$\\ldots$. | shell, the general region for the value of energy
for an electron on the orbital |\n| angular momentum or
azimuthal quantum number | 1 | $0 \\leq 1 \\leq n$
-1 | subshell, the shape of the orbital |\n| magnetic quantum number | $m_{l}$ | $-1 \\leq m_{l} \\leq$
$l$ | orientation of the orbital |"} {"id": "textbook_736", "title": "235. The Pauli Exclusion Principle - ", "content": "Quantum Numbers, Their Properties, and Significance"} {"id": "textbook_737", "title": "235. The Pauli Exclusion Principle - ", "content": "| | Symbol | Allowed
values | Physical meaning |\n| :--- | :--- | :--- | :--- |\n| spin quantum number | $m_{S}$ | $\\frac{1}{2},-\\frac{1}{2}$ | direction of the intrinsic quantum \"spinning\" of
the electron |"} {"id": "textbook_738", "title": "235. The Pauli Exclusion Principle - ", "content": "TABLE 3.1\n"} {"id": "textbook_739", "title": "236. EXAMPLE 3.7 - ", "content": ""} {"id": "textbook_740", "title": "237. Working with Shells and Subshells - ", "content": "\nIndicate the number of subshells, the number of orbitals in each subshell, and the values of 1 and $m_{l}$ for the orbitals in the $n=4$ shell of an atom.\n"} {"id": "textbook_741", "title": "238. Solution - ", "content": "\nFor $n=4, l$ can have values of $0,1,2$, and 3 . Thus, $s, p, d$, and $f$ subshells are found in the $n=4$ shell of an atom. For $l=0$ (the $s$ subshell), $m_{l}$ can only be 0 . Thus, there is only one $4 s$ orbital. For $l=1$ ( $p$-type orbitals), $m$ can have values of $-1,0,+1$, so we find three $4 p$ orbitals. For $l=2$ ( $d$-type orbitals), $m_{l}$ can have values of $-2,-1,0$, $+1,+2$, so we have five $4 d$ orbitals. When $l=3$ ( $f$-type orbitals), $m_{l}$ can have values of $-3,-2,-1,0,+1,+2,+3$, and we can have seven $4 f$ orbitals. Thus, we find a total of 16 orbitals in the $n=4$ shell of an atom.\n"} {"id": "textbook_742", "title": "239. Check Your Learning - ", "content": "\nIdentify the subshell in which electrons with the following quantum numbers are found: (a) $n=3,1=1$; (b) $n=$ $5, l=3$; (c) $n=2, l=0$.\n"} {"id": "textbook_743", "title": "240. Answer: - ", "content": "\n(a) $3 p$ (b) $5 f$ (c) 2 s\n"} {"id": "textbook_744", "title": "241. EXAMPLE 3.8 - ", "content": ""} {"id": "textbook_745", "title": "242. Maximum Number of Electrons - ", "content": "\nCalculate the maximum number of electrons that can occupy a shell with (a) $n=2$, (b) $n=5$, and (c) $n$ as a variable. Note you are only looking at the orbitals with the specified $n$ value, not those at lower energies.\n"} {"id": "textbook_746", "title": "243. Solution - ", "content": "\n(a) When $n=2$, there are four orbitals (a single $2 s$ orbital, and three orbitals labeled $2 p$ ). These four orbitals can contain eight electrons.\n(b) When $n=5$, there are five subshells of orbitals that we need to sum:"} {"id": "textbook_747", "title": "243. Solution - ", "content": "1 orbital labeled $5 s$\n3 orbitals labeled 5p\n5 orbitals labeled $5 d$\n7 orbitals labeled $5 f$\n+9 orbitals labeled $5 g$\n25 orbitals total"} {"id": "textbook_748", "title": "243. Solution - ", "content": "Again, each orbital holds two electrons, so 50 electrons can fit in this shell.\n(c) The number of orbitals in any shell $n$ will equal $n^{2}$. There can be up to two electrons in each orbital, so the maximum number of electrons will be $2 \\times n^{2}$.\n"} {"id": "textbook_749", "title": "244. Check Your Learning - ", "content": "\nIf a shell contains a maximum of 32 electrons, what is the principal quantum number, $n$ ?\n"} {"id": "textbook_750", "title": "245. Answer: - ", "content": "\n$n=4$\n"} {"id": "textbook_751", "title": "246. EXAMPLE 3.9 - ", "content": ""} {"id": "textbook_752", "title": "247. Working with Quantum Numbers - ", "content": "\nComplete the following table for atomic orbitals:"} {"id": "textbook_753", "title": "247. Working with Quantum Numbers - ", "content": "| Orbital | $\\boldsymbol{n}$ | $\\boldsymbol{I}$ | $\\boldsymbol{m}_{\\boldsymbol{l}}$ degeneracy | Radial nodes (no.) |\n| :--- | :--- | :--- | :--- | :--- |\n| $4 f$ | | | | |\n| | 4 | 1 | | |\n| | 7 | | 7 | 3 |\n| $5 d$ | | | | |\n"} {"id": "textbook_754", "title": "248. Solution - ", "content": "\nThe table can be completed using the following rules:"} {"id": "textbook_755", "title": "248. Solution - ", "content": "- The orbital designation is $n l$, where $l=0,1,2,3,4,5, \\ldots$ is mapped to the letter sequence $s, p, d, f, g, h, \\ldots$,\n- The $m_{l}$ degeneracy is the number of orbitals within an 1 subshell, and so is $2 l+1$ (there is one $s$ orbital, three $p$ orbitals, five $d$ orbitals, seven $f$ orbitals, and so forth).\n- The number of radial nodes is equal to $n-1-1$.\n\n| Orbital | $\\boldsymbol{n}$ | $\\boldsymbol{l}$ | $\\boldsymbol{m}_{\\boldsymbol{l}}$ degeneracy | Radial nodes (no.) |\n| :--- | :--- | :--- | :--- | :--- |\n| $4 f$ | 4 | 3 | 7 | 0 |\n| $4 p$ | 4 | 1 | 3 | 2 |\n| $7 f$ | 7 | 3 | 7 | 3 |\n| $5 d$ | 5 | 2 | 5 | 2 |\n"} {"id": "textbook_756", "title": "249. Check Your Learning - ", "content": "\nHow many orbitals have $l=2$ and $n=3$ ?\n"} {"id": "textbook_757", "title": "250. Answer: - ", "content": "\nThe five degenerate $3 d$ orbitals\n"} {"id": "textbook_758", "title": "250. Answer: - 250.1. Electronic Structure of Atoms (Electron Configurations)", "content": ""} {"id": "textbook_759", "title": "251. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_760", "title": "251. LEARNING OBJECTIVES - ", "content": "- Derive the predicted ground-state electron configurations of atoms\n- Identify and explain exceptions to predicted electron configurations for atoms and ions\n- Relate electron configurations to element classifications in the periodic table"} {"id": "textbook_761", "title": "251. LEARNING OBJECTIVES - ", "content": "Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom.\n"} {"id": "textbook_762", "title": "252. Orbital Energies and Atomic Structure - ", "content": "\nThe energy of atomic orbitals increases as the principal quantum number, $n$, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of 1 differ so that the energy of the orbitals increases within a shell in the order $s
Chemical Bonding and Molecular Geometry - ", "content": "\n"} {"id": "textbook_957", "title": "320. CHAPTER 4
Chemical Bonding and Molecular Geometry - ", "content": "Figure 4.1 Nicknamed \"buckyballs,\" buckminsterfullerene molecules ( $\\mathrm{C}_{60}$ ) contain only carbon atoms (left) arranged to form a geometric framework of hexagons and pentagons, similar to the pattern on a soccer ball (center). This molecular structure is named after architect R. Buckminster Fuller, whose innovative designs combined simple geometric shapes to create large, strong structures such as this weather radar dome near Tucson, Arizona (right). (credit middle: modification of work by \"Petey21\"/Wikimedia Commons; credit right: modification of work by Bill Morrow)\n"} {"id": "textbook_958", "title": "321. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_959", "title": "321. CHAPTER OUTLINE - 321.1. Ionic Bonding", "content": "\n4.2 Covalent Bonding\n4.3 Chemical Nomenclature\n4.4 Lewis Symbols and Structures\n4.5 Formal Charges and Resonance\n4.6 Molecular Structure and Polarity"} {"id": "textbook_960", "title": "321. CHAPTER OUTLINE - 321.1. Ionic Bonding", "content": "INTRODUCTION It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized:\nbuckminsterfullerene. This molecule was named after the architect and inventor R. Buckminster Fuller (1895-1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, $\\mathrm{C}_{60}$, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory-the topic of this chapter-which explains how individual atoms connect to form more complex structures.\n"} {"id": "textbook_961", "title": "321. CHAPTER OUTLINE - 321.2. Ionic Bonding", "content": ""} {"id": "textbook_962", "title": "322. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_963", "title": "322. LEARNING OBJECTIVES - ", "content": "- Explain the formation of cations, anions, and ionic compounds\n- Predict the charge of common metallic and nonmetallic elements, and write their electron configurations"} {"id": "textbook_964", "title": "322. LEARNING OBJECTIVES - ", "content": "As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell."} {"id": "textbook_965", "title": "322. LEARNING OBJECTIVES - ", "content": "Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason-the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely."} {"id": "textbook_966", "title": "322. LEARNING OBJECTIVES - ", "content": "Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, $\\mathrm{Cl}_{2}$, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure 4.2). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.\n\n\nFIGURE 4.2 (a) Sodium is a soft metal that must be stored in mineral oil to prevent reaction with air or water. (b) Chlorine is a pale yellow-green gas. (c) When combined, they form white crystals of sodium chloride (table salt). (credit a: modification of work by \"Jurii\"/Wikimedia Commons)\n"} {"id": "textbook_967", "title": "323. The Formation of Ionic Compounds - ", "content": "\nBinary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table."} {"id": "textbook_968", "title": "323. The Formation of Ionic Compounds - ", "content": "As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, indicates that this ionic compound contains two aluminum cations, $\\mathrm{Al}^{3+}$, for every three oxide anions, $\\mathrm{O}^{2-}$ [thus, $(2 \\times+3)+(3 \\times-2)=0$ ]."} {"id": "textbook_969", "title": "323. The Formation of Ionic Compounds - ", "content": "It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride ( NaCl ) \"molecule\" because there is not a\nsingle ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic-the same in all directions-meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, threedimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of $\\mathrm{Na}^{+}$cations and $\\mathrm{Cl}^{-}$anions (Figure 4.3).\n"} {"id": "textbook_970", "title": "323. The Formation of Ionic Compounds - ", "content": "FIGURE 4.3 The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is \"bonded\" to all of the surrounding ions-six in this case."} {"id": "textbook_971", "title": "323. The Formation of Ionic Compounds - ", "content": "The strong electrostatic attraction between $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions holds them tightly together in solid NaCl . It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions:\n\n$$\n\\mathrm{NaCl}(s) \\longrightarrow \\mathrm{Na}^{+}(g)+\\mathrm{Cl}^{-}(g) \\quad \\Delta H=769 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_972", "title": "324. Electronic Structures of Cations - ", "content": "\nWhen forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}$. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a $2+$ charge, and an electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$. The $\\mathrm{Ca}^{2+}$ ion is therefore isoelectronic with the noble gas Ar.\n\nFor groups 13-17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full $d$ subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms $3+$ ions $\\left(\\mathrm{Al}^{3+}\\right)$."} {"id": "textbook_973", "title": "324. Electronic Structures of Cations - ", "content": "Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions $\\mathrm{Tl}^{3+}, \\mathrm{Sn}^{4+}, \\mathrm{Pb}^{4+}$, and $\\mathrm{Bi}^{5+}$, a partial loss of these atoms' valence shell electrons can also lead to the formation of $\\mathrm{Tl}^{+}, \\mathrm{Sn}^{2+}, \\mathrm{Pb}^{2+}$, and $\\mathrm{Bi}^{3+}$ ions. The formation of these $1+, 2+$, and $3+$ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence $s$-electron pair for atoms of the heavy elements of groups 13,14 , and 15 . Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, $\\mathrm{Hg}_{2}{ }^{2+}$ (an ion formed from two mercury atoms, with an $\\mathrm{Hg}-\\mathrm{Hg}$ bond), in addition to the expected monatomic ion $\\mathrm{Hg}^{2+}$ (formed from only one mercury atom)."} {"id": "textbook_974", "title": "324. Electronic Structures of Cations - ", "content": "Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have $2+$ or $3+$ charges that result from the loss of their outermost $s$ electron(s) first, sometimes followed by the loss of one or two $d$ electrons from the next-to-outermost shell. For example, iron $\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}\\right)$ forms the ion $\\mathrm{Fe}^{2+}\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}\\right)$ by the loss of the $4 s$ electrons and the ion $\\mathrm{Fe}^{3+}\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}\\right)$ by the loss of the $4 s$ electron and one of the $3 d$ electrons. Although the $d$ orbitals of\nthe transition elements are-according to the Aufbau principle-the last to fill when building up electron configurations, the outermost $s$ electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a $3+$ charge, resulting from the loss of their outermost $s$ electrons and a $d$ or $f$ electron.\n"} {"id": "textbook_975", "title": "325. EXAMPLE 4.1 - ", "content": ""} {"id": "textbook_976", "title": "326. Determining the Electronic Structures of Cations - ", "content": "\nThere are at least 14 elements categorized as \"essential trace elements\" for the human body. They are called \"essential\" because they are required for healthy bodily functions, \"trace\" because they are required only in small amounts, and \"elements\" in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as $\\mathrm{Cr}^{3+}$ and $\\mathrm{Zn}^{2+}$. Write the electron configurations of these cations.\n"} {"id": "textbook_977", "title": "327. Solution - ", "content": "\nFirst, write the electron configuration for the neutral atoms:\n$\\mathrm{Zn}:[\\mathrm{Ar}] 3 \\mathrm{~d}^{10} 4 s^{2}$\nCr: $[\\mathrm{Ar}] 3 d^{5} 4 s^{1}$\nNext, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the $s$ orbital first and then from the $d$ orbital. For the $p$-block elements, electrons are removed from the $p$ orbitals and then from the $s$ orbital. Zinc is a member of group 12, so it should have a charge of $2+$, and thus loses only the two electrons in its $S$ orbital. Chromium is a transition element and should lose its $s$ electrons and then its $d$ electrons when forming a cation. Thus, we find the following electron configurations of the ions:\n$\\mathrm{Zn}^{2+}:[\\mathrm{Ar}] 3 d^{10}$\n$\\mathrm{Cr}^{3+}:[\\mathrm{Ar}] 3 d^{3}$\n"} {"id": "textbook_978", "title": "328. Check Your Learning - ", "content": "\nPotassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.\n"} {"id": "textbook_979", "title": "329. Answer: - ", "content": "\n$\\mathrm{K}^{+}$: $[\\mathrm{Ar}], \\mathrm{Mg}^{2+}$ : $[\\mathrm{Ne}]$\n"} {"id": "textbook_980", "title": "330. Electronic Structures of Anions - ", "content": "\nMost monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer $s$ and $p$ orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the $s$ and $p$ orbitals of the parent atom. Oxygen, for example, has the electron configuration $1 s^{2} 2 s^{2} 2 p^{4}$, whereas the oxygen anion has the electron configuration of the noble gas neon ( Ne ), $1 s^{2} 2 s^{2} 2 p^{6}$. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of $2-\\left(\\mathrm{O}^{2-}\\right)$.\n"} {"id": "textbook_981", "title": "331. EXAMPLE 4.2 - ", "content": ""} {"id": "textbook_982", "title": "332. Determining the Electronic Structure of Anions - ", "content": "\nSelenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.\n"} {"id": "textbook_983", "title": "333. Solution - ", "content": "\n$\\mathrm{Se}^{2-}:[\\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{6}$\n$I^{-}:[\\mathrm{Kr}] 4 d^{10} 5 s^{2} 5 p^{6}$\n"} {"id": "textbook_984", "title": "334. Check Your Learning - ", "content": "\nWrite the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.\n"} {"id": "textbook_985", "title": "335. Answer: - ", "content": "\nP: [Ne] $3 s^{2} 3 p^{3} ; \\mathrm{P}^{3-}:[\\mathrm{Ne}] 3 s^{2} 3 p^{6}$\n"} {"id": "textbook_986", "title": "335. Answer: - 335.1. Covalent Bonding", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_987", "title": "335. Answer: - 335.1. Covalent Bonding", "content": "- Describe the formation of covalent bonds\n- Define electronegativity and assess the polarity of covalent bonds"} {"id": "textbook_988", "title": "335. Answer: - 335.1. Covalent Bonding", "content": "Ionic bonding results from the electrostatic attraction of oppositely charged ions that are typically produced by the transfer of electrons between metallic and nonmetallic atoms. A different type of bonding results from the mutual attraction of atoms for a \"shared\" pair of electrons. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an $\\mathrm{H}_{2}$ molecule; each hydrogen atom in the $\\mathrm{H}_{2}$ molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He ."} {"id": "textbook_989", "title": "335. Answer: - 335.1. Covalent Bonding", "content": "Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.\n"} {"id": "textbook_990", "title": "336. Formation of Covalent Bonds - ", "content": "\nNonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, $\\mathrm{H}_{2}$, contains a covalent bond between its two hydrogen atoms. Figure 4.4 illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the $x$-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the $x$-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.\n\n\nFIGURE 4.4 The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved."} {"id": "textbook_991", "title": "336. Formation of Covalent Bonds - ", "content": "It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:"} {"id": "textbook_992", "title": "336. Formation of Covalent Bonds - ", "content": "$$\n\\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{H}(\\mathrm{~g}) \\quad \\Delta H=436 \\mathrm{~kJ}\n$$"} {"id": "textbook_993", "title": "336. Formation of Covalent Bonds - ", "content": "Conversely, the same amount of energy is released when one mole of $\\mathrm{H}_{2}$ molecules forms from two moles of H atoms:"} {"id": "textbook_994", "title": "336. Formation of Covalent Bonds - ", "content": "$$\n2 \\mathrm{H}(\\mathrm{~g}) \\longrightarrow \\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta H=-436 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_995", "title": "337. Pure vs. Polar Covalent Bonds - ", "content": "\nIf the atoms that form a covalent bond are identical, as in $\\mathrm{H}_{2}, \\mathrm{Cl}_{2}$, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in\npure covalent bonds have an equal probability of being near each nucleus.\nIn the case of $\\mathrm{Cl}_{2}$, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:"} {"id": "textbook_996", "title": "337. Pure vs. Polar Covalent Bonds - ", "content": "$$\n\\mathrm{Cl}+\\mathrm{Cl} \\longrightarrow \\mathrm{Cl}_{2}\n$$"} {"id": "textbook_997", "title": "337. Pure vs. Polar Covalent Bonds - ", "content": "The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, $\\mathrm{Cl}_{2}$ also features a pure covalent bond."} {"id": "textbook_998", "title": "337. Pure vs. Polar Covalent Bonds - ", "content": "When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the $\\mathrm{H}-\\mathrm{Cl}$ bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure 4.5 shows the distribution of electrons in the $\\mathrm{H}-\\mathrm{Cl}$ bond. Note that the shaded area around Cl is much larger than it is around H . Compare this to Figure 4.4, which shows the even distribution of electrons in the $\\mathrm{H}_{2}$ nonpolar bond.\n\nWe sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter \"delta,\" $\\delta$, with a plus sign or minus sign to indicate whether the atom has a partial positive charge ( $\\delta+$ ) or a partial negative charge ( $\\delta-$ ). This symbolism is shown for the $\\mathrm{H}-\\mathrm{Cl}$ molecule in Figure 4.5.\n\n(a)\n$\\delta+\\delta-$\n$\\mathrm{H}-\\mathrm{Cl}$\n(b)"} {"id": "textbook_999", "title": "337. Pure vs. Polar Covalent Bonds - ", "content": "FIGURE 4.5 (a) The distribution of electron density in the HCl molecule is uneven. The electron density is greater around the chlorine nucleus. The small, black dots indicate the location of the hydrogen and chlorine nuclei in the molecule. (b) Symbols $\\delta+$ and $\\delta$ - indicate the polarity of the $\\mathrm{H}-\\mathrm{Cl}$ bond.\n"} {"id": "textbook_1000", "title": "338. Electronegativity - ", "content": "\nWhether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms."} {"id": "textbook_1001", "title": "338. Electronegativity - ", "content": "Figure 4.6 shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure 4.7). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as $\\mathrm{XeO}_{2}$ do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)\n"} {"id": "textbook_1002", "title": "338. Electronegativity - ", "content": "FIGURE 4.6 The electronegativity values derived by Pauling follow predictable periodic trends, with the higher electronegativities toward the upper right of the periodic table.\n"} {"id": "textbook_1003", "title": "339. Electronegativity versus Electron Affinity - ", "content": "\nWe must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in $\\mathrm{kJ} / \\mathrm{mol}$. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4 .\n"} {"id": "textbook_1004", "title": "340. Portrait of a Chemist - ", "content": ""} {"id": "textbook_1005", "title": "341. Linus Pauling - ", "content": "\nLinus Pauling, shown in Figure 4.7, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.\n"} {"id": "textbook_1006", "title": "341. Linus Pauling - ", "content": "FIGURE 4.7 Linus Pauling (1901-1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons."} {"id": "textbook_1007", "title": "341. Linus Pauling - ", "content": "Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease-the presence of a genetically inherited abnormal protein in the blood-and paved the way for the field of molecular genetics. His work was also pivotal in curbing the\ntesting of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.\n"} {"id": "textbook_1008", "title": "342. Electronegativity and Bond Type - ", "content": "\nThe absolute value of the difference in electronegativity ( $\\Delta \\mathrm{EN}$ ) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds $\\mathrm{H}-\\mathrm{H}, \\mathrm{H}-\\mathrm{Cl}$, and $\\mathrm{Na}-\\mathrm{Cl}$ are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 4.8 shows the relationship between electronegativity difference and bond type.\n\n\nFIGURE 4.8 As the electronegativity difference increases between two atoms, the bond becomes more ionic.\nA rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure 4.8. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in $\\mathrm{NH}_{3}$ a difference of 0.9 , yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in $\\mathrm{MnI}_{2}$ have a difference of 1.0, yet both of these substances form ionic compounds."} {"id": "textbook_1009", "title": "342. Electronegativity and Bond Type - ", "content": "The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic."} {"id": "textbook_1010", "title": "342. Electronegativity and Bond Type - ", "content": "Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as $\\mathrm{OH}^{-}, \\mathrm{NO}_{3}{ }^{-}$, and $\\mathrm{NH}_{4}{ }^{+}$, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, $\\mathrm{KNO}_{3}$, contains the $\\mathrm{K}^{+}$cation and the polyatomic $\\mathrm{NO}_{3}{ }^{-}$anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions $\\mathrm{K}^{+}$and $\\mathrm{NO}_{3}{ }^{-}$, as well as covalent between the nitrogen and oxygen atoms in $\\mathrm{NO}_{3}{ }^{-}$.\n"} {"id": "textbook_1011", "title": "343. EXAMPLE 4.3 - ", "content": ""} {"id": "textbook_1012", "title": "344. Electronegativity and Bond Polarity - ", "content": "\nBond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure 4.6, arrange the following covalent bonds-all commonly found in amino acids-in order of increasing polarity. Then designate the positive and negative atoms using the symbols $\\delta+$ and $\\delta-$ :"} {"id": "textbook_1013", "title": "344. Electronegativity and Bond Polarity - ", "content": "C-H, C-N, C-O, N-H, O-H, S-H\n"} {"id": "textbook_1014", "title": "345. Solution - ", "content": "\nThe polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the $\\delta$ - designation is the more electronegative of the two. Table 4.1 shows these bonds in order of increasing polarity."} {"id": "textbook_1015", "title": "345. Solution - ", "content": "Bond Polarity and Electronegativity Difference"} {"id": "textbook_1016", "title": "345. Solution - ", "content": "| Bond | $\\Delta \\mathrm{EN}$ | Polarity |\n| :---: | :---: | :---: |\n| C-H | 0.4 | $\\stackrel{\\delta-}{\\mathrm{C}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |\n| S-H | 0.4 | $\\stackrel{\\delta-}{S-}-\\stackrel{\\delta+}{H}$ |\n| C-N | 0.5 | $\\stackrel{\\delta+}{\\mathrm{C}}-\\stackrel{\\delta-}{\\mathrm{N}}$ |\n| N-H | 0.9 | $\\stackrel{\\delta-}{\\mathrm{N}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |\n| C-O | 1.0 | $\\stackrel{\\delta+}{\\mathrm{C}}-\\stackrel{\\delta-}{\\mathrm{O}}$ |\n| O-H | 1.4 | $\\stackrel{\\delta-}{\\mathrm{O}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |"} {"id": "textbook_1017", "title": "345. Solution - ", "content": "TABLE 4.1\n"} {"id": "textbook_1018", "title": "346. Check Your Learning - ", "content": "\nSilicones are polymeric compounds containing, among others, the following types of covalent bonds: $\\mathrm{Si}-\\mathrm{O}$, $\\mathrm{Si}-\\mathrm{C}, \\mathrm{C}-\\mathrm{H}$, and $\\mathrm{C}-\\mathrm{C}$. Using the electronegativity values in Figure 4.6, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols $\\delta+$ and $\\delta-$."} {"id": "textbook_1019", "title": "346. Check Your Learning - ", "content": "Answer:"} {"id": "textbook_1020", "title": "346. Check Your Learning - ", "content": "| Bond | Electronegativity Difference | Polarity |\n| :--- | :--- | :--- |\n| $\\mathrm{C}-\\mathrm{C}$ | 0.0 | nonpolar |\n| $\\mathrm{C}-\\mathrm{H}$ | 0.4 | $\\delta-\\stackrel{\\delta+}{\\mathrm{C}}-\\mathrm{H}$ |\n| $\\mathrm{Si}-\\mathrm{C}$ | 0.7 | $\\delta+\\delta-$
$\\mathrm{Si}-\\mathrm{C}$ |\n| $\\mathrm{Si}-\\mathrm{O}$ | 1.7 | $\\delta+\\delta-$
$\\mathrm{Si}-\\mathrm{O}$ |\n"} {"id": "textbook_1021", "title": "346. Check Your Learning - 346.1. Chemical Nomenclature", "content": ""} {"id": "textbook_1022", "title": "347. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1023", "title": "347. LEARNING OBJECTIVES - ", "content": "- Derive names for common types of inorganic compounds using a systematic approach"} {"id": "textbook_1024", "title": "347. LEARNING OBJECTIVES - ", "content": "Nomenclature, a collection of rules for naming things, is important in science and in many other situations."} {"id": "textbook_1025", "title": "347. LEARNING OBJECTIVES - ", "content": "This module describes an approach that is used to name simple ionic and molecular compounds, such as $\\mathrm{NaCl}, \\mathrm{CaCO}_{3}$, and $\\mathrm{N}_{2} \\mathrm{O}_{4}$. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids (subsequent chapters in this text will focus on these compounds in great detail). We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry.\n"} {"id": "textbook_1026", "title": "348. Ionic Compounds - ", "content": "\nTo name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly.\n"} {"id": "textbook_1027", "title": "349. Compounds Containing Only Monatomic Ions - ", "content": "\nThe name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix -ide). Some examples are given in Table 4.2."} {"id": "textbook_1028", "title": "349. Compounds Containing Only Monatomic Ions - ", "content": "| Names of Some Ionic Compounds | |\n| :--- | :--- |\n| NaCl, sodium chloride | $\\mathrm{Na}_{2} \\mathrm{O}$, sodium oxide |\n| KBr , potassium bromide | CdS, cadmium sulfide |\n| $\\mathrm{CaI}_{2}$, calcium iodide | $\\mathrm{Mg}_{3} \\mathrm{~N}_{2}$, magnesium nitride |\n| CsF, cesium fluoride | $\\mathrm{Ca}_{3} \\mathrm{P}_{2}$, calcium phosphide |\n| LiCl, lithium chloride | $\\mathrm{Al}_{4} \\mathrm{C}_{3}$, aluminum carbide |"} {"id": "textbook_1029", "title": "349. Compounds Containing Only Monatomic Ions - ", "content": "TABLE 4.2\n"} {"id": "textbook_1030", "title": "350. Compounds Containing Polyatomic lons - ", "content": "\nCompounds containing polyatomic ions are named similarly to those containing only monatomic ions, i.e., by naming first the cation and then the anion. Examples are shown in Table 4.3."} {"id": "textbook_1031", "title": "350. Compounds Containing Polyatomic lons - ", "content": "| Names of Some Polyatomic lonic Compounds | |\n| :--- | :--- |\n| $\\mathrm{KC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$, potassium acetate | $\\mathrm{NH}_{4} \\mathrm{Cl}$, ammonium chloride |\n| $\\mathrm{NaHCO}_{3}$, sodium bicarbonate | $\\mathrm{CaSO}_{4}$, calcium sulfate |\n| $\\mathrm{Al}_{2}\\left(\\mathrm{CO}_{3}\\right)_{3}$, aluminum carbonate | $\\mathrm{Mg}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, magnesium phosphate |"} {"id": "textbook_1032", "title": "350. Compounds Containing Polyatomic lons - ", "content": "TABLE 4.3\n"} {"id": "textbook_1033", "title": "351. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_1034", "title": "352. Ionic Compounds in Your Cabinets - ", "content": "\nEvery day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table 4.4. Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula."} {"id": "textbook_1035", "title": "352. Ionic Compounds in Your Cabinets - ", "content": "Everyday Ionic Compounds"} {"id": "textbook_1036", "title": "352. Ionic Compounds in Your Cabinets - ", "content": "| Ionic Compound | Use |\n| :---: | :---: |\n| NaCl , sodium chloride | ordinary table salt |\n| KI, potassium iodide | added to \"iodized\" salt for thyroid health |\n| NaF, sodium fluoride | ingredient in toothpaste |\n| $\\mathrm{NaHCO}_{3}$, sodium bicarbonate | baking soda; used in cooking (and as antacid) |\n| $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, sodium carbonate | washing soda; used in cleaning agents |\n| NaOCl , sodium hypochlorite | active ingredient in household bleach |\n| $\\mathrm{CaCO}_{3}$ calcium carbonate | ingredient in antacids |\n| $\\mathrm{Mg}(\\mathrm{OH})_{2}$, magnesium hydroxide | ingredient in antacids |\n| $\\mathrm{Al}(\\mathrm{OH})_{3}$, aluminum hydroxide | ingredient in antacids |\n| NaOH , sodium hydroxide | lye; used as drain cleaner |\n| $\\mathrm{K}_{3} \\mathrm{PO}_{4}$, potassium phosphate | food additive (many purposes) |\n| $\\mathrm{MgSO}_{4}$, magnesium sulfate | added to purified water |\n| $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$, sodium hydrogen phosphate | anti-caking agent; used in powdered products |\n| $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$, sodium sulfite | preservative |"} {"id": "textbook_1037", "title": "352. Ionic Compounds in Your Cabinets - ", "content": "TABLE 4.4\n"} {"id": "textbook_1038", "title": "353. Compounds Containing a Metal Ion with a Variable Charge - ", "content": "\nMost of the transition metals and some main group metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or $3+$ (see Figure 3.40), and the two corresponding compound formulas are $\\mathrm{FeCl}_{2}$ and $\\mathrm{FeCl}_{3}$. The simplest name, \"iron chloride,\" will, in this case, be ambiguous, as it does not distinguish between\nthese two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table 4.5."} {"id": "textbook_1039", "title": "353. Compounds Containing a Metal Ion with a Variable Charge - ", "content": "| Some Ionic Compounds with Variably Charged Metal lons | |\n| :--- | :--- |\n| Compound | Name |\n| $\\mathrm{FeCl}_{2}$ | iron(II) chloride |\n| $\\mathrm{FeCl}_{3}$ | iron(III) chloride |\n| $\\mathrm{Hg}_{2} \\mathrm{O}$ | mercury(I) oxide |\n| HgO | mercury(II) oxide |\n| $\\mathrm{SnF}_{2}$ | tin(II) fluoride |\n| $\\mathrm{SnF}_{4}$ | tin(IV) fluoride |"} {"id": "textbook_1040", "title": "353. Compounds Containing a Metal Ion with a Variable Charge - ", "content": "TABLE 4.5"} {"id": "textbook_1041", "title": "353. Compounds Containing a Metal Ion with a Variable Charge - ", "content": "Out-of-date nomenclature used the suffixes -ic and -ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, $\\mathrm{FeCl}_{3}$, was previously called ferric chloride, and iron(II) chloride, $\\mathrm{FeCl}_{2}$, was known as ferrous chloride. Though this naming convention has been largely abandoned by the scientific community, it remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula $\\mathrm{SnF}_{2}$, which is more properly named tin(II) fluoride. The other fluoride of tin is $\\mathrm{SnF}_{4}$, which was previously called stannic fluoride but is now named tin(IV) fluoride.\n"} {"id": "textbook_1042", "title": "354. Ionic Hydrates - ", "content": "\nIonic compounds that contain water molecules as integral components of their crystals are called hydrates. The name for an ionic hydrate is derived by adding a term to the name for the anhydrous (meaning \"not hydrated\") compound that indicates the number of water molecules associated with each formula unit of the compound. The added word begins with a Greek prefix denoting the number of water molecules (see Table 4.6) and ends with \"hydrate.\" For example, the anhydrous compound copper(II) sulfate also exists as a hydrate containing five water molecules and named copper(II) sulfate pentahydrate. Washing soda is the common name for a hydrate of sodium carbonate containing 10 water molecules; the systematic name is sodium carbonate decahydrate."} {"id": "textbook_1043", "title": "354. Ionic Hydrates - ", "content": "Formulas for ionic hydrates are written by appending a vertically centered dot, a coefficient representing the number of water molecules, and the formula for water. The two examples mentioned in the previous paragraph are represented by the formulas"} {"id": "textbook_1044", "title": "354. Ionic Hydrates - ", "content": "> copper(II) sulfate pentahydrate $\\mathrm{CuSO}_{4} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}$\n> sodium carbonate decahydrate $\\mathrm{Na}_{2} \\mathrm{CO}_{3} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$"} {"id": "textbook_1045", "title": "354. Ionic Hydrates - ", "content": "Nomenclature Prefixes"} {"id": "textbook_1046", "title": "354. Ionic Hydrates - ", "content": "| Number | | Prefix | | Number | | Prefix |\n| :--- | :--- | :--- | :--- | :--- | :---: | :---: |\n| 1 (sometimes omitted) | mono- | 6 | hexa- | | | |\n| 2 | di- | | 7 | hepta- | | |\n| 2 | tri- | | 8 | octa- | | |\n| 3 | tetra- | | 9 | nona- | | |\n| 4 | penta- | | 10 | deca- | | |\n| 5 | | | | | | |"} {"id": "textbook_1047", "title": "354. Ionic Hydrates - ", "content": "TABLE 4.6\n"} {"id": "textbook_1048", "title": "355. (8) EXAMPLE 4.4 - ", "content": ""} {"id": "textbook_1049", "title": "356. Naming Ionic Compounds - ", "content": "\nName the following ionic compounds:\n(a) $\\mathrm{Fe}_{2} \\mathrm{~S}_{3}$\n(b) CuSe\n(c) GaN\n(d) $\\mathrm{MgSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{Ti}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n"} {"id": "textbook_1050", "title": "357. Solution - ", "content": "\nThe anions in these compounds have a fixed negative charge ( $\\mathrm{S}^{2-}, \\mathrm{Se}^{2-}, \\mathrm{N}^{3-}$, and $\\mathrm{SO}_{4}{ }^{2-}$ ), and the compounds must be neutral. Because the total number of positive charges in each compound must equal the total number of negative charges, the positive ions must be $\\mathrm{Fe}^{3+}, \\mathrm{Cu}^{2+}, \\mathrm{Ga}^{3+}, \\mathrm{Mg}^{2+}$, and $\\mathrm{Ti}^{3+}$. These charges are used in the names of the metal ions:\n(a) iron(III) sulfide\n(b) copper(II) selenide\n(c) gallium(III) nitride\n(d) magnesium sulfate heptahydrate\n(e) titanium(III) sulfate\n"} {"id": "textbook_1051", "title": "358. Check Your Learning - ", "content": "\nWrite the formulas of the following ionic compounds:\n(a) chromium(III) phosphide\n(b) mercury(II) sulfide\n(c) manganese(II) phosphate\n(d) copper(I) oxide\n(e) iron(III) chloride dihydrate\n"} {"id": "textbook_1052", "title": "359. Answer: - ", "content": "\n(a) CrP ; (b) HgS ; (c) $\\mathrm{Mn}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$; (d) $\\mathrm{Cu}_{2} \\mathrm{O}$; (e) $\\mathrm{FeCl}_{3} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$\n"} {"id": "textbook_1053", "title": "360. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_1054", "title": "361. Erin Brokovich and Chromium Contamination - ", "content": "\nIn the early 1990s, legal file clerk Erin Brockovich (Figure 4.9) discovered a high rate of serious illnesses in the small town of Hinckley, California. Her investigation eventually linked the illnesses to groundwater contaminated by $\\mathrm{Cr}(\\mathrm{VI})$ used by Pacific Gas \\& Electric (PG\\&E) to fight corrosion in a nearby natural gas pipeline. As dramatized in the film Erin Brokovich (for which Julia Roberts won an Oscar), Erin and lawyer Edward Masry sued PG\\&E for contaminating the water near Hinckley in 1993. The settlement they won in 1996-\\$333 million-was the largest amount ever awarded for a direct-action lawsuit in the US at that time.\n"} {"id": "textbook_1055", "title": "361. Erin Brokovich and Chromium Contamination - ", "content": "FIGURE 4.9 (a) Erin Brockovich found that $\\mathrm{Cr}(\\mathrm{VI})$, used by PG\\&E, had contaminated the Hinckley, California, water supply. (b) The $\\mathrm{Cr}(\\mathrm{VI})$ ion is often present in water as the polyatomic ions chromate, $\\mathrm{CrO}_{4}{ }^{2-}$ (left), and dichromate, $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ (right)."} {"id": "textbook_1056", "title": "361. Erin Brokovich and Chromium Contamination - ", "content": "Chromium compounds are widely used in industry, such as for chrome plating, in dye-making, as preservatives, and to prevent corrosion in cooling tower water, as occurred near Hinckley. In the environment, chromium exists primarily in either the Cr (III) or Cr (VI) forms. Cr (III), an ingredient of many vitamin and nutritional supplements, forms compounds that are not very soluble in water, and it has low toxicity. But $\\mathrm{Cr}(\\mathrm{VI})$ is much more toxic and forms compounds that are reasonably soluble in water.\nExposure to small amounts of $\\mathrm{Cr}(\\mathrm{VI})$ can lead to damage of the respiratory, gastrointestinal, and immune systems, as well as the kidneys, liver, blood, and skin."} {"id": "textbook_1057", "title": "361. Erin Brokovich and Chromium Contamination - ", "content": "Despite cleanup efforts, $\\mathrm{Cr}(\\mathrm{VI})$ groundwater contamination remains a problem in Hinckley and other locations across the globe. A 2010 study by the Environmental Working Group found that of 35 US cities tested, 31 had higher levels of $\\mathrm{Cr}(\\mathrm{VI})$ in their tap water than the public health goal of 0.02 parts per billion set by the California Environmental Protection Agency.\n"} {"id": "textbook_1058", "title": "362. Molecular (Covalent) Compounds - ", "content": "\nThe bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios."} {"id": "textbook_1059", "title": "362. Molecular (Covalent) Compounds - ", "content": "Compounds Composed of Two Elements\nWhen two nonmetallic elements form a molecular compound, several combination ratios are often possible."} {"id": "textbook_1060", "title": "362. Molecular (Covalent) Compounds - ", "content": "For example, carbon and oxygen can form the compounds CO and $\\mathrm{CO}_{2}$. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix -ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table 4.6.\n\nWhen only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, CO is named carbon monoxide, and $\\mathrm{CO}_{2}$ is called carbon dioxide. When two vowels are adjacent, the $a$ in the Greek prefix is usually dropped. Some other examples are shown in Table 4.7."} {"id": "textbook_1061", "title": "362. Molecular (Covalent) Compounds - ", "content": "Names of Some Molecular Compounds Composed of Two Elements"} {"id": "textbook_1062", "title": "362. Molecular (Covalent) Compounds - ", "content": "| Compound | | | | Name | | Compound | | Name |\n| :--- | :--- | :--- | :--- | :--- | :---: | :---: | :---: | :---: |\n| $\\mathrm{SO}_{2}$ | sulfur dioxide | | $\\mathrm{BCl}_{3}$ | boron trichloride | | | | |\n| $\\mathrm{SO}_{3}$ | sulfur trioxide | | $\\mathrm{SF}_{6}$ | sulfur hexafluoride | | | | |\n| $\\mathrm{NO}_{2}$ | nitrogen dioxide | | $\\mathrm{PF}_{5}$ | phosphorus pentafluoride | | | | |\n| $\\mathrm{N}_{2} \\mathrm{O}_{4}$ | dinitrogen tetroxide | | $\\mathrm{P}_{4} \\mathrm{O}_{10}$ | tetraphosphorus decaoxide | | | | |\n| $\\mathrm{N}_{2} \\mathrm{O}_{5}$ | dinitrogen pentoxide | | $\\mathrm{IF}_{7}$ | iodine heptafluoride | | | | |"} {"id": "textbook_1063", "title": "362. Molecular (Covalent) Compounds - ", "content": "TABLE 4.7"} {"id": "textbook_1064", "title": "362. Molecular (Covalent) Compounds - ", "content": "There are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, $\\mathrm{N}_{2} \\mathrm{O}$ is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And $\\mathrm{H}_{2} \\mathrm{O}$ is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them.\n"} {"id": "textbook_1065", "title": "363. EXAMPLE 4.5 - ", "content": ""} {"id": "textbook_1066", "title": "364. Naming Covalent Compounds - ", "content": "\nName the following covalent compounds:\n(a) $\\mathrm{SF}_{6}$\n(b) $\\mathrm{N}_{2} \\mathrm{O}_{3}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{O}_{7}$\n(d) $\\mathrm{P}_{4} \\mathrm{O}_{6}$\n"} {"id": "textbook_1067", "title": "365. Solution - ", "content": "\nBecause these compounds consist solely of nonmetals, we use prefixes to designate the number of atoms of each element:\n(a) sulfur hexafluoride\n(b) dinitrogen trioxide\n(c) dichlorine heptoxide\n(d) tetraphosphorus hexoxide\n"} {"id": "textbook_1068", "title": "366. Check Your Learning - ", "content": "\nWrite the formulas for the following compounds:\n(a) phosphorus pentachloride\n(b) dinitrogen monoxide\n(c) iodine heptafluoride\n(d) carbon tetrachloride\n"} {"id": "textbook_1069", "title": "367. Answer: - ", "content": "\n(a) $\\mathrm{PCl}_{5}$; (b) $\\mathrm{N}_{2} \\mathrm{O}$; (c) $\\mathrm{IF}_{7}$; (d) $\\mathrm{CCl}_{4}$\n"} {"id": "textbook_1070", "title": "368. LINK TO LEARNING - ", "content": "\nThe following website (http://openstax.org/l/16chemcompname) provides practice with naming chemical compounds and writing chemical formulas. You can choose binary, polyatomic, and variable charge ionic compounds, as well as molecular compounds.\n"} {"id": "textbook_1071", "title": "369. Binary Acids - ", "content": "\nSome compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, $\\mathrm{H}^{+}$, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound's name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):"} {"id": "textbook_1072", "title": "369. Binary Acids - ", "content": "1. The word \"hydrogen\" is changed to the prefix hydro-\n2. The other nonmetallic element name is modified by adding the suffix -ic\n3. The word \"acid\" is added as a second word"} {"id": "textbook_1073", "title": "369. Binary Acids - ", "content": "For example, when the gas HCl (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table 4.8."} {"id": "textbook_1074", "title": "369. Binary Acids - ", "content": "Names of Some Simple Acids"} {"id": "textbook_1075", "title": "369. Binary Acids - ", "content": "| Name of Gas | Name of Acid |\n| :--- | :--- |\n| $\\mathrm{HF}(g)$, hydrogen fluoride | $\\mathrm{HF}(a q)$, hydrofluoric acid |\n| $\\mathrm{HCl}(g)$, hydrogen chloride | $\\mathrm{HCl}(a q)$, hydrochloric acid |\n| $\\mathrm{HBr}(g)$, hydrogen bromide | $\\mathrm{HBr}(\\mathrm{aq})$, hydrobromic acid |\n| $\\mathrm{HI}(g)$, hydrogen iodide | $\\mathrm{HI}(\\mathrm{aq})$, hydroiodic acid |\n| $\\mathrm{H}_{2} \\mathrm{~S}(g)$, hydrogen sulfide | $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq})$, hydrosulfuric acid |"} {"id": "textbook_1076", "title": "369. Binary Acids - ", "content": "TABLE 4.8\n"} {"id": "textbook_1077", "title": "370. Oxyacids - ", "content": "\nMany compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly\ndiscuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:"} {"id": "textbook_1078", "title": "370. Oxyacids - ", "content": "1. Omit \"hydrogen\"\n2. Start with the root name of the anion\n3. Replace -ate with -ic, or -ite with -ous\n4. Add \"acid\""} {"id": "textbook_1079", "title": "370. Oxyacids - ", "content": "For example, consider $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ (which you might be tempted to call \"hydrogen carbonate\"). To name this correctly, \"hydrogen\" is omitted; the -ate of carbonate is replace with -ic; and acid is added-so its name is carbonic acid. Other examples are given in Table 4.9. There are some exceptions to the general naming method (e.g., $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is called sulfuric acid, not sulfic acid, and $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ is sulfurous, not sulfous, acid)."} {"id": "textbook_1080", "title": "370. Oxyacids - ", "content": "Names of Common Oxyacids"} {"id": "textbook_1081", "title": "370. Oxyacids - ", "content": "| Formula | Anion Name | Acid Name |\n| :--- | :--- | :--- |\n| $\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ | acetate | acetic acid |\n| $\\mathrm{HNO}_{3}$ | nitrate | nitric acid |\n| $\\mathrm{HNO}_{2}$ | nitrite | nitrous acid |\n| $\\mathrm{HClO}_{4}$ | perchlorate | perchloric acid |\n| $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ | carbonate | carbonic acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfate | sulfuric acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ | sulfite | sulfurous acid |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ | phosphate | phosphoric acid |"} {"id": "textbook_1082", "title": "370. Oxyacids - ", "content": "TABLE 4.9\n"} {"id": "textbook_1083", "title": "370. Oxyacids - 370.1. Lewis Symbols and Structures", "content": ""} {"id": "textbook_1084", "title": "371. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1085", "title": "371. LEARNING OBJECTIVES - ", "content": "- Write Lewis symbols for neutral atoms and ions\n- Draw Lewis structures depicting the bonding in simple molecules"} {"id": "textbook_1086", "title": "371. LEARNING OBJECTIVES - ", "content": "Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.\n"} {"id": "textbook_1087", "title": "372. Lewis Symbols - ", "content": "\nWe use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:"} {"id": "textbook_1088", "title": "372. Lewis Symbols - ", "content": "- Ca"} {"id": "textbook_1089", "title": "372. Lewis Symbols - ", "content": "Figure 4.10 shows the Lewis symbols for the elements of the third period of the periodic table."} {"id": "textbook_1090", "title": "372. Lewis Symbols - ", "content": "| Atoms | Electronic Configuration | Lewis Symbol |\n| :--- | :--- | :--- |\n| sodium | $[\\mathrm{Ne}] 3 s^{1}$ | $\\mathrm{Na} \\cdot$ |\n| magnesium | $[\\mathrm{Ne}] 3 s^{2}$ | $\\cdot \\mathrm{Mg} \\cdot$ |\n| aluminum | $[\\mathrm{Ne}] 3 s^{2} 3 p^{1}$ | $\\cdot \\dot{\\mathrm{Al} \\cdot} \\cdot$ |\n| silicon | $[\\mathrm{Ne}] 3 s^{2} 3 p^{2}$ | $\\cdot \\dot{\\mathrm{Si}} \\cdot$ |\n| phosphorus | $[\\mathrm{Ne}] 3 s^{2} 3 p^{3}$ | $\\ddot{\\mathrm{P}} \\cdot$ |\n| sulfur | $[\\mathrm{Ne}] 3 s^{2} 3 p^{4}$ | $\\ddot{\\mathrm{~S}} \\cdot$ |\n| chlorine | $[\\mathrm{Ne}] 3 s^{2} 3 p^{5}$ | $: \\ddot{\\mathrm{Cl}} \\cdot$ |\n| argon | $[\\mathrm{Ne}] 3 s^{2} 3 p^{6}$ | $: \\ddot{\\mathrm{Ar}}:$ |"} {"id": "textbook_1091", "title": "372. Lewis Symbols - ", "content": "FIGURE 4.10 Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table."} {"id": "textbook_1092", "title": "372. Lewis Symbols - ", "content": "Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:\n\\$\\underset{\\substack{sodium
\natom}}{\\mathrm{Na} \\cdot \\underset{\\substack{sodium
"} {"id": "textbook_1093", "title": "372. Lewis Symbols - ", "content": "cation}}{\\mathrm{Na}^{+}}+\\mathrm{e}^{-}}\\$| calcium |\n| :---: |\n| atom |\\$>\\underset{\\substack{calcium
"} {"id": "textbook_1094", "title": "372. Lewis Symbols - ", "content": "cation}}{\\mathrm{Ca}^{2+}}+\\mathrm{Ce}^{-}\\$"} {"id": "textbook_1095", "title": "372. Lewis Symbols - ", "content": "Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:\n"} {"id": "textbook_1096", "title": "372. Lewis Symbols - ", "content": "Figure 4.11 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds."} {"id": "textbook_1097", "title": "372. Lewis Symbols - ", "content": "| Metal | | Nonmetal | Ionic Compound |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Na} \\cdot$
sodium atom | $+$ | $\\ddot{\\mathrm{Cl}}$
chlorine atom | $\\longrightarrow \\quad \\mathrm{Na}^{+}[: \\stackrel{\\ddot{\\mathrm{C}} \\cdot:}{\\cdot .}]^{-}$
sodium chloride (sodium ion and chloride ion) |\n| -Mg\u2022
magnesium atom | + | $\\ddot{O}$
oxygen atom | $\\longrightarrow \\quad \\mathrm{Mg}^{2+}[: \\ddot{\\mathrm{O}}:]^{2-}$
magnesium oxide (magnesium ion and oxide ion) |\n| - Ca\u2022
calcium atom | $+$ | $2: \\ddot{F}$
fluorine atoms | $\\longrightarrow \\quad \\mathrm{Ca}^{2+}[: \\ddot{\\mathrm{F}}:]_{2}^{-}$
calcium fluoride (calcium ion and two fluoride ions) |"} {"id": "textbook_1098", "title": "372. Lewis Symbols - ", "content": "FIGURE 4.11 Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change.\n"} {"id": "textbook_1099", "title": "373. Lewis Structures - ", "content": "\nWe also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:\n"} {"id": "textbook_1100", "title": "373. Lewis Structures - ", "content": "The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:\n\n\nA single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.\n"} {"id": "textbook_1101", "title": "374. The Octet Rule - ", "content": "\nThe other halogen molecules ( $\\mathrm{F}_{2}, \\mathrm{Br}_{2}, \\mathrm{I}_{2}$, and $\\mathrm{At}_{2}$ ) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule."} {"id": "textbook_1102", "title": "374. The Octet Rule - ", "content": "The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table ( $\\mathrm{C}, \\mathrm{N}, \\mathrm{O}$, and F ). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in $\\mathrm{CCl}_{4}$ (carbon tetrachloride) and silicon in $\\mathrm{SiH}_{4}$ (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:\n"} {"id": "textbook_1103", "title": "374. The Octet Rule - ", "content": "Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in $\\mathrm{NH}_{3}$ (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:\n"} {"id": "textbook_1104", "title": "374. The Octet Rule - ", "content": "Double and Triple Bonds\nAs previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in $\\mathrm{CH}_{2} \\mathrm{O}$ (formaldehyde) and between the two carbon atoms in $\\mathrm{C}_{2} \\mathrm{H}_{4}$ (ethylene):\n"} {"id": "textbook_1105", "title": "374. The Octet Rule - ", "content": "A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion $\\left(\\mathrm{CN}^{-}\\right)$:"} {"id": "textbook_1106", "title": "374. The Octet Rule - ", "content": "```\n:C:::O: or :C\\equivO: :C:::N\\overline{* or :C\\equivN:}\n```"} {"id": "textbook_1107", "title": "374. The Octet Rule - ", "content": "carbon monoxide cyanide ion\n"} {"id": "textbook_1108", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "\nFor very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:\n"} {"id": "textbook_1109", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:"} {"id": "textbook_1110", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.\n2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).\n3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an\noctet around each atom.\n4. Place all remaining electrons on the central atom.\n5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible."} {"id": "textbook_1111", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "Let us determine the Lewis structures of $\\mathrm{SiH}_{4}, \\mathrm{CHO}_{2}{ }^{-}, \\mathrm{NO}^{+}$, and $\\mathrm{OF}_{2}$ as examples in following this procedure:"} {"id": "textbook_1112", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "1. Determine the total number of valence (outer shell) electrons in the molecule or ion."} {"id": "textbook_1113", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- For a molecule, we add the number of valence electrons on each atom in the molecule:"} {"id": "textbook_1114", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "> | $\\mathrm{SiH}_{4}$ |\n| :--- |\n| $\\mathrm{Si}: 4$ valence electrons/atom $\\times 1$ atom $=4$ |\n| $+\\mathrm{H}: 1$ valence electron $/$ atom $\\times 4$ atoms $=4$ |"} {"id": "textbook_1115", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "$$\n=8 \\text { valence electrons }\n$$"} {"id": "textbook_1116", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- For a negative ion, such as $\\mathrm{CHO}_{2}{ }^{-}$, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):"} {"id": "textbook_1117", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{CHO}_{2}{ }^{-} \\\\\n& \\mathrm{C}: 4 \\text { valence electrons/atom } \\times 1 \\text { atom }=4 \\\\\n& \\mathrm{H}: 1 \\text { valence electron/atom } \\times 1 \\text { atom }=1 \\\\\n& \\mathrm{O}: 6 \\text { valence electrons/atom } \\times 2 \\text { atoms }=12 \\\\\n& +\\quad 1 \\text { additional electron }=1 \\\\\n& \\hline\n\\end{aligned}\n$$"} {"id": "textbook_1118", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- For a positive ion, such as $\\mathrm{NO}^{+}$, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:\n\n$$\n\\begin{aligned}\n& \\mathrm{NO}^{+} \\\\\n& \\mathrm{N}: 5 \\text { valence electrons/atom } \\times 1 \\text { atom }=5 \\\\\n& \\\\\n& \\begin{aligned}\n\\mathrm{O}: 6 \\text { valence electron/atom } \\times 1 \\text { atom } & =6 \\\\\n+-1 \\text { electron (positive charge) } & =-1 \\\\\n\\hline & =10 \\text { valence electrons }\n\\end{aligned}\n\\end{aligned}\n$$"} {"id": "textbook_1119", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- Since $\\mathrm{OF}_{2}$ is a neutral molecule, we simply add the number of valence electrons:\n\n$$\n\\begin{aligned}\n& \\mathrm{OF}_{2} \\\\\n& \\mathrm{O}: 6 \\text { valence electrons/atom } \\times 1 \\text { atom }=6 \\\\\n&+\\mathrm{F}: 7 \\text { valence electrons/atom } \\times 2 \\text { atoms }=14 \\\\\n& \\hline=20 \\text { valence electrons }\n\\end{aligned}\n$$"} {"id": "textbook_1120", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)\n"} {"id": "textbook_1121", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "When several arrangements of atoms are possible, as for $\\mathrm{CHO}_{2}{ }^{-}$, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In $\\mathrm{CHO}_{2}{ }^{-}$, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in $\\mathrm{POCl}_{3}, \\mathrm{~S}$ in $\\mathrm{SO}_{2}$, and Cl in $\\mathrm{ClO}_{4}{ }^{-}$. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.\n3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons."} {"id": "textbook_1122", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- There are no remaining electrons on $\\mathrm{SiH}_{4}$, so it is unchanged:\n\n"} {"id": "textbook_1123", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "\n4. Place all remaining electrons on the central atom."} {"id": "textbook_1124", "title": "375. Writing Lewis Structures with the Octet Rule - ", "content": "- For $\\mathrm{SiH}_{4}, \\mathrm{CHO}_{2}{ }^{-}$, and $\\mathrm{NO}^{+}$, there are no remaining electrons; we already placed all of the electrons determined in Step 1.\n- For $\\mathrm{OF}_{2}$, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:\n
regions | Two regions of
high electron
density (bonds
and/or
unshared
pairs) | Three regions
of high
electron
density (bonds
and/or
unshared
pairs) | Four regions
of high
electron
density (bonds
and/or
unshared
pairs) | Five regions of
high electron
density (bonds
and/or
unshared
pairs) | Six regions of
high electron
density (bonds
and/or
unshared
pairs) |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| Spatial
arrangement | | | | | |"} {"id": "textbook_1221", "title": "412. VSEPR Theory - ", "content": "FIGURE 4.16 The basic electron-pair geometries predicted by VSEPR theory maximize the space around any region of electron density (bonds or lone pairs).\n"} {"id": "textbook_1222", "title": "413. Electron-pair Geometry versus Molecular Structure - ", "content": "\nIt is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure 4.16 describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.\n\nWe differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom."} {"id": "textbook_1223", "title": "413. Electron-pair Geometry versus Molecular Structure - ", "content": "For example, the methane molecule, $\\mathrm{CH}_{4}$, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure 4.17). On the other hand, the ammonia molecule, $\\mathrm{NH}_{3}$, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure 4.18).\n
Linear | | | | |\n| 3 | 
Trigonal planar | 
Bent or angular | | | |\n| 4 |
Tetrahedral | Trigonal pyramid | Bent or angular | | |\n| 5 | 
Trigonal bipyramid | 
Sawhorse or seesaw | 
T-shape | 
Linear | |\n| 6 | 
Octahedral |  | 
Square planar | 
T-shape | Linear |"} {"id": "textbook_1232", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "FIGURE 4.19 The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry."} {"id": "textbook_1233", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "According to VSEPR theory, the terminal atom locations (Xs in Figure 4.19) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct $X$ positions, as shown in Figure 4.20: an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure 4.19, the axial position is surrounded by bond angles of $90^{\\circ}$, whereas the equatorial position has more space available because of the $120^{\\circ}$ bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs."} {"id": "textbook_1234", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the $\\mathrm{ClF}_{3}$ molecule (Figure 4.20). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.\n\n\nFIGURE 4.20 (a) In a trigonal bipyramid, the two axial positions are located directly across from one another, whereas the three equatorial positions are located in a triangular arrangement. (b-d) The two lone pairs (red lines) in $\\mathrm{ClF}_{3}$ have several possible arrangements, but the T -shaped molecular structure (b) is the one actually observed, consistent with the larger lone pairs both occupying equatorial positions."} {"id": "textbook_1235", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electronpair geometry. The two lone pairs are on opposite sides of the octahedron ( $180^{\\circ}$ apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure 4.19)."} {"id": "textbook_1236", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "Predicting Electron Pair Geometry and Molecular Structure\nThe following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:"} {"id": "textbook_1237", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "1. Write the Lewis structure of the molecule or polyatomic ion.\n2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.\n3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure 4.19, first column).\n4. Use the number of lone pairs to determine the molecular structure (Figure 4.19). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom."} {"id": "textbook_1238", "title": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - ", "content": "The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry.\n"} {"id": "textbook_1239", "title": "415. EXAMPLE 4.11 - ", "content": ""} {"id": "textbook_1240", "title": "416. Predicting Electron-pair Geometry and Molecular Structure: $\\mathbf{C O}_{\\mathbf{2}}$ and $\\mathbf{B C l}_{\\mathbf{3}}$ - ", "content": "\nPredict the electron-pair geometry and molecular structure for each of the following:\n(a) carbon dioxide, $\\mathrm{CO}_{2}$, a molecule produced by the combustion of fossil fuels\n(b) boron trichloride, $\\mathrm{BCl}_{3}$, an important industrial chemical\n"} {"id": "textbook_1241", "title": "417. Solution - ", "content": "\n(a) We write the Lewis structure of $\\mathrm{CO}_{2}$ as:\n"} {"id": "textbook_1242", "title": "417. Solution - ", "content": "This shows us two regions of high electron density around the carbon atom-each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of $180^{\\circ}$. The\nelectron-pair geometry and molecular structure are identical, and $\\mathrm{CO}_{2}$ molecules are linear.\n(b) We write the Lewis structure of $\\mathrm{BCl}_{3}$ as:\n
Advanced Theories of Bonding - ", "content": "\n"} {"id": "textbook_1341", "title": "463. CHAPTER 5
Advanced Theories of Bonding - ", "content": "Figure 5.1 Oxygen molecules orient randomly most of the time, as shown in the top magnified view. However, when we pour liquid oxygen through a magnet, the molecules line up with the magnetic field, and the attraction allows them to stay suspended between the poles of the magnet where the magnetic field is strongest. Other diatomic molecules (like $\\mathrm{N}_{2}$ ) flow past the magnet. The detailed explanation of bonding described in this chapter allows us to understand this phenomenon. (credit: modification of work by Jefferson Lab)\n"} {"id": "textbook_1342", "title": "464. CHAPTER OUTLINE - ", "content": "\n5.1 Valence Bond Theory\n5.2 Hybrid Atomic Orbitals\n5.3 Multiple Bonds\n5.4 Molecular Orbital Theory"} {"id": "textbook_1343", "title": "464. CHAPTER OUTLINE - ", "content": "INTRODUCTION We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of $s, p$, and $d$ atomic orbitals with molecular shapes that show angles like $120^{\\circ}$ and $109.5^{\\circ}$ ? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both $\\mathrm{N}_{2}$ and $\\mathrm{O}_{2}$ have fairly similar Lewis structures that contain lone pairs of electrons."} {"id": "textbook_1344", "title": "464. CHAPTER OUTLINE - ", "content": "```\n:N\u4e09N: :\n```"} {"id": "textbook_1345", "title": "464. CHAPTER OUTLINE - ", "content": "Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen (shown in Figure 5.1) is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations.\n"} {"id": "textbook_1346", "title": "464. CHAPTER OUTLINE - 464.1. Valence Bond Theory", "content": ""} {"id": "textbook_1347", "title": "465. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1348", "title": "465. LEARNING OBJECTIVES - ", "content": "- Describe the formation of covalent bonds in terms of atomic orbital overlap\n- Define and give examples of $\\sigma$ and $\\pi$ bonds"} {"id": "textbook_1349", "title": "465. LEARNING OBJECTIVES - ", "content": "As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts threedimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding.\n"} {"id": "textbook_1350", "title": "466. Atomic Orbital Overlap - ", "content": "\nThere are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an $s$ orbital, a dumbbell shape for a $p$ orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization."} {"id": "textbook_1351", "title": "466. Atomic Orbital Overlap - ", "content": "Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms' positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap."} {"id": "textbook_1352", "title": "466. Atomic Orbital Overlap - ", "content": "The energy of the system depends on how much the orbitals overlap. Figure 5.2 illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure 5.2.\n"} {"id": "textbook_1353", "title": "466. Atomic Orbital Overlap - ", "content": "FIGURE 5.2 (a) The interaction of two hydrogen atoms changes as a function of distance. (b) The energy of the system changes as the atoms interact. The lowest (most stable) energy occurs at a distance of 74 pm , which is the bond length observed for the $\\mathrm{H}_{2}$ molecule."} {"id": "textbook_1354", "title": "466. Atomic Orbital Overlap - ", "content": "In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two $s$ orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure 5.3 illustrates this for two $p$ orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.\n\n(a)\n\n(b)\n\nFIGURE 5.3 (a) The overlap of two $p$ orbitals is greatest when the orbitals are directed end to end. (b) Any other arrangement results in less overlap. The dots indicate the locations of the nuclei."} {"id": "textbook_1355", "title": "466. Atomic Orbital Overlap - ", "content": "The overlap of two $s$ orbitals (as in $\\mathrm{H}_{2}$ ), the overlap of an $s$ orbital and a $p$ orbital (as in HCl ), and the end-to-end overlap of two $p$ orbitals (as in $\\mathrm{Cl}_{2}$ ) all produce sigma bonds ( $\\boldsymbol{\\sigma}$ bonds), as illustrated in Figure 5.4. A $\\sigma$ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as $\\sigma$ bonds in valence bond theory.\n\n\nFIGURE 5.4 Sigma ( $\\sigma$ ) bonds form from the overlap of the following: (a) two $s$ orbitals, (b) an $s$ orbital and a $p$ orbital, and (c) two $p$ orbitals. The dots indicate the locations of the nuclei."} {"id": "textbook_1356", "title": "466. Atomic Orbital Overlap - ", "content": "A pi bond ( $\\boldsymbol{\\pi}$ bond) is a type of covalent bond that results from the side-by-side overlap of two $p$ orbitals, as illustrated in Figure 5.5. In a $\\pi$ bond, the regions of orbital overlap lie on opposite sides of the internuclear axis."} {"id": "textbook_1357", "title": "466. Atomic Orbital Overlap - ", "content": "Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.\n"} {"id": "textbook_1358", "title": "466. Atomic Orbital Overlap - ", "content": "FIGURE 5.5 $\\mathrm{Pi}(\\pi)$ bonds form from the side-by-side overlap of two $p$ orbitals. The dots indicate the location of the nuclei."} {"id": "textbook_1359", "title": "466. Atomic Orbital Overlap - ", "content": "While all single bonds are $\\sigma$ bonds, multiple bonds consist of both $\\sigma$ and $\\pi$ bonds. As the Lewis structures below suggest, $\\mathrm{O}_{2}$ contains a double bond, and $\\mathrm{N}_{2}$ contains a triple bond. The double bond consists of one $\\sigma$ bond and one $\\pi$ bond, and the triple bond consists of one $\\sigma$ bond and two $\\pi$ bonds. Between any two atoms, the first bond formed will always be a $\\sigma$ bond, but there can only be one $\\sigma$ bond in any one location. In any multiple bond, there will be one $\\sigma$ bond, and the remaining one or two bonds will be $\\pi$ bonds. These bonds are described in more detail later in this chapter.\n
atoms | considers electrons delocalized throughout the
entire molecule |\n| creates bonds from overlap of atomic orbitals ( $s, p, d . .)$.
and hybrid orbitals $\\left(s p, s p^{2}, s p^{3} \\ldots\\right)$ | combines atomic orbitals to form molecular
orbitals $\\left(\\sigma, \\sigma^{*}, \\pi, \\pi^{*}\\right)$ |\n| forms $\\sigma$ or $\\pi$ bonds | creates bonding and antibonding interactions
based on which orbitals are filled |\n| predicts molecular shape based on the number of
regions of electron density | predicts the arrangement of electrons in
molecules |\n| needs multiple structures to describe resonance | |"} {"id": "textbook_1466", "title": "503. LINK TO LEARNING - ", "content": "TABLE 5.1"} {"id": "textbook_1467", "title": "503. LINK TO LEARNING - ", "content": "Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, $\\Psi$, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital ( $\\Psi^{2}$ ). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin."} {"id": "textbook_1468", "title": "503. LINK TO LEARNING - ", "content": "We will consider the molecular orbitals in molecules composed of two identical atoms ( $\\mathrm{H}_{2}$ or $\\mathrm{Cl}_{2}$, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur."} {"id": "textbook_1469", "title": "503. LINK TO LEARNING - ", "content": "The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure 5.28). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.\n"} {"id": "textbook_1470", "title": "503. LINK TO LEARNING - ", "content": "FIGURE 5.28 (a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude."} {"id": "textbook_1471", "title": "503. LINK TO LEARNING - ", "content": "There are two types of molecular orbitals that can form from the overlap of two atomic $s$ orbitals on adjacent atoms. The two types are illustrated in Figure 5.29. The in-phase combination produces a lower energy $\\boldsymbol{\\sigma}_{\\boldsymbol{s}}$\nmolecular orbital (read as \"sigma-s\") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $\\sigma_{s}^{*}$ molecular orbital (read as \"sigma-s-star\") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a $\\sigma_{S}$ orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $\\sigma_{s}^{*}$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.\n\n\nFIGURE 5.29 Sigma ( $\\sigma$ ) and sigma-star ( $\\sigma^{*}$ ) molecular orbitals are formed by the combination of two $s$ atomic orbitals. The dots $(\\cdot)$ indicate the locations of nuclei.\n"} {"id": "textbook_1472", "title": "504. LINK TO LEARNING - ", "content": "\nYou can watch animations (http://openstax.org/l/16molecorbital) visualizing the calculated atomic orbitals combining to form various molecular orbitals at the Orbitron website."} {"id": "textbook_1473", "title": "504. LINK TO LEARNING - ", "content": "In $p$ orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a twodimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When $p$ orbitals overlap end to end, they create $\\sigma$ and $\\sigma^{*}$ orbitals (Figure 5.30). If two atoms are located along the $x$-axis in a Cartesian coordinate system, the two $p_{x}$ orbitals overlap end to end and form $\\sigma_{p x}$ (bonding) and $\\sigma_{p x}^{*}$ (antibonding) (read as \"sigma-p-x\" and \"sigma-p-x star,\" respectively). Just as with $s$-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.\n\n\nFIGURE 5.30 Combining wave functions of two $p$ atomic orbitals along the internuclear axis creates two molecular orbitals, $\\sigma_{p}$ and $\\sigma_{p}^{*}$.\nThe side-by-side overlap of two $p$ orbitals gives rise to a pi ( $\\boldsymbol{\\pi}$ ) bonding molecular orbital and a $\\boldsymbol{\\pi}^{*}$ *\nantibonding molecular orbital, as shown in Figure 5.31. In valence bond theory, we describe $\\pi$ bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the $p$ orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the $\\pi$ orbital by this same\nshape, and a $\\pi$ bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.\n\n\nFIGURE 5.31 Side-by-side overlap of each two $p$ orbitals results in the formation of two $\\pi$ molecular orbitals. Combining the out-of-phase orbitals results in an antibonding molecular orbital with two nodes. One contains the internuclear axis, and one is perpendicular to the axis. Combining the in-phase orbitals results in a bonding orbital. There is a node (blue) containing the internuclear axis with the two lobes of the orbital located above and below this node.\n\nIn the molecular orbitals of diatomic molecules, each atom also has two sets of $p$ orbitals oriented side by side ( $p_{y}$ and $p_{z}$ ), so these four atomic orbitals combine pairwise to create two $\\pi$ orbitals and two $\\pi^{*}$ orbitals. The $\\pi_{p y}$ and $\\pi_{p y}^{*}$ orbitals are oriented at right angles to the $\\pi_{p z}$ and $\\pi_{p z}^{*}$ orbitals. Except for their orientation, the $\\pi_{p y}$ and $\\pi_{p z}$ orbitals are identical and have the same energy; they are degenerate orbitals. The $\\pi_{p y}^{*}$ and $\\pi_{p z}^{*}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic $p$ orbitals in two atoms: $\\sigma_{p x}$ and $\\sigma_{p x}^{*}, \\pi_{p y}$ and $\\pi_{p y}^{*}, \\pi_{p z}$ and $\\pi_{p z}^{*}$.\n"} {"id": "textbook_1474", "title": "505. EXAMPLE 5.6 - ", "content": ""} {"id": "textbook_1475", "title": "506. Molecular Orbitals - ", "content": "\nPredict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.\n\n"} {"id": "textbook_1476", "title": "507. Solution - ", "content": "\n(a) is an in-phase combination, resulting in a $\\sigma_{3 p}$ orbital\n(b) will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.\n(c) is an out-of-phase combination, resulting in a $\\pi_{3 p}^{*}$ orbital.\n"} {"id": "textbook_1477", "title": "508. Check Your Learning - ", "content": "\nLabel the molecular orbital shown as $\\sigma$ or $\\pi$, bonding or antibonding and indicate where the node occurs.\n\n"} {"id": "textbook_1478", "title": "509. Answer: - ", "content": "\nThe orbital is located along the internuclear axis, so it is a $\\sigma$ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.\n"} {"id": "textbook_1479", "title": "510. Portrait of a Chemist - ", "content": ""} {"id": "textbook_1480", "title": "511. Walter Kohn: Nobel Laureate - ", "content": "\nWalter Kohn (Figure 5.32) is a theoretical physicist who studies the electronic structure of solids. His work combines the principles of quantum mechanics with advanced mathematical techniques. This technique, called density functional theory, makes it possible to compute properties of molecular orbitals, including their shape and energies. Kohn and mathematician John Pople were awarded the Nobel Prize in Chemistry in 1998 for their contributions to our understanding of electronic structure. Kohn also made significant contributions to the physics of semiconductors.\n"} {"id": "textbook_1481", "title": "511. Walter Kohn: Nobel Laureate - ", "content": "FIGURE 5.32 Walter Kohn developed methods to describe molecular orbitals. (credit: image courtesy of Walter Kohn)"} {"id": "textbook_1482", "title": "511. Walter Kohn: Nobel Laureate - ", "content": "Kohn's biography has been remarkable outside the realm of physical chemistry as well. He was born in Austria, and during World War II he was part of the Kindertransport program that rescued 10,000 children from the Nazi regime. His summer jobs included discovering gold deposits in Canada and helping Polaroid explain how its instant film worked. Dr. Kohn passed away in 2016 at the age of 93.\n"} {"id": "textbook_1483", "title": "512. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_1484", "title": "513. Computational Chemistry in Drug Design - ", "content": "\nWhile the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (see Figure 5.33). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.\n"} {"id": "textbook_1485", "title": "513. Computational Chemistry in Drug Design - ", "content": "FIGURE 5.33 The molecule shown, HIV-1 protease, is an important target for pharmaceutical research. By designing molecules that bind to this protein, scientists are able to drastically inhibit the progress of the disease.\n"} {"id": "textbook_1486", "title": "514. Molecular Orbital Energy Diagrams - ", "content": "\nThe relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure 5.34). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six $2 p$ atomic orbitals results in three bonding orbitals (one $\\sigma$ and two $\\pi$ ) and three antibonding orbitals (one $\\sigma^{*}$ and two $\\pi^{*}$ )."} {"id": "textbook_1487", "title": "514. Molecular Orbital Energy Diagrams - ", "content": "We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure 5.34). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\\mathrm{Be}_{2}{ }^{+}$) would have the molecular electron configuration $\\left(\\sigma_{1 s}\\right)^{2}\\left(\\sigma_{1 s}^{*}\\right)^{2}\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{1}$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.\n"} {"id": "textbook_1488", "title": "514. Molecular Orbital Energy Diagrams - ", "content": "FIGURE 5.34 This is the molecular orbital diagram for the homonuclear diatomic $\\mathrm{Be}_{2}{ }^{+}$, showing the molecular orbitals of the valence shell only. The molecular orbitals are filled in the same manner as atomic orbitals, using the Aufbau principle and Hund's rule.\n"} {"id": "textbook_1489", "title": "515. Bond Order - ", "content": "\nThe filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons."} {"id": "textbook_1490", "title": "515. Bond Order - ", "content": "When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1 , a double bond has a bond order of 2 , and a triple bond has a bond order of 3 . We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.\n\nIn the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:\n\n$$\n\\text { bond order }=\\frac{\\text { (number of bonding electrons) }- \\text { (number of antibonding electrons) }}{2}\n$$"} {"id": "textbook_1491", "title": "515. Bond Order - ", "content": "The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.\n"} {"id": "textbook_1492", "title": "516. Bonding in Diatomic Molecules - ", "content": "\nA dihydrogen molecule $\\left(\\mathrm{H}_{2}\\right)$ forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the $\\sigma_{1 s}$ bonding orbital. A dihydrogen molecule, $\\mathrm{H}_{2}$, readily forms because the energy of a $\\mathrm{H}_{2}$ molecule is lower than that of two H atoms. The $\\sigma_{1 s}$ orbital that contains both electrons is lower in energy than either of the two 1 s atomic orbitals."} {"id": "textbook_1493", "title": "516. Bonding in Diatomic Molecules - ", "content": "A molecular orbital can hold two electrons, so both electrons in the $\\mathrm{H}_{2}$ molecule are in the $\\sigma_{1 s}$ bonding orbital; the electron configuration is $\\left(\\sigma_{1 s}\\right)^{2}$. We represent this configuration by a molecular orbital energy diagram (Figure 5.35) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.\n"} {"id": "textbook_1494", "title": "516. Bonding in Diatomic Molecules - ", "content": "FIGURE 5.35 The molecular orbital energy diagram predicts that $\\mathrm{H}_{2}$ will be a stable molecule with lower energy than the separated atoms."} {"id": "textbook_1495", "title": "516. Bonding in Diatomic Molecules - ", "content": "A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have"} {"id": "textbook_1496", "title": "516. Bonding in Diatomic Molecules - ", "content": "$$\n\\text { bond order in } \\mathrm{H}_{2}=\\frac{(2-0)}{2}=1\n$$"} {"id": "textbook_1497", "title": "516. Bonding in Diatomic Molecules - ", "content": "Because the bond order for the $\\mathrm{H}-\\mathrm{H}$ bond is equal to 1 , the bond is a single bond.\nA helium atom has two electrons, both of which are in its $1 s$ orbital. Two helium atoms do not combine to form a dihelium molecule, $\\mathrm{He}_{2}$, with four electrons, because the stabilizing effect of the two electrons in the lowerenergy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of $\\mathrm{He}_{2}$ as $\\left(\\sigma_{1 s}\\right)^{2}\\left(\\sigma_{1 s}^{*}\\right)^{2}$ as in Figure 5.36. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.\n\n$$\n\\text { bond order in } \\mathrm{He}_{2}=\\frac{(2-2)}{2}=0\n$$"} {"id": "textbook_1498", "title": "516. Bonding in Diatomic Molecules - ", "content": "A bond order of zero indicates that no bond is formed between two atoms.\n"} {"id": "textbook_1499", "title": "516. Bonding in Diatomic Molecules - ", "content": "FIGURE 5.36 The molecular orbital energy diagram predicts that $\\mathrm{He}_{2}$ will not be a stable molecule, since it has equal numbers of bonding and antibonding electrons."} {"id": "textbook_1500", "title": "516. Bonding in Diatomic Molecules - ", "content": "The Diatomic Molecules of the Second Period\nEight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: $\\mathrm{Li}_{2}, \\mathrm{Be}_{2}, \\mathrm{~B}_{2}, \\mathrm{C}_{2}, \\mathrm{~N}_{2}, \\mathrm{O}_{2}, \\mathrm{~F}_{2}$, and $\\mathrm{Ne}_{2}$. However, we can predict that the $\\mathrm{Be}_{2}$ molecule and the $\\mathrm{Ne}_{2}$ molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table 5.2)."} {"id": "textbook_1501", "title": "516. Bonding in Diatomic Molecules - ", "content": "We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies."} {"id": "textbook_1502", "title": "516. Bonding in Diatomic Molecules - ", "content": "Consistent with Hund's rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place."} {"id": "textbook_1503", "title": "516. Bonding in Diatomic Molecules - ", "content": "As we saw in valence bond theory, $\\sigma$ bonds are generally more stable than $\\pi$ bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, $\\sigma$ orbitals are usually more stable than $\\pi$ orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure 5.37. Looking at $\\mathrm{Ne}_{2}$ molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the $p$ orbitals (Li through N) we observe a different pattern, in which the $\\sigma_{p}$ orbital is higher in energy than the $\\pi_{p}$ set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.\n"} {"id": "textbook_1504", "title": "516. Bonding in Diatomic Molecules - ", "content": "FIGURE 5.37 This shows the MO diagrams for each homonuclear diatomic molecule in the second period. The orbital energies decrease across the period as the effective nuclear charge increases and atomic radius decreases. Between $\\mathrm{N}_{2}$ and $\\mathrm{O}_{2}$, the order of the orbitals changes."} {"id": "textbook_1505", "title": "516. Bonding in Diatomic Molecules - ", "content": "This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The $\\sigma_{\\mathrm{s}}$ wavefunction mathematically combines with the $\\sigma_{p}$ wavefunction, with the result that the $\\sigma_{s}$ orbital becomes more stable, and the $\\sigma_{p}$ orbital becomes less stable (Figure 5.38). Similarly, the antibonding orbitals also undergo s-p mixing, with the $\\sigma_{\\mathrm{s}^{*}}$ becoming more stable and the $\\sigma_{\\mathrm{p}^{*}}$ becoming less stable.\n\n\nFIGURE 5.38 Without mixing, the MO pattern occurs as expected, with the $\\sigma_{p}$ orbital lower in energy than the $\\pi_{p}$ orbitals. When s-p mixing occurs, the orbitals shift as shown, with the $\\sigma_{p}$ orbital higher in energy than the $\\pi_{p}$ orbitals. $s$-p mixing occurs when the $s$ and $p$ orbitals have similar energies. The energy difference between $2 s$ and $2 p$ orbitals in $\\mathrm{O}, \\mathrm{F}$, and Ne is greater than that in Li, Be, B, C, and N. Because of this, $\\mathrm{O}_{2}, \\mathrm{~F}_{2}$, and $\\mathrm{Ne}_{2}$ exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure 5.37. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the $\\sigma_{p}$ orbital is raised above the $\\pi_{p}$ set."} {"id": "textbook_1506", "title": "516. Bonding in Diatomic Molecules - ", "content": "Using the MO diagrams shown in Figure 5.37, we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table 5.2, $\\mathrm{Be}_{2}$ and $\\mathrm{Ne}_{2}$ molecules would have a bond order of 0 , and these molecules do not exist."} {"id": "textbook_1507", "title": "516. Bonding in Diatomic Molecules - ", "content": "Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements"} {"id": "textbook_1508", "title": "516. Bonding in Diatomic Molecules - ", "content": "| Molecule | Electron Configuration | Bond Order |\n| :--- | :--- | :--- |\n| $\\mathrm{Li}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}$ | 1 |\n| $\\mathrm{Be}_{2}$ (unstable) | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}$ | 0 |\n| $\\mathrm{~B}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{2}$ | 1 |\n| $\\mathrm{C}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}$ | 2 |"} {"id": "textbook_1509", "title": "516. Bonding in Diatomic Molecules - ", "content": "TABLE 5.2"} {"id": "textbook_1510", "title": "516. Bonding in Diatomic Molecules - ", "content": "| Molecule | Electron Configuration | Bond Order |\n| :--- | :--- | :--- |\n| $\\mathrm{N}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\sigma_{2 p x}\\right)^{2}$ | 3 |\n| $\\mathrm{O}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{2}$ | 2 |\n| $\\mathrm{~F}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{4}$ | 1 |\n| $\\mathrm{Ne}_{2}$ (unstable) | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{4}\\left(\\sigma_{2 p x}^{*}\\right)^{2}$ | 0 |"} {"id": "textbook_1511", "title": "516. Bonding in Diatomic Molecules - ", "content": "TABLE 5.2"} {"id": "textbook_1512", "title": "516. Bonding in Diatomic Molecules - ", "content": "The combination of two lithium atoms to form a lithium molecule, $\\mathrm{Li}_{2}$, is analogous to the formation of $\\mathrm{H}_{2}$, but the atomic orbitals involved are the valence $2 s$ orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the $\\sigma_{2 s}$ bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the $\\mathrm{Li}_{2}$ molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table 5.2 with a bond order greater than zero are also known."} {"id": "textbook_1513", "title": "516. Bonding in Diatomic Molecules - ", "content": "The $\\mathrm{O}_{2}$ molecule has enough electrons to half fill the $\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)$ level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for $\\mathrm{O}_{2}$ is in accord with the fact that the oxygen molecule has two unpaired electrons (Figure 5.40). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.\n"} {"id": "textbook_1514", "title": "517. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_1515", "title": "518. Band Theory - ", "content": "\nWhen two identical atomic orbitals on different atoms combine, two molecular orbitals result (see Figure 5.29). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are inphase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase."} {"id": "textbook_1516", "title": "518. Band Theory - ", "content": "In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically $>10^{23}$ atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When $N$ valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, $N / 2$ (filled) bonding orbitals and $N / 2$ (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure 5.39 shows the bands for three important classes of materials: insulators, semiconductors, and conductors.\n\n\nFIGURE 5.39 Molecular orbitals in solids are so closely spaced that they are described as bands. The valence band is lower in energy and the conduction band is higher in energy. The type of solid is determined by the size of the \"band gap\" between the valence and conduction bands. Only a very small amount of energy is required to move electrons from the valance band to the conduction band in a conductor, and so they conduct electricity well. In an insulator, the band gap is large, so that very few electrons move, and they are poor conductors of electricity. Semiconductors are in between: they conduct electricity better than insulators, but not as well as conductors.\n\nIn order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is \"easy\" to overcome, so they are good conductors of electricity. In an insulator, the band gap is so \"large\" that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when \"moderate\" amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics."} {"id": "textbook_1517", "title": "518. Band Theory - ", "content": "Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to $46 \\%$ of the energy in sunlight could be converted into electricity using solar cells.\n"} {"id": "textbook_1518", "title": "519. EXAMPLE 5.7 - ", "content": "\nMolecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons\nDraw the molecular orbital diagram for the oxygen molecule, $\\mathrm{O}_{2}$. From this diagram, calculate the bond order for $\\mathrm{O}_{2}$. How does this diagram account for the paramagnetism of $\\mathrm{O}_{2}$ ?\n"} {"id": "textbook_1519", "title": "520. Solution - ", "content": "\nWe draw a molecular orbital energy diagram similar to that shown in Figure 5.37. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure 5.40.\n"} {"id": "textbook_1520", "title": "520. Solution - ", "content": "FIGURE 5.40 The molecular orbital energy diagram for $\\mathrm{O}_{2}$ predicts two unpaired electrons.\nWe calculate the bond order as"} {"id": "textbook_1521", "title": "520. Solution - ", "content": "$$\n\\mathrm{O}_{2}=\\frac{(8-4)}{2}=2\n$$"} {"id": "textbook_1522", "title": "520. Solution - ", "content": "Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the ( $\\Pi_{2 p y}, \\Pi_{2 p z}$ )* molecular orbitals.\n"} {"id": "textbook_1523", "title": "521. Check Your Learning - ", "content": "\nThe main component of air is $N_{2}$. From the molecular orbital diagram of $N_{2}$, predict its bond order and whether it is diamagnetic or paramagnetic.\n"} {"id": "textbook_1524", "title": "522. Answer: - ", "content": "\n$\\mathrm{N}_{2}$ has a bond order of 3 and is diamagnetic.\n"} {"id": "textbook_1525", "title": "523. EXAMPLE 5.8 - ", "content": ""} {"id": "textbook_1526", "title": "524. Ion Predictions with MO Diagrams - ", "content": "\nGive the molecular orbital configuration for the valence electrons in $\\mathrm{C}_{2}{ }^{2-}$. Will this ion be stable?\n"} {"id": "textbook_1527", "title": "525. Solution - ", "content": "\nLooking at the appropriate MO diagram, we see that the $\\pi$ orbitals are lower in energy than the $\\sigma_{p}$ orbital. The valence electron configuration for $\\mathrm{C}_{2}$ is $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}$. Adding two more electrons to generate the $\\mathrm{C}_{2}{ }^{2-}$ anion will give a valence electron configuration of $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\sigma_{2 p x}\\right)^{2}$. Since this has six more bonding electrons than antibonding, the bond order will be 3 , and the ion should be stable.\n"} {"id": "textbook_1528", "title": "526. Check Your Learning - ", "content": "\nHow many unpaired electrons would be present on a $\\mathrm{Be}_{2}{ }^{2-}$ ion? Would it be paramagnetic or diamagnetic?\n"} {"id": "textbook_1529", "title": "527. Answer: - ", "content": "\ntwo, paramagnetic\n"} {"id": "textbook_1530", "title": "528. LINK TO LEARNING - ", "content": "\nCreating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings (http://openstax.org/l/16orbitaldiag) of the molecular orbitals for $\\mathrm{C}_{6} \\mathrm{H}_{6}$.\n"} {"id": "textbook_1531", "title": "529. Key Terms - ", "content": "\nantibonding orbital molecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule\nbond order number of pairs of electrons between two atoms; it can be found by the number of bonds in a Lewis structure or by the difference between the number of bonding and antibonding electrons divided by two\nbonding orbital molecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule\ndegenerate orbitals orbitals that have the same energy\ndiamagnetism phenomenon in which a material is not magnetic itself but is repelled by a magnetic field; it occurs when there are only paired electrons present\nhomonuclear diatomic molecule molecule consisting of two identical atoms\nhybrid orbital orbital created by combining atomic orbitals on a central atom\nhybridization model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound\nlinear combination of atomic orbitals technique for combining atomic orbitals to create molecular orbitals\nmolecular orbital region of space in which an electron has a high probability of being found in a molecule\nmolecular orbital diagram visual representation of the relative energy levels of molecular orbitals\nmolecular orbital theory model that describes the behavior of electrons delocalized throughout a molecule in terms of the combination of atomic wave functions\nnode plane separating different lobes of orbitals, where the probability of finding an electron is zero\noverlap coexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond\nparamagnetism phenomenon in which a material is not magnetic itself but is attracted to a magnetic field; it occurs when there are unpaired electrons present\npi bond ( $\\boldsymbol{\\pi}$ bond) covalent bond formed by side-byside overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis\ns-p mixing change that causes $\\sigma_{p}$ orbitals to be less stable than $\\pi_{p}$ orbitals due to the mixing of $S$ and $p$-based molecular orbitals of similar energies.\nsigma bond ( $\\sigma$ bond) covalent bond formed by overlap of atomic orbitals along the internuclear axis\n$\\boldsymbol{s p}$ hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one $s$ and one $p$ orbital\n$\\boldsymbol{s} \\boldsymbol{p}^{\\mathbf{2}}$ hybrid orbital one of a set of three orbitals with a trigonal planar arrangement that results from combining one $s$ and two $p$ orbitals\n$\\boldsymbol{s p} \\boldsymbol{p}^{\\mathbf{3}}$ hybrid orbital one of a set of four orbitals with a tetrahedral arrangement that results from combining one $s$ and three $p$ orbitals\n$\\boldsymbol{s p}^{\\mathbf{3}} \\boldsymbol{d}$ hybrid orbital one of a set of five orbitals with a trigonal bipyramidal arrangement that results from combining one $s$, three $p$, and one $d$ orbital\n$\\boldsymbol{s p}^{\\mathbf{3}} \\boldsymbol{d}^{\\mathbf{2}}$ hybrid orbital one of a set of six orbitals with an octahedral arrangement that results from combining one $s$, three $p$, and two $d$ orbitals\nvalence bond theory description of bonding that involves atomic orbitals overlapping to form $\\sigma$ or $\\pi$ bonds, within which pairs of electrons are shared\n$\\boldsymbol{\\pi}$ bonding orbital molecular orbital formed by side-by-side overlap of atomic orbitals, in which the electron density is found on opposite sides of the internuclear axis\n$\\boldsymbol{\\pi}$ * bonding orbital antibonding molecular orbital formed by out of phase side-by-side overlap of atomic orbitals, in which the electron density is found on both sides of the internuclear axis, and there is a node between the nuclei\n$\\boldsymbol{\\sigma}$ bonding orbital molecular orbital in which the electron density is found along the axis of the bond\n$\\boldsymbol{\\sigma}^{*}$ bonding orbital antibonding molecular orbital formed by out-of-phase overlap of atomic orbital along the axis of the bond, generating a node between the nuclei\n"} {"id": "textbook_1532", "title": "530. Key Equations - ", "content": "\nbond order $=\\frac{(\\text { number of bonding electron) }-(\\text { number of antibonding electrons })}{2}$\n"} {"id": "textbook_1533", "title": "531. Summary - ", "content": ""} {"id": "textbook_1534", "title": "531. Summary - 531.1. Valence Bond Theory", "content": "\nValence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a $\\sigma$ bond. When they overlap in a fashion that creates a node along this axis, they form a $\\pi$ bond. Dipole moments can be used to determine partial separations of charges between atoms.\n"} {"id": "textbook_1535", "title": "531. Summary - 531.2. Hybrid Atomic Orbitals", "content": "\nWe can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma ( $\\sigma$ ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply $s p$ hybridization; three, $s p^{2}$ hybridization; four, $s p^{3}$ hybridization; five, $s p^{3} d$ hybridization; and six, $s p^{3} d^{2}$ hybridization. $\\operatorname{Pi}(\\pi)$ bonds are formed from unhybridized atomic orbitals ( $p$ or $d$ orbitals).\n"} {"id": "textbook_1536", "title": "531. Summary - 531.3. Multiple Bonds", "content": "\nMultiple bonds consist of a $\\sigma$ bond located along the axis between two atoms and one or two $\\pi$ bonds. The $\\sigma$ bonds are usually formed by the overlap of hybridized atomic orbitals, while the $\\pi$ bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of $\\pi$ bonds\ncan vary.\n"} {"id": "textbook_1537", "title": "531. Summary - 531.4. Molecular Orbital Theory", "content": "\nMolecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called $\\sigma$ MOs. They can be formed from $s$ orbitals or from $p$ orbitals oriented in an end-to-end fashion. Molecular orbitals formed from $p$ orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called $\\pi$ orbitals."} {"id": "textbook_1538", "title": "531. Summary - 531.4. Molecular Orbital Theory", "content": "We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund's rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory.\n"} {"id": "textbook_1539", "title": "532. Exercises - ", "content": ""} {"id": "textbook_1540", "title": "532. Exercises - 532.1. Valence Bond Theory", "content": "\n1. Explain how $\\sigma$ and $\\pi$ bonds are similar and how they are different.\n2. Use valence bond theory to explain the bonding in $\\mathrm{F}_{2}$, HF , and ClBr . Sketch the overlap of the atomic orbitals involved in the bonds.\n3. Use valence bond theory to explain the bonding in $\\mathrm{O}_{2}$. Sketch the overlap of the atomic orbitals involved in the bonds in $\\mathrm{O}_{2}$.\n4. How many $\\sigma$ and $\\pi$ bonds are present in the molecule HCN?\n5. A friend tells you $N_{2}$ has three $\\pi$ bonds due to overlap of the three $p$-orbitals on each $N$ atom. Do you agree?\n6. Draw the Lewis structures for $\\mathrm{CO}_{2}$ and CO , and predict the number of $\\sigma$ and $\\pi$ bonds for each molecule.\n(a) $\\mathrm{CO}_{2}$\n(b) CO\n"} {"id": "textbook_1541", "title": "532. Exercises - 532.2. Hybrid Atomic Orbitals", "content": "\n7. Why is the concept of hybridization required in valence bond theory?\n8. Give the shape that describes each hybrid orbital set:\n(a) $s p^{2}$\n(b) $s p^{3} d$\n(c) $s p$\n(d) $s p^{3} d^{2}$\n9. Explain why a carbon atom cannot form five bonds using $s p^{3} d$ hybrid orbitals.\n10. What is the hybridization of the central atom in each of the following?\n(a) $\\mathrm{BeH}_{2}$\n(b) $\\mathrm{SF}_{6}$\n(c) $\\mathrm{PO}_{4}{ }^{3-}$\n(d) $\\mathrm{PCl}_{5}$\n11. A molecule with the formula $A B_{3}$ could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.\n12. Methionine, $\\mathrm{CH}_{3} \\mathrm{SCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?\n\n13. Sulfuric acid is manufactured by a series of reactions represented by the following equations:\n$\\mathrm{S}_{8}(\\mathrm{~s})+8 \\mathrm{O}_{2}(g) \\longrightarrow 8 \\mathrm{SO}_{2}(g)$\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g)$\n$\\mathrm{SO}_{3}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{4}(l)$\nDraw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:\n(a) circular $\\mathrm{S}_{8}$ molecule\n(b) $\\mathrm{SO}_{2}$ molecule\n(c) $\\mathrm{SO}_{3}$ molecule\n(d) $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ molecule (the hydrogen atoms are bonded to oxygen atoms)\n14. Two important industrial chemicals, ethene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, and propene, $\\mathrm{C}_{3} \\mathrm{H}_{6}$, are produced by the steam (or thermal) cracking process:"} {"id": "textbook_1542", "title": "532. Exercises - 532.2. Hybrid Atomic Orbitals", "content": "$$\n2 \\mathrm{C}_{3} \\mathrm{H}_{8}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{C}_{3} \\mathrm{H}_{6}(g)+\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_1543", "title": "532. Exercises - 532.2. Hybrid Atomic Orbitals", "content": "For each of the four carbon compounds, do the following:\n(a) Draw a Lewis structure.\n(b) Predict the geometry about the carbon atom.\n(c) Determine the hybridization of each type of carbon atom.\n15. Analysis of a compound indicates that it contains $77.55 \\% \\mathrm{Xe}$ and $22.45 \\% \\mathrm{~F}$ by mass.\n(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).\n(b) Write a Lewis structure for the compound.\n(c) Predict the shape of the molecules of the compound.\n(d) What hybridization is consistent with the shape you predicted?\n16. Consider nitrous acid, $\\mathrm{HNO}_{2}$ (HONO).\n(a) Write a Lewis structure.\n(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the $\\mathrm{HNO}_{2}$ molecule?\n(c) What is the hybridization on the internal oxygen and nitrogen atoms in $\\mathrm{HNO}_{2}$ ?\n17. Strike-anywhere matches contain a layer of $\\mathrm{KClO}_{3}$ and a layer of $\\mathrm{P}_{4} \\mathrm{~S}_{3}$. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. $\\mathrm{KClO}_{3}$ contains the $\\mathrm{ClO}_{3}{ }^{-}$ion. $\\mathrm{P}_{4} \\mathrm{~S}_{3}$ is an unusual molecule with the skeletal structure.\n
Composition of Substances and Solutions - ", "content": "\n"} {"id": "textbook_1551", "title": "533. CHAPTER 6
Composition of Substances and Solutions - ", "content": "Figure 6.1 The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat)\n"} {"id": "textbook_1552", "title": "534. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_1553", "title": "534. CHAPTER OUTLINE - 534.1. Formula Mass", "content": "\n6.2 Determining Empirical and Molecular Formulas\n6.3 Molarity\n6.4 Other Units for Solution Concentrations"} {"id": "textbook_1554", "title": "534. CHAPTER OUTLINE - 534.1. Formula Mass", "content": "INTRODUCTION Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. Proper pool maintenance requires regular additions of various chemical compounds in carefully measured amounts. For example, the relative amount of calcium ion, $\\mathrm{Ca}^{2+}$, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also contains anions; thus, it is necessary to know both the relative amount of $\\mathrm{Ca}^{2+}$ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter.\n"} {"id": "textbook_1555", "title": "534. CHAPTER OUTLINE - 534.2. Formula Mass", "content": ""} {"id": "textbook_1556", "title": "535. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1557", "title": "535. LEARNING OBJECTIVES - ", "content": "- Calculate formula masses for covalent and ionic compounds"} {"id": "textbook_1558", "title": "535. LEARNING OBJECTIVES - ", "content": "Many argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton's atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.\n"} {"id": "textbook_1559", "title": "536. Formula Mass - ", "content": "\nAn earlier chapter of this text described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance's formula.\n"} {"id": "textbook_1560", "title": "537. Formula Mass for Covalent Substances - ", "content": "\nFor covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform $\\left(\\mathrm{CHCl}_{3}\\right)$, a covalent compound once used as a surgical anesthetic and now primarily used in the production of the \"anti-stick\" polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 6.2 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu ."} {"id": "textbook_1561", "title": "537. Formula Mass for Covalent Substances - ", "content": "| Element | Quantity | Average atomic
mass (amu) | Subtotal
$(\\mathbf{a m u})$ | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 1 | $\\times$ | 12.01 | $=$ | 12.01 |\n| H | 1 | $\\times$ | 1.008 | $=$ | 1.008 |\n| Cl | 3 | $\\times$ | 35.45 | $=$ | 106.35 |\n| Molecular mass | | | | | 119.37 |"} {"id": "textbook_1562", "title": "537. Formula Mass for Covalent Substances - ", "content": "FIGURE 6.2 The average mass of a chloroform molecule, $\\mathrm{CHCl}_{3}$, is 119.37 amu , which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform."} {"id": "textbook_1563", "title": "537. Formula Mass for Covalent Substances - ", "content": "Likewise, the molecular mass of an aspirin molecule, $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure 6.3)."} {"id": "textbook_1564", "title": "537. Formula Mass for Covalent Substances - ", "content": "| Element | Quantity | Average atomic
mass (amu) | Subtotal
$(\\mathbf{a m u})$ | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 9 | $\\times$ | 12.01 | $=$ | 108.09 |\n| H | 8 | $\\times$ | 1.008 | $=$ | 8.064 |\n| O | 4 | $\\times$ | 16.00 | $=$ | 64.00 |\n| Molecular mass | | | | | 180.15 |"} {"id": "textbook_1565", "title": "537. Formula Mass for Covalent Substances - ", "content": "FIGURE 6.3 The average mass of an aspirin molecule is 180.15 amu . The model shows the molecular structure of aspirin, $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$.\n"} {"id": "textbook_1566", "title": "538. EXAMPLE 6.1 - ", "content": ""} {"id": "textbook_1567", "title": "539. Computing Molecular Mass for a Covalent Compound - ", "content": "\nIbuprofen, $\\mathrm{C}_{13} \\mathrm{H}_{18} \\mathrm{O}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?\n"} {"id": "textbook_1568", "title": "540. Solution - ", "content": "\nMolecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:"} {"id": "textbook_1569", "title": "540. Solution - ", "content": "| Element | Quantity | Average atomic
mass (amu) | Subtotal
(amu) | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 13 | $\\times$ | 12.01 | $=$ | 156.13 |\n| H | 18 | $\\times$ | 1.008 | $=$ | 18.144 |\n| O | 2 | $\\times$ | 16.00 | $=$ | 32.00 |\n| Molecular mass | | | | | 206.27 |\n"} {"id": "textbook_1570", "title": "541. Check Your Learning - ", "content": "\nAcetaminophen, $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{NO}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?\n"} {"id": "textbook_1571", "title": "542. Answer: - ", "content": "\n151.16 amu\n"} {"id": "textbook_1572", "title": "543. Formula Mass for Ionic Compounds - ", "content": "\nIonic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound's formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the \"molecular mass.\""} {"id": "textbook_1573", "title": "543. Formula Mass for Ionic Compounds - ", "content": "As an example, consider sodium chloride, NaCl , the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, $\\mathrm{Na}^{+}$, and chloride anions, $\\mathrm{Cl}^{-}$, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (see Figure 6.4)."} {"id": "textbook_1574", "title": "543. Formula Mass for Ionic Compounds - ", "content": "| Element | Quantity | | Average atomic
mass (amu) | Subtotal | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Na | 1 | $\\times$ | 22.99 | $=$ | 22.99 |\n| Cl | 1 | $\\times$ | 35.45 | $=$ | 35.45 |\n| | | | | | |\n| Formula mass | | | | | 58.44 |"} {"id": "textbook_1575", "title": "543. Formula Mass for Ionic Compounds - ", "content": "FIGURE 6.4 Table salt, NaCl , contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu ."} {"id": "textbook_1576", "title": "543. Formula Mass for Ionic Compounds - ", "content": "Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing\nor additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.\n"} {"id": "textbook_1577", "title": "544. EXAMPLE 6.2 - ", "content": ""} {"id": "textbook_1578", "title": "545. Computing Formula Mass for an Ionic Compound - ", "content": "\nAluminum sulfate, $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass ( amu ) of this compound?\n"} {"id": "textbook_1579", "title": "546. Solution - ", "content": "\nThe formula for this compound indicates it contains $\\mathrm{Al}^{3+}$ and $\\mathrm{SO}_{4}{ }^{2-}$ ions combined in a $2: 3$ ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, $\\mathrm{Al}_{2} \\mathrm{~S}_{3} \\mathrm{O}_{12}$. Following the approach outlined above, the formula mass for this compound is calculated as follows:"} {"id": "textbook_1580", "title": "546. Solution - ", "content": "| Element | Quantity | Average atomic
mass (amu) | Subtotal
$(\\mathbf{a m u})$ | | |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Al | 2 | $\\times$ | 26.98 | $=$ | 53.96 |\n| S | 3 | $\\times$ | 32.06 | $=$ | 96.18 |\n| O | 12 | $\\times$ | 16.00 | $=$ | 192.00 |\n| Molecular mass 342.14 | | | | | |\n"} {"id": "textbook_1581", "title": "547. Check Your Learning - ", "content": "\nCalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?\n"} {"id": "textbook_1582", "title": "548. Answer: - ", "content": "\n310.18 amu\n"} {"id": "textbook_1583", "title": "548. Answer: - 548.1. Determining Empirical and Molecular Formulas", "content": "\nA previous chapter of this text discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.\n"} {"id": "textbook_1584", "title": "549. Percent Composition - ", "content": "\nThe elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound's formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound's percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:"} {"id": "textbook_1585", "title": "549. Percent Composition - ", "content": "$$\n\\begin{aligned}\n& \\% \\mathrm{H}=\\frac{\\text { mass } \\mathrm{H}}{\\text { mass compound }} \\times 100 \\% \\\\\n& \\% \\mathrm{C}=\\frac{\\text { mass } \\mathrm{C}}{\\text { mass compound }} \\times 100 \\%\n\\end{aligned}\n$$"} {"id": "textbook_1586", "title": "549. Percent Composition - ", "content": "If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C , the percent composition would be calculated to be $25 \\% \\mathrm{H}$ and $75 \\% \\mathrm{C}$ :\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{H}=\\frac{2.5 \\mathrm{~g} \\mathrm{H}}{10.0 \\mathrm{~g} \\text { compound }} \\times 100 \\%=25 \\% \\\\\n& \\% \\mathrm{C}=\\frac{7.5 \\mathrm{~g} \\mathrm{C}}{10.0 \\mathrm{~g} \\text { compound }} \\times 100 \\%=75 \\%\n\\end{aligned}\n$$\n"} {"id": "textbook_1587", "title": "550. EXAMPLE 6.3 - ", "content": ""} {"id": "textbook_1588", "title": "551. Calculation of Percent Composition - ", "content": "\nAnalysis of a $12.04-\\mathrm{g}$ sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain $7.34 \\mathrm{~g} \\mathrm{C}, 1.85 \\mathrm{~g} \\mathrm{H}$, and 2.85 g N . What is the percent composition of this compound?\n"} {"id": "textbook_1589", "title": "552. Solution - ", "content": "\nTo calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:"} {"id": "textbook_1590", "title": "552. Solution - ", "content": "$$\n\\begin{aligned}\n& \\% \\mathrm{C}=\\frac{7.34 \\mathrm{~g} \\mathrm{C}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=61.0 \\% \\\\\n& \\% \\mathrm{H}=\\frac{1.85 \\mathrm{~g} \\mathrm{H}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=15.4 \\% \\\\\n& \\% \\mathrm{~N}=\\frac{2.85 \\mathrm{~g} \\mathrm{~N}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=23.7 \\%\n\\end{aligned}\n$$\n\nThe analysis results indicate that the compound is $61.0 \\% \\mathrm{C}, 15.4 \\% \\mathrm{H}$, and $23.7 \\% \\mathrm{~N}$ by mass.\n"} {"id": "textbook_1591", "title": "553. Check Your Learning - ", "content": "\nA 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain $3.01 \\mathrm{~g} \\mathrm{C}, 4.00 \\mathrm{~g} \\mathrm{O}$, and 17.81 g Cl . What is this compound's percent composition?\n"} {"id": "textbook_1592", "title": "554. Answer: - ", "content": "\n12.1\\% C, 16.1\\% O, 71.79\\% Cl\n"} {"id": "textbook_1593", "title": "555. Determining Percent Composition from Molecular or Empirical Formulas - ", "content": "\nPercent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia $\\left(\\mathrm{NH}_{3}\\right)$, ammonium nitrate $\\left(\\mathrm{NH}_{4} \\mathrm{NO}_{3}\\right)$, and urea $\\left(\\mathrm{CH}_{4} \\mathrm{~N}_{2} \\mathrm{O}\\right)$. The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of $\\mathrm{NH}_{3}$ contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 $\\times 1.008 \\mathrm{amu})=3.024 \\mathrm{amu}$. The formula mass of ammonia is therefore $(14.01 \\mathrm{amu}+3.024 \\mathrm{amu})=17.03 \\mathrm{amu}$, and its percent composition is:"} {"id": "textbook_1594", "title": "555. Determining Percent Composition from Molecular or Empirical Formulas - ", "content": "$$\n\\begin{aligned}\n& \\% \\mathrm{~N}=\\frac{14.01 \\mathrm{amu} \\mathrm{~N}}{17.03 \\mathrm{amu} \\mathrm{NH}} 3 \\mathrm{am} \\\\\n& \\% \\mathrm{H}=\\frac{3.024 \\mathrm{amu} \\mathrm{H}}{17.03 \\mathrm{amu} \\mathrm{NH}_{3} \\times 100 \\%=82.27 \\%}=17.76 \\%\n\\end{aligned}\n$$"} {"id": "textbook_1595", "title": "555. Determining Percent Composition from Molecular or Empirical Formulas - ", "content": "This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 6.4. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements.\n"} {"id": "textbook_1596", "title": "556. EXAMPLE 6.4 - ", "content": ""} {"id": "textbook_1597", "title": "557. Determining Percent Composition from a Molecular Formula - ", "content": "\nAspirin is a compound with the molecular formula $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$. What is its percent composition?\n"} {"id": "textbook_1598", "title": "558. Solution - ", "content": "\nTo calculate the percent composition, the masses of $\\mathrm{C}, \\mathrm{H}$, and O in a known mass of $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$ are needed. It is convenient to consider 1 mol of $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$ and use its molar mass ( $180.159 \\mathrm{~g} / \\mathrm{mole}$, determined from the chemical formula) to calculate the percentages of each of its elements:\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{C}=\\frac{9 \\mathrm{~mol} \\mathrm{C} \\times \\text { molar mass } \\mathrm{C}}{\\mathrm{molar} \\text { mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{9 \\times 12.01 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{108.09 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\\n& \\% \\mathrm{C}=60.00 \\% \\mathrm{C} \\\\\n& \\% \\mathrm{H}=\\frac{8 \\mathrm{~mol} \\mathrm{H} \\times \\text { molar mass } \\mathrm{H}}{\\mathrm{molar} \\text { mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{8 \\times 1.008 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{8.064 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\\n& \\% \\mathrm{H}=4.476 \\% \\mathrm{H} \\\\\n& \\% \\mathrm{O}=\\frac{4 \\text { mol O } \\times \\text { molar mass } \\mathrm{O}}{\\text { molar mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{4 \\times 16.00 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{64.00 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\\n& \\% \\mathrm{O}=35.52 \\%\n\\end{aligned}\n$$"} {"id": "textbook_1599", "title": "558. Solution - ", "content": "Note that these percentages sum to equal $100.00 \\%$ when appropriately rounded.\nCheck Your Learning\nTo three significant digits, what is the mass percentage of iron in the compound $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ ?\n"} {"id": "textbook_1600", "title": "559. Answer: - ", "content": "\n69.9\\% Fe\n"} {"id": "textbook_1601", "title": "560. Determination of Empirical Formulas - ", "content": "\nAs previously mentioned, the most common approach to determining a compound's chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H . The corresponding numbers of atoms (in moles) are:"} {"id": "textbook_1602", "title": "560. Determination of Empirical Formulas - ", "content": "$$\n\\begin{aligned}\n& 1.71 \\mathrm{~g} \\mathrm{C} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}=0.142 \\mathrm{~mol} \\mathrm{C} \\\\\n& 0.287 \\mathrm{~g} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.008 \\mathrm{~g} \\mathrm{H}}=0.284 \\mathrm{~mol} \\mathrm{H}\n\\end{aligned}\n$$"} {"id": "textbook_1603", "title": "560. Determination of Empirical Formulas - ", "content": "Thus, this compound may be represented by the formula $\\mathrm{C}_{0.142} \\mathrm{H}_{0.284}$. Per convention, formulas contain wholenumber subscripts, which can be achieved by dividing each subscript by the smaller subscript:"} {"id": "textbook_1604", "title": "560. Determination of Empirical Formulas - ", "content": "$$\n\\mathrm{C}_{\\frac{0.142}{0.142}} \\mathrm{H}_{\\frac{0.284}{0.142}} \\text { or } \\mathrm{CH}_{2}\n$$"} {"id": "textbook_1605", "title": "560. Determination of Empirical Formulas - ", "content": "(Recall that subscripts of \" 1 \" are not written but rather assumed if no other number is present.)\nThe empirical formula for this compound is thus $\\mathrm{CH}_{2}$. This may or not be the compound's molecular formula\nas well; however, additional information is needed to make that determination (as discussed later in this section)."} {"id": "textbook_1606", "title": "560. Determination of Empirical Formulas - ", "content": "Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O . Following the same approach yields a tentative empirical formula of:"} {"id": "textbook_1607", "title": "560. Determination of Empirical Formulas - ", "content": "$$\n\\mathrm{Cl}_{0.150} \\mathrm{O}_{0.525}=\\mathrm{Cl}_{\\frac{0.150}{0.150}} \\mathrm{O}_{\\frac{0.525}{0.150}}=\\mathrm{ClO}_{3.5}\n$$"} {"id": "textbook_1608", "title": "560. Determination of Empirical Formulas - ", "content": "In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding $\\mathrm{Cl}_{2} \\mathrm{O}_{7}$ as the final empirical formula."} {"id": "textbook_1609", "title": "560. Determination of Empirical Formulas - ", "content": "In summary, empirical formulas are derived from experimentally measured element masses by:"} {"id": "textbook_1610", "title": "560. Determination of Empirical Formulas - ", "content": "1. Deriving the number of moles of each element from its mass\n2. Dividing each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula\n3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained"} {"id": "textbook_1611", "title": "560. Determination of Empirical Formulas - ", "content": "Figure 6.5 outlines this procedure in flow chart fashion for a substance containing elements A and X.\n"} {"id": "textbook_1612", "title": "560. Determination of Empirical Formulas - ", "content": "FIGURE 6.5 The empirical formula of a compound can be derived from the masses of all elements in the sample.\n"} {"id": "textbook_1613", "title": "561. EXAMPLE 6.5 - ", "content": ""} {"id": "textbook_1614", "title": "562. Determining a Compound's Empirical Formula from the Masses of Its Elements - ", "content": "\nA sample of the black mineral hematite (Figure 6.6), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?\n"} {"id": "textbook_1615", "title": "562. Determining a Compound's Empirical Formula from the Masses of Its Elements - ", "content": "FIGURE 6.6 Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)\n"} {"id": "textbook_1616", "title": "563. Solution - ", "content": "\nThis problem provides the mass in grams of each element. Begin by finding the moles of each:"} {"id": "textbook_1617", "title": "563. Solution - ", "content": "$$\n\\begin{aligned}\n& 34.97 \\mathrm{~g} \\mathrm{Fe}\\left(\\frac{\\mathrm{~mol} \\mathrm{Fe}}{55.85 \\mathrm{~g}}\\right)=0.6261 \\mathrm{~mol} \\mathrm{Fe} \\\\\n& 15.03 \\mathrm{~g} \\mathrm{O}\\left(\\frac{\\mathrm{~mol} \\mathrm{O}}{16.00 \\mathrm{~g}}\\right)=0.9394 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$"} {"id": "textbook_1618", "title": "563. Solution - ", "content": "Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:"} {"id": "textbook_1619", "title": "563. Solution - ", "content": "$$\n\\begin{aligned}\n& \\frac{0.6261}{0.6261}=1.000 \\mathrm{~mol} \\mathrm{Fe} \\\\\n& \\frac{0.9394}{0.6261}=1.500 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$"} {"id": "textbook_1620", "title": "563. Solution - ", "content": "The ratio is 1.000 mol of iron to 1.500 mol of oxygen $\\left(\\mathrm{Fe}_{1} \\mathrm{O}_{1.5}\\right)$. Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:"} {"id": "textbook_1621", "title": "563. Solution - ", "content": "$$\n2\\left(\\mathrm{Fe}_{1} \\mathrm{O}_{1.5}\\right)=\\mathrm{Fe}_{2} \\mathrm{O}_{3}\n$$"} {"id": "textbook_1622", "title": "563. Solution - ", "content": "The empirical formula is $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$.\n"} {"id": "textbook_1623", "title": "564. Check Your Learning - ", "content": "\nWhat is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?\n"} {"id": "textbook_1624", "title": "565. Answer: - ", "content": "\n$\\mathrm{N}_{2} \\mathrm{O}_{5}$\n"} {"id": "textbook_1625", "title": "566. LINK TO LEARNING - ", "content": "\nFor additional worked examples illustrating the derivation of empirical formulas, watch the brief video (http://openstax.org/l/16empforms) clip.\n"} {"id": "textbook_1626", "title": "567. Deriving Empirical Formulas from Percent Composition - ", "content": "\nFinally, with regard to deriving empirical formulas, consider instances in which a compound's percent composition is available rather than the absolute masses of the compound's constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.\n"} {"id": "textbook_1627", "title": "568. EXAMPLE 6.6 - ", "content": ""} {"id": "textbook_1628", "title": "569. Determining an Empirical Formula from Percent Composition - ", "content": "\nThe bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29\\% C and $72.71 \\% \\mathrm{O}$ (Figure 6.7). What is the empirical formula for this gas?\n"} {"id": "textbook_1629", "title": "569. Determining an Empirical Formula from Percent Composition - ", "content": "FIGURE 6.7 An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \"Dual Freq\"/Wikimedia Commons)\n"} {"id": "textbook_1630", "title": "570. Solution - ", "content": "\nSince the scale for percentages is 100 , it is most convenient to calculate the mass of elements present in a sample weighing 100 g . The calculation is \"most convenient\" because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element's mass percentage. This numerical equivalence results from the definition of the \"percentage\" unit, whose name is derived from the Latin phrase per centum meaning \"by the hundred.\" Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:"} {"id": "textbook_1631", "title": "570. Solution - ", "content": "$$\n\\begin{aligned}\n27.29 \\% \\mathrm{C} & =\\frac{27.29 \\mathrm{~g} \\mathrm{C}}{100 \\mathrm{~g} \\text { compound }} \\\\\n72.71 \\% \\mathrm{O} & =\\frac{72.71 \\mathrm{~g} \\mathrm{O}}{100 \\mathrm{~g} \\text { compound }}\n\\end{aligned}\n$$"} {"id": "textbook_1632", "title": "570. Solution - ", "content": "The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element's mass by its molar mass:"} {"id": "textbook_1633", "title": "570. Solution - ", "content": "$$\n\\begin{aligned}\n& 27.29 \\mathrm{~g} \\mathrm{C}\\left(\\frac{\\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g}}\\right)=2.272 \\mathrm{~mol} \\mathrm{C} \\\\\n& 72.71 \\mathrm{~g} \\mathrm{O}\\left(\\frac{\\mathrm{~mol} \\mathrm{O}}{16.00 \\mathrm{~g}}\\right)=4.544 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$"} {"id": "textbook_1634", "title": "570. Solution - ", "content": "Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:"} {"id": "textbook_1635", "title": "570. Solution - ", "content": "$$\n\\begin{aligned}\n& \\frac{2.272 \\mathrm{~mol} \\mathrm{C}}{2.272}=1 \\\\\n& \\frac{4.544 \\mathrm{~mol} \\mathrm{O}}{2.272}=2\n\\end{aligned}\n$$"} {"id": "textbook_1636", "title": "570. Solution - ", "content": "Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is $\\mathrm{CO}_{2}$.\n"} {"id": "textbook_1637", "title": "571. Check Your Learning - ", "content": "\nWhat is the empirical formula of a compound containing $40.0 \\% \\mathrm{C}, 6.71 \\% \\mathrm{H}$, and $53.28 \\% \\mathrm{O}$ ?\n"} {"id": "textbook_1638", "title": "572. Answer: - ", "content": "\n$\\mathrm{CH}_{2} \\mathrm{O}$\n"} {"id": "textbook_1639", "title": "573. Derivation of Molecular Formulas - ", "content": "\nRecall that empirical formulas are symbols representing the relative numbers of a compound's elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in a previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.\n\nMolecular formulas are derived by comparing the compound's molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n):"} {"id": "textbook_1640", "title": "573. Derivation of Molecular Formulas - ", "content": "$$\n\\frac{\\text { molecular or molar mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{~mol}}\\right)}{\\text { empirical formula mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{~mol}}\\right)}=n \\text { formula units/molecule }\n$$"} {"id": "textbook_1641", "title": "573. Derivation of Molecular Formulas - ", "content": "The molecular formula is then obtained by multiplying each subscript in the empirical formula by $n$, as shown by the generic empirical formula $\\mathrm{A}_{\\mathrm{x}} \\mathrm{B}_{\\mathrm{y}}$ :"} {"id": "textbook_1642", "title": "573. Derivation of Molecular Formulas - ", "content": "$$\n\\left(A_{x} B_{y}\\right)_{n}=A_{n x} B_{n y}\n$$"} {"id": "textbook_1643", "title": "573. Derivation of Molecular Formulas - ", "content": "For example, consider a covalent compound whose empirical formula is determined to be $\\mathrm{CH}_{2} \\mathrm{O}$. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one 0 atom). If the compound's molecular mass is determined to be 180 amu , this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:"} {"id": "textbook_1644", "title": "573. Derivation of Molecular Formulas - ", "content": "$$\n\\frac{180 \\mathrm{amu} / \\mathrm{molecule}}{30 \\frac{\\mathrm{amu}}{\\text { formula unit }}}=6 \\text { formula units } / \\mathrm{molec} \\text {. }\n$$"} {"id": "textbook_1645", "title": "573. Derivation of Molecular Formulas - ", "content": "Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:"} {"id": "textbook_1646", "title": "573. Derivation of Molecular Formulas - ", "content": "$$\n\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)_{6}=\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\n$$"} {"id": "textbook_1647", "title": "573. Derivation of Molecular Formulas - ", "content": "Note that this same approach may be used when the molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, as opposed to single units and molecules.\n"} {"id": "textbook_1648", "title": "574. EXAMPLE 6.7 - ", "content": ""} {"id": "textbook_1649", "title": "575. Determination of the Molecular Formula for Nicotine - ", "content": "\nNicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains $74.02 \\% \\mathrm{C}, 8.710 \\% \\mathrm{H}$, and $17.27 \\% \\mathrm{~N}$. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?\n"} {"id": "textbook_1650", "title": "576. Solution - ", "content": "\nDetermining the molecular formula from the provided data will require comparison of the compound's empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound's empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:"} {"id": "textbook_1651", "title": "576. Solution - ", "content": "$$\n\\begin{aligned}\n& (74.02 \\mathrm{~g} \\mathrm{C})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}\\right)=6.163 \\mathrm{~mol} \\mathrm{C} \\\\\n& (8.710 \\mathrm{~g} \\mathrm{H})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.01 \\mathrm{~g} \\mathrm{H}}\\right)=8.624 \\mathrm{~mol} \\mathrm{H} \\\\\n& (17.27 \\mathrm{~g} \\mathrm{~N})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{~N}}{14.01 \\mathrm{~g} \\mathrm{~N}}\\right)=1.233 \\mathrm{~mol} \\mathrm{~N}\n\\end{aligned}\n$$"} {"id": "textbook_1652", "title": "576. Solution - ", "content": "Next, calculate the molar ratios of these elements relative to the least abundant element, N ."} {"id": "textbook_1653", "title": "576. Solution - ", "content": "$$\n\\begin{gathered}\n6.163 \\mathrm{~mol} \\mathrm{C} / 1.233 \\mathrm{~mol} \\mathrm{~N}=5 \\\\\n8.264 \\mathrm{~mol} \\mathrm{H} / 1.233 \\mathrm{~mol} \\mathrm{~N}=7 \\\\\n1.233 \\mathrm{~mol} \\mathrm{~N} / 1.233 \\mathrm{~mol} \\mathrm{~N}=1 \\\\\n\\frac{1.233}{1.233}=1.000 \\mathrm{~mol} \\mathrm{~N} \\\\\n\\frac{6.163}{1.233}=4.998 \\mathrm{~mol} \\mathrm{C} \\\\\n\\frac{8.624}{1.233}=6.994 \\mathrm{~mol} \\mathrm{H}\n\\end{gathered}\n$$"} {"id": "textbook_1654", "title": "576. Solution - ", "content": "The C-to- N and H -to- N molar ratios are adequately close to whole numbers, and so the empirical formula is $\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{~N}$. The empirical formula mass for this compound is therefore $81.13 \\mathrm{amu} /$ formula unit, or $81.13 \\mathrm{~g} / \\mathrm{mol}$ formula unit."} {"id": "textbook_1655", "title": "576. Solution - ", "content": "Calculate the molar mass for nicotine from the given mass and molar amount of compound:"} {"id": "textbook_1656", "title": "576. Solution - ", "content": "$$\n\\frac{40.57 \\mathrm{~g} \\text { nicotine }}{0.2500 \\mathrm{~mol} \\text { nicotine }}=\\frac{162.3 \\mathrm{~g}}{\\mathrm{~mol}}\n$$"} {"id": "textbook_1657", "title": "576. Solution - ", "content": "Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:"} {"id": "textbook_1658", "title": "576. Solution - ", "content": "$$\n\\frac{162.3 \\mathrm{~g} / \\mathrm{mol}}{81.13 \\frac{\\mathrm{~g}}{\\text { formula unit }}}=2 \\text { formula units } / \\text { molecule }\n$$"} {"id": "textbook_1659", "title": "576. Solution - ", "content": "Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:"} {"id": "textbook_1660", "title": "576. Solution - ", "content": "$$\n\\left(\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{~N}\\right)_{2}=\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}\n$$\n"} {"id": "textbook_1661", "title": "577. Check Your Learning - ", "content": "\nWhat is the molecular formula of a compound with a percent composition of $49.47 \\% \\mathrm{C}, 5.201 \\% \\mathrm{H}, 28.84 \\% \\mathrm{~N}$, and $16.48 \\% \\mathrm{O}$, and a molecular mass of 194.2 amu ?\n"} {"id": "textbook_1662", "title": "578. Answer: - ", "content": "\n$\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\n"} {"id": "textbook_1663", "title": "578. Answer: - 578.1. Molarity", "content": ""} {"id": "textbook_1664", "title": "579. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1665", "title": "579. LEARNING OBJECTIVES - ", "content": "- Describe the fundamental properties of solutions\n- Calculate solution concentrations using molarity\n- Perform dilution calculations using the dilution equation"} {"id": "textbook_1666", "title": "579. LEARNING OBJECTIVES - ", "content": "Preceding sections of this chapter focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures-samples of matter containing two or more substances physically combined-are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet's atmosphere determines its ability to sustain aerobic\nlife. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an \"alloy\") determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 6.8). This section will describe one of the most common ways in which the relative compositions of mixtures may be quantified.\n"} {"id": "textbook_1667", "title": "579. LEARNING OBJECTIVES - ", "content": "FIGURE 6.8 Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage's sweetness. (credit: Jane Whitney)\n"} {"id": "textbook_1668", "title": "580. Solutions - ", "content": "\nSolutions have previously been defined as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. A more thorough treatment of solution properties is provided in the chapter on solutions and colloids, but provided here is an introduction to some of the basic properties of solutions."} {"id": "textbook_1669", "title": "580. Solutions - ", "content": "The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution."} {"id": "textbook_1670", "title": "580. Solutions - ", "content": "A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration)."} {"id": "textbook_1671", "title": "580. Solutions - ", "content": "Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity ( $\\boldsymbol{M}$ ) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter ( 1 L ) of the solution:"} {"id": "textbook_1672", "title": "580. Solutions - ", "content": "$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}\n$$\n"} {"id": "textbook_1673", "title": "581. EXAMPLE 6.8 - ", "content": ""} {"id": "textbook_1674", "title": "582. Calculating Molar Concentrations - ", "content": "\nA $355-\\mathrm{mL}$ soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?\n"} {"id": "textbook_1675", "title": "583. Solution - ", "content": "\nSince the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L :"} {"id": "textbook_1676", "title": "583. Solution - ", "content": "$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}=\\frac{0.133 \\mathrm{~mol}}{355 \\mathrm{~mL} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}}=0.375 \\mathrm{M}\n$$\n"} {"id": "textbook_1677", "title": "584. Check Your Learning - ", "content": "\nA teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL ?\n"} {"id": "textbook_1678", "title": "585. Answer: - ", "content": ""} {"id": "textbook_1679", "title": "585. Answer: - 585.1. M", "content": ""} {"id": "textbook_1680", "title": "586. EXAMPLE 6.9 - ", "content": ""} {"id": "textbook_1681", "title": "587. Deriving Moles and Volumes from Molar Concentrations - ", "content": "\nHow much sugar (mol) is contained in a modest sip ( $\\sim 10 \\mathrm{~mL}$ ) of the soft drink from Example 6.8?\n"} {"id": "textbook_1682", "title": "588. Solution - ", "content": "\nRearrange the definition of molarity to isolate the quantity sought, moles of sugar, then substitute the value for molarity derived in Example 6.8, 0.375 M :"} {"id": "textbook_1683", "title": "588. Solution - ", "content": "$$\n\\begin{gathered}\n\\qquad \\begin{array}{c}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }} \\\\\n\\text { mol solute }\n\\end{array}=M \\times \\mathrm{L} \\text { solution } \\\\\n\\text { mol solute }=0.375 \\frac{\\mathrm{~mol} \\text { sugar }}{\\mathrm{L}} \\times\\left(10 \\mathrm{~mL} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}\\right)=0.004 \\mathrm{~mol} \\text { sugar }\n\\end{gathered}\n$$\n"} {"id": "textbook_1684", "title": "589. Check Your Learning - ", "content": "\nWhat volume (mL) of the sweetened tea described in Example 6.8 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?\n"} {"id": "textbook_1685", "title": "590. Answer: - ", "content": "\n80 mL\n"} {"id": "textbook_1686", "title": "591. EXAMPLE 6.10 - ", "content": ""} {"id": "textbook_1687", "title": "592. Calculating Molar Concentrations from the Mass of Solute - ", "content": "\nDistilled white vinegar (Figure 6.9) is a solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, in water. A 0.500 - L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?\n"} {"id": "textbook_1688", "title": "592. Calculating Molar Concentrations from the Mass of Solute - ", "content": "FIGURE 6.9 Distilled white vinegar is a solution of acetic acid in water.\n"} {"id": "textbook_1689", "title": "593. Solution - ", "content": "\nAs in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. Since the mass of solute is provided instead of its molar amount, use the solute's molar mass to obtain the amount of solute in moles:"} {"id": "textbook_1690", "title": "593. Solution - ", "content": "$$\n\\begin{gathered}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }=} \\frac{25.2 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}{60.052 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}}{0.500 \\mathrm{~L} \\text { solution }}=0.839 \\mathrm{M} \\\\\n\\begin{array}{c}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}=0.839 \\mathrm{M} \\\\\nM=\\frac{.0839 \\text { mol solute }}{1.00 \\mathrm{~L} \\text { solution }}\n\\end{array}\n\\end{gathered}\n$$\n"} {"id": "textbook_1691", "title": "594. Check Your Learning - ", "content": "\nCalculate the molarity of 6.52 g of $\\mathrm{CoCl}_{2}(128.9 \\mathrm{~g} / \\mathrm{mol})$ dissolved in an aqueous solution with a total volume of 75.0 mL .\n"} {"id": "textbook_1692", "title": "595. Answer: - ", "content": "\n0.674 M\n"} {"id": "textbook_1693", "title": "596. EXAMPLE 6.11 - ", "content": ""} {"id": "textbook_1694", "title": "597. Determining the Mass of Solute in a Given Volume of Solution - ", "content": "\nHow many grams of NaCl are contained in 0.250 L of a $5.30-\\mathrm{M}$ solution?\n"} {"id": "textbook_1695", "title": "598. Solution - ", "content": "\nThe volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 6.9:"} {"id": "textbook_1696", "title": "598. Solution - ", "content": "$$\n\\begin{gathered}\n\\qquad M=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }} \\\\\n\\text { mol solute }=M \\times \\mathrm{L} \\text { solution } \\\\\n\\text { mol solute }=5.30 \\frac{\\mathrm{~mol} \\mathrm{NaCl}}{\\mathrm{~L}} \\times 0.250 \\mathrm{~L}=1.325 \\mathrm{~mol} \\mathrm{NaCl}\n\\end{gathered}\n$$"} {"id": "textbook_1697", "title": "598. Solution - ", "content": "Finally, this molar amount is used to derive the mass of NaCl :"} {"id": "textbook_1698", "title": "598. Solution - ", "content": "$$\n1.325 \\mathrm{~mol} \\mathrm{NaCl} \\times \\frac{58.44 \\mathrm{~g} \\mathrm{NaCl}}{\\mathrm{~mol} \\mathrm{NaCl}}=77.4 \\mathrm{~g} \\mathrm{NaCl}\n$$\n"} {"id": "textbook_1699", "title": "599. Check Your Learning - ", "content": "\nHow many grams of $\\mathrm{CaCl}_{2}(110.98 \\mathrm{~g} / \\mathrm{mol})$ are contained in 250.0 mL of a $0.200-M$ solution of calcium chloride?\n"} {"id": "textbook_1700", "title": "600. Answer: - ", "content": "\n$5.55 \\mathrm{~g} \\mathrm{CaCl}_{2}$"} {"id": "textbook_1701", "title": "600. Answer: - ", "content": "When performing calculations stepwise, as in Example 6.11, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 6.11, the molar amount of NaCl computed in the first step, 1.325 mol , would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If the guard digit had not been retained, the final calculation for the mass of NaCl would have been 77.1 g , a difference of 0.3 g ."} {"id": "textbook_1702", "title": "600. Answer: - ", "content": "In addition to retaining a guard digit for intermediate calculations, rounding errors may also be avoided by performing computations in a single step (see Example 6.12). This eliminates intermediate steps so that only the final result is rounded.\n"} {"id": "textbook_1703", "title": "601. EXAMPLE 6.12 - ", "content": ""} {"id": "textbook_1704", "title": "602. Determining the Volume of Solution Containing a Given Mass of Solute - ", "content": "\nIn Example 6.10, the concentration of acetic acid in white vinegar was determined to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid?\n"} {"id": "textbook_1705", "title": "603. Solution - ", "content": "\nFirst, use the molar mass to calculate moles of acetic acid from the given mass:"} {"id": "textbook_1706", "title": "603. Solution - ", "content": "$$\n\\mathrm{g} \\text { solute } \\times \\frac{\\text { mol solute }}{\\mathrm{g} \\text { solute }}=\\text { mol solute }\n$$"} {"id": "textbook_1707", "title": "603. Solution - ", "content": "Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:"} {"id": "textbook_1708", "title": "603. Solution - ", "content": "$$\n\\text { mol solute } \\times \\frac{\\mathrm{L} \\text { solution }}{\\text { mol solute }}=\\mathrm{L} \\text { solution }\n$$"} {"id": "textbook_1709", "title": "603. Solution - ", "content": "Combining these two steps into one yields:"} {"id": "textbook_1710", "title": "603. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{g} \\text { solute } \\times \\frac{\\text { mol solute }}{\\mathrm{g} \\text { solute }} \\times \\frac{\\mathrm{L} \\text { solution }}{\\text { mol solute }}=\\mathrm{L} \\text { solution } \\\\\n75.6 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\left(\\frac{\\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}{60.05 \\mathrm{~g}}\\right)\\left(\\frac{\\mathrm{L} \\text { solution }}{0.839 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}\\right)=1.50 \\mathrm{~L} \\text { solution }\n\\end{gathered}\n$$\n"} {"id": "textbook_1711", "title": "604. Check Your Learning - ", "content": "\nWhat volume of a $1.50-M \\mathrm{KBr}$ solution contains 66.0 g KBr ?\n"} {"id": "textbook_1712", "title": "605. Answer: - ", "content": "\n0.370 L\n"} {"id": "textbook_1713", "title": "606. Dilution of Solutions - ", "content": "\nDilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 6.10).\n"} {"id": "textbook_1714", "title": "606. Dilution of Solutions - ", "content": "FIGURE 6.10 Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)"} {"id": "textbook_1715", "title": "606. Dilution of Solutions - ", "content": "Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, a solution of lesser concentration may be prepared.\nFor example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents."} {"id": "textbook_1716", "title": "606. Dilution of Solutions - ", "content": "A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the number of moles of solute in a solution $(n)$ is equal to the product of the solution's molarity $(M)$ and its volume in liters $(L)$ :"} {"id": "textbook_1717", "title": "606. Dilution of Solutions - ", "content": "$$\nn=M L\n$$"} {"id": "textbook_1718", "title": "606. Dilution of Solutions - ", "content": "Expressions like these may be written for a solution before and after it is diluted:"} {"id": "textbook_1719", "title": "606. Dilution of Solutions - ", "content": "$$\n\\begin{aligned}\n& n_{1}=M_{1} L_{1} \\\\\n& n_{2}=M_{2} L_{2}\n\\end{aligned}\n$$"} {"id": "textbook_1720", "title": "606. Dilution of Solutions - ", "content": "where the subscripts \" 1 \" and \" 2 \" refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, $n_{1}=n_{2}$. Thus, these two equations may be set equal to one another:"} {"id": "textbook_1721", "title": "606. Dilution of Solutions - ", "content": "$$\nM_{1} L_{1}=M_{2} L_{2}\n$$"} {"id": "textbook_1722", "title": "606. Dilution of Solutions - ", "content": "This relation is commonly referred to as the dilution equation. Although this equation uses molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used as long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:"} {"id": "textbook_1723", "title": "606. Dilution of Solutions - ", "content": "$$\nC_{1} V_{1}=C_{2} V_{2}\n$$"} {"id": "textbook_1724", "title": "606. Dilution of Solutions - ", "content": "where $C$ and $V$ are concentration and volume, respectively.\n"} {"id": "textbook_1725", "title": "607. LINK TO LEARNING - ", "content": "\nUse the simulation (http://openstax.org/l/16Phetsolvents) to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.\n"} {"id": "textbook_1726", "title": "608. EXAMPLE 6.13 - ", "content": ""} {"id": "textbook_1727", "title": "609. Determining the Concentration of a Diluted Solution - ", "content": "\nIf 0.850 L of a $5.00-M$ solution of copper nitrate, $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?\n"} {"id": "textbook_1728", "title": "610. Solution - ", "content": "\nThe stock concentration, $C_{1}$, and volume, $V_{1}$, are provided as well as the volume of the diluted solution, $V_{2}$. Rearrange the dilution equation to isolate the unknown property, the concentration of the diluted solution, $C_{2}$ :"} {"id": "textbook_1729", "title": "610. Solution - ", "content": "$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nC_{2}=\\frac{C_{1} V_{1}}{V_{2}}\n\\end{gathered}\n$$"} {"id": "textbook_1730", "title": "610. Solution - ", "content": "Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L ), the diluted solution's concentration is expected to be less than one-half 5 M . This ballpark estimate will be compared to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:"} {"id": "textbook_1731", "title": "610. Solution - ", "content": "$$\nC_{2}=\\frac{0.850 \\mathrm{~L} \\times 5.00 \\frac{\\mathrm{~mol}}{\\mathrm{~L}}}{1.80 \\mathrm{~L}}=2.36 M\n$$"} {"id": "textbook_1732", "title": "610. Solution - ", "content": "This result compares well to our ballpark estimate (it's a bit less than one-half the stock concentration, 5 M ).\n"} {"id": "textbook_1733", "title": "611. Check Your Learning - ", "content": "\nWhat is the concentration of the solution that results from diluting 25.0 mL of a $2.04-\\mathrm{M}$ solution of $\\mathrm{CH}_{3} \\mathrm{OH}$ to 500.0 mL ?\n"} {"id": "textbook_1734", "title": "612. Answer: - ", "content": "\n$0.102 \\mathrm{MCH}_{3} \\mathrm{OH}$\n"} {"id": "textbook_1735", "title": "613. EXAMPLE 6.14 - ", "content": ""} {"id": "textbook_1736", "title": "614. Volume of a Diluted Solution - ", "content": "\nWhat volume of 0.12 MHBr can be prepared from $11 \\mathrm{~mL}(0.011 \\mathrm{~L})$ of 0.45 M HBr ?\n"} {"id": "textbook_1737", "title": "615. Solution - ", "content": "\nProvided are the volume and concentration of a stock solution, $V_{1}$ and $C_{1}$, and the concentration of the resultant diluted solution, $C_{2}$. Find the volume of the diluted solution, $V_{2}$ by rearranging the dilution equation to isolate $V_{2}$ :"} {"id": "textbook_1738", "title": "615. Solution - ", "content": "$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nV_{2}=\\frac{C_{1} V_{1}}{C_{2}}\n\\end{gathered}\n$$"} {"id": "textbook_1739", "title": "615. Solution - ", "content": "Since the diluted concentration $(0.12 M)$ is slightly more than one-fourth the original concentration $(0.45 M)$,\nthe volume of the diluted solution is expected to be roughly four times the original volume, or around 44 mL . Substituting the given values and solving for the unknown volume yields:"} {"id": "textbook_1740", "title": "615. Solution - ", "content": "$$\n\\begin{gathered}\nV_{2}=\\frac{(0.45 M)(0.011 \\mathrm{~L})}{(0.12 M)} \\\\\nV_{2}=0.041 \\mathrm{~L}\n\\end{gathered}\n$$"} {"id": "textbook_1741", "title": "615. Solution - ", "content": "The volume of the $0.12-M$ solution is $0.041 \\mathrm{~L}(41 \\mathrm{~mL})$. The result is reasonable and compares well with the rough estimate.\n"} {"id": "textbook_1742", "title": "616. Check Your Learning - ", "content": "\nA laboratory experiment calls for $0.125 \\mathrm{M} \\mathrm{HNO}_{3}$. What volume of $0.125 \\mathrm{M} \\mathrm{HNO}_{3}$ can be prepared from 0.250 L of $1.88 \\mathrm{M} \\mathrm{HNO}_{3}$ ?\n"} {"id": "textbook_1743", "title": "617. Answer: - ", "content": "\n3.76 L\n"} {"id": "textbook_1744", "title": "618. EXAMPLE 6.15 - ", "content": ""} {"id": "textbook_1745", "title": "619. Volume of a Concentrated Solution Needed for Dilution - ", "content": "\nWhat volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH ?\n"} {"id": "textbook_1746", "title": "620. Solution - ", "content": "\nGiven are the concentration of a stock solution, $C_{1}$, and the volume and concentration of the resultant diluted solution, $V_{2}$ and $C_{2}$. Find the volume of the stock solution, $V_{1}$ by rearranging the dilution equation to isolate $V_{1}$ :"} {"id": "textbook_1747", "title": "620. Solution - ", "content": "$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nV_{1}=\\frac{C_{2} V_{2}}{C_{1}}\n\\end{gathered}\n$$"} {"id": "textbook_1748", "title": "620. Solution - ", "content": "Since the concentration of the diluted solution $0.100 M$ is roughly one-sixteenth that of the stock solution (1.59 $M$ ), the volume of the stock solution is expected to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:"} {"id": "textbook_1749", "title": "620. Solution - ", "content": "$$\n\\begin{gathered}\nV_{1}=\\frac{(0.100 M)(5.00 \\mathrm{~L})}{1.59 M} \\\\\nV_{1}=0.314 \\mathrm{~L}\n\\end{gathered}\n$$"} {"id": "textbook_1750", "title": "620. Solution - ", "content": "Thus, 0.314 L of the $1.59-M$ solution is needed to prepare the desired solution. This result is consistent with the rough estimate.\n"} {"id": "textbook_1751", "title": "621. Check Your Learning - ", "content": "\nWhat volume of a $0.575-M$ solution of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, can be prepared from 50.00 mL of a $3.00-\\mathrm{M}$ glucose solution?\n"} {"id": "textbook_1752", "title": "622. Answer: - ", "content": "\n0.261 L\n"} {"id": "textbook_1753", "title": "622. Answer: - 622.1. Other Units for Solution Concentrations", "content": ""} {"id": "textbook_1754", "title": "623. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1755", "title": "623. LEARNING OBJECTIVES - ", "content": "- Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-permillion (ppm), and parts-per-billion (ppb)\n- Perform computations relating a solution's concentration and its components' volumes and/or masses using these units"} {"id": "textbook_1756", "title": "623. LEARNING OBJECTIVES - ", "content": "The previous section introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. This section will describe some other units of concentration that are commonly used in various applications, either for convenience or by convention.\n"} {"id": "textbook_1757", "title": "624. Mass Percentage - ", "content": "\nEarlier in this chapter, percent composition was introduced as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component's mass to the solution's mass, expressed as a percentage:"} {"id": "textbook_1758", "title": "624. Mass Percentage - ", "content": "$$\n\\text { mass percentage }=\\frac{\\text { mass of component }}{\\text { mass of solution }} \\times 100 \\%\n$$"} {"id": "textbook_1759", "title": "624. Mass Percentage - ", "content": "Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, \\%, although more detailed symbols are often used including \\%mass, \\%weight, and (w/w)\\%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).\nMass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 6.11) cites the concentration of its active ingredient, sodium hypochlorite ( NaOCl ), as being $7.4 \\%$. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl .\n"} {"id": "textbook_1760", "title": "624. Mass Percentage - ", "content": "FIGURE 6.11 Liquid bleach is an aqueous solution of sodium hypochlorite ( NaOCl ). This brand has a concentration of $7.4 \\% \\mathrm{NaOCl}$ by mass.\n"} {"id": "textbook_1761", "title": "625. Calculation of Percent by Mass - ", "content": "\nA $5.0-\\mathrm{g}$ sample of spinal fluid contains $3.75 \\mathrm{mg}(0.00375 \\mathrm{~g})$ of glucose. What is the percent by mass of glucose in spinal fluid?\n"} {"id": "textbook_1762", "title": "626. Solution - ", "content": "\nThe spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about $0.1 \\%$. Substituting the given masses into the equation defining mass percentage yields:\n\n$$\n\\% \\text { glucose }=\\frac{3.75 \\mathrm{mg} \\text { glucose } \\times \\frac{1 \\mathrm{~g}}{1000 \\mathrm{mg}}}{5.0 \\mathrm{~g} \\text { spinal fluid }}=0.075 \\%\n$$"} {"id": "textbook_1763", "title": "626. Solution - ", "content": "The computed mass percentage agrees with our rough estimate (it's a bit less than 0.1\\%).\nNote that while any mass unit may be used to compute a mass percentage ( $\\mathrm{mg}, \\mathrm{g}, \\mathrm{kg}, \\mathrm{oz}$, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, the solute mass unit in the numerator was converted from mg to g to match the units in the denominator. Alternatively, the spinal fluid mass unit in the denominator could have been converted from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.\n"} {"id": "textbook_1764", "title": "627. Check Your Learning - ", "content": "\nA bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?\n"} {"id": "textbook_1765", "title": "628. Answer: - ", "content": "\n14.8\\%\n"} {"id": "textbook_1766", "title": "629. EXAMPLE 6.17 - ", "content": ""} {"id": "textbook_1767", "title": "630. Calculations using Mass Percentage - ", "content": "\n\"Concentrated\" hydrochloric acid is an aqueous solution of $37.2 \\% \\mathrm{HCl}$ that is commonly used as a laboratory reagent. The density of this solution is $1.19 \\mathrm{~g} / \\mathrm{mL}$. What mass of HCl is contained in 0.500 L of this solution?\n"} {"id": "textbook_1768", "title": "631. Solution - ", "content": "\nThe HCl concentration is near $40 \\%$, so a 100-g portion of this solution would contain about 40 g of HCl . Since the solution density isn't greatly different from that of water ( $1 \\mathrm{~g} / \\mathrm{mL}$ ), a reasonable estimate of the HCl mass in $500 \\mathrm{~g}(0.5 \\mathrm{~L})$ of the solution is about five times greater than that in a 100 g portion, or $5 \\times 40=200 \\mathrm{~g}$. In order to derive the mass of solute in a solution from its mass percentage, the mass of the solution must be known. Using the solution density given, convert the solution's volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:\n\n\nFor proper unit cancellation, the $0.500-\\mathrm{L}$ volume is converted into 500 mL , and the mass percentage is expressed as a ratio, $37.2 \\mathrm{~g} \\mathrm{HCl} / \\mathrm{g}$ solution:"} {"id": "textbook_1769", "title": "631. Solution - ", "content": "$$\n500 \\mathrm{~mL} \\text { solution }\\left(\\frac{1.19 \\mathrm{~g} \\text { solution }}{\\mathrm{mL} \\text { solution }}\\right)\\left(\\frac{37.2 \\mathrm{~g} \\mathrm{HCl}}{100 \\mathrm{~g} \\text { solution }}\\right)=221 \\mathrm{~g} \\mathrm{HCl}\n$$"} {"id": "textbook_1770", "title": "631. Solution - ", "content": "This mass of HCl is consistent with our rough estimate of approximately 200 g .\nCheck Your Learning\nWhat volume of concentrated HCl solution contains 125 g of HCl ?\n"} {"id": "textbook_1771", "title": "632. Answer: - ", "content": "\n282 mL\n"} {"id": "textbook_1772", "title": "633. Volume Percentage - ", "content": "\nLiquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, $\\% \\mathrm{vol}$ or $(\\mathrm{v} / \\mathrm{v}) \\%$ :"} {"id": "textbook_1773", "title": "633. Volume Percentage - ", "content": "$$\n\\text { volume percentage }=\\frac{\\text { volume solute }}{\\text { volume solution }} \\times 100 \\%\n$$\n"} {"id": "textbook_1774", "title": "634. EXAMPLE 6.18 - ", "content": ""} {"id": "textbook_1775", "title": "635. Calculations using Volume Percentage - ", "content": "\nRubbing alcohol (isopropanol) is usually sold as a 70\\%vol aqueous solution. If the density of isopropyl alcohol is $0.785 \\mathrm{~g} / \\mathrm{mL}$, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?\n"} {"id": "textbook_1776", "title": "636. Solution - ", "content": "\nPer the definition of volume percentage, the isopropanol volume is $70 \\%$ of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:\n$\\left(355 \\mathrm{~mL}\\right.$ solution) $\\left(\\frac{70 \\mathrm{~mL} \\text { isopropyl alcohol }}{100 \\mathrm{~mL} \\text { solution }}\\right)\\left(\\frac{0.785 \\mathrm{~g} \\text { isopropyl alcohol }}{1 \\mathrm{~mL} \\text { isopropyl alcohol }}\\right)=195 \\mathrm{~g}$ isopropyl alchol\n"} {"id": "textbook_1777", "title": "637. Check Your Learning - ", "content": "\nWine is approximately $12 \\%$ ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$ by volume. Ethanol has a molar mass of $46.06 \\mathrm{~g} / \\mathrm{mol}$ and a density $0.789 \\mathrm{~g} / \\mathrm{mL}$. How many moles of ethanol are present in a $750-\\mathrm{mL}$ bottle of wine?\n"} {"id": "textbook_1778", "title": "638. Answer: - ", "content": "\n1.5 mol ethanol\n"} {"id": "textbook_1779", "title": "639. Mass-Volume Percentage - ", "content": "\n\"Mixed\" percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute's mass to the solution's volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of $0.9 \\%$ mass $/$ volume ( $\\mathrm{m} / \\mathrm{v}$ ), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as \"blood sugar\") is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter ( 100 mL ) of blood (Figure 6.12).\n\n\nFIGURE 6.12 \"Mixed\" mass-volume units are commonly encountered in medical settings. (a) The NaCl concentration of physiological saline is $0.9 \\%(\\mathrm{~m} / \\mathrm{v})$. (b) This device measures glucose levels in a sample of blood. The normal range for glucose concentration in blood (fasting) is around $70-100 \\mathrm{mg} / \\mathrm{dL}$. (credit a: modification of work by \"The National Guard\"/Flickr; credit b: modification of work by Biswarup Ganguly)\n"} {"id": "textbook_1780", "title": "640. Parts per Million and Parts per Billion - ", "content": "\nVery low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (\"part per hundred\") units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules."} {"id": "textbook_1781", "title": "640. Parts per Million and Parts per Billion - ", "content": "The mass-based definitions of ppm and ppb are given here:"} {"id": "textbook_1782", "title": "640. Parts per Million and Parts per Billion - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{ppm}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{6} \\mathrm{ppm} \\\\\n& \\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb}\n\\end{aligned}\n$$"} {"id": "textbook_1783", "title": "640. Parts per Million and Parts per Billion - ", "content": "Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm . Inline water filters are designed to reduce the concentration of fluoride and several other tracelevel contaminants in tap water (Figure 6.13).\n"} {"id": "textbook_1784", "title": "640. Parts per Million and Parts per Billion - ", "content": "FIGURE 6.13 (a) In some areas, trace-level concentrations of contaminants can render unfiltered tap water unsafe for drinking and cooking. (b) Inline water filters reduce the concentration of solutes in tap water. (credit a: modification of work by Jenn Durfey; credit b: modification of work by \"vastateparkstaff\"/Wikimedia commons)\n"} {"id": "textbook_1785", "title": "641. EXAMPLE 6.19 - ", "content": ""} {"id": "textbook_1786", "title": "642. Calculation of Parts per Million and Parts per Billion Concentrations - ", "content": "\nAccording to the EPA, when the concentration of lead in tap water reaches 15 ppb , certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead ( $\\mu \\mathrm{g}$ ) would be contained in a typical glass of water ( 300 mL )?\n"} {"id": "textbook_1787", "title": "643. Solution - ", "content": "\nThe definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm . Comparing these two unit definitions shows that ppm is 1000 times greater than $\\mathrm{ppb}\\left(1 \\mathrm{ppm}=10^{3} \\mathrm{ppb}\\right)$. Thus:"} {"id": "textbook_1788", "title": "643. Solution - ", "content": "$$\n15 \\mathrm{ppb} \\times \\frac{1 \\mathrm{ppm}}{10^{3} \\mathrm{ppb}}=0.015 \\mathrm{ppm}\n$$"} {"id": "textbook_1789", "title": "643. Solution - ", "content": "The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. Since the volume of solution ( 300 mL ) is given, its density must be used to derive the corresponding mass. Assume the density of tap water to be roughly the same as that of pure water ( $\\sim 1.00 \\mathrm{~g} / \\mathrm{mL}$ ), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:"} {"id": "textbook_1790", "title": "643. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb} \\\\\n\\text { mass solute }=\\frac{\\mathrm{ppb} \\times \\text { mass solution }}{10^{9} \\mathrm{ppb}} \\\\\n\\text { mass solute }=\\frac{15 \\mathrm{ppb} \\times 300 \\mathrm{~mL} \\times \\frac{1.00 \\mathrm{~g}}{\\mathrm{~mL}}}{10^{9} \\mathrm{ppb}}=4.5 \\times 10^{-6} \\mathrm{~g}\n\\end{gathered}\n$$"} {"id": "textbook_1791", "title": "643. Solution - ", "content": "Finally, convert this mass to the requested unit of micrograms:"} {"id": "textbook_1792", "title": "643. Solution - ", "content": "$$\n4.5 \\times 10^{-6} \\mathrm{~g} \\times \\frac{1 \\mu \\mathrm{~g}}{10^{-6} \\mathrm{~g}}=4.5 \\mu \\mathrm{~g}\n$$\n"} {"id": "textbook_1793", "title": "644. Check Your Learning - ", "content": "\nA 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units.\n"} {"id": "textbook_1794", "title": "645. Answer: - ", "content": "\n9.6 ppm, 9600 ppb\n"} {"id": "textbook_1795", "title": "646. Key Terms - ", "content": "\naqueous solution solution for which water is the solvent\nAvogadro's number ( $\\boldsymbol{N}_{\\boldsymbol{A}}$ ) experimentally determined value of the number of entities comprising 1 mole of substance, equal to $6.022 \\times$ $10^{23} \\mathrm{~mol}^{-1}$\nconcentrated qualitative term for a solution containing solute at a relatively high concentration\nconcentration quantitative measure of the relative amounts of solute and solvent present in a solution\ndilute qualitative term for a solution containing solute at a relatively low concentration\ndilution process of adding solvent to a solution in order to lower the concentration of solutes\ndissolved describes the process by which solute components are dispersed in a solvent\nempirical formula mass sum of average atomic masses for all atoms represented in an empirical formula\nformula mass sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass\nmass percentage ratio of solute-to-solution mass\nexpressed as a percentage\nmass-volume percent ratio of solute mass to solution volume, expressed as a percentage\nmolar mass mass in grams of 1 mole of a substance\nmolarity (M) unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution\nmole amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ${ }^{12} \\mathrm{C}$\nparts per billion (ppb) ratio of solute-to-solution mass multiplied by $10^{9}$\nparts per million (ppm) ratio of solute-to-solution mass multiplied by $10^{6}$\npercent composition percentage by mass of the various elements in a compound\nsolute solution component present in a concentration less than that of the solvent\nsolvent solution component present in a concentration that is higher relative to other components\nvolume percentage ratio of solute-to-solution volume expressed as a percentage\n"} {"id": "textbook_1796", "title": "647. Key Equations - ", "content": "\n$\\% \\mathrm{X}=\\frac{\\text { mass } \\mathrm{X}}{\\text { mass compound }} \\times 100 \\%$\n$\\frac{\\text { molecular or molar mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{mol}}\\right)}{\\text { empirical formula mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{mol}}\\right)}=n$ formula units/molecule\n$\\left(\\mathrm{A}_{\\mathrm{x}} \\mathrm{B}_{\\mathrm{y}}\\right)_{\\mathrm{n}}=\\mathrm{A}_{\\mathrm{nx}} \\mathrm{B}_{\\mathrm{ny}}$\n$M=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}$\n$C_{1} V_{1}=C_{2} V_{2}$\nPercent by mass $=\\frac{\\text { mass of solute }}{\\text { mass of solution }} \\times 100$\n$\\mathrm{ppm}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{6} \\mathrm{ppm}$\n$\\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb}$\n"} {"id": "textbook_1797", "title": "648. Summary - ", "content": ""} {"id": "textbook_1798", "title": "649. 1 Formula Mass - ", "content": "\nThe formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula\nmass (amu) and molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) for any substance are numerically equivalent (for example, one $\\mathrm{H}_{2} \\mathrm{O}$ molecule weighs approximately18 amu and 1 mole of $\\mathrm{H}_{2} \\mathrm{O}$ molecules weighs approximately 18 g).\n"} {"id": "textbook_1799", "title": "649. 1 Formula Mass - 649.1. Determining Empirical and Molecular", "content": ""} {"id": "textbook_1800", "title": "650. Formulas - ", "content": "\nThe chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound's percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound's empirical formula. The empirical formula mass of a covalent compound may be compared to the compound's molecular or molar mass to derive a molecular formula.\n"} {"id": "textbook_1801", "title": "650. Formulas - 650.1. Molarity", "content": "\nSolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given\namount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.\n"} {"id": "textbook_1802", "title": "650. Formulas - 650.2. Other Units for Solution Concentrations", "content": "\nIn addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components' masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.\n"} {"id": "textbook_1803", "title": "651. Exercises - ", "content": ""} {"id": "textbook_1804", "title": "651. Exercises - 651.1. Formula Mass", "content": "\n1. What is the total mass (amu) of carbon in each of the following molecules?\n(a) $\\mathrm{CH}_{4}$\n(b) $\\mathrm{CHCl}_{3}$\n(c) $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{O}_{6}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n2. What is the total mass of hydrogen in each of the molecules?\n(a) $\\mathrm{CH}_{4}$\n(b) $\\mathrm{CHCl}_{3}$\n(c) $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{O}_{6}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n3. Calculate the molecular or formula mass of each of the following:\n(a) $\\mathrm{P}_{4}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ (acetic acid)\n(e) $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ (sucrose, cane sugar)\n4. Determine the molecular mass of the following compounds:\n(a)\n
Stoichiometry of Chemical Reactions - ", "content": "\n"} {"id": "textbook_1809", "title": "652. CHAPTER 7
Stoichiometry of Chemical Reactions - ", "content": "Figure 7.1 Many modern rocket fuels are solid mixtures of substances combined in carefully measured amounts and ignited to yield a thrust-generating chemical reaction. (credit: modification of work by NASA)\n"} {"id": "textbook_1810", "title": "653. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_1811", "title": "653. CHAPTER OUTLINE - 653.1. Writing and Balancing Chemical Equations", "content": "\n7.2 Classifying Chemical Reactions\n"} {"id": "textbook_1812", "title": "653. CHAPTER OUTLINE - 653.2. Reaction Stoichiometry", "content": "\n7.4 Reaction Yields\n7.5 Quantitative Chemical Analysis"} {"id": "textbook_1813", "title": "653. CHAPTER OUTLINE - 653.2. Reaction Stoichiometry", "content": "INTRODUCTION Solid-fuel rockets are a central feature in the world's space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure 7.1). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions-that is, the reaction stoichiometry.\n"} {"id": "textbook_1814", "title": "653. CHAPTER OUTLINE - 653.3. Writing and Balancing Chemical Equations", "content": ""} {"id": "textbook_1815", "title": "654. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1816", "title": "654. LEARNING OBJECTIVES - ", "content": "- Derive chemical equations from narrative descriptions of chemical reactions.\n- Write and balance chemical equations in molecular, total ionic, and net ionic formats."} {"id": "textbook_1817", "title": "654. LEARNING OBJECTIVES - ", "content": "An earlier chapter of this text introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule $\\left(\\mathrm{CH}_{4}\\right)$ and two diatomic oxygen molecules $\\left(\\mathrm{O}_{2}\\right)$ to produce one carbon dioxide molecule $\\left(\\mathrm{CO}_{2}\\right)$ and two water molecules $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$. The chemical equation representing this process is provided in the upper half of Figure 7.2, with space-filling molecular models shown in the lower half of the figure.\n"} {"id": "textbook_1818", "title": "654. LEARNING OBJECTIVES - ", "content": "FIGURE 7.2 The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top)."} {"id": "textbook_1819", "title": "654. LEARNING OBJECTIVES - ", "content": "This example illustrates the fundamental aspects of any chemical equation:"} {"id": "textbook_1820", "title": "654. LEARNING OBJECTIVES - ", "content": "1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.\n2. The substances generated by the reaction are called products, and their formulas are placed on the right side of the equation.\n3. Plus signs (+) separate individual reactant and product formulas, and an arrow $(\\longrightarrow)$ separates the reactant and product (left and right) sides of the equation.\n4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted."} {"id": "textbook_1821", "title": "654. LEARNING OBJECTIVES - ", "content": "It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 7.3). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:"} {"id": "textbook_1822", "title": "654. LEARNING OBJECTIVES - ", "content": "- One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.\n- One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.\n- One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide\nmolecules and 2 moles of water molecules.\nMixture before reaction\n"} {"id": "textbook_1823", "title": "654. LEARNING OBJECTIVES - ", "content": "Mixture after reaction"} {"id": "textbook_1824", "title": "654. LEARNING OBJECTIVES - ", "content": "FIGURE 7.3 Regardless of the absolute numbers of molecules involved, the ratios between numbers of molecules of each species that react (the reactants) and molecules of each species that form (the products) are the same and are given by the chemical reaction equation.\n"} {"id": "textbook_1825", "title": "655. Balancing Equations - ", "content": "\nThe chemical equation described in section 4.1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element's subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is"} {"id": "textbook_1826", "title": "655. Balancing Equations - ", "content": "$$\n\\left(1 \\mathrm{CO}_{2} \\text { molecule } \\times \\frac{2 \\mathrm{O} \\text { atoms }}{\\mathrm{CO}_{2} \\text { molecule }}\\right)+\\left(2 \\mathrm{H}_{2} \\mathrm{O} \\text { molecules } \\times \\frac{1 \\mathrm{O} \\text { atom }}{\\mathrm{H}_{2} \\mathrm{O} \\text { molecule }}\\right)=4 \\mathrm{O} \\text { atoms }\n$$"} {"id": "textbook_1827", "title": "655. Balancing Equations - ", "content": "The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:"} {"id": "textbook_1828", "title": "655. Balancing Equations - ", "content": "$$\n\\mathrm{CH}_{4}+2 \\mathrm{O}_{2} \\longrightarrow \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_1829", "title": "655. Balancing Equations - ", "content": "| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| C | $1 \\times 1=1$ | $1 \\times 1=1$ | $1=1$, yes |\n| H | $4 \\times 1=4$ | $2 \\times 2=4$ | $4=4$, yes |\n| O | $2 \\times 2=4$ | $(1 \\times 2)+(2 \\times 1)=4$ | $4=4$, yes |"} {"id": "textbook_1830", "title": "655. Balancing Equations - ", "content": "A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:"} {"id": "textbook_1831", "title": "655. Balancing Equations - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2} \\quad \\text { (unbalanced) }\n$$"} {"id": "textbook_1832", "title": "655. Balancing Equations - ", "content": "Comparing the number of H and O atoms on either side of this equation confirms its imbalance:"} {"id": "textbook_1833", "title": "655. Balancing Equations - ", "content": "| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| H | $1 \\times 2=2$ | $1 \\times 2=2$ | $2=2$, yes |\n| O | $1 \\times 1=1$ | $1 \\times 2=2$ | $1 \\neq 2$, no |"} {"id": "textbook_1834", "title": "655. Balancing Equations - ", "content": "The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from $\\mathrm{H}_{2} \\mathrm{O}$ to $\\mathrm{H}_{2} \\mathrm{O}_{2}$ would yield balance in the number of atoms, but doing so also changes the reactant's identity (it's now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for $\\mathrm{H}_{2} \\mathrm{O}$ to 2 ."} {"id": "textbook_1835", "title": "655. Balancing Equations - ", "content": "| $2 \\mathrm{H}_{2} \\mathrm{O}$ | | $\\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2}$ | (unbalanced) |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| H | $\\mathbf{2} \\times 2=4$ | $1 \\times 2=2$ | $4 \\neq 2$, no |\n| O | $2 \\times 1=2$ | $1 \\times 2=2$ | $2=2$, yes |"} {"id": "textbook_1836", "title": "655. Balancing Equations - ", "content": "The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the $\\mathrm{H}_{2}$ product to 2 ."} {"id": "textbook_1837", "title": "655. Balancing Equations - ", "content": "| $2 \\mathrm{H}_{2} \\mathrm{O}$ | | $\\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2}$ | (balanced) |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| H | $2 \\times 2=4$ | $\\mathbf{2} \\times 2=4$ | $4=4$, yes |\n| O | $2 \\times 1=2$ | $1 \\times 2=2$ | $2=2$, yes |"} {"id": "textbook_1838", "title": "655. Balancing Equations - ", "content": "These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:"} {"id": "textbook_1839", "title": "655. Balancing Equations - ", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 2 \\mathrm{H}_{2}+\\mathrm{O}_{2}\n$$\n"} {"id": "textbook_1840", "title": "656. EXAMPLE 7.1 - ", "content": ""} {"id": "textbook_1841", "title": "657. Balancing Chemical Equations - ", "content": "\nWrite a balanced equation for the reaction of molecular nitrogen $\\left(\\mathrm{N}_{2}\\right)$ and oxygen $\\left(\\mathrm{O}_{2}\\right)$ to form dinitrogen pentoxide.\n"} {"id": "textbook_1842", "title": "658. Solution - ", "content": "\nFirst, write the unbalanced equation."} {"id": "textbook_1843", "title": "658. Solution - ", "content": "$$\n\\mathrm{N}_{2}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{~N}_{2} \\mathrm{O}_{5} \\quad \\text { (unbalanced) }\n$$"} {"id": "textbook_1844", "title": "658. Solution - ", "content": "Next, count the number of each type of atom present in the unbalanced equation."} {"id": "textbook_1845", "title": "658. Solution - ", "content": "| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| N | $1 \\times 2=2$ | $1 \\times 2=2$ | $2=2$, yes |\n| O | $1 \\times 2=2$ | $1 \\times 5=5$ | $2 \\neq 5$, no |"} {"id": "textbook_1846", "title": "658. Solution - ", "content": "Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}_{5}$ to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas)."} {"id": "textbook_1847", "title": "658. Solution - ", "content": "| $\\mathrm{N}_{2}+\\mathbf{5} \\mathrm{O}_{2} \\longrightarrow \\mathbf{2} \\mathrm{~N}_{2} \\mathrm{O}_{5}$ | | (unbalanced) | |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| N | $1 \\times 2=2$ | $\\mathbf{2} \\times 2=4$ | $2 \\neq 4$, no |\n| O | $\\mathbf{5} \\times 2=10$ | $\\mathbf{2} \\times 5=10$ | $10=10$, yes |"} {"id": "textbook_1848", "title": "658. Solution - ", "content": "The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant $\\mathrm{N}_{2}$ to 2 ."} {"id": "textbook_1849", "title": "658. Solution - ", "content": "| $2 \\mathrm{~N}_{2}+5 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{~N}_{2} \\mathrm{O}_{5}$ | | | |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| N | $2 \\times 2=4$ | $2 \\times 2=4$ | $4=4$, yes |\n| O | $5 \\times 2=10$ | $2 \\times 5=10$ | $10=10$, yes |"} {"id": "textbook_1850", "title": "658. Solution - ", "content": "The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.\n"} {"id": "textbook_1851", "title": "659. Check Your Learning - ", "content": "\nWrite a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)\n"} {"id": "textbook_1852", "title": "660. Answer: - ", "content": "\n$2 \\mathrm{NH}_{4} \\mathrm{NO}_{3} \\longrightarrow 2 \\mathrm{~N}_{2}+\\mathrm{O}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$"} {"id": "textbook_1853", "title": "660. Answer: - ", "content": "It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation's coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane $\\left(\\mathrm{C}_{2} \\mathrm{H}_{6}\\right)$ with oxygen to yield $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$, represented by the unbalanced equation:"} {"id": "textbook_1854", "title": "660. Answer: - ", "content": "$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\quad \\text { (unbalanced) }\n$$"} {"id": "textbook_1855", "title": "660. Answer: - ", "content": "Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:"} {"id": "textbook_1856", "title": "660. Answer: - ", "content": "$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{CO}_{2} \\quad \\text { (unbalanced) }\n$$"} {"id": "textbook_1857", "title": "660. Answer: - ", "content": "This results in seven $O$ atoms on the product side of the equation, an odd number-no integer coefficient can be used with the $\\mathrm{O}_{2}$ reactant to yield an odd number, so a fractional coefficient, $\\frac{7}{2}$, is used instead to yield a\nprovisional balanced equation:"} {"id": "textbook_1858", "title": "660. Answer: - ", "content": "$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\frac{7}{2} \\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{CO}_{2}\n$$"} {"id": "textbook_1859", "title": "660. Answer: - ", "content": "A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2 :"} {"id": "textbook_1860", "title": "660. Answer: - ", "content": "$$\n2 \\mathrm{C}_{2} \\mathrm{H}_{6}+7 \\mathrm{O}_{2} \\longrightarrow 6 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{CO}_{2}\n$$"} {"id": "textbook_1861", "title": "660. Answer: - ", "content": "Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,"} {"id": "textbook_1862", "title": "660. Answer: - ", "content": "$$\n3 \\mathrm{~N}_{2}+9 \\mathrm{H}_{2} \\longrightarrow 6 \\mathrm{NH}_{3}\n$$"} {"id": "textbook_1863", "title": "660. Answer: - ", "content": "the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:"} {"id": "textbook_1864", "title": "660. Answer: - ", "content": "$$\n\\mathrm{N}_{2}+3 \\mathrm{H}_{2} \\longrightarrow 2 \\mathrm{NH}_{3}\n$$\n"} {"id": "textbook_1865", "title": "661. LINK TO LEARNING - ", "content": "\nUse this interactive tutorial (http://openstax.org/l/16BalanceEq) for additional practice balancing equations.\n"} {"id": "textbook_1866", "title": "662. Additional Information in Chemical Equations - ", "content": "\nThe physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include $s$ for solids, $l$ for liquids, $g$ for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here:"} {"id": "textbook_1867", "title": "662. Additional Information in Chemical Equations - ", "content": "$$\n2 \\mathrm{Na}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_1868", "title": "662. Additional Information in Chemical Equations - ", "content": "This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water)."} {"id": "textbook_1869", "title": "662. Additional Information in Chemical Equations - ", "content": "Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation's arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta ( $\\Delta$ ) over the arrow."} {"id": "textbook_1870", "title": "662. Additional Information in Chemical Equations - ", "content": "$$\n\\mathrm{CaCO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_1871", "title": "662. Additional Information in Chemical Equations - ", "content": "Other examples of these special conditions will be encountered in more depth in later chapters.\n"} {"id": "textbook_1872", "title": "663. Equations for Ionic Reactions - ", "content": "\nGiven the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of $\\mathrm{CaCl}_{2}$ and $\\mathrm{AgNO}_{3}$ are mixed, a reaction takes place producing aqueous $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and solid AgCl :"} {"id": "textbook_1873", "title": "663. Equations for Ionic Reactions - ", "content": "$$\n\\mathrm{CaCl}_{2}(a q)+2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{AgCl}(s)\n$$"} {"id": "textbook_1874", "title": "663. Equations for Ionic Reactions - ", "content": "This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn't explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:"} {"id": "textbook_1875", "title": "663. Equations for Ionic Reactions - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{CaCl}_{2}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\\\\n& 2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\\\\n& \\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)\n\\end{aligned}\n$$"} {"id": "textbook_1876", "title": "663. Equations for Ionic Reactions - ", "content": "Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, $s$."} {"id": "textbook_1877", "title": "663. Equations for Ionic Reactions - ", "content": "Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:"} {"id": "textbook_1878", "title": "663. Equations for Ionic Reactions - ", "content": "$$\n\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)+2 \\mathrm{Ag} \\mathrm{Cl}(s)\n$$"} {"id": "textbook_1879", "title": "663. Equations for Ionic Reactions - ", "content": "Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, $\\mathrm{Ca}^{2+}(a q)$ and $\\mathrm{NO}_{3}{ }^{-}(a q)$. These spectator ions-ions whose presence is required to maintain charge neutrality-are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:"} {"id": "textbook_1880", "title": "663. Equations for Ionic Reactions - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)+2 \\mathrm{AgCl}(s) \\\\\n2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow 2 \\mathrm{AgCl}(s)\n\\end{gathered}\n$$"} {"id": "textbook_1881", "title": "663. Equations for Ionic Reactions - ", "content": "Following the convention of using the smallest possible integers as coefficients, this equation is then written:"} {"id": "textbook_1882", "title": "663. Equations for Ionic Reactions - ", "content": "$$\n\\mathrm{Cl}^{-}(a q)+\\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{AgCl}(s)\n$$"} {"id": "textbook_1883", "title": "663. Equations for Ionic Reactions - ", "content": "This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of $\\mathrm{Cl}^{-}$and $\\mathrm{Ag}^{+}$.\n"} {"id": "textbook_1884", "title": "664. EXAMPLE 7.2 - ", "content": ""} {"id": "textbook_1885", "title": "665. Ionic and Molecular Equations - ", "content": "\nWhen carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.\n"} {"id": "textbook_1886", "title": "666. Solution - ", "content": "\nBegin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:"} {"id": "textbook_1887", "title": "666. Solution - ", "content": "$$\n\\mathrm{CO}_{2}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\text { (unbalanced) }\n$$"} {"id": "textbook_1888", "title": "666. Solution - ", "content": "Balance is achieved easily in this case by changing the coefficient for NaOH to 2 , resulting in the molecular equation for this reaction:"} {"id": "textbook_1889", "title": "666. Solution - ", "content": "$$\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_1890", "title": "666. Solution - ", "content": "The two dissolved ionic compounds, NaOH and $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, can be represented as dissociated ions to yield the complete ionic equation:"} {"id": "textbook_1891", "title": "666. Solution - ", "content": "$$\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_1892", "title": "666. Solution - ", "content": "Finally, identify the spectator ion(s), in this case $\\mathrm{Na}^{+}(a q)$, and remove it from each side of the equation to generate the net ionic equation:"} {"id": "textbook_1893", "title": "666. Solution - ", "content": "$$\n\\begin{aligned}\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q) & \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{OH}^{-}(a q) & \\longrightarrow \\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{aligned}\n$$\n"} {"id": "textbook_1894", "title": "667. Check Your Learning - ", "content": "\nDiatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:"} {"id": "textbook_1895", "title": "667. Check Your Learning - ", "content": "$$\n\\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { electricity }} \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g)\n$$"} {"id": "textbook_1896", "title": "667. Check Your Learning - ", "content": "Write balanced molecular, complete ionic, and net ionic equations for this process.\n"} {"id": "textbook_1897", "title": "668. Answer: - ", "content": "\n$2 \\mathrm{NaCl}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (molecular)\n$2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (complete ionic)\n$2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{OH}^{-}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (net ionic)\n"} {"id": "textbook_1898", "title": "668. Answer: - 668.1. Classifying Chemical Reactions", "content": ""} {"id": "textbook_1899", "title": "669. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_1900", "title": "669. LEARNING OBJECTIVES - ", "content": "- Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)\n- Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations\n- Identify common acids and bases\n- Predict the solubility of common inorganic compounds by using solubility rules\n- Compute the oxidation states for elements in compounds"} {"id": "textbook_1901", "title": "669. LEARNING OBJECTIVES - ", "content": "Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction.\n"} {"id": "textbook_1902", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "\nA precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter)."} {"id": "textbook_1903", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in a later chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that\nhave been observed for many ionic compounds (Table 7.1)."} {"id": "textbook_1904", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "| Soluble Ionic Compounds | contain these ions | exceptions |\n| :---: | :---: | :---: |\n| | $\\mathrm{NH}_{4}{ }^{+}$
group I cations:
$\\mathrm{Li}^{+}$
$\\mathrm{Na}^{+}$
$\\mathrm{K}^{+}$
$\\mathrm{Rb}^{+}$
$\\mathrm{Cs}^{+}$ | none |\n| | $\\begin{aligned} & \\mathrm{Cl}^{-} \\\\ & \\mathrm{Br}^{-} \\\\ & \\mathrm{I}^{-} \\end{aligned}$ | compounds with $\\mathrm{Ag}^{+}, \\mathrm{Hg}_{2}{ }^{2+}$, and $\\mathrm{Pb}^{2+}$ |\n| | $\\mathrm{F}^{-}$ | compounds with group 2 metal cations, $\\mathrm{Pb}^{2+}$, and $\\mathrm{Fe}^{3+}$ |\n| | $\\begin{aligned} & \\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}^{-} \\\\ & \\mathrm{HCO}_{3}^{-} \\\\ & \\mathrm{NO}_{3}^{-} \\\\ & \\mathrm{ClO}_{3}^{-} \\end{aligned}$ | none |\n| | $\\mathrm{SO}_{4}{ }^{2-}$ | compounds with $\\mathrm{Ag}^{+}, \\mathrm{Ba}^{2+}, \\mathrm{Ca}^{2+}, \\mathrm{Hg}_{2}{ }^{2+}, \\mathrm{Pb}^{2+}$ and $\\mathrm{Sr}^{2+}$ |\n| Insoluble Ionic Compounds | contain these ions | exceptions |\n| | $\\begin{aligned} & \\mathrm{CO}_{3}{ }^{2-} \\\\ & \\mathrm{CrO}_{4}{ }^{2-} \\\\ & \\mathrm{PO}_{4}{ }^{3-} \\\\ & \\mathrm{S}^{2-} \\end{aligned}$ | compounds with group 1 cations and $\\mathrm{NH}_{4}{ }^{+}$ |\n| | $\\mathrm{OH}^{-}$ | compounds with group 1 cations and $\\mathrm{Ba}^{2+}$ |"} {"id": "textbook_1905", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "TABLE 7.1"} {"id": "textbook_1906", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:"} {"id": "textbook_1907", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "$$\n2 \\mathrm{KI}(a q)+\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{PbI}_{2}(s)+2 \\mathrm{KNO}_{3}(a q)\n$$"} {"id": "textbook_1908", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts."} {"id": "textbook_1909", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "The net ionic equation representing this reaction is:"} {"id": "textbook_1910", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "$$\n\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{PbI}_{2}(s)\n$$"} {"id": "textbook_1911", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "Lead iodide is a bright yellow solid that was formerly used as an artist's pigment known as iodine yellow (Figure 7.4). The properties of pure $\\mathrm{PbI}_{2}$ crystals make them useful for fabrication of X-ray and gamma ray detectors.\n"} {"id": "textbook_1912", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "FIGURE 7.4 A precipitate of $\\mathrm{PbI}_{2}$ forms when solutions containing $\\mathrm{Pb}^{2+}$ and $\\mathrm{I}^{-}$are mixed. (credit: Der Kreole/ Wikimedia Commons)"} {"id": "textbook_1913", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "The solubility guidelines in Table 7.1 may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium chloride will yield a solution containing $\\mathrm{Ag}^{+}, \\mathrm{NO}_{3}{ }^{-}$, $\\mathrm{Na}^{+}$, and $\\mathrm{Cl}^{-}$ions. Aside from the two ionic compounds originally present in the solutions, $\\mathrm{AgNO}_{3}$ and NaCl , two additional ionic compounds may be derived from this collection of ions: $\\mathrm{NaNO}_{3}$ and AgCl . The solubility guidelines indicate all nitrate salts are soluble but that AgCl is one of insoluble. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:"} {"id": "textbook_1914", "title": "670. Precipitation Reactions and Solubility Rules - ", "content": "$$\n\\begin{gathered}\n\\mathrm{NaCl}(a q)+\\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{NaNO}_{3}(a q) \\quad \\text { (molecular) } \\\\\n\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s) \\quad \\text { (net ionic) }\n\\end{gathered}\n$$\n"} {"id": "textbook_1915", "title": "671. EXAMPLE 7.3 - ", "content": ""} {"id": "textbook_1916", "title": "672. Predicting Precipitation Reactions - ", "content": "\nPredict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.\n(a) potassium sulfate and barium nitrate\n(b) lithium chloride and silver acetate\n(c) lead nitrate and ammonium carbonate\n"} {"id": "textbook_1917", "title": "673. Solution - ", "content": "\n(a) The two possible products for this combination are $\\mathrm{KNO}_{3}$ and $\\mathrm{BaSO}_{4}$. The solubility guidelines indicate\n$\\mathrm{BaSO}_{4}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is"} {"id": "textbook_1918", "title": "673. Solution - ", "content": "$$\n\\mathrm{Ba}^{2+}(a q)+\\mathrm{SO}_{4}^{2-}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)\n$$"} {"id": "textbook_1919", "title": "673. Solution - ", "content": "(b) The two possible products for this combination are $\\mathrm{LiC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ and AgCl . The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is"} {"id": "textbook_1920", "title": "673. Solution - ", "content": "$$\n\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s)\n$$"} {"id": "textbook_1921", "title": "673. Solution - ", "content": "(c) The two possible products for this combination are $\\mathrm{PbCO}_{3}$ and $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$. The solubility guidelines indicate $\\mathrm{PbCO}_{3}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is"} {"id": "textbook_1922", "title": "673. Solution - ", "content": "$$\n\\mathrm{Pb}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\longrightarrow \\mathrm{PbCO}_{3}(s)\n$$\n"} {"id": "textbook_1923", "title": "674. Check Your Learning - ", "content": "\nWhich solution could be used to precipitate the barium ion, $\\mathrm{Ba}^{2+}$, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?\n"} {"id": "textbook_1924", "title": "675. Answer: - ", "content": "\nsodium sulfate, $\\mathrm{BaSO}_{4}$\n"} {"id": "textbook_1925", "title": "676. Acid-Base Reactions - ", "content": "\nAn acid-base reaction is one in which a hydrogen ion, $\\mathrm{H}^{+}$, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text."} {"id": "textbook_1926", "title": "676. Acid-Base Reactions - ", "content": "For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, $\\mathrm{H}_{3} \\mathrm{O}^{+}$. As an example, consider the equation shown here:"} {"id": "textbook_1927", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\mathrm{HCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) \\longrightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$"} {"id": "textbook_1928", "title": "676. Acid-Base Reactions - ", "content": "The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, $\\mathrm{H}_{3} \\mathrm{O}^{+}$ions are produced by a chemical reaction in which $\\mathrm{H}^{+}$ions are transferred from HCl molecules to $\\mathrm{H}_{2} \\mathrm{O}$ molecules (Figure 7.5).\n\n\nFIGURE 7.5 When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions)."} {"id": "textbook_1929", "title": "676. Acid-Base Reactions - ", "content": "The nature of HCl is such that its reaction with water as just described is essentially $100 \\%$ efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 7.2). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:"} {"id": "textbook_1930", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$"} {"id": "textbook_1931", "title": "676. Acid-Base Reactions - ", "content": "When dissolved in water under typical conditions, only about $1 \\%$ of acetic acid molecules are present in the ionized form, $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$(Figure 7.6). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)\n\n(a)\n\n(b)"} {"id": "textbook_1932", "title": "676. Acid-Base Reactions - ", "content": "FIGURE 7.6 (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Br\u00fccke-Osteuropa/Wikimedia Commons)"} {"id": "textbook_1933", "title": "676. Acid-Base Reactions - ", "content": "Common Strong Acids"} {"id": "textbook_1934", "title": "676. Acid-Base Reactions - ", "content": "| Compound Formula | Name in Aqueous Solution |\n| :--- | :--- |\n| HBr | hydrobromic acid |\n| HCl | hydrochloric acid |\n| HI | hydroiodic acid |\n| $\\mathrm{HNO}_{3}$ | nitric acid |\n| $\\mathrm{HClO}_{4}$ | perchloric acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfuric acid |"} {"id": "textbook_1935", "title": "676. Acid-Base Reactions - ", "content": "TABLE 7.2"} {"id": "textbook_1936", "title": "676. Acid-Base Reactions - ", "content": "A base is a substance that will dissolve in water to yield hydroxide ions, $\\mathrm{OH}^{-}$. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion-for example, NaOH and $\\mathrm{Ca}(\\mathrm{OH})_{2}$. Unlike the acid compounds discussed previously, these compounds do not react chemically with water; instead they dissolve and dissociate, releasing hydroxide ions directly into the solution. For example, KOH and $\\mathrm{Ba}(\\mathrm{OH})_{2}$ dissolve in water and dissociate completely to produce cations ( $\\mathrm{K}^{+}$and $\\mathrm{Ba}^{2+}$, respectively) and hydroxide ions, $\\mathrm{OH}^{-}$. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases."} {"id": "textbook_1937", "title": "676. Acid-Base Reactions - ", "content": "Consider as an example the dissolution of lye (sodium hydroxide) in water:"} {"id": "textbook_1938", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\mathrm{NaOH}(s) \\longrightarrow \\mathrm{Na}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_1939", "title": "676. Acid-Base Reactions - ", "content": "This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield $\\mathrm{Na}^{+}$and $\\mathrm{OH}^{-}$ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.\nUnlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 7.7). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:"} {"id": "textbook_1940", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_1941", "title": "676. Acid-Base Reactions - ", "content": "This is, by definition, an acid-base reaction, in this case involving the transfer of $\\mathrm{H}^{+}$ions from water molecules to ammonia molecules. Under typical conditions, only about $1 \\%$ of the dissolved ammonia is present as $\\mathrm{NH}_{4}{ }^{+}$ ions.\n"} {"id": "textbook_1942", "title": "676. Acid-Base Reactions - ", "content": "FIGURE 7.7 Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)"} {"id": "textbook_1943", "title": "676. Acid-Base Reactions - ", "content": "A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base (but not water), and the products are often a salt and water"} {"id": "textbook_1944", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\text { acid + base } \\longrightarrow \\text { salt + water }\n$$"} {"id": "textbook_1945", "title": "676. Acid-Base Reactions - ", "content": "To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ) is ingested to ease symptoms associated with excess stomach acid ( HCl ):"} {"id": "textbook_1946", "title": "676. Acid-Base Reactions - ", "content": "$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_1947", "title": "676. Acid-Base Reactions - ", "content": "Note that in addition to water, this reaction produces a salt, magnesium chloride.\n"} {"id": "textbook_1948", "title": "677. EXAMPLE 7.4 - ", "content": ""} {"id": "textbook_1949", "title": "678. Writing Equations for Acid-Base Reactions - ", "content": "\nWrite balanced chemical equations for the acid-base reactions described here:\n(a) the weak acid hydrogen hypochlorite reacts with water\n(b) a solution of barium hydroxide is neutralized with a solution of nitric acid\n"} {"id": "textbook_1950", "title": "679. Solution - ", "content": "\n(a) The two reactants are provided, HOCl and $\\mathrm{H}_{2} \\mathrm{O}$. Since the substance is reported to be an acid, its reaction with water will involve the transfer of $\\mathrm{H}^{+}$from HOCl to $\\mathrm{H}_{2} \\mathrm{O}$ to generate hydronium ions, $\\mathrm{H}_{3} \\mathrm{O}^{+}$and hypochlorite ions, $\\mathrm{OCl}^{-}$."} {"id": "textbook_1951", "title": "679. Solution - ", "content": "$$\n\\mathrm{HOCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{OCl}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$"} {"id": "textbook_1952", "title": "679. Solution - ", "content": "A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.\n(b) The two reactants are provided, $\\mathrm{Ba}(\\mathrm{OH})_{2}$ and $\\mathrm{HNO}_{3}$. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide $\\left(\\mathrm{Ba}^{2+}\\right)$ and the anion generated when the acid transfers its hydrogen ion $\\left(\\mathrm{NO}_{3}{ }^{-}\\right)$."} {"id": "textbook_1953", "title": "679. Solution - ", "content": "$$\n\\mathrm{Ba}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_1954", "title": "679. Solution - ", "content": "Check Your Learning\nWrite the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.)\n"} {"id": "textbook_1955", "title": "680. Answer: - ", "content": "\n$\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n"} {"id": "textbook_1956", "title": "681. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_1957", "title": "682. Stomach Antacids - ", "content": "\nOur stomachs contain a solution of roughly 0.03 M HCl , which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, $\\mathrm{CaCO}_{3}$. The reaction,"} {"id": "textbook_1958", "title": "682. Stomach Antacids - ", "content": "$$\n\\mathrm{CaCO}_{3}(s)+2 \\mathrm{HCl}(a q) \\rightleftharpoons \\mathrm{CaCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_1959", "title": "682. Stomach Antacids - ", "content": "not only neutralizes stomach acid, it also produces $\\mathrm{CO}_{2}(g)$, which may result in a satisfying belch.\nMilk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, $\\mathrm{Mg}(\\mathrm{OH})_{2}$. It works according to the reaction:"} {"id": "textbook_1960", "title": "682. Stomach Antacids - ", "content": "$$\n\\operatorname{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_1961", "title": "682. Stomach Antacids - ", "content": "The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that:"} {"id": "textbook_1962", "title": "682. Stomach Antacids - ", "content": "$$\n\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{OH}^{-} \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_1963", "title": "682. Stomach Antacids - ", "content": "This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.\n"} {"id": "textbook_1964", "title": "683. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_1965", "title": "684. Culinary Aspects of Chemistry - ", "content": "\nExamples of acid-base chemistry are abundant in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. $\\mathrm{NaHCO}_{3}$ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter \"rises.\" Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter."} {"id": "textbook_1966", "title": "684. Culinary Aspects of Chemistry - ", "content": "Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 7.8). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a \"sour\" taste that we seem to enjoy.\n"} {"id": "textbook_1967", "title": "684. Culinary Aspects of Chemistry - ", "content": "FIGURE 7.8 A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish."} {"id": "textbook_1968", "title": "684. Culinary Aspects of Chemistry - ", "content": "Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.\n"} {"id": "textbook_1969", "title": "685. LINK TO LEARNING - ", "content": "\nExplore the microscopic view (http://openstax.org/l/16AcidsBases) of strong and weak acids and bases.\n"} {"id": "textbook_1970", "title": "686. Oxidation-Reduction Reactions - ", "content": "\nEarth's atmosphere contains about $20 \\%$ molecular oxygen, $\\mathrm{O}_{2}$, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving $\\mathrm{O}_{2}$, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification."} {"id": "textbook_1971", "title": "686. Oxidation-Reduction Reactions - ", "content": "Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:"} {"id": "textbook_1972", "title": "686. Oxidation-Reduction Reactions - ", "content": "$$\n2 \\mathrm{Na}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NaCl}(s)\n$$"} {"id": "textbook_1973", "title": "686. Oxidation-Reduction Reactions - ", "content": "It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:"} {"id": "textbook_1974", "title": "686. Oxidation-Reduction Reactions - ", "content": "$$\n\\begin{aligned}\n& 2 \\mathrm{Na}(s) \\longrightarrow 2 \\mathrm{Na}^{+}(s)+2 \\mathrm{e}^{-} \\\\\n& \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}(s)\n\\end{aligned}\n$$"} {"id": "textbook_1975", "title": "686. Oxidation-Reduction Reactions - ", "content": "These equations show that Na atoms lose electrons while Cl atoms (in the $\\mathrm{Cl}_{2}$ molecule) gain electrons, the \" s \" subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:"} {"id": "textbook_1976", "title": "686. Oxidation-Reduction Reactions - ", "content": "```\noxidation = loss of electrons\nreduction = gain of electrons\n```"} {"id": "textbook_1977", "title": "686. Oxidation-Reduction Reactions - ", "content": "In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium."} {"id": "textbook_1978", "title": "686. Oxidation-Reduction Reactions - ", "content": "$$\n\\begin{aligned}\n\\text { reducing agent } & =\\text { species that is oxidized } \\\\\n\\text { oxidizing agent } & =\\text { species that is reduced }\n\\end{aligned}\n$$"} {"id": "textbook_1979", "title": "686. Oxidation-Reduction Reactions - ", "content": "Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl :"} {"id": "textbook_1980", "title": "686. Oxidation-Reduction Reactions - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$"} {"id": "textbook_1981", "title": "686. Oxidation-Reduction Reactions - ", "content": "The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound were ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion."} {"id": "textbook_1982", "title": "686. Oxidation-Reduction Reactions - ", "content": "1. The oxidation number of an atom in an elemental substance is zero.\n2. The oxidation number of a monatomic ion is equal to the ion's charge.\n3. Oxidation numbers for common nonmetals are usually assigned as follows:"} {"id": "textbook_1983", "title": "686. Oxidation-Reduction Reactions - ", "content": "- Hydrogen: +1 when combined with nonmetals, -1 when combined with metals\n- Oxygen: - 2 in most compounds, sometimes -1 (so-called peroxides, $\\mathrm{O}_{2}{ }^{2-}$ ), very rarely $-\\frac{1}{2}$ (so-called superoxides, $\\mathrm{O}_{2}{ }^{-}$), positive values when combined with F (values vary)\n- Halogens: -1 for $F$ always, -1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)"} {"id": "textbook_1984", "title": "686. Oxidation-Reduction Reactions - ", "content": "4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion."} {"id": "textbook_1985", "title": "686. Oxidation-Reduction Reactions - ", "content": "Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., $2+$ ), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2 ). This convention aims to emphasize the distinction between these two related properties.\n"} {"id": "textbook_1986", "title": "687. EXAMPLE 7.5 - ", "content": ""} {"id": "textbook_1987", "title": "688. Assigning Oxidation Numbers - ", "content": "\nFollow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:\n(a) $\\mathrm{H}_{2} \\mathrm{~S}$\n(b) $\\mathrm{SO}_{3}{ }^{2-}$\n(c) $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$\n"} {"id": "textbook_1988", "title": "689. Solution - ", "content": "\n(a) According to guideline 3 , the oxidation number for H is +1 ."} {"id": "textbook_1989", "title": "689. Solution - ", "content": "Using this oxidation number and the compound's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:"} {"id": "textbook_1990", "title": "689. Solution - ", "content": "$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{H}_{2} \\mathrm{~S}=0=(2 \\times+1)+(1 \\times x) \\\\\nx=0-(2 \\times+1)=-2\n\\end{gathered}\n$$"} {"id": "textbook_1991", "title": "689. Solution - ", "content": "(b) Guideline 3 suggests the oxidation number for oxygen is -2 ."} {"id": "textbook_1992", "title": "689. Solution - ", "content": "Using this oxidation number and the ion's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:"} {"id": "textbook_1993", "title": "689. Solution - ", "content": "$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{SO}_{3}{ }^{2-}=-2=(3 \\times-2)+(1 \\times x) \\\\\nx=-2-(3 \\times-2)=+4\n\\end{gathered}\n$$"} {"id": "textbook_1994", "title": "689. Solution - ", "content": "(c) For ionic compounds, it's convenient to assign oxidation numbers for the cation and anion separately."} {"id": "textbook_1995", "title": "689. Solution - ", "content": "According to guideline 2, the oxidation number for sodium is +1 .\nAssuming the usual oxidation number for oxygen ( -2 per guideline 3 ), the oxidation number for sulfur is calculated as directed by guideline 4:"} {"id": "textbook_1996", "title": "689. Solution - ", "content": "$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{SO}_{4}{ }^{2-}=-2=(4 \\times-2)+(1 \\times x) \\\\\nx=-2-(4 \\times-2)=+6\n\\end{gathered}\n$$\n"} {"id": "textbook_1997", "title": "690. Check Your Learning - ", "content": "\nAssign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:\n(a) $\\mathrm{KNO}_{3}$\n(b) $\\mathrm{AlH}_{3}$\n(c) $\\mathrm{NH}_{4}{ }^{+}$\n(d) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$"} {"id": "textbook_1998", "title": "690. Check Your Learning - ", "content": "Answer:\n(a) N, +5; (b) Al, +3; (c) N, -3; (d) P, +5"} {"id": "textbook_1999", "title": "690. Check Your Learning - ", "content": "Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 7.6.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:"} {"id": "textbook_2000", "title": "690. Check Your Learning - ", "content": "$$\n\\begin{aligned}\n\\text { oxidation } & =\\text { increase in oxidation number } \\\\\n\\text { reduction } & =\\text { decrease in oxidation number }\n\\end{aligned}\n$$"} {"id": "textbook_2001", "title": "690. Check Your Learning - ", "content": "Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl ) and chlorine is reduced (its oxidation number decreases from 0 in $\\mathrm{Cl}_{2}$ to -1 in NaCl ). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in $\\mathrm{H}_{2}$ to +1 in HCl ) and chlorine is reduced (its oxidation number decreases from 0 in\n$\\mathrm{Cl}_{2}$ to -1 in HCl ).\nSeveral subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted in Figure 7.1 are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:"} {"id": "textbook_2002", "title": "690. Check Your Learning - ", "content": "$$\n10 \\mathrm{Al}(s)+6 \\mathrm{NH}_{4} \\mathrm{ClO}_{4}(s) \\longrightarrow 4 \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{AlCl}_{3}(s)+12 \\mathrm{H}_{2} \\mathrm{O}(g)+3 \\mathrm{~N}_{2}(g)\n$$\n"} {"id": "textbook_2003", "title": "691. LINK TO LEARNING - ", "content": "\nWatch a brief video (http://openstax.org/1/16hybridrocket) showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at\n3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture."} {"id": "textbook_2004", "title": "691. LINK TO LEARNING - ", "content": "Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:"} {"id": "textbook_2005", "title": "691. LINK TO LEARNING - ", "content": "$$\n\\mathrm{Zn}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{ZnCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_2006", "title": "691. LINK TO LEARNING - ", "content": "Metallic elements may also be oxidized by solutions of other metal salts; for example:"} {"id": "textbook_2007", "title": "691. LINK TO LEARNING - ", "content": "$$\n\\mathrm{Cu}(s)+2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{Ag}(s)\n$$"} {"id": "textbook_2008", "title": "691. LINK TO LEARNING - ", "content": "This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting $\\mathrm{Cu}^{2+}$ ions dissolve in the solution to yield a characteristic blue color (Figure 7.9).\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 7.9 (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)\n"} {"id": "textbook_2009", "title": "692. EXAMPLE 7.6 - ", "content": ""} {"id": "textbook_2010", "title": "693. Describing Redox Reactions - ", "content": "\nIdentify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.\n(a) $\\mathrm{ZnCO}_{3}(s) \\longrightarrow \\mathrm{ZnO}(s)+\\mathrm{CO}_{2}(g)$\n(b) $2 \\mathrm{Ga}(l)+3 \\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{GaBr}_{3}(s)$\n(c) $2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)$\n(d) $\\mathrm{BaCl}_{2}(a q)+\\mathrm{K}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)+2 \\mathrm{KCl}(a q)$\n(e) $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$\n"} {"id": "textbook_2011", "title": "694. Solution - ", "content": "\nRedox reactions are identified per definition if one or more elements undergo a change in oxidation number.\n(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.\n(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in $\\mathrm{Ga}(I)$ to +3 in $\\operatorname{GaBr}_{3}(S)$. The reducing agent is $\\mathrm{Ga}(I)$. Bromine is reduced, its oxidation number decreasing from 0 in $\\mathrm{Br}_{2}(I)$ to -1 in $\\operatorname{GaBr}_{3}(S)$. The oxidizing agent is $\\mathrm{Br}_{2}(I)$.\n(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from -1 in $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{aq})$ to 0 in $\\mathrm{O}_{2}(\\mathrm{~g})$. Oxygen is also reduced, its oxidation number decreasing from -1 in $\\mathrm{H}_{2} \\mathrm{O}_{2}(a q)$ to -2 in $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$. For disproportionation reactions, the same substance functions as an oxidant and a reductant.\n(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.\n(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from -2 in $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ to +4 in $\\mathrm{CO}_{2}(\\mathrm{~g})$. The reducing agent (fuel) is $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$. Oxygen is reduced, its oxidation number decreasing from 0 in $\\mathrm{O}_{2}(\\mathrm{~g})$ to -2 in $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$. The oxidizing agent is $\\mathrm{O}_{2}(\\mathrm{~g})$.\n"} {"id": "textbook_2012", "title": "695. Check Your Learning - ", "content": "\nThis equation describes the production of tin(II) chloride:"} {"id": "textbook_2013", "title": "695. Check Your Learning - ", "content": "$$\n\\mathrm{Sn}(s)+2 \\mathrm{HCl}(g) \\longrightarrow \\mathrm{SnCl}_{2}(s)+\\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_2014", "title": "695. Check Your Learning - ", "content": "Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.\n"} {"id": "textbook_2015", "title": "696. Answer: - ", "content": "\nYes, a single-replacement reaction. $\\mathrm{Sn}(s)$ is the reductant, $\\mathrm{HCl}(g)$ is the oxidant.\n"} {"id": "textbook_2016", "title": "697. Balancing Redox Reactions via the Half-Reaction Method - ", "content": "\nRedox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:"} {"id": "textbook_2017", "title": "697. Balancing Redox Reactions via the Half-Reaction Method - ", "content": "1. Write the two half-reactions representing the redox process.\n2. Balance all elements except oxygen and hydrogen.\n3. Balance oxygen atoms by adding $\\mathrm{H}_{2} \\mathrm{O}$ molecules.\n4. Balance hydrogen atoms by adding $\\mathrm{H}^{+}$ions.\n5. Balance charge by adding electrons.\n6. If necessary, multiply each half-reaction's coefficients by the smallest possible integers to yield equal numbers of electrons in each.\n7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.\n8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:\na. Add $\\mathrm{OH}^{-}$ions to both sides of the equation in numbers equal to the number of $\\mathrm{H}^{+}$ions.\nb. On the side of the equation containing both $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$ions, combine these ions to yield water molecules.\nc. Simplify the equation by removing any redundant water molecules.\n9. Finally, check to see that both the number of atoms and the total charges ${ }^{\\frac{1}{}}$ are balanced.\n"} {"id": "textbook_2018", "title": "698. EXAMPLE 7.7 - ", "content": ""} {"id": "textbook_2019", "title": "699. Balancing Redox Reactions in Acidic Solution - ", "content": "\nWrite a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution."} {"id": "textbook_2020", "title": "699. Balancing Redox Reactions in Acidic Solution - ", "content": "$$\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Cr}^{3+}+\\mathrm{Fe}^{3+}\n$$\n"} {"id": "textbook_2021", "title": "700. Solution - ", "content": "\nStep 1.\nWrite the two half-reactions.\nEach half-reaction will contain one reactant and one product with one element in common."} {"id": "textbook_2022", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow \\mathrm{Cr}^{3+}\n\\end{gathered}\n$$"} {"id": "textbook_2023", "title": "700. Solution - ", "content": "Step 2.\nBalance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms."} {"id": "textbook_2024", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow 2 \\mathrm{Cr}^{3+}\n\\end{gathered}\n$$"} {"id": "textbook_2025", "title": "700. Solution - ", "content": "Step 3.\nBalance oxygen atoms by adding $\\mathrm{H}_{2} \\mathrm{O}$ molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side."} {"id": "textbook_2026", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$"} {"id": "textbook_2027", "title": "700. Solution - ", "content": "Step 4.\nBalance hydrogen atoms by adding $\\mathrm{H}^{+}$ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side."} {"id": "textbook_2028", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$"} {"id": "textbook_2029", "title": "700. Solution - ", "content": "Step 5.\nBalance charge by adding electrons. The iron half-reaction shows a total charge of $2+$ on the left side (1 $\\mathrm{Fe}^{2+}$ ion) and $3+$ on the right side ( $1 \\mathrm{Fe}^{3+}$ ion). Adding one electron to the right side brings that side's total charge to $(3+)+(1-)=2+$, and charge balance is achieved.\n\nThe chromium half-reaction shows a total charge of $(1 \\times 2-)+(14 \\times 1+)=12+$ on the left side $\\left(1 \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right.$ ion and $14 \\mathrm{H}^{+}$ions). The total charge on the right side is $(2 \\times 3+)=6+\\left(2 \\mathrm{Cr}^{3+}\\right.$ ions $)$. Adding six electrons to\nthe left side will bring that side's total charge to $(12++6-)=6+$, and charge balance is achieved."} {"id": "textbook_2030", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+}+\\mathrm{e}^{-} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$"} {"id": "textbook_2031", "title": "700. Solution - ", "content": "Step 6.\nMultiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction's coefficient must be multiplied by 6 ."} {"id": "textbook_2032", "title": "700. Solution - ", "content": "$$\n\\begin{gathered}\n6 \\mathrm{Fe}^{2+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+6 \\mathrm{e}^{-} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$"} {"id": "textbook_2033", "title": "700. Solution - ", "content": "Step 7.\nAdd the balanced half-reactions and cancel species that appear on both sides of the equation."} {"id": "textbook_2034", "title": "700. Solution - ", "content": "$$\n6 \\mathrm{Fe}^{2+}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+6 \\mathrm{e}^{-}+2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_2035", "title": "700. Solution - ", "content": "Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:"} {"id": "textbook_2036", "title": "700. Solution - ", "content": "$$\n6 \\mathrm{Fe}^{2+}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_2037", "title": "700. Solution - ", "content": "A final check of atom and charge balance confirms the equation is balanced."} {"id": "textbook_2038", "title": "700. Solution - ", "content": "| | Reactants | Products |\n| :--- | :--- | :--- |\n| Fe | 6 | 6 |\n| Cr | 2 | 2 |\n| O | 7 | 7 |\n| H | 14 | 14 |\n| charge | $24+$ | $24+$ |\n"} {"id": "textbook_2039", "title": "701. Check Your Learning - ", "content": "\nIn basic solution, molecular chlorine, $\\mathrm{Cl}_{2}$, reacts with hydroxide ions, $\\mathrm{OH}^{-}$, to yield chloride ions, $\\mathrm{Cl}^{-}$. and chlorate ions, $\\mathrm{ClO}_{3}{ }^{-}$. HINT: This is a disproportionation reaction in which the element chlorine is both oxidized and reduced. Write a balanced equation for this reaction.\n"} {"id": "textbook_2040", "title": "702. Answer: - ", "content": "\n$3 \\mathrm{Cl}_{2}(a q)+6 \\mathrm{OH}^{-}(a q) \\longrightarrow 5 \\mathrm{Cl}^{-}(a q)+\\mathrm{ClO}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n"} {"id": "textbook_2041", "title": "702. Answer: - 702.1. Reaction Stoichiometry", "content": ""} {"id": "textbook_2042", "title": "703. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2043", "title": "703. LEARNING OBJECTIVES - ", "content": "- Explain the concept of stoichiometry as it pertains to chemical reactions\n- Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products\n- Perform stoichiometric calculations involving mass, moles, and solution molarity"} {"id": "textbook_2044", "title": "703. LEARNING OBJECTIVES - ", "content": "A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing\nclassification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction's stoichiometry, a term derived from the Greek words stoicheion (meaning \"element\") and metron (meaning \"measure\"). In this module, the use of balanced chemical equations for various stoichiometric applications is explored."} {"id": "textbook_2045", "title": "703. LEARNING OBJECTIVES - ", "content": "The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $\\frac{3}{4}$ cup milk, and one egg. The \"equation\" representing the preparation of pancakes per this recipe is"} {"id": "textbook_2046", "title": "703. LEARNING OBJECTIVES - ", "content": "$$\n1 \\text { cup mix }+\\frac{3}{4} \\text { cup milk }+1 \\text { egg } \\longrightarrow 8 \\text { pancakes }\n$$"} {"id": "textbook_2047", "title": "703. LEARNING OBJECTIVES - ", "content": "If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is"} {"id": "textbook_2048", "title": "703. LEARNING OBJECTIVES - ", "content": "$$\n24 \\text { pancakes } \\times \\frac{1 \\mathrm{egg}}{8 \\text { pancakes- }}=3 \\mathrm{eggs}\n$$"} {"id": "textbook_2049", "title": "703. LEARNING OBJECTIVES - ", "content": "Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:"} {"id": "textbook_2050", "title": "703. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)\n$$"} {"id": "textbook_2051", "title": "703. LEARNING OBJECTIVES - ", "content": "This equation shows ammonia molecules are produced from hydrogen molecules in a $2: 3$ ratio, and stoichiometric factors may be derived using any amount (number) unit:\n\n$$\n\\frac{2 \\mathrm{NH}_{3} \\text { molecules }}{3 \\mathrm{H}_{2} \\text { molecules }} \\text { or } \\frac{2 \\text { doz } \\mathrm{NH}_{3} \\text { molecules }}{3 \\text { doz } \\mathrm{H}_{2} \\text { molecules }} \\text { or } \\frac{2 \\mathrm{~mol} \\mathrm{NH}_{3} \\text { molecules }}{3 \\mathrm{~mol} \\mathrm{H}_{2} \\text { molecules }}\n$$"} {"id": "textbook_2052", "title": "703. LEARNING OBJECTIVES - ", "content": "These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.\n"} {"id": "textbook_2053", "title": "704. EXAMPLE 7.8 - ", "content": ""} {"id": "textbook_2054", "title": "705. Moles of Reactant Required in a Reaction - ", "content": "\nHow many moles of $\\mathrm{I}_{2}$ are required to react with 0.429 mol of Al according to the following equation (see Figure 7.10)?"} {"id": "textbook_2055", "title": "705. Moles of Reactant Required in a Reaction - ", "content": "$$\n2 \\mathrm{Al}+3 \\mathrm{I}_{2} \\longrightarrow 2 \\mathrm{AlI}_{3}\n$$"} {"id": "textbook_2056", "title": "705. Moles of Reactant Required in a Reaction - ", "content": ""} {"id": "textbook_2057", "title": "705. Moles of Reactant Required in a Reaction - ", "content": "FIGURE 7.10 Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)\n"} {"id": "textbook_2058", "title": "706. Solution - ", "content": "\nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest\nis $\\frac{3 \\mathrm{~mol} \\mathrm{I}_{2}}{2 \\mathrm{~mol} \\mathrm{Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:\n"} {"id": "textbook_2059", "title": "706. Solution - ", "content": "$$\n\\begin{aligned}\n\\mathrm{mol} \\mathrm{I}_{2} & =0.429 \\mathrm{~mol} \\mathrm{At} \\times \\frac{3 \\mathrm{~mol} \\mathrm{I}_{2}}{2 \\mathrm{~mol} \\mathrm{At}} \\\\\n& =0.644 \\mathrm{~mol} \\mathrm{I}_{2}\n\\end{aligned}\n$$\n"} {"id": "textbook_2060", "title": "707. Check Your Learning - ", "content": "\nHow many moles of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ are required to react with 1.36 mol of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ to produce $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ according to the equation $3 \\mathrm{Ca}(\\mathrm{OH})_{2}+2 \\mathrm{H}_{3} \\mathrm{PO}_{4} \\longrightarrow \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}+6 \\mathrm{H}_{2} \\mathrm{O}$ ?\n"} {"id": "textbook_2061", "title": "708. Answer: - ", "content": "\n2.04 mol\n"} {"id": "textbook_2062", "title": "709. EXAMPLE 7.9 - ", "content": ""} {"id": "textbook_2063", "title": "710. Number of Product Molecules Generated by a Reaction - ", "content": "\nHow many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?"} {"id": "textbook_2064", "title": "710. Number of Product Molecules Generated by a Reaction - ", "content": "$$\n\\mathrm{C}_{3} \\mathrm{H}_{8}+5 \\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{CO}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}\n$$\n"} {"id": "textbook_2065", "title": "711. Solution - ", "content": "\nThe approach here is the same as for Example 7.8, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro's number."} {"id": "textbook_2066", "title": "711. Solution - ", "content": "The balanced equation shows that carbon dioxide is produced from propane in a $3: 1$ ratio:"} {"id": "textbook_2067", "title": "711. Solution - ", "content": "$$\n\\frac{3 \\mathrm{~mol} \\mathrm{CO}_{2}}{1 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{8}}\n$$"} {"id": "textbook_2068", "title": "711. Solution - ", "content": "Using this stoichiometric factor, the provided molar amount of propane, and Avogadro's number,\n\n"} {"id": "textbook_2069", "title": "712. Check Your Learning - ", "content": "\nHow many $\\mathrm{NH}_{3}$ molecules are produced by the reaction of 4.0 mol of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ according to the following equation:"} {"id": "textbook_2070", "title": "712. Check Your Learning - ", "content": "$$\n\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}+\\mathrm{Ca}(\\mathrm{OH})_{2} \\longrightarrow 2 \\mathrm{NH}_{3}+\\mathrm{CaSO}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$\n"} {"id": "textbook_2071", "title": "713. Answer: - ", "content": "\n$4.8 \\times 10^{24} \\mathrm{NH}_{3}$ molecules"} {"id": "textbook_2072", "title": "713. Answer: - ", "content": "These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an\neasy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.\n"} {"id": "textbook_2073", "title": "714. EXAMPLE 7.10 - ", "content": ""} {"id": "textbook_2074", "title": "715. Relating Masses of Reactants and Products - ", "content": "\nWhat mass of sodium hydroxide, NaOH , would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ] by the following reaction?"} {"id": "textbook_2075", "title": "715. Relating Masses of Reactants and Products - ", "content": "$$\n\\mathrm{MgCl}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{NaCl}(a q)\n$$\n"} {"id": "textbook_2076", "title": "716. Solution - ", "content": "\nThe approach used previously in Example 7.8 and Example 7.9 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:\n\n"} {"id": "textbook_2077", "title": "717. Check Your Learning - ", "content": "\nWhat mass of gallium oxide, $\\mathrm{Ga}_{2} \\mathrm{O}_{3}$, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $4 \\mathrm{Ga}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{Ga}_{2} \\mathrm{O}_{3}$.\n"} {"id": "textbook_2078", "title": "718. Answer: - ", "content": "\n39.0 g\n"} {"id": "textbook_2079", "title": "719. EXAMPLE 7.11 - ", "content": ""} {"id": "textbook_2080", "title": "720. Relating Masses of Reactants - ", "content": "\nWhat mass of oxygen gas, $\\mathrm{O}_{2}$, from the air is consumed in the combustion of 702 g of octane, $\\mathrm{C}_{8} \\mathrm{H}_{18}$, one of the principal components of gasoline?"} {"id": "textbook_2081", "title": "720. Relating Masses of Reactants - ", "content": "$$\n2 \\mathrm{C}_{8} \\mathrm{H}_{18}+25 \\mathrm{O}_{2} \\longrightarrow 16 \\mathrm{CO}_{2}+18 \\mathrm{H}_{2} \\mathrm{O}\n$$\n"} {"id": "textbook_2082", "title": "721. Solution - ", "content": "\nThe approach required here is the same as for the Example 7.10, differing only in that the provided and requested masses are both for reactant species.\n\n"} {"id": "textbook_2083", "title": "722. Check Your Learning - ", "content": "\nWhat mass of CO is required to react with 25.13 g of $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ according to the equation\n$\\mathrm{Fe}_{2} \\mathrm{O}_{3}+3 \\mathrm{CO} \\longrightarrow 2 \\mathrm{Fe}+3 \\mathrm{CO}_{2}$ ?\n"} {"id": "textbook_2084", "title": "723. Answer: - ", "content": "\n13.22 g"} {"id": "textbook_2085", "title": "723. Answer: - ", "content": "These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 7.11 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.\n\n\nFIGURE 7.11 The flowchart depicts the various computational steps involved in most reaction stoichiometry\ncalculations.\n"} {"id": "textbook_2086", "title": "724. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2087", "title": "725. Airbags - ", "content": "\nAirbags (Figure 7.12) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, $\\mathrm{NaN}_{3}$. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of $\\mathrm{NaN}_{3}$ to initiate its decomposition:"} {"id": "textbook_2088", "title": "725. Airbags - ", "content": "$$\n2 \\mathrm{NaN}_{3}(s) \\longrightarrow 3 \\mathrm{~N}_{2}(g)+2 \\mathrm{Na}(s)\n$$"} {"id": "textbook_2089", "title": "725. Airbags - ", "content": "This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second ( $\\sim 0.03-0.1 \\mathrm{~s}$ ). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass $(\\sim 100 \\mathrm{~g})$ of $\\mathrm{NaN}_{3}$ will generate approximately 50 L of $\\mathrm{N}_{2}$.\n"} {"id": "textbook_2090", "title": "725. Airbags - ", "content": "FIGURE 7.12 Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)\n"} {"id": "textbook_2091", "title": "725. Airbags - 725.1. Reaction Yields", "content": ""} {"id": "textbook_2092", "title": "726. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2093", "title": "726. LEARNING OBJECTIVES - ", "content": "- Explain the concepts of theoretical yield and limiting reactants/reagents.\n- Derive the theoretical yield for a reaction under specified conditions.\n- Calculate the percent yield for a reaction."} {"id": "textbook_2094", "title": "726. LEARNING OBJECTIVES - ", "content": "The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.\n"} {"id": "textbook_2095", "title": "727. Limiting Reactant - ", "content": "\nConsider another food analogy, making grilled cheese sandwiches (Figure 7.13):\n1 slice of cheese +2 slices of bread $\\longrightarrow 1$ sandwich\nStoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio."} {"id": "textbook_2096", "title": "727. Limiting Reactant - ", "content": "Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.\n"} {"id": "textbook_2097", "title": "727. Limiting Reactant - ", "content": "FIGURE 7.13 Sandwich making can illustrate the concepts of limiting and excess reactants.\nConsider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:"} {"id": "textbook_2098", "title": "727. Limiting Reactant - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$"} {"id": "textbook_2099", "title": "727. Limiting Reactant - ", "content": "The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of $\\mathrm{H}_{2}$ and 2 moles of $\\mathrm{Cl}_{2}$. This represents a $3: 2$ (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine ( 2 mol ) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted."} {"id": "textbook_2100", "title": "727. Limiting Reactant - ", "content": "An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction's stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield"} {"id": "textbook_2101", "title": "727. Limiting Reactant - ", "content": "$$\n\\text { mol } \\mathrm{HCl} \\text { produced }=3 \\mathrm{~mol} \\mathrm{H}_{2} \\times \\frac{2 \\mathrm{~mol} \\mathrm{HCl}}{1 \\mathrm{~mol} \\mathrm{H}_{2}}=6 \\mathrm{~mol} \\mathrm{HCl}\n$$"} {"id": "textbook_2102", "title": "727. Limiting Reactant - ", "content": "Complete reaction of the provided chlorine would produce"} {"id": "textbook_2103", "title": "727. Limiting Reactant - ", "content": "$$\n\\text { mol } \\mathrm{HCl} \\text { produced }=2 \\mathrm{~mol} \\mathrm{Cl}_{2} \\times \\frac{2 \\mathrm{~mol} \\mathrm{HCl}}{1 \\mathrm{~mol} \\mathrm{Cl}_{2}}=4 \\mathrm{~mol} \\mathrm{HCl}\n$$"} {"id": "textbook_2104", "title": "727. Limiting Reactant - ", "content": "The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl , there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 7.14).\n\n\nFIGURE 7.14 When $\\mathrm{H}_{2}$ and $\\mathrm{Cl}_{2}$ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.\n"} {"id": "textbook_2105", "title": "728. LINK TO LEARNING - ", "content": "\nView this interactive simulation (http://openstax.org/l/16reactantprod) illustrating the concepts of limiting and excess reactants.\n"} {"id": "textbook_2106", "title": "729. EXAMPLE 7.12 - ", "content": ""} {"id": "textbook_2107", "title": "730. Identifying the Limiting Reactant - ", "content": "\nSilicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:"} {"id": "textbook_2108", "title": "730. Identifying the Limiting Reactant - ", "content": "$$\n3 \\mathrm{Si}(s)+2 \\mathrm{~N}_{2}(g) \\longrightarrow \\mathrm{Si}_{3} \\mathrm{~N}_{4}(s)\n$$"} {"id": "textbook_2109", "title": "730. Identifying the Limiting Reactant - ", "content": "Which is the limiting reactant when 2.00 g of Si and 1.50 g of $\\mathrm{N}_{2}$ react?\n"} {"id": "textbook_2110", "title": "731. Solution - ", "content": "\nCompute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant."} {"id": "textbook_2111", "title": "731. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{mol} \\mathrm{Si}=2.00 \\mathrm{~g} \\mathrm{Si} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}}{28.09 \\mathrm{~g} \\mathrm{Si}}=0.0712 \\mathrm{~mol} \\mathrm{Si} \\\\\n\\mathrm{~mol} \\mathrm{~N}_{2}=1.50-\\mathrm{g}_{\\mathrm{g}}^{\\mathrm{g} \\mathrm{~N}_{2}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{~N}_{2}}{28.02-\\frac{\\mathrm{g} \\mathrm{~N}_{2}}{}}=0.0535 \\mathrm{~mol} \\mathrm{~N}_{2}\n\\end{gathered}\n$$"} {"id": "textbook_2112", "title": "731. Solution - ", "content": "The provided $\\mathrm{Si}: \\mathrm{N}_{2}$ molar ratio is:"} {"id": "textbook_2113", "title": "731. Solution - ", "content": "$$\n\\frac{0.0712 \\mathrm{~mol} \\mathrm{Si}}{0.0535 \\mathrm{~mol} \\mathrm{~N}_{2}}=\\frac{1.33 \\mathrm{~mol} \\mathrm{Si}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\n$$"} {"id": "textbook_2114", "title": "731. Solution - ", "content": "The stoichiometric Si: $\\mathrm{N}_{2}$ ratio is:"} {"id": "textbook_2115", "title": "731. Solution - ", "content": "$$\n\\frac{3 \\mathrm{~mol} \\mathrm{Si}}{2 \\mathrm{~mol} \\mathrm{~N}_{2}}=\\frac{1.5 \\mathrm{~mol} \\mathrm{Si}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\n$$"} {"id": "textbook_2116", "title": "731. Solution - ", "content": "Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant."} {"id": "textbook_2117", "title": "731. Solution - ", "content": "Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield"} {"id": "textbook_2118", "title": "731. Solution - ", "content": "$$\n\\mathrm{mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4} \\text { produced }=0.0712 \\mathrm{~mol} \\mathrm{Si} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}}{3 \\mathrm{~mol} \\mathrm{Si}}=0.0237 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}\n$$"} {"id": "textbook_2119", "title": "731. Solution - ", "content": "while the 0.0535 moles of nitrogen would produce"} {"id": "textbook_2120", "title": "731. Solution - ", "content": "$$\n\\mathrm{mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4} \\text { produced }=0.0535 \\mathrm{~mol} \\mathrm{~N}_{2} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}}{2 \\mathrm{~mol} \\mathrm{~N}_{2}}=0.0268 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}\n$$"} {"id": "textbook_2121", "title": "731. Solution - ", "content": "Since silicon yields the lesser amount of product, it is the limiting reactant.\n"} {"id": "textbook_2122", "title": "732. Check Your Learning - ", "content": "\nWhich is the limiting reactant when 5.00 g of $\\mathrm{H}_{2}$ and 10.0 g of $\\mathrm{O}_{2}$ react and form water?\n"} {"id": "textbook_2123", "title": "733. Answer: - ", "content": "\n$\\mathrm{O}_{2}$\n"} {"id": "textbook_2124", "title": "734. Percent Yield - ", "content": "\nThe amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction's theoretical yield is achieved is commonly expressed as its percent yield:"} {"id": "textbook_2125", "title": "734. Percent Yield - ", "content": "$$\n\\text { percent yield }=\\frac{\\text { actual yield }}{\\text { theoretical yield }} \\times 100 \\%\n$$"} {"id": "textbook_2126", "title": "734. Percent Yield - ", "content": "Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.\n"} {"id": "textbook_2127", "title": "735. EXAMPLE 7.13 - ", "content": ""} {"id": "textbook_2128", "title": "736. Calculation of Percent Yield - ", "content": "\nUpon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:"} {"id": "textbook_2129", "title": "736. Calculation of Percent Yield - ", "content": "$$\n\\mathrm{CuSO}_{4}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Cu}(s)+\\mathrm{ZnSO}_{4}(a q)\n$$"} {"id": "textbook_2130", "title": "736. Calculation of Percent Yield - ", "content": "What is the percent yield?\n"} {"id": "textbook_2131", "title": "737. Solution - ", "content": "\nThe provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:"} {"id": "textbook_2132", "title": "737. Solution - ", "content": "$$\n1.274-\\mathrm{g} \\mathrm{CuSO}_{4} \\times \\frac{1 \\mathrm{molCuSO}_{4}}{159.62-\\mathrm{gCuSO}_{4}} \\times \\frac{1 \\mathrm{molCu}}{1 \\mathrm{molCuSO}_{4}} \\times \\frac{63.55 \\mathrm{~g} \\mathrm{Cu}}{1 \\mathrm{~mol} \\mathrm{Cu}}=0.5072 \\mathrm{~g} \\mathrm{Cu}\n$$"} {"id": "textbook_2133", "title": "737. Solution - ", "content": "Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be"} {"id": "textbook_2134", "title": "737. Solution - ", "content": "$$\n\\begin{aligned}\n\\text { percent yield } & =\\left(\\frac{\\text { actual yield }}{\\text { theoretical yield }}\\right) \\times 100 \\\\\n\\text { percent yield } & =\\left(\\frac{0.392 \\mathrm{~g} \\mathrm{Cu}}{0.5072 \\mathrm{~g} \\mathrm{Cu}}\\right) \\times 100 \\\\\n& =77.3 \\%\n\\end{aligned}\n$$\n"} {"id": "textbook_2135", "title": "738. Check Your Learning - ", "content": "\n excess HF?"} {"id": "textbook_2136", "title": "738. Check Your Learning - ", "content": "$$\n\\mathrm{CCl}_{4}+2 \\mathrm{HF} \\longrightarrow \\mathrm{CF}_{2} \\mathrm{Cl}_{2}+2 \\mathrm{HCl}\n$$\n"} {"id": "textbook_2137", "title": "739. Answer: - ", "content": "\n48.3\\%\n"} {"id": "textbook_2138", "title": "740. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_2139", "title": "741. Green Chemistry and Atom Economy - ", "content": "\nThe purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the \"Twelve Principles of Green Chemistry\" (see details at this website (http://openstax.org/l/16greenchem)). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:"} {"id": "textbook_2140", "title": "741. Green Chemistry and Atom Economy - ", "content": "$$\n\\text { atom economy }=\\frac{\\text { mass of product }}{\\text { mass of reactants }} \\times 100 \\%\n$$"} {"id": "textbook_2141", "title": "741. Green Chemistry and Atom Economy - ", "content": "Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry."} {"id": "textbook_2142", "title": "741. Green Chemistry and Atom Economy - ", "content": "The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 7.15). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole ( 206 g ) of ibuprofen, an atom economy of $40 \\%$. In the 1990 s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of $\\sim 80 \\%$, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency's Greener Synthetic Pathways Award in 1997.\n\n(a)\n
Gases - ", "content": "\n"} {"id": "textbook_2213", "title": "763. CHAPTER 8
Gases - ", "content": "Figure 8.1 The hot air inside these balloons is less dense than the surrounding cool air. This results in a buoyant force that causes the balloons to rise when their guy lines are untied. (credit: modification of work by Anthony Quintano)\n"} {"id": "textbook_2214", "title": "764. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_2215", "title": "764. CHAPTER OUTLINE - 764.1. Gas Pressure", "content": "\n8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n"} {"id": "textbook_2216", "title": "764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases", "content": "\n8.5 The Kinetic-Molecular Theory\n8.6 Non-Ideal Gas Behavior"} {"id": "textbook_2217", "title": "764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases", "content": "INTRODUCTION We are surrounded by an ocean of gas-the atmosphere-and many of the properties of gases are familiar to us from our daily activities. Heated gases expand, which can make a hot air balloon rise (Figure 8.1) or cause a blowout in a bicycle tire left in the sun on a hot day."} {"id": "textbook_2218", "title": "764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases", "content": "Gases have played an important part in the development of chemistry. In the seventeenth and eighteenth centuries, many scientists investigated gas behavior, providing the first mathematical descriptions of the behavior of matter."} {"id": "textbook_2219", "title": "764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases", "content": "In this chapter, we will examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it.\n"} {"id": "textbook_2220", "title": "764. CHAPTER OUTLINE - 764.3. Gas Pressure", "content": ""} {"id": "textbook_2221", "title": "765. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2222", "title": "765. LEARNING OBJECTIVES - ", "content": "- Define the property of pressure\n- Define and convert among the units of pressure measurements\n- Describe the operation of common tools for measuring gas pressure\n- Calculate pressure from manometer data"} {"id": "textbook_2223", "title": "765. LEARNING OBJECTIVES - ", "content": "The earth's atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes-for example, when your ears \"pop\" during takeoff and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure 8.2). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.\n"} {"id": "textbook_2224", "title": "765. LEARNING OBJECTIVES - ", "content": "FIGURE 8.2 The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail.\n"} {"id": "textbook_2225", "title": "766. LINK TO LEARNING - ", "content": "\nA dramatic illustration (http://openstax.org/l/16atmospressur1) of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased."} {"id": "textbook_2226", "title": "766. LINK TO LEARNING - ", "content": "A smaller scale demonstration (http://openstax.org/l/16atmospressur2) of this phenomenon is briefly explained."} {"id": "textbook_2227", "title": "766. LINK TO LEARNING - ", "content": "Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant."} {"id": "textbook_2228", "title": "766. LINK TO LEARNING - ", "content": "In general, pressure is defined as the force exerted on a given area: $P=\\frac{F}{A}$. Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area."} {"id": "textbook_2229", "title": "766. LINK TO LEARNING - ", "content": "Let's apply this concept to determine which exerts a greater pressure in Figure 8.3-the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft\n(footprint area of $250 \\mathrm{in}^{2}$ ), so the pressure exerted by each foot is about $14 \\mathrm{lb} / \\mathrm{in}^{2}$ :"} {"id": "textbook_2230", "title": "766. LINK TO LEARNING - ", "content": "$$\n\\text { pressure per elephant foot }=14,000 \\frac{\\mathrm{lb}}{\\text { elephant }} \\times \\frac{1 \\text { elephant }}{4 \\text { feet }} \\times \\frac{1 \\text { foot }}{250 \\mathrm{in}^{2}}=14 \\mathrm{lb} / \\mathrm{in}^{2}\n$$"} {"id": "textbook_2231", "title": "766. LINK TO LEARNING - ", "content": "The figure skater weighs about 120 lbs , supported on two skate blades, each with an area of about $2 \\mathrm{in}^{2}$, so the pressure exerted by each blade is about $30 \\mathrm{lb} / \\mathrm{in}^{2}$ :\n\n$$\n\\text { pressure per skate blade }=120 \\frac{\\mathrm{lb}}{\\text { skater }} \\times \\frac{1 \\text { skater }}{2 \\text { blades }} \\times \\frac{1 \\text { blade }}{2 \\mathrm{in}^{2}}=30 \\mathrm{lb} / \\mathrm{in}^{2}\n$$"} {"id": "textbook_2232", "title": "766. LINK TO LEARNING - ", "content": "Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure. On the other hand, if the skater removes their skates and stands with bare feet (or regular footwear) on the ice, the larger area over which their weight is applied greatly reduces the pressure exerted:"} {"id": "textbook_2233", "title": "766. LINK TO LEARNING - ", "content": "$$\n\\text { pressure per human foot }=120 \\frac{\\mathrm{lb}}{\\text { skater }} \\times \\frac{1 \\text { skater }}{2 \\text { feet }} \\times \\frac{1 \\text { foot }}{30 \\mathrm{in}^{2}}=2 \\mathrm{lb} / \\mathrm{in}^{2}\n$$"} {"id": "textbook_2234", "title": "766. LINK TO LEARNING - ", "content": ""} {"id": "textbook_2235", "title": "766. LINK TO LEARNING - ", "content": "FIGURE 8.3 Although (a) an elephant's weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of the skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi)"} {"id": "textbook_2236", "title": "766. LINK TO LEARNING - ", "content": "The SI unit of pressure is the pascal ( $\\mathbf{P a}$ ), with $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$, where N is the newton, a unit of force defined as 1 $\\mathrm{kg} \\mathrm{m} / \\mathrm{s}^{2}$. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal ( $1 \\mathrm{kPa}=$ 1000 Pa ) or bar ( $1 \\mathrm{bar}=100,000 \\mathrm{~Pa}$ ). In the United States, pressure is often measured in pounds of force on an area of one square inch-pounds per square inch (psi)-for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris $\\left(45^{\\circ}\\right)$. Table 8.1 provides some information on these and a few other common units for pressure measurements"} {"id": "textbook_2237", "title": "766. LINK TO LEARNING - ", "content": "Pressure Units"} {"id": "textbook_2238", "title": "766. LINK TO LEARNING - ", "content": "| Unit Name and Abbreviation | Definition or Relation to Other Unit |\n| :--- | :--- |\n| pascal (Pa) | $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$
recommended IUPAC unit |\n| kilopascal (kPa) | $1 \\mathrm{kPa}=1000 \\mathrm{~Pa}$ |\n| pounds per square inch (psi) | air pressure at sea level is ~14.7 psi |\n| atmosphere (atm) | 1 atm $=101,325 \\mathrm{~Pa}=760$ torr
air pressure at sea level is $\\sim 1 \\mathrm{~atm}$ |"} {"id": "textbook_2239", "title": "766. LINK TO LEARNING - ", "content": "TABLE 8.1"} {"id": "textbook_2240", "title": "766. LINK TO LEARNING - ", "content": "| Unit Name and Abbreviation | Definition or Relation to Other Unit |\n| :--- | :--- |\n| bar (bar, or b) | 1 bar $=100,000 \\mathrm{~Pa}$ (exactly)
commonly used in meteorology |\n| millibar (mbar, or mb) | $1000 \\mathrm{mbar}=1 \\mathrm{bar}$ |\n| inches of mercury (in. Hg) | $1 \\mathrm{in} . \\mathrm{Hg}=3386 \\mathrm{~Pa}$
used by aviation industry, also some weather reports |\n| torr | 1 torr $=\\frac{1}{760}$ atm
named after Evangelista Torricelli, inventor of the barometer |\n| millimeters of mercury (mm Hg) | $1 \\mathrm{~mm} \\mathrm{Hg} \\mathrm{\\sim 1} \\mathrm{torr}$ |"} {"id": "textbook_2241", "title": "766. LINK TO LEARNING - ", "content": "TABLE 8.1\n"} {"id": "textbook_2242", "title": "767. EXAMPLE 8.1 - ", "content": ""} {"id": "textbook_2243", "title": "768. Conversion of Pressure Units - ", "content": "\nThe United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in . Hg into:\n(a) torr\n(b) atm\n(c) kPa\n(d) mbar\n"} {"id": "textbook_2244", "title": "769. Solution - ", "content": "\nThis is a unit conversion problem. The relationships between the various pressure units are given in Table 8.1.\n(a) $29.2 \\mathrm{im} \\mathrm{Hg} \\times \\frac{25.4 \\mathrm{~mm}}{1 \\mathrm{im}} \\times \\frac{1 \\text { torr }}{1 \\mathrm{mmHg}}=742$ torr\n(b) $742 \\mathrm{tOrf} \\times \\frac{1 \\mathrm{~atm}}{760 \\text { tort }}=0.976 \\mathrm{~atm}$\n(c) 742 torf $\\times \\frac{101.325 \\mathrm{kPa}}{760 \\text { tort }}=98.9 \\mathrm{kPa}$\n(d) $98.9 \\mathrm{kPa} \\times \\frac{1000 \\mathrm{~Pa}}{1 \\mathrm{kPa}} \\times \\frac{1 \\text { bar }}{100,000 \\mathrm{~Pa}} \\times \\frac{1000 \\mathrm{mbar}}{1 \\text { bar }}=989 \\mathrm{mbar}$\n"} {"id": "textbook_2245", "title": "770. Check Your Learning - ", "content": "\nA typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?\n"} {"id": "textbook_2246", "title": "771. Answer: - ", "content": "\n0.974 atm; $740 \\mathrm{~mm} \\mathrm{Hg} ; 98.7 \\mathrm{kPa} ; 0.987$ bar"} {"id": "textbook_2247", "title": "771. Answer: - ", "content": "We can measure atmospheric pressure, the force exerted by the atmosphere on the earth's surface, with a barometer (Figure 8.4). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore\nproportional to the pressure exerted by the atmosphere.\n\n\nFIGURE 8.4 In a barometer, the height, $h$, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall.\n\nIf the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury ( Hg ) is about 13.6-times denser than water, a mercury barometer only needs to be $\\frac{1}{13.6}$ as tall as a water barometer-a more suitable size. Standard atmospheric pressure of 1 atm at sea level ( $101,325 \\mathrm{~Pa}$ ) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, $p$ :"} {"id": "textbook_2248", "title": "771. Answer: - ", "content": "$$\np=h \\rho g\n$$"} {"id": "textbook_2249", "title": "771. Answer: - ", "content": "where $h$ is the height of the fluid, $\\rho$ is the density of the fluid, and $g$ is acceleration due to gravity.\n"} {"id": "textbook_2250", "title": "772. EXAMPLE 8.2 - ", "content": ""} {"id": "textbook_2251", "title": "773. Calculation of Barometric Pressure - ", "content": "\nShow the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury $=13.6 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n"} {"id": "textbook_2252", "title": "774. Solution - ", "content": "\nThe hydrostatic pressure is given by $p=h \\rho g$, with $h=760 \\mathrm{~mm}, \\rho=13.6 \\mathrm{~g} / \\mathrm{cm}^{3}$, and $g=9.81 \\mathrm{~m} / \\mathrm{s}^{2}$. Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of $\\sim 101,325 \\mathrm{~Pa}$.)"} {"id": "textbook_2253", "title": "774. Solution - ", "content": "$$\n\\begin{gathered}\n101,325 \\mathrm{~N} / \\mathrm{m}^{2}=101,325 \\frac{\\mathrm{~kg} \\cdot \\mathrm{~m} / \\mathrm{s}^{2}}{\\mathrm{~m}^{2}}=101,325 \\frac{\\mathrm{~kg}}{\\mathrm{~m} \\cdot \\mathrm{~s}^{2}} \\\\\np=\\left(760 \\mathrm{~mm} \\times \\frac{1 \\mathrm{~m}}{1000 \\mathrm{~mm}}\\right) \\times\\left(\\frac{13.6 \\mathrm{~g}}{1 \\mathrm{~cm}^{3}} \\times \\frac{1 \\mathrm{~kg}}{1000 \\mathrm{~g}} \\times \\frac{(100 \\mathrm{~cm})^{3}}{(1 \\mathrm{~m})^{3}}\\right) \\times\\left(\\frac{9.81 \\mathrm{~m}}{1 \\mathrm{~s}^{2}}\\right) \\\\\n=(0.760 \\mathrm{~m})\\left(13,600 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)\\left(9.81 \\mathrm{~m} / \\mathrm{s}^{2}\\right)=1.01 \\times 10^{5} \\mathrm{~kg} / \\mathrm{ms}^{2}=1.01 \\times 10^{5} \\mathrm{~N} / \\mathrm{m}^{2}\n\\end{gathered}\n$$"} {"id": "textbook_2254", "title": "774. Solution - ", "content": "$$\n=1.01 \\times 10^{5} \\mathrm{~Pa}\n$$\n"} {"id": "textbook_2255", "title": "775. Check Your Learning - ", "content": "\nCalculate the height of a column of water at $25^{\\circ} \\mathrm{C}$ that corresponds to normal atmospheric pressure. The density of water at this temperature is $1.0 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n"} {"id": "textbook_2256", "title": "776. Answer: - ", "content": "\n10.3 m"} {"id": "textbook_2257", "title": "776. Answer: - ", "content": "A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( $h$ in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 8.5) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.\n"} {"id": "textbook_2258", "title": "776. Answer: - ", "content": "FIGURE 8.5 A manometer can be used to measure the pressure of a gas. The (difference in) height between the liquid levels ( $h$ ) is a measure of the pressure. Mercury is usually used because of its large density.\n"} {"id": "textbook_2259", "title": "777. EXAMPLE 8.3 - ", "content": ""} {"id": "textbook_2260", "title": "778. Calculation of Pressure Using a Closed-End Manometer - ", "content": "\nThe pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar\n\n"} {"id": "textbook_2261", "title": "779. Solution - ", "content": "\nThe pressure of the gas is equal to a column of mercury of height 26.4 cm . (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm Hg , or mercury.) We could use the equation $p=h \\rho g$ as in Example 8.2, but it is simpler to just convert between units using Table 8.1.\n(a) $26.4 \\mathrm{cmHg} \\times \\frac{10 \\mathrm{mmHz}}{1 \\mathrm{cmHg}} \\times \\frac{1 \\text { torr }}{1 \\mathrm{mmHg}}=264$ torr\n(b) 264 tori $\\times \\frac{1 \\mathrm{Am}}{760 \\text { torr }} \\times \\frac{101,325 \\mathrm{~Pa}}{1+35,200 \\mathrm{~Pa}}$\n(c) $35,200 \\mathrm{~Pa} \\times \\frac{1 \\mathrm{bar}}{100,000 \\mathrm{~Pa}}=0.352 \\mathrm{bar}$\n"} {"id": "textbook_2262", "title": "780. Check Your Learning - ", "content": "\nThe pressure of a sample of gas is measured with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar\n\n"} {"id": "textbook_2263", "title": "781. Answer: - ", "content": "\n(a) $\\sim 150$ torr; (b) $\\sim 20,000 \\mathrm{~Pa}$; (c) $\\sim 0.20$ bar\n"} {"id": "textbook_2264", "title": "782. EXAMPLE 8.4 - ", "content": ""} {"id": "textbook_2265", "title": "783. Calculation of Pressure Using an Open-End Manometer - ", "content": "\nThe pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown to the right. Determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n"} {"id": "textbook_2266", "title": "784. Solution - ", "content": "\nThe pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)\n(a) In mm Hg , this is: $137 \\mathrm{~mm} \\mathrm{Hg}+760 \\mathrm{~mm} \\mathrm{Hg}=897 \\mathrm{~mm} \\mathrm{Hg}$\n(b) $897 \\mathrm{mmHg} \\times \\frac{1 \\mathrm{~atm}}{760 \\mathrm{~mm} \\mathrm{Hg}}=1.18 \\mathrm{~atm}$\n(c) $1.18 \\mathrm{~atm} \\times \\frac{101.325 \\mathrm{kPa}}{1 \\mathrm{~atm}}=1.20 \\times 10^{2} \\mathrm{kPa}$\n"} {"id": "textbook_2267", "title": "785. Check Your Learning - ", "content": "\nThe pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown to the right. Determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n"} {"id": "textbook_2268", "title": "785. Check Your Learning - ", "content": "Answer:\n(a) 642 mm Hg ; (b) 0.845 atm ; (c) 85.6 kPa\n"} {"id": "textbook_2269", "title": "786. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2270", "title": "787. Measuring Blood Pressure - ", "content": "\nBlood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = \"pulse\"). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure 8.6). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure-the peak pressure in the cardiac cycle. When the cuff's pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart's ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure-the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury ( mm Hg ).\n\n(a)\n\n(b)\n\nFIGURE 8.6 (a) A medical technician prepares to measure a patient's blood pressure with a sphygmomanometer. (b) A typical sphygmomanometer uses a valved rubber bulb to inflate the cuff and a diaphragm gauge to measure pressure. (credit a: modification of work by Master Sgt. Jeffrey Allen)\n"} {"id": "textbook_2271", "title": "788. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_2272", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "\nThroughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth's weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure 8.7) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.\n"} {"id": "textbook_2273", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "FIGURE 8.7 Meteorologists use weather maps to describe and predict weather. Regions of high (H) and low (L) pressure have large effects on weather conditions. The gray lines represent locations of constant pressure known as isobars. (credit: modification of work by National Oceanic and Atmospheric Administration)"} {"id": "textbook_2274", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "In terms of weather, low-pressure systems occur when the earth's surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events."} {"id": "textbook_2275", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "The atmosphere is the gaseous layer that surrounds a planet. Earth's atmosphere, which is roughly 100-125 km thick, consists of roughly $78.1 \\%$ nitrogen and $21.0 \\%$ oxygen, and can be subdivided further into the regions shown in Figure 8.8: the exosphere (furthest from earth, $>700 \\mathrm{~km}$ above sea level), the thermosphere ( $80-700 \\mathrm{~km}$ ), the mesosphere ( $50-80 \\mathrm{~km}$ ), the stratosphere (second lowest level of our atmosphere, $12-50 \\mathrm{~km}$ above sea level), and the troposphere (up to 12 km above sea level, roughly $80 \\%$ of the earth's atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.\n"} {"id": "textbook_2276", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "FIGURE 8.8 Earth's atmosphere has five layers: the troposphere, the stratosphere, the mesosphere, the thermosphere, and the exosphere."} {"id": "textbook_2277", "title": "789. Meteorology, Climatology, and Atmospheric Science - ", "content": "Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.\n"} {"id": "textbook_2278", "title": "789. Meteorology, Climatology, and Atmospheric Science - 789.1. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law", "content": ""} {"id": "textbook_2279", "title": "790. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2280", "title": "790. LEARNING OBJECTIVES - ", "content": "- Identify the mathematical relationships between the various properties of gases\n- Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions"} {"id": "textbook_2281", "title": "790. LEARNING OBJECTIVES - ", "content": "During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 8.9), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today's standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas-a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation-the ideal gas law-that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.\n\n(a)\n\n(b)\n\n(c)"} {"id": "textbook_2282", "title": "790. LEARNING OBJECTIVES - ", "content": "FIGURE 8.9 In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.\n"} {"id": "textbook_2283", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "\nImagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 8.10) and the pressure increases.\n"} {"id": "textbook_2284", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "FIGURE 8.10 The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases."} {"id": "textbook_2285", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 8.11. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then $P$ and $T$ are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor."} {"id": "textbook_2286", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "| Temperature
$\\left({ }^{\\circ} \\mathbf{C}\\right)$ | Temperature
$\\mathbf{( K )}$ | Pressure
$\\mathbf{( k P a )}$ |\n| :---: | :---: | :---: |\n| -100 | 173 | 36.0 |\n| -50 | 223 | 46.4 |\n| 0 | 273 | 56.7 |\n| 50 | 323 | 67.1 |\n| 100 | 373 | 77.5 |\n| 150 | 423 | 88.0 |"} {"id": "textbook_2287", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": ""} {"id": "textbook_2288", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "FIGURE 8.11 For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at $-273^{\\circ} \\mathrm{C}$, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero."} {"id": "textbook_2289", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas ( $\\sim 1700$ ), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the $P$ - $T$ relationship for gases is known as either Amontons's law or Gay-Lussac's law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:"} {"id": "textbook_2290", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "$$\nP \\propto T \\text { or } P=\\text { constant } \\times T \\text { or } P=k \\times T\n$$"} {"id": "textbook_2291", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "where $\\alpha$ means \"is proportional to,\" and $k$ is a proportionality constant that depends on the identity, amount, and volume of the gas."} {"id": "textbook_2292", "title": "791. Pressure and Temperature: Amontons's Law - ", "content": "For a confined, constant volume of gas, the ratio $\\frac{P}{T}$ is therefore constant (i.e., $\\frac{P}{T}=k$ ). If the gas is initially in \"Condition 1\" (with $P=P_{1}$ and $T=T_{1}$ ), and then changes to \"Condition 2\" (with $P=P_{2}$ and $T=T_{2}$ ), we have that $\\frac{P_{1}}{T_{1}}=k$ and $\\frac{P_{2}}{T_{2}}=k$, which reduces to $\\frac{P_{1}}{T_{1}}=\\frac{P_{2}}{T_{2}}$. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations ( 0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)\n"} {"id": "textbook_2293", "title": "792. EXAMPLE 8.5 - ", "content": ""} {"id": "textbook_2294", "title": "793. Predicting Change in Pressure with Temperature - ", "content": "\nA can of hair spray is used until it is empty except for the propellant, isobutane gas.\n(a) On the can is the warning \"Store only at temperatures below $120^{\\circ} \\mathrm{F}\\left(48.8^{\\circ} \\mathrm{C}\\right)$. Do not incinerate.\" Why?\n(b) The gas in the can is initially at $24^{\\circ} \\mathrm{C}$ and 360 kPa , and the can has a volume of 350 mL . If the can is left in a car that reaches $50^{\\circ} \\mathrm{C}$ on a hot day, what is the new pressure in the can?\n"} {"id": "textbook_2295", "title": "794. Solution - ", "content": "\n(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)\n(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons's/Gay-Lussac's law. Taking $P_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature where the pressure is unknown and $P_{2}$ as the unknown pressure, and converting ${ }^{\\circ} \\mathrm{C}$ to K , we have:"} {"id": "textbook_2296", "title": "794. Solution - ", "content": "$$\n\\frac{P_{1}}{T_{1}}=\\frac{P_{2}}{T_{2}} \\text { which means that } \\frac{360 \\mathrm{kPa}}{297 \\mathrm{~K}}=\\frac{P_{2}}{323 \\mathrm{~K}}\n$$\n\nRearranging and solving gives: $P_{2}=\\frac{360 \\mathrm{kPa} \\times 323 \\mathrm{~K}}{297 \\mathrm{~K}}=390 \\mathrm{kPa}$\n"} {"id": "textbook_2297", "title": "795. Check Your Learning - ", "content": "\nA sample of nitrogen, $\\mathrm{N}_{2}$, occupies 45.0 mL at $27^{\\circ} \\mathrm{C}$ and 600 torr. What pressure will it have if cooled to $-73^{\\circ} \\mathrm{C}$ while the volume remains constant?\n"} {"id": "textbook_2298", "title": "796. Answer: - ", "content": "\n400 torr\n"} {"id": "textbook_2299", "title": "797. Volume and Temperature: Charles's Law - ", "content": "\nIf we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let's say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.\n"} {"id": "textbook_2300", "title": "798. LINK TO LEARNING - ", "content": "\nThis video (http://openstax.org/l/16CharlesLaw) shows how cooling and heating a gas causes its volume to decrease or increase, respectively."} {"id": "textbook_2301", "title": "798. LINK TO LEARNING - ", "content": "These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 8.12."} {"id": "textbook_2302", "title": "798. LINK TO LEARNING - ", "content": "| Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Temperature (K) | Volume (L) |\n| :---: | :---: | :---: |\n| -3 | 270 | 22 |\n| -23 | 250 | 21 |\n| -53 | 220 | 18 |\n| -162 | 111 | 9 |"} {"id": "textbook_2303", "title": "798. LINK TO LEARNING - ", "content": ""} {"id": "textbook_2304", "title": "798. LINK TO LEARNING - ", "content": "FIGURE 8.12 The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph's origin, representing a temperature of absolute zero."} {"id": "textbook_2305", "title": "798. LINK TO LEARNING - ", "content": "The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles's law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre C\u00e9sar Charles. Charles's law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant."} {"id": "textbook_2306", "title": "798. LINK TO LEARNING - ", "content": "Mathematically, this can be written as:"} {"id": "textbook_2307", "title": "798. LINK TO LEARNING - ", "content": "$$\nV \\alpha T \\text { or } V=\\text { constant } \\cdot T \\text { or } V=k \\cdot T \\text { or } V_{1} / T_{1}=V_{2} / T_{2}\n$$\n\nwith $k$ being a proportionality constant that depends on the amount and pressure of the gas.\nFor a confined, constant pressure gas sample, $\\frac{V}{T}$ is constant (i.e., the ratio $=k$ ), and as seen with the $P-T$ relationship, this leads to another form of Charles's law: $\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}}$.\n"} {"id": "textbook_2308", "title": "799. EXAMPLE 8.6 - ", "content": ""} {"id": "textbook_2309", "title": "800. Predicting Change in Volume with Temperature - ", "content": "\nA sample of carbon dioxide, $\\mathrm{CO}_{2}$, occupies 0.300 L at $10^{\\circ} \\mathrm{C}$ and 750 torr. What volume will the gas have at $30^{\\circ} \\mathrm{C}$ and 750 torr?\n"} {"id": "textbook_2310", "title": "801. Solution - ", "content": "\nBecause we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\\circ} \\mathrm{C}$ into K we have:"} {"id": "textbook_2311", "title": "801. Solution - ", "content": "$$\n\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}} \\text { which means that } \\frac{0.300 \\mathrm{~L}}{283 \\mathrm{~K}}=\\frac{V_{2}}{303 \\mathrm{~K}}\n$$\n\nRearranging and solving gives: $V_{2}=\\frac{0.300 \\mathrm{~L} \\times 303 \\mathrm{~K}}{283 \\mathrm{~K}}=0.321 \\mathrm{~L}$\nThis answer supports our expectation from Charles's law, namely, that raising the gas temperature (from 283 K to 303 K ) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L ).\n"} {"id": "textbook_2312", "title": "802. Check Your Learning - ", "content": "\nA sample of oxygen, $\\mathrm{O}_{2}$, occupies 32.2 mL at $30^{\\circ} \\mathrm{C}$ and 452 torr. What volume will it occupy at $-70^{\\circ} \\mathrm{C}$ and the same pressure?\n"} {"id": "textbook_2313", "title": "803. Answer: - ", "content": "\n21.6 mL\n"} {"id": "textbook_2314", "title": "804. EXAMPLE 8.7 - ", "content": ""} {"id": "textbook_2315", "title": "805. Measuring Temperature with a Volume Change - ", "content": "\nTemperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of $150.0 \\mathrm{~cm}^{3}$ when immersed in a mixture of ice and water $\\left(0.00^{\\circ} \\mathrm{C}\\right)$. When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is $131.7 \\mathrm{~cm}^{3}$. Find the temperature of boiling ammonia on the kelvin and Celsius scales.\n"} {"id": "textbook_2316", "title": "806. Solution - ", "content": "\nA volume change caused by a temperature change at constant pressure means we should use Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\\circ} \\mathrm{C}$ into K we have:"} {"id": "textbook_2317", "title": "806. Solution - ", "content": "$$\n\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}} \\text { which means that } \\frac{150.0 \\mathrm{~cm}^{3}}{273.15 \\mathrm{~K}}=\\frac{131.7 \\mathrm{~cm}^{3}}{T_{2}}\n$$"} {"id": "textbook_2318", "title": "806. Solution - ", "content": "Rearrangement gives $T_{2}=\\frac{131.7 \\mathrm{~cm}^{3} \\times 273.15 \\mathrm{~K}}{150.0 \\mathrm{~cm}^{3}}=239.8 \\mathrm{~K}$\nSubtracting 273.15 from 239.8 K , we find that the temperature of the boiling ammonia on the Celsius scale is $-33.4^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2319", "title": "807. Check Your Learning - ", "content": "\nWhat is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?\n"} {"id": "textbook_2320", "title": "808. Answer: - ", "content": "\n635 mL\n"} {"id": "textbook_2321", "title": "809. Volume and Pressure: Boyle's Law - ", "content": "\nIf we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say $25^{\\circ} \\mathrm{C}$. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 8.13.\n\n\n\n\nFIGURE 8.13 When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since $P$ and $V$ are inversely proportional, a graph of $\\frac{1}{P}$ vs. $V$ is linear."} {"id": "textbook_2322", "title": "809. Volume and Pressure: Boyle's Law - ", "content": "Unlike the $P-T$ and $V-T$ relationships, pressure and volume are not directly proportional to each other. Instead, $P$ and $V$ exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:"} {"id": "textbook_2323", "title": "809. Volume and Pressure: Boyle's Law - ", "content": "$$\nP \\propto 1 / V \\text { or } P=k \\cdot 1 / V \\text { or } P \\cdot V=k \\text { or } P_{1} V_{1}=P_{2} V_{2}\n$$"} {"id": "textbook_2324", "title": "809. Volume and Pressure: Boyle's Law - ", "content": "with $k$ being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure ( $\\frac{1}{P}$ ) versus the volume $(V)$, or the inverse of volume $\\left(\\frac{1}{V}\\right)$ versus the pressure $(P)$. Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to \"linearize\" their data. If we plot $P$ versus $V$, we obtain a hyperbola (see Figure 8.14).\n\n\nFIGURE 8.14 The relationship between pressure and volume is inversely proportional. (a) The graph of $P$ vs. $V$ is a hyperbola, whereas (b) the graph of $\\left(\\frac{1}{P}\\right)$ vs. $V$ is linear."} {"id": "textbook_2325", "title": "809. Volume and Pressure: Boyle's Law - ", "content": "The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle's law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.\n"} {"id": "textbook_2326", "title": "810. EXAMPLE 8.8 - ", "content": ""} {"id": "textbook_2327", "title": "811. Volume of a Gas Sample - ", "content": "\nThe sample of gas in Figure 8.13 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL , using:\n(a) the $P$ - $V$ graph in Figure 8.13\n(b) the $\\frac{1}{P}$ vs. $V$ graph in Figure 8.13\n(c) the Boyle's law equation"} {"id": "textbook_2328", "title": "811. Volume of a Gas Sample - ", "content": "Comment on the likely accuracy of each method.\n"} {"id": "textbook_2329", "title": "812. Solution - ", "content": "\n(a) Estimating from the $P$ - $V$ graph gives a value for $P$ somewhere around 27 psi .\n(b) Estimating from the $\\frac{1}{P}$ versus $V$ graph give a value of about 26 psi .\n(c) From Boyle's law, we know that the product of pressure and volume ( $P V$ ) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have $P_{1} V_{1}=k$ and $P_{2} V_{2}=k$ which means that $P_{1} V_{1}=P_{2} V_{2}$."} {"id": "textbook_2330", "title": "812. Solution - ", "content": "Using $P_{1}$ and $V_{1}$ as the known values 13.0 psi and $15.0 \\mathrm{~mL}, P_{2}$ as the pressure at which the volume is unknown, and $V_{2}$ as the unknown volume, we have:\n\n$$\nP_{1} V_{1}=P_{2} V_{2} \\text { or } 13.0 \\mathrm{psi} \\times 15.0 \\mathrm{~mL}=P_{2} \\times 7.5 \\mathrm{~mL}\n$$"} {"id": "textbook_2331", "title": "812. Solution - ", "content": "Solving:"} {"id": "textbook_2332", "title": "812. Solution - ", "content": "$$\nP_{2}=\\frac{13.0 \\mathrm{psi} \\times 15.0 \\mathrm{~mL}}{7.5 \\mathrm{~mL}}=26 \\mathrm{psi}\n$$"} {"id": "textbook_2333", "title": "812. Solution - ", "content": "It was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.\n"} {"id": "textbook_2334", "title": "813. Check Your Learning - ", "content": "\nThe sample of gas in Figure 8.13 has a volume of 30.0 mL at a pressure of 6.5 psi . Determine the volume of the gas at a pressure of 11.0 psi , using:\n(a) the $P$ - $V$ graph in Figure 8.13\n(b) the $\\frac{1}{P}$ vs. $V$ graph in Figure 8.13\n(c) the Boyle's law equation"} {"id": "textbook_2335", "title": "813. Check Your Learning - ", "content": "Comment on the likely accuracy of each method.\n"} {"id": "textbook_2336", "title": "814. Answer: - ", "content": "\n(a) about $17-18 \\mathrm{~mL}$; (b) $\\sim 18 \\mathrm{~mL}$; (c) 17.7 mL ; it was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow\n"} {"id": "textbook_2337", "title": "815. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2338", "title": "816. Breathing and Boyle's Law - ", "content": "\nWhat do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle's law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle's law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle's law cycle for the rest of your life (Figure 8.15).\n"} {"id": "textbook_2339", "title": "816. Breathing and Boyle's Law - ", "content": "FIGURE 8.15 Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs.\n"} {"id": "textbook_2340", "title": "817. Moles of Gas and Volume: Avogadro's Law - ", "content": "\nThe Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro's law: For a confined gas, the volume ( $V$ ) and number of moles (n) are directly proportional if the pressure and temperature both remain constant."} {"id": "textbook_2341", "title": "817. Moles of Gas and Volume: Avogadro's Law - ", "content": "In equation form, this is written as:"} {"id": "textbook_2342", "title": "817. Moles of Gas and Volume: Avogadro's Law - ", "content": "$$\nV \\propto n \\text { or } V=k \\times n \\text { or } \\frac{V_{1}}{n_{1}}=\\frac{V_{2}}{n_{2}}\n$$"} {"id": "textbook_2343", "title": "817. Moles of Gas and Volume: Avogadro's Law - ", "content": "Mathematical relationships can also be determined for the other variable pairs, such as $P$ versus $n$, and $n$ versus $T$.\n"} {"id": "textbook_2344", "title": "818. LINK TO LEARNING - ", "content": "\nVisit this interactive PhET simulation (http://openstax.org/l/16IdealGasLaw) to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).\n"} {"id": "textbook_2345", "title": "819. The Ideal Gas Law - ", "content": "\nTo this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:"} {"id": "textbook_2346", "title": "819. The Ideal Gas Law - ", "content": "- Boyle's law: $P V=$ constant at constant $T$ and $n$\n- Amontons's law: $\\frac{P}{T}=$ constant at constant $V$ and $n$\n- Charles's law: $\\frac{V}{T}=$ constant at constant $P$ and $n$\n- Avogadro's law: $\\frac{V}{n}=$ constant at constant $P$ and $T$"} {"id": "textbook_2347", "title": "819. The Ideal Gas Law - ", "content": "Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:"} {"id": "textbook_2348", "title": "819. The Ideal Gas Law - ", "content": "$$\nP V=n R T\n$$"} {"id": "textbook_2349", "title": "819. The Ideal Gas Law - ", "content": "where $P$ is the pressure of a gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its temperature on the kelvin scale, and $R$ is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being $0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$ and 8.314 kPa L $\\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$."} {"id": "textbook_2350", "title": "819. The Ideal Gas Law - ", "content": "Gases whose properties of $P, V$, and $T$ are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures."} {"id": "textbook_2351", "title": "819. The Ideal Gas Law - ", "content": "The ideal gas equation contains five terms, the gas constant $R$ and the variable properties $P, V, n$, and $T$. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.\n"} {"id": "textbook_2352", "title": "820. EXAMPLE 8.9 - ", "content": ""} {"id": "textbook_2353", "title": "821. Using the Ideal Gas Law - ", "content": "\nMethane, $\\mathrm{CH}_{4}$, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of $\\mathrm{CH}_{4}$. What is the volume of this much methane at $25^{\\circ} \\mathrm{C}$ and 745 torr?\n"} {"id": "textbook_2354", "title": "822. Solution - ", "content": "\nWe must rearrange $P V=n R T$ to solve for $V: V=\\frac{n R T}{P}$\nIf we choose to use $R=0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm."} {"id": "textbook_2355", "title": "822. Solution - ", "content": "Converting into the \"right\" units:"} {"id": "textbook_2356", "title": "822. Solution - ", "content": "$$\n\\begin{gathered}\nn=655 \\mathrm{gCH}_{4} \\times \\frac{1 \\mathrm{~mol}}{16.043-\\mathrm{g}_{4} \\mathrm{CH}_{4}}=40.8 \\mathrm{~mol} \\\\\nT=25^{\\circ} \\mathrm{C}+273=298 \\mathrm{~K} \\\\\nP=745 \\mathrm{tmm} \\times \\frac{1 \\mathrm{~atm}}{760 \\mathrm{tmf}}=0.980 \\mathrm{~atm} \\\\\nV=\\frac{n R T}{P}=\\frac{(40.8 \\mathrm{mor})\\left(0.08206 \\mathrm{~L} \\mathrm{atmol}^{-1} \\mathrm{~K}^{-1}\\right)(298 \\mathrm{~K})}{0.980 \\mathrm{~atm}}=1.02 \\times 10^{3} \\mathrm{~L}\n\\end{gathered}\n$$"} {"id": "textbook_2357", "title": "822. Solution - ", "content": "It would require 1020 L ( 269 gal ) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.\n"} {"id": "textbook_2358", "title": "823. Check Your Learning - ", "content": "\nCalculate the pressure in bar of 2520 moles of hydrogen gas stored at $27^{\\circ} \\mathrm{C}$ in the $180-\\mathrm{L}$ storage tank of a modern hydrogen-powered car.\n"} {"id": "textbook_2359", "title": "824. Answer: - ", "content": "\n350 bar"} {"id": "textbook_2360", "title": "824. Answer: - ", "content": "If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\\frac{P_{1} V_{1}}{T_{1}}=\\frac{P_{2} V_{2}}{T_{2}}$ using units of atm, L, and K. Both sets of conditions are equal to the product of $n \\times R$ (where $n=$ the number of moles of the gas and $R$ is the ideal gas law constant).\n"} {"id": "textbook_2361", "title": "825. EXAMPLE 8.10 - ", "content": ""} {"id": "textbook_2362", "title": "826. Using the Combined Gas Law - ", "content": "\nWhen filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8.16). If the water temperature is $27^{\\circ} \\mathrm{C}$, how many liters of air will such a tank provide to a diver's lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm ?\n"} {"id": "textbook_2363", "title": "826. Using the Combined Gas Law - ", "content": "FIGURE 8.16 Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)"} {"id": "textbook_2364", "title": "826. Using the Combined Gas Law - ", "content": "Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is $37^{\\circ} \\mathrm{C}$, we have:"} {"id": "textbook_2365", "title": "826. Using the Combined Gas Law - ", "content": "$$\n\\frac{P_{1} V_{1}}{T_{1}}=\\frac{P_{2} V_{2}}{T_{2}} \\rightarrow \\frac{(153 \\mathrm{~atm})(13.2 \\mathrm{~L})}{(300 \\mathrm{~K})}=\\frac{(3.13 \\mathrm{~atm})\\left(V_{2}\\right)}{(310 \\mathrm{~K})}\n$$"} {"id": "textbook_2366", "title": "826. Using the Combined Gas Law - ", "content": "Solving for $V_{2}$ :"} {"id": "textbook_2367", "title": "826. Using the Combined Gas Law - ", "content": "$$\nV_{2}=\\frac{(153 \\mathrm{~atm})(13.2 \\mathrm{~L})(310 \\mathrm{~K})}{(300 \\mathrm{~K})(3.13 \\mathrm{~atm})}=667 \\mathrm{~L}\n$$"} {"id": "textbook_2368", "title": "826. Using the Combined Gas Law - ", "content": "(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good \"ballpark\" estimate.)\n"} {"id": "textbook_2369", "title": "827. Check Your Learning - ", "content": "\nA sample of ammonia is found to occupy 0.250 L under laboratory conditions of $27^{\\circ} \\mathrm{C}$ and 0.850 atm . Find the volume of this sample at $0^{\\circ} \\mathrm{C}$ and 1.00 atm .\n"} {"id": "textbook_2370", "title": "828. Answer: - ", "content": "\n0.193 L\n"} {"id": "textbook_2371", "title": "829. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2372", "title": "830. The Interdependence between Ocean Depth and Pressure in Scuba Diving - ", "content": "\nWhether scuba diving at the Great Barrier Reef in Australia (shown in Figure 8.17) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.\n\n\nFIGURE 8.17 Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)"} {"id": "textbook_2373", "title": "830. The Interdependence between Ocean Depth and Pressure in Scuba Diving - ", "content": "Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as \"atmospheres absolute\" or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body's air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle's law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.\n"} {"id": "textbook_2374", "title": "831. Standard Conditions of Temperature and Pressure - ", "content": "\nWe have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 $\\mathrm{kPa}) .{ }^{\\frac{1}{2}}$ At STP, one mole of an ideal gas has a volume of about $22.4 \\mathrm{~L}-$ this is referred to as the standard molar volume (Figure 8.18)."} {"id": "textbook_2375", "title": "831. Standard Conditions of Temperature and Pressure - ", "content": "[^4]"} {"id": "textbook_2376", "title": "831. Standard Conditions of Temperature and Pressure - ", "content": "FIGURE 8.18 Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of $\\sim 22.4 \\mathrm{~L}$ at STP.\n"} {"id": "textbook_2377", "title": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2378", "title": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES", "content": "- Use the ideal gas law to compute gas densities and molar masses\n- Perform stoichiometric calculations involving gaseous substances\n- State Dalton's law of partial pressures and use it in calculations involving gaseous mixtures"} {"id": "textbook_2379", "title": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES", "content": "The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the \"father of modern chemistry,\" changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer JosephLouis Lagrange said, \"It took the mob only a moment to remove his head; a century will not suffice to reproduce it.\" ${ }^{2}$ Much of the knowledge we do have about Lavoisier's contributions is due to his wife, MarieAnne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study."} {"id": "textbook_2380", "title": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES", "content": "As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask \"How much?\" The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles ( $n$ ). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed."} {"id": "textbook_2381", "title": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES", "content": "[^5]"} {"id": "textbook_2382", "title": "832. Gas Density and Molar Mass - ", "content": "\nThe ideal gas law described previously in this chapter relates the properties of pressure $P$, volume $V$, temperature $T$, and molar amount $n$. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:"} {"id": "textbook_2383", "title": "832. Gas Density and Molar Mass - ", "content": "$$\nP V=n R T\n$$"} {"id": "textbook_2384", "title": "832. Gas Density and Molar Mass - ", "content": "The density $d$ of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance."} {"id": "textbook_2385", "title": "832. Gas Density and Molar Mass - ", "content": "$$\nd=\\frac{m}{V}\n$$"} {"id": "textbook_2386", "title": "832. Gas Density and Molar Mass - ", "content": "Rearranging the ideal gas equation to isolate $V$ and substituting into the density equation yields"} {"id": "textbook_2387", "title": "832. Gas Density and Molar Mass - ", "content": "$$\nd=\\frac{m P}{n R T}=\\left(\\frac{m}{n}\\right) \\frac{P}{R T}\n$$"} {"id": "textbook_2388", "title": "832. Gas Density and Molar Mass - ", "content": "The ratio $m / n$ is the definition of molar mass, $\\mathcal{M}$ :"} {"id": "textbook_2389", "title": "832. Gas Density and Molar Mass - ", "content": "$$\n\\mathscr{M}=\\frac{m}{n}\n$$"} {"id": "textbook_2390", "title": "832. Gas Density and Molar Mass - ", "content": "The density equation can then be written"} {"id": "textbook_2391", "title": "832. Gas Density and Molar Mass - ", "content": "$$\nd=\\frac{\\mathscr{M} P}{R T}\n$$"} {"id": "textbook_2392", "title": "832. Gas Density and Molar Mass - ", "content": "This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 8.11.\n"} {"id": "textbook_2393", "title": "833. EXAMPLE 8.11 - ", "content": ""} {"id": "textbook_2394", "title": "834. Measuring Gas Density - ", "content": "\nWhat is the density of molecular nitrogen gas at STP?\n"} {"id": "textbook_2395", "title": "835. Solution - ", "content": "\nThe molar mass of molecular nitrogen, $\\mathrm{N}_{2}$, is $28.01 \\mathrm{~g} / \\mathrm{mol}$. Substituting this value along with standard temperature and pressure into the gas density equation yields"} {"id": "textbook_2396", "title": "835. Solution - ", "content": "$$\nd=\\frac{\\mathscr{M} P}{R T}=\\frac{(28.01 \\mathrm{~g} / \\mathrm{mol})(1.00 \\mathrm{~atm})}{\\left(0.0821 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(273 \\mathrm{~K})}=1.25 \\mathrm{~g} / \\mathrm{L}\n$$\n"} {"id": "textbook_2397", "title": "836. Check Your Learning - ", "content": "\nWhat is the density of molecular hydrogen gas at $17.0^{\\circ} \\mathrm{C}$ and a pressure of 760 torr?\n"} {"id": "textbook_2398", "title": "837. Answer: - ", "content": "\n$\\mathrm{d}=0.0847 \\mathrm{~g} / \\mathrm{L}$"} {"id": "textbook_2399", "title": "837. Answer: - ", "content": "When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation"} {"id": "textbook_2400", "title": "837. Answer: - ", "content": "$$\nP V=n R T\n$$"} {"id": "textbook_2401", "title": "837. Answer: - ", "content": "and the definition of molar mass"} {"id": "textbook_2402", "title": "837. Answer: - ", "content": "$$\n\\mathscr{M}=\\frac{m}{n}\n$$"} {"id": "textbook_2403", "title": "837. Answer: - ", "content": "yields the following equation:"} {"id": "textbook_2404", "title": "837. Answer: - ", "content": "$$\n\\mathscr{M}=\\frac{m R T}{P V}\n$$"} {"id": "textbook_2405", "title": "837. Answer: - ", "content": "Determining the molar mass of a gas via this approach is demonstrated in Example 8.12.\n"} {"id": "textbook_2406", "title": "838. EXAMPLE 8.12 - ", "content": ""} {"id": "textbook_2407", "title": "839. Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula - ", "content": "\nCyclopropane, a gas once used with oxygen as a general anesthetic, is composed of $85.7 \\%$ carbon and $14.3 \\%$ hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and $50^{\\circ} \\mathrm{C}$, what is the molecular formula for cyclopropane?\n"} {"id": "textbook_2408", "title": "840. Solution - ", "content": "\nFirst determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the $100-\\mathrm{g}$ sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:"} {"id": "textbook_2409", "title": "840. Solution - ", "content": "$$\n\\begin{array}{lll}\n85.7 \\mathrm{~g} \\mathrm{C} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}=7.136 \\mathrm{~mol} \\mathrm{C} & \\frac{7.136}{7.136}=1.00 \\mathrm{~mol} \\mathrm{C} \\\\\n14.3 \\mathrm{~g} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.01 \\mathrm{~g} \\mathrm{H}}=14.158 \\mathrm{~mol} \\mathrm{H} & \\frac{14.158}{7.136}=1.98 \\mathrm{~mol} \\mathrm{H}\n\\end{array}\n$$"} {"id": "textbook_2410", "title": "840. Solution - ", "content": "Empirical formula is $\\mathrm{CH}_{2}$ [empirical mass (EM) of $14.03 \\mathrm{~g} / \\mathrm{emp}$ irical unit].\nNext, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:"} {"id": "textbook_2411", "title": "840. Solution - ", "content": "$$\n\\mathscr{M}=\\frac{m R T}{P V}=\\frac{(1.56 \\mathrm{~g})\\left(0.0821 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(323 \\mathrm{~K})}{(0.984 \\mathrm{~atm})(1.00 \\mathrm{~L})}=42.0 \\mathrm{~g} / \\mathrm{mol}\n$$"} {"id": "textbook_2412", "title": "840. Solution - ", "content": "Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:"} {"id": "textbook_2413", "title": "840. Solution - ", "content": "$$\n\\frac{M}{E M}=\\frac{42.0 \\mathrm{~g} / \\mathrm{mol}}{14.0 \\mathrm{~g} / \\mathrm{mol}}=3\n$$"} {"id": "textbook_2414", "title": "840. Solution - ", "content": "The molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:"} {"id": "textbook_2415", "title": "840. Solution - ", "content": "$$\n\\left(\\mathrm{CH}_{2}\\right)_{3}=\\mathrm{C}_{3} \\mathrm{H}_{6}\n$$\n"} {"id": "textbook_2416", "title": "841. Check Your Learning - ", "content": "\nAcetylene, a fuel used welding torches, is composed of $92.3 \\% \\mathrm{C}$ and $7.7 \\% \\mathrm{H}$ by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and $59.5^{\\circ} \\mathrm{C}$, what is the molecular formula for acetylene?\n"} {"id": "textbook_2417", "title": "842. Answer: - ", "content": "\nEmpirical formula, CH; Molecular formula, $\\mathrm{C}_{2} \\mathrm{H}_{2}$\n"} {"id": "textbook_2418", "title": "843. EXAMPLE 8.13 - ", "content": ""} {"id": "textbook_2419", "title": "844. Determining the Molar Mass of a Volatile Liquid - ", "content": "\nThe approximate molar mass of a volatile liquid can be determined by:"} {"id": "textbook_2420", "title": "844. Determining the Molar Mass of a Volatile Liquid - ", "content": "1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that\nmay escape through the hole\n2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure\n3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample's mass (see Figure 8.19)\n"} {"id": "textbook_2421", "title": "844. Determining the Molar Mass of a Volatile Liquid - ", "content": "FIGURE 8.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At $t_{l \\longrightarrow g}$, the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)"} {"id": "textbook_2422", "title": "844. Determining the Molar Mass of a Volatile Liquid - ", "content": "Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 $\\mathrm{cm}^{3}$ at $99.6^{\\circ} \\mathrm{C}$ when the atmospheric pressure is 742.1 mm Hg . What is the approximate molar mass of chloroform?\n"} {"id": "textbook_2423", "title": "845. Solution - ", "content": "\nSince $\\mathcal{M}=\\frac{m}{n}$ and $n=\\frac{P V}{R T}$, substituting and rearranging gives $\\mathcal{M}=\\frac{m R T}{P V}$,\nthen"} {"id": "textbook_2424", "title": "845. Solution - ", "content": "$$\n\\mathcal{M}=\\frac{m R T}{P V}=\\frac{(0.494 \\mathrm{~g}) \\times 0.08206 \\mathrm{~L} \\cdot \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K} \\times 372.8 \\mathrm{~K}}{0.976 \\mathrm{~atm} \\times 0.129 \\mathrm{~L}}=120 \\mathrm{~g} / \\mathrm{mol} .\n$$\n"} {"id": "textbook_2425", "title": "846. Check Your Learning - ", "content": "\nA sample of phosphorus that weighs $3.243 \\times 10^{-2} \\mathrm{~g}$ exerts a pressure of 31.89 kPa in a $56.0-\\mathrm{mL}$ bulb at $550^{\\circ} \\mathrm{C}$. What are the molar mass and molecular formula of phosphorus vapor?\n"} {"id": "textbook_2426", "title": "847. Answer: - ", "content": "\n$124 \\mathrm{~g} / \\mathrm{mol} \\mathrm{P}_{4}$\n"} {"id": "textbook_2427", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "\nUnless they chemically react with each other, the individual gases in a mixture of gases do not affect each other's pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 8.20). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton's law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:"} {"id": "textbook_2428", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "$$\nP_{\\text {Total }}=P_{A}+P_{B}+P_{C}+\\ldots=\\Sigma_{\\mathrm{i}} P_{\\mathrm{i}}\n$$"} {"id": "textbook_2429", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "In the equation $P_{\\text {Total }}$ is the total pressure of a mixture of gases, $P_{A}$ is the partial pressure of gas $A ; P_{B}$ is the partial pressure of gas $B ; P_{C}$ is the partial pressure of gas $C$; and so on.\n"} {"id": "textbook_2430", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "FIGURE 8.20 If equal-volume cylinders containing gasses at pressures of $300 \\mathrm{kPa}, 450 \\mathrm{kPa}$, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa ."} {"id": "textbook_2431", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction ( $\\boldsymbol{X}$ ), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:"} {"id": "textbook_2432", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "$$\nP_{A}=X_{A} \\times P_{\\text {Total }} \\quad \\text { where } \\quad X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}\n$$"} {"id": "textbook_2433", "title": "848. The Pressure of a Mixture of Gases: Dalton's Law - ", "content": "where $P_{A}, X_{A}$, and $n_{A}$ are the partial pressure, mole fraction, and number of moles of gas A, respectively, and $n_{\\text {Total }}$ is the number of moles of all components in the mixture.\n"} {"id": "textbook_2434", "title": "849. EXAMPLE 8.14 - ", "content": ""} {"id": "textbook_2435", "title": "850. The Pressure of a Mixture of Gases - ", "content": "\nA 10.0-L vessel contains $2.50 \\times 10^{-3} \\mathrm{~mol}$ of $\\mathrm{H}_{2}, 1.00 \\times 10^{-3} \\mathrm{~mol}$ of He , and $3.00 \\times 10^{-4} \\mathrm{~mol}$ of Ne at $35^{\\circ} \\mathrm{C}$.\n(a) What are the partial pressures of each of the gases?\n(b) What is the total pressure in atmospheres?\n"} {"id": "textbook_2436", "title": "851. Solution - ", "content": "\nThe gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\\frac{n R T}{V}$ :"} {"id": "textbook_2437", "title": "851. Solution - ", "content": "$$\n\\begin{aligned}\n& P_{\\mathrm{H}_{2}}=\\frac{\\left(2.50 \\times 10^{-3} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(308 \\mathrm{~K})}{10.0 \\mathrm{t}}=6.32 \\times 10^{-3} \\mathrm{~atm} \\\\\n& P_{\\mathrm{He}}=\\frac{\\left(1.00 \\times 10^{-3} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(308 \\mathrm{~K})}{10.0 \\mathrm{t}}=2.53 \\times 10^{-3} \\mathrm{~atm} \\\\\n& \\left.P_{\\mathrm{Ne}}=\\frac{\\left(3.00 \\times 10^{-4} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{mmol}^{-1} \\mathrm{~K}^{-1}\\right)}{10.0 \\mathrm{t}}(308 \\mathrm{~K})\\right)=7.58 \\times 10^{-4} \\mathrm{~atm}\n\\end{aligned}\n$$"} {"id": "textbook_2438", "title": "851. Solution - ", "content": "The total pressure is given by the sum of the partial pressures:"} {"id": "textbook_2439", "title": "851. Solution - ", "content": "$$\nP_{\\mathrm{T}}=P_{\\mathrm{H}_{2}}+P_{\\mathrm{He}}+P_{\\mathrm{Ne}}=(0.00632+0.00253+0.00076) \\mathrm{atm}=9.61 \\times 10^{-3} \\mathrm{~atm}\n$$\n"} {"id": "textbook_2440", "title": "852. Check Your Learning - ", "content": "\nA 5.73-L flask at $25^{\\circ} \\mathrm{C}$ contains 0.0388 mol of $\\mathrm{N}_{2}, 0.147 \\mathrm{~mol}$ of CO , and $0.0803 \\mathrm{~mol} \\mathrm{of} \\mathrm{H}_{2}$. What is the total pressure in the flask in atmospheres?\n"} {"id": "textbook_2441", "title": "853. Answer: - ", "content": "\n1.137 atm"} {"id": "textbook_2442", "title": "853. Answer: - ", "content": "Here is another example of this concept, but dealing with mole fraction calculations.\n"} {"id": "textbook_2443", "title": "854. EXAMPLE 8.15 - ", "content": ""} {"id": "textbook_2444", "title": "855. The Pressure of a Mixture of Gases - ", "content": "\nA gas mixture used for anesthesia contains 2.83 mol oxygen, $\\mathrm{O}_{2}$, and 8.41 mol nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$. The total pressure of the mixture is 192 kPa .\n(a) What are the mole fractions of $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}$ ?\n(b) What are the partial pressures of $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}$ ?\n"} {"id": "textbook_2445", "title": "856. Solution - ", "content": "\nThe mole fraction is given by $X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}$ and the partial pressure is $P_{A}=X_{A} \\times P_{\\text {Total }}$.\nFor $\\mathrm{O}_{2}$,"} {"id": "textbook_2446", "title": "856. Solution - ", "content": "$$\nX_{O_{2}}=\\frac{n_{O_{2}}}{n_{\\text {Total }}}=\\frac{2.83 \\mathrm{~mol}}{(2.83+8.41) \\mathrm{mol}}=0.252\n$$"} {"id": "textbook_2447", "title": "856. Solution - ", "content": "and $P_{O_{2}}=X_{O_{2}} \\times P_{\\text {Total }}=0.252 \\times 192 \\mathrm{kPa}=48.4 \\mathrm{kPa}$\nFor $\\mathrm{N}_{2} \\mathrm{O}$,"} {"id": "textbook_2448", "title": "856. Solution - ", "content": "$$\nX_{N_{2} O}=\\frac{n_{N_{2} O}}{n_{\\text {Total }}}=\\frac{8.41 \\mathrm{~mol}}{(2.83+8.41) \\mathrm{mol}}=0.748\n$$\n\nand\n$P_{N_{2} O}=X_{N_{2} O} \\times P_{\\text {Total }}=0.748 \\times 192 \\mathrm{kPa}=144 \\mathrm{kPa}$\n"} {"id": "textbook_2449", "title": "857. Check Your Learning - ", "content": "\nWhat is the pressure of a mixture of 0.200 g of $\\mathrm{H}_{2}, 1.00 \\mathrm{~g}$ of $\\mathrm{N}_{2}$, and 0.820 g of Ar in a container with a volume of 2.00 L at $20^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_2450", "title": "858. Answer: - ", "content": "\n1.87 atm\n"} {"id": "textbook_2451", "title": "859. Collection of Gases over Water - ", "content": "\nA simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 8.21), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.\n\n\nFIGURE 8.21 When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers)."} {"id": "textbook_2452", "title": "859. Collection of Gases over Water - ", "content": "However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor-this is referred to as the \"dry\" gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 8.22); more detailed information on the temperature dependence of water vapor can be found in Table 8.2, and vapor pressure will be discussed in more detail in the chapter on liquids.\n"} {"id": "textbook_2453", "title": "859. Collection of Gases over Water - ", "content": "FIGURE 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature."} {"id": "textbook_2454", "title": "859. Collection of Gases over Water - ", "content": "Vapor Pressure of Ice and Water in Various Temperatures at Sea Level"} {"id": "textbook_2455", "title": "859. Collection of Gases over Water - ", "content": "| Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Pressure (torr) | Temperature ( ${ }^{\\circ} \\mathrm{C}$ ) | Pressure (torr) | Temperature ( ${ }^{\\circ} \\mathrm{C}$ ) | Pressure (torr) |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| -10 | 1.95 | 18 | 15.5 | 30 | 31.8 |\n| -5 | 3.0 | 19 | 16.5 | 35 | 42.2 |\n| -2 | 3.9 | 20 | 17.5 | 40 | 55.3 |\n| 0 | 4.6 | 21 | 18.7 | 50 | 92.5 |\n| 2 | 5.3 | 22 | 19.8 | 60 | 149.4 |\n| 4 | 6.1 | 23 | 21.1 | 70 | 233.7 |\n| 6 | 7.0 | 24 | 22.4 | 80 | 355.1 |\n| 8 | 8.0 | 25 | 23.8 | 90 | 525.8 |\n| 10 | 9.2 | 26 | 25.2 | 95 | 633.9 |\n| 12 | 10.5 | 27 | 26.7 | 99 | 733.2 |\n| 14 | 12.0 | 28 | 28.3 | 100.0 | 760.0 |\n| 16 | 13.6 | 29 | 30.0 | 101.0 | 787.6 |"} {"id": "textbook_2456", "title": "859. Collection of Gases over Water - ", "content": "TABLE 8.2\n"} {"id": "textbook_2457", "title": "860. EXAMPLE 8.16 - ", "content": ""} {"id": "textbook_2458", "title": "861. Pressure of a Gas Collected Over Water - ", "content": "\nIf 0.200 L of argon is collected over water at a temperature of $26^{\\circ} \\mathrm{C}$ and a pressure of 750 torr in a system like that shown in Figure 8.21, what is the partial pressure of argon?\n"} {"id": "textbook_2459", "title": "862. Solution - ", "content": "\nAccording to Dalton's law, the total pressure in the bottle ( 750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:"} {"id": "textbook_2460", "title": "862. Solution - ", "content": "$$\nP_{\\mathrm{T}}=P_{\\mathrm{Ar}}+P_{\\mathrm{H}_{2} \\mathrm{O}}\n$$"} {"id": "textbook_2461", "title": "862. Solution - ", "content": "Rearranging this equation to solve for the pressure of argon gives:"} {"id": "textbook_2462", "title": "862. Solution - ", "content": "$$\nP_{\\mathrm{Ar}}=P_{\\mathrm{T}}-P_{\\mathrm{H}_{2} \\mathrm{O}}\n$$"} {"id": "textbook_2463", "title": "862. Solution - ", "content": "The pressure of water vapor above a sample of liquid water at $26^{\\circ} \\mathrm{C}$ is 25.2 torr (Appendix E), so:"} {"id": "textbook_2464", "title": "862. Solution - ", "content": "$$\nP_{\\mathrm{Ar}}=750 \\text { torr }-25.2 \\text { torr }=725 \\text { torr }\n$$\n"} {"id": "textbook_2465", "title": "863. Check Your Learning - ", "content": "\nA sample of oxygen collected over water at a temperature of $29.0^{\\circ} \\mathrm{C}$ and a pressure of 764 torr has a volume of 0.560 L . What volume would the dry oxygen have under the same conditions of temperature and pressure?\n"} {"id": "textbook_2466", "title": "864. Answer: - ", "content": "\n0.583 L\n"} {"id": "textbook_2467", "title": "865. Chemical Stoichiometry and Gases - ", "content": "\nChemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions."} {"id": "textbook_2468", "title": "865. Chemical Stoichiometry and Gases - ", "content": "We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.\n"} {"id": "textbook_2469", "title": "866. Avogadro's Law Revisited - ", "content": "\nSometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure."} {"id": "textbook_2470", "title": "866. Avogadro's Law Revisited - ", "content": "We can extend Avogadro's law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)$, a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.\n\nThe explanation for this is illustrated in Figure 8.23. According to Avogadro's law, equal volumes of gaseous $\\mathrm{N}_{2}$, $\\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$, at the same temperature and pressure, contain the same number of molecules. Because one molecule of $\\mathrm{N}_{2}$ reacts with three molecules of $\\mathrm{H}_{2}$ to produce two molecules of $\\mathrm{NH}_{3}$, the volume of $\\mathrm{H}_{2}$ required is three times the volume of $\\mathrm{N}_{2}$, and the volume of $\\mathrm{NH}_{3}$ produced is two times the volume of $\\mathrm{N}_{2}$.\n\n\nFIGURE 8.23 One volume of $\\mathrm{N}_{2}$ combines with three volumes of $\\mathrm{H}_{2}$ to form two volumes of $\\mathrm{NH}_{3}$.\n"} {"id": "textbook_2471", "title": "867. EXAMPLE 8.17 - ", "content": ""} {"id": "textbook_2472", "title": "868. Reaction of Gases - ", "content": "\nPropane, $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$, is used in gas grills to provide the heat for cooking. What volume of $\\mathrm{O}_{2}(\\mathrm{~g})$ measured at $25{ }^{\\circ} \\mathrm{C}$ and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.\n"} {"id": "textbook_2473", "title": "869. Solution - ", "content": "\nThe ratio of the volumes of $\\mathrm{C}_{3} \\mathrm{H}_{8}$ and $\\mathrm{O}_{2}$ will be equal to the ratio of their coefficients in the balanced equation for the reaction:\n\n$$\n\\begin{array}{lll}\n\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})+5 \\mathrm{O}_{2}(\\mathrm{~g}) & \\longrightarrow & 3 \\mathrm{CO}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\\\\n1 \\text { volume }+5 \\text { volumes } & & 3 \\text { volumes }+4 \\text { volumes }\n\\end{array}\n$$\n\nFrom the equation, we see that one volume of $\\mathrm{C}_{3} \\mathrm{H}_{8}$ will react with five volumes of $\\mathrm{O}_{2}$ :"} {"id": "textbook_2474", "title": "869. Solution - ", "content": "$$\n2.7 \\mathrm{LC}_{3} \\mathrm{H}_{8}-\\frac{5 \\mathrm{LO}_{2}}{1 \\mathrm{LC}_{3} \\mathrm{H}_{8}}=13.5 \\mathrm{~L} \\mathrm{O}_{2}\n$$"} {"id": "textbook_2475", "title": "869. Solution - ", "content": "A volume of 13.5 L of $\\mathrm{O}_{2}$ will be required to react with $2.7 \\mathrm{~L}^{\\text {of }} \\mathrm{C}_{3} \\mathrm{H}_{8}$.\n"} {"id": "textbook_2476", "title": "870. Check Your Learning - ", "content": "\nAn acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, at $0^{\\circ} \\mathrm{C}$ and 1 atm. How many tanks of oxygen, each providing $7.00 \\times 10^{3} \\mathrm{~L}$ of $\\mathrm{O}_{2}$ at $0^{\\circ} \\mathrm{C}$ and 1 atm , will be required to burn the acetylene?"} {"id": "textbook_2477", "title": "870. Check Your Learning - ", "content": "$$\n2 \\mathrm{C}_{2} \\mathrm{H}_{2}+5 \\mathrm{O}_{2} \\longrightarrow 4 \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$\n"} {"id": "textbook_2478", "title": "871. Answer: - ", "content": "\n3.34 tanks ( $2.34 \\times 10^{4} \\mathrm{~L}$ )\n"} {"id": "textbook_2479", "title": "872. EXAMPLE 8.18 - ", "content": ""} {"id": "textbook_2480", "title": "873. Volumes of Reacting Gases - ", "content": "\nAmmonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at $25^{\\circ} \\mathrm{C}$ and 1 atm , was manufactured. What volume of $\\mathrm{H}_{2}(\\mathrm{~g})$, measured under the same conditions, was required to prepare this amount of ammonia by reaction with $\\mathrm{N}_{2}$ ?"} {"id": "textbook_2481", "title": "873. Volumes of Reacting Gases - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)\n$$\n"} {"id": "textbook_2482", "title": "874. Solution - ", "content": "\nBecause equal volumes of $\\mathrm{H}_{2}$ and $\\mathrm{NH}_{3}$ contain equal numbers of molecules and each three molecules of $\\mathrm{H}_{2}$ that react produce two molecules of $\\mathrm{NH}_{3}$, the ratio of the volumes of $\\mathrm{H}_{2}$ and $\\mathrm{NH}_{3}$ will be equal to $3: 2$. Two volumes of $\\mathrm{NH}_{3}$, in this case in units of billion $\\mathrm{ft}^{3}$, will be formed from three volumes of $\\mathrm{H}_{2}$ :"} {"id": "textbook_2483", "title": "874. Solution - ", "content": "$$\n683 \\text { billion } \\mathrm{ft}^{3} \\mathrm{NH}_{3} \\times \\frac{3 \\text { billion } \\mathrm{ft}^{3} \\mathrm{H}_{2}}{2 \\text { bitlion } \\mathrm{ft}^{3} \\mathrm{NH}_{3}}=1.02 \\times 10^{3} \\text { billion } \\mathrm{ft}^{3} \\mathrm{H}_{2}\n$$"} {"id": "textbook_2484", "title": "874. Solution - ", "content": "The manufacture of 683 billion $\\mathrm{ft}^{3}$ of $\\mathrm{NH}_{3}$ required 1020 billion $\\mathrm{ft}^{3}$ of $\\mathrm{H}_{2}$. (At $25^{\\circ} \\mathrm{C}$ and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)\n"} {"id": "textbook_2485", "title": "875. Check Your Learning - ", "content": "\nWhat volume of $\\mathrm{O}_{2}(\\mathrm{~g})$ measured at $25{ }^{\\circ} \\mathrm{C}$ and 760 torr is required to react with 17.0 L of ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$, measured under the same conditions of temperature and pressure? The products are $\\mathrm{CO}_{2}$ and water vapor.\n"} {"id": "textbook_2486", "title": "876. Answer: - ", "content": "\n51.0 L\n"} {"id": "textbook_2487", "title": "877. EXAMPLE 8.19 - ", "content": ""} {"id": "textbook_2488", "title": "878. Volume of Gaseous Product - ", "content": "\nWhat volume of hydrogen at $27^{\\circ} \\mathrm{C}$ and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?"} {"id": "textbook_2489", "title": "878. Volume of Gaseous Product - ", "content": "$$\n2 \\mathrm{Ga}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{GaCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$\n"} {"id": "textbook_2490", "title": "879. Solution - ", "content": "\nConvert the provided mass of the limiting reactant, Ga , to moles of hydrogen produced:"} {"id": "textbook_2491", "title": "879. Solution - ", "content": "$$\n8.88 \\mathrm{~g} \\mathrm{Ga} \\times \\frac{1 \\mathrm{molGa}}{69.723 \\mathrm{~g} \\mathrm{ga}} \\times \\frac{3 \\mathrm{~mol} \\mathrm{H}_{2}}{2 \\mathrm{molGa}}=0.191 \\mathrm{~mol} \\mathrm{H}_{2}\n$$"} {"id": "textbook_2492", "title": "879. Solution - ", "content": "Convert the provided temperature and pressure values to appropriate units ( K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:"} {"id": "textbook_2493", "title": "879. Solution - ", "content": "$$\nV=\\left(\\frac{n R T}{P}\\right)=\\frac{0.191 \\mathrm{mot} \\times 0.08206 \\mathrm{~L} \\mathrm{atmmol}^{-1} \\mathrm{~K}^{-1} \\times 300 \\mathrm{~K}}{0.951 \\mathrm{~atm}}=4.94 \\mathrm{~L}\n$$\n"} {"id": "textbook_2494", "title": "880. Check Your Learning - ", "content": "\nSulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of $\\mathrm{SO}_{2}$ at $343{ }^{\\circ} \\mathrm{C}$ and 1.21 atm is produced by burning 1.00 kg of sulfur in excess oxygen?\n"} {"id": "textbook_2495", "title": "881. Answer: - ", "content": "\n$1.30 \\times 10^{3} \\mathrm{~L}$\n"} {"id": "textbook_2496", "title": "882. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_2497", "title": "883. Greenhouse Gases and Climate Change - ", "content": "\nThe thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than $0.5 \\%$ of the air molecules. Of the energy from the sun that reaches the earth, almost $\\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most of this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than $30^{\\circ} \\mathrm{C}$ (nearly $60^{\\circ} \\mathrm{F}$ ). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth's climate (Figure 8.24).\n"} {"id": "textbook_2498", "title": "883. Greenhouse Gases and Climate Change - ", "content": "FIGURE 8.24 Greenhouse gases trap enough of the sun's energy to make the planet habitable-this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events."} {"id": "textbook_2499", "title": "883. Greenhouse Gases and Climate Change - ", "content": "There is strong evidence from multiple sources that higher atmospheric levels of $\\mathrm{CO}_{2}$ are caused by human activity, with fossil fuel burning accounting for about $\\frac{3}{4}$ of the recent increase in $\\mathrm{CO}_{2}$. Reliable data from ice cores reveals that $\\mathrm{CO}_{2}$ concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the $\\mathrm{CO}_{2}$ concentration has increased from preindustrial levels of $\\sim 280 \\mathrm{ppm}$ to more than 400 ppm today (Figure 8.25)."} {"id": "textbook_2500", "title": "883. Greenhouse Gases and Climate Change - ", "content": "Carbon Dioxide in the Atmosphere\n"} {"id": "textbook_2501", "title": "883. Greenhouse Gases and Climate Change - ", "content": "FIGURE $8.25 \\mathrm{CO}_{2}$ levels over the past 700,000 years were typically from 200-300 ppm, with a steep, unprecedented increase over the past 50 years.\n"} {"id": "textbook_2502", "title": "884. LINK TO LEARNING - ", "content": "\nClick here (http://openstax.org/l/16GlobalWarming) to see a 2-minute video explaining greenhouse gases and global warming.\n"} {"id": "textbook_2503", "title": "885. Portrait of a Chemist - ", "content": ""} {"id": "textbook_2504", "title": "886. Susan Solomon - ", "content": "\nAtmospheric and climate scientist Susan Solomon (Figure 8.26) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine's 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change."} {"id": "textbook_2505", "title": "886. Susan Solomon - ", "content": "She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA."} {"id": "textbook_2506", "title": "886. Susan Solomon - ", "content": "For more information, watch this video (http://openstax.org/l/16SusanSolomon) about Susan Solomon.\n"} {"id": "textbook_2507", "title": "886. Susan Solomon - ", "content": "FIGURE 8.26 Susan Solomon's research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)\n"} {"id": "textbook_2508", "title": "886. Susan Solomon - 886.1. Effusion and Diffusion of Gases", "content": ""} {"id": "textbook_2509", "title": "887. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2510", "title": "887. LEARNING OBJECTIVES - ", "content": "- Define and explain effusion and diffusion\n- State Graham's law and use it to compute relevant gas properties"} {"id": "textbook_2511", "title": "887. LEARNING OBJECTIVES - ", "content": "If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule"} {"id": "textbook_2512", "title": "887. LEARNING OBJECTIVES - ", "content": "In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 8.27). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly-regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 8.27. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).\n"} {"id": "textbook_2513", "title": "887. LEARNING OBJECTIVES - ", "content": "FIGURE 8.27 (a) Two gases, $\\mathrm{H}_{2}$ and $\\mathrm{O}_{2}$, are initially separated. (b) When the stopcock is opened, they mix together."} {"id": "textbook_2514", "title": "887. LEARNING OBJECTIVES - ", "content": "The lighter gas, $\\mathrm{H}_{2}$, passes through the opening faster than $\\mathrm{O}_{2}$, so just after the stopcock is opened, more $\\mathrm{H}_{2}$ molecules move to the $\\mathrm{O}_{2}$ side than $\\mathrm{O}_{2}$ molecules move to the $\\mathrm{H}_{2}$ side. (c) After a short time, both the slowermoving $\\mathrm{O}_{2}$ molecules and the faster-moving $\\mathrm{H}_{2}$ molecules have distributed themselves evenly on both sides of the vessel."} {"id": "textbook_2515", "title": "887. LEARNING OBJECTIVES - ", "content": "We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:"} {"id": "textbook_2516", "title": "887. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate of diffusion }=\\frac{\\text { amount of gas passing through an area }}{\\text { unit of time }}\n$$"} {"id": "textbook_2517", "title": "887. LEARNING OBJECTIVES - ", "content": "The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation."} {"id": "textbook_2518", "title": "887. LEARNING OBJECTIVES - ", "content": "A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 8.28). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.\n"} {"id": "textbook_2519", "title": "887. LEARNING OBJECTIVES - ", "content": "Diffusion\n"} {"id": "textbook_2520", "title": "887. LEARNING OBJECTIVES - ", "content": "Effusion"} {"id": "textbook_2521", "title": "887. LEARNING OBJECTIVES - ", "content": "FIGURE 8.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion."} {"id": "textbook_2522", "title": "887. LEARNING OBJECTIVES - ", "content": "If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 8.29). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham's law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:"} {"id": "textbook_2523", "title": "887. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate of effusion } \\propto \\frac{1}{\\sqrt{\\mathcal{M}}}\n$$"} {"id": "textbook_2524", "title": "887. LEARNING OBJECTIVES - ", "content": "This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:"} {"id": "textbook_2525", "title": "887. LEARNING OBJECTIVES - ", "content": "$$\n\\frac{\\text { rate of effusion of } A}{\\text { rate of effusion of } B}=\\frac{\\sqrt{\\mathcal{N}_{B}}}{\\sqrt{\\mathcal{N}_{A}}}\n$$"} {"id": "textbook_2526", "title": "887. LEARNING OBJECTIVES - ", "content": ""} {"id": "textbook_2527", "title": "887. LEARNING OBJECTIVES - ", "content": "FIGURE 8.29 The left photograph shows two balloons inflated with different gases, helium (orange) and argon (blue).The right-side photograph shows the balloons approximately 12 hours after being filled, at which time the helium balloon has become noticeably more deflated than the argon balloon, due to the greater effusion rate of the lighter helium gas. (credit: modification of work by Paul Flowers)\n"} {"id": "textbook_2528", "title": "888. EXAMPLE 8.20 - ", "content": ""} {"id": "textbook_2529", "title": "889. Applying Graham's Law to Rates of Effusion - ", "content": "\nCalculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.\n"} {"id": "textbook_2530", "title": "890. Solution - ", "content": "\nFrom Graham's law, we have:"} {"id": "textbook_2531", "title": "890. Solution - ", "content": "$$\n\\frac{\\text { rate of effusion of hydrogen }}{\\text { rate of effusion of oxygen }}=\\frac{\\sqrt{32 \\mathrm{gmol}^{-1}}}{\\sqrt{2 \\mathrm{gmol}^{-1}}}=\\frac{\\sqrt{16}}{\\sqrt{1}}=\\frac{4}{1}\n$$\n\nHydrogen effuses four times as rapidly as oxygen.\n"} {"id": "textbook_2532", "title": "891. Check Your Learning - ", "content": "\nAt a particular pressure and temperature, nitrogen gas effuses at the rate of $79 \\mathrm{~mL} / \\mathrm{s}$. Under the same conditions, at what rate will sulfur dioxide effuse?\n"} {"id": "textbook_2533", "title": "892. Answer: - ", "content": "\n$52 \\mathrm{~mL} / \\mathrm{s}$\n"} {"id": "textbook_2534", "title": "893. EXAMPLE 8.21 - ", "content": ""} {"id": "textbook_2535", "title": "894. Effusion Time Calculations - ", "content": "\nIt takes 243 s for $4.46 \\times 10^{-5} \\mathrm{~mol}$ Xe to effuse through a tiny hole. Under the same conditions, how long will it take $4.46 \\times 10^{-5} \\mathrm{~mol} \\mathrm{Ne}$ to effuse?\n"} {"id": "textbook_2536", "title": "895. Solution - ", "content": "\nIt is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:"} {"id": "textbook_2537", "title": "895. Solution - ", "content": "$$\n\\text { rate of effusion }=\\frac{\\text { amount of gas transferred }}{\\text { time }}\n$$"} {"id": "textbook_2538", "title": "895. Solution - ", "content": "and combine it with Graham's law:"} {"id": "textbook_2539", "title": "895. Solution - ", "content": "$$\n\\frac{\\text { rate of effusion of gas } \\mathrm{Xe}}{\\text { rate of effusion of gas Ne }}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$"} {"id": "textbook_2540", "title": "895. Solution - ", "content": "To get:"} {"id": "textbook_2541", "title": "895. Solution - ", "content": "$$\n\\frac{\\frac{\\text { amount of Xe transferred }}{\\text { time for Xe }}}{\\frac{\\text { amount of Ne transferred }}{\\text { time for Ne }}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$"} {"id": "textbook_2542", "title": "895. Solution - ", "content": "Noting that amount of $A=$ amount of $B$, and solving for time for $N e$ :"} {"id": "textbook_2543", "title": "895. Solution - ", "content": "$$\n\\frac{\\frac{\\text { amount of } \\mathrm{Xe}}{\\text { time for } \\mathrm{Xe}}}{\\frac{\\text { amount of Ne }}{\\text { time for } \\mathrm{Ne}}}=\\frac{\\text { time for } \\mathrm{Ne}}{\\text { time for } \\mathrm{Xe}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$"} {"id": "textbook_2544", "title": "895. Solution - ", "content": "and substitute values:"} {"id": "textbook_2545", "title": "895. Solution - ", "content": "$$\n\\frac{\\text { time for } \\mathrm{Ne}}{243 \\mathrm{~s}}=\\sqrt{\\frac{20.2 \\mathrm{~g} \\mathrm{~mol}}{131.3 \\mathrm{~m} \\mathrm{~mol}}}=0.392\n$$"} {"id": "textbook_2546", "title": "895. Solution - ", "content": "Finally, solve for the desired quantity:"} {"id": "textbook_2547", "title": "895. Solution - ", "content": "$$\n\\text { time for } \\mathrm{Ne}=0.392 \\times 243 \\mathrm{~s}=95.3 \\mathrm{~s}\n$$"} {"id": "textbook_2548", "title": "895. Solution - ", "content": "Note that this answer is reasonable: Since Ne is lighter than Xe , the effusion rate for Ne will be larger than that for Xe , which means the time of effusion for Ne will be smaller than that for Xe .\n"} {"id": "textbook_2549", "title": "896. Check Your Learning - ", "content": "\nA party balloon filled with helium deflates to $\\frac{2}{3}$ of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air ( $\\mathcal{M}=28.2 \\mathrm{~g} / \\mathrm{mol}$ ) to deflate to $\\frac{1}{2}$ of its original volume?\n"} {"id": "textbook_2550", "title": "897. Answer: - ", "content": "\n32 h\n"} {"id": "textbook_2551", "title": "898. EXAMPLE 8.22 - ", "content": ""} {"id": "textbook_2552", "title": "899. Determining Molar Mass Using Graham's Law - ", "content": "\nAn unknown gas effuses 1.66 times more rapidly than $\\mathrm{CO}_{2}$. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?\n"} {"id": "textbook_2553", "title": "900. Solution - ", "content": "\nFrom Graham's law, we have:"} {"id": "textbook_2554", "title": "900. Solution - ", "content": "$$\n\\frac{\\text { rate of effusion of Unknown }}{\\text { rate of effusion of } \\mathrm{CO}_{2}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{CO}_{2}}}}{\\sqrt{\\mathcal{M}_{\\text {Unknown }}}}\n$$"} {"id": "textbook_2555", "title": "900. Solution - ", "content": "Plug in known data:"} {"id": "textbook_2556", "title": "900. Solution - ", "content": "$$\n\\frac{1.66}{1}=\\frac{\\sqrt{44.0 \\mathrm{~g} / \\mathrm{mol}}}{\\sqrt{\\mathcal{M}_{\\text {Unknown }}}}\n$$"} {"id": "textbook_2557", "title": "900. Solution - ", "content": "Solve:"} {"id": "textbook_2558", "title": "900. Solution - ", "content": "$$\n\\mathcal{M}_{\\text {Unknown }}=\\frac{44.0 \\mathrm{~g} / \\mathrm{mol}}{(1.66)^{2}}=16.0 \\mathrm{~g} / \\mathrm{mol}\n$$"} {"id": "textbook_2559", "title": "900. Solution - ", "content": "The gas could well be $\\mathrm{CH}_{4}$, the only gas with this molar mass.\n"} {"id": "textbook_2560", "title": "901. Check Your Learning - ", "content": "\nHydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.\n"} {"id": "textbook_2561", "title": "902. Answer: - ", "content": "\n$163 \\mathrm{~g} / \\mathrm{mol}$\n"} {"id": "textbook_2562", "title": "903. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_2563", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - ", "content": "\nGaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only $0.72 \\%$ of ${ }^{235} \\mathrm{U}$, the kind of uranium that is \"fissile,\" that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is $2-5 \\%{ }^{235} \\mathrm{U}$, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham's law. In a gaseous diffusion enrichment plant, uranium hexafluoride ( $\\mathrm{UF}_{6}$, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The ${ }^{235} \\mathrm{UF}_{6}$ molecules have a higher average speed and diffuse through the barrier a little faster than the heavier ${ }^{238} \\mathrm{UF}_{6}$ molecules. The gas that has passed through the barrier is slightly enriched in ${ }^{235} \\mathrm{UF}_{6}$ and the residual gas is slightly depleted. The small difference in molecular weights between ${ }^{235} \\mathrm{UF}_{6}$ and ${ }^{238} \\mathrm{UF}_{6}$ only about $0.4 \\%$ enrichment, is achieved in one diffuser (Figure 8.30). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.\n\n\nFIGURE 8.30 In a diffuser, gaseous $U F_{6}$ is pumped through a porous barrier, which partially separates ${ }^{235} \\mathrm{UF}_{6}$ from ${ }^{238} \\mathrm{UF}_{6}$ The $\\mathrm{UF}_{6}$ must pass through many large diffuser units to achieve sufficient enrichment in ${ }^{235} \\mathrm{U}$."} {"id": "textbook_2564", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - ", "content": "The large scale separation of gaseous ${ }^{235} \\mathrm{UF}_{6}$ from ${ }^{238} \\mathrm{UF}_{6}$ was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about $10^{-6} \\mathrm{~cm}$ in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive $\\mathrm{UF}_{6}$."} {"id": "textbook_2565", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - ", "content": "Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high\npressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.\n"} {"id": "textbook_2566", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "\nLEARNING OBJECTIVES"} {"id": "textbook_2567", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "- State the postulates of the kinetic-molecular theory\n- Use this theory's postulates to explain the gas laws"} {"id": "textbook_2568", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm . Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships."} {"id": "textbook_2569", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term \"molecule\" will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)"} {"id": "textbook_2570", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.\n2. The molecules composing the gas are negligibly small compared to the distances between them.\n3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.\n4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).\n5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas."} {"id": "textbook_2571", "title": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory", "content": "The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle's, Charles's, Amontons's, Avogadro's, and Dalton's laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham's law.\n"} {"id": "textbook_2572", "title": "905. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I - ", "content": "\nRecalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:"} {"id": "textbook_2573", "title": "905. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I - ", "content": "- Amontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 8.31).\n- Charles's law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.\n- Boyle's law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is, if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with\nthe container wall will therefore occur more frequently and the pressure exerted by the gas will increase (Figure 8.31).\n- Avogadro's law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 8.31).\n- Dalton's Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.\n"} {"id": "textbook_2574", "title": "905. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I - ", "content": "FIGURE 8.31 (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.\n"} {"id": "textbook_2575", "title": "906. molecular speeds and Kinetic Energy - ", "content": "\nThe previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample."} {"id": "textbook_2576", "title": "906. molecular speeds and Kinetic Energy - ", "content": "In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 8.32).\n"} {"id": "textbook_2577", "title": "906. molecular speeds and Kinetic Energy - ", "content": "FIGURE 8.32 The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, $v_{p}$, is a little less than $400 \\mathrm{~m} / \\mathrm{s}$, while the root mean square speed, $u_{\\mathrm{rms}}$, is closer to $500 \\mathrm{~m} / \\mathrm{s}$."} {"id": "textbook_2578", "title": "906. molecular speeds and Kinetic Energy - ", "content": "The kinetic energy (KE) of a particle of mass ( $m$ ) and speed $(u)$ is given by:"} {"id": "textbook_2579", "title": "906. molecular speeds and Kinetic Energy - ", "content": "$$\n\\mathrm{KE}=\\frac{1}{2} m u^{2}\n$$"} {"id": "textbook_2580", "title": "906. molecular speeds and Kinetic Energy - ", "content": "Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules ( $\\mathrm{J}=\\mathrm{kg}$ $\\mathrm{m}^{2} \\mathrm{~s}^{-2}$ ). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, $\\mathbf{u}_{\\mathbf{r m s}}$, is defined as the square root of the average of the squares of the speeds with $n=$ the number of particles:"} {"id": "textbook_2581", "title": "906. molecular speeds and Kinetic Energy - ", "content": "$$\nu_{\\mathrm{rms}}=\\sqrt{\\overline{u^{2}}}=\\sqrt{\\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\\ldots}{n}}\n$$"} {"id": "textbook_2582", "title": "906. molecular speeds and Kinetic Energy - ", "content": "The average kinetic energy for a mole of particles, $\\mathrm{KE}_{\\text {avg }}$, is then equal to:"} {"id": "textbook_2583", "title": "906. molecular speeds and Kinetic Energy - ", "content": "$$\n\\mathrm{KE}_{\\mathrm{avg}}=\\frac{1}{2} M u_{\\mathrm{rms}}^{2}\n$$"} {"id": "textbook_2584", "title": "906. molecular speeds and Kinetic Energy - ", "content": "where $M$ is the molar mass expressed in units of $\\mathrm{kg} / \\mathrm{mol}$. The $\\mathrm{KE}_{\\text {avg }}$ of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:"} {"id": "textbook_2585", "title": "906. molecular speeds and Kinetic Energy - ", "content": "$$\n\\mathrm{KE}_{\\mathrm{avg}}=\\frac{3}{2} R T\n$$"} {"id": "textbook_2586", "title": "906. molecular speeds and Kinetic Energy - ", "content": "where $R$ is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is $8.314 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}\\left(8.314 \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-2} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)$. These two separate equations for $\\mathrm{KE}_{\\text {avg }}$ may be combined and rearranged to yield a relation between molecular speed and temperature:"} {"id": "textbook_2587", "title": "906. molecular speeds and Kinetic Energy - ", "content": "$$\n\\begin{gathered}\n\\frac{1}{2} M u_{\\mathrm{rms}}^{2}=\\frac{3}{2} R T \\\\\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}}\n\\end{gathered}\n$$\n"} {"id": "textbook_2588", "title": "907. EXAMPLE 8.23 - ", "content": ""} {"id": "textbook_2589", "title": "908. Calculation of $u_{\\text {rms }}$ - ", "content": "\nCalculate the root-mean-square speed for a nitrogen molecule at $30^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2590", "title": "909. Solution - ", "content": "\nConvert the temperature into Kelvin:"} {"id": "textbook_2591", "title": "909. Solution - ", "content": "$$\n30^{\\circ} \\mathrm{C}+273=303 \\mathrm{~K}\n$$"} {"id": "textbook_2592", "title": "909. Solution - ", "content": "Determine the molar mass of nitrogen in kilograms:"} {"id": "textbook_2593", "title": "909. Solution - ", "content": "$$\n\\frac{28.0 \\frac{\\mathrm{~g}}{\\mathrm{o}}}{1 \\mathrm{~mol}} \\times \\frac{1 \\mathrm{~kg}}{1000 \\frac{\\mathrm{~g}}{\\mathrm{o}}}=0.028 \\mathrm{~kg} / \\mathrm{mol}\n$$"} {"id": "textbook_2594", "title": "909. Solution - ", "content": "Replace the variables and constants in the root-mean-square speed equation, replacing Joules with the equivalent $\\mathrm{kg} \\mathrm{m}{ }^{2} \\mathrm{~s}^{-2}$ :"} {"id": "textbook_2595", "title": "909. Solution - ", "content": "$$\n\\begin{gathered}\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}} \\\\\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3(8.314 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K})(303 \\mathrm{~K})}{(0.028 \\mathrm{~kg} / \\mathrm{mol})}}=\\sqrt{2.70 \\times 10^{5} \\mathrm{~m}^{2} \\mathrm{~s}^{-2}}=519 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$\n"} {"id": "textbook_2596", "title": "910. Check Your Learning - ", "content": "\nCalculate the root-mean-square speed for a mole of oxygen molecules at $-23^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2597", "title": "911. Answer: - ", "content": "\n441 m/s"} {"id": "textbook_2598", "title": "911. Answer: - ", "content": "If the temperature of a gas increases, its $\\mathrm{KE}_{\\text {avg }}$ increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, $\\mathrm{KE}_{\\text {avg }}$ decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 8.33.\n\n\nFIGURE 8.33 The molecular speed distribution for nitrogen gas $\\left(N_{2}\\right)$ shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases."} {"id": "textbook_2599", "title": "911. Answer: - ", "content": "At a given temperature, all gases have the same $\\mathrm{KE}_{\\text {avg }}$ for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher $u_{r m s}$, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower $u_{r m s}$, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in Figure 8.34.\n"} {"id": "textbook_2600", "title": "911. Answer: - ", "content": "FIGURE 8.34 molecular speed is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.\n"} {"id": "textbook_2601", "title": "912. LINK TO LEARNING - ", "content": "\nThe gas simulator (http://openstax.org/l/16MolecVelocity) may be used to examine the effect of temperature on molecular speeds. Examine the simulator's \"energy histograms\" (molecular speed distributions) and \"species information\" (which gives average speed values) for molecules of different masses at various temperatures."} {"id": "textbook_2602", "title": "912. LINK TO LEARNING - ", "content": "The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II\nAccording to Graham's law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates."} {"id": "textbook_2603", "title": "912. LINK TO LEARNING - ", "content": "The rate of effusion of a gas depends directly on the (average) speed of its molecules:\neffusion rate $\\propto u_{\\text {rms }}$\nUsing this relation, and the equation relating molecular speed to mass, Graham's law may be easily derived as shown here:"} {"id": "textbook_2604", "title": "912. LINK TO LEARNING - ", "content": "$$\n\\begin{gathered}\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}} \\\\\nM=\\frac{3 R T}{u_{\\mathrm{rms}}^{2}}=\\frac{3 R T}{\\bar{u}^{2}} \\\\\n\\frac{\\text { effusion rate } \\mathrm{A}}{\\text { effusion rate } \\mathrm{B}}=\\frac{u_{\\mathrm{rms} \\mathrm{~A}}}{u_{\\mathrm{rms} \\mathrm{~B}}}=\\frac{\\sqrt{\\frac{3 R T}{M_{\\mathrm{A}}}}}{\\sqrt{\\frac{3 R T}{M_{\\mathrm{B}}}}}=\\sqrt{\\frac{M_{\\mathrm{B}}}{M_{\\mathrm{A}}}}\n\\end{gathered}\n$$"} {"id": "textbook_2605", "title": "912. LINK TO LEARNING - ", "content": "The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham's law.\n"} {"id": "textbook_2606", "title": "912. LINK TO LEARNING - 912.1. Non-Ideal Gas Behavior", "content": ""} {"id": "textbook_2607", "title": "913. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2608", "title": "913. LEARNING OBJECTIVES - ", "content": "- Describe the physical factors that lead to deviations from ideal gas behavior\n- Explain how these factors are represented in the van der Waals equation\n- Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior\n- Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation"} {"id": "textbook_2609", "title": "913. LEARNING OBJECTIVES - ", "content": "Thus far, the ideal gas law, $P V=n R T$, has been applied to a variety of different types of problems, ranging from\nreaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered."} {"id": "textbook_2610", "title": "913. LEARNING OBJECTIVES - ", "content": "One way in which the accuracy of $P V=n R T$ can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, $V_{\\mathrm{m}}$ ) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor ( $\\mathbf{Z}$ ) with:\n\n$$\n\\mathrm{Z}=\\frac{\\text { molar volume of gas at same } T \\text { and } P}{\\text { molar volume of ideal gas at same } T \\text { and } P}=\\left(\\frac{P V_{m}}{R T}\\right)_{\\text {measured }}\n$$"} {"id": "textbook_2611", "title": "913. LEARNING OBJECTIVES - ", "content": "Ideal gas behavior is therefore indicated when this ratio is equal to 1 , and any deviation from 1 is an indication of non-ideal behavior. Figure 8.35 shows plots of Z over a large pressure range for several common gases.\n\n\nFIGURE 8.35 A graph of the compressibility factor $(Z)$ vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law."} {"id": "textbook_2612", "title": "913. LEARNING OBJECTIVES - ", "content": "As is apparent from Figure 8.35, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas."} {"id": "textbook_2613", "title": "913. LEARNING OBJECTIVES - ", "content": "Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle's law.\n\nAt relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure 8.36). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.\n\n\nFIGURE 8.36 (a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted at constant volume compared to an ideal gas."} {"id": "textbook_2614", "title": "913. LEARNING OBJECTIVES - ", "content": "There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them."} {"id": "textbook_2615", "title": "913. LEARNING OBJECTIVES - ", "content": "$$\nP V=n R T \\longrightarrow\\left(P+\\frac{a n^{2}}{V^{2}}\\right)(V-n b)=n R T\n$$"} {"id": "textbook_2616", "title": "913. LEARNING OBJECTIVES - ", "content": "The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant $b$ corresponds to the size of the molecules of a particular gas. The \"correction\" to the pressure term in the ideal gas law is $\\frac{n^{2} a}{V^{2}}$, and the \"correction\" to the volume is $n b$. Note that when $V$ is relatively large and $n$ is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, $P V=n R T$. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table 8.3."} {"id": "textbook_2617", "title": "913. LEARNING OBJECTIVES - ", "content": "Values of van der Waals Constants for Some Common Gases"} {"id": "textbook_2618", "title": "913. LEARNING OBJECTIVES - ", "content": "| Gas | $\\boldsymbol{a}\\left(\\mathrm{L}^{2} \\mathrm{~atm} / \\mathrm{mol}^{2}\\right)$ | $\\boldsymbol{b}(\\mathrm{L} / \\mathrm{mol})$ |\n| :--- | :--- | :--- |\n| $\\mathrm{N}_{2}$ | 1.39 | 0.0391 |\n| $\\mathrm{O}_{2}$ | 1.36 | 0.0318 |\n| $\\mathrm{CO}_{2}$ | 3.59 | 0.0427 |\n| $\\mathrm{H}_{2} \\mathrm{O}$ | 5.46 | 0.0305 |\n| He | 0.0342 | 0.0237 |\n| $\\mathrm{CCl}_{4}$ | 20.4 | 0.1383 |"} {"id": "textbook_2619", "title": "913. LEARNING OBJECTIVES - ", "content": "TABLE 8.3"} {"id": "textbook_2620", "title": "913. LEARNING OBJECTIVES - ", "content": "At low pressures, the correction for intermolecular attraction, $a$, is more important than the one for molecular volume, $b$. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by $P V=n R T$ over a small range of pressures. This behavior is reflected by the \"dips\" in several of the compressibility curves shown in Figure 8.35. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised ( Z decreases with increasing $P$ ). At very high pressures, the gas becomes less compressible ( Z increases with $P$ ), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume."} {"id": "textbook_2621", "title": "913. LEARNING OBJECTIVES - ", "content": "Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded-this is, however, very often not the case.\n"} {"id": "textbook_2622", "title": "914. EXAMPLE 8.24 - ", "content": ""} {"id": "textbook_2623", "title": "915. Comparison of Ideal Gas Law and van der Waals Equation - ", "content": "\nA 4.25-L flask contains $3.46 \\mathrm{~mol} \\mathrm{CO}_{2}$ at $229^{\\circ} \\mathrm{C}$. Calculate the pressure of this sample of $\\mathrm{CO}_{2}$ :\n(a) from the ideal gas law\n(b) from the van der Waals equation\n(c) Explain the reason(s) for the difference.\n"} {"id": "textbook_2624", "title": "916. Solution - ", "content": "\n(a) From the ideal gas law:"} {"id": "textbook_2625", "title": "916. Solution - ", "content": "$$\nP=\\frac{n R T}{V}=\\frac{3.46 \\mathrm{mot} \\times 0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\times 502 \\mathrm{~K}}{4.25 \\mathrm{t}}=33.5 \\mathrm{~atm}\n$$"} {"id": "textbook_2626", "title": "916. Solution - ", "content": "(b) From the van der Waals equation:"} {"id": "textbook_2627", "title": "916. Solution - ", "content": "$$\n\\begin{gathered}\n\\left(P+\\frac{n^{2} a}{V^{2}}\\right) \\times(V-n b)=n R T \\rightarrow P=\\frac{n R T}{(V-n b)}-\\frac{n^{2} a}{V^{2}} \\\\\nP=\\frac{3.46 \\mathrm{~mol} \\times 0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}}{}=1 \\mathrm{~K}^{-1} \\times 502 \\mathrm{~K} \\\\\n\\left(4.25 \\mathrm{~L}-3.46 \\mathrm{~mol} \\times 0.0427 \\mathrm{~L} \\mathrm{~mol}^{-1}\\right)\n\\end{gathered} \\frac{(3.46 \\mathrm{~mol})^{2} \\times 3.59 \\mathrm{~L}^{2} \\mathrm{~atm} \\mathrm{~mol}^{2}}{(4.25 \\mathrm{~L})^{2}} .\n$$"} {"id": "textbook_2628", "title": "916. Solution - ", "content": "This finally yields $P=32.4 \\mathrm{~atm}$.\n(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because $\\mathrm{CO}_{2}$ molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.\n"} {"id": "textbook_2629", "title": "917. Check your Learning - ", "content": "\nA $560-\\mathrm{mL}$ flask contains $21.3 \\mathrm{~g} \\mathrm{~N} \\mathrm{~N}_{2}$ at $145^{\\circ} \\mathrm{C}$. Calculate the pressure of $\\mathrm{N}_{2}$ :\n(a) from the ideal gas law\n(b) from the van der Waals equation\n(c) Explain the reason(s) for the difference.\n"} {"id": "textbook_2630", "title": "918. Answer: - ", "content": "\n(a) 46.562 atm ; (b) 46.594 atm ; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions.\n"} {"id": "textbook_2631", "title": "919. Key Terms - ", "content": "\nabsolute zero temperature at which the volume of a gas would be zero according to Charles's law.\nAmontons's law (also, Gay-Lussac's law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant\natmosphere (atm) unit of pressure; 1 atm = $101,325 \\mathrm{~Pa}$\nAvogadro's law volume of a gas at constant temperature and pressure is proportional to the number of gas molecules\nbar (bar or b) unit of pressure; $1 \\mathrm{bar}=100,000 \\mathrm{~Pa}$\nbarometer device used to measure atmospheric pressure\nBoyle's law volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured\nCharles's law volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant\ncompressibility factor ( $Z$ ) ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation\nDalton's law of partial pressures total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.\ndiffusion movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)\neffusion transfer of gaseous atoms or molecules from a container to a vacuum through very small openings\nGraham's law of effusion rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses\nhydrostatic pressure pressure exerted by a fluid due to gravity\nideal gas hypothetical gas whose physical properties are perfectly described by the gas laws\nideal gas constant ( $\\boldsymbol{R}$ ) constant derived from the\n $8.314 \\mathrm{~L} \\mathrm{kPa} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\nideal gas law relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws\nkinetic molecular theory theory based on simple principles and assumptions that effectively explains ideal gas behavior\nmanometer device used to measure the pressure of a gas trapped in a container\nmean free path average distance a molecule travels between collisions\nmole fraction ( $\\boldsymbol{X}$ ) concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components\npartial pressure pressure exerted by an individual gas in a mixture\npascal (Pa) SI unit of pressure; $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$\npounds per square inch (psi) unit of pressure common in the US\npressure force exerted per unit area\nrate of diffusion amount of gas diffusing through a given area over a given time\nroot mean square speed ( $\\boldsymbol{u}_{\\text {rms }}$ ) measure of average speed for a group of particles calculated as the square root of the average squared speed\nstandard conditions of temperature and pressure (STP) $273.15 \\mathrm{~K}\\left(0^{\\circ} \\mathrm{C}\\right)$ and $1 \\mathrm{~atm}(101.325 \\mathrm{kPa})$\nstandard molar volume volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally\ntorr unit of pressure; 1 torr $=\\frac{1}{760} \\mathrm{~atm}$\nvan der Waals equation modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior\nvapor pressure of water pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature\n"} {"id": "textbook_2632", "title": "920. Key Equations - ", "content": "\n$$\nP=\\frac{F}{A}\n$$"} {"id": "textbook_2633", "title": "920. Key Equations - ", "content": "$p=h \\rho g$\n$P V=n R T$\n$P_{\\text {Total }}=P_{A}+P_{B}+P_{C}+\\ldots=\\Sigma_{\\mathrm{i}} P_{\\mathrm{i}}$\n$P_{A}=X_{A} P_{\\text {Total }}$\n$X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}$\nrate of diffusion $=\\frac{\\text { amount of gas passing through an area }}{\\text { unit of time }}$\n$\\frac{\\text { rate of effusion of gas } \\mathrm{A}}{\\text { rate of effusion of gas } \\mathrm{B}}=\\frac{\\sqrt{m_{B}}}{\\sqrt{m_{A}}}=\\frac{\\sqrt{\\mathcal{N}_{B}}}{\\sqrt{\\mathcal{N}_{A}}}$\n$u_{\\mathrm{rms}}=\\sqrt{\\overline{u^{2}}}=\\sqrt{\\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\\ldots}{n}}$\n$\\mathrm{KE}_{\\mathrm{avg}}=\\frac{3}{2} R T$\n$u_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}}$\n$\\mathrm{Z}=\\frac{\\text { molar volume of gas at same } T \\text { and } P}{\\text { molar volume of ideal gas at same } T \\text { and } P}=\\left(\\frac{P \\times V_{m}}{R \\times T}\\right)_{\\text {measured }}$\n$\\left(P+\\frac{n^{2} a}{V^{2}}\\right) \\times(V-n b)=n R T$\n"} {"id": "textbook_2634", "title": "921. Summary - ", "content": ""} {"id": "textbook_2635", "title": "921. Summary - 921.1. Gas Pressure", "content": "\nGases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.\n"} {"id": "textbook_2636", "title": "921. Summary - 921.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law", "content": "\nThe behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons's law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro's law)."} {"id": "textbook_2637", "title": "921. Summary - 921.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law", "content": "The equations describing these laws are special cases of the ideal gas law, $P V=n R T$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its kelvin temperature, and $R$ is the ideal (universal) gas constant.\n"} {"id": "textbook_2638", "title": "921. Summary - 921.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions", "content": "\nThe ideal gas law can be used to derive a number of convenient equations relating directly measured\nquantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton's law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro's law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.\n"} {"id": "textbook_2639", "title": "921. Summary - 921.4. Effusion and Diffusion of Gases", "content": "\nGaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules' masses (Graham's law).\n"} {"id": "textbook_2640", "title": "921. Summary - 921.5. The Kinetic-Molecular Theory", "content": "\nThe kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the\ngas and the mass of its molecules.\n"} {"id": "textbook_2641", "title": "921. Summary - 921.6. Non-Ideal Gas Behavior", "content": "\nGas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate\ndescription of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.\n"} {"id": "textbook_2642", "title": "922. Exercises - ", "content": ""} {"id": "textbook_2643", "title": "922. Exercises - 922.1. Gas Pressure", "content": "\n1. Why are sharp knives more effective than dull knives? (Hint: Think about the definition of pressure.)\n2. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?\n3. Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?\n4. A typical barometric pressure in Redding, California, is about 750 mm Hg . Calculate this pressure in atm and kPa.\n5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg . What is this pressure in atmospheres and kilopascals?\n6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?\n7. Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?\n8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars ( $1 \\mathrm{bar}=0.987 \\mathrm{~atm}$ ). What is that pressure in torr and kPa ?\n9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.\n10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa . What is the pressure of this gas in atmospheres and torr?\n11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.\n(a) What was the pressure in kPa ?\n(b) The pressure near the seacoast in the northeastern United States is usually reported near $30.0 \\mathrm{in} . \\mathrm{Hg}$. During a hurricane, the pressure may fall to near 28.0 in . Hg. Calculate the drop in pressure in torr.\n12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?\n13. The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar\n\n14. The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in . Hg, determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar\n\n15. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg , determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n16. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg , determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n17. How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?\n"} {"id": "textbook_2644", "title": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law", "content": "\n18. Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?\n19. Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8.16) change as they rise to the surface, assuming that they remain intact.\n20. One way to state Boyle's law is \"All other things being equal, the pressure of a gas is inversely proportional to its volume.\" (a) What is the meaning of the term \"inversely proportional?\" (b) What are the \"other things\" that must be equal?\n21. An alternate way to state Avogadro's law is \"All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.\" (a) What is the meaning of the term \"directly proportional?\" (b) What are the \"other things\" that must be equal?\n22. How would the graph in Figure 8.12 change if the number of moles of gas in the sample used to determine the curve were doubled?\n23. How would the graph in Figure 8.13 change if the number of moles of gas in the sample used to determine the curve were doubled?\n24. In addition to the data found in Figure 8.13, what other information do we need to find the mass of the sample of air used to determine the graph?\n25. Determine the volume of 1 mol of $\\mathrm{CH}_{4}$ gas at 150 K and 1 atm, using Figure 8.12.\n26. Determine the pressure of the gas in the syringe shown in Figure 8.13 when its volume is 12.5 mL , using:\n(a) the appropriate graph\n(b) Boyle's law\n27. A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 ${ }^{\\circ} \\mathrm{C}$. If the can is thrown into a fire $\\left(\\mathrm{T}=475^{\\circ} \\mathrm{C}\\right)$, what will be the pressure in the hot can?\n28. What is the temperature of an 11.2 - L sample of carbon monoxide, CO , at 744 torr if it occupies 13.3 L at 55 ${ }^{\\circ} \\mathrm{C}$ and 744 torr?\n29. A $2.50-\\mathrm{L}$ volume of hydrogen measured at $-196^{\\circ} \\mathrm{C}$ is warmed to $100^{\\circ} \\mathrm{C}$. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.\n30. A balloon inflated with three breaths of air has a volume of 1.7 L . At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?\n31. A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of $25^{\\circ} \\mathrm{C}$ at ground level. What is the volume of the balloon under these conditions?\n\n32. The volume of an automobile air bag was 66.8 L when inflated at $25^{\\circ} \\mathrm{C}$ with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa ?\n33. How many moles of gaseous boron trifluoride, $\\mathrm{BF}_{3}$, are contained in a $4.3410-\\mathrm{L}$ bulb at 788.0 K if the pressure is 1.220 atm ? How many grams of $\\mathrm{BF}_{3}$ ?\n34. Iodine, $\\mathrm{I}_{2}$, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a $73.3-\\mathrm{mL}$ bulb that contains 0.292 g of $\\mathrm{I}_{2}$ vapor at a pressure of 0.462 atm ?\n35. How many grams of gas are present in each of the following cases?\n(a) 0.100 L of $\\mathrm{CO}_{2}$ at 307 torr and $26^{\\circ} \\mathrm{C}$\n(b) $8.75 \\mathrm{~L}^{\\text {of }} \\mathrm{C}_{2} \\mathrm{H}_{4}$, at 378.3 kPa and 483 K\n(c) 221 mL of Ar at 0.23 torr and $-54^{\\circ} \\mathrm{C}$\n36. A high altitude balloon is filled with $1.41 \\times 10^{4} \\mathrm{~L}$ of hydrogen at a temperature of $21^{\\circ} \\mathrm{C}$ and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km , where the temperature is $-48^{\\circ} \\mathrm{C}$ and the pressure is 63.1 torr?\n37. A cylinder of medical oxygen has a volume of 35.4 L , and contains $\\mathrm{O}_{2}$ at a pressure of 151 atm and a temperature of $25^{\\circ} \\mathrm{C}$. What volume of $\\mathrm{O}_{2}$ does this correspond to at normal body conditions, that is, 1 atm and $37^{\\circ} \\mathrm{C}$ ?\n38. A large scuba tank (Figure 8.16) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at $20^{\\circ} \\mathrm{C}$ and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at $20^{\\circ} \\mathrm{C}$ ?\n39. A 20.0-L cylinder containing 11.34 kg of butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm , and a temperature of $27^{\\circ} \\mathrm{C}$.\n40. While resting, the average $70-\\mathrm{kg}$ human male consumes 14 L of pure $\\mathrm{O}_{2}$ per hour at $25^{\\circ} \\mathrm{C}$ and 100 kPa . How many moles of $\\mathrm{O}_{2}$ are consumed by a 70 kg man while resting for 1.0 h ?\n41. For a given amount of gas showing ideal behavior, draw labeled graphs of:\n(a) the variation of $P$ with $V$\n(b) the variation of $V$ with $T$\n(c) the variation of $P$ with $T$\n(d) the variation of $\\frac{1}{P}$ with $V$\n42. A liter of methane gas, $\\mathrm{CH}_{4}$, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, $\\mathrm{H}_{2}$, at STP. Using Avogadro's law as a starting point, explain why.\n43. The effect of chlorofluorocarbons (such as $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ ) on the depletion of the ozone layer is well known. The use of substitutes, such as $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~F}(g)$, for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:\n(a) $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~F}(g)$\n44. As 1 g of the radioactive element radium decays over 1 year, it produces $1.16 \\times 10^{18}$ alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of $25^{\\circ} \\mathrm{C}$ ?\n45. A balloon with a volume of 100.21 L at $21^{\\circ} \\mathrm{C}$ and 0.981 atm is released and just barely clears the top of Mount Crumpit in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 ${ }^{\\circ} \\mathrm{C}$, what is the pressure experienced by the balloon as it clears Mount Crumpet?\n46. If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?\n47. If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?\n"} {"id": "textbook_2645", "title": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions", "content": "\n48. What is the density of laughing gas, dinitrogen monoxide, $\\mathrm{N}_{2} \\mathrm{O}$, at a temperature of 325 K and a pressure of 113.0 kPa ?\n49. Calculate the density of Freon $12, \\mathrm{CF}_{2} \\mathrm{Cl}_{2}$, at $30.0^{\\circ} \\mathrm{C}$ and 0.954 atm.\n50. Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.\n51. A cylinder of $\\mathrm{O}_{2}(\\mathrm{~g})$ used in breathing by patients with emphysema has a volume of 3.00 L at a pressure of 10.0 atm . If the temperature of the cylinder is $28.0^{\\circ} \\mathrm{C}$, what mass of oxygen is in the cylinder?\n52. What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature $26^{\\circ} \\mathrm{C}$ and a pressure of 307 torr?\n53. What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature $126^{\\circ} \\mathrm{C}$ and a pressure of 777 torr?\n54. How could you show experimentally that the molecular formula of propene is $\\mathrm{C}_{3} \\mathrm{H}_{6}$, not $\\mathrm{CH}_{2}$ ?\n55. The density of a certain gaseous fluoride of phosphorus is $3.93 \\mathrm{~g} / \\mathrm{L}$ at STP. Calculate the molar mass of this fluoride and determine its molecular formula.\n56. Consider this question: What is the molecular formula of a compound that contains $39 \\% \\mathrm{C}, 45 \\% \\mathrm{~N}$, and $16 \\% \\mathrm{H}$ if 0.157 g of the compound occupies 125 mL with a pressure of 99.5 kPa at $22^{\\circ} \\mathrm{C}$ ?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n57. A 36.0-L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 $\\mathrm{g} \\mathrm{CO}_{2}, 805 \\mathrm{~g} \\mathrm{O}_{2}$, and $4,880 \\mathrm{~g} \\mathrm{~N}_{2}$. At 25 degrees C , what is the pressure in the cylinder in atmospheres?\n58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains $5.0 \\% \\mathrm{CO}_{2}, 12.0 \\% \\mathrm{O}_{2}$, and the remainder $\\mathrm{N}_{2}$ at a total pressure of 146 atm . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)\n59. A sample of gas isolated from unrefined petroleum contains $90.0 \\% \\mathrm{CH}_{4}, 8.9 \\% \\mathrm{C}_{2} \\mathrm{H}_{6}$, and $1.1 \\% \\mathrm{C}_{3} \\mathrm{H}_{8}$ at a total pressure of 307.2 kPa . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)\n60. A mixture of 0.200 g of $\\mathrm{H}_{2}, 1.00 \\mathrm{~g}$ of $\\mathrm{N}_{2}$, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.\n61. Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than $3.0 \\% \\mathrm{O}_{2}$ is not. If enough $\\mathrm{O}_{2}$ is added to a cylinder of $\\mathrm{H}_{2}$ at 33.2 atm to bring the total pressure to 34.5 atm , is the mixture explosive?\n62. A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as $2 \\times 10^{-6} \\mathrm{mg} / \\mathrm{L}$ of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at $26^{\\circ} \\mathrm{C}$ ?\n63. A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of $18{ }^{\\circ} \\mathrm{C}$. What is the pressure of the carbon monoxide? (See Table 8.2 for the vapor pressure of water.)\n64. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of $27^{\\circ} \\mathrm{C}$. The mass of the gas was 0.472 g . What was the molar mass of the gas?\n65. Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO :\n$2 \\mathrm{HgO}(s) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{O}_{2}(g)$\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{O}_{2}$ at $23{ }^{\\circ} \\mathrm{C}$ and 0.975 atm is produced by the decomposition of 5.36 g of HgO ?\n(b) Answer the question.\n66. Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:\n$4 \\mathrm{H}_{2} \\mathrm{O}(g)+3 \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+4 \\mathrm{H}_{2}(g)$\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{H}_{2}$ at a pressure of 745 torr and a temperature of $20^{\\circ} \\mathrm{C}$ can be prepared from the reaction of 15.0 g of $\\mathrm{H}_{2} \\mathrm{O}$ ?\n(b) Answer the question.\n67. The chlorofluorocarbon $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ can be recycled into a different compound by reaction with hydrogen to produce $\\mathrm{CH}_{2} \\mathrm{~F}_{2}(\\mathrm{~g})$, a compound useful in chemical manufacturing:\n$\\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)+4 \\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{CH}_{2} \\mathrm{~F}_{2}(g)+2 \\mathrm{HCl}(g)$\n(a) Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and $35.5^{\\circ} \\mathrm{C}$ would be required to react with 1 ton $\\left(1.000 \\times 10^{3} \\mathrm{~kg}\\right)$ of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ ?\n(b) Answer the question.\n68. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide $\\left(\\mathrm{NaN}_{3}\\right)$. The other product is sodium metal. Calculate the volume of nitrogen gas at $27{ }^{\\circ} \\mathrm{C}$ and 756 torr formed by the decomposition of 125 g of sodium azide.\n69. Lime, CaO , is produced by heating calcium carbonate, $\\mathrm{CaCO}_{3}$; carbon dioxide is the other product.\n(a) Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton $\\left(1.000 \\times 10^{3} \\mathrm{~kg}\\right)$ of calcium carbonate?\n(b) Answer the question.\n70. Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, and solid calcium hydroxide were formed by the reaction of calcium carbide, $\\mathrm{CaC}_{2}$, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{C}_{2} \\mathrm{H}_{2}$ at 1.005 atm and $12.2{ }^{\\circ} \\mathrm{C}$ is formed by the reaction of 15.48 g of $\\mathrm{CaC}_{2}$ with water?\n(b) Answer the question.\n71. Calculate the volume of oxygen required to burn 12.00 L of ethane gas, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, to produce carbon dioxide and water, if the volumes of $\\mathrm{C}_{2} \\mathrm{H}_{6}$ and $\\mathrm{O}_{2}$ are measured under the same conditions of temperature and pressure.\n72. What volume of $\\mathrm{O}_{2}$ at STP is required to oxidize 8.0 L of NO at STP to $\\mathrm{NO}_{2}$ ? What volume of $\\mathrm{NO}_{2}$ is produced at STP?\n73. Consider the following questions:\n(a) What is the total volume of the $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $600^{\\circ} \\mathrm{C}$ and 0.888 atm produced by the combustion of 1.00 L of $\\mathrm{C}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ measured at STP?\n(b) What is the partial pressure of $\\mathrm{H}_{2} \\mathrm{O}$ in the product gases?\n74. Methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, is produced industrially by the following reaction:\n$\\mathrm{CO}(g)+2 \\mathrm{H}_{2}(g) \\xrightarrow{\\text { copper catalyst } 300^{\\circ} \\mathrm{C}, 300 \\mathrm{~atm}} \\mathrm{CH}_{3} \\mathrm{OH}(g)$\nAssuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.\n75. What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of $\\mathrm{BaO}_{2}$ to BaO and $\\mathrm{O}_{2}$ ?\n76. A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of $\\mathrm{N}_{2}$ and 1.25 L of $\\mathrm{O}_{2}$ at STP. What is the colorless gas?\n77. Ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is produced industrially from ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, by the following sequence of reactions:\n$3 \\mathrm{C}_{2} \\mathrm{H}_{4}+2 \\mathrm{H}_{2} \\mathrm{SO}_{4} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{HSO}_{4}+\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{SO}_{4}$\n$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{HSO}_{4}+\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{SO}_{4}+3 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 3 \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nWhat volume of ethylene at STP is required to produce 1.000 metric ton ( 1000 kg ) of ethanol if the overall yield of ethanol is $90.1 \\%$ ?\n78. One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature $\\left(37^{\\circ} \\mathrm{C}\\right)$ and a pressure of 743 torr, what is the molar mass of hemoglobin?\n79. A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH , the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)\n80. One method of analyzing amino acids is the van Slyke method. The characteristic amino groups $\\left(-\\mathrm{NH}_{2}\\right)$ in protein material are allowed to react with nitrous acid, $\\mathrm{HNO}_{2}$, to form $\\mathrm{N}_{2}$ gas. From the volume of the gas, the amount of amino acid can be determined. A $0.0604-\\mathrm{g}$ sample of a biological sample containing glycine, $\\mathrm{CH}_{2}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{COOH}$, was analyzed by the van Slyke method and yielded 3.70 mL of $\\mathrm{N}_{2}$ collected over water at a pressure of 735 torr and $29^{\\circ} \\mathrm{C}$. What was the percentage of glycine in the sample?\n$\\mathrm{CH}_{2}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{HNO}_{2} \\longrightarrow \\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{N}_{2}$\n"} {"id": "textbook_2646", "title": "922. Exercises - 922.4. Effusion and Diffusion of Gases", "content": "\n81. A balloon filled with helium gas takes 6 hours to deflate to $50 \\%$ of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by $50 \\%$ ?\n82. Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 8.27.\n83. Starting with the definition of rate of effusion and Graham's finding relating rate and molar mass, show how to derive the Graham's law equation, relating the relative rates of effusion for two gases to their molecular masses.\n84. Heavy water, $\\mathrm{D}_{2} \\mathrm{O}$ (molar mass $=20.03 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ), can be separated from ordinary water, $\\mathrm{H}_{2} \\mathrm{O}$ (molar mass $=$ 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{D}_{2} \\mathrm{O}$.\n85. Which of the following gases diffuse more slowly than oxygen? $\\mathrm{F}_{2}, \\mathrm{Ne}, \\mathrm{N}_{2} \\mathrm{O}, \\mathrm{C}_{2} \\mathrm{H}_{2}, \\mathrm{NO}, \\mathrm{Cl}_{2}, \\mathrm{H}_{2} \\mathrm{~S}$\n86. During the discussion of gaseous diffusion for enriching uranium, it was claimed that ${ }^{235} \\mathrm{UF}_{6}$ diffuses $0.4 \\%$ faster than ${ }^{238} \\mathrm{UF}_{6}$. Show the calculation that supports this value. The molar mass of ${ }^{235} \\mathrm{UF}_{6}=$ $235.043930+6 \\times 18.998403=349.034348 \\mathrm{~g} / \\mathrm{mol}$, and the molar mass of ${ }^{238} \\mathrm{UF}_{6}=238.050788+6 \\times$ $18.998403=352.041206 \\mathrm{~g} / \\mathrm{mol}$.\n87. Calculate the relative rate of diffusion of ${ }^{1} \\mathrm{H}_{2}$ (molar mass $2.0 \\mathrm{~g} / \\mathrm{mol}$ ) compared with ${ }^{2} \\mathrm{H}_{2}$ (molar mass 4.0 $\\mathrm{g} / \\mathrm{mol}$ ) and the relative rate of diffusion of $\\mathrm{O}_{2}$ (molar mass $32 \\mathrm{~g} / \\mathrm{mol}$ ) compared with $\\mathrm{O}_{3}$ (molar mass $48 \\mathrm{~g} /$ mol ).\n88. A gas of unknown identity diffuses at a rate of $83.3 \\mathrm{~mL} / \\mathrm{s}$ in a diffusion apparatus in which carbon dioxide diffuses at the rate of $102 \\mathrm{~mL} / \\mathrm{s}$. Calculate the molecular mass of the unknown gas.\n89. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of $\\mathrm{NH}_{4} \\mathrm{Cl}$ forms where gaseous $\\mathrm{NH}_{3}$ and gaseous HCl first come into contact. $\\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g) \\longrightarrow \\mathrm{NH}_{4} \\mathrm{Cl}(s)$ At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both $\\mathrm{NH}_{3}$ and HCl , and find out how much faster $\\mathrm{NH}_{3}$ diffuses than HCl .)\n"} {"id": "textbook_2647", "title": "922. Exercises - 922.5. The Kinetic-Molecular Theory", "content": "\n90. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.\n91. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.\n92. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:\n(a) The pressure of the gas is increased by reducing the volume at constant temperature.\n(b) The pressure of the gas is increased by increasing the temperature at constant volume.\n(c) The average speed of the molecules is increased by a factor of 2.\n93. The distribution of molecular speeds in a sample of helium is shown in Figure 8.34. If the sample is cooled, will the distribution of speeds look more like that of $\\mathrm{H}_{2}$ or of $\\mathrm{H}_{2} \\mathrm{O}$ ? Explain your answer.\n94. What is the ratio of the average kinetic energy of a $\\mathrm{SO}_{2}$ molecule to that of an $\\mathrm{O}_{2}$ molecule in a mixture of two gases? What is the ratio of the root mean square speeds, $u_{\\text {rms }}$, of the two gases?\n95. A 1-L sample of CO initially at STP is heated to 546 K , and its volume is increased to 2 L .\n(a) What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?\n(b) What is the effect on the average kinetic energy of the molecules?\n(c) What is the effect on the root mean square speed of the molecules?\n96. The root mean square speed of $\\mathrm{H}_{2}$ molecules at $25^{\\circ} \\mathrm{C}$ is about $1.6 \\mathrm{~km} / \\mathrm{s}$. What is the root mean square speed of a $\\mathrm{N}_{2}$ molecule at $25^{\\circ} \\mathrm{C}$ ?\n97. Answer the following questions:\n(a) Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?\n(b) Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?\n(c) At a pressure of 1 atm and a temperature of $20^{\\circ} \\mathrm{C}$, dry air has a density of $1.2256 \\mathrm{~g} / \\mathrm{L}$. What is the (average) molar mass of dry air?\n(d) The average temperature of the gas in a hot-air balloon is $1.30 \\times 10^{2}{ }^{\\circ} \\mathrm{F}$. Calculate its density, assuming the molar mass equals that of dry air.\n(e) The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?\n(f) An average balloon has a diameter of 60 feet and a volume of $1.1 \\times 10^{5} \\mathrm{ft}^{3}$. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?\n(g) A balloon carries 40.0 gallons of liquid propane (density $0.5005 \\mathrm{~g} / \\mathrm{L}$ ). What volume of $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ gas is produced by the combustion of this propane?\n(h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in $\\mathrm{kJ} / \\mathrm{min}$ ) from the hot air in the bag during the flight?\n98. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas $2, \\frac{R_{1}}{R_{2}}$, is the same at $0^{\\circ} \\mathrm{C}$ and $100^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2648", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "\n99. Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?\n"} {"id": "textbook_2649", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "Temperature\n"} {"id": "textbook_2650", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "PV/RT\n"} {"id": "textbook_2651", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "Temperature\n\n\n"} {"id": "textbook_2652", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "Temperature\n100. Explain why the plot of $P V$ for $\\mathrm{CO}_{2}$ differs from that of an ideal gas.\n\n101. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.\n(a) high pressure, small volume\n(b) high temperature, low pressure\n(c) low temperature, high pressure\n102. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.\n103. For which of the following gases should the correction for the molecular volume be largest: $\\mathrm{CO}, \\mathrm{CO}_{2}, \\mathrm{H}_{2}, \\mathrm{He}, \\mathrm{NH}_{3}, \\mathrm{SF}_{6}$ ?\n104. A 0.245 -L flask contains $0.467 \\mathrm{~mol} \\mathrm{CO}_{2}$ at $159{ }^{\\circ} \\mathrm{C}$. Calculate the pressure:\n(a) using the ideal gas law\n(b) using the van der Waals equation\n(c) Explain the reason for the difference.\n(d) Identify which correction (that for P or V ) is dominant and why.\n105. Answer the following questions:\n(a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like?\n(b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified?\n(c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.\n(d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.\n(e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?"} {"id": "textbook_2653", "title": "922. Exercises - 922.6. Non-Ideal Gas Behavior", "content": "418 8\u2022Exercises\n"} {"id": "textbook_2654", "title": "923. CHAPTER 9 Thermochemistry - ", "content": "\n"} {"id": "textbook_2655", "title": "923. CHAPTER 9 Thermochemistry - ", "content": "Figure 9.1 Sliding a match head along a rough surface initiates a combustion reaction that produces energy in the form of heat and light. (credit: modification of work by Laszlo Ilyes)\n"} {"id": "textbook_2656", "title": "924. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_2657", "title": "924. CHAPTER OUTLINE - 924.1. Energy Basics", "content": ""} {"id": "textbook_2658", "title": "924. CHAPTER OUTLINE - 924.2. Calorimetry", "content": ""} {"id": "textbook_2659", "title": "924. CHAPTER OUTLINE - 924.3. Enthalpy", "content": "\n9.4 Strengths of Ionic and Covalent Bonds"} {"id": "textbook_2660", "title": "924. CHAPTER OUTLINE - 924.3. Enthalpy", "content": "INTRODUCTION Chemical reactions, such as those that occur when you light a match, involve changes in energy as well as matter. Societies at all levels of development could not function without the energy released by chemical reactions. In 2012, about $85 \\%$ of US energy consumption came from the combustion of petroleum products, coal, wood, and garbage. We use this energy to produce electricity ( $38 \\%$ ); to transport food, raw materials, manufactured goods, and people (27\\%); for industrial production (21\\%); and to heat and power our homes and businesses (10\\%). ${ }^{\\frac{1}{2}}$ While these combustion reactions help us meet our essential energy needs, they are also recognized by the majority of the scientific community as a major contributor to global climate change."} {"id": "textbook_2661", "title": "924. CHAPTER OUTLINE - 924.3. Enthalpy", "content": "Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy."} {"id": "textbook_2662", "title": "924. CHAPTER OUTLINE - 924.3. Enthalpy", "content": "[^6]The chemical bond is simply another form of energy, and its strength is indicated in exactly the same units as any other process involving energy. That is, in joules (or kilojoules) - or, when taken on a molar basis, in J/mol or $\\mathrm{kJ} / \\mathrm{mol}$. All of the principles involving energy we have examined up to now apply to bond energies exactly as they do to all other forms of energy: that is, energy may change form or be absorbed or released, but it cannot be created or destroyed. The energy of a chemical bond is indicated by the bond enthalpy, taken by convention to be the energy required to break a chemical bond. The same energy is released when a chemical bond is formed. Given that a chemical reaction involves breaking some bonds and making others, the enthalpy change for a chemical reaction can be estimated by analyzing the bonds broken and formed during the reaction. While this procedure gives an estimate of the overall enthalpy, it should be noted that a far more precise method involves using enthalpies of formation; the fact that there can be significant differences between, for example, an $\\mathrm{O}-\\mathrm{H}$ bond in water and that in acetic acid accounts for the discrepancy in calculating $\\Delta H$ using bond enthalpies versus the use of enthalpies of formation.\n"} {"id": "textbook_2663", "title": "924. CHAPTER OUTLINE - 924.4. Energy Basics", "content": ""} {"id": "textbook_2664", "title": "925. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2665", "title": "925. LEARNING OBJECTIVES - ", "content": "- Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes\n- Distinguish the related properties of heat, thermal energy, and temperature\n- Define and distinguish specific heat and heat capacity, and describe the physical implications of both\n- Perform calculations involving heat, specific heat, and temperature change"} {"id": "textbook_2666", "title": "925. LEARNING OBJECTIVES - ", "content": "Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure 9.2). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges.\n\n\nFIGURE 9.2 The energy involved in chemical changes is important to our daily lives: (a) A cheeseburger for lunch provides the energy you need to get through the rest of the day; (b) the combustion of gasoline provides the energy that moves your car (and you) between home, work, and school; and (c) coke, a processed form of coal, provides the energy needed to convert iron ore into iron, which is essential for making many of the products we use daily. (credit a: modification of work by \"Pink Sherbet Photography\"/Flickr; credit b: modification of work by Jeffery Turner)"} {"id": "textbook_2667", "title": "925. LEARNING OBJECTIVES - ", "content": "Over $90 \\%$ of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world's energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter."} {"id": "textbook_2668", "title": "925. LEARNING OBJECTIVES - ", "content": "This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes-an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science.\n"} {"id": "textbook_2669", "title": "926. Energy - ", "content": "\nEnergy can be defined as the capacity to supply heat or do work. One type of work ( $\\boldsymbol{w}$ ) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire-we move matter (the air in the pump) against the opposing force of the air already in the tire."} {"id": "textbook_2670", "title": "926. Energy - ", "content": "Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy, the energy an object has because of its relative position, composition, or condition, and kinetic energy, the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure 9.3). A battery has potential energy because the chemicals within it can produce electricity that can do work.\n\n(a)\n\n(b)"} {"id": "textbook_2671", "title": "926. Energy - ", "content": "FIGURE 9.3 (a) Water at a higher elevation, for example, at the top of Victoria Falls, has a higher potential energy than water at a lower elevation. As the water falls, some of its potential energy is converted into kinetic energy. (b) If the water flows through generators at the bottom of a dam, such as the Hoover Dam shown here, its kinetic energy is converted into electrical energy. (credit a: modification of work by Steve Jurvetson; credit b: modification of work by \"curimedia\"/Wikimedia commons)"} {"id": "textbook_2672", "title": "926. Energy - ", "content": "Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.)"} {"id": "textbook_2673", "title": "926. Energy - ", "content": "When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car's engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic\nenergy) when they move the cylinders' pistons.\nAccording to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry.\n"} {"id": "textbook_2674", "title": "927. Thermal Energy, Temperature, and Heat - ", "content": "\nThermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of \"hot\" or \"cold.\" When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is \"hot.\" When the atoms and molecules are moving slowly, they have lower average KE, and we say that the object is \"cold\" (Figure 9.4). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.\n"} {"id": "textbook_2675", "title": "927. Thermal Energy, Temperature, and Heat - ", "content": "Hot water\n(a)\n"} {"id": "textbook_2676", "title": "927. Thermal Energy, Temperature, and Heat - ", "content": "Cold water\n(b)"} {"id": "textbook_2677", "title": "927. Thermal Energy, Temperature, and Heat - ", "content": "FIGURE 9.4 (a) The molecules in a sample of hot water move more rapidly than (b) those in a sample of cold water.\n"} {"id": "textbook_2678", "title": "928. LINK TO LEARNING - ", "content": "\nClick on this interactive simulation (http://openstax.org/l/16PHETtempFX) to view the effects of temperature on molecular motion."} {"id": "textbook_2679", "title": "928. LINK TO LEARNING - ", "content": "Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure 9.5. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.\n"} {"id": "textbook_2680", "title": "928. LINK TO LEARNING - ", "content": "FIGURE 9.5 (a) In an alcohol or mercury thermometer, the liquid (dyed red for visibility) expands when heated and contracts when cooled, much more so than the glass tube that contains the liquid. (b) In a bimetallic thermometer, two different metals (such as brass and steel) form a two-layered strip. When heated or cooled, one of the metals (brass) expands or contracts more than the other metal (steel), causing the strip to coil or uncoil. Both types of thermometers have a calibrated scale that indicates the temperature. (credit a: modification of work by \"dwstucke\"/Flickr)\n"} {"id": "textbook_2681", "title": "929. LINK TO LEARNING - ", "content": "\nThe following demonstration (http://openstax.org/l/16Bimetallic) allows one to view the effects of heating and cooling a coiled bimetallic strip."} {"id": "textbook_2682", "title": "929. LINK TO LEARNING - ", "content": "Heat ( $\\boldsymbol{q}$ ) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance ( L ). The atoms and molecules in H have a higher average KE than those in L . If we place substance H in contact with substance L , the thermal energy will flow spontaneously from substance H to substance L . The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance $L$ will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure 9.6).\n"} {"id": "textbook_2683", "title": "929. LINK TO LEARNING - ", "content": "FIGURE 9.6 (a) Substances H and L are initially at different temperatures, and their atoms have different average kinetic energies. (b) When they contact each other, collisions between the molecules result in the transfer of kinetic (thermal) energy from the hotter to the cooler matter. (c) The two objects reach \"thermal equilibrium\" when both substances are at the same temperature and their molecules have the same average kinetic energy.\n"} {"id": "textbook_2684", "title": "930. LINK TO LEARNING - ", "content": "\nClick on the PhET simulation (http://openstax.org/l/16PHETenergy) to explore energy forms and changes. Visit\nthe Energy Systems tab to create combinations of energy sources, transformation methods, and outputs. Click on Energy Symbols to visualize the transfer of energy."} {"id": "textbook_2685", "title": "930. LINK TO LEARNING - ", "content": "Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process-this process also releases energy in the form of light as evidenced by the torch's flame (Figure 9.7). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.\n\n(a)\n\n(b)"} {"id": "textbook_2686", "title": "930. LINK TO LEARNING - ", "content": "FIGURE 9.7 (a) An oxyacetylene torch produces heat by the combustion of acetylene in oxygen. The energy released by this exothermic reaction heats and then melts the metal being cut. The sparks are tiny bits of the molten metal flying away. (b) A cold pack uses an endothermic process to create the sensation of cold. (credit a: modification of work by \"Skatebiker\"/Wikimedia commons)"} {"id": "textbook_2687", "title": "930. LINK TO LEARNING - ", "content": "Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to $1 \\mathrm{~kg} \\mathrm{~m}{ }^{2} / \\mathrm{s}^{2}$, which is also called 1 newton-meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules."} {"id": "textbook_2688", "title": "930. LINK TO LEARNING - ", "content": "We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat ( $q$ ) it absorbs or releases when it experiences a temperature change $(\\Delta T)$ of 1 degree Celsius (or equivalently, 1 kelvin):"} {"id": "textbook_2689", "title": "930. LINK TO LEARNING - ", "content": "$$\nC=\\frac{q}{\\Delta T}\n$$"} {"id": "textbook_2690", "title": "930. LINK TO LEARNING - ", "content": "Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property-its value is proportional to the amount of the substance."} {"id": "textbook_2691", "title": "930. LINK TO LEARNING - ", "content": "For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast\niron frying pan is found by observing that it takes $18,150 \\mathrm{~J}$ of energy to raise the temperature of the pan by $50.0^{\\circ} \\mathrm{C}$ :"} {"id": "textbook_2692", "title": "930. LINK TO LEARNING - ", "content": "$$\nC_{\\text {small pan }}=\\frac{18,140 \\mathrm{~J}}{50.0^{\\circ} \\mathrm{C}}=363 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2693", "title": "930. LINK TO LEARNING - ", "content": "The larger cast iron frying pan, while made of the same substance, requires $90,700 \\mathrm{~J}$ of energy to raise its temperature by $50.0^{\\circ} \\mathrm{C}$. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:"} {"id": "textbook_2694", "title": "930. LINK TO LEARNING - ", "content": "$$\nC_{\\text {large pan }}=\\frac{90,700 \\mathrm{~J}}{50.0^{\\circ} \\mathrm{C}}=1814 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2695", "title": "930. LINK TO LEARNING - ", "content": "The specific heat capacity (c) of a substance, commonly called its \"specific heat,\" is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):"} {"id": "textbook_2696", "title": "930. LINK TO LEARNING - ", "content": "$$\nc=\\frac{q}{\\mathrm{~m} \\Delta \\mathrm{~T}}\n$$"} {"id": "textbook_2697", "title": "930. LINK TO LEARNING - ", "content": "Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property-the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g . The specific heat of iron (the material used to make the pan) is therefore:"} {"id": "textbook_2698", "title": "930. LINK TO LEARNING - ", "content": "$$\nc_{\\text {iron }}=\\frac{18,140 \\mathrm{~J}}{(808 \\mathrm{~g})\\left(50.0^{\\circ} \\mathrm{C}\\right)}=0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2699", "title": "930. LINK TO LEARNING - ", "content": "The large frying pan has a mass of 4040 g . Using the data for this pan, we can also calculate the specific heat of iron:"} {"id": "textbook_2700", "title": "930. LINK TO LEARNING - ", "content": "$$\nc_{\\text {iron }}=\\frac{90,700 \\mathrm{~J}}{(4040 \\mathrm{~g})\\left(50.0^{\\circ} \\mathrm{C}\\right)}=0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2701", "title": "930. LINK TO LEARNING - ", "content": "Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of $\\mathrm{J} / \\mathrm{mol}{ }^{\\circ} \\mathrm{C}$ (Figure 9.8).\n"} {"id": "textbook_2702", "title": "930. LINK TO LEARNING - ", "content": "FIGURE 9.8 Because of its larger mass, a large frying pan has a larger heat capacity than a small frying pan. Because they are made of the same material, both frying pans have the same specific heat. (credit: Mark Blaser)\nWater has a relatively high specific heat (about $4.2 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ for the liquid and $2.09 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ for the solid); most metals have much lower specific heats (usually less than $1 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ ). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table 9.1."} {"id": "textbook_2703", "title": "930. LINK TO LEARNING - ", "content": "Specific Heats of Common Substances at $25^{\\circ} \\mathrm{C}$ and 1 bar"} {"id": "textbook_2704", "title": "930. LINK TO LEARNING - ", "content": "| Substance | Symbol (state) | Specific Heat ( $\\mathrm{J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ ) |\n| :---: | :---: | :---: |\n| helium | $\\mathrm{He}(\\mathrm{g})$ | 5.193 |\n| water | $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$ | 4.184 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}(\\mathrm{I})$ | 2.376 |\n| ice | $\\mathrm{H}_{2} \\mathrm{O}(s)$ | 2.093 (at $\\left.-10^{\\circ} \\mathrm{C}\\right)$ |\n| water vapor | $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | 1.864 |\n| nitrogen | $\\mathrm{N}_{2}(\\mathrm{~g})$ | 1.040 |\n| air | | 1.007 |\n| oxygen | $\\mathrm{O}_{2}(\\mathrm{~g})$ | 0.918 |\n| aluminum | $\\mathrm{Al}(\\mathrm{s})$ | 0.897 |\n| carbon dioxide | $\\mathrm{CO}_{2}(\\mathrm{~g})$ | 0.853 |\n| argon | $\\operatorname{Ar}(\\mathrm{g})$ | 0.522 |\n| iron | $\\mathrm{Fe}(s)$ | 0.449 |\n| copper | $\\mathrm{Cu}(s)$ | 0.385 |\n| lead | $\\mathrm{Pb}(s)$ | 0.130 |\n| gold | $\\mathrm{Au}(s)$ | 0.129 |\n| silicon | Si(s) | 0.712 |"} {"id": "textbook_2705", "title": "930. LINK TO LEARNING - ", "content": "TABLE 9.1"} {"id": "textbook_2706", "title": "930. LINK TO LEARNING - ", "content": "If we know the mass of a substance and its specific heat, we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:"} {"id": "textbook_2707", "title": "930. LINK TO LEARNING - ", "content": "$$\n\\begin{aligned}\n& q=(\\text { specific heat }) \\times(\\text { mass of substance }) \\times(\\text { temperature change }) \\\\\n& q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)\n\\end{aligned}\n$$\n\nIn this equation, $c$ is the specific heat of the substance, $m$ is its mass, and $\\Delta T$ (which is read \"delta $T$ \") is the temperature change, $T_{\\text {final }}-T_{\\text {initial }}$. If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, $T_{\\text {final }}-T_{\\text {initial }}$ has a positive value, and the value of $q$ is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, $T_{\\text {final }}-T_{\\text {initial }}$ has a negative value, and the value of $q$ is negative.\n"} {"id": "textbook_2708", "title": "931. EXAMPLE 9.1 - ", "content": ""} {"id": "textbook_2709", "title": "932. Measuring Heat - ", "content": "\nA flask containing $8.0 \\times 10^{2} \\mathrm{~g}$ of water is heated, and the temperature of the water increases from $21^{\\circ} \\mathrm{C}$ to 85 ${ }^{\\circ} \\mathrm{C}$. How much heat did the water absorb?\n"} {"id": "textbook_2710", "title": "933. Solution - ", "content": "\nTo answer this question, consider these factors:"} {"id": "textbook_2711", "title": "933. Solution - ", "content": "- the specific heat of the substance being heated (in this case, water)\n- the amount of substance being heated (in this case, $8.0 \\times 10^{2} \\mathrm{~g}$ )\n- the magnitude of the temperature change (in this case, from $21^{\\circ} \\mathrm{C}$ to $85^{\\circ} \\mathrm{C}$ )."} {"id": "textbook_2712", "title": "933. Solution - ", "content": "The specific heat of water is $4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, so to heat 1 g of water by $1^{\\circ} \\mathrm{C}$ requires 4.184 J . We note that since 4.184 J is required to heat 1 g of water by $1^{\\circ} \\mathrm{C}$, we will need 800 times as much to heat $8.0 \\times 10^{2} \\mathrm{~g}$ of water by 1 ${ }^{\\circ} \\mathrm{C}$. Finally, we observe that since 4.184 J are required to heat 1 g of water by $1^{\\circ} \\mathrm{C}$, we will need 64 times as much to heat it by $64^{\\circ} \\mathrm{C}$ (that is, from $21^{\\circ} \\mathrm{C}$ to $85^{\\circ} \\mathrm{C}$ )."} {"id": "textbook_2713", "title": "933. Solution - ", "content": "This can be summarized using the equation:"} {"id": "textbook_2714", "title": "933. Solution - ", "content": "$$\n\\begin{aligned}\n& q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right) \\\\\n= & \\left(4.184 \\mathrm{~J} / \\frac{\\mathrm{g}}{}{ }^{\\circ} \\mathrm{C}\\right) \\times\\left(8.0 \\times 10^{2} \\frac{\\mathrm{~g}}{\\mathrm{~g}}\\right) \\times(85-21)^{\\circ} \\mathrm{C} \\\\\n= & \\left(4.184 \\mathrm{~J} / \\frac{\\mathrm{g}}{}{ }^{\\circ} \\mathrm{C}\\right) \\times\\left(8.0 \\times 10^{2} \\frac{\\mathrm{~g}}{\\mathrm{C}}\\right) \\times(64)^{\\circ} \\mathrm{C} \\\\\n= & 210,000 \\mathrm{~J}\\left(=2.1 \\times 10^{2} \\mathrm{~kJ}\\right)\n\\end{aligned}\n$$"} {"id": "textbook_2715", "title": "933. Solution - ", "content": "Because the temperature increased, the water absorbed heat and $q$ is positive.\n"} {"id": "textbook_2716", "title": "934. Check Your Learning - ", "content": "\nHow much heat, in joules, must be added to a 502 g iron skillet to increase its temperature from $25^{\\circ} \\mathrm{C}$ to 250 ${ }^{\\circ} \\mathrm{C}$ ? The specific heat of iron is $0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2717", "title": "935. Answer: - ", "content": "\n$5.07 \\times 10^{4} \\mathrm{~J}$"} {"id": "textbook_2718", "title": "935. Answer: - ", "content": "Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced.\n"} {"id": "textbook_2719", "title": "936. EXAMPLE 9.2 - ", "content": ""} {"id": "textbook_2720", "title": "937. Determining Other Quantities - ", "content": "\nA piece of unknown metal weighs 348 g . When the metal piece absorbs 6.64 kJ of heat, its temperature increases from $22.4^{\\circ} \\mathrm{C}$ to $43.6^{\\circ} \\mathrm{C}$. Determine the specific heat of this metal (which might provide a clue to its identity).\n"} {"id": "textbook_2721", "title": "938. Solution - ", "content": "\nSince mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:"} {"id": "textbook_2722", "title": "938. Solution - ", "content": "$$\nq=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)\n$$"} {"id": "textbook_2723", "title": "938. Solution - ", "content": "Substituting the known values:"} {"id": "textbook_2724", "title": "938. Solution - ", "content": "$$\n6640 \\mathrm{~J}=c \\times(348 \\mathrm{~g}) \\times(43.6-22.4)^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2725", "title": "938. Solution - ", "content": "Solving:"} {"id": "textbook_2726", "title": "938. Solution - ", "content": "$$\nc=\\frac{6640 \\mathrm{~J}}{(348 \\mathrm{~g}) \\times\\left(21.2^{\\circ} \\mathrm{C}\\right)}=0.900 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2727", "title": "938. Solution - ", "content": "Comparing this value with the values in Table 9.1, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.\n"} {"id": "textbook_2728", "title": "939. Check Your Learning - ", "content": "\nA piece of unknown metal weighs 217 g . When the metal piece absorbs 1.43 kJ of heat, its temperature increases from $24.5^{\\circ} \\mathrm{C}$ to $39.1^{\\circ} \\mathrm{C}$. Determine the specific heat of this metal, and predict its identity.\n"} {"id": "textbook_2729", "title": "940. Answer: - ", "content": "\n$c=0.451 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$; the metal is likely to be iron\n"} {"id": "textbook_2730", "title": "941. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2731", "title": "942. Solar Thermal Energy Power Plants - ", "content": "\nThe sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat \"stored\" in that substance is then converted into electricity."} {"id": "textbook_2732", "title": "942. Solar Thermal Energy Power Plants - ", "content": "The Solana Generating Station in Arizona's Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure 9.9). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities.\n"} {"id": "textbook_2733", "title": "942. Solar Thermal Energy Power Plants - ", "content": "FIGURE 9.9 This solar thermal plant uses parabolic trough mirrors to concentrate sunlight. (credit a: modification of work by Bureau of Land Management)"} {"id": "textbook_2734", "title": "942. Solar Thermal Energy Power Plants - ", "content": "The 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure 9.10). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over $538^{\\circ} \\mathrm{C}$ that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization.\n\n(a)\n\n(b)\n\nFIGURE 9.10 (a) The Ivanpah solar thermal plant uses 170,000 mirrors to concentrate sunlight on water-filled towers. (b) It covers 4000 acres of public land near the Mojave Desert and the California-Nevada border. (credit a: modification of work by Craig Dietrich; credit b: modification of work by \"USFWS Pacific Southwest Region\"/Flickr)\n"} {"id": "textbook_2735", "title": "942. Solar Thermal Energy Power Plants - 942.1. Calorimetry", "content": ""} {"id": "textbook_2736", "title": "943. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2737", "title": "943. LEARNING OBJECTIVES - ", "content": "- Explain the technique of calorimetry\n- Calculate and interpret heat and related properties using typical calorimetry data"} {"id": "textbook_2738", "title": "943. LEARNING OBJECTIVES - ", "content": "One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study. The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system).\n\nA calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure 9.11). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.\n"} {"id": "textbook_2739", "title": "943. LEARNING OBJECTIVES - ", "content": "FIGURE 9.11 In a calorimetric determination, either (a) an exothermic process occurs and heat, $q$, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, $q$, is positive, indicating that thermal energy is transferred from the surroundings to the system."} {"id": "textbook_2740", "title": "943. LEARNING OBJECTIVES - ", "content": "Calorimetry measurements are important in understanding the heat transferred in reactions involving everything from microscopic proteins to massive machines. During her time at the National Bureau of Standards, research chemist Reatha Clark King performed calorimetric experiments to understand the precise heats of various flourine compounds. Her work was important to NASA in their quest for better rocket fuels."} {"id": "textbook_2741", "title": "943. LEARNING OBJECTIVES - ", "content": "Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment, which effectively limits the \"surroundings\" to the nonsystem components with the calorimeter (and the calorimeter itself). This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure 9.12). These easy-to-use \"coffee cup\" calorimeters allow more heat exchange with the outside environment, and therefore produce less accurate energy values.\n"} {"id": "textbook_2742", "title": "943. LEARNING OBJECTIVES - ", "content": "FIGURE 9.12 A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture."} {"id": "textbook_2743", "title": "943. LEARNING OBJECTIVES - ", "content": "Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure 9.13).\n\n\nFIGURE 9.13 Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research."} {"id": "textbook_2744", "title": "943. LEARNING OBJECTIVES - ", "content": "Before discussing the calorimetry of chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal $(M)$, and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W . The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature-that is, when they reach thermal equilibrium (Figure 9.14). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either its external environment. Under these ideal circumstances, the net heat change is zero:"} {"id": "textbook_2745", "title": "943. LEARNING OBJECTIVES - ", "content": "$$\nq_{\\text {substance } \\mathrm{M}}+q_{\\text {substance } \\mathrm{W}}=0\n$$"} {"id": "textbook_2746", "title": "943. LEARNING OBJECTIVES - ", "content": "This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:"} {"id": "textbook_2747", "title": "943. LEARNING OBJECTIVES - ", "content": "$$\nq_{\\text {substance } \\mathrm{M}}=-q_{\\text {substance } \\mathrm{W}}\n$$"} {"id": "textbook_2748", "title": "943. LEARNING OBJECTIVES - ", "content": "The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{\\text {substance } \\mathrm{M}}$ and $q_{\\text {substance W }}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either $q$ value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, $q_{\\text {substance } M}$ is a negative value and $q_{\\text {substance } \\mathrm{W}}$ is positive, since heat is transferred from M to W .\n"} {"id": "textbook_2749", "title": "943. LEARNING OBJECTIVES - ", "content": "FIGURE 9.14 In a simple calorimetry process, (a) heat, $q$, is transferred from the hot metal, $M$, to the cool water, W, until (b) both are at the same temperature.\n"} {"id": "textbook_2750", "title": "944. EXAMPLE 9.3 - ", "content": ""} {"id": "textbook_2751", "title": "945. Heat Transfer between Substances at Different Temperatures - ", "content": "\nA 360.0-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at $24.0^{\\circ} \\mathrm{C}$. The final temperature of the water was measured as $42.7^{\\circ} \\mathrm{C}$. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table 9.1), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).\n"} {"id": "textbook_2752", "title": "946. Solution - ", "content": "\nThe temperature of the water increases from $24.0^{\\circ} \\mathrm{C}$ to $42.7^{\\circ} \\mathrm{C}$, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat \"lost\" to the outside environment, then heat given off by rebar = -heat taken in by water, or:"} {"id": "textbook_2753", "title": "946. Solution - ", "content": "$$\nq_{\\text {rebar }}=-q_{\\text {water }}\n$$"} {"id": "textbook_2754", "title": "946. Solution - ", "content": "Since we know how heat is related to other measurable quantities, we have:"} {"id": "textbook_2755", "title": "946. Solution - ", "content": "$$\n(c \\times m \\times \\Delta T)_{\\mathrm{rebar}}=-(c \\times m \\times \\Delta T)_{\\mathrm{water}}\n$$"} {"id": "textbook_2756", "title": "946. Solution - ", "content": "Letting $\\mathrm{f}=$ final and $\\mathrm{i}=$ initial, in expanded form, this becomes:"} {"id": "textbook_2757", "title": "946. Solution - ", "content": "$$\nc_{\\text {rebar }} \\times m_{\\text {rebar }} \\times\\left(T_{\\mathrm{f}, \\text { rebar }}-T_{\\mathrm{i}, \\text { rebar }}\\right)=-c_{\\text {water }} \\times m_{\\text {water }} \\times\\left(T_{\\mathrm{f}, \\text { water }}-T_{\\mathrm{i}, \\text { water }}\\right)\n$$"} {"id": "textbook_2758", "title": "946. Solution - ", "content": "The density of water is $1.0 \\mathrm{~g} / \\mathrm{mL}$, so 425 mL of water $=425 \\mathrm{~g}$. Noting that the final temperature of both the rebar and water is $42.7^{\\circ} \\mathrm{C}$, substituting known values yields:"} {"id": "textbook_2759", "title": "946. Solution - ", "content": "$$\n\\begin{gathered}\n\\left(0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(360.0 \\mathrm{~g})\\left(42.7{ }^{\\circ} \\mathrm{C}-T_{\\mathrm{i}, \\text { rebar }}\\right)=-\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(425 \\mathrm{~g})\\left(42.7^{\\circ} \\mathrm{C}-24.0^{\\circ} \\mathrm{C}\\right) \\\\\nT_{i, r e b a r}=\\frac{\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(425 \\mathrm{~g})\\left(42.7^{\\circ} \\mathrm{C}-24.0^{\\circ} \\mathrm{C}\\right)}{\\left(0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(360.0 \\mathrm{~g})}+42.7^{\\circ} \\mathrm{C}\n\\end{gathered}\n$$\n\nSolving this gives $T_{i, \\text { rebar }}=248^{\\circ} \\mathrm{C}$, so the initial temperature of the rebar was $248{ }^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2760", "title": "947. Check Your Learning - ", "content": "\nA $248-\\mathrm{g}$ piece of copper is dropped into 390 mL of water at $22.6^{\\circ} \\mathrm{C}$. The final temperature of the water was\nmeasured as $39.9^{\\circ} \\mathrm{C}$. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.\n"} {"id": "textbook_2761", "title": "948. Answer: - ", "content": "\nThe initial temperature of the copper was $335.6^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2762", "title": "949. Check Your Learning - ", "content": "\nA 248 -g piece of copper initially at $314^{\\circ} \\mathrm{C}$ is dropped into 390 mL of water initially at $22.6^{\\circ} \\mathrm{C}$. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.\n"} {"id": "textbook_2763", "title": "950. Answer: - ", "content": "\nThe final temperature (reached by both copper and water) is $38.7^{\\circ} \\mathrm{C}$."} {"id": "textbook_2764", "title": "950. Answer: - ", "content": "This method can also be used to determine other quantities, such as the specific heat of an unknown metal.\n"} {"id": "textbook_2765", "title": "951. EXAMPLE 9.4 - ", "content": ""} {"id": "textbook_2766", "title": "952. Identifying a Metal by Measuring Specific Heat - ", "content": "\nA 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at $22.0^{\\circ} \\mathrm{C}$. The final temperature is $28.5^{\\circ} \\mathrm{C}$. Use these data to determine the specific heat of the metal. Use this result to identify the metal.\n"} {"id": "textbook_2767", "title": "953. Solution - ", "content": "\nAssuming perfect heat transfer, heat given off by metal $=-$ heat taken in by water, or:"} {"id": "textbook_2768", "title": "953. Solution - ", "content": "$$\nq_{\\text {metal }}=-q_{\\text {water }}\n$$"} {"id": "textbook_2769", "title": "953. Solution - ", "content": "In expanded form, this is:"} {"id": "textbook_2770", "title": "953. Solution - ", "content": "$$\nc_{\\text {metal }} \\times m_{\\text {metal }} \\times\\left(T_{\\mathrm{f}, \\text { metal }}-T_{\\mathrm{i}, \\text { metal }}\\right)=-c_{\\text {water }} \\times m_{\\text {water }} \\times\\left(T_{\\mathrm{f}, \\text { water }}-T_{\\mathrm{i}, \\text { water }}\\right)\n$$"} {"id": "textbook_2771", "title": "953. Solution - ", "content": "Noting that since the metal was submerged in boiling water, its initial temperature was $100.0^{\\circ} \\mathrm{C}$; and that for water, $60.0 \\mathrm{~mL}=60.0 \\mathrm{~g}$; we have:"} {"id": "textbook_2772", "title": "953. Solution - ", "content": "$$\n\\left(c_{\\text {metal }}\\right)(59.7 \\mathrm{~g})\\left(28.5^{\\circ} \\mathrm{C}-100.0^{\\circ} \\mathrm{C}\\right)=-\\left(4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(60.0 \\mathrm{~g})\\left(28.5^{\\circ} \\mathrm{C}-22.0^{\\circ} \\mathrm{C}\\right)\n$$"} {"id": "textbook_2773", "title": "953. Solution - ", "content": "Solving this:"} {"id": "textbook_2774", "title": "953. Solution - ", "content": "$$\nc_{\\text {metal }}=\\frac{-\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(60.0 \\mathrm{~g})\\left(6.5^{\\circ} \\mathrm{C}\\right)}{(59.7 \\mathrm{~g})\\left(-71.5^{\\circ} \\mathrm{C}\\right)}=0.38 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_2775", "title": "953. Solution - ", "content": "Comparing this with values in Table 9.1, our experimental specific heat is closest to the value for copper (0.39 $\\mathrm{J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ), so we identify the metal as copper.\n"} {"id": "textbook_2776", "title": "954. Check Your Learning - ", "content": "\nA 92.9-g piece of a silver/gray metal is heated to $178.0^{\\circ} \\mathrm{C}$, and then quickly transferred into 75.0 mL of water initially at $24.0^{\\circ} \\mathrm{C}$. After 5 minutes, both the metal and the water have reached the same temperature: $29.7^{\\circ} \\mathrm{C}$. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).\n"} {"id": "textbook_2777", "title": "955. Answer: - ", "content": "\n$c_{\\text {metal }}=0.13 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$\nThis specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead."} {"id": "textbook_2778", "title": "955. Answer: - ", "content": "When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have\nbeen discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the outside environment. Because energy is neither created nor destroyed during a chemical reaction, the heat produced or consumed in the reaction (the \"system\"), $q_{\\text {reaction }}$, plus the heat absorbed or lost by the solution (the \"surroundings\"), $q_{\\text {solution }}$, must add up to zero:"} {"id": "textbook_2779", "title": "955. Answer: - ", "content": "$$\nq_{\\text {reaction }}+q_{\\text {solution }}=0\n$$"} {"id": "textbook_2780", "title": "955. Answer: - ", "content": "This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:"} {"id": "textbook_2781", "title": "955. Answer: - ", "content": "$$\nq_{\\text {reaction }}=-q_{\\text {solution }}\n$$"} {"id": "textbook_2782", "title": "955. Answer: - ", "content": "This concept lies at the heart of all calorimetry problems and calculations.\n"} {"id": "textbook_2783", "title": "956. EXAMPLE 9.5 - ", "content": ""} {"id": "textbook_2784", "title": "957. Heat Produced by an Exothermic Reaction - ", "content": "\nWhen 50.0 mL of $1.00 \\mathrm{M} \\mathrm{HCl}(\\mathrm{aq})$ and 50.0 mL of $1.00 \\mathrm{M} \\mathrm{NaOH}(\\mathrm{aq})$, both at $22.0^{\\circ} \\mathrm{C}$, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of $28.9^{\\circ} \\mathrm{C}$. What is the approximate amount of heat produced by this reaction?"} {"id": "textbook_2785", "title": "957. Heat Produced by an Exothermic Reaction - ", "content": "$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n"} {"id": "textbook_2786", "title": "958. Solution - ", "content": "\nTo visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at $22.0^{\\circ} \\mathrm{C}$. The HCl and NaOH then react until the solution temperature reaches $28.9^{\\circ} \\mathrm{C}$."} {"id": "textbook_2787", "title": "958. Solution - ", "content": "The heat given off by the reaction is equal to that taken in by the solution. Therefore:"} {"id": "textbook_2788", "title": "958. Solution - ", "content": "$$\nq_{\\text {reaction }}=-q_{\\text {solution }}\n$$"} {"id": "textbook_2789", "title": "958. Solution - ", "content": "(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)"} {"id": "textbook_2790", "title": "958. Solution - ", "content": "Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:"} {"id": "textbook_2791", "title": "958. Solution - ", "content": "$$\nq_{\\text {solution }}=(c \\times m \\times \\Delta T)_{\\text {solution }}\n$$"} {"id": "textbook_2792", "title": "958. Solution - ", "content": "To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately $1.0 \\mathrm{~g} / \\mathrm{mL}$, so 100.0 mL has a mass of about $1.0 \\times 10^{2} \\mathrm{~g}$ (two significant figures). The specific heat of water is approximately $4.184 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$, so we use that for the specific heat of the solution. Substituting these values gives:"} {"id": "textbook_2793", "title": "958. Solution - ", "content": "$$\nq_{\\text {solution }}=\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)\\left(1.0 \\times 10^{2} \\mathrm{~g}\\right)\\left(28.9^{\\circ} \\mathrm{C}-22.0^{\\circ} \\mathrm{C}\\right)=2.9 \\times 10^{3} \\mathrm{~J}\n$$"} {"id": "textbook_2794", "title": "958. Solution - ", "content": "Finally, since we are trying to find the heat of the reaction, we have:"} {"id": "textbook_2795", "title": "958. Solution - ", "content": "$$\nq_{\\text {reaction }}=-q_{\\text {solution }}=-2.9 \\times 10^{3} \\mathrm{~J}\n$$"} {"id": "textbook_2796", "title": "958. Solution - ", "content": "The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.\n"} {"id": "textbook_2797", "title": "959. Check Your Learning - ", "content": "\nWhen 100 mL of $0.200 \\mathrm{M} \\mathrm{NaCl}(\\mathrm{aq})$ and 100 mL of $0.200 \\mathrm{M} \\mathrm{AgNO}_{3}(\\mathrm{aq})$, both at $21.9^{\\circ} \\mathrm{C}$, are mixed in a coffee cup calorimeter, the temperature increases to $23.5^{\\circ} \\mathrm{C}$ as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?\n"} {"id": "textbook_2798", "title": "960. Answer: - ", "content": "\n$1.34 \\times 1.3 \\mathrm{~kJ}$; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water\n"} {"id": "textbook_2799", "title": "961. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2800", "title": "962. Thermochemistry of Hand Warmers - ", "content": "\nWhen working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure 9.15). A common reusable hand warmer contains a supersaturated solution of $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail)."} {"id": "textbook_2801", "title": "962. Thermochemistry of Hand Warmers - ", "content": "The process $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(a q) \\longrightarrow \\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(s)$ is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ redissolves and can be reused.\n"} {"id": "textbook_2802", "title": "962. Thermochemistry of Hand Warmers - ", "content": "FIGURE 9.15 Chemical hand warmers produce heat that warms your hand on a cold day. In this one, you can see the metal disc that initiates the exothermic precipitation reaction. (credit: modification of work by Science Buddies TV/YouTube)"} {"id": "textbook_2803", "title": "962. Thermochemistry of Hand Warmers - ", "content": "Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is\n$2 \\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$. Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.\n"} {"id": "textbook_2804", "title": "963. LINK TO LEARNING - ", "content": "\nThis link (http://openstax.org/l/16Handwarmer) shows the precipitation reaction that occurs when the disk in a chemical hand warmer is flexed.\n"} {"id": "textbook_2805", "title": "964. EXAMPLE 9.6 - ", "content": ""} {"id": "textbook_2806", "title": "965. Heat Flow in an Instant Ice Pack - ", "content": "\nWhen solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an \"instant ice pack\" (Figure 9.16). When 3.21 g of solid $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ dissolves in 50.0 g of water at $24.9^{\\circ} \\mathrm{C}$ in a calorimeter, the temperature decreases to $20.3^{\\circ} \\mathrm{C}$."} {"id": "textbook_2807", "title": "965. Heat Flow in an Instant Ice Pack - ", "content": "Calculate the value of $q$ for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.\n"} {"id": "textbook_2808", "title": "965. Heat Flow in an Instant Ice Pack - ", "content": "FIGURE 9.16 An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water. The cold pack then removes thermal energy from your body.\n"} {"id": "textbook_2809", "title": "966. Solution - ", "content": "\nWe assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:"} {"id": "textbook_2810", "title": "966. Solution - ", "content": "$$\nq_{\\mathrm{rxn}}=-q_{\\mathrm{soln}}\n$$"} {"id": "textbook_2811", "title": "966. Solution - ", "content": "with \"rxn\" and \"soln\" used as shorthand for \"reaction\" and \"solution,\" respectively.\nAssuming also that the specific heat of the solution is the same as that for water, we have:"} {"id": "textbook_2812", "title": "966. Solution - ", "content": "$$\n\\begin{aligned}\n& q_{\\mathrm{rxn}}=-q_{\\mathrm{soln}}=-(c \\times m \\times \\Delta T)_{\\text {soln }} \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(53.2 \\mathrm{~g}) \\times\\left(20.3^{\\circ} \\mathrm{C}-24.9^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(53.2 \\mathrm{~g}) \\times\\left(-4.6^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& +1.0 \\times 10^{3} \\mathrm{~J}=+1.0 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_2813", "title": "966. Solution - ", "content": "The positive sign for $q$ indicates that the dissolution is an endothermic process.\n"} {"id": "textbook_2814", "title": "967. Check Your Learning - ", "content": "\nWhen a $3.00-\\mathrm{g}$ sample of KCl was added to $3.00 \\times 10^{2} \\mathrm{~g}$ of water in a coffee cup calorimeter, the temperature decreased by $1.05^{\\circ} \\mathrm{C}$. How much heat is involved in the dissolution of the KCl ? What assumptions did you make?\n"} {"id": "textbook_2815", "title": "968. Answer: - ", "content": "\n1.33 kJ ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water"} {"id": "textbook_2816", "title": "968. Answer: - ", "content": "If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter."} {"id": "textbook_2817", "title": "968. Answer: - ", "content": "The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term \"bomb\" comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the \"bomb\") that contains the reactants and is itself submerged in water (Figure 9.17). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is absorbed by the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known $q$, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.\n\n\nFIGURE 9.17 (a) A bomb calorimeter is used to measure heat produced by reactions involving gaseous reactants or products, such as combustion. (b) The reactants are contained in the gas-tight \"bomb,\" which is submerged in water and surrounded by insulating materials. (credit a: modification of work by \"Harbor1\"/Wikimedia commons)\n"} {"id": "textbook_2818", "title": "969. LINK TO LEARNING - ", "content": "\nClick on this link (http://openstax.org/l/16BombCal) to view how a bomb calorimeter is prepared for action. This site (http://openstax.org/l/16Calorcalcs) shows calorimetric calculations using sample data.\n"} {"id": "textbook_2819", "title": "970. EXAMPLE 9.7 - ", "content": ""} {"id": "textbook_2820", "title": "971. Bomb Calorimetry - ", "content": "\nWhen 3.12 g of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases from $23.8^{\\circ} \\mathrm{C}$ to $35.6^{\\circ} \\mathrm{C}$. The calorimeter contains 775 g of water, and the bomb itself has a heat\ncapacity of $893 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$. How much heat was produced by the combustion of the glucose sample?\n"} {"id": "textbook_2821", "title": "972. Solution - ", "content": "\nThe combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)"} {"id": "textbook_2822", "title": "972. Solution - ", "content": "The heat produced by the reaction is absorbed by the water and the bomb:"} {"id": "textbook_2823", "title": "972. Solution - ", "content": "$$\n\\begin{aligned}\n& q_{\\mathrm{rxn}}=-\\left(q_{\\text {water }}+q_{\\text {bomb }}\\right) \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(775 \\mathrm{~g}) \\times\\left(35.6^{\\circ} \\mathrm{C}-23.8{ }^{\\circ} \\mathrm{C}\\right)+893 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C} \\times\\left(35.6^{\\circ} \\mathrm{C}-23.8{ }^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& =-(38,300 \\mathrm{~J}+10,500 \\mathrm{~J}) \\\\\n& =-48,800 \\mathrm{~J}=-48.8 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_2824", "title": "972. Solution - ", "content": "This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.\n"} {"id": "textbook_2825", "title": "973. Check Your Learning - ", "content": "\nWhen 0.963 g of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases by $8.39^{\\circ} \\mathrm{C}$. The bomb has a heat capacity of $784 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$ and is submerged in 925 mL of water. How much heat was produced by the combustion of the benzene sample?\n"} {"id": "textbook_2826", "title": "974. Answer: - ", "content": "\n$\\mathrm{q}_{\\mathrm{rx}}=-39.0 \\mathrm{~kJ}$ (the reaction produced 39.0 kJ of heat)"} {"id": "textbook_2827", "title": "974. Answer: - ", "content": "Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. ${ }^{2}$ These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes."} {"id": "textbook_2828", "title": "974. Answer: - ", "content": "For example Carla Prado's team at University of Alberta undertook whole-body calorimetry to understand the energy expenditures of women who had recently given birth. Studies like this help develop better recommendations and regimens for nutrition, exercise, and general wellbeing during this period of significant physiological change. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories ( 1 kcal ), the amount of energy needed to heat 1 kg of water by $1^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_2829", "title": "975. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2830", "title": "976. Measuring Nutritional Calories - ", "content": "\nIn your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) $=1000 \\mathrm{cal}$, or 1 kcal . (This is approximately the amount of energy needed to heat 1 kg of water by $1^{\\circ} \\mathrm{C}$.)"} {"id": "textbook_2831", "title": "976. Measuring Nutritional Calories - ", "content": "The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/ g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure 9.18)."} {"id": "textbook_2832", "title": "976. Measuring Nutritional Calories - ", "content": "[^7]"} {"id": "textbook_2833", "title": "976. Measuring Nutritional Calories - ", "content": "FIGURE 9.18 (a) Macaroni and cheese contain energy in the form of the macronutrients in the food. (b) The food's nutritional information is shown on the package label. In the US, the energy content is given in Calories (per serving); the rest of the world usually uses kilojoules. (credit a: modification of work by \"Rex Roof\"/Flickr)"} {"id": "textbook_2834", "title": "976. Measuring Nutritional Calories - ", "content": "For the example shown in (b), the total energy per 228-g portion is calculated by:\n$(5 \\mathrm{~g}$ protein $\\times 4$ Calories $/ \\mathrm{g})+(31 \\mathrm{~g}$ carb $\\times 4$ Calories $/ \\mathrm{g})+(12 \\mathrm{~g}$ fat $\\times 9$ Calories $/ \\mathrm{g})=252$ Calories\nSo, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freezedried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.\n\nToday, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.\n"} {"id": "textbook_2835", "title": "977. LINK TO LEARNING - ", "content": "\nClick on this link (http://openstax.org/l/16USDA) to access the US Department of Agriculture (USDA) National Nutrient Database, containing nutritional information on over 8000 foods.\n"} {"id": "textbook_2836", "title": "977. LINK TO LEARNING - 977.1. Enthalpy", "content": ""} {"id": "textbook_2837", "title": "978. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_2838", "title": "978. LEARNING OBJECTIVES - ", "content": "- State the first law of thermodynamics\n- Define enthalpy and explain its classification as a state function\n- Write and balance thermochemical equations\n- Calculate enthalpy changes for various chemical reactions\n- Explain Hess's law and use it to compute reaction enthalpies"} {"id": "textbook_2839", "title": "978. LEARNING OBJECTIVES - ", "content": "Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics."} {"id": "textbook_2840", "title": "978. LEARNING OBJECTIVES - ", "content": "Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy ( $\\boldsymbol{U}$ ), sometimes symbolized as $E$."} {"id": "textbook_2841", "title": "978. LEARNING OBJECTIVES - ", "content": "As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat $(q)$ from the surroundings or when the surroundings do work ( $w$ ) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire's temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings."} {"id": "textbook_2842", "title": "978. LEARNING OBJECTIVES - ", "content": "The relationship between internal energy, heat, and work can be represented by the equation:"} {"id": "textbook_2843", "title": "978. LEARNING OBJECTIVES - ", "content": "$$\n\\Delta U=q+w\n$$"} {"id": "textbook_2844", "title": "978. LEARNING OBJECTIVES - ", "content": "as shown in Figure 9.19. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive $q$ is heat flow in; negative $q$ is heat flow out) or work done on or by the system. The work, $w$, is positive if it is done on the system and negative if it is done by the system.\n\n\nFIGURE 9.19 The internal energy, $U$, of a system can be changed by heat flow and work. If heat flows into the system, $q_{i n}$, or work is done on the system, $w_{\\text {on }}$, its internal energy increases, $\\Delta U>0$. If heat flows out of the system, $q_{\\text {out }}$, or work is done by the system, $w_{\\text {by }}$, its internal energy decreases, $\\Delta U<0$.\nA type of work called expansion work (or pressure-volume work) occurs when a system pushes back the\nsurroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.\n"} {"id": "textbook_2845", "title": "979. LINK TO LEARNING - ", "content": "\nThis view of an internal combustion engine (http://openstax.org/l/16combustion) illustrates the conversion of energy produced by the exothermic combustion reaction of a fuel such as gasoline into energy of motion."} {"id": "textbook_2846", "title": "979. LINK TO LEARNING - ", "content": "As discussed, the relationship between internal energy, heat, and work can be represented as $\\Delta U=q+w$. Internal energy is an example of a state function (or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value does depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m , and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (Figure 9.20). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).\n"} {"id": "textbook_2847", "title": "979. LINK TO LEARNING - ", "content": "FIGURE 9.20 Paths $X$ and $Y$ represent two different routes to the summit of Mt. Kilimanjaro. Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)"} {"id": "textbook_2848", "title": "979. LINK TO LEARNING - ", "content": "Chemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system's internal energy $(U)$ and the mathematical product of its pressure ( $P$ ) and volume ( $V$ ):"} {"id": "textbook_2849", "title": "979. LINK TO LEARNING - ", "content": "$$\nH=U+P V\n$$"} {"id": "textbook_2850", "title": "979. LINK TO LEARNING - ", "content": "Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change ( $\\boldsymbol{\\Delta H} \\boldsymbol{H})$ is:"} {"id": "textbook_2851", "title": "979. LINK TO LEARNING - ", "content": "$$\n\\Delta H=\\Delta U+P \\Delta V\n$$"} {"id": "textbook_2852", "title": "979. LINK TO LEARNING - ", "content": "The mathematical product $P \\Delta V$ represents work ( $w$ ), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of $\\Delta V$ and $w$ will always be opposite:"} {"id": "textbook_2853", "title": "979. LINK TO LEARNING - ", "content": "$$\nP \\Delta V=-w\n$$"} {"id": "textbook_2854", "title": "979. LINK TO LEARNING - ", "content": "Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:"} {"id": "textbook_2855", "title": "979. LINK TO LEARNING - ", "content": "$$\n\\begin{aligned}\n& \\Delta H=\\Delta U+P \\Delta V \\\\\n& =q_{\\mathrm{p}}+w-w \\\\\n& =q_{\\mathrm{p}}\n\\end{aligned}\n$$"} {"id": "textbook_2856", "title": "979. LINK TO LEARNING - ", "content": "where $q_{p}$ is the heat of reaction under conditions of constant pressure.\nAnd so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow $\\left(q_{p}\\right)$ and enthalpy change $(\\Delta H)$ for the process are equal."} {"id": "textbook_2857", "title": "979. LINK TO LEARNING - ", "content": "The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter (Figure 9.17) is not equal to $\\Delta H$ because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with $q=$ $\\Delta H$, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions."} {"id": "textbook_2858", "title": "979. LINK TO LEARNING - ", "content": "The following conventions apply when using $\\Delta H$ :"} {"id": "textbook_2859", "title": "979. LINK TO LEARNING - ", "content": "- A negative value of an enthalpy change, $\\Delta H<0$, indicates an exothermic reaction; a positive value, $\\Delta H>0$, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its $\\Delta H$ is changed (a process that is endothermic in one direction is exothermic in the opposite direction).\n- Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a $\\Delta H$ value following the equation for the reaction. This $\\Delta H$ value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. For example, consider this equation:"} {"id": "textbook_2860", "title": "979. LINK TO LEARNING - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-286 \\mathrm{~kJ}\n$$"} {"id": "textbook_2861", "title": "979. LINK TO LEARNING - ", "content": "This equation indicates that when 1 mole of hydrogen gas and $\\frac{1}{2}$ mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor ( $\\Delta H$ is an extensive property):\n(two-fold increase in amounts)"} {"id": "textbook_2862", "title": "979. LINK TO LEARNING - ", "content": "$$\n\\begin{array}{lr}\n2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l) & \\Delta H=2 \\times(-286 \\mathrm{~kJ})=-572 \\mathrm{~kJ} \\\\\n\\text { (two-fold decrease in amounts) } & \\\\\n\\frac{1}{2} \\mathrm{H}_{2}(g)+\\frac{1}{4} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(l) & \\Delta H=\\frac{1}{2} \\times(-286 \\mathrm{~kJ})=-143 \\mathrm{~kJ}\n\\end{array}\n$$"} {"id": "textbook_2863", "title": "979. LINK TO LEARNING - ", "content": "- The enthalpy change of a reaction depends on the physical states of the reactants and products, so these must be shown. For example, when 1 mole of hydrogen gas and $\\frac{1}{2}$ mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released."} {"id": "textbook_2864", "title": "979. LINK TO LEARNING - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=-242 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2865", "title": "980. EXAMPLE 9.8 - ", "content": ""} {"id": "textbook_2866", "title": "981. Writing Thermochemical Equations - ", "content": "\nWhen 0.0500 mol of $\\mathrm{HCl}(\\mathrm{aq})$ reacts with 0.0500 mol of $\\mathrm{NaOH}(\\mathrm{aq})$ to form 0.0500 mol of $\\mathrm{NaCl}(\\mathrm{aq}), 2.9 \\mathrm{~kJ}$ of heat are produced. Write a balanced thermochemical equation for the reaction of one mole of HCl ."} {"id": "textbook_2867", "title": "981. Writing Thermochemical Equations - ", "content": "$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n"} {"id": "textbook_2868", "title": "982. Solution - ", "content": "\nFor the reaction of 0.0500 mol acid $(\\mathrm{HCl}), q=-2.9 \\mathrm{~kJ}$. The reactants are provided in stoichiometric amounts (same molar ratio as in the balanced equation), and so the amount of acid may be used to calculate a molar enthalpy change. Since $\\Delta H$ is an extensive property, it is proportional to the amount of acid neutralized:"} {"id": "textbook_2869", "title": "982. Solution - ", "content": "$$\n\\Delta H=1 \\mathrm{molHCt} \\times \\frac{-2.9 \\mathrm{~kJ}}{0.0500 \\mathrm{molHCt}}=-58 \\mathrm{~kJ}\n$$"} {"id": "textbook_2870", "title": "982. Solution - ", "content": "The thermochemical equation is then"} {"id": "textbook_2871", "title": "982. Solution - ", "content": "$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-58 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2872", "title": "983. Check Your Learning - ", "content": "\nWhen $1.34 \\mathrm{~g} \\mathrm{Zn}(s)$ reacts with 60.0 mL of $0.750 \\mathrm{M} \\mathrm{HCl}(\\mathrm{aq}), 3.14 \\mathrm{~kJ}$ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:"} {"id": "textbook_2873", "title": "983. Check Your Learning - ", "content": "$$\n\\mathrm{Zn}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{ZnCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$\n"} {"id": "textbook_2874", "title": "984. Answer: - ", "content": "\n$\\Delta H=-153 \\mathrm{~kJ}$"} {"id": "textbook_2875", "title": "984. Answer: - ", "content": "Be sure to take both stoichiometry and limiting reactants into account when determining the $\\Delta H$ for a chemical reaction.\n"} {"id": "textbook_2876", "title": "985. EXAMPLE 9.9 - ", "content": ""} {"id": "textbook_2877", "title": "986. Writing Thermochemical Equations - ", "content": "\nA gummy bear contains 2.67 g sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. When it reacts with 7.19 g potassium chlorate, $\\mathrm{KClO}_{3}, 43.7$ kJ of heat are produced. Write a thermochemical equation for the reaction of one mole of sucrose:"} {"id": "textbook_2878", "title": "986. Writing Thermochemical Equations - ", "content": "$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(a q)+8 \\mathrm{KClO}_{3}(a q) \\longrightarrow 12 \\mathrm{CO}_{2}(g)+11 \\mathrm{H}_{2} \\mathrm{O}(l)+8 \\mathrm{KCl}(a q) .\n$$\n"} {"id": "textbook_2879", "title": "987. Solution - ", "content": "\nUnlike the previous example exercise, this one does not involve the reaction of stoichiometric amounts of reactants, and so the limiting reactant must be identified (it limits the yield of the reaction and the amount of thermal energy produced or consumed)."} {"id": "textbook_2880", "title": "987. Solution - ", "content": "The provided amounts of the two reactants are"} {"id": "textbook_2881", "title": "987. Solution - ", "content": "$$\n\\begin{aligned}\n& (2.67 \\mathrm{~g})(1 \\mathrm{~mol} / 342.3 \\mathrm{~g})=0.00780 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11} \\\\\n& (7.19 \\mathrm{~g})(1 \\mathrm{~mol} / 122.5 \\mathrm{~g})=0.0587 \\mathrm{~mol} \\mathrm{KCIO}\n\\end{aligned}\n$$"} {"id": "textbook_2882", "title": "987. Solution - ", "content": "The provided molar ratio of perchlorate-to-sucrose is then"} {"id": "textbook_2883", "title": "987. Solution - ", "content": "$$\n0.0587 \\mathrm{~mol} \\mathrm{KCIO}_{3} / 0.00780 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}=7.52\n$$"} {"id": "textbook_2884", "title": "987. Solution - ", "content": "The balanced equation indicates $8 \\mathrm{~mol}_{\\mathrm{KClO}}^{3}$ are required for reaction with $1 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. Since the provided amount of $\\mathrm{KClO}_{3}$ is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:"} {"id": "textbook_2885", "title": "987. Solution - ", "content": "$$\n\\triangle \\mathrm{H}=-43.7 \\mathrm{~kJ} / 0.0587 \\mathrm{~mol} \\mathrm{KCIO}_{3}=744 \\mathrm{~kJ} / \\mathrm{mol} \\mathrm{KCIO}_{3}\n$$"} {"id": "textbook_2886", "title": "987. Solution - ", "content": "Because the equation, as written, represents the reaction of $8 \\mathrm{~mol}_{\\mathrm{KClO}}^{3}$, the enthalpy change is"} {"id": "textbook_2887", "title": "987. Solution - ", "content": "$$\n(744 \\mathrm{~kJ} / \\mathrm{mol} \\mathrm{KCIO} 3)\\left(8 \\mathrm{~mol} \\mathrm{KCIO}_{3}\\right)=5960 \\mathrm{~kJ}\n$$"} {"id": "textbook_2888", "title": "987. Solution - ", "content": "The enthalpy change for this reaction is -5960 kJ , and the thermochemical equation is:"} {"id": "textbook_2889", "title": "987. Solution - ", "content": "$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}+8 \\mathrm{KClO}_{3} \\longrightarrow 12 \\mathrm{CO}_{2}+11 \\mathrm{H}_{2} \\mathrm{O}+8 \\mathrm{KCl} \\quad \\Delta H=-5960 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2890", "title": "988. Check Your Learning - ", "content": "\nWhen 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of $\\mathrm{FeCl}_{2}(s)$ and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of $\\mathrm{FeCl}_{2}(s)$ is produced?\n"} {"id": "textbook_2891", "title": "989. Answer: - ", "content": "\n$\\Delta H=-338 \\mathrm{~kJ}$"} {"id": "textbook_2892", "title": "989. Answer: - ", "content": "Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A standard state is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M , and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm . Because the $\\Delta H$ of a reaction changes very little with such small changes in pressure ( $1 \\mathrm{bar}=0.987 \\mathrm{~atm}$ ), $\\Delta H$ values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted \" o \" in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K , this temperature will be assumed unless some other temperature is specified. Thus, the symbol $\\left(\\Delta H^{\\circ}\\right)$ is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol $\\Delta H$ is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)"} {"id": "textbook_2893", "title": "989. Answer: - ", "content": "The enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the extensive nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the $\\Delta H$ for specific amounts of reactants). However, we often find it more useful to divide one extensive property $(\\Delta H)$ by another (amount of substance), and report a per-amount intensive value of $\\Delta H$, often \"normalized\" to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)\n"} {"id": "textbook_2894", "title": "990. Standard Enthalpy of Combustion - ", "content": "\nStandard enthalpy of combustion ( $\\Delta \\boldsymbol{H}_{C}^{\\circ}$ ) is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called \"heat of combustion.\" For example, the enthalpy of combustion of ethanol, $-1366.8 \\mathrm{~kJ} / \\mathrm{mol}$, is the amount of heat produced when one mole of ethanol undergoes complete combustion at $25^{\\circ} \\mathrm{C}$ and 1 atmosphere pressure, yielding products also at $25^{\\circ} \\mathrm{C}$ and 1 atm ."} {"id": "textbook_2895", "title": "990. Standard Enthalpy of Combustion - ", "content": "$$\n\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-1366.8 \\mathrm{~kJ}\n$$"} {"id": "textbook_2896", "title": "990. Standard Enthalpy of Combustion - ", "content": "Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 9.2. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline."} {"id": "textbook_2897", "title": "990. Standard Enthalpy of Combustion - ", "content": "| Substance | Combustion Reaction | Enthalpy of Combustion, $\\Delta H_{c}^{\\circ}$
$\\left(\\frac{\\mathrm{kJ}}{\\mathrm{mol}}\\right.$ at $\\left.25^{\\circ} \\mathrm{C}\\right)$ |\n| :--- | :--- | :--- |\n| carbon | $\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$ | -393.5 |\n| hydrogen | $\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)$ | -285.8 |\n| magnesium | $\\mathrm{Mg}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{MgO}(s)$ | -601.6 |\n| sulfur | $\\mathrm{S}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{SO}_{2}(g)$ | -296.8 |\n| carbon
monoxide | $\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$ | -283.0 |\n| methane | $\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -890.8 |\n| acetylene | $\\mathrm{C}_{2} \\mathrm{H}_{2}(g)+\\frac{5}{2} \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$ | -1301.1 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -1366.8 |\n| methanol | $\\mathrm{CH}_{3} \\mathrm{OH}(l)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -726.1 |\n| isooctane | $\\mathrm{C}_{8} \\mathrm{H}_{18}(l)+\\frac{25}{2} \\mathrm{O}_{2}(g) \\longrightarrow 8 \\mathrm{CO}_{2}(g)+9 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -5461 |"} {"id": "textbook_2898", "title": "990. Standard Enthalpy of Combustion - ", "content": "TABLE 9.2\n"} {"id": "textbook_2899", "title": "991. EXAMPLE 9.10 - ", "content": ""} {"id": "textbook_2900", "title": "992. Using Enthalpy of Combustion - ", "content": "\nAs Figure 9.21 suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 $\\mathrm{g} / \\mathrm{mL}$.\n"} {"id": "textbook_2901", "title": "992. Using Enthalpy of Combustion - ", "content": "FIGURE 9.21 The combustion of gasoline is very exothermic. (credit: modification of work by \"AlexEagle\"/Flickr)\n"} {"id": "textbook_2902", "title": "993. Solution - ", "content": "\nStarting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the\nnecessary conversions. Table 9.2 gives this value as -5460 kJ per 1 mole of isooctane $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18}\\right)$.\nUsing these data,\n\n\nThe combustion of 1.00 L of isooctane produces $33,100 \\mathrm{~kJ}$ of heat. (This amount of energy is enough to melt 99.2 kg , or about 218 lbs , of ice.)"} {"id": "textbook_2903", "title": "993. Solution - ", "content": "Note: If you do this calculation one step at a time, you would find:"} {"id": "textbook_2904", "title": "993. Solution - ", "content": "$$\n\\begin{aligned}\n& 1.00 \\mathrm{LC}_{8} \\mathrm{H}_{18} \\longrightarrow 1.00 \\times 10^{3} \\mathrm{~mL} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 1.00 \\times 10^{3} \\mathrm{mLC}_{8} \\mathrm{H}_{18} \\longrightarrow 692 \\mathrm{~g} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 692 \\mathrm{~g} \\mathrm{C}_{8} \\mathrm{H}_{18} \\longrightarrow 6.07 \\mathrm{~mol} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 6.07 \\mathrm{~mol} \\mathrm{C}_{8} \\mathrm{H}_{18} \\longrightarrow-3.31 \\times 10^{4} \\mathrm{~kJ}\n\\end{aligned}\n$$\n"} {"id": "textbook_2905", "title": "994. Check Your Learning - ", "content": "\nHow much heat is produced by the combustion of 125 g of acetylene?\n"} {"id": "textbook_2906", "title": "995. Answer: - ", "content": "\n$6.25 \\times 10^{3} \\mathrm{~kJ}$\n"} {"id": "textbook_2907", "title": "996. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_2908", "title": "997. Emerging Algae-Based Energy Technologies (Biofuels) - ", "content": "\nAs reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (Figure 9.22). The species of algae used are nontoxic, biodegradable, and among the world's fastest growing organisms. About $50 \\%$ of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of biofuel per hectare-much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.\n"} {"id": "textbook_2909", "title": "997. Emerging Algae-Based Energy Technologies (Biofuels) - ", "content": "FIGURE 9.22 (a) Tiny algal organisms can be (b) grown in large quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)"} {"id": "textbook_2910", "title": "997. Emerging Algae-Based Energy Technologies (Biofuels) - ", "content": "According to the US Department of Energy, only 39,000 square kilometers (about 0.4\\% of the land mass of the US or less than $\\frac{1}{7}$ of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive-for instance, the US Air Force is producing jet fuel from algae at a total cost of under $\\$ 5$ per gallon. ${ }^{\\frac{3}{3}}$ The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and $\\mathrm{CO}_{2}$ as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (Figure 9.23).\n"} {"id": "textbook_2911", "title": "997. Emerging Algae-Based Energy Technologies (Biofuels) - ", "content": "FIGURE 9.23 Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.\n"} {"id": "textbook_2912", "title": "998. LINK TO LEARNING - ", "content": "\nClick here (http://openstax.org/l/16biofuel) to learn more about the process of creating algae biofuel.\n"} {"id": "textbook_2913", "title": "999. Standard Enthalpy of Formation - ", "content": "\nA standard enthalpy of formation $\\Delta H_{\\mathrm{f}}^{\\circ}$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess's law."} {"id": "textbook_2914", "title": "999. Standard Enthalpy of Formation - ", "content": "The standard enthalpy of formation of $\\mathrm{CO}_{2}(\\mathrm{~g})$ is $-393.5 \\mathrm{~kJ} / \\mathrm{mol}$. This is the enthalpy change for the exothermic reaction:"} {"id": "textbook_2915", "title": "999. Standard Enthalpy of Formation - ", "content": "$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=\\Delta H^{\\circ}=-393.5 \\mathrm{~kJ}\n$$"} {"id": "textbook_2916", "title": "999. Standard Enthalpy of Formation - ", "content": "starting with the reactants at a pressure of 1 atm and $25^{\\circ} \\mathrm{C}$ (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of $\\mathrm{CO}_{2}$, also at 1 atm and $25^{\\circ} \\mathrm{C}$. For nitrogen dioxide, $\\mathrm{NO}_{2}(\\mathrm{~g}), \\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ is $33.2 \\mathrm{~kJ} / \\mathrm{mol}$. This is the enthalpy change for the reaction:"} {"id": "textbook_2917", "title": "999. Standard Enthalpy of Formation - ", "content": "$$\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=\\Delta H^{\\circ}=+33.2 \\mathrm{~kJ}\n$$"} {"id": "textbook_2918", "title": "999. Standard Enthalpy of Formation - ", "content": "A reaction equation with $\\frac{1}{2}$ mole of $\\mathrm{N}_{2}$ and 1 mole of $\\mathrm{O}_{2}$ is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, $\\mathrm{NO}_{2}(g)$."} {"id": "textbook_2919", "title": "999. Standard Enthalpy of Formation - ", "content": "You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as $-2984 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of $\\mathrm{P}_{4} \\mathrm{O}_{10}$ ) to strongly endothermic (such as $+226.7 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$ ). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under"} {"id": "textbook_2920", "title": "999. Standard Enthalpy of Formation - ", "content": "3 For more on algal fuel, see http://www.theguardian.com/environment/2010/feb/13/algae-solve-pentagon-fuel-problem.\nstandard conditions, which is 1 atm for gases and 1 M for solutions.\n"} {"id": "textbook_2921", "title": "1000. EXAMPLE 9.11 - ", "content": ""} {"id": "textbook_2922", "title": "1001. Evaluating an Enthalpy of Formation - ", "content": "\nOzone, $\\mathrm{O}_{3}(\\mathrm{~g})$, forms from oxygen, $\\mathrm{O}_{2}(\\mathrm{~g})$, by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ of ozone from the following information:"} {"id": "textbook_2923", "title": "1001. Evaluating an Enthalpy of Formation - ", "content": "$$\n3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{O}_{3}(g) \\quad \\Delta H^{\\circ}=+286 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2924", "title": "1002. Solution - ", "content": "\n$\\Delta H_{\\mathrm{f}}^{\\circ}$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{O}_{3}(g)$ is the enthalpy change for the reaction:"} {"id": "textbook_2925", "title": "1002. Solution - ", "content": "$$\n\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{O}_{3}(g)\n$$"} {"id": "textbook_2926", "title": "1002. Solution - ", "content": "For the formation of 2 mol of $\\mathrm{O}_{3}(\\mathrm{~g}), \\Delta H^{\\circ}=+286 \\mathrm{~kJ}$. This ratio, $\\left(\\frac{286 \\mathrm{~kJ}}{2 \\mathrm{~mol} \\mathrm{O}_{3}}\\right)$, can be used as a conversion factor to find the heat produced when 1 mole of $\\mathrm{O}_{3}(\\mathrm{~g})$ is formed, which is the enthalpy of formation for $\\mathrm{O}_{3}(\\mathrm{~g})$ :\n\n$$\n\\Delta H^{\\circ} \\text { for } 1 \\text { mole of } \\mathrm{O}_{3}(g)=1 \\mathrm{mel}_{3} \\times \\frac{286 \\mathrm{~kJ}}{2 \\mathrm{~mol} \\Theta_{3}}=143 \\mathrm{~kJ}\n$$\n\nTherefore, $\\Delta H_{\\mathrm{f}}^{\\circ}\\left[\\mathrm{O}_{3}(g)\\right]=+143 \\mathrm{~kJ} / \\mathrm{mol}$.\n"} {"id": "textbook_2927", "title": "1003. Check Your Learning - ", "content": "\nHydrogen gas, $\\mathrm{H}_{2}$, reacts explosively with gaseous chlorine, $\\mathrm{Cl}_{2}$, to form hydrogen chloride, $\\mathrm{HCl}(g)$. What is the enthalpy change for the reaction of 1 mole of $\\mathrm{H}_{2}(\\mathrm{~g})$ with 1 mole of $\\mathrm{Cl}_{2}(\\mathrm{~g})$ if both the reactants and products are at standard state conditions? The standard enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g})$ is $-92.3 \\mathrm{~kJ} / \\mathrm{mol}$.\n"} {"id": "textbook_2928", "title": "1004. Answer: - ", "content": "\nFor the reaction $\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H^{\\circ}=-184.6 \\mathrm{~kJ}$\n"} {"id": "textbook_2929", "title": "1005. EXAMPLE 9.12 - ", "content": ""} {"id": "textbook_2930", "title": "1006. Writing Reaction Equations for $\\Delta H_{\\mathrm{f}}^{\\circ}$ - ", "content": "\nWrite the heat of formation reaction equations for:\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I})$\n(b) $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$\n"} {"id": "textbook_2931", "title": "1007. Solution - ", "content": "\nRemembering that $\\Delta H_{\\mathrm{f}}^{\\circ}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:\n(a) $2 \\mathrm{C}(s$, graphite $)+3 \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)$\n(b) $3 \\mathrm{Ca}(s)+\\frac{1}{2} \\mathrm{P}_{4}(s)+4 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$"} {"id": "textbook_2932", "title": "1007. Solution - ", "content": "Note: The standard state of carbon is graphite, and phosphorus exists as $\\mathrm{P}_{4}$.\n"} {"id": "textbook_2933", "title": "1008. Check Your Learning - ", "content": "\nWrite the heat of formation reaction equations for:\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}($ I $)$\n(b) $\\mathrm{Na}_{2} \\mathrm{CO}_{3}(s)$\n"} {"id": "textbook_2934", "title": "1009. Answer: - ", "content": "\n(a) $4 \\mathrm{C}\\left(s\\right.$, graphite) $+5 \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(l)$; (b)\n$2 \\mathrm{Na}(s)+\\mathrm{C}(s$, graphite $)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(s)$\n"} {"id": "textbook_2935", "title": "1010. Hess's Law - ", "content": "\nThere are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment."} {"id": "textbook_2936", "title": "1010. Hess's Law - ", "content": "This type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess's law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:"} {"id": "textbook_2937", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H^{\\circ}=-394 \\mathrm{~kJ}\n$$"} {"id": "textbook_2938", "title": "1010. Hess's Law - ", "content": "In the two-step process, first carbon monoxide is formed:"} {"id": "textbook_2939", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) \\quad \\Delta H^{\\circ}=-111 \\mathrm{~kJ}\n$$"} {"id": "textbook_2940", "title": "1010. Hess's Law - ", "content": "Then, carbon monoxide reacts further to form carbon dioxide:"} {"id": "textbook_2941", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H^{\\circ}=-283 \\mathrm{~kJ}\n$$"} {"id": "textbook_2942", "title": "1010. Hess's Law - ", "content": "The equation describing the overall reaction is the sum of these two chemical changes:"} {"id": "textbook_2943", "title": "1010. Hess's Law - ", "content": "$$\n\\begin{aligned}\n& \\text { Step 1: } \\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) \\\\\n& \\text { Step 2: } \\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\\\\n& \\text { Sum: } \\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{CO}_{2}(g)\n\\end{aligned}\n$$"} {"id": "textbook_2944", "title": "1010. Hess's Law - ", "content": "Because the CO produced in Step 1 is consumed in Step 2, the net change is:"} {"id": "textbook_2945", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_2946", "title": "1010. Hess's Law - ", "content": "According to Hess's law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps."} {"id": "textbook_2947", "title": "1010. Hess's Law - ", "content": "$$\n\\begin{array}{ll}\n\\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) & \\Delta H^{\\circ}=-111 \\mathrm{~kJ} \\\\\n\\frac{\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)}{\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)} & \\frac{\\Delta H^{\\circ}=-283 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-394 \\mathrm{~kJ}}\n\\end{array}\n$$"} {"id": "textbook_2948", "title": "1010. Hess's Law - ", "content": "The result is shown in Figure 9.24. We see that $\\Delta H$ of the overall reaction is the same whether it occurs in one step or two. This finding (overall $\\Delta H$ for the reaction = sum of $\\Delta H$ values for reaction \"steps\" in the overall reaction) is true in general for chemical and physical processes.\n"} {"id": "textbook_2949", "title": "1010. Hess's Law - ", "content": "FIGURE 9.24 The formation of $\\mathrm{CO}_{2}(g)$ from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess's law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants."} {"id": "textbook_2950", "title": "1010. Hess's Law - ", "content": "Before we further practice using Hess's law, let us recall two important features of $\\Delta H$."} {"id": "textbook_2951", "title": "1010. Hess's Law - ", "content": "1. $\\Delta H$ is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of $\\mathrm{NO}_{2}(g)$ is +33.2 kJ :"} {"id": "textbook_2952", "title": "1010. Hess's Law - ", "content": "$$\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) \\quad \\Delta H=+33.2 \\mathrm{~kJ}\n$$"} {"id": "textbook_2953", "title": "1010. Hess's Law - ", "content": "When 2 moles of $\\mathrm{NO}_{2}$ (twice as much) are formed, the $\\Delta H$ will be twice as large:"} {"id": "textbook_2954", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H=+66.4 \\mathrm{~kJ}\n$$"} {"id": "textbook_2955", "title": "1010. Hess's Law - ", "content": "In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.\n2. $\\Delta H$ for a reaction in one direction is equal in magnitude and opposite in sign to $\\Delta H$ for the reaction in the reverse direction. For example, given that:"} {"id": "textbook_2956", "title": "1010. Hess's Law - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H=-184.6 \\mathrm{~kJ}\n$$"} {"id": "textbook_2957", "title": "1010. Hess's Law - ", "content": "Then, for the \"reverse\" reaction, the enthalpy change is also \"reversed\":"} {"id": "textbook_2958", "title": "1010. Hess's Law - ", "content": "$$\n2 \\mathrm{HCl}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad \\Delta H=+184.6 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2959", "title": "1011. EXAMPLE 9.13 - ", "content": ""} {"id": "textbook_2960", "title": "1012. Stepwise Calculation of $\\Delta H_{\\mathrm{f}}^{\\circ}$ Using Hess's Law - ", "content": "\nDetermine the enthalpy of formation, $\\Delta H_{f}^{\\circ}$, of $\\mathrm{FeCl}_{3}(s)$ from the enthalpy changes of the following two-step process that occurs under standard state conditions:"} {"id": "textbook_2961", "title": "1012. Stepwise Calculation of $\\Delta H_{\\mathrm{f}}^{\\circ}$ Using Hess's Law - ", "content": "$$\n\\begin{array}{cr}\n\\mathrm{Fe}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{2}(s) & \\Delta H^{\\circ}=-341.8 \\mathrm{~kJ} \\\\\n\\mathrm{FeCl}_{2}(s)+\\frac{1}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s) & \\Delta H^{\\circ}=-57.7 \\mathrm{~kJ}\n\\end{array}\n$$\n"} {"id": "textbook_2962", "title": "1013. Solution - ", "content": "\nWe are trying to find the standard enthalpy of formation of $\\mathrm{FeCl}_{3}(s)$, which is equal to $\\Delta H^{\\circ}$ for the reaction:\n\n$$\n\\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=?\n$$"} {"id": "textbook_2963", "title": "1013. Solution - ", "content": "Looking at the reactions, we see that the reaction for which we want to find $\\Delta H^{\\circ}$ is the sum of the two reactions with known $\\Delta H$ values, so we must sum their $\\Delta H \\mathrm{~s}$ :"} {"id": "textbook_2964", "title": "1013. Solution - ", "content": "$$\n\\begin{array}{ll}\n\\mathrm{Fe}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{2}(s) & \\Delta H^{\\circ}=-341.8 \\mathrm{~kJ} \\\\\n\\frac{\\mathrm{FeCl}_{2}(s)+\\frac{1}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s)}{\\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s)} & \\frac{\\Delta H^{\\circ}=-57.7 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-399.5 \\mathrm{~kJ}}\n\\end{array}\n$$"} {"id": "textbook_2965", "title": "1013. Solution - ", "content": "The enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, of $\\mathrm{FeCl}_{3}(s)$ is $-399.5 \\mathrm{~kJ} / \\mathrm{mol}$.\n"} {"id": "textbook_2966", "title": "1014. Check Your Learning - ", "content": "\nCalculate $\\Delta H$ for the process:"} {"id": "textbook_2967", "title": "1014. Check Your Learning - ", "content": "$$\n\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g)\n$$"} {"id": "textbook_2968", "title": "1014. Check Your Learning - ", "content": "from the following information:"} {"id": "textbook_2969", "title": "1014. Check Your Learning - ", "content": "$$\n\\begin{aligned}\n\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g) & \\Delta H=180.5 \\mathrm{~kJ} \\\\\n\\mathrm{NO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) & \\Delta H=-57.06 \\mathrm{~kJ}\n\\end{aligned}\n$$\n"} {"id": "textbook_2970", "title": "1015. Answer: - ", "content": "\n66.4 kJ"} {"id": "textbook_2971", "title": "1015. Answer: - ", "content": "Here is a less straightforward example that illustrates the thought process involved in solving many Hess's law problems. It shows how we can find many standard enthalpies of formation (and other values of $\\Delta H$ ) if they are difficult to determine experimentally.\n"} {"id": "textbook_2972", "title": "1016. EXAMPLE 9.14 - ", "content": ""} {"id": "textbook_2973", "title": "1017. A More Challenging Problem Using Hess's Law - ", "content": "\nChlorine monofluoride can react with fluorine to form chlorine trifluoride:\n(i) $\\mathrm{ClF}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)$\n$\\Delta H^{\\circ}=$ ?"} {"id": "textbook_2974", "title": "1017. A More Challenging Problem Using Hess's Law - ", "content": "Use the reactions here to determine the $\\Delta H^{\\circ}$ for reaction (i):"} {"id": "textbook_2975", "title": "1017. A More Challenging Problem Using Hess's Law - ", "content": "$$\n\\text { (ii) } 2 \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{O}_{2}(g)+2 \\mathrm{~F}_{2}(g) \\quad \\Delta H_{(i i)}^{\\circ}=-49.4 \\mathrm{~kJ}\n$$\n\n$$\n\\text { (iii) } 2 \\mathrm{ClF}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\mathrm{OF}_{2}(g) \\quad \\Delta H_{(i i i)}^{\\circ}=+214.0 \\mathrm{~kJ}\n$$\n\n$$\n\\text { (iv) } \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\quad \\Delta H_{(i v)}^{\\circ}=+236.2 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2976", "title": "1018. Solution - ", "content": "\nOur goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that $\\operatorname{ClF}(g)$ is needed as a reactant. This can be obtained by multiplying reaction (iii) by $\\frac{1}{2}$, which means that the $\\Delta H^{\\circ}$ change is also multiplied by $\\frac{1}{2}$ :"} {"id": "textbook_2977", "title": "1018. Solution - ", "content": "$$\n\\mathrm{ClF}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{1}{2} \\mathrm{OF}_{2}(g) \\quad \\Delta H^{\\circ}=\\frac{1}{2}(214.0)=+107.0 \\mathrm{~kJ}\n$$"} {"id": "textbook_2978", "title": "1018. Solution - ", "content": "Next, we see that $\\mathrm{F}_{2}$ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the $\\Delta H^{\\circ}$ changes sign and is halved:"} {"id": "textbook_2979", "title": "1018. Solution - ", "content": "$$\n\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{OF}_{2}(g) \\quad \\Delta H^{\\circ}=+24.7 \\mathrm{~kJ}\n$$"} {"id": "textbook_2980", "title": "1018. Solution - ", "content": "To get $\\mathrm{ClF}_{3}$ as a product, reverse (iv), changing the sign of $\\Delta H^{\\circ}$ :"} {"id": "textbook_2981", "title": "1018. Solution - ", "content": "$$\n\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) \\quad \\Delta H^{\\circ}=-236.2 \\mathrm{~kJ}\n$$"} {"id": "textbook_2982", "title": "1018. Solution - ", "content": "Now check to make sure that these reactions add up to the reaction we want:"} {"id": "textbook_2983", "title": "1018. Solution - ", "content": "$$\n\\begin{array}{ll}\n\\mathrm{ClF}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{1}{2} \\mathrm{OF}_{2}(g) & \\Delta H^{\\circ}=+107.0 \\mathrm{~kJ} \\\\\n\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{OF}_{2}(g) & \\Delta H^{\\circ}=+24.7 \\mathrm{~kJ} \\\\\n\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) & \\frac{\\Delta H^{\\circ}=-236.2 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-104.5 \\mathrm{~kJ}}\n\\end{array}\n$$\n\nReactants $\\frac{1}{2} \\mathrm{O}_{2}$ and $\\frac{1}{2} \\mathrm{O}_{2}$ cancel out product $\\mathrm{O}_{2}$; product $\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}$ cancels reactant $\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}$; and reactant $\\frac{3}{2} \\mathrm{OF}_{2}$ is cancelled by products $\\frac{1}{2} \\mathrm{OF}_{2}$ and $\\mathrm{OF}_{2}$. This leaves only reactants $\\mathrm{ClF}(g)$ and $\\mathrm{F}_{2}(g)$ and product $\\mathrm{ClF}_{3}(g)$, which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified $\\Delta H^{\\circ}$ values will give the desired $\\Delta H^{\\circ}$ :"} {"id": "textbook_2984", "title": "1018. Solution - ", "content": "$$\n\\Delta H^{\\circ}=(+107.0 \\mathrm{~kJ})+(24.7 \\mathrm{~kJ})+(-236.2 \\mathrm{~kJ})=-104.5 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_2985", "title": "1019. Check Your Learning - ", "content": "\nAluminum chloride can be formed from its elements:\n(i) $2 \\mathrm{Al}(s)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{AlCl}_{3}(s) \\quad \\Delta H^{\\circ}=$ ?"} {"id": "textbook_2986", "title": "1019. Check Your Learning - ", "content": "Use the reactions here to determine the $\\Delta H^{\\circ}$ for reaction (i):\n(ii) $\\mathrm{HCl}(g) \\longrightarrow \\mathrm{HCl}(a q) \\quad \\Delta H_{(i i)}^{\\circ}=-74.8 \\mathrm{~kJ}$\n(iii) $\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H_{(i i i)}^{\\circ}=-185 \\mathrm{~kJ}$\n(iv) $\\mathrm{AlCl}_{3}(a q) \\longrightarrow \\mathrm{AlCl}_{3}(s) \\quad \\Delta H_{\\text {(iv) }}^{\\circ}=+323 \\mathrm{~kJ} / \\mathrm{mol}$\n(v) $2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g) \\quad \\Delta H_{(v)}^{\\circ}=-1049 \\mathrm{~kJ}$\n"} {"id": "textbook_2987", "title": "1020. Answer: - ", "content": "\n-1407 kJ"} {"id": "textbook_2988", "title": "1020. Answer: - ", "content": "We also can use Hess's law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with $\\sum$ representing \"the sum of\" and $n$ standing for the stoichiometric coefficients:"} {"id": "textbook_2989", "title": "1020. Answer: - ", "content": "$$\n\\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { reactants })\n$$"} {"id": "textbook_2990", "title": "1020. Answer: - ", "content": "The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest."} {"id": "textbook_2991", "title": "1020. Answer: - ", "content": "EXAMPLE 9.15\n"} {"id": "textbook_2992", "title": "1021. Using Hess's Law - ", "content": "\nWhat is the standard enthalpy change for the reaction:"} {"id": "textbook_2993", "title": "1021. Using Hess's Law - ", "content": "$$\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g) \\quad \\Delta H^{\\circ}=?\n$$\n"} {"id": "textbook_2994", "title": "1022. Solution: Using the Equation - ", "content": "\nUse the special form of Hess's law given previously, and values from Appendix G:"} {"id": "textbook_2995", "title": "1022. Solution: Using the Equation - ", "content": "$$\n\\left.\\left.\\begin{array}{rl}\n& \\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { reactants }) \\\\\n= & {[2 \\mathrm{molHNO}(\\mathrm{qq})}\n\\end{array} \\times \\frac{-207.4 \\mathrm{~kJ}}{m+1 \\mathrm{mNO}(\\mathrm{~mol})}+1 \\mathrm{NO}(\\mathrm{~g})-\\times \\frac{+90.2 \\mathrm{~kJ}}{\\mathrm{molNO}(\\mathrm{~g})}\\right]\\right] .\n$$\n"} {"id": "textbook_2996", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "\nAlternatively, we can write this reaction as the sum of the decompositions of $3 \\mathrm{NO}_{2}(g)$ and $1 \\mathrm{H}_{2} \\mathrm{O}(I)$ into their constituent elements, and the formation of $2 \\mathrm{HNO}_{3}(\\mathrm{aq})$ and $1 \\mathrm{NO}(g)$ from their constituent elements. Writing out these reactions, and noting their relationships to the $\\Delta H_{\\mathrm{f}}^{\\circ}$ values for these compounds (from Appendix G), we have:"} {"id": "textbook_2997", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "$$\n\\begin{gathered}\n3 \\mathrm{NO}_{2}(g) \\longrightarrow 3 / 2 \\mathrm{~N}_{2}(g)+3 \\mathrm{O}_{2}(g) \\quad \\Delta H_{1}^{\\circ}=-99.6 \\mathrm{~kJ} \\\\\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\quad \\Delta H_{2}^{\\circ}=+285.8 \\mathrm{~kJ}\\left[-1 \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right] \\\\\n\\mathrm{H}_{2}(g)+\\mathrm{N}_{2}(g)+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q) \\quad \\Delta H_{3}^{\\circ}=-414.8 \\mathrm{~kJ}\\left[2 \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{HNO}_{3}\\right)\\right] \\\\\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{NO}(\\mathrm{~g})\n\\end{gathered} \\Delta H_{4}^{\\circ}=+90.2 \\mathrm{~kJ}[1 \\times(\\mathrm{NO})] \\mathrm{l}\n$$"} {"id": "textbook_2998", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "Summing these reaction equations gives the reaction we are interested in:"} {"id": "textbook_2999", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "$$\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g)\n$$"} {"id": "textbook_3000", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "Summing their enthalpy changes gives the value we want to determine:"} {"id": "textbook_3001", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "$$\n\\begin{aligned}\n\\Delta H_{\\mathrm{rxn}}^{\\circ} & =\\Delta H_{1}^{\\circ}+\\Delta H_{2}^{\\circ}+\\Delta H_{3}^{\\circ}+\\Delta H_{4}^{\\circ}=(-99.6 \\mathrm{~kJ})+(+285.8 \\mathrm{~kJ})+(-414.8 \\mathrm{~kJ})+(+90.2 \\mathrm{~kJ}) \\\\\n& =-138.4 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_3002", "title": "1023. Solution: Supporting Why the General Equation Is Valid - ", "content": "So the standard enthalpy change for this reaction is $\\Delta H^{\\circ}=-138.4 \\mathrm{~kJ}$.\nNote that this result was obtained by (1) multiplying the $\\Delta H_{\\mathrm{f}}^{\\circ}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\\Delta H_{\\mathrm{f}}^{\\circ}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.\n"} {"id": "textbook_3003", "title": "1024. Check Your Learning - ", "content": "\nCalculate the heat of combustion of 1 mole of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(I)$, when $\\mathrm{H}_{2} \\mathrm{O}(I)$ and $\\mathrm{CO}_{2}(g)$ are formed. Use the following enthalpies of formation: $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I}),-278 \\mathrm{~kJ} / \\mathrm{mol} ; \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I}),-286 \\mathrm{~kJ} / \\mathrm{mol}$; and $\\mathrm{CO}_{2}(\\mathrm{~g}),-394 \\mathrm{~kJ} / \\mathrm{mol}$.\n"} {"id": "textbook_3004", "title": "1025. Answer: - ", "content": "\n-1368 kJ/mol\n"} {"id": "textbook_3005", "title": "1025. Answer: - 1025.1. Strengths of Ionic and Covalent Bonds", "content": ""} {"id": "textbook_3006", "title": "1026. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3007", "title": "1026. LEARNING OBJECTIVES - ", "content": "- Describe the energetics of covalent and ionic bond formation and breakage\n- Use the Born-Haber cycle to compute lattice energies for ionic compounds\n- Use average covalent bond energies to estimate enthalpies of reaction"} {"id": "textbook_3008", "title": "1026. LEARNING OBJECTIVES - ", "content": "A bond's strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms."} {"id": "textbook_3009", "title": "1026. LEARNING OBJECTIVES - ", "content": "It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:"} {"id": "textbook_3010", "title": "1026. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{H}(\\mathrm{~g}) \\quad \\text { bond energy }=436 \\mathrm{~kJ}\n$$"} {"id": "textbook_3011", "title": "1026. LEARNING OBJECTIVES - ", "content": "Conversely, the same amount of energy is released when one mole of $\\mathrm{H}_{2}$ molecules forms from two moles of H atoms:"} {"id": "textbook_3012", "title": "1026. LEARNING OBJECTIVES - ", "content": "$$\n2 \\mathrm{H}(g) \\longrightarrow \\mathrm{H}_{2}(g) \\quad \\text { bond energy }=-436 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_3013", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "\nStable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 4.4). The stronger a bond, the greater the energy required to break it."} {"id": "textbook_3014", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, $\\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}$, is defined as the standard enthalpy change for the endothermic reaction:"} {"id": "textbook_3015", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\mathrm{XY}(g) \\longrightarrow \\mathrm{X}(g)+\\mathrm{Y}(g) \\quad \\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}=\\Delta H^{\\circ}\n$$"} {"id": "textbook_3016", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "For example, the bond energy of the pure covalent $\\mathrm{H}-\\mathrm{H}$ bond, $\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}$, is 436 kJ per mole of $\\mathrm{H}-\\mathrm{H}$ bonds broken:"} {"id": "textbook_3017", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{H}(g) \\quad \\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}=\\Delta H^{\\circ}=436 \\mathrm{~kJ}\n$$"} {"id": "textbook_3018", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four $\\mathrm{C}-\\mathrm{H}$ bond energies in $\\mathrm{CH}_{4}, 1660 \\mathrm{~kJ}$, is equal to the standard enthalpy change of the reaction:\n\n\nThe average $\\mathrm{C}-\\mathrm{H}$ bond energy, $\\mathrm{D}_{\\mathrm{C}-\\mathrm{H}}$, is $1660 / 4=415 \\mathrm{~kJ} / \\mathrm{mol}$ because there are four moles of $\\mathrm{C}-\\mathrm{H}$ bonds broken per mole of the reaction. Although the four $\\mathrm{C}-\\mathrm{H}$ bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires $439 \\mathrm{~kJ} / \\mathrm{mol}$ ), the remaining bonds are easier to break. The $415 \\mathrm{~kJ} / \\mathrm{mol}$ value is the average, not the exact value required to break any one bond."} {"id": "textbook_3019", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 9.3, and a comparison of bond lengths and bond strengths for some common bonds appears in"} {"id": "textbook_3020", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "Table 9.4. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, $\\mathrm{C}-\\mathrm{F}$ is $439 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{C}-\\mathrm{Cl}$ is $330 \\mathrm{~kJ} / \\mathrm{mol}$, and $\\mathrm{C}-\\mathrm{Br}$ is $275 \\mathrm{~kJ} / \\mathrm{mol}$."} {"id": "textbook_3021", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "Bond Energies (kJ/mol)"} {"id": "textbook_3022", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{H}$ | 436 | C-S | 260 | $\\mathrm{F}-\\mathrm{Cl}$ | 255 |\n| H-C | 415 | $\\mathrm{C}-\\mathrm{Cl}$ | 330 | $\\mathrm{F}-\\mathrm{Br}$ | 235 |\n| $\\mathrm{H}-\\mathrm{N}$ | 390 | $\\mathrm{C}-\\mathrm{Br}$ | 275 | Si-Si | 230 |\n| H-O | 464 | C-I | 240 | Si-P | 215 |\n| H-F | 569 | $\\mathrm{N}-\\mathrm{N}$ | 160 | Si-S | 225 |\n| H-Si | 395 | $\\mathrm{N}=\\mathrm{N}$ | 418 | $\\mathrm{Si}-\\mathrm{Cl}$ | 359 |\n| $\\mathrm{H}-\\mathrm{P}$ | 320 | $\\mathrm{N} \\equiv \\mathrm{N}$ | 946 | $\\mathrm{Si}-\\mathrm{Br}$ | 290 |\n| H-S | 340 | $\\mathrm{N}-\\mathrm{O}$ | 200 | Si-I | 215 |\n| $\\mathrm{H}-\\mathrm{Cl}$ | 432 | N-F | 270 | P-P | 215 |\n| $\\mathrm{H}-\\mathrm{Br}$ | 370 | N-P | 210 | P-S | 230 |\n| H-I | 295 | $\\mathrm{N}-\\mathrm{Cl}$ | 200 | $\\mathrm{P}-\\mathrm{Cl}$ | 330 |\n| C-C | 345 | $\\mathrm{N}-\\mathrm{Br}$ | 245 | $\\mathrm{P}-\\mathrm{Br}$ | 270 |\n| $\\mathrm{C}=\\mathrm{C}$ | 611 | O-O | 140 | P-I | 215 |\n| $\\mathrm{C} \\equiv \\mathrm{C}$ | 837 | $\\mathrm{O}=\\mathrm{O}$ | 498 | S-S | 215 |\n| $\\mathrm{C}-\\mathrm{N}$ | 290 | O-F | 160 | $\\mathrm{S}-\\mathrm{Cl}$ | 250 |\n| $\\mathrm{C}=\\mathrm{N}$ | 615 | O-Si | 370 | $\\mathrm{S}-\\mathrm{Br}$ | 215 |\n| $\\mathrm{C} \\equiv \\mathrm{N}$ | 891 | O-P | 350 | $\\mathrm{Cl}-\\mathrm{Cl}$ | 243 |\n| $\\mathrm{C}-\\mathrm{O}$ | 350 | $\\mathrm{O}-\\mathrm{Cl}$ | 205 | $\\mathrm{Cl}-\\mathrm{Br}$ | 220 |\n| $\\mathrm{C}=\\mathrm{O}$ | 741 | O-I | 200 | Cl-I | 210 |\n| $\\mathrm{C} \\equiv \\mathrm{O}$ | 1080 | F-F | 160 | $\\mathrm{Br}-\\mathrm{Br}$ | 190 |\n| C-F | 439 | F-Si | 540 | $\\mathrm{Br}-\\mathrm{I}$ | 180 |\n| $\\mathrm{C}-\\mathrm{Si}$ | 360 | F-P | 489 | I-I | 150 |"} {"id": "textbook_3023", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "TABLE 9.3"} {"id": "textbook_3024", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C-P | 265 | F-S | 285 | | |"} {"id": "textbook_3025", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "TABLE 9.3"} {"id": "textbook_3026", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "Average Bond Lengths and Bond Energies for Some Common Bonds"} {"id": "textbook_3027", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "| Bond | Bond Length (\u00c5) | Bond Energy (kJ/mol) |\n| :--- | :--- | :--- |\n| $\\mathrm{C}-\\mathrm{C}$ | 1.54 | 345 |\n| $\\mathrm{C}=\\mathrm{C}$ | 1.34 | 611 |\n| $\\mathrm{C} \\equiv \\mathrm{C}$ | 1.20 | 837 |\n| $\\mathrm{C}-\\mathrm{N}$ | 1.43 | 290 |\n| $\\mathrm{C}=\\mathrm{N}$ | 1.38 | 615 |\n| $\\mathrm{C} \\equiv \\mathrm{N}$ | 1.16 | 891 |\n| $\\mathrm{C}-\\mathrm{O}$ | 1.43 | 741 |\n| $\\mathrm{C}=\\mathrm{O}$ | 1.23 | 1.13 |"} {"id": "textbook_3028", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "TABLE 9.4"} {"id": "textbook_3029", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the $\\mathrm{H}_{2}$ molecule shown in Figure 5.2, at the bond distance of 74 pm the system is $7.24 \\times 10^{-19} \\mathrm{~J}$ lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, as we will learn in more detail later, bond energies are often discussed on a per-mole basis. For example, it requires $7.24 \\times 10^{-19} \\mathrm{~J}$ to break one $\\mathrm{H}-\\mathrm{H}$ bond, but it takes $4.36 \\times 10^{5} \\mathrm{~J}$ to break 1 mole of $\\mathrm{H}-\\mathrm{H}$ bonds. A comparison of some bond lengths and energies is shown in Figure 5.2 and Table 9.3. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first $\\mathrm{C}-\\mathrm{H}$ bond in $\\mathrm{CH}_{4}$ requires $439.3 \\mathrm{~kJ} / \\mathrm{mol}$, while breaking the first $\\mathrm{C}-\\mathrm{H}$ bond in $\\mathrm{H}-\\mathrm{CH}_{2} \\mathrm{C}_{6} \\mathrm{H}_{5}$ (a common paint thinner) requires $375.5 \\mathrm{~kJ} / \\mathrm{mol}$."} {"id": "textbook_3030", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "As seen in Table 9.3 and Table 9.4, an average carbon-carbon single bond is $347 \\mathrm{~kJ} / \\mathrm{mol}$, while in a carboncarbon double bond, the $\\pi$ bond increases the bond strength by $267 \\mathrm{~kJ} / \\mathrm{mol}$. Adding an additional $\\pi$ bond causes a further increase of $225 \\mathrm{~kJ} / \\mathrm{mol}$. We can see a similar pattern when we compare other $\\sigma$ and $\\pi$ bonds. Thus, each individual $\\pi$ bond is generally weaker than a corresponding $\\sigma$ bond between the same two atoms. In a $\\sigma$ bond, there is a greater degree of orbital overlap than in a $\\pi$ bond."} {"id": "textbook_3031", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction ( $\\Delta H$ negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction ( $\\Delta H$ positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.\n\nThe enthalpy change, $\\Delta H$, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy \"in\", positive sign) plus the energy released when all bonds are formed in the products (energy \"out,\" negative sign). This can be expressed mathematically in the following way:"} {"id": "textbook_3032", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\Delta H=\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }}\n$$"} {"id": "textbook_3033", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "In this expression, the symbol $\\Sigma$ means \"the sum of\" and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 9.4) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction."} {"id": "textbook_3034", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "Consider the following reaction:"} {"id": "textbook_3035", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$"} {"id": "textbook_3036", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "or"} {"id": "textbook_3037", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\mathrm{H}-\\mathrm{H}(g)+\\mathrm{Cl}-\\mathrm{Cl}(g) \\longrightarrow 2 \\mathrm{H}-\\mathrm{Cl}(g)\n$$"} {"id": "textbook_3038", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "To form two moles of HCl , one mole of $\\mathrm{H}-\\mathrm{H}$ bonds and one mole of $\\mathrm{Cl}-\\mathrm{Cl}$ bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the $\\mathrm{H}-\\mathrm{H}$ bond $(436 \\mathrm{~kJ} / \\mathrm{mol})$ and the $\\mathrm{Cl}-\\mathrm{Cl}$ bond ( $243 \\mathrm{~kJ} / \\mathrm{mol}$ ). During the reaction, two moles of $\\mathrm{H}-\\mathrm{Cl}$ bonds are formed (bond energy $=432 \\mathrm{~kJ} / \\mathrm{mol}$ ), releasing 2 $\\times 432 \\mathrm{~kJ}$; or 864 kJ . Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:"} {"id": "textbook_3039", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "$$\n\\begin{aligned}\n\\Delta H & =\\sum \\mathrm{D}_{\\text {bonds broken }}-\\sum \\mathrm{D}_{\\text {bonds formed }} \\\\\n\\Delta H & =\\left[\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}+\\mathrm{D}_{\\mathrm{Cl}-\\mathrm{Cl}}\\right]-2 \\mathrm{D}_{\\mathrm{H}-\\mathrm{Cl}} \\\\\n& =[436+243]-2(432)=-185 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_3040", "title": "1027. Bond Strength: Covalent Bonds - ", "content": "This excess energy is released as heat, so the reaction is exothermic. Appendix $G$ gives a value for the standard molar enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g}), \\Delta H_{\\mathrm{f}}^{\\circ}$, of $-92.307 \\mathrm{~kJ} / \\mathrm{mol}$. Twice that value is -184.6 kJ , which agrees well with the answer obtained earlier for the formation of two moles of HCl .\n"} {"id": "textbook_3041", "title": "1028. EXAMPLE 9.16 - ", "content": ""} {"id": "textbook_3042", "title": "1029. Using Bond Energies to Calculate Approximate Enthalpy Changes - ", "content": "\nMethanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO , and hydrogen, $\\mathrm{H}_{2}$, from which methanol can be produced. Using the bond energies in Table 9.4, calculate the approximate enthalpy change, $\\Delta H$, for the reaction here:"} {"id": "textbook_3043", "title": "1029. Using Bond Energies to Calculate Approximate Enthalpy Changes - ", "content": "$$\n\\mathrm{CO}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{~g})\n$$\n"} {"id": "textbook_3044", "title": "1030. Solution - ", "content": "\nFirst, we need to write the Lewis structures of the reactants and the products:\n"} {"id": "textbook_3045", "title": "1030. Solution - ", "content": "From this, we see that $\\Delta H$ for this reaction involves the energy required to break a $\\mathrm{C}-\\mathrm{O}$ triple bond and two $\\mathrm{H}-\\mathrm{H}$ single bonds, as well as the energy produced by the formation of three $\\mathrm{C}-\\mathrm{H}$ single bonds, a $\\mathrm{C}-\\mathrm{O}$ single bond, and an $\\mathrm{O}-\\mathrm{H}$ single bond. We can express this as follows:\n\n$$\n\\begin{aligned}\n\\Delta H & =\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }} \\\\\n\\Delta H & =\\left[\\mathrm{D}_{\\mathrm{C} \\equiv \\mathrm{O}}+2\\left(\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}\\right)\\right]-\\left[3\\left(\\mathrm{D}_{\\mathrm{C}-\\mathrm{H}}\\right)+\\mathrm{D}_{\\mathrm{C}-\\mathrm{O}}+\\mathrm{D}_{\\mathrm{O}-\\mathrm{H}}\\right]\n\\end{aligned}\n$$"} {"id": "textbook_3046", "title": "1030. Solution - ", "content": "Using the bond energy values in Table 9.4, we obtain:"} {"id": "textbook_3047", "title": "1030. Solution - ", "content": "$$\n\\begin{aligned}\n\\Delta H & =[1080+2(436)]-[3(415)+350+464] \\\\\n& =-107 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_3048", "title": "1030. Solution - ", "content": "We can compare this value to the value calculated based on $\\Delta H_{\\mathrm{f}}^{\\circ}$ data from Appendix G :"} {"id": "textbook_3049", "title": "1030. Solution - ", "content": "$$\n\\begin{aligned}\n\\Delta H & =\\left[\\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{CH}_{3} \\mathrm{OH}(g)\\right]-\\left[\\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{CO}(g)+2 \\times \\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{H}_{2}\\right] \\\\\n& =[-201.0]-[-110.52+2 \\times 0] \\\\\n& =-90.5 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_3050", "title": "1030. Solution - ", "content": "Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.\n"} {"id": "textbook_3051", "title": "1031. Check Your Learning - ", "content": "\nEthyl alcohol, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:\n"} {"id": "textbook_3052", "title": "1031. Check Your Learning - ", "content": "Using the bond energies in Table 9.4, calculate an approximate enthalpy change, $\\Delta H$, for this reaction.\n"} {"id": "textbook_3053", "title": "1032. Answer: - ", "content": "\n-35 kJ\n"} {"id": "textbook_3054", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "\nAn ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ( $\\boldsymbol{\\Delta} \\boldsymbol{H}_{\\text {lattice }}$ ) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:"} {"id": "textbook_3055", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "$$\n\\mathrm{MX}(s) \\longrightarrow \\mathrm{M}^{n+}(g)+\\mathrm{X}^{n-}(g) \\quad \\Delta H_{\\text {lattice }}\n$$"} {"id": "textbook_3056", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, $\\Delta H_{\\text {lattice }}=769 \\mathrm{~kJ}$. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous $\\mathrm{Na}^{+}$ and $\\mathrm{Cl}^{-}$ions. When one mole each of gaseous $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions form solid $\\mathrm{NaCl}, 769 \\mathrm{~kJ}$ of heat is released."} {"id": "textbook_3057", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "The lattice energy $\\Delta H_{\\text {lattice }}$ of an ionic crystal can be expressed by the following equation (derived from Coulomb's law, governing the forces between electric charges):"} {"id": "textbook_3058", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "$$\n\\Delta H_{\\text {lattice }}=\\frac{\\mathrm{C}\\left(\\mathrm{Z}^{+}\\right)\\left(\\mathrm{Z}^{-}\\right)}{\\mathrm{R}_{\\mathrm{o}}}\n$$"} {"id": "textbook_3059", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "in which C is a constant that depends on the type of crystal structure; $\\mathrm{Z}^{+}$and $\\mathrm{Z}^{-}$are the charges on the ions; and $R_{0}$ is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy\nof an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of $\\mathrm{LiF}\\left(\\mathrm{Z}^{+}\\right.$and $\\mathrm{Z}^{-}=1$ ) is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, whereas that of $\\mathrm{MgO}\\left(\\mathrm{Z}^{+}\\right.$and $\\mathrm{Z}^{-}=2$ ) is $3900 \\mathrm{~kJ} / \\mathrm{mol}\\left(\\mathrm{R}_{\\mathrm{o}}\\right.$ is nearly the same-about 200 pm for both compounds)."} {"id": "textbook_3060", "title": "1033. Ionic Bond Strength and Lattice Energy - ", "content": "Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of $\\mathrm{MgF}_{2}(2957 \\mathrm{~kJ} / \\mathrm{mol})$ to that of $\\mathrm{MgI}_{2}(2327 \\mathrm{~kJ} / \\mathrm{mol})$ to observe the effect on lattice energy of the smaller ionic size of $\\mathrm{F}^{-}$as compared to $\\mathrm{I}^{-}$.\n"} {"id": "textbook_3061", "title": "1034. EXAMPLE 9.17 - ", "content": ""} {"id": "textbook_3062", "title": "1035. Lattice Energy Comparisons - ", "content": "\nThe precious gem ruby is aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, containing traces of $\\mathrm{Cr}^{3+}$. The compound $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$ is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ or $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$ ?\n"} {"id": "textbook_3063", "title": "1036. Solution - ", "content": "\nIn these two ionic compounds, the charges $\\mathrm{Z}^{+}$and $\\mathrm{Z}^{-}$are the same, so the difference in lattice energy will depend upon $\\mathrm{R}_{0}$. The $\\mathrm{O}^{2-}$ ion is smaller than the $\\mathrm{Se}^{2-}$ ion. Thus, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ would have a shorter interionic distance than $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$, and $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ would have the larger lattice energy.\n"} {"id": "textbook_3064", "title": "1037. Check Your Learning - ", "content": "\nZinc oxide, ZnO , is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl ?\n"} {"id": "textbook_3065", "title": "1038. Answer: - ", "content": "\nZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl .\n"} {"id": "textbook_3066", "title": "1039. The Born-Haber Cycle - ", "content": "\nIt is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess's law that breaks down the formation of an ionic solid into a series of individual steps:"} {"id": "textbook_3067", "title": "1039. The Born-Haber Cycle - ", "content": "- $\\Delta H_{\\mathrm{f}}^{\\circ}$, the standard enthalpy of formation of the compound\n- IE, the ionization energy of the metal\n- $E A$, the electron affinity of the nonmetal\n- $\\Delta H_{s}^{\\circ}$, the enthalpy of sublimation of the metal\n- $D$, the bond dissociation energy of the nonmetal\n- $\\Delta H_{\\text {lattice }}$, the lattice energy of the compound"} {"id": "textbook_3068", "title": "1039. The Born-Haber Cycle - ", "content": "Figure 9.25 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.\n"} {"id": "textbook_3069", "title": "1039. The Born-Haber Cycle - ", "content": "FIGURE 9.25 The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states."} {"id": "textbook_3070", "title": "1039. The Born-Haber Cycle - ", "content": "We begin with the elements in their most common states, $\\operatorname{Cs}(s)$ and $\\mathrm{F}_{2}(g)$. The $\\Delta H_{s}^{\\circ}$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the $F-F$ bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the $y$-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, of the compound from its elements. In this case, the overall change is exothermic."} {"id": "textbook_3071", "title": "1039. The Born-Haber Cycle - ", "content": "Hess's law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 9.5 shows this for cesium fluoride, CsF."} {"id": "textbook_3072", "title": "1039. The Born-Haber Cycle - ", "content": "| Enthalpy of
sublimation of
$\\mathrm{Cs}(s)$ | $\\mathrm{Cs}(s) \\longrightarrow \\mathrm{Cs}(g)$ | $\\Delta H=\\Delta H_{s}^{\\circ}=76.5 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| :--- | :--- | :--- |\n| One-half of the
bond energy of
$\\mathrm{F}_{2}$ | $\\frac{1}{2} \\mathrm{~F}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{F}(\\mathrm{g})$ | $\\Delta H=\\frac{1}{2} D=79.4 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Ionization
energy of $\\mathrm{Cs}(g)$ | $\\mathrm{Cs}(g) \\longrightarrow \\mathrm{Cs}^{+}(g)+\\mathrm{e}^{-}$ | $\\Delta H=I E=375.7 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Electron
affinity of F | $\\mathrm{F}(g)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{F}^{-}(g)$ | $\\Delta H=E A=-328.2 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Negative of the
lattice energy
of CsF $(s)$ | $\\mathrm{Cs}{ }^{+}(g)+\\mathrm{F}^{-}(g) \\longrightarrow \\mathrm{CsF}(s)$ | $\\Delta H=-\\Delta H_{\\text {lattice }}=?$ |\n| Enthalpy of
formation of
CsF $(s)$, add
steps $1-5$ | $\\Delta H=\\Delta H_{f}^{\\circ}=\\Delta H_{s}^{\\circ}+\\frac{1}{2} D+I E+(E A)+\\left(-\\Delta H_{\\mathrm{lattice}}\\right)$ | $\\Delta H=-553.5 \\mathrm{~kJ} / \\mathrm{mol}$ |\n"} {"id": "textbook_3073", "title": "1040. TABLE 9.5 - ", "content": "\nThus, the lattice energy can be calculated from other values. For cesium fluoride, using this data, the lattice energy is:"} {"id": "textbook_3074", "title": "1040. TABLE 9.5 - ", "content": "$$\n\\Delta H_{\\text {lattice }}=76.5+79.4+375.7+(-328.2)-(-553.5)=756.9 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3075", "title": "1040. TABLE 9.5 - ", "content": "The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $\\Delta H_{s}^{\\circ}$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy $\\Delta H_{\\text {lattice, }}$, and standard enthalpy of formation $\\Delta H_{\\mathrm{f}}^{\\circ}$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom."} {"id": "textbook_3076", "title": "1040. TABLE 9.5 - ", "content": "Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600-4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between $150-400 \\mathrm{~kJ} / \\mathrm{mol}$ for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.\n"} {"id": "textbook_3077", "title": "1041. Key Terms - ", "content": "\nbomb calorimeter device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products\nbond energy (also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance\nBorn-Haber cycle thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements\ncalorie (cal) unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J\ncalorimeter device used to measure the amount of heat absorbed or released in a chemical or physical process\ncalorimetry process of measuring the amount of heat involved in a chemical or physical process\nchemical thermodynamics area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes\nendothermic process chemical reaction or physical change that absorbs heat\nenergy capacity to supply heat or do work\nenthalpy ( $\\boldsymbol{H}$ ) sum of a system's internal energy and the mathematical product of its pressure and volume\nenthalpy change ( $\\mathbf{\\Delta H}$ ) heat released or absorbed by a system under constant pressure during a chemical or physical process\nexothermic process chemical reaction or physical change that releases heat\nexpansion work (pressure-volume work) work done as a system expands or contracts against external pressure\nfirst law of thermodynamics internal energy of a system changes due to heat flow in or out of the system or work done on or by the system\nheat (q) transfer of thermal energy between two bodies\nheat capacity (C) extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin)\nHess's law if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps\nhydrocarbon compound composed only of hydrogen and carbon; the major component of fossil fuels\ninternal energy ( $\\boldsymbol{U}$ ) total of all possible kinds of energy present in a substance or substances\njoule (J) SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J $=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}$ and $4.184 \\mathrm{~J}=1 \\mathrm{cal}$\nkinetic energy energy of a moving body, in joules, equal to $\\frac{1}{2} m v^{2}$ (where $m=$ mass and $v=$ velocity)\nlattice energy ( $\\boldsymbol{\\Delta} \\boldsymbol{H}_{\\text {lattice }}$ ) energy required to separate one mole of an ionic solid into its component gaseous ions\nnutritional calorie (Calorie) unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal\npotential energy energy of a particle or system of particles derived from relative position, composition, or condition\nspecific heat capacity (c) intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin)\nstandard enthalpy of combustion $\\left(\\Delta H_{c}^{\\circ}\\right)$ heat released when one mole of a compound undergoes complete combustion under standard conditions\nstandard enthalpy of formation $\\left(\\Delta H_{\\mathrm{f}}^{\\circ}\\right)$ enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions\nstandard state set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations, usually at a temperature of 298.15 K\nstate function property depending only on the state of a system, and not the path taken to reach that state\nsurroundings all matter other than the system being studied\nsystem portion of matter undergoing a chemical or physical change being studied\ntemperature intensive property of matter that is a quantitative measure of \"hotness\" and \"coldness\"\nthermal energy kinetic energy associated with the random motion of atoms and molecules\nthermochemistry study of measuring the amount\nof heat absorbed or released during a chemical reaction or a physical change\nwork (w) energy transfer due to changes in\nexternal, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force\n"} {"id": "textbook_3078", "title": "1042. Key Equations - ", "content": "\n$q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)$\n$\\Delta U=q+w$\n$\\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}($ products $)-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}($ reactants $)$\nBond energy for a diatomic molecule: XY $(g) \\longrightarrow \\mathrm{X}(g)+\\mathrm{Y}(g)$"} {"id": "textbook_3079", "title": "1042. Key Equations - ", "content": "$$\n\\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}=\\Delta H^{\\circ}\n$$"} {"id": "textbook_3080", "title": "1042. Key Equations - ", "content": "Enthalpy change: $\\Delta H=\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }}$\nLattice energy for a solid MX: MX $(s) \\longrightarrow \\mathrm{M}^{n+}(g)+\\mathrm{X}^{n-}(g) \\quad \\Delta H_{\\text {lattice }}$\nLattice energy for an ionic crystal: $\\Delta H_{\\text {lattice }}=\\frac{\\mathrm{C}\\left(\\mathrm{Z}^{+}\\right)\\left(\\mathrm{Z}^{-}\\right)}{\\mathrm{R}_{\\mathrm{o}}}$\n"} {"id": "textbook_3081", "title": "1043. Summary - ", "content": ""} {"id": "textbook_3082", "title": "1043. Summary - 1043.1. Energy Basics", "content": "\nEnergy is the capacity to supply heat or do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics)."} {"id": "textbook_3083", "title": "1043. Summary - 1043.1. Energy Basics", "content": "The thermal energy of matter is due to the kinetic energies of its constituent atoms or molecules. Temperature is an intensive property of matter reflecting hotness or coldness that increases as the average kinetic energy increases. Heat is the transfer of thermal energy between objects at different temperatures. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule ( J )."} {"id": "textbook_3084", "title": "1043. Summary - 1043.1. Energy Basics", "content": "Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes.\n"} {"id": "textbook_3085", "title": "1043. Summary - 1043.2. Calorimetry", "content": "\nCalorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed\nin the process using known mathematical relations.\nCalorimeters are designed to minimize energy exchange between their contents and the external environment. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.\n"} {"id": "textbook_3086", "title": "1043. Summary - 1043.3. Enthalpy", "content": "\nIf a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, $q$ for the change is called the enthalpy change with the symbol $\\Delta H$, or $\\Delta H^{\\circ}$ for reactions occurring under standard state conditions at 298 K . The value of $\\Delta H$ for a reaction in one direction is equal in magnitude, but opposite in sign, to $\\Delta H$ for the reaction in the opposite direction, and $\\Delta H$ is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess's law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.\n"} {"id": "textbook_3087", "title": "1043. Summary - 1043.4. Strengths of Ionic and Covalent Bonds", "content": "\nThe strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than\nsingle bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase\nions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.\n"} {"id": "textbook_3088", "title": "1044. Exercises - ", "content": ""} {"id": "textbook_3089", "title": "1044. Exercises - 1044.1. Energy Basics", "content": "\n1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?\n2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile.\n3. Explain the difference between heat capacity and specific heat of a substance.\n4. Calculate the heat capacity, in joules and in calories per degree, of the following:\n(a) 28.4 g of water\n(b) 1.00 oz of lead\n5. Calculate the heat capacity, in joules and in calories per degree, of the following:\n(a) 45.8 g of nitrogen gas\n(b) 1.00 pound of aluminum metal\n6. How much heat, in joules and in calories, must be added to a $75.0-\\mathrm{g}$ iron block with a specific heat of $0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ to increase its temperature from $25^{\\circ} \\mathrm{C}$ to its melting temperature of $1535{ }^{\\circ} \\mathrm{C}$ ?\n7. How much heat, in joules and in calories, is required to heat a $28.4-\\mathrm{g}(1-\\mathrm{oz})$ ice cube from $-23.0^{\\circ} \\mathrm{C}$ to -1.0 ${ }^{\\circ} \\mathrm{C}$ ?\n8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?\n9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?\n10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from $23.2^{\\circ} \\mathrm{C}$ to $89.6^{\\circ} \\mathrm{C}$.\n(a) What is the specific heat of the substance?\n(b) If it is one of the substances found in Table 9.1, what is its likely identity?\n11. A piece of unknown solid substance weighs 437.2 g , and requires 8460 J to increase its temperature from $19.3^{\\circ} \\mathrm{C}$ to $68.9^{\\circ} \\mathrm{C}$.\n(a) What is the specific heat of the substance?\n(b) If it is one of the substances found in Table 9.1, what is its likely identity?\n12. An aluminum kettle weighs 1.05 kg .\n(a) What is the heat capacity of the kettle?\n(b) How much heat is required to increase the temperature of this kettle from $23.0^{\\circ} \\mathrm{C}$ to $99.0^{\\circ} \\mathrm{C}$ ?\n(c) How much heat is required to heat this kettle from $23.0^{\\circ} \\mathrm{C}$ to $99.0^{\\circ} \\mathrm{C}$ if it contains 1.25 L of water (density of $0.997 \\mathrm{~g} / \\mathrm{mL}$ and a specific heat of $4.184 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ )?\n13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about $85^{\\circ} \\mathrm{F}$. Unless it is heated, a waterbed that contains 892 L of water cools from $85^{\\circ} \\mathrm{F}$ to $72^{\\circ} \\mathrm{F}$ in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh , to keep the bed from cooling. Note that 1 kilowatt-hour $(\\mathrm{kWh})=3.6 \\times 10^{6} \\mathrm{~J}$, and assume that the density of water is $1.0 \\mathrm{~g} / \\mathrm{mL}$ (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield \"positive\" or \"negative\" errors)?\n"} {"id": "textbook_3090", "title": "1044. Exercises - 1044.2. Calorimetry", "content": "\n14. A $500-\\mathrm{mL}$ bottle of water at room temperature and a $2-\\mathrm{L}$ bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the $500-\\mathrm{mL}$ bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the $500-\\mathrm{mL}$ bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.\n15. Would the amount of heat measured for the reaction in Example 9.5 be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.\n16. Would the amount of heat absorbed by the dissolution in Example 9.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.\n17. Would the amount of heat absorbed by the dissolution in Example 9.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer.\n18. How many milliliters of water at $23^{\\circ} \\mathrm{C}$ with a density of $1.00 \\mathrm{~g} / \\mathrm{mL}$ must be mixed with 180 mL (about 6 oz ) of coffee at $95^{\\circ} \\mathrm{C}$ so that the resulting combination will have a temperature of $60^{\\circ} \\mathrm{C}$ ? Assume that coffee and water have the same density and the same specific heat.\n19. How much will the temperature of a cup ( 180 g ) of coffee at $95^{\\circ} \\mathrm{C}$ be reduced when a 45 g silver spoon (specific heat $0.24 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ) at $25^{\\circ} \\mathrm{C}$ is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.\n20. A $45-\\mathrm{g}$ aluminum spoon (specific heat $0.88 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ) at $24^{\\circ} \\mathrm{C}$ is placed in $180 \\mathrm{~mL}(180 \\mathrm{~g})$ of coffee at $85^{\\circ} \\mathrm{C}$ and the temperature of the two become equal.\n(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.\n(b) The first time a student solved this problem she got an answer of $88^{\\circ} \\mathrm{C}$. Explain why this is clearly an incorrect answer.\n21. The temperature of the cooling water as it leaves the hot engine of an automobile is $240^{\\circ} \\mathrm{F}$. After it passes through the radiator it has a temperature of $175^{\\circ} \\mathrm{F}$. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of $4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$.\n22. A $70.0-\\mathrm{g}$ piece of metal at $80.0^{\\circ} \\mathrm{C}$ is placed in 100 g of water at $22.0^{\\circ} \\mathrm{C}$ contained in a calorimeter like that shown in Figure 9.12. The metal and water come to the same temperature at $24.6^{\\circ} \\mathrm{C}$. How much heat did the metal give up to the water? What is the specific heat of the metal?\n23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at $26.5^{\\circ} \\mathrm{C}$ in a calorimeter like that in Figure 9.12, what is the resulting temperature of the water?\n24. A $0.500-\\mathrm{g}$ sample of KCl is added to 50.0 g of water in a calorimeter (Figure 9.12). If the temperature decreases by $1.05^{\\circ} \\mathrm{C}$, what is the approximate amount of heat involved in the dissolution of the KCl , assuming the specific heat of the resulting solution is $4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ? Is the reaction exothermic or endothermic?\n25. Dissolving 3.0 g of $\\mathrm{CaCl}_{2}(s)$ in 150.0 g of water in a calorimeter (Figure 9.12) at $22.4^{\\circ} \\mathrm{C}$ causes the temperature to rise to $25.8^{\\circ} \\mathrm{C}$. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is $4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ? Is the reaction exothermic or endothermic?\n26. When 50.0 g of $0.200 \\mathrm{M} \\mathrm{NaCl}(\\mathrm{aq})$ at $24.1^{\\circ} \\mathrm{C}$ is added to 100.0 g of $0.100 \\mathrm{M} \\mathrm{AgNO}_{3}(\\mathrm{aq})$ at $24.1^{\\circ} \\mathrm{C}$ in a calorimeter, the temperature increases to $25.2^{\\circ} \\mathrm{C}$ as $\\mathrm{AgCl}(s)$ forms. Assuming the specific heat of the solution and products is $4.20 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, calculate the approximate amount of heat in joules produced.\n27. The addition of 3.15 g of $\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ to a solution of 1.52 g of $\\mathrm{NH}_{4} \\mathrm{SCN}$ in 100 g of water in a calorimeter caused the temperature to fall by $3.1^{\\circ} \\mathrm{C}$. Assuming the specific heat of the solution and products is 4.20 $\\mathrm{J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:\n$\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}(s)+2 \\mathrm{NH}_{4} \\mathrm{SCN}(\\mathrm{aq}) \\longrightarrow \\mathrm{Ba}(\\mathrm{SCN})_{2}(\\mathrm{aq})+2 \\mathrm{NH}_{3}(\\mathrm{aq})+10 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$\n28. The reaction of 50 mL of acid and 50 mL of base described in Example 9.5 increased the temperature of the solution by $6.9^{\\circ} \\mathrm{C}$. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of $22.0^{\\circ} \\mathrm{C}$ ? Explain your answer.\n29. If the 3.21 g of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ in Example 9.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer.\n30. When 1.0 g of fructose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)$, a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by $1.58^{\\circ} \\mathrm{C}$. If the heat capacity of the calorimeter and its contents is $9.90 \\mathrm{~kJ} /{ }^{\\circ} \\mathrm{C}$, what is $q$ for this combustion?\n31. When a $0.740-\\mathrm{g}$ sample of trinitrotoluene (TNT), $\\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{~N}_{2} \\mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature increases from $23.4^{\\circ} \\mathrm{C}$ to $26.9^{\\circ} \\mathrm{C}$. The heat capacity of the calorimeter is $534 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?\n32. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter (Figure 9.17), the temperature increases by $1.48{ }^{\\circ} \\mathrm{C}$. If the heat capacity of the calorimeter is $21.6 \\mathrm{~kJ} /{ }^{\\circ} \\mathrm{C}$, determine the heat produced by combustion of a ton of coal $\\left(2.000 \\times 10^{3}\\right.$ pounds).\n33. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g . What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g?\n34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories ( 16 kcal ). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g?\n35. What is the maximum mass of carbohydrate in a $6-$ oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g?\n36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra $1.1 \\times 10^{3}$ Calories if the average number of Calories for fat is 9.1 Calories/g?\n37. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories $/ \\mathrm{g}$, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g?\n38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $\\$ 4.23$, or a liter of isooctane (density, $0.6919 \\mathrm{~g} / \\mathrm{mL}$ ) that costs $\\$ 0.45$ ? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories.\n"} {"id": "textbook_3091", "title": "1044. Exercises - 1044.3. Enthalpy", "content": "\n39. Explain how the heat measured in Example 9.5 differs from the enthalpy change for the exothermic reaction described by the following equation:\n$\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n40. Using the data in the check your learning section of Example 9.5, calculate $\\Delta H$ in $\\mathrm{kJ} / \\mathrm{mol}$ of $\\mathrm{AgNO}_{3}(\\mathrm{aq})$ for the reaction: $\\mathrm{NaCl}(a q)+\\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{NaNO}_{3}(a q)$\n41. Calculate the enthalpy of solution ( $\\Delta H$ for the dissolution) per mole of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ under the conditions described in Example 9.6.\n42. Calculate $\\Delta H$ for the reaction described by the equation. (Hint: Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.)\n$\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}(s)+2 \\mathrm{NH}_{4} \\mathrm{SCN}(a q) \\longrightarrow \\mathrm{Ba}(\\mathrm{SCN})_{2}(a q)+2 \\mathrm{NH}_{3}(a q)+10 \\mathrm{H}_{2} \\mathrm{O}(l)$\n43. Calculate the enthalpy of solution ( $\\Delta H$ for the dissolution) per mole of $\\mathrm{CaCl}_{2}$ (refer to Exercise 9.25).\n44. Although the gas used in an oxyacetylene torch (Figure 9.7) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 9.2. Considering the conditions for which the tabulated data are reported, suggest an explanation.\n45. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?\n46. How much heat is produced by combustion of 125 g of methanol under standard state conditions?\n47. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?\n48. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?\n49. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?\n50. How much heat is produced when 100 mL of 0.250 M HCl (density, $1.00 \\mathrm{~g} / \\mathrm{mL}$ ) and 200 mL of 0.150 M NaOH (density, $1.00 \\mathrm{~g} / \\mathrm{mL}$ ) are mixed?\n$\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-58 \\mathrm{~kJ}$\nIf both solutions are at the same temperature and the specific heat of the products is $4.19 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, how much will the temperature increase? What assumption did you make in your calculation?\n51. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from $26.74^{\\circ} \\mathrm{C}$ to $27.93^{\\circ} \\mathrm{C}$. What is the heat capacity of the calorimeter and its contents?\n52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, $6.00 \\mathrm{kcal} / \\mathrm{mol}$ ) was used in household refrigerators. What mass of $\\mathrm{SO}_{2}$ must be evaporated to remove as much heat as evaporation of 1.00 kg of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ (enthalpy of vaporization is $17.4 \\mathrm{~kJ} / \\mathrm{mol}$ )?\nThe vaporization reactions for $\\mathrm{SO}_{2}$ and $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ are $\\mathrm{SO}_{2}(l) \\longrightarrow \\mathrm{SO}_{2}(g)$ and $\\mathrm{CCl}_{2} \\mathrm{~F}(l) \\longrightarrow \\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)$, respectively.\n53. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to $35.0^{\\circ} \\mathrm{C}$, as the heat provided when 100 g of steam is cooled from $110^{\\circ} \\mathrm{C}$ to $100^{\\circ} \\mathrm{C}$.\n54. Which of the enthalpies of combustion in Table 9.2 the table are also standard enthalpies of formation?\n55. Does the standard enthalpy of formation of $\\mathrm{H}_{2} \\mathrm{O}(g)$ differ from $\\Delta H^{\\circ}$ for the reaction\n$2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) ?$\n56. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red $\\mathrm{HgO}(s)$ to $\\mathrm{Hg}(\\mathrm{I})$ and $\\mathrm{O}_{2}(g)$ under standard conditions?\n57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn , is burned to form $\\mathrm{Mn}_{3} \\mathrm{O}_{4}(s)$ at standard state conditions?\n58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe , is burned to form $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$ at standard state conditions?\n59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:\n$4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-907 \\mathrm{~kJ}$\n$2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H=-113 \\mathrm{~kJ}$\n$3 \\mathrm{NO}_{2}+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g) \\quad \\Delta H=-139 \\mathrm{~kJ}$\nDetermine the total energy change for the production of one mole of aqueous nitric acid by this process.\n60. Both graphite and diamond burn.\n$\\mathrm{C}(s$, diamond $)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$\nFor the conversion of graphite to diamond:\n$\\mathrm{C}(s$, graphite $) \\longrightarrow \\mathrm{C}(s$, diamond $) \\quad \\Delta H^{\\circ}=1.90 \\mathrm{~kJ}$\nWhich produces more heat, the combustion of graphite or the combustion of diamond?\n61. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: $\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)$\n62. Which produces more heat?\n$\\mathrm{Os}(s) \\longrightarrow 2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{OsO}_{4}(s)$\nor\n$\\mathrm{Os}(s) \\longrightarrow 2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{OsO}_{4}(g)$\nfor the phase change $\\mathrm{OsO}_{4}(s) \\longrightarrow \\mathrm{OsO}_{4}(g) \\quad \\Delta H=56.4 \\mathrm{~kJ}$\n63. Calculate $\\Delta \\boldsymbol{H}^{\\circ}$ for the process\n$\\mathrm{Sb}(s)+\\frac{5}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{5}(s)$\nfrom the following information:\n$\\mathrm{Sb}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{3}(s) \\quad \\Delta H^{\\circ}=-314 \\mathrm{~kJ}$\n$\\mathrm{SbCl}_{3}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{5}(s) \\quad \\Delta H^{\\circ}=-80 \\mathrm{~kJ}$\n64. Calculate $\\Delta H^{\\circ}$ for the process $\\mathrm{Zn}(s)+\\mathrm{S}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{ZnSO}_{4}(s)$\nfrom the following information:\n$\\mathrm{Zn}(s)+\\mathrm{S}(s) \\longrightarrow \\mathrm{ZnS}(s) \\quad \\Delta H^{\\circ}=-206.0 \\mathrm{~kJ}$\n$\\mathrm{ZnS}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{ZnSO}_{4}(s) \\quad \\Delta H^{\\circ}=-776.8 \\mathrm{~kJ}$\n65. Calculate $\\Delta H$ for the process $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{Cl}_{2}(g)$ from the following information:"} {"id": "textbook_3092", "title": "1044. Exercises - 1044.3. Enthalpy", "content": "$$\n\\begin{array}{lr}\n\\mathrm{Hg}(l)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{HgCl}_{2}(s) & \\Delta H=-224 \\mathrm{~kJ} \\\\\n\\mathrm{Hg}(l)+\\mathrm{HgCl}_{2}(s) \\longrightarrow \\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) & \\Delta H=-41.2 \\mathrm{~kJ}\n\\end{array}\n$$"} {"id": "textbook_3093", "title": "1044. Exercises - 1044.3. Enthalpy", "content": "66. Calculate $\\Delta H^{\\circ}$ for the process $\\mathrm{Co}_{3} \\mathrm{O}_{4}(s) \\longrightarrow 3 \\mathrm{Co}(s)+2 \\mathrm{O}_{2}(g)$ from the following information:\n\n$$\n\\begin{array}{lr}\n\\mathrm{Co}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CoO}(s) & \\Delta H^{\\circ}=-237.9 \\mathrm{~kJ} \\\\\n3 \\mathrm{CoO}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Co}_{3} \\mathrm{O}_{4}(s) & \\Delta H^{\\circ}=-177.5 \\mathrm{~kJ}\n\\end{array}\n$$"} {"id": "textbook_3094", "title": "1044. Exercises - 1044.3. Enthalpy", "content": "67. Calculate the standard molar enthalpy of formation of $\\mathrm{NO}(g)$ from the following data:\n$\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H^{\\circ}=66.4 \\mathrm{~kJ}$\n$2 \\mathrm{NO}(g)+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H^{\\circ}=-114.1 \\mathrm{~kJ}$\n68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:\n(a) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$\n(b) $\\mathrm{Si}(s)+2 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SiCl}_{4}(g)$\n(c) $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(d) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:\n(a) $\\mathrm{Si}(s)+2 \\mathrm{~F}_{2}(g) \\longrightarrow \\mathrm{SiF}_{4}(g)$\n(b) $2 \\mathrm{C}(s)+2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(l)$\n(c) $\\mathrm{CH}_{4}(g)+\\mathrm{N}_{2}(g) \\longrightarrow \\mathrm{HCN}(g)+\\mathrm{NH}_{3}(g)$;\n(d) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g)$\n70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.\n(a) $2 \\mathrm{Ag}_{2} \\mathrm{O}(s) \\longrightarrow 4 \\mathrm{Ag}(s)+\\mathrm{O}_{2}(g)$\n(b) $\\mathrm{SnO}(s)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{Sn}(s)+\\mathrm{CO}_{2}(g)$\n(c) $\\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{Cr}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(d) $2 \\mathrm{Al}(s)+\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s) \\longrightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{Fe}(s)$\n71. The decomposition of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ under standard conditions.\n$2 \\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{l}) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{O}_{2}(g)$\n72. Calculate the enthalpy of combustion of propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$, for the formation of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ and $\\mathrm{CO}_{2}(\\mathrm{~g})$. The enthalpy of formation of propane is $-104 \\mathrm{~kJ} / \\mathrm{mol}$.\n73. Calculate the enthalpy of combustion of butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}(g)$ for the formation of $\\mathrm{H}_{2} \\mathrm{O}(g)$ and $\\mathrm{CO}_{2}(g)$. The enthalpy of formation of butane is $-126 \\mathrm{~kJ} / \\mathrm{mol}$.\n74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned?\n75. The white pigment $\\mathrm{TiO}_{2}$ is prepared by the reaction of titanium tetrachloride, $\\mathrm{TiCl}_{4}$, with water vapor in the gas phase: $\\mathrm{TiCl}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{TiO}_{2}(s)+4 \\mathrm{HCl}(g)$.\nHow much heat is evolved in the production of exactly 1 mole of $\\mathrm{TiO}_{2}(s)$ under standard state conditions?\n76. Water gas, a mixture of $\\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: $\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{H}_{2}(g)$.\n(a) Assuming that coke has the same enthalpy of formation as graphite, calculate $\\Delta H^{\\circ}$ for this reaction.\n(b) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst:\n$2 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH}(g)$.\nUnder the conditions of the reaction, methanol forms as a gas. Calculate $\\Delta H^{\\circ}$ for this reaction and for the condensation of gaseous methanol to liquid methanol.\n(c) Calculate the heat of combustion of 1 mole of liquid methanol to $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ and $\\mathrm{CO}_{2}(\\mathrm{~g})$.\n77. In the early days of automobiles, illumination at night was provided by burning acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, $\\mathrm{CaC}_{2}$ :\n$\\mathrm{CaC}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s)+\\mathrm{C}_{2} \\mathrm{H}_{2}(g)$.\nCalculate the standard enthalpy of the reaction. The $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{CaC}_{2}$ is $-15.14 \\mathrm{kcal} / \\mathrm{mol}$.\n78. From the data in Table 9.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: $\\mathrm{CO}(g), \\mathrm{CH}_{4}(g)$, or $\\mathrm{C}_{2} \\mathrm{H}_{2}(g)$.\n79. The enthalpy of combustion of hard coal averages $-35 \\mathrm{~kJ} / \\mathrm{g}$, that of gasoline, $1.28 \\times 10^{5} \\mathrm{~kJ} / \\mathrm{gal}$. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is $0.692 \\mathrm{~g} / \\mathrm{mL}$ (the same as the density of isooctane).\n80. Ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is used as a fuel for motor vehicles, particularly in Brazil.\n(a) Write the balanced equation for the combustion of ethanol to $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.\n(b) The density of ethanol is $0.7893 \\mathrm{~g} / \\mathrm{mL}$. Calculate the enthalpy of combustion of exactly 1 L of ethanol.\n(c) Assuming that an automobile's mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n -octane, $\\mathrm{C}_{8} \\mathrm{H}_{18}$\n$\\left(\\Delta H_{\\mathrm{f}}^{\\circ}=-208.4 \\mathrm{~kJ} / \\mathrm{mol}\\right.$; density $\\left.=0.7025 \\mathrm{~g} / \\mathrm{mL}\\right)$.\n81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane $\\left[\\mathrm{B}_{2} \\mathrm{H}_{6}\\right.$, produces $\\mathrm{B}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$, methane $\\left[\\mathrm{CH}_{4}\\right.$, produces $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$, and hydrazine [ $\\mathrm{N}_{2} \\mathrm{H}_{4}$, produces $\\mathrm{N}_{2}(\\mathrm{~g})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{B}_{2} \\mathrm{H}_{6}(\\mathrm{~g}), \\mathrm{CH}_{4}(\\mathrm{~g})$, and $\\mathrm{N}_{2} \\mathrm{H}_{4}(I)$ may be found in Appendix G .\n82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?\n83. Ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About $80 \\%$ of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst.\n$\\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)$\nUsing the data in the table in Appendix G, calculate $\\Delta H^{\\circ}$ for the reaction.\n84. The oxidation of the sugar glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, is described by the following equation:\n$\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)+6 \\mathrm{O}_{2}(g) \\longrightarrow 6 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-2816 \\mathrm{~kJ}$\nThe metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.\n(a) How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?\n(b) How many Calories can be produced by the metabolism of 1.0 g of glucose?\n85. Propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}$, is a hydrocarbon that is commonly used as a fuel.\n(a) Write a balanced equation for the complete combustion of propane gas.\n(b) Calculate the volume of air at $25^{\\circ} \\mathrm{C}$ and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent $\\mathrm{O}_{2}$ by volume. (Hint: We will see how to do this calculation in a later chapter on gases-for now use the information that 1.00 L of air at $25^{\\circ} \\mathrm{C}$ and 1.00 atm contains 0.275 g of $\\mathrm{O}_{2}$.)\n(c) The heat of combustion of propane is $-2,219.2 \\mathrm{~kJ} / \\mathrm{mol}$. Calculate the heat of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ of propane given that $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})=-285.8 \\mathrm{~kJ} / \\mathrm{mol}$ and $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{CO}_{2}(\\mathrm{~g})=-393.5 \\mathrm{~kJ} / \\mathrm{mol}$.\n(d) Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00\nkilograms of water, calculate the increase in temperature of the water.\n86. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with $89 \\%$ efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).\n(a) Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was $56^{\\circ} \\mathrm{F}$; at this temperature and a pressure of 1 atm, natural gas has a density of $0.681 \\mathrm{~g} / \\mathrm{L}$.\n(b) How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [ $\\mathrm{C}_{3} \\mathrm{H}_{8}$ : density, $0.5318 \\mathrm{~g} / \\mathrm{mL}$; enthalpy of combustion, $2219 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of $\\mathrm{CO}_{2}(g)$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I}]$ and the furnace used to burn the LPG has the same efficiency as the gas furnace.\n(c) What mass of carbon dioxide is produced by combustion of the methane used to heat the house?\n(d) What mass of water is produced by combustion of the methane used to heat the house?\n(e) What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains $23 \\%$ oxygen by mass. The average density of air during the month was $1.22 \\mathrm{~g} / \\mathrm{L}$. (f) How many kilowatt-hours ( $1 \\mathrm{kWh}=3.6 \\times 10^{6} \\mathrm{~J}$ ) of electricity would be required to provide the heat necessary to heat the house? Note electricity is $100 \\%$ efficient in producing heat inside a house.\n(g) Although electricity is $100 \\%$ efficient in producing heat inside a house, production and distribution of electricity is not $100 \\%$ efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about $40 \\%$. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is $40 \\%$ ?\n"} {"id": "textbook_3095", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "\n87. Which bond in each of the following pairs of bonds is the strongest?\n(a) $\\mathrm{C}-\\mathrm{C}$ or $\\mathrm{C}=\\mathrm{C}$\n(b) $\\mathrm{C}-\\mathrm{N}$ or $\\mathrm{C} \\equiv \\mathrm{N}$\n(c) $\\mathrm{C} \\equiv \\mathrm{O}$ or $\\mathrm{C}=\\mathrm{O}$\n(d) $\\mathrm{H}-\\mathrm{F}$ or $\\mathrm{H}-\\mathrm{Cl}$\n(e) $\\mathrm{C}-\\mathrm{H}$ or $\\mathrm{O}-\\mathrm{H}$\n(f) $\\mathrm{C}-\\mathrm{N}$ or $\\mathrm{C}-\\mathrm{O}$\n88. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:\n(a) $\\mathrm{H}_{2}(g)+\\mathrm{Br}_{2}(g) \\longrightarrow 2 \\mathrm{HBr}(g)$\n(b) $\\mathrm{CH}_{4}(g)+\\mathrm{I}_{2}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{I}(g)+\\mathrm{HI}(g)$\n(c) $\\mathrm{C}_{2} \\mathrm{H}_{4}(g)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n89. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:\n(a) $\\mathrm{Cl}_{2}(g)+3 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{ClF}_{3}(g)$\n(b) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}(g)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{H}_{3} \\mathrm{CCH}_{3}(g)$\n(c) $2 \\mathrm{C}_{2} \\mathrm{H}_{6}(g)+7 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n90. Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.\n(a) Use the bond energy found in Table 9.3 and Table 9.4 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)\n(b) Use the enthalpy of reaction and the bond energies for $\\mathrm{H}_{2}$ and $\\mathrm{Cl}_{2}$ to solve for the energy of one mole of HCl bonds."} {"id": "textbook_3096", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{HCl}(g) \\quad \\Delta H_{\\mathrm{rxn}}^{\\circ}=-184.7 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3097", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "91. Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.\n92. When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:\n
mol) |\n| :---: | :---: | :---: |\n| K | 419 | 3050 |\n| Ca | 590 | 1140 |"} {"id": "textbook_3101", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "(d) The first ionization energy of Mg is $738 \\mathrm{~kJ} / \\mathrm{mol}$ and that of Al is $578 \\mathrm{~kJ} / \\mathrm{mol}$. Account for this difference."} {"id": "textbook_3102", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "[^8]101. The lattice energy of LiF is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, and the Li-F distance is 200.8 pm . NaF crystallizes in the same structure as LiF but with a $\\mathrm{Na}-\\mathrm{F}$ distance of 231 pm . Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175 , or $4090 \\mathrm{~kJ} / \\mathrm{mol}$ ? Explain your choice.\n102. For which of the following substances is the least energy required to convert one mole of the solid into separate ions?\n(a) MgO\n(b) SrO\n(c) KF\n(d) CsF\n(e) $\\mathrm{MgF}_{2}$\n103. The reaction of a metal, $M$, with a halogen, $X_{2}$, proceeds by an exothermic reaction as indicated by this equation: $\\mathrm{M}(s)+\\mathrm{X}_{2}(g) \\longrightarrow \\mathrm{MX}_{2}(s)$. For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.\n(a) a large radius vs. a small radius for $\\mathrm{M}^{+2}$\n(b) a high ionization energy vs. a low ionization energy for $M$\n(c) an increasing bond energy for the halogen\n(d) a decreasing electron affinity for the halogen\n(e) an increasing size of the anion formed by the halogen\n104. The lattice energy of LiF is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, and the Li-F distance is 201 pm . MgO crystallizes in the same structure as LiF but with a $\\mathrm{Mg}-\\mathrm{O}$ distance of 205 pm . Which of the following values most closely approximates the lattice energy of MgO: $256 \\mathrm{~kJ} / \\mathrm{mol}, 512 \\mathrm{~kJ} / \\mathrm{mol}, 1023 \\mathrm{~kJ} / \\mathrm{mol}, 2046 \\mathrm{~kJ} / \\mathrm{mol}$, or $4008 \\mathrm{~kJ} /$ mol? Explain your choice.\n105. Which compound in each of the following pairs has the larger lattice energy? Note: $\\mathrm{Mg}^{2+}$ and $\\mathrm{Li}^{+}$have similar radii; $\\mathrm{O}^{2-}$ and $\\mathrm{F}^{-}$have similar radii. Explain your choices.\n(a) MgO or MgSe\n(b) LiF or MgO\n(c) $\\mathrm{Li}_{2} \\mathrm{O}$ or LiCl\n(d) $\\mathrm{Li}_{2} \\mathrm{Se}$ or MgO\n106. Which compound in each of the following pairs has the larger lattice energy? Note: $\\mathrm{Ba}^{2+}$ and $\\mathrm{K}^{+}$have similar radii; $\\mathrm{S}^{2-}$ and $\\mathrm{Cl}^{-}$have similar radii. Explain your choices.\n(a) $\\mathrm{K}_{2} \\mathrm{O}$ or $\\mathrm{Na}_{2} \\mathrm{O}$\n(b) $\\mathrm{K}_{2} \\mathrm{~S}$ or BaS\n(c) KCl or BaS\n(d) BaS or $\\mathrm{BaCl}_{2}$\n107. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n(a) MgO\n(b) SrO\n(c) KF\n(d) CsF\n(e) $\\mathrm{MgF}_{2}$\n108. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n(a) $\\mathrm{K}_{2} \\mathrm{~S}$\n(b) $\\mathrm{K}_{2} \\mathrm{O}$\n(c) CaS\n(d) $\\mathrm{Cs}_{2} \\mathrm{~S}$\n(e) CaO\n109. The lattice energy of $K F$ is $794 \\mathrm{~kJ} / \\mathrm{mol}$, and the interionic distance is 269 pm . The $\\mathrm{Na}-\\mathrm{F}$ distance in NaF, which has the same structure as KF, is 231 pm . Which of the following values is the closest approximation of the lattice energy of NaF: $682 \\mathrm{~kJ} / \\mathrm{mol}, 794 \\mathrm{~kJ} / \\mathrm{mol}, 924 \\mathrm{~kJ} / \\mathrm{mol}, 1588 \\mathrm{~kJ} / \\mathrm{mol}$, or 3175 kJ/mol? Explain your answer."} {"id": "textbook_3103", "title": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds", "content": "474 9\u2022Exercises\n"} {"id": "textbook_3104", "title": "1045. CHAPTER 10
Liquids and Solids - ", "content": "\n"} {"id": "textbook_3105", "title": "1045. CHAPTER 10
Liquids and Solids - ", "content": "Figure 10.1 Solid carbon dioxide (\"dry ice\", left) sublimes vigorously when placed in a liquid (right), cooling the liquid and generating a dense mist of water above the cylinder. (credit: modification of work by Paul Flowers)\n"} {"id": "textbook_3106", "title": "1046. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_3107", "title": "1046. CHAPTER OUTLINE - 1046.1. Intermolecular Forces", "content": "\n10.2 Properties of Liquids\n10.3 Phase Transitions\n10.4 Phase Diagrams\n10.5 The Solid State of Matter\n10.6 Lattice Structures in Crystalline Solids"} {"id": "textbook_3108", "title": "1046. CHAPTER OUTLINE - 1046.1. Intermolecular Forces", "content": "INTRODUCTION Leprosy has been a devastating disease throughout much of human history. Aside from the symptoms and complications of the illness, its social stigma led sufferers to be cast out of communities and isolated in colonies; in some regions this practice lasted well into the twentieth century. At that time, the best potential treatment for leprosy was oil from the chaulmoogra tree, but the oil was extremely thick, causing blisters and making usage painful and ineffective. Healthcare professionals seeking a better application contacted Alice Ball, a young chemist at the University of Hawaii, who had focused her masters thesis on a similar plant. Ball initiated a sequence of procedures (repeated acidification and purification to change the characteristics of the oil and isolate the active substances (esters, discussed later in this text). The \"Ball Method\" as it later came to be called, became the standard treatment for leprosy for decades. In the liquid and solid states, atomic and molecular interactions are of considerable strength and play an important role in determining a number of physical properties of the substance. For example, the thickness, or viscosity, of the chaulmoogra oil was due to its intermolecular forces. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.\n"} {"id": "textbook_3109", "title": "1046. CHAPTER OUTLINE - 1046.2. Intermolecular Forces", "content": ""} {"id": "textbook_3110", "title": "1047. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3111", "title": "1047. LEARNING OBJECTIVES - ", "content": "- Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)\n- Identify the types of intermolecular forces experienced by specific molecules based on their structures\n- Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state"} {"id": "textbook_3112", "title": "1047. LEARNING OBJECTIVES - ", "content": "As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase \"intermolecular attraction\" to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions."} {"id": "textbook_3113", "title": "1047. LEARNING OBJECTIVES - ", "content": "Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:"} {"id": "textbook_3114", "title": "1047. LEARNING OBJECTIVES - ", "content": "- Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.\n- Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide."} {"id": "textbook_3115", "title": "1047. LEARNING OBJECTIVES - ", "content": "The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles' KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure 10.2 illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE , of a given substance.\n"} {"id": "textbook_3116", "title": "1047. LEARNING OBJECTIVES - ", "content": "FIGURE 10.2 Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)"} {"id": "textbook_3117", "title": "1047. LEARNING OBJECTIVES - ", "content": "As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between $\\mathrm{H}_{2} \\mathrm{O}$ molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid $\\mathrm{H}_{2} \\mathrm{O}$. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure 10.3.\n"} {"id": "textbook_3118", "title": "1047. LEARNING OBJECTIVES - ", "content": "FIGURE 10.3 Condensation forms when water vapor in the air is cooled enough to form liquid water, such as (a) on the outside of a cold beverage glass or (b) in the form of fog. (credit a: modification of work by Jenny Downing; credit b: modification of work by Cory Zanker)"} {"id": "textbook_3119", "title": "1047. LEARNING OBJECTIVES - ", "content": "We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter's fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure 10.4.\n\n\nFIGURE 10.4 Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state. (credit: modification of work by \"Sam-Cat\"/Flickr)"} {"id": "textbook_3120", "title": "1047. LEARNING OBJECTIVES - ", "content": "Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter.\n"} {"id": "textbook_3121", "title": "1048. LINK TO LEARNING - ", "content": "\nAccess this interactive simulation (http://openstax.org/l/16phetvisual) on states of matter, phase transitions, and intermolecular forces. This simulation is useful for visualizing concepts introduced throughout this chapter.\n"} {"id": "textbook_3122", "title": "1049. Forces between Molecules - ", "content": "\nUnder appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure 10.5 illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy-430 kilojoules.\n"} {"id": "textbook_3123", "title": "1049. Forces between Molecules - ", "content": "FIGURE 10.5 Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance's properties."} {"id": "textbook_3124", "title": "1049. Forces between Molecules - ", "content": "All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.\n"} {"id": "textbook_3125", "title": "1050. Dispersion Forces - ", "content": "\nOne of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron's location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species-a so-called dispersion force like that illustrated in Figure 10.6.\n"} {"id": "textbook_3126", "title": "1050. Dispersion Forces - ", "content": "FIGURE 10.6 Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules."} {"id": "textbook_3127", "title": "1050. Dispersion Forces - ", "content": "Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close."} {"id": "textbook_3128", "title": "1050. Dispersion Forces - ", "content": "Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. $\\mathrm{F}_{2}$ and $\\mathrm{Cl}_{2}$ are gases at room temperature (reflecting weaker attractive forces); $\\mathrm{Br}_{2}$ is a liquid, and $\\mathrm{I}_{2}$ is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 10.1."} {"id": "textbook_3129", "title": "1050. Dispersion Forces - ", "content": "Melting and Boiling Points of the Halogens"} {"id": "textbook_3130", "title": "1050. Dispersion Forces - ", "content": "| Halogen | Molar Mass | Atomic Radius | Melting Point | Boiling Point |\n| :---: | :---: | :---: | :---: | :---: |\n| fluorine, $\\mathrm{F}_{2}$ | $38 \\mathrm{~g} / \\mathrm{mol}$ | 72 pm | 53 K | 85 K |\n| chlorine, $\\mathrm{Cl}_{2}$ | $71 \\mathrm{~g} / \\mathrm{mol}$ | 99 pm | 172 K | 238 K |\n| bromine, $\\mathrm{Br}_{2}$ | $160 \\mathrm{~g} / \\mathrm{mol}$ | 114 pm | 266 K | 332 K |\n| iodine, $\\mathrm{I}_{2}$ | $254 \\mathrm{~g} / \\mathrm{mol}$ | 133 pm | 387 K | 457 K |\n| astatine, $\\mathrm{At}_{2}$ | $420 \\mathrm{~g} / \\mathrm{mol}$ | 150 pm | 575 K | 610 K |"} {"id": "textbook_3131", "title": "1050. Dispersion Forces - ", "content": "TABLE 10.1"} {"id": "textbook_3132", "title": "1050. Dispersion Forces - ", "content": "The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule's charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.\n"} {"id": "textbook_3133", "title": "1051. EXAMPLE 10.1 - ", "content": ""} {"id": "textbook_3134", "title": "1052. London Forces and Their Effects - ", "content": "\nOrder the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: $\\mathrm{CH}_{4}$, $\\mathrm{SiH}_{4}, \\mathrm{GeH}_{4}$, and $\\mathrm{SnH}_{4}$. Explain your reasoning.\n"} {"id": "textbook_3135", "title": "1053. Solution - ", "content": "\nApplying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of $\\mathrm{CH}_{4}, \\mathrm{SiH}_{4}, \\mathrm{GeH}_{4}$, and $\\mathrm{SnH}_{4}$ are approximately $16 \\mathrm{~g} / \\mathrm{mol}, 32 \\mathrm{~g} / \\mathrm{mol}, 77 \\mathrm{~g} /$ mol, and $123 \\mathrm{~g} / \\mathrm{mol}$, respectively. Therefore, $\\mathrm{CH}_{4}$ is expected to have the lowest boiling point and $\\mathrm{SnH}_{4}$ the highest boiling point. The ordering from lowest to highest boiling point is expected to be $\\mathrm{CH}_{4}<\\mathrm{SiH}_{4}<\\mathrm{GeH}_{4}<$ $\\mathrm{SnH}_{4}$."} {"id": "textbook_3136", "title": "1053. Solution - ", "content": "A graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:\n\n"} {"id": "textbook_3137", "title": "1054. Check Your Learning - ", "content": "\nOrder the following hydrocarbons from lowest to highest boiling point: $\\mathrm{C}_{2} \\mathrm{H}_{6}, \\mathrm{C}_{3} \\mathrm{H}_{8}$, and $\\mathrm{C}_{4} \\mathrm{H}_{10}$.\n"} {"id": "textbook_3138", "title": "1055. Answer: - ", "content": "\n$\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}<\\mathrm{C}_{4} \\mathrm{H}_{10}$. All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore $\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}<\\mathrm{C}_{4} \\mathrm{H}_{10}$."} {"id": "textbook_3139", "title": "1055. Answer: - ", "content": "The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers $n$-pentane, isopentane, and neopentane (shown in Figure 10.7) are $36^{\\circ} \\mathrm{C}, 27^{\\circ} \\mathrm{C}$, and $9.5^{\\circ} \\mathrm{C}$, respectively. Even though these compounds are composed of molecules with the same chemical formula, $\\mathrm{C}_{5} \\mathrm{H}_{12}$, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for $n$-pentane and least for neopentane. The elongated shape of $n$-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip's contact, the stronger the connection.\n\n\nFIGURE 10.7 The strength of the dispersion forces increases with the contact area between molecules, as demonstrated by the boiling points of these pentane isomers.\n"} {"id": "textbook_3140", "title": "1056. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3141", "title": "1057. Geckos and Intermolecular Forces - ", "content": "\nGeckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos' feet to behave this way.\n\nGeckos' toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure 10.8, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces-weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko's weight."} {"id": "textbook_3142", "title": "1057. Geckos and Intermolecular Forces - ", "content": "In 2014, two scientists developed a model to explain how geckos can rapidly transition from \"sticky\" to \"non-sticky.\" Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos' feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Later research led by Alyssa Stark at University of Akron showed that geckos can maintain their hold on hydrophobic surfaces (similar to the leaves in their habitats) equally well whether the surfaces were wet or dry. Stark's experiment used a ribbon to gently pull the geckos until they slipped, so that the researchers could determine the geckos' ability to hold various surfaces under wet and dry conditions. Further investigations may eventually lead to the development of better adhesives and other applications.\n"} {"id": "textbook_3143", "title": "1057. Geckos and Intermolecular Forces - ", "content": "FIGURE 10.8 Geckos' toes contain large numbers of tiny hairs (setae), which branch into many triangular tips (spatulae). Geckos adhere to surfaces because of van der Waals attractions between the surface and a gecko's millions of spatulae. By changing how the spatulae contact the surface, geckos can turn their stickiness \"on\" and \"off.\" (credit photo: modification of work by \"JC*+A!\"/Flickr)\n"} {"id": "textbook_3144", "title": "1058. LINK TO LEARNING - ", "content": "\nWatch this video (http://openstax.org/l/16kellaraut) to learn more about Kellar Autumn's research that determined that van der Waals forces are responsible for a gecko's ability to cling and climb.\n"} {"id": "textbook_3145", "title": "1059. Dipole-Dipole Attractions - ", "content": "\nRecall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule-a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl . In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction-the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure 10.9.\n\n\nFIGURE 10.9 This image shows two arrangements of polar molecules, such as HCl , that allow an attraction between the partial negative end of one molecule and the partial positive end of another."} {"id": "textbook_3146", "title": "1059. Dipole-Dipole Attractions - ", "content": "The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar $\\mathrm{F}_{2}$ molecules. Both HCl and $\\mathrm{F}_{2}$ consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K , molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to \"stick together\" to form a liquid, whereas the relatively weaker dispersion forces between nonpolar $\\mathrm{F}_{2}$ molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl ( 188 K ) compared to $\\mathrm{F}_{2}(85 \\mathrm{~K})$ is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar $\\mathrm{F}_{2}$ molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.\n"} {"id": "textbook_3147", "title": "1060. Dipole-Dipole Forces and Their Effects - ", "content": "\nPredict which will have the higher boiling point: $\\mathrm{N}_{2}$ or CO. Explain your reasoning.\n"} {"id": "textbook_3148", "title": "1061. Solution - ", "content": "\nCO and $\\mathrm{N}_{2}$ are both diatomic molecules with masses of about 28 amu , so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because $\\mathrm{N}_{2}$ is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar $\\mathrm{N}_{2}$ molecules, so CO is expected to have the higher boiling point.\n"} {"id": "textbook_3149", "title": "1062. Check Your Learning - ", "content": "\nPredict which will have the higher boiling point: ICl or $\\mathrm{Br}_{2}$. Explain your reasoning.\n"} {"id": "textbook_3150", "title": "1063. Answer: - ", "content": "\nICl. ICl and $\\mathrm{Br}_{2}$ have similar masses ( $\\sim 160 \\mathrm{amu}$ ) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; $\\mathrm{Br}_{2}$ is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.\n"} {"id": "textbook_3151", "title": "1064. Hydrogen Bonding - ", "content": "\nNitrosyl fluoride (ONF, molecular mass 49 amu ) is a gas at room temperature. Water ( $\\mathrm{H}_{2} \\mathrm{O}$, molecular mass 18 amu ) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded ( 4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of $\\mathrm{F}, \\mathrm{O}$, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with $\\mathrm{F}-\\mathrm{H}, \\mathrm{O}-\\mathrm{H}$, or $\\mathrm{N}-\\mathrm{H}$ moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF $\\cdots \\mathrm{HF}, \\mathrm{H}_{2} \\mathrm{O} \\cdots \\mathrm{HOH}$, and $\\mathrm{H}_{3} \\mathrm{~N} \\cdots \\mathrm{HNH}_{2}$, in which the hydrogen bonds are denoted by dots. Figure 10.10 illustrates hydrogen bonding between water molecules.\n"} {"id": "textbook_3152", "title": "1064. Hydrogen Bonding - ", "content": "FIGURE 10.10 Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules."} {"id": "textbook_3153", "title": "1064. Hydrogen Bonding - ", "content": "Despite use of the word \"bond,\" keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to $10 \\%$ as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces."} {"id": "textbook_3154", "title": "1064. Hydrogen Bonding - ", "content": "Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group $15\\left(\\mathrm{NH}_{3}, \\mathrm{PH}_{3}, \\mathrm{AsH}_{3}\\right.$, and $\\left.\\mathrm{SbH}_{3}\\right)$, group 16 hydrides $\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{~S}, \\mathrm{H}_{2} \\mathrm{Se}\\right.$, and $\\mathrm{H}_{2} \\mathrm{Te}$ ), and group 17 hydrides ( $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$, and HI ). The boiling points of the heaviest three hydrides for each group are plotted in Figure 10.11. As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.\n"} {"id": "textbook_3155", "title": "1064. Hydrogen Bonding - ", "content": "FIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3,4 , and 5 ."} {"id": "textbook_3156", "title": "1064. Hydrogen Bonding - ", "content": "If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect $\\mathrm{NH}_{3}$ to boil at about $-120^{\\circ} \\mathrm{C}, \\mathrm{H}_{2} \\mathrm{O}$ to boil at about $-80^{\\circ} \\mathrm{C}$, and HF to boil at about $-110^{\\circ} \\mathrm{C}$. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure 10.12. The stark contrast between our na\u00efve predictions and reality provides compelling evidence for the strength of hydrogen bonding.\n"} {"id": "textbook_3157", "title": "1064. Hydrogen Bonding - ", "content": "FIGURE 10.12 In comparison to periods 3-5, the binary hydrides of period 2 elements in groups 17, 16 and 15 (F, O and N , respectively) exhibit anomalously high boiling points due to hydrogen bonding.\n"} {"id": "textbook_3158", "title": "1065. EXAMPLE 10.3 - ", "content": ""} {"id": "textbook_3159", "title": "1066. Effect of Hydrogen Bonding on Boiling Points - ", "content": "\nConsider the compounds dimethylether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$, and propane $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right)$. Their boiling points, not necessarily in order, are $-42.1^{\\circ} \\mathrm{C},-24.8^{\\circ} \\mathrm{C}$, and $78.4^{\\circ} \\mathrm{C}$. Match each compound with its boiling point. Explain your reasoning.\n"} {"id": "textbook_3160", "title": "1067. Solution - ", "content": "\nThe VSEPR-predicted shapes of $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}, \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, and $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ are similar, as are their molar masses ( $46 \\mathrm{~g} / \\mathrm{mol}, 46 \\mathrm{~g} / \\mathrm{mol}$, and $44 \\mathrm{~g} / \\mathrm{mol}$, respectively), so they will exhibit similar dispersion forces. Since $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ is nonpolar, it may exhibit only dispersion forces. Because $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ is polar, it will also experience dipole-dipole attractions. Finally, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ has an - OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}<\\mathrm{CH}_{3} \\mathrm{OCH}_{3}<\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$. The boiling point of propane is -42.1 ${ }^{\\circ} \\mathrm{C}$, the boiling point of dimethylether is $-24.8^{\\circ} \\mathrm{C}$, and the boiling point of ethanol is $78.5^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_3161", "title": "1068. Check Your Learning - ", "content": "\nEthane $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{3}\\right)$ has a melting point of $-183^{\\circ} \\mathrm{C}$ and a boiling point of $-89^{\\circ} \\mathrm{C}$. Predict the melting and boiling points for methylamine $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{2}\\right)$. Explain your reasoning.\n"} {"id": "textbook_3162", "title": "1069. Answer: - ", "content": "\nThe melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ and $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ are similar in size and mass, but methylamine possesses an - NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of $-93^{\\circ} \\mathrm{C}$ and a boiling point of $-6^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_3163", "title": "1070. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_3164", "title": "1071. Hydrogen Bonding and DNA - ", "content": "\nDeoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism's characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism's offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure 10.13.\n"} {"id": "textbook_3165", "title": "1071. Hydrogen Bonding and DNA - ", "content": "FIGURE 10.13 Two separate DNA molecules form a double-stranded helix in which the molecules are held together via hydrogen bonding. (credit: modification of work by Jerome Walker, Dennis Myts)"} {"id": "textbook_3166", "title": "1071. Hydrogen Bonding and DNA - ", "content": "Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and $T$ share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure 10.14.\n"} {"id": "textbook_3167", "title": "1071. Hydrogen Bonding and DNA - ", "content": "FIGURE 10.14 The geometries of the base molecules result in maximum hydrogen bonding between adenine and thymine (AT) and between guanine and cytosine (GC), so-called \"complementary base pairs.\""} {"id": "textbook_3168", "title": "1071. Hydrogen Bonding and DNA - ", "content": "The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily \"unzip\" down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication.\n"} {"id": "textbook_3169", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_3170", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "- Distinguish between adhesive and cohesive forces\n- Define viscosity, surface tension, and capillary rise\n- Describe the roles of intermolecular attractive forces in each of these properties/phenomena"} {"id": "textbook_3171", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure 10.15, have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).\n"} {"id": "textbook_3172", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.15 (a) Honey and (b) motor oil are examples of liquids with high viscosities; they flow slowly. (credit a: modification of work by Scott Bauer; credit b: modification of work by David Nagy)"} {"id": "textbook_3173", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table 10.2 shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases."} {"id": "textbook_3174", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "Viscosities of Common Substances at $25^{\\circ} \\mathrm{C}$"} {"id": "textbook_3175", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "| Substance | Formula | Viscosity (mPa\u2022s) |\n| :---: | :---: | :---: |\n| water | $\\mathrm{H}_{2} \\mathrm{O}$ | 0.890 |\n| mercury | Hg | 1.526 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | 1.074 |\n| octane | $\\mathrm{C}_{8} \\mathrm{H}_{18}$ | 0.508 |\n| ethylene glycol | $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2}(\\mathrm{OH})$ | 16.1 |\n| honey | variable | $\\sim 2,000-10,000$ |\n| motor oil | variable | $\\sim 50-500$ |"} {"id": "textbook_3176", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "TABLE 10.2"} {"id": "textbook_3177", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "The various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface-that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure 10.16, because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.\n"} {"id": "textbook_3178", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.16 Attractive forces result in a spherical water drop that minimizes surface area; cohesive forces hold the sphere together; adhesive forces keep the drop attached to the web. (credit photo: modification of work by \"OliBac\"/Flickr)"} {"id": "textbook_3179", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table 10.3. Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively \"tough skin\" that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure 10.17, even though they are denser than water, move on its surface because they are supported by the surface tension."} {"id": "textbook_3180", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "Surface Tensions of Common Substances at $25^{\\circ} \\mathrm{C}$"} {"id": "textbook_3181", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "| Substance | Formula | Surface Tension (mN/m) |\n| :---: | :---: | :---: |\n| water | $\\mathrm{H}_{2} \\mathrm{O}$ | 71.99 |\n| mercury | Hg | 458.48 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | 21.97 |\n| octane | $\\mathrm{C}_{8} \\mathrm{H}_{18}$ | 21.14 |\n| ethylene glycol | $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2}(\\mathrm{OH})$ | 47.99 |"} {"id": "textbook_3182", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "TABLE 10.3\n"} {"id": "textbook_3183", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.17 Surface tension (right) prevents this insect, a \"water strider,\" from sinking into the water.\nSurface tension is affected by a variety of variables, including the introduction of additional substances on the surface. In the late 1800s, Agnes Pockels, who was initially blocked from pursuing a scientific career but\nstudied on her own, began investigating the impact and characteristics of soapy and greasy films in water. Using homemade materials, she developed an instrument known as a trough for measuring surface contaminants and their effects. With the support of renowned scientist Lord Rayleigh, her 1891 paper showed that surface contamination significantly reduces surface tension, and also that changing the characteristics of the surface (compressing or expanding it) also affects surface tension. Decades later, Irving Langmuir and Katharine Blodgett built on Pockels' work in their own trough and important advances in surface chemistry. Langmuir pioneered methods for producing single-molecule layers of film; Blodgett applied these to the development of non-reflective glass (critical for film-making and other applications), and also studied methods related to cleaning surfaces, which are important in semiconductor fabrication."} {"id": "textbook_3184", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "The IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not \"wet\" the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure 10.18).\n"} {"id": "textbook_3185", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.18 Differences in the relative strengths of cohesive and adhesive forces result in different meniscus shapes for mercury (left) and water (right) in glass tubes. (credit: Mark Ott)"} {"id": "textbook_3186", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "If you place one end of a paper towel in spilled wine, as shown in Figure 10.19, the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action-when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.\n"} {"id": "textbook_3187", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.19 Wine wicks up a paper towel (left) because of the strong attractions of water (and ethanol) molecules to the -OH groups on the towel's cellulose fibers and the strong attractions of water molecules to other water (and ethanol) molecules (right). (credit photo: modification of work by Mark Blaser)"} {"id": "textbook_3188", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many -OH groups. Water molecules are attracted to these -OH groups and form hydrogen bonds with them, which draws the $\\mathrm{H}_{2} \\mathrm{O}$ molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers."} {"id": "textbook_3189", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure 10.20. If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.\n"} {"id": "textbook_3190", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "FIGURE 10.20 Depending upon the relative strengths of adhesive and cohesive forces, a liquid may rise (such as water) or fall (such as mercury) in a glass capillary tube. The extent of the rise (or fall) is directly proportional to the surface tension of the liquid and inversely proportional to the density of the liquid and the radius of the tube."} {"id": "textbook_3191", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the\nfollowing equation:"} {"id": "textbook_3192", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "$$\nh=\\frac{2 T \\cos \\theta}{r \\rho g}\n$$"} {"id": "textbook_3193", "title": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids", "content": "In this equation, $h$ is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, $T$ is the surface tension of the liquid, $\\theta$ is the contact angle between the liquid and the tube, $r$ is the radius of the tube, $\\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity, $9.8 \\mathrm{~m} / \\mathrm{s}^{2}$. When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of $0^{\\circ}$. This is the situation for water rising in a glass tube.\n"} {"id": "textbook_3194", "title": "1072. EXAMPLE 10.4 - ", "content": ""} {"id": "textbook_3195", "title": "1073. Capillary Rise - ", "content": "\nAt $25^{\\circ} \\mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm ?\nFor water, $T=71.99 \\mathrm{mN} / \\mathrm{m}$ and $\\rho=1.0 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n"} {"id": "textbook_3196", "title": "1074. Solution - ", "content": "\nThe liquid will rise to a height $h$ given by: $h=\\frac{2 T \\cos \\theta}{r \\rho g}$\nThe Newton is defined as a $\\mathrm{kg} \\mathrm{m} / \\mathrm{s}^{2}$, and so the provided surface tension is equivalent to $0.07199 \\mathrm{~kg} / \\mathrm{s}^{2}$. The provided density must be converted into units that will cancel appropriately: $\\rho=1000 \\mathrm{~kg} / \\mathrm{m}^{3}$. The diameter of the tube in meters is 0.00025 m , so the radius is 0.000125 m . For a glass tube immersed in water, the contact angle is $\\theta=0^{\\circ}$, so $\\cos \\theta=1$. Finally, acceleration due to gravity on the earth is $g=9.8 \\mathrm{~m} / \\mathrm{s}^{2}$. Substituting these values into the equation, and cancelling units, we have:"} {"id": "textbook_3197", "title": "1074. Solution - ", "content": "$$\nh=\\frac{2\\left(0.07199 \\mathrm{~kg} / \\mathrm{s}^{2}\\right)}{(0.000125 \\mathrm{~m})\\left(1000 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)\\left(9.8 \\mathrm{~m} / \\mathrm{s}^{2}\\right)}=0.12 \\mathrm{~m}=12 \\mathrm{~cm}\n$$\n"} {"id": "textbook_3198", "title": "1075. Check Your Learning - ", "content": "\nWater rises in a glass capillary tube to a height of 8.4 cm . What is the diameter of the capillary tube?\n"} {"id": "textbook_3199", "title": "1076. Answer: - ", "content": "\ndiameter $=0.36 \\mathrm{~mm}$\n"} {"id": "textbook_3200", "title": "1077. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3201", "title": "1078. Biomedical Applications of Capillary Action - ", "content": "\nMany medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure 10.21. When your finger is pricked, a drop of blood forms and holds together due to surface tension-the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrowdiameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising.\n"} {"id": "textbook_3202", "title": "1078. Biomedical Applications of Capillary Action - ", "content": "FIGURE 10.21 Blood is collected for medical analysis by capillary action, which draws blood into a small diameter glass tube. (credit: modification of work by Centers for Disease Control and Prevention)\n"} {"id": "textbook_3203", "title": "1078. Biomedical Applications of Capillary Action - 1078.1. Phase Transitions", "content": ""} {"id": "textbook_3204", "title": "1079. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3205", "title": "1079. LEARNING OBJECTIVES - ", "content": "- Define phase transitions and phase transition temperatures\n- Explain the relation between phase transition temperatures and intermolecular attractive forces\n- Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes"} {"id": "textbook_3206", "title": "1079. LEARNING OBJECTIVES - ", "content": "We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth's water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.\n"} {"id": "textbook_3207", "title": "1080. Vaporization and Condensation - ", "content": "\nWhen a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid's vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 10.22 , and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.\n\n(a)\n\n(b)\n\n(c)"} {"id": "textbook_3208", "title": "1080. Vaporization and Condensation - ", "content": "FIGURE 10.22 In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from\nthe liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue."} {"id": "textbook_3209", "title": "1080. Vaporization and Condensation - ", "content": "The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring \"recapture\" of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.\n"} {"id": "textbook_3210", "title": "1081. EXAMPLE 10.5 - ", "content": ""} {"id": "textbook_3211", "title": "1082. Explaining Vapor Pressure in Terms of IMFs - ", "content": "\nGiven the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:\n\n"} {"id": "textbook_3212", "title": "1083. Solution - ", "content": "\nDiethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two -OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.\n"} {"id": "textbook_3213", "title": "1084. Check Your Learning - ", "content": "\nAt $20^{\\circ} \\mathrm{C}$, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:"} {"id": "textbook_3214", "title": "1084. Check Your Learning - ", "content": "| Compound | methanol $\\mathrm{CH}_{3} \\mathrm{OH}$ | ethanol $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | propanol $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}$ | butanol $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{OH}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| Vapor Pressure at $20^{\\circ} \\mathrm{C}$ | 11.9 kPa | 5.95 kPa | 2.67 kPa | 0.56 kPa |\n"} {"id": "textbook_3215", "title": "1085. Answer: - ", "content": "\nAll these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:\n$\\mathrm{P}_{\\text {methanol }}>\\mathrm{P}_{\\text {ethanol }}>\\mathrm{P}_{\\text {propanol }}>\\mathrm{P}_{\\text {butanol }}$."} {"id": "textbook_3216", "title": "1085. Answer: - ", "content": "As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic\nenergies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 10.23. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.\n"} {"id": "textbook_3217", "title": "1085. Answer: - ", "content": "FIGURE 10.23 Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, $K E$, to escape from the liquid into the gas phase.\n"} {"id": "textbook_3218", "title": "1086. Boiling Points - ", "content": "\nWhen the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth's atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to $1 \\mathrm{~atm}(101.3 \\mathrm{kPa})$. Figure 10.24 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid's boiling point on surrounding pressure.\n\n\nFIGURE 10.24 The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)\n"} {"id": "textbook_3219", "title": "1087. EXAMPLE 10.6 - ", "content": ""} {"id": "textbook_3220", "title": "1088. A Boiling Point at Reduced Pressure - ", "content": "\nA typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa . Use the graph in Figure 10.24 to determine the boiling point of water at this elevation.\n"} {"id": "textbook_3221", "title": "1089. Solution - ", "content": "\nThe graph of the vapor pressure of water versus temperature in Figure 10.24 indicates that the vapor pressure of water is 68 kPa at about $90^{\\circ} \\mathrm{C}$. Thus, at about $90^{\\circ} \\mathrm{C}$, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.\n"} {"id": "textbook_3222", "title": "1090. Check Your Learning - ", "content": "\nThe boiling point of ethyl ether was measured to be $10^{\\circ} \\mathrm{C}$ at a base camp on the slopes of Mount Everest. Use Figure 10.24 to determine the approximate atmospheric pressure at the camp.\n"} {"id": "textbook_3223", "title": "1091. Answer: - ", "content": "\nApproximately 40 kPa (0.4 atm)"} {"id": "textbook_3224", "title": "1091. Answer: - ", "content": "The quantitative relation between a substance's vapor pressure and its temperature is described by the Clausius-Clapeyron equation:"} {"id": "textbook_3225", "title": "1091. Answer: - ", "content": "$$\nP=A e^{-\\Delta H_{\\mathrm{vap}} / R T}\n$$"} {"id": "textbook_3226", "title": "1091. Answer: - ", "content": "where $\\Delta H_{\\text {vap }}$ is the enthalpy of vaporization for the liquid, $R$ is the gas constant, and $A$ is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:"} {"id": "textbook_3227", "title": "1091. Answer: - ", "content": "$$\n\\ln P=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T}+\\ln A\n$$"} {"id": "textbook_3228", "title": "1091. Answer: - ", "content": "This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature $T_{1}$, the vapor pressure is $\\mathrm{P}_{1}$, and at temperature $\\mathrm{T}_{2}$, the vapor pressure is $\\mathrm{P}_{2}$, the corresponding linear equations are:"} {"id": "textbook_3229", "title": "1091. Answer: - ", "content": "$$\n\\ln P_{1}=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{1}}+\\ln A \\quad \\text { and } \\quad \\ln P_{2}=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{2}}+\\ln A\n$$"} {"id": "textbook_3230", "title": "1091. Answer: - ", "content": "Since the constant, $A$, is the same, these two equations may be rearranged to isolate $\\ln A$ and then set them equal to one another:"} {"id": "textbook_3231", "title": "1091. Answer: - ", "content": "$$\n\\ln P_{1}+\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{1}}=\\ln P_{2}+\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{2}}\n$$"} {"id": "textbook_3232", "title": "1091. Answer: - ", "content": "which can be combined into:"} {"id": "textbook_3233", "title": "1091. Answer: - ", "content": "$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$\n"} {"id": "textbook_3234", "title": "1092. EXAMPLE 10.7 - ", "content": ""} {"id": "textbook_3235", "title": "1093. Estimating Enthalpy of Vaporization - ", "content": "\nIsooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At $34.0^{\\circ} \\mathrm{C}$, the vapor pressure of isooctane is 10.0 kPa , and at $98.8^{\\circ} \\mathrm{C}$, its vapor pressure is 100.0 kPa . Use this information to estimate the enthalpy of vaporization for isooctane.\n"} {"id": "textbook_3236", "title": "1094. Solution - ", "content": "\nThe enthalpy of vaporization, $\\Delta H_{\\text {vap }}$, can be determined by using the Clausius-Clapeyron equation:"} {"id": "textbook_3237", "title": "1094. Solution - ", "content": "$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$"} {"id": "textbook_3238", "title": "1094. Solution - ", "content": "Since we have two vapor pressure-temperature values ( $T_{1}=34.0^{\\circ} \\mathrm{C}=307.2 \\mathrm{~K}, P_{1}=10.0 \\mathrm{kPa}$ and $T_{2}=98.8^{\\circ} \\mathrm{C}=$ $372.0 \\mathrm{~K}, P_{2}=100 \\mathrm{kPa}$ ), we can substitute them into this equation and solve for $\\Delta H_{\\text {vap }}$. Rearranging the Clausius-Clapeyron equation and solving for $\\Delta H_{\\text {vap }}$ yields:"} {"id": "textbook_3239", "title": "1094. Solution - ", "content": "$$\n\\Delta H_{\\mathrm{vap}}=\\frac{R \\cdot \\ln \\left(\\frac{P_{2}}{P_{1}}\\right)}{\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)}=\\frac{(8.3145 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K}) \\cdot \\ln \\left(\\frac{100 \\mathrm{kPa}}{10.0 \\mathrm{kPa}}\\right)}{\\left(\\frac{1}{307.2 \\mathrm{~K}}-\\frac{1}{372.0 \\mathrm{~K}}\\right)}=33,800 \\mathrm{~J} / \\mathrm{mol}=33.8 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3240", "title": "1094. Solution - ", "content": "Note that the pressure can be in any units, so long as they agree for both $P$ values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.\n"} {"id": "textbook_3241", "title": "1095. Check Your Learning - ", "content": "\nAt $20.0^{\\circ} \\mathrm{C}$, the vapor pressure of ethanol is 5.95 kPa , and at $63.5^{\\circ} \\mathrm{C}$, its vapor pressure is 53.3 kPa . Use this information to estimate the enthalpy of vaporization for ethanol.\n"} {"id": "textbook_3242", "title": "1096. Answer: - ", "content": "\n$41,360 \\mathrm{~J} / \\mathrm{mol}$ or $41.4 \\mathrm{~kJ} / \\mathrm{mol}$\n"} {"id": "textbook_3243", "title": "1097. EXAMPLE 10.8 - ", "content": ""} {"id": "textbook_3244", "title": "1098. Estimating Temperature (or Vapor Pressure) - ", "content": "\nFor benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{6}\\right)$, the normal boiling point is $80.1^{\\circ} \\mathrm{C}$ and the enthalpy of vaporization is $30.8 \\mathrm{~kJ} / \\mathrm{mol}$. What is the boiling point of benzene in Denver, where atmospheric pressure $=83.4 \\mathrm{kPa}$ ?\n"} {"id": "textbook_3245", "title": "1099. Solution - ", "content": "\nIf the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, $\\Delta H_{\\text {vap, }}$, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:"} {"id": "textbook_3246", "title": "1099. Solution - ", "content": "$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$"} {"id": "textbook_3247", "title": "1099. Solution - ", "content": "Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value ( $T_{1}=80.1^{\\circ} \\mathrm{C}=353.3 \\mathrm{~K}, P_{1}=101.3 \\mathrm{kPa}, \\Delta H_{\\text {vap }}=30.8$ $\\mathrm{kJ} / \\mathrm{mol}$ ) and want to find the temperature ( $T_{2}$ ) that corresponds to vapor pressure $P_{2}=83.4 \\mathrm{kPa}$. We can substitute these values into the Clausius-Clapeyron equation and then solve for $T_{2}$. Rearranging the ClausiusClapeyron equation and solving for $T_{2}$ yields:\n$T_{2}=\\left(\\frac{-R \\cdot \\ln \\left(\\frac{P_{2}}{P_{1}}\\right)}{\\Delta H_{\\text {vap }}}+\\frac{1}{T_{1}}\\right)^{-1}=\\left(\\frac{-(8.3145 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}) \\cdot \\ln \\left(\\frac{83.4 \\mathrm{kPa}}{101.3 \\mathrm{kPa}}\\right)}{30,800 \\mathrm{~J} / \\mathrm{mol}}+\\frac{1}{353.3 \\mathrm{~K}}\\right)^{-1}=346.9 \\mathrm{~K}$ or $73.8^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3248", "title": "1100. Check Your Learning - ", "content": "\nFor acetone $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CO}$, the normal boiling point is $56.5^{\\circ} \\mathrm{C}$ and the enthalpy of vaporization is $31.3 \\mathrm{~kJ} / \\mathrm{mol}$. What is the vapor pressure of acetone at $25.0^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3249", "title": "1101. Answer: - ", "content": "\n30.1 kPa\n"} {"id": "textbook_3250", "title": "1102. Enthalpy of Vaporization - ", "content": "\nVaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or\na shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, $\\Delta H_{\\text {vap }}$. For example, the vaporization of water at standard temperature is represented by:"} {"id": "textbook_3251", "title": "1102. Enthalpy of Vaporization - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H_{\\mathrm{vap}}=44.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3252", "title": "1102. Enthalpy of Vaporization - ", "content": "As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:"} {"id": "textbook_3253", "title": "1102. Enthalpy of Vaporization - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H_{\\mathrm{con}}=-\\Delta H_{\\mathrm{vap}}=-44.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n"} {"id": "textbook_3254", "title": "1103. EXAMPLE 10.9 - ", "content": ""} {"id": "textbook_3255", "title": "1104. Using Enthalpy of Vaporization - ", "content": "\nOne way our body is cooled is by evaporation of the water in sweat (Figure 10.25). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water ( 1.5 kg ) at $T=37{ }^{\\circ} \\mathrm{C}$ (normal body temperature); $\\Delta H_{\\text {vap }}=43.46 \\mathrm{~kJ} / \\mathrm{mol}$ at $37^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_3256", "title": "1104. Using Enthalpy of Vaporization - ", "content": "FIGURE 10.25 Evaporation of sweat helps cool the body. (credit: \"Kullez\"/Flickr)\n"} {"id": "textbook_3257", "title": "1105. Solution - ", "content": "\nWe start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:"} {"id": "textbook_3258", "title": "1105. Solution - ", "content": "$$\n1.5 \\mathrm{~L} \\times \\frac{1000 \\mathrm{~g}}{1 \\mathrm{~g}} \\times \\frac{1 \\mathrm{mot}}{18 \\frac{\\mathrm{~g}}{\\mathrm{o}}} \\times \\frac{43.46 \\mathrm{~kJ}}{1 \\mathrm{mot}}=3.6 \\times 10^{3} \\mathrm{~kJ}\n$$"} {"id": "textbook_3259", "title": "1105. Solution - ", "content": "Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.\n"} {"id": "textbook_3260", "title": "1106. Check Your Learning - ", "content": "\nHow much heat is required to evaporate 100.0 g of liquid ammonia, $\\mathrm{NH}_{3}$, at its boiling point if its enthalpy of vaporization is $4.8 \\mathrm{~kJ} / \\mathrm{mol}$ ?\n"} {"id": "textbook_3261", "title": "1107. Answer: - ", "content": "\n28 kJ\n"} {"id": "textbook_3262", "title": "1108. Melting and Freezing - ", "content": "\nWhen we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued\nheating increase the temperature of the liquid (Figure 10.26).\n"} {"id": "textbook_3263", "title": "1108. Melting and Freezing - ", "content": "FIGURE 10.26 (a) This beaker of ice has a temperature of $-12.0^{\\circ} \\mathrm{C}$. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to $0^{\\circ} \\mathrm{C}$. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still $0^{\\circ} \\mathrm{C}$. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to $22.2^{\\circ} \\mathrm{C}$. (credit: modification of work by Mark Ott)"} {"id": "textbook_3264", "title": "1108. Melting and Freezing - ", "content": "If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing)."} {"id": "textbook_3265", "title": "1108. Melting and Freezing - ", "content": "The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures."} {"id": "textbook_3266", "title": "1108. Melting and Freezing - ", "content": "The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, $\\Delta \\mathrm{H}_{\\text {fus }}$ of the substance. The enthalpy of fusion of ice is $6.0 \\mathrm{~kJ} / \\mathrm{mol}$ at $0^{\\circ} \\mathrm{C}$. Fusion (melting) is an endothermic process:"} {"id": "textbook_3267", "title": "1108. Melting and Freezing - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H_{\\text {fus }}=6.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3268", "title": "1108. Melting and Freezing - ", "content": "The reciprocal process, freezing, is an exothermic process whose enthalpy change is $-6.0 \\mathrm{~kJ} / \\mathrm{mol}$ at $0^{\\circ} \\mathrm{C}$ :"} {"id": "textbook_3269", "title": "1108. Melting and Freezing - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(s)\n$$"} {"id": "textbook_3270", "title": "1108. Melting and Freezing - ", "content": "$$\n\\Delta H_{\\mathrm{frz}}=-\\Delta H_{\\mathrm{fus}}=-6.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n"} {"id": "textbook_3271", "title": "1109. Sublimation and Deposition - ", "content": "\nSome solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid $\\mathrm{CO}_{2}$ ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 10.27). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.\n"} {"id": "textbook_3272", "title": "1109. Sublimation and Deposition - ", "content": "FIGURE 10.27 Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)"} {"id": "textbook_3273", "title": "1109. Sublimation and Deposition - ", "content": "Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, $\\Delta \\mathrm{H}_{\\text {sub }}$, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:"} {"id": "textbook_3274", "title": "1109. Sublimation and Deposition - ", "content": "$$\n\\mathrm{CO}_{2}(s) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H_{\\text {sub }}=26.1 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3275", "title": "1109. Sublimation and Deposition - ", "content": "Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:"} {"id": "textbook_3276", "title": "1109. Sublimation and Deposition - ", "content": "$$\n\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(s) \\quad \\Delta H_{\\mathrm{dep}}=-\\Delta H_{\\mathrm{sub}}=-26.1 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_3277", "title": "1109. Sublimation and Deposition - ", "content": "Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess's Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 10.28. For example:\n"} {"id": "textbook_3278", "title": "1109. Sublimation and Deposition - ", "content": "FIGURE 10.28 For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.\n"} {"id": "textbook_3279", "title": "1110. Heating and Cooling Curves - ", "content": "\nIn the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, $q$, and its accompanying temperature change, $\\Delta T$, was introduced:"} {"id": "textbook_3280", "title": "1110. Heating and Cooling Curves - ", "content": "$$\nq=m c \\Delta T\n$$"} {"id": "textbook_3281", "title": "1110. Heating and Cooling Curves - ", "content": "where $m$ is the mass of the substance and $c$ is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure 10.29 shows a typical heating curve."} {"id": "textbook_3282", "title": "1110. Heating and Cooling Curves - ", "content": "Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water's temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.\n"} {"id": "textbook_3283", "title": "1110. Heating and Cooling Curves - ", "content": "FIGURE 10.29 A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.\n"} {"id": "textbook_3284", "title": "1111. EXAMPLE 10.10 - ", "content": "\nTotal Heat Needed to Change Temperature and Phase for a Substance\nHow much heat is required to convert 135 g of ice at $-15^{\\circ} \\mathrm{C}$ into water vapor at $120^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3285", "title": "1112. Solution - ", "content": "\nThe transition described involves the following steps:"} {"id": "textbook_3286", "title": "1112. Solution - ", "content": "1. Heat ice from $-15^{\\circ} \\mathrm{C}$ to $0^{\\circ} \\mathrm{C}$\n2. Melt ice\n3. Heat water from $0^{\\circ} \\mathrm{C}$ to $100^{\\circ} \\mathrm{C}$\n4. Boil water\n5. Heat steam from $100^{\\circ} \\mathrm{C}$ to $120^{\\circ} \\mathrm{C}$"} {"id": "textbook_3287", "title": "1112. Solution - ", "content": "The heat needed to change the temperature of a given substance (with no change in phase) is: $q=m \\times c \\times \\Delta T$ (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by $q=$ $n \\times \\Delta H$."} {"id": "textbook_3288", "title": "1112. Solution - ", "content": "Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:"} {"id": "textbook_3289", "title": "1112. Solution - ", "content": "$$\nq_{\\text {total }}=(m \\cdot c \\cdot \\Delta T)_{\\text {ice }}+n \\cdot \\Delta H_{\\text {fus }}+(m \\cdot c \\cdot \\Delta T)_{\\text {water }}+n \\cdot \\Delta H_{\\mathrm{vap}}+(m \\cdot c \\cdot \\Delta T)_{\\text {steam }}\n$$"} {"id": "textbook_3290", "title": "1112. Solution - ", "content": "$$\n\\begin{aligned}\n& =\\left(135 \\mathrm{~g} \\cdot 2.09 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 15^{\\circ} \\mathrm{C}\\right)+\\left(135 \\cdot \\frac{1 \\mathrm{~mol}}{18.02 \\mathrm{~g}} \\cdot 6.01 \\mathrm{~kJ} / \\mathrm{mol}\\right) \\\\\n& +\\left(135 \\mathrm{~g} \\cdot 4.18 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 100^{\\circ} \\mathrm{C}\\right)+\\left(135 \\mathrm{~g} \\cdot \\frac{1 \\mathrm{~mol}}{18.02 \\mathrm{~g}} \\cdot 40.67 \\mathrm{~kJ} / \\mathrm{mol}\\right) \\\\\n& +\\left(135 \\mathrm{~g} \\cdot 1.84 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 20^{\\circ} \\mathrm{C}\\right) \\\\\n& =4230 \\mathrm{~J}+45.0 \\mathrm{~kJ}+56,500 \\mathrm{~J}+305 \\mathrm{~kJ}+4970 \\mathrm{~J}\n\\end{aligned}\n$$"} {"id": "textbook_3291", "title": "1112. Solution - ", "content": "Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:"} {"id": "textbook_3292", "title": "1112. Solution - ", "content": "$$\n=4.23 \\mathrm{~kJ}+45.0 \\mathrm{~kJ}+56.5 \\mathrm{~kJ}+305 \\mathrm{~kJ}+4.97 \\mathrm{~kJ}=416 \\mathrm{~kJ}\n$$\n"} {"id": "textbook_3293", "title": "1113. Check Your Learning - ", "content": "\nWhat is the total amount of heat released when 94.0 g water at $80.0^{\\circ} \\mathrm{C}$ cools to form ice at $-30.0^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3294", "title": "1114. Answer: - ", "content": "\n68.7 kJ\n"} {"id": "textbook_3295", "title": "1114. Answer: - 1114.1. Phase Diagrams", "content": ""} {"id": "textbook_3296", "title": "1115. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3297", "title": "1115. LEARNING OBJECTIVES - ", "content": "- Explain the construction and use of a typical phase diagram\n- Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties\n- Describe the supercritical fluid phase of matter"} {"id": "textbook_3298", "title": "1115. LEARNING OBJECTIVES - ", "content": "In the previous module, the variation of a liquid's equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance's melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure 10.30.\n"} {"id": "textbook_3299", "title": "1115. LEARNING OBJECTIVES - ", "content": "FIGURE 10.30 The physical state of a substance and its phase-transition temperatures are represented graphically in a phase diagram."} {"id": "textbook_3300", "title": "1115. LEARNING OBJECTIVES - ", "content": "To illustrate the utility of these plots, consider the phase diagram for water shown in Figure 10.31.\n"} {"id": "textbook_3301", "title": "1115. LEARNING OBJECTIVES - ", "content": "FIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties."} {"id": "textbook_3302", "title": "1115. LEARNING OBJECTIVES - ", "content": "We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of $-10^{\\circ} \\mathrm{C}$ correspond to the region of the diagram labeled \"ice.\" Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of $50^{\\circ} \\mathrm{C}$ correspond to the \"water\" region-here, water exists only as a liquid. At 25 kPa and $200^{\\circ} \\mathrm{C}$, water exists only in the gaseous state. Note that on the $\\mathrm{H}_{2} \\mathrm{O}$ phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here."} {"id": "textbook_3303", "title": "1115. LEARNING OBJECTIVES - ", "content": "The curve BC in Figure 10.31 is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This \"liquid-vapor\" curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm , the boiling point is $100^{\\circ} \\mathrm{C}$. Notice that the liquid-vapor curve terminates at a temperature of $374^{\\circ} \\mathrm{C}$ and a pressure of 218 atm ,\nindicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module."} {"id": "textbook_3304", "title": "1115. LEARNING OBJECTIVES - ", "content": "The solid-vapor curve, labeled AB in Figure 10.31, indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure 10.31, we would see that ice has a vapor pressure of about 0.20 kPa at $-10^{\\circ} \\mathrm{C}$. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa , ice will sublime. This is the basis for the \"freeze-drying\" process often used to preserve foods, such as the ice cream shown in Figure 10.32.\n"} {"id": "textbook_3305", "title": "1115. LEARNING OBJECTIVES - ", "content": "FIGURE 10.32 Freeze-dried foods, like this ice cream, are dehydrated by sublimation at pressures below the triple point for water. (credit: \"lwao\"/Flickr)"} {"id": "textbook_3306", "title": "1115. LEARNING OBJECTIVES - ", "content": "The solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure 10.33. The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.\n"} {"id": "textbook_3307", "title": "1115. LEARNING OBJECTIVES - ", "content": "FIGURE 10.33 The immense pressures beneath glaciers result in partial melting to produce a layer of water that provides lubrication to assist glacial movement. This satellite photograph shows the advancing edge of the Perito Moreno glacier in Argentina. (credit: NASA)"} {"id": "textbook_3308", "title": "1115. LEARNING OBJECTIVES - ", "content": "The point of intersection of all three curves is labeled B in Figure 10.31. At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data\npair is called the triple point. At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature.\n"} {"id": "textbook_3309", "title": "1116. EXAMPLE 10.11 - ", "content": ""} {"id": "textbook_3310", "title": "1117. Determining the State of Water - ", "content": "\nUsing the phase diagram for water given in Figure 10.31, determine the state of water at the following temperatures and pressures:\n(a) $-10^{\\circ} \\mathrm{C}$ and 50 kPa\n(b) $25^{\\circ} \\mathrm{C}$ and 90 kPa\n(c) $50^{\\circ} \\mathrm{C}$ and 40 kPa\n(d) $80^{\\circ} \\mathrm{C}$ and 5 kPa\n(e) $-10^{\\circ} \\mathrm{C}$ and 0.3 kPa\n(f) $50^{\\circ} \\mathrm{C}$ and 0.3 kPa\n"} {"id": "textbook_3311", "title": "1118. Solution - ", "content": "\nUsing the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas.\n"} {"id": "textbook_3312", "title": "1119. Check Your Learning - ", "content": "\nWhat phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa ? If the pressure is held at 50 kPa ?\n"} {"id": "textbook_3313", "title": "1120. Answer: - ", "content": "\nAt $0.3 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow \\mathrm{g}$ at $-58^{\\circ} \\mathrm{C}$. At $50 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow 1$ at $0^{\\circ} \\mathrm{C}, \\mathrm{l} \\longrightarrow \\mathrm{g}$ at $78^{\\circ} \\mathrm{C}$"} {"id": "textbook_3314", "title": "1120. Answer: - ", "content": "Consider the phase diagram for carbon dioxide shown in Figure 10.34 as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for $\\mathrm{CO}_{2}$ increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous $\\mathrm{CO}_{2}$. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water.\n"} {"id": "textbook_3315", "title": "1120. Answer: - ", "content": "FIGURE 10.34 A phase diagram for carbon dioxide is shown. The pressure axis is plotted on a logarithmic scale to accommodate the large range of values.\n"} {"id": "textbook_3316", "title": "1121. Determining the State of Carbon Dioxide - ", "content": "\nUsing the phase diagram for carbon dioxide shown in Figure 10.34, determine the state of $\\mathrm{CO}_{2}$ at the following temperatures and pressures:\n(a) $-30^{\\circ} \\mathrm{C}$ and 2000 kPa\n(b) $-90^{\\circ} \\mathrm{C}$ and 1000 kPa\n(c) $-60^{\\circ} \\mathrm{C}$ and 100 kPa\n(d) $-40^{\\circ} \\mathrm{C}$ and 1500 kPa\n(e) $0^{\\circ} \\mathrm{C}$ and 100 kPa\n(f) $20^{\\circ} \\mathrm{C}$ and 100 kPa\n"} {"id": "textbook_3317", "title": "1122. Solution - ", "content": "\nUsing the phase diagram for carbon dioxide provided, we can determine that the state of $\\mathrm{CO}_{2}$ at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas.\n"} {"id": "textbook_3318", "title": "1123. Check Your Learning - ", "content": "\nIdentify the phase changes that carbon dioxide will undergo as its temperature is increased from $-100^{\\circ} \\mathrm{C}$ while holding its pressure constant at 1500 kPa . At 50 kPa . At what approximate temperatures do these phase changes occur?\n"} {"id": "textbook_3319", "title": "1124. Answer: - ", "content": "\nat $1500 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow 1$ at $-55^{\\circ} \\mathrm{C}, 1 \\longrightarrow \\mathrm{~g}$ at $-10^{\\circ} \\mathrm{C}$;\nat $50 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow \\mathrm{g}$ at $-60^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3320", "title": "1125. Supercritical Fluids - ", "content": "\nIf we place a sample of water in a sealed container at $25^{\\circ} \\mathrm{C}$, remove the air, and let the vaporizationcondensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm . A distinct boundary between the more dense liquid and the less dense gas is clearly\nobserved. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure 10.31), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of $374^{\\circ} \\mathrm{C}$, the vapor pressure has risen to 218 atm , and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (Figure 10.35). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in the following table."} {"id": "textbook_3321", "title": "1125. Supercritical Fluids - ", "content": "| Substance | Critical Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Critical Pressure (kPa) |\n| :--- | :--- | :--- |\n| hydrogen | -240.0 | 1300 |\n| nitrogen | -147.2 | 3400 |\n| oxygen | -118.9 | 5000 |\n| carbon dioxide | 31.1 | 7400 |\n| ammonia | 132.4 | 11,300 |\n| sulfur dioxide | 157.2 | 7800 |\n| water | 374.0 | 22,000 |"} {"id": "textbook_3322", "title": "1125. Supercritical Fluids - ", "content": ""} {"id": "textbook_3323", "title": "1125. Supercritical Fluids - ", "content": "FIGURE 10.35 (a) A sealed container of liquid carbon dioxide slightly below its critical point is heated, resulting in (b) the formation of the supercritical fluid phase. Cooling the supercritical fluid lowers its temperature and pressure below the critical point, resulting in the reestablishment of separate liquid and gaseous phases (c and d). Colored floats illustrate differences in density between the liquid, gaseous, and supercritical fluid states. (credit: modification of work by \"mrmrobin\"/YouTube)\n"} {"id": "textbook_3324", "title": "1126. LINK TO LEARNING - ", "content": "\nObserve the liquid-to-supercritical fluid transition (http://openstax.org/l/16supercrit) for carbon dioxide."} {"id": "textbook_3325", "title": "1126. LINK TO LEARNING - ", "content": "Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus\noils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the $\\mathrm{CO}_{2}$ can be easily recovered by reducing the pressure and collecting the resulting gas.\n"} {"id": "textbook_3326", "title": "1127. EXAMPLE 10.13 - ", "content": ""} {"id": "textbook_3327", "title": "1128. The Critical Temperature of Carbon Dioxide - ", "content": "\nIf we shake a carbon dioxide fire extinguisher on a cool day ( $18{ }^{\\circ} \\mathrm{C}$ ), we can hear liquid $\\mathrm{CO}_{2}$ sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day ( $35^{\\circ} \\mathrm{C}$ ). Explain these observations.\n"} {"id": "textbook_3328", "title": "1129. Solution - ", "content": "\nOn the cool day, the temperature of the $\\mathrm{CO}_{2}$ is below the critical temperature of $\\mathrm{CO}_{2}, 304 \\mathrm{~K}$ or $31^{\\circ} \\mathrm{C}$, so liquid $\\mathrm{CO}_{2}$ is present in the cylinder. On the hot day, the temperature of the $\\mathrm{CO}_{2}$ is greater than its critical temperature of $31^{\\circ} \\mathrm{C}$. Above this temperature no amount of pressure can liquefy $\\mathrm{CO}_{2}$ so no liquid $\\mathrm{CO}_{2}$ exists in the fire extinguisher.\n"} {"id": "textbook_3329", "title": "1130. Check Your Learning - ", "content": "\nAmmonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?\n"} {"id": "textbook_3330", "title": "1131. Answer: - ", "content": "\nThe critical temperature of ammonia is 405.5 K , which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.\n"} {"id": "textbook_3331", "title": "1132. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3332", "title": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - ", "content": "\nCoffee is the world's second most widely traded commodity, following only petroleum. Across the globe, people love coffee's aroma and taste. Many of us also depend on one component of coffee-caffeine-to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee's stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening."} {"id": "textbook_3333", "title": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - ", "content": "Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee's taste and aroma also dissolve in $\\mathrm{H}_{2} \\mathrm{O}$, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane $\\left(\\mathrm{CH}_{2} \\mathrm{Cl}_{2}\\right)$ and ethyl acetate $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}\\right)$ have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee."} {"id": "textbook_3334", "title": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - ", "content": "Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure 10.36). At temperatures above 304.2 K and pressures above $7376 \\mathrm{kPa}, \\mathrm{CO}_{2}$ is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes $97-99 \\%$ of the caffeine, leaving coffee's flavor and aroma compounds intact. Because $\\mathrm{CO}_{2}$ is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs.\n"} {"id": "textbook_3335", "title": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - ", "content": "FIGURE 10.36 (a) Caffeine molecules have both polar and nonpolar regions, making it soluble in solvents of varying polarities. (b) The schematic shows a typical decaffeination process involving supercritical carbon dioxide.\n"} {"id": "textbook_3336", "title": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - 1133.1. The Solid State of Matter", "content": ""} {"id": "textbook_3337", "title": "1134. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3338", "title": "1134. LEARNING OBJECTIVES - ", "content": "- Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids\n- Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids\n- Explain the ways in which crystal defects can occur in a solid"} {"id": "textbook_3339", "title": "1134. LEARNING OBJECTIVES - ", "content": "When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure 10.37).\n"} {"id": "textbook_3340", "title": "1134. LEARNING OBJECTIVES - ", "content": "FIGURE 10.37 The entities of a solid phase may be arranged in a regular, repeating pattern (crystalline solids) or\nrandomly (amorphous).\nMetals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in Figure 10.38), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.\n\n(a)\n\n(b)"} {"id": "textbook_3341", "title": "1134. LEARNING OBJECTIVES - ", "content": "FIGURE 10.38 (a) Silicon dioxide, $\\mathrm{SiO}_{2}$, is abundant in nature as one of several crystalline forms of the mineral quartz. (b) Rapid cooling of molten $\\mathrm{SiO}_{2}$ yields an amorphous solid known as \"fused silica\"."} {"id": "textbook_3342", "title": "1134. LEARNING OBJECTIVES - ", "content": "Crystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.\n"} {"id": "textbook_3343", "title": "1135. Ionic Solids - ", "content": "\nIonic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure 10.39). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions-in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.\n"} {"id": "textbook_3344", "title": "1135. Ionic Solids - ", "content": "FIGURE 10.39 Sodium chloride is an ionic solid.\n"} {"id": "textbook_3345", "title": "1136. Metallic Solids - ", "content": "\nMetallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure 10.40. The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a \"sea\" of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their\nmalleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below $200^{\\circ} \\mathrm{C}$. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above $1000^{\\circ} \\mathrm{C}$. These differences reflect differences in strengths of metallic bonding among the metals.\n"} {"id": "textbook_3346", "title": "1136. Metallic Solids - ", "content": "FIGURE 10.40 Copper is a metallic solid.\n"} {"id": "textbook_3347", "title": "1137. Covalent Network Solid - ", "content": "\nCovalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure 10.41. To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above $3500^{\\circ} \\mathrm{C}$.\n\n\nFIGURE 10.41 A covalent crystal contains a three-dimensional network of covalent bonds, as illustrated by the structures of diamond, silicon dioxide, silicon carbide, and graphite. Graphite is an exceptional example, composed of planar sheets of covalent crystals that are held together in layers by noncovalent forces. Unlike typical covalent solids, graphite is very soft and electrically conductive.\n"} {"id": "textbook_3348", "title": "1138. Molecular Solid - ", "content": "\nMolecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure 10.42, are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as $\\mathrm{H}_{2}, \\mathrm{~N}_{2}, \\mathrm{O}_{2}$, and $\\mathrm{F}_{2}$, have weak attractive forces and form molecular solids with very low melting points (below $-200^{\\circ} \\mathrm{C}$ ). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar\nmolecules) melt at still higher temperatures. Examples include ice (melting point, $0^{\\circ} \\mathrm{C}$ ) and table sugar (melting point, $185^{\\circ} \\mathrm{C}$ ).\n\ncarbon dioxide\n\niodine\n\nFIGURE 10.42 Carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$ consists of small, nonpolar molecules and forms a molecular solid with a melting point of $-78^{\\circ} \\mathrm{C}$. Iodine $\\left(\\mathrm{I}_{2}\\right)$ consists of larger, nonpolar molecules and forms a molecular solid that melts at $114^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_3349", "title": "1139. Properties of Solids - ", "content": "\nA crystalline solid, like those listed in Table 10.4, has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures."} {"id": "textbook_3350", "title": "1139. Properties of Solids - ", "content": "Types of Crystalline Solids and Their Properties"} {"id": "textbook_3351", "title": "1139. Properties of Solids - ", "content": "| Type of Solid | Type of Particles | Type of Attractions | Properties | Examples |\n| :---: | :---: | :---: | :---: | :---: |\n| ionic | ions | ionic
bonds | hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points | $\\begin{aligned} & \\mathrm{NaCl}, \\\\ & \\mathrm{Al}_{2} \\mathrm{O}_{3} \\end{aligned}$ |\n| metallic | atoms of electropositive elements | metallic
bonds | shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature | $\\begin{aligned} & \\mathrm{Cu}, \\mathrm{Fe}, \\mathrm{Ti}, \\\\ & \\mathrm{~Pb}, \\mathrm{U} \\end{aligned}$ |\n| covalent network | atoms of electronegative elements | covalent bonds | very hard, not conductive, very high melting points | C
(diamond), $\\mathrm{SiO}_{2}, \\mathrm{SiC}$ |\n| molecular | molecules (or atoms) | IMFs | variable hardness, variable brittleness, not conductive, low melting points | $\\begin{aligned} & \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{CO}_{2}, \\\\ & \\mathrm{I}_{2}, \\\\ & \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11} \\end{aligned}$ |"} {"id": "textbook_3352", "title": "1139. Properties of Solids - ", "content": "TABLE 10.4\n"} {"id": "textbook_3353", "title": "1140. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_3354", "title": "1141. Graphene: Material of the Future - ", "content": "\nCarbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure 10.43. You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.\n"} {"id": "textbook_3355", "title": "1141. Graphene: Material of the Future - ", "content": "FIGURE 10.43 Diamond is extremely hard because of the strong bonding between carbon atoms in all directions. Graphite (in pencil lead) rubs off onto paper due to the weak attractions between the carbon layers. An image of a graphite surface shows the distance between the centers of adjacent carbon atoms. (credit left photo: modification of work by Steve Jurvetson; credit middle photo: modification of work by United States Geological Survey)"} {"id": "textbook_3356", "title": "1141. Graphene: Material of the Future - ", "content": "You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure 10.44, is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.\nGraphene sheet\n\n"} {"id": "textbook_3357", "title": "1141. Graphene: Material of the Future - ", "content": "Buckyball\n"} {"id": "textbook_3358", "title": "1141. Graphene: Material of the Future - ", "content": "Nanotube\n"} {"id": "textbook_3359", "title": "1141. Graphene: Material of the Future - ", "content": "Stacked sheets"} {"id": "textbook_3360", "title": "1141. Graphene: Material of the Future - ", "content": "FIGURE 10.44 Graphene sheets can be formed into buckyballs, nanotubes, and stacked layers.\n"} {"id": "textbook_3361", "title": "1142. Crystal Defects - ", "content": "\nIn a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure 10.45. Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.\n"} {"id": "textbook_3362", "title": "1142. Crystal Defects - ", "content": "FIGURE 10.45 Types of crystal defects include vacancies, interstitial atoms, and substitutions impurities.\n"} {"id": "textbook_3363", "title": "1142. Crystal Defects - 1142.1. Lattice Structures in Crystalline Solids", "content": ""} {"id": "textbook_3364", "title": "1143. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3365", "title": "1143. LEARNING OBJECTIVES - ", "content": "- Describe the arrangement of atoms and ions in crystalline structures\n- Compute ionic radii using unit cell dimensions\n- Explain the use of X-ray diffraction measurements in determining crystalline structures"} {"id": "textbook_3366", "title": "1143. LEARNING OBJECTIVES - ", "content": "Over $90 \\%$ of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally.\n"} {"id": "textbook_3367", "title": "1144. The Structures of Metals - ", "content": "\nWe will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow.\n"} {"id": "textbook_3368", "title": "1145. Unit Cells of Metals - ", "content": "\nThe structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure 10.46.\n\n\nFIGURE 10.46 A unit cell shows the locations of lattice points repeating in all directions.\nLet us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure 10.47. This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.\n"} {"id": "textbook_3369", "title": "1145. Unit Cells of Metals - ", "content": "FIGURE 10.47 When metal atoms are arranged with spheres in one layer directly above or below spheres in another layer, the lattice structure is called simple cubic. Note that the spheres are in contact."} {"id": "textbook_3370", "title": "1145. Unit Cells of Metals - ", "content": "In a simple cubic structure, the spheres are not packed as closely as they could be, and they only \"fill\" about $52 \\%$ of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure 10.48, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.\n"} {"id": "textbook_3371", "title": "1145. Unit Cells of Metals - ", "content": "FIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six."} {"id": "textbook_3372", "title": "1145. Unit Cells of Metals - ", "content": "In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure 10.49. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight \"corners,\" there is $8 \\times \\frac{1}{8}=1$ atom within one simple cubic unit cell.\n\n"} {"id": "textbook_3373", "title": "1145. Unit Cells of Metals - ", "content": "8 corners"} {"id": "textbook_3374", "title": "1145. Unit Cells of Metals - ", "content": "Simple cubic lattice cell\nFIGURE 10.49 A simple cubic lattice unit cell contains one-eighth of an atom at each of its eight corners, so it contains one atom total.\n"} {"id": "textbook_3375", "title": "1146. EXAMPLE 10.14 - ", "content": ""} {"id": "textbook_3376", "title": "1147. Calculation of Atomic Radius and Density for Metals, Part 1 - ", "content": "\nThe edge length of the unit cell of alpha polonium is 336 pm .\n(a) Determine the radius of a polonium atom.\n(b) Determine the density of alpha polonium.\n"} {"id": "textbook_3377", "title": "1148. Solution - ", "content": "\nAlpha polonium crystallizes in a simple cubic unit cell:\n\n(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $1=$ $2 r$. Therefore, the radius of Po is $r=\\frac{1}{2}=\\frac{336 \\mathrm{pm}}{2}=168 \\mathrm{pm}$.\n(b) Density is given by density $=\\frac{\\text { mass }}{\\text { volume }}$. The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom."} {"id": "textbook_3378", "title": "1148. Solution - ", "content": "The mass of a Po unit cell can be found by:"} {"id": "textbook_3379", "title": "1148. Solution - ", "content": "$$\n1 \\text { Po unit cell } \\times \\frac{1 \\text { Po atom }}{1 \\text { Po unit cell }} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Po}}{6.022 \\times 10^{23} \\text { Po atoms }} \\times \\frac{208.998 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Po}}=3.47 \\times 10^{-22} \\mathrm{~g}\n$$"} {"id": "textbook_3380", "title": "1148. Solution - ", "content": "The volume of a Po unit cell can be found by:"} {"id": "textbook_3381", "title": "1148. Solution - ", "content": "$$\nV=l^{3}=\\left(336 \\times 10^{-10} \\mathrm{~cm}\\right)^{3}=3.79 \\times 10^{-23} \\mathrm{~cm}^{3}\n$$"} {"id": "textbook_3382", "title": "1148. Solution - ", "content": "(Note that the edge length was converted from pm to cm to get the usual volume units for density.)\nTherefore, the density of $\\mathrm{Po}=\\frac{3.471 \\times 10^{-22} \\mathrm{~g}}{3.79 \\times 10^{-23} \\mathrm{~cm}^{3}}=9.16 \\mathrm{~g} / \\mathrm{cm}^{3}$\n"} {"id": "textbook_3383", "title": "1149. Check Your Learning - ", "content": "\nThe edge length of the unit cell for nickel is 0.3524 nm . The density of Ni is $8.90 \\mathrm{~g} / \\mathrm{cm}^{3}$. Does nickel crystallize in a simple cubic structure? Explain.\n"} {"id": "textbook_3384", "title": "1150. Answer: - ", "content": "\nNo. If Ni was simple cubic, its density would be given by:\n1 Ni atom $\\times \\frac{1 \\mathrm{~mol} \\mathrm{Ni}}{6.022 \\times 10^{23} \\mathrm{Ni} \\text { atoms }} \\times \\frac{58.693 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Ni}}=9.746 \\times 10^{-23} \\mathrm{~g}$\n$V=l^{3}=\\left(3.524 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}=4.376 \\times 10^{-23} \\mathrm{~cm}^{3}$\nThen the density of Ni would be $=\\frac{9.746 \\times 10^{-23} \\mathrm{~g}}{4.376 \\times 10^{-23} \\mathrm{~cm}^{3}}=2.23 \\mathrm{~g} / \\mathrm{cm}^{3}$\nSince the actual density of Ni is not close to this, Ni does not form a simple cubic structure."} {"id": "textbook_3385", "title": "1150. Answer: - ", "content": "Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell-all of which are illustrated in Figure 10.50. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)\n"} {"id": "textbook_3386", "title": "1151. Lattice point locations - ", "content": "\n"} {"id": "textbook_3387", "title": "1151. Lattice point locations - ", "content": "FIGURE 10.50 Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell."} {"id": "textbook_3388", "title": "1151. Lattice point locations - ", "content": "Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure 10.51. This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.\n\n\nBody-centered cubic structure\nFIGURE 10.51 In a body-centered cubic structure, atoms in a specific layer do not touch each other. Each atom touches four atoms in the layer above it and four atoms in the layer below it."} {"id": "textbook_3389", "title": "1151. Lattice point locations - ", "content": "Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about $68 \\%$ of the total volume. Isomorphous metals with a BCC structure include $\\mathrm{K}, \\mathrm{Ba}, \\mathrm{Cr}, \\mathrm{Mo}, \\mathrm{W}$, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)"} {"id": "textbook_3390", "title": "1151. Lattice point locations - ", "content": "Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure 10.52. This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ( $6 \\times \\frac{1}{2}=3$ atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.\n\n\nFace-centered cubic structure\nFIGURE 10.52 A face-centered cubic solid has atoms at the corners and, as the name implies, at the centers of the faces of its unit cells."} {"id": "textbook_3391", "title": "1151. Lattice point locations - ", "content": "Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying $74 \\%$ of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12 . The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure 10.53.\n"} {"id": "textbook_3392", "title": "1151. Lattice point locations - ", "content": "Side view\n\n\nTop view\n"} {"id": "textbook_3393", "title": "1151. Lattice point locations - ", "content": "Rotated view"} {"id": "textbook_3394", "title": "1151. Lattice point locations - ", "content": "Cubic closest packed structure\nFIGURE 10.53 A CCP arrangement consists of three repeating layers (ABCABC...) of hexagonally arranged atoms. Atoms in a CCP structure have a coordination number of 12 because they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below. By rotating our perspective, we can see that a CCP structure has a unit cell with a face containing an atom from layer A at one corner, atoms from layer B across a diagonal (at two corners and in the middle of the face), and an atom from layer C at the remaining corner. This is the same as a face-centered cubic arrangement.\n\nBecause closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure 10.54. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB $\\cdot \\cdot$ ). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type $A$, type $B$, and type $C$ close-packed layers (i.e., ABCABCABC\u2026). About two-thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include $\\mathrm{Cd}, \\mathrm{Co}, \\mathrm{Li}, \\mathrm{Mg}, \\mathrm{Na}$, and Zn , and metals that crystallize in a CCP structure include $\\mathrm{Ag}, \\mathrm{Al}, \\mathrm{Ca}$, $\\mathrm{Cu}, \\mathrm{Ni}, \\mathrm{Pb}$, and Pt .\n"} {"id": "textbook_3395", "title": "1151. Lattice point locations - ", "content": "FIGURE 10.54 In both types of closest packing, atoms are packed as compactly as possible. Hexagonal closest packing consists of two alternating layers (ABABAB...). Cubic closest packing consists of three alternating layers (ABCABCABC...).\n"} {"id": "textbook_3396", "title": "1152. EXAMPLE 10.15 - ", "content": ""} {"id": "textbook_3397", "title": "1153. Calculation of Atomic Radius and Density for Metals, Part 2 - ", "content": "\nCalcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm .\n(a) What is the atomic radius of Ca in this structure?\n(b) Calculate the density of Ca.\n"} {"id": "textbook_3398", "title": "1154. Solution - ", "content": "\n\n(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii $(\\mathrm{d}=4 r)$. Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:"} {"id": "textbook_3399", "title": "1154. Solution - ", "content": "$$\na^{2}+a^{2}=d^{2} \\longrightarrow(558.8 \\mathrm{pm})^{2}+(558.5 \\mathrm{pm})^{2}=(4 r)^{2}\n$$"} {"id": "textbook_3400", "title": "1154. Solution - ", "content": "Solving this gives $r=\\sqrt{\\frac{(558.8 \\mathrm{pm})^{2}+(558.5 \\mathrm{pm})^{2}}{16}}=197.6 \\mathrm{pm}$ for a Ca radius.\n(b) Density is given by density $=\\frac{\\text { mass }}{\\text { volume }}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A facecentered Ca unit cell has one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom $)$ and one-half of an atom on each of the six faces $6 \\times \\frac{1}{2}=3$ atoms), for a total of four atoms in the unit cell."} {"id": "textbook_3401", "title": "1154. Solution - ", "content": "The mass of the unit cell can be found by:"} {"id": "textbook_3402", "title": "1154. Solution - ", "content": "$$\n1 \\mathrm{Ca} \\text { unit cell } \\times \\frac{4 \\mathrm{Ca} \\text { atoms }}{1 \\mathrm{Ca} \\text { unit cell }} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Ca}}{6.022 \\times 10^{23} \\mathrm{Ca} \\text { atoms }} \\times \\frac{40.078 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Ca}}=2.662 \\times 10^{-22} \\mathrm{~g}\n$$"} {"id": "textbook_3403", "title": "1154. Solution - ", "content": "The volume of a Ca unit cell can be found by:"} {"id": "textbook_3404", "title": "1154. Solution - ", "content": "$$\nV=a^{3}=\\left(558.8 \\times 10^{-10} \\mathrm{~cm}\\right)^{3}=1.745 \\times 10^{-22} \\mathrm{~cm}^{3}\n$$"} {"id": "textbook_3405", "title": "1154. Solution - ", "content": "(Note that the edge length was converted from pm to cm to get the usual volume units for density.)\nThen, the density of $\\mathrm{Ca}=\\frac{2.662 \\times 10^{-22} \\mathrm{~g}}{1.745 \\times 10^{-22} \\mathrm{~cm}^{3}}=1.53 \\mathrm{~g} / \\mathrm{cm}^{3}$\n"} {"id": "textbook_3406", "title": "1155. Check Your Learning - ", "content": "\nSilver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm .\n(a) What is the atomic radius of Ag in this structure?\n(b) Calculate the density of Ag.\n"} {"id": "textbook_3407", "title": "1156. Answer: - ", "content": "\n(a) 144 pm ; (b) $10.5 \\mathrm{~g} / \\mathrm{cm}^{3}$"} {"id": "textbook_3408", "title": "1156. Answer: - ", "content": "In general, a unit cell is defined by the lengths of three axes ( $a, b$, and $c$ ) and the angles ( $\\alpha, \\beta$, and $\\gamma$ ) between them, as illustrated in Figure 10.55. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.\n\n\nFIGURE 10.55 A unit cell is defined by the lengths of its three axes ( $a, b$, and $c$ ) and the angles ( $\\alpha, \\beta$, and $\\gamma$ ) between the axes."} {"id": "textbook_3409", "title": "1156. Answer: - ", "content": "There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure 10.56."} {"id": "textbook_3410", "title": "1156. Answer: - ", "content": "| System/Axes/Angles | Unit Cells |\n| :---: | :---: |\n| Cubic $\\begin{gathered} a=b=c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple
Face-centered
Body-centered |\n| Tetragonal $\\begin{gathered} a=b \\neq c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple
Body-centered |\n| Orthorhombic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple
Body-centered
Base-centered
Face-centered |\n| Monoclinic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha=\\gamma=90^{\\circ} ; \\beta \\neq 90^{\\circ} \\end{gathered}$ | Simple
Base-centered |\n| Triclinic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha \\neq \\beta \\neq \\gamma \\neq 90^{\\circ} \\end{gathered}$ |  |\n| Hexagonal $\\begin{gathered} a=b \\neq c \\\\ \\alpha=\\beta=90^{\\circ} ; \\gamma=120^{\\circ} \\end{gathered}$ |  |\n| Rhombohedral $\\begin{gathered} a=b=c \\\\ \\alpha=\\beta=\\gamma \\neq 90^{\\circ} \\end{gathered}$ |  |"} {"id": "textbook_3411", "title": "1156. Answer: - ", "content": "FIGURE 10.56 There are seven different lattice systems and 14 different unit cells.\n"} {"id": "textbook_3412", "title": "1157. The Structures of Ionic Crystals - ", "content": "\nIonic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size."} {"id": "textbook_3413", "title": "1157. The Structures of Ionic Crystals - ", "content": "Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are\nsurrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound."} {"id": "textbook_3414", "title": "1157. The Structures of Ionic Crystals - ", "content": "In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. Figure 10.57 illustrates both of these types of holes.\n\n\nFIGURE 10.57 Cations may occupy two types of holes between anions: octahedral holes or tetrahedral holes.\nDepending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure 10.58. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing.\n\n\nFIGURE 10.58 A cation's size and the shape of the hole occupied by the compound are directly related.\nThere are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of $2: 1$; all of the tetrahedral holes are filled at this ratio. Examples include $\\mathrm{Li}_{2} \\mathrm{O}, \\mathrm{Na}_{2} \\mathrm{O}, \\mathrm{Li}_{2} \\mathrm{~S}$,\nand $\\mathrm{Na}_{2} \\mathrm{~S}$. Compounds with a ratio of less than $2: 1$ may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant.\n"} {"id": "textbook_3415", "title": "1158. EXAMPLE 10.16 - ", "content": ""} {"id": "textbook_3416", "title": "1159. Occupancy of Tetrahedral Holes - ", "content": "\nZinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?\n"} {"id": "textbook_3417", "title": "1160. Solution - ", "content": "\nBecause there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\\frac{1}{2} \\times 2$, or 1 , zinc ion per sulfide ion. Thus, the formula is ZnS .\n"} {"id": "textbook_3418", "title": "1161. Check Your Learning - ", "content": "\nLithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?\n"} {"id": "textbook_3419", "title": "1162. Answer: - ", "content": "\n$\\mathrm{Li}_{2} \\mathrm{Se}$"} {"id": "textbook_3420", "title": "1162. Answer: - ", "content": "The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO , $\\mathrm{MnS}, \\mathrm{NaCl}$, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty.\n"} {"id": "textbook_3421", "title": "1163. EXAMPLE 10.17 - ", "content": ""} {"id": "textbook_3422", "title": "1164. Stoichiometry of Ionic Compounds - ", "content": "\nSapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?\n"} {"id": "textbook_3423", "title": "1165. Solution - ", "content": "\nBecause there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\\frac{2}{3}: 1$, which would give $\\mathrm{Al}_{2 / 3} \\mathrm{O}$. The simplest whole number ratio is $2: 3$, so the formula is $\\mathrm{Al}_{2} \\mathrm{O}_{3}$.\n"} {"id": "textbook_3424", "title": "1166. Check Your Learning - ", "content": "\nThe white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?\n"} {"id": "textbook_3425", "title": "1167. Answer: - ", "content": "\n$\\mathrm{TiO}_{2}$"} {"id": "textbook_3426", "title": "1167. Answer: - ", "content": "In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl , and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in $\\mathrm{SrH}_{2}, \\mathrm{UO}_{2}, \\mathrm{SrCl}_{2}$, and $\\mathrm{CaF}_{2}$."} {"id": "textbook_3427", "title": "1167. Answer: - ", "content": "Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar.\n"} {"id": "textbook_3428", "title": "1168. Unit Cells of Ionic Compounds - ", "content": "\nMany ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures."} {"id": "textbook_3429", "title": "1168. Unit Cells of Ionic Compounds - ", "content": "When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl , (illustrated in Figure 10.59) is an example of this, with $\\mathrm{Cs}^{+}$and $\\mathrm{Cl}^{-}$having radii of 174 pm and 181 pm , respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by $\\mathrm{Cs}^{+}$ions overlapping unit cells formed by $\\mathrm{Cl}^{-}$ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.\n\n\nSimple cubic structure\nFIGURE 10.59 Ionic compounds with similar-sized cations and anions, such as CsCl , usually form a simple cubic structure. They can be described by unit cells with either cations at the corners or anions at the corners.\nWe have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures."} {"id": "textbook_3430", "title": "1168. Unit Cells of Ionic Compounds - ", "content": "When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure 10.60. Sodium chloride, NaCl , is an example of this, with $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$having radii of 102 pm and 181 pm , respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl .\n\n\nFace-centered simple cubic structure\nFIGURE 10.60 Ionic compounds with anions that are much larger than cations, such as NaCl , usually form an FCC structure. They can be described by FCC unit cells with cations in the octahedral holes."} {"id": "textbook_3431", "title": "1168. Unit Cells of Ionic Compounds - ", "content": "The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure 10.61. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about $40 \\%$ of the radius of a sulfide ion, so these small $\\mathrm{Zn}^{2+}$ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS .\n\n\nZnS face-centered unit cell\nFIGURE 10.61 ZnS , zinc sulfide (or zinc blende) forms an FCC unit cell with sulfide ions at the lattice points and much smaller zinc ions occupying half of the tetrahedral holes in the structure."} {"id": "textbook_3432", "title": "1168. Unit Cells of Ionic Compounds - ", "content": "A calcium fluoride unit cell, like that shown in Figure 10.62, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, $\\mathrm{CaF}_{2}$. Close examination of Figure 10.62 will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.\n\n$\\mathrm{CaF}_{2}$ face-centered unit cell\nFIGURE 10.62 Calcium fluoride, $\\mathrm{CaF}_{2}$, forms an FCC unit cell with calcium ions (green) at the lattice points and fluoride ions (red) occupying all of the tetrahedral sites between them.\n"} {"id": "textbook_3433", "title": "1169. Calculation of Ionic Radii - ", "content": "\nIf we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.\n"} {"id": "textbook_3434", "title": "1170. EXAMPLE 10.18 - ", "content": ""} {"id": "textbook_3435", "title": "1171. Calculation of Ionic Radii - ", "content": "\nThe edge length of the unit cell of LiCl ( NaCl -like structure, FCC ) is 0.514 nm or $5.14 \\AA$. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in Figure 10.60, calculate the ionic radius for the chloride ion."} {"id": "textbook_3436", "title": "1171. Calculation of Ionic Radii - ", "content": "Note: The length unit angstrom, $\\AA$, is often used to represent atomic-scale dimensions and is equivalent to $10^{-10} \\mathrm{~m}$.\n"} {"id": "textbook_3437", "title": "1172. Solution - ", "content": "\nOn the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:\n"} {"id": "textbook_3438", "title": "1172. Solution - ", "content": "Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter-which equals two radii-from the chloride ion in the center of the face), so $d=4 r$. From the Pythagorean theorem, we have:"} {"id": "textbook_3439", "title": "1172. Solution - ", "content": "$$\na^{2}+a^{2}=d^{2}\n$$"} {"id": "textbook_3440", "title": "1172. Solution - ", "content": "which yields:"} {"id": "textbook_3441", "title": "1172. Solution - ", "content": "$$\n(0.514 \\mathrm{~nm})^{2}+(0.514 \\mathrm{~nm})^{2}=(4 r)^{2}=16 r^{2}\n$$"} {"id": "textbook_3442", "title": "1172. Solution - ", "content": "Solving this gives:"} {"id": "textbook_3443", "title": "1172. Solution - ", "content": "$$\nr=\\sqrt{\\frac{(0.514 \\mathrm{~nm})^{2}+(0.514 \\mathrm{~nm})^{2}}{16}}=0.182 \\mathrm{~nm}(1.82 \\AA) \\text { for a } \\mathrm{Cl}^{-} \\text {radius. }\n$$\n"} {"id": "textbook_3444", "title": "1173. Check Your Learning - ", "content": "\nThe edge length of the unit cell of KCl ( NaCl -like structure, FCC ) is 6.28 \u00c5. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is $1.82 \\AA$.\n"} {"id": "textbook_3445", "title": "1174. Answer: - ", "content": "\nThe radius of the potassium ion is $1.33 \\AA$."} {"id": "textbook_3446", "title": "1174. Answer: - ", "content": "It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.\n"} {"id": "textbook_3447", "title": "1175. X-Ray Crystallography - ", "content": "\nThe size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few $\\AA$ )."} {"id": "textbook_3448", "title": "1175. X-Ray Crystallography - ", "content": "When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves' maxima are separated (see Figure 10.63).\n"} {"id": "textbook_3449", "title": "1175. X-Ray Crystallography - ", "content": "Constructive interface\n(a)\n"} {"id": "textbook_3450", "title": "1175. X-Ray Crystallography - ", "content": "Destructive interface\n(b)"} {"id": "textbook_3451", "title": "1175. X-Ray Crystallography - ", "content": "FIGURE 10.63 Light waves occupying the same space experience interference, combining to yield waves of greater (a) or lesser (b) intensity, depending upon the separation of their maxima and minima."} {"id": "textbook_3452", "title": "1175. X-Ray Crystallography - ", "content": "When X-rays of a certain wavelength, $\\lambda$, are scattered by atoms in adjacent crystal planes separated by a distance, $d$, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, $n$, of the wavelength. This condition is satisfied when the angle of the diffracted beam, $\\theta$, is related to the wavelength and interatomic distance by the equation:"} {"id": "textbook_3453", "title": "1175. X-Ray Crystallography - ", "content": "$$\nn \\lambda=2 d \\sin \\theta\n$$"} {"id": "textbook_3454", "title": "1175. X-Ray Crystallography - ", "content": "This relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. Figure 10.64 illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference.\n"} {"id": "textbook_3455", "title": "1175. X-Ray Crystallography - ", "content": "FIGURE 10.64 The diffraction of X-rays scattered by the atoms within a crystal permits the determination of the distance between the atoms. The top image depicts constructive interference between two scattered waves and a resultant diffracted wave of high intensity. The bottom image depicts destructive interference and a low intensity diffracted wave.\n"} {"id": "textbook_3456", "title": "1176. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16bragg) for more details on the Bragg equation and a simulator that allows you to explore the effect of each variable on the intensity of the diffracted wave."} {"id": "textbook_3457", "title": "1176. LINK TO LEARNING - ", "content": "An X-ray diffractometer, such as the one illustrated in Figure 10.65, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise.\n"} {"id": "textbook_3458", "title": "1176. LINK TO LEARNING - ", "content": "FIGURE 10.65 (a) In a diffractometer, a beam of X-rays strikes a crystalline material, producing (b) an X-ray diffraction pattern that can be analyzed to determine the crystal structure.\n"} {"id": "textbook_3459", "title": "1177. EXAMPLE 10.19 - ", "content": ""} {"id": "textbook_3460", "title": "1178. Using the Bragg Equation - ", "content": "\nIn a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction $(n=1)$ occurred at an angle $\\theta=25.25^{\\circ}$. Determine the spacing between the diffracting planes in copper.\n"} {"id": "textbook_3461", "title": "1179. Solution - ", "content": "\nThe distance between the planes is found by solving the Bragg equation, $n \\lambda=2 d \\sin \\theta$, for $d$.\nThis gives: $d=\\frac{n \\lambda}{2 \\sin \\theta}=\\frac{1(0.1315 \\mathrm{~nm})}{2 \\sin \\left(25.25^{\\circ}\\right)}=0.154 \\mathrm{~nm}$\n"} {"id": "textbook_3462", "title": "1180. Check Your Learning - ", "content": "\nA crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm . What is the angle for the first order diffraction?\n"} {"id": "textbook_3463", "title": "1181. Answer: - ", "content": "\n$10.8^{\\circ}$\n"} {"id": "textbook_3464", "title": "1182. Portrait of a Chemist - ", "content": ""} {"id": "textbook_3465", "title": "1183. X-ray Crystallographer Rosalind Franklin - ", "content": "\nThe discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins, who provided experimental proof of DNA's structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin's research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type \"B\") and a short, wide fiber formed when dried (type \"A\"). Her X-ray diffraction images of DNA (Figure 10.66) provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued\nto work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo.\n"} {"id": "textbook_3466", "title": "1183. X-ray Crystallographer Rosalind Franklin - ", "content": "FIGURE 10.66 This illustration shows an X-ray diffraction image similar to the one Franklin found in her research. (credit: National Institutes of Health)\n"} {"id": "textbook_3467", "title": "1184. Key Terms - ", "content": "\nadhesive force force of attraction between molecules of different chemical identities\namorphous solid (also, noncrystalline solid) solid in which the particles lack an ordered internal structure\nbody-centered cubic (BCC) solid crystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell\nbody-centered cubic unit cell simplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube\nboiling point temperature at which the vapor pressure of a liquid equals the pressure of the gas above it\nBragg equation equation that relates the angles at which X-rays are diffracted by the atoms within a crystal\ncapillary action flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules\nClausius-Clapeyron equation mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance\ncohesive force force of attraction between identical molecules\ncondensation change from a gaseous to a liquid state\ncoordination number number of atoms closest to any given atom in a crystal or to the central metal atom in a complex\ncovalent network solid solid whose particles are held together by covalent bonds\ncritical point temperature and pressure above which a gas cannot be condensed into a liquid\ncrystalline solid solid in which the particles are arranged in a definite repeating pattern\ncubic closest packing (CCP) crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations ( ABC )\ndeposition change from a gaseous state directly to a solid state\ndiffraction redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions\ndipole-dipole attraction intermolecular attraction between two permanent dipoles\ndispersion force (also, London dispersion force) attraction between two rapidly fluctuating,\ntemporary dipoles; significant only when particles are very close together\ndynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates\nface-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face\nface-centered cubic unit cell simplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face\nfreezing change from a liquid state to a solid state\nfreezing point temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point\nhexagonal closest packing (HCP) crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)\nhole (also, interstice) space between atoms within a crystal\nhydrogen bonding occurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N\ninduced dipole temporary dipole formed when the electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule\ninstantaneous dipole temporary dipole that occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically\nintermolecular force noncovalent attractive force between atoms, molecules, and/or ions\ninterstitial sites spaces between the regular particle positions in any array of atoms or ions\nionic solid solid composed of positive and negative ions held together by strong electrostatic attractions\nisomorphous possessing the same crystalline structure\nmelting change from a solid state to a liquid state\nmelting point temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point\nmetallic solid solid composed of metal atoms\nmolecular solid solid composed of neutral molecules held together by intermolecular forces of attraction\nnormal boiling point temperature at which a\nliquid's vapor pressure equals 1 atm ( 760 torr)\noctahedral hole open space in a crystal at the center of six particles located at the corners of an octahedron\nphase diagram pressure-temperature graph summarizing conditions under which the phases of a substance can exist\npolarizability measure of the ability of a charge to distort a molecule's charge distribution (electron cloud)\nsimple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners\nsimple cubic unit cell (also, primitive cubic unit cell) unit cell in the simple cubic structure\nspace lattice all points within a crystal that have identical environments\nsublimation change from solid state directly to gaseous state\nsupercritical fluid substance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states\nsurface tension energy required to increase the area, or length, of a liquid surface by a given amount\n"} {"id": "textbook_3468", "title": "1185. Key Equations - ", "content": "\n$h=\\frac{2 T \\cos \\theta}{r \\rho g}$\n$P=A e^{-\\Delta H_{\\mathrm{vap}} / R T}$\n$\\ln P=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T}+\\ln A$\n$\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)$\n$n \\lambda=2 d \\sin \\theta$"} {"id": "textbook_3469", "title": "1185. Key Equations - ", "content": "Summary\n"} {"id": "textbook_3470", "title": "1185. Key Equations - 1185.1. Intermolecular Forces", "content": "\nThe physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient KE to move past each other."} {"id": "textbook_3471", "title": "1185. Key Equations - 1185.1. Intermolecular Forces", "content": "Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one polar molecule for the partial positive end of another. The temporary dipole that results from the\ntetrahedral hole tetrahedral space formed by four atoms or ions in a crystal\ntriple point temperature and pressure at which the vapor, liquid, and solid phases of a substance are in equilibrium\nunit cell smallest portion of a space lattice that is repeated in three dimensions to form the entire lattice\nvacancy defect that occurs when a position that should contain an atom or ion is vacant\nvan der Waals force attractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules\nvapor pressure (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature\nvaporization change from liquid state to gaseous state\nviscosity measure of a liquid's resistance to flow\nX-ray crystallography experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal\nmotion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: $\\mathrm{F}, \\mathrm{O}$, or N .\n"} {"id": "textbook_3472", "title": "1185. Key Equations - 1185.2. Properties of Liquids", "content": "\nThe intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid's viscosity (resistance to flow) and surface tension\n(elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.\n"} {"id": "textbook_3473", "title": "1185. Key Equations - 1185.3. Phase Transitions", "content": "\nPhase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.\n"} {"id": "textbook_3474", "title": "1185. Key Equations - 1185.4. Phase Diagrams", "content": "\nThe temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phasetransition temperatures and pressures. The point of intersection of all three curves represents the substance's triple point-the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance's critical point, the pressure and temperature above which a liquid phase cannot exist.\n"} {"id": "textbook_3475", "title": "1185. Key Equations - 1185.5. The Solid State of Matter", "content": "\nSome substances form crystalline solids consisting\nof particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.\n"} {"id": "textbook_3476", "title": "1185. Key Equations - 1185.6. Lattice Structures in Crystalline Solids", "content": "\nThe structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closestpacked structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements.\n"} {"id": "textbook_3477", "title": "1186. Exercises - ", "content": ""} {"id": "textbook_3478", "title": "1186. Exercises - 1186.1. Intermolecular Forces", "content": "\n1. In terms of their bulk properties, how do liquids and solids differ? How are they similar?\n2. In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?\n3. In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?\n4. Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.\n5. What is the evidence that all neutral atoms and molecules exert attractive forces on each other?\n6. Open the PhET States of Matter Simulation (http://openstax.org/l/16phetvisual) to answer the following questions:\n(a) Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?\n(b) For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.\n(c) Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.\n7. Define the following and give an example of each:\n(a) dispersion force\n(b) dipole-dipole attraction\n(c) hydrogen bond\n8. The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?\n9. Why do the boiling points of the noble gases increase in the order $\\mathrm{He}<\\mathrm{Ne}<\\mathrm{Ar}<\\mathrm{Kr}<\\mathrm{Xe}$ ?\n10. Neon and HF have approximately the same molecular masses.\n(a) Explain why the boiling points of Neon and HF differ.\n(b) Compare the change in the boiling points of $\\mathrm{Ne}, \\mathrm{Ar}, \\mathrm{Kr}$, and Xe with the change of the boiling points of $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$, and HI , and explain the difference between the changes with increasing atomic or molecular mass.\n11. Arrange each of the following sets of compounds in order of increasing boiling point temperature:\n(a) $\\mathrm{HCl}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{SiH}_{4}$\n(b) $\\mathrm{F}_{2}, \\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$\n(c) $\\mathrm{CH}_{4}, \\mathrm{C}_{2} \\mathrm{H}_{6}, \\mathrm{C}_{3} \\mathrm{H}_{8}$\n(d) $\\mathrm{O}_{2}, \\mathrm{NO}, \\mathrm{N}_{2}$\n12. The molecular mass of butanol, $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{OH}$, is 74.14 ; that of ethylene glycol, $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}$, is 62.08 , yet their boiling points are $117.2^{\\circ} \\mathrm{C}$ and $174^{\\circ} \\mathrm{C}$, respectively. Explain the reason for the difference.\n13. On the basis of intermolecular attractions, explain the differences in the boiling points of $n$-butane $\\left(-1^{\\circ} \\mathrm{C}\\right)$ and chloroethane $\\left(12^{\\circ} \\mathrm{C}\\right)$, which have similar molar masses.\n14. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone $\\left(56.2^{\\circ} \\mathrm{C}\\right)$ and 1-propanol $\\left(97.4^{\\circ} \\mathrm{C}\\right)$, which have similar molar masses.\n15. The melting point of $\\mathrm{H}_{2} \\mathrm{O}(s)$ is $0^{\\circ} \\mathrm{C}$. Would you expect the melting point of $\\mathrm{H}_{2} \\mathrm{~S}(s)$ to be $-85^{\\circ} \\mathrm{C}, 0^{\\circ} \\mathrm{C}$, or 185 ${ }^{\\circ} \\mathrm{C}$ ? Explain your answer.\n16. Silane $\\left(\\mathrm{SiH}_{4}\\right)$, phosphine $\\left(\\mathrm{PH}_{3}\\right)$, and hydrogen sulfide $\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$ melt at $-185^{\\circ} \\mathrm{C},-133^{\\circ} \\mathrm{C}$, and $-85^{\\circ} \\mathrm{C}$, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?\n17. Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules.\n18. Under certain conditions, molecules of acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$, form \"dimers,\" pairs of acetic acid molecules held together by strong intermolecular attractions:\n
$(\\mathbf{m N / m})$ | Viscosity
$(\\mathbf{m P a} \\mathbf{s})$ |\n| :---: | :---: | :---: | :---: |\n| diethyl ether
$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}$ | | 17 | 0.22 |\n| acetone
$\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ | | 23 | 0.31 |\n| ethanol
$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | | 22 | 1.07 |\n| ethylene glycol
$\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}(\\mathrm{OH})$ | | 48 | 16.1 |"} {"id": "textbook_3483", "title": "1186. Exercises - 1186.2. Properties of Liquids", "content": "(a) Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.\n(b) Explain their differences in surface tension in terms of the size and shape of their molecules and their"} {"id": "textbook_3484", "title": "1186. Exercises - 1186.2. Properties of Liquids", "content": "IMFs:\n25. You may have heard someone use the figure of speech \"slower than molasses in winter\" to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.\n26. It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this.\n27. The surface tension and viscosity of water at several different temperatures are given in this table."} {"id": "textbook_3485", "title": "1186. Exercises - 1186.2. Properties of Liquids", "content": "| Water | Surface Tension (mN/m) | Viscosity (mPa s) |\n| :---: | :---: | :---: |\n| $0{ }^{\\circ} \\mathrm{C}$ | 75.6 | 1.79 |\n| $20^{\\circ} \\mathrm{C}$ | 72.8 | 1.00 |\n| $60^{\\circ} \\mathrm{C}$ | 66.2 | 0.47 |\n| $100^{\\circ} \\mathrm{C}$ | 58.9 | 0.28 |"} {"id": "textbook_3486", "title": "1186. Exercises - 1186.2. Properties of Liquids", "content": "(a) As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.\n(b) As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.\n28. At $25^{\\circ} \\mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm ? Refer to Example 10.4 for the required information.\n29. Water rises in a glass capillary tube to a height of 17 cm . What is the diameter of the capillary tube?\n"} {"id": "textbook_3487", "title": "1186. Exercises - 1186.3. Phase Transitions", "content": "\n30. Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?\n31. Heat is added to ice at $0^{\\circ} \\mathrm{C}$. Explain why the temperature of the ice does not change. What does change?\n32. What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container?\n33. Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate?\n34. Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime?\n35. What is the relationship between the intermolecular forces in a liquid and its vapor pressure?\n36. What is the relationship between the intermolecular forces in a solid and its melting temperature?\n37. Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day?\n38. Carbon tetrachloride, $\\mathrm{CCl}_{4}$, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At $57.8^{\\circ} \\mathrm{C}$, the vapor pressure of $\\mathrm{CCl}_{4}$ is 54.0 kPa , and its enthalpy of vaporization is 33.05 $\\mathrm{kJ} / \\mathrm{mol}$. Use this information to estimate the normal boiling point for $\\mathrm{CCl}_{4}$.\n39. When is the boiling point of a liquid equal to its normal boiling point?\n40. How does the boiling of a liquid differ from its evaporation?\n41. Use the information in Figure 10.24 to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa .\n42. A syringe at a temperature of $20^{\\circ} \\mathrm{C}$ is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?\n43. Explain the following observations:\n(a) It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level).\n(b) Perspiring is a mechanism for cooling the body.\n44. The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.\n45. Explain why the molar enthalpies of vaporization of the following substances increase in the order $\\mathrm{CH}_{4}<$ $\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}$, even though the type of IMF (dispersion) is the same.\n46. Explain why the enthalpies of vaporization of the following substances increase in the order $\\mathrm{CH}_{4}<\\mathrm{NH}_{3}<$ $\\mathrm{H}_{2} \\mathrm{O}$, even though all three substances have approximately the same molar mass.\n47. The enthalpy of vaporization of $\\mathrm{CO}_{2}(I)$ is $9.8 \\mathrm{~kJ} / \\mathrm{mol}$. Would you expect the enthalpy of vaporization of $\\mathrm{CS}_{2}(\\mathrm{I})$ to be $28 \\mathrm{~kJ} / \\mathrm{mol}, 9.8 \\mathrm{~kJ} / \\mathrm{mol}$, or $-8.4 \\mathrm{~kJ} / \\mathrm{mol}$ ? Discuss the plausibility of each of these answers.\n48. The hydrogen fluoride molecule, HF , is more polar than a water molecule, $\\mathrm{H}_{2} \\mathrm{O}$ (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.\n49. Ethyl chloride (boiling point, $13^{\\circ} \\mathrm{C}$ ) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride.\n50. Which contains the compounds listed correctly in order of increasing boiling points?\n(a) $\\mathrm{N}_{2}<\\mathrm{CS}_{2}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{KCl}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{N}_{2}<\\mathrm{CS}_{2}<\\mathrm{KCl}$\n(c) $\\mathrm{N}_{2}<\\mathrm{KCl}<\\mathrm{CS}_{2}<\\mathrm{H}_{2} \\mathrm{O}$\n(d) $\\mathrm{CS}_{2}<\\mathrm{N}_{2}<\\mathrm{KCl}<\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{KCl}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{CS}_{2}<\\mathrm{N}_{2}$\n51. How much heat is required to convert 422 g of liquid $\\mathrm{H}_{2} \\mathrm{O}$ at $23.5^{\\circ} \\mathrm{C}$ into steam at $150^{\\circ} \\mathrm{C}$ ?\n52. Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20 -ounce bottle of water that had been in the refrigerator at $3.8^{\\circ} \\mathrm{C}$, how much heat is needed to convert all of that water into sweat and then to vapor? (Note: Your body temperature is $36.6^{\\circ} \\mathrm{C}$. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)\n53. Titanium tetrachloride, $\\mathrm{TiCl}_{4}$, has a melting point of $-23.2^{\\circ} \\mathrm{C}$ and has a $\\Delta H_{\\text {fusion }}=9.37 \\mathrm{~kJ} / \\mathrm{mol}$.\n(a) How much energy is required to melt 263.1 g TiCl 4 ?\n(b) For $\\mathrm{TiCl}_{4}$, which will likely have the larger magnitude: $\\Delta H_{\\text {fusion }}$ or $\\Delta H_{\\text {vaporization }}$ ? Explain your reasoning.\n"} {"id": "textbook_3488", "title": "1186. Exercises - 1186.4. Phase Diagrams", "content": "\n54. From the phase diagram for water (Figure 10.31), determine the state of water at:\n(a) $35^{\\circ} \\mathrm{C}$ and 85 kPa\n(b) $-15^{\\circ} \\mathrm{C}$ and 40 kPa\n(c) $-15^{\\circ} \\mathrm{C}$ and 0.1 kPa\n(d) $75^{\\circ} \\mathrm{C}$ and 3 kPa\n(e) $40^{\\circ} \\mathrm{C}$ and 0.1 kPa\n(f) $60^{\\circ} \\mathrm{C}$ and 50 kPa\n55. What phase changes will take place when water is subjected to varying pressure at a constant temperature of $0.005^{\\circ} \\mathrm{C}$ ? At $40^{\\circ} \\mathrm{C}$ ? At $-40^{\\circ} \\mathrm{C}$ ?\n56. Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm . What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning.\n57. From the phase diagram for carbon dioxide in Figure 10.34, determine the state of $\\mathrm{CO}_{2}$ at:\n(a) $20^{\\circ} \\mathrm{C}$ and 1000 kPa\n(b) $10^{\\circ} \\mathrm{C}$ and 2000 kPa\n(c) $10^{\\circ} \\mathrm{C}$ and 100 kPa\n(d) $-40^{\\circ} \\mathrm{C}$ and 500 kPa\n(e) $-80^{\\circ} \\mathrm{C}$ and 1500 kPa\n(f) $-80^{\\circ} \\mathrm{C}$ and 10 kPa\n58. Determine the phase changes that carbon dioxide undergoes as pressure is increased at a constant temperature of (a) $-50^{\\circ} \\mathrm{C}$ and (b) $50^{\\circ} \\mathrm{C}$. If the temperature is held at $-40^{\\circ} \\mathrm{C}$ ? At $20^{\\circ} \\mathrm{C}$ ? (See the phase diagram in Figure 10.34.)\n59. Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of $20^{\\circ} \\mathrm{C}$. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.\n60. Dry ice, $\\mathrm{CO}_{2}(s)$, does not melt at atmospheric pressure. It sublimes at a temperature of $-78{ }^{\\circ} \\mathrm{C}$. What is the lowest pressure at which $\\mathrm{CO}_{2}(s)$ will melt to give $\\mathrm{CO}_{2}(I)$ ? At approximately what temperature will this occur? (See Figure 10.34 for the phase diagram.)\n61. If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer.\n62. Is it possible to liquefy nitrogen at room temperature (about $25^{\\circ} \\mathrm{C}$ )? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers.\n63. Elemental carbon has one gas phase, one liquid phase, and two different solid phases, as shown in the phase diagram:\n\n(a) On the phase diagram, label the gas and liquid regions.\n(b) Graphite is the most stable phase of carbon at normal conditions. On the phase diagram, label the graphite phase.\n(c) If graphite at normal conditions is heated to 2500 K while the pressure is increased to $10^{10} \\mathrm{~Pa}$, it is converted into diamond. Label the diamond phase.\n(d) Circle each triple point on the phase diagram.\n(e) In what phase does carbon exist at 5000 K and $10^{8} \\mathrm{~Pa}$ ?\n(f) If the temperature of a sample of carbon increases from 3000 K to 5000 K at a constant pressure of $10^{6}$ Pa , which phase transition occurs, if any?\n"} {"id": "textbook_3489", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "\n64. What types of liquids typically form amorphous solids?\n65. At very low temperatures oxygen, $\\mathrm{O}_{2}$, freezes and forms a crystalline solid. Which best describes these crystals?\n(a) ionic\n(b) covalent network\n(c) metallic\n(d) amorphous\n(e) molecular crystals\n66. As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid?\n(a) ionic\n(b) covalent network\n(c) metallic\n(d) amorphous\n(e) molecular crystals\n67. Explain why ice, which is a crystalline solid, has a melting temperature of $0^{\\circ} \\mathrm{C}$, whereas butter, which is an amorphous solid, softens over a range of temperatures.\n68. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:\n(a) $\\mathrm{SiO}_{2}$\n(b) KCl\n(c) Cu\n(d) $\\mathrm{CO}_{2}$\n(e) C (diamond)\n(f) $\\mathrm{BaSO}_{4}$\n(g) $\\mathrm{NH}_{3}$\n(h) $\\mathrm{NH}_{4} \\mathrm{~F}$\n(i) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n69. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:\n(a) $\\mathrm{CaCl}_{2}$\n(b) SiC\n(c) $\\mathrm{N}_{2}$\n(d) Fe\n(e) C (graphite)\n(f) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(g) HCl\n(h) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n(i) $\\mathrm{K}_{3} \\mathrm{PO}_{4}$\n70. Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:"} {"id": "textbook_3490", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "| Substance | Appearance | Melting Point | Electrical
Conductivity | Solubility in Water |\n| :---: | :---: | :---: | :---: | :---: |\n| X | lustrous, malleable | $1500^{\\circ} \\mathrm{C}$ | high | insoluble |\n| Y | soft, yellow | $113^{\\circ} \\mathrm{C}$ | none | insoluble |\n| Z | hard, white | $800^{\\circ} \\mathrm{C}$ | only if melted/
dissolved | soluble |"} {"id": "textbook_3491", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "71. Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:"} {"id": "textbook_3492", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "| Appearance | | Melting Point | Electrical
Conductivity | Solubility in Water |\n| :---: | :---: | :---: | :---: | :---: |\n| X | brittle, white | $800^{\\circ} \\mathrm{C}$ | only if melted/
dissolved | soluble |\n| Y | shiny, malleable | $1100^{\\circ} \\mathrm{C}$ | high | insoluble |\n| Z | hard, colorless | $3550^{\\circ} \\mathrm{C}$ | none | insoluble |"} {"id": "textbook_3493", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "72. Identify the following substances as ionic, metallic, covalent network, or molecular solids:"} {"id": "textbook_3494", "title": "1186. Exercises - 1186.5. The Solid State of Matter", "content": "Substance A is malleable, ductile, conducts electricity well, and has a melting point of $1135^{\\circ} \\mathrm{C}$. Substance $B$ is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072\n${ }^{\\circ} \\mathrm{C}$. Substance C is very hard, does not conduct electricity, and has a melting point of $3440^{\\circ} \\mathrm{C}$. Substance D is soft, does not conduct electricity, and has a melting point of $185^{\\circ} \\mathrm{C}$.\n73. Substance A is shiny, conducts electricity well, and melts at $975^{\\circ} \\mathrm{C}$. Substance A is likely a(n):\n(a) ionic solid\n(b) metallic solid\n(c) molecular solid\n(d) covalent network solid\n74. Substance B is hard, does not conduct electricity, and melts at $1200^{\\circ} \\mathrm{C}$. Substance $B$ is likely a(n):\n(a) ionic solid\n(b) metallic solid\n(c) molecular solid\n(d) covalent network solid\n"} {"id": "textbook_3495", "title": "1186. Exercises - 1186.6. Lattice Structures in Crystalline Solids", "content": "\n75. Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell.\n76. Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell.\n77. What is the coordination number of a chromium atom in the body-centered cubic structure of chromium?\n78. What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum?\n79. Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom?\n80. Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom?\n81. Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 \u00c5.\n(a) What is the atomic radius of tungsten in this structure?\n(b) Calculate the density of tungsten.\n82. Platinum (atomic radius $=1.38 \\AA$ \u00c5) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.\n83. Barium crystallizes in a body-centered cubic unit cell with an edge length of $5.025 \\AA$\n(a) What is the atomic radius of barium in this structure?\n(b) Calculate the density of barium.\n84. Aluminum (atomic radius $=1.43 \\AA$ ) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.\n85. The density of aluminum is $2.7 \\mathrm{~g} / \\mathrm{cm}^{3}$; that of silicon is $2.3 \\mathrm{~g} / \\mathrm{cm}^{3}$. Explain why Si has the lower density even though it has heavier atoms.\n86. The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?\n87. Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer.\n88. A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer.\n89. What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions?\n90. A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound?\n91. A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer.\n92. Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closestpacked array with all of the octahedral holes occupied: $\\mathrm{Li}, \\mathrm{Na}, \\mathrm{Be}, \\mathrm{Ca}$, or Al ?\n93. What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium?\n94. Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure.\n95. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation.\n96. Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound?\n97. One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?\n98. NaH crystallizes with the same crystal structure as NaCl . The edge length of the cubic unit cell of NaH is 4.880 \u00c5.\n(a) Calculate the ionic radius of $\\mathrm{H}^{-}$. (The ionic radius of $\\mathrm{Li}^{+}$is 0.0.95 $\\AA$.)\n(b) Calculate the density of NaH .\n99. Thallium(I) iodide crystallizes with the same structure as CsCl . The edge length of the unit cell of TlI is $4.20 \\AA$. Calculate the ionic radius of $\\mathrm{TI}^{+}$. (The ionic radius of $\\mathrm{I}^{-}$is $2.16 \\AA$.)\n100. A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.\n(a) What is the empirical formula of this compound? Explain your answer.\n(b) What is the coordination number of the $\\mathrm{Mn}^{3+}$ ion?\n(c) Calculate the edge length of the unit cell if the radius of a $\\mathrm{Mn}^{3+}$ ion is 0.65 A .\n(d) Calculate the density of the compound.\n101. What is the spacing between crystal planes that diffract $X$-rays with a wavelength of 1.541 nm at an angle $\\theta$ of $15.55^{\\circ}$ (first order reflection)?\n102. A diffractometer using $X$-rays with a wavelength of 0.2287 nm produced first order diffraction peak for a crystal angle $\\theta=16.21^{\\circ}$. Determine the spacing between the diffracting planes in this crystal.\n103. A metal with spacing between planes equal to 0.4164 nm diffracts $X$-rays with a wavelength of 0.2879 nm . What is the diffraction angle for the first order diffraction peak?\n104. Gold crystallizes in a face-centered cubic unit cell. The second-order reflection ( $n=2$ ) of $X$-rays for the planes that make up the tops and bottoms of the unit cells is at $\\theta=22.20^{\\circ}$. The wavelength of the X -rays is $1.54 \\AA$. What is the density of metallic gold?\n105. When an electron in an excited molybdenum atom falls from the $L$ to the $K$ shell, an $X$-ray is emitted. These X-rays are diffracted at an angle of $7.75^{\\circ}$ by planes with a separation of $2.64 \\AA$. What is the difference in energy between the $K$ shell and the $L$ shell in molybdenum assuming a first order diffraction?\n"} {"id": "textbook_3496", "title": "1187. CHAPTER 11 - ", "content": "\nSolutions and Colloids\n"} {"id": "textbook_3497", "title": "1187. CHAPTER 11 - ", "content": "Figure 11.1 Coral reefs, such as this one at the Palmyra Atoll National Wildlife Refuge, are vital to the ecosystem of earth's oceans. The health of coral reefs and all marine life depends on the specific chemical composition of the complex mixture known as seawater. (credit: modification of work by \"USFWS - Pacific Region\"/Wikimedia Commons)\n"} {"id": "textbook_3498", "title": "1188. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_3499", "title": "1188. CHAPTER OUTLINE - 1188.1. The Dissolution Process", "content": "\n11.2 Electrolytes\n11.3 Solubility\n11.4 Colligative Properties\n11.5 Colloids"} {"id": "textbook_3500", "title": "1188. CHAPTER OUTLINE - 1188.1. The Dissolution Process", "content": "INTRODUCTION Coral reefs are home to about $25 \\%$ of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution known as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans."} {"id": "textbook_3501", "title": "1188. CHAPTER OUTLINE - 1188.1. The Dissolution Process", "content": "Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. This chapter considers the nature of solutions and examines factors that determine whether a solution will form and what properties it may have. The properties of colloids-mixtures containing dispersed particles larger than the molecules and ions of typical solutions-are also discussed.\n"} {"id": "textbook_3502", "title": "1188. CHAPTER OUTLINE - 1188.2. The Dissolution Process", "content": ""} {"id": "textbook_3503", "title": "1189. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3504", "title": "1189. LEARNING OBJECTIVES - ", "content": "- Describe the basic properties of solutions and how they form\n- Predict whether a given mixture will yield a solution based on molecular properties of its components\n- Explain why some solutions either produce or absorb heat when they form"} {"id": "textbook_3505", "title": "1189. LEARNING OBJECTIVES - ", "content": "An earlier chapter of this text introduced solutions, defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:"} {"id": "textbook_3506", "title": "1189. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(s) \\longrightarrow \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(a q)\n$$"} {"id": "textbook_3507", "title": "1189. LEARNING OBJECTIVES - ", "content": "The subscript \"aq\" in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to \"settle out\" over time."} {"id": "textbook_3508", "title": "1189. LEARNING OBJECTIVES - ", "content": "Potassium dichromate, $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$, is an ionic compound composed of colorless potassium ions, $\\mathrm{K}^{+}$, and orange dichromate ions, $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$. When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure 11.2), as indicated in this equation:"} {"id": "textbook_3509", "title": "1189. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(s) \\longrightarrow 2 \\mathrm{~K}^{+}(a q)+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(a q)\n$$"} {"id": "textbook_3510", "title": "1189. LEARNING OBJECTIVES - ", "content": "As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.\n\n\nFIGURE 11.2 When potassium dichromate $\\left(\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}\\right)$ is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott)\n"} {"id": "textbook_3511", "title": "1190. LINK TO LEARNING - ", "content": "\nVisit this virtual lab (http://openstax.org/l/16Phetsugar) to view simulations of the dissolution of common covalent and ionic substances (sugar and salt) in water."} {"id": "textbook_3512", "title": "1190. LINK TO LEARNING - ", "content": "Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table 11.1 gives examples of several different solutions and the phases of the solutes and solvents."} {"id": "textbook_3513", "title": "1190. LINK TO LEARNING - ", "content": "| Solution | Solute | Solvent |\n| :--- | :--- | :--- |\n| air | $\\mathrm{O}_{2}(g)$ | $\\mathrm{N}_{2}(g)$ |\n| soft drinks ${ }^{\\underline{1}}$ | $\\mathrm{CO}_{2}(g)$ | $\\mathrm{H}_{2} \\mathrm{O}(I)$ |\n| hydrogen in palladium | $\\mathrm{H}_{2}(g)$ | $\\mathrm{Pd}(s)$ |\n| rubbing alcohol | $\\mathrm{H}_{2} \\mathrm{O}(I)$ | $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}(I)(2$-propanol $)$ |\n| saltwater | $\\mathrm{NaCl}^{(s)}$ | $\\mathrm{H}_{2} \\mathrm{O}(I)$ |\n| brass | $\\mathrm{Zn}(s)$ | $\\mathrm{Cu}(s)$ |"} {"id": "textbook_3514", "title": "1190. LINK TO LEARNING - ", "content": "TABLE 11.1\nSolutions exhibit these defining traits:"} {"id": "textbook_3515", "title": "1190. LINK TO LEARNING - ", "content": "- They are homogeneous; after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).\n- The physical state of a solution-solid, liquid, or gas-is typically the same as that of the solvent, as demonstrated by the examples in Table 11.1.\n- The components of a solution are dispersed on a molecular scale; they consist of a mixture of separated solute particles (molecules, atoms, and/or ions) each closely surrounded by solvent species.\n- The dissolved solute in a solution will not settle out or separate from the solvent.\n- The composition of a solution, or the concentrations of its components, can be varied continuously (within limits determined by the solubility of the components, discussed in detail later in this chapter)."} {"id": "textbook_3516", "title": "1190. LINK TO LEARNING - ", "content": ""} {"id": "textbook_3517", "title": "1191. The Formation of Solutions - ", "content": "\nThe formation of a solution is an example of a spontaneous process, a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes a mixture is stirred to speed up the dissolution process, but this is not necessary; a homogeneous solution will form eventually. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter's discussion, it will suffice to consider two criteria that favor, but do not guarantee, the spontaneous formation of a solution:"} {"id": "textbook_3518", "title": "1191. The Formation of Solutions - ", "content": "1. a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)\n2. an increased dispersal of matter in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)"} {"id": "textbook_3519", "title": "1191. The Formation of Solutions - ", "content": "In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in matter dispersal always results when a solution forms from the uniform distribution of solute molecules throughout a solvent."} {"id": "textbook_3520", "title": "1191. The Formation of Solutions - ", "content": "When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution. A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions."} {"id": "textbook_3521", "title": "1191. The Formation of Solutions - ", "content": "1 If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution."} {"id": "textbook_3522", "title": "1191. The Formation of Solutions - ", "content": "When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure 11.3). The formation of this solution clearly involves an increase in matter dispersal, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.\n"} {"id": "textbook_3523", "title": "1191. The Formation of Solutions - ", "content": "Valve closed\n"} {"id": "textbook_3524", "title": "1191. The Formation of Solutions - ", "content": "Valve open"} {"id": "textbook_3525", "title": "1191. The Formation of Solutions - ", "content": "FIGURE 11.3 Samples of helium and argon spontaneously mix to give a solution.\nIdeal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ and ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ form ideal solutions, as do mixtures of the hydrocarbons pentane, $\\mathrm{C}_{5} \\mathrm{H}_{12}$, and hexane, $\\mathrm{C}_{6} \\mathrm{H}_{14}$. Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure 11.3 will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how increased matter dispersal alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution."} {"id": "textbook_3526", "title": "1191. The Formation of Solutions - ", "content": "Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solventsolvent, and solute-solvent. As illustrated in Figure 11.4, the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.\n"} {"id": "textbook_3527", "title": "1191. The Formation of Solutions - ", "content": "FIGURE 11.4 This schematic representation of dissolution shows a stepwise process involving the endothermic separation of solute and solvent species (Steps 1 and 2) and exothermic solvation (Step 3)."} {"id": "textbook_3528", "title": "1191. The Formation of Solutions - ", "content": "Consider the example of an ionic compound dissolving in water. Formation of the solution requires the electrostatic forces between the cations and anions of the compound (solute-solute) be overcome completely as attractive forces are established between these ions and water molecules (solute-solvent). Hydrogen bonding between a relatively small fraction of the water molecules must also be overcome to accommodate any dissolved solute. If the solute's electrostatic forces are significantly greater than the solvation forces, the dissolution process is significantly endothermic and the compound may not dissolve to an appreciable extent. Calcium carbonate, the major component of coral reefs, is one example of such an \"insoluble\" ionic compound (see Figure 11.1). On the other hand, if the solvation forces are much stronger than the compound's electrostatic forces, the dissolution is significantly exothermic and the compound may be highly soluble. A common example of this type of ionic compound is sodium chloride, commonly known as table salt."} {"id": "textbook_3529", "title": "1191. The Formation of Solutions - ", "content": "As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate $\\left(\\mathrm{NH}_{4} \\mathrm{NO}_{3}\\right)$ is one such example and is used to make instant cold packs, like the one pictured in Figure 11.5, which are used for treating injuries. A thin-walled plastic bag of water is sealed inside a larger bag with solid $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$. When the smaller bag is broken, a solution of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.\n\n\nFIGURE 11.5 An instant cold pack gets cold when certain salts, such as ammonium nitrate, dissolve in water-an endothermic process.\n"} {"id": "textbook_3530", "title": "1192. LINK TO LEARNING - ", "content": "\nWatch this brief video (http://openstax.org/l/16endoexo) illustrating endothermic and exothermic dissolution processes.\n"} {"id": "textbook_3531", "title": "1192. LINK TO LEARNING - 1192.1. Electrolytes", "content": ""} {"id": "textbook_3532", "title": "1193. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3533", "title": "1193. LEARNING OBJECTIVES - ", "content": "- Define and give examples of electrolytes\n- Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes\n- Relate electrolyte strength to solute-solvent attractive forces"} {"id": "textbook_3534", "title": "1193. LEARNING OBJECTIVES - ", "content": "When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100\\% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte."} {"id": "textbook_3535", "title": "1193. LEARNING OBJECTIVES - ", "content": "Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit\n(Figure 11.6).\n\n\nacetic acid solution Low conductivity"} {"id": "textbook_3536", "title": "1193. LEARNING OBJECTIVES - ", "content": "FIGURE 11.6 Solutions of nonelectrolytes such as ethanol do not contain dissolved ions and cannot conduct electricity. Solutions of electrolytes contain ions that permit the passage of electricity. The conductivity of an electrolyte solution is related to the strength of the electrolyte.\n"} {"id": "textbook_3537", "title": "1194. Ionic Electrolytes - ", "content": "\nWater and other polar molecules are attracted to ions, as shown in Figure 11.7. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.\n"} {"id": "textbook_3538", "title": "1194. Ionic Electrolytes - ", "content": "FIGURE 11.7 As potassium chloride ( KCl ) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the $\\mathrm{K}^{+}$and $\\mathrm{Cl}^{-}$ions. Water molecules in front of and behind the ions are not shown."} {"id": "textbook_3539", "title": "1194. Ionic Electrolytes - ", "content": "When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Even sparingly, soluble ionic compounds are strong electrolytes, since the small amount that does dissolve will dissociate completely."} {"id": "textbook_3540", "title": "1194. Ionic Electrolytes - ", "content": "Consider what happens at the microscopic level when solid KCl is added to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules surround individual $\\mathrm{K}^{+}$and $\\mathrm{Cl}^{-}$ions, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure 11.7 shows. Overcoming the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution as the ions transition from fixed positions in the undissolved compound to widely dispersed, solvated ions in solution.\n"} {"id": "textbook_3541", "title": "1195. Covalent Electrolytes - ", "content": "\nPure water is an extremely poor conductor of electricity because it is only very slightly ionized-only about two out of every 1 billion molecules ionize at $25^{\\circ} \\mathrm{C}$. Water ionizes when one molecule of water gives up a proton $\\left(\\mathrm{H}^{+}\\right.$ ion) to another molecule of water, yielding hydronium and hydroxide ions."} {"id": "textbook_3542", "title": "1195. Covalent Electrolytes - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_3543", "title": "1195. Covalent Electrolytes - ", "content": "In some cases, solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, an aqueous solution of HCl is a very good conductor, indicating that an appreciable concentration of ions exists within the solution."} {"id": "textbook_3544", "title": "1195. Covalent Electrolytes - ", "content": "Because HCl is an acid, its molecules react with water, transferring $\\mathrm{H}^{+}$ions to form hydronium ions $\\left(\\mathrm{H}_{3} \\mathrm{O}^{+}\\right)$and\nchloride ions $\\left(\\mathrm{Cl}^{-}\\right)$:\n"} {"id": "textbook_3545", "title": "1195. Covalent Electrolytes - ", "content": "This reaction is essentially $100 \\%$ complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry.\n"} {"id": "textbook_3546", "title": "1195. Covalent Electrolytes - 1195.1. Solubility", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_3547", "title": "1195. Covalent Electrolytes - 1195.1. Solubility", "content": "- Describe the effects of temperature and pressure on solubility\n- State Henry's law and use it in calculations involving the solubility of a gas in a liquid\n- Explain the degrees of solubility possible for liquid-liquid solutions"} {"id": "textbook_3548", "title": "1195. Covalent Electrolytes - 1195.1. Solubility", "content": "Imagine adding a small amount of sugar to a glass of water, stirring until all the sugar has dissolved, and then adding a bit more. You can repeat this process until the sugar concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solventsolvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved sugar remains. The concentration of sugar in the solution at this point is known as its solubility.\nThe solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.\nWhen a solute's concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute's concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.\n"} {"id": "textbook_3549", "title": "1196. LINK TO LEARNING - ", "content": "\nUse this interactive simulation (http://openstax.org/l/16Phetsoluble) to prepare various saturated solutions."} {"id": "textbook_3550", "title": "1196. LINK TO LEARNING - ", "content": "Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states (a detailed treatment of this important concept is provided in the text chapters on equilibrium). For example, the carbonated beverage in an open container that has not yet \"gone flat\" is supersaturated with carbon dioxide gas; given time, the $\\mathrm{CO}_{2}$ concentration will decrease until it reaches its solubility.\n"} {"id": "textbook_3551", "title": "1197. LINK TO LEARNING - ", "content": "\nWatch this impressive video (http://openstax.org/1/16NaAcetate) showing the precipitation of sodium acetate from a supersaturated solution.\n"} {"id": "textbook_3552", "title": "1198. Solutions of Gases in Liquids - ", "content": "\nAs for any solution, the solubility of a gas in a liquid is affected by the intermolecular attractive forces between solute and solvent species. Unlike solid and liquid solutes, however, there is no solute-solute intermolecular attraction to overcome when a gaseous solute dissolves in a liquid solvent (see Figure 11.4) since the atoms or molecules comprising a gas are far separated and experience negligible interactions. Consequently, solutesolvent interactions are the sole energetic factor affecting solubility. For example, the water solubility of\noxygen is approximately three times greater than that of helium (there are greater dispersion forces between water and the larger oxygen molecules) but 100 times less than the solubility of chloromethane, $\\mathrm{CHCl}_{3}$ (polar chloromethane molecules experience dipole-dipole attraction to polar water molecules). Likewise note the solubility of oxygen in hexane, $\\mathrm{C}_{6} \\mathrm{H}_{14}$, is approximately 20 times greater than it is in water because greater dispersion forces exist between oxygen and the larger hexane molecules."} {"id": "textbook_3553", "title": "1198. Solutions of Gases in Liquids - ", "content": "Temperature is another factor affecting solubility, with gas solubility typically decreasing as temperature increases (Figure 11.8). This inverse relation between temperature and dissolved gas concentration is responsible for one of the major impacts of thermal pollution in natural waters.\n\n\nFIGURE 11.8 The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa ( 1 atm ) of gas above the solutions."} {"id": "textbook_3554", "title": "1198. Solutions of Gases in Liquids - ", "content": "When the temperature of a river, lake, or stream is raised, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water's ecosystems and, in severe cases, can result in large-scale fish kills (Figure 11.9).\n\n(a)\n\n(b)"} {"id": "textbook_3555", "title": "1198. Solutions of Gases in Liquids - ", "content": "FIGURE 11.9 (a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service)"} {"id": "textbook_3556", "title": "1198. Solutions of Gases in Liquids - ", "content": "The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the\nsolution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with $\\mathrm{CO}_{2}$ at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure 11.10). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become \"flat.\"\n\n\nFIGURE 11.10 Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of $\\mathrm{CO}_{2}$ is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. (credit: modification of work by Derrick Coetzee)"} {"id": "textbook_3557", "title": "1198. Solutions of Gases in Liquids - ", "content": "For many gaseous solutes, the relation between solubility, $C_{\\mathrm{g}}$, and partial pressure, $P_{\\mathrm{g}}$, is a proportional one:"} {"id": "textbook_3558", "title": "1198. Solutions of Gases in Liquids - ", "content": "$$\nC_{\\mathrm{g}}=k P_{\\mathrm{g}}\n$$"} {"id": "textbook_3559", "title": "1198. Solutions of Gases in Liquids - ", "content": "where $k$ is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry's law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.\n"} {"id": "textbook_3560", "title": "1199. EXAMPLE 11.1 - ", "content": ""} {"id": "textbook_3561", "title": "1200. Application of Henry's Law - ", "content": "\nAt $20^{\\circ} \\mathrm{C}$, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa is $1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$. Use Henry's law to determine the solubility of oxygen when its partial pressure is 20.7 kPa , the approximate pressure of oxygen in earth's atmosphere.\n"} {"id": "textbook_3562", "title": "1201. Solution - ", "content": "\nAccording to Henry's law, for an ideal solution the solubility, $C_{\\mathrm{g}}$, of a gas ( $1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$, in this case) is directly proportional to the pressure, $P_{\\mathrm{g}}$, of the undissolved gas above the solution ( 101.3 kPa in this case). Because both $C_{\\mathrm{g}}$ and $P_{\\mathrm{g}}$ are known, this relation can be rearragned and used to solve for $k$.\n\n$$\n\\begin{aligned}\nC_{\\mathrm{g}} & =k P_{\\mathrm{g}} \\\\\nk & =\\frac{C_{\\mathrm{g}}}{P_{\\mathrm{g}}} \\\\\n& =\\frac{1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}}{101.3 \\mathrm{kPa}} \\\\\n& =1.36 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{kPa}^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_3563", "title": "1201. Solution - ", "content": "Now, use $k$ to find the solubility at the lower pressure."} {"id": "textbook_3564", "title": "1201. Solution - ", "content": "$$\n\\begin{aligned}\nC_{\\mathrm{g}}= & k P_{\\mathrm{g}} \\\\\n& 1.36 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{kPa}^{-1} \\times 20.7 \\mathrm{kPa} \\\\\n= & 2.82 \\times 10^{-4} \\mathrm{~mol} \\mathrm{~L}^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_3565", "title": "1201. Solution - ", "content": "Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.\n"} {"id": "textbook_3566", "title": "1202. Check Your Learning - ", "content": "\nExposing a 100.0 mL sample of water at $0^{\\circ} \\mathrm{C}$ to an atmosphere containing a gaseous solute at 152 torr resulted in the dissolution of $1.45 \\times 10^{-3} \\mathrm{~g}$ of the solute. Use Henry's law to determine the solubility of this gaseous solute when its pressure is 760 torr.\n"} {"id": "textbook_3567", "title": "1203. Answer: - ", "content": "\n$7.25 \\times 10^{-3}$ in 100.0 mL or $0.0725 \\mathrm{~g} / \\mathrm{L}$\n"} {"id": "textbook_3568", "title": "1204. EXAMPLE 11.2 - ", "content": ""} {"id": "textbook_3569", "title": "1205. Thermal Pollution and Oxygen Solubility - ", "content": "\nA certain species of freshwater trout requires a dissolved oxygen concentration of $7.5 \\mathrm{mg} / \\mathrm{L}$. Could these fish thrive in a thermally polluted mountain stream (water temperature is $30.0^{\\circ} \\mathrm{C}$, partial pressure of atmospheric oxygen is 0.17 atm )? Use the data in Figure 11.8 to estimate a value for the Henry's law constant at this temperature.\n"} {"id": "textbook_3570", "title": "1206. Solution - ", "content": "\nFirst, estimate the Henry's law constant for oxygen in water at the specified temperature of $30.0^{\\circ} \\mathrm{C}$ (Figure 11.8 indicates the solubility at this temperature is approximately $\\sim 1.2 \\mathrm{~mol} / \\mathrm{L})$."} {"id": "textbook_3571", "title": "1206. Solution - ", "content": "$$\nk=\\frac{C_{\\mathrm{g}}}{P_{\\mathrm{g}}}=1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} / 1.00 \\mathrm{~atm}=1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} \\mathrm{~atm}\n$$"} {"id": "textbook_3572", "title": "1206. Solution - ", "content": "Then, use this $k$ value to compute the oxygen solubility at the specified oxygen partial pressure, 0.17 atm ."} {"id": "textbook_3573", "title": "1206. Solution - ", "content": "$$\nC_{g}=k P_{g}=\\left(1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} \\mathrm{~atm}\\right)(0.17 \\mathrm{~atm})=2.0 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}\n$$"} {"id": "textbook_3574", "title": "1206. Solution - ", "content": "Finally, convert this dissolved oxygen concentration from mol/L to mg/L."} {"id": "textbook_3575", "title": "1206. Solution - ", "content": "$$\n\\left(2.0 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}\\right)(32.0 \\mathrm{~g} / 1 \\mathrm{~mol})(1000 \\mathrm{mg} / \\mathrm{g})=6.4 \\mathrm{mg} / \\mathrm{L}\n$$"} {"id": "textbook_3576", "title": "1206. Solution - ", "content": "This concentration is lesser than the required minimum value of $7.5 \\mathrm{mg} / \\mathrm{L}$, and so these trout would likely not thrive in the polluted stream.\n"} {"id": "textbook_3577", "title": "1207. Check Your Learning - ", "content": "\nWhat dissolved oxygen concentration is expected for the stream above when it returns to a normal summer time temperature of $15^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3578", "title": "1208. Answer: - ", "content": "\n$8.2 \\mathrm{mg} / \\mathrm{L}$\n"} {"id": "textbook_3579", "title": "1209. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3580", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "\nDecompression sickness (DCS), or \"the bends,\" is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver's blood are proportionally higher per Henry's law."} {"id": "textbook_3581", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver's blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds ( 10 or $20 \\mathrm{~m} / \\mathrm{min}$ ) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure 11.11). Researchers are also investigating related body reactions and defenses in order to develop better testing and treatment for decompression sicknetss. For example, Ingrid Eftedal, a barophysiologist specializing in bodily reactions to diving, has shown that white blood cells undergo chemical and genetic changes as a result of the condition; these can potentially be used to create biomarker tests and other methods to manage decompression sickness.\n"} {"id": "textbook_3582", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "FIGURE 11.11 (a) US Navy divers undergo training in a recompression chamber. (b) Divers receive hyperbaric oxygen therapy."} {"id": "textbook_3583", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "Deviations from Henry's law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water increases more rapidly with increasing pressure than predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.\n"} {"id": "textbook_3584", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure 11.12), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed\nthat the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved $\\mathrm{CO}_{2}$ were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.\n"} {"id": "textbook_3585", "title": "1210. Decompression Sickness or \"The Bends\" - ", "content": "FIGURE 11.12 (a) It is believed that the 1986 disaster that killed more than 1700 people near Lake Nyos in Cameroon resulted when a large volume of carbon dioxide gas was released from the lake. (b) $\\mathrm{A} \\mathrm{CO}_{2}$ vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans)\n"} {"id": "textbook_3586", "title": "1211. Solutions of Liquids in Liquids - ", "content": "\nSome liquids may be mixed in any proportions to yield solutions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure 11.13) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline, mixtures of which are used as lubricating fuels for various types of outdoor power equipment (chainsaws, leaf blowers, and so on).\n"} {"id": "textbook_3587", "title": "1211. Solutions of Liquids in Liquids - ", "content": "FIGURE 11.13 Water and antifreeze are miscible; mixtures of the two are homogeneous in all proportions. (credit: \"dno1967\"/Wikimedia commons)"} {"id": "textbook_3588", "title": "1211. Solutions of Liquids in Liquids - ", "content": "Miscible liquids are typically those with very similar polarities. Consider, for example, liquids that are polar or capable of hydrogen bonding. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solventsolvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom \"like dissolves like.\""} {"id": "textbook_3589", "title": "1211. Solutions of Liquids in Liquids - ", "content": "Two liquids that do not mix to an appreciable extent are called immiscible. Separate layers are formed when immiscible liquids are poured into the same container. Gasoline, oil (Figure 11.14), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. Relatively weak attractive forces between the polar water molecules and the nonpolar liquid molecules are not adequate to overcome much stronger hydrogen bonding between water molecules. The distinction between immiscibility and miscibility is really one of extent, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.\n"} {"id": "textbook_3590", "title": "1211. Solutions of Liquids in Liquids - ", "content": "FIGURE 11.14 Water and oil are immiscible. Mixtures of these two substances will form two separate layers with the less dense oil floating on top of the water. (credit: \"Yortw\"/Flickr)"} {"id": "textbook_3591", "title": "1211. Solutions of Liquids in Liquids - ", "content": "Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure 11.15).\n\n\nFIGURE 11.15 Bromine (the deep orange liquid on the left) and water (the clear liquid in the middle) are partially miscible. The top layer in the mixture on the right is a saturated solution of bromine in water; the bottom layer is a saturated solution of water in bromine. (credit: Paul Flowers)\n"} {"id": "textbook_3592", "title": "1212. Solutions of Solids in Liquids - ", "content": "\nThe dependence of solubility on temperature for a number of solids in water is shown by the solubility curves in Figure 11.16. Reviewing these data indicates a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.\n\n\nFIGURE 11.16 This graph shows how the solubility of several solids changes with temperature."} {"id": "textbook_3593", "title": "1212. Solutions of Solids in Liquids - ", "content": "The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure 11.17, take advantage of this behavior.\n"} {"id": "textbook_3594", "title": "1212. Solutions of Solids in Liquids - ", "content": "FIGURE 11.17 This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is \"clicked.\" (credit: modification of work by \"Velela\"/Wikimedia Commons)\n"} {"id": "textbook_3595", "title": "1213. LINK TO LEARNING - ", "content": "\nThis video (http://openstax.org/l/16handwarmer) shows the crystallization process occurring in a hand warmer.\n"} {"id": "textbook_3596", "title": "1213. LINK TO LEARNING - 1213.1. Colligative Properties", "content": ""} {"id": "textbook_3597", "title": "1214. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3598", "title": "1214. LEARNING OBJECTIVES - ", "content": "- Express concentrations of solution components using mole fraction and molality\n- Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)\n- Perform calculations using the mathematical equations that describe these various colligative effects\n- Describe the process of distillation and its practical applications\n- Explain the process of osmosis and describe how it is applied industrially and in nature"} {"id": "textbook_3599", "title": "1214. LEARNING OBJECTIVES - ", "content": "The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.\n"} {"id": "textbook_3600", "title": "1215. Mole Fraction and Molality - ", "content": "\nSeveral units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity $(M)$ is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:"} {"id": "textbook_3601", "title": "1215. Mole Fraction and Molality - ", "content": "$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}\n$$"} {"id": "textbook_3602", "title": "1215. Mole Fraction and Molality - ", "content": "Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different\nat different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality."} {"id": "textbook_3603", "title": "1215. Mole Fraction and Molality - ", "content": "The mole fraction, $X$, of a component is the ratio of its molar amount to the total number of moles of all solution components:"} {"id": "textbook_3604", "title": "1215. Mole Fraction and Molality - ", "content": "$$\nX_{\\mathrm{A}}=\\frac{\\mathrm{mol} \\mathrm{~A}}{\\text { total mol of all components }}\n$$"} {"id": "textbook_3605", "title": "1215. Mole Fraction and Molality - ", "content": "By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one."} {"id": "textbook_3606", "title": "1215. Mole Fraction and Molality - ", "content": "Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:"} {"id": "textbook_3607", "title": "1215. Mole Fraction and Molality - ", "content": "$$\nm=\\frac{\\mathrm{mol} \\text { solute }}{\\mathrm{kg} \\text { solvent }}\n$$"} {"id": "textbook_3608", "title": "1215. Mole Fraction and Molality - ", "content": "Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.\n"} {"id": "textbook_3609", "title": "1216. EXAMPLE 11.3 - ", "content": ""} {"id": "textbook_3610", "title": "1217. Calculating Mole Fraction and Molality - ", "content": "\nThe antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}$, in a solution prepared from $2.22 \\times 10^{3} \\mathrm{~g}$ of ethylene glycol and $2.00 \\times 10^{3} \\mathrm{~g}$ of water (approximately 2 L of glycol and 2 L of water)?\n"} {"id": "textbook_3611", "title": "1218. Solution - ", "content": "\n(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the definition of mole fraction."} {"id": "textbook_3612", "title": "1218. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}=2.22 \\times 10^{3} \\mathrm{~g} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{62.07 \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}=35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2} \\\\\n\\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}=2.00 \\times 10^{3} \\mathrm{~g} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}{18.02 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}}=111 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O} \\\\\nX_{\\text {ethylene glycol }}=\\frac{35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{(35.8+111) \\mathrm{mol} \\text { total }}=0.244\n\\end{gathered}\n$$"} {"id": "textbook_3613", "title": "1218. Solution - ", "content": "Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).\n(b) Derive moles of solute and mass of solvent (in kg)."} {"id": "textbook_3614", "title": "1218. Solution - ", "content": "First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:"} {"id": "textbook_3615", "title": "1218. Solution - ", "content": "$$\n2.22 \\times 10^{3} \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}\\left(\\frac{\\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{2}(\\mathrm{OH})_{2}}{62.07 \\mathrm{~g}}\\right)=35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}\n$$"} {"id": "textbook_3616", "title": "1218. Solution - ", "content": "Then, convert the mass of the water from grams to kilograms:"} {"id": "textbook_3617", "title": "1218. Solution - ", "content": "$$\n2.00 \\times 10^{3} \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}\\left(\\frac{1 \\mathrm{~kg}}{1000 \\mathrm{~g}}\\right)=2.00 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_3618", "title": "1218. Solution - ", "content": "Finally, calculate molality per its definition:"} {"id": "textbook_3619", "title": "1218. Solution - ", "content": "$$\n\\begin{aligned}\n& \\text { molality }=\\frac{\\text { mol solute }}{\\mathrm{kg} \\text { solvent }} \\\\\n& \\text { molality }=\\frac{35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{2 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& \\text { molality }=17.9 \\mathrm{~m}\n\\end{aligned}\n$$\n"} {"id": "textbook_3620", "title": "1219. Check Your Learning - ", "content": "\nWhat are the mole fraction and molality of a solution that contains 0.850 g of ammonia, $\\mathrm{NH}_{3}$, dissolved in 125 g of water?\n"} {"id": "textbook_3621", "title": "1220. Answer: - ", "content": "\n$7.14 \\times 10^{-3} ; 0.399 \\mathrm{~m}$\n"} {"id": "textbook_3622", "title": "1221. EXAMPLE 11.4 - ", "content": ""} {"id": "textbook_3623", "title": "1222. Converting Mole Fraction and Molal Concentrations - ", "content": "\nCalculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.\n"} {"id": "textbook_3624", "title": "1223. Solution - ", "content": "\nConverting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:"} {"id": "textbook_3625", "title": "1223. Solution - ", "content": "$$\n\\frac{3.0 \\mathrm{~mol} \\mathrm{NaCl}}{1 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}}\n$$\n\nThe numerator for this solution's mole fraction is, therefore, 3.0 mol NaCl . The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg"} {"id": "textbook_3626", "title": "1223. Solution - ", "content": "$$\n1.0 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}\\left(\\frac{1000 \\mathrm{~g}}{1 \\mathrm{~kg}}\\right)\\left(\\frac{\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}}{18.02 \\mathrm{~g}}\\right)=55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_3627", "title": "1223. Solution - ", "content": "and then substituting these molar amounts into the definition for mole fraction."} {"id": "textbook_3628", "title": "1223. Solution - ", "content": "$$\n\\begin{aligned}\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=\\frac{\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}}{\\mathrm{~mol} \\mathrm{NaCl}+\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=\\frac{55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}{3.0 \\mathrm{~mol} \\mathrm{NaCl}+55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=0.95 \\\\\n& X_{\\mathrm{NaCl}}=\\frac{\\mathrm{mol} \\mathrm{NaCl}}{\\mathrm{~mol} \\mathrm{NaCl}+\\mathrm{mol} \\mathrm{H} \\mathrm{O}} \\\\\n& X_{\\mathrm{NaCl}}\n\\end{aligned}=\\frac{3.0 \\mathrm{~mol} \\mathrm{NaCl}}{3.0 \\mathrm{~mol} \\mathrm{NaCl}+55 \\mathrm{~mol} \\mathrm{H}}{ }_{2} \\mathrm{O} .0 .052 .\n$$\n"} {"id": "textbook_3629", "title": "1224. Check Your Learning - ", "content": "\nThe mole fraction of iodine, $\\mathrm{I}_{2}$, dissolved in dichloromethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$, is 0.115 . What is the molal concentration, $m$, of iodine in this solution?\n"} {"id": "textbook_3630", "title": "1225. Answer: - ", "content": "\n1.50 m\n"} {"id": "textbook_3631", "title": "1226. EXAMPLE 11.5 - ", "content": ""} {"id": "textbook_3632", "title": "1227. Molality and Molarity Conversions - ", "content": "\nIntravenous infusion of a 0.556 M aqueous solution of glucose (density of $1.04 \\mathrm{~g} / \\mathrm{mL}$ ) is part of some postoperative recovery therapies. What is the molal concentration of glucose in this solution?\n"} {"id": "textbook_3633", "title": "1228. Solution - ", "content": "\nThe provided molal concentration may be explicitly written as:"} {"id": "textbook_3634", "title": "1228. Solution - ", "content": "$$\nM=0.556 \\mathrm{~mol} \\text { glucose } / 1 \\mathrm{~L} \\text { solution }\n$$"} {"id": "textbook_3635", "title": "1228. Solution - ", "content": "Consider the definition of molality:"} {"id": "textbook_3636", "title": "1228. Solution - ", "content": "$$\nm=\\text { mol solute } / \\mathrm{kg} \\text { solvent }\n$$"} {"id": "textbook_3637", "title": "1228. Solution - ", "content": "The amount of glucose in 1-L of this solution is 0.556 mol , so the mass of water in this volume of solution is needed."} {"id": "textbook_3638", "title": "1228. Solution - ", "content": "First, compute the mass of 1.00 L of the solution:"} {"id": "textbook_3639", "title": "1228. Solution - ", "content": "$$\n(1.0 \\mathrm{~L} \\text { soln })(1.04 \\mathrm{~g} / \\mathrm{mL})(1000 \\mathrm{~mL} / 1 \\mathrm{~L})(1 \\mathrm{~kg} / 1000 \\mathrm{~g})=1.04 \\mathrm{~kg} \\text { soln }\n$$"} {"id": "textbook_3640", "title": "1228. Solution - ", "content": "This is the mass of both the water and its solute, glucose, and so the mass of glucose must be subtracted. Compute the mass of glucose from its molar amount:"} {"id": "textbook_3641", "title": "1228. Solution - ", "content": "$$\n(0.556 \\mathrm{~mol} \\text { glucose })(180.2 \\mathrm{~g} / 1 \\mathrm{~mol})=100.2 \\mathrm{~g} \\text { or } 0.1002 \\mathrm{~kg}\n$$"} {"id": "textbook_3642", "title": "1228. Solution - ", "content": "Subtracting the mass of glucose yields the mass of water in the solution:"} {"id": "textbook_3643", "title": "1228. Solution - ", "content": "$$\n1.04 \\mathrm{~kg} \\text { solution }-0.1002 \\mathrm{~kg} \\text { glucose }=0.94 \\mathrm{~kg} \\text { water }\n$$"} {"id": "textbook_3644", "title": "1228. Solution - ", "content": "Finally, the molality of glucose in this solution is computed as:"} {"id": "textbook_3645", "title": "1228. Solution - ", "content": "$$\nm=0.556 \\mathrm{~mol} \\text { glucose } / 0.94 \\mathrm{~kg} \\text { water }=0.59 \\mathrm{~m}\n$$\n"} {"id": "textbook_3646", "title": "1229. Check Your Learning - ", "content": "\nNitric acid, $\\mathrm{HNO}_{3}(\\mathrm{aq})$, is commercially available as a 33.7 m aqueous solution (density $=1.35 \\mathrm{~g} / \\mathrm{mL}$ ). What is the molarity of this solution?\n"} {"id": "textbook_3647", "title": "1230. Answer: - ", "content": "\n14.6 M\n"} {"id": "textbook_3648", "title": "1231. Vapor Pressure Lowering - ", "content": "\nAs described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:"} {"id": "textbook_3649", "title": "1231. Vapor Pressure Lowering - ", "content": "$$\n\\text { liquid } \\rightleftharpoons \\text { gas }\n$$"} {"id": "textbook_3650", "title": "1231. Vapor Pressure Lowering - ", "content": "Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid's vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure 11.18). While this interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the more dispersed nature of matter in a solution, compared to separate solvent and solute phases, serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly\nhigher boiling point as described in the next section of this module.\n"} {"id": "textbook_3651", "title": "1231. Vapor Pressure Lowering - ", "content": "FIGURE 11.18 The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules."} {"id": "textbook_3652", "title": "1231. Vapor Pressure Lowering - ", "content": "The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult's law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution."} {"id": "textbook_3653", "title": "1231. Vapor Pressure Lowering - ", "content": "$$\nP_{\\mathrm{A}}=X_{\\mathrm{A}} P_{\\mathrm{A}}^{*}\n$$"} {"id": "textbook_3654", "title": "1231. Vapor Pressure Lowering - ", "content": "where $P_{\\mathrm{A}}$ is the partial pressure exerted by component A in the solution, $P_{\\mathrm{A}}^{*}$ is the vapor pressure of pure A, and $X_{\\mathrm{A}}$ is the mole fraction of A in the solution."} {"id": "textbook_3655", "title": "1231. Vapor Pressure Lowering - ", "content": "Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton's law of partial pressures), the total vapor pressure exerted by a solution containing $i$ components is"} {"id": "textbook_3656", "title": "1231. Vapor Pressure Lowering - ", "content": "$$\nP_{\\text {solution }}=\\sum_{i} P_{i}=\\sum_{i} X_{i} P_{i}^{*}\n$$"} {"id": "textbook_3657", "title": "1231. Vapor Pressure Lowering - ", "content": "A nonvolatile substance is one whose vapor pressure is negligible ( $P^{*} \\approx 0$ ), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:"} {"id": "textbook_3658", "title": "1231. Vapor Pressure Lowering - ", "content": "$$\nP_{\\text {solution }}=X_{\\text {solvent }} P_{\\text {solvent }}^{*}\n$$"} {"id": "textbook_3659", "title": "1231. Vapor Pressure Lowering - ", "content": "ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, at $40^{\\circ} \\mathrm{C}$. The vapor pressure of pure ethanol is 0.178 atm at $40^{\\circ} \\mathrm{C}$. Glycerin is essentially nonvolatile at this temperature.\n"} {"id": "textbook_3660", "title": "1232. Solution - ", "content": "\nSince the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult's law as:"} {"id": "textbook_3661", "title": "1232. Solution - ", "content": "$$\nP_{\\text {solution }}=X_{\\text {solvent }} P_{\\text {solvent }}^{*}\n$$"} {"id": "textbook_3662", "title": "1232. Solution - ", "content": "First, calculate the molar amounts of each solution component using the provided mass data.\n$92.1-\\mathrm{g}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}}{92.094-\\frac{\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}}{}}=1.00 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}$\n$184.4-\\mathrm{EC}_{2} \\mathrm{H}_{5} \\mathrm{OH} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}}{46.069-\\mathrm{gC}_{2} \\mathrm{H}_{5} \\mathrm{OH}}=4.000 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\nNext, calculate the mole fraction of the solvent (ethanol) and use Raoult's law to compute the solution's vapor pressure."} {"id": "textbook_3663", "title": "1232. Solution - ", "content": "$$\n\\begin{aligned}\n& X_{\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}}=\\frac{4.000 \\mathrm{~mol}}{(1.00 \\mathrm{~mol}+4.000 \\mathrm{~mol})}=0.800 \\\\\n& P_{\\text {solv }}=X_{\\text {solv }} P_{\\text {solv }}^{*}=0.800 \\times 0.178 \\mathrm{~atm}=0.142 \\mathrm{~atm}\n\\end{aligned}\n$$\n"} {"id": "textbook_3664", "title": "1233. Check Your Learning - ", "content": "\nA solution contains 5.00 g of urea, $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$ (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at $25^{\\circ} \\mathrm{C}$ is 23.7 torr, what is the vapor pressure of the solution assuming ideal behavior?\n"} {"id": "textbook_3665", "title": "1234. Answer: - ", "content": "\n23.4 torr\n"} {"id": "textbook_3666", "title": "1235. Distillation of Solutions - ", "content": "\nSolutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult's law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected-effectively separating it from component B."} {"id": "textbook_3667", "title": "1235. Distillation of Solutions - ", "content": "Distillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in Figure 11.19.\n"} {"id": "textbook_3668", "title": "1235. Distillation of Solutions - ", "content": "FIGURE 11.19 A typical laboratory distillation unit is shown in (a) a photograph and (b) a schematic diagram of the components. (credit a: modification of work by \"Rifleman82\"/Wikimedia commons; credit b: modification of work by \"Slashme\"/Wikimedia Commons)"} {"id": "textbook_3669", "title": "1235. Distillation of Solutions - ", "content": "Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure 11.20.\n"} {"id": "textbook_3670", "title": "1235. Distillation of Solutions - ", "content": "Small molecules:"} {"id": "textbook_3671", "title": "1235. Distillation of Solutions - ", "content": "- Low boiling point\n- Very volatile\n- Flows easily\n- Ignites easily\n"} {"id": "textbook_3672", "title": "1235. Distillation of Solutions - ", "content": "Large molecules:"} {"id": "textbook_3673", "title": "1235. Distillation of Solutions - ", "content": "- High boiling point\n- Not very volatile\n- Does not flow easily\n- Does not ignite easily"} {"id": "textbook_3674", "title": "1235. Distillation of Solutions - ", "content": "FIGURE 11.20 Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures.\n"} {"id": "textbook_3675", "title": "1236. Boiling Point Elevation - ", "content": "\nAs described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution's boiling point will subsequently be increased. Vapor pressure increases with temperature, and so a solution will require a higher temperature than will pure solvent to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, $\\Delta T_{\\mathrm{b}}$, is called boiling point elevation and is directly proportional to the molal concentration of solute species:"} {"id": "textbook_3676", "title": "1236. Boiling Point Elevation - ", "content": "$$\n\\Delta T_{\\mathrm{b}}=K_{\\mathrm{b}} m\n$$"} {"id": "textbook_3677", "title": "1236. Boiling Point Elevation - ", "content": "where $K_{\\mathrm{b}}$ is the boiling point elevation constant, or the ebullioscopic constant and $m$ is the molal concentration (molality) of all solute species.\n\nBoiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of $K_{\\mathrm{b}}$ for several solvents are listed in Table 11.2."} {"id": "textbook_3678", "title": "1236. Boiling Point Elevation - ", "content": "Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents"} {"id": "textbook_3679", "title": "1236. Boiling Point Elevation - ", "content": "| Solvent | Boiling Point ( ${ }^{\\circ} \\mathrm{C}$ at 1 atm$)$ | $K_{\\mathrm{b}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ | Freezing Point $\\left({ }^{\\circ} \\mathrm{C}\\right.$ at 1 atm$)$ | $K_{\\mathrm{f}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| water | 100.0 | 0.512 | 0.0 | 1.86 |"} {"id": "textbook_3680", "title": "1236. Boiling Point Elevation - ", "content": "TABLE 11.2"} {"id": "textbook_3681", "title": "1236. Boiling Point Elevation - ", "content": "| Solvent | Boiling Point ( ${ }^{\\circ} \\mathrm{C}$ at $\\left.\\mathbf{1} \\mathbf{~ a t m}\\right)$ | $K_{\\mathrm{b}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ | Freezing Point ( ${ }^{\\circ} \\mathrm{C}$ at 1 atm$)$ | $K_{\\mathrm{f}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| hydrogen acetate | 118.1 | 3.07 | 16.6 | 3.9 |\n| benzene | 80.1 | 2.53 | 5.5 | 5.12 |\n| chloroform | 61.26 | 3.63 | -63.5 | 4.68 |\n| nitrobenzene | 210.9 | 5.24 | 5.67 | 8.1 |"} {"id": "textbook_3682", "title": "1236. Boiling Point Elevation - ", "content": "TABLE 11.2"} {"id": "textbook_3683", "title": "1236. Boiling Point Elevation - ", "content": "The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose ( $342 \\mathrm{~g} / \\mathrm{mol}$ ) and a 1 m aqueous solution of ethylene glycol ( $62 \\mathrm{~g} / \\mathrm{mol}$ ) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.\n"} {"id": "textbook_3684", "title": "1237. EXAMPLE 11.7 - ", "content": ""} {"id": "textbook_3685", "title": "1238. Calculating the Boiling Point of a Solution - ", "content": "\nAssuming ideal solution behavior, what is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?\n"} {"id": "textbook_3686", "title": "1239. Solution - ", "content": "\nUse the equation relating boiling point elevation to solute molality to solve this problem in two steps.\n"} {"id": "textbook_3687", "title": "1239. Solution - ", "content": "Step 1. Calculate the change in boiling point."} {"id": "textbook_3688", "title": "1239. Solution - ", "content": "$$\n\\Delta T_{\\mathrm{b}}=K_{\\mathrm{b}} m=2.53{ }^{\\circ} \\mathrm{Cm}^{-1} \\times 0.33 m=0.83^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_3689", "title": "1239. Solution - ", "content": "Step 2. Add the boiling point elevation to the pure solvent's boiling point.\nBoiling temperature $=80.1^{\\circ} \\mathrm{C}+0.83{ }^{\\circ} \\mathrm{C}=80.9^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3690", "title": "1240. Check Your Learning - ", "content": "\nAssuming ideal solution behavior, what is the boiling point of the antifreeze described in Example 11.3?\n"} {"id": "textbook_3691", "title": "1241. Answer: - ", "content": "\n$109.2^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3692", "title": "1242. EXAMPLE 11.8 - ", "content": ""} {"id": "textbook_3693", "title": "1243. The Boiling Point of an Iodine Solution - ", "content": "\nFind the boiling point of a solution of 92.1 g of iodine, $\\mathrm{I}_{2}$, in 800.0 g of chloroform, $\\mathrm{CHCl}_{3}$, assuming that the iodine is nonvolatile and that the solution is ideal.\n"} {"id": "textbook_3694", "title": "1244. Solution - ", "content": "\nA four-step approach to solving this problem is outlined below.\n"} {"id": "textbook_3695", "title": "1244. Solution - ", "content": "Step 1. Convert from grams to moles of $\\mathrm{I}_{2}$ using the molar mass of $\\mathrm{I}_{2}$ in the unit conversion factor. Result: 0.363 mol\nStep 2. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.\nResult: 0.454 m\nStep 3. Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.\nResult: $1.65^{\\circ} \\mathrm{C}$\nStep 4. Determine the new boiling point from the boiling point of the pure solvent and the change.\nResult: $62.91{ }^{\\circ} \\mathrm{C}$\nCheck each result as a self-assessment.\n"} {"id": "textbook_3696", "title": "1245. Check Your Learning - ", "content": "\nWhat is the boiling point of a solution of 1.0 g of glycerin, $\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}$, in 47.8 g of water? Assume an ideal solution.\n"} {"id": "textbook_3697", "title": "1246. Answer: - ", "content": "\n$100.12{ }^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3698", "title": "1247. Freezing Point Depression - ", "content": "\nSolutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in \"de-icing\" schemes that use salt (Figure 11.21), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an \"antifreeze\" in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below $0^{\\circ} \\mathrm{C}$ (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).\n"} {"id": "textbook_3699", "title": "1247. Freezing Point Depression - ", "content": "FIGURE 11.21 Rock salt $(\\mathrm{NaCl})$, calcium chloride $\\left(\\mathrm{CaCl}_{2}\\right)$, or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)"} {"id": "textbook_3700", "title": "1247. Freezing Point Depression - ", "content": "The decrease in freezing point of a dilute solution compared to that of the pure solvent, $\\Delta T_{\\mathrm{f}}$, is called the freezing point depression and is directly proportional to the molal concentration of the solute"} {"id": "textbook_3701", "title": "1247. Freezing Point Depression - ", "content": "$$\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m\n$$"} {"id": "textbook_3702", "title": "1247. Freezing Point Depression - ", "content": "where $m$ is the molal concentration of the solute and $K_{\\mathrm{f}}$ is called the freezing point depression constant (or cryoscopic constant). Just as for boiling point elevation constants, these are characteristic properties whose\nvalues depend on the chemical identity of the solvent. Values of $K_{\\mathrm{f}}$ for several solvents are listed in Table 11.2.\n"} {"id": "textbook_3703", "title": "1248. EXAMPLE 11.9 - ", "content": ""} {"id": "textbook_3704", "title": "1249. Calculation of the Freezing Point of a Solution - ", "content": "\nAssuming ideal solution behavior, what is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example 11.4?\n"} {"id": "textbook_3705", "title": "1250. Solution - ", "content": "\nUse the equation relating freezing point depression to solute molality to solve this problem in two steps.\n"} {"id": "textbook_3706", "title": "1250. Solution - ", "content": "Step 1. Calculate the change in freezing point."} {"id": "textbook_3707", "title": "1250. Solution - ", "content": "$$\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m=5.12^{\\circ} \\mathrm{C} \\mathrm{~m}^{-1} \\times 0.33 \\mathrm{~m}=1.7^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_3708", "title": "1250. Solution - ", "content": "Step 2. Subtract the freezing point change observed from the pure solvent's freezing point.\nFreezing Temperature $=5.5^{\\circ} \\mathrm{C}-1.7^{\\circ} \\mathrm{C}=3.8^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3709", "title": "1251. Check Your Learning - ", "content": "\nAssuming ideal solution behavior, what is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?\n"} {"id": "textbook_3710", "title": "1252. Answer: - ", "content": "\n$-9.3^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3711", "title": "1253. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3712", "title": "1254. Colligative Properties and De-Icing - ", "content": "\nSodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than $0{ }^{\\circ} \\mathrm{C}$, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (\"rock salt\") for use on roads, since they tend to be somewhat less corrosive than the NaCl , and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride."} {"id": "textbook_3713", "title": "1254. Colligative Properties and De-Icing - ", "content": "Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Figure 11.22).\n\n(a)\n\n(b)"} {"id": "textbook_3714", "title": "1254. Colligative Properties and De-Icing - ", "content": "FIGURE 11.22 Freezing point depression is exploited to remove ice from (a) roadways and (b) the control surfaces of aircraft.\n"} {"id": "textbook_3715", "title": "1255. Phase Diagram for a Solution - ", "content": "\nThe colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid (Figure 11.23).\n"} {"id": "textbook_3716", "title": "1255. Phase Diagram for a Solution - ", "content": "FIGURE 11.23 Phase diagrams for a pure solvent (solid curves) and a solution formed by dissolving nonvolatile solute in the solvent (dashed curves)."} {"id": "textbook_3717", "title": "1255. Phase Diagram for a Solution - ", "content": "The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, $\\Delta P$, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution's boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, $\\Delta T_{\\mathrm{b}}$, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, $\\Delta T_{\\mathrm{f}}$, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that\nundergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed of solvent only, and so transitions between these phases are not subject to colligative effects.\n"} {"id": "textbook_3718", "title": "1256. Osmosis and Osmotic Pressure of Solutions - ", "content": "\nA number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes."} {"id": "textbook_3719", "title": "1256. Osmosis and Osmotic Pressure of Solutions - ", "content": "Consider the apparatus illustrated in Figure 11.24, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.\n\n\nFIGURE 11.24 (a) A solution and pure solvent are initially separated by an osmotic membrane. (b) Net transfer of solvent molecules to the solution occurs until its osmotic pressure yields equal rates of transfer in both directions."} {"id": "textbook_3720", "title": "1256. Osmosis and Osmotic Pressure of Solutions - ", "content": "When osmosis is carried out in an apparatus like that shown in Figure 11.24, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ( $\\boldsymbol{\\Pi}$ ) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, $M$, and absolute temperature, $T$, according to the equation"} {"id": "textbook_3721", "title": "1256. Osmosis and Osmotic Pressure of Solutions - ", "content": "$$\n\\Pi=M R T\n$$"} {"id": "textbook_3722", "title": "1256. Osmosis and Osmotic Pressure of Solutions - ", "content": "where $R$ is the universal gas constant.\n"} {"id": "textbook_3723", "title": "1257. EXAMPLE 11.10 - ", "content": ""} {"id": "textbook_3724", "title": "1258. Calculation of Osmotic Pressure - ", "content": "\nAssuming ideal solution behavior, what is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, $37^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3725", "title": "1259. Solution - ", "content": "\nFind the osmotic pressure, $\\Pi$, using the formula $\\Pi=M R T$, where $T$ is on the Kelvin scale ( 310 K ) and the value of $R$ is expressed in appropriate units ( $0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{K}$ )."} {"id": "textbook_3726", "title": "1259. Solution - ", "content": "$$\n\\begin{aligned}\n\\Pi & =M R T \\\\\n& =0.30 \\mathrm{~mol} / \\mathrm{L} \\times 0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K} \\times 310 \\mathrm{~K} \\\\\n& =7.6 \\mathrm{~atm}\n\\end{aligned}\n$$\n"} {"id": "textbook_3727", "title": "1260. Check Your Learning - ", "content": "\nAssuming ideal solution behavior, what is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, in water at $37^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3728", "title": "1261. Answer: - ", "content": "\n5.3 atm"} {"id": "textbook_3729", "title": "1261. Answer: - ", "content": "If a solution is placed in an apparatus like the one shown in Figure 11.25, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.\n\n\nFIGURE 11.25 Applying a pressure greater than the osmotic pressure of a solution will reverse osmosis. Solvent molecules from the solution are pushed into the pure solvent.\n"} {"id": "textbook_3730", "title": "1262. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_3731", "title": "1263. Reverse Osmosis Water Purification - ", "content": "\nIn the process of osmosis, diffusion serves to move water through a semipermeable membrane from a less concentrated solution to a more concentrated solution. Osmotic pressure is the amount of pressure that must be applied to the more concentrated solution to cause osmosis to stop. If greater pressure is applied, the water will go from the more concentrated solution to a less concentrated (more pure) solution. This is called reverse osmosis. Reverse osmosis (RO) is used to purify water in many applications, from desalination plants in coastal cities, to water-purifying machines in grocery stores (Figure 11.26), and smaller reverse-osmosis household units. With a hand-operated pump, small RO units can be used in third-world countries, disaster areas, and in lifeboats. Our military forces have a variety of generatoroperated RO units that can be transported in vehicles to remote locations.\n"} {"id": "textbook_3732", "title": "1263. Reverse Osmosis Water Purification - ", "content": "FIGURE 11.26 Reverse osmosis systems for purifying drinking water are shown here on (a) small and (b) large scales. (credit a: modification of work by Jerry Kirkhart; credit b: modification of work by Willard J. Lathrop)"} {"id": "textbook_3733", "title": "1263. Reverse Osmosis Water Purification - ", "content": "Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm . Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure 11.27.\n"} {"id": "textbook_3734", "title": "1263. Reverse Osmosis Water Purification - ", "content": "FIGURE 11.27 Red blood cell membranes are water permeable and will (a) swell and possibly rupture in a hypotonic solution; (b) maintain normal volume and shape in an isotonic solution; and (c) shrivel and possibly die in a hypertonic solution. (credit a/b/c: modifications of work by \"LadyofHats\"/Wikimedia commons)\n"} {"id": "textbook_3735", "title": "1264. Determination of Molar Masses - ", "content": "\nOsmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the number of solute species present in a given amount of solution. Consequently, measuring one of these properties for a solution prepared using a known mass of solute permits determination of the solute's molar mass.\n"} {"id": "textbook_3736", "title": "1265. EXAMPLE 11.11 - ", "content": ""} {"id": "textbook_3737", "title": "1266. Determination of a Molar Mass from a Freezing Point Depression - ", "content": "\nA solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at $2.32^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of this compound?\n"} {"id": "textbook_3738", "title": "1267. Solution - ", "content": "\nSolve this problem using the following steps.\n"} {"id": "textbook_3739", "title": "1267. Solution - ", "content": "Step 1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 11.2)."} {"id": "textbook_3740", "title": "1267. Solution - ", "content": "$$\n\\Delta T_{\\mathrm{f}}=5.5^{\\circ} \\mathrm{C}-2.32^{\\circ} \\mathrm{C}=3.2^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_3741", "title": "1267. Solution - ", "content": "Step 2. Determine the molal concentration from $K_{\\mathrm{f}}$, the freezing point depression constant for benzene (Table 11.2), and $\\Delta T_{\\mathrm{f}}$."} {"id": "textbook_3742", "title": "1267. Solution - ", "content": "$$\n\\begin{gathered}\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m \\\\\nm=\\frac{\\Delta T_{\\mathrm{f}}}{K_{\\mathrm{f}}}=\\frac{3.2^{\\circ} \\mathrm{C}}{5.12^{\\circ} \\mathrm{C} m^{-1}}=0.63 \\mathrm{~m}\n\\end{gathered}\n$$"} {"id": "textbook_3743", "title": "1267. Solution - ", "content": "Step 3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution."} {"id": "textbook_3744", "title": "1267. Solution - ", "content": "$$\n\\text { Moles of solute }=\\frac{0.63 \\mathrm{~mol} \\text { solute }}{1.00 \\mathrm{~kg} \\text { solvent }} \\times 0.0550 \\mathrm{~kg} \\text { solvent }=0.035 \\mathrm{~mol}\n$$"} {"id": "textbook_3745", "title": "1267. Solution - ", "content": "Step 4. Determine the molar mass from the mass of the solute and the number of moles in that mass."} {"id": "textbook_3746", "title": "1267. Solution - ", "content": "$$\n\\text { Molar mass }=\\frac{4.00 \\mathrm{~g}}{0.035 \\mathrm{~mol}}=1.1 \\times 10^{2} \\mathrm{~g} / \\mathrm{mol}\n$$\n"} {"id": "textbook_3747", "title": "1268. Check Your Learning - ", "content": "\nA solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of $64.5^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of this compound?\n"} {"id": "textbook_3748", "title": "1269. Answer: - ", "content": "\n$1.8 \\times 10^{2} \\mathrm{~g} / \\mathrm{mol}$\n"} {"id": "textbook_3749", "title": "1270. EXAMPLE 11.12 - ", "content": ""} {"id": "textbook_3750", "title": "1271. Determination of a Molar Mass from Osmotic Pressure - ", "content": "\nA 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at $22{ }^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of hemoglobin?\n"} {"id": "textbook_3751", "title": "1272. Solution - ", "content": "\nHere is one set of steps that can be used to solve the problem:\n"} {"id": "textbook_3752", "title": "1272. Solution - ", "content": "Step 1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure."} {"id": "textbook_3753", "title": "1272. Solution - ", "content": "$$\n\\begin{gathered}\n\\Pi=\\frac{5.9 \\mathrm{torr} \\times 1 \\mathrm{~atm}}{760 \\mathrm{torr}}=7.8 \\times 10^{-3} \\mathrm{~atm} \\\\\n\\Pi=M R T \\\\\nM=\\frac{\\Pi}{R T}=\\frac{7.8 \\times 10^{-3} \\mathrm{~atm}}{(0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K})(295 \\mathrm{~K})}=3.2 \\times 10^{-4} \\mathrm{M}\n\\end{gathered}\n$$"} {"id": "textbook_3754", "title": "1272. Solution - ", "content": "Step 2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution."} {"id": "textbook_3755", "title": "1272. Solution - ", "content": "$$\n\\text { moles of hemoglobin }=\\frac{3.2 \\times 10^{-4} \\mathrm{~mol}}{1 \\mathrm{Lsolttion}} \\times 0.500 \\mathrm{Lsolution}=1.6 \\times 10^{-4} \\mathrm{~mol}\n$$"} {"id": "textbook_3756", "title": "1272. Solution - ", "content": "Step 3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass."} {"id": "textbook_3757", "title": "1272. Solution - ", "content": "$$\n\\text { molar mass }=\\frac{10.0 \\mathrm{~g}}{1.6 \\times 10^{-4} \\mathrm{~mol}}=6.2 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}\n$$\n"} {"id": "textbook_3758", "title": "1273. Check Your Learning - ", "content": "\nAssuming ideal solution behavior, what is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at $25^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_3759", "title": "1274. Answer: - ", "content": "\n$3 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}$\n"} {"id": "textbook_3760", "title": "1275. Colligative Properties of Electrolytes - ", "content": "\nAs noted previously in this module, the colligative properties of a solution depend only on the number, not on the identity, of solute species dissolved. The concentration terms in the equations for various colligative properties (freezing point depression, boiling point elevation, osmotic pressure) pertain to all solute species present in the solution. For the solutions considered thus far in this chapter, the solutes have been nonelectrolytes that dissolve physically without dissociation or any other accompanying process. Each molecule that dissolves yields one dissolved solute molecule. The dissolution of an electroyte, however, is not this simple, as illustrated by the two common examples below:"} {"id": "textbook_3761", "title": "1275. Colligative Properties of Electrolytes - ", "content": "$$\n\\begin{array}{ll}\n\\text { dissociation } & \\mathrm{NaCl}(\\mathrm{~s}) \\longrightarrow \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) \\\\\n\\text { ionization } & \\mathrm{HCl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\longrightarrow \\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})\n\\end{array}\n$$"} {"id": "textbook_3762", "title": "1275. Colligative Properties of Electrolytes - ", "content": "Considering the first of these examples, and assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions ( 1.0 mol Na and $1.0 \\mathrm{~mol} \\mathrm{Cl}^{-}$) per each kilogram of water, and its freezing point depression is expected to be"} {"id": "textbook_3763", "title": "1275. Colligative Properties of Electrolytes - ", "content": "$$\n\\Delta T_{\\mathrm{f}}=2.0 \\mathrm{~mol} \\text { ions } / \\mathrm{kg} \\text { water } \\times 1.86^{\\circ} \\mathrm{C} \\mathrm{~kg} \\text { water } / \\mathrm{mol} \\text { ion }=3.7^{\\circ} \\mathrm{C} .\n$$"} {"id": "textbook_3764", "title": "1275. Colligative Properties of Electrolytes - ", "content": "When this solution is actually prepared and its freezing point depression measured, however, a value of $3.4^{\\circ} \\mathrm{C}$ is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution."} {"id": "textbook_3765", "title": "1275. Colligative Properties of Electrolytes - ", "content": "To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van't Hoff is used. The van't Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:"} {"id": "textbook_3766", "title": "1275. Colligative Properties of Electrolytes - ", "content": "$$\ni=\\frac{\\text { moles of particles in solution }}{\\text { moles of formula units dissolved }}\n$$"} {"id": "textbook_3767", "title": "1275. Colligative Properties of Electrolytes - ", "content": "Values for measured van't Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table 11.3."} {"id": "textbook_3768", "title": "1275. Colligative Properties of Electrolytes - ", "content": "Predicated and Measured van't Hoff Factors for Several 0.050 m Aqueous Solutions"} {"id": "textbook_3769", "title": "1275. Colligative Properties of Electrolytes - ", "content": "| Formula unit | Classification | Dissolution products | $\\boldsymbol{i}$ (predicted) | $\\boldsymbol{i}$ (measured) |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ (glucose) | Nonelectrolyte | $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ | 1 | 1.0 |\n| NaCl | Strong electrolyte | $\\mathrm{Na}^{+}, \\mathrm{Cl}^{-}$ | 2 | 1.9 |\n| HCl | Strong electrolyte (acid) | $\\mathrm{H}_{3} \\mathrm{O}^{+}, \\mathrm{Cl}^{-}$ | 2 | 1.9 |\n| $\\mathrm{MgSO}_{4}$ | Strong electrolyte | $\\mathrm{Mg}^{2+}, \\mathrm{SO}_{4}{ }^{2-}$, | 2 | 1.3 |"} {"id": "textbook_3770", "title": "1275. Colligative Properties of Electrolytes - ", "content": "TABLE 11.3"} {"id": "textbook_3771", "title": "1275. Colligative Properties of Electrolytes - ", "content": "| Formula unit | Classification | Dissolution products | $\\boldsymbol{i}$ (predicted) | $\\boldsymbol{i}$ (measured) |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{MgCl}_{2}$ | Strong electrolyte | $\\mathrm{Mg}^{2+}, 2 \\mathrm{Cl}^{-}$ | 3 | 2.7 |\n| $\\mathrm{FeCl}_{3}$ | Strong electrolyte | $\\mathrm{Fe}^{3+}, 3 \\mathrm{Cl}^{-}$ | 4 | 3.4 |"} {"id": "textbook_3772", "title": "1275. Colligative Properties of Electrolytes - ", "content": "TABLE 11.3"} {"id": "textbook_3773", "title": "1275. Colligative Properties of Electrolytes - ", "content": "In 1923, the chemists Peter Debye and Erich H\u00fcckel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure 11.28). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van't Hoff factors for the electrolytes in Table 11.3 are for 0.05 m solutions, at which concentration the value of $i$ for NaCl is 1.9 , as opposed to an ideal value of 2 .\n"} {"id": "textbook_3774", "title": "1275. Colligative Properties of Electrolytes - ", "content": "FIGURE 11.28 Dissociation of ionic compounds in water is not always complete due to the formation of ion pairs.\n"} {"id": "textbook_3775", "title": "1276. EXAMPLE 11.13 - ", "content": ""} {"id": "textbook_3776", "title": "1277. The Freezing Point of a Solution of an Electrolyte - ", "content": "\nThe concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Use this information and a predicted value for the van't Hoff factor (Table 11.3) to\ndetermine the freezing temperature the solution (assume ideal solution behavior).\n"} {"id": "textbook_3777", "title": "1278. Solution - ", "content": "\nSolve this problem using the following series of steps.\n"} {"id": "textbook_3778", "title": "1278. Solution - ", "content": "Step 1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl\nStep 2. Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor ( 2 mol ions $/ 1 \\mathrm{~mol} \\mathrm{NaCl}$ ).\nResult: 0.14 mol ions\nStep 3. Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms.\nResult: 1.2 m\nStep 4. Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes.\nResult: $2.1^{\\circ} \\mathrm{C}$\nStep 5. Determine the new freezing point from the freezing point of the pure solvent and the change.\nResult: $-2.1^{\\circ} \\mathrm{C}$\nCheck each result as a self-assessment, taking care to avoid rounding errors by retaining guard digits in each step's result for computing the next step's result.\n"} {"id": "textbook_3779", "title": "1279. Check Your Learning - ", "content": "\nAssuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of 0.724 g of $\\mathrm{CaCl}_{2}$ in 175 g of water.\n"} {"id": "textbook_3780", "title": "1280. Answer: - ", "content": "\n$-0.208{ }^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_3781", "title": "1280. Answer: - 1280.1. Colloids", "content": ""} {"id": "textbook_3782", "title": "1281. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3783", "title": "1281. LEARNING OBJECTIVES - ", "content": "- Describe the composition and properties of colloidal dispersions\n- List and explain several technological applications of colloids"} {"id": "textbook_3784", "title": "1281. LEARNING OBJECTIVES - ", "content": "As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, a solution is a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. Another class of mixtures called colloids (or colloidal dispersions) exhibit properties intermediate between those of suspensions and solutions (Figure 11.29). The particles in a colloid are larger than most simple molecules;\nhowever, colloidal particles are small enough that they do not settle out upon standing.\n\n\nFIGURE 11.29 (a) A solution is a homogeneous mixture that appears clear, such as the saltwater in this aquarium. (b) In a colloid, such as milk, the particles are much larger but remain dispersed and do not settle. (c) A suspension, such as mud, is a heterogeneous mixture of suspended particles that appears cloudy and in which the particles can settle. (credit a photo: modification of work by Adam Wimsatt; credit b photo: modification of work by Melissa Wiese; credit c photo: modification of work by Peter Burgess)"} {"id": "textbook_3785", "title": "1281. LEARNING OBJECTIVES - ", "content": "The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure 11.30. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.\n"} {"id": "textbook_3786", "title": "1281. LEARNING OBJECTIVES - ", "content": "FIGURE 11.30 The paths of searchlight beams are made visible when light is scattered by colloidal-size particles in the air (fog, smoke, etc.). (credit: \"Bahman\"/Wikimedia Commons)"} {"id": "textbook_3787", "title": "1281. LEARNING OBJECTIVES - ", "content": "The term \"colloid\"-from the Greek words kolla, meaning \"glue,\" and eidos, meaning \"like\"-was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units."} {"id": "textbook_3788", "title": "1281. LEARNING OBJECTIVES - ", "content": "Analogous to the identification of solution components as \"solute\" and \"solvent,\" the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium. Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table 11.4."} {"id": "textbook_3789", "title": "1281. LEARNING OBJECTIVES - ", "content": "Examples of Colloidal Systems"} {"id": "textbook_3790", "title": "1281. LEARNING OBJECTIVES - ", "content": "| Dispersed Phase | Dispersion Medium | Common Examples | Name |\n| :--- | :--- | :--- | :--- |\n| solid | gas | smoke, dust | - |\n| solid | liquid | starch in water, some inks, paints, milk of magnesia | sol |\n| solid | solid | some colored gems, some alloys | - |\n| liquid | liquid | clouds, fogs, mists, sprays | aerosol |\n| liquid | solid | milk, mayonnaise, butter | emulsion |\n| liquid | jellies, gels, pearl, opal $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right.$ in $\\left.\\mathrm{SiO}_{2}\\right)$ | gel | |"} {"id": "textbook_3791", "title": "1281. LEARNING OBJECTIVES - ", "content": "TABLE 11.4"} {"id": "textbook_3792", "title": "1281. LEARNING OBJECTIVES - ", "content": "| Dispersed Phase | Dispersion Medium | Common Examples | Name |\n| :--- | :--- | :--- | :--- |\n| gas | liquid | foams, whipped cream, beaten egg whites | foam |\n| gas | solid | pumice, floating soaps | - |"} {"id": "textbook_3793", "title": "1281. LEARNING OBJECTIVES - ", "content": "TABLE 11.4\n"} {"id": "textbook_3794", "title": "1282. Preparation of Colloidal Systems - ", "content": "\nColloids are prepared by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods:"} {"id": "textbook_3795", "title": "1282. Preparation of Colloidal Systems - ", "content": "1. Dispersion methods: breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.\n2. Condensation methods: growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets."} {"id": "textbook_3796", "title": "1282. Preparation of Colloidal Systems - ", "content": "A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air."} {"id": "textbook_3797", "title": "1282. Preparation of Colloidal Systems - ", "content": "An emulsion may be prepared by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent, a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein serving as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents."} {"id": "textbook_3798", "title": "1282. Preparation of Colloidal Systems - ", "content": "Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. A red colloidal suspension of iron(III) hydroxide may be prepared by mixing a concentrated solution of iron(III) chloride with hot water:"} {"id": "textbook_3799", "title": "1282. Preparation of Colloidal Systems - ", "content": "$$\n\\mathrm{Fe}^{3+}(\\mathrm{aq})+3 \\mathrm{Cl}^{-}(\\mathrm{aq})+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Fe}(\\mathrm{OH})_{3}(s)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+3 \\mathrm{Cl}^{-}(a q) .\n$$"} {"id": "textbook_3800", "title": "1282. Preparation of Colloidal Systems - ", "content": "A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate:"} {"id": "textbook_3801", "title": "1282. Preparation of Colloidal Systems - ", "content": "$$\n\\mathrm{Au}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}\n$$"} {"id": "textbook_3802", "title": "1282. Preparation of Colloidal Systems - ", "content": "Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids.\n"} {"id": "textbook_3803", "title": "1283. Soaps and Detergents - ", "content": "\nPioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, $\\mathrm{K}_{2} \\mathrm{CO}_{3}$, from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula $\\mathrm{C}_{17} \\mathrm{H}_{35} \\mathrm{CO}_{2} \\mathrm{Na}$ and contains an uncharged nonpolar hydrocarbon chain, the $\\mathrm{C}_{17} \\mathrm{H}_{35}$ - unit, and an ionic carboxylate group, the $-\\mathrm{CO}_{2}{ }^{-}$unit (Figure 11.31).\n\n
Thermodynamics - ", "content": "\n"} {"id": "textbook_3836", "title": "1293. CHAPTER 12
Thermodynamics - ", "content": "Figure 12.1 Geysers are a dramatic display of thermodynamic principles in nature. Water deep within the underground channels of the geyser is under high pressure and heated to high temperature by magma. When a pocket of water near the surface reaches boiling point and is expelled, the resulting drop in pressure causes larger volumes of water to flash boil, forcefully ejecting steam and water in an impressive eruption. (credit: modification of work by Yellowstone National Park)\n"} {"id": "textbook_3837", "title": "1294. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_3838", "title": "1294. CHAPTER OUTLINE - 1294.1. Spontaneity", "content": "\n12.2 Entropy\n12.3 The Second and Third Laws of Thermodynamics\n12.4 Free Energy"} {"id": "textbook_3839", "title": "1294. CHAPTER OUTLINE - 1294.1. Spontaneity", "content": "INTRODUCTION Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions.\n"} {"id": "textbook_3840", "title": "1294. CHAPTER OUTLINE - 1294.2. Spontaneity", "content": ""} {"id": "textbook_3841", "title": "1295. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3842", "title": "1295. LEARNING OBJECTIVES - ", "content": "- Distinguish between spontaneous and nonspontaneous processes\n- Describe the dispersal of matter and energy that accompanies certain spontaneous processes"} {"id": "textbook_3843", "title": "1295. LEARNING OBJECTIVES - ", "content": "Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth's atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is \"driven\" by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze."} {"id": "textbook_3844", "title": "1295. LEARNING OBJECTIVES - ", "content": "The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure 12.2).\n"} {"id": "textbook_3845", "title": "1295. LEARNING OBJECTIVES - ", "content": "FIGURE 12.2 Both U-238 and Tc-99m undergo spontaneous radioactive decay, but at drastically different rates. Over the course of one week, essentially all of a Tc-99m sample and none of a U-238 sample will have decayed."} {"id": "textbook_3846", "title": "1295. LEARNING OBJECTIVES - ", "content": "As another example, consider the conversion of diamond into graphite (Figure 12.3)."} {"id": "textbook_3847", "title": "1295. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{C}(s, \\text { diamond }) \\longrightarrow \\mathrm{C}(s, \\text { graphite })\n$$"} {"id": "textbook_3848", "title": "1295. LEARNING OBJECTIVES - ", "content": "The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow; so, for all practical purposes diamonds are indeed \"forever.\" Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.\n"} {"id": "textbook_3849", "title": "1295. LEARNING OBJECTIVES - ", "content": "FIGURE 12.3 The conversion of carbon from the diamond allotrope to the graphite allotrope is spontaneous at ambient pressure, but its rate is immeasurably slow at low to moderate temperatures. This process is known as graphitization, and its rate can be increased to easily measurable values at temperatures in the 1000-2000 K range. (credit \"diamond\" photo: modification of work by \"Fancy Diamonds\"/Flickr; credit \"graphite\" photo: modification of work by images-of-elements.com/carbon.php)\n"} {"id": "textbook_3850", "title": "1296. Dispersal of Matter and Energy - ", "content": "\nExtending the discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas in one flask and the other flask is empty $(P=0)$. (Figure 12.4). When the valve is opened, the gas spontaneously expands to fill both flasks equally. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero."} {"id": "textbook_3851", "title": "1296. Dispersal of Matter and Energy - ", "content": "$$\nw=-P \\Delta V=0 \\quad(P=0 \\text { in a vacuum })\n$$"} {"id": "textbook_3852", "title": "1296. Dispersal of Matter and Energy - ", "content": "Note as well that since the system is isolated, no heat has been exchanged with the surroundings ( $q=0$ ). The first law of thermodynamics confirms that there has been no change in the system's internal energy as a result of this process."} {"id": "textbook_3853", "title": "1296. Dispersal of Matter and Energy - ", "content": "$$\n\\Delta U=q+w=0+0=0\n$$"} {"id": "textbook_3854", "title": "1296. Dispersal of Matter and Energy - ", "content": "The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous expansion took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).\n"} {"id": "textbook_3855", "title": "1296. Dispersal of Matter and Energy - ", "content": "FIGURE 12.4 An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a\nsecond flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks."} {"id": "textbook_3856", "title": "1296. Dispersal of Matter and Energy - ", "content": "Now consider two objects at different temperatures: object X at temperature $T_{\\mathrm{X}}$ and object Y at temperature $T_{\\mathrm{Y}}$, with $T_{\\mathrm{X}}>T_{\\mathrm{Y}}$ (Figure 12.5). When these objects come into contact, heat spontaneously flows from the hotter object $(\\mathrm{X})$ to the colder one $(\\mathrm{Y})$. This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y."} {"id": "textbook_3857", "title": "1296. Dispersal of Matter and Energy - ", "content": "$$\nq_{\\mathrm{X}}<0 \\quad \\text { and } \\quad q_{\\mathrm{Y}}=-q_{\\mathrm{X}}>0\n$$"} {"id": "textbook_3858", "title": "1296. Dispersal of Matter and Energy - ", "content": "From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.\n"} {"id": "textbook_3859", "title": "1296. Dispersal of Matter and Energy - ", "content": "FIGURE 12.5 When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object."} {"id": "textbook_3860", "title": "1296. Dispersal of Matter and Energy - ", "content": "As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.\n"} {"id": "textbook_3861", "title": "1297. EXAMPLE 12.1 - ", "content": ""} {"id": "textbook_3862", "title": "1298. Redistribution of Matter during a Spontaneous Process - ", "content": "\nDescribe how matter is redistributed when the following spontaneous processes take place:\n(a) A solid sublimes.\n(b) A gas condenses.\n(c) A drop of food coloring added to a glass of water forms a solution with uniform color.\n"} {"id": "textbook_3863", "title": "1299. Solution - ", "content": "\n"} {"id": "textbook_3864", "title": "1299. Solution - ", "content": "FIGURE 12.6 (credit a: modification of work by Jenny Downing; credit b: modification of work by \"Fuzzy Gerdes\"/Flickr; credit c: modification of work by Paul A. Flowers)\n(a) Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition.\n(b) Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after\nthe gas-to-liquid transition.\n(c) The process in question is diffusion. This process yields a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop of dye, zero in the water), and the final state of the system contains a single dye concentration throughout.\n"} {"id": "textbook_3865", "title": "1300. Check Your Learning - ", "content": "\nDescribe how energy is redistributed when a spoon at room temperature is placed in a cup of hot coffee.\n"} {"id": "textbook_3866", "title": "1301. Answer: - ", "content": "\nHeat will spontaneously flow from the hotter object (coffee) to the colder object (spoon), resulting in a more uniform distribution of thermal energy as the spoon warms and the coffee cools.\n"} {"id": "textbook_3867", "title": "1301. Answer: - 1301.1. Entropy", "content": ""} {"id": "textbook_3868", "title": "1302. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_3869", "title": "1302. LEARNING OBJECTIVES - ", "content": "- Define entropy\n- Explain the relationship between entropy and the number of microstates\n- Predict the sign of the entropy change for chemical and physical processes"} {"id": "textbook_3870", "title": "1302. LEARNING OBJECTIVES - ", "content": "In 1824, at the age of 28, Nicolas L\u00e9onard Sadi Carnot (Figure 12.7) published the results of an extensive study regarding the efficiency of steam heat engines. A later review of Carnot's findings by Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat ( $q_{\\mathrm{rev}}$ ) and the kelvin temperature ( $T$ ). In thermodynamics, a reversible process is one that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be \"reversed\") by an infinitesimally small change in some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible.\n\n(a)\n\n(b)\n\nFIGURE 12.7 (a) Nicholas L\u00e9onard Sadi Carnot's research into steam-powered machinery and (b) Rudolf Clausius's later study of those findings led to groundbreaking discoveries about spontaneous heat flow processes."} {"id": "textbook_3871", "title": "1302. LEARNING OBJECTIVES - ", "content": "Similar to other thermodynamic properties, this new quantity is a state function, so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property entropy (S) and defined its change for any process as the following:"} {"id": "textbook_3872", "title": "1302. LEARNING OBJECTIVES - ", "content": "$$\n\\Delta S=\\frac{q_{\\mathrm{rev}}}{T}\n$$"} {"id": "textbook_3873", "title": "1302. LEARNING OBJECTIVES - ", "content": "The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states.\n"} {"id": "textbook_3874", "title": "1303. Entropy and Microstates - ", "content": "\nFollowing the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates ( $W$ ) possible for the system. A microstate is a specific configuration of all the locations and energies of the atoms or molecules that make up a system. The relation between a system's entropy and the number of possible microstates is"} {"id": "textbook_3875", "title": "1303. Entropy and Microstates - ", "content": "$$\nS=k \\ln W\n$$"} {"id": "textbook_3876", "title": "1303. Entropy and Microstates - ", "content": "where $k$ is the Boltzmann constant, $1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K}$.\nAs for other state functions, the change in entropy for a process is the difference between its final ( $S_{\\mathrm{f}}$ ) and initial $\\left(S_{\\mathrm{i}}\\right)$ values:"} {"id": "textbook_3877", "title": "1303. Entropy and Microstates - ", "content": "$$\n\\Delta S=S_{\\mathrm{f}}-S_{\\mathrm{i}}=k \\ln W_{\\mathrm{f}}-k \\ln W_{\\mathrm{i}}=k \\ln \\frac{W_{\\mathrm{f}}}{W_{\\mathrm{i}}}\n$$"} {"id": "textbook_3878", "title": "1303. Entropy and Microstates - ", "content": "For processes involving an increase in the number of microstates, $W_{\\mathrm{f}}>W_{\\mathrm{i}}$, the entropy of the system increases and $\\Delta S>0$. Conversely, processes that reduce the number of microstates, $W_{\\mathrm{f}}
Fundamental Equilibrium Concepts - ", "content": "\n"} {"id": "textbook_4062", "title": "1356. CHAPTER 13
Fundamental Equilibrium Concepts - ", "content": "Figure 13.1 Transport of carbon dioxide in the body involves several reversible chemical reactions, including hydrolysis and acid ionization (among others).\n"} {"id": "textbook_4063", "title": "1357. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_4064", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "\n13.2 Equilibrium Constants\n13.3 Shifting Equilibria: Le Ch\u00e2telier's Principle\n13.4 Equilibrium Calculations"} {"id": "textbook_4065", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "INTRODUCTION Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure 13.1). This chapter provides a thorough introduction to the essential aspects of chemical equilibria."} {"id": "textbook_4066", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "We now have a good understanding of chemical and physical change that allow us to determine, for any given process:"} {"id": "textbook_4067", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "1. Whether the process is endothermic or exothermic\n2. Whether the process is accompanied by an increase of decrease in entropy\n3. Whether a process will be spontaneous, non-spontaneous, or what we have called an equilibrium process"} {"id": "textbook_4068", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "Recall that when the value $\\Delta G$ for a reaction is zero, we consider there to be no free energy change-that is, no\nfree energy available to do useful work. Does this mean a reaction where $\\Delta G=0$ comes to a complete halt? No, it does not. Just as a liquid exists in equilibrium with its vapor in a closed container, where the rates of evaporation and condensation are equal, there is a connection to the state of equilibrium for a phase change or a chemical reaction. That is, at equilibrium, the forward and reverse rates of reaction are equal. We will develop that concept and extend it to a relationship between equilibrium and free energy later in this chapter."} {"id": "textbook_4069", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "In the explanation that follows, we will use the term $Q$ to refer to any reactant or product concentration or pressure. When the concentrations or pressure of reactants and products are at equilibrium, the term $K$ will be used. This will be more clearly explained as we go along in this chapter."} {"id": "textbook_4070", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "Now we will consider the connection between the free energy change and the equilibrium constant. The fundamental relationship is:\n$\\Delta G^{\\circ}=-\\mathrm{R} T \\ln K$-this can be for $K_{c}$ or $K_{p}$ (and we will see later, any equilibrium constant we encounter).\nWe also know that the form of $K$ can be used in non-equilibrium conditions as the reaction quotient, $Q$. The defining relationship here is"} {"id": "textbook_4071", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$"} {"id": "textbook_4072", "title": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria", "content": "Without the superscript, the value of $\\Delta G$ can be calculated for any set of concentrations.\nNote that since $Q$ is a mass-action reaction of productions/reactants, as a reaction proceeds from left to right, product concentrations increase as reactant concentrations decrease, until $Q=K$, and at which time $\\Delta G$ becomes zero:\n$0=\\Delta G^{\\circ}+R T \\ln K$, a relationship that reduces to our defining connection between $Q$ and $K$.\nThus, we can see clearly that as a reaction moves toward equilibrium, the value of $\\Delta G$ goes to zero.\nNow, think back to the connection between the signs of $\\Delta G^{\\circ}$ and $\\Delta H^{\\circ}$\n$\\Delta H^{\\circ} \\quad \\Delta S^{\\circ} \\quad$ Result\nNegative Positive Always spontaneous\nPositive Negative Never spontaneous\nPositive Positive Spontaneous at high temperatures\nNegative Negative Spontaneous at low temperatures\nOnly in the last two cases is there a point at which the process swings from spontaneous to non-spontaneous (or the reverse); in these cases, the process must pass through equilibrium when the change occurs. The concept of the connection between the free energy change and the equilibrium constant is an important one that we will expand upon in future sections. The fact that the change in free energy for an equilibrium process is zero, and that displacement of a process from that zero point results in a drive to re-establish equilibrium is fundamental to understanding the behavior of chemical reactions and phase changes.\n"} {"id": "textbook_4073", "title": "1357. CHAPTER OUTLINE - 1357.2. Chemical Equilibria", "content": ""} {"id": "textbook_4074", "title": "1358. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4075", "title": "1358. LEARNING OBJECTIVES - ", "content": "- Describe the nature of equilibrium systems\n- Explain the dynamic nature of a chemical equilibrium"} {"id": "textbook_4076", "title": "1358. LEARNING OBJECTIVES - ", "content": "The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of \"reactant\" and \"product,\" a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both."} {"id": "textbook_4077", "title": "1358. LEARNING OBJECTIVES - ", "content": "Figure 13.2 illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:"} {"id": "textbook_4078", "title": "1358. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$"} {"id": "textbook_4079", "title": "1358. LEARNING OBJECTIVES - ", "content": "Note that a special double arrow is used to emphasize the reversible nature of the reaction.\n"} {"id": "textbook_4080", "title": "1358. LEARNING OBJECTIVES - ", "content": "FIGURE 13.2 (a) A sealed tube containing colorless $\\mathrm{N}_{2} \\mathrm{O}_{4}$ darkens as it decomposes to yield brown $\\mathrm{NO}_{2}$. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal."} {"id": "textbook_4081", "title": "1358. LEARNING OBJECTIVES - ", "content": "For this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:"} {"id": "textbook_4082", "title": "1358. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n\\operatorname{rate}_{f} & =k_{f}\\left[\\mathrm{~N}_{-2} \\mathrm{O}_{4}\\right] \\\\\n\\operatorname{rate}_{r} & =k_{r}\\left[\\mathrm{NO}_{2}\\right]^{2}\n\\end{aligned}\n$$"} {"id": "textbook_4083", "title": "1358. LEARNING OBJECTIVES - ", "content": "As the reaction begins ( $t=0$ ), the concentration of the $\\mathrm{N}_{2} \\mathrm{O}_{4}$ reactant is finite and that of the $\\mathrm{NO}_{2}$ product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N.\n${ }_{2} \\mathrm{O}_{4}$ is consumed and its concentration falls, while $\\mathrm{NO}_{2}$ is produced and its concentration increases (Figure $13.2 \\mathbf{b})$. The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure 13.2c). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure $13.2 \\mathbf{b}$ and Figure 13.2c). It's important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not \"stopped,\" but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.\n"} {"id": "textbook_4084", "title": "1358. LEARNING OBJECTIVES - ", "content": "FIGURE 13.3 A two-person juggling act illustrates the dynamic aspect of chemical equilibria. Each person is throwing and catching clubs at the same rate, and each holds a (approximately) constant number of clubs."} {"id": "textbook_4085", "title": "1358. LEARNING OBJECTIVES - ", "content": "Physical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:"} {"id": "textbook_4086", "title": "1358. LEARNING OBJECTIVES - ", "content": "$$\n\\operatorname{Br}_{2}(l) \\rightleftharpoons \\mathrm{Br}_{2}(g)\n$$"} {"id": "textbook_4087", "title": "1358. LEARNING OBJECTIVES - ", "content": "When liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in Figure 13.4.\n"} {"id": "textbook_4088", "title": "1358. LEARNING OBJECTIVES - ", "content": "FIGURE 13.4 A sealed tube containing an equilibrium mixture of liquid and gaseous bromine. (credit: http://images-of-elements.com/bromine.php)\n"} {"id": "textbook_4089", "title": "1358. LEARNING OBJECTIVES - 1358.1. Equilibrium Constants", "content": ""} {"id": "textbook_4090", "title": "1359. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4091", "title": "1359. LEARNING OBJECTIVES - ", "content": "- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions\n- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures\n- Relate the magnitude of an equilibrium constant to properties of the chemical system"} {"id": "textbook_4092", "title": "1359. LEARNING OBJECTIVES - ", "content": "The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( $\\boldsymbol{Q}$ ). For a reversible reaction described by"} {"id": "textbook_4093", "title": "1359. LEARNING OBJECTIVES - ", "content": "$$\nm \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D}\n$$"} {"id": "textbook_4094", "title": "1359. LEARNING OBJECTIVES - ", "content": "the reaction quotient is derived directly from the stoichiometry of the balanced equation as"} {"id": "textbook_4095", "title": "1359. LEARNING OBJECTIVES - ", "content": "$$\nQ_{c}=\\frac{[\\mathrm{C}]^{x}[\\mathrm{D}]^{y}}{[\\mathrm{~A}]^{m}[\\mathrm{~B}]^{n}}\n$$"} {"id": "textbook_4096", "title": "1359. LEARNING OBJECTIVES - ", "content": "where the subscript $c$ denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:"} {"id": "textbook_4097", "title": "1359. LEARNING OBJECTIVES - ", "content": "$$\nQ_{p}=\\frac{P_{\\mathrm{C}}^{x} P_{\\mathrm{D}}^{y}}{P_{\\mathrm{A}}^{m} P_{\\mathrm{B}}^{n}}\n$$"} {"id": "textbook_4098", "title": "1359. LEARNING OBJECTIVES - ", "content": "Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing $Q$. In most cases, this will introduce only modest errors in calculations involving reaction quotients.\n"} {"id": "textbook_4099", "title": "1360. EXAMPLE 13.1 - ", "content": ""} {"id": "textbook_4100", "title": "1361. Writing Reaction Quotient Expressions - ", "content": "\nWrite the concentration-based reaction quotient expression for each of the following reactions:\n(a) $3 \\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{O}_{3}(g)$\n(b) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(c) $4 \\mathrm{NH}_{3}(g)+7 \\mathrm{O}_{2}(g) \\rightleftharpoons 4 \\mathrm{NO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n"} {"id": "textbook_4101", "title": "1362. Solution - ", "content": "\n(a) $Q_{c}=\\frac{\\left[\\mathrm{O}_{3}\\right]^{2}}{\\left[\\mathrm{O}_{2}\\right]^{3}}$\n(b) $Q_{c}=\\frac{\\left[\\mathrm{NH}_{3}\\right]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]^{3}}$\n(c) $Q_{c}=\\frac{\\left[\\mathrm{NO}_{2}\\right]^{4}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{6}}{\\left[\\mathrm{NH}_{3}\\right]^{4}\\left[\\mathrm{O}_{2}\\right]^{7}}$\n"} {"id": "textbook_4102", "title": "1363. Check Your Learning - ", "content": "\nWrite the concentration-based reaction quotient expression for each of the following reactions:\n(a) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{8}(g) \\rightleftharpoons 2 \\mathrm{C}_{2} \\mathrm{H}_{4}(g)$\n(c) $2 \\mathrm{C}_{4} \\mathrm{H}_{10}(g)+13 \\mathrm{O}_{2}(g) \\rightleftharpoons 8 \\mathrm{CO}_{2}(g)+10 \\mathrm{H}_{2} \\mathrm{O}(g)$\n"} {"id": "textbook_4103", "title": "1364. Answer: - ", "content": "\n(a) $Q_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}$; (b) $Q_{c}=\\frac{\\left[\\mathrm{C}_{2} \\mathrm{H}_{4}\\right]^{2}}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]} ;$ (c) $Q_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]^{8}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]{ }^{10}}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{10}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]^{13}}$\n"} {"id": "textbook_4104", "title": "1364. Answer: - ", "content": "FIGURE 13.5 Changes in concentrations and $Q_{C}$ for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only."} {"id": "textbook_4105", "title": "1364. Answer: - ", "content": "The numerical value of $Q$ varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction's status. To illustrate this point, consider the oxidation of sulfur dioxide:"} {"id": "textbook_4106", "title": "1364. Answer: - ", "content": "$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)\n$$"} {"id": "textbook_4107", "title": "1364. Answer: - ", "content": "Two different experimental scenarios are depicted in Figure 13.5, one in which this reaction is initiated with a mixture of reactants only, $\\mathrm{SO}_{2}$ and $\\mathrm{O}_{2}$, and another that begins with only product, $\\mathrm{SO}_{3}$. For the reaction that begins with a mixture of reactants only, $Q$ is initially equal to zero:"} {"id": "textbook_4108", "title": "1364. Answer: - ", "content": "$$\nQ_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=\\frac{0^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=0\n$$"} {"id": "textbook_4109", "title": "1364. Answer: - ", "content": "As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of $Q_{c}$ ), product concentration increases (as does the numerator of $Q_{c}$ ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of $Q_{c}$."} {"id": "textbook_4110", "title": "1364. Answer: - ", "content": "If the reaction begins with only product present, the value of $Q_{c}$ is initially undefined (immeasurably large, or infinite):"} {"id": "textbook_4111", "title": "1364. Answer: - ", "content": "$$\nQ_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{0} \\rightarrow \\infty\n$$"} {"id": "textbook_4112", "title": "1364. Answer: - ", "content": "In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of $Q_{c}$ decrease with time, the reactant concentrations and the denominator of $Q_{c}$ increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium."} {"id": "textbook_4113", "title": "1364. Answer: - ", "content": "The constant value of $Q$ exhibited by a system at equilibrium is called the equilibrium constant, $\\boldsymbol{K}$ :"} {"id": "textbook_4114", "title": "1364. Answer: - ", "content": "$$\nK \\equiv Q \\text { at equilibrium }\n$$"} {"id": "textbook_4115", "title": "1364. Answer: - ", "content": "Comparison of the data plots in Figure 13.5 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant.\n"} {"id": "textbook_4116", "title": "1365. EXAMPLE 13.2 - ", "content": ""} {"id": "textbook_4117", "title": "1366. Evaluating a Reaction Quotient - ", "content": "\nGaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:"} {"id": "textbook_4118", "title": "1366. Evaluating a Reaction Quotient - ", "content": "$$\n2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g)\n$$"} {"id": "textbook_4119", "title": "1366. Evaluating a Reaction Quotient - ", "content": "When $0.10 \\mathrm{~mol} \\mathrm{NO}_{2}$ is added to a 1.0-L flask at $25^{\\circ} \\mathrm{C}$, the concentration changes so that at equilibrium, $\\left[\\mathrm{NO}_{2}\\right]=$ 0.016 M and $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0.042 \\mathrm{M}$.\n(a) What is the value of the reaction quotient before any reaction occurs?\n(b) What is the value of the equilibrium constant for the reaction?\n"} {"id": "textbook_4120", "title": "1367. Solution - ", "content": "\nAs for all equilibrium calculations in this text, use the simplified equations for $Q$ and $K$ and disregard any concentration or pressure units, as noted previously in this section.\n(a) Before any product is formed, $\\left[\\mathrm{NO}_{2}\\right]=\\frac{0.10 \\mathrm{~mol}}{1.0 \\mathrm{~L}}=0.10 M$, and $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0 \\mathrm{M}$. Thus,"} {"id": "textbook_4121", "title": "1367. Solution - ", "content": "$$\nQ_{c}=\\frac{\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]}{\\left[\\mathrm{NO}_{2}\\right]^{2}}=\\frac{0}{0.10^{2}}=0\n$$"} {"id": "textbook_4122", "title": "1367. Solution - ", "content": "(b) At equilibrium, $K_{c}=Q_{c}=\\frac{\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]}{\\left[\\mathrm{NO}_{2}\\right]^{2}}=\\frac{0.042}{0.016^{2}}=1.6 \\times 10^{2}$. The equilibrium constant is $1.6 \\times 10^{2}$.\n"} {"id": "textbook_4123", "title": "1368. Check Your Learning - ", "content": "\nFor the reaction $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$, the concentrations at equilibrium are $\\left[\\mathrm{SO}_{2}\\right]=0.90 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=0.35$ $M$, and $\\left[\\mathrm{SO}_{3}\\right]=1.1 \\mathrm{M}$. What is the value of the equilibrium constant, $K_{c}$ ?\n"} {"id": "textbook_4124", "title": "1369. Answer: - ", "content": "\n$K_{c}=4.3$"} {"id": "textbook_4125", "title": "1369. Answer: - ", "content": "By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large $K$ will reach equilibrium when most of the reactant has been converted to product, whereas a small $K$ indicates the reaction achieves equilibrium after very little reactant has been converted. It's important to keep in mind that the magnitude of $K$ does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer."} {"id": "textbook_4126", "title": "1369. Answer: - ", "content": "The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing $Q$ to $K$ for\nan equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below:"} {"id": "textbook_4127", "title": "1369. Answer: - ", "content": "$$\n\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\quad K_{c}=0.640 \\quad \\mathrm{~T}=800^{\\circ} \\mathrm{C}\n$$"} {"id": "textbook_4128", "title": "1369. Answer: - ", "content": "The bar charts in Figure 13.6 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction's equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.\n"} {"id": "textbook_4129", "title": "1369. Answer: - ", "content": "FIGURE 13.6 Compositions of three mixtures before $\\left(Q_{c} \\neq K_{c}\\right)$ and after $\\left(Q_{c}=K_{c}\\right)$ equilibrium is established for the reaction $\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g)$.\n"} {"id": "textbook_4130", "title": "1370. EXAMPLE 13.3 - ", "content": ""} {"id": "textbook_4131", "title": "1371. Predicting the Direction of Reaction - ", "content": "\nGiven here are the starting concentrations of reactants and products for three experiments involving this reaction:"} {"id": "textbook_4132", "title": "1371. Predicting the Direction of Reaction - ", "content": "$$\n\\begin{gathered}\n\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\\\\nK_{c}=0.64\n\\end{gathered}\n$$"} {"id": "textbook_4133", "title": "1371. Predicting the Direction of Reaction - ", "content": "Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown."} {"id": "textbook_4134", "title": "1371. Predicting the Direction of Reaction - ", "content": "| Reactants/Products | | Experiment 1 | Experiment 2 |\n| :--- | :--- | :--- | :--- |\n| $[\\mathrm{CO}]_{\\mathrm{i}}$ | 0.020 M | 0.011 M | 0.0094 M |\n| $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]_{\\mathrm{i}}$ | 0.020 M | 0.0011 M | 0.0025 M |\n| $\\left[\\mathrm{CO}_{2}\\right]_{\\mathrm{i}}$ | 0.0040 M | 0.037 M | 0.0015 M |\n| $\\left[\\mathrm{H}_{2}\\right]_{\\mathrm{i}}$ | 0.0040 M | 0.046 M | 0.0076 M |\n"} {"id": "textbook_4135", "title": "1372. Solution - ", "content": "\nExperiment 1:"} {"id": "textbook_4136", "title": "1372. Solution - ", "content": "$$\nQ_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]}{[\\mathrm{CO}]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}=\\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040\n$$"} {"id": "textbook_4137", "title": "1372. Solution - ", "content": "$Q_{c}
(increase in entropy) | $\\Delta \\mathrm{G}<0$ at high temperature $\\Delta G>0$ at low temperature Process is spontaneous at high temperature | $\\Delta G<0$ at any temperature
Process is spontaneous at any temperature |\n| $\\Delta \\mathrm{S}<0$ (decrease in entropy) | $\\Delta \\mathrm{G}>0$ at any temperature
Process is nonspontaneous at any temperature | $\\Delta \\mathrm{G}<0$ at low temperature $\\Delta G>0$ at high temperature Process is spontaneous at low temperature |"} {"id": "textbook_4331", "title": "1418. Temperature Dependence of Spontaneity - ", "content": "FIGURE 13.8 There are four possibilities regarding the signs of enthalpy and entropy changes.\n"} {"id": "textbook_4332", "title": "1419. EXAMPLE 13.10 - ", "content": ""} {"id": "textbook_4333", "title": "1420. Predicting the Temperature Dependence of Spontaneity - ", "content": "\nThe incomplete combustion of carbon is described by the following equation:"} {"id": "textbook_4334", "title": "1420. Predicting the Temperature Dependence of Spontaneity - ", "content": "$$\n2 \\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}(g)\n$$"} {"id": "textbook_4335", "title": "1420. Predicting the Temperature Dependence of Spontaneity - ", "content": "How does the spontaneity of this process depend upon temperature?\n"} {"id": "textbook_4336", "title": "1421. Solution - ", "content": "\nCombustion processes are exothermic ( $\\Delta H<0$ ). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, $\\Delta S>0$ ). The reaction is therefore spontaneous $(\\Delta G<0)$ at all temperatures.\n"} {"id": "textbook_4337", "title": "1422. Check Your Learning - ", "content": "\nPopular chemical hand warmers generate heat by the air-oxidation of iron:"} {"id": "textbook_4338", "title": "1422. Check Your Learning - ", "content": "$$\n4 \\mathrm{Fe}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)\n$$"} {"id": "textbook_4339", "title": "1422. Check Your Learning - ", "content": "How does the spontaneity of this process depend upon temperature?\n"} {"id": "textbook_4340", "title": "1423. Answer: - ", "content": "\n$\\Delta H$ and $\\Delta S$ are negative; the reaction is spontaneous at low temperatures."} {"id": "textbook_4341", "title": "1423. Answer: - ", "content": "When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms \"high\" and \"low\" mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in \"spontaneity\" (as reflected by its $\\Delta G$ ) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which $\\Delta G$ is plotted on the $y$ axis versus $T$ on the $x$ axis:"} {"id": "textbook_4342", "title": "1423. Answer: - ", "content": "$$\n\\begin{gathered}\n\\Delta G=\\Delta H-T \\Delta S \\\\\ny=b+m x\n\\end{gathered}\n$$"} {"id": "textbook_4343", "title": "1423. Answer: - ", "content": "Such a plot is shown in Figure 13.9. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative $\\Delta G$ ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the $x$-intercept of the line, that is, the value of $T$ for which $\\Delta G$ is zero:"} {"id": "textbook_4344", "title": "1423. Answer: - ", "content": "$$\n\\Delta G=0=\\Delta H-T \\Delta S\n$$"} {"id": "textbook_4345", "title": "1423. Answer: - ", "content": "$$\nT=\\frac{\\Delta H}{\\Delta S}\n$$"} {"id": "textbook_4346", "title": "1423. Answer: - ", "content": "So, saying a process is spontaneous at \"high\" or \"low\" temperatures means the temperature is above or below, respectively, that temperature at which $\\Delta G$ for the process is zero. As noted earlier, the condition of $\\Delta \\mathrm{G}=0$ describes a system at equilibrium.\n"} {"id": "textbook_4347", "title": "1423. Answer: - ", "content": "FIGURE 13.9 These plots show the variation in $\\Delta G$ with temperature for the four possible combinations of arithmetic sign for $\\Delta H$ and $\\Delta S$.\n"} {"id": "textbook_4348", "title": "1424. EXAMPLE 13.11 - ", "content": ""} {"id": "textbook_4349", "title": "1425. Equilibrium Temperature for a Phase Transition - ", "content": "\nAs defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.\n"} {"id": "textbook_4350", "title": "1426. Solution - ", "content": "\nThe process of interest is the following phase change:"} {"id": "textbook_4351", "title": "1426. Solution - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)\n$$"} {"id": "textbook_4352", "title": "1426. Solution - ", "content": "When this process is at equilibrium, $\\Delta G=0$, so the following is true:"} {"id": "textbook_4353", "title": "1426. Solution - ", "content": "$$\n0=\\Delta H^{\\circ}-T \\Delta S^{\\circ} \\quad \\text { or } \\quad T=\\frac{\\Delta H^{\\circ}}{\\Delta S^{\\circ}}\n$$"} {"id": "textbook_4354", "title": "1426. Solution - ", "content": "Using the standard thermodynamic data from Appendix G,"} {"id": "textbook_4355", "title": "1426. Solution - ", "content": "$$\n\\begin{aligned}\n\\Delta H^{\\circ} & =1 \\mathrm{~mol} \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-1 \\mathrm{~mol} \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n& =(1 \\mathrm{~mol})-241.82 \\mathrm{~kJ} / \\mathrm{mol}-(1 \\mathrm{~mol})(-241.82 \\mathrm{~kJ} / \\mathrm{mol})=44.01 \\mathrm{~kJ} \\\\\n\\Delta S^{\\circ} & =1 \\mathrm{~mol} \\times \\Delta S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-1 \\mathrm{~mol} \\times \\Delta S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n& =(1 \\mathrm{~mol}) 188.8 \\mathrm{~J} / \\mathrm{K} \\cdot \\mathrm{~mol}-(1 \\mathrm{~mol}) 70.0 \\mathrm{~J} / \\mathrm{K} \\cdot \\mathrm{~mol}=118.8 \\mathrm{~J} / \\mathrm{K} \\\\\nT & =\\frac{\\Delta H^{\\circ}}{\\Delta S^{\\circ}}=\\frac{44.01 \\times 10^{3} \\mathrm{~J}}{118.8 \\mathrm{~J} / \\mathrm{K}}=370.5 \\mathrm{~K}=97.3^{\\circ} \\mathrm{C}\n\\end{aligned}\n$$"} {"id": "textbook_4356", "title": "1426. Solution - ", "content": "The accepted value for water's normal boiling point is $373.2 \\mathrm{~K}\\left(100.0^{\\circ} \\mathrm{C}\\right)$, and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.\n"} {"id": "textbook_4357", "title": "1427. Check Your Learning - ", "content": "\nUse the information in Appendix G to estimate the boiling point of $\\mathrm{CS}_{2}$.\n"} {"id": "textbook_4358", "title": "1428. Answer: - ", "content": "\n313 K (accepted value 319 K )\n"} {"id": "textbook_4359", "title": "1429. Free Energy and Equilibrium - ", "content": "\nThe free energy change for a process may be viewed as a measure of its driving force. A negative value for $\\Delta G$ represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When $\\Delta G$ is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium)."} {"id": "textbook_4360", "title": "1429. Free Energy and Equilibrium - ", "content": "In the section on equilibrium, the reaction quotient, $Q$, was introduced as a convenient measure of the status of an equilibrium system. Recall that $Q$ is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When $Q$ is lesser than the equilibrium constant, $K$, the reaction will proceed in the forward direction until equilibrium is reached and $Q=K$. Conversely, if $Q>K$, the process will proceed in the reverse direction until equilibrium is achieved."} {"id": "textbook_4361", "title": "1429. Free Energy and Equilibrium - ", "content": "The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M ) is related to the standard free energy change, according to this equation:"} {"id": "textbook_4362", "title": "1429. Free Energy and Equilibrium - ", "content": "$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$"} {"id": "textbook_4363", "title": "1429. Free Energy and Equilibrium - ", "content": "$R$ is the gas constant ( $8.314 \\mathrm{~J} / \\mathrm{K} \\mathrm{mol}$ ), $T$ is the kelvin or absolute temperature, and $Q$ is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.12.\n"} {"id": "textbook_4364", "title": "1430. EXAMPLE 13.12 - ", "content": ""} {"id": "textbook_4365", "title": "1431. Calculating $\\boldsymbol{\\Delta} \\boldsymbol{G}$ under Nonstandard Conditions - ", "content": "\nWhat is the free energy change for the process shown here under the specified conditions?"} {"id": "textbook_4366", "title": "1431. Calculating $\\boldsymbol{\\Delta} \\boldsymbol{G}$ under Nonstandard Conditions - ", "content": "$$\n\\begin{aligned}\nT=25^{\\circ} \\mathrm{C}, P_{\\mathrm{N}_{2}}=0.870 \\mathrm{~atm}, P_{\\mathrm{H}_{2}} & =0.250 \\mathrm{~atm}, \\text { and } P_{\\mathrm{NH}_{3}}=12.9 \\mathrm{~atm} \\\\\n2 \\mathrm{NH}_{3}(g) & \\longrightarrow 3 \\mathrm{H}_{2}(g)+\\mathrm{N}_{2}(g) \\quad \\Delta G^{\\circ}=33.0 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$\n"} {"id": "textbook_4367", "title": "1432. Solution - ", "content": "\nThe equation relating free energy change to standard free energy change and reaction quotient may be used directly:"} {"id": "textbook_4368", "title": "1432. Solution - ", "content": "$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q=33.0 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}}+\\left(8.314 \\frac{\\mathrm{~J}}{\\mathrm{~mol} \\mathrm{~K}} \\times 298 \\mathrm{~K} \\times \\ln \\frac{\\left(0.250^{3}\\right) \\times 0.870}{12.9^{2}}\\right)=9680 \\frac{\\mathrm{~J}}{\\mathrm{~mol}} \\text { or } 9.68 \\mathrm{~kJ} / \\mathrm{mol}\n$$"} {"id": "textbook_4369", "title": "1432. Solution - ", "content": "Since the computed value for $\\Delta G$ is positive, the reaction is nonspontaneous under these conditions.\n"} {"id": "textbook_4370", "title": "1433. Check Your Learning - ", "content": "\nCalculate the free energy change for this same reaction at $875^{\\circ} \\mathrm{C}$ in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?\n"} {"id": "textbook_4371", "title": "1434. Answer: - ", "content": "\n$\\Delta G=-47 \\mathrm{~kJ} / \\mathrm{mol}$; yes"} {"id": "textbook_4372", "title": "1434. Answer: - ", "content": "For a system at equilibrium, $Q=K$ and $\\Delta G=0$, and the previous equation may be written as"} {"id": "textbook_4373", "title": "1434. Answer: - ", "content": "$$\n\\begin{array}{rlrr}\n0 & =\\Delta G^{\\circ}+R T \\ln K & & \\text { (at equilibrium) } \\\\\n\\Delta G^{\\circ} & =-R T \\ln K & \\text { or } & K=e^{-\\frac{\\Delta G^{\\circ}}{R T}}\n\\end{array}\n$$"} {"id": "textbook_4374", "title": "1434. Answer: - ", "content": "This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1."} {"id": "textbook_4375", "title": "1434. Answer: - ", "content": "Relations between Standard Free Energy Changes and Equilibrium Constants"} {"id": "textbook_4376", "title": "1434. Answer: - ", "content": "| $\\boldsymbol{K}$ | $\\Delta \\boldsymbol{G}^{\\circ}$ | Composition of an Equilibrium Mixture |\n| :--- | :--- | :--- |\n| $>1$ | $<0$ | Products are more abundant |\n| $<1$ | $>0$ | Reactants are more abundant |\n| $=1$ | $=0$ | Reactants and products are comparably abundant |"} {"id": "textbook_4377", "title": "1434. Answer: - ", "content": "TABLE 13.1\n"} {"id": "textbook_4378", "title": "1435. EXAMPLE 13.13 - ", "content": ""} {"id": "textbook_4379", "title": "1436. Calculating an Equilibrium Constant using Standard Free Energy Change - ", "content": "\nGiven that the standard free energies of formation of $\\mathrm{Ag}^{+}(\\mathrm{aq}), \\mathrm{Cl}^{-}(\\mathrm{aq})$, and $\\mathrm{AgCl}(s)$ are $77.1 \\mathrm{~kJ} / \\mathrm{mol},-131.2 \\mathrm{~kJ} /$ mol , and $-109.8 \\mathrm{~kJ} / \\mathrm{mol}$, respectively, calculate the solubility product, $K_{\\mathrm{sp}}$, for AgCl .\n"} {"id": "textbook_4380", "title": "1437. Solution - ", "content": "\nThe reaction of interest is the following:"} {"id": "textbook_4381", "title": "1437. Solution - ", "content": "$$\n\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]\n$$"} {"id": "textbook_4382", "title": "1437. Solution - ", "content": "The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:"} {"id": "textbook_4383", "title": "1437. Solution - ", "content": "$$\n\\begin{aligned}\n& \\Delta G^{\\circ}=\\left[\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{Ag}^{+}(a q)\\right)+\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{Cl}^{-}(a q)\\right)\\right]-\\left[\\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{AgCl}(s))\\right] \\\\\n& =[77.1 \\mathrm{~kJ} / \\mathrm{mol}-131.2 \\mathrm{~kJ} / \\mathrm{mol}]-[-109.8 \\mathrm{~kJ} / \\mathrm{mol}]=55.7 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$"} {"id": "textbook_4384", "title": "1437. Solution - ", "content": "The equilibrium constant for the reaction may then be derived from its standard free energy change:"} {"id": "textbook_4385", "title": "1437. Solution - ", "content": "$$\nK_{\\mathrm{sp}}=e^{-\\frac{\\Delta G^{\\circ}}{R T}}=\\exp \\left(-\\frac{\\Delta G^{\\circ}}{R T}\\right)=\\exp \\left(-\\frac{55.7 \\times 10^{3} \\mathrm{~J} / \\mathrm{mol}}{8.314 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K} \\times 298.15 \\mathrm{~K}}\\right)=\\exp (-22.470)=e^{-22.470}=1.74 \\times 10^{-10}\n$$"} {"id": "textbook_4386", "title": "1437. Solution - ", "content": "This result is in reasonable agreement with the value provided in Appendix J.\n"} {"id": "textbook_4387", "title": "1438. Check Your Learning - ", "content": "\nUse the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at $25^{\\circ} \\mathrm{C}$."} {"id": "textbook_4388", "title": "1438. Check Your Learning - ", "content": "$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n"} {"id": "textbook_4389", "title": "1439. Answer: - ", "content": "\n$K=0.32$"} {"id": "textbook_4390", "title": "1439. Answer: - ", "content": "To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of $Q$ ), equilibrium is established when the system's free energy is minimized (Figure 13.10). If a system consists of reactants and products in nonequilibrium amounts $(Q \\neq K)$, the reaction will proceed spontaneously in the direction necessary to establish equilibrium.\n"} {"id": "textbook_4391", "title": "1439. Answer: - ", "content": "FIGURE 13.10 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.\n"} {"id": "textbook_4392", "title": "1440. Key Terms - ", "content": "\nequilibrium state of a reversible reaction in which the forward and reverse processes occur at equal rates\nequilibrium constant ( $\\boldsymbol{K}$ ) value of the reaction quotient for a system at equilibrium; may be expressed using concentrations ( $K_{c}$ ) or partial pressures ( $K_{p}$ )\nheterogeneous equilibria equilibria in which reactants and products occupy two or more different phases\nhomogeneous equilibria equilibria in which all reactants and products occupy the same phase\nlaw of mass action when a reversible reaction has\nattained equilibrium at a given temperature, the reaction quotient remains constant\nLe Ch\u00e2telier's principle an equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium\nreaction quotient $(Q)$ mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations $\\left(Q_{c}\\right)$ or pressures $\\left(Q_{p}\\right)$\nreversible reaction chemical reaction that can proceed in both the forward and reverse directions under given conditions\n"} {"id": "textbook_4393", "title": "1441. Key Equations - ", "content": "\n$$\n\\begin{array}{lr}\nQ_{c}=\\frac{[\\mathrm{C}]^{x}[\\mathrm{D}]^{y}}{[\\mathrm{~A}]^{m}[\\mathrm{~B}]^{n}} & \\text { for the reaction } m \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D} \\\\\nQ_{P}=\\frac{\\left(P_{C}\\right)^{x}\\left(P_{D}\\right)^{y}}{\\left(P_{A}\\right)^{m}\\left(P_{B}\\right)^{n}} & \\text { for the reaction } m \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D} \\\\\nP=M R T & \\\\\nK_{c}=Q_{c} \\text { at equilibrium } & \\\\\nK_{p}=Q_{p} \\text { at equilibrium } & \\\\\nK_{P}=K_{c}(R T)^{\\Delta n} &\n\\end{array}\n$$\n"} {"id": "textbook_4394", "title": "1442. Summary - ", "content": ""} {"id": "textbook_4395", "title": "1442. Summary - 1442.1. Chemical Equilibria", "content": "\nA reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species.\n"} {"id": "textbook_4396", "title": "1442. Summary - 1442.2. Equilibrium Constants", "content": "\nThe composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, $Q$. For a reaction at equilibrium, the composition is constant, and $Q$ is called the equilibrium constant, $K$."} {"id": "textbook_4397", "title": "1442. Summary - 1442.2. Equilibrium Constants", "content": "A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.\n"} {"id": "textbook_4398", "title": "1442. Summary - 1442.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "\nSystems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system's response to\nthese disturbances is described by Le Ch\u00e2telier's principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture's composition ( $K$ does not change).\n"} {"id": "textbook_4399", "title": "1442. Summary - 1442.4. Equilibrium Calculations", "content": "\nCalculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. We also learned that a negative value for $\\Delta G$ indicates a spontaneous process; a positive $\\Delta G$ indicates a nonspontaneous process; and a $\\Delta G$ of zero indicates that the system is at equilibrium. We also saw how free energy, spontaneity, and equilibrium relate.\n"} {"id": "textbook_4400", "title": "1443. Exercises - ", "content": ""} {"id": "textbook_4401", "title": "1443. Exercises - 1443.1. Chemical Equilibria", "content": "\n1. What does it mean to describe a reaction as \"reversible\"?\n2. When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?\n3. If a reaction is reversible, when can it be said to have reached equilibrium?\n4. Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?\n5. If the concentrations of products and reactants are equal, is the system at equilibrium?\n"} {"id": "textbook_4402", "title": "1443. Exercises - 1443.2. Equilibrium Constants", "content": "\n6. Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.\n7. Explain why an equilibrium between $\\mathrm{Br}_{2}(I)$ and $\\mathrm{Br}_{2}(g)$ would not be established if the container were not a closed vessel shown in Figure 13.4.\n8. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure $\\mathrm{NO}_{2}$ or with pure $\\mathrm{N}_{2} \\mathrm{O}_{4}$ ? $2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g)$\n9. Among the solubility rules previously discussed is the statement: All chlorides are soluble except $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$, $\\mathrm{AgCl}, \\mathrm{PbCl}_{2}$, and CuCl .\n(a) Write the expression for the equilibrium constant for the reaction represented by the equation\n$\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n(b) Write the expression for the equilibrium constant for the reaction represented by the equation $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\rightleftharpoons \\mathrm{PbCl}_{2}(s)$. Is $K_{c}>1$, <1, or $\\approx 1$ ? Explain your answer.\n10. Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates-except those of the ammonium ion and the alkali metals-are insoluble.\n(a) Write the expression for the equilibrium constant for the reaction represented by the equation\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n(b) Write the expression for the equilibrium constant for the reaction represented by the equation $3 \\mathrm{Ba}^{2+}(a q)+2 \\mathrm{PO}_{4}{ }^{3-}(a q) \\rightleftharpoons \\mathrm{Ba}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n11. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $3 \\mathrm{C}_{2} \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{C}_{6} \\mathrm{H}_{6}(g)$. Which value of $K_{c}$ would make this reaction most useful commercially? $K_{c} \\approx 0.01, K_{c} \\approx 1$, or $K_{c} \\approx 10$. Explain your answer.\n12. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $\\mathrm{KI}(a q)+\\mathrm{I}_{2}(a q) \\rightleftharpoons \\mathrm{KI}_{3}(a q)$ give the same expression for the reaction quotient. $\\mathrm{KI}_{3}$ is composed of the ions $\\mathrm{K}^{+}$and $\\mathrm{I}_{3}{ }^{-}$.\n13. For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be $100 \\%$. Is $K_{c}>1,<1$, or $\\approx 1$ for a titration reaction?\n14. For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is $K_{c}>1,<1$, or $\\approx 1$ for a useful precipitation reaction?\n15. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:\n(a) $\\mathrm{CH}_{4}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{Cl}(g)+\\mathrm{HCl}(g)$\n(b) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n(c) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(d) $\\mathrm{BaSO}_{3}(s) \\rightleftharpoons \\mathrm{BaO}(s)+\\mathrm{SO}_{2}(g)$\n(e) $\\mathrm{P}_{4}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{P}_{4} \\mathrm{O}_{10}(s)$\n(f) $\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{Br}(g)$\n(g) $\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(h) $\\mathrm{CuSO}_{4} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CuSO}_{4}(s)+5 \\mathrm{H}_{2} \\mathrm{O}(g)$\n16. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(b) $4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(c) $\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)$\n(d) $\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n(e) $\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)$\n(f) $2 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(s) \\rightleftharpoons 2 \\mathrm{PbO}(s)+4 \\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(g)$\n(g) $2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(h) $\\mathrm{S}_{8}(g) \\rightleftharpoons 8 \\mathrm{~S}(g)$\n17. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{c}=17 ;\\left[\\mathrm{NH}_{3}\\right]=0.20 M,\\left[\\mathrm{~N}_{2}\\right]=1.00 M,\\left[\\mathrm{H}_{2}\\right]=1.00 M$\n(b) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{P}=6.8 \\times 10^{4} ; \\mathrm{NH}_{3}=3.0 \\mathrm{~atm}, \\mathrm{~N}_{2}=2.0 \\mathrm{~atm}, \\mathrm{H}_{2}=1.0 \\mathrm{~atm}$\n(c) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{c}=0.230 ;\\left[\\mathrm{SO}_{3}\\right]=0.00 \\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=1.00 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=1.00 \\mathrm{M}$\n(d) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$ $K_{P}=16.5 ; \\mathrm{SO}_{3}=1.00 \\mathrm{~atm}, \\mathrm{SO}_{2}=1.00 \\mathrm{~atm}, \\mathrm{O}_{2}=1.00 \\mathrm{~atm}$\n(e) $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$ $K_{c}=4.6 \\times 10^{4} ;[\\mathrm{NO}]=1.00 M,\\left[\\mathrm{Cl}_{2}\\right]=1.00 M,[\\mathrm{NOCl}]=0 M$\n(f) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$K_{P}=0.050 ; \\mathrm{NO}=10.0 \\mathrm{~atm}, \\mathrm{~N}_{2}=\\mathrm{O}_{2}=5 \\mathrm{~atm}$\n18. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\quad K_{c}=17 ;\\left[\\mathrm{NH}_{3}\\right]=0.50 \\mathrm{M},\\left[\\mathrm{N}_{2}\\right]=0.15 \\mathrm{M},\\left[\\mathrm{H}_{2}\\right]=0.12 \\mathrm{M}$\n(b) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{P}=6.8 \\times 10^{4} ; \\mathrm{NH}_{3}=2.00 \\mathrm{~atm}, \\mathrm{~N}_{2}=10.00 \\mathrm{~atm}, \\mathrm{H}_{2}=10.00 \\mathrm{~atm}$\n(c) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{c}=0.230 ;\\left[\\mathrm{SO}_{3}\\right]=2.00 \\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=2.00 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=2.00 \\mathrm{M}$\n(d) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{P}=6.5 \\mathrm{~atm} ; \\mathrm{SO}_{2}=1.00 \\mathrm{~atm}, \\mathrm{O}_{2}=1.130 \\mathrm{~atm}, \\mathrm{SO}_{3}=0 \\mathrm{~atm}$\n(e) $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$\n$K_{P}=2.5 \\times 10^{3} ; \\mathrm{NO}=1.00 \\mathrm{~atm}, \\mathrm{Cl}_{2}=1.00 \\mathrm{~atm}, \\mathrm{NOCl}=0 \\mathrm{~atm}$\n(f) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$K_{c}=0.050 ;\\left[\\mathrm{N}_{2}\\right]=0.100 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=0.200 \\mathrm{M},[\\mathrm{NO}]=1.00 \\mathrm{M}$\n19. The following reaction has $K_{P}=4.50 \\times 10^{-5}$ at 720 K .\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\nIf a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? $P\\left(\\mathrm{NH}_{3}\\right)=93 \\mathrm{~atm}, P\\left(\\mathrm{~N}_{2}\\right)=48 \\mathrm{~atm}$, and $P\\left(\\mathrm{H}_{2}\\right)=52$ atm\n20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?\n$\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{SO}_{2}(g)+\\mathrm{Cl}_{2}(g)$\n$\\left[\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right]=0.12 \\mathrm{M},\\left[\\mathrm{Cl}_{2}\\right]=0.16 \\mathrm{M}$ and $\\left[\\mathrm{SO}_{2}\\right]=0.050 \\mathrm{M} . K_{c}$ for the reaction is 0.078 .\n21. Which of the systems described in Exercise 13.15 are homogeneous equilibria? Which are heterogeneous equilibria?\n22. Which of the systems described in Exercise 13.16 are homogeneous equilibria? Which are heterogeneous equilibria?\n23. For which of the reactions in Exercise 13.15 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?\n24. For which of the reactions in Exercise 13.16 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?\n25. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n$K_{c}=0.50$ at $400^{\\circ} \\mathrm{C}$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightleftharpoons 2 \\mathrm{HI}(g)$\n$K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n(c) $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{4}(s)+10 \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=4.08 \\times 10^{-25}$ at $25^{\\circ} \\mathrm{C}$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=0.122$ at $50^{\\circ} \\mathrm{C}$\n26. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.\n(a) $\\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{BrCl}(g)$\n$K_{c}=4.7 \\times 10^{-2}$ at $25^{\\circ} \\mathrm{C}$\n(b) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n$K_{P}=48.2$ at $500^{\\circ} \\mathrm{C}$\n(c) $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CaCl}_{2}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=5.09 \\times 10^{-44}$ at $25^{\\circ} \\mathrm{C}$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=0.196$ at $60^{\\circ} \\mathrm{C}$\n27. What is the value of the equilibrium constant expression for the change $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$ at $30^{\\circ} \\mathrm{C}$ ? (See Appendix E.)\n28. Write the expression of the reaction quotient for the ionization of HOCN in water.\n29. Write the reaction quotient expression for the ionization of $\\mathrm{NH}_{3}$ in water.\n30. What is the approximate value of the equilibrium constant $K_{P}$ for the change $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(l) \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(g)$ at $25^{\\circ} \\mathrm{C}$. (The equilibrium vapor pressure for this substance is 570 torr at $25^{\\circ} \\mathrm{C}$.)\n"} {"id": "textbook_4403", "title": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "\n31. The following equation represents a reversible decomposition:\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\nUnder what conditions will decomposition in a closed container proceed to completion so that no $\\mathrm{CaCO}_{3}$ remains?\n32. Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium.\n33. What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?\n34. The following reaction occurs when a burner on a gas stove is lit:\n$\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\nIs an equilibrium among $\\mathrm{CH}_{4}, \\mathrm{O}_{2}, \\mathrm{CO}_{2}$, and $\\mathrm{H}_{2} \\mathrm{O}$ established under these conditions? Explain your answer.\n35. A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, $\\mathrm{SO}_{3}$, from sulfur dioxide, $\\mathrm{SO}_{2}$, and oxygen, $\\mathrm{O}_{2}$, shown here. At high temperatures, the rate of formation of $\\mathrm{SO}_{3}$ is higher, but the equilibrium amount (concentration or partial pressure) of $\\mathrm{SO}_{3}$ is lower than it would be at lower temperatures.\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?\n(b) Is the reaction endothermic or exothermic?\n36. Suggest four ways in which the concentration of hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$, could be increased in an equilibrium described by the following equation:\n$\\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{4}(g) \\quad \\Delta H=95 \\mathrm{~kJ}$\n37. Suggest four ways in which the concentration of $\\mathrm{PH}_{3}$ could be increased in an equilibrium described by the following equation:"} {"id": "textbook_4404", "title": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "$$\n\\mathrm{P}_{4}(g)+6 \\mathrm{H}_{2}(g) \\rightleftharpoons 4 \\mathrm{PH}_{3}(g) \\quad \\Delta H=110.5 \\mathrm{~kJ}\n$$"} {"id": "textbook_4405", "title": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "38. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$\\Delta H=92 \\mathrm{~kJ}$\n(b) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$\\Delta H=181 \\mathrm{~kJ}$\n(c) $2 \\mathrm{O}_{3}(g) \\rightleftharpoons 3 \\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\Delta H=-285 \\mathrm{~kJ}$\n(d) $\\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{CaCO}_{3}(s)$\n$\\Delta H=-176 \\mathrm{~kJ}$\n39. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?\n(a) $2 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)$\n$\\Delta H=484 \\mathrm{~kJ}$\n(b) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n$\\Delta H=-92.2 \\mathrm{~kJ}$\n(c) $2 \\mathrm{Br}(g) \\rightleftharpoons \\mathrm{Br}_{2}(g)$\n(d) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\rightleftharpoons 2 \\mathrm{HI}(g)$"} {"id": "textbook_4406", "title": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "$$\n\\begin{aligned}\n\\Delta H=-224 \\mathrm{~kJ} \\\\\n\\Delta H=53 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_4407", "title": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle", "content": "40. Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$2 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(g) \\quad \\Delta H=-90.2 \\mathrm{~kJ}$\n(b) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if more $\\mathrm{H}_{2}$ is added?\n(c) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if CO is removed?\n(d) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if $\\mathrm{CH}_{3} \\mathrm{OH}$ is added?\n(e) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if the temperature of the system is increased?\n41. Nitrogen and oxygen react at high temperatures.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g) \\quad \\Delta H=181 \\mathrm{~kJ}$\n(b) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if more $\\mathrm{O}_{2}$ is added?\n(c) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if $\\mathrm{N}_{2}$ is removed?\n(d) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if NO is added?\n(e) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if the volume of the reaction vessel is decreased?\n(f) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if the temperature of the system is increased?\n42. Water gas, a mixture of $\\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.\n(a) Write the expression for the equilibrium constant for the reversible reaction\n$\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2}(g) \\quad \\Delta H=131.30 \\mathrm{~kJ}$\n(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?\n(c) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2} \\mathrm{O}$ is removed?\n(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?\n(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?\n43. Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{Fe}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=98.7 \\mathrm{~kJ}$\n(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?\n(c) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2} \\mathrm{O}$ is removed?\n(d) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2}$ is added?\n(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?\n(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?\n44. Ammonia is a weak base that reacts with water according to this equation:\n$\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{+}(a q)+\\mathrm{OH}^{-}(a q)$\nWill any of the following increase the percent of ammonia that is converted to the ammonium ion in water?\n(a) Addition of NaOH\n(b) Addition of HCl\n(c) Addition of $\\mathrm{NH}_{4} \\mathrm{Cl}$\n45. Acetic acid is a weak acid that reacts with water according to this equation:\n$\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)$\nWill any of the following increase the percent of acetic acid that reacts and produces $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$ion?\n(a) Addition of HCl\n(b) Addition of NaOH\n(c) Addition of $\\mathrm{NaCH}_{3} \\mathrm{CO}_{2}$\n46. Suggest two ways in which the equilibrium concentration of $\\mathrm{Ag}^{+}$can be reduced in a solution of $\\mathrm{Na}^{+}, \\mathrm{Cl}^{-}$, $\\mathrm{Ag}^{+}$, and $\\mathrm{NO}_{3}{ }^{-}$, in contact with solid AgCl .\n$\\mathrm{Na}^{+}(a q)+\\mathrm{Cl}^{-}(a q)+\\mathrm{Ag}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q) \\rightleftharpoons \\mathrm{AgCl}(s)+\\mathrm{Na}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q)$\n$\\Delta H=-65.9 \\mathrm{~kJ}$\n47. How can the pressure of water vapor be increased in the following equilibrium?\n$\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=41 \\mathrm{~kJ}$\n48. A solution is saturated with silver sulfate and contains excess solid silver sulfate:\n$\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\rightleftharpoons 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\nA small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive $\\mathrm{Ag}^{+}$ ions. Explain this observation.\n49. When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?\n(a) HCl\n(b) $\\mathrm{H}^{+}+\\mathrm{Cl}^{-}$\n(c) HOCl\n(d) $\\mathrm{H}^{+}+\\mathrm{OCl}^{-}$\n"} {"id": "textbook_4408", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "\n50. A reaction is represented by this equation: $\\mathrm{A}(a q)+2 \\mathrm{~B}(a q) \\rightleftharpoons 2 \\mathrm{C}(a q) \\quad K_{c}=1 \\times 10^{3}$\n(a) Write the mathematical expression for the equilibrium constant.\n(b) Using concentrations $\\leq 1 M$, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.\n51. A reaction is represented by this equation: $2 \\mathrm{~W}(a q) \\rightleftharpoons \\mathrm{X}(a q)+2 \\mathrm{Y}(a q)$"} {"id": "textbook_4409", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\nK_{c}=5 \\times 10^{-4}\n$$"} {"id": "textbook_4410", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "(a) Write the mathematical expression for the equilibrium constant.\n(b) Using concentrations of $\\leq 1 M$, identify two sets of concentrations that describe a mixture of $\\mathrm{W}, \\mathrm{X}$, and Y at equilibrium.\n52. What is the value of the equilibrium constant at $500^{\\circ} \\mathrm{C}$ for the formation of $\\mathrm{NH}_{3}$ according to the following equation?\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\nAn equilibrium mixture of $\\mathrm{NH}_{3}(g), \\mathrm{H}_{2}(g)$, and $\\mathrm{N}_{2}(g)$ at $500^{\\circ} \\mathrm{C}$ was found to contain $1.35 \\mathrm{MH}_{2}, 1.15 \\mathrm{M}_{2}$, and $4.12 \\times 10^{-1} \\mathrm{MNH}_{3}$.\n53. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. $\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 3 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g)$\nWhat is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: $\\mathrm{CH}_{4}, 0.126 \\mathrm{M} ; \\mathrm{H}_{2} \\mathrm{O}, 0.242 \\mathrm{M} ; \\mathrm{CO}, 0.126 \\mathrm{M} ; \\mathrm{H}_{2} 1.15 \\mathrm{M}$, at a temperature of $760^{\\circ} \\mathrm{C}$ ?\n54. A $0.72-\\mathrm{mol} \\mathrm{sample}$ of $\\mathrm{PCl}_{5}$ is put into a $1.00-\\mathrm{L}$ vessel and heated. At equilibrium, the vessel contains 0.40 mol of $\\mathrm{PCl}_{3}(g)$ and 0.40 mol of $\\mathrm{Cl}_{2}(g)$. Calculate the value of the equilibrium constant for the decomposition of $\\mathrm{PCl}_{5}$ to $\\mathrm{PCl}_{3}$ and $\\mathrm{Cl}_{2}$ at this temperature.\n55. At 1 atm and $25^{\\circ} \\mathrm{C}, \\mathrm{NO}_{2}$ with an initial concentration of 1.00 M is $0.0033 \\%$ decomposed into NO and $\\mathrm{O}_{2}$. Calculate the value of the equilibrium constant for the reaction.\n$2 \\mathrm{NO}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)$\n56. Calculate the value of the equilibrium constant $K_{P}$ for the reaction $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$ from these equilibrium pressures: NO, $0.050 \\mathrm{~atm} ; \\mathrm{Cl}_{2}, 0.30 \\mathrm{~atm} ; \\mathrm{NOCl}, 1.2 \\mathrm{~atm}$.\n57. When heated, iodine vapor dissociates according to this equation:\n$\\mathrm{I}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{I}(\\mathrm{g})$\nAt 1274 K , a sample exhibits a partial pressure of $\\mathrm{I}_{2}$ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm . Determine the value of the equilibrium constant, $K_{P}$, for the decomposition at 1274 K .\n58. A sample of ammonium chloride was heated in a closed container.\n$\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)$\nAt equilibrium, the pressure of $\\mathrm{NH}_{3}(g)$ was found to be 1.75 atm . What is the value of the equilibrium constant $K_{P}$ for the decomposition at this temperature?\n59. At a temperature of $60^{\\circ} \\mathrm{C}$, the vapor pressure of water is 0.196 atm . What is the value of the equilibrium constant $K_{P}$ for the vaporization equilibrium at $60^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n60. Complete the following partial ICE tables.\n(a)"} {"id": "textbook_4411", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_4412", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $-\\quad+x$\n(b)"} {"id": "textbook_4413", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$"} {"id": "textbook_4414", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $\\quad+x$\n(c)"} {"id": "textbook_4415", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n2 \\mathrm{CH}_{4}(g) \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{2}(g)+3 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_4416", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change\n(d)"} {"id": "textbook_4417", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) & \\rightleftharpoons \\mathrm{CO}(g)+3 \\mathrm{H}_{2}(g) \\\\\n\\text { change_ } & \\rightleftharpoons-\n\\end{array}\n$$"} {"id": "textbook_4418", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "(e)"} {"id": "textbook_4419", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)\n$$"} {"id": "textbook_4420", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $+x$\n(f)"} {"id": "textbook_4421", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\mathrm{Ni}(s)+4 \\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Ni}(\\mathrm{CO})_{4}(g)\n$$"} {"id": "textbook_4422", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $+x$\n61. Complete the following partial ICE tables.\n(a)"} {"id": "textbook_4423", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$"} {"id": "textbook_4424", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change"} {"id": "textbook_4425", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\begin{array}{llll}\n& - & +x\n\\end{array}\n$$"} {"id": "textbook_4426", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "(b)"} {"id": "textbook_4427", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\mathrm{CS}_{2}(g)+4 \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{CH}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{~S}(g)\n$$"} {"id": "textbook_4428", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $+x$\n(c)\n$\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{HCl}(g)$\nchange $+x$\n(d)\n$2 \\mathrm{NH}_{3}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(g)$\nchange $\\qquad$\n$\\qquad$ $+x$\n(e)"} {"id": "textbook_4429", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\mathrm{NH}_{4} \\mathrm{HS}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{H}_{2} \\mathrm{~S}(g)\n$$"} {"id": "textbook_4430", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change"} {"id": "textbook_4431", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n+x\n$$"} {"id": "textbook_4432", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "(f)"} {"id": "textbook_4433", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n\\mathrm{Fe}(s)+5 \\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Fe}(\\mathrm{CO})_{5}(g)\n$$"} {"id": "textbook_4434", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "change $+x$\n62. Why are there no changes specified for Ni in Exercise 13.60, part (f)? What property of Ni does change?\n63. Why are there no changes specified for $\\mathrm{NH}_{4} \\mathrm{HS}$ in Exercise 13.61, part (e)? What property of $\\mathrm{NH}_{4} \\mathrm{HS}$ does change?\n64. Analysis of the gases in a sealed reaction vessel containing $\\mathrm{NH}_{3}, \\mathrm{~N}_{2}$, and $\\mathrm{H}_{2}$ at equilibrium at $400^{\\circ} \\mathrm{C}$ established the concentration of $\\mathrm{N}_{2}$ to be 1.2 M and the concentration of $\\mathrm{H}_{2}$ to be 0.24 M .\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) \\quad K_{c}=0.50$ at $400^{\\circ} \\mathrm{C}$\nCalculate the equilibrium molar concentration of $\\mathrm{NH}_{3}$.\n65. Calculate the number of moles of HI that are at equilibrium with 1.25 mol of $\\mathrm{H}_{2}$ and $1.25 \\mathrm{~mol}^{\\text {of }} \\mathrm{I}_{2}$ in a 5.00-L flask at $448^{\\circ} \\mathrm{C}$.\n$\\mathrm{H}_{2}+\\mathrm{I}_{2} \\rightleftharpoons 2 \\mathrm{HI} \\quad K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n66. What is the pressure of BrCl in an equilibrium mixture of $\\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$, and BrCl if the pressure of $\\mathrm{Cl}_{2}$ in the mixture is 0.115 atm and the pressure of $\\mathrm{Br}_{2}$ in the mixture is 0.450 atm ?\n$\\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{BrCl}(g) \\quad K_{P}=4.7 \\times 10^{-2}$\n67. What is the pressure of $\\mathrm{CO}_{2}$ in a mixture at equilibrium that contains 0.50 atm $\\mathrm{H}_{2}, 2.0$ atm of $\\mathrm{H}_{2} \\mathrm{O}$, and 1.0 atm of CO at $990^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2}(g)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{CO}(g) \\quad K_{P}=1.6$ at $990^{\\circ} \\mathrm{C}$\n68. Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.\n$\\mathrm{CoO}(s)+\\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Co}(s)+\\mathrm{CO}_{2}(g) \\quad K_{c}=4.90 \\times 10^{2}$ at $550^{\\circ} \\mathrm{C}$\nWhat concentration of CO remains in an equilibrium mixture with $\\left[\\mathrm{CO}_{2}\\right]=0.100 \\mathrm{M}$ ?\n69. Carbon reacts with water vapor at elevated temperatures.\n$\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2}(g) \\quad K_{c}=0.2$ at $1000^{\\circ} \\mathrm{C}$\nAssuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=0.500 \\mathrm{M}$ at $1000^{\\circ} \\mathrm{C}$ ?\n70. Sodium sulfate 10 -hydrate, $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$, dehydrates according to the equation\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{4}(s)+10 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad K_{P}=4.08 \\times 10^{-25}$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of water vapor at equilibrium with a mixture of $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{NaSO}_{4}$ ?\n71. Calcium chloride 6-hydrate, $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$, dehydrates according to the equation\n$\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CaCl}_{2}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad K_{P}=5.09 \\times 10^{-44}$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of water vapor at equilibrium with a mixture of $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CaCl}_{2}$ at $25{ }^{\\circ} \\mathrm{C}$ ?\n72. A student solved the following problem and found the equilibrium concentrations to be $\\left[\\mathrm{SO}_{2}\\right]=0.590 \\mathrm{M}$, $\\left[\\mathrm{O}_{2}\\right]=0.0450 \\mathrm{M}$, and $\\left[\\mathrm{SO}_{3}\\right]=0.260 \\mathrm{M}$. How could this student check the work without reworking the problem? The problem was: For the following reaction at $600^{\\circ} \\mathrm{C}$ :\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g) \\quad K_{c}=4.32$\nWhat are the equilibrium concentrations of all species in a mixture that was prepared with $\\left[\\mathrm{SO}_{3}\\right]=0.500$ $\\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=0 \\mathrm{M}$, and $\\left[\\mathrm{O}_{2}\\right]=0.350 \\mathrm{M}$ ?\n73. A student solved the following problem and found $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0.16 \\mathrm{M}$ at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ in a mixture formed from a sample of $\\mathrm{NO}_{2}$ with a concentration of 0.10\n$M$ ?\n$2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\quad K_{c}=160$\n74. Assume that the change in concentration of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ is small enough to be neglected in the following problem.\n(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ with chloroform as the solvent.\n$\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g) \\quad K_{c}=1.07 \\times 10^{-5}$ in chloroform\n(b) Confirm that the change is small enough to be neglected.\n75. Assume that the change in concentration of $\\mathrm{COCl}_{2}$ is small enough to be neglected in the following problem.\n(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of $\\mathrm{COCl}_{2}$ with an initial concentration of 0.3166 M .\n$\\mathrm{COCl}_{2}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{Cl}_{2}(g) \\quad K_{c}=2.2 \\times 10^{-10}$\n(b) Confirm that the change is small enough to be neglected.\n76. Assume that the change in pressure of $\\mathrm{H}_{2} \\mathrm{~S}$ is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of $\\mathrm{H}_{2} \\mathrm{~S}$ with an initial pressure of 0.824 atm ."} {"id": "textbook_4435", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{~S}(g) \\rightleftharpoons 2 \\mathrm{H}_{2}(g)+\\mathrm{S}_{2}(g) \\quad K_{P}=2.2 \\times 10^{-6}\n$$"} {"id": "textbook_4436", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "(b) Confirm that the change is small enough to be neglected.\n77. What are all concentrations after a mixture that contains $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=1.00 \\mathrm{M}$ and $\\left[\\mathrm{Cl}_{2} \\mathrm{O}\\right]=1.00 \\mathrm{M}$ comes to equilibrium at $25^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{Cl}_{2} \\mathrm{O}(g) \\rightleftharpoons 2 \\mathrm{HOCl}(g) \\quad K_{c}=0.0900$\n78. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of $\\mathrm{H}_{2}$ and 63.5 g of iodine at 448 ${ }^{\\circ} \\mathrm{C}$.\n$\\mathrm{H}_{2}+\\mathrm{I}_{2} \\rightleftharpoons 2 \\mathrm{HI} \\quad K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n79. Butane exists as two isomers, $n$-butane and isobutane.\n\n$K_{P}=2.5$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?\n80. What is the minimum mass of $\\mathrm{CaCO}_{3}$ required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant $\\left(K_{c}\\right)$ is 0.50 for the decomposition reaction of $\\mathrm{CaCO}_{3}$ at that temperature?\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\n81. The equilibrium constant $\\left(K_{c}\\right)$ for this reaction is 1.60 at $990^{\\circ} \\mathrm{C}$ :\n$\\mathrm{H}_{2}(g)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{CO}(g)$\nCalculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of $\\mathrm{H}_{2}, 2.00 \\mathrm{~mol}$ of $\\mathrm{CO}_{2}, 0.750 \\mathrm{~mol}$ of $\\mathrm{H}_{2} \\mathrm{O}$, and 1.00 mol of CO to a $5.00-\\mathrm{L}$ container at $990^{\\circ} \\mathrm{C}$.\n82. In a 3.0-L vessel, the following equilibrium partial pressures are measured: $\\mathrm{N}_{2}, 190$ torr; $\\mathrm{H}_{2}, 317$ torr; $\\mathrm{NH}_{3}, 1.00 \\times 10^{3}$ torr.\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(a) How will the partial pressures of $\\mathrm{H}_{2}, \\mathrm{~N}_{2}$, and $\\mathrm{NH}_{3}$ change if $\\mathrm{H}_{2}$ is removed from the system? Will they increase, decrease, or remain the same?\n(b) Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.\n83. The equilibrium constant $\\left(K_{c}\\right)$ for this reaction is 5.0 at a given temperature.\n$\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g)$\n(a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of $\\mathrm{CO}, 0.30 \\mathrm{~mol}$ of water vapor, and 0.90 mol of $\\mathrm{H}_{2}$ in a liter. How many moles of $\\mathrm{CO}_{2}$ were there in the equilibrium mixture?\n(b) Maintaining the same temperature, additional $\\mathrm{H}_{2}$ was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of $\\mathrm{CO}, 0.30$ mol of water vapor, and 1.2 mol of $\\mathrm{H}_{2}$ in a liter. How many moles of $\\mathrm{CO}_{2}$ were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.\n84. Antimony pentachloride decomposes according to this equation:\n$\\mathrm{SbCl}_{5}(g) \\rightleftharpoons \\mathrm{SbCl}_{3}(g)+\\mathrm{Cl}_{2}(g)$\nAn equilibrium mixture in a 5.00 -L flask at $448{ }^{\\circ} \\mathrm{C}$ contains 3.85 g of $\\mathrm{SbCl}_{5}, 9.14 \\mathrm{~g}$ of $\\mathrm{SbCl}_{3}$, and 2.84 g of $\\mathrm{Cl}_{2}$. How many grams of each will be found if the mixture is transferred into a $2.00-\\mathrm{L}$ flask at the same temperature?\n85. Consider the equilibrium\n$4 \\mathrm{NO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 4 \\mathrm{NH}_{3}(g)+7 \\mathrm{O}_{2}(g)$\n(a) What is the expression for the equilibrium constant $\\left(K_{c}\\right)$ of the reaction?\n(b) How must the concentration of $\\mathrm{NH}_{3}$ change to reach equilibrium if the reaction quotient is less than the equilibrium constant?\n(c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of $\\mathrm{NO}_{2}$ ?\n(d) If the change in the pressure of $\\mathrm{NO}_{2}$ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of $\\mathrm{O}_{2}$ change?\n86. The binding of oxygen by hemoglobin $(\\mathrm{Hb})$, giving oxyhemoglobin $\\left(\\mathrm{HbO}_{2}\\right)$, is partially regulated by the concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and dissolved $\\mathrm{CO}_{2}$ in the blood. Although the equilibrium is complicated, it can be summarized as\n$\\mathrm{HbO}_{2}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}-\\mathrm{Hb}-\\mathrm{H}^{+}+\\mathrm{O}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(a) Write the equilibrium constant expression for this reaction.\n(b) Explain why the production of lactic acid and $\\mathrm{CO}_{2}$ in a muscle during exertion stimulates release of $\\mathrm{O}_{2}$ from the oxyhemoglobin in the blood passing through the muscle.\n87. Liquid $\\mathrm{N}_{2} \\mathrm{O}_{3}$ is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and $\\mathrm{NO}_{2}$. At $25^{\\circ} \\mathrm{C}$, a value of $K_{P}=1.91$ has been established for this decomposition. If 0.236 moles of $\\mathrm{N}_{2} \\mathrm{O}_{3}$ are placed in a 1.52 - L vessel at $25^{\\circ} \\mathrm{C}$, calculate the equilibrium partial pressures of $\\mathrm{N}_{2} \\mathrm{O}_{3}(\\mathrm{~g}), \\mathrm{NO}_{2}(\\mathrm{~g})$, and $\\mathrm{NO}(\\mathrm{g})$.\n88. A 1.00-L vessel at $400^{\\circ} \\mathrm{C}$ contains the following equilibrium concentrations: $\\mathrm{N}_{2}, 1.00 \\mathrm{M} ; \\mathrm{H}_{2}, 0.50 \\mathrm{M}$; and $\\mathrm{NH}_{3}, 0.25 \\mathrm{M}$. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 $M$ ? The equilibrium reaction is $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n89. Calculate the equilibrium constant at $25^{\\circ} \\mathrm{C}$ for each of the following reactions from the value of $\\Delta G^{\\circ}$ given.\n(a) $\\mathrm{I}_{2}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{ICl}(g)$\n$\\Delta G^{\\circ}=-10.88 \\mathrm{~kJ}$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow 2 \\mathrm{HI}(g)$\n$\\Delta G^{\\circ}=3.4 \\mathrm{~kJ}$\n(c) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g) \\quad \\Delta G^{\\circ}=-39 \\mathrm{~kJ}$\n(d) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g) \\quad \\Delta G^{\\circ}=-141.82 \\mathrm{~kJ}$\n(e) $\\mathrm{CS}_{2}(g) \\longrightarrow \\mathrm{CS}_{2}(l) \\quad \\Delta G^{\\circ}=-1.88 \\mathrm{~kJ}$\n90. Calculate the equilibrium constant at the temperature given.\n(a) $\\mathrm{O}_{2}(g)+2 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{~F}_{2} \\mathrm{O}(g) \\quad\\left(\\mathrm{T}=100^{\\circ} \\mathrm{C}\\right)$\n(b) $\\mathrm{I}_{2}(s)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{IBr}(g) \\quad\\left(\\mathrm{T}=0.0^{\\circ} \\mathrm{C}\\right)$\n(c) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\quad\\left(\\mathrm{T}=575^{\\circ} \\mathrm{C}\\right)$\n$(d) \\mathrm{N}_{2} \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{NO}_{2}(g) \\quad\\left(\\mathrm{T}=-10.0^{\\circ} \\mathrm{C}\\right)$\n(e) $\\mathrm{SnCl}_{4}(l) \\longrightarrow \\mathrm{SnCl}_{4}(g) \\quad\\left(\\mathrm{T}=200^{\\circ} \\mathrm{C}\\right)$\n91. Calculate the equilibrium constant at the temperature given.\n(a) $\\mathrm{I}_{2}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{ICl}(g) \\quad\\left(\\mathrm{T}=100^{\\circ} \\mathrm{C}\\right)$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow 2 \\mathrm{HI}(g) \\quad\\left(\\mathrm{T}=0.0^{\\circ} \\mathrm{C}\\right)$\n(c) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g) \\quad\\left(\\mathrm{T}=125^{\\circ} \\mathrm{C}\\right)$\n(d) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g) \\quad\\left(\\mathrm{T}=675^{\\circ} \\mathrm{C}\\right)$\n(e) $\\mathrm{CS}_{2}(g) \\longrightarrow \\mathrm{CS}_{2}(l) \\quad\\left(\\mathrm{T}=90^{\\circ} \\mathrm{C}\\right)$\n92. Consider the following reaction at 298 K :\n$\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g) \\quad K_{P}=0.142$\nWhat is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.\n93. Determine the normal boiling point (in kelvin) of dichloroethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.\n94. Under what conditions is $\\mathrm{N}_{2} \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{NO}_{2}(g)$ spontaneous?\n95. At room temperature, the equilibrium constant $\\left(K_{W}\\right)$ for the self-ionization of water is $1.00 \\times 10^{-14}$. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)\n96. Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction $2 \\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{SO}_{2}(g) \\rightleftharpoons \\frac{3}{8} \\mathrm{~S}_{8}(s$, rhombic $)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?\n97. Consider the decomposition of $\\mathrm{CaCO}_{3}(s)$ into $\\mathrm{CaO}(s)$ and $\\mathrm{CO}_{2}(g)$. What is the equilibrium partial pressure of $\\mathrm{CO}_{2}$ at room temperature?\n98. In the laboratory, hydrogen chloride $(\\mathrm{HCl}(g))$ and ammonia $\\left(\\mathrm{NH}_{3}(g)\\right)$ often escape from bottles of their solutions and react to form the ammonium chloride $\\left(\\mathrm{NH}_{4} \\mathrm{Cl}(s)\\right)$, the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and $\\mathrm{NH}_{3}$ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)\n99. Benzene can be prepared from acetylene. $3 \\mathrm{C}_{2} \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{C}_{6} \\mathrm{H}_{6}(g)$. Determine the equilibrium constant at $25^{\\circ} \\mathrm{C}$ and at $850^{\\circ} \\mathrm{C}$. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?\n100. Carbon dioxide decomposes into CO and $\\mathrm{O}_{2}$ at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at $1000{ }^{\\circ} \\mathrm{C}$ for which the initial pressure of $\\mathrm{CO}_{2}$ was 1.15 atm?\n101. Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.\n$\\mathrm{CH}_{4}(g)+4 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+4 \\mathrm{HCl}(g)$\nWhat is the equilibrium constant for the reaction at 850 K ? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?\n102. Acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, can form a dimer, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}$, in the gas phase. $2 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{g}) \\longrightarrow\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}(\\mathrm{~g})$\nThe dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer."} {"id": "textbook_4437", "title": "1443. Exercises - 1443.4. Equilibrium Calculations", "content": "\nAt $25^{\\circ} \\mathrm{C}$, the equilibrium constant for the dimerization is $1.3 \\times 10^{3}$ (pressure in atm). What is $\\Delta S^{\\circ}$ for the reaction?\n"} {"id": "textbook_4438", "title": "1444. CHAPTER 14
Acid-Base Equilibria - ", "content": "\n"} {"id": "textbook_4439", "title": "1444. CHAPTER 14
Acid-Base Equilibria - ", "content": "Figure 14.1 Sinkholes such as this are the result of reactions between acidic groundwaters and basic rock formations, like limestone. (credit: modification of work by Emil Kehnel)\n"} {"id": "textbook_4440", "title": "1445. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_4441", "title": "1445. CHAPTER OUTLINE - 1445.1. Br\u00f8nsted-Lowry Acids and Bases", "content": ""} {"id": "textbook_4442", "title": "1446. 2 pH and pOH - ", "content": "\n14.3 Relative Strengths of Acids and Bases\n14.4 Hydrolysis of Salts\n14.5 Polyprotic Acids\n14.6 Buffers\n14.7 Acid-Base Titrations"} {"id": "textbook_4443", "title": "1446. 2 pH and pOH - ", "content": "INTRODUCTION Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$, is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases). These H+ transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation (Figure 14.1) to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class.\n"} {"id": "textbook_4444", "title": "1446. 2 pH and pOH - 1446.1. Br\u00f8nsted-Lowry Acids and Bases", "content": ""} {"id": "textbook_4445", "title": "1447. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4446", "title": "1447. LEARNING OBJECTIVES - ", "content": "- Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition\n- Write equations for acid and base ionization reactions\n- Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations\n- Describe the acid-base behavior of amphiprotic substances"} {"id": "textbook_4447", "title": "1447. LEARNING OBJECTIVES - ", "content": "The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be $\\mathrm{CO}_{2}$ ), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions."} {"id": "textbook_4448", "title": "1447. LEARNING OBJECTIVES - ", "content": "Johannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, $\\mathrm{H}^{+}$. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, ${ }^{1} \\mathrm{H}$.) A compound that donates a proton to another compound is called a Br\u00f8nsted-Lowry acid, and a compound that accepts a proton is called a Br\u00f8nsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base)."} {"id": "textbook_4449", "title": "1447. LEARNING OBJECTIVES - ", "content": "The concept of conjugate pairs is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates $\\mathrm{H}^{+}$, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts $\\mathrm{H}^{+}$, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, $\\mathrm{OH}^{-}$, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, $\\mathrm{NH}_{4}{ }^{+}$, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.\n"} {"id": "textbook_4450", "title": "1447. LEARNING OBJECTIVES - ", "content": "The reaction between a Br\u00f8nsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:\n"} {"id": "textbook_4451", "title": "1447. LEARNING OBJECTIVES - ", "content": "Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, $\\mathrm{C}_{5} \\mathrm{NH}_{5}$, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:\n"} {"id": "textbook_4452", "title": "1447. LEARNING OBJECTIVES - ", "content": "The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acidbase reactions for two amphiprotic species, bicarbonate ion and water:"} {"id": "textbook_4453", "title": "1447. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\\\\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{OH}^{-}(a q)\n\\end{aligned}\n$$"} {"id": "textbook_4454", "title": "1447. LEARNING OBJECTIVES - ", "content": "The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter."} {"id": "textbook_4455", "title": "1447. LEARNING OBJECTIVES - ", "content": "In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:\n"} {"id": "textbook_4456", "title": "1447. LEARNING OBJECTIVES - ", "content": "The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at $25^{\\circ} \\mathrm{C}$, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, $\\boldsymbol{K}_{\\mathbf{w}}$ :"} {"id": "textbook_4457", "title": "1447. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$"} {"id": "textbook_4458", "title": "1447. LEARNING OBJECTIVES - ", "content": "The slight ionization of pure water is reflected in the small value of the equilibrium constant; at $25^{\\circ} \\mathrm{C}, K_{\\mathrm{w}}$ has a value of $1.0 \\times 10^{-14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at $100^{\\circ} \\mathrm{C}$, the value for $K_{\\mathrm{w}}$ is about $5.6 \\times 10^{-13}$, roughly 50 times larger than the value at $25^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_4459", "title": "1448. EXAMPLE 14.1 - ", "content": ""} {"id": "textbook_4460", "title": "1449. Ion Concentrations in Pure Water - ", "content": "\nWhat are the hydronium ion concentration and the hydroxide ion concentration in pure water at $25^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_4461", "title": "1450. Solution - ", "content": "\nThe autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=x$. At $25^{\\circ} \\mathrm{C}$ :"} {"id": "textbook_4462", "title": "1450. Solution - ", "content": "$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=(x)(x)=x^{2}=1.0 \\times 10^{-14}\n$$"} {"id": "textbook_4463", "title": "1450. Solution - ", "content": "So:"} {"id": "textbook_4464", "title": "1450. Solution - ", "content": "$$\nx=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=\\sqrt{1.0 \\times 10^{-14}}=1.0 \\times 10^{-7} \\mathrm{M}\n$$"} {"id": "textbook_4465", "title": "1450. Solution - ", "content": "The hydronium ion concentration and the hydroxide ion concentration are the same, $1.0 \\times 10^{-7} \\mathrm{M}$.\n"} {"id": "textbook_4466", "title": "1451. Check Your Learning - ", "content": "\nThe ion product of water at $80^{\\circ} \\mathrm{C}$ is $2.4 \\times 10^{-13}$. What are the concentrations of hydronium and hydroxide ions in pure water at $80^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_4467", "title": "1452. Answer: - ", "content": "\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=4.9 \\times 10^{-7} \\mathrm{M}$\n"} {"id": "textbook_4468", "title": "1453. EXAMPLE 14.2 - ", "content": ""} {"id": "textbook_4469", "title": "1454. The Inverse Relation between $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right.$] and $\\left[\\mathrm{OH}^{-}\\right.$] - ", "content": "\nA solution of an acid in water has a hydronium ion concentration of $2.0 \\times 10^{-6} \\mathrm{M}$. What is the concentration of hydroxide ion at $25^{\\circ} \\mathrm{C}$ ?\n"} {"id": "textbook_4470", "title": "1455. Solution - ", "content": "\nUse the value of the ion-product constant for water at $25^{\\circ} \\mathrm{C}$"} {"id": "textbook_4471", "title": "1455. Solution - ", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14}\n$$"} {"id": "textbook_4472", "title": "1455. Solution - ", "content": "to calculate the missing equilibrium concentration.\nRearrangement of the $K_{\\mathrm{w}}$ expression shows that $\\left[\\mathrm{OH}^{-}\\right]$is inversely proportional to $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$:"} {"id": "textbook_4473", "title": "1455. Solution - ", "content": "$$\n\\left[\\mathrm{OH}^{-}\\right]=\\frac{K_{\\mathrm{w}}}{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}=\\frac{1.0 \\times 10^{-14}}{2.0 \\times 10^{-6}}=5.0 \\times 10^{-9}\n$$"} {"id": "textbook_4474", "title": "1455. Solution - ", "content": "Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier's principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration."} {"id": "textbook_4475", "title": "1455. Solution - ", "content": "Substituting the ion concentrations into the $K_{\\mathrm{w}}$ expression confirms this calculation, resulting in the expected value:"} {"id": "textbook_4476", "title": "1455. Solution - ", "content": "$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=\\left(2.0 \\times 10^{-6}\\right)\\left(5.0 \\times 10^{-9}\\right)=1.0 \\times 10^{-14}\n$$\n"} {"id": "textbook_4477", "title": "1456. Check Your Learning - ", "content": "\nWhat is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 $M$ at $25^{\\circ} \\mathrm{C}$ ?"} {"id": "textbook_4478", "title": "1456. Check Your Learning - ", "content": "Answer:\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1 \\times 10^{-11} \\mathrm{M}$\n"} {"id": "textbook_4479", "title": "1457. EXAMPLE 14.3 - ", "content": ""} {"id": "textbook_4480", "title": "1458. Representing the Acid-Base Behavior of an Amphoteric Substance - ", "content": "\nWrite separate equations representing the reaction of $\\mathrm{HSO}_{3}{ }^{-}$\n(a) as an acid with $\\mathrm{OH}^{-}$\n(b) as a base with HI\n"} {"id": "textbook_4481", "title": "1459. Solution - ", "content": "\n(a) $\\mathrm{HSO}_{3}{ }^{-}(a q)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{SO}_{3}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(b) $\\mathrm{HSO}_{3}{ }^{-}(a q)+\\mathrm{HI}(a q) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q)+\\mathrm{I}^{-}(a q)$\n"} {"id": "textbook_4482", "title": "1460. Check Your Learning - ", "content": "\nWrite separate equations representing the reaction of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(a) as a base with HBr\n(b) as an acid with $\\mathrm{OH}^{-}$\n"} {"id": "textbook_4483", "title": "1461. Answer: - ", "content": "\n(a) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)+\\mathrm{HBr}(a q) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{Br}^{-}(a q)$; (b)\n$\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{HPO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n"} {"id": "textbook_4484", "title": "1462. 2 pH and pOH - ", "content": ""} {"id": "textbook_4485", "title": "1463. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4486", "title": "1463. LEARNING OBJECTIVES - ", "content": "- Explain the characterization of aqueous solutions as acidic, basic, or neutral\n- Express hydronium and hydroxide ion concentrations on the pH and pOH scales\n- Perform calculations relating pH and pOH"} {"id": "textbook_4487", "title": "1463. LEARNING OBJECTIVES - ", "content": "As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ( $K_{\\mathrm{w}}$ ). The concentrations of these ions in a solution are often critical determinants of the solution's properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions."} {"id": "textbook_4488", "title": "1463. LEARNING OBJECTIVES - ", "content": "A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where \" X \" is the quantity of interest and \"log\" is the base-10 logarithm:"} {"id": "textbook_4489", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{pX}=-\\log \\mathrm{X}\n$$"} {"id": "textbook_4490", "title": "1463. LEARNING OBJECTIVES - ", "content": "The $\\mathbf{p H}$ of a solution is therefore defined as shown here, where $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$is the molar concentration of hydronium ion in the solution:"} {"id": "textbook_4491", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\n$$"} {"id": "textbook_4492", "title": "1463. LEARNING OBJECTIVES - ", "content": "Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:"} {"id": "textbook_4493", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-\\mathrm{pH}}\n$$"} {"id": "textbook_4494", "title": "1463. LEARNING OBJECTIVES - ", "content": "Likewise, the hydroxide ion molarity may be expressed as a p-function, or $\\mathbf{p O H}$ :"} {"id": "textbook_4495", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]\n$$"} {"id": "textbook_4496", "title": "1463. LEARNING OBJECTIVES - ", "content": "or"} {"id": "textbook_4497", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\left[\\mathrm{OH}^{-}\\right]=10^{-\\mathrm{pOH}}\n$$"} {"id": "textbook_4498", "title": "1463. LEARNING OBJECTIVES - ", "content": "Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_{\\mathrm{w}}$ expression:"} {"id": "textbook_4499", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$"} {"id": "textbook_4500", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{gathered}\n-\\log K_{\\mathrm{w}}=-\\log \\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\\right)=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]+-\\log \\left[\\mathrm{OH}^{-}\\right] \\\\\n\\mathrm{p} K_{\\mathrm{w}}=\\mathrm{pH}+\\mathrm{pOH}\n\\end{gathered}\n$$"} {"id": "textbook_4501", "title": "1463. LEARNING OBJECTIVES - ", "content": "At $25^{\\circ} \\mathrm{C}$, the value of $K_{\\mathrm{W}}$ is $1.0 \\times 10^{-14}$, and so:"} {"id": "textbook_4502", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n14.00=\\mathrm{pH}+\\mathrm{pOH}\n$$"} {"id": "textbook_4503", "title": "1463. LEARNING OBJECTIVES - ", "content": "As was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is $1.0 \\times$ $10^{-7} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. The pH and pOH of a neutral solution at this temperature are therefore:"} {"id": "textbook_4504", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(1.0 \\times 10^{-7}\\right)=7.00 \\\\\n& \\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log \\left(1.0 \\times 10^{-7}\\right)=7.00\n\\end{aligned}\n$$"} {"id": "textbook_4505", "title": "1463. LEARNING OBJECTIVES - ", "content": "And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \\times 10^{-7} \\mathrm{M}$ and hydroxide ion molarities less than $1.0 \\times 10^{-7} \\mathrm{M}$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \\times 10^{-7} \\mathrm{M}$ and hydroxide ion molarities greater than $1.0 \\times 10^{-7} M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00)."} {"id": "textbook_4506", "title": "1463. LEARNING OBJECTIVES - ", "content": "Since the autoionization constant $K_{\\mathrm{w}}$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than $25^{\\circ} \\mathrm{C}$. For example, the \"Check Your Learning\" exercise accompanying Example 14.1 showed the hydronium molarity of pure water at $80^{\\circ} \\mathrm{C}$ is $4.9 \\times 10^{-7} \\mathrm{M}$, which corresponds to pH and pOH values of:"} {"id": "textbook_4507", "title": "1463. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(4.9 \\times 10^{-7}\\right)=6.31 \\\\\n& \\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log \\left(4.9 \\times 10^{-7}\\right)=6.31\n\\end{aligned}\n$$"} {"id": "textbook_4508", "title": "1463. LEARNING OBJECTIVES - ", "content": "At this temperature, then, neutral solutions exhibit $\\mathrm{pH}=\\mathrm{pOH}=6.31$, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31 . This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around $36-40^{\\circ} \\mathrm{C}$. Unless otherwise noted, references to pH values are presumed to be those at $25^{\\circ} \\mathrm{C}$ (Table 14.1)."} {"id": "textbook_4509", "title": "1463. LEARNING OBJECTIVES - ", "content": "| Summary of Relations for Acidic, Basic and Neutral Solutions | | |\n| :---: | :---: | :---: |\n| Classification | Relative lon Concentrations | pH at $\\mathbf{2 5}^{\\circ} \\mathrm{C}$ |\n| acidic | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}<7$ |\n| neutral | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}=7$ |\n| basic | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}>7$ |"} {"id": "textbook_4510", "title": "1463. LEARNING OBJECTIVES - ", "content": "TABLE 14.1"} {"id": "textbook_4511", "title": "1463. LEARNING OBJECTIVES - ", "content": "Figure 14.2 shows the relationships between $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for solutions classified as acidic, basic, and neutral.\n\n\nFIGURE 14.2 The pH and pOH scales represent concentrations of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{OH}^{-}$, respectively. The pH and pOH values of some common substances at $25^{\\circ} \\mathrm{C}$ are shown in this chart.\n"} {"id": "textbook_4512", "title": "1464. EXAMPLE 14.4 - ", "content": ""} {"id": "textbook_4513", "title": "1465. Calculation of pH from $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$ - ", "content": "\nWhat is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \\times 10^{-3} \\mathrm{M}$ ?\n"} {"id": "textbook_4514", "title": "1466. Solution - ", "content": "\n$$\n\\begin{gathered}\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right] \\\\\n=-\\log \\left(1.2 \\times 10^{-3}\\right) \\\\\n=-(-2.92)=2.92\n\\end{gathered}\n$$"} {"id": "textbook_4515", "title": "1466. Solution - ", "content": "(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)\n"} {"id": "textbook_4516", "title": "1467. Check Your Learning - ", "content": "\nWater exposed to air contains carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, due to the reaction between carbon dioxide and water:"} {"id": "textbook_4517", "title": "1467. Check Your Learning - ", "content": "$$\n\\mathrm{CO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)\n$$"} {"id": "textbook_4518", "title": "1467. Check Your Learning - ", "content": "Air-saturated water has a hydronium ion concentration caused by the dissolved $\\mathrm{CO}_{2}$ of $2.0 \\times 10^{-6} \\mathrm{M}$, about 20-times larger than that of pure water. Calculate the pH of the solution at $25^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_4519", "title": "1468. Answer: - ", "content": ""} {"id": "textbook_4520", "title": "1468. Answer: - 1468.1. 0", "content": ""} {"id": "textbook_4521", "title": "1469. EXAMPLE 14.5 - ", "content": ""} {"id": "textbook_4522", "title": "1470. Calculation of Hydronium Ion Concentration from pH - ", "content": "\nCalculate the hydronium ion concentration of blood, the pH of which is 7.3.\n"} {"id": "textbook_4523", "title": "1471. Solution - ", "content": "\n$$\n\\begin{gathered}\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=7.3 \\\\\n\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-7.3 \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-7.3} \\text { or }\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\text {antilog of }-7.3} \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=5 \\times 10^{-8} \\mathrm{M}}\n\\end{gathered}\n$$"} {"id": "textbook_4524", "title": "1471. Solution - ", "content": "(On a calculator take the antilog, or the \"inverse\" log, of -7.3 , or calculate $10^{-7.3}$.)\n"} {"id": "textbook_4525", "title": "1472. Check Your Learning - ", "content": "\nCalculate the hydronium ion concentration of a solution with a pH of -1.07 .\n"} {"id": "textbook_4526", "title": "1473. Answer: - ", "content": "\n12 M\n"} {"id": "textbook_4527", "title": "1474. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_4528", "title": "1475. Environmental Science - ", "content": "\nNormal rainwater has a pH between 5 and 6 due to the presence of dissolved $\\mathrm{CO}_{2}$ which forms carbonic acid:"} {"id": "textbook_4529", "title": "1475. Environmental Science - ", "content": "$$\n\\begin{gathered}\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\\\\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\rightleftharpoons \\mathrm{H}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q)\n\\end{gathered}\n$$"} {"id": "textbook_4530", "title": "1475. Environmental Science - ", "content": "Acid rain is rainwater that has a pH of less than 5 , due to a variety of nonmetal oxides, including $\\mathrm{CO}_{2}, \\mathrm{SO}_{2}, \\mathrm{SO}_{3}$, NO , and $\\mathrm{NO}_{2}$ being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:"} {"id": "textbook_4531", "title": "1475. Environmental Science - ", "content": "$$\n\\begin{gathered}\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{SO}_{3}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\\\\n\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{HSO}_{4}^{-}(a q)\n\\end{gathered}\n$$"} {"id": "textbook_4532", "title": "1475. Environmental Science - ", "content": "Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of \"roasting\" ores of metal sulfides in metalrefining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine."} {"id": "textbook_4533", "title": "1475. Environmental Science - ", "content": "Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until\nthe 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India."} {"id": "textbook_4534", "title": "1475. Environmental Science - ", "content": "For further information on acid rain, visit this website (http://openstax.org/l/16EPA) hosted by the US Environmental Protection Agency.\n"} {"id": "textbook_4535", "title": "1475. Environmental Science - ", "content": "FIGURE 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by \"Eden, Janine and Jim\"/Flickr)\n"} {"id": "textbook_4536", "title": "1476. EXAMPLE 14.6 - ", "content": ""} {"id": "textbook_4537", "title": "1477. Calculation of pOH - ", "content": "\nWhat are the pOH and the pH of a $0.0125-M$ solution of potassium hydroxide, KOH ?\n"} {"id": "textbook_4538", "title": "1478. Solution - ", "content": "\nPotassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding $\\left[\\mathrm{OH}^{-}\\right]=0.0125 \\mathrm{M}$ :"} {"id": "textbook_4539", "title": "1478. Solution - ", "content": "$$\n\\begin{aligned}\n\\mathrm{pOH}= & =-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log 0.0125 \\\\\n& =-(-1.903)=1.903\n\\end{aligned}\n$$"} {"id": "textbook_4540", "title": "1478. Solution - ", "content": "The pH can be found from the pOH :"} {"id": "textbook_4541", "title": "1478. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{pH}+\\mathrm{pOH}=14.00 \\\\\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00-1.903=12.10\n\\end{gathered}\n$$\n"} {"id": "textbook_4542", "title": "1479. Check Your Learning - ", "content": "\nThe hydronium ion concentration of vinegar is approximately $4 \\times 10^{-3} \\mathrm{M}$. What are the corresponding values of pOH and pH ?\n"} {"id": "textbook_4543", "title": "1480. Answer: - ", "content": "\n$\\mathrm{pOH}=11.6, \\mathrm{pH}=2.4$"} {"id": "textbook_4544", "title": "1480. Answer: - ", "content": "The acidity of a solution is typically assessed experimentally by measurement of its pH . The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4).\n"} {"id": "textbook_4545", "title": "1480. Answer: - ", "content": "FIGURE 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of $\\pm 0.002 \\mathrm{pH}$ units, and may cost in excess of $\\$ 1000$. (b) A portable pH meter has lower resolution ( 0.01 pH units), lower accuracy ( $\\pm 0.2 \\mathrm{pH}$ units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)"} {"id": "textbook_4546", "title": "1480. Answer: - ", "content": "The pH of a solution may also be visually estimated using colored indicators (Figure 14.5). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.\n"} {"id": "textbook_4547", "title": "1480. Answer: - ", "content": "FIGURE 14.5 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH . (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa)\n"} {"id": "textbook_4548", "title": "1480. Answer: - 1480.1. Relative Strengths of Acids and Bases", "content": ""} {"id": "textbook_4549", "title": "1481. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4550", "title": "1481. LEARNING OBJECTIVES - ", "content": "- Assess the relative strengths of acids and bases according to their ionization constants\n- Rationalize trends in acid-base strength in relation to molecular structure\n- Carry out equilibrium calculations for weak acid-base systems"} {"id": "textbook_4551", "title": "1481. LEARNING OBJECTIVES - ", "content": ""} {"id": "textbook_4552", "title": "1482. Acid and Base Ionization Constants - ", "content": "\nThe relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure 14.6."} {"id": "textbook_4553", "title": "1482. Acid and Base Ionization Constants - ", "content": "| 6 Strong Acids | | 6 Strong Bases | |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{HClO}_{4}$ | perchloric acid | LiOH | lithium hydroxide |\n| HCl | hydrochloric acid | NaOH | sodium hydroxide |\n| HBr | hydrobromic acid | KOH | potassium hydroxide |\n| HI | hydroiodic acid | $\\mathrm{Ca}(\\mathrm{OH})_{2}$ | calcium hydroxide |\n| $\\mathrm{HNO}_{3}$ | nitric acid | $\\mathrm{Sr}(\\mathrm{OH})_{2}$ | strontium hydroxide |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfuric acid | $\\mathrm{Ba}(\\mathrm{OH})_{2}$ | barium hydroxide |"} {"id": "textbook_4554", "title": "1482. Acid and Base Ionization Constants - ", "content": "FIGURE 14.6 Some of the common strong acids and bases are listed here.\nThe relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, $\\boldsymbol{K}_{\\mathbf{a}}$. For the reaction of an acid HA:"} {"id": "textbook_4555", "title": "1482. Acid and Base Ionization Constants - ", "content": "$$\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q)\n$$"} {"id": "textbook_4556", "title": "1482. Acid and Base Ionization Constants - ", "content": "the acid ionization constant is written"} {"id": "textbook_4557", "title": "1482. Acid and Base Ionization Constants - ", "content": "$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n$$"} {"id": "textbook_4558", "title": "1482. Acid and Base Ionization Constants - ", "content": "where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]$ in the equation. The larger the $K_{\\mathrm{a}}$ of an acid, the larger the concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{A}^{-}$relative to the concentration of the nonionized acid, HA , in an equilibrium mixture, and the stronger the acid. An acid is classified as \"strong\" when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large ( $K_{\\mathrm{a}} \\approx \\infty$ ). Acids that are partially ionized are called \"weak,\" and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H."} {"id": "textbook_4559", "title": "1482. Acid and Base Ionization Constants - ", "content": "To illustrate this idea, three acid ionization equations and $K_{\\mathrm{a}}$ values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}<\\mathrm{HNO}_{2}<\\mathrm{HSO}_{4}{ }^{-}$:\n\n$$\n\\begin{array}{ccc}\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}(a q) & K_{\\mathrm{a}}=1.8 \\times 10^{-5} \\\\\n\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}{ }^{-}(a q) & K_{\\mathrm{a}}=4.6 \\times 10^{-4} \\\\\n\\mathrm{HSO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) & K_{\\mathrm{a}}=1.2 \\times 10^{-2}\n\\end{array}\n$$"} {"id": "textbook_4560", "title": "1482. Acid and Base Ionization Constants - ", "content": "Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:"} {"id": "textbook_4561", "title": "1482. Acid and Base Ionization Constants - ", "content": "$$\n\\% \\text { ionization }=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{[\\mathrm{HA}]_{0}} \\times 100\n$$"} {"id": "textbook_4562", "title": "1482. Acid and Base Ionization Constants - ", "content": "where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, $\\left[\\mathrm{A}^{-}\\right]=$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$). Unlike the $K_{\\mathrm{a}}$ value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.\n"} {"id": "textbook_4563", "title": "1483. EXAMPLE 14.7 - ", "content": ""} {"id": "textbook_4564", "title": "1484. Calculation of Percent Ionization from pH - ", "content": "\nCalculate the percent ionization of a $0.125-M$ solution of nitrous acid (a weak acid), with a pH of 2.09 .\n"} {"id": "textbook_4565", "title": "1485. Solution - ", "content": "\nThe percent ionization for an acid is:"} {"id": "textbook_4566", "title": "1485. Solution - ", "content": "$$\n\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{\\left[\\mathrm{HNO}_{2}\\right]_{0}} \\times 100\n$$"} {"id": "textbook_4567", "title": "1485. Solution - ", "content": "Converting the provided pH to hydronium ion molarity yields"} {"id": "textbook_4568", "title": "1485. Solution - ", "content": "$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-2.09}=0.0081 \\mathrm{M}\n$$"} {"id": "textbook_4569", "title": "1485. Solution - ", "content": "Substituting this value and the provided initial acid concentration into the percent ionization equation gives"} {"id": "textbook_4570", "title": "1485. Solution - ", "content": "$$\n\\frac{8.1 \\times 10^{-3}}{0.125} \\times 100=6.5 \\%\n$$"} {"id": "textbook_4571", "title": "1485. Solution - ", "content": "(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)\n"} {"id": "textbook_4572", "title": "1486. Check Your Learning - ", "content": "\nCalculate the percent ionization of a $0.10-M$ solution of acetic acid with a pH of 2.89 .\n"} {"id": "textbook_4573", "title": "1487. Answer: - ", "content": "\n1.3\\% ionized\n"} {"id": "textbook_4574", "title": "1488. LINK TO LEARNING - ", "content": "\nView the simulation (http://openstax.org/l/16AcidBase) of strong and weak acids and bases at the molecular level."} {"id": "textbook_4575", "title": "1488. LINK TO LEARNING - ", "content": "Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant $\\left(K_{\\mathbf{b}}\\right)$ in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:"} {"id": "textbook_4576", "title": "1488. LINK TO LEARNING - ", "content": "$$\n\\mathrm{B}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HB}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_4577", "title": "1488. LINK TO LEARNING - ", "content": "the ionization constant is written as"} {"id": "textbook_4578", "title": "1488. LINK TO LEARNING - ", "content": "$$\nK_{\\mathrm{b}}=\\frac{\\left[\\mathrm{HB}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{B}]}\n$$"} {"id": "textbook_4579", "title": "1488. LINK TO LEARNING - ", "content": "Inspection of the data for three weak bases presented below shows the base strength increases in the order $\\mathrm{NO}_{2}{ }^{-}<\\mathrm{CH}_{2} \\mathrm{CO}_{2}^{-}<\\mathrm{NH}_{3}$."} {"id": "textbook_4580", "title": "1488. LINK TO LEARNING - ", "content": "$$\n\\begin{array}{ll}\n\\mathrm{NO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HNO}_{2}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=2.17 \\times 10^{-11} \\\\\n\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=5.6 \\times 10^{-10} \\\\\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=1.8 \\times 10^{-5}\n\\end{array}\n$$"} {"id": "textbook_4581", "title": "1488. LINK TO LEARNING - ", "content": "A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a\nbase is also reflected in its percent ionization, computed as"} {"id": "textbook_4582", "title": "1488. LINK TO LEARNING - ", "content": "$$\n\\% \\text { ionization }=\\left[\\mathrm{OH}^{-}\\right]_{e q} /[\\mathrm{B}]_{0} \\times 100 \\%\n$$"} {"id": "textbook_4583", "title": "1488. LINK TO LEARNING - ", "content": "but will vary depending on the base ionization constant and the initial concentration of the solution.\n"} {"id": "textbook_4584", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "\nBr\u00f8nsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, $K_{\\mathrm{a}}$ or $K_{\\mathrm{b}}$, which represents the extent of the acid or base ionization reaction. For the conjugate acidbase pair HA / $\\mathrm{A}^{-}$, ionization equilibrium equations and ionization constant expressions are"} {"id": "textbook_4585", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "$$\n\\begin{aligned}\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q) & K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\\\\n\\mathrm{A}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{OH}^{-}(a q)+\\mathrm{HA}(a q) & K_{\\mathrm{b}}=\\frac{[\\mathrm{HA}]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{A}^{-}\\right]}\n\\end{aligned}\n$$"} {"id": "textbook_4586", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "Adding these two chemical equations yields the equation for the autoionization for water:"} {"id": "textbook_4587", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "$$\n\\begin{gathered}\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{A}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q)+\\mathrm{OH}^{-}(a q)+\\mathrm{HA}(a q) \\\\\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n\\end{gathered}\n$$"} {"id": "textbook_4588", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so"} {"id": "textbook_4589", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "$$\nK_{\\mathrm{a}} \\times K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\times \\frac{[\\mathrm{HA}]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{A}^{-}\\right]}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=K_{\\mathrm{w}}\n$$"} {"id": "textbook_4590", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, $K_{\\mathrm{w}}$. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:"} {"id": "textbook_4591", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "$$\nK_{\\mathrm{a}}=K_{\\mathrm{w}} / K_{\\mathrm{b}} \\text { or } K_{\\mathrm{b}}=K_{\\mathrm{w}} / K_{\\mathrm{a}}\n$$"} {"id": "textbook_4592", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "The inverse proportional relation between $K_{\\mathrm{a}}$ and $K_{\\mathrm{b}}$ means the stronger the acid or base, the weaker its conjugate partner. Figure 14.7 illustrates this relation for several conjugate acid-base pairs.\n"} {"id": "textbook_4593", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "Relative conjugate base strength\n\n\nFIGURE 14.7 Relative strengths of several conjugate acid-base pairs are shown.\n\n\nFIGURE 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance."} {"id": "textbook_4594", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "The listing of conjugate acid-base pairs shown in Figure 14.8 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table's columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution."} {"id": "textbook_4595", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$, meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for \"strong\" acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be \"weak,\" and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides $\\mathrm{HCl}, \\mathrm{HBr}$, and HI are strong acids in water but weak acids in ethanol (strength increasing $\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI}$ )."} {"id": "textbook_4596", "title": "1489. Relative Strengths of Conjugate Acid-Base Pairs - ", "content": "The right column of Figure 14.8 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don't undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acidbase pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large $K_{\\mathrm{a}}$, and so its conjugate base will exhibit a $K_{\\mathrm{b}}$ that is essentially zero:\nstrong acid: $\\quad K_{\\mathrm{a}} \\approx \\infty$\nconjugate base : $K_{\\mathrm{b}}=K_{\\mathrm{W}} / K_{\\mathrm{a}}=K_{\\mathrm{W}} / \\infty \\approx 0$\nA similar approach can be used to support the observation that conjugate acids of strong bases ( $K_{\\mathrm{b}} \\approx \\infty$ ) are of negligible strength ( $K_{\\mathrm{a}} \\approx 0$ ).\n"} {"id": "textbook_4597", "title": "1490. EXAMPLE 14.8 - ", "content": ""} {"id": "textbook_4598", "title": "1491. Calculating Ionization Constants for Conjugate Acid-Base Pairs - ", "content": "\nUse the $K_{\\mathrm{b}}$ for the nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$, to calculate the $K_{\\mathrm{a}}$ for its conjugate acid.\n"} {"id": "textbook_4599", "title": "1492. Solution - ", "content": "\n$K_{\\mathrm{b}}$ for $\\mathrm{NO}_{2}{ }^{-}$is given in this section as $2.17 \\times 10^{-11}$. The conjugate acid of $\\mathrm{NO}_{2}{ }^{-}$is $\\mathrm{HNO}_{2} ; K_{\\mathrm{a}}$ for $\\mathrm{HNO}_{2}$ can be calculated using the relationship:"} {"id": "textbook_4600", "title": "1492. Solution - ", "content": "$$\nK_{\\mathrm{a}} \\times K_{\\mathrm{b}}=1.0 \\times 10^{-14}=K_{\\mathrm{w}}\n$$"} {"id": "textbook_4601", "title": "1492. Solution - ", "content": "Solving for $K_{\\mathrm{a}}$ yields"} {"id": "textbook_4602", "title": "1492. Solution - ", "content": "$$\nK_{\\mathrm{a}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{b}}}=\\frac{1.0 \\times 10^{-14}}{2.17 \\times 10^{-11}}=4.6 \\times 10^{-4}\n$$"} {"id": "textbook_4603", "title": "1492. Solution - ", "content": "This answer can be verified by finding the $K_{\\mathrm{a}}$ for $\\mathrm{HNO}_{2}$ in Appendix H.\n"} {"id": "textbook_4604", "title": "1493. Check Your Learning - ", "content": "\nDetermine the relative acid strengths of $\\mathrm{NH}_{4}{ }^{+}$and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as $4.9 \\times 10^{-10}$. The ionization constant of $\\mathrm{NH}_{4}{ }^{+}$is not listed, but the ionization constant of its conjugate base, $\\mathrm{NH}_{3}$, is listed as $1.8 \\times 10^{-5}$.\n"} {"id": "textbook_4605", "title": "1494. Answer: - ", "content": "\n$\\mathrm{NH}_{4}{ }^{+}$is the slightly stronger acid ( $K_{\\mathrm{a}}$ for $\\mathrm{NH}_{4}{ }^{+}=5.6 \\times 10^{-10}$ ).\n"} {"id": "textbook_4606", "title": "1495. Acid-Base Equilibrium Calculations - ", "content": "\nThe chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.\n\n(8) EXAMPLE 14.9"} {"id": "textbook_4607", "title": "1495. Acid-Base Equilibrium Calculations - ", "content": "Determination of $\\boldsymbol{K}_{\\mathrm{a}}$ from Equilibrium Concentrations\nAcetic acid is the principal ingredient in vinegar (Figure 14.9) that provides its sour taste. At equilibrium, a solution contains $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=0.0787 \\mathrm{M}$ and $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=0.00118 \\mathrm{M}$. What is the value of $K_{\\mathrm{a}}$ for acetic acid?\n"} {"id": "textbook_4608", "title": "1495. Acid-Base Equilibrium Calculations - ", "content": "FIGURE 14.9 Vinegar contains acetic acid, a weak acid. (credit: modification of work by \"HomeSpot HQ\"/Flickr)\n"} {"id": "textbook_4609", "title": "1496. Solution - ", "content": "\nThe relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_{\\mathrm{a}}$ for acetic acid."} {"id": "textbook_4610", "title": "1496. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q) \\\\\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]}=\\frac{(0.00118)(0.00118)}{0.0787}=1.77 \\times 10^{-5}\n\\end{gathered}\n$$\n"} {"id": "textbook_4611", "title": "1497. Check Your Learning - ", "content": "\nThe $\\mathrm{HSO}_{4}{ }^{-}$ion, weak acid used in some household cleansers:"} {"id": "textbook_4612", "title": "1497. Check Your Learning - ", "content": "$$\n\\mathrm{HSO}_{4}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)\n$$"} {"id": "textbook_4613", "title": "1497. Check Your Learning - ", "content": "What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.027 \\mathrm{M} ;\\left[\\mathrm{HSO}_{4}{ }^{-}\\right]=0.29 \\mathrm{M}$; and $\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=0.13 \\mathrm{M}$ ?\n"} {"id": "textbook_4614", "title": "1498. Answer: - ", "content": "\n$K_{\\mathrm{a}}$ for $\\mathrm{HSO}_{4}^{-}=1.2 \\times 10^{-2}$\n"} {"id": "textbook_4615", "title": "1499. EXAMPLE 14.10 - ", "content": ""} {"id": "textbook_4616", "title": "1500. Determination of $\\boldsymbol{K}_{\\mathbf{b}}$ from Equilibrium Concentrations - ", "content": "\nCaffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$ is a weak base. What is the value of $K_{\\mathrm{b}}$ for caffeine if a solution at equilibrium has $\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}\\right]=0.050 \\mathrm{M},\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}\\right]=5.0 \\times 10^{-3} \\mathrm{M}$, and $\\left[\\mathrm{OH}^{-}\\right]=2.5 \\times 10^{-3} \\mathrm{M}$ ?\n"} {"id": "textbook_4617", "title": "1501. Solution - ", "content": "\nThe relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_{\\mathrm{b}}$ for caffeine."} {"id": "textbook_4618", "title": "1501. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\\\\nK_{\\mathrm{b}}=\\frac{\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}\\right]}=\\frac{\\left(5.0 \\times 10^{-3}\\right)\\left(2.5 \\times 10^{-3}\\right)}{0.050}=2.5 \\times 10^{-4}\n\\end{gathered}\n$$\n"} {"id": "textbook_4619", "title": "1502. Check Your Learning - ", "content": "\nWhat is the equilibrium constant for the ionization of the $\\mathrm{HPO}_{4}{ }^{2-}$ ion, a weak base"} {"id": "textbook_4620", "title": "1502. Check Your Learning - ", "content": "$$\n\\mathrm{HPO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_4621", "title": "1502. Check Your Learning - ", "content": "if the composition of an equilibrium mixture is as follows: $\\left[\\mathrm{OH}^{-}\\right]=1.3 \\times 10^{-6} \\mathrm{M}$; $\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}\\right]=0.042 \\mathrm{M}$; and $\\left[\\mathrm{HPO}_{4}{ }^{2-}\\right]=0.341 \\mathrm{M}$ ?"} {"id": "textbook_4622", "title": "1502. Check Your Learning - ", "content": "Answer:\n$K_{\\mathrm{b}}$ for $\\mathrm{HPO}_{4}{ }^{2-}=1.6 \\times 10^{-7}$\n"} {"id": "textbook_4623", "title": "1503. EXAMPLE 14.11 - ", "content": ""} {"id": "textbook_4624", "title": "1504. Determination of $\\boldsymbol{K}_{\\mathbf{a}}$ or $\\boldsymbol{K}_{\\mathbf{b}}$ from $\\mathbf{p H}$ - ", "content": "\nThe pH of a $0.0516-M$ solution of nitrous acid, $\\mathrm{HNO}_{2}$, is 2.34 . What is its $K_{\\mathrm{a}}$ ?"} {"id": "textbook_4625", "title": "1504. Determination of $\\boldsymbol{K}_{\\mathbf{a}}$ or $\\boldsymbol{K}_{\\mathbf{b}}$ from $\\mathbf{p H}$ - ", "content": "$$\n\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}^{-}(a q)\n$$\n"} {"id": "textbook_4626", "title": "1505. Solution - ", "content": "\nThe nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as \"initial\" values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$is present $\\left(1 \\times 10^{-7} \\mathrm{M}\\right)$ due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected."} {"id": "textbook_4627", "title": "1505. Solution - ", "content": "The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an \"equilibrium\" value for the ICE table:"} {"id": "textbook_4628", "title": "1505. Solution - ", "content": "$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-2.34}=0.0046 \\mathrm{M}\n$$"} {"id": "textbook_4629", "title": "1505. Solution - ", "content": "The ICE table for this system is then"} {"id": "textbook_4630", "title": "1505. Solution - ", "content": "| | $\\mathrm{HNO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathbf{H}_{3} \\mathrm{O}^{+}+\\mathrm{NO}_{2}^{-}$ | | | |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.0516 | | $\\sim 0$ | 0 |\n| Change $(M)$ | -0.0046 | | +0.0046 | +0.0046 |\n| Equilibrium concentration $(M)$ | 0.0470 | | 0.0046 | 0.0046 |"} {"id": "textbook_4631", "title": "1505. Solution - ", "content": "Finally, calculate the value of the equilibrium constant using the data in the table:"} {"id": "textbook_4632", "title": "1505. Solution - ", "content": "$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{NO}_{2}^{-}\\right]}{\\left[\\mathrm{HNO}_{2}\\right]}=\\frac{(0.0046)(0.0046)}{(0.0470)}=4.6 \\times 10^{-4}\n$$\n"} {"id": "textbook_4633", "title": "1506. Check Your Learning. - ", "content": "\nThe pH of a solution of household ammonia, a $0.950-M$ solution of $\\mathrm{NH}_{3}$, is 11.612 . What is $K_{\\mathrm{b}}$ for $\\mathrm{NH}_{3}$.\n"} {"id": "textbook_4634", "title": "1507. Answer: - ", "content": "\n$K_{\\mathrm{b}}=1.8 \\times 10^{-5}$\n"} {"id": "textbook_4635", "title": "1508. Calculating Equilibrium Concentrations in a Weak Acid Solution - ", "content": "\nFormic acid, $\\mathrm{HCO}_{2} \\mathrm{H}$, is one irritant that causes the body's reaction to some ant bites and stings (Figure 14.10).\n"} {"id": "textbook_4636", "title": "1508. Calculating Equilibrium Concentrations in a Weak Acid Solution - ", "content": "FIGURE 14.10 The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)\nWhat is the concentration of hydronium ion and the pH of a $0.534-M$ solution of formic acid?"} {"id": "textbook_4637", "title": "1508. Calculating Equilibrium Concentrations in a Weak Acid Solution - ", "content": "$$\n\\mathrm{HCO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-4}\n$$\n"} {"id": "textbook_4638", "title": "1509. Solution - ", "content": "\nThe ICE table for this system is\n"} {"id": "textbook_4639", "title": "1509. Solution - ", "content": "Substituting the equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression gives"} {"id": "textbook_4640", "title": "1509. Solution - ", "content": "$$\n\\begin{aligned}\nK_{\\mathrm{a}}= & 1.8 \\times 10^{-4}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{HCO}_{2}^{-}\\right]}{\\left[\\mathrm{HCO}_{2} \\mathrm{H}\\right]} \\\\\n& =\\frac{(x)(x)}{0.534-x}=1.8 \\times 10^{-4}\n\\end{aligned}\n$$"} {"id": "textbook_4641", "title": "1509. Solution - ", "content": "The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that $x$ will be much lesser than 0.534 , and so the equation becomes"} {"id": "textbook_4642", "title": "1509. Solution - ", "content": "$$\nK_{\\mathrm{a}}=1.8 \\times 10^{-4}=\\frac{x^{2}}{0.534}\n$$"} {"id": "textbook_4643", "title": "1509. Solution - ", "content": "Solving the equation for $x$ yields"} {"id": "textbook_4644", "title": "1509. Solution - ", "content": "$$\n\\begin{gathered}\nx^{2}=0.534 \\times\\left(1.8 \\times 10^{-4}\\right)=9.6 \\times 10^{-5} \\\\\nx=\\sqrt{9.6 \\times 10^{-5}}\n\\end{gathered}\n$$"} {"id": "textbook_4645", "title": "1509. Solution - ", "content": "$$\n=9.8 \\times 10^{-3} M\n$$"} {"id": "textbook_4646", "title": "1509. Solution - ", "content": "To check the assumption that $x$ is small compared to 0.534 , its relative magnitude can be estimated:"} {"id": "textbook_4647", "title": "1509. Solution - ", "content": "$$\n\\frac{x}{0.534}=\\frac{9.8 \\times 10^{-3}}{0.534}=1.8 \\times 10^{-2}(1.8 \\% \\text { of } 0.534)\n$$"} {"id": "textbook_4648", "title": "1509. Solution - ", "content": "Because $x$ is less than $5 \\%$ of the initial concentration, the assumption is valid.\nAs defined in the ICE table, $x$ is equal to the equilibrium concentration of hydronium ion:"} {"id": "textbook_4649", "title": "1509. Solution - ", "content": "$$\nx=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.0098 \\mathrm{M}\n$$"} {"id": "textbook_4650", "title": "1509. Solution - ", "content": "Finally, the pH is calculated to be"} {"id": "textbook_4651", "title": "1509. Solution - ", "content": "$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log (0.0098)=2.01\n$$\n"} {"id": "textbook_4652", "title": "1510. Check Your Learning - ", "content": "\nOnly a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a $0.100-M$ solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ ?"} {"id": "textbook_4653", "title": "1510. Check Your Learning - ", "content": "$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-5}\n$$\n"} {"id": "textbook_4654", "title": "1511. Answer: - ", "content": "\npercent ionization $=1.3 \\%$\n"} {"id": "textbook_4655", "title": "1512. EXAMPLE 14.13 - ", "content": ""} {"id": "textbook_4656", "title": "1513. Calculating Equilibrium Concentrations in a Weak Base Solution - ", "content": "\nFind the concentration of hydroxide ion, the pOH , and the pH of a $0.25-M$ solution of trimethylamine, a weak base:"} {"id": "textbook_4657", "title": "1513. Calculating Equilibrium Concentrations in a Weak Base Solution - ", "content": "$$\n\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=6.3 \\times 10^{-5}\n$$\n"} {"id": "textbook_4658", "title": "1514. Solution - ", "content": "\nThe ICE table for this system is"} {"id": "textbook_4659", "title": "1514. Solution - ", "content": "| | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathbf{N}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}+\\mathrm{OH}^{-}$ | | | |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.25 | | 0 | $\\sim 0$ |\n| Change $(M)$ | $-x$ | | $x$ | $x$ |\n| Equilibrium concentration $(M)$ | $0.25+(-x)$ | | $0+x$ | $\\sim 0+x$ |"} {"id": "textbook_4660", "title": "1514. Solution - ", "content": "Substituting the equilibrium concentration terms into the $K_{\\mathrm{b}}$ expression gives"} {"id": "textbook_4661", "title": "1514. Solution - ", "content": "$$\nK_{\\mathrm{b}}=\\frac{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]}=\\frac{(x)(x)}{0.25-x}=6.3 \\times 10^{-5}\n$$"} {"id": "textbook_4662", "title": "1514. Solution - ", "content": "Assuming $x \\ll 0.25$ and solving for $x$ yields"} {"id": "textbook_4663", "title": "1514. Solution - ", "content": "$$\nx=4.0 \\times 10^{-3} M\n$$"} {"id": "textbook_4664", "title": "1514. Solution - ", "content": "This value is less than $5 \\%$ of the initial concentration ( 0.25 ), so the assumption is justified."} {"id": "textbook_4665", "title": "1514. Solution - ", "content": "As defined in the ICE table, $x$ is equal to the equilibrium concentration of hydroxide ion:"} {"id": "textbook_4666", "title": "1514. Solution - ", "content": "$$\n\\begin{aligned}\n{\\left[\\mathrm{OH}^{-}\\right]=} & \\sim 0+x=x=4.0 \\times 10^{-3} \\mathrm{M} \\\\\n& =4.0 \\times 10^{-3} \\mathrm{M}\n\\end{aligned}\n$$"} {"id": "textbook_4667", "title": "1514. Solution - ", "content": "The pOH is calculated to be"} {"id": "textbook_4668", "title": "1514. Solution - ", "content": "$$\n\\mathrm{pOH}=-\\log \\left(4.0 \\times 10^{-3}\\right)=2.40\n$$"} {"id": "textbook_4669", "title": "1514. Solution - ", "content": "Using the relation introduced in the previous section of this chapter:"} {"id": "textbook_4670", "title": "1514. Solution - ", "content": "$$\n\\mathrm{pH}+\\mathrm{pOH}=\\mathrm{p} K_{\\mathrm{w}}=14.00\n$$"} {"id": "textbook_4671", "title": "1514. Solution - ", "content": "permits the computation of pH :"} {"id": "textbook_4672", "title": "1514. Solution - ", "content": "$$\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00-2.40=11.60\n$$\n"} {"id": "textbook_4673", "title": "1515. Check Your Learning - ", "content": "\nCalculate the hydroxide ion concentration and the percent ionization of a $0.0325-M$ solution of ammonia, a weak base with a $K_{\\mathrm{b}}$ of $1.76 \\times 10^{-5}$.\n"} {"id": "textbook_4674", "title": "1516. Answer: - ", "content": "\n$7.56 \\times 10^{-4} \\mathrm{M}, 2.33 \\%$"} {"id": "textbook_4675", "title": "1516. Answer: - ", "content": "In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that $x$ is negligible cannot be made. Calculations of this sort are demonstrated in Example 14.14 below.\n"} {"id": "textbook_4676", "title": "1517. EXAMPLE 14.14 - ", "content": ""} {"id": "textbook_4677", "title": "1518. Calculating Equilibrium Concentrations without Simplifying Assumptions - ", "content": "\nSodium bisulfate, $\\mathrm{NaHSO}_{4}$, is used in some household cleansers as a source of the $\\mathrm{HSO}_{4}{ }^{-}$ion, a weak acid. What is the pH of a $0.50-M$ solution of $\\mathrm{HSO}_{4}{ }^{-}$?"} {"id": "textbook_4678", "title": "1518. Calculating Equilibrium Concentrations without Simplifying Assumptions - ", "content": "$$\n\\mathrm{HSO}_{4}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\quad K_{\\mathrm{a}}=1.2 \\times 10^{-2}\n$$\n"} {"id": "textbook_4679", "title": "1519. Solution - ", "content": "\nThe ICE table for this system is"} {"id": "textbook_4680", "title": "1519. Solution - ", "content": "| | $\\mathrm{HSO}_{4}{ }^{-}+\\mathbf{H}_{\\mathbf{2}} \\mathrm{O} \\rightleftharpoons \\mathbf{H}_{3} \\mathbf{O}^{+}+\\mathbf{S O}_{4}{ }^{\\mathbf{2 -}}$ | | | |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.50 | | $\\sim 0$ | 0 |\n| Change $(M)$ | $-x$ | | $+x$ | $+x$ |\n| Equilibrium
concentration $(M)$ | $0.50-x$ | | $x$ | $x$ |"} {"id": "textbook_4681", "title": "1519. Solution - ", "content": "Substituting the equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression gives"} {"id": "textbook_4682", "title": "1519. Solution - ", "content": "$$\nK_{\\mathrm{a}}=1.2 \\times 10^{-2}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]}{\\left[\\mathrm{HSO}_{4}^{-}\\right]}=\\frac{(x)(x)}{0.50-x}\n$$"} {"id": "textbook_4683", "title": "1519. Solution - ", "content": "If the assumption that $x \\ll 0.5$ is made, simplifying and solving the above equation yields"} {"id": "textbook_4684", "title": "1519. Solution - ", "content": "$$\nx=0.077 M\n$$"} {"id": "textbook_4685", "title": "1519. Solution - ", "content": "This value of $x$ is clearly not significantly less than $0.50 M$; rather, it is approximately $15 \\%$ of the initial concentration:\nWhen we check the assumption, we calculate:"} {"id": "textbook_4686", "title": "1519. Solution - ", "content": "$$\n\\begin{gathered}\n\\frac{x}{\\left[\\mathrm{HSO}_{4}^{-}\\right]_{\\mathrm{i}}} \\\\\n\\frac{x}{0.50}=\\frac{7.7 \\times 10^{-2}}{0.50}=0.15(15 \\%)\n\\end{gathered}\n$$"} {"id": "textbook_4687", "title": "1519. Solution - ", "content": "Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:"} {"id": "textbook_4688", "title": "1519. Solution - ", "content": "$$\nK_{\\mathrm{a}}=1.2 \\times 10^{-2}=\\frac{(x)(x)}{0.50-x}\n$$"} {"id": "textbook_4689", "title": "1519. Solution - ", "content": "Rearranging this equation yields"} {"id": "textbook_4690", "title": "1519. Solution - ", "content": "$$\n6.0 \\times 10^{-3}-1.2 \\times 10^{-2} x=x^{2}\n$$"} {"id": "textbook_4691", "title": "1519. Solution - ", "content": "Writing the equation in quadratic form gives"} {"id": "textbook_4692", "title": "1519. Solution - ", "content": "$$\nx^{2}+1.2 \\times 10^{-2} x-6.0 \\times 10^{-3}=0\n$$"} {"id": "textbook_4693", "title": "1519. Solution - ", "content": "Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to $x$. As defined in the ICE table, $x$ is equal to the hydronium concentration."} {"id": "textbook_4694", "title": "1519. Solution - ", "content": "$$\n\\begin{aligned}\n& x=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.072 M \\\\\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log (0.072)=1.14\n\\end{aligned}\n$$\n"} {"id": "textbook_4695", "title": "1520. Check Your Learning - ", "content": "\nCalculate the pH in a $0.010-M$ solution of caffeine, a weak base:"} {"id": "textbook_4696", "title": "1520. Check Your Learning - ", "content": "$$\n\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=2.5 \\times 10^{-4}\n$$\n"} {"id": "textbook_4697", "title": "1521. Answer: - ", "content": "\npH 11.16\n"} {"id": "textbook_4698", "title": "1522. Effect of Molecular Structure on Acid-Base Strength - ", "content": ""} {"id": "textbook_4699", "title": "1523. Binary Acids and Bases - ", "content": "\nIn the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the $\\mathrm{H}-\\mathrm{A}$ bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is $\\mathrm{HF}<\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI}$. Likewise, for group 16, the order of increasing acid strength is $\\mathrm{H}_{2} \\mathrm{O}<$ $\\mathrm{H}_{2} \\mathrm{~S}<\\mathrm{H}_{2} \\mathrm{Se}<\\mathrm{H}_{2} \\mathrm{Te}$.\n\nAcross a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the $\\mathrm{H}-\\mathrm{A}$ bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\\mathrm{CH}_{4}<\\mathrm{NH}_{3}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{HF}$; across the third row, it is $\\mathrm{SiH}_{4}<\\mathrm{PH}_{3}<\\mathrm{H}_{2} \\mathrm{~S}<\\mathrm{HCl}$ (see Figure 14.11).\n"} {"id": "textbook_4700", "title": "1523. Binary Acids and Bases - ", "content": "FIGURE 14.11 The figure shows trends in the strengths of binary acids and bases.\n"} {"id": "textbook_4701", "title": "1524. Ternary Acids and Bases - ", "content": "\nTernary compounds composed of hydrogen, oxygen, and some third element (\"E\") may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula $\\mathrm{O}_{\\mathrm{m}} \\mathrm{E}(\\mathrm{OH})_{\\mathrm{n}}$. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, $\\mathrm{O}_{2} \\mathrm{~S}(\\mathrm{OH})_{2}$, sulfurous acid, $\\mathrm{OS}(\\mathrm{OH})_{2}$, nitric acid, $\\mathrm{O}_{2} \\mathrm{NOH}$, perchloric acid, $\\mathrm{O}_{3} \\mathrm{ClOH}$, aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, and potassium hydroxide, KOH :\n
Added (mL) | Moles of NaOH
Added | pH Values 0.100 M
$\\mathrm{HCl}^{1}$ | pH Values 0.100 M
$\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}^{2}$ |\n| :--- | :--- | :--- | :--- |\n| 28.0 | 0.00280 | 11.75 | 11.75 |\n| 30.0 | 0.00300 | 11.96 | 11.96 |\n| 35.0 | 0.00350 | 12.22 | 12.22 |\n| 40.0 | 0.00400 | 12.36 | 12.36 |\n| 45.0 | 0.00450 | 12.46 | 12.46 |\n| 50.0 | 0.00500 | 12.52 | 12.52 |"} {"id": "textbook_4932", "title": "1583. Answer: - ", "content": "TABLE 14.2\n"} {"id": "textbook_4933", "title": "1583. Answer: - ", "content": "FIGURE 14.18 (a) The titration curve for the titration of 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 7.00 pH . (b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH .\n"} {"id": "textbook_4934", "title": "1584. Acid-Base Indicators - ", "content": "\nCertain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than $5.0 \\times 10^{-9} M(\\mathrm{pH}<8.3)$. In more basic solutions where the hydronium ion concentration is less than $5.0 \\times 10^{-9} M(\\mathrm{pH}>8.3)$, it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acidbase indicators are either weak organic acids or weak organic bases."} {"id": "textbook_4935", "title": "1584. Acid-Base Indicators - ", "content": "The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:"} {"id": "textbook_4936", "title": "1584. Acid-Base Indicators - ", "content": "$$\n\\begin{aligned}\n& \\underset{\\text { red }}{\\mathrm{HIn}(a q)}+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{In}^{-}(a q) \\\\\n& \\text { yellow } \\\\\n& \\quad K_{a}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{In}^{-}\\right]}{[\\mathrm{HIn}]}=4.0 \\times 10^{-4}\n\\end{aligned}\n$$"} {"id": "textbook_4937", "title": "1584. Acid-Base Indicators - ", "content": "The anion of methyl orange, $\\mathrm{In}^{-}$, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Ch\u00e2telier's principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers."} {"id": "textbook_4938", "title": "1584. Acid-Base Indicators - ", "content": "The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species $\\mathrm{In}^{-}$and HIn. If most of the indicator (typically about $60-90 \\%$ or more) is present as $\\mathrm{In}^{-}$, the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The HendersonHasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):"} {"id": "textbook_4939", "title": "1584. Acid-Base Indicators - ", "content": "$$\n\\mathrm{pH}=\\mathrm{p} K \\mathrm{a}+\\log \\left(\\frac{\\left[\\mathrm{In}^{-}\\right]}{[\\mathrm{HIn}]}\\right)\n$$"} {"id": "textbook_4940", "title": "1584. Acid-Base Indicators - ", "content": "In solutions where $\\mathrm{pH}>\\mathrm{p} K_{\\mathrm{a}}$, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When $\\mathrm{pH}<\\mathrm{p} K_{\\mathrm{a}}$, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa , appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately $\\mathrm{p} K_{\\mathrm{a}} \\pm 1$."} {"id": "textbook_4941", "title": "1584. Acid-Base Indicators - ", "content": "There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure 14.19 presents several indicators, their colors, and their color-change intervals.\n"} {"id": "textbook_4942", "title": "1584. Acid-Base Indicators - ", "content": "FIGURE 14.19 This chart illustrates the color change intervals for several acid-base indicators.\n"} {"id": "textbook_4943", "title": "1584. Acid-Base Indicators - ", "content": "FIGURE 14.20 Titration curves for strong and weak acids illustrating the proper choice of acid-base indicator. Any of the three indicators will exhibit a reasonably sharp color change at the equivalence point of the strong acid titration, but only phenolphthalein is suitable for use in the weak acid titration.\n\nThe titration curves shown in Figure 14.20 illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of $\\sim 24 \\mathrm{~mL}$ of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.\n\nThe weak acid titration curve in Figure 14.20 shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-toyellow color change over a relatively large volume interval ( $0-6 \\mathrm{~mL}$ ), completing the color change well before the equivalence point ( 25 mL ) has been reached. Use of litmus would show a color change that begins after adding $7-8 \\mathrm{~mL}$ of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.\n"} {"id": "textbook_4944", "title": "1585. Key Terms - ", "content": "\nacid ionization reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid\nacid ionization constant ( $\\boldsymbol{K}_{\\mathrm{a}}$ ) equilibrium constant for an acid ionization reaction\nacid-base indicator weak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH\nacidic a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]$\namphiprotic species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction\namphoteric species that can act as either an acid or a base\nautoionization reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions\nbase ionization reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base\nbase ionization constant ( $\\boldsymbol{K}_{\\mathbf{b}}$ ) equilibrium constant for a base ionization reaction\nbasic a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]$\nBr\u00f8nsted-Lowry acid proton donor\nBr\u00f8nsted-Lowry base proton acceptor\nbuffer mixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added\nbuffer capacity amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)\ncolor-change interval range in pH over which the color change of an indicator is observed\nconjugate acid substance formed when a base gains a proton\nconjugate base substance formed when an acid\n"} {"id": "textbook_4945", "title": "1586. Key Equations - ", "content": "\n$K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14}\\left(\\right.$ at $\\left.25^{\\circ} \\mathrm{C}\\right)$\n$\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$\n$\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-\\mathrm{pH}}$\n$\\left[\\mathrm{OH}^{-}\\right]=10^{-\\mathrm{pOH}}$\n$\\mathrm{pH}+\\mathrm{pOH}=\\mathrm{p} K_{\\mathrm{w}}=14.00$ at $25^{\\circ} \\mathrm{C}$\n$K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}$\n$K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{HB}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{B}]}$\nloses a proton\ndiprotic acid acid containing two ionizable hydrogen atoms per molecule\ndiprotic base base capable of accepting two protons\nHenderson-Hasselbalch equation logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions\nion-product constant for water ( $\\boldsymbol{K}_{\\mathrm{w}}$ ) equilibrium constant for the autoionization of water\nleveling effect observation that acid-base strength of solutes in a given solvent is limited to that of the solvent's characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)\nmonoprotic acid acid containing one ionizable hydrogen atom per molecule\nneutral describes a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=$ [ $\\mathrm{OH}^{-}$]\noxyacid ternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom\npercent ionization ratio of the concentration of ionized acid to initial acid concentration expressed as a percentage\npH logarithmic measure of the concentration of hydronium ions in a solution\n$\\mathbf{p O H}$ logarithmic measure of the concentration of hydroxide ions in a solution\nstepwise ionization process in which a polyprotic acid is ionized by losing protons sequentially\ntitration curve plot of some sample property (such as pH ) versus volume of added titrant\ntriprotic acid acid that contains three ionizable hydrogen atoms per molecule\n$K_{\\mathrm{a}} \\times K_{\\mathrm{b}}=1.0 \\times 10^{-14}=K_{\\mathrm{w}}$\nPercent ionization $=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{[\\mathrm{HA}]_{0}} \\times 100$\n$\\mathrm{p} K_{\\mathrm{a}}=-\\log K_{\\mathrm{a}}$\n$\\mathrm{p} K_{\\mathrm{b}}=-\\log K_{\\mathrm{b}}$\n$\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}$\n"} {"id": "textbook_4946", "title": "1587. Summary - ", "content": ""} {"id": "textbook_4947", "title": "1587. Summary - 1587.1. Br\u00f8nsted-Lowry Acids and Bases", "content": "\nA compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nstedLowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, $\\mathrm{H}_{3} \\mathrm{O}^{+}$, and the hydroxide ion, $\\mathrm{OH}^{-}$when it undergoes autoionization:"} {"id": "textbook_4948", "title": "1587. Summary - 1587.1. Br\u00f8nsted-Lowry Acids and Bases", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"} {"id": "textbook_4949", "title": "1587. Summary - 1587.1. Br\u00f8nsted-Lowry Acids and Bases", "content": "The ion product of water, $K_{\\mathrm{w}}$ is the equilibrium constant for the autoionization reaction:"} {"id": "textbook_4950", "title": "1587. Summary - 1587.1. Br\u00f8nsted-Lowry Acids and Bases", "content": "$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14} \\text { at } 25^{\\circ} \\mathrm{C}\n$$\n"} {"id": "textbook_4951", "title": "1588. 2 pH and pOH - ", "content": "\nConcentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At $25^{\\circ} \\mathrm{C}$, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]\\right)$, basic $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]\\right)$, or neutral $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]\\right)$. At $25^{\\circ} \\mathrm{C}$, a pH $<7$ indicates an acidic solution, a $\\mathrm{pH}>7$ a basic solution, and a $\\mathrm{pH}=7$ a neutral solution.\n"} {"id": "textbook_4952", "title": "1588. 2 pH and pOH - 1588.1. Relative Strengths of Acids and Bases", "content": "\nThe relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists\nbetween the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table $\\left(\\mathrm{CH}_{4}<\\mathrm{NH}_{3}\\right.$ $<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{HF}$ ), and they increase down a group ( $\\mathrm{HF}<$ $\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI})$. The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases $\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}<\\right.$ $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ ). The strengths of oxyacids also increase as the electronegativity of the central element increases $\\left[\\mathrm{H}_{2} \\mathrm{SeO}_{4}<\\mathrm{H}_{2} \\mathrm{SO}_{4}\\right]$.\n"} {"id": "textbook_4953", "title": "1588. 2 pH and pOH - 1588.2. Hydrolysis of Salts", "content": "\nThe ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.\n"} {"id": "textbook_4954", "title": "1588. 2 pH and pOH - 1588.3. Polyprotic Acids", "content": "\nAn acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.\n"} {"id": "textbook_4955", "title": "1588. 2 pH and pOH - 1588.4. Buffers", "content": "\nSolutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one\nconjugate partner and preventing further buffering action.\n"} {"id": "textbook_4956", "title": "1588. 2 pH and pOH - 1588.5. Acid-Base Titrations", "content": "\nThe titration curve for an acid-base titration is\ntypically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points.\n"} {"id": "textbook_4957", "title": "1589. Exercises - ", "content": ""} {"id": "textbook_4958", "title": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases", "content": "\n1. Write equations that show $\\mathrm{NH}_{3}$ as both a conjugate acid and a conjugate base.\n2. Write equations that show $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$acting both as an acid and as a base.\n3. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:\n(a) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) HCl\n(c) $\\mathrm{NH}_{3}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(e) $\\mathrm{NH}_{4}{ }^{+}$\n(f) $\\mathrm{HSO}_{4}{ }^{-}$\n4. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:\n(a) $\\mathrm{HNO}_{3}$\n(b) $\\mathrm{PH}_{4}^{+}$\n(c) $\\mathrm{H}_{2} \\mathrm{~S}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}$\n(e) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(f) $\\mathrm{HS}^{-}$\n5. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:\n(a) $\\mathrm{H}_{2} \\mathrm{O}$\n(b) $\\mathrm{OH}^{-}$\n(c) $\\mathrm{NH}_{3}$\n(d) $\\mathrm{CN}^{-}$\n(e) $\\mathrm{S}^{2-}$\n(f) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n6. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:\n(a) $\\mathrm{HS}^{-}$\n(b) $\\mathrm{PO}_{4}{ }^{3-}$\n(c) $\\mathrm{NH}_{2}{ }^{-}$\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(e) $\\mathrm{O}^{2-}$\n(f) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n7. What is the conjugate acid of each of the following? What is the conjugate base of each?\n(a) $\\mathrm{OH}^{-}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{HCO}_{3}{ }^{-}$\n(d) $\\mathrm{NH}_{3}$\n(e) $\\mathrm{HSO}_{4}^{-}$\n(f) $\\mathrm{H}_{2} \\mathrm{O}_{2}$\n(g) $\\mathrm{HS}^{-}$\n(h) $\\mathrm{H}_{5} \\mathrm{~N}_{2}{ }^{+}$\n8. What is the conjugate acid of each of the following? What is the conjugate base of each?\n(a) $\\mathrm{H}_{2} \\mathrm{~S}$\n(b) $\\mathrm{H}_{2} \\mathrm{PO}_{4}-$\n(c) $\\mathrm{PH}_{3}$\n(d) $\\mathrm{HS}^{-}$\n(e) $\\mathrm{HSO}_{3}{ }^{-}$\n(f) $\\mathrm{H}_{3} \\mathrm{O}_{2}{ }^{+}$\n(g) $\\mathrm{H}_{4} \\mathrm{~N}_{2}$\n(h) $\\mathrm{CH}_{3} \\mathrm{OH}$\n9. Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:\n(a) $\\mathrm{HNO}_{3}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{NO}_{3}{ }^{-}$\n(b) $\\mathrm{CN}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HCN}+\\mathrm{OH}^{-}$\n(c) $\\mathrm{H}_{2} \\mathrm{SO}_{4}+\\mathrm{Cl}^{-} \\longrightarrow \\mathrm{HCl}+\\mathrm{HSO}_{4}^{-}$\n(d) $\\mathrm{HSO}_{4}{ }^{-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{O}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 2 \\mathrm{OH}^{-}$\n(f) $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})\\right]^{+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+} \\longrightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}$\n(g) $\\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{NH}_{2}{ }^{-} \\longrightarrow \\mathrm{HS}^{-}+\\mathrm{NH}_{3}$\n10. Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:\n(a) $\\mathrm{NO}_{2}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HNO}_{2}+\\mathrm{OH}^{-}$\n(b) $\\mathrm{HBr}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{Br}^{-}$\n(c) $\\mathrm{HS}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{OH}^{-}$\n(d) $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{HPO}_{4}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}+\\mathrm{HCl} \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}+\\mathrm{Cl}^{-}$\n(f) $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+} \\longrightarrow\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}$\n(g) $\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{H}^{-} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{O}^{-}+\\mathrm{H}_{2}$\n11. What are amphiprotic species? Illustrate with suitable equations.\n12. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:\n(a) $\\mathrm{H}_{2} \\mathrm{O}$\n(b) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(c) $\\mathrm{S}^{2-}$\n(d) $\\mathrm{CO}_{3}{ }^{2-}$\n(e) $\\mathrm{HSO}_{4}{ }^{-}$\n13. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.\n(a) $\\mathrm{NH}_{3}$\n(b) $\\mathrm{HPO}_{4}^{-}$\n(c) $\\mathrm{Br}^{-}$\n(d) $\\mathrm{NH}_{4}{ }^{+}$\n(e) $\\mathrm{ASO}_{4}{ }^{3-}$\n14. Is the self-ionization of water endothermic or exothermic? The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $2.9 \\times$ $10^{-14}$ at $40^{\\circ} \\mathrm{C}$ and $9.3 \\times 10^{-14}$ at $60^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_4959", "title": "1590. 2 pH and pOH - ", "content": "\n15. Explain why a sample of pure water at $40^{\\circ} \\mathrm{C}$ is neutral even though $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.7 \\times 10^{-7} \\mathrm{M}$. $K_{\\mathrm{w}}$ is $2.9 \\times 10^{-14}$ at $40^{\\circ} \\mathrm{C}$.\n16. The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $2.9 \\times 10^{-14}$ at $40^{\\circ} \\mathrm{C}$. Calculate $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for pure water at $40^{\\circ} \\mathrm{C}$.\n17. The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $9.311 \\times 10^{-14}$ at $60^{\\circ} \\mathrm{C}$. Calculate $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for pure water at $60^{\\circ} \\mathrm{C}$.\n18. Calculate the pH and the pOH of each of the following solutions at $25^{\\circ} \\mathrm{C}$ for which the substances ionize completely:\n(a) 0.200 M HCl\n(b) 0.0143 M NaOH\n(c) $3.0 \\mathrm{M} \\mathrm{HNO}_{3}$\n(d) $0.0031 \\mathrm{M} \\mathrm{Ca}(\\mathrm{OH})_{2}$\n19. Calculate the pH and the pOH of each of the following solutions at $25^{\\circ} \\mathrm{C}$ for which the substances ionize completely:\n(a) $0.000259 \\mathrm{M} \\mathrm{HClO}_{4}$\n(b) 0.21 M NaOH\n(c) $0.000071 \\mathrm{MBa}(\\mathrm{OH})_{2}$\n(d) 2.5 M KOH\n20. What are the pH and pOH of a solution of 2.0 M HCl , which ionizes completely?\n21. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52 ?\n22. Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH . See Figure 14.2 for useful information.\n23. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH . See Figure 14.2 for useful information.\n24. The hydronium ion concentration in a sample of rainwater is found to be $1.7 \\times 10^{-6} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. What is the concentration of hydroxide ions in the rainwater?\n25. The hydroxide ion concentration in household ammonia is $3.2 \\times 10^{-3} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. What is the concentration of hydronium ions in the solution?\n"} {"id": "textbook_4960", "title": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases", "content": "\n26. Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.\n27. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.\n28. Use this list of important industrial compounds (and Figure 14.8) to answer the following questions regarding: $\\mathrm{Ca}(\\mathrm{OH})_{2}, \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}, \\mathrm{HCl}, \\mathrm{H}_{2} \\mathrm{CO}_{3}$, $\\mathrm{HF}, \\mathrm{HNO}_{2}, \\mathrm{HNO}_{3}, \\mathrm{H}_{3} \\mathrm{PO}_{4}, \\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{NH}_{3}, \\mathrm{NaOH}, \\mathrm{Na}_{2} \\mathrm{CO}_{3}$. (a) Identify the strong Br\u00f8nsted-Lowry acids and strong Br\u00f8nsted-Lowry bases.\n(b) Identify the compounds that can behave as Br\u00f8nsted-Lowry acids with strengths lying between those of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{H}_{2} \\mathrm{O}$.\n(c) Identify the compounds that can behave as Br\u00f8nsted-Lowry bases with strengths lying between those of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{OH}^{-}$.\n29. The odor of vinegar is due to the presence of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.\n30. Household ammonia is a solution of the weak base $\\mathrm{NH}_{3}$ in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.\n31. Explain why the ionization constant, $K_{\\mathrm{a}}$, for $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is larger than the ionization constant for $\\mathrm{H}_{2} \\mathrm{SO}_{3}$.\n32. Explain why the ionization constant, $K_{\\mathrm{a}}$, for HI is larger than the ionization constant for HF.\n33. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.\n34. Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of $\\mathrm{HNO}_{3}$ with CuO .\n35. What is the ionization constant at $25^{\\circ} \\mathrm{C}$ for the weak acid $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$, the conjugate acid of the weak base $\\mathrm{CH}_{3} \\mathrm{NH}_{2}, K_{\\mathrm{b}}=4.4 \\times 10^{-4}$.\n36. What is the ionization constant at $25^{\\circ} \\mathrm{C}$ for the weak acid $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$, the conjugate acid of the weak base $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}, K_{\\mathrm{b}}=5.9 \\times 10^{-4}$ ?\n37. Which base, $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ or $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$, is the stronger base? Which conjugate acid, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$or $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$, is the stronger acid?\n38. Which is the stronger acid, $\\mathrm{NH}_{4}{ }^{+}$or HBrO ?\n39. Which is the stronger base, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}^{2}$ or $\\mathrm{H}_{2} \\mathrm{BO}_{3}-$ ?\n40. Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.\n(a) $\\mathrm{H}_{2} \\mathrm{O}$ or HF\n(b) $\\mathrm{B}(\\mathrm{OH})_{3}$ or $\\mathrm{Al}(\\mathrm{OH})_{3}$\n(c) $\\mathrm{HSO}_{3}{ }^{-}$or $\\mathrm{HSO}_{4}^{-}$\n(d) $\\mathrm{NH}_{3}$ or $\\mathrm{H}_{2} \\mathrm{~S}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}$ or $\\mathrm{H}_{2} \\mathrm{Te}$\n41. Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.\n(a) $\\mathrm{HSO}_{4}^{-}$or $\\mathrm{HSeO}_{4}^{-}$\n(b) $\\mathrm{NH}_{3}$ or $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{PH}_{3}$ or HI\n(d) $\\mathrm{NH}_{3}$ or $\\mathrm{PH}_{3}$\n(e) $\\mathrm{H}_{2} \\mathrm{~S}$ or HBr\n42. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n(a) acidity: $\\mathrm{HCl}, \\mathrm{HBr}, \\mathrm{HI}$\n(b) basicity: $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{OH}^{-}, \\mathrm{H}^{-}, \\mathrm{Cl}^{-}$\n(c) basicity: $\\mathrm{Mg}(\\mathrm{OH})_{2}, \\mathrm{Si}(\\mathrm{OH})_{4}, \\mathrm{ClO}_{3}(\\mathrm{OH})$ (Hint: Formula could also be written as $\\mathrm{HClO}_{4}$.)\n(d) acidity: $\\mathrm{HF}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NH}_{3}, \\mathrm{CH}_{4}$\n43. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n(a) acidity: $\\mathrm{NaHSO}_{3}, \\mathrm{NaHSeO}_{3}, \\mathrm{NaHSO}_{4}$\n(b) basicity: $\\mathrm{BrO}_{2}{ }^{-}, \\mathrm{ClO}_{2}{ }^{-}, \\mathrm{IO}_{2}{ }^{-}$\n(c) acidity: $\\mathrm{HOCl}, \\mathrm{HOBr}, \\mathrm{HOI}$\n(d) acidity: $\\mathrm{HOCl}, \\mathrm{HOClO}, \\mathrm{HOClO}_{2}, \\mathrm{HOClO}_{3}$\n(e) basicity: $\\mathrm{NH}_{2}{ }^{-}, \\mathrm{HS}^{-}, \\mathrm{HTe}^{-}, \\mathrm{PH}_{2}{ }^{-}$\n(f) basicity: $\\mathrm{BrO}^{-}, \\mathrm{BrO}_{2}^{-}, \\mathrm{BrO}_{3}^{-}, \\mathrm{BrO}_{4}^{-}$\n44. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, $\\mathrm{F}^{-}$or $\\mathrm{CN}^{-}$, is the stronger base?\n45. The active ingredient formed by aspirin in the body is salicylic acid, $\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)$. The carboxyl group $\\left(-\\mathrm{CO}_{2} \\mathrm{H}\\right)$ acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a $0.001-M$ aqueous solution of $\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)$.\n46. Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.\n47. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?\n48. Which of the following will increase the percent of $\\mathrm{NH}_{3}$ that is converted to the ammonium ion in water?\n(a) addition of NaOH\n(b) addition of HCl\n(c) addition of $\\mathrm{NH}_{4} \\mathrm{Cl}$\n49. Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?\n(a) addition of NaOH\n(b) addition of HCl\n(c) addition of NaF\n50. What is the effect on the concentrations of $\\mathrm{NO}_{2}{ }^{-}, \\mathrm{HNO}_{2}$, and $\\mathrm{OH}^{-}$when the following are added to a solution of $\\mathrm{KNO}_{2}$ in water:\n(a) HCl\n(b) $\\mathrm{HNO}_{2}$\n(c) NaOH\n(d) NaCl\n(e) KNO\n51. What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?\n(a) HCl\n(b) KF\n(c) NaCl\n(d) KOH\n(e) HF\n52. Why is the hydronium ion concentration in a solution that is 0.10 Min HCl and 0.10 Min HCOOH determined by the concentration of HCl ?\n53. From the equilibrium concentrations given, calculate $K_{\\mathrm{a}}$ for each of the weak acids and $K_{\\mathrm{b}}$ for each of the weak bases.\n(a) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}:\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.34 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=1.34 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=9.866 \\times 10^{-2} \\mathrm{M}$;\n(b) $\\mathrm{ClO}^{-}:\\left[\\mathrm{OH}^{-}\\right]=4.0 \\times 10^{-4} \\mathrm{M}$;\n[ HClO ] $=2.38 \\times 10^{-4} \\mathrm{M}$;\n$\\left[\\mathrm{ClO}^{-}\\right]=0.273 \\mathrm{M}$;\n(c) $\\mathrm{HCO}_{2} \\mathrm{H}:\\left[\\mathrm{HCO}_{2} \\mathrm{H}\\right]=0.524 \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=9.8 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{HCO}_{2}{ }^{-}\\right]=9.8 \\times 10^{-3} \\mathrm{M}$;\n(d) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}:\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}\\right]=0.233 \\mathrm{M}$;\n$\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\right]=2.3 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=2.3 \\times 10^{-3} \\mathrm{M}$\n54. From the equilibrium concentrations given, calculate $K_{\\mathrm{a}}$ for each of the weak acids and $K_{\\mathrm{b}}$ for each of the weak bases.\n(a) $\\mathrm{NH}_{3}:\\left[\\mathrm{OH}^{-}\\right]=3.1 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{4}{ }^{+}\\right]=3.1 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{3}\\right]=0.533 \\mathrm{M}$;\n(b) $\\mathrm{HNO}_{2}:\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.011 \\mathrm{M}$;\n$\\left[\\mathrm{NO}_{2}{ }^{-}\\right]=0.0438 \\mathrm{M}$;\n$\\left[\\mathrm{HNO}_{2}\\right]=1.07 \\mathrm{M}$;\n(c) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}:\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]=0.25 \\mathrm{M}$;\n$\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]=4.3 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{OH}^{-}\\right]=3.7 \\times 10^{-3} \\mathrm{M}$;\n(d) $\\mathrm{NH}_{4}{ }^{+}:\\left[\\mathrm{NH}_{4}{ }^{+}\\right]=0.100 \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{3}\\right]=7.5 \\times 10^{-6} \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=7.5 \\times 10^{-6} \\mathrm{M}$\n55. Determine $K_{\\mathrm{b}}$ for the nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$. In a $0.10-M$ solution this base is $0.0015 \\%$ ionized.\n56. Determine $K_{\\mathrm{a}}$ for hydrogen sulfate ion, $\\mathrm{HSO}_{4}{ }^{-}$. In a $0.10-M$ solution the acid is $29 \\%$ ionized.\n57. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n(a) $\\mathrm{F}^{-}$\n(b) $\\mathrm{NH}_{4}{ }^{+}$\n(c) $\\mathrm{AsO}_{4}{ }^{3-}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$\n(e) $\\mathrm{NO}_{2}{ }^{-}$\n(f) $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$(as a base)\n58. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n(a) $\\mathrm{HTe}^{-}$(as a base)\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}$\n(c) $\\mathrm{HAsO}_{4}{ }^{2-}$ (as a base)\n(d) $\\mathrm{HO}_{2}{ }^{-}$(as a base)\n(e) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}+$\n(f) $\\mathrm{HSO}_{3}{ }^{-}$(as a base)\n59. Using the $K_{\\mathrm{a}}$ value of $1.4 \\times 10^{-5}$, place $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}$ in the correct location in Figure 14.7.\n60. Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.\n(a) 0.0092 M HClO , a weak acid\n(b) $0.0784 \\mathrm{MC}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}$, a weak base\n(c) 0.0810 M HCN , a weak acid\n(d) $0.11 \\mathrm{M}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}$, a weak base\n(e) $0.120 \\mathrm{MFe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{2+}$ a weak acid, $K_{\\mathrm{a}}=1.6 \\times 10^{-7}$\n61. Propionic acid, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}\\left(K_{\\mathrm{a}}=1.34 \\times 10^{-5}\\right)$, is used in the manufacture of calcium propionate, a food preservative. What is the pH of a $0.698-M$ solution of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$ ?\n62. White vinegar is a $5.0 \\%$ by mass solution of acetic acid in water. If the density of white vinegar is $1.007 \\mathrm{~g} /$ $\\mathrm{cm}^{3}$, what is the pH ?\n63. The ionization constant of lactic acid, $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CO}_{2} \\mathrm{H}$, an acid found in the blood after strenuous exercise, is $1.36 \\times 10^{-4}$. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L , what is the concentration of hydronium ion in the solution?\n64. Nicotine, $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}$, is a base that will accept two protons ( $K_{\\mathrm{b} 1}=7 \\times 10^{-7}, K_{\\mathrm{b} 2}=1.4 \\times 10^{-11}$ ). What is the concentration of each species present in a $0.050-M$ solution of nicotine?\n65. The pH of a $0.23-M$ solution of HF is 1.92 . Determine $K_{\\mathrm{a}}$ for HF from these data.\n66. The pH of a $0.15-M$ solution of $\\mathrm{HSO}_{4}{ }^{-}$is 1.43. Determine $K_{\\mathrm{a}}$ for $\\mathrm{HSO}_{4}{ }^{-}$from these data.\n67. The pH of a $0.10-M$ solution of caffeine is 11.70 . Determine $K_{\\mathrm{b}}$ for caffeine from these data: $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q)$\n68. The pH of a solution of household ammonia, a 0.950 M solution of $\\mathrm{NH}_{3}$, is 11.612 . Determine $K_{\\mathrm{b}}$ for $\\mathrm{NH}_{3}$ from these data.\n"} {"id": "textbook_4961", "title": "1590. 2 pH and pOH - 1590.2. Hydrolysis of Salts", "content": "\n69. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) $\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}$\n(b) RbI\n(c) $\\mathrm{KHCO}_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{NH}_{3} \\mathrm{Br}$\n70. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) $\\mathrm{FeCl}_{3}$\n(b) $\\mathrm{K}_{2} \\mathrm{CO}_{3}$\n(c) $\\mathrm{NH}_{4} \\mathrm{Br}$\n(d) $\\mathrm{KClO}_{4}$\n71. Novocaine, $\\mathrm{C}_{13} \\mathrm{H}_{21} \\mathrm{O}_{2} \\mathrm{~N}_{2} \\mathrm{Cl}$, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is $7 \\times 10^{-6}$. Is a solution of novocaine acidic or basic? What are $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right]$, and pH of a $2.0 \\%$ solution by mass of novocaine, assuming that the density of the solution is $1.0 \\mathrm{~g} / \\mathrm{mL}$.\n"} {"id": "textbook_4962", "title": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids", "content": "\n72. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a $0.134-M$ solution of $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, a diprotic acid: $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right],\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right],\\left[\\mathrm{HCO}_{3}{ }^{-}\\right]$, $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$ ? No calculations are needed to answer this question.\n73. Calculate the concentration of each species present in a $0.050-M$ solution of $\\mathrm{H}_{2} \\mathrm{~S}$.\n74. Calculate the concentration of each species present in a $0.010-M$ solution of phthalic acid, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}$."} {"id": "textbook_4963", "title": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids", "content": "$$\n\\begin{array}{lr}\n\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2}\\right)^{-}(a q) & K_{\\mathrm{a}}=1.1 \\times 10^{-3} \\\\\n\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2}\\right)(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2}\\right)_{2}^{2-}(a q) & K_{\\mathrm{a}}=3.9 \\times 10^{-6}\n\\end{array}\n$$"} {"id": "textbook_4964", "title": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids", "content": "75. Salicylic acid, $\\mathrm{HOC}_{6} \\mathrm{H}_{4} \\mathrm{CO}_{2} \\mathrm{H}$, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.\n(a) Both functional groups of salicylic acid ionize in water, with $K_{\\mathrm{a}}=1.0 \\times 10^{-3}$ for the $-\\mathrm{CO}_{2} \\mathrm{H}$ group and $4.2 \\times 10^{-13}$ for the -OH group. What is the pH of a saturated solution of the acid (solubility $=1.8 \\mathrm{~g} / \\mathrm{L}$ ).\n(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{CO}_{2} \\mathrm{H}$. The $-\\mathrm{CO}_{2} \\mathrm{H}$ functional group is still present, but its acidity is reduced, $K_{\\mathrm{a}}=3.0 \\times 10^{-4}$. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).\n76. The ion $\\mathrm{HTe}^{-}$is an amphiprotic species; it can act as either an acid or a base.\n(a) What is $K_{\\mathrm{a}}$ for the acid reaction of $\\mathrm{HTe}^{-}$with $\\mathrm{H}_{2} \\mathrm{O}$ ?\n(b) What is $K_{\\mathrm{b}}$ for the reaction in which $\\mathrm{HTe}^{-}$functions as a base in water?\n(c) Demonstrate whether or not the second ionization of $\\mathrm{H}_{2} \\mathrm{Te}$ can be neglected in the calculation of [ $\\mathrm{HTe}^{-}$] in a 0.10 M solution of $\\mathrm{H}_{2} \\mathrm{Te}$.\n"} {"id": "textbook_4965", "title": "1590. 2 pH and pOH - 1590.4. Buffers", "content": "\n77. Explain why a buffer can be prepared from a mixture of $\\mathrm{NH}_{4} \\mathrm{Cl}$ and NaOH but not from $\\mathrm{NH}_{3}$ and NaOH .\n78. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and a salt of its conjugate base $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$.\n79. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base $\\mathrm{NH}_{3}$ and a salt of its conjugate acid $\\mathrm{NH}_{4} \\mathrm{Cl}$.\n80. What is $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$in a solution of $0.25 \\mathrm{MCH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ and $0.030 \\mathrm{M} \\mathrm{NaCH} \\mathrm{NO}_{2}$ ? $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-5}$\n81. What is $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$in a solution of $0.075 \\mathrm{M} \\mathrm{HNO}_{2}$ and $0.030 M \\mathrm{NaNO}_{2}$ ?\n$\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=4.5 \\times 10^{-5}$\n82. What is $\\left[\\mathrm{OH}^{-}\\right]$in a solution of $0.125 \\mathrm{MCH}_{3} \\mathrm{NH}_{2}$ and $0.130 \\mathrm{MCH}_{3} \\mathrm{NH}_{3} \\mathrm{Cl}$ ?\n$\\mathrm{CH}_{3} \\mathrm{NH}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{NH}_{3}+(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=4.4 \\times 10^{-4}$\n83. What is $\\left[\\mathrm{OH}^{-}\\right]$in a solution of $1.25 \\mathrm{MNH}_{3}$ and $0.78 \\mathrm{MNH}_{4} \\mathrm{NO}_{3}$ ?\n$\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{( }(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=1.8 \\times 10^{-5}$\n84. What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:\n(a) HCl\n(b) $\\mathrm{KCH}_{3} \\mathrm{CO}_{2}$\n(c) NaCl\n(d) KOH\n(e) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n85. What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:\n(a) KI\n(b) $\\mathrm{NH}_{3}$\n(c) HI\n(d) NaOH\n(e) $\\mathrm{NH}_{4} \\mathrm{Cl}$\n86. What will be the pH of a buffer solution prepared from $0.20 \\mathrm{~mol} \\mathrm{NH}_{3}, 0.40 \\mathrm{~mol} \\mathrm{NH}_{4} \\mathrm{NO}_{3}$, and just enough water to give 1.00 L of solution?\n87. Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of $\\mathrm{KH}_{2} \\mathrm{PO}_{4}$, and enough water to make 0.500 L of solution.\n88. How much solid $\\mathrm{NaCH}_{3} \\mathrm{CO}_{2} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$ must be added to 0.300 L of a $0.50-M$ acetic acid solution to give a buffer with a pH of 5.00 ? (Hint: Assume a negligible change in volume as the solid is added.)\n89. What mass of $\\mathrm{NH}_{4} \\mathrm{Cl}$ must be added to 0.750 L of a $0.100-\\mathrm{M}$ solution of $\\mathrm{NH}_{3}$ to give a buffer solution with a pH of 9.26 ? (Hint: Assume a negligible change in volume as the solid is added.)\n90. A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use $1.80 \\times 10^{-5}$ as $K_{\\mathrm{a}}$ for acetic acid.\n(a) What is the pH of the solution?\n(b) Is the solution acidic or basic?\n(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to 0.200 L of the original buffer?\n91. A $5.36-\\mathrm{g}$ sample of $\\mathrm{NH}_{4} \\mathrm{Cl}$ was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L .\n(a) What is the pH of this buffer solution?\n(b) Is the solution acidic or basic?\n(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to the solution?\n"} {"id": "textbook_4966", "title": "1590. 2 pH and pOH - 1590.5. Acid-Base Titrations", "content": "\n92. Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.\n93. Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH .\n94. Calculate the pH at the following points in a titration of $40 \\mathrm{~mL}(0.040 \\mathrm{~L})$ of 0.100 M barbituric acid $\\left(K_{\\mathrm{a}}=9.8\\right.$ $\\times 10^{-5}$ ) with 0.100 M KOH .\n(a) no KOH added\n(b) 20 mL of KOH solution added\n(c) 39 mL of KOH solution added\n(d) 40 mL of KOH solution added\n(e) 41 mL of KOH solution added\n95. The indicator dinitrophenol is an acid with a $K_{\\mathrm{a}}$ of $1.1 \\times 10^{-4}$. In a $1.0 \\times 10^{-4}-M$ solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from $10 \\%$ ionized (colorless) to $90 \\%$ ionized (yellow)."} {"id": "textbook_4967", "title": "1590. 2 pH and pOH - 1590.5. Acid-Base Titrations", "content": "724 14\u2022Exercises\n"} {"id": "textbook_4968", "title": "1591. CHAPTER 15 Equilibria of Other Reaction Classes - ", "content": "\n"} {"id": "textbook_4969", "title": "1591. CHAPTER 15 Equilibria of Other Reaction Classes - ", "content": "Figure 15.1 The mineral fluorite $\\left(\\mathrm{CaF}_{2}\\right)$ is formed when dissolved calcium and fluoride ions precipitate from groundwater within the Earth's crust. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metal ions in the crystal.\n"} {"id": "textbook_4970", "title": "1592. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_4971", "title": "1592. CHAPTER OUTLINE - 1592.1. Precipitation and Dissolution", "content": "\n15.2 Lewis Acids and Bases\n15.3 Coupled Equilibria"} {"id": "textbook_4972", "title": "1592. CHAPTER OUTLINE - 1592.1. Precipitation and Dissolution", "content": "INTRODUCTION The mineral fluorite, $\\mathrm{CaF}_{2}$ Figure 15.1, is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble $\\mathrm{CaF}_{2}$ in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH , and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions.\n"} {"id": "textbook_4973", "title": "1592. CHAPTER OUTLINE - 1592.2. Precipitation and Dissolution", "content": ""} {"id": "textbook_4974", "title": "1593. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_4975", "title": "1593. LEARNING OBJECTIVES - ", "content": "- Write chemical equations and equilibrium expressions representing solubility equilibria\n- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations"} {"id": "textbook_4976", "title": "1593. LEARNING OBJECTIVES - ", "content": "Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal\nrates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.\n"} {"id": "textbook_4977", "title": "1594. The Solubility Product - ", "content": "\nRecall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established."} {"id": "textbook_4978", "title": "1594. The Solubility Product - ", "content": "In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous $\\mathrm{Ag}^{+}$and $\\mathrm{Cl}^{-}$ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (Figure 15.2). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.\n\n\nFIGURE 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of $\\mathrm{Ag}^{+}$and $\\mathrm{Cl}^{-}$ions in equilibrium with undissolved silver chloride."} {"id": "textbook_4979", "title": "1594. The Solubility Product - ", "content": "The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, $\\boldsymbol{K}_{\\mathbf{s p}}$, in this case"} {"id": "textbook_4980", "title": "1594. The Solubility Product - ", "content": "$$\n\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}(a q)\\right]\\left[\\mathrm{Cl}^{-}(a q)\\right]\n$$"} {"id": "textbook_4981", "title": "1594. The Solubility Product - ", "content": "Recall that only gases and solutes are represented in equilibrium constant expressions, so the $K_{\\text {sp }}$ does not include a term for the undissolved AgCl . A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.\n"} {"id": "textbook_4982", "title": "1595. EXAMPLE 15.1 - ", "content": ""} {"id": "textbook_4983", "title": "1596. Writing Equations and Solubility Products - ", "content": "\nWrite the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:\n(a) AgI, silver iodide, a solid with antiseptic properties\n(b) $\\mathrm{CaCO}_{3}$, calcium carbonate, the active ingredient in many over-the-counter chewable antacids\n(c) $\\mathrm{Mg}(\\mathrm{OH})_{2}$, magnesium hydroxide, the active ingredient in Milk of Magnesia\n(d) $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}$, magnesium ammonium phosphate, an essentially insoluble substance used in tests for\nmagnesium\n(e) $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{OH}$, the mineral apatite, a source of phosphate for fertilizers"} {"id": "textbook_4984", "title": "1596. Writing Equations and Solubility Products - ", "content": "Solution\n(a) $\\mathrm{AgI}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{I}^{-}(a q) \\quad K_{\\text {sp }}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{I}^{-}\\right]$\n(b) $\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$\n(c) $\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}$\n(d) $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+\\mathrm{NH}_{4}{ }^{+}(a q)+\\mathrm{PO}_{4}{ }^{3-}(a q)$\n$K_{\\text {sp }}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{NH}_{4}^{+}\\right]\\left[\\mathrm{PO}_{4}{ }^{3-}\\right]$\n(e) $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right) 3 \\mathrm{OH}(s) \\rightleftharpoons 5 \\mathrm{Ca}^{2+}(a q)+3 \\mathrm{PO}_{4}{ }^{3-}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]^{5}\\left[\\mathrm{PO}_{4}^{3-}\\right]^{3}\\left[\\mathrm{OH}^{-}\\right]$"} {"id": "textbook_4985", "title": "1596. Writing Equations and Solubility Products - ", "content": "Check Your Learning\nWrite the dissolution equation and the solubility product for each of the following slightly soluble compounds:\n(a) $\\mathrm{BaSO}_{4}$\n(b) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(c) $\\mathrm{Al}(\\mathrm{OH})_{3}$\n(d) $\\mathrm{Pb}(\\mathrm{OH}) \\mathrm{Cl}$\n"} {"id": "textbook_4986", "title": "1597. Answer: - ", "content": "\n(a) $\\mathrm{BaSO}_{4}(s) \\rightleftharpoons \\mathrm{Ba}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ba}^{2+}\\right]\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]$;\n(b) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\rightleftharpoons 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]^{2}\\left[\\mathrm{SO}_{4}{ }^{2-}\\right] ;$\n(c) $\\mathrm{Al}(\\mathrm{OH})_{3}(s) \\rightleftharpoons \\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Al}^{3+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{3}$;\n(d) $\\mathrm{Pb}(\\mathrm{OH}) \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+\\mathrm{OH}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]\\left[\\mathrm{Cl}^{-}\\right]$\n"} {"id": "textbook_4987", "title": "1598. $K_{\\text {sp }}$ and Solubility - ", "content": "\nThe $K_{\\text {sp }}$ of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:"} {"id": "textbook_4988", "title": "1598. $K_{\\text {sp }}$ and Solubility - ", "content": "$$\n\\mathrm{M}_{p} \\mathrm{X}_{q}(s) \\rightleftharpoons p \\mathrm{M}^{\\mathrm{m}+}(a q)+q \\mathrm{X}^{\\mathrm{n}-}(a q)\n$$"} {"id": "textbook_4989", "title": "1598. $K_{\\text {sp }}$ and Solubility - ", "content": "For cases such as these, one may derive $K_{\\text {sp }}$ values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound's molar solubility, measured as moles of dissolved solute per liter of saturated solution.\n"} {"id": "textbook_4990", "title": "1599. EXAMPLE 15.2 - ", "content": ""} {"id": "textbook_4991", "title": "1600. Calculation of $\\boldsymbol{K}_{\\text {sp }}$ from Equilibrium Concentrations - ", "content": "\nFluorite, $\\mathrm{CaF}_{2}$, is a slightly soluble solid that dissolves according to the equation:"} {"id": "textbook_4992", "title": "1600. Calculation of $\\boldsymbol{K}_{\\text {sp }}$ from Equilibrium Concentrations - ", "content": "$$\n\\mathrm{CaF}_{2}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{~F}^{-}(a q)\n$$"} {"id": "textbook_4993", "title": "1600. Calculation of $\\boldsymbol{K}_{\\text {sp }}$ from Equilibrium Concentrations - ", "content": "The concentration of $\\mathrm{Ca}^{2+}$ in a saturated solution of $\\mathrm{CaF}_{2}$ is $2.15 \\times 10^{-4} \\mathrm{M}$. What is the solubility product of fluorite?\n"} {"id": "textbook_4994", "title": "1601. Solution - ", "content": "\nAccording to the stoichiometry of the dissolution equation, the fluoride ion molarity of a $\\mathrm{CaF}_{2}$ solution is equal to twice its calcium ion molarity:"} {"id": "textbook_4995", "title": "1601. Solution - ", "content": "$$\n\\left[\\mathrm{F}^{-}\\right]=\\left(2 \\mathrm{~mol} \\mathrm{~F}^{-} / 1 \\mathrm{~mol} \\mathrm{Ca}^{2+}\\right)=(2)\\left(2.15 \\times 10^{-4} M\\right)=4.30 \\times 10^{-4} M\n$$"} {"id": "textbook_4996", "title": "1601. Solution - ", "content": "Substituting the ion concentrations into the $K_{\\mathrm{sp}}$ expression gives"} {"id": "textbook_4997", "title": "1601. Solution - ", "content": "$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}=\\left(2.15 \\times 10^{-4}\\right)\\left(4.30 \\times 10^{-4}\\right)^{2}=3.98 \\times 10^{-11}\n$$\n"} {"id": "textbook_4998", "title": "1602. Check Your Learning - ", "content": "\nIn a saturated solution of $\\mathrm{Mg}(\\mathrm{OH})_{2}$, the concentration of $\\mathrm{Mg}^{2+}$ is $1.31 \\times 10^{-4} \\mathrm{M}$. What is the solubility product for $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ?"} {"id": "textbook_4999", "title": "1602. Check Your Learning - ", "content": "$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)\n$$\n"} {"id": "textbook_5000", "title": "1603. Answer: - ", "content": "\n$8.99 \\times 10^{-12}$\n"} {"id": "textbook_5001", "title": "1604. EXAMPLE 15.3 - ", "content": ""} {"id": "textbook_5002", "title": "1605. Determination of Molar Solubility from $\\boldsymbol{K}_{\\mathbf{s p}}$ - ", "content": "\nThe $K_{\\mathrm{sp}}$ of copper(I) bromide, CuBr , is $6.3 \\times 10^{-9}$. Calculate the molar solubility of copper bromide.\n"} {"id": "textbook_5003", "title": "1606. Solution - ", "content": "\nThe dissolution equation and solubility product expression are"} {"id": "textbook_5004", "title": "1606. Solution - ", "content": "$$\n\\begin{gathered}\n\\mathrm{CuBr}(s) \\rightleftharpoons \\mathrm{Cu}^{+}(a q)+\\mathrm{Br}^{-}(a q) \\\\\nK_{\\mathrm{sp}}=\\left[\\mathrm{Cu}^{+}\\right]\\left[\\mathrm{Br}^{-}\\right]\n\\end{gathered}\n$$"} {"id": "textbook_5005", "title": "1606. Solution - ", "content": "Following the ICE approach to this calculation yields the table"} {"id": "textbook_5006", "title": "1606. Solution - ", "content": "| | $\\mathrm{CuBr}(s) \\rightleftharpoons \\mathrm{Cu}^{+}(a q) \\quad+\\mathrm{Br}^{-}(\\mathrm{aq})$ | | |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | | 0 | 0 |\n| Change $(M)$ | $+x$ | $+x$ | |\n| Equilibrium concentration $(M)$ | | $x$ | $x$ |"} {"id": "textbook_5007", "title": "1606. Solution - ", "content": "Substituting the equilibrium concentration terms into the solubility product expression and solving for $x$ yields"} {"id": "textbook_5008", "title": "1606. Solution - ", "content": "$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Cu}^{+}\\right]\\left[\\mathrm{Br}^{-}\\right] \\\\\n6.3 \\times 10^{-9}=(x)(x)=x^{2} \\\\\nx=\\sqrt{\\left(6.3 \\times 10^{-9}\\right)}=7.9 \\times 10^{-5} M\n\\end{gathered}\n$$"} {"id": "textbook_5009", "title": "1606. Solution - ", "content": "Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is $7.9 \\times 10^{-5} \\mathrm{M}$.\n"} {"id": "textbook_5010", "title": "1607. Check Your Learning - ", "content": "\nThe $K_{\\mathrm{sp}}$ of AgI is $1.5 \\times 10^{-16}$. Calculate the molar solubility of silver iodide.\n"} {"id": "textbook_5011", "title": "1608. Answer: - ", "content": "\n$1.2 \\times 10^{-8} \\mathrm{M}$\n"} {"id": "textbook_5012", "title": "1609. EXAMPLE 15.4 - ", "content": ""} {"id": "textbook_5013", "title": "1610. Determination of Molar Solubility from $\\boldsymbol{K}_{\\text {sp }}$ - ", "content": "\nThe $K_{\\mathrm{sp}}$ of calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, is $1.3 \\times 10^{-6}$. Calculate the molar solubility of calcium hydroxide.\n"} {"id": "textbook_5014", "title": "1611. Solution - ", "content": "\nThe dissolution equation and solubility product expression are"} {"id": "textbook_5015", "title": "1611. Solution - ", "content": "$$\n\\begin{aligned}\n\\mathrm{Ca}(\\mathrm{OH})_{2}(s) & \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\\\\nK_{\\text {sp }} & =\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}\n\\end{aligned}\n$$"} {"id": "textbook_5016", "title": "1611. Solution - ", "content": "The ICE table for this system is"} {"id": "textbook_5017", "title": "1611. Solution - ", "content": "| | $\\mathbf{C a}(\\mathbf{O H})_{2}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(\\mathrm{aq}) \\mathbf{+} \\mathbf{2 O H}^{-}(\\mathrm{aq})$ | | |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | | 0 | 0 |\n| Change $(M)$ | | $+x$ | $+2 x$ |\n| Equilibrium concentration $(M)$ | | $x$ | $2 x$ |"} {"id": "textbook_5018", "title": "1611. Solution - ", "content": "Substituting terms for the equilibrium concentrations into the solubility product expression and solving for $x$ gives"} {"id": "textbook_5019", "title": "1611. Solution - ", "content": "$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2} \\\\\n1.3 \\times 10^{-6}=(x)(2 x)^{2}=(x)\\left(4 x^{2}\\right)=4 x^{3} \\\\\nx=\\sqrt[3]{\\frac{1.3 \\times 10^{-6}}{4}}=6.9 \\times 10^{-3} M\n\\end{gathered}\n$$"} {"id": "textbook_5020", "title": "1611. Solution - ", "content": "As defined in the ICE table, $x$ is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is $6.9 \\times 10^{-3} \\mathrm{M}$.\n"} {"id": "textbook_5021", "title": "1612. Check Your Learning - ", "content": "\nThe $K_{\\mathrm{sp}}$ of $\\mathrm{PbI}_{2}$ is $1.4 \\times 10^{-8}$. Calculate the molar solubility of lead(II) iodide.\n"} {"id": "textbook_5022", "title": "1613. Answer: - ", "content": "\n$1.5 \\times 10^{-3} \\mathrm{M}$\n"} {"id": "textbook_5023", "title": "1614. EXAMPLE 15.5 - ", "content": ""} {"id": "textbook_5024", "title": "1615. Determination of $\\boldsymbol{K}_{\\text {sp }}$ from Gram Solubility - ", "content": "\nMany of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist's pigment chrome yellow, $\\mathrm{PbCrO}_{4}$, is $4.6 \\times 10^{-6} \\mathrm{~g} / \\mathrm{L}$. Determine the solubility product for $\\mathrm{PbCrO}_{4}$.\n"} {"id": "textbook_5025", "title": "1615. Determination of $\\boldsymbol{K}_{\\text {sp }}$ from Gram Solubility - ", "content": "FIGURE 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow $\\left(\\mathrm{PbCrO}_{4}\\right)$, examples include Prussian blue $\\left(\\mathrm{Fe}_{7}(\\mathrm{CN})_{18}\\right)$, the reddish-orange color vermilion $(\\mathrm{HgS})$, and green color veridian $\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{3}\\right)$. (credit: Sonny Abesamis)\n"} {"id": "textbook_5026", "title": "1616. Solution - ", "content": "\nBefore calculating the solubility product, the provided solubility must be converted to molarity:"} {"id": "textbook_5027", "title": "1616. Solution - ", "content": "$$\n\\begin{gathered}\n{\\left[\\mathrm{PbCrO}_{4}\\right]=\\frac{4.6 \\times 10^{-6} \\mathrm{~g} \\mathrm{PbCrO}_{4}}{1 \\mathrm{~L}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{PbCrO}_{4}}{323.2 \\mathrm{~g} \\mathrm{PbCrO}_{4}}} \\\\\n=\\frac{1.4 \\times 10^{-8} \\mathrm{~mol} \\mathrm{PbCrO}_{4}}{1 \\mathrm{~L}} \\\\\n=1.4 \\times 10^{-8} \\mathrm{M}\n\\end{gathered}\n$$"} {"id": "textbook_5028", "title": "1616. Solution - ", "content": "The dissolution equation for this compound is"} {"id": "textbook_5029", "title": "1616. Solution - ", "content": "$$\n\\mathrm{PbCrO}_{4}(s) \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+\\mathrm{CrO}_{4}^{2-}(a q)\n$$"} {"id": "textbook_5030", "title": "1616. Solution - ", "content": "The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both $\\left[\\mathrm{Pb}^{2+}\\right]$ and $\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]$ are equal to the molar solubility of $\\mathrm{PbCrO}_{4}$ :"} {"id": "textbook_5031", "title": "1616. Solution - ", "content": "$$\n\\left[\\mathrm{Pb}^{2+}\\right]=\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]=1.4 \\times 10^{-8} \\mathrm{M}\n$$\n\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]=\\left(1.4 \\times 10^{-8}\\right)\\left(1.4 \\times 10^{-8}\\right)=2.0 \\times 10^{-16}$\n"} {"id": "textbook_5032", "title": "1617. Check Your Learning - ", "content": "\nThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at $20^{\\circ} \\mathrm{C}$. What is its solubility product?\n"} {"id": "textbook_5033", "title": "1618. Answer: - ", "content": "\n$1.69 \\times 10^{-4}$\n"} {"id": "textbook_5034", "title": "1619. EXAMPLE 15.6 - ", "content": ""} {"id": "textbook_5035", "title": "1620. Calculating the Solubility of $\\mathrm{Hg}_{2} \\mathbf{C l}_{\\mathbf{2}}$ - ", "content": "\nCalomel, $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$, is a compound composed of the diatomic ion of mercury(I), $\\mathrm{Hg}_{2}{ }^{2+}$, and chloride ions, $\\mathrm{Cl}^{-}$. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small $K_{\\mathrm{sp}}$ :"} {"id": "textbook_5036", "title": "1620. Calculating the Solubility of $\\mathrm{Hg}_{2} \\mathbf{C l}_{\\mathbf{2}}$ - ", "content": "$$\n\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) \\rightleftharpoons \\mathrm{Hg}_{2}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=1.1 \\times 10^{-18}\n$$"} {"id": "textbook_5037", "title": "1620. Calculating the Solubility of $\\mathrm{Hg}_{2} \\mathbf{C l}_{\\mathbf{2}}$ - ", "content": "Calculate the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n"} {"id": "textbook_5038", "title": "1621. Solution - ", "content": "\nThe dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ is equal to the concentration of $\\mathrm{Hg}_{2}{ }^{2+}$ ions"} {"id": "textbook_5039", "title": "1621. Solution - ", "content": "Following the ICE approach results in"} {"id": "textbook_5040", "title": "1621. Solution - ", "content": "| | $\\mathbf{H g}_{2} \\mathrm{Cl}_{2}(\\mathbf{s}) \\rightleftharpoons \\mathbf{H g}_{2}{ }^{2+}(a q) \\mathbf{+} \\mathbf{2 C l}(a q)$ | | |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | | 0 | 0 |\n| Change $(M)$ | | $+x$ | $+2 x$ |\n| Equilibrium concentration $(M)$ | $x$ | $2 x$ | |"} {"id": "textbook_5041", "title": "1621. Solution - ", "content": "Substituting the equilibrium concentration terms into the solubility product expression and solving for $x$ gives"} {"id": "textbook_5042", "title": "1621. Solution - ", "content": "$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Hg}_{2}^{2+}\\right]\\left[\\mathrm{Cl}^{-}\\right]^{2} \\\\\n1.1 \\times 10^{-18}=(x)(2 x)^{2} \\\\\n4 x^{3}=1.1 \\times 10^{-18} \\\\\nx=\\sqrt[3]{\\left(\\frac{1.1 \\times 10^{-18}}{4}\\right)}=6.5 \\times 10^{-7} M \\\\\n{\\left[\\mathrm{Hg}_{2}^{2+}\\right]=6.5 \\times 10^{-7} M=6.5 \\times 10^{-7} M} \\\\\n{\\left[\\mathrm{Cl}^{-}\\right]=2 x=2\\left(6.5 \\times 10^{-7}\\right)=1.3 \\times 10^{-6} M}\n\\end{gathered}\n$$"} {"id": "textbook_5043", "title": "1621. Solution - ", "content": "The dissolution stoichiometry shows the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ is equal to $\\left[\\mathrm{Hg}_{2}{ }^{2+}\\right]$, or $6.5 \\times 10^{-7} \\mathrm{M}$.\n"} {"id": "textbook_5044", "title": "1622. Check Your Learning - ", "content": "\nDetermine the molar solubility of $\\mathrm{MgF}_{2}$ from its solubility product: $K_{\\mathrm{sp}}=6.4 \\times 10^{-9}$.\n"} {"id": "textbook_5045", "title": "1623. Answer: - ", "content": "\n$1.2 \\times 10^{-3} \\mathrm{M}$\n"} {"id": "textbook_5046", "title": "1624. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_5047", "title": "1625. Using Barium Sulfate for Medical Imaging - ", "content": "\nVarious types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the $K_{\\text {sp }}$ of barium sulfate is $2.3 \\times 10^{-8}$, very little of it dissolves as it coats the lining of the patient's intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).\n\n\nFIGURE 15.4 A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by \"glitzy queen00\"/Wikimedia Commons)"} {"id": "textbook_5048", "title": "1625. Using Barium Sulfate for Medical Imaging - ", "content": "Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn's disease, and ulcers in addition to other conditions."} {"id": "textbook_5049", "title": "1625. Using Barium Sulfate for Medical Imaging - ", "content": "Visit this website (http://openstax.org/l/16barium) for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.\n"} {"id": "textbook_5050", "title": "1626. Predicting Precipitation - ", "content": "\nThe equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:"} {"id": "textbook_5051", "title": "1626. Predicting Precipitation - ", "content": "$$\n\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\quad K_{s p}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{CO}_{3}^{2-}\\right]=8.7 \\times 10^{-9}\n$$"} {"id": "textbook_5052", "title": "1626. Predicting Precipitation - ", "content": "It is important to realize that this equilibrium is established in any aqueous solution containing $\\mathrm{Ca}^{2+}$ and $\\mathrm{CO}_{3}{ }^{2-}$ ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, $Q_{s p}$, that exceeds the solubility product, $K_{\\text {sp }}$, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established ( $Q_{s p}=K_{\\mathrm{sp}}$ ). The comparison of $Q_{s p}$ to $K_{\\mathrm{sp}}$ to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:\n$Q_{s p}
Electrochemistry - ", "content": "Figure 16.1 Electric vehicles are powered by batteries, devices that harness the energy of spontaneous redox reactions. (credit: modification of work by Robert Couse-Baker)\n"} {"id": "textbook_5265", "title": "1691. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_5266", "title": "1691. CHAPTER OUTLINE - 1691.1. Review of Redox Chemistry", "content": "\n16.2 Galvanic Cells\n16.3 Electrode and Cell Potentials\n16.4 Potential, Free Energy, and Equilibrium\n16.5 Batteries and Fuel Cells\n16.6 Corrosion\n16.7 Electrolysis"} {"id": "textbook_5267", "title": "1691. CHAPTER OUTLINE - 1691.1. Review of Redox Chemistry", "content": "INTRODUCTION Another chapter in this text introduced the chemistry of reduction-oxidation (redox) reactions. This important reaction class is defined by changes in oxidation states for one or more reactant elements, and it includes a subset of reactions involving the transfer of electrons between reactant species. Around the turn of the nineteenth century, chemists began exploring ways these electrons could be transferred indirectly via an external circuit rather than directly via intimate contact of redox reactants. In the two centuries since, the field of electrochemistry has evolved to yield significant insights on the fundamental aspects of redox chemistry as well as a wealth of technologies ranging from industrial-scale metallurgical processes to robust, rechargeable batteries for electric vehicles (Figure 16.1). In this chapter, the essential concepts of electrochemistry will be addressed.\n"} {"id": "textbook_5268", "title": "1691. CHAPTER OUTLINE - 1691.2. Review of Redox Chemistry", "content": ""} {"id": "textbook_5269", "title": "1692. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5270", "title": "1692. LEARNING OBJECTIVES - ", "content": "- Describe defining traits of redox chemistry\n- Identify the oxidant and reductant of a redox reaction\n- Balance chemical equations for redox reactions using the half-reaction method"} {"id": "textbook_5271", "title": "1692. LEARNING OBJECTIVES - ", "content": "Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier text chapter (see chapter on reaction stoichiometry). Readers wishing additional review are referred to the text chapter on reaction stoichiometry.\n"} {"id": "textbook_5272", "title": "1693. Oxidation Numbers - ", "content": "\nBy definition, a redox reaction is one that entails changes in oxidation number (or oxidation state) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description, the oxidation number of an atom in an element is equal to zero. For an atom in a compound, the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic. Consequential to these rules, the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered."} {"id": "textbook_5273", "title": "1693. Oxidation Numbers - ", "content": "Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements' oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl , is comprised of $\\mathrm{Na}^{+}$cations and $\\mathrm{Cl}^{-}$anions, and so oxidation numbers for sodium and chlorine are, +1 and -1 , respectively. Calcium fluoride, $\\mathrm{CaF}_{2}$, is comprised of $\\mathrm{Ca}^{2+}$ cations and $\\mathrm{F}^{-}$anions, and so oxidation numbers for calcium and fluorine are, +2 and -1 , respectively."} {"id": "textbook_5274", "title": "1693. Oxidation Numbers - ", "content": "Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent $\\mathrm{O}-\\mathrm{H}$ bonds. The shared electrons comprising an $\\mathrm{O}-\\mathrm{H}$ bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral."} {"id": "textbook_5275", "title": "1693. Oxidation Numbers - ", "content": "Imagine that the polarization of shared electrons within the $\\mathrm{O}-\\mathrm{H}$ bonds of water were $100 \\%$ complete-the result would be transfer of electrons from H to O , and water would be an ionic compound comprised of $\\mathrm{O}^{2-}$ anions and $\\mathrm{H}^{+}$cations. And so, the oxidations numbers for oxygen and hydrogen in water are -2 and +1 , respectively. Applying this same logic to carbon tetrachloride, $\\mathrm{CCl}_{4}$, yields oxidation numbers of +4 for carbon and -1 for chlorine. In the nitrate ion, $\\mathrm{NO}_{3}^{-}$, the oxidation number for nitrogen is +5 and that for oxygen is -2 , summing to equal the 1 - charge on the molecule:"} {"id": "textbook_5276", "title": "1693. Oxidation Numbers - ", "content": "$$\n(1 \\mathrm{~N} \\text { atom })\\left(\\frac{+5}{\\mathrm{~N} \\text { atom }}\\right)+(3 \\mathrm{O} \\text { atoms })\\left(\\frac{-2}{\\mathrm{O} \\text { atom }}\\right)=+5+-6=-1\n$$\n"} {"id": "textbook_5277", "title": "1694. Balancing Redox Equations - ", "content": "\nThe unbalanced equation below describes the decomposition of molten sodium chloride:"} {"id": "textbook_5278", "title": "1694. Balancing Redox Equations - ", "content": "$$\n\\mathrm{NaCl}(l) \\longrightarrow \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g) \\quad \\text { unbalanced }\n$$"} {"id": "textbook_5279", "title": "1694. Balancing Redox Equations - ", "content": "This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes reduction) and that for Cl is increased from -1 to 0 (it undergoes oxidation). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na :"} {"id": "textbook_5280", "title": "1694. Balancing Redox Equations - ", "content": "$$\n2 \\mathrm{NaCl}(l) \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g) \\quad \\text { balanced }\n$$"} {"id": "textbook_5281", "title": "1694. Balancing Redox Equations - ", "content": "Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, $\\mathrm{H}^{+}(a q)$ and $\\mathrm{OH}^{-}(a q)$, as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the half-reaction method is helpful. This approach involves the following steps:"} {"id": "textbook_5282", "title": "1694. Balancing Redox Equations - ", "content": "1. Write skeletal equations for the oxidation and reduction half-reactions.\n2. Balance each half-reaction for all elements except H and O .\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\n4. Balance each half-reaction for H by adding $\\mathrm{H}^{+}$.\n5. Balance each half-reaction for charge by adding electrons.\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\n7. Add the two half-reactions and simplify.\n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify."} {"id": "textbook_5283", "title": "1694. Balancing Redox Equations - ", "content": "The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.\n"} {"id": "textbook_5284", "title": "1695. EXAMPLE 16.1 - ", "content": ""} {"id": "textbook_5285", "title": "1696. Balancing Equations for Redox Reactions in Acidic Solutions - ", "content": "\nWrite the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.\n"} {"id": "textbook_5286", "title": "1697. Solution - ", "content": "\nFollowing the steps of the half-reaction method:"} {"id": "textbook_5287", "title": "1697. Solution - ", "content": "1. Write skeletal equations for the oxidation and reduction half-reactions."} {"id": "textbook_5288", "title": "1697. Solution - ", "content": "$$\n\\begin{array}{ll}\n\\text { oxidation: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q) \\\\\n\\text { reduction: } & \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)\n\\end{array}\n$$\n\n2. Balance each half-reaction for all elements except $H$ and $O$.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)$\nreduction: $\\quad \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)$\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)$\nreduction: $\\quad \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+\\mathbf{2 H}_{\\mathbf{2}} \\mathbf{O}(l)$\n4. Balance each half-reaction for $H$ by adding $H^{+}$.\n\n$$\n\\begin{array}{ll}\n\\text { oxidation: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q) \\\\\n\\text { reduction: } & \\mathbf{3 H}^{+}(a q)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{array}\n$$\n\n5. Balance each half-reaction for charge by adding electrons.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathbf{e}^{-}$\nreduction: $\\quad 3 \\mathrm{e}^{-}+3 \\mathrm{H}^{+}(a q)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\noxidation $(\\times 3): \\quad 3 \\mathrm{Cu}(s) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+62 \\mathrm{e}^{-}$\nreduction $(\\times 2): \\quad \\quad 63-\\mathrm{e}^{-}+63-\\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{NO}(g)+42 \\mathrm{H}_{2} \\mathrm{O}(l)$\n7. Add the two half-reactions and simplify.\n$3 \\mathrm{Cu}(s)+\\mathbf{6} \\mathrm{e}^{-}+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+\\mathbf{6} \\mathrm{e}^{-}+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$3 \\mathrm{Cu}(s)+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$\n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify."} {"id": "textbook_5289", "title": "1697. Solution - ", "content": "This step not necessary since the solution is stipulated to be acidic.\nThe balanced equation for the reaction in an acidic solution is then"} {"id": "textbook_5290", "title": "1697. Solution - ", "content": "$$\n3 \\mathrm{Cu}(s)+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n"} {"id": "textbook_5291", "title": "1698. Check Your Learning - ", "content": "\nThe reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction."} {"id": "textbook_5292", "title": "1698. Check Your Learning - ", "content": "Answer:\n$\\mathrm{Cu}(s)+2 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n"} {"id": "textbook_5293", "title": "1699. EXAMPLE 16.2 - ", "content": ""} {"id": "textbook_5294", "title": "1700. Balancing Equations for Redox Reactions in Basic Solutions - ", "content": "\nWrite the balanced equation representing reaction between aqueous permanganate ion, $\\mathrm{MnO}_{4}{ }^{-}$, and solid chromium(III) hydroxide, $\\mathrm{Cr}(\\mathrm{OH})_{3}$, to yield solid manganese(IV) oxide, $\\mathrm{MnO}_{2}$, and aqueous chromate ion, $\\mathrm{CrO}_{4}{ }^{2-}$ The reaction takes place in a basic solution.\n"} {"id": "textbook_5295", "title": "1701. Solution - ", "content": "\nFollowing the steps of the half-reaction method:"} {"id": "textbook_5296", "title": "1701. Solution - ", "content": "1. Write skeletal equations for the oxidation and reduction half-reactions.\noxidation: $\\quad \\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n2. Balance each half-reaction for all elements except $H$ and $O$.\noxidation: $\\quad \\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\noxidation: $\\quad \\mathbf{H}_{2} \\mathbf{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathbf{H}_{\\mathbf{2}} \\mathbf{O}(l)$\n4. Balance each half-reaction for $H$ by adding $H^{+}$.\noxidation: $\\quad \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathbf{5 H}^{+}(a q)$\nreduction: $\\quad 4 \\mathbf{H}^{+}(a q)+\\mathrm{MnO}_{4}^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n5. Balance each half-reaction for charge by adding electrons.\noxidation: $\\quad \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+5 \\mathrm{H}^{+}(a q)+\\mathbf{3 e}^{-}$\nreduction: $\\quad 3 \\mathbf{e}^{-}+4 \\mathrm{H}^{+}(a q)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\nThis step is not necessary since the number of electrons is already in balance.\n7. Add the two half-reactions and simplify.\n$\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}(\\boldsymbol{(})+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathbf{3} \\mathbf{e}^{-}+\\mathbf{- 4 \\mathbf { H } ^ { + }}(\\mathrm{aq})+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathbf{5}^{-} \\mathrm{H}^{+}(a q)$ $+3 \\mathrm{e}^{-}+\\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{H}^{+}(a q)+\\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.\n$\\mathbf{O H}^{-}(a q)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{H}^{+}(a q)+\\mathbf{O H}^{-}(a q)+\\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n$\\mathrm{OH}^{-}(a q)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n"} {"id": "textbook_5297", "title": "1702. Check Your Learning - ", "content": "\nAqueous permanganate ion may also be reduced using aqueous bromide ion, $\\mathrm{Br}^{-}$, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, $\\mathrm{BrO}_{3}{ }^{-}$. Write the balanced equation for this\nreaction occurring in a basic medium.\nAnswer:\n$\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{MnO}_{4}^{-}(a q)+\\mathrm{Br}^{-}(a q) \\longrightarrow 2 \\mathrm{MnO}_{2}(s)+\\mathrm{BrO}_{3}^{-}(a q)+2 \\mathrm{OH}^{-}(a q)$\n"} {"id": "textbook_5298", "title": "1702. Check Your Learning - 1702.1. Galvanic Cells", "content": ""} {"id": "textbook_5299", "title": "1703. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5300", "title": "1703. LEARNING OBJECTIVES - ", "content": "- Describe the function of a galvanic cell and its components\n- Use cell notation to symbolize the composition and construction of galvanic cells"} {"id": "textbook_5301", "title": "1703. LEARNING OBJECTIVES - ", "content": "As demonstration of spontaneous chemical change, Figure 16.2 shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.\n\n\nFIGURE 16.2 A copper wire and an aqueous solution of silver nitrate (left) are brought into contact (center) and a spontaneous transfer of electrons occurs, creating blue $\\mathrm{Cu}^{2+}(a q)$ and gray $\\operatorname{Ag}(s)$ (right).\n\nThese observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, $\\mathrm{Cu}^{2+}(a q)$, which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:"} {"id": "textbook_5302", "title": "1703. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{array}{ll}\n\\text { overall reaction: } & \\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s) \\\\\n\\text { oxidation half-reaction: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n\\text {reduction half-reaction: } & 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Ag}(s)\n\\end{array}\n$$"} {"id": "textbook_5303", "title": "1703. LEARNING OBJECTIVES - ", "content": "Consider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells, and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells)."} {"id": "textbook_5304", "title": "1703. LEARNING OBJECTIVES - ", "content": "A galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure 16.3. The cell is comprised of two half-cells, each containing the redox conjugate pair (\"couple\") of a single reactant. The half-cell shown at the left contains the $\\mathrm{Cu}(0) / \\mathrm{Cu}(\\mathrm{II})$ couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the $\\mathrm{Ag}(\\mathrm{I}) / \\mathrm{Ag}(0)$ couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode. By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell's reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge. The spontaneous reaction in this cell produces $\\mathrm{Cu}^{2+}$ cations in the anode half-cell and consumes $\\mathrm{Ag}^{+}$ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of $\\mathrm{Cu}^{2+}$ in the anode half-cell are balanced by an influx of $\\mathrm{NO}_{3}{ }^{-}$from the salt bridge, while a flow of $\\mathrm{Na}^{+}$into the cathode half-cell compensates for the decreasing $\\mathrm{Ag}^{+}$concentration.\n\n\nFIGURE 16.3 A galvanic cell based on the spontaneous reaction between copper and silver(I) ions.\n"} {"id": "textbook_5305", "title": "1704. Cell Notation - ", "content": "\nAbbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics, and they are written following a few guidelines:"} {"id": "textbook_5306", "title": "1704. Cell Notation - ", "content": "- The relevant components of each half-cell are represented by their chemical formulas or element symbols\n- All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas\n- By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode"} {"id": "textbook_5307", "title": "1704. Cell Notation - ", "content": "A verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure 16.3 consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:"} {"id": "textbook_5308", "title": "1704. Cell Notation - ", "content": "$$\n\\mathrm{Cu}(s)\\left|1 M \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\| 1 M \\mathrm{AgNO}_{3}(a q)\\right| \\mathrm{Ag}(s)\n$$"} {"id": "textbook_5309", "title": "1704. Cell Notation - ", "content": "Consider a different galvanic cell (see Figure 16.4) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:\nnet cell reaction:\noxidation half-reaction:\nreduction half-reaction:"} {"id": "textbook_5310", "title": "1704. Cell Notation - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{Mg}(s)+2 \\mathrm{Fe}^{3+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{Fe}^{2+}(a q) \\\\\n& \\operatorname{Mg}(s) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n& 2 \\mathrm{Fe}^{3+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Fe}^{2+}(a q)\n\\end{aligned}\n$$"} {"id": "textbook_5311", "title": "1704. Cell Notation - ", "content": "In this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II)\nchloride, immersed in which is a platinum cathode. The cell schematic is then written as"} {"id": "textbook_5312", "title": "1704. Cell Notation - ", "content": "$$\n\\operatorname{Mg}(s)\\left|0.1 M \\mathrm{MgCl}_{2}(a q) \\| 0.2 M \\mathrm{FeCl}_{3}(a q), 0.3 M \\mathrm{FeCl}_{2}(a q)\\right| \\operatorname{Pt}(s)\n$$"} {"id": "textbook_5313", "title": "1704. Cell Notation - ", "content": "Notice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance ( Pt ) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell's redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes.\n"} {"id": "textbook_5314", "title": "1704. Cell Notation - ", "content": "FIGURE 16.4 A galvanic cell based on the spontaneous reaction between magnesium and iron(III) ions.\n"} {"id": "textbook_5315", "title": "1705. EXAMPLE 16.3 - ", "content": ""} {"id": "textbook_5316", "title": "1706. Writing Galvanic Cell Schematics - ", "content": "\nA galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a $1 \\mathrm{M} \\mathrm{CrCl}_{3}$ solution and another in which a copper wire is immersed in $1 \\mathrm{M} \\mathrm{CuCl}_{2}$. Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode halfreaction, the cathode half-reaction, and the overall cell reaction.\n"} {"id": "textbook_5317", "title": "1707. Solution - ", "content": "\nSince the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:"} {"id": "textbook_5318", "title": "1707. Solution - ", "content": "$$\n\\mathrm{Cr}(s)\\left|1 M \\mathrm{CrCl}_{3}(a q) \\| 1 M \\mathrm{CuCl}_{2}(a q)\\right| \\mathrm{Cu}(s)\n$$"} {"id": "textbook_5319", "title": "1707. Solution - ", "content": "The half-reactions for this cell are"} {"id": "textbook_5320", "title": "1707. Solution - ", "content": "$$\n\\begin{array}{ll}\n\\text { anode (oxidation): } & \\mathrm{Cr}(s) \\longrightarrow \\mathrm{Cr}^{3+}(a q)+3 \\mathrm{e}^{-} \\\\\n\\text {cathode (reduction): } & \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)\n\\end{array}\n$$"} {"id": "textbook_5321", "title": "1707. Solution - ", "content": "Multiplying to make the number of electrons lost by Cr and gained by $\\mathrm{Cu}^{2+}$ equal yields"} {"id": "textbook_5322", "title": "1707. Solution - ", "content": "$$\n\\begin{array}{ll}\n\\text { anode (oxidation): } & 2 \\mathrm{Cr}(s) \\longrightarrow 2 \\mathrm{Cr}^{3+}(a q)+6 \\mathrm{e}^{-} \\\\\n\\text {cathode (reduction): } & 3 \\mathrm{Cu}^{2+}(a q)+6 \\mathrm{e}^{-} \\longrightarrow 3 \\mathrm{Cu}(s)\n\\end{array}\n$$"} {"id": "textbook_5323", "title": "1707. Solution - ", "content": "Adding the half-reaction equations and simplifying yields an equation for the cell reaction:"} {"id": "textbook_5324", "title": "1707. Solution - ", "content": "$$\n2 \\mathrm{Cr}(s)+3 \\mathrm{Cu}^{2+}(a q) \\longrightarrow 2 \\mathrm{Cr}^{3+}(a q)+3 \\mathrm{Cu}(s)\n$$\n"} {"id": "textbook_5325", "title": "1708. Check Your Learning - ", "content": "\nOmitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below."} {"id": "textbook_5326", "title": "1708. Check Your Learning - ", "content": "$$\n\\mathrm{Sn}^{4+}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Sn}^{2+}(a q)+\\mathrm{Zn}^{2+}(a q)\n$$\n"} {"id": "textbook_5327", "title": "1709. Answer: - ", "content": "\n$\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q) \\| \\mathrm{Sn}^{4+}(a q), \\mathrm{Sn}^{2+}(a q)\\right| \\operatorname{Pt}(s)$\n"} {"id": "textbook_5328", "title": "1709. Answer: - 1709.1. Electrode and Cell Potentials", "content": ""} {"id": "textbook_5329", "title": "1710. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5330", "title": "1710. LEARNING OBJECTIVES - ", "content": "- Describe and relate the definitions of electrode and cell potentials\n- Interpret electrode potentials in terms of relative oxidant and reductant strengths\n- Calculate cell potentials and predict redox spontaneity using standard electrode potentials"} {"id": "textbook_5331", "title": "1710. LEARNING OBJECTIVES - ", "content": "Unlike the spontaneous oxidation of copper by aqueous silver(I) ions described in section 17.2, immersing a copper wire in an aqueous solution of lead(II) ions yields no reaction. The two species, $\\mathrm{Ag}^{+}(\\mathrm{aq})$ and $\\mathrm{Pb}^{2+}(\\mathrm{aq})$, thus show a distinct difference in their redox activity towards copper: the silver ion spontaneously oxidized copper, but the lead ion did not. Electrochemical cells permit this relative redox activity to be quantified by an easily measured property, potential. This property is more commonly called voltage when referenced in regard to electrical applications, and it is a measure of energy accompanying the transfer of charge. Potentials are measured in the volt unit, defined as one joule of energy per one coulomb of charge, $\\mathrm{V}=\\mathrm{J} / \\mathrm{C}$."} {"id": "textbook_5332", "title": "1710. LEARNING OBJECTIVES - ", "content": "When measured for purposes of electrochemistry, a potential reflects the driving force for a specific type of charge transfer process, namely, the transfer of electrons between redox reactants. Considering the nature of potential in this context, it is clear that the potential of a single half-cell or a single electrode can't be measured; \"transfer\" of electrons requires both a donor and recipient, in this case a reductant and an oxidant, respectively. Instead, a half-cell potential may only be assessed relative to that of another half-cell. It is only the difference in potential between two half-cells that may be measured, and these measured potentials are called\ncell potentials, E $_{\\text {cell }}$, defined as"} {"id": "textbook_5333", "title": "1710. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{E}_{\\text {cell }}=\\mathrm{E}_{\\text {cathode }}-\\mathrm{E}_{\\text {anode }}\n$$"} {"id": "textbook_5334", "title": "1710. LEARNING OBJECTIVES - ", "content": "where $\\mathrm{E}_{\\text {cathode }}$ and $\\mathrm{E}_{\\text {anode }}$ are the potentials of two different half-cells functioning as specified in the subscripts. As for other thermodynamic quantities, the standard cell potential, $\\mathbf{E}^{\\circ}{ }_{\\text {cell }}$, is a cell potential measured when both half-cells are under standard-state conditions (1 M concentrations, 1 bar pressures, 298 K ):"} {"id": "textbook_5335", "title": "1710. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{E}_{\\text {cell }}^{\\circ}=\\mathrm{E}_{\\text {cathode }}^{\\circ}-\\mathrm{E}_{\\text {anode }}^{\\circ}\n$$"} {"id": "textbook_5336", "title": "1710. LEARNING OBJECTIVES - ", "content": "To simplify the collection and sharing of potential data for half-reactions, the scientific community has designated one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of exactly 0 V . This half-cell is the standard hydrogen electrode (SHE) and it is based on halfreaction below:"} {"id": "textbook_5337", "title": "1710. LEARNING OBJECTIVES - ", "content": "$$\n2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_5338", "title": "1710. LEARNING OBJECTIVES - ", "content": "A typical SHE contains an inert platinum electrode immersed in precisely $1 M$ aqueous $\\mathrm{H}^{+}$and a stream of bubbling $\\mathrm{H}_{2}$ gas at 1 bar pressure, all maintained at a temperature of 298 K (see Figure 16.5).\n"} {"id": "textbook_5339", "title": "1710. LEARNING OBJECTIVES - ", "content": "FIGURE 16.5 A standard hydrogen electrode (SHE).\nThe assigned potential of the SHE permits the definition of a conveniently measured potential for a single halfcell. The electrode potential ( $\\mathbf{E}_{\\mathbf{X}}$ ) for a half-cell $X$ is defined as the potential measured for a cell comprised of $X$ acting as cathode and the SHE acting as anode:"} {"id": "textbook_5340", "title": "1710. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n& E_{\\text {cell }}=E_{X}-E_{\\text {SHE }} \\\\\n& E_{S H E}=0 V(\\text { defined }) \\\\\n& E_{\\text {cell }}=E_{X}\n\\end{aligned}\n$$"} {"id": "textbook_5341", "title": "1710. LEARNING OBJECTIVES - ", "content": "When the half-cell $X$ is under standard-state conditions, its potential is the standard electrode potential, $\\mathbf{E}^{\\circ} \\mathbf{x}$. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials."} {"id": "textbook_5342", "title": "1710. LEARNING OBJECTIVES - ", "content": "This approach to measuring electrode potentials is illustrated in Figure 16.6, which depicts a cell comprised of an SHE connected to a copper(II)/copper(0) half-cell under standard-state conditions. A voltmeter in the external circuit allows measurement of the potential difference between the two half-cells. Since the Cu halfcell is designated as the cathode in the definition of cell potential, it is connected to the red (positive) input of the voltmeter, while the designated SHE anode is connected to the black (negative) input. These connections insure that the sign of the measured potential will be consistent with the sign conventions of electrochemistry per the various definitions discussed above. A cell potential of +0.337 V is measured, and so"} {"id": "textbook_5343", "title": "1710. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{E}_{\\text {cell }}^{\\circ}=\\mathrm{E}_{\\mathrm{Cu}}^{\\circ}=+0.337 \\mathrm{~V}\n$$"} {"id": "textbook_5344", "title": "1710. LEARNING OBJECTIVES - ", "content": "Tabulations of $\\mathrm{E}^{\\circ}$ values for other half-cells measured in a similar fashion are available as reference literature to permit calculations of cell potentials and the prediction of the spontaneity of redox processes.\n"} {"id": "textbook_5345", "title": "1710. LEARNING OBJECTIVES - ", "content": "FIGURE 16.6 A cell permitting experimental measurement of the standard electrode potential for the half-reaction $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)$"} {"id": "textbook_5346", "title": "1710. LEARNING OBJECTIVES - ", "content": "Table 16.1 provides a listing of standard electrode potentials for a selection of half-reactions in numerical order, and a more extensive alphabetical listing is given in Appendix L."} {"id": "textbook_5347", "title": "1710. LEARNING OBJECTIVES - ", "content": "Selected Standard Reduction Potentials at $25^{\\circ} \\mathrm{C}$"} {"id": "textbook_5348", "title": "1710. LEARNING OBJECTIVES - ", "content": "| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :--- | :--- |\n| $\\mathrm{F}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{~F}^{-}(a q)$ | +2.866 |\n| $\\mathrm{PbO}_{2}(s)+\\mathrm{SO}_{4}{ }^{2-}(a q)+4 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.69 |\n| $\\mathrm{MnO}_{4}^{-}(a q)+8 \\mathrm{H}^{+}(a q)+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.507 |\n| $\\mathrm{Au}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}(s)$ | +1.498 |\n| $\\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}(a q)$ | +1.35827 |\n| $\\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.229 |\n| $\\mathrm{Pt}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pt}(s)$ | +1.20 |\n| $\\mathrm{Br}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}(a q)$ | +1.0873 |\n| $\\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)$ | +0.7996 |"} {"id": "textbook_5349", "title": "1710. LEARNING OBJECTIVES - ", "content": "TABLE 16.1"} {"id": "textbook_5350", "title": "1710. LEARNING OBJECTIVES - ", "content": "| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}(l)$ | +0.7973 |\n| $\\mathrm{Fe}^{3+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}^{2+}(a q)$ | +0.771 |\n| $\\mathrm{MnO}_{4}{ }^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{MnO}_{2}(s)+4 \\mathrm{OH}^{-}(a q)$ | +0.558 |\n| $\\mathrm{I}_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{I}^{-}(a q)$ | +0.5355 |\n| $\\mathrm{NiO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(\\mathrm{OH})_{2}(s)+2 \\mathrm{OH}^{-}(a q)$ | +0.49 |\n| $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)$ | +0.34 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}(l)+2 \\mathrm{Cl}^{-}(a q)$ | +0.26808 |\n| $\\mathrm{AgCl}(s)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)+\\mathrm{Cl}^{-}(a q)$ | +0.22233 |\n| $\\mathrm{Sn}^{4+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+}(a q)$ | +0.151 |\n| $2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)$ | 0.00 |\n| $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)$ | -0.1262 |\n| $\\mathrm{Sn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}(s)$ | -0.1375 |\n| $\\mathrm{Ni}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(s)$ | -0.257 |\n| $\\mathrm{Co}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}(s)$ | -0.28 |\n| $\\mathrm{PbSO}_{4}(s)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)+\\mathrm{SO}_{4}{ }^{2-}(a q)$ | -0.3505 |\n| $\\mathrm{Cd}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}(s)$ | -0.4030 |\n| $\\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}(s)$ | -0.447 |\n| $\\mathrm{Cr}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}(s)$ | -0.744 |\n| $\\mathrm{Mn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}(s)$ | -1.185 |\n| $\\mathrm{Zn}(\\mathrm{OH})_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)+2 \\mathrm{OH}^{-}(a q)$ | -1.245 |\n| $\\mathrm{Zn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)$ | -0.7618 |\n| $\\mathrm{Al}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}(s)$ | -1.662 |\n| $\\mathrm{Mg}^{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mg}(s)$ | -2.372 |\n| $\\mathrm{Na}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(s)$ | -2.71 |"} {"id": "textbook_5351", "title": "1710. LEARNING OBJECTIVES - ", "content": "TABLE 16.1"} {"id": "textbook_5352", "title": "1710. LEARNING OBJECTIVES - ", "content": "| Half-Reaction | $E^{0}(\\mathrm{~V})$ |\n| :--- | :--- |\n| $\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ca}(s)$ | -2.868 |\n| $\\mathrm{Ba}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ba}(s)$ | -2.912 |\n| $\\mathrm{~K}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{K}(s)$ | -2.931 |\n| $\\mathrm{Li}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Li}(s)$ | -3.04 |"} {"id": "textbook_5353", "title": "1710. LEARNING OBJECTIVES - ", "content": "TABLE 16.1\n"} {"id": "textbook_5354", "title": "1711. EXAMPLE 16.4 - ", "content": ""} {"id": "textbook_5355", "title": "1712. Calculating Standard Cell Potentials - ", "content": "\nWhat is the standard potential of the galvanic cell shown in Figure 16.3?\n"} {"id": "textbook_5356", "title": "1713. Solution - ", "content": "\nThe cell in Figure 16.3 is galvanic, the spontaneous cell reaction involving oxidation of its copper anode and reduction of silver(I) ions at its silver cathode:\ncell reaction:\nanode half-reaction:\ncathode half-reaction:"} {"id": "textbook_5357", "title": "1713. Solution - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s) \\\\\n& \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n& 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Ag}(s)\n\\end{aligned}\n$$"} {"id": "textbook_5358", "title": "1713. Solution - ", "content": "The standard cell potential computed as"} {"id": "textbook_5359", "title": "1713. Solution - ", "content": "$$\n\\begin{aligned}\n\\mathrm{E}_{\\text {cell }}^{\\circ} & =\\mathrm{E}_{\\text {cathode }}^{\\circ}-\\mathrm{E}_{\\text {anode }}^{\\circ} \\\\\n& =\\mathrm{E}_{\\mathrm{Ag}}^{\\circ}-\\mathrm{E}_{\\mathrm{Cu}}^{\\circ} \\\\\n& =0.7996 \\mathrm{~V}-0.34 \\mathrm{~V} \\\\\n& =+0.46 \\mathrm{~V}\n\\end{aligned}\n$$\n"} {"id": "textbook_5360", "title": "1714. Check Your Learning - ", "content": "\nWhat is the standard cell potential expected if the silver cathode half-cell in Figure 16.3 is replaced with a lead half-cell: $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)$ ?\n"} {"id": "textbook_5361", "title": "1715. Answer: - ", "content": "\n-0. 47 V\n"} {"id": "textbook_5362", "title": "1716. Intrepreting Electrode and Cell Potentials - ", "content": "\nThinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations in Example 16.4 have just shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness (\"strength\") with which aqueous $\\mathrm{Ag}^{+}$and $\\mathrm{Pb}^{2+}$ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, $E^{\\circ}$. Since by convention electrode potentials are for reduction processes, an increased value of $E^{\\circ}$ corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.\n\nApplying this logic to the numerically ordered listing of standard electrode potentials in Table 16.1 shows this listing to be likewise in order of the oxidizing strength of the half-reaction's reactant species, decreasing from strongest oxidant (most positive $E^{\\circ}$ ) to weakest oxidant (most negative $E^{\\circ}$ ). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, $E^{\\circ}{ }_{\\text {cell }}$ is positive when $E^{\\circ}{ }_{\\text {cathode }}>E^{\\circ}{ }_{\\text {anode }}$, and so any redox reaction in which the oxidant's entry is above the reductant's entry is predicted to be spontaneous."} {"id": "textbook_5363", "title": "1716. Intrepreting Electrode and Cell Potentials - ", "content": "Reconsideration of the two redox reactions in Example 16.4 provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by $\\mathrm{Ag}^{+}$is predicted to be spontaneous ( $E^{\\circ}{ }_{\\text {cathode }}>E^{\\circ}$ anode and so $E^{\\circ}{ }_{\\text {cell }}>0$ ). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by $\\mathrm{Pb}^{2+}$ is nonspontaneous ( $E^{\\circ}{ }_{\\text {cathode }}
Products more abundant at equilibrium |\n| $<1$ | $>0$ | $<0$ | Reaction is non-spontaneous under standard conditions
Reactants more abundant at equilibrium |\n| $=1$ | $=0$ | $=0$ | Reaction is at equilibrium under standard conditions
Reactants and products equally abundant |"} {"id": "textbook_5393", "title": "1724. $E^{\\circ}$ and $K$ - ", "content": "TABLE 16.2\n"} {"id": "textbook_5394", "title": "1725. EXAMPLE 16.6 - ", "content": ""} {"id": "textbook_5395", "title": "1726. Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes - ", "content": "\nUse data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at $25^{\\circ} \\mathrm{C}$. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products."} {"id": "textbook_5396", "title": "1726. Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes - ", "content": "$$\n2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Fe}(s) \\rightleftharpoons 2 \\mathrm{Ag}(s)+\\mathrm{Fe}^{2+}(a q)\n$$\n"} {"id": "textbook_5397", "title": "1727. Solution - ", "content": "\nThe reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L."} {"id": "textbook_5398", "title": "1727. Solution - ", "content": "$$\n\\begin{array}{llrl}\n\\text { anode (oxidation): } & \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Fe} 2+/ \\mathrm{Fe}}^{\\circ}=-0.447 \\mathrm{~V} \\\\\n\\text { cathode (reduction): } & 2 \\times\\left(\\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)\\right) & E_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{\\circ}=0.7996 \\mathrm{~V} \\\\\n& E_{\\text {cell }}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}= & E_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{\\circ}-E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=+1.247 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5399", "title": "1727. Solution - ", "content": "With $n=2$, the equilibrium constant is then"} {"id": "textbook_5400", "title": "1727. Solution - ", "content": "$$\n\\begin{aligned}\nE_{\\mathrm{cell}}^{\\circ} & =\\frac{0.0592 \\mathrm{~V}}{n} \\log K \\\\\nK & =10^{n \\times E_{\\mathrm{cell}}^{\\circ} / 0.0592 \\mathrm{~V}} \\\\\nK & =10^{2 \\times 1.247 \\mathrm{~V} / 0.0592 \\mathrm{~V}} \\\\\nK & =10^{42.128} \\\\\nK & =1.3 \\times 10^{42}\n\\end{aligned}\n$$"} {"id": "textbook_5401", "title": "1727. Solution - ", "content": "The standard free energy is then"} {"id": "textbook_5402", "title": "1727. Solution - ", "content": "$$\n\\begin{aligned}\n& \\Delta G^{\\circ}=-n F E_{\\mathrm{cell}}^{\\circ} \\\\\n& \\Delta G^{\\circ}=-2 \\times 96,485 \\frac{\\mathrm{C}}{\\mathrm{~mol}} \\times 1.247 \\frac{\\mathrm{~J}}{\\mathrm{C}}=-240.6 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}}\n\\end{aligned}\n$$"} {"id": "textbook_5403", "title": "1727. Solution - ", "content": "The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The $K$ value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.\n"} {"id": "textbook_5404", "title": "1728. Check Your Learning - ", "content": "\nWhat is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?"} {"id": "textbook_5405", "title": "1728. Check Your Learning - ", "content": "$$\n\\mathrm{Sn}(s)+2 \\mathrm{Cu}^{2+}(a q) \\rightleftharpoons \\mathrm{Sn}^{2+}(a q)+2 \\mathrm{Cu}^{+}(a q)\n$$"} {"id": "textbook_5406", "title": "1728. Check Your Learning - ", "content": "Answer:\nSpontaneous; $n=2 ; E_{\\text {cell }}^{\\circ}=+0.291 \\mathrm{~V} ; \\Delta G^{\\circ}=-56.2 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}} ; K=6.8 \\times 10^{9}$.\n"} {"id": "textbook_5407", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "\nMost of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose."} {"id": "textbook_5408", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$"} {"id": "textbook_5409", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "Notice the reaction quotient, $Q$, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:"} {"id": "textbook_5410", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "$$\n\\begin{gathered}\n-n F E_{\\mathrm{cell}}=-n F E_{\\mathrm{cell}}^{\\circ}+R T \\ln Q \\\\\nE_{\\mathrm{cell}}=E_{\\mathrm{cell}}^{\\circ}-\\frac{R T}{n F} \\ln Q\n\\end{gathered}\n$$"} {"id": "textbook_5411", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, $n$, the temperature, $T$, and the reaction mixture composition as reflected in $Q$. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants ( $R$ and $F$ ) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:"} {"id": "textbook_5412", "title": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - ", "content": "$$\nE_{\\mathrm{cell}}=E_{\\mathrm{cell}}^{\\circ}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q\n$$\n"} {"id": "textbook_5413", "title": "1730. EXAMPLE 16.7 - ", "content": ""} {"id": "textbook_5414", "title": "1731. Predicting Redox Spontaneity Under Nonstandard Conditions - ", "content": "\nUse the Nernst equation to predict the spontaneity of the redox reaction shown below."} {"id": "textbook_5415", "title": "1731. Predicting Redox Spontaneity Under Nonstandard Conditions - ", "content": "$$\n\\mathrm{Co}(s)+\\mathrm{Fe}^{2+}(a q, 1.94 M) \\longrightarrow \\mathrm{Co}^{2+}(a q, 0.15 M)+\\mathrm{Fe}(s)\n$$\n"} {"id": "textbook_5416", "title": "1732. Solution - ", "content": "\nCollecting information from Appendix L and the problem,"} {"id": "textbook_5417", "title": "1732. Solution - ", "content": "$$\n\\begin{array}{llrl}\n\\text { Anode (oxidation): } & \\mathrm{Co}(s) \\longrightarrow \\mathrm{Co}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Co}^{2+} / \\mathrm{Co}}^{\\circ}=-0.28 \\mathrm{~V} \\\\\n\\text { Cathode (reduction): } & \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}(s) & E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=-0.447 \\mathrm{~V} \\\\\n& E_{\\mathrm{cell}}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}=-0.447 \\mathrm{~V}-(-0.28 \\mathrm{~V})=-0.17 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5418", "title": "1732. Solution - ", "content": "Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:"} {"id": "textbook_5419", "title": "1732. Solution - ", "content": "$$\n\\begin{aligned}\nQ & =\\frac{\\left[\\mathrm{Co}^{2+}\\right]}{\\left[\\mathrm{Fe}^{2+}\\right]}=\\frac{0.15 M}{1.94 M}=0.077 \\\\\nE_{\\text {cell }} & =E_{\\text {cell }}^{\\circ}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q \\\\\nE_{\\text {cell }} & =-0.17 \\mathrm{~V}-\\frac{0.0592 \\mathrm{~V}}{2} \\log 0.077 \\\\\nE_{\\text {cell }} & =-0.17 \\mathrm{~V}+0.033 \\mathrm{~V}=-0.14 \\mathrm{~V}\n\\end{aligned}\n$$"} {"id": "textbook_5420", "title": "1732. Solution - ", "content": "The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.\n"} {"id": "textbook_5421", "title": "1733. Check Your Learning - ", "content": "\nFor the cell schematic below, identify values for $n$ and $Q$, and calculate the cell potential, $E_{\\text {cell }}$ -"} {"id": "textbook_5422", "title": "1733. Check Your Learning - ", "content": "$$\n\\mathrm{Al}(s)\\left|\\mathrm{Al}^{3+}(a q, 0.15 M) \\| \\mathrm{Cu}^{2+}(a q, 0.025 M)\\right| \\mathrm{Cu}(s)\n$$\n"} {"id": "textbook_5423", "title": "1734. Answer: - ", "content": "\n$n=6 ; Q=1440 ; E_{\\text {cell }}=+1.97 \\mathrm{~V}$, spontaneous."} {"id": "textbook_5424", "title": "1734. Answer: - ", "content": "A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same halfreaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.\n"} {"id": "textbook_5425", "title": "1735. EXAMPLE 16.8 - ", "content": ""} {"id": "textbook_5426", "title": "1736. Concentration Cells - ", "content": "\nWhat is the cell potential of the concentration cell described by"} {"id": "textbook_5427", "title": "1736. Concentration Cells - ", "content": "$$\n\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q, 0.10 M) \\| \\mathrm{Zn}^{2+}(a q, 0.50 M)\\right| \\mathrm{Zn}(s)\n$$\n"} {"id": "textbook_5428", "title": "1737. Solution - ", "content": "\nFrom the information given:"} {"id": "textbook_5429", "title": "1737. Solution - ", "content": "| Anode: | $\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q, 0.10 M)+2 \\mathrm{e}^{-}$ | $E_{\\text {anode }}^{\\circ}$ | $=-0.7618 \\mathrm{~V}$ |\n| :--- | :--- | :--- | :--- |\n| Cathode: | $\\mathrm{Zn}^{2+}(a q, 0.50 M)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)$ | $E_{\\text {cathode }}^{\\circ}$ | $=-0.7618 \\mathrm{~V}$ |\n| Overall: | $\\mathrm{Zn}^{2+}(a q, 0.50 M) \\longrightarrow \\mathrm{Zn}^{2+}(a q, 0.10 M)$ | $E_{\\text {cell }}^{\\circ}$ | $=0.000 \\mathrm{~V}$ |"} {"id": "textbook_5430", "title": "1737. Solution - ", "content": "Substituting into the Nernst equation,"} {"id": "textbook_5431", "title": "1737. Solution - ", "content": "$$\nE_{\\text {cell }}=0.000 \\mathrm{~V}-\\frac{0.0592 \\mathrm{~V}}{2} \\log \\frac{0.10}{0.50}=+0.021 \\mathrm{~V}\n$$"} {"id": "textbook_5432", "title": "1737. Solution - ", "content": "The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater ( $E_{\\text {cathode }}>E_{\\text {anode }}$ ).\n"} {"id": "textbook_5433", "title": "1738. Check Your Learning - ", "content": "\nThe concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?\n"} {"id": "textbook_5434", "title": "1739. Answer: - ", "content": "\n$E_{\\text {cell }}=0.000 \\mathrm{~V} ;\\left[\\mathrm{Zn}^{2+}\\right]_{\\text {cathode }}=\\left[\\mathrm{Zn}^{2+}\\right]_{\\text {anode }}=0.30 \\mathrm{M}$\n"} {"id": "textbook_5435", "title": "1739. Answer: - 1739.1. Batteries and Fuel Cells", "content": ""} {"id": "textbook_5436", "title": "1740. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5437", "title": "1740. LEARNING OBJECTIVES - ", "content": "- Describe the electrochemistry associated with several common batteries\n- Distinguish the operation of a fuel cell from that of a battery"} {"id": "textbook_5438", "title": "1740. LEARNING OBJECTIVES - ", "content": "There are many technological products associated with the past two centuries of electrochemistry research, none more immediately obvious than the battery. A battery is a galvanic cell that has been specially designed and constructed in a way that best suits its intended use a source of electrical power for specific applications. Among the first successful batteries was the Daniell cell, which relied on the spontaneous oxidation of zinc by copper(II) ions (Figure 16.8):"} {"id": "textbook_5439", "title": "1740. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{Zn}(s)+\\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{Cu}(s)\n$$"} {"id": "textbook_5440", "title": "1740. LEARNING OBJECTIVES - ", "content": ""} {"id": "textbook_5441", "title": "1740. LEARNING OBJECTIVES - ", "content": "FIGURE 16.8 Illustration of a Daniell cell taken from a 1904 journal publication (left) along with a simplified illustration depicting the electrochemistry of the cell (right). The 1904 design used a porous clay pot to both contain one of the half-cell's content and to serve as a salt bridge to the other half-cell."} {"id": "textbook_5442", "title": "1740. LEARNING OBJECTIVES - ", "content": "Modern batteries exist in a multitude of forms to accommodate various applications, from tiny button batteries that provide the modest power needs of a wristwatch to the very large batteries used to supply backup energy to municipal power grids. Some batteries are designed for single-use applications and cannot be recharged\n(primary cells), while others are based on conveniently reversible cell reactions that allow recharging by an external power source (secondary cells). This section will provide a summary of the basic electrochemical aspects of several batteries familiar to most consumers, and will introduce a related electrochemical device called a fuel cell that can offer improved performance in certain applications.\n"} {"id": "textbook_5443", "title": "1741. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16batteries) to learn more about batteries.\n"} {"id": "textbook_5444", "title": "1742. Single-Use Batteries - ", "content": "\nA common primary battery is the dry cell, which uses a zinc can as both container and anode (\"-\" terminal) and a graphite rod as the cathode (\"+\" terminal). The Zn can is filled with an electrolyte paste containing manganese(IV) oxide, zinc(II) chloride, ammonium chloride, and water. A graphite rod is immersed in the electrolyte paste to complete the cell. The spontaneous cell reaction involves the oxidation of zinc:"} {"id": "textbook_5445", "title": "1742. Single-Use Batteries - ", "content": "$$\n\\text { anode reaction: } \\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+2 \\mathrm{e}^{-}\n$$"} {"id": "textbook_5446", "title": "1742. Single-Use Batteries - ", "content": "and the reduction of manganese(IV)\nreduction reaction: $2 \\mathrm{MnO}_{2}(s)+2 \\mathrm{NH}_{4} \\mathrm{Cl}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-}$\nwhich together yield the cell reaction:\ncell reaction: $2 \\mathrm{MnO}_{2}(s)+2 \\mathrm{NH}_{4} \\mathrm{Cl}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-} E_{\\text {cell }} \\sim 1.5 \\mathrm{~V}$\nThe voltage (cell potential) of a dry cell is approximately 1.5 V . Dry cells are available in various sizes (e.g., D, C, AA, AAA). All sizes of dry cells comprise the same components, and so they exhibit the same voltage, but larger cells contain greater amounts of the redox reactants and therefore are capable of transferring correspondingly greater amounts of charge. Like other galvanic cells, dry cells may be connected in series to yield batteries with greater voltage outputs, if needed.\n"} {"id": "textbook_5447", "title": "1742. Single-Use Batteries - ", "content": "FIGURE 16.9 A schematic diagram shows a typical dry cell.\n"} {"id": "textbook_5448", "title": "1743. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16zinccarbon) to learn more about zinc-carbon batteries."} {"id": "textbook_5449", "title": "1743. LINK TO LEARNING - ", "content": "Alkaline batteries (Figure 16.10) were developed in the 1950s to improve on the performance of the dry cell,\nand they were designed around the same redox couples. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are"} {"id": "textbook_5450", "title": "1743. LINK TO LEARNING - ", "content": "| $\\mathrm{Zn}(s)+2 \\mathrm{OH}^{-}(a q)$
anode:
cathode: $2 \\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{OH}^{-}(a q)$ |\n| :--- |\n| cell: |"} {"id": "textbook_5451", "title": "1743. LINK TO LEARNING - ", "content": "An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so they should be removed from devices for longterm storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.\n"} {"id": "textbook_5452", "title": "1743. LINK TO LEARNING - ", "content": "FIGURE 16.10 Alkaline batteries were designed as improved replacements for zinc-carbon (dry cell) batteries.\n"} {"id": "textbook_5453", "title": "1744. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16alkaline) to learn more about alkaline batteries.\n"} {"id": "textbook_5454", "title": "1745. Rechargeable (Secondary) Batteries - ", "content": "\nNickel-cadmium, or NiCd, batteries (Figure 16.11) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a \"jelly-roll\" design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are"} {"id": "textbook_5455", "title": "1745. Rechargeable (Secondary) Batteries - ", "content": "$$\n\\left.\\begin{array}{l}\n\\text { anode: } \\quad \\mathrm{Cd}(s)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Cd}(\\mathrm{OH})_{2}(s)+2 \\mathrm{e}^{-} \\\\\n\\text {cathode: } \\mathrm{NiO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(\\mathrm{OH})_{2}(s)+2 \\mathrm{OH}^{-}(a q)\n\\end{array}\\right]\n$$"} {"id": "textbook_5456", "title": "1745. Rechargeable (Secondary) Batteries - ", "content": "When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be ruptured or incinerated, and they should be disposed of in accordance with relevant toxic waste guidelines.\n"} {"id": "textbook_5457", "title": "1745. Rechargeable (Secondary) Batteries - ", "content": "FIGURE 16.11 NiCd batteries use a \"jelly-roll\" design that significantly increases the amount of current the battery can deliver as compared to a similar-sized alkaline battery.\n"} {"id": "textbook_5458", "title": "1746. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16NiCdrecharge) for more information about nickel cadmium rechargeable batteries."} {"id": "textbook_5459", "title": "1746. LINK TO LEARNING - ", "content": "Lithium ion batteries (Figure 16.12) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are"} {"id": "textbook_5460", "title": "1746. LINK TO LEARNING - ", "content": "$$\n\\begin{array}{lll}\n\\text { anode: } & \\mathrm{LiCoO}_{2} \\rightleftharpoons \\mathrm{Li}_{1-x} \\mathrm{CoO}_{2}+x \\mathrm{Li}^{+}+x \\mathrm{e}^{-} \\\\\n\\text {cathode: } x \\mathrm{Li}^{+}+x \\mathrm{e}^{-}+x \\mathrm{C}_{6} \\rightleftharpoons x \\mathrm{LiC}_{6} & \\\\\n\\hline \\text { cell: } & \\mathrm{LiCoO}_{2}+x \\mathrm{C}_{6} \\rightleftharpoons \\mathrm{Li}_{1-x} \\mathrm{CoO}_{2}+x \\mathrm{LiC}_{6} & E_{\\text {cell }} \\sim 3.7 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5461", "title": "1746. LINK TO LEARNING - ", "content": "The variable stoichiometry of the cell reaction leads to variation in cell voltages, but for typical conditions, $x$ is usually no more than 0.5 and the cell voltage is approximately 3.7 V . Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.\n"} {"id": "textbook_5462", "title": "1746. LINK TO LEARNING - ", "content": "FIGURE 16.12 In a lithium ion battery, charge flows as the lithium ions are transferred between the anode and cathode.\n"} {"id": "textbook_5463", "title": "1747. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/l/16lithiumion) for more information about lithium ion batteries."} {"id": "textbook_5464", "title": "1747. LINK TO LEARNING - ", "content": "The lead acid battery (Figure 16.13) is the type of secondary battery commonly used in automobiles. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are"} {"id": "textbook_5465", "title": "1747. LINK TO LEARNING - ", "content": "$$\n\\begin{aligned}\n& \\text { anode: } \\mathrm{Pb}(s)+\\mathrm{HSO}_{4}^{-}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s)+\\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\\\\n& \\text {cathode: } \\mathrm{PbO}_{2}(s)+\\mathrm{HSO}_{4}^{-}(a q)+3 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n& \\hline \\text { cell: }\n\\end{aligned} \\quad \\mathrm{Pb}(s)+\\mathrm{PbO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow 2 \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad E_{\\text {cell }} \\sim 2 \\mathrm{~V} \\text { lel }\n$$"} {"id": "textbook_5466", "title": "1747. LINK TO LEARNING - ", "content": "Each cell produces 2 V , so six cells are connected in series to produce a $12-\\mathrm{V}$ car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, $\\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$, but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.\n"} {"id": "textbook_5467", "title": "1747. LINK TO LEARNING - ", "content": "FIGURE 16.13 The lead acid battery in your automobile consists of six cells connected in series to give 12 V .\n"} {"id": "textbook_5468", "title": "1748. LINK TO LEARNING - ", "content": "\nVisit this site (http://openstax.org/1/16leadacid) for more information about lead acid batteries.\n"} {"id": "textbook_5469", "title": "1749. Fuel Cells - ", "content": "\nA fuel cell is a galvanic cell that uses traditional combustive fuels, most often hydrogen or methane, that are continuously fed into the cell along with an oxidant. (An alternative, but not very popular, name for a fuel cell is a flow battery.) Within the cell, fuel and oxidant undergo the same redox chemistry as when they are combusted, but via a catalyzed electrochemical that is significantly more efficient. For example, a typical hydrogen fuel cell uses graphite electrodes embedded with platinum-based catalysts to accelerate the two halfcell reactions:\n"} {"id": "textbook_5470", "title": "1749. Fuel Cells - ", "content": "FIGURE 16.14 In this hydrogen fuel cell, oxygen from the air reacts with hydrogen, producing water and electricity."} {"id": "textbook_5471", "title": "1749. Fuel Cells - ", "content": "$$\n\\begin{aligned}\n& \\text { Anode: } \\quad 2 \\mathrm{H}_{2}(g) \\longrightarrow 4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\\\\n& \\text {Cathode: } \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) \\\\\n& \\hline \\text { Cell: } \\quad 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\\\\n& \\hline \\text { cell } \\sim 1.2 \\mathrm{~V}\n\\end{aligned}\n$$"} {"id": "textbook_5472", "title": "1749. Fuel Cells - ", "content": "These types of fuel cells generally produce voltages of approximately 1.2 V . Compared to an internal combustion engine, the energy efficiency of a fuel cell using the same redox reaction is typically more than double ( $\\sim 20 \\%-25 \\%$ for an engine versus $\\sim 50 \\%-75 \\%$ for a fuel cell). Hydrogen fuel cells are commonly used on extended space missions, and prototypes for personal vehicles have been developed, though the technology remains relatively immature.\n"} {"id": "textbook_5473", "title": "1750. LINK TO LEARNING - ", "content": "\nCheck out this link (http://openstax.org/l/16fuelcells) to learn more about fuel cells.\n"} {"id": "textbook_5474", "title": "1750. LINK TO LEARNING - 1750.1. Corrosion", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_5475", "title": "1750. LINK TO LEARNING - 1750.1. Corrosion", "content": "- Define corrosion\n- List some of the methods used to prevent or slow corrosion"} {"id": "textbook_5476", "title": "1750. LINK TO LEARNING - 1750.1. Corrosion", "content": "Corrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year.\n"} {"id": "textbook_5477", "title": "1751. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_5478", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "\nThe Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure 16.15). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper \"skin.\" So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide ( $\\mathrm{Cu}_{2} \\mathrm{O}$ ), which is red, and then to copper(II) oxide, which is black"} {"id": "textbook_5479", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "$$\n\\begin{aligned}\n2 \\mathrm{Cu}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) & \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{O}(s) \\\\\n\\mathrm{Cu}_{2} \\mathrm{O}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) & \\longrightarrow 2 \\mathrm{CuO}(s)\n\\end{aligned}\n$$"} {"id": "textbook_5480", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the CuO"} {"id": "textbook_5481", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "$$\n\\begin{gathered}\n2 \\mathrm{CuO}(s)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{CO}_{3}(\\mathrm{OH})_{2}(s) \\\\\n3 \\mathrm{CuO}(s)+2 \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{2}\\left(\\mathrm{CO}_{3}\\right)_{2}(\\mathrm{OH})_{2}(s) \\\\\n4 \\mathrm{CuO}(s)+\\mathrm{SO}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{4} \\mathrm{SO}_{4}(\\mathrm{OH})_{6}(s)\n\\end{gathered}\n$$"} {"id": "textbook_5482", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "These three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation, a phenomenon discussed further in another chapter of this text.\n"} {"id": "textbook_5483", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "FIGURE 16.15 (a) The Statue of Liberty is covered with a copper skin, and was originally brown, as shown in this painting. (b) Exposure to the elements has resulted in the formation of the blue-green patina seen today."} {"id": "textbook_5484", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure 16.15. The relevant redox reactions are described by the following equations:"} {"id": "textbook_5485", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "$$\n\\begin{array}{llr}\n\\text { anode: } & \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=-0.44 \\mathrm{~V} \\\\\n\\text { cathode: } & \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l) & E_{\\mathrm{O}_{2} / \\mathrm{O}^{2}}^{\\circ}=+1.23 \\mathrm{~V} \\\\\n\\text { overall: } & 2 \\mathrm{Fe}(s)+\\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q) \\longrightarrow 2 \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) & E_{\\mathrm{cell}}^{\\circ}=+1.67 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5486", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:"} {"id": "textbook_5487", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "$$\n4 \\mathrm{Fe}^{2+}(a q)+\\mathrm{O}_{2}(g)+(4+2 x) \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3} \\cdot x \\mathrm{H}_{2} \\mathrm{O}(s)+8 \\mathrm{H}^{+}(a q)\n$$"} {"id": "textbook_5488", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "The stoichiometry of the hydrate varies, as indicated by the use of $x$ in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.\n"} {"id": "textbook_5489", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "FIGURE 16.16 Corrosion can occur when a painted iron or steel surface is exposed to the environment by a scratch through the paint. A galvanic cell results that may be approximated by the simplified cell schematic $\\mathrm{Fe}(s) \\mid \\mathrm{Fe}^{2+}(\\mathrm{aq})$ $\\| \\mathrm{O}_{2}(\\mathrm{aq}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{D}) \\mid \\mathrm{Fe}(\\mathrm{s})$."} {"id": "textbook_5490", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion."} {"id": "textbook_5491", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Other strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron."} {"id": "textbook_5492", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Iron and other metals may also be protected from corrosion by galvanization, a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph."} {"id": "textbook_5493", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure 16.17). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.\n"} {"id": "textbook_5494", "title": "1752. Statue of Liberty: Changing Colors - ", "content": "FIGURE 16.17 Cathodic protection is a useful approach to electrochemically preventing corrosion of underground storage tanks.\n"} {"id": "textbook_5495", "title": "1752. Statue of Liberty: Changing Colors - 1752.1. Electrolysis", "content": ""} {"id": "textbook_5496", "title": "1753. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5497", "title": "1753. LEARNING OBJECTIVES - ", "content": "- Describe the process of electrolysis\n- Compare the operation of electrolytic cells with that of galvanic cells\n- Perform stoichiometric calculations for electrolytic processes"} {"id": "textbook_5498", "title": "1753. LEARNING OBJECTIVES - ", "content": "Electrochemical cells in which spontaneous redox reactions take place (galvanic cells) have been the topic of discussion so far in this chapter. In these cells, electrical work is done by a redox system on its surroundings as electrons produced by the redox reaction are transferred through an external circuit. This final section of the chapter will address an alternative scenario in which an external circuit does work on a redox system by imposing a voltage sufficient to drive an otherwise nonspontaneous reaction, a process known as electrolysis. A familiar example of electrolysis is recharging a battery, which involves use of an external power source to drive the spontaneous (discharge) cell reaction in the reverse direction, restoring to some extent the composition of the half-cells and the voltage of the battery. Perhaps less familiar is the use of electrolysis in the refinement of metallic ores, the manufacture of commodity chemicals, and the electroplating of metallic coatings on various products (e.g., jewelry, utensils, auto parts). To illustrate the essential concepts of electrolysis, a few specific processes will be considered.\n"} {"id": "textbook_5499", "title": "1754. The Electrolysis of Molten Sodium Chloride - ", "content": "\nMetallic sodium, Na , and chlorine gas, $\\mathrm{Cl}_{2}$, are used in numerous applications, and their industrial production relies on the large-scale electrolysis of molten sodium chloride, $\\mathrm{NaCl}(1)$. The industrial process typically uses a Downs cell similar to the simplified illustration shown in Figure 16.18. The reactions associated with this process are:"} {"id": "textbook_5500", "title": "1754. The Electrolysis of Molten Sodium Chloride - ", "content": "| anode: | $2 \\mathrm{Cl}^{-}(l) \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-}$ |\n| :--- | :---: |\n| cathode: | $\\mathrm{Na}^{+}(l)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(l)$ |\n| cell: | $2 \\mathrm{Na}^{+}(l)+2 \\mathrm{Cl}^{-}(l) \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)$ |"} {"id": "textbook_5501", "title": "1754. The Electrolysis of Molten Sodium Chloride - ", "content": "The cell potential for the above process is negative, indicating the reaction as written (decomposition of liquid $\\mathrm{NaCl})$ is not spontaneous. To force this reaction, a positive potential of magnitude greater than the negative cell potential must be applied to the cell.\n"} {"id": "textbook_5502", "title": "1754. The Electrolysis of Molten Sodium Chloride - ", "content": "FIGURE 16.18 Cells of this sort (a cell for the electrolysis of molten sodium chloride) are used in the Downs process for production of sodium and chlorine, and they typically use iron cathodes and carbon anodes.\n"} {"id": "textbook_5503", "title": "1755. The Electrolysis of Water - ", "content": "\nWater may be electrolytically decomposed in a cell similar to the one illustrated in Figure 16.19. To improve electrical conductivity without introducing a different redox species, the hydrogen ion concentration of the water is typically increased by addition of a strong acid. The redox processes associated with this cell are"} {"id": "textbook_5504", "title": "1755. The Electrolysis of Water - ", "content": "| anode: | $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-}$ | $E_{\\text {anode }}^{\\circ}$ |\n| :--- | ---: | :--- |\n| cathode: $2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)$ | $E_{\\text {cathode }}^{\\circ}$ | $=0 \\mathrm{~V}$ |\n| cell: | $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 229 \\mathrm{~V}$ | |\n| | | $E_{\\text {cell }}^{\\circ}$ |"} {"id": "textbook_5505", "title": "1755. The Electrolysis of Water - ", "content": "Again, the cell potential as written is negative, indicating a nonspontaneous cell reaction that must be driven by imposing a cell voltage greater than +1.229 V . Keep in mind that standard electrode potentials are used to inform thermodynamic predictions here, though the cell is not operating under standard state conditions. Therefore, at best, calculated cell potentials should be considered ballpark estimates.\n"} {"id": "textbook_5506", "title": "1755. The Electrolysis of Water - ", "content": "FIGURE 16.19 The electrolysis of water produces stoichiometric amounts of oxygen gas at the anode and hydrogen at the anode.\n"} {"id": "textbook_5507", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "\nWhen aqueous solutions of ionic compounds are electrolyzed, the anode and cathode half-reactions may involve the electrolysis of either water species $\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}^{+}, \\mathrm{OH}^{-}\\right)$or solute species (the cations and anions of the compound). As an example, the electrolysis of aqueous sodium chloride could involve either of these two anode reactions:"} {"id": "textbook_5508", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "$$\n\\begin{array}{ll}\n\\text { (i) } 2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} & E_{\\text {anode }}^{\\circ}=+1.35827 \\mathrm{~V} \\\\\n\\text { (ii) } 2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} & E_{\\text {anode }}^{\\circ}=+1.229 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5509", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "The standard electrode (reduction) potentials of these two half-reactions indicate water may be oxidized at a less negative/more positive potential $(-1.229 \\mathrm{~V})$ than chloride ion $(-1.358 \\mathrm{~V})$. Thermodynamics thus predicts that water would be more readily oxidized, though in practice it is observed that both water and chloride ion are oxidized under typical conditions, producing a mixture of oxygen and chlorine gas."} {"id": "textbook_5510", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "Turning attention to the cathode, the possibilities for reduction are:"} {"id": "textbook_5511", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "$$\n\\begin{array}{ll}\n\\text { (iii) } 2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g) & E_{\\text {cathode }}^{\\circ}=0 \\mathrm{~V} \\\\\n\\text { (iv) } 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)+2 \\mathrm{OH}^{-}(a q) & E_{\\text {cathode }}^{\\circ}=-0.8277 \\mathrm{~V} \\\\\n\\text { (v) } \\mathrm{Na}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(s) & E_{\\text {cathode }}^{\\circ}=-2.71 \\mathrm{~V}\n\\end{array}\n$$"} {"id": "textbook_5512", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "Comparison of these standard half-reaction potentials suggests the reduction of hydrogen ion is\nthermodynamically favored. However, in a neutral aqueous sodium chloride solution, the concentration of hydrogen ion is far below the standard state value of 1 M (approximately $10^{-7} \\mathrm{M}$ ), and so the observed cathode reaction is actually reduction of water. The net cell reaction in this case is then"} {"id": "textbook_5513", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "$$\n\\text { cell: } 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g)+2 \\mathrm{OH}^{-}(a q) \\quad E_{\\mathrm{cell}}^{\\circ}=-2.186 \\mathrm{~V}\n$$"} {"id": "textbook_5514", "title": "1756. The Electrolysis of Aqueous Sodium Chloride - ", "content": "This electrolysis reaction is part of the chlor-alkali process used by industry to produce chlorine and sodium hydroxide (lye).\n"} {"id": "textbook_5515", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "\nAn important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. The silver plating of eating utensils is used here to illustrate the process. (Figure 16.20).\n"} {"id": "textbook_5516", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "FIGURE 16.20 This schematic shows an electrolytic cell for silver plating eating utensils.\nIn the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. Applying a sufficient potential results in the oxidation of the silver anode"} {"id": "textbook_5517", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "$$\n\\text { anode }: \\mathrm{Ag}(s) \\longrightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-}\n$$"} {"id": "textbook_5518", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "and reduction of silver ion at the (spoon) cathode:"} {"id": "textbook_5519", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "$$\n\\text { cathode: } \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)\n$$"} {"id": "textbook_5520", "title": "1757. Chemistry in Everyday Life Electroplating - ", "content": "The net result is the transfer of silver metal from the anode to the cathode. Several experimental factors must be carefully controlled to obtain high-quality silver coatings, including the exact composition of the electrolyte solution, the cell voltage applied, and the rate of the electrolysis reaction (electrical current).\n"} {"id": "textbook_5521", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "\nElectrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second ( $\\mathrm{A}=1 \\mathrm{C} / \\mathrm{s}$ ). The charge transferred, $Q$, by passage of a constant current, $I$, over a specified time interval, $t$, is then given by the simple mathematical product"} {"id": "textbook_5522", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "$$\nQ=I t\n$$"} {"id": "textbook_5523", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "When electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process"} {"id": "textbook_5524", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "$$\n\\mathrm{M}^{\\mathrm{n}+}(a q)+\\mathrm{ne}^{-} \\longrightarrow \\mathrm{M}(s)\n$$"} {"id": "textbook_5525", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "involves the transfer of $n$ mole of electrons. The charge transferred is, therefore,"} {"id": "textbook_5526", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "$$\nQ=n F\n$$"} {"id": "textbook_5527", "title": "1758. Quantitative Aspects of Electrolysis - ", "content": "where $F$ is Faraday's constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction.\n"} {"id": "textbook_5528", "title": "1759. EXAMPLE 16.9 - ", "content": ""} {"id": "textbook_5529", "title": "1760. Converting Current to Moles of Electrons - ", "content": "\nIn one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?\n"} {"id": "textbook_5530", "title": "1761. Solution - ", "content": "\nFaraday's constant can be used to convert the charge ( $Q$ ) into moles of electrons (n). The charge is the current (I) multiplied by the time"} {"id": "textbook_5531", "title": "1761. Solution - ", "content": "$$\nn=\\frac{Q}{F}=\\frac{\\frac{10.23 \\mathrm{C}}{\\mathrm{~s}} \\times 1 \\mathrm{hr} \\times \\frac{60 \\mathrm{~min}}{\\mathrm{hr}} \\times \\frac{60 \\mathrm{~s}}{\\mathrm{~min}}}{96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}}=\\frac{36,830 \\mathrm{C}}{96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}}=0.3817 \\mathrm{~mol} \\mathrm{e}^{-}\n$$"} {"id": "textbook_5532", "title": "1761. Solution - ", "content": "From the problem, the solution contains $\\mathrm{AgNO}_{3}$, so the reaction at the cathode involves 1 mole of electrons for each mole of silver"} {"id": "textbook_5533", "title": "1761. Solution - ", "content": "$$\n\\text { cathode: } \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\operatorname{Ag}(s)\n$$"} {"id": "textbook_5534", "title": "1761. Solution - ", "content": "The atomic mass of silver is $107.9 \\mathrm{~g} / \\mathrm{mol}$, so"} {"id": "textbook_5535", "title": "1761. Solution - ", "content": "$$\n\\text { mass } \\mathrm{Ag}=0.3817 \\mathrm{~mol} \\mathrm{e} ~-~ \\times \\frac{1 \\mathrm{~mol} \\mathrm{Ag}}{1 \\mathrm{~mol} \\mathrm{e}^{-}} \\times \\frac{107.9 \\mathrm{~g} \\mathrm{Ag}}{1 \\mathrm{~mol} \\mathrm{Ag}}=41.19 \\mathrm{~g} \\mathrm{Ag}\n$$\n"} {"id": "textbook_5536", "title": "1762. Check Your Learning - ", "content": "\nAluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?"} {"id": "textbook_5537", "title": "1762. Check Your Learning - ", "content": "Answer:\n$\\mathrm{Al}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}(s) ; 0.0777 \\mathrm{~mol} \\mathrm{Al}=2.10 \\mathrm{~g} \\mathrm{Al}$.\n"} {"id": "textbook_5538", "title": "1763. EXAMPLE 16.10 - ", "content": ""} {"id": "textbook_5539", "title": "1764. Time Required for Deposition - ", "content": "\nIn one application, a $0.010-\\mathrm{mm}$ layer of chromium must be deposited on a part with a total surface area of 3.3 $\\mathrm{m}^{2}$ from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is $7.19 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n"} {"id": "textbook_5540", "title": "1765. Solution - ", "content": "\nFirst, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):"} {"id": "textbook_5541", "title": "1765. Solution - ", "content": "$$\n\\text { volume }=\\left(0.010 \\mathrm{~mm} \\times \\frac{1 \\mathrm{~cm}}{10 \\mathrm{~mm}}\\right) \\times\\left(3.3 \\mathrm{~m}^{2} \\times\\left(\\frac{10,000 \\mathrm{~cm}^{2}}{1 \\mathrm{~m}^{2}}\\right)\\right)=33 \\mathrm{~cm}^{3}\n$$"} {"id": "textbook_5542", "title": "1765. Solution - ", "content": "Use the computed volume and the provided density to calculate the molar amount of chromium required:"} {"id": "textbook_5543", "title": "1765. Solution - ", "content": "$$\n\\begin{gathered}\n\\text { mass }=\\text { volume } \\times \\text { density }=33 \\mathrm{~cm}^{3} \\times \\frac{7.19 \\mathrm{~g}}{\\mathrm{~cm}^{3}}=237 \\mathrm{~g} \\mathrm{Cr} \\\\\n\\\\\n\\mathrm{~mol} \\mathrm{Cr}=237 \\mathrm{~g} \\mathrm{Cr} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Cr}}{52.00 \\mathrm{~g} \\mathrm{Cr}}=4.56 \\mathrm{~mol} \\mathrm{Cr}\n\\end{gathered}\n$$"} {"id": "textbook_5544", "title": "1765. Solution - ", "content": "The stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:"} {"id": "textbook_5545", "title": "1765. Solution - ", "content": "$$\nQ=4.56 \\mathrm{~mol} \\mathrm{Cr} \\times \\frac{3 \\mathrm{~mol} \\mathrm{e}^{-}}{1 \\mathrm{~mol} \\mathrm{Cr}} \\times \\frac{96485 \\mathrm{C}}{\\mathrm{~mol} \\mathrm{e}^{-}}=1.32 \\times 10^{6} \\mathrm{C}\n$$"} {"id": "textbook_5546", "title": "1765. Solution - ", "content": "Finally, if this charge is passed at a rate of $33.46 \\mathrm{C} / \\mathrm{s}$, the required time is:"} {"id": "textbook_5547", "title": "1765. Solution - ", "content": "$$\nt=\\frac{Q}{I}=\\frac{1.32 \\times 10^{6} \\mathrm{C}}{33.46 \\mathrm{C} / \\mathrm{s}}=3.95 \\times 10^{4} \\mathrm{~s}=11.0 \\mathrm{hr}\n$$\n"} {"id": "textbook_5548", "title": "1766. Check Your Learning - ", "content": "\nWhat mass of zinc is required to galvanize the top of a $3.00 \\mathrm{~m} \\times 5.50 \\mathrm{~m}$ sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of $\\mathrm{Zn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and the current is 25.5 A , how long will it take to galvanize the top of the iron? The density of zinc is $7.140 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n"} {"id": "textbook_5549", "title": "1767. Answer: - ", "content": "\n11.8 kg Zn requires 382 hours.\n"} {"id": "textbook_5550", "title": "1768. Key Terms - ", "content": "\nactive electrode electrode that participates as a reactant or product in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidationreduction reaction\nalkaline battery primary battery similar to a dry cell that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an improved replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell\nanode electrode in an electrochemical cell at which oxidation occurs\nbattery single or series of galvanic cells designed for use as a source of electrical power\ncathode electrode in an electrochemical cell at which reduction occurs\ncathodic protection approach to preventing corrosion of a metal object by connecting it to a sacrificial anode composed of a more readily oxidized metal\ncell notation (schematic) symbolic representation of the components and reactions in an electrochemical cell\ncell potential ( $\\boldsymbol{E}_{\\text {cell }}$ ) difference in potential of the cathode and anode half-cells\nconcentration cell galvanic cell comprising halfcells of identical composition but for the concentration of one redox reactant or product\ncorrosion degradation of metal via a natural electrochemical process\ndry cell primary battery, also called a zinc-carbon battery, based on the spontaneous oxidation of zinc by manganese(IV)\nelectrode potential ( $E_{\\mathrm{X}}$ ) the potential of a cell in which the half-cell of interest acts as a cathode when connected to the standard hydrogen electrode\nelectrolysis process using electrical energy to cause a nonspontaneous process to occur\nelectrolytic cell electrochemical cell in which an external source of electrical power is used to drive an otherwise nonspontaneous process\nFaraday's constant (F) charge on 1 mol of electrons; $F=96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}$\nfuel cell devices similar to galvanic cells that require a continuous feed of redox reactants; also called a flow battery\ngalvanic (voltaic) cell electrochemical cell in\nwhich a spontaneous redox reaction takes place; also called a voltaic cell\ngalvanization method of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.\nhalf cell component of a cell that contains the redox conjugate pair (\"couple\") of a single reactant\ninert electrode electrode that conducts electrons to and from the reactants in a half-cell but that is not itself oxidized or reduced\nlead acid battery rechargeable battery commonly used in automobiles; it typically comprises six galvanic cells based on Pb half-reactions in acidic solution\nlithium ion battery widely used rechargeable battery commonly used in portable electronic devices, based on lithium ion transfer between the anode and cathode\nNernst equation relating the potential of a redox system to its composition\nnickel-cadmium battery rechargeable battery based on Ni/Cd half-cells with applications similar to those of lithium ion batteries\nprimary cell nonrechargeable battery, suitable for single use only\nsacrificial anode electrode constructed from an easily oxidized metal, often magnesium or zinc, used to prevent corrosion of metal objects via cathodic protection\nsalt bridge tube filled with inert electrolyte solution\nsecondary cell battery designed to allow recharging\nstandard cell potential ( $E_{\\text {cell }}^{\\circ}$ ) the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K\nstandard electrode potential $\\left(\\left(E_{X}^{\\circ}\\right)\\right)$ electrode potential measured under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K\nstandard hydrogen electrode (SHE) half-cell based on hydrogen ion production, assigned a potential of exactly 0 V under standard state conditions, used as the universal reference for measuring electrode potential\n"} {"id": "textbook_5551", "title": "1769. Key Equations - ", "content": "\n$E_{\\text {cell }}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}$\n$E_{\\text {cell }}^{\\circ}=\\frac{R T}{n F} \\ln K$\n$E_{\\text {cell }}^{\\circ}=\\frac{0.0257 \\mathrm{~V}}{n} \\ln K=\\frac{0.0592 \\mathrm{~V}}{n} \\log K$\n$E_{\\text {cell }}=E_{\\text {cell }}^{\\circ}-\\frac{R T}{n F} \\ln Q \\quad$ (Nernst equation)\n$E_{\\text {cell }}=E^{\\circ}{ }_{\\text {cell }}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q$\n$\\Delta G=-n F E_{\\text {cell }}$\n$\\Delta G^{\\circ}=-n F E_{\\text {cell }}^{\\circ}$\n$w_{\\text {ele }}=w_{\\text {max }}=-n F E_{\\text {cell }}$\n$Q=I \\times t=n \\times F$\n"} {"id": "textbook_5552", "title": "1770. Summary - ", "content": ""} {"id": "textbook_5553", "title": "1770. Summary - 1770.1. Review of Redox Chemistry", "content": "\nRedox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.\n"} {"id": "textbook_5554", "title": "1770. Summary - 1770.2. Galvanic Cells", "content": "\nGalvanic cells are devices in which a spontaneous redox reaction occurs indirectly, with the oxidant and reductant redox couples contained in separate half-cells. Electrons are transferred from the reductant (in the anode half-cell) to the oxidant (in the cathode half-cell) through an external circuit, and inert solution phase ions are transferred between half-cells, through a salt bridge, to maintain charge neutrality. The construction and composition of a galvanic cell may be succinctly represented using chemical formulas and others symbols in the form of a cell schematic (cell notation).\n"} {"id": "textbook_5555", "title": "1770. Summary - 1770.3. Electrode and Cell Potentials", "content": "\nThe property of potential, $E$, is the energy associated with the separation/transfer of charge. In electrochemistry, the potentials of cells and halfcells are thermodynamic quantities that reflect the driving force or the spontaneity of their redox processes. The cell potential of an electrochemical cell is the difference in between its cathode and anode. To permit easy sharing of half-cell potential data, the standard hydrogen electrode (SHE) is\nassigned a potential of exactly 0 V and used to define a single electrode potential for any given half-cell. The electrode potential of a half-cell, $E_{X}$, is the cell potential of said half-cell acting as a cathode when connected to a SHE acting as an anode. When the half-cell is operating under standard state conditions, its potential is the standard electrode potential, $E^{\\circ}{ }_{\\mathrm{X}}$. Standard electrode potentials reflect the relative oxidizing strength of the half-reaction's reactant, with stronger oxidants exhibiting larger (more positive) $E^{o}{ }_{X}$ values. Tabulations of standard electrode potentials may be used to compute standard cell potentials, $E^{o}$ cell, for many redox reactions. The arithmetic sign of a cell potential indicates the spontaneity of the cell reaction, with positive values for spontaneous reactions and negative values for nonspontaneous reactions (spontaneous in the reverse direction).\n"} {"id": "textbook_5556", "title": "1770. Summary - 1770.4. Potential, Free Energy, and Equilibrium", "content": "\nPotential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction's cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of $K$ ). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient, $Q$, as\ndescribed by the Nernst equation.\n"} {"id": "textbook_5557", "title": "1770. Summary - 1770.5. Batteries and Fuel Cells", "content": "\nGalvanic cells designed specifically to function as electrical power supplies are called batteries. A variety of both single-use batteries (primary cells) and rechargeable batteries (secondary cells) are commercially available to serve a variety of applications, with important specifications including voltage, size, and lifetime. Fuel cells, sometimes called flow batteries, are devices that harness the energy of spontaneous redox reactions normally associated with combustion processes. Like batteries, fuel cells enable the reaction's electron transfer via an external circuit, but they require continuous input of the redox reactants (fuel and oxidant) from an external reservoir. Fuel cells are typically much more efficient in converting the energy released by the reaction to useful work in comparison to internal combustion engines.\n"} {"id": "textbook_5558", "title": "1770. Summary - 1770.6. Corrosion", "content": "\nSpontaneous oxidation of metals by natural\nelectrochemical processes is called corrosion, familiar examples including the rusting of iron and the tarnishing of silver. Corrosion process involve the creation of a galvanic cell in which different sites on the metal object function as anode and cathode, with the corrosion taking place at the anodic site. Approaches to preventing corrosion of metals include use of a protective coating of zinc (galvanization) and the use of sacrificial anodes connected to the metal object (cathodic protection).\n"} {"id": "textbook_5559", "title": "1770. Summary - 1770.7. Electrolysis", "content": "\nNonspontaneous redox processes may be forced to occur in electrochemical cells by the application of an appropriate potential using an external power source-a process known as electrolysis. Electrolysis is the basis for certain ore refining processes, the industrial production of many chemical commodities, and the electroplating of metal coatings on various products. Measurement of the current flow during electrolysis permits stoichiometric calculations.\n"} {"id": "textbook_5560", "title": "1771. Exercises - ", "content": ""} {"id": "textbook_5561", "title": "1771. Exercises - 1771.1. Review of Redox Chemistry", "content": "\n1. Identify each half-reaction below as either oxidation or reduction.\n(a) $\\mathrm{Fe}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}$\n(b) $\\mathrm{Cr} \\longrightarrow \\mathrm{Cr}^{3+}+3 \\mathrm{e}^{-}$\n(c) $\\mathrm{MnO}_{4}{ }^{2-} \\longrightarrow \\mathrm{MnO}_{4}^{-}+\\mathrm{e}^{-}$\n(d) $\\mathrm{Li}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Li}$\n2. Identify each half-reaction below as either oxidation or reduction.\n(a) $\\mathrm{Cl}^{-} \\longrightarrow \\mathrm{Cl}_{2}$\n(b) $\\mathrm{Mn}^{2+} \\longrightarrow \\mathrm{MnO}_{2}$\n(c) $\\mathrm{H}_{2} \\longrightarrow \\mathrm{H}^{+}$\n(d) $\\mathrm{NO}_{3}{ }^{-} \\longrightarrow \\mathrm{NO}$\n3. Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.\n(a) $\\mathrm{Ca} \\longrightarrow \\mathrm{Ca}^{2+}+2 \\mathrm{e}^{-}, \\mathrm{F}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{~F}^{-}$\n(b) $\\mathrm{Li} \\longrightarrow \\mathrm{Li}^{+}+\\mathrm{e}^{-}, \\mathrm{Cl}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}$\n(c) $\\mathrm{Fe} \\longrightarrow \\mathrm{Fe}^{3+}+3 \\mathrm{e}^{-}, \\mathrm{Br}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}^{-}$\n(d) $\\mathrm{Ag} \\longrightarrow \\mathrm{Ag}^{+}+\\mathrm{e}^{-}, \\mathrm{MnO}_{4}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\n4. Balance the equations below assuming they occur in an acidic solution.\n(a) $\\mathrm{H}_{2} \\mathrm{O}_{2}+\\mathrm{Sn}^{2+} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Sn}^{4+}$\n(b) $\\mathrm{PbO}_{2}+\\mathrm{Hg} \\longrightarrow \\mathrm{Hg}_{2}{ }^{2+}+\\mathrm{Pb}^{2+}$\n(c) $\\mathrm{Al}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow \\mathrm{Al}^{3+}+\\mathrm{Cr}^{3+}$\n5. Identify the oxidant and reductant of each reaction of the previous exercise.\n6. Balance the equations below assuming they occur in a basic solution.\n(a) $\\mathrm{SO}_{3}{ }^{2-}(a q)+\\mathrm{Cu}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}(a q)+\\mathrm{Cu}(\\mathrm{OH})(s)$\n(b) $\\mathrm{O}_{2}(g)+\\mathrm{Mn}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n(c) $\\mathrm{NO}_{3}{ }^{-}(a q)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{NO}(g)$\n(d) $\\mathrm{Al}(s)+\\mathrm{CrO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{Cr}(\\mathrm{OH})_{4}{ }^{-}(a q)$\n7. Identify the oxidant and reductant of each reaction of the previous exercise.\n8. Why don't hydroxide ions appear in equations for half-reactions occurring in acidic solution?\n9. Why don't hydrogen ions appear in equations for half-reactions occurring in basic solution?\n10. Why must the charge balance in oxidation-reduction reactions?\n"} {"id": "textbook_5562", "title": "1771. Exercises - 1771.2. Galvanic Cells", "content": "\n11. Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.\n(a) $\\mathrm{Mg}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s)$\n(c) $\\mathrm{Mn}(s)+\\mathrm{Sn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Sn}(s)$\n(d) $3 \\mathrm{CuNO}_{3}(a q)+\\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Au}(s)$\n12. Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.\n(a) $\\mathrm{Mg}(s)\\left|\\mathrm{Mg}^{2+}{ }^{2+}(a q) \\| \\mathrm{Cu}^{2+}(a q)\\right| \\mathrm{Cu}(s)$\n(b) $\\mathrm{Ni}(s)\\left|\\mathrm{Ni}^{2+}(a q) \\| \\mathrm{Ag}^{+}(a q)\\right| \\operatorname{Ag}(s)$\n13. Write a balanced equation for the cell reaction of each cell in the previous exercise.\n14. Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.\n(a) $\\mathrm{Al}(s)+\\mathrm{Zr}^{4+}(a q) \\longrightarrow \\mathrm{Al}^{3+}(a q)+\\mathrm{Zr}(s)$\n(b) $\\mathrm{Ag}^{+}(a q)+\\mathrm{NO}(g) \\longrightarrow \\mathrm{Ag}(s)+\\mathrm{NO}_{3}{ }^{-}(a q) \\quad$ (acidic solution)\n(c) $\\mathrm{SiO}_{3}{ }^{2-}(a q)+\\mathrm{Mg}(s) \\longrightarrow \\mathrm{Si}(s)+\\mathrm{Mg}(\\mathrm{OH})_{2}(s)$\n(basic solution)\n(d) $\\mathrm{ClO}_{3}^{-}(a q)+\\mathrm{MnO}_{2}(s) \\longrightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{MnO}_{4}^{-}(a q) \\quad$ (basic solution)\n15. Identify the oxidant and reductant in each reaction of the previous exercise.\n16. From the information provided, use cell notation to describe the following systems:\n(a) In one half-cell, a solution of $\\mathrm{Pt}\\left(\\mathrm{NO}_{3}\\right)_{2}$ forms Pt metal, while in the other half-cell, Cu metal goes into a\n$\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution with all solute concentrations 1 M .\n(b) The cathode consists of a gold electrode in a $0.55 \\mathrm{M} \\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}$ solution and the anode is a magnesium electrode in $0.75 \\mathrm{M} \\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n(c) One half-cell consists of a silver electrode in a $1 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution, and in the other half-cell, a copper electrode in $1 \\mathrm{MCu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ is oxidized.\n17. Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?\n18. An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.\n19. An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.\n20. The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode.\n"} {"id": "textbook_5563", "title": "1771. Exercises - 1771.3. Electrode and Cell Potentials", "content": "\n21. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.\n(a) $\\mathrm{Mg}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s)$\n(c) $\\mathrm{Mn}(s)+\\mathrm{Sn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Sn}(s)$\n(d) $3 \\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q) \\longrightarrow 3 \\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q)+\\mathrm{Au}(s)$\n22. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.\n(a) $\\mathrm{Mn}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Al}(s) \\longrightarrow 2 \\mathrm{Al}^{3+}(a q)+3 \\mathrm{Cu}(s)$\n(c) $\\mathrm{Na}(s)+\\mathrm{LiNO}_{3}(a q) \\longrightarrow \\mathrm{NaNO}_{3}(a q)+\\mathrm{Li}(s)$\n(d) $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Ba}(s) \\longrightarrow \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Ca}(s)$\n23. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.\n$\\mathrm{Cu}(s)\\left|\\mathrm{Cu}^{2+}(a q) \\| \\mathrm{Au}^{3+}(a q)\\right| \\mathrm{Au}(s)$\n24. Determine the cell reaction and standard cell potential at $25^{\\circ} \\mathrm{C}$ for a cell made from a cathode half-cell consisting of a silver electrode in $1 M$ silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?\n25. Determine the cell reaction and standard cell potential at $25^{\\circ} \\mathrm{C}$ for a cell made from an anode half-cell containing a cadmium electrode in 1 M cadmium nitrate and a cathode half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?\n26. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.\n$\\operatorname{Pt}(s)\\left|\\mathrm{H}_{2}(g)\\right| \\mathrm{H}^{+}(a q) \\| \\mathrm{Br}_{2}(a q), \\mathrm{Br}^{-}(a q) \\mid \\operatorname{Pt}(s)$\n"} {"id": "textbook_5564", "title": "1771. Exercises - 1771.4. Potential, Free Energy, and Equilibrium", "content": "\n27. For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).\n(a) $0.000 \\mathrm{~V}, \\mathrm{n}=2$\n(b) $+0.434 \\mathrm{~V}, \\mathrm{n}=2$\n(c) $-2.439 \\mathrm{~V}, \\mathrm{n}=1$\n28. For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.\n(a) $12 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{n}=3$\n(b) $-45 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{n}=1$\n29. Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K .\n(a) $\\mathrm{Hg}(l)+\\mathrm{S}^{2-}(a q, 0.10 M)+2 \\mathrm{Ag}^{+}(a q, 0.25 M) \\longrightarrow 2 \\mathrm{Ag}(s)+\\mathrm{HgS}(s)$\n(b) The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.\n(c) The cell comprised of a half-cell in which aqueous bromine $(1.0 \\mathrm{M})$ is being oxidized to bromide ion $(0.11 M)$ and a half-cell in which $\\mathrm{Al}^{3+}(0.023 M)$ is being reduced to aluminum metal.\n30. Determine $\\Delta G$ and $\\Delta G^{\\circ}$ for each of the reactions in the previous problem.\n31. Use the data in Appendix $L$ to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.\n(a) $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$\n(b) $\\mathrm{CdS}(s) \\rightleftharpoons \\mathrm{Cd}^{2+}(a q)+\\mathrm{S}^{2-}(a q) \\quad$ at 377 K\n(c) $\\mathrm{Hg}^{2+}(a q)+4 \\mathrm{Br}^{-}(a q) \\rightleftharpoons\\left[\\mathrm{HgBr}_{4}\\right]^{2-}(a q)$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad$ at $25^{\\circ} \\mathrm{C}$\n"} {"id": "textbook_5565", "title": "1771. Exercises - 1771.5. Batteries and Fuel Cells", "content": "\n32. Consider a battery made from one half-cell that consists of a copper electrode in $1 \\mathrm{MCuSO}_{4}$ solution and another half-cell that consists of a lead electrode in $1 \\mathrm{M} \\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n(a) What is the standard cell potential for the battery?\n(b) What are the reactions at the anode, cathode, and the overall reaction?\n(c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.\n(d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some $\\mathrm{PbSO}_{4}(s)$ forms. Would the cell potential increase, decrease, or remain the same?\n33. Consider a battery with the overall reaction: $\\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow 2 \\mathrm{Ag}(s)+\\mathrm{Cu}^{2+}(a q)$.\n(a) What is the reaction at the anode and cathode?\n(b) A battery is \"dead\" when its cell potential is zero. What is the value of $Q$ when this battery is dead?\n(c) If a particular dead battery was found to have $\\left[\\mathrm{Cu}^{2+}\\right]=0.11 \\mathrm{M}$, what was the concentration of silver ion?\n34. Why do batteries go dead, but fuel cells do not?\n35. Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.\n36. Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.\n"} {"id": "textbook_5566", "title": "1771. Exercises - 1771.6. Corrosion", "content": "\n37. Which member of each pair of metals is more likely to corrode (oxidize)?\n(a) Mg or Ca\n(b) Au or Hg\n(c) Fe or Zn\n(d) Ag or Pt\n38. Consider the following metals: $\\mathrm{Ag}, \\mathrm{Au}, \\mathrm{Mg}, \\mathrm{Ni}$, and Zn . Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use -0.447 V as the standard reduction potential for steel.\n39. Aluminum $\\left(E_{\\mathrm{Al}^{3+} / \\mathrm{Al}}^{\\circ}=-2.07 \\mathrm{~V}\\right)$ is more easily oxidized than iron $\\left(E_{\\mathrm{Fe}^{3+} / \\mathrm{Fe}}^{\\circ}=-0.477 \\mathrm{~V}\\right)$, and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. What might explain this observation?\n40. If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.\n41. Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?\n42. Why would a sacrificial anode made of lithium metal be a bad choice\n"} {"id": "textbook_5567", "title": "1771. Exercises - 1771.7. Electrolysis", "content": "\n43. If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?\n44. For the scenario in the previous question, how many electrons moved through the circuit?\n45. Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.\n(a) $\\mathrm{CaCl}_{2}$\n(b) LiH\n(c) $\\mathrm{AlCl}_{3}$\n(d) $\\mathrm{CrBr}_{3}$\n46. What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of $3.33 \\times 10^{5} \\mathrm{C}$ passes through each cell?\n47. How long would it take to reduce 1 mole of each of the following ions using the current indicated?\n(a) $\\mathrm{Al}^{3+}, 1.234 \\mathrm{~A}$\n(b) $\\mathrm{Ca}^{2+}, 22.2 \\mathrm{~A}$\n(c) $\\mathrm{Cr}^{5+}, 37.45 \\mathrm{~A}$\n(d) $\\mathrm{Au}^{3+}, 3.57 \\mathrm{~A}$\n48. A current of 2.345 A passes through the cell shown in Figure 16.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm ? (Hint: Is hydrogen the only gas present above the water?)\n49. An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a $\\mathrm{Zn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123 -mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is $7.140 \\mathrm{~g} / \\mathrm{cm}^{3}$."} {"id": "textbook_5568", "title": "1771. Exercises - 1771.7. Electrolysis", "content": "79816 \u2022 Exercises\n"} {"id": "textbook_5569", "title": "1772. CHAPTER 17 Kinetics - ", "content": "\n"} {"id": "textbook_5570", "title": "1772. CHAPTER 17 Kinetics - ", "content": "Figure 17.1 An agama lizard basks in the sun. As its body warms, the chemical reactions of its metabolism speed up.\n"} {"id": "textbook_5571", "title": "1773. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_5572", "title": "1773. CHAPTER OUTLINE - 1773.1. Chemical Reaction Rates", "content": "\n17.2 Factors Affecting Reaction Rates\n"} {"id": "textbook_5573", "title": "1773. CHAPTER OUTLINE - 1773.2. Rate Laws", "content": "\n17.4 Integrated Rate Laws\n17.5 Collision Theory\n17.6 Reaction Mechanisms\n17.7 Catalysis"} {"id": "textbook_5574", "title": "1773. CHAPTER OUTLINE - 1773.2. Rate Laws", "content": "INTRODUCTION The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun's rays is critical to the lizard's survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators."} {"id": "textbook_5575", "title": "1773. CHAPTER OUTLINE - 1773.2. Rate Laws", "content": "From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: \"Will the reaction produce the desired products in useful quantities?\" The second question is: \"How rapidly will the reaction occur?\" A third question is often asked when investigating reactions in greater detail: \"What specific molecular-level processes take place as the reaction occurs?\" Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled."} {"id": "textbook_5576", "title": "1773. CHAPTER OUTLINE - 1773.2. Rate Laws", "content": "The study of chemical kinetics concerns the second and third questions-that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors\nthat influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur.\n"} {"id": "textbook_5577", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_5578", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "- Define chemical reaction rate\n- Derive rate expressions from the balanced equation for a given chemical reaction\n- Calculate reaction rates from experimental data"} {"id": "textbook_5579", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.\n\nThe rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution's conductivity."} {"id": "textbook_5580", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, in an aqueous solution changes slowly over time as it decomposes according to the equation:"} {"id": "textbook_5581", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_5582", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:\n\n$$\n\\text { rate of decomposition of } \\begin{aligned}\n\\mathrm{H}_{2} \\mathrm{O}_{2} & =-\\frac{\\text { change in concentration of reactant }}{\\text { time interval }} \\\\\n& =-\\frac{\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{2}}-\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{1}}}{t_{2}-t_{1}} \\\\\n& =-\\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}\n\\end{aligned}\n$$\n\nThis mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta ( $\\Delta$ ) indicates \"change in.\" Thus, $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{1}}$ represents the molar concentration of hydrogen peroxide at some time $t_{1}$; likewise, $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{2}}$ represents the molar concentration of hydrogen peroxide at a later time $t_{2}$; and $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ represents the change in molar concentration of hydrogen peroxide during the time interval $\\Delta t$ (that is, $t_{2}-t_{1}$ ). Since the reactant concentration decreases as the reaction proceeds, $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by -1 . Figure 17.2 provides an example of data collected during the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$."} {"id": "textbook_5583", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "| Time (h) | $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\Delta t(\\mathrm{~h})$ | $\\begin{aligned} & \\text { Rate of Decomposition, } \\\\ & \\left(\\mathrm{mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\\right) \\end{aligned}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 0.00 | 1.000 | | | |\n| | | $\\geq-0.500$ | 6.00 | 0.0833 |\n| 6.00 | $0.500=$ | | | |\n| | | $\\geq-0.250$ | 6.00 | 0.0417 |\n| 12.00 | $0.250=$ | | | |\n| | - | $\\geq-0.125$ | 6.00 | 0.0208 |\n| 18.00 | $0.125=$ | | | |\n| 24.00 | - | $\\geq-0.062$ | 6.00 | 0.010 |"} {"id": "textbook_5584", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "FIGURE 17.2 The rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ in an aqueous solution decreases as the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases."} {"id": "textbook_5585", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of $40^{\\circ} \\mathrm{C}$. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:"} {"id": "textbook_5586", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "$$\n\\frac{-\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{-(0.500 \\mathrm{~mol} / \\mathrm{L}-1.000 \\mathrm{~mol} / \\mathrm{L})}{(6.00 \\mathrm{~h}-0.00 \\mathrm{~h})}=0.0833 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$"} {"id": "textbook_5587", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:"} {"id": "textbook_5588", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "$$\n\\frac{-\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{-(0.0625 \\mathrm{~mol} / \\mathrm{L}-0.125 \\mathrm{~mol} / \\mathrm{L})}{(24.00 \\mathrm{~h}-18.00 \\mathrm{~h})}=0.010 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$"} {"id": "textbook_5589", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at \"time zero,\" when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle's initial rate-analogous to the beginning of a chemical reaction-would be the speedometer reading at the moment the driver begins pressing the brakes ( $t_{0}$ ). A few moments later, the instantaneous rate at a specific moment-call it $t_{1}$-would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car's average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop ( $\\Delta t$ ). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates."} {"id": "textbook_5590", "title": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates", "content": "The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at any time $t$ is given by the slope of a straight line that is tangent to the curve at that time (Figure 17.3). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.\n\n\nFIGURE 17.3 This graph shows a plot of concentration versus time for a 1.000 M solution of $\\mathrm{H}_{2} \\mathrm{O}_{2}$. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at $t=0 \\mathrm{~h}$ (\"initial rate\") and at $t=12 \\mathrm{~h}$ (\"instantaneous rate\" at 12 h ).\n"} {"id": "textbook_5591", "title": "1774. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_5592", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "\nPhysicians often use disposable test strips to measure the amounts of various substances in a patient's urine (Figure 17.4). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations."} {"id": "textbook_5593", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:"} {"id": "textbook_5594", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "$$\n\\begin{gathered}\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+\\mathrm{O}_{2} \\xrightarrow{\\text { catalyst }} \\mathrm{C}_{6} \\mathrm{H}_{10} \\mathrm{O}_{6}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\\\\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{I}^{-} \\xrightarrow{\\text { catalyst }} \\mathrm{I}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\n\\end{gathered}\n$$"} {"id": "textbook_5595", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change."} {"id": "textbook_5596", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.\n"} {"id": "textbook_5597", "title": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - ", "content": "FIGURE 17.4 Test strips are commonly used to detect the presence of specific substances in a person's urine. Many test strips have several pads containing various reagents to permit the detection of multiple substances on a single strip. (credit: Iqbal Osman)\n"} {"id": "textbook_5598", "title": "1776. Relative Rates of Reaction - ", "content": "\nThe rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction"} {"id": "textbook_5599", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n\\mathrm{aA} \\longrightarrow \\mathrm{bB}\n$$"} {"id": "textbook_5600", "title": "1776. Relative Rates of Reaction - ", "content": "can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction:"} {"id": "textbook_5601", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n\\text { rate }=-\\left(\\frac{1}{\\mathrm{a}}\\right)\\left(\\frac{\\Delta \\mathrm{A}}{\\Delta t}\\right)=\\left(\\frac{1}{\\mathrm{~b}}\\right)\\left(\\frac{\\Delta \\mathrm{B}}{\\Delta t}\\right)\n$$"} {"id": "textbook_5602", "title": "1776. Relative Rates of Reaction - ", "content": "Consider the reaction represented by the following equation:"} {"id": "textbook_5603", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n2 \\mathrm{NH}_{3}(g) \\longrightarrow \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_5604", "title": "1776. Relative Rates of Reaction - ", "content": "The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:"} {"id": "textbook_5605", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n-\\frac{\\Delta \\mathrm{mol} \\mathrm{NH}_{3}}{\\Delta t} \\times \\frac{1 \\mathrm{~mol} \\mathrm{~N}_{2}}{2 \\mathrm{~mol} \\mathrm{NH}_{3}}=\\frac{\\Delta \\mathrm{mol} \\mathrm{~N}_{2}}{\\Delta t}\n$$"} {"id": "textbook_5606", "title": "1776. Relative Rates of Reaction - ", "content": "This may be represented in an abbreviated format by omitting the units of the stoichiometric factor:"} {"id": "textbook_5607", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n-\\frac{1}{2} \\frac{\\Delta \\mathrm{~mol} \\mathrm{NH}}{3} \\text { } \\quad \\Delta t \\quad=\\frac{\\Delta \\mathrm{mol} \\mathrm{~N}_{2}}{\\Delta t}\n$$"} {"id": "textbook_5608", "title": "1776. Relative Rates of Reaction - ", "content": "Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:"} {"id": "textbook_5609", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{NH}_{3}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{N}_{2}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5610", "title": "1776. Relative Rates of Reaction - ", "content": "Similarly, the rate of formation of $\\mathrm{H}_{2}$ is three times the rate of formation of $\\mathrm{N}_{2}$ because three moles of $\\mathrm{H}_{2}$ are produced for each mole of $\\mathrm{N}_{2}$ produced."} {"id": "textbook_5611", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{H}_{2}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{N}_{2}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5612", "title": "1776. Relative Rates of Reaction - ", "content": "Figure 17.5 illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen\nand hydrogen at $1100^{\\circ} \\mathrm{C}$. Slopes of the tangent lines at $t=500 \\mathrm{~s}$ show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:"} {"id": "textbook_5613", "title": "1776. Relative Rates of Reaction - ", "content": "$$\n\\frac{2.91 \\times 10^{-6} \\mathrm{M} / \\mathrm{s}}{9.70 \\times 10^{-7} \\mathrm{M} / \\mathrm{s}} \\approx 3\n$$"} {"id": "textbook_5614", "title": "1776. Relative Rates of Reaction - ", "content": ""} {"id": "textbook_5615", "title": "1776. Relative Rates of Reaction - ", "content": "FIGURE 17.5 Changes in concentrations of the reactant and products for the reaction $2 \\mathrm{NH}_{3} \\longrightarrow \\mathrm{~N}_{2}+3 \\mathrm{H}_{2}$. The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at $t=500 \\mathrm{~s}$."} {"id": "textbook_5616", "title": "1776. Relative Rates of Reaction - ", "content": "EXAMPLE 17.1\n"} {"id": "textbook_5617", "title": "1777. Expressions for Relative Reaction Rates - ", "content": "\nThe first step in the production of nitric acid is the combustion of ammonia:"} {"id": "textbook_5618", "title": "1777. Expressions for Relative Reaction Rates - ", "content": "$$\n4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$"} {"id": "textbook_5619", "title": "1777. Expressions for Relative Reaction Rates - ", "content": "Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.\n"} {"id": "textbook_5620", "title": "1778. Solution - ", "content": "\nConsidering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:"} {"id": "textbook_5621", "title": "1778. Solution - ", "content": "$$\n-\\frac{1}{4} \\frac{\\Delta\\left[\\mathrm{NH}_{3}\\right]}{\\Delta t}=-\\frac{1}{5} \\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{1}{4} \\frac{\\Delta[\\mathrm{NO}]}{\\Delta t}=\\frac{1}{6} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5622", "title": "1778. Solution - ", "content": "Check Your Learning\nThe rate of formation of $\\mathrm{Br}_{2}$ is $6.0 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}$ in a reaction described by the following net ionic equation:"} {"id": "textbook_5623", "title": "1778. Solution - ", "content": "$$\n5 \\mathrm{Br}^{-}+\\mathrm{BrO}_{3}^{-}+6 \\mathrm{H}^{+} \\longrightarrow 3 \\mathrm{Br}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}\n$$"} {"id": "textbook_5624", "title": "1778. Solution - ", "content": "Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.\n"} {"id": "textbook_5625", "title": "1779. Answer: - ", "content": "\n$-\\frac{1}{5} \\frac{\\Delta\\left[\\mathrm{Br}^{-}\\right]}{\\Delta t}=-\\frac{\\Delta\\left[\\mathrm{BrO}_{3}{ }^{-}\\right]}{\\Delta t}=-\\frac{1}{6} \\frac{\\Delta\\left[\\mathrm{H}^{+}\\right]}{\\Delta t}=\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{Br}_{2}\\right]}{\\Delta t}=\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}$\n"} {"id": "textbook_5626", "title": "1780. EXAMPLE 17.2 - ", "content": ""} {"id": "textbook_5627", "title": "1781. Reaction Rate Expressions for Decomposition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}_{\\mathbf{2}}$ - ", "content": "\nThe graph in Figure 17.3 shows the rate of the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ over time:"} {"id": "textbook_5628", "title": "1781. Reaction Rate Expressions for Decomposition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}_{\\mathbf{2}}$ - ", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\n$$"} {"id": "textbook_5629", "title": "1781. Reaction Rate Expressions for Decomposition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}_{\\mathbf{2}}$ - ", "content": "Based on these data, the instantaneous rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at $t=11.1 \\mathrm{~h}$ is determined to be $3.20 \\times 10^{-2} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{h}$, that is:\n\n$$\n-\\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=3.20 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$"} {"id": "textbook_5630", "title": "1781. Reaction Rate Expressions for Decomposition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}_{\\mathbf{2}}$ - ", "content": "What is the instantaneous rate of production of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{O}_{2}$ ?\n"} {"id": "textbook_5631", "title": "1782. Solution - ", "content": "\nThe reaction stoichiometry shows that"} {"id": "textbook_5632", "title": "1782. Solution - ", "content": "$$\n-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5633", "title": "1782. Solution - ", "content": "Therefore:"} {"id": "textbook_5634", "title": "1782. Solution - ", "content": "$$\n\\frac{1}{2} \\times 3.20 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}=\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5635", "title": "1782. Solution - ", "content": "and"} {"id": "textbook_5636", "title": "1782. Solution - ", "content": "$$\n\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}=1.60 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$\n"} {"id": "textbook_5637", "title": "1783. Check Your Learning - ", "content": "\nIf the rate of decomposition of ammonia, $\\mathrm{NH}_{3}$, at 1150 K is $2.10 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}$, what is the rate of production of nitrogen and hydrogen?\n"} {"id": "textbook_5638", "title": "1784. Answer: - ", "content": "\n$1.05 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}, \\mathrm{N}_{2}$ and $3.15 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}, \\mathrm{H}_{2}$.\n"} {"id": "textbook_5639", "title": "1784. Answer: - 1784.1. Factors Affecting Reaction Rates", "content": ""} {"id": "textbook_5640", "title": "1785. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5641", "title": "1785. LEARNING OBJECTIVES - ", "content": "- Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates"} {"id": "textbook_5642", "title": "1785. LEARNING OBJECTIVES - ", "content": "The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.\n"} {"id": "textbook_5643", "title": "1786. The Chemical Nature of the Reacting Substances - ", "content": "\nThe rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air\novernight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.\n"} {"id": "textbook_5644", "title": "1787. The Physical States of the Reactants - ", "content": "\nA chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder (Figure 17.6). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.\n\n(a)\n\n(b)"} {"id": "textbook_5645", "title": "1787. The Physical States of the Reactants - ", "content": "FIGURE 17.6 (a) Iron powder reacts rapidly with dilute hydrochloric acid and produces bubbles of hydrogen gas: $2 \\mathrm{Fe}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{FeCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)$. (b) An iron nail reacts more slowly because the surface area exposed to the acid is much less.\n"} {"id": "textbook_5646", "title": "1788. LINK TO LEARNING - ", "content": "\nWatch this video (http://openstax.org/l/16cesium) to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.\n"} {"id": "textbook_5647", "title": "1789. Temperature of the Reactants - ", "content": "\nChemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by $10^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_5648", "title": "1790. Concentrations of the Reactants - ", "content": "\nThe rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate $\\left(\\mathrm{CaCO}_{3}\\right)$ deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure 17.7). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:"} {"id": "textbook_5649", "title": "1790. Concentrations of the Reactants - ", "content": "$$\n\\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q)\n$$"} {"id": "textbook_5650", "title": "1790. Concentrations of the Reactants - ", "content": "Calcium carbonate reacts with sulfurous acid as follows:"} {"id": "textbook_5651", "title": "1790. Concentrations of the Reactants - ", "content": "$$\n\\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{3}(a q) \\longrightarrow \\mathrm{CaSO}_{3}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_5652", "title": "1790. Concentrations of the Reactants - ", "content": "In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about $20 \\%$ oxygen.\n"} {"id": "textbook_5653", "title": "1790. Concentrations of the Reactants - ", "content": "FIGURE 17.7 Statues made from carbonate compounds such as limestone and marble typically weather slowly over time due to the actions of water, and thermal expansion and contraction. However, pollutants like sulfur dioxide can accelerate weathering. As the concentration of air pollutants increases, deterioration of limestone occurs more rapidly. (credit: James P Fisher III)\n"} {"id": "textbook_5654", "title": "1791. LINK TO LEARNING - ", "content": "\nPhosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this video (http://openstax.org/l/16phosphor) to see an example.\n"} {"id": "textbook_5655", "title": "1792. The Presence of a Catalyst - ", "content": "\nRelatively dilute aqueous solutions of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:"} {"id": "textbook_5656", "title": "1792. The Presence of a Catalyst - ", "content": "$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(\\mathrm{~g})\n$$"} {"id": "textbook_5657", "title": "1792. The Presence of a Catalyst - ", "content": "Under typical conditions, this decomposition occurs very slowly. When dilute $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{aq})$ is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound's exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter.\n"} {"id": "textbook_5658", "title": "1793. (C) LINK TO LEARNING - ", "content": "\nChemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions \\& Rates interactive (http://openstax.org/l/16PHETreaction) to explore how temperature, concentration, and the nature of the reactants affect reaction rates.\n"} {"id": "textbook_5659", "title": "1793. (C) LINK TO LEARNING - 1793.1. Rate Laws", "content": ""} {"id": "textbook_5660", "title": "1794. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5661", "title": "1794. LEARNING OBJECTIVES - ", "content": "- Explain the form and function of a rate law\n- Use rate laws to calculate reaction rates\n- Use rate and concentration data to identify reaction orders and derive rate laws"} {"id": "textbook_5662", "title": "1794. LEARNING OBJECTIVES - ", "content": "As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants.\nRate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation"} {"id": "textbook_5663", "title": "1794. LEARNING OBJECTIVES - ", "content": "$$\na A+b B \\longrightarrow \\text { products }\n$$"} {"id": "textbook_5664", "title": "1794. LEARNING OBJECTIVES - ", "content": "where $a$ and $b$ are stoichiometric coefficients. The rate law for this reaction is written as:"} {"id": "textbook_5665", "title": "1794. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate }=k[A]^{m}[B]^{n}\n$$"} {"id": "textbook_5666", "title": "1794. LEARNING OBJECTIVES - ", "content": "in which $[A]$ and $[B]$ represent the molar concentrations of reactants, and $k$ is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents $m$ and $n$ are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant $k$ and the reaction orders $m$ and $n$ must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant $k$ is independent of the reactant concentrations, but it does vary with temperature."} {"id": "textbook_5667", "title": "1794. LEARNING OBJECTIVES - ", "content": "The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is $m$ order with respect to $A$ and $n$ order with respect to $B$. For example, if $m=1$ and $n=2$, the reaction is first order in $A$ and second order in $B$. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall $(1+2=3)$. A few specific examples are shown below to further illustrate this concept."} {"id": "textbook_5668", "title": "1794. LEARNING OBJECTIVES - ", "content": "The rate law:"} {"id": "textbook_5669", "title": "1794. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\n$$"} {"id": "textbook_5670", "title": "1794. LEARNING OBJECTIVES - ", "content": "describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:"} {"id": "textbook_5671", "title": "1794. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]^{2}\n$$"} {"id": "textbook_5672", "title": "1794. LEARNING OBJECTIVES - ", "content": "describes a reaction that is second order in $\\mathrm{C}_{4} \\mathrm{H}_{6}$ and second order overall. The rate law:"} {"id": "textbook_5673", "title": "1794. LEARNING OBJECTIVES - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$"} {"id": "textbook_5674", "title": "1794. LEARNING OBJECTIVES - ", "content": "describes a reaction that is first order in $\\mathrm{H}^{+}$, first order in $\\mathrm{OH}^{-}$, and second order overall.\n"} {"id": "textbook_5675", "title": "1795. EXAMPLE 17.3 - ", "content": ""} {"id": "textbook_5676", "title": "1796. Writing Rate Laws from Reaction Orders - ", "content": "\nAn experiment shows that the reaction of nitrogen dioxide with carbon monoxide:"} {"id": "textbook_5677", "title": "1796. Writing Rate Laws from Reaction Orders - ", "content": "$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_5678", "title": "1796. Writing Rate Laws from Reaction Orders - ", "content": "is second order in $\\mathrm{NO}_{2}$ and zero order in CO at $100^{\\circ} \\mathrm{C}$. What is the rate law for the reaction?\n"} {"id": "textbook_5679", "title": "1797. Solution - ", "content": "\nThe reaction will have the form:"} {"id": "textbook_5680", "title": "1797. Solution - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{m}[\\mathrm{CO}]^{n}\n$$"} {"id": "textbook_5681", "title": "1797. Solution - ", "content": "The reaction is second order in $\\mathrm{NO}_{2}$; thus $m=2$. The reaction is zero order in CO; thus $n=0$. The rate law is:\n\n$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2}[\\mathrm{CO}]^{0}=k\\left[\\mathrm{NO}_{2}\\right]^{2}\n$$"} {"id": "textbook_5682", "title": "1797. Solution - ", "content": "Remember that a number raised to the zero power is equal to 1 , thus $[\\mathrm{CO}]^{0}=1$, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of $\\mathrm{NO}_{2}$. A later chapter section on reaction mechanisms will explain how a reactant's concentration can have no effect on a reaction rate despite being involved in the reaction.\n"} {"id": "textbook_5683", "title": "1798. Check Your Learning - ", "content": "\nThe rate law for the reaction:"} {"id": "textbook_5684", "title": "1798. Check Your Learning - ", "content": "$$\n\\mathrm{H}_{2}(g)+2 \\mathrm{NO}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$"} {"id": "textbook_5685", "title": "1798. Check Your Learning - ", "content": "has been determined to be rate $=k[N O]^{2}\\left[\\mathrm{H}_{2}\\right]$. What are the orders with respect to each reactant, and what is the\noverall order of the reaction?\n"} {"id": "textbook_5686", "title": "1799. Answer: - ", "content": "\norder in $\\mathrm{NO}=2$; order in $\\mathrm{H}_{2}=1$; overall order $=3$\n"} {"id": "textbook_5687", "title": "1800. Check Your Learning - ", "content": "\nIn a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ and ethyl acetate $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3}\\right)$ as a sample reaction before studying the chemical reactions that produce biodiesel:"} {"id": "textbook_5688", "title": "1800. Check Your Learning - ", "content": "$$\n\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OCOCH}_{3}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\n$$"} {"id": "textbook_5689", "title": "1800. Check Your Learning - ", "content": "The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:"} {"id": "textbook_5690", "title": "1800. Check Your Learning - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]\n$$"} {"id": "textbook_5691", "title": "1800. Check Your Learning - ", "content": "What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?\n"} {"id": "textbook_5692", "title": "1801. Answer: - ", "content": "\norder in $\\mathrm{CH}_{3} \\mathrm{OH}=1$; order in $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3}=0$; overall order $=1$"} {"id": "textbook_5693", "title": "1801. Answer: - ", "content": "A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.\n"} {"id": "textbook_5694", "title": "1802. EXAMPLE 17.4 - ", "content": ""} {"id": "textbook_5695", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "\nOzone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure 17.8). One such reaction is the combination of nitric oxide, NO, with ozone, $\\mathrm{O}_{3}$ :"} {"id": "textbook_5696", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "FIGURE 17.8 A contour map showing stratospheric ozone concentration and the \"ozone hole\" that occurs over Antarctica during its spring months. (credit: modification of work by NASA)"} {"id": "textbook_5697", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "$$\n\\mathrm{NO}(\\mathrm{~g})+\\mathrm{O}_{3}(\\mathrm{~g}) \\longrightarrow \\mathrm{NO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})\n$$"} {"id": "textbook_5698", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "This reaction has been studied in the laboratory, and the following rate data were determined at $25^{\\circ} \\mathrm{C}$."} {"id": "textbook_5699", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "| Trial | $\\mathrm{NO}$ | $\\left\\mathrm{O}_{3}\\right$ |\n| :--- | :--- | :--- |\n| 1 | $\\frac{\\Delta\\left[\\mathrm{NO}_{2}\\right]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ | |\n| $1.00 \\times 10^{-6}$ | $3.00 \\times 10^{-6}$ | $6.60 \\times 10^{-5}$ |\n| 2 | $1.00 \\times 10^{-6}$ | $6.00 \\times 10^{-6}$ |\n| 3 | $1.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |\n| 4 | $2.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |\n| 5 | $3.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |"} {"id": "textbook_5700", "title": "1803. Determining a Rate Law from Initial Rates - ", "content": "Determine the rate law and the rate constant for the reaction at $25^{\\circ} \\mathrm{C}$.\n"} {"id": "textbook_5701", "title": "1804. Solution - ", "content": "\nThe rate law will have the form:"} {"id": "textbook_5702", "title": "1804. Solution - ", "content": "$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{O}_{3}\\right]^{n}\n$$"} {"id": "textbook_5703", "title": "1804. Solution - ", "content": "Determine the values of $m, n$, and $k$ from the experimental data using the following three-part process:\nStep 1.\nDetermine the value of m from the data in which [ NO ] varies and $\\left[\\mathrm{O}_{3}\\right.$ ] is constant. In the last three experiments, [ NO ] varies while $\\left[\\mathrm{O}_{3}\\right]$ remains constant. When [NO] doubles from trial 3 to 4 , the rate doubles, and when [ NO ] triples from trial 3 to 5 , the rate also triples. Thus, the rate is also directly proportional to [NO], and $m$ in the rate law is equal to 1 .\n\nStep 2.\nDetermine the value of n from data in which $\\left[\\mathrm{O}_{3}\\right]$ varies and [NO] is constant. In the first three experiments, [ NO ] is constant and $\\left[\\mathrm{O}_{3}\\right]$ varies. The reaction rate changes in direct proportion to the change in [ $\\mathrm{O}_{3}$ ]. When $\\left[\\mathrm{O}_{3}\\right]$ doubles from trial 1 to 2, the rate doubles; when $\\left[\\mathrm{O}_{3}\\right]$ triples from trial 1 to 3 , the rate increases also triples. Thus, the rate is directly proportional to $\\left[\\mathrm{O}_{3}\\right]$, and $n$ is equal to 1.The rate law is thus:"} {"id": "textbook_5704", "title": "1804. Solution - ", "content": "$$\n\\text { rate }=k[\\mathrm{NO}]^{1}\\left[\\mathrm{O}_{3}\\right]^{1}=k[\\mathrm{NO}]\\left[\\mathrm{O}_{3}\\right]\n$$"} {"id": "textbook_5705", "title": "1804. Solution - ", "content": "Step 3.\nDetermine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:"} {"id": "textbook_5706", "title": "1804. Solution - ", "content": "$$\n\\begin{aligned}\n& k=\\frac{\\text { rate }}{[\\mathrm{NO}]\\left[\\mathrm{O}_{3}\\right]} \\\\\n&=\\frac{6.60 \\times 10^{-5} \\overline{\\mathrm{molL}}^{-1}}{\\mathrm{~s}}-1 \\\\\n&\\left(1.00 \\times 10^{-6} \\mathrm{molL}^{-1}\\right)\\left(3.00 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)\n\\end{aligned}\n$$\n"} {"id": "textbook_5707", "title": "1805. Check Your Learning - ", "content": "\nAcetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:"} {"id": "textbook_5708", "title": "1805. Check Your Learning - ", "content": "$$\n\\mathrm{CH}_{3} \\mathrm{CHO}(\\mathrm{~g}) \\longrightarrow \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{~g})\n$$"} {"id": "textbook_5709", "title": "1805. Check Your Learning - ", "content": "Determine the rate law and the rate constant for the reaction from the following experimental data:"} {"id": "textbook_5710", "title": "1805. Check Your Learning - ", "content": "| Trial | $\\left\\mathrm{CH}_{3} \\mathrm{CHO}\\right$ | $-\\frac{\\Delta\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ |\n| :--- | :--- | :--- |\n| 1 | $1.75 \\times 10^{-3}$ | $2.06 \\times 10^{-11}$ |\n| 2 | $3.50 \\times 10^{-3}$ | $8.24 \\times 10^{-11}$ |\n| 3 | $7.00 \\times 10^{-3}$ | $3.30 \\times 10^{-10}$ |\n"} {"id": "textbook_5711", "title": "1806. Answer: - ", "content": "\nrate $=k\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]^{2}$ with $k=6.73 \\times 10^{-6} \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$\n"} {"id": "textbook_5712", "title": "1807. EXAMPLE 17.5 - ", "content": ""} {"id": "textbook_5713", "title": "1808. Determining Rate Laws from Initial Rates - ", "content": "\nUsing the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:"} {"id": "textbook_5714", "title": "1808. Determining Rate Laws from Initial Rates - ", "content": "$$\n2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)\n$$"} {"id": "textbook_5715", "title": "1808. Determining Rate Laws from Initial Rates - ", "content": "| Trial | $\\mathrm{NO}$ | $\\left\\mathrm{Cl}_{2}\\right$ | $-\\frac{\\Delta[\\mathrm{NO}]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n| 1 | 0.10 | 0.10 | 0.00300 |\n| 2 | 0.10 | 0.15 | 0.00450 |\n| 3 | 0.15 | 0.10 | 0.00675 |\n"} {"id": "textbook_5716", "title": "1809. Solution - ", "content": "\nThe rate law for this reaction will have the form:"} {"id": "textbook_5717", "title": "1809. Solution - ", "content": "$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{Cl}_{2}\\right]^{n}\n$$"} {"id": "textbook_5718", "title": "1809. Solution - ", "content": "As in Example 17.4, approach this problem in a stepwise fashion, determining the values of $m$ and $n$ from the experimental data and then using these values to determine the value of $k$. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of $m$ and $n$ :\n"} {"id": "textbook_5719", "title": "1810. Step 1. - ", "content": "\nDetermine the value of m from the data in which $[\\mathrm{NO}]$ varies and $\\left[\\mathrm{Cl}_{2}\\right]$ is constant. Write the ratios with the subscripts $x$ and $y$ to indicate data from two different trials:"} {"id": "textbook_5720", "title": "1810. Step 1. - ", "content": "$$\n\\frac{\\operatorname{rate}_{x}}{\\operatorname{rate}_{y}}=\\frac{k[\\mathrm{NO}]_{x}^{m}\\left[\\mathrm{Cl}_{2}\\right]_{x}^{n}}{k[\\mathrm{NO}]_{y}^{m}\\left[\\mathrm{Cl}_{2}\\right]_{y}^{n}}\n$$"} {"id": "textbook_5721", "title": "1810. Step 1. - ", "content": "Using the third trial and the first trial, in which $\\left[\\mathrm{Cl}_{2}\\right]$ does not vary, gives:"} {"id": "textbook_5722", "title": "1810. Step 1. - ", "content": "$$\n\\frac{\\text { rate } 3}{\\text { rate } 1}=\\frac{0.00675}{0.00300}=\\frac{k(0.15)^{m}(0.10)^{n}}{k(0.10)^{m}(0.10)^{n}}\n$$"} {"id": "textbook_5723", "title": "1810. Step 1. - ", "content": "Canceling equivalent terms in the numerator and denominator leaves:"} {"id": "textbook_5724", "title": "1810. Step 1. - ", "content": "$$\n\\frac{0.00675}{0.00300}=\\frac{(0.15)^{m}}{(0.10)^{m}}\n$$"} {"id": "textbook_5725", "title": "1810. Step 1. - ", "content": "which simplifies to:"} {"id": "textbook_5726", "title": "1810. Step 1. - ", "content": "$$\n2.25=(1.5)^{m}\n$$"} {"id": "textbook_5727", "title": "1810. Step 1. - ", "content": "Use logarithms to determine the value of the exponent $m$ :"} {"id": "textbook_5728", "title": "1810. Step 1. - ", "content": "$$\n\\begin{aligned}\n\\ln (2.25) & =m \\ln (1.5) \\\\\n\\frac{\\ln (2.25)}{\\ln (1.5)} & =m \\\\\n2 & =m\n\\end{aligned}\n$$"} {"id": "textbook_5729", "title": "1810. Step 1. - ", "content": "Confirm the result"} {"id": "textbook_5730", "title": "1810. Step 1. - ", "content": "$$\n1.5^{2}=2.25\n$$"} {"id": "textbook_5731", "title": "1810. Step 1. - ", "content": "Step 2.\nDetermine the value of n from data in which [ $\\mathrm{Cl}_{2}$ ] varies and [ NO ] is constant."} {"id": "textbook_5732", "title": "1810. Step 1. - ", "content": "$$\n\\frac{\\text { rate } 2}{\\text { rate } 1}=\\frac{0.00450}{0.00300}=\\frac{k(0.10)^{m}(0.15)^{n}}{k(0.10)^{m}(0.10)^{n}}\n$$"} {"id": "textbook_5733", "title": "1810. Step 1. - ", "content": "Cancelation gives:"} {"id": "textbook_5734", "title": "1810. Step 1. - ", "content": "$$\n\\frac{0.0045}{0.0030}=\\frac{(0.15)^{n}}{(0.10)^{n}}\n$$"} {"id": "textbook_5735", "title": "1810. Step 1. - ", "content": "which simplifies to:"} {"id": "textbook_5736", "title": "1810. Step 1. - ", "content": "$$\n1.5=(1.5)^{n}\n$$"} {"id": "textbook_5737", "title": "1810. Step 1. - ", "content": "Thus $n$ must be 1 , and the form of the rate law is:"} {"id": "textbook_5738", "title": "1810. Step 1. - ", "content": "$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{Cl}_{2}\\right]^{n}=k[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n$$"} {"id": "textbook_5739", "title": "1810. Step 1. - ", "content": "Step 3.\nDetermine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are $\\mathrm{mol} / \\mathrm{L} / \\mathrm{s}$. The units for $k$ are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are $\\mathrm{mol}^{3} / \\mathrm{L}^{3}$. The units for $k$ should be $\\mathrm{mol}^{-2} \\mathrm{~L}^{2} / \\mathrm{s}$ so that the rate is in terms of $\\mathrm{mol} / \\mathrm{L} / \\mathrm{s}$."} {"id": "textbook_5740", "title": "1810. Step 1. - ", "content": "To determine the value of $k$ once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for $k$ :"} {"id": "textbook_5741", "title": "1810. Step 1. - ", "content": "$$\n\\begin{aligned}\n0.00300 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1} & =k\\left(0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)^{2}\\left(0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)^{1} \\\\\nk & =3.0 \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n"} {"id": "textbook_5742", "title": "1811. Check Your Learning - ", "content": "\nUse the provided initial rate data to derive the rate law for the reaction whose equation is:"} {"id": "textbook_5743", "title": "1811. Check Your Learning - ", "content": "| $\\mathrm{OCl}^{-}(a q)+\\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{OI}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$ | | | |\n| :--- | :--- | :--- | :--- |\n| Trial | $\\left\\mathrm{OCl}^{-}\\right$ | $\\left\\mathrm{I}^{-}\\right$ | Initial Rate (mol/L/s) |\n| 1 | 0.0040 | 0.0020 | 0.00184 |\n| 2 | 0.0020 | 0.0040 | 0.00092 |\n| 3 | 0.0020 | 0.0020 | 0.00046 |"} {"id": "textbook_5744", "title": "1811. Check Your Learning - ", "content": "Determine the rate law expression and the value of the rate constant $k$ with appropriate units for this reaction.\n"} {"id": "textbook_5745", "title": "1812. Answer: - ", "content": "\n$$\n\\begin{aligned}\n& \\frac{\\text { rate } 2}{\\text { rate } 3}=\\frac{0.00092}{0.00046}=\\frac{k(0.0020)^{x}(0.0040)^{y}}{k(0.0020)^{x}(0.0020)^{y}} \\\\\n& 2.00=2.00^{y} \\\\\n& y=1 \\\\\n& \\frac{\\text { rate } 1}{\\text { rate } 2}=\\frac{0.00184}{0.00092}=\\frac{k(0.0040)^{x}(0.0020)^{y}}{k(0.0020)^{x}(0.0040)^{y}} \\\\\n& 2.00=\\frac{2^{x}}{2^{y}} \\\\\n& 2.00=\\frac{2^{x}}{2^{1}} \\\\\n& 4.00=2^{x} \\\\\n& x=2\n\\end{aligned}\n$$"} {"id": "textbook_5746", "title": "1812. Answer: - ", "content": "Substituting the concentration data from trial 1 and solving for $k$ yields:"} {"id": "textbook_5747", "title": "1812. Answer: - ", "content": "$$\n\\begin{aligned}\n\\text { rate } & =k\\left[\\mathrm{OCl}^{-}\\right]^{2}\\left[\\mathrm{I}^{-}\\right]^{1} \\\\\n0.00184 & =k(0.0040)^{2}(0.0020)^{1} \\\\\nk & =5.75 \\times 10^{4} \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n"} {"id": "textbook_5748", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "\nIn some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the\nchemical equation for the reaction. This is merely a coincidence and very often not the case.\nRate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:"} {"id": "textbook_5749", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "$$\n\\begin{array}{lc}\n\\mathrm{NO}_{2}+\\mathrm{CO} \\longrightarrow \\mathrm{NO}+\\mathrm{CO}_{2} & \\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2} \\\\\n\\mathrm{CH}_{3} \\mathrm{CHO} \\longrightarrow \\mathrm{CH}_{4}+\\mathrm{CO} & \\text { rate }=k\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]^{2} \\\\\n2 \\mathrm{~N}_{2} \\mathrm{O}_{5} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2} & \\text { rate }=k\\left[\\mathrm{~N}_{2} \\mathrm{O}_{5}\\right] \\\\\n2 \\mathrm{NO}_{2}+\\mathrm{F}_{2} \\longrightarrow 2 \\mathrm{NO}_{2} \\mathrm{~F} & \\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]\\left[\\mathrm{F}_{2}\\right] \\\\\n2 \\mathrm{NO}_{2} \\mathrm{Cl} \\longrightarrow 2 \\mathrm{NO}_{2}+\\mathrm{Cl}_{2} & \\text { rate }=k\\left[\\mathrm{NO}_{2} \\mathrm{Cl}\\right]\n\\end{array}\n$$"} {"id": "textbook_5750", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry."} {"id": "textbook_5751", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example 17.4 was determined to be $\\mathrm{L} \\mathrm{mol}^{-1} \\mathrm{~s}^{-1}$. For the third-order reaction described in Example 17.5, the unit for $k$ was derived to be $\\mathrm{L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$. Dimensional analysis requires the rate constant unit for a reaction whose overall order is $x$ to be $\\mathrm{L}^{x-1} \\mathrm{~mol}^{1-x} \\mathrm{~s}^{-1}$. Table 17.1 summarizes the rate constant units for common reaction orders."} {"id": "textbook_5752", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "Rate Constant Units for Common Reaction Orders"} {"id": "textbook_5753", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "| Overall Reaction Order $(\\boldsymbol{x})$ | Rate Constant Unit ( $\\mathrm{L}^{\\boldsymbol{x - 1}} \\mathrm{mol}^{1-x} \\mathrm{~s}^{-1}$ ) |\n| :--- | :--- |\n| 0 (zero) | $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}$ |\n| 1 (first) | $\\mathrm{s}^{-1}$ |\n| 2 (second) | $\\mathrm{L} \\mathrm{mol}^{-1} \\mathrm{~s}^{-1}$ |\n| 3 (third) | $\\mathrm{L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$ |"} {"id": "textbook_5754", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "TABLE 17.1"} {"id": "textbook_5755", "title": "1813. Reaction Order and Rate Constant Units - ", "content": "Note that the units in this table were derived using specific units for concentration ( $\\mathrm{mol} / \\mathrm{L}$ ) and time ( s ), though any valid units for these two properties may be used.\n"} {"id": "textbook_5756", "title": "1813. Reaction Order and Rate Constant Units - 1813.1. Integrated Rate Laws", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_5757", "title": "1813. Reaction Order and Rate Constant Units - 1813.1. Integrated Rate Laws", "content": "- Explain the form and function of an integrated rate law\n- Perform integrated rate law calculations for zero-, first-, and second-order reactions\n- Define half-life and carry out related calculations\n- Identify the order of a reaction from concentration/time data"} {"id": "textbook_5758", "title": "1813. Reaction Order and Rate Constant Units - 1813.1. Integrated Rate Laws", "content": "The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.\n\nUsing calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first, second-, and zero-order reactions.\n"} {"id": "textbook_5759", "title": "1814. First-Order Reactions - ", "content": "\nIntegration of the rate law for a simple first-order reaction (rate $=k[A]$ ) results in an equation describing how the reactant concentration varies with time:"} {"id": "textbook_5760", "title": "1814. First-Order Reactions - ", "content": "$$\n[A]_{t}=[A]_{0} e^{-k t}\n$$"} {"id": "textbook_5761", "title": "1814. First-Order Reactions - ", "content": "where $[A] t$ is the concentration of $A$ at any time $t,[A]_{0}$ is the initial concentration of $A$, and $k$ is the first-order rate constant."} {"id": "textbook_5762", "title": "1814. First-Order Reactions - ", "content": "For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:"} {"id": "textbook_5763", "title": "1814. First-Order Reactions - ", "content": "$$\n\\ln \\left(\\frac{[A]_{t}}{[A]_{0}}\\right)=-k t \\quad \\text { or } \\quad \\ln \\left(\\frac{[A]_{0}}{[A]_{t}}\\right)=k t\n$$"} {"id": "textbook_5764", "title": "1814. First-Order Reactions - ", "content": "and a format showing a linear dependence of concentration in time:"} {"id": "textbook_5765", "title": "1814. First-Order Reactions - ", "content": "$$\n\\ln [A]_{t}=\\ln [A]_{0}-k t\n$$\n"} {"id": "textbook_5766", "title": "1815. EXAMPLE 17.6 - ", "content": ""} {"id": "textbook_5767", "title": "1816. The Integrated Rate Law for a First-Order Reaction - ", "content": "\nThe rate constant for the first-order decomposition of cyclobutane, $\\mathrm{C}_{4} \\mathrm{H}_{8}$ at $500^{\\circ} \\mathrm{C}$ is $9.2 \\times 10^{-3} \\mathrm{~s}^{-1}$ :"} {"id": "textbook_5768", "title": "1816. The Integrated Rate Law for a First-Order Reaction - ", "content": "$$\n\\mathrm{C}_{4} \\mathrm{H}_{8} \\longrightarrow 2 \\mathrm{C}_{2} \\mathrm{H}_{4}\n$$"} {"id": "textbook_5769", "title": "1816. The Integrated Rate Law for a First-Order Reaction - ", "content": "How long will it take for $80.0 \\%$ of a sample of $\\mathrm{C}_{4} \\mathrm{H}_{8}$ to decompose?\n"} {"id": "textbook_5770", "title": "1817. Solution - ", "content": "\nSince the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:"} {"id": "textbook_5771", "title": "1817. Solution - ", "content": "$$\n\\ln \\left(\\frac{[A]_{0}}{[A]_{t}}\\right)=k t\n$$"} {"id": "textbook_5772", "title": "1817. Solution - ", "content": "The initial concentration of $\\mathrm{C}_{4} \\mathrm{H}_{8},[A]_{0}$, is not provided, but the provision that $80.0 \\%$ of the sample has decomposed is enough information to solve this problem. Let $x$ be the initial concentration, in which case the concentration after $80.0 \\%$ decomposition is $20.0 \\%$ of $x$ or $0.200 x$. Rearranging the rate law to isolate $t$ and substituting the provided quantities yields:"} {"id": "textbook_5773", "title": "1817. Solution - ", "content": "$$\n\\begin{aligned}\nt & =\\ln \\frac{[x]}{[0.200 x]} \\times \\frac{1}{k} \\\\\n& =\\ln 5 \\times \\frac{1}{9.2 \\times 10^{-3} \\mathrm{~s}^{-1}} \\\\\n& =1.609 \\times \\frac{1}{9.2 \\times 10^{-3} \\mathrm{~s}^{-1}} \\\\\n& =1.7 \\times 10^{2} \\mathrm{~s}\n\\end{aligned}\n$$\n"} {"id": "textbook_5774", "title": "1818. Check Your Learning - ", "content": "\nIodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:"} {"id": "textbook_5775", "title": "1818. Check Your Learning - ", "content": "$$\n\\mathrm{I}-131 \\longrightarrow \\mathrm{Xe}-131+\\text { electron }\n$$"} {"id": "textbook_5776", "title": "1818. Check Your Learning - ", "content": "The decay is first-order with a rate constant of $0.138 \\mathrm{~d}^{-1}$. How many days will it take for $90 \\%$ of the iodine -131 in a 0.500 M solution of this substance to decay to Xe-131?\n"} {"id": "textbook_5777", "title": "1819. Answer: - ", "content": "\n16.7 days"} {"id": "textbook_5778", "title": "1819. Answer: - ", "content": "In the next example exercise, a linear format for the integrated rate law will be convenient:"} {"id": "textbook_5779", "title": "1819. Answer: - ", "content": "$$\n\\begin{aligned}\n\\ln [A]_{t} & =(-k)(t)+\\ln [A]_{0} \\\\\ny & =m x+b\n\\end{aligned}\n$$"} {"id": "textbook_5780", "title": "1819. Answer: - ", "content": "A plot of $\\ln [A]_{t}$ versus $t$ for a first-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $\\ln [A]_{0}$. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.\n"} {"id": "textbook_5781", "title": "1820. EXAMPLE 17.7 - ", "content": ""} {"id": "textbook_5782", "title": "1821. Graphical Determination of Reaction Order and Rate Constant - ", "content": "\nShow that the data in Figure 17.2 can be represented by a first-order rate law by graphing $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ versus time. Determine the rate constant for the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ from these data.\n"} {"id": "textbook_5783", "title": "1822. Solution - ", "content": "\nThe data from Figure 17.2 are tabulated below, and a plot of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ is shown in Figure 17.9."} {"id": "textbook_5784", "title": "1822. Solution - ", "content": "| Trial | | Time (h) | $\\left\\mathrm{H}_{2} \\mathrm{O}_{2}\\right$ |\n| :--- | :--- | :--- | :--- |\n| 1 | 0.00 | 1.000 | 0.000 |\n| 2 | 6.00 | 0.500 | -0.693 |\n| 3 | 12.00 | 0.250 | -1.386 |\n| 4 | 18.00 | 0.125 | -2.079 |\n| 5 | 24.00 | 0.0625 | -2.772 |"} {"id": "textbook_5785", "title": "1822. Solution - ", "content": ""} {"id": "textbook_5786", "title": "1822. Solution - ", "content": "FIGURE 17.9 A linear relationship between $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ and time suggests the decomposition of hydrogen peroxide is a first-order reaction.\n\nThe plot of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ versus time is linear, indicating that the reaction may be described by a first-order rate law."} {"id": "textbook_5787", "title": "1822. Solution - ", "content": "According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot's slope."} {"id": "textbook_5788", "title": "1822. Solution - ", "content": "$$\n\\text { slope }=\\frac{\\text { change in } y}{\\text { change in } x}=\\frac{\\Delta y}{\\Delta x}=\\frac{\\Delta \\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}\n$$"} {"id": "textbook_5789", "title": "1822. Solution - ", "content": "The slope of this line may be derived from two values of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ at different values of $t$ (one near each end of the line is preferable). For example, the value of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ when $t$ is 0.00 h is 0.000 ; the value when $t=24.00 \\mathrm{~h}$ is $-2.772$"} {"id": "textbook_5790", "title": "1822. Solution - ", "content": "$$\n\\begin{aligned}\n\\text { slope } & =\\frac{-2.772-0.000}{24.00-0.00 \\mathrm{~h}} \\\\\n& =\\frac{-2.772}{24.00 \\mathrm{~h}} \\\\\n& =-0.116 \\mathrm{~h}^{-1} \\\\\nk & =- \\text { slope }=-\\left(-0.116 \\mathrm{~h}^{-1}\\right)=0.116 \\mathrm{~h}^{-1}\n\\end{aligned}\n$$\n"} {"id": "textbook_5791", "title": "1823. Check Your Learning - ", "content": "\nGraph the following data to determine whether the reaction $A \\longrightarrow B+C$ is first order."} {"id": "textbook_5792", "title": "1823. Check Your Learning - ", "content": "| Trial | | Time (s) |\n| :--- | :--- | :--- |\n| $A]$ | | |\n| 1 | 4.0 | 0.220 |\n| 2 | 8.0 | 0.144 |\n| 3 | 12.0 | 0.110 |\n| 4 | 16.0 | 0.088 |\n| 5 | 20.0 | 0.074 |\n"} {"id": "textbook_5793", "title": "1824. Answer: - ", "content": "\nThe plot of $\\ln [A]_{t}$ vs. $t$ is not linear, indicating the reaction is not first order:\n\n"} {"id": "textbook_5794", "title": "1825. Second-Order Reactions - ", "content": "\nThe equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:"} {"id": "textbook_5795", "title": "1825. Second-Order Reactions - ", "content": "$$\n\\text { rate }=k[A]^{2}\n$$"} {"id": "textbook_5796", "title": "1825. Second-Order Reactions - ", "content": "For these second-order reactions, the integrated rate law is:"} {"id": "textbook_5797", "title": "1825. Second-Order Reactions - ", "content": "$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$"} {"id": "textbook_5798", "title": "1825. Second-Order Reactions - ", "content": "where the terms in the equation have their usual meanings as defined earlier.\n"} {"id": "textbook_5799", "title": "1826. EXAMPLE 17.8 - ", "content": "\nThe Integrated Rate Law for a Second-Order Reaction\nThe reaction of butadiene gas $\\left(\\mathrm{C}_{4} \\mathrm{H}_{6}\\right)$ to yield $\\mathrm{C}_{8} \\mathrm{H}_{12}$ gas is described by the equation:"} {"id": "textbook_5800", "title": "1826. EXAMPLE 17.8 - ", "content": "$$\n2 \\mathrm{C}_{4} \\mathrm{H}_{6}(g) \\longrightarrow \\mathrm{C}_{8} \\mathrm{H}_{12}(g)\n$$"} {"id": "textbook_5801", "title": "1826. EXAMPLE 17.8 - ", "content": "This \"dimerization\" reaction is second order with a rate constant equal to $5.76 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}$ under certain conditions. If the initial concentration of butadiene is 0.200 M , what is the concentration after 10.0 min?\n"} {"id": "textbook_5802", "title": "1827. Solution - ", "content": "\nFor a second-order reaction, the integrated rate law is written"} {"id": "textbook_5803", "title": "1827. Solution - ", "content": "$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$"} {"id": "textbook_5804", "title": "1827. Solution - ", "content": "We know three variables in this equation: $[A]_{0}=0.200 \\mathrm{~mol} / \\mathrm{L}, k=5.76 \\times 10^{-2} \\mathrm{~L} / \\mathrm{mol} / \\mathrm{min}$, and $t=10.0 \\mathrm{~min}$. Therefore, we can solve for $[A]$, the fourth variable:\n\n$$\n\\begin{aligned}\n\\frac{1}{[A]_{t}} & =\\left(5.76 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}\\right)(10 \\mathrm{~min})+\\frac{1}{0.200 \\mathrm{~mol}^{-1}} \\\\\n\\frac{1}{[A]_{t}} & =\\left(5.76 \\times 10^{-1} \\mathrm{~L} \\mathrm{~mol}^{-1}\\right)+5.00 \\mathrm{~L} \\mathrm{~mol}^{-1} \\\\\n\\frac{1}{[A]_{t}} & =5.58 \\mathrm{~L} \\mathrm{~mol}^{-1} \\\\\n{[A]_{t} } & =1.79 \\times 10^{-1} \\mathrm{~mol} \\mathrm{~L}^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_5805", "title": "1827. Solution - ", "content": "Therefore $0.179 \\mathrm{~mol} / \\mathrm{L}$ of butadiene remain at the end of 10.0 min , compared to the $0.200 \\mathrm{~mol} / \\mathrm{L}$ that was originally present.\n"} {"id": "textbook_5806", "title": "1828. Check Your Learning - ", "content": "\nIf the initial concentration of butadiene is 0.0200 M , what is the concentration remaining after 20.0 min?\n"} {"id": "textbook_5807", "title": "1829. Answer: - ", "content": "\n$0.0195 \\mathrm{~mol} / \\mathrm{L}$"} {"id": "textbook_5808", "title": "1829. Answer: - ", "content": "The integrated rate law for second-order reactions has the form of the equation of a straight line:"} {"id": "textbook_5809", "title": "1829. Answer: - ", "content": "$$\n\\begin{aligned}\n\\frac{1}{[A]_{t}} & =k t+\\frac{1}{[A]_{0}} \\\\\ny & =m x+b\n\\end{aligned}\n$$"} {"id": "textbook_5810", "title": "1829. Answer: - ", "content": "A plot of $\\frac{1}{[A]_{t}}$ versus $t$ for a second-order reaction is a straight line with a slope of $k$ and a $y$-intercept of $\\frac{1}{[A]_{0}}$. If the plot is not a straight line, then the reaction is not second order.\n"} {"id": "textbook_5811", "title": "1830. EXAMPLE 17.9 - ", "content": ""} {"id": "textbook_5812", "title": "1831. Graphical Determination of Reaction Order and Rate Constant - ", "content": "\nThe data below are for the same reaction described in Example 17.8. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.\n"} {"id": "textbook_5813", "title": "1832. Solution - ", "content": "\n| Trial | Time (s) | $\\left\\mathrm{C}_{4} \\mathrm{H}_{6}\\right$ |\n| :--- | :--- | :--- |\n| 1 | 0 | $1.00 \\times 10^{-2}$ |\n| 2 | 1600 | $5.04 \\times 10^{-3}$ |\n| 3 | 3200 | $3.37 \\times 10^{-3}$ |\n| 4 | 4800 | $2.53 \\times 10^{-3}$ |\n| 5 | 6200 | $2.08 \\times 10^{-3}$ |"} {"id": "textbook_5814", "title": "1832. Solution - ", "content": "In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}$ versus $t$ and compare it to a plot of $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}}$ versus $t$. The values needed for these plots follow."} {"id": "textbook_5815", "title": "1832. Solution - ", "content": "| Time (s) | $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]}\\left(M^{-1}\\right)$ | $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]$ |\n| :--- | :--- | :--- |\n| 0 | 100 | -4.605 |\n| 1600 | 198 | -5.289 |\n| 3200 | 296 | -5.692 |\n| 4800 | 395 | -5.978 |\n| 6200 | 481 | -6.175 |"} {"id": "textbook_5816", "title": "1832. Solution - ", "content": "The plots are shown in Figure 17.10, which clearly shows the plot of $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}$ versus $t$ is not linear, therefore the reaction is not first order. The plot of $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}}$ versus $t$ is linear, indicating that the reaction is second order.\n\n"} {"id": "textbook_5817", "title": "1832. Solution - ", "content": "FIGURE 17.10 These two graphs show first- and second-order plots for the dimerization of $\\mathrm{C}_{4} \\mathrm{H}_{6}$. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics."} {"id": "textbook_5818", "title": "1832. Solution - ", "content": "According to the second-order integrated rate law, the rate constant is equal to the slope of the $\\frac{1}{[A]_{t}}$ versus $t$ plot. Using the data for $t=0 \\mathrm{~s}$ and $t=6200 \\mathrm{~s}$, the rate constant is estimated as follows:"} {"id": "textbook_5819", "title": "1832. Solution - ", "content": "$$\nk=\\text { slope }=\\frac{\\left(481 M^{-1}-100 M^{-1}\\right)}{(6200 \\mathrm{~s}-0 \\mathrm{~s})}=0.0614 \\mathrm{M}^{-1} \\mathrm{~s}^{-1}\n$$\n"} {"id": "textbook_5820", "title": "1833. Check Your Learning - ", "content": "\nDo the following data fit a second-order rate law?"} {"id": "textbook_5821", "title": "1833. Check Your Learning - ", "content": "| Trial | Time (s) | $A$ |\n| :--- | :--- | :--- |\n| 1 | 5 | 0.952 |\n| 2 | 10 | 0.625 |\n| 3 | 15 | 0.465 |\n| 4 | 20 | 0.370 |\n| 5 | 25 | 0.308 |\n| 6 | 35 | 0.230 |\n"} {"id": "textbook_5822", "title": "1834. Answer: - ", "content": "\nYes. The plot of $\\frac{1}{[A]_{t}}$ vs. $t$ is linear:\n\n"} {"id": "textbook_5823", "title": "1835. Zero-Order Reactions - ", "content": "\nFor zero-order reactions, the differential rate law is:"} {"id": "textbook_5824", "title": "1835. Zero-Order Reactions - ", "content": "$$\n\\text { rate }=k\n$$"} {"id": "textbook_5825", "title": "1835. Zero-Order Reactions - ", "content": "A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can't be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren't met, and for this reason the more prudent term pseudo-zero-order is sometimes used."} {"id": "textbook_5826", "title": "1835. Zero-Order Reactions - ", "content": "The integrated rate law for a zero-order reaction is a linear function:"} {"id": "textbook_5827", "title": "1835. Zero-Order Reactions - ", "content": "$$\n\\begin{aligned}\n{[A]_{t} } & =-k t+[A]_{0} \\\\\ny & =m x+b\n\\end{aligned}\n$$"} {"id": "textbook_5828", "title": "1835. Zero-Order Reactions - ", "content": "A plot of $[A]$ versus $t$ for a zero-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $[A]_{0}$. Figure 17.11 shows a plot of $\\left[\\mathrm{NH}_{3}\\right]$ versus $t$ for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz $\\left(\\mathrm{SiO}_{2}\\right)$ surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.\n"} {"id": "textbook_5829", "title": "1836. EXAMPLE 17.10 - ", "content": ""} {"id": "textbook_5830", "title": "1837. Graphical Determination of Zero-Order Rate Constant - ", "content": "\nUse the data plot in Figure 17.11 to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.\n"} {"id": "textbook_5831", "title": "1838. Solution - ", "content": "\nThe integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, $[A]_{t}$, versus time, $t$, with a slope equal to the negative of the rate constant, $-k$. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W ) is estimated from the graph. Using the ammonia concentrations at $t=0$ and $t=1000 \\mathrm{~s}$ :"} {"id": "textbook_5832", "title": "1838. Solution - ", "content": "$$\nk=- \\text { slope }=-\\frac{\\left(0.0015 \\mathrm{~mol} \\mathrm{~L}^{-1}-0.0028 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)}{(1000 \\mathrm{~s}-0 \\mathrm{~s})}=1.3 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\n$$\n"} {"id": "textbook_5833", "title": "1839. Check Your Learning - ", "content": "\nThe zero-order plot in Figure 17.11 shows an initial ammonia concentration of $0.0028 \\mathrm{~mol} \\mathrm{~L}^{-1}$ decreasing linearly with time for 1000 s . Assuming no change in this zero-order behavior, at what time (min) will the concentration reach $0.0001 \\mathrm{~mol} \\mathrm{~L}^{-1}$ ?\n"} {"id": "textbook_5834", "title": "1840. Answer: - ", "content": "\n35 min\n"} {"id": "textbook_5835", "title": "1840. Answer: - ", "content": "FIGURE 17.11 The decomposition of $\\mathrm{NH}_{3}$ on a tungsten $(\\mathrm{W})$ surface is a zero-order reaction, whereas on a quartz $\\left(\\mathrm{SiO}_{2}\\right)$ surface, the reaction is first order."} {"id": "textbook_5836", "title": "1840. Answer: - ", "content": "The Half-Life of a Reaction\nThe half-life of a reaction ( $\\boldsymbol{t}_{\\mathbf{1} / \\mathbf{2}}$ ) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure 17.2) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases from 1.000 M to 0.500 M . During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M ; during the third half-life, it decreases from 0.250 M to 0.125 M . The concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.\n"} {"id": "textbook_5837", "title": "1841. First-Order Reactions - ", "content": "\nAn equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:"} {"id": "textbook_5838", "title": "1841. First-Order Reactions - ", "content": "$$\n\\begin{aligned}\n\\ln \\frac{[A]_{0}}{[A]_{t}} & =k t \\\\\nt & =\\ln \\frac{[A]_{0}}{[A]_{t}} \\times \\frac{1}{k}\n\\end{aligned}\n$$"} {"id": "textbook_5839", "title": "1841. First-Order Reactions - ", "content": "Invoking the definition of half-life, symbolized $t_{1 / 2}$, requires that the concentration of $A$ at this point is one-half its initial concentration: $t=t_{1 / 2},[A]_{t}=\\frac{1}{2}[A]_{0}$."} {"id": "textbook_5840", "title": "1841. First-Order Reactions - ", "content": "Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for halflife:"} {"id": "textbook_5841", "title": "1841. First-Order Reactions - ", "content": "$$\n\\begin{aligned}\nt_{1 / 2} & =\\ln \\frac{[A]_{0}}{\\frac{1}{2}[A]_{0}} \\times \\frac{1}{k} \\\\\n& =\\ln 2 \\times \\frac{1}{k}=0.693 \\times \\frac{1}{k} \\\\\nt_{1 / 2} & =\\frac{0.693}{k}\n\\end{aligned}\n$$"} {"id": "textbook_5842", "title": "1841. First-Order Reactions - ", "content": "This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.\n"} {"id": "textbook_5843", "title": "1842. EXAMPLE 17.11 - ", "content": ""} {"id": "textbook_5844", "title": "1843. Calculation of a First-order Rate Constant using Half-Life - ", "content": "\nCalculate the rate constant for the first-order decomposition of hydrogen peroxide in water at $40^{\\circ} \\mathrm{C}$, using the data given in Figure 17.12.\n"} {"id": "textbook_5845", "title": "1843. Calculation of a First-order Rate Constant using Half-Life - ", "content": "FIGURE 17.12 The decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}\\left(2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\\right)$ at $40^{\\circ} \\mathrm{C}$ is illustrated. The intensity of the color symbolizes the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at the indicated times; $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is actually colorless.\n"} {"id": "textbook_5846", "title": "1844. Solution - ", "content": "\nInspecting the concentration/time data in Figure 17.12 shows the half-life for the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is $2.16 \\times 10^{4} \\mathrm{~s}$ :"} {"id": "textbook_5847", "title": "1844. Solution - ", "content": "$$\n\\begin{aligned}\nt_{1 / 2} & =\\frac{0.693}{k} \\\\\nk & =\\frac{0.693}{t_{1 / 2}}=\\frac{0.693}{2.16 \\times 10^{4} \\mathrm{~s}}=3.21 \\times 10^{-5} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n"} {"id": "textbook_5848", "title": "1845. Check Your Learning - ", "content": "\nThe first-order radioactive decay of iodine-131 exhibits a rate constant of $0.138 \\mathrm{~d}^{-1}$. What is the half-life for this decay?"} {"id": "textbook_5849", "title": "1845. Check Your Learning - ", "content": "Answer:\n5.02 d.\n"} {"id": "textbook_5850", "title": "1846. Second-Order Reactions - ", "content": "\nFollowing the same approach as used for first-order reactions, an equation relating the half-life of a secondorder reaction to its rate constant and initial concentration may be derived from its integrated rate law:"} {"id": "textbook_5851", "title": "1846. Second-Order Reactions - ", "content": "$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$"} {"id": "textbook_5852", "title": "1846. Second-Order Reactions - ", "content": "or"} {"id": "textbook_5853", "title": "1846. Second-Order Reactions - ", "content": "$$\n\\frac{1}{[A]}-\\frac{1}{[A]_{0}}=k t\n$$"} {"id": "textbook_5854", "title": "1846. Second-Order Reactions - ", "content": "Restrict $t$ to $t_{1 / 2}$"} {"id": "textbook_5855", "title": "1846. Second-Order Reactions - ", "content": "$$\nt=t_{1 / 2}\n$$"} {"id": "textbook_5856", "title": "1846. Second-Order Reactions - ", "content": "define $[A]_{t}$ as one-half $[A]_{0}$\n\n$$\n[A]_{t}=\\frac{1}{2}[A]_{0}\n$$"} {"id": "textbook_5857", "title": "1846. Second-Order Reactions - ", "content": "and then substitute into the integrated rate law and simplify:"} {"id": "textbook_5858", "title": "1846. Second-Order Reactions - ", "content": "$$\n\\begin{aligned}\n\\frac{1}{\\frac{1}{2}[A]_{0}}-\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\n\\frac{2}{[A]_{0}}-\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\n\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\nt_{1 / 2} & =\\frac{1}{k[A]_{0}}\n\\end{aligned}\n$$"} {"id": "textbook_5859", "title": "1846. Second-Order Reactions - ", "content": "For a second-order reaction, $t_{1 / 2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.\n"} {"id": "textbook_5860", "title": "1847. Zero-Order Reactions - ", "content": "\nAs for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law:"} {"id": "textbook_5861", "title": "1847. Zero-Order Reactions - ", "content": "$$\n[A]=-k t+[A]_{0}\n$$"} {"id": "textbook_5862", "title": "1847. Zero-Order Reactions - ", "content": "Restricting the time and concentrations to those defined by half-life: $t=t_{1 / 2}$ and $[A]=\\frac{[A]_{0}}{2}$. Substituting these terms into the zero-order integrated rate law yields:"} {"id": "textbook_5863", "title": "1847. Zero-Order Reactions - ", "content": "$$\n\\begin{aligned}\n\\frac{[\\mathrm{A}]_{0}}{2} & =-k t_{1 / 2}+[\\mathrm{A}]_{0} \\\\\nk t_{1 / 2} & =\\frac{[\\mathrm{A}]_{0}}{2} \\\\\nt_{1 / 2} & =\\frac{[A]_{0}}{2 k}\n\\end{aligned}\n$$"} {"id": "textbook_5864", "title": "1847. Zero-Order Reactions - ", "content": "As for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases."} {"id": "textbook_5865", "title": "1847. Zero-Order Reactions - ", "content": "Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 17.2."} {"id": "textbook_5866", "title": "1847. Zero-Order Reactions - ", "content": "Summary of Rate Laws for Zero-, First-, and Second-Order Reactions"} {"id": "textbook_5867", "title": "1847. Zero-Order Reactions - ", "content": "| Zero-Order | | First-Order | Second-Order |\n| :--- | :--- | :--- | :--- |\n| rate law | rate $=k$ | rate $=k[A]$ | rate $=k[A]^{2}$ |\n| units of rate constant | $M s^{-1}$ | $\\mathrm{~s}^{-1}$ | $M^{-1} \\mathrm{~s}^{-1}$ |\n| integrated rate law | $[A]=-k t+[A]_{0}$ | $\\ln [A]=-k t+\\ln [A]_{0}$ | $\\frac{1}{[A]}=k t+\\left(\\frac{1}{[A]_{0}}\\right)$ |"} {"id": "textbook_5868", "title": "1847. Zero-Order Reactions - ", "content": "TABLE 17.2"} {"id": "textbook_5869", "title": "1847. Zero-Order Reactions - ", "content": "| | Zero-Order | First-Order | Second-Order |\n| :--- | :--- | :--- | :--- |\n| plot needed for linear fit of rate data | $[A]$ vs. $t$ | $\\ln [A]$ vs. $t$ | $\\frac{1}{[A]}$ vs. $t$ |\n| relationship between slope of linear
plot and rate constant | $k=-$ slope | $k=-$ slope | $k=$ slope |\n| half-life | $t_{1 / 2}=\\frac{[A]_{0}}{2 k}$ | $t_{1 / 2}=\\frac{0.693}{k}$ | $t_{1 / 2}=\\frac{1}{[A]_{0} k}$ |"} {"id": "textbook_5870", "title": "1847. Zero-Order Reactions - ", "content": "TABLE 17.2\n"} {"id": "textbook_5871", "title": "1848. EXAMPLE 17.12 - ", "content": ""} {"id": "textbook_5872", "title": "1849. Half-Life for Zero-Order and Second-Order Reactions - ", "content": "\nWhat is the half-life for the butadiene dimerization reaction described in Example 17.8?\n"} {"id": "textbook_5873", "title": "1850. Solution - ", "content": "\nThe reaction in question is second order, is initiated with a $0.200 \\mathrm{~mol} \\mathrm{~L}^{-1}$ reactant solution, and exhibits a rate constant of $0.0576 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}$. Substituting these quantities into the second-order half-life equation:"} {"id": "textbook_5874", "title": "1850. Solution - ", "content": "$$\nt_{1 / 2}=\\frac{1}{\\left[\\left(0.0576 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}\\right)\\left(0.200 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)\\right]}=18 \\mathrm{~min}\n$$\n"} {"id": "textbook_5875", "title": "1851. Check Your Learning - ", "content": "\nWhat is the half-life (min) for the thermal decomposition of ammonia on tungsten (see Figure 17.11)?\n"} {"id": "textbook_5876", "title": "1852. Answer: - ", "content": "\n87 min\n"} {"id": "textbook_5877", "title": "1852. Answer: - 1852.1. Collision Theory", "content": ""} {"id": "textbook_5878", "title": "1853. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5879", "title": "1853. LEARNING OBJECTIVES - ", "content": "- Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates\n- Define the concepts of activation energy and transition state\n- Use the Arrhenius equation in calculations relating rate constants to temperature"} {"id": "textbook_5880", "title": "1853. LEARNING OBJECTIVES - ", "content": "We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates."} {"id": "textbook_5881", "title": "1853. LEARNING OBJECTIVES - ", "content": "Collision theory is based on the following postulates:"} {"id": "textbook_5882", "title": "1853. LEARNING OBJECTIVES - ", "content": "1. The rate of a reaction is proportional to the rate of reactant collisions:"} {"id": "textbook_5883", "title": "1853. LEARNING OBJECTIVES - ", "content": "$$\n\\text { reaction rate } \\propto \\frac{\\# \\text { collisions }}{\\text { time }}\n$$"} {"id": "textbook_5884", "title": "1853. LEARNING OBJECTIVES - ", "content": "2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.\n3. The collision must occur with adequate energy to permit mutual penetration of the reacting species' valence shells so that the electrons can rearrange and form new bonds (and new chemical species)."} {"id": "textbook_5885", "title": "1853. LEARNING OBJECTIVES - ", "content": "We can see the importance of the two physical factors noted in postulates 2 and 3 , the orientation and energy\nof collisions, when we consider the reaction of carbon monoxide with oxygen:"} {"id": "textbook_5886", "title": "1853. LEARNING OBJECTIVES - ", "content": "$$\n2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_5887", "title": "1853. LEARNING OBJECTIVES - ", "content": "Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure."} {"id": "textbook_5888", "title": "1853. LEARNING OBJECTIVES - ", "content": "The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:"} {"id": "textbook_5889", "title": "1853. LEARNING OBJECTIVES - ", "content": "$$\n\\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{O}(g)\n$$"} {"id": "textbook_5890", "title": "1853. LEARNING OBJECTIVES - ", "content": "Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 17.13. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms $(\\mathrm{O}=\\mathrm{C}=\\mathrm{O})$. This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.\n\n\nFIGURE 17.13 Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur."} {"id": "textbook_5891", "title": "1853. LEARNING OBJECTIVES - ", "content": "If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states."} {"id": "textbook_5892", "title": "1853. LEARNING OBJECTIVES - ", "content": "Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.\n"} {"id": "textbook_5893", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "\nThe minimum energy necessary to form a product during a collision between reactants is called the activation energy $\\left(\\boldsymbol{E}_{\\mathrm{a}}\\right)$. How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly."} {"id": "textbook_5894", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "Figure 17.14 shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation"} {"id": "textbook_5895", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "$$\nA+B \\longrightarrow C+D\n$$"} {"id": "textbook_5896", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, $A+B$. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, $C+D$. The diagram depicts the reaction's activation energy, $E_{a}$, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, $\\Delta H$, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic $(\\Delta H<0)$ since it yields a decrease in system enthalpy.\n"} {"id": "textbook_5897", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "FIGURE 17.14 Reaction diagram for the exothermic reaction $A+B \\longrightarrow C+D$.\nThe Arrhenius equation relates the activation energy and the rate constant, $k$, for many chemical reactions:"} {"id": "textbook_5898", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "$$\nk=A e^{-E_{\\mathrm{a}} / R T}\n$$"} {"id": "textbook_5899", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "In this equation, $R$ is the ideal gas constant, which has a value $8.314 \\mathrm{~J} / \\mathrm{mol} / \\mathrm{K}, \\mathrm{T}$ is temperature on the Kelvin scale, $E_{\\mathrm{a}}$ is the activation energy in joules per mole, $e$ is the constant 2.7183 , and $A$ is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules."} {"id": "textbook_5900", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, $A$, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for $A$ and faster reaction rates.\nThe exponential term, $e^{-E a / R T}$, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure 17.15(a). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy $(R T)$ to overcome the activation barriers $\\left(E_{a}\\right)$. A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.\n\nThe exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy ( $R T$ ) to overcome the activation barrier $\\left(E_{a}\\right)$, as shown in Figure $17.15(\\mathbf{b})$. This yields a greater value for the rate constant and a correspondingly faster reaction rate.\n"} {"id": "textbook_5901", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "FIGURE 17.15 Molecular energy distributions showing numbers of molecules with energies exceeding (a) two different activation energies at a given temperature, and (b) a given activation energy at two different temperatures."} {"id": "textbook_5902", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "A convenient approach for determining $E_{\\mathrm{a}}$ for a reaction involves the measurement of $k$ at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation"} {"id": "textbook_5903", "title": "1854. Activation Energy and the Arrhenius Equation - ", "content": "$$\n\\begin{aligned}\n\\ln k & =\\left(\\frac{-E_{\\mathrm{a}}}{R}\\right)\\left(\\frac{1}{T}\\right)+\\ln A \\\\\ny & =m x+b\n\\end{aligned}\n$$\n\nA plot of $\\ln k$ versus $\\frac{1}{T}$ is linear with a slope equal to $\\frac{-E_{\\mathrm{a}}}{R}$ and a $y$-intercept equal to $\\ln A$.\n"} {"id": "textbook_5904", "title": "1855. Determination of $E_{\\mathrm{a}}$ - ", "content": "\nThe variation of the rate constant with temperature for the decomposition of $\\mathrm{HI}(g)$ to $\\mathrm{H}_{2}(g)$ and $\\mathrm{I}_{2}(g)$ is given here. What is the activation energy for the reaction?"} {"id": "textbook_5905", "title": "1855. Determination of $E_{\\mathrm{a}}$ - ", "content": "$$\n2 \\mathrm{HI}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g)\n$$"} {"id": "textbook_5906", "title": "1855. Determination of $E_{\\mathrm{a}}$ - ", "content": "| $\\boldsymbol{T}(\\mathrm{K})$ | $k(\\mathrm{~L} / \\mathrm{mol} / \\mathrm{s})$ |\n| :---: | :---: |\n| 555 | $3.52 \\times 10^{-7}$ |\n| 575 | $1.22 \\times 10^{-6}$ |\n| 645 | $8.59 \\times 10^{-5}$ |\n| 700 | $1.16 \\times 10^{-3}$ |\n| 781 | $3.95 \\times 10^{-2}$ |\n"} {"id": "textbook_5907", "title": "1856. Solution - ", "content": "\nUse the provided data to derive values of $\\frac{1}{T}$ and $\\ln k$ :"} {"id": "textbook_5908", "title": "1856. Solution - ", "content": "| $\\frac{1}{\\mathrm{~T}}\\left(\\mathrm{~K}^{-1}\\right)$ | $\\ln \\boldsymbol{k}$ |\n| :--- | :--- |\n| $1.80 \\times 10^{-3}$ | -14.860 |\n| $1.74 \\times 10^{-3}$ | -13.617 |\n\n\n| $\\frac{1}{\\mathrm{~T}}\\left(\\mathrm{~K}^{-1}\\right)$ | $\\ln k$ |\n| :--- | :--- |\n| $1.55 \\times 10^{-3}$ | -9.362 |\n| $1.43 \\times 10^{-3}$ | -6.759 |\n| $1.28 \\times 10^{-3}$ | -3.231 |"} {"id": "textbook_5909", "title": "1856. Solution - ", "content": "Figure 17.16 is a graph of $\\ln k$ versus $\\frac{1}{T}$. In practice, the equation of the line (slope and $y$-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.\n"} {"id": "textbook_5910", "title": "1856. Solution - ", "content": "FIGURE 17.16 This graph shows the linear relationship between $\\ln k$ and $\\frac{1}{T}$ for the reaction $2 \\mathrm{HI} \\longrightarrow \\mathrm{H}_{2}+\\mathrm{I}_{2}$ according to the Arrhenius equation."} {"id": "textbook_5911", "title": "1856. Solution - ", "content": "$$\n\\begin{aligned}\n\\text { Slope }= & \\frac{\\Delta(\\ln k)}{\\Delta\\left(\\frac{1}{T}\\right)} \\\\\n= & \\frac{(-14.860)-(-3.231)}{\\left(1.80 \\times 10^{-3} \\mathrm{~K}^{-1}\\right)-\\left(1.28 \\times 10^{-3} \\mathrm{~K}^{-1}\\right)} \\\\\n= & \\frac{-11.629}{0.52 \\times 10^{-3} \\mathrm{~K}^{-1}}=-2.2 \\times 10^{4} \\mathrm{~K} \\\\\n= & -\\frac{E_{\\mathrm{a}}}{R} \\\\\nE_{\\mathrm{a}}= & - \\text { slope } \\times R=-\\left(-2.2 \\times 10^{4} \\mathrm{~K} \\times 8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right) \\\\\n& 1.8 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1} \\text { or } 180 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_5912", "title": "1856. Solution - ", "content": "Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:"} {"id": "textbook_5913", "title": "1856. Solution - ", "content": "$$\n\\ln \\frac{k_{1}}{k_{2}}=\\frac{E_{\\mathrm{a}}}{R}\\left(\\frac{1}{T_{2}}-\\frac{1}{T_{1}}\\right)\n$$"} {"id": "textbook_5914", "title": "1856. Solution - ", "content": "Rearranging this equation to isolate activation energy yields:"} {"id": "textbook_5915", "title": "1856. Solution - ", "content": "$$\nE_{\\mathrm{a}}=-R\\left(\\frac{\\ln k_{2}-\\ln k_{1}}{\\left(\\frac{1}{T_{2}}\\right)-\\left(\\frac{1}{T_{1}}\\right)}\\right)\n$$"} {"id": "textbook_5916", "title": "1856. Solution - ", "content": "Any two data pairs may be substituted into this equation-for example, the first and last entries from the above data table:"} {"id": "textbook_5917", "title": "1856. Solution - ", "content": "$$\nE_{\\mathrm{a}}=-8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\left(\\frac{-3.231-(-14.860)}{1.28 \\times 10^{-3} \\mathrm{~K}^{-1}-1.80 \\times 10^{-3} \\mathrm{~K}^{-1}}\\right)\n$$"} {"id": "textbook_5918", "title": "1856. Solution - ", "content": "and the result is $E_{\\mathrm{a}}=1.8 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1}$ or $180 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nThis approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.\n"} {"id": "textbook_5919", "title": "1857. Check Your Learning - ", "content": "\nThe rate constant for the rate of decomposition of $\\mathrm{N}_{2} \\mathrm{O}_{5}$ to NO and $\\mathrm{O}_{2}$ in the gas phase is $1.66 \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$ at 650 K and $7.39 \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$ at 700 K :"} {"id": "textbook_5920", "title": "1857. Check Your Learning - ", "content": "$$\n2 \\mathrm{~N}_{2} \\mathrm{O}_{5}(g) \\longrightarrow 4 \\mathrm{NO}(g)+3 \\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_5921", "title": "1857. Check Your Learning - ", "content": "Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.\n"} {"id": "textbook_5922", "title": "1858. Answer: - ", "content": "\n$1.1 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1}$ or $110 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n"} {"id": "textbook_5923", "title": "1858. Answer: - 1858.1. Reaction Mechanisms", "content": ""} {"id": "textbook_5924", "title": "1859. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_5925", "title": "1859. LEARNING OBJECTIVES - ", "content": "- Distinguish net reactions from elementary reactions (steps)\n- Identify the molecularity of elementary reactions\n- Write a balanced chemical equation for a process given its reaction mechanism\n- Derive the rate law consistent with a given reaction mechanism"} {"id": "textbook_5926", "title": "1859. LEARNING OBJECTIVES - ", "content": "Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs."} {"id": "textbook_5927", "title": "1859. LEARNING OBJECTIVES - ", "content": "The decomposition of ozone, for example, appears to follow a mechanism with two steps:"} {"id": "textbook_5928", "title": "1859. LEARNING OBJECTIVES - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{O}_{2}(g)+\\mathrm{O} \\\\\n& \\mathrm{O}+\\mathrm{O}_{3}(g) \\longrightarrow 2 \\mathrm{O}_{2}(g)\n\\end{aligned}\n$$"} {"id": "textbook_5929", "title": "1859. LEARNING OBJECTIVES - ", "content": "Each of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:"} {"id": "textbook_5930", "title": "1859. LEARNING OBJECTIVES - ", "content": "$$\n2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_5931", "title": "1859. LEARNING OBJECTIVES - ", "content": "Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates."} {"id": "textbook_5932", "title": "1859. LEARNING OBJECTIVES - ", "content": "While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react\nto give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one $\\mathrm{O}_{3}$ decomposes to yield $\\mathrm{O}_{2}$ and an oxygen atom, and a second $\\mathrm{O}_{3}$ molecule subsequently reacts with the oxygen atom to yield two additional $\\mathrm{O}_{2}$ molecules."} {"id": "textbook_5933", "title": "1859. LEARNING OBJECTIVES - ", "content": "Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction's equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.\n"} {"id": "textbook_5934", "title": "1860. Unimolecular Elementary Reactions - ", "content": "\nThe molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product:"} {"id": "textbook_5935", "title": "1860. Unimolecular Elementary Reactions - ", "content": "$$\nA \\longrightarrow \\text { products }\n$$"} {"id": "textbook_5936", "title": "1860. Unimolecular Elementary Reactions - ", "content": "The rate law for a unimolecular reaction is first order:"} {"id": "textbook_5937", "title": "1860. Unimolecular Elementary Reactions - ", "content": "$$\n\\text { rate }=k[A]\n$$"} {"id": "textbook_5938", "title": "1860. Unimolecular Elementary Reactions - ", "content": "A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:"} {"id": "textbook_5939", "title": "1860. Unimolecular Elementary Reactions - ", "content": "$$\n\\mathrm{O}_{3} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{O}\n$$"} {"id": "textbook_5940", "title": "1860. Unimolecular Elementary Reactions - ", "content": "illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an \"overall\" reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, $\\mathrm{C}_{4} \\mathrm{H}_{8}$, to ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, is represented by the following chemical equation:\n"} {"id": "textbook_5941", "title": "1860. Unimolecular Elementary Reactions - ", "content": "This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing firstorder behavior:"} {"id": "textbook_5942", "title": "1860. Unimolecular Elementary Reactions - ", "content": "$$\n\\text { rate }=-\\frac{\\Delta\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]}{\\Delta t}=k\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]\n$$"} {"id": "textbook_5943", "title": "1860. Unimolecular Elementary Reactions - ", "content": "This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.\n"} {"id": "textbook_5944", "title": "1861. Bimolecular Elementary Reactions - ", "content": "\nA bimolecular reaction involves two reactant species, for example:"} {"id": "textbook_5945", "title": "1861. Bimolecular Elementary Reactions - ", "content": "$$\n\\begin{gathered}\nA+B \\longrightarrow \\text { products } \\\\\n\\text { and } \\\\\n2 A \\longrightarrow \\text { products }\n\\end{gathered}\n$$"} {"id": "textbook_5946", "title": "1861. Bimolecular Elementary Reactions - ", "content": "For the first type, in which the two reactant molecules are different, the rate law is first-order in $A$ and first order in $B$ (second-order overall):"} {"id": "textbook_5947", "title": "1861. Bimolecular Elementary Reactions - ", "content": "$$\n\\text { rate }=kA\n$$"} {"id": "textbook_5948", "title": "1861. Bimolecular Elementary Reactions - ", "content": "For the second type, in which two identical molecules collide and react, the rate law is second order in $A$ :"} {"id": "textbook_5949", "title": "1861. Bimolecular Elementary Reactions - ", "content": "$$\n\\text { rate }=kA=k[A]^{2}\n$$"} {"id": "textbook_5950", "title": "1861. Bimolecular Elementary Reactions - ", "content": "Some chemical reactions occur by mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:"} {"id": "textbook_5951", "title": "1861. Bimolecular Elementary Reactions - ", "content": "$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)\n$$"} {"id": "textbook_5952", "title": "1861. Bimolecular Elementary Reactions - ", "content": "(see Figure 17.17)\n"} {"id": "textbook_5953", "title": "1861. Bimolecular Elementary Reactions - ", "content": "FIGURE 17.17 The probable mechanism for the reaction between $\\mathrm{NO}_{2}$ and CO to yield NO and $\\mathrm{CO}_{2}$.\nBimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:"} {"id": "textbook_5954", "title": "1861. Bimolecular Elementary Reactions - ", "content": "$$\n\\mathrm{O}(\\mathrm{~g})+\\mathrm{O}_{3}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{O}_{2}(\\mathrm{~g})\n$$\n"} {"id": "textbook_5955", "title": "1862. Termolecular Elementary Reactions - ", "content": "\nAn elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:"} {"id": "textbook_5956", "title": "1862. Termolecular Elementary Reactions - ", "content": "$$\n\\begin{aligned}\n& 2 \\mathrm{NO}+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2} \\\\\n& \\text { rate }=k[\\mathrm{NO}]^{2}\\left[\\mathrm{O}_{2}\\right]\n\\end{aligned}\n$$"} {"id": "textbook_5957", "title": "1862. Termolecular Elementary Reactions - ", "content": "Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:"} {"id": "textbook_5958", "title": "1862. Termolecular Elementary Reactions - ", "content": "$$\n\\begin{aligned}\n& 2 \\mathrm{NO}+\\mathrm{Cl}_{2} \\longrightarrow 2 \\mathrm{NOCl} \\\\\n& \\text { rate }=k[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n\\end{aligned}\n$$\n"} {"id": "textbook_5959", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "\nIt's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure 17.18.\n"} {"id": "textbook_5960", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "FIGURE 17.18 A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)"} {"id": "textbook_5961", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of $\\mathrm{NO}_{2}$ and CO provides an illustrative example:"} {"id": "textbook_5962", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)\n$$"} {"id": "textbook_5963", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "For temperatures above $225^{\\circ} \\mathrm{C}$, the rate law has been found to be:"} {"id": "textbook_5964", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\text { rate }=k\\left\\mathrm{NO}_{2}\\right\n$$"} {"id": "textbook_5965", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "The reaction is first order with respect to $\\mathrm{NO}_{2}$ and first-order with respect to CO . This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures."} {"id": "textbook_5966", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "At temperatures below $225^{\\circ} \\mathrm{C}$, the reaction is described by a rate law that is second order with respect to $\\mathrm{NO}_{2}$ :"} {"id": "textbook_5967", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2}\n$$"} {"id": "textbook_5968", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "This rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:"} {"id": "textbook_5969", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{NO}_{2}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{NO}_{3}(g)+\\mathrm{NO}(g) \\text { (slow) } \\\\\n& \\mathrm{NO}_{3}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}_{2}(g)+\\mathrm{CO}_{2}(g) \\text { (fast) }\n\\end{aligned}\n$$"} {"id": "textbook_5970", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "The rate-determining (slower) step gives a rate law showing second-order dependence on the $\\mathrm{NO}_{2}$ concentration, and the sum of the two equations gives the net overall reaction."} {"id": "textbook_5971", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive."} {"id": "textbook_5972", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "As discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species $\\mathrm{N}_{2} \\mathrm{O}_{2}$. When this reaction is at equilibrium:"} {"id": "textbook_5973", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{NO}+\\mathrm{NO} \\rightleftharpoons \\mathrm{~N}_{2} \\mathrm{O}_{2} \\\\\n& \\mathrm{rate}_{\\text {forward }}=\\text { rate }_{\\text {reverse }} \\\\\n& k_{1}[\\mathrm{NO}]^{2}=k_{-1}\\left[\\mathrm{~N}_{2} \\mathrm{O}_{2}\\right]\n\\end{aligned}\n$$"} {"id": "textbook_5974", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:"} {"id": "textbook_5975", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "$$\n\\left(\\frac{\\mathrm{k}_{1}[\\mathrm{NO}]^{2}}{\\mathrm{k}_{-1}}\\right)=\\left[\\mathrm{N}_{2} \\mathrm{O}_{2}\\right]\n$$"} {"id": "textbook_5976", "title": "1863. Relating Reaction Mechanisms to Rate Laws - ", "content": "Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.\n"} {"id": "textbook_5977", "title": "1864. EXAMPLE 17.14 - ", "content": ""} {"id": "textbook_5978", "title": "1865. Deriving a Rate Law from a Reaction Mechanism - ", "content": "\nThe two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:"} {"id": "textbook_5979", "title": "1865. Deriving a Rate Law from a Reaction Mechanism - ", "content": "$$\n\\begin{array}{ll}\n\\text { Step 1: } \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{NOCl}_{2}(g) & \\text { fast } \\\\\n\\text { Step 2: } \\mathrm{NOCl}_{2}(g)+\\mathrm{NO}(g) \\longrightarrow 2 \\mathrm{NOCl}(g) & \\text { slow }\n\\end{array}\n$$"} {"id": "textbook_5980", "title": "1865. Deriving a Rate Law from a Reaction Mechanism - ", "content": "Use this mechanism to derive the equation and predicted rate law for the overall reaction.\n"} {"id": "textbook_5981", "title": "1866. Solution - ", "content": "\nThe equation for the overall reaction is obtained by adding the two elementary reactions:"} {"id": "textbook_5982", "title": "1866. Solution - ", "content": "$$\n2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)\n$$"} {"id": "textbook_5983", "title": "1866. Solution - ", "content": "To derive a rate law from this mechanism, first write rates laws for each of the two steps."} {"id": "textbook_5984", "title": "1866. Solution - ", "content": "$$\n\\begin{aligned}\n& \\operatorname{rate}_{1}=k_{1}[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right] \\text { for the forward reaction of step } 1 \\\\\n& \\text { rate }_{-1}=k_{-1}\\left[\\mathrm{NOCl}_{2}\\right] \\text { for the reverse reaction of step } 1 \\\\\n& \\operatorname{rate}_{2}=k_{2}\\left\\mathrm{NOCl}_{2}\\right \\text { for step } 2\n\\end{aligned}\n$$"} {"id": "textbook_5985", "title": "1866. Solution - ", "content": "Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [ $\\mathrm{NOCl}_{2}$ ]. To remedy this, use the first step's rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations."} {"id": "textbook_5986", "title": "1866. Solution - ", "content": "Assuming step 1 is at equilibrium:"} {"id": "textbook_5987", "title": "1866. Solution - ", "content": "$$\n\\begin{aligned}\n\\text { rate }_{1} & =\\text { rate }_{-1} \\\\\nk_{1}[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right] & =k_{-1}\\left[\\mathrm{NOCl}_{2}\\right] \\\\\n{\\left[\\mathrm{NOCl}_{2}\\right] } & =\\left(\\frac{k_{1}}{k_{-1}}\\right)[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right]\n\\end{aligned}\n$$"} {"id": "textbook_5988", "title": "1866. Solution - ", "content": "Substituting this expression into the rate law for step 2 yields:"} {"id": "textbook_5989", "title": "1866. Solution - ", "content": "$$\n\\operatorname{rate}_{2}=\\operatorname{rate}_{\\text {overall }}=\\left(\\frac{k_{2} k_{1}}{k_{-1}}\\right)[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n$$\n"} {"id": "textbook_5990", "title": "1867. Check Your Learning - ", "content": "\nThe first step of a proposed multistep mechanism is:"} {"id": "textbook_5991", "title": "1867. Check Your Learning - ", "content": "$$\n\\mathrm{F}_{2}(g) \\rightleftharpoons 2 \\mathrm{~F}(g) \\text { fast }\n$$"} {"id": "textbook_5992", "title": "1867. Check Your Learning - ", "content": "Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.\n"} {"id": "textbook_5993", "title": "1868. Answer: - ", "content": "\n$[\\mathrm{F}]=\\left(\\frac{k_{1}\\left[\\mathrm{~F}_{2}\\right]}{k_{-1}}\\right)^{1 / 2}$\n"} {"id": "textbook_5994", "title": "1868. Answer: - 1868.1. Catalysis", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_5995", "title": "1868. Answer: - 1868.1. Catalysis", "content": "- Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams\n- List examples of catalysis in natural and industrial processes"} {"id": "textbook_5996", "title": "1868. Answer: - 1868.1. Catalysis", "content": ""} {"id": "textbook_5997", "title": "1869. Catalysts Do Not Affect Equilibrium - ", "content": "\nA catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations."} {"id": "textbook_5998", "title": "1869. Catalysts Do Not Affect Equilibrium - ", "content": "The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation"} {"id": "textbook_5999", "title": "1869. Catalysts Do Not Affect Equilibrium - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$"} {"id": "textbook_6000", "title": "1869. Catalysts Do Not Affect Equilibrium - ", "content": "A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year."} {"id": "textbook_6001", "title": "1869. Catalysts Do Not Affect Equilibrium - ", "content": "Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.\n"} {"id": "textbook_6002", "title": "1870. Portrait of a Chemist - ", "content": ""} {"id": "textbook_6003", "title": "1871. Fritz Haber - ", "content": "\nIn the early 20th century, German chemist Fritz Haber (Figure 17.19) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb."} {"id": "textbook_6004", "title": "1871. Fritz Haber - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$"} {"id": "textbook_6005", "title": "1871. Fritz Haber - ", "content": "The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for $78 \\%$ of air, diatomic nitrogen $\\left(\\mathrm{N}_{2}\\right)$ is nutritionally unavailable due the tremendous stability of the nitrogennitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation)."} {"id": "textbook_6006", "title": "1871. Fritz Haber - ", "content": "Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.\n"} {"id": "textbook_6007", "title": "1871. Fritz Haber - ", "content": "FIGURE 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery."} {"id": "textbook_6008", "title": "1871. Fritz Haber - ", "content": "In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, \"During peace time a scientist belongs to the World, but during war time he belongs to his country.\" ${ }^{1}$ Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science."} {"id": "textbook_6009", "title": "1871. Fritz Haber - ", "content": "Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995."} {"id": "textbook_6010", "title": "1871. Fritz Haber - ", "content": "It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood."} {"id": "textbook_6011", "title": "1871. Fritz Haber - ", "content": "To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which $\\mathrm{N}_{2}, \\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$ are at equilibrium or are coming to equilibrium."} {"id": "textbook_6012", "title": "1871. Fritz Haber - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$"} {"id": "textbook_6013", "title": "1871. Fritz Haber - ", "content": "The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure."} {"id": "textbook_6014", "title": "1871. Fritz Haber - ", "content": "Although increasing the pressure of a mixture of $\\mathrm{N}_{2}, \\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$ will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of $\\mathrm{N}_{2}$ and $\\mathrm{H}_{2}$, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:"} {"id": "textbook_6015", "title": "1871. Fritz Haber - ", "content": "$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g) \\quad \\Delta H=-92.2 \\mathrm{~kJ}\n$$"} {"id": "textbook_6016", "title": "1871. Fritz Haber - ", "content": "Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature."} {"id": "textbook_6017", "title": "1871. Fritz Haber - ", "content": "Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly."} {"id": "textbook_6018", "title": "1871. Fritz Haber - ", "content": "[^10]In the commercial production of ammonia, conditions of about $500^{\\circ} \\mathrm{C}, 150-900 \\mathrm{~atm}$, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 17.20).\n"} {"id": "textbook_6019", "title": "1871. Fritz Haber - ", "content": "FIGURE 17.20 Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant."} {"id": "textbook_6020", "title": "1871. Fritz Haber - ", "content": "Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function."} {"id": "textbook_6021", "title": "1871. Fritz Haber - ", "content": "Figure 17.21 shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn't necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower $E_{\\mathrm{a}}$ ).\n\n\nFIGURE 17.21 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states).\n"} {"id": "textbook_6022", "title": "1872. EXAMPLE 17.15 - ", "content": ""} {"id": "textbook_6023", "title": "1873. Reaction Diagrams for Catalyzed Reactions - ", "content": "\nThe two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst.\n\n(a)\n\n(b)\n"} {"id": "textbook_6024", "title": "1874. Solution - ", "content": "\nActivation energies are calculated by subtracting the reactant energy from the transition state energy."} {"id": "textbook_6025", "title": "1874. Solution - ", "content": "$$\n\\begin{aligned}\n& \\operatorname{diagram}(\\mathrm{a}): E_{\\mathrm{a}}=32 \\mathrm{~kJ}-6 \\mathrm{~kJ}=26 \\mathrm{~kJ} \\\\\n& \\operatorname{diagram}(\\mathrm{~b}): E_{\\mathrm{a}}=20 \\mathrm{~kJ}-6 \\mathrm{~kJ}=14 \\mathrm{~kJ}\n\\end{aligned}\n$$"} {"id": "textbook_6026", "title": "1874. Solution - ", "content": "The catalyzed reaction is the one with lesser activation energy, in this case represented by diagram (b).\n"} {"id": "textbook_6027", "title": "1875. Check Your Learning - ", "content": "\nReaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare?\n\n"} {"id": "textbook_6028", "title": "1876. Answer: - ", "content": "\nFor the first step, $E_{\\mathrm{a}}=80 \\mathrm{~kJ}$ for (a) and 70 kJ for (b), so diagram (b) depicts the catalyzed reaction. Activation energies for the second steps of both mechanisms are the same, 20 kJ .\n"} {"id": "textbook_6029", "title": "1877. Homogeneous Catalysts - ", "content": "\nA homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product."} {"id": "textbook_6030", "title": "1877. Homogeneous Catalysts - ", "content": "As an important illustration of homogeneous catalysis, consider the earth's ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:"} {"id": "textbook_6031", "title": "1877. Homogeneous Catalysts - ", "content": "$$\n3 \\mathrm{O}_{2}(\\mathrm{~g}) \\xrightarrow{h v} 2 \\mathrm{O}_{3}(\\mathrm{~g})\n$$"} {"id": "textbook_6032", "title": "1877. Homogeneous Catalysts - ", "content": "Ozone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following two-step mechanism:"} {"id": "textbook_6033", "title": "1877. Homogeneous Catalysts - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{O}_{3} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{O} \\\\\n& \\mathrm{O}+\\mathrm{O}_{3} \\longrightarrow 2 \\mathrm{O}_{2}\n\\end{aligned}\n$$"} {"id": "textbook_6034", "title": "1877. Homogeneous Catalysts - ", "content": "A number of substances can catalyze the decomposition of ozone. For example, the nitric oxide -catalyzed decomposition of ozone is believed to occur via the following three-step mechanism:"} {"id": "textbook_6035", "title": "1877. Homogeneous Catalysts - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{NO}(g)+\\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(g) \\\\\n& \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{O}_{2}(g)+\\mathrm{O}(g) \\\\\n& \\mathrm{NO}_{2}(g)+\\mathrm{O}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)\n\\end{aligned}\n$$"} {"id": "textbook_6036", "title": "1877. Homogeneous Catalysts - ", "content": "As required, the overall reaction is the same for both the two-step uncatalyzed mechanism and the three-step NO-catalyzed mechanism:"} {"id": "textbook_6037", "title": "1877. Homogeneous Catalysts - ", "content": "$$\n2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_6038", "title": "1877. Homogeneous Catalysts - ", "content": "Notice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction.\n"} {"id": "textbook_6039", "title": "1878. Portrait of a Chemist - ", "content": ""} {"id": "textbook_6040", "title": "1879. Mario J. Molina - ", "content": "\nThe 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure 17.22), and F. Sherwood Rowland \"for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone.\" ${ }^{2}$ Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).\n\n\nFIGURE 17.22 (a) Mexican chemist Mario Molina (1943 -) shared the Nobel Prize in Chemistry in 1995 for his research on (b) the Antarctic ozone hole. (credit a: courtesy of Mario Molina; credit b: modification of work by NASA)"} {"id": "textbook_6041", "title": "1879. Mario J. Molina - ", "content": "In 1974, Molina and Rowland published a paper in the journal Nature detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth's upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable \"hole\" forms above Antarctica, and an increase in the amount of solar ultraviolet radiation - strongly linked to the prevalence of skin cancers-reaches earth's surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction."} {"id": "textbook_6042", "title": "1879. Mario J. Molina - ", "content": "Molina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons-once widely used as refrigerants and propellants-are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:"} {"id": "textbook_6043", "title": "1879. Mario J. Molina - ", "content": "$$\n\\mathrm{CH}_{3} \\mathrm{Cl}+\\mathrm{OH} \\longrightarrow \\mathrm{Cl}+\\text { other products }\n$$"} {"id": "textbook_6044", "title": "1879. Mario J. Molina - ", "content": "Chlorine radicals break down ozone and are regenerated by the following catalytic cycle:"} {"id": "textbook_6045", "title": "1879. Mario J. Molina - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{Cl}+\\mathrm{O}_{3} \\longrightarrow \\mathrm{ClO}+\\mathrm{O}_{2} \\\\\n& \\mathrm{ClO}+\\mathrm{O} \\longrightarrow \\mathrm{Cl}+\\mathrm{O}_{2} \\\\\n& \\text { overall Reaction: } \\mathrm{O}_{3}+\\mathrm{O} \\longrightarrow 2 \\mathrm{O}_{2}\n\\end{aligned}\n$$"} {"id": "textbook_6046", "title": "1879. Mario J. Molina - ", "content": "A single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms $\\mathrm{Cl}_{2}$ and $\\mathrm{ClONO}_{2}$."} {"id": "textbook_6047", "title": "1879. Mario J. Molina - ", "content": "[^11]Since receiving his portion of the Nobel Prize, Molina has continued his work in atmospheric chemistry at MIT.\n"} {"id": "textbook_6048", "title": "1880. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_6049", "title": "1881. Glucose-6-Phosphate Dehydrogenase Deficiency - ", "content": "\nEnzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure 17.23, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (Figure 17.24).\n"} {"id": "textbook_6050", "title": "1881. Glucose-6-Phosphate Dehydrogenase Deficiency - ", "content": "FIGURE 17.23 Glucose-6-phosphate dehydrogenase is a rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells."} {"id": "textbook_6051", "title": "1881. Glucose-6-Phosphate Dehydrogenase Deficiency - ", "content": "A disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.\n"} {"id": "textbook_6052", "title": "1881. Glucose-6-Phosphate Dehydrogenase Deficiency - ", "content": "FIGURE 17.24 In the mechanism for the pentose phosphate pathway, G6PD catalyzes the reaction that regulates NADPH, a co-enzyme that regulates glutathione, an antioxidant that protects red blood cells and other cells from oxidative damage.\n"} {"id": "textbook_6053", "title": "1882. Heterogeneous Catalysts - ", "content": "\nA heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase."} {"id": "textbook_6054", "title": "1882. Heterogeneous Catalysts - ", "content": "Heterogeneous catalysis typically involves the following processes:"} {"id": "textbook_6055", "title": "1882. Heterogeneous Catalysts - ", "content": "1. Adsorption of the reactant(s) onto the surface of the catalyst\n2. Activation of the adsorbed reactant(s)\n3. Reaction of the adsorbed reactant(s)\n4. Desorption of product(s) from the surface of the catalyst"} {"id": "textbook_6056", "title": "1882. Heterogeneous Catalysts - ", "content": "Figure 17.25 illustrates the steps of a mechanism for the reaction of compounds containing a carbon-carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon-carbon double bonds) to produce saturated fats and oils (which contain only carbon-carbon single bonds).\n"} {"id": "textbook_6057", "title": "1882. Heterogeneous Catalysts - ", "content": "FIGURE 17.25 Mechanism for the Ni-catalyzed reaction $\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{6}$. (a) Hydrogen is adsorbed on the surface, breaking the $\\mathrm{H}-\\mathrm{H}$ bonds and forming $\\mathrm{Ni}-\\mathrm{H}$ bonds. (b) Ethylene is adsorbed on the surface, breaking the $\\mathrm{C}-\\mathrm{C}$ $\\pi$-bond and forming $\\mathrm{Ni}-\\mathrm{C}$ bonds. (c) Atoms diffuse across the surface and form new $\\mathrm{C}-\\mathrm{H}$ bonds when they collide.\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{6}$ molecules desorb from the Ni surface."} {"id": "textbook_6058", "title": "1882. Heterogeneous Catalysts - ", "content": "Many important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure 17.26).\n"} {"id": "textbook_6059", "title": "1883. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_6060", "title": "1884. Automobile Catalytic Converters - ", "content": "\nScientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure 17.26).\n\n\nFIGURE 17.26 A catalytic converter allows for the combustion of all carbon-containing compounds to carbon dioxide, while at the same time reducing the output of nitrogen oxide and other pollutants in emissions from gasoline-burning engines."} {"id": "textbook_6061", "title": "1884. Automobile Catalytic Converters - ", "content": "Most modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion of nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:"} {"id": "textbook_6062", "title": "1884. Automobile Catalytic Converters - ", "content": "$$\n\\begin{aligned}\n& 2 \\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\\\\n& 2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{CO}_{2}(\\mathrm{~g}) \\\\\n& 2 \\mathrm{C}_{8} \\mathrm{H}_{18}(g)+25 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 16 \\mathrm{CO}_{2}(\\mathrm{~g})+18 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g})\n\\end{aligned}\n$$"} {"id": "textbook_6063", "title": "1884. Automobile Catalytic Converters - ", "content": "In order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures.\n"} {"id": "textbook_6064", "title": "1885. LINK TO LEARNING - ", "content": "\nThe University of California at Davis' \"ChemWiki\" provides a thorough explanation (http://openstax.org/l/ 16 catconvert) of how catalytic converters work.\n"} {"id": "textbook_6065", "title": "1886. HOW SCIENCES INTERCONNECT - ", "content": ""} {"id": "textbook_6066", "title": "1887. Enzyme Structure and Function - ", "content": "\nThe study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table 17.3."} {"id": "textbook_6067", "title": "1887. Enzyme Structure and Function - ", "content": "Classes of Enzymes and Their Functions"} {"id": "textbook_6068", "title": "1887. Enzyme Structure and Function - ", "content": "| Class | Function |\n| :--- | :--- |\n| oxidoreductases | redox reactions |\n| transferases | transfer of functional groups |\n| hydrolases | hydrolysis reactions |\n| lyases | group elimination to form double bonds |\n| isomerases | isomerization |\n| ligases | bond formation with ATP hydrolysis |"} {"id": "textbook_6069", "title": "1887. Enzyme Structure and Function - ", "content": "TABLE 17.3"} {"id": "textbook_6070", "title": "1887. Enzyme Structure and Function - ", "content": "Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme's active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure 17.27).\n\n\nFIGURE 17.27 (a) According to the lock-and-key model, the shape of an enzyme's active site is a perfect fit for the substrate. (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate.\n"} {"id": "textbook_6071", "title": "1888. LINK TO LEARNING - ", "content": "\nThe Royal Society of Chemistry (http://openstax.org/l/16enzymes) provides an excellent introduction to enzymes for students and teachers."} {"id": "textbook_6072", "title": "1888. LINK TO LEARNING - ", "content": "The connection between the rate of a reaction and its equilibrium constant is one we can easily determine with just a bit of algebraic substitution. For a reaction where substance A forms B (and the reverse)"} {"id": "textbook_6073", "title": "1888. LINK TO LEARNING - ", "content": "$$\nA \\rightleftharpoons B\n$$"} {"id": "textbook_6074", "title": "1888. LINK TO LEARNING - ", "content": "The rate of the forward reaction is"} {"id": "textbook_6075", "title": "1888. LINK TO LEARNING - ", "content": "$$\n\\operatorname{Rate}(f)=k(f)[A]\n$$"} {"id": "textbook_6076", "title": "1888. LINK TO LEARNING - ", "content": "And the rate of the reverse reaction is"} {"id": "textbook_6077", "title": "1888. LINK TO LEARNING - ", "content": "$$\n\\operatorname{Rate}(r)=k(r)[B]\n$$"} {"id": "textbook_6078", "title": "1888. LINK TO LEARNING - ", "content": "Once equilibrium is established, the rates of the forward and reverse reactions are equal:"} {"id": "textbook_6079", "title": "1888. LINK TO LEARNING - ", "content": "$$\n\\operatorname{Rate}(f)=\\operatorname{Rate}(r)=k(f)[A]=k(r)[B]\n$$"} {"id": "textbook_6080", "title": "1888. LINK TO LEARNING - ", "content": "Rearranging a bit, we get"} {"id": "textbook_6081", "title": "1888. LINK TO LEARNING - ", "content": "$$\n\\operatorname{Rate}_{f}=\\operatorname{Rate}_{r} \\text { so } k(f)[A]=k(r)[B]\n$$"} {"id": "textbook_6082", "title": "1888. LINK TO LEARNING - ", "content": "Also recall that the equilibrium constant is simply the ratio of product to reactant concentration at equilibrium:"} {"id": "textbook_6083", "title": "1888. LINK TO LEARNING - ", "content": "$$\n\\begin{gathered}\n\\frac{k(f)}{k(r)}=\\frac{[B]}{[A]} \\\\\nK=\\frac{[B]}{[A]}\n\\end{gathered}\n$$"} {"id": "textbook_6084", "title": "1888. LINK TO LEARNING - ", "content": "So the equilibrium constant turns out to be the ratio of the forward to the reverse rate constants. This relationship also helps cement our understanding of the nature of a catalyst. That is, a catalyst does not change the fundamental equilibrium (or the underlying thermodynamics) of a reaction. Rather, what it does is alter the rate constant for the reaction - that is, both rate constants, forward and reverse, equally. In doing so, catalysts usually speed up the rate at which reactions attain equilibrium (though they can be used to slow down the rate of reaction as well!).\n"} {"id": "textbook_6085", "title": "1889. Key Terms - ", "content": "\nactivated complex (also, transition state) unstable combination of reactant species formed during a chemical reaction\nactivation energy $\\left(\\boldsymbol{E}_{\\mathrm{a}}\\right)$ minimum energy necessary in order for a reaction to take place\nArrhenius equation mathematical relationship between a reaction's rate constant, activation energy, and temperature\naverage rate rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred\nbimolecular reaction elementary reaction involving two reactant species\ncatalyst substance that increases the rate of a reaction without itself being consumed by the reaction\ncollision theory model that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics\nelementary reaction reaction that takes place in a single step, precisely as depicted in its chemical equation\nfrequency factor (A) proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation\nhalf-life of a reaction ( $\\boldsymbol{t}_{\\mathbf{l} / \\mathbf{2}}$ ) time required for half of a given amount of reactant to be consumed\nheterogeneous catalyst catalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur\nhomogeneous catalyst catalyst present in the same phase as the reactants\ninitial rate instantaneous rate of a chemical reaction at $t=0 \\mathrm{~s}$ (immediately after the reaction has begun)\ninstantaneous rate rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time\nintegrated rate law equation that relates the\nconcentration of a reactant to elapsed time of reaction\nintermediate species produced in one step of a reaction mechanism and consumed in a subsequent step\nmethod of initial rates common experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations\nmolecularity number of reactant species involved in an elementary reaction\noverall reaction order sum of the reaction orders for each substance represented in the rate law\nrate constant (k) proportionality constant in a rate law\nrate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time\nrate law (also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants\nrate of reaction measure of the speed at which a chemical reaction takes place\nrate-determining step (also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction\nreaction diagram used in chemical kinetics to illustrate various properties of a reaction\nreaction mechanism stepwise sequence of elementary reactions by which a chemical change takes place\nreaction order value of an exponent in a rate law (for example, zero order for 0 , first order for 1 , second order for 2 , and so on)\ntermolecular reaction elementary reaction involving three reactant species\nunimolecular reaction elementary reaction involving a single reactant species\n"} {"id": "textbook_6086", "title": "1890. Key Equations - ", "content": "\nrelative reaction rates for $a \\mathrm{~A} \\longrightarrow b \\mathrm{~B}=-\\frac{1}{a} \\frac{\\Delta[\\mathrm{~A}]}{\\Delta t}=\\frac{1}{b} \\frac{\\Delta[\\mathrm{~B}]}{\\Delta t}$\nintegrated rate law for zero-order reactions: $[A]_{t}=-k t+[A]_{0}$,\nhalf-life for a zero-order reaction $t_{1 / 2}=\\frac{[A]_{0}}{2 k}$\nintegrated rate law for first-order reactions: $\\ln [A]_{t}=-k t+\\ln [A]_{0}$,\nhalf-life for a first-order reaction $t_{1 / 2}=\\frac{0.693}{k}$\nintegrated rate law for second-order reactions: $\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}$,\nhalf-life for a second-order reaction $t_{1 / 2}=\\frac{1}{[A]_{0} k}$\n$k=A e^{-E_{\\mathrm{a}} / R T}$\n$\\ln k=\\left(\\frac{-E_{\\mathrm{a}}}{R}\\right)\\left(\\frac{1}{T}\\right)+\\ln A$\n$\\ln \\frac{k_{1}}{k_{2}}=\\frac{E_{\\mathrm{a}}}{R}\\left(\\frac{1}{T_{2}}-\\frac{1}{T_{1}}\\right)$\n"} {"id": "textbook_6087", "title": "1891. Summary - ", "content": ""} {"id": "textbook_6088", "title": "1891. Summary - 1891.1. Chemical Reaction Rates", "content": "\nThe rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.\n"} {"id": "textbook_6089", "title": "1891. Summary - 1891.2. Factors Affecting Reaction Rates", "content": "\nThe rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy.\n"} {"id": "textbook_6090", "title": "1891. Summary - 1891.3. Rate Laws", "content": "\nRate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.\n"} {"id": "textbook_6091", "title": "1891. Summary - 1891.4. Integrated Rate Laws", "content": "\nIntegrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations."} {"id": "textbook_6092", "title": "1891. Summary - 1891.4. Integrated Rate Laws", "content": "The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction's half-life varies with rate constant and,\nfor some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.\n"} {"id": "textbook_6093", "title": "1891. Summary - 1891.5. Collision Theory", "content": "\nChemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction's rate constant, activation energy, temperature, and dependence on collision orientation.\n"} {"id": "textbook_6094", "title": "1891. Summary - 1891.6. Reaction Mechanisms", "content": "\nThe sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.\n"} {"id": "textbook_6095", "title": "1891. Summary - 1891.7. Catalysis", "content": "\nCatalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy, but they do not affect equilibrium. Catalysts\ncan be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than\nthe reactants).\n"} {"id": "textbook_6096", "title": "1892. Exercises - ", "content": ""} {"id": "textbook_6097", "title": "1892. Exercises - 1892.1. Chemical Reaction Rates", "content": "\n1. What is the difference between average rate, initial rate, and instantaneous rate?\n2. Ozone decomposes to oxygen according to the equation $2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\\mathrm{O}_{3}$ and the formation of oxygen.\n3. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction $\\mathrm{Cl}_{2}(g)+3 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{ClF}_{3}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\\mathrm{Cl}_{2}$ and $\\mathrm{F}_{2}$ and the formation of $\\mathrm{ClF}_{3}$.\n4. A study of the rate of dimerization of $\\mathrm{C}_{4} \\mathrm{H}_{6}$ gave the data shown in the table:\n$2 \\mathrm{C}_{4} \\mathrm{H}_{6} \\longrightarrow \\mathrm{C}_{8} \\mathrm{H}_{12}$"} {"id": "textbook_6098", "title": "1892. Exercises - 1892.1. Chemical Reaction Rates", "content": "| Time (s) | 0 | 1600 | 3200 | 4800 | 6200 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{C}_{4} \\mathrm{H}_{6}\\right$ | $1.00 \\times 10^{-2}$ | $5.04 \\times 10^{-3}$ | $3.37 \\times 10^{-3}$ | $2.53 \\times 10^{-3}$ | $2.08 \\times 10^{-3}$ |"} {"id": "textbook_6099", "title": "1892. Exercises - 1892.1. Chemical Reaction Rates", "content": "(a) Determine the average rate of dimerization between 0 s and 1600 s , and between 1600 s and 3200 s .\n(b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus $\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]$. What are the units of this rate?\n(c) Determine the average rate of formation of $\\mathrm{C}_{8} \\mathrm{H}_{12}$ at 1600 s and the instantaneous rate of formation at 3200 $s$ from the rates found in parts (a) and (b).\n5. A study of the rate of the reaction represented as $2 A \\longrightarrow B$ gave the following data:"} {"id": "textbook_6100", "title": "1892. Exercises - 1892.1. Chemical Reaction Rates", "content": "| Time (s) | 0.0 | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 35.0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $A$ | 1.00 | 0.775 | 0.625 | 0.465 | 0.360 | 0.285 | 0.230 |"} {"id": "textbook_6101", "title": "1892. Exercises - 1892.1. Chemical Reaction Rates", "content": "(a) Determine the average rate of disappearance of $A$ between 0.0 s and 10.0 s , and between 10.0 s and 20.0 s .\n(b) Estimate the instantaneous rate of disappearance of $A$ at 15.0 s from a graph of time versus [ $A$ ]. What are the units of this rate?\n(c) Use the rates found in parts (a) and (b) to determine the average rate of formation of $B$ between 0.00 s and 10.0 s , and the instantaneous rate of formation of $B$ at 15.0 s .\n6. Consider the following reaction in aqueous solution:\n$5 \\mathrm{Br}^{-}(a q)+\\mathrm{BrO}_{3}{ }^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\longrightarrow 3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\nIf the rate of disappearance of $\\mathrm{Br}^{-}(\\mathrm{aq})$ at a particular moment during the reaction is $3.5 \\times 10^{-4} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$, what is the rate of appearance of $\\mathrm{Br}_{2}(a q)$ at that moment?\n"} {"id": "textbook_6102", "title": "1892. Exercises - 1892.2. Factors Affecting Reaction Rates", "content": "\n7. Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.\n8. Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)\n9. Go to the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen ( O ) and carbon monoxide ( CO ) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on \"Reload Launcher\" and change to \"Angled shot\" to see the difference.\n(a) What happens when the angle of the collision is changed?\n(b) Explain how this is relevant to rate of reaction.\n10. In the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive, use the \"Many Collisions\" tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select \"Show bonds\" under Options. How is the rate of the reaction affected by concentration and temperature?\n11. In the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select \"Show Bonds\" under Options.\n(a) Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?\n(b) Click \"Pause\" and then \"Reset All,\" and then enter 15 molecules of A and 10 molecules of BC once again. Select \"Show Bonds\" under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.\n"} {"id": "textbook_6103", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "\n12. How do the rate of a reaction and its rate constant differ?\n13. Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:\n(a) What is the order of the reaction with respect to that reactant?\n(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?\n14. Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:\n(a) What is the order of the reaction with respect to that reactant?\n(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold."} {"id": "textbook_6104", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "What is the order of the reaction with respect to that reactant?\n15. How will the rate of reaction change for the process: $\\mathrm{CO}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\\left[\\mathrm{NO}_{2}\\right]^{2}$ ?\n(a) Decreasing the pressure of $\\mathrm{NO}_{2}$ from 0.50 atm to 0.250 atm .\n(b) Increasing the concentration of CO from 0.01 M to 0.03 M .\n16. How will each of the following affect the rate of the reaction: $\\mathrm{CO}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\\left\\mathrm{NO}_{2}\\right$ ?\n(a) Increasing the pressure of $\\mathrm{NO}_{2}$ from 0.1 atm to 0.3 atm\n(b) Increasing the concentration of CO from 0.02 M to 0.06 M .\n17. Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $\\mathrm{NO}+\\mathrm{O}_{3} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$ is first order with respect to both NO and $\\mathrm{O}_{3}$ with a rate constant of $2.20 \\times 10^{7}$ $\\mathrm{L} / \\mathrm{mol} / \\mathrm{s}$. What is the instantaneous rate of disappearance of NO when $[\\mathrm{NO}]=3.3 \\times 10^{-6} \\mathrm{M}$ and $\\left[\\mathrm{O}_{3}\\right]=5.9 \\times$ $10^{-7} \\mathrm{M}$ ?\n18. Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:\n${ }_{15}^{32} \\mathrm{P} \\longrightarrow{ }_{16}^{32} \\mathrm{~S}+\\mathrm{e}^{-}$\nrate $=4.85 \\times 10^{-2}$ day $^{-1}\\left[{ }^{32} \\mathrm{P}\\right]$\nWhat is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M ?\n19. The rate constant for the radioactive decay of ${ }^{14} \\mathrm{C}$ is $1.21 \\times 10^{-4}$ year ${ }^{-1}$. The products of the decay are nitrogen atoms and electrons (beta particles):\n${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+\\mathrm{e}^{-}$\nrate $=k\\left[{ }_{6}^{14} \\mathrm{C}\\right]$\nWhat is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of $6.5 \\times$ $10^{-9} \\mathrm{M}$ ?\n20. The decomposition of acetaldehyde is a second order reaction with a rate constant of $4.71 \\times 10^{-8} \\mathrm{~L} \\mathrm{~mol}^{-1}$ $\\mathrm{s}^{-1}$. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of $5.55 \\times 10^{-4} \\mathrm{M}$ ?\n21. Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by $25-30 \\%$. Women metabolize alcohol a little more slowly than men:"} {"id": "textbook_6105", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $\\left\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right$ | $4.4 \\times 10^{-2}$ | $3.3 \\times 10^{-2}$ | $2.2 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $2.0 \\times 10^{-2}$ | $2.0 \\times 10^{-2}$ | $2.0 \\times 10^{-2}$ |"} {"id": "textbook_6106", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "Determine the rate law, the rate constant, and the overall order for this reaction.\n22. Under certain conditions the decomposition of ammonia on a metal surface gives the following data:"} {"id": "textbook_6107", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $\\left\\mathrm{NH}_{3}\\right$ | $1.0 \\times 10^{-3}$ | $2.0 \\times 10^{-3}$ | $3.0 \\times 10^{-3}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $1.5 \\times 10^{-6}$ | $1.5 \\times 10^{-6}$ | $1.5 \\times 10^{-6}$ |"} {"id": "textbook_6108", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "Determine the rate law, the rate constant, and the overall order for this reaction.\n23. Nitrosyl chloride, NOCl , decomposes to NO and $\\mathrm{Cl}_{2}$.\n$2 \\mathrm{NOCl}(g) \\longrightarrow 2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g)$\nDetermine the rate law, the rate constant, and the overall order for this reaction from the following data:"} {"id": "textbook_6109", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $\\mathrm{NOCl}$ | 0.10 | 0.20 | 0.30 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $8.0 \\times 10^{-10}$ | $3.2 \\times 10^{-9}$ | $7.2 \\times 10^{-9}$ |"} {"id": "textbook_6110", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "24. From the following data, determine the rate law, the rate constant, and the order with respect to $A$ for the reaction $A \\longrightarrow 2 C$."} {"id": "textbook_6111", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $A$ | $1.33 \\times 10^{-2}$ | $2.66 \\times 10^{-2}$ | $3.99 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $3.80 \\times 10^{-7}$ | $1.52 \\times 10^{-6}$ | $3.42 \\times 10^{-6}$ |"} {"id": "textbook_6112", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "25. Nitrogen monoxide reacts with chlorine according to the equation:\n$2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)$\nThe following initial rates of reaction have been observed for certain reactant concentrations:"} {"id": "textbook_6113", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| [NO] (mol/L) | $\\left\\mathrm{Cl}_{2}\\right$ | |\n| :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | | |\n| 0.50 | 0.50 | 1.14 |\n| 1.00 | 0.50 | 4.56 |\n| 1.00 | 1.00 | 9.12 |"} {"id": "textbook_6114", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "What is the rate law that describes the rate's dependence on the concentrations of NO and $\\mathrm{Cl}_{2}$ ? What is the rate constant? What are the orders with respect to each reactant?\n26. Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: $\\mathrm{H}_{2}(g)+2 \\mathrm{NO}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\nDetermine the rate law, the rate constant, and the orders with respect to each reactant from the following data:"} {"id": "textbook_6115", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $\\mathrm{NO}$ | 0.30 | 0.60 | 0.60 |\n| :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{H}_{2}\\right$ | 0.35 | 0.35 | 0.70 |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $2.835 \\times 10^{-3}$ | $1.134 \\times 10^{-2}$ | $2.268 \\times 10^{-2}$ |"} {"id": "textbook_6116", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "27. For the reaction $A \\longrightarrow B+C$, the following data were obtained at $30^{\\circ} \\mathrm{C}$ :"} {"id": "textbook_6117", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $A$ | 0.230 | 0.356 | 0.557 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $4.17 \\times 10^{-4}$ | $9.99 \\times 10^{-4}$ | $2.44 \\times 10^{-3}$ |"} {"id": "textbook_6118", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "(a) What is the order of the reaction with respect to [ $A$ ], and what is the rate law?\n(b) What is the rate constant?\n28. For the reaction $Q \\longrightarrow W+X$, the following data were obtained at $30^{\\circ} \\mathrm{C}$ :"} {"id": "textbook_6119", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "| $[Q]_{\\text {initial }}(M)$ | 0.170 | 0.212 | 0.357 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $6.68 \\times 10^{-3}$ | $1.04 \\times 10^{-2}$ | $2.94 \\times 10^{-2}$ |"} {"id": "textbook_6120", "title": "1892. Exercises - 1892.3. Rate Laws", "content": "(a) What is the order of the reaction with respect to [ $Q$ ], and what is the rate law?\n(b) What is the rate constant?\n29. The rate constant for the first-order decomposition at $45^{\\circ} \\mathrm{C}$ of dinitrogen pentoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$, dissolved in chloroform, $\\mathrm{CHCl}_{3}$, is $6.2 \\times 10^{-4} \\mathrm{~min}^{-1}$.\n$2 \\mathrm{~N}_{2} \\mathrm{O}_{5} \\longrightarrow 4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\nWhat is the rate of the reaction when $\\left[\\mathrm{N}_{2} \\mathrm{O}_{5}\\right]=0.40 \\mathrm{M}$ ?\n30. The annual production of $\\mathrm{HNO}_{3}$ in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.\n(a) $4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(b) $2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g)$\n(c) $3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g)$\n\nThe first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in $\\mathrm{O}_{2}$, what is the rate of formation of $\\mathrm{NO}_{2}$ when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M ? The rate constant for the reaction is $5.8 \\times 10^{-6} \\mathrm{~L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$.\n31. The following data have been determined for the reaction:\n$\\mathrm{I}^{-}+\\mathrm{OCl}^{-} \\longrightarrow \\mathrm{IO}^{-}+\\mathrm{Cl}^{-}$\n"} {"id": "textbook_6121", "title": "1893. 1 - ", "content": "\n2\n3"} {"id": "textbook_6122", "title": "1893. 1 - ", "content": "| $\\left[\\mathrm{I}^{-}\\right]_{\\text {initial }}(M)$ | 0.10 | 0.20 | 0.30 |\n| :---: | :---: | :---: | :---: |\n| $\\left[\\mathrm{OCl}^{-}\\right]_{\\text {initial }}(M)$ | 0.050 | 0.050 | 0.010 |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $3.05 \\times 10^{-4}$ | $6.20 \\times 10^{-4}$ | $1.83 \\times 10^{-4}$ |"} {"id": "textbook_6123", "title": "1893. 1 - ", "content": "Determine the rate law and the rate constant for this reaction.\n"} {"id": "textbook_6124", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "\n32. Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of $A$ at varying times.\n33. Use the data provided to graphically determine the order and rate constant of the following reaction:\n$\\mathrm{SO}_{2} \\mathrm{Cl}_{2} \\longrightarrow \\mathrm{SO}_{2}+\\mathrm{Cl}_{2}$"} {"id": "textbook_6125", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "| Time (s) | 0 | $5.00 \\times 10^{3}$ | $1.00 \\times 10^{4}$ | $1.50 \\times 10^{4}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right$ | 0.100 | 0.0896 | 0.0802 | 0.0719 |\n| Time (s) | $2.50 \\times 10^{4}$ | $3.00 \\times 10^{4}$ | $4.00 \\times 10^{4}$ | |\n| $\\left\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right$ | 0.0577 | 0.0517 | 0.0415 | |"} {"id": "textbook_6126", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "34. Pure ozone decomposes slowly to oxygen, $2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)$. Use the data provided in a graphical method and determine the order and rate constant of the reaction."} {"id": "textbook_6127", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "| Time (h) | 0 | $2.0 \\times 10^{3}$ | $7.6 \\times 10^{3}$ | $1.00 \\times 10^{4}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{O}_{3}\\right$ | $1.00 \\times 10^{-5}$ | $4.98 \\times 10^{-6}$ | $2.07 \\times 10^{-6}$ | $1.66 \\times 10^{-6}$ |\n| Time (h) | $1.23 \\times 10^{4}$ | $1.43 \\times 10^{4}$ | $1.70 \\times 10^{4}$ | |\n| $\\left\\mathrm{O}_{3}\\right$ | $1.39 \\times 10^{-6}$ | $1.22 \\times 10^{-6}$ | $1.05 \\times 10^{-6}$ | |"} {"id": "textbook_6128", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "35. From the given data, use a graphical method to determine the order and rate constant of the following reaction:\n$2 X \\longrightarrow Y+Z$"} {"id": "textbook_6129", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "| $2 X \\longrightarrow Y+$ Time (s) | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 30.0 | 35.0 | 40.0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $X$ | 0.0990 | 0.0497 | 0.0332 | 0.0249 | 0.0200 | 0.0166 | 0.0143 | 0.0125 |"} {"id": "textbook_6130", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "36. What is the half-life for the first-order decay of phosphorus- 32 ? $\\left({ }_{15}^{32} \\mathrm{P} \\longrightarrow{ }_{16}^{32} \\mathrm{~S}+\\mathrm{e}^{-}\\right)$The rate constant for the decay is $4.85 \\times 10^{-2}$ day $^{-1}$.\n37. What is the half-life for the first-order decay of carbon-14? $\\left({ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+\\mathrm{e}^{-}\\right)$The rate constant for the decay is $1.21 \\times 10^{-4}$ year $^{-1}$.\n38. What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M ? The rate constant for this second-order reaction is $8.0 \\times 10^{-8} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$.\n39. What is the half-life for the decomposition of $\\mathrm{O}_{3}$ when the concentration of $\\mathrm{O}_{3}$ is $2.35 \\times 10^{-6} \\mathrm{M}$ ? The rate constant for this second-order reaction is $50.4 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~h}^{-1}$.\n40. The reaction of compound $A$ to give compounds $C$ and $D$ was found to be second-order in $A$. The rate constant for the reaction was determined to be $2.42 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$. If the initial concentration is $0.500 \\mathrm{~mol} / \\mathrm{L}$, what is the value of $t_{1 / 2}$ ?\n41. The half-life of a reaction of compound $A$ to give compounds $D$ and $E$ is 8.50 min when the initial concentration of $A$ is 0.150 M . How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to $A$ or (b) second order with respect to $A$ ?\n42. Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of $3 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}$ that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing $0.15 \\mu \\mathrm{~g}\\left(0.15 \\times 10^{-6} \\mathrm{~g}\\right)$ of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant."} {"id": "textbook_6131", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "| [Penicillin] (M) | Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathbf{m i n}^{-1}\\right)$ | |\n| :--- | :--- | :--- |\n| $2.0 \\times 10^{-6}$ | $1.0 \\times 10^{-10}$ | |\n| $3.0 \\times 10^{-6}$ | $1.5 \\times 10^{-10}$ | |\n| $4.0 \\times 10^{-6}$ | $2.0 \\times 10^{-10}$ | |"} {"id": "textbook_6132", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "43. Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h , respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days ( 48 h )?\n44. There are two molecules with the formula $\\mathrm{C}_{3} \\mathrm{H}_{6}$. Propene, $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:\n
$\\left[\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{~N}_{3} \\mathrm{O}_{9}\\right]$
$(M)$ | 4.88 | 3.52 | 2.29 | 1.81 | 5.33 | 4.05 | 2.95 | 1.72 |\n| $t(\\mathrm{~s})$ | 300 | 300 | 300 | 300 | 180 | 180 | 180 | 180 |\n| $\\%$
Decomposed | 52.0 | 52.9 | 53.2 | 53.9 | 34.6 | 35.9 | 36.0 | 35.4 |"} {"id": "textbook_6135", "title": "1893. 1 - 1893.1. Integrated Rate Laws", "content": "49. For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene $\\left(\\mathrm{CH}_{2}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}\\right)$ has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:\n
$[\\mathrm{NO}]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | Initial Concentration,
$\\left[\\mathrm{H}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~min}^{-1}\\right)$ | Initial Rate of Formation
of $\\mathrm{N}_{2}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0060 | 0.0010 | $1.8 \\times 10^{-4}$ |\n| 2 | 0.0060 | 0.0020 | $3.6 \\times 10^{-4}$ |\n| 3 | 0.0010 | 0.0060 | $0.30 \\times 10^{-4}$ |\n| 4 | 0.0020 | 0.0060 | $1.2 \\times 10^{-4}$ |"} {"id": "textbook_6144", "title": "1893. 1 - 1893.3. Reaction Mechanisms", "content": "Consider the following questions:\n(a) Determine the order for each of the reactants, NO and $\\mathrm{H}_{2}$, from the data given and show your reasoning.\n(b) Write the overall rate law for the reaction.\n(c) Calculate the value of the rate constant, $k$, for the reaction. Include units.\n(d) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of $\\mathrm{H}_{2}$ had been consumed.\n(e) The following sequence of elementary steps is a proposed mechanism for the reaction."} {"id": "textbook_6145", "title": "1893. 1 - 1893.3. Reaction Mechanisms", "content": "Step 1: $\\mathrm{NO}+\\mathrm{NO} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{2}$\nStep 2: $\\mathrm{N}_{2} \\mathrm{O}_{2}+\\mathrm{H}_{2} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{N}_{2} \\mathrm{O}$\nStep 3: $\\mathrm{N}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\rightleftharpoons \\mathrm{~N}_{2}+\\mathrm{H}_{2} \\mathrm{O}$\nBased on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.\n75. The reaction of CO with $\\mathrm{Cl}_{2}$ gives phosgene $\\left(\\mathrm{COCl}_{2}\\right)$, a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:\n$\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{Cl}(g)$ (fast, $\\mathrm{k}_{1}$ represents the forward rate constant, $k_{-1}$ the reverse rate constant)\n$\\mathrm{CO}(g)+\\mathrm{Cl}(g) \\longrightarrow \\mathrm{COCl}(g)$ (slow, $k_{2}$ the rate constant)\n$\\mathrm{COCl}(g)+\\mathrm{Cl}(g) \\longrightarrow \\mathrm{COCl}_{2}(g)$ (fast, $k_{3}$ the rate constant)\n(a) Write the overall reaction.\n(b) Identify all intermediates.\n(c) Write the rate law for each elementary reaction.\n(d) Write the overall rate law expression.\n"} {"id": "textbook_6146", "title": "1893. 1 - 1893.4. Catalysis", "content": "\n76. Account for the increase in reaction rate brought about by a catalyst.\n77. Compare the functions of homogeneous and heterogeneous catalysts.\n78. Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:\n$\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$\n$\\mathrm{O}_{3}+\\mathrm{Cl} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{ClO}$\n$\\mathrm{ClO}+\\mathrm{O} \\longrightarrow \\mathrm{Cl}+\\mathrm{O}_{2}$\n(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:\n$2 \\mathrm{O}_{3} \\longrightarrow 3 \\mathrm{O}_{2}$\n(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism:\n$\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$\n$\\mathrm{O}_{3}+\\mathrm{NO} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\n$\\mathrm{NO}_{2}+\\mathrm{O} \\longrightarrow \\mathrm{NO}+\\mathrm{O}_{2}$\nIs NO a catalyst for the decomposition? Explain your answer.\n79. Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: $\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{C}(s) \\rightleftharpoons \\mathrm{H}_{2}(g)+\\mathrm{CO}(g)$. Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst. What will happen to the concentrations of $\\mathrm{H}_{2}$, CO , and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if more catalyst is added?\n80. Nitrogen and oxygen react at high temperatures. What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if a catalyst is added?\n81. For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:\n(a)\n\n(a)\n(b)\n\n(a)\n\n(b)\n\n(b)\n82. For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:\n(a)\n\n83. For each of the following reaction diagrams, estimate the activation energy $\\left(E_{\\mathrm{a}}\\right)$ of the reaction:\n(a)\n\n(b)\n\n84. For each of the following reaction diagrams, estimate the activation energy $\\left(E_{\\mathrm{a}}\\right)$ of the reaction:\n(a)\n\n(b)\n\n85. Assuming the diagrams in Exercise 17.83 represent different mechanisms for the same reaction, which of the reactions has the faster rate?\n86. Consider the similarities and differences in the two reaction diagrams shown in Exercise 17.84. Do these diagrams represent two different overall reactions, or do they represent the same overall reaction taking place by two different mechanisms? Explain your answer.\n"} {"id": "textbook_6147", "title": "1894. CHAPTER 18
Representative Metals, Metalloids, and Nonmetals - ", "content": "\n"} {"id": "textbook_6148", "title": "1894. CHAPTER 18
Representative Metals, Metalloids, and Nonmetals - ", "content": "Figure 18.1 Purity is extremely important when preparing silicon wafers. Technicians in a cleanroom prepare silicon without impurities (left). The CEO of VLSI Research, Don Hutcheson, shows off a pure silicon wafer (center). A silicon wafer covered in Pentium chips is an enlarged version of the silicon wafers found in many electronics used today (right). (credit middle: modification of work by \"Intel Free Press\"/Flickr; credit right: modification of work by Naotake Murayama)\n"} {"id": "textbook_6149", "title": "1895. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_6150", "title": "1895. CHAPTER OUTLINE - 1895.1. Periodicity", "content": "\n18.2 Occurrence and Preparation of the Representative Metals\n18.3 Structure and General Properties of the Metalloids\n18.4 Structure and General Properties of the Nonmetals\n18.5 Occurrence, Preparation, and Compounds of Hydrogen\n18.6 Occurrence, Preparation, and Properties of Carbonates\n18.7 Occurrence, Preparation, and Properties of Nitrogen\n18.8 Occurrence, Preparation, and Properties of Phosphorus\n18.9 Occurrence, Preparation, and Compounds of Oxygen\n18.10 Occurrence, Preparation, and Properties of Sulfur\n18.11 Occurrence, Preparation, and Properties of Halogens\n18.12 Occurrence, Preparation, and Properties of the Noble Gases"} {"id": "textbook_6151", "title": "1895. CHAPTER OUTLINE - 1895.1. Periodicity", "content": "INTRODUCTION The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table.\n"} {"id": "textbook_6152", "title": "1895. CHAPTER OUTLINE - 1895.2. Periodicity", "content": ""} {"id": "textbook_6153", "title": "1896. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_6154", "title": "1896. LEARNING OBJECTIVES - ", "content": "- Classify elements\n- Make predictions about the periodicity properties of the representative elements"} {"id": "textbook_6155", "title": "1896. LEARNING OBJECTIVES - ", "content": "We begin this section by examining the behaviors of representative metals in relation to their positions in the periodic table. The primary focus of this section will be the application of periodicity to the representative metals."} {"id": "textbook_6156", "title": "1896. LEARNING OBJECTIVES - ", "content": "It is possible to divide elements into groups according to their electron configurations. The representative elements are elements where the $s$ and $p$ orbitals are filling. The transition elements are elements where the $d$ orbitals (groups 3-11 on the periodic table) are filling, and the inner transition metals are the elements where the $f$ orbitals are filling. The $d$ orbitals fill with the elements in group 11; therefore, the elements in group 12 qualify as representative elements because the last electron enters an $s$ orbital. Metals among the representative elements are the representative metals. Metallic character results from an element's ability to lose its outer valence electrons and results in high thermal and electrical conductivity, among other physical and chemical properties. There are 20 nonradioactive representative metals in groups $1,2,3,12,13,14$, and 15 of the periodic table (the elements shaded in yellow in Figure 18.2). The radioactive elements copernicium, flerovium, polonium, and livermorium are also metals but are beyond the scope of this chapter."} {"id": "textbook_6157", "title": "1896. LEARNING OBJECTIVES - ", "content": "In addition to the representative metals, some of the representative elements are metalloids. A metalloid is an element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors."} {"id": "textbook_6158", "title": "1896. LEARNING OBJECTIVES - ", "content": "The remaining representative elements are nonmetals. Unlike metals, which typically form cations and ionic compounds (containing ionic bonds), nonmetals tend to form anions or molecular compounds. In general, the combination of a metal and a nonmetal produces a salt. A salt is an ionic compound consisting of cations and anions.\n"} {"id": "textbook_6159", "title": "1896. LEARNING OBJECTIVES - ", "content": "FIGURE 18.2 The location of the representative metals is shown in the periodic table. Nonmetals are shown in green, metalloids in purple, and the transition metals and inner transition metals in blue."} {"id": "textbook_6160", "title": "1896. LEARNING OBJECTIVES - ", "content": "Most of the representative metals do not occur naturally in an uncombined state because they readily react with water and oxygen in the air. However, it is possible to isolate elemental beryllium, magnesium, zinc, cadmium, mercury, aluminum, tin, and lead from their naturally occurring minerals and use them because they react very slowly with air. Part of the reason why these elements react slowly is that these elements react with air to form a protective coating. The formation of this protective coating is passivation. The coating is a nonreactive film of oxide or some other compound. Elemental magnesium, aluminum, zinc, and tin are important in the fabrication of many familiar items, including wire, cookware, foil, and many household and personal objects. Although beryllium, cadmium, mercury, and lead are readily available, there are limitations in their use because of their toxicity.\n"} {"id": "textbook_6161", "title": "1897. Group 1: The Alkali Metals - ", "content": "\nThe alkali metals lithium, sodium, potassium, rubidium, cesium, and francium constitute group 1 of the periodic table. Although hydrogen is in group 1 (and also in group 17), it is a nonmetal and deserves separate consideration later in this chapter. The name alkali metal is in reference to the fact that these metals and their oxides react with water to form very basic (alkaline) solutions."} {"id": "textbook_6162", "title": "1897. Group 1: The Alkali Metals - ", "content": "The properties of the alkali metals are similar to each other as expected for elements in the same family. The alkali metals have the largest atomic radii and the lowest first ionization energy in their periods. This combination makes it very easy to remove the single electron in the outermost (valence) shell of each. The easy loss of this valence electron means that these metals readily form stable cations with a charge of $1+$. Their\nreactivity increases with increasing atomic number due to the ease of losing the lone valence electron (decreasing ionization energy). Since oxidation is so easy, the reverse, reduction, is difficult, which explains why it is hard to isolate the elements. The solid alkali metals are very soft; lithium, shown in Figure 18.3, has the lowest density of any metal $\\left(0.5 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)$."} {"id": "textbook_6163", "title": "1897. Group 1: The Alkali Metals - ", "content": "The alkali metals all react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. This means they are easier to oxidize than is hydrogen. As an example, the reaction of lithium with water is:\n"} {"id": "textbook_6164", "title": "1897. Group 1: The Alkali Metals - ", "content": "FIGURE 18.3 Lithium floats in paraffin oil because its density is less than the density of paraffin oil.\nAlkali metals react directly with all the nonmetals (except the noble gases) to yield binary ionic compounds containing $1+$ metal ions. These metals are so reactive that it is necessary to avoid contact with both moisture and oxygen in the air. Therefore, they are stored in sealed containers under mineral oil, as shown in Figure 18.4 , to prevent contact with air and moisture. The pure metals never exist free (uncombined) in nature due to their high reactivity. In addition, this high reactivity makes it necessary to prepare the metals by electrolysis of alkali metal compounds.\n"} {"id": "textbook_6165", "title": "1897. Group 1: The Alkali Metals - ", "content": "FIGURE 18.4 To prevent contact with air and water, potassium for laboratory use comes as sticks or beads stored under kerosene or mineral oil, or in sealed containers. (credit: http://images-of-elements.com/potassium.php)"} {"id": "textbook_6166", "title": "1897. Group 1: The Alkali Metals - ", "content": "Unlike many other metals, the reactivity and softness of the alkali metals make these metals unsuitable for structural applications. However, there are applications where the reactivity of the alkali metals is an advantage. For example, the production of metals such as titanium and zirconium relies, in part, on the ability of sodium to reduce compounds of these metals. The manufacture of many organic compounds, including\ncertain dyes, drugs, and perfumes, utilizes reduction by lithium or sodium.\nSodium and its compounds impart a bright yellow color to a flame, as seen in Figure 18.5. Passing an electrical discharge through sodium vapor also produces this color. In both cases, this is an example of an emission spectrum as discussed in the chapter on electronic structure. Streetlights sometime employ sodium vapor lights because the sodium vapor penetrates fog better than most other light. This is because the fog does not scatter yellow light as much as it scatters white light. The other alkali metals and their salts also impart color to a flame. Lithium creates a bright, crimson color, whereas the others create a pale, violet color.\n"} {"id": "textbook_6167", "title": "1897. Group 1: The Alkali Metals - ", "content": "FIGURE 18.5 Dipping a wire into a solution of a sodium salt and then heating the wire causes emission of a bright yellow light, characteristic of sodium.\n"} {"id": "textbook_6168", "title": "1898. LINK TO LEARNING - ", "content": "\nThis video (http://openstax.org/l/16alkalih2o) demonstrates the reactions of the alkali metals with water.\n"} {"id": "textbook_6169", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "\nThe alkaline earth metals (beryllium, magnesium, calcium, strontium, barium, and radium) constitute group 2 of the periodic table. The name alkaline metal comes from the fact that the oxides of the heavier members of the group react with water to form alkaline solutions. The nuclear charge increases when going from group 1 to group 2. Because of this charge increase, the atoms of the alkaline earth metals are smaller and have higher first ionization energies than the alkali metals within the same period. The higher ionization energy makes the alkaline earth metals less reactive than the alkali metals; however, they are still very reactive elements. Their reactivity increases, as expected, with increasing size and decreasing ionization energy. In chemical reactions, these metals readily lose both valence electrons to form compounds in which they exhibit an oxidation state of $2+$. Due to their high reactivity, it is common to produce the alkaline earth metals, like the alkali metals, by\nelectrolysis. Even though the ionization energies are low, the two metals with the highest ionization energies (beryllium and magnesium) do form compounds that exhibit some covalent characters. Like the alkali metals, the heavier alkaline earth metals impart color to a flame. As in the case of the alkali metals, this is part of the emission spectrum of these elements. Calcium and strontium produce shades of red, whereas barium produces a green color."} {"id": "textbook_6170", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "Magnesium is a silver-white metal that is malleable and ductile at high temperatures. Passivation decreases the reactivity of magnesium metal. Upon exposure to air, a tightly adhering layer of magnesium oxycarbonate forms on the surface of the metal and inhibits further reaction. (The carbonate comes from the reaction of carbon dioxide in the atmosphere.) Magnesium is the lightest of the widely used structural metals, which is why most magnesium production is for lightweight alloys."} {"id": "textbook_6171", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "Magnesium (shown in Figure 18.6), calcium, strontium, and barium react with water and air. At room temperature, barium shows the most vigorous reaction. The products of the reaction with water are hydrogen and the metal hydroxide. The formation of hydrogen gas indicates that the heavier alkaline earth metals are better reducing agents (more easily oxidized) than is hydrogen. As expected, these metals react with both acids and nonmetals to form ionic compounds. Unlike most salts of the alkali metals, many of the common salts of the alkaline earth metals are insoluble in water because of the high lattice energies of these compounds, containing a divalent metal ion.\n"} {"id": "textbook_6172", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "FIGURE 18.6 From left to right: $\\mathrm{Mg}(s)$, warm water at pH 7 , and the resulting solution with a pH greater than 7 , as indicated by the pink color of the phenolphthalein indicator. (credit: modification of work by Sahar Atwa)"} {"id": "textbook_6173", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "The potent reducing power of hot magnesium is useful in preparing some metals from their oxides. Indeed, magnesium's affinity for oxygen is so great that burning magnesium reacts with carbon dioxide, producing elemental carbon:"} {"id": "textbook_6174", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "$$\n2 \\mathrm{Mg}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow 2 \\mathrm{MgO}(s)+\\mathrm{C}(s)\n$$"} {"id": "textbook_6175", "title": "1899. Group 2: The Alkaline Earth Metals - ", "content": "For this reason, $\\mathrm{CO}_{2}$ fire extinguisher will not extinguish a magnesium fire. Additionally, the brilliant white light emitted by burning magnesium makes it useful in flares and fireworks.\n"} {"id": "textbook_6176", "title": "1900. Group 12 - ", "content": "\nThe elements in group 12 are transition elements; however, the last electron added is not a $d$ electron, but an $s$ electron. Since the last electron added is an $s$ electron, these elements qualify as representative metals, or post-transition metals. The group 12 elements behave more like the alkaline earth metals than transition metals. Group 12 contains the four elements zinc, cadmium, mercury, and copernicium. Each of these elements has two electrons in its outer shell $\\left(n s^{2}\\right)$. When atoms of these metals form cations with a charge of $2+$, where the two outer electrons are lost, they have pseudo-noble gas electron configurations. Mercury is sometimes an exception because it also exhibits an oxidation state of $1+i n$ compounds that contain a diatomic $\\mathrm{Hg}_{2}{ }^{2+}$ ion. In their elemental forms and in compounds, cadmium and mercury are both toxic."} {"id": "textbook_6177", "title": "1900. Group 12 - ", "content": "Zinc is the most reactive in group 12, and mercury is the least reactive. (This is the reverse of the reactivity\ntrend of the metals of groups 1 and 2, in which reactivity increases down a group. The increase in reactivity with increasing atomic number only occurs for the metals in groups 1 and 2.) The decreasing reactivity is due to the formation of ions with a pseudo-noble gas configuration and to other factors that are beyond the scope of this discussion. The chemical behaviors of zinc and cadmium are quite similar to each other but differ from that of mercury."} {"id": "textbook_6178", "title": "1900. Group 12 - ", "content": "Zinc and cadmium have lower reduction potentials than hydrogen, and, like the alkali metals and alkaline earth metals, they will produce hydrogen gas when they react with acids. The reaction of zinc with hydrochloric acid, shown in Figure 18.7, is:\n"} {"id": "textbook_6179", "title": "1900. Group 12 - ", "content": "FIGURE 18.7 Zinc is an active metal. It dissolves in hydrochloric acid, forming a solution of colorless $\\mathrm{Zn}^{2+}$ ions, $\\mathrm{Cl}^{-}$ ions, and hydrogen gas."} {"id": "textbook_6180", "title": "1900. Group 12 - ", "content": "Zinc is a silvery metal that quickly tarnishes to a blue-gray appearance. This change in color is due to an adherent coating of a basic carbonate, $\\mathrm{Zn}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}$, which passivates the metal to inhibit further corrosion. Dry cell and alkaline batteries contain a zinc anode. Brass ( Cu and Zn ) and some bronze ( Cu , Sn , and sometimes Zn ) are important zinc alloys. About half of zinc production serves to protect iron and other metals from corrosion. This protection may take the form of a sacrificial anode (also known as a galvanic anode, which is a means of providing cathodic protection for various metals) or as a thin coating on the protected metal. Galvanized steel is steel with a protective coating of zinc.\n"} {"id": "textbook_6181", "title": "1901. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_6182", "title": "1902. Sacrificial Anodes - ", "content": "\nA sacrificial anode, or galvanic anode, is a means of providing cathodic protection of various metals. Cathodic protection refers to the prevention of corrosion by converting the corroding metal into a cathode. As a cathode, the metal resists corrosion, which is an oxidation process. Corrosion occurs at the sacrificial anode instead of at the cathode."} {"id": "textbook_6183", "title": "1902. Sacrificial Anodes - ", "content": "The construction of such a system begins with the attachment of a more active metal (more negative reduction potential) to the metal needing protection. Attachment may be direct or via a wire. To complete the circuit, a salt bridge is necessary. This salt bridge is often seawater or ground water. Once the circuit is complete, oxidation (corrosion) occurs at the anode and not the cathode."} {"id": "textbook_6184", "title": "1902. Sacrificial Anodes - ", "content": "The commonly used sacrificial anodes are magnesium, aluminum, and zinc. Magnesium has the most negative reduction potential of the three and serves best when the salt bridge is less efficient due to a low electrolyte concentration such as in freshwater. Zinc and aluminum work better in saltwater than does magnesium. Aluminum is lighter than zinc and has a higher capacity; however, an oxide coating may passivate the aluminum. In special cases, other materials are useful. For example, iron will protect copper."} {"id": "textbook_6185", "title": "1902. Sacrificial Anodes - ", "content": "Mercury is very different from zinc and cadmium. Mercury is the only metal that is liquid at $25^{\\circ} \\mathrm{C}$. Many metals dissolve in mercury, forming solutions called amalgams (see the feature on Amalgams), which are alloys of mercury with one or more other metals. Mercury, shown in Figure 18.8, is a nonreactive element that is more difficult to oxidize than hydrogen. Thus, it does not displace hydrogen from acids; however, it will react with strong oxidizing acids, such as nitric acid:"} {"id": "textbook_6186", "title": "1902. Sacrificial Anodes - ", "content": "$$\n\\begin{gathered}\n\\mathrm{Hg}(l)+\\mathrm{HCl}(a q) \\longrightarrow \\text { no reaction } \\\\\n3 \\mathrm{Hg}(l)+8 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{NO}(g)\n\\end{gathered}\n$$"} {"id": "textbook_6187", "title": "1902. Sacrificial Anodes - ", "content": "The clear NO initially formed quickly undergoes further oxidation to the reddish brown $\\mathrm{NO}_{2}$.\n"} {"id": "textbook_6188", "title": "1902. Sacrificial Anodes - ", "content": "FIGURE 18.8 From left to right: $\\mathrm{Hg}\\left(\\mathrm{l}, \\mathrm{Hg}+\\right.$ concentrated $\\mathrm{HCl}, \\mathrm{Hg}+$ concentrated $\\mathrm{HNO}_{3}$. (credit: Sahar Atwa)\nMost mercury compounds decompose when heated. Most mercury compounds contain mercury with a 2+oxidation state. When there is a large excess of mercury, it is possible to form compounds containing the $\\mathrm{Hg}_{2}{ }^{2+}$ ion. All mercury compounds are toxic, and it is necessary to exercise great care in their synthesis.\n"} {"id": "textbook_6189", "title": "1903. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_6190", "title": "1904. Amalgams - ", "content": "\nAn amalgam is an alloy of mercury with one or more other metals. This is similar to considering steel to be an alloy of iron with other metals. Most metals will form an amalgam with mercury, with the main exceptions being iron, platinum, tungsten, and tantalum."} {"id": "textbook_6191", "title": "1904. Amalgams - ", "content": "Due to toxicity issues with mercury, there has been a significant decrease in the use of amalgams. Historically, amalgams were important in electrolytic cells and in the extraction of gold. Amalgams of the alkali metals still find use because they are strong reducing agents and easier to handle than the pure alkali metals."} {"id": "textbook_6192", "title": "1904. Amalgams - ", "content": "Prospectors had a problem when they found finely divided gold. They learned that adding mercury to their pans collected the gold into the mercury to form an amalgam for easier collection. Unfortunately, losses of small amounts of mercury over the years left many streams in California polluted with mercury."} {"id": "textbook_6193", "title": "1904. Amalgams - ", "content": "Dentists use amalgams containing silver and other metals to fill cavities. There are several reasons to use an amalgam including low cost, ease of manipulation, and longevity compared to alternate materials. Dental amalgams are approximately $50 \\%$ mercury by weight, which, in recent years, has become a concern due to the toxicity of mercury."} {"id": "textbook_6194", "title": "1904. Amalgams - ", "content": "After reviewing the best available data, the Food and Drug Administration (FDA) considers amalgam-based fillings to be safe for adults and children over six years of age. Even with multiple fillings, the mercury levels in the patients remain far below the lowest levels associated with harm. Clinical studies have found no link between dental amalgams and health problems. Health issues may not be the same in cases of\nchildren under six or pregnant women. The FDA conclusions are in line with the opinions of the Environmental Protection Agency (EPA) and Centers for Disease Control (CDC). The only health consideration noted is that some people are allergic to the amalgam or one of its components.\n"} {"id": "textbook_6195", "title": "1905. Group 13 - ", "content": "\nGroup 13 contains the metalloid boron and the metals aluminum, gallium, indium, and thallium. The lightest element, boron, is semiconducting, and its binary compounds tend to be covalent and not ionic. The remaining elements of the group are metals, but their oxides and hydroxides change characters. The oxides and hydroxides of aluminum and gallium exhibit both acidic and basic behaviors. A substance, such as these two, that will react with both acids and bases is amphoteric. This characteristic illustrates the combination of nonmetallic and metallic behaviors of these two elements. Indium and thallium oxides and hydroxides exhibit only basic behavior, in accordance with the clearly metallic character of these two elements. The melting point of gallium is unusually low (about $30^{\\circ} \\mathrm{C}$ ) and will melt in your hand."} {"id": "textbook_6196", "title": "1905. Group 13 - ", "content": "Aluminum is amphoteric because it will react with both acids and bases. A typical reaction with an acid is:"} {"id": "textbook_6197", "title": "1905. Group 13 - ", "content": "$$\n2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_6198", "title": "1905. Group 13 - ", "content": "The products of the reaction of aluminum with a base depend upon the reaction conditions, with the following being one possibility:"} {"id": "textbook_6199", "title": "1905. Group 13 - ", "content": "$$\n2 \\mathrm{Al}(s)+2 \\mathrm{NaOH}(a q)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right+3 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_6200", "title": "1905. Group 13 - ", "content": "With both acids and bases, the reaction with aluminum generates hydrogen gas.\nThe group 13 elements have a valence shell electron configuration of $n s^{2} n p^{1}$. Aluminum normally uses all of its valence electrons when it reacts, giving compounds in which it has an oxidation state of $3+$. Although many of these compounds are covalent, others, such as $\\mathrm{AlF}_{3}$ and $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$, are ionic. Aqueous solutions of aluminum salts contain the cation $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$, abbreviated as $\\mathrm{Al}^{3+}(a q)$. Gallium, indium, and thallium also form ionic compounds containing $\\mathrm{M}^{3+}$ ions. These three elements exhibit not only the expected oxidation state of $3+$ from the three valence electrons but also an oxidation state (in this case, $1+$ ) that is two below the expected value. This phenomenon, the inert pair effect, refers to the formation of a stable ion with an oxidation state two lower than expected for the group. The pair of electrons is the valence $s$ orbital for those elements. In general, the inert pair effect is important for the lower $p$-block elements. In an aqueous solution, the $\\mathrm{Tl}^{+}(a q)$ ion is more stable than is $\\mathrm{Tl}^{3+}(a q)$. In general, these metals will react with air and water to form $3+\\mathrm{ions}$; however, thallium reacts to give thallium(I) derivatives. The metals of group 13 all react directly with nonmetals such as sulfur, phosphorus, and the halogens, forming binary compounds."} {"id": "textbook_6201", "title": "1905. Group 13 - ", "content": "The metals of group 13 ( $\\mathrm{Al}, \\mathrm{Ga}, \\mathrm{In}$, and Tl ) are all reactive. However, passivation occurs as a tough, hard, thin film of the metal oxide forms upon exposure to air. Disruption of this film may counter the passivation, allowing the metal to react. One way to disrupt the film is to expose the passivated metal to mercury. Some of the metal dissolves in the mercury to form an amalgam, which sheds the protective oxide layer to expose the metal to further reaction. The formation of an amalgam allows the metal to react with air and water.\n"} {"id": "textbook_6202", "title": "1906. LINK TO LEARNING - ", "content": "\nAlthough easily oxidized, the passivation of aluminum makes it very useful as a strong, lightweight building material. Because of the formation of an amalgam, mercury is corrosive to structural materials made of aluminum. This video (http://openstax.org/l/16aluminumhg) demonstrates how the integrity of an aluminum beam can be destroyed by the addition of a small amount of elemental mercury."} {"id": "textbook_6203", "title": "1906. LINK TO LEARNING - ", "content": "The most important uses of aluminum are in the construction and transportation industries, and in the manufacture of aluminum cans and aluminum foil. These uses depend on the lightness, toughness, and strength of the metal, as well as its resistance to corrosion. Because aluminum is an excellent conductor of heat and resists corrosion, it is useful in the manufacture of cooking utensils."} {"id": "textbook_6204", "title": "1906. LINK TO LEARNING - ", "content": "Aluminum is a very good reducing agent and may replace other reducing agents in the isolation of certain metals from their oxides. Although more expensive than reduction by carbon, aluminum is important in the isolation of $\\mathrm{Mo}, \\mathrm{W}$, and Cr from their oxides.\n"} {"id": "textbook_6205", "title": "1907. Group 14 - ", "content": "\nThe metallic members of group 14 are tin, lead, and flerovium. Carbon is a typical nonmetal. The remaining elements of the group, silicon and germanium, are examples of semimetals or metalloids. Tin and lead form the stable divalent cations, $\\mathrm{Sn}^{2+}$ and $\\mathrm{Pb}^{2+}$, with oxidation states two below the group oxidation state of $4+$. The stability of this oxidation state is a consequence of the inert pair effect. Tin and lead also form covalent compounds with a formal $4+$-oxidation state. For example, $\\mathrm{SnCl}_{4}$ and $\\mathrm{PbCl}_{4}$ are low-boiling covalent liquids.\n"} {"id": "textbook_6206", "title": "1907. Group 14 - ", "content": "FIGURE 18.9 (a) Tin(II) chloride is an ionic solid; (b) tin(IV) chloride is a covalent liquid.\nTin reacts readily with nonmetals and acids to form tin(II) compounds (indicating that it is more easily oxidized than hydrogen) and with nonmetals to form either tin(II) or tin(IV) compounds (shown in Figure 18.9), depending on the stoichiometry and reaction conditions. Lead is less reactive. It is only slightly easier to oxidize than hydrogen, and oxidation normally requires a hot concentrated acid."} {"id": "textbook_6207", "title": "1907. Group 14 - ", "content": "Many of these elements exist as allotropes. Allotropes are two or more forms of the same element in the same physical state with different chemical and physical properties. There are two common allotropes of tin. These allotropes are grey (brittle) tin and white tin. As with other allotropes, the difference between these forms of tin is in the arrangement of the atoms. White tin is stable above $13.2^{\\circ} \\mathrm{C}$ and is malleable like other metals. At low temperatures, gray tin is the more stable form. Gray tin is brittle and tends to break down to a powder. Consequently, articles made of tin will disintegrate in cold weather, particularly if the cold spell is lengthy. The change progresses slowly from the spot of origin, and the gray tin that is first formed catalyzes further change. In a way, this effect is similar to the spread of an infection in a plant or animal body, leading people to call this process tin disease or tin pest."} {"id": "textbook_6208", "title": "1907. Group 14 - ", "content": "The principal use of tin is in the coating of steel to form tin plate-sheet iron, which constitutes the tin in tin cans. Important tin alloys are bronze ( Cu and Sn ) and solder ( Sn and Pb ). Lead is important in the lead storage batteries in automobiles.\n"} {"id": "textbook_6209", "title": "1908. Group 15 - ", "content": "\nBismuth, the heaviest member of group 15, is a less reactive metal than the other representative metals. It readily gives up three of its five valence electrons to active nonmetals to form the tri-positive ion, $\\mathrm{Bi}^{3+}$. It forms compounds with the group oxidation state of $5+$ only when treated with strong oxidizing agents. The stability of the $3+$-oxidation state is another example of the inert pair effect.\n"} {"id": "textbook_6210", "title": "1908. Group 15 - 1908.1. Occurrence and Preparation of the Representative Metals", "content": ""} {"id": "textbook_6211", "title": "1909. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_6212", "title": "1909. LEARNING OBJECTIVES - ", "content": "- Identify natural sources of representative metals\n- Describe electrolytic and chemical reduction processes used to prepare these elements from natural sources"} {"id": "textbook_6213", "title": "1909. LEARNING OBJECTIVES - ", "content": "Because of their reactivity, we do not find most representative metals as free elements in nature. However, compounds that contain ions of most representative metals are abundant. In this section, we will consider the two common techniques used to isolate the metals from these compounds-electrolysis and chemical reduction."} {"id": "textbook_6214", "title": "1909. LEARNING OBJECTIVES - ", "content": "These metals primarily occur in minerals, with lithium found in silicate or phosphate minerals, and sodium and potassium found in salt deposits from evaporation of ancient seas and in silicates. The alkaline earth metals occur as silicates and, with the exception of beryllium, as carbonates and sulfates. Beryllium occurs as the mineral beryl, $\\mathrm{Be}_{3} \\mathrm{Al}_{2} \\mathrm{Si}_{6} \\mathrm{O}_{18}$, which, with certain impurities, may be either the gemstone emerald or aquamarine. Magnesium is in seawater and, along with the heavier alkaline earth metals, occurs as silicates, carbonates, and sulfates. Aluminum occurs abundantly in many types of clay and in bauxite, an impure aluminum oxide hydroxide. The principle tin ore is the oxide cassiterite, $\\mathrm{SnO}_{2}$, and the principle lead and thallium ores are the sulfides or the products of weathering of the sulfides. The remaining representative metals occur as impurities in zinc or aluminum ores.\n"} {"id": "textbook_6215", "title": "1910. Electrolysis - ", "content": "\nIons of metals in of groups 1 and 2, along with aluminum, are very difficult to reduce; therefore, it is necessary to prepare these elements by electrolysis, an important process discussed in the chapter on electrochemistry. Briefly, electrolysis involves using electrical energy to drive unfavorable chemical reactions to completion; it is useful in the isolation of reactive metals in their pure forms. Sodium, aluminum, and magnesium are typical examples."} {"id": "textbook_6216", "title": "1910. Electrolysis - ", "content": "The Preparation of Sodium\nThe most important method for the production of sodium is the electrolysis of molten sodium chloride; the set-up is a Downs cell, shown in Figure 18.10. The reaction involved in this process is:"} {"id": "textbook_6217", "title": "1910. Electrolysis - ", "content": "$$\n2 \\mathrm{NaCl}(l) \\xrightarrow[600^{\\circ} \\mathrm{C}]{\\text { electrolysis }} 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)\n$$"} {"id": "textbook_6218", "title": "1910. Electrolysis - ", "content": "The electrolysis cell contains molten sodium chloride (melting point $801^{\\circ} \\mathrm{C}$ ), to which calcium chloride has been added to lower the melting point to $600^{\\circ} \\mathrm{C}$ (a colligative effect). The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Chloride ions migrate to the positively charged anode, lose electrons, and undergo oxidation to chlorine gas. The overall cell reaction comes from adding the following reactions:"} {"id": "textbook_6219", "title": "1910. Electrolysis - ", "content": "$$\n\\begin{aligned}\n& \\text { at the cathode: } 2 \\mathrm{Na}^{+}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Na}(l) \\\\\n& \\text { at the anode: } 2 \\mathrm{Cl}^{-} \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\\\\n& \\text {overall change: } 2 \\mathrm{Na}^{+}+2 \\mathrm{Cl}^{-} \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)\n\\end{aligned}\n$$"} {"id": "textbook_6220", "title": "1910. Electrolysis - ", "content": "Separation of the molten sodium and chlorine prevents recombination. The liquid sodium, which is less dense than molten sodium chloride, floats to the surface and flows into a collector. The gaseous chlorine goes to storage tanks. Chlorine is also a valuable product.\n"} {"id": "textbook_6221", "title": "1910. Electrolysis - ", "content": "FIGURE 18.10 Pure sodium metal is isolated by electrolysis of molten sodium chloride using a Downs cell. It is not possible to isolate sodium by electrolysis of aqueous solutions of sodium salts because hydrogen ions are more easily reduced than are sodium ions; as a result, hydrogen gas forms at the cathode instead of the desired sodium metal. The high temperature required to melt NaCl means that liquid sodium metal forms."} {"id": "textbook_6222", "title": "1910. Electrolysis - ", "content": "The Preparation of Aluminum\nThe preparation of aluminum utilizes a process invented in 1886 by Charles M. Hall, who began to work on the problem while a student at Oberlin College in Ohio. Paul L. T. H\u00e9roult discovered the process independently a month or two later in France. In honor to the two inventors, this electrolysis cell is known as the Hall-H\u00e9roult cell. The Hall-H\u00e9roult cell is an electrolysis cell for the production of aluminum. Figure 18.11 illustrates the Hall-H\u00e9roult cell."} {"id": "textbook_6223", "title": "1910. Electrolysis - ", "content": "The production of aluminum begins with the purification of bauxite, the most common source of aluminum. The reaction of bauxite, $\\mathrm{AlO}(\\mathrm{OH})$, with hot sodium hydroxide forms soluble sodium aluminate, while clay and other impurities remain undissolved:"} {"id": "textbook_6224", "title": "1910. Electrolysis - ", "content": "$$\n\\mathrm{AlO}(\\mathrm{OH})(s)+\\mathrm{NaOH}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right\n$$"} {"id": "textbook_6225", "title": "1910. Electrolysis - ", "content": "After the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:"} {"id": "textbook_6226", "title": "1910. Electrolysis - ", "content": "$$\n\\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{Na}^{+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"} {"id": "textbook_6227", "title": "1910. Electrolysis - ", "content": "The next step is to remove the precipitated aluminum hydroxide by filtration. Heating the hydroxide produces aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, which dissolves in a molten mixture of cryolite, $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$, and calcium fluoride, $\\mathrm{CaF}_{2}$. Electrolysis of this solution takes place in a cell like that shown in Figure 18.11. Reduction of aluminum ions to the metal occurs at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode.\n\n\nFIGURE 18.11 An electrolytic cell is used for the production of aluminum. The electrolysis of a solution of cryolite and calcium fluoride results in aluminum metal at the cathode, and oxygen, carbon monoxide, and carbon dioxide at the anode."} {"id": "textbook_6228", "title": "1910. Electrolysis - ", "content": "The Preparation of Magnesium\nMagnesium is the other metal that is isolated in large quantities by electrolysis. Seawater, which contains approximately $0.5 \\%$ magnesium chloride, serves as the major source of magnesium. Addition of calcium hydroxide to seawater precipitates magnesium hydroxide. The addition of hydrochloric acid to magnesium hydroxide, followed by evaporation of the resultant aqueous solution, leaves pure magnesium chloride. The electrolysis of molten magnesium chloride forms liquid magnesium and chlorine gas:"} {"id": "textbook_6229", "title": "1910. Electrolysis - ", "content": "$$\n\\begin{gathered}\n\\mathrm{MgCl}_{2}(a q)+\\mathrm{Ca}(\\mathrm{OH})_{2}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+\\mathrm{CaCl}_{2}(a q) \\\\\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{MgCl}_{2}(l) \\longrightarrow \\mathrm{Mg}(l)+\\mathrm{Cl}_{2}(g)\n\\end{gathered}\n$$"} {"id": "textbook_6230", "title": "1910. Electrolysis - ", "content": "Some production facilities have moved away from electrolysis completely. In the next section, we will see how the Pidgeon process leads to the chemical reduction of magnesium.\n"} {"id": "textbook_6231", "title": "1911. Chemical Reduction - ", "content": "\nIt is possible to isolate many of the representative metals by chemical reduction using other elements as reducing agents. In general, chemical reduction is much less expensive than electrolysis, and for this reason, chemical reduction is the method of choice for the isolation of these elements. For example, it is possible to produce potassium, rubidium, and cesium by chemical reduction, as it is possible to reduce the molten chlorides of these metals with sodium metal. This may be surprising given that these metals are more reactive than sodium; however, the metals formed are more volatile than sodium and can be distilled for collection. The removal of the metal vapor leads to a shift in the equilibrium to produce more metal (see how reactions can be driven in the discussions of Le Ch\u00e2telier's principle in the chapter on fundamental equilibrium concepts)."} {"id": "textbook_6232", "title": "1911. Chemical Reduction - ", "content": "The production of magnesium, zinc, and tin provide additional examples of chemical reduction."} {"id": "textbook_6233", "title": "1911. Chemical Reduction - ", "content": "The Preparation of Magnesium\nThe Pidgeon process involves the reaction of magnesium oxide with elemental silicon at high temperatures to form pure magnesium:"} {"id": "textbook_6234", "title": "1911. Chemical Reduction - ", "content": "$$\n\\mathrm{Si}(s)+2 \\mathrm{MgO}(s) \\xrightarrow{\\Delta} \\mathrm{SiO}_{2}(s)+2 \\mathrm{Mg}(g)\n$$"} {"id": "textbook_6235", "title": "1911. Chemical Reduction - ", "content": "Although this reaction is unfavorable in terms of thermodynamics, the removal of the magnesium vapor produced takes advantage of Le Ch\u00e2telier's principle to continue the forward progress of the reaction. Over $75 \\%$ of the world's production of magnesium, primarily in China, comes from this process."} {"id": "textbook_6236", "title": "1911. Chemical Reduction - ", "content": "The Preparation of Zinc\nZinc ores usually contain zinc sulfide, zinc oxide, or zinc carbonate. After separation of these compounds from the ores, heating in air converts the ore to zinc oxide by one of the following reactions:"} {"id": "textbook_6237", "title": "1911. Chemical Reduction - ", "content": "$$\n\\begin{gathered}\n2 \\mathrm{ZnS}(s)+3 \\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} 2 \\mathrm{ZnO}(s)+2 \\mathrm{SO}_{2}(g) \\\\\n\\mathrm{ZnCO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{ZnO}(s)+\\mathrm{CO}_{2}(g)\n\\end{gathered}\n$$"} {"id": "textbook_6238", "title": "1911. Chemical Reduction - ", "content": "Carbon, in the form of coal, reduces the zinc oxide to form zinc vapor:"} {"id": "textbook_6239", "title": "1911. Chemical Reduction - ", "content": "$$\n\\mathrm{ZnO}(s)+\\mathrm{C}(s) \\longrightarrow \\mathrm{Zn}(g)+\\mathrm{CO}(g)\n$$"} {"id": "textbook_6240", "title": "1911. Chemical Reduction - ", "content": "The zinc can be distilled (boiling point $907^{\\circ} \\mathrm{C}$ ) and condensed. This zinc contains impurities of cadmium ( 767 $\\left.{ }^{\\circ} \\mathrm{C}\\right)$, iron $\\left(2862^{\\circ} \\mathrm{C}\\right)$, lead $\\left(1750^{\\circ} \\mathrm{C}\\right)$, and arsenic $\\left(613^{\\circ} \\mathrm{C}\\right)$. Careful redistillation produces pure zinc. Arsenic and cadmium are distilled from the zinc because they have lower boiling points. At higher temperatures, the zinc is distilled from the other impurities, mainly lead and iron.\n"} {"id": "textbook_6241", "title": "1912. The Preparation of Tin - ", "content": "\nThe ready reduction of tin(IV) oxide by the hot coals of a campfire accounts for the knowledge of tin in the ancient world. In the modern process, the roasting of tin ores containing $\\mathrm{SnO}_{2}$ removes contaminants such as arsenic and sulfur as volatile oxides. Treatment of the remaining material with hydrochloric acid removes the oxides of other metals. Heating the purified ore with carbon at temperature above $1000{ }^{\\circ} \\mathrm{C}$ produces tin:"} {"id": "textbook_6242", "title": "1912. The Preparation of Tin - ", "content": "$$\n\\mathrm{SnO}_{2}(s)+2 \\mathrm{C}(s) \\xrightarrow{\\Delta} \\mathrm{Sn}(s)+2 \\mathrm{CO}(g)\n$$"} {"id": "textbook_6243", "title": "1912. The Preparation of Tin - ", "content": "The molten tin collects at the bottom of the furnace and is drawn off and cast into blocks.\n"} {"id": "textbook_6244", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:"} {"id": "textbook_6245", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "- Describe the general preparation, properties, and uses of the metalloids\n- Describe the preparation, properties, and compounds of boron and silicon"} {"id": "textbook_6246", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "A series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements-boron and silicon-in more detail."} {"id": "textbook_6247", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "The metalloid boron exhibits many similarities to its neighbor carbon and its diagonal neighbor silicon. All three elements form covalent compounds. However, boron has one distinct difference in that its $2 s^{2} 2 p^{1}$ outer electron structure gives it one less valence electron than it has valence orbitals. Although boron exhibits an oxidation state of $3+$ in most of its stable compounds, this electron deficiency provides boron with the ability to form other, sometimes fractional, oxidation states, which occur, for example, in the boron hydrides."} {"id": "textbook_6248", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "Silicon has the valence shell electron configuration $3 s^{2} 3 p^{2}$, and it commonly forms tetrahedral structures in which it is $s p^{3}$ hybridized with a formal oxidation state of $4+$. The major differences between the chemistry of carbon and silicon result from the relative strength of the carbon-carbon bond, carbon's ability to form stable bonds to itself, and the presence of the empty $3 d$ valence-shell orbitals in silicon. Silicon's empty $d$ orbitals and boron's empty $p$ orbital enable tetrahedral silicon compounds and trigonal planar boron compounds to act as Lewis acids. Carbon, on the other hand, has no available valence shell orbitals; tetrahedral carbon compounds cannot act as Lewis acids. Germanium is very similar to silicon in its chemical behavior."} {"id": "textbook_6249", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "Arsenic and antimony generally form compounds in which an oxidation state of $3+$ or $5+$ is exhibited; however, arsenic can form arsenides with an oxidation state of $3-$. These elements tarnish only slightly in dry air but readily oxidize when warmed."} {"id": "textbook_6250", "title": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids", "content": "Tellurium combines directly with most elements. The most stable tellurium compounds are the tellurides-salts of $\\mathrm{Te}^{2-}$ formed with active metals and lanthanides-and compounds with oxygen, fluorine, and chlorine, in which tellurium normally exhibits an oxidation state $2+$ or $4+$. Although tellurium(VI) compounds are known (for example, $\\mathrm{TeF}_{6}$ ), there is a marked resistance to oxidation to this maximum group oxidation state.\n"} {"id": "textbook_6251", "title": "1913. Structures of the Metalloids - ", "content": "\nCovalent bonding is the key to the crystal structures of the metalloids. In this regard, these elements resemble nonmetals in their behavior."} {"id": "textbook_6252", "title": "1913. Structures of the Metalloids - ", "content": "Elemental silicon, germanium, arsenic, antimony, and tellurium are lustrous, metallic-looking solids. Silicon and germanium crystallize with a diamond structure. Each atom within the crystal has covalent bonds to four neighboring atoms at the corners of a regular tetrahedron. Single crystals of silicon and germanium are giant, three-dimensional molecules. There are several allotropes of arsenic with the most stable being layer like and containing puckered sheets of arsenic atoms. Each arsenic atom forms covalent bonds to three other atoms within the sheet. The crystal structure of antimony is similar to that of arsenic, both shown in Figure 18.12. The structures of arsenic and antimony are similar to the structure of graphite, covered later in this chapter. Tellurium forms crystals that contain infinite spiral chains of tellurium atoms. Each atom in the chain bonds to two other atoms.\n"} {"id": "textbook_6253", "title": "1914. LINK TO LEARNING - ", "content": "\nExplore a cubic diamond (http://openstax.org/l/16crystal) crystal structure.\n"} {"id": "textbook_6254", "title": "1914. LINK TO LEARNING - ", "content": "FIGURE 18.12 (a) Arsenic and (b) antimony have a layered structure similar to that of (c) graphite, except that the layers are puckered rather than planar. (d) Elemental tellurium forms spiral chains."} {"id": "textbook_6255", "title": "1914. LINK TO LEARNING - ", "content": "Pure crystalline boron is transparent. The crystals consist of icosahedra, as shown in Figure 18.13, with a boron atom at each corner. In the most common form of boron, the icosahedra pack together in a manner similar to the cubic closest packing of spheres. All boron-boron bonds within each icosahedron are identical and are approximately 176 pm in length. In the different forms of boron, there are different arrangements and connections between the icosahedra.\n"} {"id": "textbook_6256", "title": "1914. LINK TO LEARNING - ", "content": "FIGURE 18.13 An icosahedron is a symmetrical, solid shape with 20 faces, each of which is an equilateral triangle. The faces meet at 12 corners."} {"id": "textbook_6257", "title": "1914. LINK TO LEARNING - ", "content": "The name silicon is derived from the Latin word for flint, silex. The metalloid silicon readily forms compounds containing Si-O-Si bonds, which are of prime importance in the mineral world. This bonding capability is in contrast to the nonmetal carbon, whose ability to form carbon-carbon bonds gives it prime importance in the plant and animal worlds."} {"id": "textbook_6258", "title": "1914. LINK TO LEARNING - ", "content": "Occurrence, Preparation, and Compounds of Boron and Silicon\nBoron constitutes less than $0.001 \\%$ by weight of the earth's crust. In nature, it only occurs in compounds with oxygen. Boron is widely distributed in volcanic regions as boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$, and in dry lake regions, including the desert areas of California, as borates and salts of boron oxyacids, such as borax, $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$."} {"id": "textbook_6259", "title": "1914. LINK TO LEARNING - ", "content": "Elemental boron is chemically inert at room temperature, reacting with only fluorine and oxygen to form boron trifluoride, $\\mathrm{BF}_{3}$, and boric oxide, $\\mathrm{B}_{2} \\mathrm{O}_{3}$, respectively. At higher temperatures, boron reacts with all nonmetals, except tellurium and the noble gases, and with nearly all metals; it oxidizes to $\\mathrm{B}_{2} \\mathrm{O}_{3}$ when heated with concentrated nitric or sulfuric acid. Boron does not react with nonoxidizing acids. Many boron compounds react readily with water to give boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$ (sometimes written as $\\mathrm{H}_{3} \\mathrm{BO}_{3}$ )."} {"id": "textbook_6260", "title": "1914. LINK TO LEARNING - ", "content": "Reduction of boric oxide with magnesium powder forms boron ( $95-98.5 \\%$ pure) as a brown, amorphous powder:"} {"id": "textbook_6261", "title": "1914. LINK TO LEARNING - ", "content": "$$\n\\mathrm{B}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{Mg}(s) \\longrightarrow 2 \\mathrm{~B}(s)+3 \\mathrm{MgO}(s)\n$$"} {"id": "textbook_6262", "title": "1914. LINK TO LEARNING - ", "content": "An amorphous substance is a material that appears to be a solid, but does not have a long-range order like a true solid. Treatment with hydrochloric acid removes the magnesium oxide. Further purification of the boron begins with conversion of the impure boron into boron trichloride. The next step is to heat a mixture of boron trichloride and hydrogen:"} {"id": "textbook_6263", "title": "1914. LINK TO LEARNING - ", "content": "$$\n2 \\mathrm{BCl}_{3}(g)+3 \\mathrm{H}_{2}(g) \\xrightarrow{1500^{\\circ} \\mathrm{C}} 2 \\mathrm{~B}(s)+6 \\mathrm{HCl}(g) \\quad \\Delta H^{\\circ}=253.7 \\mathrm{~kJ}\n$$"} {"id": "textbook_6264", "title": "1914. LINK TO LEARNING - ", "content": "Silicon makes up nearly one-fourth of the mass of the earth's crust-second in abundance only to oxygen. The crust is composed almost entirely of minerals in which the silicon atoms are at the center of the silicon-oxygen tetrahedron, which connect in a variety of ways to produce, among other things, chains, layers, and threedimensional frameworks. These minerals constitute the bulk of most common rocks, soil, and clays. In addition, materials such as bricks, ceramics, and glasses contain silicon compounds."} {"id": "textbook_6265", "title": "1914. LINK TO LEARNING - ", "content": "It is possible to produce silicon by the high-temperature reduction of silicon dioxide with strong reducing agents, such as carbon and magnesium:"} {"id": "textbook_6266", "title": "1914. LINK TO LEARNING - ", "content": "$$\n\\begin{aligned}\n& \\mathrm{SiO}_{2}(s)+2 \\mathrm{C}(s) \\xrightarrow{\\Delta} \\mathrm{Si}(s)+2 \\mathrm{CO}(g) \\\\\n& \\mathrm{SiO}_{2}(s)+2 \\mathrm{Mg}(s) \\xrightarrow{\\Delta} \\mathrm{Si}(s)+2 \\mathrm{MgO}(s)\n\\end{aligned}\n$$"} {"id": "textbook_6267", "title": "1914. LINK TO LEARNING - ", "content": "Extremely pure silicon is necessary for the manufacture of semiconductor electronic devices. This process begins with the conversion of impure silicon into silicon tetrahalides, or silane ( $\\mathrm{SiH}_{4}$ ), followed by\ndecomposition at high temperatures. Zone refining, illustrated in Figure 18.14, completes the purification. In this method, a rod of silicon is heated at one end by a heat source that produces a thin cross-section of molten silicon. Slowly lowering the rod through the heat source moves the molten zone from one end of the rod to other. As this thin, molten region moves, impurities in the silicon dissolve in the liquid silicon and move with the molten region. Ultimately, the impurities move to one end of the rod, which is then cut off.\n\n\nFIGURE 18.14 A zone-refining apparatus used to purify silicon.\nThis highly purified silicon, containing no more than one part impurity per million parts of silicon, is the most important element in the computer industry. Pure silicon is necessary in semiconductor electronic devices such as transistors, computer chips, and solar cells."} {"id": "textbook_6268", "title": "1914. LINK TO LEARNING - ", "content": "Like some metals, passivation of silicon occurs due the formation of a very thin film of oxide (primarily silicon dioxide, $\\mathrm{SiO}_{2}$ ). Silicon dioxide is soluble in hot aqueous base; thus, strong bases destroy the passivation. Removal of the passivation layer allows the base to dissolve the silicon, forming hydrogen gas and silicate anions. For example:"} {"id": "textbook_6269", "title": "1914. LINK TO LEARNING - ", "content": "$$\n\\mathrm{Si}(s)+4 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{SiO}_{4}{ }^{4-}(a q)+2 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_6270", "title": "1914. LINK TO LEARNING - ", "content": "Silicon reacts with halogens at high temperatures, forming volatile tetrahalides, such as $\\operatorname{SiF}_{4}$.\nUnlike carbon, silicon does not readily form double or triple bonds. Silicon compounds of the general formula $\\mathrm{SiX}_{4}$, where X is a highly electronegative group, can act as Lewis acids to form six-coordinate silicon. For example, silicon tetrafluoride, $\\mathrm{SiF}_{4}$, reacts with sodium fluoride to yield $\\mathrm{Na}_{2}\\left[\\mathrm{SiF}_{6}\\right]$, which contains the octahedral $\\left[\\mathrm{SiF}_{6}\\right]^{2-}$ ion in which silicon is $s p^{3} d^{2}$ hybridized:"} {"id": "textbook_6271", "title": "1914. LINK TO LEARNING - ", "content": "$$\n2 \\mathrm{NaF}(s)+\\mathrm{SiF}_{4}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{SiF}_{6}(s)\n$$"} {"id": "textbook_6272", "title": "1914. LINK TO LEARNING - ", "content": "Antimony reacts readily with stoichiometric amounts of fluorine, chlorine, bromine, or iodine, yielding trihalides or, with excess fluorine or chlorine, forming the pentahalides $\\mathrm{SbF}_{5}$ and $\\mathrm{SbCl}_{5}$. Depending on the stoichiometry, it forms antimony(III) sulfide, $\\mathrm{Sb}_{2} \\mathrm{~S}_{3}$, or antimony(V) sulfide when heated with sulfur. As expected, the metallic nature of the element is greater than that of arsenic, which lies immediately above it in group 15.\n"} {"id": "textbook_6273", "title": "1915. Boron and Silicon Halides - ", "content": "\nBoron trihalides $-\\mathrm{BF}_{3}, \\mathrm{BCl}_{3}, \\mathrm{BBr}_{3}$, and $\\mathrm{BI}_{3}-$ can be prepared by the direct reaction of the elements. These nonpolar molecules contain boron with $s p^{2}$ hybridization and a trigonal planar molecular geometry. The fluoride and chloride compounds are colorless gasses, the bromide is a liquid, and the iodide is a white crystalline solid."} {"id": "textbook_6274", "title": "1915. Boron and Silicon Halides - ", "content": "Except for boron trifluoride, the boron trihalides readily hydrolyze in water to form boric acid and the corresponding hydrohalic acid. Boron trichloride reacts according to the equation:"} {"id": "textbook_6275", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{BCl}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{B}(\\mathrm{OH})_{3}(a q)+3 \\mathrm{HCl}(a q)\n$$"} {"id": "textbook_6276", "title": "1915. Boron and Silicon Halides - ", "content": "Boron trifluoride reacts with hydrofluoric acid, to yield a solution of fluoroboric acid, $\\mathrm{HBF}_{4}$ :"} {"id": "textbook_6277", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{BF}_{3}(a q)+\\mathrm{HF}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{BF}_{4}^{-}(a q)\n$$"} {"id": "textbook_6278", "title": "1915. Boron and Silicon Halides - ", "content": "In this reaction, the $\\mathrm{BF}_{3}$ molecule acts as the Lewis acid (electron pair acceptor) and accepts a pair of electrons from a fluoride ion:\n"} {"id": "textbook_6279", "title": "1915. Boron and Silicon Halides - ", "content": "All the tetrahalides of silicon, $\\mathrm{SiX}_{4}$, have been prepared. Silicon tetrachloride can be prepared by direct chlorination at elevated temperatures or by heating silicon dioxide with chlorine and carbon:"} {"id": "textbook_6280", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{SiO}_{2}(s)+2 \\mathrm{C}(s)+2 \\mathrm{Cl}_{2}(g) \\xrightarrow{\\Delta} \\mathrm{SiCl}_{4}(g)+2 \\mathrm{CO}(g)\n$$"} {"id": "textbook_6281", "title": "1915. Boron and Silicon Halides - ", "content": "Silicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling $\\left(57^{\\circ} \\mathrm{C}\\right)$, colorless liquid."} {"id": "textbook_6282", "title": "1915. Boron and Silicon Halides - ", "content": "It is possible to prepare silicon tetrafluoride by the reaction of silicon dioxide with hydrofluoric acid:"} {"id": "textbook_6283", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{SiO}_{2}(s)+4 \\mathrm{HF}(g) \\longrightarrow \\mathrm{SiF}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-191.2 \\mathrm{~kJ}\n$$"} {"id": "textbook_6284", "title": "1915. Boron and Silicon Halides - ", "content": "Hydrofluoric acid is the only common acid that will react with silicon dioxide or silicates. This reaction occurs because the silicon-fluorine bond is the only bond that silicon forms that is stronger than the silicon-oxygen bond. For this reason, it is possible to store all common acids, other than hydrofluoric acid, in glass containers."} {"id": "textbook_6285", "title": "1915. Boron and Silicon Halides - ", "content": "Except for silicon tetrafluoride, silicon halides are extremely sensitive to water. Upon exposure to water, $\\mathrm{SiCl}_{4}$ reacts rapidly with hydroxide groups, replacing all four chlorine atoms to produce unstable orthosilicic acid, $\\mathrm{Si}(\\mathrm{OH})_{4}$ or $\\mathrm{H}_{4} \\mathrm{SiO}_{4}$, which slowly decomposes into $\\mathrm{SiO}_{2}$."} {"id": "textbook_6286", "title": "1915. Boron and Silicon Halides - ", "content": "Boron and Silicon Oxides and Derivatives\nBoron burns at $700^{\\circ} \\mathrm{C}$ in oxygen, forming boric oxide, $\\mathrm{B}_{2} \\mathrm{O}_{3}$. Boric oxide is necessary for the production of heat-resistant borosilicate glass, like that shown in Figure 18.15 and certain optical glasses. Boric oxide dissolves in hot water to form boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$ :"} {"id": "textbook_6287", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{B}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{~B}(\\mathrm{OH})_{3}(a q)\n$$"} {"id": "textbook_6288", "title": "1915. Boron and Silicon Halides - ", "content": ""} {"id": "textbook_6289", "title": "1915. Boron and Silicon Halides - ", "content": "FIGURE 18.15 Laboratory glassware, such as Pyrex and Kimax, is made of borosilicate glass because it does not break when heated. The inclusion of borates in the glass helps to mediate the effects of thermal expansion and contraction. This reduces the likelihood of thermal shock, which causes silicate glass to crack upon rapid heating or cooling. (credit: \"Tweenk\"/Wikimedia Commons)"} {"id": "textbook_6290", "title": "1915. Boron and Silicon Halides - ", "content": "The boron atom in $\\mathrm{B}(\\mathrm{OH})_{3}$ is $s p^{2}$ hybridized and is located at the center of an equilateral triangle with oxygen atoms at the corners. In solid $\\mathrm{B}(\\mathrm{OH})_{3}$, hydrogen bonding holds these triangular units together. Boric acid, shown in Figure 18.16, is a very weak acid that does not act as a proton donor but rather as a Lewis acid, accepting an unshared pair of electrons from the Lewis base $\\mathrm{OH}^{-}$:"} {"id": "textbook_6291", "title": "1915. Boron and Silicon Halides - ", "content": "$$\n\\mathrm{B}(\\mathrm{OH})_{3}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{B}(\\mathrm{OH})_{4}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\quad K_{\\mathrm{a}}=5.8 \\times 10^{-10}\n$$"} {"id": "textbook_6292", "title": "1915. Boron and Silicon Halides - ", "content": ""} {"id": "textbook_6293", "title": "1915. Boron and Silicon Halides - ", "content": "FIGURE 18.16 Boric acid has a planar structure with three -OH groups spread out equally at $120^{\\circ}$ angles from each other."} {"id": "textbook_6294", "title": "1915. Boron and Silicon Halides - ", "content": "Heating boric acid to $100^{\\circ} \\mathrm{C}$ causes molecules of water to split out between pairs of adjacent -OH groups to form metaboric acid, $\\mathrm{HBO}_{2}$. At about $150^{\\circ} \\mathrm{C}$, additional B-O-B linkages form, connecting the $\\mathrm{BO}_{3}$ groups together with shared oxygen atoms to form tetraboric acid, $\\mathrm{H}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7}$. Complete water loss, at still higher temperatures, results in boric oxide."} {"id": "textbook_6295", "title": "1915. Boron and Silicon Halides - ", "content": "Borates are salts of the oxyacids of boron. Borates result from the reactions of a base with an oxyacid or from the fusion of boric acid or boric oxide with a metal oxide or hydroxide. Borate anions range from the simple trigonal planar $\\mathrm{BO}_{3}{ }^{3-}$ ion to complex species containing chains and rings of three- and four-coordinated boron atoms. The structures of the anions found in $\\mathrm{CaB}_{2} \\mathrm{O}_{4}, \\mathrm{~K}\\left[\\mathrm{~B}_{5} \\mathrm{O}_{6}(\\mathrm{OH})_{4}\\right] \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ (commonly written $\\mathrm{KB}_{5} \\mathrm{O}_{8} \\cdot 4 \\mathrm{H}_{2} \\mathrm{O}$ ) and $\\mathrm{Na}_{2}\\left[\\mathrm{~B}_{4} \\mathrm{O}_{5}(\\mathrm{OH})_{4}\\right] \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ (commonly written $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ ) are shown in Figure 18.17. Commercially, the most important borate is borax, $\\mathrm{Na}_{2}\\left[\\mathrm{~B}_{4} \\mathrm{O}_{5}(\\mathrm{OH})_{4}\\right] \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$, which is an important component of some laundry detergents. Most of the supply of borax comes directly from dry lakes, such as Searles Lake in California, or is prepared from kernite, $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 4 \\mathrm{H}_{2} \\mathrm{O}$.\n\n(a)\n
Nuclear Chemistry - ", "content": "\n"} {"id": "textbook_7026", "title": "2049. CHAPTER 20
Nuclear Chemistry - ", "content": "Figure 20.1 Nuclear chemistry provides the basis for many useful diagnostic and therapeutic methods in medicine, such as these positron emission tomography (PET) scans. The PET/computed tomography scan on the left shows muscle activity. The brain scans in the center show chemical differences in dopamine signaling in the brains of addicts and nonaddicts. The images on the right show an oncological application of PET scans to identify lymph node metastasis.\n"} {"id": "textbook_7027", "title": "2050. CHAPTER OUTLINE - ", "content": "\n20.1 Nuclear Structure and Stability\n20.2 Nuclear Equations\n20.3 Radioactive Decay\n20.4 Transmutation and Nuclear Energy\n20.5 Uses of Radioisotopes\n20.6 Biological Effects of Radiation"} {"id": "textbook_7028", "title": "2050. CHAPTER OUTLINE - ", "content": "INTRODUCTION The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions."} {"id": "textbook_7029", "title": "2050. CHAPTER OUTLINE - ", "content": "This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas.\n"} {"id": "textbook_7030", "title": "2050. CHAPTER OUTLINE - 2050.1. Nuclear Structure and Stability", "content": ""} {"id": "textbook_7031", "title": "2051. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7032", "title": "2051. LEARNING OBJECTIVES - ", "content": "- Describe nuclear structure in terms of protons, neutrons, and electrons\n- Calculate mass defect and binding energy for nuclei\n- Explain trends in the relative stability of nuclei"} {"id": "textbook_7033", "title": "2051. LEARNING OBJECTIVES - ", "content": "Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of ${ }_{1}^{1} \\mathrm{H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ( Z ) of the element, and the sum of the number of protons and the number of neutrons is the mass number (A). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation ${ }_{Z}^{A} \\mathrm{X}$, where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, ${ }_{6}^{14} \\mathrm{C}$ ). Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, ${ }_{6}^{14} \\mathrm{C}$ is called \"carbon-14.\""} {"id": "textbook_7034", "title": "2051. LEARNING OBJECTIVES - ", "content": "Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about $10^{-15}$ meters, a nucleus is quite small compared to the radius of the entire atom, which is about $10^{-10}$ meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \\times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of $22.6 \\mathrm{~g} / \\mathrm{cm}^{3}$. If the earth's density were equal to the average nuclear density, the earth's radius would be only about 200 meters (earth's actual radius is approximately $6.4 \\times 10^{6}$ meters, 30,000 times larger). Example 20.1 demonstrates just how great nuclear densities can be in the natural world.\n"} {"id": "textbook_7035", "title": "2052. EXAMPLE 20.1 - ", "content": ""} {"id": "textbook_7036", "title": "2053. Density of a Neutron Star - ", "content": "\nNeutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star's outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses ( 1 solar mass $=M_{\\odot}=$ mass of the sun $=1.99 \\times 10^{30} \\mathrm{~kg}$ ) and a diameter of 26 km .\n(a) What is the density of this neutron star?\n(b) How does this neutron star's density compare to the density of a uranium nucleus, which has a diameter of about $15 \\mathrm{fm}\\left(1 \\mathrm{fm}=10^{-15} \\mathrm{~m}\\right)$ ?\n"} {"id": "textbook_7037", "title": "2054. Solution - ", "content": "\nWe can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:"} {"id": "textbook_7038", "title": "2054. Solution - ", "content": "$$\nd=\\frac{m}{V} \\quad \\text { with } \\quad V=\\frac{4}{3} \\pi r^{3}\n$$"} {"id": "textbook_7039", "title": "2054. Solution - ", "content": "(a) The radius of the neutron star is $\\frac{1}{2} \\times 26 \\mathrm{~km}=\\frac{1}{2} \\times 2.6 \\times 10^{4} \\mathrm{~m}=1.3 \\times 10^{4} \\mathrm{~m}$, so the density of the neutron star is:"} {"id": "textbook_7040", "title": "2054. Solution - ", "content": "$$\nd=\\frac{m}{V}=\\frac{m}{\\frac{4}{3} \\pi r^{3}}=\\frac{2.4\\left(1.99 \\times 10^{30} \\mathrm{~kg}\\right)}{\\frac{4}{3} \\pi\\left(1.3 \\times 10^{4} \\mathrm{~m}\\right)^{3}}=5.2 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}\n$$"} {"id": "textbook_7041", "title": "2054. Solution - ", "content": "(b) The radius of the $\\mathrm{U}-235$ nucleus is $\\frac{1}{2} \\times 15 \\times 10^{-15} \\mathrm{~m}=7.5 \\times 10^{-15} \\mathrm{~m}$, so the density of the U-235 nucleus is:"} {"id": "textbook_7042", "title": "2054. Solution - ", "content": "$$\nd=\\frac{m}{V}=\\frac{m}{\\frac{4}{3} \\pi r^{3}}=\\frac{235 \\mathrm{amu}\\left(\\frac{1.66 \\times 10^{-27} \\mathrm{~kg}}{1 \\mathrm{amu}}\\right)}{\\frac{4}{3} \\pi\\left(7.5 \\times 10^{-15} \\mathrm{~m}\\right)^{3}}=2.2 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}\n$$"} {"id": "textbook_7043", "title": "2054. Solution - ", "content": "These values are fairly similar (same order of magnitude), but the neutron star is more than twice as dense as the U-235 nucleus.\n"} {"id": "textbook_7044", "title": "2055. Check Your Learning - ", "content": "\nFind the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km , and compare it to the density of a hydrogen nucleus, which has a diameter of $1.75 \\mathrm{fm}\\left(1 \\mathrm{fm}=1 \\times 10^{-15} \\mathrm{~m}\\right)$.\n"} {"id": "textbook_7045", "title": "2056. Answer: - ", "content": "\nThe density of the neutron star is $3.4 \\times 10^{18} \\mathrm{~kg} / \\mathrm{m}^{3}$. The density of a hydrogen nucleus is $6.0 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}$. The neutron star is 5.7 times denser than the hydrogen nucleus."} {"id": "textbook_7046", "title": "2056. Answer: - ", "content": "To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than $10^{-15}$ meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.\n"} {"id": "textbook_7047", "title": "2057. LINK TO LEARNING - ", "content": "\nVisit this website (http://openstax.org/1/16fourfund) for more information about the four fundamental forces.\n"} {"id": "textbook_7048", "title": "2058. Nuclear Binding Energy - ", "content": "\nAs a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:"} {"id": "textbook_7049", "title": "2058. Nuclear Binding Energy - ", "content": "$$\n(2 \\times \\underset{\\text { protons }}{1.0073 \\mathrm{amu}})+\\underset{\\text { neutrons }}{(2 \\times 1.0087 \\mathrm{amu})}+\\underset{\\text { electrons }}{1 \\times 0.00055 \\mathrm{amu})}=4.0 \\begin{aligned}\n& 4.0331 \\mathrm{amu} \\\\\n& (2 \\times 0\n\\end{aligned}\n$$"} {"id": "textbook_7050", "title": "2058. Nuclear Binding Energy - ", "content": "However, mass spectrometric measurements reveal that the mass of an ${ }_{2}^{4} \\mathrm{He}$ atom is 4.0026 amu , less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a \"loss\" in mass of $4.0331 \\mathrm{amu}-4.0026 \\mathrm{amu}=0.0305 \\mathrm{amu}$. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms' nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions."} {"id": "textbook_7051", "title": "2058. Nuclear Binding Energy - ", "content": "The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:"} {"id": "textbook_7052", "title": "2058. Nuclear Binding Energy - ", "content": "$$\nE=m c^{2}\n$$"} {"id": "textbook_7053", "title": "2058. Nuclear Binding Energy - ", "content": "where $E$ is energy, $m$ is mass of the matter being converted, and $c$ is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this\nmass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example 20.2. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making $1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}$.\n"} {"id": "textbook_7054", "title": "2059. EXAMPLE 20.2 - ", "content": ""} {"id": "textbook_7055", "title": "2060. Calculation of Nuclear Binding Energy - ", "content": "\nDetermine the binding energy for the nuclide ${ }_{2}^{4} \\mathrm{He}$ in:\n(a) joules per mole of nuclei\n(b) joules per nucleus\n(c) MeV per nucleus\n"} {"id": "textbook_7056", "title": "2061. Solution - ", "content": "\nThe mass defect for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is 0.0305 amu , as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}^{2}$ ).\n(a) First, express the mass defect in $\\mathrm{g} / \\mathrm{mol}$. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore $0.0305 \\mathrm{~g} / \\mathrm{mol}$. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg , since $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}^{2}$. Converting grams into kilograms yields a mass defect of $3.05 \\times 10^{-5} \\mathrm{~kg} /$ mol. Substituting this quantity into the mass-energy equivalence equation yields:"} {"id": "textbook_7057", "title": "2061. Solution - ", "content": "$$\n\\begin{aligned}\n& E=m c^{2}=\\frac{3.05 \\times 10^{-5} \\mathrm{~kg}}{\\mathrm{~mol}} \\times\\left(\\frac{2.998 \\times 10^{8} \\mathrm{~m}}{\\mathrm{~s}}\\right)^{2}=2.74 \\times 10^{12} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-2} \\mathrm{~mol}^{-1} \\\\\n& =2.74 \\times 10^{12} \\mathrm{~J} \\mathrm{~mol}^{-1}=2.74 \\mathrm{TJ} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_7058", "title": "2061. Solution - ", "content": "Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg , roughly the mass of typical drop of water).\n(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro's number:"} {"id": "textbook_7059", "title": "2061. Solution - ", "content": "$$\nE=2.74 \\times 10^{12} \\mathrm{~J} \\mathrm{~mol}^{-1} \\times \\frac{1 \\mathrm{~mol}}{6.022 \\times 10^{23} \\text { nuclei }}=4.55 \\times 10^{-12} \\mathrm{~J}=4.55 \\mathrm{pJ}\n$$"} {"id": "textbook_7060", "title": "2061. Solution - ", "content": "(c) Recall that $1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}$. Using the binding energy computed in part (b):"} {"id": "textbook_7061", "title": "2061. Solution - ", "content": "$$\nE=4.55 \\times 10^{-12} \\mathrm{~J} \\times \\frac{1 \\mathrm{eV}}{1.602 \\times 10^{-19} \\mathrm{~J}}=2.84 \\times 10^{7} \\mathrm{eV}=28.4 \\mathrm{MeV}\n$$\n"} {"id": "textbook_7062", "title": "2062. Check Your Learning - ", "content": "\nWhat is the binding energy for the nuclide ${ }_{9}^{19} \\mathrm{~F}$ (atomic mass: 18.9984 amu ) in MeV per nucleus?\n"} {"id": "textbook_7063", "title": "2063. Answer: - ", "content": "\n148.4 MeV"} {"id": "textbook_7064", "title": "2063. Answer: - ", "content": "Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit\nenthalpies on the order of thousands of $\\mathrm{kJ} / \\mathrm{mol}$, which is equivalent to mass differences in the nanogram range $\\left(10^{-9} \\mathrm{~g}\\right)$. On the other hand, nuclear binding energies are typically on the order of billions of $\\mathrm{kJ} / \\mathrm{mol}$, corresponding to mass differences in the milligram range $\\left(10^{-3} \\mathrm{~g}\\right)$.\nNuclear Stability\nA nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure 20.2 represents nuclei that have a 1:1 ratio of protons to neutrons ( $\\mathrm{n}: \\mathrm{p}$ ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead- 207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.\n"} {"id": "textbook_7065", "title": "2063. Answer: - ", "content": "FIGURE 20.2 This plot shows the nuclides that are known to exist and those that are stable. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. Note that all isotopes of elements with atomic numbers greater than 83 are unstable. The solid line is the line where $\\mathrm{n}=\\mathrm{Z}$."} {"id": "textbook_7066", "title": "2063. Answer: - ", "content": "The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter."} {"id": "textbook_7067", "title": "2063. Answer: - ", "content": "Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table 20.1). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons $(2,8,20,28,50,82$, and 126 ) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of\nboth protons and neutrons, such as ${ }_{2}^{4} \\mathrm{He},{ }_{8}^{16} \\mathrm{O},{ }_{20}^{40} \\mathrm{Ca}$, and ${ }_{82}^{208} \\mathrm{~Pb}$, are called \"double magic\" and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter."} {"id": "textbook_7068", "title": "2063. Answer: - ", "content": "Stable Nuclear Isotopes"} {"id": "textbook_7069", "title": "2063. Answer: - ", "content": "| Number of Stable Isotopes | Proton Number | Neutron Number |\n| :--- | :--- | :--- |\n| 157 | even | even |\n| 53 | even | odd |\n| 50 | odd | even |\n| 5 | odd | odd |"} {"id": "textbook_7070", "title": "2063. Answer: - ", "content": "TABLE 20.1"} {"id": "textbook_7071", "title": "2063. Answer: - ", "content": "The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example 20.2 that the binding energy for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is 28.4 MeV . The binding energy per nucleon for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is therefore:"} {"id": "textbook_7072", "title": "2063. Answer: - ", "content": "$$\n\\frac{28.4 \\mathrm{MeV}}{4 \\text { nucleons }}=7.10 \\mathrm{MeV} / \\text { nucleon }\n$$"} {"id": "textbook_7073", "title": "2063. Answer: - ", "content": "In Example 20.3, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure 20.3.\n"} {"id": "textbook_7074", "title": "2063. Answer: - ", "content": "FIGURE 20.3 The binding energy per nucleon is largest for nuclides with mass number of approximately 56.\n"} {"id": "textbook_7075", "title": "2064. EXAMPLE 20.3 - ", "content": ""} {"id": "textbook_7076", "title": "2065. Calculation of Binding Energy per Nucleon - ", "content": "\nThe iron nuclide ${ }_{26}^{56} \\mathrm{Fe}$ lies near the top of the binding energy curve (Figure 20.3) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide ${ }_{26}^{56} \\mathrm{Fe}$ (atomic mass of 55.9349 amu )?\n"} {"id": "textbook_7077", "title": "2066. Solution - ", "content": "\nAs in Example 20.2, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an ${ }_{26}^{56} \\mathrm{Fe}$ atom:"} {"id": "textbook_7078", "title": "2066. Solution - ", "content": "$$\n\\begin{aligned}\n& \\text { Mass defect }=[(26 \\times 1.0073 \\mathrm{amu})+(30 \\times 1.0087 \\mathrm{amu})+(26 \\times 0.00055 \\mathrm{amu})]-55.9349 \\mathrm{amu} \\\\\n& =56.4651 \\mathrm{amu}-55.9349 \\mathrm{amu} \\\\\n& =0.5302 \\mathrm{amu}\n\\end{aligned}\n$$"} {"id": "textbook_7079", "title": "2066. Solution - ", "content": "We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:"} {"id": "textbook_7080", "title": "2066. Solution - ", "content": "$$\n\\begin{aligned}\n& E=m c^{2}=0.5302 \\mathrm{amu} \\times \\frac{1.6605 \\times 10^{-27} \\mathrm{~kg}}{1 \\mathrm{amu}} \\times\\left(2.998 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)^{2} \\\\\n& =7.913 \\times 10^{-11} \\mathrm{~kg} \\cdot \\mathrm{~m} / \\mathrm{s}^{2} \\\\\n& =7.913 \\times 10^{-11} \\mathrm{~J}\n\\end{aligned}\n$$"} {"id": "textbook_7081", "title": "2066. Solution - ", "content": "We then convert the binding energy in joules per nucleus into units of MeV per nuclide:"} {"id": "textbook_7082", "title": "2066. Solution - ", "content": "$$\n7.913 \\times 10^{-11} \\mathrm{~J} \\times \\frac{1 \\mathrm{MeV}}{1.602 \\times 10^{-13} \\mathrm{~J}}=493.9 \\mathrm{MeV}\n$$"} {"id": "textbook_7083", "title": "2066. Solution - ", "content": "Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:"} {"id": "textbook_7084", "title": "2066. Solution - ", "content": "$$\n\\text { Binding energy per nucleon }=\\frac{493.9 \\mathrm{MeV}}{56}=8.820 \\mathrm{MeV} / \\text { nucleon }\n$$"} {"id": "textbook_7085", "title": "2066. Solution - ", "content": "Note that this is almost $25 \\%$ larger than the binding energy per nucleon for ${ }_{2}^{4} \\mathrm{He}$.\n(Note also that this is the same process as in Example 20.1, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)\n"} {"id": "textbook_7086", "title": "2067. Check Your Learning - ", "content": "\nWhat is the binding energy per nucleon in ${ }_{9}^{19} \\mathrm{~F}$ (atomic mass, 18.9984 amu )?\n"} {"id": "textbook_7087", "title": "2068. Answer: - ", "content": "\n7.810 MeV/nucleon\n"} {"id": "textbook_7088", "title": "2068. Answer: - 2068.1. Nuclear Equations", "content": ""} {"id": "textbook_7089", "title": "2069. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7090", "title": "2069. LEARNING OBJECTIVES - ", "content": "- Identify common particles and energies involved in nuclear reactions\n- Write and balance nuclear equations"} {"id": "textbook_7091", "title": "2069. LEARNING OBJECTIVES - ", "content": "Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.\n"} {"id": "textbook_7092", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "\nMany entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure 20.4. Protons ( ${ }_{1}^{1} \\mathrm{p}$, also represented by the symbol $\\left.{ }_{1}^{1} \\mathrm{H}\\right)$ and neutrons $\\left({ }_{0}^{1} \\mathrm{n}\\right)$ are the constituents of atomic nuclei, and have been described previously. Alpha particles $\\left({ }_{2}^{4} \\mathrm{He}\\right.$, also represented by the symbol $\\left.{ }_{2}^{4} \\alpha\\right)$ are high-energy helium nuclei. Beta particles $\\left({ }_{-1}^{0} \\beta\\right.$, also represented by the symbol $\\left.{ }_{-1}^{0} \\mathrm{e}\\right)$ are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons $\\left({ }_{+1}^{0} \\mathrm{e}\\right.$, also represented by the symbol $\\left.{ }_{+1}^{0} \\beta\\right)$ are positively charged electrons (\"anti-electrons\"). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4 , so it is symbolized ${ }_{2}^{4} \\mathrm{He}$. This works because, in general, the ion charge is not important in the balancing of nuclear equations."} {"id": "textbook_7093", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "| Name | Symbol(s) | Representation | Description |\n| :---: | :---: | :---: | :---: |\n| Alpha particle | ${ }_{2}^{4} \\mathrm{He}$ or ${ }_{2}^{4} \\alpha$ |  | (High-energy) helium nuclei consisting of two protons and two neutrons |\n| Beta particle | ${ }_{-1}^{0} \\mathrm{e}$ or ${ }_{-1}^{0} \\beta$ | $\\bigcirc$ | (High-energy) electrons |\n| Positron | ${ }_{+1}^{0} \\mathrm{e}$ or ${ }_{+1}^{0} \\beta$ | ( | Particles with the same mass as an electron but with 1 unit of positive charge |\n| Proton | ${ }_{1}^{1} \\mathrm{H}$ or ${ }_{1}^{1} \\mathrm{p}$ | $\\pm$ | Nuclei of hydrogen atoms |\n| Neutron | ${ }_{0}^{1} n$ |  | Particles with a mass approximately equal to that of a proton but with no charge |\n| Gamma ray | $\\gamma$ | $\\sim \\sim \\sim \\gamma$ | Very high-energy electromagnetic radiation |"} {"id": "textbook_7094", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "FIGURE 20.4 Although many species are encountered in nuclear reactions, this table summarizes the names, symbols, representations, and descriptions of the most common of these."} {"id": "textbook_7095", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays $(\\boldsymbol{\\gamma})$-and other much smaller subnuclear particles, which are beyond the scope of this chapter-according to the mass-energy equivalence equation $E=m c^{2}$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:"} {"id": "textbook_7096", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "$$\n{ }_{-1}^{0} \\mathrm{e}+{ }_{+1}^{0} \\mathrm{e} \\longrightarrow \\gamma+\\gamma\n$$"} {"id": "textbook_7097", "title": "2070. Types of Particles in Nuclear Reactions - ", "content": "As seen in the chapter discussing light and electromagnetic radiation, gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays that can behave as particles in the wave-particle duality sense. Gamma rays are a type of high energy electromagnetic radiation produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions.\n"} {"id": "textbook_7098", "title": "2071. Balancing Nuclear Reactions - ", "content": "\nA balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of nucleons (subatomic particles within the atoms' nuclei) rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:"} {"id": "textbook_7099", "title": "2071. Balancing Nuclear Reactions - ", "content": "1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.\n2. The sum of the charges of the reactants equals the sum of the charges of the products."} {"id": "textbook_7100", "title": "2071. Balancing Nuclear Reactions - ", "content": "If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that ${ }_{8}^{17} \\mathrm{O}$ is a product of the nuclear reaction of ${ }_{7}^{14} \\mathrm{~N}$ and ${ }_{2}^{4} \\mathrm{He}$ if we knew that a proton, ${ }_{1}^{1} \\mathrm{H}$, was one of the two products. Example 20.4 shows how we can identify a nuclide by balancing the nuclear reaction.\n"} {"id": "textbook_7101", "title": "2072. EXAMPLE 20.4 - ", "content": ""} {"id": "textbook_7102", "title": "2073. Balancing Equations for Nuclear Reactions - ", "content": "\nThe reaction of an $\\alpha$ particle with magnesium- $25\\left({ }_{12}^{25} \\mathrm{Mg}\\right)$ produces a proton and a nuclide of another element. Identify the new nuclide produced.\n"} {"id": "textbook_7103", "title": "2074. Solution - ", "content": "\nThe nuclear reaction can be written as:"} {"id": "textbook_7104", "title": "2074. Solution - ", "content": "$$\n{ }_{12}^{25} \\mathrm{Mg}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{1}^{1} \\mathrm{H}+{ }_{\\mathrm{Z}}^{\\mathrm{A}} \\mathrm{X}\n$$"} {"id": "textbook_7105", "title": "2074. Solution - ", "content": "where $A$ is the mass number and $Z$ is the atomic number of the new nuclide, $X$. Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:"} {"id": "textbook_7106", "title": "2074. Solution - ", "content": "$$\n25+4=\\mathrm{A}+1, \\text { or } \\mathrm{A}=28\n$$"} {"id": "textbook_7107", "title": "2074. Solution - ", "content": "Similarly, the charges must balance, so:"} {"id": "textbook_7108", "title": "2074. Solution - ", "content": "$$\n12+2=\\mathrm{Z}+1, \\text { and } \\mathrm{Z}=13\n$$"} {"id": "textbook_7109", "title": "2074. Solution - ", "content": "Check the periodic table: The element with nuclear charge $=+13$ is aluminum. Thus, the product is ${ }_{13}^{28} \\mathrm{Al}$.\n"} {"id": "textbook_7110", "title": "2075. Check Your Learning - ", "content": "\nThe nuclide ${ }_{53}^{125}$ I combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?\n"} {"id": "textbook_7111", "title": "2076. Answer: - ", "content": "\n${ }_{53}^{125} \\mathrm{I}+{ }_{-1}^{0} \\mathrm{e} \\longrightarrow{ }_{52}^{125} \\mathrm{Te}$"} {"id": "textbook_7112", "title": "2076. Answer: - ", "content": "Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:"} {"id": "textbook_7113", "title": "2076. Answer: - ", "content": "- The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting $\\alpha$ particles:"} {"id": "textbook_7114", "title": "2076. Answer: - ", "content": "$$\n{ }_{84}^{212} \\mathrm{Po} \\longrightarrow{ }_{82}^{208} \\mathrm{~Pb}+{ }_{2}^{4} \\mathrm{He}\n$$"} {"id": "textbook_7115", "title": "2076. Answer: - ", "content": "- The first nuclide to be prepared by artificial means was an isotope of oxygen, ${ }^{17} \\mathrm{O}$. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with $\\alpha$ particles:"} {"id": "textbook_7116", "title": "2076. Answer: - ", "content": "$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}\n$$"} {"id": "textbook_7117", "title": "2076. Answer: - ", "content": "- James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with ${ }^{12} \\mathrm{C}$ by the nuclear reaction between ${ }^{9} \\mathrm{Be}$ and ${ }^{4} \\mathrm{He}$ :"} {"id": "textbook_7118", "title": "2076. Answer: - ", "content": "$$\n{ }_{4}^{9} \\mathrm{Be}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{6}^{12} \\mathrm{C}+{ }_{0}^{1} \\mathrm{n}\n$$"} {"id": "textbook_7119", "title": "2076. Answer: - ", "content": "- The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, ${ }_{1}^{2} \\mathrm{H}$ ), by Emilio Segre and Carlo Perrier in 1937:"} {"id": "textbook_7120", "title": "2076. Answer: - ", "content": "$$\n{ }_{1}^{2} \\mathrm{H}+{ }_{42}^{97} \\mathrm{Mo} \\longrightarrow 2{ }_{0}^{1} \\mathrm{n}+{ }_{43}^{97} \\mathrm{Tc}\n$$"} {"id": "textbook_7121", "title": "2076. Answer: - ", "content": "- The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was:"} {"id": "textbook_7122", "title": "2076. Answer: - ", "content": "$$\n{ }_{92}^{235} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{35}^{87} \\mathrm{Br}+{ }_{57}^{146} \\mathrm{La}+3{ }_{0}^{1} \\mathrm{n}\n$$\n"} {"id": "textbook_7123", "title": "2076. Answer: - 2076.1. Radioactive Decay", "content": ""} {"id": "textbook_7124", "title": "2077. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7125", "title": "2077. LEARNING OBJECTIVES - ", "content": "- Recognize common modes of radioactive decay\n- Identify common particles and energies involved in nuclear decay reactions\n- Write and balance nuclear decay equations\n- Calculate kinetic parameters for decay processes, including half-life\n- Describe common radiometric dating techniques"} {"id": "textbook_7126", "title": "2077. LEARNING OBJECTIVES - ", "content": "Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences-chemistry and physics), who was the first to coin the term \"radioactivity,\" and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed."} {"id": "textbook_7127", "title": "2077. LEARNING OBJECTIVES - ", "content": "The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure 20.5).\n"} {"id": "textbook_7128", "title": "2077. LEARNING OBJECTIVES - ", "content": "FIGURE 20.5 A nucleus of uranium-238 (the parent nuclide) undergoes $\\alpha$ decay to form thorium-234 (the daughter nuclide). The alpha particle removes two protons (green) and two neutrons (gray) from the uranium-238 nucleus.\n"} {"id": "textbook_7129", "title": "2078. LINK TO LEARNING - ", "content": "\nAlthough the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here (http://openstax.org/l/16cloudchamb) to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.\n"} {"id": "textbook_7130", "title": "2079. Types of Radioactive Decay - ", "content": "\nErnest Rutherford's experiments involving the interaction of radiation with a magnetic or electric field (Figure 20.6) helped him determine that one type of radiation consisted of positively charged and relatively massive $\\alpha$ particles; a second type was made up of negatively charged and much less massive $\\beta$ particles; and a third was uncharged electromagnetic waves, $\\gamma$ rays. We now know that $\\alpha$ particles are high-energy helium nuclei, $\\beta$ particles are high-energy electrons, and $\\gamma$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.\n"} {"id": "textbook_7131", "title": "2079. Types of Radioactive Decay - ", "content": "FIGURE 20.6 Alpha particles, which are attracted to the negative plate and deflected by a relatively small amount, must be positively charged and relatively massive. Beta particles, which are attracted to the positive plate and deflected a relatively large amount, must be negatively charged and relatively light. Gamma rays, which are unaffected by the electric field, must be uncharged."} {"id": "textbook_7132", "title": "2079. Types of Radioactive Decay - ", "content": "Alpha ( $\\alpha$ ) decay is the emission of an $\\alpha$ particle from the nucleus. For example, polonium-210 undergoes $\\alpha$ decay:"} {"id": "textbook_7133", "title": "2079. Types of Radioactive Decay - ", "content": "$$\n{ }_{84}^{210} \\mathrm{Po} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{82}^{206} \\mathrm{~Pb} \\quad \\text { or } \\quad{ }_{84}^{210} \\mathrm{Po} \\longrightarrow{ }_{2}^{4} \\alpha+{ }_{82}^{206} \\mathrm{~Pb}\n$$"} {"id": "textbook_7134", "title": "2079. Types of Radioactive Decay - ", "content": "Alpha decay occurs primarily in heavy nuclei ( $\\mathrm{A}>200, \\mathrm{Z}>83$ ). Because the loss of an $\\alpha$ particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing $\\alpha$ decay lies below the band of stability (refer to Figure 20.2), the daughter nuclide will lie closer to the band."} {"id": "textbook_7135", "title": "2079. Types of Radioactive Decay - ", "content": "Beta ( $\\boldsymbol{\\beta}$ ) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes $\\beta$ decay:"} {"id": "textbook_7136", "title": "2079. Types of Radioactive Decay - ", "content": "$$\n{ }_{53}^{131} \\mathrm{I} \\longrightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{54}^{131} \\mathrm{Xe} \\quad \\text { or } \\quad{ }_{53}^{131} \\mathrm{I} \\longrightarrow{ }_{-1}^{0} \\beta+{ }_{54}^{131} \\mathrm{Xe}\n$$"} {"id": "textbook_7137", "title": "2079. Types of Radioactive Decay - ", "content": "Beta decay, which can be thought of as the conversion of a neutron into a proton and a $\\beta$ particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide."} {"id": "textbook_7138", "title": "2079. Types of Radioactive Decay - ", "content": "Gamma emission ( $\\boldsymbol{\\gamma}$ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a $\\gamma$ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits $\\gamma$ radiation and is used in many applications including cancer treatment:"} {"id": "textbook_7139", "title": "2079. Types of Radioactive Decay - ", "content": "$$\n{ }_{27}^{60} \\mathrm{Co}^{*} \\longrightarrow{ }_{0}^{0} \\gamma+{ }_{27}^{60} \\mathrm{Co}\n$$"} {"id": "textbook_7140", "title": "2079. Types of Radioactive Decay - ", "content": "There is no change in mass number or atomic number during the emission of a $\\gamma$ ray unless the $\\gamma$ emission accompanies one of the other modes of decay."} {"id": "textbook_7141", "title": "2079. Types of Radioactive Decay - ", "content": "Positron emission ( $\\boldsymbol{\\beta}^{+}$decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:"} {"id": "textbook_7142", "title": "2079. Types of Radioactive Decay - ", "content": "$$\n{ }_{8}^{15} \\mathrm{O} \\longrightarrow{ }_{+1}^{0} \\mathrm{e}+{ }_{7}^{15} \\mathrm{~N} \\quad \\text { or } \\quad{ }_{8}^{15} \\mathrm{O} \\longrightarrow{ }_{+1}^{0} \\beta+{ }_{7}^{15} \\mathrm{~N}\n$$"} {"id": "textbook_7143", "title": "2079. Types of Radioactive Decay - ", "content": "Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide."} {"id": "textbook_7144", "title": "2079. Types of Radioactive Decay - ", "content": "Electron capture occurs when one of the inner electrons in an atom is captured by the atom's nucleus. For example, potassium-40 undergoes electron capture:"} {"id": "textbook_7145", "title": "2079. Types of Radioactive Decay - ", "content": "$$\n{ }_{19}^{40} \\mathrm{~K}+{ }_{-1}^{0} \\mathrm{e} \\longrightarrow{ }_{18}^{40} \\mathrm{Ar}\n$$"} {"id": "textbook_7146", "title": "2079. Types of Radioactive Decay - ", "content": "Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for \"proton-rich\" nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur."} {"id": "textbook_7147", "title": "2079. Types of Radioactive Decay - ", "content": "Figure 20.7 summarizes these types of decay, along with their equations and changes in atomic and mass numbers."} {"id": "textbook_7148", "title": "2079. Types of Radioactive Decay - ", "content": "| Type | Nuclear equation | Representation | Change in mass/atomic numbers |\n| :---: | :---: | :---: | :---: |\n| Alpha decay | ${ }_{\\mathrm{Z}}^{\\mathrm{A}} \\mathrm{X} \\rightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{\\mathrm{Z}-2}^{\\mathrm{A}-4} \\mathrm{Y}$ |  | A: decrease by 4
Z: decrease by 2 |\n| Beta decay | ${ }_{\\mathrm{Z}} \\mathrm{A} X \\rightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{\\mathrm{Z}+1}{ }^{\\text {a }} \\mathrm{Y}$ |  | A: unchanged
Z: increase by 1 |\n| Gamma decay | ${ }_{Z}^{A} X \\rightarrow{ }_{0}^{0} \\gamma+{ }_{Z}^{A} Y$ | Excited nuclear state | A: unchanged
Z: unchanged |\n| Positron emission | ${ }_{\\mathrm{Z}} \\mathrm{X} \\mathrm{X} \\rightarrow{ }_{+1}^{0} \\mathrm{e}+{ }_{Y-1}{ }_{1}^{\\text {P }} \\mathrm{Y}$ | $8 \\bigcirc \\underset{\\oplus}{\\downarrow} \\longrightarrow$ | A: unchanged
Z: decrease by 1 |\n| Electron capture | ${ }_{\\mathrm{Z}}{ }^{\\text {P }} \\rightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{\\mathrm{Y}-1}{ }_{1} \\mathrm{Y}$ |  | A: unchanged
Z: decrease by 1 |"} {"id": "textbook_7149", "title": "2079. Types of Radioactive Decay - ", "content": "FIGURE 20.7 This table summarizes the type, nuclear equation, representation, and any changes in the mass or atomic numbers for various types of decay.\n"} {"id": "textbook_7150", "title": "2080. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_7151", "title": "2081. PET Scan - ", "content": "\nPositron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient's body function (Figure 20.8). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This \"tagged\" compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 20.8 A PET scanner (a) uses radiation to provide an image of how part of a patient's body functions. The scans it produces can be used to image a healthy brain (b) or can be used for diagnosing medical conditions such as Alzheimer's disease (c). (credit a: modification of work by Jens Maus)\nFor example, $\\mathrm{F}-18$ is produced by proton bombardment of ${ }^{18} \\mathrm{O}\\left({ }_{8}^{18} \\mathrm{O}+{ }_{1}^{1} \\mathrm{p} \\longrightarrow{ }_{9}^{18} \\mathrm{~F}+{ }_{0}^{1} \\mathrm{n}\\right)$ and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The ${ }^{18} \\mathrm{~F}$ emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient's body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer's disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and Xrays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.\n"} {"id": "textbook_7152", "title": "2082. Radioactive Decay Series - ", "content": "\nThe naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product-that is, a nuclide on the band of stability (Figure 20.9). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.\n"} {"id": "textbook_7153", "title": "2082. Radioactive Decay Series - ", "content": "FIGURE 20.9 Uranium-238 undergoes a radioactive decay series consisting of 14 separate steps before producing stable lead-206. This series consists of eight $\\alpha$ decays and $\\operatorname{six} \\beta$ decays.\n"} {"id": "textbook_7154", "title": "2083. Radioactive Half-Lives - ", "content": "\nRadioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life $\\left(t_{1 / 2}\\right)$, the time required for half of the atoms in a sample to decay. An isotope's half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem."} {"id": "textbook_7155", "title": "2083. Radioactive Half-Lives - ", "content": "For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure 20.10). In a given cobalt-60 source, since half of the ${ }_{27}^{60}$ Co nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.\n\n\nFIGURE 20.10 For cobalt-60, which has a half-life of 5.27 years, $50 \\%$ remains after 5.27 years (one half-life), 25\\% remains after 10.54 years (two half-lives), $12.5 \\%$ remains after 15.81 years (three half-lives), and so on."} {"id": "textbook_7156", "title": "2083. Radioactive Half-Lives - ", "content": "Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for firstorder chemical reactions. We generally substitute the number of nuclei, $N$, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:\ndecay rate $=\\lambda N$ with $\\lambda=$ the decay constant for the particular radioisotope\nThe decay constant, $\\lambda$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, $t_{1 / 2}$ :"} {"id": "textbook_7157", "title": "2083. Radioactive Half-Lives - ", "content": "$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{t_{1 / 2}} \\quad \\text { or } \\quad t_{1 / 2}=\\frac{\\ln 2}{\\lambda}=\\frac{0.693}{\\lambda}\n$$"} {"id": "textbook_7158", "title": "2083. Radioactive Half-Lives - ", "content": "The first-order equations relating amount, $N$, and time are:"} {"id": "textbook_7159", "title": "2083. Radioactive Half-Lives - ", "content": "$$\nN_{t}=N_{0} e^{-\\lambda t} \\quad \\text { or } \\quad t=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)\n$$"} {"id": "textbook_7160", "title": "2083. Radioactive Half-Lives - ", "content": "where $N_{0}$ is the initial number of nuclei or moles of the isotope, and $N_{t}$ is the number of nuclei/moles remaining at time $t$. Example 20.5 applies these calculations to find the rates of radioactive decay for specific nuclides.\n"} {"id": "textbook_7161", "title": "2084. EXAMPLE 20.5 - ", "content": ""} {"id": "textbook_7162", "title": "2085. Rates of Radioactive Decay - ", "content": "\n${ }_{27}^{60} \\mathrm{Co}$ decays with a half-life of 5.27 years to produce ${ }_{28}^{60} \\mathrm{Ni}$.\n(a) What is the decay constant for the radioactive disintegration of cobalt-60?\n(b) Calculate the fraction of a sample of the ${ }_{27}^{60}$ Co isotope that will remain after 15 years.\n(c) How long does it take for a sample of ${ }_{27}^{60} \\mathrm{Co}$ to disintegrate to the extent that only $2.0 \\%$ of the original amount remains?\n"} {"id": "textbook_7163", "title": "2086. Solution - ", "content": "\n(a) The value of the rate constant is given by:"} {"id": "textbook_7164", "title": "2086. Solution - ", "content": "$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{5.27 \\mathrm{y}}=0.132 \\mathrm{y}^{-1}\n$$"} {"id": "textbook_7165", "title": "2086. Solution - ", "content": "(b) The fraction of ${ }_{27}^{60} \\mathrm{Co}$ that is left after time $t$ is given by $\\frac{N_{t}}{N_{0}}$. Rearranging the first-order relationship $N_{t}=$ $N_{0} \\mathrm{e}^{-\\lambda t}$ to solve for this ratio yields:"} {"id": "textbook_7166", "title": "2086. Solution - ", "content": "$$\n\\frac{N_{t}}{N_{0}}=e^{-\\lambda t}=e^{-(0.132 / \\mathrm{y})(15 \\times \\mathrm{y})}=0.138\n$$"} {"id": "textbook_7167", "title": "2086. Solution - ", "content": "The fraction of ${ }_{27}^{60}$ Co that will remain after 15.0 years is 0.138 . Or put another way, $13.8 \\%$ of the ${ }_{27}^{60} \\mathrm{Co}$ originally present will remain after 15 years.\n(c) $2.00 \\%$ of the original amount of ${ }_{27}^{60} \\mathrm{Co}$ is equal to $0.0200 \\times N_{0}$. Substituting this into the equation for time for first-order kinetics, we have:"} {"id": "textbook_7168", "title": "2086. Solution - ", "content": "$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)=-\\frac{1}{0.132 \\mathrm{y}^{-1}} \\ln \\left(\\frac{0.0200 \\times N_{0}}{N_{0}}\\right)=29.6 \\mathrm{y}\n$$\n"} {"id": "textbook_7169", "title": "2087. Check Your Learning - ", "content": "\nRadon-222, ${ }_{86}^{222} \\mathrm{Rn}$, has a half-life of 3.823 days. How long will it take a sample of radon- 222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?\n"} {"id": "textbook_7170", "title": "2088. Answer: - ", "content": "\n11.1 days"} {"id": "textbook_7171", "title": "2088. Answer: - ", "content": "Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of ${ }_{83}^{209} \\mathrm{Bi}$ is $1.9 \\times 10^{19}$ years; ${ }_{94}^{239} \\mathrm{Ra}$ is 24,000 years; ${ }_{86}^{222} \\mathrm{Rn}$ is 3.82 days; and element- 111 ( Rg for roentgenium) is $1.5 \\times$ $10^{-3}$ seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table 20.2, and others are listed in Appendix M."} {"id": "textbook_7172", "title": "2088. Answer: - ", "content": "Half-lives of Radioactive Isotopes Important to Medicine"} {"id": "textbook_7173", "title": "2088. Answer: - ", "content": "| Type $^{\\mathbf{1}}$ | Decay Mode | Half-Life | Uses |\n| :--- | :--- | :--- | :--- |\n| F-18 | $\\beta^{+}$decay | 110. minutes | PET scans |\n| Co-60 | $\\beta$ decay, $\\gamma$ decay | 5.27 years | cancer treatment |\n| Tc-99m | $\\gamma$ decay | 8.01 hours | scans of brain, lung, heart, bone |\n| I-131 | $\\beta$ decay | 8.02 days | thyroid scans and treatment |\n| Tl-201 | electron capture | 73 hours | heart and arteries scans; cardiac stress tests |"} {"id": "textbook_7174", "title": "2088. Answer: - ", "content": "TABLE 20.2\n"} {"id": "textbook_7175", "title": "2089. Radiometric Dating - ", "content": "\nSeveral radioisotopes have half-lives and other properties that make them useful for purposes of \"dating\" the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about"} {"id": "textbook_7176", "title": "2089. Radiometric Dating - ", "content": "[^12]the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type."} {"id": "textbook_7177", "title": "2089. Radiometric Dating - ", "content": "Radioactive Dating Using Carbon-14\nThe radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.\nNaturally occurring carbon consists of three isotopes: ${ }_{6}^{12} \\mathrm{C}$, which constitutes about $99 \\%$ of the carbon on earth; ${ }_{6}^{13} \\mathrm{C}$, about $1 \\%$ of the total; and trace amounts of ${ }_{6}^{14} \\mathrm{C}$. Carbon- 14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:"} {"id": "textbook_7178", "title": "2089. Radiometric Dating - ", "content": "$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{6}^{14} \\mathrm{C}+{ }_{1}^{1} \\mathrm{H}\n$$"} {"id": "textbook_7179", "title": "2089. Radiometric Dating - ", "content": "All isotopes of carbon react with oxygen to produce $\\mathrm{CO}_{2}$ molecules. The ratio of ${ }_{6}^{14} \\mathrm{CO}_{2}$ to ${ }_{6}^{12} \\mathrm{CO}_{2}$ depends on the ratio of ${ }_{6}^{14} \\mathrm{CO}$ to ${ }_{6}^{12} \\mathrm{CO}$ in the atmosphere. The natural abundance of ${ }_{6}^{14} \\mathrm{CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of ${ }_{6}^{14} \\mathrm{C}_{6}^{14} \\mathrm{CO}_{2}$ and ${ }_{6}^{12} \\mathrm{CO}_{2}$ into plants is a regular part of the photosynthesis process, which means that the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio found in a living plant is the same as the ${ }_{6}^{14} \\mathrm{C}$ : ${ }_{6}^{12} \\mathrm{C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because ${ }_{6}^{12} \\mathrm{C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by $\\beta$ emission with a half-life of 5730 years:"} {"id": "textbook_7180", "title": "2089. Radiometric Dating - ", "content": "$$\n{ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+{ }_{-1}^{0} \\mathrm{e}\n$$"} {"id": "textbook_7181", "title": "2089. Radiometric Dating - ", "content": "Thus, the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure 20.11 visually depicts this process.\n\n\nFIGURE 20.11 Along with stable carbon-12, radioactive carbon-14 is taken in by plants and animals, and remains at a constant level within them while they are alive. After death, the $\\mathrm{C}-14$ decays and the $\\mathrm{C}-14: \\mathrm{C}-12$ ratio in the remains decreases. Comparing this ratio to the $\\mathrm{C}-14: \\mathrm{C}-12$ ratio in living organisms allows us to determine how long ago the organism lived (and died).\nFor example, with the half-life of ${ }_{6}^{14} \\mathrm{C}$ being 5730 years, if the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.\n"} {"id": "textbook_7182", "title": "2090. LINK TO LEARNING - ", "content": "\nVisit this website (http://openstax.org/l/16phetradiom) to perform simulations of radiometric dating.\n"} {"id": "textbook_7183", "title": "2091. EXAMPLE 20.6 - ", "content": ""} {"id": "textbook_7184", "title": "2092. Radiocarbon Dating - ", "content": "\nA tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations $/ \\mathrm{min} / \\mathrm{g}$ of C, estimate the age of the Dead Sea Scrolls.\n"} {"id": "textbook_7185", "title": "2093. Solution - ", "content": "\nThe rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive $\\mathrm{C}-14$ left in the paper, so we can substitute the rates for the amounts, $N$, in the relationship:"} {"id": "textbook_7186", "title": "2093. Solution - ", "content": "$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right) \\rightarrow t=-\\frac{1}{\\lambda} \\ln \\left(\\frac{\\text { Rate }_{t}}{\\text { Rate }_{0}}\\right)\n$$"} {"id": "textbook_7187", "title": "2093. Solution - ", "content": "where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript $t$ represents the current time."} {"id": "textbook_7188", "title": "2093. Solution - ", "content": "The decay constant can be determined from the half-life of C-14, 5730 years:"} {"id": "textbook_7189", "title": "2093. Solution - ", "content": "$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{5730 \\mathrm{y}}=1.21 \\times 10^{-4} \\mathrm{y}^{-1}\n$$"} {"id": "textbook_7190", "title": "2093. Solution - ", "content": "Substituting and solving, we have:"} {"id": "textbook_7191", "title": "2093. Solution - ", "content": "$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{\\text { Rate }_{t}}{\\text { Rate }_{0}}\\right)=-\\frac{1}{1.21 \\times 10^{-4} \\mathrm{y}^{-1}} \\ln \\left(\\frac{10.8 \\mathrm{dis} / \\mathrm{min} / \\mathrm{g} \\mathrm{C}}{13.6 \\mathrm{dis} / \\mathrm{min} / \\mathrm{g} \\mathrm{C}}\\right)=1910 \\mathrm{y}\n$$"} {"id": "textbook_7192", "title": "2093. Solution - ", "content": "Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure 20.12).\n"} {"id": "textbook_7193", "title": "2093. Solution - ", "content": "FIGURE 20.12 Carbon-14 dating has shown that these pages from the Dead Sea Scrolls were written or copied on paper made from plants that died between 100 BC and AD 50.\n"} {"id": "textbook_7194", "title": "2094. Check Your Learning - ", "content": "\nMore accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun's tomb have a C-14 decay rate of 9.07 disintegrations $/ \\mathrm{min} / \\mathrm{g}$ of C . How long ago did King Tut's reign come to an end?\n"} {"id": "textbook_7195", "title": "2095. Answer: - ", "content": "\nabout 3350 years ago, or approximately 1340 BC"} {"id": "textbook_7196", "title": "2095. Answer: - ", "content": "There have been some significant, well-documented changes to the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio. The accuracy of a straightforward application of this technique depends on the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of $\\mathrm{CO}_{2}$ molecules (largely ${ }_{6}^{12} \\mathrm{CO}_{2}$ ) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the ${ }_{6}^{14} \\mathrm{C}$ has decayed), the ratio of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ in the atmosphere may be changing. This manmade increase in ${ }_{6}^{12} \\mathrm{CO}_{2}$ in the atmosphere causes the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.\n"} {"id": "textbook_7197", "title": "2096. Radioactive Dating Using Nuclides Other than Carbon-14 - ", "content": "\nRadioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original $\\mathrm{U}-238$ to decay into $\\mathrm{Pb}-206$. In a sample of rock that does not contain appreciable amounts of $\\mathrm{Pb}-208$, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form $\\mathrm{Ar}-40$ with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40\nratio yields the age of the rock. Other methods, such as rubidium-strontium dating ( $\\mathrm{Rb}-87$ decays into $\\mathrm{Sr}-87$ with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth's age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.\n"} {"id": "textbook_7198", "title": "2097. EXAMPLE 20.7 - ", "content": ""} {"id": "textbook_7199", "title": "2098. Radioactive Dating of Rocks - ", "content": "\nAn igneous rock contains $9.58 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{U}-238$ and $2.51 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{Pb}-206$, and much, much smaller amounts of $\\mathrm{Pb}-208$. Determine the approximate time at which the rock formed.\n"} {"id": "textbook_7200", "title": "2099. Solution - ", "content": "\nThe sample of rock contains very little $\\mathrm{Pb}-208$, the most common isotope of lead, so we can safely assume that all the $\\mathrm{Pb}-206$ in the rock was produced by the radioactive decay of $\\mathrm{U}-238$. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay."} {"id": "textbook_7201", "title": "2099. Solution - ", "content": "The amount of $\\mathrm{U}-238$ currently in the rock is:"} {"id": "textbook_7202", "title": "2099. Solution - ", "content": "$$\n9.58 \\times 10^{-5} \\frac{\\mathrm{~g} U}{\\delta} \\times\\left(\\frac{1 \\mathrm{~mol} \\mathrm{U}}{238 \\mathrm{~g} U}\\right)=4.03 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$"} {"id": "textbook_7203", "title": "2099. Solution - ", "content": "Because when one mole of U-238 decays, it produces one mole of $\\mathrm{Pb}-206$, the amount of $\\mathrm{U}-238$ that has undergone radioactive decay since the rock was formed is:"} {"id": "textbook_7204", "title": "2099. Solution - ", "content": "$$\n2.51 \\times 10^{-5} \\underset{\\S}{\\mathrm{~g} P b} \\times\\left(\\frac{1 \\mathrm{molPb}}{206 \\frac{\\mathrm{~g} P \\mathrm{~Pb}}{}}\\right) \\times\\left(\\frac{1 \\mathrm{~mol} \\mathrm{U}}{1 \\mathrm{molPb}}\\right)=1.22 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$"} {"id": "textbook_7205", "title": "2099. Solution - ", "content": "The total amount of U-238 originally present in the rock is therefore:"} {"id": "textbook_7206", "title": "2099. Solution - ", "content": "$$\n4.03 \\times 10^{-7} \\mathrm{~mol}+1.22 \\times 10^{-7} \\mathrm{~mol}=5.25 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$"} {"id": "textbook_7207", "title": "2099. Solution - ", "content": "The amount of time that has passed since the formation of the rock is given by:"} {"id": "textbook_7208", "title": "2099. Solution - ", "content": "$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)\n$$"} {"id": "textbook_7209", "title": "2099. Solution - ", "content": "with $N_{0}$ representing the original amount of $\\mathrm{U}-238$ and $N_{t}$ representing the present amount of $\\mathrm{U}-238$.\n$\\mathrm{U}-238$ decays into $\\mathrm{Pb}-206$ with a half-life of $4.5 \\times 10^{9} \\mathrm{y}$, so the decay constant $\\lambda$ is:"} {"id": "textbook_7210", "title": "2099. Solution - ", "content": "$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{4.5 \\times 10^{9} \\mathrm{y}}=1.54 \\times 10^{-10} \\mathrm{y}^{-1}\n$$"} {"id": "textbook_7211", "title": "2099. Solution - ", "content": "Substituting and solving, we have:"} {"id": "textbook_7212", "title": "2099. Solution - ", "content": "$$\nt=-\\frac{1}{1.54 \\times 10^{-10} \\mathrm{y}^{-1}} \\ln \\left(\\frac{4.03 \\times 10^{-7} \\mathrm{~mol} U}{5.25 \\times 10^{-7} \\mathrm{molU}}\\right)=1.7 \\times 10^{9} \\mathrm{y}\n$$"} {"id": "textbook_7213", "title": "2099. Solution - ", "content": "Therefore, the rock is approximately 1.7 billion years old.\n"} {"id": "textbook_7214", "title": "2100. Check Your Learning - ", "content": "\nA sample of rock contains $6.14 \\times 10^{-4} \\mathrm{~g}$ of $\\mathrm{Rb}-87$ and $3.51 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{Sr}-87$. Calculate the age of the rock. (The half-life of the $\\beta$ decay of $\\mathrm{Rb}-87$ is $4.7 \\times 10^{10} \\mathrm{y}$.)\n"} {"id": "textbook_7215", "title": "2101. Answer: - ", "content": "\n$3.7 \\times 10^{9} \\mathrm{y}$\n"} {"id": "textbook_7216", "title": "2101. Answer: - 2101.1. Transmutation and Nuclear Energy", "content": ""} {"id": "textbook_7217", "title": "2102. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7218", "title": "2102. LEARNING OBJECTIVES - ", "content": "- Describe the synthesis of transuranium nuclides\n- Explain nuclear fission and fusion processes\n- Relate the concepts of critical mass and nuclear chain reactions\n- Summarize basic requirements for nuclear fission and fusion reactors"} {"id": "textbook_7219", "title": "2102. LEARNING OBJECTIVES - ", "content": "After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Scientists learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society's needs during peace.\n"} {"id": "textbook_7220", "title": "2103. Synthesis of Nuclides - ", "content": "\nNuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford's laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed $\\alpha$ particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:"} {"id": "textbook_7221", "title": "2103. Synthesis of Nuclides - ", "content": "$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}\n$$"} {"id": "textbook_7222", "title": "2103. Synthesis of Nuclides - ", "content": "The ${ }_{8}^{17} \\mathrm{O}$ and ${ }_{1}^{1} \\mathrm{H}$ nuclei that are produced are stable, so no further (nuclear) changes occur.\nTo reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news.\n"} {"id": "textbook_7223", "title": "2104. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_7224", "title": "2105. CERN Particle Accelerator - ", "content": "\nLocated near Geneva, the CERN (\"Conseil Europ\u00e9en pour la Recherche Nucl\u00e9aire,\" or European Council for Nuclear Research) Laboratory is the world's premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer ( 17 mile ) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure 20.13). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.\n"} {"id": "textbook_7225", "title": "2105. CERN Particle Accelerator - ", "content": "FIGURE 20.13 A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere)\nIn 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This longanticipated discovery made worldwide news and resulted in the awarding of the 2013 Nobel Prize in Physics to Fran\u00e7ois Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.\n"} {"id": "textbook_7226", "title": "2106. LINK TO LEARNING - ", "content": "\nFamous physicist Brian Cox talks about his work on the Large Hadron Collider at CERN, providing an entertaining and engaging tour (http://openstax.org/l/16tedCERN) of this massive project and the physics behind it.\nView a short video (http://openstax.org/l/16CERNvideo) from CERN, describing the basics of how its particle accelerators work."} {"id": "textbook_7227", "title": "2106. LINK TO LEARNING - ", "content": "Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92 . Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these-element 93, neptunium ( Np ) -was first made in 1940 by McMillan and Abelson by bombarding uranium- 238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:"} {"id": "textbook_7228", "title": "2106. LINK TO LEARNING - ", "content": "$$\n\\begin{aligned}\n& { }_{92}^{238} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{92}^{239} \\mathrm{U} \\\\\n& \\begin{array}{r}\n239 \\\\\n92 \\\\\n239 \\\\\n\\end{array} \\longrightarrow{ }_{93}^{239} \\mathrm{~Np}+{ }_{-1}^{0} \\mathrm{e} \\\\\n& \\text { half-life }=23.5 \\mathrm{~min} \\\\\n& { }_{93}^{239} \\mathrm{~Np} \\longrightarrow{ }_{94}^{239} \\mathrm{Pu}+{ }_{-1}^{0} \\mathrm{e} \\\\\n& \\text { half-life }=2.36 \\text { days }\n\\end{aligned}\n$$"} {"id": "textbook_7229", "title": "2106. LINK TO LEARNING - ", "content": "Plutonium is now mostly formed in nuclear reactors as a byproduct during the fission of $\\mathrm{U}-235$. Additional neutrons are released during this fission process (see the next section), some of which combine with U-238 nuclei to form uranium-239; this undergoes $\\beta$ decay to form neptunium- 239 , which in turn undergoes $\\beta$ decay to form plutonium-239 as illustrated in the preceding three equations. These processes are summarized in the equation:"} {"id": "textbook_7230", "title": "2106. LINK TO LEARNING - ", "content": "$$\n{ }_{92}^{238} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{92}^{239} \\mathrm{U} \\xrightarrow{\\beta^{-}}{ }_{93}^{239} \\mathrm{~Np} \\xrightarrow{\\beta^{-}}{ }_{94}^{239} \\mathrm{Pu}\n$$"} {"id": "textbook_7231", "title": "2106. LINK TO LEARNING - ", "content": "Heavier isotopes of plutonium-Pu-240, $\\mathrm{Pu}-241$, and $\\mathrm{Pu}-242-$ are also produced when lighter plutonium\nnuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years."} {"id": "textbook_7232", "title": "2106. LINK TO LEARNING - ", "content": "Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses."} {"id": "textbook_7233", "title": "2106. LINK TO LEARNING - ", "content": "The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table 20.3."} {"id": "textbook_7234", "title": "2106. LINK TO LEARNING - ", "content": "Preparation of Some of the Transuranium Elements"} {"id": "textbook_7235", "title": "2106. LINK TO LEARNING - ", "content": "| Name | Symbol | Atomic Number | Reaction |\n| :---: | :---: | :---: | :---: |\n| americium | Am | 95 | ${ }_{94}^{239} \\mathrm{Pu}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{95}^{240} \\mathrm{Am}+{ }_{-1}^{0} \\mathrm{e}$ |\n| curium | Cm | 96 | ${ }_{94}^{239} \\mathrm{Pu}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{96}^{242} \\mathrm{Cm}+{ }_{0}^{1} \\mathrm{n}$ |\n| californium | Cf | 98 | ${ }_{96}^{242} \\mathrm{Cm}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{98}^{245} \\mathrm{Cf}+{ }_{0}^{1} \\mathrm{n}$ |\n| einsteinium | Es | 99 | ${ }_{92}^{238} \\mathrm{U}+15_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{99}^{253} \\mathrm{Es}+7{ }_{-1}^{0} \\mathrm{e}$ |\n| mendelevium | Md | 101 | ${ }_{99}^{253} \\mathrm{Es}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{101}^{256} \\mathrm{Md}+{ }_{0}^{1} \\mathrm{n}$ |\n| nobelium | No | 102 | ${ }_{96}^{246} \\mathrm{Cm}+{ }_{6}^{12} \\mathrm{C} \\longrightarrow{ }_{102}^{254} \\mathrm{No}+4{ }_{0}^{1} \\mathrm{n}$ |\n| rutherfordium | Rf | 104 | ${ }_{98}^{249} \\mathrm{Cf}+{ }_{6}^{12} \\mathrm{C} \\longrightarrow{ }_{104}^{257} \\mathrm{Rf}+4{ }_{0}^{1} \\mathrm{n}$ |\n| seaborgium | Sg | 106 | $\\begin{array}{r} { }_{82}^{206} \\mathrm{~Pb}+{ }_{24}^{54} \\mathrm{Cr} \\longrightarrow{ }_{106}^{257} \\mathrm{Sg}+3{ }_{0}^{1} \\mathrm{n} \\\\ { }_{98}^{249} \\mathrm{Cf}+{ }_{8}^{18} \\mathrm{O} \\longrightarrow{ }_{106}^{263} \\mathrm{Sg}+4{ }_{0}^{1} \\mathrm{n} \\end{array}$ |\n| meitnerium | Mt | 107 | ${ }_{83}^{209} \\mathrm{Bi}+{ }_{26}^{58} \\mathrm{Fe} \\longrightarrow{ }_{109}^{266} \\mathrm{Mt}+{ }_{0}^{1} \\mathrm{n}$ |"} {"id": "textbook_7236", "title": "2106. LINK TO LEARNING - ", "content": "TABLE 20.3\n"} {"id": "textbook_7237", "title": "2107. Nuclear Fission - ", "content": "\nMany heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon-that is, mass numbers and binding energies per nucleon that are closer to the \"peak\" of the binding energy graph near 56 (see Figure 20.3). Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure 20.14.\n"} {"id": "textbook_7238", "title": "2107. Nuclear Fission - ", "content": "FIGURE 20.14 When a slow neutron hits a fissionable U-235 nucleus, it is absorbed and forms an unstable U-236 nucleus. The U-236 nucleus then rapidly breaks apart into two smaller nuclei (in this case, Ba-141 and Kr-92) along with several neutrons (usually two or three), and releases a very large amount of energy."} {"id": "textbook_7239", "title": "2107. Nuclear Fission - ", "content": "Among the products of Meitner, Hahn, and Strassman's fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for $\\mathrm{U}-235$, and a graph showing the distribution of its fission products and their yields, are shown in Figure 20.15. Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium."} {"id": "textbook_7240", "title": "2107. Nuclear Fission - ", "content": "$$\n\\begin{aligned}\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{98}^{90} \\mathrm{Sr}+{ }_{54}^{144} \\mathrm{Xe}+2{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{35}^{87} \\mathrm{Br}+{ }_{57}^{146} \\mathrm{La}+3{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{37}^{96} R b+{ }_{55}^{137} \\mathrm{Cs}+3{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{52}^{137} \\mathrm{Te}+{ }_{40}^{97} \\mathrm{Zr}+2{ }_{0}^{1} n \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{56}^{141} \\mathrm{Ba}+{ }_{36}^{92} \\mathrm{Kr}+3{ }_{0}^{1} \\mathrm{n}\n\\end{aligned}\n$$"} {"id": "textbook_7241", "title": "2107. Nuclear Fission - ", "content": "(a)\n\n(b)"} {"id": "textbook_7242", "title": "2107. Nuclear Fission - ", "content": "FIGURE 20.15 (a) Nuclear fission of U-235 produces a range of fission products. (b) The larger fission products of $\\mathrm{U}-235$ are typically one isotope with a mass number around $85-105$, and another isotope with a mass number that is about $50 \\%$ larger, that is, about 130-150.\n"} {"id": "textbook_7243", "title": "2108. LINK TO LEARNING - ", "content": "\nView this link (http://openstax.org/l/16fission) to see a simulation of nuclear fission.\n$\\mathrm{U}-235$ undergoes fission, the products weigh about 0.2 grams less than the reactants; this \"lost\" mass is converted into a very large amount of energy, about $1.8 \\times 10^{10} \\mathrm{~kJ}$ per mole of $\\mathrm{U}-235$. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal."} {"id": "textbook_7244", "title": "2108. LINK TO LEARNING - ", "content": "As described earlier, when undergoing fission U-235 produces two \"medium-sized\" nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (see Figure 20.16). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.\n\n\nFIGURE 20.16 The fission of a large nucleus, such as U-235, produces two or three neutrons, each of which is capable of causing fission of another nucleus by the reactions shown. If this process continues, a nuclear chain reaction occurs."} {"id": "textbook_7245", "title": "2108. LINK TO LEARNING - ", "content": "Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure 20.17)."} {"id": "textbook_7246", "title": "2108. LINK TO LEARNING - ", "content": "Sub-critical mass\n\n(a)\n\nCritical mass\n\n(b)"} {"id": "textbook_7247", "title": "2108. LINK TO LEARNING - ", "content": "FIGURE 20.17 (a) In a subcritical mass, the fissile material is too small and allows too many neutrons to escape the material, so a chain reaction does not occur. (b) In a critical mass, a large enough number of neutrons in the fissile material induce fission to create a chain reaction."} {"id": "textbook_7248", "title": "2108. LINK TO LEARNING - ", "content": "An atomic bomb (Figure 20.18) contains several pounds of fissionable material, ${ }_{92}^{235} \\mathrm{U}$ or ${ }_{94}^{239} \\mathrm{Pu}$, a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.\n\n\nFIGURE 20.18 (a) The nuclear fission bomb that destroyed Hiroshima on August 6, 1945, consisted of two subcritical masses of U-235, where conventional explosives were used to fire one of the subcritical masses into the other, creating the critical mass for the nuclear explosion. (b) The plutonium bomb that destroyed Nagasaki on August 9,1945 , consisted of a hollow sphere of plutonium that was rapidly compressed by conventional explosives. This led to a concentration of plutonium in the center that was greater than the critical mass necessary for the nuclear explosion.\n"} {"id": "textbook_7249", "title": "2109. Fission Reactors - ", "content": "\nChain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure 20.19). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.\n"} {"id": "textbook_7250", "title": "2109. Fission Reactors - ", "content": "FIGURE 20.19 (a) The Diablo Canyon Nuclear Power Plant near San Luis Obispo is the only nuclear power plant currently in operation in California. The domes are the containment structures for the nuclear reactors, and the brown building houses the turbine where electricity is generated. Ocean water is used for cooling. (b) The Diablo Canyon uses a pressurized water reactor, one of a few different fission reactor designs in use around the world, to produce electricity. Energy from the nuclear fission reactions in the core heats water in a closed, pressurized system. Heat from this system produces steam that drives a turbine, which in turn produces electricity. (credit a: modification of work by \"Mike\" Michael L. Baird; credit b: modification of work by the Nuclear Regulatory Commission)\n"} {"id": "textbook_7251", "title": "2110. Nuclear Fuels - ", "content": "\nNuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from $0.05-0.3 \\%$ of the uranium oxide $\\mathrm{U}_{3} \\mathrm{O}_{8}$; the uranium in the ore is about $99.3 \\%$ nonfissionable $\\mathrm{U}-238$ with only $0.7 \\%$ fissionable $U-235$. Nuclear reactors require a fuel with a higher concentration of $U-235$ than is found in nature; it is normally enriched to have about $5 \\%$ of uranium mass as $\\mathrm{U}-235$. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.\n\nIn the gaseous diffusion enrichment plant where U - 235 fuel is prepared, $\\mathrm{UF}_{6}$ (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for $\\mathrm{UF}_{6}$ to pass through. The slightly lighter ${ }^{235} \\mathrm{UF}_{6}$ molecules diffuse through the barrier slightly faster than the heavier ${ }^{238} \\mathrm{UF}_{6}$ molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of ${ }^{235} \\mathrm{UF}_{6}$ to the level needed by the nuclear reactor. The basis for this process, Graham's law, is described in the chapter on gases. The enriched $\\mathrm{UF}_{6}$ gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramicencased, enriched uranium (usually $\\mathrm{UO}_{2}$ ) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.\n"} {"id": "textbook_7252", "title": "2111. Nuclear Moderators - ", "content": "\nNeutrons produced by nuclear reactions move too fast to cause fission (refer back to Figure 20.17). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used highpurity graphite as a moderator. Modern reactors in the US exclusively use heavy water $\\left({ }_{1}^{2} \\mathrm{H}_{2} \\mathrm{O}\\right)$ or light water (ordinary $\\mathrm{H}_{2} \\mathrm{O}$ ), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.\n"} {"id": "textbook_7253", "title": "2112. Reactor Coolants - ", "content": "\nA nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.\n"} {"id": "textbook_7254", "title": "2113. Control Rods - ", "content": "\nNuclear reactors use control rods (Figure 20.20) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:"} {"id": "textbook_7255", "title": "2113. Control Rods - ", "content": "$$\n{ }_{5}^{10} \\mathrm{~B}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{3}^{7} \\mathrm{Li}+{ }_{2}^{4} \\mathrm{He}\n$$"} {"id": "textbook_7256", "title": "2113. Control Rods - ", "content": "When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.\n"} {"id": "textbook_7257", "title": "2113. Control Rods - ", "content": "FIGURE 20.20 The nuclear reactor core shown in (a) contains the fuel and control rod assembly shown in (b).\n(credit: modification of work by E. Generalic, http://glossary.periodni.com/glossary.php?en=control+rod)\n"} {"id": "textbook_7258", "title": "2114. Shield and Containment System - ", "content": "\nDuring its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:"} {"id": "textbook_7259", "title": "2114. Shield and Containment System - ", "content": "1. The reactor vessel, a steel shell that is 3-20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor\n2. A main shield of $1-3$ meters of high-density concrete\n3. A personnel shield of lighter materials that protects operators from $\\gamma$ rays and X-rays"} {"id": "textbook_7260", "title": "2114. Shield and Containment System - ", "content": "In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.\n"} {"id": "textbook_7261", "title": "2115. LINK TO LEARNING - ", "content": "\nClick here to watch a 3-minute video (http://openstax.org/l/16nucreactors) from the Nuclear Energy Institute on how nuclear reactors work."} {"id": "textbook_7262", "title": "2115. LINK TO LEARNING - ", "content": "Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.\n"} {"id": "textbook_7263", "title": "2116. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_7264", "title": "2117. Nuclear Accidents - ", "content": "\nThe importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima)."} {"id": "textbook_7265", "title": "2117. Nuclear Accidents - ", "content": "In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least $2200^{\\circ} \\mathrm{C}$, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:"} {"id": "textbook_7266", "title": "2117. Nuclear Accidents - ", "content": "$$\n\\mathrm{Zr}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{ZrO}_{2}(s)+2 \\mathrm{H}_{2}(g)\n$$"} {"id": "textbook_7267", "title": "2117. Nuclear Accidents - ", "content": "The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process."} {"id": "textbook_7268", "title": "2117. Nuclear Accidents - ", "content": "Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure 20.21).\n"} {"id": "textbook_7269", "title": "2117. Nuclear Accidents - ", "content": "FIGURE 20.21 (a) In this 2010 photo of Three Mile Island, the remaining structures from the damaged Unit 2 reactor are seen on the left, whereas the separate Unit 1 reactor, unaffected by the accident, continues generating power to this day (right). (b) President Jimmy Carter visited the Unit 2 control room a few days after the accident in 1979."} {"id": "textbook_7270", "title": "2117. Nuclear Accidents - ", "content": "Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland."} {"id": "textbook_7271", "title": "2117. Nuclear Accidents - ", "content": "In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events."} {"id": "textbook_7272", "title": "2117. Nuclear Accidents - ", "content": "An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan's nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan's atomic energy program is still stalled (Figure 20.22).\n"} {"id": "textbook_7273", "title": "2117. Nuclear Accidents - ", "content": "FIGURE 20.22 (a) After the accident, contaminated waste had to be removed, and (b) an evacuation zone was set up around the plant in areas that received heavy doses of radioactive fallout. (credit a: modification of work by \"Live Action Hero\"/Flickr)"} {"id": "textbook_7274", "title": "2117. Nuclear Accidents - ", "content": "The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about $0.1 \\%$ of the mass of the fuel is converted into energy. The other $99.9 \\%$ remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate."} {"id": "textbook_7275", "title": "2117. Nuclear Accidents - ", "content": "Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60 , some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed.\n"} {"id": "textbook_7276", "title": "2118. LINK TO LEARNING - ", "content": "\nExplore the information in this link (http://openstax.org/l/16wastemgmt) to learn about the approaches to nuclear waste management.\n"} {"id": "textbook_7277", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "\nThe process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:"} {"id": "textbook_7278", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "$$\n4{ }_{1}^{1} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+2{ }_{+1}^{0} \\mathrm{e}^{+}\n$$"} {"id": "textbook_7279", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "A helium nucleus has a mass that is $0.7 \\%$ less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about $3.6 \\times 10^{11} \\mathrm{~kJ}$ of energy per mole of ${ }_{2}^{4} \\mathrm{He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 ( $1.8 \\times 10^{10} \\mathrm{~kJ}$ ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane ( 5471 kJ )."} {"id": "textbook_7280", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, ${ }_{1}^{2} \\mathrm{H}$ and a triton, ${ }_{1}^{3} \\mathrm{H}$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:"} {"id": "textbook_7281", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "$$\n{ }_{1}^{2} \\mathrm{H}+{ }_{1}^{3} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{0}^{1} \\mathrm{n}\n$$"} {"id": "textbook_7282", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "This change proceeds with a mass loss of 0.0188 amu , corresponding to the release of $1.69 \\times 10^{9}$ kilojoules per mole of ${ }_{2}^{4} \\mathrm{He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide."} {"id": "textbook_7283", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "Useful fusion reactions require very high temperatures for their initiation-about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe-stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur."} {"id": "textbook_7284", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure 20.23). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.\n\n(a)\n\n(b)"} {"id": "textbook_7285", "title": "2119. Nuclear Fusion and Fusion Reactors - ", "content": "FIGURE 20.23 (a) This model is of the International Thermonuclear Experimental Reactor (ITER) reactor. Currently under construction in the south of France with an expected completion date of 2027, the ITER will be the world's largest experimental Tokamak nuclear fusion reactor with a goal of achieving large-scale sustained energy production. (b) In 2012, the National Ignition Facility at Lawrence Livermore National Laboratory briefly produced over 500,000,000,000 watts ( 500 terawatts, or 500 TW ) of peak power and delivered 1,850,000 joules (1.85 MJ) of energy, the largest laser energy ever produced and 1000 times the power usage of the entire United States in any given moment. Although lasting only a few billionths of a second, the 192 lasers attained the conditions needed for nuclear fusion ignition. This image shows the target prior to the laser shot. (credit a: modification of work by Stephan Mosel)\n"} {"id": "textbook_7286", "title": "2119. Nuclear Fusion and Fusion Reactors - 2119.1. Uses of Radioisotopes", "content": ""} {"id": "textbook_7287", "title": "2120. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7288", "title": "2120. LEARNING OBJECTIVES - ", "content": "- List common applications of radioactive isotopes"} {"id": "textbook_7289", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more."} {"id": "textbook_7290", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioimmunossays (RIA), for example, rely on radioisotopes to detect the presence and/or concentration of certain antigens. Developed by Rosalyn Sussman Yalow and Solomon Berson in the 1950s, the technique is known for extreme sensitivity, meaning that it can detect and measure very small quantities of a substance. Prior to its discovery, most similar detection relied on large enough quantities to produce visible outcomes. RIA revolutionized and expanded entire fields of study, most notably endocrinology, and is commonly used in narcotics detection, blood bank screening, early cancer screening, hormone measurement, and allergy diagnosis. Based on her significant contribution to medicine, Yalow received a Nobel Prize, making her the second woman to be awarded the prize for medicine."} {"id": "textbook_7291", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioisotopes have revolutionized medical practice (see Appendix M), where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\\left({ }_{43}^{99} \\mathrm{Tc}\\right)$, thallium-201 $\\left({ }_{81}^{201} \\mathrm{Tl}\\right)$, iodine-131 ( $\\left.{ }_{53}^{131} \\mathrm{I}\\right)$, and sodium-24 $\\left({ }_{11}^{24} \\mathrm{Na}\\right)$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the $\\gamma$ rays emitted by the Tc-99 isotope. Thallium-201 (Figure 20.24) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave's disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.\n"} {"id": "textbook_7292", "title": "2120. LEARNING OBJECTIVES - ", "content": "FIGURE 20.24 Administering thallium-201 to a patient and subsequently performing a stress test offer medical professionals an opportunity to visually analyze heart function and blood flow. (credit: modification of work by \"BlueOctane\"/Wikimedia Commons)"} {"id": "textbook_7293", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioisotopes used in medicine typically have short half-lives-for example, the ubiquitous Tc-99m has a halflife of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes $\\beta$ decay with a half-life of 66 hours, and the $\\mathrm{Tc}-99$ is then chemically extracted (Figure 20.25). The parent nuclide Mo-99 is part of a molybdate ion, $\\mathrm{MoO}_{4}{ }^{2-}$; when it decays, it forms the pertechnetate ion, $\\mathrm{TcO}_{4}{ }^{-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.\n"} {"id": "textbook_7294", "title": "2120. LEARNING OBJECTIVES - ", "content": "FIGURE 20.25 (a) The first Tc-99m generator (circa 1958) is used to separate Tc-99 from Mo-99. The $\\mathrm{MoO}_{4}{ }^{2-}$ is retained by the matrix in the column, whereas the $\\mathrm{TcO}_{4}{ }^{-}$passes through and is collected. (b) Tc-99 was used in this scan of the neck of a patient with Grave's disease. The scan shows the location of high concentrations of Tc-99. (credit a: modification of work by the Department of Energy; credit b: modification of work by \"MBq\"/Wikimedia Commons)"} {"id": "textbook_7295", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure 20.26). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that\nhas been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.\n"} {"id": "textbook_7296", "title": "2120. LEARNING OBJECTIVES - ", "content": "FIGURE 20.26 The cartoon in (a) shows a cobalt-60 machine used in the treatment of cancer. The diagram in (b) shows how the gantry of the Co-60 machine swings through an arc, focusing radiation on the targeted region (tumor) and minimizing the amount of radiation that passes through nearby regions."} {"id": "textbook_7297", "title": "2120. LEARNING OBJECTIVES - ", "content": "Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes $\\beta$ decay to form Ni-60, along with the emission of $\\gamma$ radiation. The overall process is:"} {"id": "textbook_7298", "title": "2120. LEARNING OBJECTIVES - ", "content": "$$\n{ }_{27}^{59} \\mathrm{Co}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{27}^{60} \\mathrm{Co} \\longrightarrow{ }_{28}^{60} \\mathrm{Ni}+{ }_{-1}^{0} \\beta+2{ }_{0}^{0} \\gamma\n$$"} {"id": "textbook_7299", "title": "2120. LEARNING OBJECTIVES - ", "content": "The overall decay scheme for this is shown graphically in Figure 20.27.\n"} {"id": "textbook_7300", "title": "2120. LEARNING OBJECTIVES - ", "content": "FIGURE 20.27 Co-60 undergoes a series of radioactive decays. The $\\gamma$ emissions are used for radiation therapy."} {"id": "textbook_7301", "title": "2120. LEARNING OBJECTIVES - ", "content": "Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants."} {"id": "textbook_7302", "title": "2120. LEARNING OBJECTIVES - ", "content": "For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:"} {"id": "textbook_7303", "title": "2120. LEARNING OBJECTIVES - ", "content": "$$\n6 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)+6 \\mathrm{O}_{2}(g)\n$$"} {"id": "textbook_7304", "title": "2120. LEARNING OBJECTIVES - ", "content": "but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to $\\mathrm{CO}_{2}$ containing a high concentration of ${ }_{6}^{14} \\mathrm{C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction."} {"id": "textbook_7305", "title": "2120. LEARNING OBJECTIVES - ", "content": "Commercial applications of radioactive materials are equally diverse (Figure 20.28). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with $\\gamma$ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.\n"} {"id": "textbook_7306", "title": "2120. LEARNING OBJECTIVES - ", "content": "FIGURE 20.28 Common commercial uses of radiation include (a) X-ray examination of luggage at an airport and (b) preservation of food. (credit a: modification of work by the Department of the Navy; credit b: modification of work by the US Department of Agriculture)"} {"id": "textbook_7307", "title": "2120. LEARNING OBJECTIVES - ", "content": "Americium-241, an $\\alpha$ emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure 20.29). The $\\alpha$ emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.\n\n\nFIGURE 20.29 Inside a smoke detector, Am-241 emits $\\alpha$ particles that ionize the air, creating a small electric current. During a fire, smoke particles impede the flow of ions, reducing the current and triggering an alarm. (credit a: modification of work by \"Muffet\"/Wikimedia Commons)\n"} {"id": "textbook_7308", "title": "2120. LEARNING OBJECTIVES - 2120.1. Biological Effects of Radiation", "content": ""} {"id": "textbook_7309", "title": "2121. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7310", "title": "2121. LEARNING OBJECTIVES - ", "content": "- Describe the biological impact of ionizing radiation\n- Define units for measuring radiation exposure\n- Explain the operation of common tools for detecting radioactivity\n- List common sources of radiation exposure in the US"} {"id": "textbook_7311", "title": "2121. LEARNING OBJECTIVES - ", "content": "The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism's repair mechanisms and possibly causing illness or even death (Figure 20.30).\n"} {"id": "textbook_7312", "title": "2121. LEARNING OBJECTIVES - ", "content": "FIGURE 20.30 Radiation can harm biological systems by damaging the DNA of cells. If this damage is not properly repaired, the cells may divide in an uncontrolled manner and cause cancer.\n"} {"id": "textbook_7313", "title": "2122. Ionizing and Nonionizing Radiation - ", "content": "\nThere is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, $\\alpha$ and $\\beta$ particles, $\\gamma$ rays, X-rays, and high-energy ultraviolet radiation) (Figure 20.31).\n"} {"id": "textbook_7314", "title": "2122. Ionizing and Nonionizing Radiation - ", "content": "FIGURE 20.31 Lower frequency, lower-energy electromagnetic radiation is nonionizing, and higher frequency, higher-energy electromagnetic radiation is ionizing."} {"id": "textbook_7315", "title": "2122. Ionizing and Nonionizing Radiation - ", "content": "Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing $\\mathrm{H}_{2} \\mathrm{O}$ (the most abundant molecule in living organisms), which forms a $\\mathrm{H}_{2} \\mathrm{O}^{+}$ion that reacts with water, forming a hydronium ion and a hydroxyl radical:\n"} {"id": "textbook_7316", "title": "2122. Ionizing and Nonionizing Radiation - ", "content": "Because the hydroxyl radical has an unpaired electron, it is highly reactive. (This is true of any substance with unpaired electrons, known as a free radical.) This hydroxyl radical can react with all kinds of biological molecules (DNA, proteins, enzymes, and so on), causing damage to the molecules and disrupting physiological processes. Examples of direct and indirect damage are shown in Figure 20.32.\n"} {"id": "textbook_7317", "title": "2122. Ionizing and Nonionizing Radiation - ", "content": "FIGURE 20.32 Ionizing radiation can (a) directly damage a biomolecule by ionizing it or breaking its bonds, or (b) create an $\\mathrm{H}_{2} \\mathrm{O}^{+}$ion, which reacts with $\\mathrm{H}_{2} \\mathrm{O}$ to form a hydroxyl radical, which in turn reacts with the biomolecule,\ncausing damage indirectly.\n"} {"id": "textbook_7318", "title": "2123. Biological Effects of Exposure to Radiation - ", "content": "\nRadiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy."} {"id": "textbook_7319", "title": "2123. Biological Effects of Exposure to Radiation - ", "content": "Different types of radiation have differing abilities to pass through material (Figure 20.33). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of $\\beta$ particles, and about 20 times that of $\\gamma$ rays and X-rays.\n"} {"id": "textbook_7320", "title": "2123. Biological Effects of Exposure to Radiation - ", "content": "FIGURE 20.33 The ability of different types of radiation to pass through material is shown. From least to most penetrating, they are alpha < beta < neutron < gamma.\n"} {"id": "textbook_7321", "title": "2124. Chemistry in Everyday Life - ", "content": ""} {"id": "textbook_7322", "title": "2125. Radon Exposure - ", "content": "\nFor many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an $\\alpha$ emitter with a half-life of 3.82 days. It is one of the products of the radioactive decay series of U-238 (Figure 20.9), which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure 20.34).\n\n\nFIGURE 20.34 Radon-222 seeps into houses and other buildings from rocks that contain uranium-238, a radon emitter. The radon enters through cracks in concrete foundations and basement floors, stone or porous cinderblock foundations, and openings for water and gas pipes."} {"id": "textbook_7323", "title": "2125. Radon Exposure - ", "content": "Radon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US ( $1.25 \\mathrm{pCi} / \\mathrm{L}$ ) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one's risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.\n"} {"id": "textbook_7324", "title": "2126. Measuring Radiation Exposure - ", "content": "\nSeveral different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure 20.35). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-M\u00fcller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-M\u00fcller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator-a material that emits light (luminesces) when excited by ionizing radiation-and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.\n\n\nFIGURE 20.35 Devices such as (a) Geiger counters, (b) scintillators, and (c) dosimeters can be used to measure radiation. (credit c: modification of work by \"osaMu\"/Wikimedia commons)"} {"id": "textbook_7325", "title": "2126. Measuring Radiation Exposure - ", "content": "A variety of units are used to measure various aspects of radiation (Figure 20.36). The SI unit for rate of radioactive decay is the becquerel ( $\\mathbf{B q}$ ), with $1 \\mathrm{~Bq}=1$ disintegration per second. The curie ( $\\mathbf{C i}$ ) and millicurie $(\\mathbf{m C i})$ are much larger units and are frequently used in medicine ( 1 curie $=1 \\mathrm{Ci}=3.7 \\times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with $1 \\mathrm{~Gy}=1 \\mathrm{~J}$ of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used ( $1 \\mathrm{rad}=0.01$ Gy; 1 rad results in the absorption of $0.01 \\mathrm{~J} / \\mathrm{kg}$ of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose. The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine ( $100 \\mathrm{rem}=1 \\mathrm{~Sv}$ ). Note that the tissue damage units (rem or Sv ) includes the energy of the radiation dose (rad or Gy) along with a biological factor referred to as the RBE (for relative biological effectiveness) that is an approximate measure of the relative damage done by the radiation. These are related by:"} {"id": "textbook_7326", "title": "2126. Measuring Radiation Exposure - ", "content": "$$\n\\text { number of rems }=\\mathrm{RBE} \\times \\text { number of rads }\n$$"} {"id": "textbook_7327", "title": "2126. Measuring Radiation Exposure - ", "content": "with RBE approximately 10 for $\\alpha$ radiation, $2(+)$ for protons and neutrons, and 1 for $\\beta$ and $\\gamma$ radiation.\n"} {"id": "textbook_7328", "title": "2126. Measuring Radiation Exposure - ", "content": "FIGURE 20.36 Different units are used to measure the rate of emission from a radioactive source, the energy that is absorbed from the source, and the amount of damage the absorbed radiation does.\n"} {"id": "textbook_7329", "title": "2127. Units of Radiation Measurement - ", "content": "\nTable 20.4 summarizes the units used for measuring radiation."} {"id": "textbook_7330", "title": "2127. Units of Radiation Measurement - ", "content": "| Measurement Purpose | Unit | Quantity Measured | Description |\n| :---: | :---: | :---: | :---: |\n| activity of source | becquerel (Bq) | radioactive decays or emissions | amount of sample that undergoes 1 decay/second |\n| | curie (Ci) | | amount of sample that undergoes 3.7 $\\times 10^{10}$ decays/second |\n| absorbed dose | gray (Gy) | energy absorbed per kg of tissue | $1 \\mathrm{~Gy}=1 \\mathrm{~J} / \\mathrm{kg}$ tissue |\n| | radiation absorbed dose (rad) | | $1 \\mathrm{rad}=0.01 \\mathrm{~J} / \\mathrm{kg}$ tissue |\n| biologically effective dose | sievert (Sv) | tissue damage | $\\mathrm{Sv}=\\mathrm{RBE} \\times \\mathrm{Gy}$ |\n| | roentgen equivalent for man (rem) | | $\\mathrm{Rem}=\\mathrm{RBE} \\times \\mathrm{rad}$ |"} {"id": "textbook_7331", "title": "2127. Units of Radiation Measurement - ", "content": "TABLE 20.4\n"} {"id": "textbook_7332", "title": "2128. EXAMPLE 20.8 - ", "content": ""} {"id": "textbook_7333", "title": "2129. Amount of Radiation - ", "content": "\nCobalt-60 ( $t_{1 / 2}=5.26 \\mathrm{y}$ ) is used in cancer therapy since the $\\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.\n(a) What is its activity in Bq?\n(b) What is its activity in Ci?\n"} {"id": "textbook_7334", "title": "2130. Solution - ", "content": "\nThe activity is given by:"} {"id": "textbook_7335", "title": "2130. Solution - ", "content": "$$\n\\text { Activity }=\\lambda N=\\left(\\frac{\\ln 2}{t_{1 / 2}}\\right) N=\\left(\\frac{\\ln 2}{5.26 \\mathrm{y}}\\right) \\times 5.00 \\mathrm{~g}=0.659 \\frac{\\mathrm{~g}}{\\mathrm{y}} \\text { of } \\mathrm{Co}-60 \\text { that decay }\n$$"} {"id": "textbook_7336", "title": "2130. Solution - ", "content": "And to convert this to decays per second:"} {"id": "textbook_7337", "title": "2130. Solution - ", "content": "$$\n\\begin{aligned}\n& 0.659 \\frac{\\mathrm{~g}}{\\mathrm{y}} \\times \\frac{1 \\mathrm{y}}{365 \\mathrm{~d}} \\times \\frac{1 \\mathrm{~d}}{24 \\mathrm{~h}} \\times \\frac{1 \\mathrm{~h}}{3600 \\mathrm{~s}} \\times \\frac{1 \\mathrm{~mol}}{59.9 \\mathrm{~g}} \\times \\frac{6.02 \\times 10^{23} \\mathrm{atoms}}{1 \\mathrm{~mol}} \\times \\frac{1 \\text { decay }}{1 \\text { atom }} \\\\\n& =2.10 \\times 10^{14 \\frac{\\text { decay }}{\\mathrm{s}}}\n\\end{aligned}\n$$"} {"id": "textbook_7338", "title": "2130. Solution - ", "content": "(a) Since $1 \\mathrm{~Bq}=\\frac{1 \\text { decay }}{\\mathrm{s}}$, the activity in Becquerel $(\\mathrm{Bq})$ is:"} {"id": "textbook_7339", "title": "2130. Solution - ", "content": "$$\n2.10 \\times 10^{14} \\frac{\\text { decay }}{s} \\times\\left(\\frac{1 \\mathrm{~Bq}}{1 \\frac{\\text { decay }}{\\mathrm{s}}}\\right)=2.10 \\times 10^{14} \\mathrm{~Bq}\n$$"} {"id": "textbook_7340", "title": "2130. Solution - ", "content": "(b) Since $1 \\mathrm{Ci}=\\frac{3.7 \\times 10^{11} \\text { decay }}{\\mathrm{s}}$, the activity in curie (Ci) is:"} {"id": "textbook_7341", "title": "2130. Solution - ", "content": "$$\n2.10 \\times 10^{14} \\frac{\\text { decay }}{s} \\times\\left(\\frac{1 \\mathrm{Ci}}{\\frac{3.7 \\times 10^{11} \\text { decay }}{\\mathrm{s}}}\\right)=5.7 \\times 10^{2} \\mathrm{Ci}\n$$\n"} {"id": "textbook_7342", "title": "2131. Check Your Learning - ", "content": "\nTritium is a radioactive isotope of hydrogen ( $t_{1 / 2}=12.32 \\mathrm{y}$ ) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu . What is the activity of a sample containing 1.00 mg of tritium (a) in Bq and (b) in Ci ?\n"} {"id": "textbook_7343", "title": "2132. Answer: - ", "content": "\n(a) $3.56 \\times 10^{11} \\mathrm{~Bq}$; (b) 0.962 Ci\n"} {"id": "textbook_7344", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "\nThe effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure 20.37, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).\n"} {"id": "textbook_7345", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "FIGURE 20.37 The total annual radiation exposure for a person in the US is about 620 mrem . The various sources and their relative amounts are shown in this bar graph. (source: U.S. Nuclear Regulatory Commission)"} {"id": "textbook_7346", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from\nchanges in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a $50 \\%$ probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person's lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table 20.5."} {"id": "textbook_7347", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "Health Effects of Radiation ${ }^{\\underline{2}}$"} {"id": "textbook_7348", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "| Exposure (rem) | Health Effect | Time to Onset (without treatment) |\n| :--- | :--- | :--- |\n| $5-10$ | changes in blood chemistry | - |\n| 50 | nausea | hours |\n| 55 | fatigue | - |\n| 70 | vomiting | - |\n| 75 | hair loss | $2-3$ weeks |\n| 90 | hemorrhage | - |\n| 100 | possible death | - |\n| 400 | destruction of intestinal lining | - |\n| | internal bleeding | - |\n| | death | within 2 months |\n| 1000 | damage to central nervous system | - |\n| | loss of consciousness; | minutes |"} {"id": "textbook_7349", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "TABLE 20.5"} {"id": "textbook_7350", "title": "2133. Effects of Long-term Radiation Exposure on the Human Body - ", "content": "It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.\n"} {"id": "textbook_7351", "title": "2134. Key Terms - ", "content": "\nalpha ( $\\alpha$ ) decay loss of an alpha particle during radioactive decay\nalpha particle ( $\\alpha$ or $\\mathbf{2}_{\\mathbf{2}}^{\\mathbf{H}} \\mathbf{H e}$ or ${ }_{\\mathbf{2}}^{\\mathbf{4}} \\boldsymbol{\\alpha}$ ) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons\nantimatter particles with the same mass but opposite properties (such as charge) of ordinary particles\nband of stability (also, belt of stability, zone of stability, or valley of stability) region of graph of number of protons versus number of neutrons containing stable (nonradioactive) nuclides\nbecquerel (Bq) SI unit for rate of radioactive decay; $1 \\mathrm{~Bq}=1$ disintegration/s\nbeta ( $\\boldsymbol{\\beta}$ ) decay breakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle\nbeta particle $\\left(\\boldsymbol{\\beta}\\right.$ or ${ }_{-\\mathbf{1}}^{\\mathbf{0}} \\mathbf{e}$ or $\\left.\\underset{-\\mathbf{1}}{\\mathbf{0}} \\boldsymbol{\\beta}\\right)$ high-energy electron\nbinding energy per nucleon total binding energy for the nucleus divided by the number of nucleons in the nucleus\nchain reaction repeated fission caused when the neutrons released in fission bombard other atoms\nchemotherapy similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells\ncontainment system (also, shield) a three-part structure of materials that protects the exterior of a nuclear fission reactor and operating personnel from the high temperatures, pressures, and radiation levels inside the reactor\ncontrol rod material inserted into the fuel assembly that absorbs neutrons and can be raised or lowered to adjust the rate of a fission reaction\ncritical mass amount of fissionable material that will support a self-sustaining (nuclear fission) chain reaction\ncurie ( $\\mathbf{C i}$ ) larger unit for rate of radioactive decay frequently used in medicine; $1 \\mathrm{Ci}=3.7 \\times 10^{10}$ disintegrations/s\ndaughter nuclide nuclide produced by the radioactive decay of another nuclide; may be stable or may decay further\nelectron capture combination of a core electron with a proton to yield a neutron within the nucleus\nelectron volt (eV) measurement unit of nuclear binding energies, with 1 eV equaling the amount energy due to the moving an electron across an electric potential difference of 1 volt\nexternal beam radiation therapy radiation delivered by a machine outside the body\nfissile (or fissionable) when a material is capable of sustaining a nuclear fission reaction\nfission splitting of a heavier nucleus into two or more lighter nuclei, usually accompanied by the conversion of mass into large amounts of energy\nfusion combination of very light nuclei into heavier nuclei, accompanied by the conversion of mass into large amounts of energy\nfusion reactor nuclear reactor in which fusion reactions of light nuclei are controlled\ngamma ( $\\boldsymbol{\\gamma}$ ) emission decay of an excited-state nuclide accompanied by emission of a gamma ray\ngamma ray ( $\\boldsymbol{\\gamma}$ or $\\mathbf{0}_{\\mathbf{0}}^{\\boldsymbol{0}} \\boldsymbol{\\gamma}$ ) short wavelength, highenergy electromagnetic radiation that exhibits wave-particle duality\nGeiger counter instrument that detects and measures radiation via the ionization produced in a Geiger-M\u00fcller tube\ngray (Gy) SI unit for measuring radiation dose; 1 Gy = 1 J absorbed $/ \\mathrm{kg}$ tissue\nhalf-life ( $\\boldsymbol{t}_{\\mathbf{1} / \\mathbf{2}}$ ) time required for half of the atoms in a radioactive sample to decay\ninternal radiation therapy (also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells\nionizing radiation radiation that can cause a molecule to lose an electron and form an ion\nmagic number nuclei with specific numbers of nucleons that are within the band of stability\nmass defect difference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass \"lost\" when nucleons are brought together to form a nucleus)\nmass-energy equivalence equation Albert Einstein's relationship showing that mass and energy are equivalent\nmillicurie ( $\\mathbf{m C i}$ ) larger unit for rate of radioactive decay frequently used in medicine; $1 \\mathrm{Ci}=3.7 \\times$ $10^{10}$ disintegrations/s\nnonionizing radiation radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules\nnuclear binding energy energy lost when an\natom's nucleons are bound together (or the energy needed to break a nucleus into its constituent protons and neutrons)\nnuclear chemistry study of the structure of atomic nuclei and processes that change nuclear structure\nnuclear fuel fissionable isotope present in sufficient quantities to provide a self-sustaining chain reaction in a nuclear reactor\nnuclear moderator substance that slows neutrons to a speed low enough to cause fission\nnuclear reaction change to a nucleus resulting in changes in the atomic number, mass number, or energy state\nnuclear reactor environment that produces energy via nuclear fission in which the chain reaction is controlled and sustained without explosion\nnuclear transmutation conversion of one nuclide into another nuclide\nnucleon collective term for protons and neutrons in a nucleus\nnuclide nucleus of a particular isotope\nparent nuclide unstable nuclide that changes spontaneously into another (daughter) nuclide\nparticle accelerator device that uses electric and magnetic fields to increase the kinetic energy of nuclei used in transmutation reactions\npositron $\\left({ }_{+1}^{0} \\beta\\right.$ or $\\left.{ }_{+1}^{0} \\mathrm{e}\\right)$ antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge\npositron emission (also, $\\beta^{+}$decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted\nradiation absorbed dose (rad) SI unit for measuring radiation dose, frequently used in medical applications; $1 \\mathrm{rad}=0.01$ Gy\nradiation dosimeter device that measures ionizing radiation and is used to determine personal radiation exposure\nradiation therapy use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing\nradioactive decay spontaneous decay of an unstable nuclide into another nuclide\nradioactive decay series chains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product\nradioactive tracer (also, radioactive label)\nradioisotope used to track or follow a substance by monitoring its radioactive emissions\nradioactivity phenomenon exhibited by an unstable nucleon that spontaneously undergoes change into a nucleon that is more stable; an unstable nucleon is said to be radioactive\nradiocarbon dating highly accurate means of dating objects 30,000-50,000 years old that were derived from once-living matter; achieved by calculating the ratio of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ in the object vs. the ratio of ${ }_{6}^{14} \\mathrm{C}$ : ${ }_{6}^{12} \\mathrm{C}$ in the present-day atmosphere\nradioisotope isotope that is unstable and undergoes conversion into a different, more stable isotope\nradiometric dating use of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations\nreactor coolant assembly used to carry the heat produced by fission in a reactor to an external boiler and turbine where it is transformed into electricity\nrelative biological effectiveness (RBE) measure of the relative damage done by radiation\nroentgen equivalent man (rem) unit for radiation damage, frequently used in medicine; 100 rem = 1 Sv\nscintillation counter instrument that uses a scintillator-a material that emits light when excited by ionizing radiation-to detect and measure radiation\nsievert (Sv) SI unit measuring tissue damage caused by radiation; takes into account energy and biological effects of radiation\nstrong nuclear force force of attraction between nucleons that holds a nucleus together\nsubcritical mass amount of fissionable material that cannot sustain a chain reaction; less than a critical mass\nsupercritical mass amount of material in which there is an increasing rate of fission\ntransmutation reaction bombardment of one type of nuclei with other nuclei or neutrons\ntransuranium element element with an atomic number greater than 92; these elements do not occur in nature\n"} {"id": "textbook_7352", "title": "2135. Key Equations - ", "content": "\n$E=m c^{2}$\ndecay rate $=\\lambda N$\n$t_{1 / 2}=\\frac{\\ln 2}{\\lambda}=\\frac{0.693}{\\lambda}$\nrem $=\\mathrm{RBE} \\times \\mathrm{rad}$\n$\\mathrm{Sv}=\\mathrm{RBE} \\times \\mathrm{Gy}$\n"} {"id": "textbook_7353", "title": "2136. Summary - ", "content": ""} {"id": "textbook_7354", "title": "2136. Summary - 2136.1. Nuclear Structure and Stability", "content": "\nAn atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This \"missing\" mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein's mass-energy equivalence equation, $E=m c^{2}$. Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56 ; these are the most stable nuclei.\n"} {"id": "textbook_7355", "title": "2136. Summary - 2136.2. Nuclear Equations", "content": "\nNuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha ( $\\alpha$ ) particles (which are highenergy helium nuclei), beta ( $\\beta$ ) particles (which are high-energy electrons), and gamma ( $\\gamma$ ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.\n"} {"id": "textbook_7356", "title": "2136. Summary - 2136.3. Radioactive Decay", "content": "\nNuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are $\\alpha$ decay, $\\beta$ decay, $\\gamma$ emission, positron emission, and electron capture. Nuclear reactions also often involve $\\gamma$ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some\nsubstances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.\n"} {"id": "textbook_7357", "title": "2136. Summary - 2136.4. Transmutation and Nuclear Energy", "content": "\nIt is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way."} {"id": "textbook_7358", "title": "2136. Summary - 2136.4. Transmutation and Nuclear Energy", "content": "Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production.\n"} {"id": "textbook_7359", "title": "2136. Summary - 2136.5. Uses of Radioisotopes", "content": "\nCompounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are\nperformed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.\n"} {"id": "textbook_7360", "title": "2136. Summary - 2136.6. Biological Effects of Radiation", "content": "\nWe are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to\nrapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating."} {"id": "textbook_7361", "title": "2136. Summary - 2136.6. Biological Effects of Radiation", "content": "Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, and including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source, and limiting time of exposure.\n"} {"id": "textbook_7362", "title": "2137. Exercises - ", "content": ""} {"id": "textbook_7363", "title": "2137. Exercises - 2137.1. Nuclear Structure and Stability", "content": "\n1. Write the following isotopes in hyphenated form (e.g., \"carbon-14\")\n(a) ${ }_{11}^{24} \\mathrm{Na}$\n(b) ${ }_{13}^{29} \\mathrm{Al}$\n(c) ${ }_{36}^{73} \\mathrm{Kr}$\n(d) ${ }_{77}^{194} \\mathrm{Ir}$\n2. Write the following isotopes in nuclide notation (e.g., \" ${ }_{6}^{14} \\mathrm{C}$ \")\n(a) oxygen-14\n(b) copper-70\n(c) tantalum-175\n(d) francium-217\n3. For the following isotopes that have missing information, fill in the missing information to complete the notation\n(a) ${ }_{14}^{34} \\mathrm{X}$\n(b) ${ }_{X}^{36} \\mathrm{P}$\n(c) ${ }_{\\mathrm{X}}{ }^{57} \\mathrm{Mn}$\n(d) ${ }_{56}^{121} \\mathrm{X}$\n4. For each of the isotopes in Exercise 20.1, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.\n5. Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:\n(a) 25 protons, 20 neutrons, 24 electrons\n(b) 45 protons, 24 neutrons, 43 electrons\n(c) 53 protons, 89 neutrons, 54 electrons\n(d) 97 protons, 146 neutrons, 97 electrons\n6. Calculate the density of the ${ }_{12}^{24} \\mathrm{Mg}$ nucleus in $\\mathrm{g} / \\mathrm{mL}$, assuming that it has the typical nuclear diameter of $1 \\times$ $10^{-13} \\mathrm{~cm}$ and is spherical in shape.\n7. What are the two principal differences between nuclear reactions and ordinary chemical changes?\n8. The mass of the atom ${ }_{11}^{23} \\mathrm{Na}$ is 22.9898 amu .\n(a) Calculate its binding energy per atom in millions of electron volts.\n(b) Calculate its binding energy per nucleon.\n9. Which of the following nuclei lie within the band of stability shown in Figure 20.2?\n(a) chlorine-37\n(b) calcium-40\n(c) ${ }^{204} \\mathrm{Bi}$\n(d) ${ }^{56} \\mathrm{Fe}$\n(e) ${ }^{206} \\mathrm{~Pb}$\n(f) ${ }^{211} \\mathrm{~Pb}$\n(g) ${ }^{222} \\mathrm{Rn}$\n(h) carbon-14\n10. Which of the following nuclei lie within the band of stability shown in Figure 20.2?\n(a) argon-40\n(b) oxygen-16\n(c) ${ }^{122} \\mathrm{Ba}$\n(d) ${ }^{58} \\mathrm{Ni}$\n(e) ${ }^{205} \\mathrm{Tl}$\n(f) ${ }^{210} \\mathrm{Tl}$\n(g) ${ }^{226} \\mathrm{Ra}$\n(h) magnesium-24\n"} {"id": "textbook_7364", "title": "2137. Exercises - 2137.2. Nuclear Equations", "content": "\n11. Write a brief description or definition of each of the following:\n(a) nucleon\n(b) $\\alpha$ particle\n(c) $\\beta$ particle\n(d) positron\n(e) $\\gamma$ ray\n(f) nuclide\n(g) mass number\n(h) atomic number\n12. Which of the various particles ( $\\alpha$ particles, $\\beta$ particles, and so on) that may be produced in a nuclear reaction are actually nuclei?\n13. Complete each of the following equations by adding the missing species:\n(a) ${ }_{13}^{27} \\mathrm{Al}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow$ ? $+{ }_{0}^{1} \\mathrm{n}$\n(b) ${ }_{94}^{239} \\mathrm{Pu}+$ ? $\\longrightarrow{ }_{96}^{242} \\mathrm{Cm}+{ }_{0}^{1} \\mathrm{n}$\n(c) ${ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow$ ? $+{ }_{1}^{1} \\mathrm{H}$\n(d) ${ }_{92}^{235} \\mathrm{U} \\longrightarrow ?+{ }_{55}^{135} \\mathrm{Cs}+4{ }_{0}^{1} \\mathrm{n}$\n14. Complete each of the following equations:\n(a) ${ }_{3}^{7} \\mathrm{Li}+? \\longrightarrow 2{ }_{2}^{4} \\mathrm{He}$\n(b) ${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+$ ?\n(c) ${ }_{13}^{27} \\mathrm{Al}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow ?+{ }_{0}^{1} \\mathrm{n}$\n(d) ${ }_{96}^{250} \\mathrm{Cm} \\longrightarrow ?+{ }_{38}^{98} \\mathrm{Sr}+4{ }_{0}^{1} \\mathrm{n}$\n15. Write a balanced equation for each of the following nuclear reactions:\n(a) the production of ${ }^{17} \\mathrm{O}$ from ${ }^{14} \\mathrm{~N}$ by $\\alpha$ particle bombardment\n(b) the production of ${ }^{14} \\mathrm{C}$ from ${ }^{14} \\mathrm{~N}$ by neutron bombardment\n(c) the production of ${ }^{233} \\mathrm{Th}$ from ${ }^{232} \\mathrm{Th}$ by neutron bombardment\n(d) the production of ${ }^{239} \\mathrm{U}$ from ${ }^{238} \\mathrm{U}$ by ${ }_{1}^{2} \\mathrm{H}$ bombardment\n16. Technetium-99 is prepared from ${ }^{98} \\mathrm{Mo}$. Molybdenum- 98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a $\\beta$ particle to yield an excited form of technetium-99, represented as ${ }^{99} \\mathrm{Tc}$ \". This excited nucleus relaxes to the ground state, represented as ${ }^{99} \\mathrm{Tc}$, by emitting a $\\gamma$ ray. The ground state of ${ }^{99} \\mathrm{Tc}$ then emits a $\\beta$ particle. Write the equations for each of these nuclear reactions.\n17. The mass of the atom ${ }_{9}^{19} \\mathrm{~F}$ is 18.99840 amu .\n(a) Calculate its binding energy per atom in millions of electron volts.\n(b) Calculate its binding energy per nucleon.\n18. For the reaction ${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+$ ?, if 100.0 g of carbon reacts, what volume of nitrogen gas $\\left(\\mathrm{N}_{2}\\right)$ is produced at 273 K and 1 atm ?\n"} {"id": "textbook_7365", "title": "2137. Exercises - 2137.3. Radioactive Decay", "content": "\n19. What are the types of radiation emitted by the nuclei of radioactive elements?\n20. What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?\n(a) an $\\alpha$ particle is emitted\n(b) a $\\beta$ particle is emitted\n(c) $\\gamma$ radiation is emitted\n(d) a positron is emitted\n(e) an electron is captured\n21. What is the change in the nucleus that results from the following decay scenarios?\n(a) emission of a $\\beta$ particle\n(b) emission of a $\\beta^{+}$particle\n(c) capture of an electron\n22. Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include $\\alpha$ particles.\n23. Why is electron capture accompanied by the emission of an X-ray?\n24. Explain, in terms of Figure 20.2, how unstable heavy nuclides (atomic number $>83$ ) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.\n25. Which of the following nuclei is most likely to decay by positron emission? Explain your choice.\n(a) chromium-53\n(b) manganese-51\n(c) iron-59\n26. The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.\n(a) ${ }_{15}^{34} \\mathrm{P}$\n(b) ${ }_{92}^{239} \\mathrm{U}$\n(c) ${ }_{20}^{38} \\mathrm{Ca}$\n(d) ${ }_{1}^{3} \\mathrm{H}$\n(e) ${ }_{94}^{245} \\mathrm{Pu}$\n27. The following nuclei do not lie in the band of stability. How would they be expected to decay?\n(a) ${ }_{15}^{28} \\mathrm{P}$\n(b) ${ }_{92}^{235} \\mathrm{U}$\n(c) ${ }_{20}^{37} \\mathrm{Ca}$\n(d) ${ }_{3}^{9} \\mathrm{Li}$\n(e) ${ }_{96}^{245} \\mathrm{Cm}$\n28. Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:\n(a) ${ }_{2}^{6} \\mathrm{He}$\n(b) ${ }_{30}^{60} \\mathrm{Zn}$\n(c) ${ }_{91}^{235} \\mathrm{~Pa}$\n(d) ${ }_{94}^{241} \\mathrm{~Np}$\n(e) ${ }^{18} \\mathrm{~F}$\n(f) ${ }^{129} \\mathrm{Ba}$\n(g) ${ }^{237} \\mathrm{Pu}$\n29. Write a nuclear reaction for each step in the formation of ${ }_{84}^{218} \\operatorname{Po}$ from ${ }_{98}^{238} \\mathrm{U}$, which proceeds by a series of decay reactions involving the step-wise emission of $\\alpha, \\beta, \\beta, \\alpha, \\alpha, \\alpha$ particles, in that order.\n30. Write a nuclear reaction for each step in the formation of ${ }_{82}^{208} \\mathrm{~Pb}$ from ${ }_{90}^{228} \\mathrm{Th}$, which proceeds by a series of decay reactions involving the step-wise emission of $\\alpha, \\alpha, \\alpha, \\alpha, \\beta, \\beta, \\alpha$ particles, in that order.\n31. Define the term half-life and illustrate it with an example.\n32. A $1.00 \\times 10^{-6}$-g sample of nobelium, ${ }_{102}^{254} \\mathrm{No}$, has a half-life of 55 seconds after it is formed. What is the percentage of ${ }_{102}^{254}$ No remaining at the following times?\n(a) 5.0 min after it forms\n(b) 1.0 h after it forms\n33. ${ }^{239} \\mathrm{Pu}$ is a nuclear waste byproduct with a half-life of $24,000 \\mathrm{y}$. What fraction of the ${ }^{239} \\mathrm{Pu}$ present today will be present in 1000 y ?\n34. The isotope ${ }^{208} \\mathrm{Tl}$ undergoes $\\beta$ decay with a half-life of 3.1 min .\n(a) What isotope is produced by the decay?\n(b) How long will it take for $99.0 \\%$ of a sample of pure ${ }^{208} \\mathrm{Tl}$ to decay?\n(c) What percentage of a sample of pure ${ }^{208} \\mathrm{Tl}$ remains un-decayed after 1.0 h ?\n35. If 1.000 g of ${ }_{88}^{226} \\mathrm{Ra}$ produces 0.0001 mL of the gas ${ }_{86}^{222} \\mathrm{Rn}$ at STP (standard temperature and pressure) in 24 h , what is the half-life of ${ }^{226} \\mathrm{Ra}$ in years?\n36. The isotope ${ }_{38}^{90} \\mathrm{Sr}$ is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a $0.500-\\mathrm{g}$ sample diminishes to 0.393 g in 10.0 y . Calculate the half-life.\n37. Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h . Calculate the rate constant for the decay of ${ }_{43}^{99} \\mathrm{Tc}$.\n38. What is the age of mummified primate skin that contains $8.25 \\%$ of the original quantity of ${ }^{14} \\mathrm{C}$ ?\n39. A sample of rock was found to contain 8.23 mg of rubidium -87 and 0.47 mg of strontium- 87 .\n(a) Calculate the age of the rock if the half-life of the decay of rubidium by $\\beta$ emission is $4.7 \\times 10^{10} \\mathrm{y}$.\n(b) If some ${ }_{38}^{87} \\mathrm{Sr}$ was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.\n40. A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of ${ }_{92}^{238} \\mathrm{U}$ and 2.52 mg of ${ }_{82}^{206} \\mathrm{~Pb}$. Calculate the age of the ore. The half-life of ${ }_{92}^{238} \\mathrm{U}$ is $4.5 \\times 10^{9} \\mathrm{yr}$.\n41. Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this ${ }^{239} \\mathrm{Pu}$ was the capture of neutrons by ${ }^{238} \\mathrm{U}$ nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed $4.7 \\times 10^{9}$ years ago?\n42. $\\mathrm{A}{ }_{4}^{7} \\mathrm{Be}$ atom (mass $=7.0169 \\mathrm{amu}$ ) decays into a ${ }_{3}^{7} \\mathrm{Li}$ atom (mass $=7.0160 \\mathrm{amu}$ ) by electron capture. How much energy (in millions of electron volts, MeV ) is produced by this reaction?\n43. $\\mathrm{A}{ }_{5}^{8} \\mathrm{~B}$ atom (mass $=8.0246 \\mathrm{amu}$ ) decays into a ${ }_{4}^{8} \\mathrm{~B}$ atom (mass $=8.0053 \\mathrm{amu}$ ) by loss of a $\\beta^{+}$particle (mass $=0.00055 \\mathrm{amu}$ ) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?\n44. Isotopes such as ${ }^{26} \\mathrm{Al}$ (half-life: $7.2 \\times 10^{5}$ years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.\n(a) ${ }^{26} \\mathrm{Al}$ decays by $\\beta^{+}$emission or electron capture. Write the equations for these two nuclear transformations.\n(b) The earth was formed about $4.7 \\times 10^{9}$ ( 4.7 billion) years ago. How old was the earth when $99.999999 \\%$ of the ${ }^{26} \\mathrm{Al}$ originally present had decayed?\n45. Write a balanced equation for each of the following nuclear reactions:\n(a) bismuth-212 decays into polonium-212\n(b) beryllium-8 and a positron are produced by the decay of an unstable nucleus\n(c) neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239\n(d) strontium-90 decays into yttrium-90\n46. Write a balanced equation for each of the following nuclear reactions:\n(a) mercury-180 decays into platinum-176\n(b) zirconium-90 and an electron are produced by the decay of an unstable nucleus\n(c) thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium- 228 by beta decay\n(d) neon-19 decays into fluorine-19\n"} {"id": "textbook_7366", "title": "2137. Exercises - 2137.4. Transmutation and Nuclear Energy", "content": "\n47. Write the balanced nuclear equation for the production of the following transuranium elements:\n(a) berkelium-244, made by the reaction of Am-241 and $\\mathrm{He}-4$\n(b) fermium-254, made by the reaction of $\\mathrm{Pu}-239$ with a large number of neutrons\n(c) lawrencium-257, made by the reaction of Cf-250 and B-11\n(d) dubnium-260, made by the reaction of Cf-249 and N-15\n48. How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic?\n49. Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?\n50. Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.\n51. Describe the components of a nuclear reactor.\n52. In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.\n53. Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.\n54. The mass of a hydrogen atom $\\left({ }_{1}^{1} \\mathrm{H}\\right)$ is 1.007825 amu ; that of a tritium atom $\\left({ }_{1}^{3} \\mathrm{H}\\right)$ is 3.01605 amu ; and that of an $\\alpha$ particle is 4.00150 amu . How much energy in kilojoules per mole of ${ }_{2}^{4} \\mathrm{He}$ produced is released by the following fusion reaction: ${ }_{1}^{1} \\mathrm{H}+{ }_{1}^{3} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}$.\n"} {"id": "textbook_7367", "title": "2137. Exercises - 2137.5. Uses of Radioisotopes", "content": "\n55. How can a radioactive nuclide be used to show that the equilibrium: $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$ is a dynamic equilibrium?\n56. Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium- 99 m is safe to leave the hospital once $75 \\%$ of the dose has decayed, when is the patient allowed to leave?\n57. Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I- 131 has a half-life of 8.70 days and decays by $\\beta^{-}$ emission.\n(a) Write an equation for the decay.\n(b) How long will it take for $95.0 \\%$ of a dose of I-131 to decay?\n"} {"id": "textbook_7368", "title": "2137. Exercises - 2137.6. Biological Effects of Radiation", "content": "\n58. If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:\n(a) cobalt-60 (a strong $\\gamma$ emitter used for irradiation)\n(b) molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)\n59. Based on what is known about Radon- 222 's primary decay method, why is inhalation so dangerous?\n60. Given specimens uranium-232 ( $t_{1 / 2}=68.9 \\mathrm{y}$ ) and uranium-233 ( $\\left.t_{1 / 2}=159,200 \\mathrm{y}\\right)$ of equal mass, which one would have greater activity and why?\n61. A scientist is studying a 2.234 g sample of thorium-229 ( $\\left.t_{1 / 2}=7340 \\mathrm{y}\\right)$ in a laboratory.\n(a) What is its activity in Bq?\n(b) What is its activity in Ci ?\n62. Given specimens neon-24 ( $t_{1 / 2}=3.38 \\mathrm{~min}$ ) and bismuth-211 ( $\\left.t_{1 / 2}=2.14 \\mathrm{~min}\\right)$ of equal mass, which one would have greater activity and why?\n"} {"id": "textbook_7369", "title": "2137. Exercises - 2137.6. Biological Effects of Radiation", "content": "Figure 21.1 All organic compounds contain carbon and most are formed by living things, although they are also formed by geological and artificial processes. (credit left: modification of work by Jon Sullivan; credit left middle: modification of work by Deb Tremper; credit right middle: modification of work by \"annszyp\"/Wikimedia Commons; credit right: modification of work by George Shuklin)\n"} {"id": "textbook_7370", "title": "2138. CHAPTER OUTLINE - ", "content": ""} {"id": "textbook_7371", "title": "2138. CHAPTER OUTLINE - 2138.1. Hydrocarbons", "content": "\n21.2 Alcohols and Ethers\n21.3 Aldehydes, Ketones, Carboxylic Acids, and Esters\n21.4 Amines and Amides"} {"id": "textbook_7372", "title": "2138. CHAPTER OUTLINE - 2138.1. Hydrocarbons", "content": "INTRODUCTION All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet \"carbon-based\" life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and $\\mathrm{CO}_{2}$. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms."} {"id": "textbook_7373", "title": "2138. CHAPTER OUTLINE - 2138.1. Hydrocarbons", "content": "Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings.\n"} {"id": "textbook_7374", "title": "2138. CHAPTER OUTLINE - 2138.2. Hydrocarbons", "content": ""} {"id": "textbook_7375", "title": "2139. LEARNING OBJECTIVES - ", "content": "\nBy the end of this section, you will be able to:"} {"id": "textbook_7376", "title": "2139. LEARNING OBJECTIVES - ", "content": "- Explain the importance of hydrocarbons and the reason for their diversity\n- Name saturated and unsaturated hydrocarbons, and molecules derived from them\n- Describe the reactions characteristic of saturated and unsaturated hydrocarbons\n- Identify structural and geometric isomers of hydrocarbons"} {"id": "textbook_7377", "title": "2139. LEARNING OBJECTIVES - ", "content": "The largest database ${ }^{\\frac{1}{1}}$ of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated ${ }^{2}$ at $10^{60}$-an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities."} {"id": "textbook_7378", "title": "2139. LEARNING OBJECTIVES - ", "content": "The simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals.\n"} {"id": "textbook_7379", "title": "2140. Alkanes - ", "content": "\nAlkanes, or saturated hydrocarbons, contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has $s p^{3}$ hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure 21.2. Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the $s p^{3}$ hybridization, the bond angles in carbon chains are close to $109.5^{\\circ}$, giving such chains in an alkane a zigzag shape."} {"id": "textbook_7380", "title": "2140. Alkanes - ", "content": "The structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, condensed formulas). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure 21.2, and several additional examples are provided in the exercises at the end of this chapter."} {"id": "textbook_7381", "title": "2140. Alkanes - ", "content": "[^13]
Formula | Melting Point
$\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Boiling Point
$\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Phase at
STP | Number of Structural
Isomers |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| methane | $\\mathrm{CH}_{4}$ | -182.5 | -161.5 | gas | 1 |\n| ethane | $\\mathrm{C}_{2} \\mathrm{H}_{6}$ | -183.3 | -88.6 | gas | 1 |\n| propane | $\\mathrm{C}_{3} \\mathrm{H}_{8}$ | -187.7 | -42.1 | gas | 1 |\n| putane | $\\mathrm{C}_{4} \\mathrm{H}_{10}$ | -138.3 | -0.5 | gas | 2 |\n| pentane | $\\mathrm{C}_{5} \\mathrm{H}_{12}$ | -129.7 | 36.1 | liquid | 3 |\n| hepane | $\\mathrm{C}_{6} \\mathrm{H}_{14}$ | -95.3 | 68.7 | liquid | 5 |\n| octane | $\\mathrm{C}_{7} \\mathrm{H}_{16} \\mathrm{H}_{18}$ | -90.6 | 98.4 | liquid | 9 |\n| nonane | $\\mathrm{C}_{9} \\mathrm{H}_{20}$ | -56.8 | 125.7 | liquid | 18 |\n| decane | $\\mathrm{C}_{10} \\mathrm{H}_{22}$ | -53.6 | -29.7 | 150.8 | liquid |\n| tetradecane | $\\mathrm{C}_{14} \\mathrm{H}_{30}$ | 5.9 | liquid | 75 | |\n| octadecane | $\\mathrm{C}_{18} \\mathrm{H}_{38}$ | 28.2 | solid | 1858 | |"} {"id": "textbook_7400", "title": "2149. Properties of Some Alkanes ${ }^{\\text {\u00b3 }}$ - ", "content": "[^14]Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula $\\mathrm{C}_{4} \\mathrm{H}_{10}$ : They are called $n$-butane and 2-methylpropane (or isobutane), and have the following Lewis structures:\n
V | Food trays, plastic wrap, bottles for mineral water and shampoo |\n| low density polyethylene (LDPE) | Shopping bags and garbage bags |\n| polypropylene (PP)
PP | Margarine tubs, microwaveable food trays |\n| polystyrene (PS) | Yogurt tubs, foam meat trays, egg cartons, vending cups, plastic cutlery, packaging for electronics and toys |\n| any other plastics (OTHER) | Plastics that do not fall into any of the above categories One example is melamine resin (plastic plates, plastic cups) |"} {"id": "textbook_7452", "title": "2164. Recycling Plastics - ", "content": "FIGURE 21.9 Each type of recyclable plastic is imprinted with a code for easy identification."} {"id": "textbook_7453", "title": "2164. Recycling Plastics - ", "content": "The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:\n
acid |
35.45
chine | $\\stackrel{\\substack{18 \\\\ \\text { 3r } \\\\ \\text { 3agan } \\\\ \\text { arg }}}{ }$ |\n|  | $\\left\\lvert\\, \\begin{gathered} \\begin{array}{c} 26 \\\\ \\mathrm{Fe} \\\\ 5 \\text { Fe. } \\\\ \\text { 1ron } \\end{array} \\\\ \\hline \\end{gathered}\\right.$ | $\\underset{\\substack{27 \\\\ \\hline 50.83 \\\\ \\text { cobat }}}{27}$ | $\\left\\lvert\\, \\begin{gathered} 28 \\\\ \\begin{array}{c} \\text { Ni } \\\\ 58.69 \\\\ \\text { nokel } \\end{array} \\\\ \\hline \\end{gathered}\\right.$ |  | $\\left\\lvert\\, \\begin{gathered} 30 \\\\ \\mathrm{Zn} \\\\ \\text { En } \\\\ \\text { zinc } \\\\ \\hline \\end{gathered}\\right.$ |  | $\\begin{gathered} 32 \\mathbf{G e} \\\\ 72.63 \\\\ \\text { gemanum } \\end{gathered}$ | $\\underbrace{}_{\\substack{33 \\\\ \\text { As } \\\\ \\text { atsenc } \\\\ \\text { anc }}}$ | $\\underset{\\substack{34 \\\\ \\text { seen } \\\\ \\text { senenum }}}{ }$ | $\\begin{gathered} 35 \\mathrm{Br} \\\\ \\text { Bran } \\\\ \\text { bomine } \\end{gathered}$ | $\\underset{\\substack{36 \\\\ \\hline 8.3, ~ \\\\ \\text { krpipon }}}{ }$ |\n| ${ }^{43} \\mathbf{T C}$ |  |  |  |  |  | $\\begin{gathered} \\ln ^{114.8} \\\\ 1104 \\\\ \\hline \\text { nourum } \\end{gathered}$ | ${\\underset{S}{50}}_{\\substack{518.7 \\\\ \\text { in }}}$ | $\\underset{\\substack{\\text { Sxili.8 } \\\\ \\text { and } \\\\ \\text { and }}}{ }$ |  | $\\left\\lvert\\, \\begin{gathered} 53 \\\\ \\text { I } \\\\ \\text { 126.9 } \\\\ \\text { iodine } \\end{gathered}\\right.$ | $\\left\\lvert\\, \\begin{gathered} 54 \\\\ \\mathrm{Xe}^{131.3} \\\\ \\text { xenon } \\end{gathered}\\right.$ |\n| $\\left\\begin{array}{c}75 \\\\ \\mathbf{R e} \\\\ \\text { 186.2. } \\\\ \\text { thenium }\\end{array}\\right.$ | ${ }^{76} \\mathrm{Os}$
190. | $\\left\\lvert\\, \\begin{array}{c\\|l} 77 \\\\ \\text { Ir } \\\\ \\text { 1922. } \\\\ \\text { indium } \\end{array}\\right.$ | ![ | $\\begin{array}{\\|c} 79 \\\\ \\text { Au } \\\\ \\text { 197.0 } \\\\ \\text { gold } \\end{array}$ | $\\begin{array}{\\|c} 80 \\\\ \\begin{array}{c} 80 \\\\ 200.6 \\\\ \\text { mercury } \\end{array} \\\\ \\hline \\end{array}$ | $\\begin{gathered} 81 \\\\ \\mathrm{TI} \\\\ \\text { 204.4. } \\\\ \\text { Raluium } \\end{gathered}$ | 82
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207.2
lead
lead | $\\begin{gathered} 83 \\\\ \\begin{array}{c} 8 i \\\\ \\text { 2ision } \\\\ \\text { biswnt } \\end{array} \\end{gathered}$ | $\\begin{array}{\\|c} 84 \\\\ \\mathbf{P O}_{0} \\\\ \\text { [209] } \\\\ \\text { polonium } \\end{array}$ |  |  |\n|  | $\\stackrel{108}{\\mathrm{Hs}}$ [277] |  | ${ }^{110} \\mathrm{DS}_{2}$ |  |  | ${\\underset{c}{123}}_{\\substack{\\text { N285] } \\\\ \\text { nhtonim }}}$ |  |  |  | $\\left.\\right\\|_{117} ^{11} \\text { TS }$ | ${ }^{118} \\mathbf{O g}$ |\n\n\n|  | ${ }^{58} \\mathrm{Ce}$ 140.1 |  | ${ }^{60} \\mathrm{Nd}$ |  | $\\sqrt[\\substack{62 \\\\ \\text { Sm } \\\\ \\text { singati }}]{ }$ | ${ }^{63} \\mathrm{Eu}$ | ${ }^{64} \\mathbf{G d} \\text { Gd. }$ | ${ }^{65} \\mathrm{~Tb}$ 158.9 |  | $\\left\\lvert\\, \\begin{gathered} 67 \\mathrm{Ho} \\\\ \\text { He4.9 } \\\\ \\text { nomium } \\end{gathered}\\right.$ | ${ }^{68}$ Er | ${ }_{\\substack{\\text { a }}}^{\\substack{\\text { Tm } \\\\ \\text { 168.9 } \\\\ \\text { thuium }}}$ |  | ${ }^{71} \\mathrm{Lu}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | $\\begin{gathered} 90 \\\\ \\hline 2320 \\\\ \\hline \\end{gathered}$ |  | ${ }_{\\substack{92 \\\\ \\hline \\\\ 238.0}}$ |  | ${ }^{94} \\mathrm{Pu}$ |  |  |  | ${ }^{98} \\mathbf{C f}$ | ${ }^{99} \\mathrm{Es}$ | $\\underset{\\substack{100 \\\\ \\text { Fm } \\\\ \\text { fetemiun }}}{ }$ |  | $\\stackrel{1}{102}_{\\substack{\\text { No } \\\\ \\text { nosediun }}}$ | $\\int_{103}^{\\mathrm{Lr}_{1}}$ |"} {"id": "textbook_7614", "title": "2216. The Periodic Table - ", "content": "Color Code\n"} {"id": "textbook_7615", "title": "2216. The Periodic Table - ", "content": "FIGURE A1"} {"id": "textbook_7616", "title": "2216. The Periodic Table - ", "content": "1082 A \u2022 The Periodic Table\n"} {"id": "textbook_7617", "title": "2217. APPENDIX B - ", "content": ""} {"id": "textbook_7618", "title": "2218. Essential Mathematics - ", "content": ""} {"id": "textbook_7619", "title": "2219. Exponential Arithmetic - ", "content": "\nExponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not equal to or greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:"} {"id": "textbook_7620", "title": "2219. Exponential Arithmetic - ", "content": "$$\n\\begin{aligned}\n1000 & =1 \\times 10^{3} \\\\\n100 & =1 \\times 10^{2} \\\\\n10 & =1 \\times 10^{1} \\\\\n1 & =1 \\times 10^{0} \\\\\n0.1 & =1 \\times 10^{-1} \\\\\n0.001 & =1 \\times 10^{-3} \\\\\n2386 & =2.386 \\times 1000=2.386 \\times 10^{3} \\\\\n0.123 & =1.23 \\times 0.1=1.23 \\times 10^{-1}\n\\end{aligned}\n$$"} {"id": "textbook_7621", "title": "2219. Exponential Arithmetic - ", "content": "The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for very large and very small numbers. For example, $1,230,000,000=1.23 \\times 10^{9}$, and $0.00000000036=3.6 \\times 10^{-10}$.\n"} {"id": "textbook_7622", "title": "2220. Addition of Exponentials - ", "content": "\nConvert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.\n"} {"id": "textbook_7623", "title": "2221. EXAMPLE B1 - ", "content": ""} {"id": "textbook_7624", "title": "2222. Adding Exponentials - ", "content": "\nAdd $5.00 \\times 10^{-5}$ and $3.00 \\times 10^{-3}$.\n"} {"id": "textbook_7625", "title": "2223. Solution - ", "content": "\n$$\n\\begin{aligned}\n3.00 \\times 10^{-3} & =300 \\times 10^{-5} \\\\\n\\left(5.00 \\times 10^{-5}\\right)+\\left(300 \\times 10^{-5}\\right) & =305 \\times 10^{-5}=3.05 \\times 10^{-3}\n\\end{aligned}\n$$\n"} {"id": "textbook_7626", "title": "2224. Subtraction of Exponentials - ", "content": "\nConvert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.\n"} {"id": "textbook_7627", "title": "2225. Subtracting Exponentials - ", "content": "\nSubtract $4.0 \\times 10^{-7}$ from $5.0 \\times 10^{-6}$.\n"} {"id": "textbook_7628", "title": "2226. Solution - ", "content": "\n$$\n\\begin{aligned}\n& 4.0 \\times 10^{-7}=0.40 \\times 10^{-6} \\\\\n& \\left(5.0 \\times 10^{-6}\\right)-\\left(0.40 \\times 10^{-6}\\right)=4.6 \\times 10^{-6}\n\\end{aligned}\n$$\n"} {"id": "textbook_7629", "title": "2227. Multiplication of Exponentials - ", "content": "\nMultiply the digit terms in the usual way and add the exponents of the exponential terms.\n"} {"id": "textbook_7630", "title": "2228. EXAMPLE B3 - ", "content": ""} {"id": "textbook_7631", "title": "2229. Multiplying Exponentials - ", "content": "\nMultiply $4.2 \\times 10^{-8}$ by $2.0 \\times 10^{3}$.\n"} {"id": "textbook_7632", "title": "2230. Solution - ", "content": "\n$$\n\\left(4.2 \\times 10^{-8}\\right) \\times\\left(2.0 \\times 10^{3}\\right)=(4.2 \\times 2.0) \\times 10^{(-8)+(+3)}=8.4 \\times 10^{-5}\n$$\n"} {"id": "textbook_7633", "title": "2231. Division of Exponentials - ", "content": "\nDivide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.\n"} {"id": "textbook_7634", "title": "2232. EXAMPLE B4 - ", "content": ""} {"id": "textbook_7635", "title": "2233. Dividing Exponentials - ", "content": "\nDivide $3.6 \\times 10^{-5}$ by $6.0 \\times 10^{-4}$.\n"} {"id": "textbook_7636", "title": "2234. Solution - ", "content": "\n$$\n\\frac{3.6 \\times 10^{-5}}{6.0 \\times 10^{-4}}=\\left(\\frac{3.6}{6.0}\\right) \\times 10^{(-5)-(-4)}=0.60 \\times 10^{-1}=6.0 \\times 10^{-2}\n$$\n"} {"id": "textbook_7637", "title": "2235. Squaring of Exponentials - ", "content": "\nSquare the digit term in the usual way and multiply the exponent of the exponential term by 2.\n"} {"id": "textbook_7638", "title": "2236. EXAMPLE B5 - ", "content": ""} {"id": "textbook_7639", "title": "2237. Squaring Exponentials - ", "content": "\nSquare the number $4.0 \\times 10^{-6}$.\n"} {"id": "textbook_7640", "title": "2238. Solution - ", "content": "\n$$\n\\left(4.0 \\times 10^{-6}\\right)^{2}=4 \\times 4 \\times 10^{2 \\times(-6)}=16 \\times 10^{-12}=1.6 \\times 10^{-11}\n$$\n"} {"id": "textbook_7641", "title": "2239. Cubing of Exponentials - ", "content": "\nCube the digit term in the usual way and multiply the exponent of the exponential term by 3.\n"} {"id": "textbook_7642", "title": "2240. Cubing Exponentials - ", "content": "\nCube the number $2 \\times 10^{4}$.\n"} {"id": "textbook_7643", "title": "2241. Solution - ", "content": "\n$\\left(2 \\times 10^{4}\\right)^{3}=2 \\times 2 \\times 2 \\times 10^{3 \\times 4}=8 \\times 10^{12}$\n"} {"id": "textbook_7644", "title": "2242. Taking Square Roots of Exponentials - ", "content": "\nIf necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2 . Extract the square root of the digit term and divide the exponential term by 2.\n"} {"id": "textbook_7645", "title": "2243. EXAMPLE B7 - ", "content": ""} {"id": "textbook_7646", "title": "2244. Finding the Square Root of Exponentials - ", "content": "\nFind the square root of $1.6 \\times 10^{-7}$.\n"} {"id": "textbook_7647", "title": "2245. Solution - ", "content": "\n$$\n\\begin{aligned}\n1.6 \\times 10^{-7} & =16 \\times 10^{-8} \\\\\n\\sqrt{16 \\times 10^{-8}}=\\sqrt{16} \\times \\sqrt{10^{-8}} & =\\sqrt{16} \\times 10^{-\\frac{8}{2}}=4.0 \\times 10^{-4}\n\\end{aligned}\n$$\n"} {"id": "textbook_7648", "title": "2246. Significant Figures - ", "content": "\nA beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more reasonable if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no useful meaning in this situation. When reporting quantities, use only as many significant figures as the accuracy of the measurement warrants."} {"id": "textbook_7649", "title": "2246. Significant Figures - ", "content": "The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example)."} {"id": "textbook_7650", "title": "2246. Significant Figures - ", "content": "EXAMPLE B8\n"} {"id": "textbook_7651", "title": "2247. Addition and Subtraction with Significant Figures - ", "content": "\nAdd 4.383 g and 0.0023 g .\n"} {"id": "textbook_7652", "title": "2248. Solution - ", "content": "\n$$\n\\begin{gathered}\n4.38 \\underline{\\mathrm{~g}} \\\\\n\\frac{0.002 \\underline{\\mathrm{3}} \\mathrm{~g}}{4.38 \\mathrm{~g}}\n\\end{gathered}\n$$"} {"id": "textbook_7653", "title": "2248. Solution - ", "content": "In multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.\n"} {"id": "textbook_7654", "title": "2249. Multiplication and Division with Significant Figures - ", "content": "\nMultiply 0.6238 by 6.6 .\n"} {"id": "textbook_7655", "title": "2250. Solution - ", "content": "\n$0.623 \\underline{8} \\times 6 . \\underline{6}=4 . \\underline{1}$"} {"id": "textbook_7656", "title": "2250. Solution - ", "content": "When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (\"round up\"). Do not change the retained digit if the digits that follow are less than 5 (\"round down\"). If the retained digit is followed by 5 , round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).\n"} {"id": "textbook_7657", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "\nThe common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2 , because 10 must be raised to the second power to equal 100 . Additional examples follow."} {"id": "textbook_7658", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "Logarithms and Exponential Numbers"} {"id": "textbook_7659", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "| Number | Number Expressed Exponentially | Common Logarithm |\n| :--- | :--- | :--- |\n| 1000 | $10^{3}$ | 3 |\n| 10 | $10^{1}$ | 1 |\n| 1 | $10^{0}$ | 0 |\n| 0.1 | $10^{-1}$ | -1 |\n| 0.001 | $10^{-3}$ | -3 |"} {"id": "textbook_7660", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "TABLE B1"} {"id": "textbook_7661", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "What is the common logarithm of 60 ? Because 60 lies between 10 and 100 , which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782 ; that is,"} {"id": "textbook_7662", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n60=10^{1.7782}\n$$"} {"id": "textbook_7663", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "The common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is -1.4069 , or"} {"id": "textbook_7664", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n0.03918=10^{-1.4069}=\\frac{1}{10^{1.4069}}\n$$"} {"id": "textbook_7665", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "To obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate $10^{x}$ (where $x$ is the logarithm of the number)."} {"id": "textbook_7666", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "The natural logarithm of a number ( $\\ln$ ) is the power to which e must be raised to equal the number; $e$ is the constant 2.7182818 . For example, the natural logarithm of 10 is 2.303 ; that is,"} {"id": "textbook_7667", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n10=e^{2.303}=2.7182818^{2.303}\n$$"} {"id": "textbook_7668", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "To obtain the natural logarithm of a number, use the $\\ln$ button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate $e^{x}$ (where $x$ is the natural logarithm of the number)."} {"id": "textbook_7669", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "Logarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving\nexponents."} {"id": "textbook_7670", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers."} {"id": "textbook_7671", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n\\log x y=\\log x+\\log y, \\text { and } \\ln x y=\\ln x+\\ln y\n$$"} {"id": "textbook_7672", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers."} {"id": "textbook_7673", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n\\log \\frac{x}{y}=\\log x-\\log y, \\text { and } \\ln \\frac{x}{y}=\\ln x-\\ln y\n$$"} {"id": "textbook_7674", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number."} {"id": "textbook_7675", "title": "2251. The Use of Logarithms and Exponential Numbers - ", "content": "$$\n\\log x^{n}=n \\log x \\text { and } \\ln x^{n}=n \\ln x\n$$\n"} {"id": "textbook_7676", "title": "2252. The Solution of Quadratic Equations - ", "content": "\nMathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions."} {"id": "textbook_7677", "title": "2252. The Solution of Quadratic Equations - ", "content": "$$\na x^{2}+b x+c=0\n$$"} {"id": "textbook_7678", "title": "2252. The Solution of Quadratic Equations - ", "content": "The solution or roots for any quadratic equation can be calculated using the following formula:"} {"id": "textbook_7679", "title": "2252. The Solution of Quadratic Equations - ", "content": "$$\nx=\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\n$$\n"} {"id": "textbook_7680", "title": "2253. EXAMPLE B10 - ", "content": ""} {"id": "textbook_7681", "title": "2254. Solving Quadratic Equations - ", "content": "\nSolve the quadratic equation $3 x^{2}+13 x-10=0$.\n"} {"id": "textbook_7682", "title": "2255. Solution - ", "content": "\nSubstituting the values $a=3, b=13, c=-10$ in the formula, we obtain"} {"id": "textbook_7683", "title": "2255. Solution - ", "content": "$$\n\\begin{gathered}\nx=\\frac{-13 \\pm \\sqrt{(13)^{2}-4 \\times 3 \\times(-10)}}{2 \\times 3} \\\\\nx=\\frac{-13 \\pm \\sqrt{169+120}}{6}=\\frac{-13 \\pm \\sqrt{289}}{6}=\\frac{-13 \\pm 17}{6}\n\\end{gathered}\n$$"} {"id": "textbook_7684", "title": "2255. Solution - ", "content": "The two roots are therefore"} {"id": "textbook_7685", "title": "2255. Solution - ", "content": "$$\nx=\\frac{-13+17}{6}=\\frac{2}{3} \\text { and } x=\\frac{-13-17}{6}=-5\n$$"} {"id": "textbook_7686", "title": "2255. Solution - ", "content": "Quadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance.\n"} {"id": "textbook_7687", "title": "2256. Two-Dimensional ( $x-y$ ) Graphing - ", "content": "\nThe relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled ( $x$ ), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured ( $y$ )."} {"id": "textbook_7688", "title": "2256. Two-Dimensional ( $x-y$ ) Graphing - ", "content": "When the value of $y$ is changing as a function of $x$ (that is, different values of $x$ correspond to different values of $y$ ), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for $(x, y)$ data pairs.\n"} {"id": "textbook_7689", "title": "2257. Graphing the Dependence of $y$ on $x$ - ", "content": "\n"} {"id": "textbook_7690", "title": "2257. Graphing the Dependence of $y$ on $x$ - ", "content": "This table contains the following points: $(1,5),(2,10),(3,7)$, and $(4,14)$. Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of $y$ on $x$.\n"} {"id": "textbook_7691", "title": "2257. Graphing the Dependence of $y$ on $x$ - ", "content": "If the function that describes the dependence of $y$ on $x$ is known, it may be used to compute $x, y$ data pairs that may subsequently be plotted.\n"} {"id": "textbook_7692", "title": "2258. Plotting Data Pairs - ", "content": "\nIf we know that $y=x^{2}+2$, we can produce a table of a few $(x, y)$ values and then plot the line based on the data shown here.\n"} {"id": "textbook_7693", "title": "2258. Plotting Data Pairs - ", "content": "| $\\boldsymbol{x}$ | $\\boldsymbol{y}=\\boldsymbol{x}^{2}+2$ |\n| :--- | :--- |\n| 3 | 11 |\n| 4 | 18 |"} {"id": "textbook_7694", "title": "2258. Plotting Data Pairs - ", "content": "\n"} {"id": "textbook_7695", "title": "2259. APPENDIX C - ", "content": ""} {"id": "textbook_7696", "title": "2260. Units and Conversion Factors - ", "content": "\n| Units of Length | |\n| :--- | :--- |\n| meter (m) | $=39.37$ inches (in.) |\n| $=1.094$ yards (yd) | |$]$| centimeter (cm) | $=0.01 \\mathrm{~m}$ (exact, definition) |\n| :--- | :--- |\n| millimeter (mm) | $=0.001 \\mathrm{~m}$ (exact, definition) |\n| kilometer (km) | $=1000 \\mathrm{~m}$ (exact, definition) |\n| angstrom (\u00c5) | $=10^{-8} \\mathrm{~cm}$ (exact, definition) |\n| yard (yd) | $=0.9144 \\mathrm{~m}$ |\n| inch (in.) | $=2.54 \\mathrm{~cm}$ (exact, definition) |\n| mile (US) | $=1.60934 \\mathrm{~km}$ |"} {"id": "textbook_7697", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C1\n\\$\\left.\\begin{array}{l|l} \\& Units of Volume
\nliter (L) \\& =0.001 \\mathrm{~m}^{3} (exact, definition)
\n=1000 \\mathrm{~cm}^{3} (exact, definition)
"} {"id": "textbook_7698", "title": "2260. Units and Conversion Factors - ", "content": "=1.057 (US) quarts\\end{array}\\right]\\$\\begin{tabular}{ll}"} {"id": "textbook_7699", "title": "2260. Units and Conversion Factors - ", "content": "milliliter (mL) \\& | $=0.001 \\mathrm{~L}$ (exact, definition) |\n| :--- |\n| $=1 \\mathrm{~cm}^{3}$ (exact, definition) |
\n\n\\hline microliter ( $\\mu \\mathrm{L}$ ) \\& | $=10^{-6} \\mathrm{~L}$ (exact, definition) |\n| :--- |\n| $=10^{-3} \\mathrm{~cm}^{3}$ (exact, definition) |
"} {"id": "textbook_7700", "title": "2260. Units and Conversion Factors - ", "content": "\n\\hline liquid quart (US) \\& | $=32(\\mathrm{US})$ liquid ounces (exact, definition) |\n| :--- |\n| $=0.25$ (US) gallon (exact, definition) |\n| $=0.9463 \\mathrm{~L}$ |
"} {"id": "textbook_7701", "title": "2260. Units and Conversion Factors - ", "content": "\\hline dry quart \\& $=1.1012 \\mathrm{~L}$
\n\\hline cubic foot (US) \\& $=28.316 \\mathrm{~L}$
\n\\hline\n\\end{tabular}"} {"id": "textbook_7702", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C2"} {"id": "textbook_7703", "title": "2260. Units and Conversion Factors - ", "content": "| gram (g) | $=0.001 \\mathrm{~kg}$ (exact, definition) |\n| :--- | :--- |\n| milligram (mg) | $=0.001 \\mathrm{~g}$ (exact, definition) |\n| kilogram (kg) | $=1000 \\mathrm{~g}$ (exact, definition)
$=2.205 \\mathrm{lb}$ |\n| ton (metric) | $=1000 \\mathrm{~kg}$ (exact, definition)
$=2204.62 \\mathrm{lb}$ |\n| ounce (oz) | $=28.35 \\mathrm{~g}$ |\n| pound (lb) | $=0.4535924 \\mathrm{~kg}$ |\n| ton (short) | $=2000 \\mathrm{lb}$ (exact, definition)
$=907.185 \\mathrm{~kg}$ |\n| ton (long) | $=2240 \\mathrm{lb}$ (exact, definition)
$=1.016$ metric ton |"} {"id": "textbook_7704", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C3"} {"id": "textbook_7705", "title": "2260. Units and Conversion Factors - ", "content": "Units of Energy"} {"id": "textbook_7706", "title": "2260. Units and Conversion Factors - ", "content": "| 4.184 joule (J) | $=1$ thermochemical calorie (cal) |\n| :--- | :--- |\n| 1 thermochemical calorie (cal) | $=4.184 \\times 10^{7} \\mathrm{erg}$ |\n| erg | $=10^{-7} \\mathrm{~J}($ exact, definition) |\n| electron-volt (eV) | $=1.60218 \\times 10^{-19} \\mathrm{~J}=23.061 \\mathrm{kcal} \\mathrm{mol}{ }^{-1}$ |\n| liter $\\bullet$ atmosphere | $=24.217 \\mathrm{cal}=101.325 \\mathrm{~J}$ (exact, definition) |\n| nutritional calorie (Cal) | $=1000 \\mathrm{cal}($ exact, definition) $=4184 \\mathrm{~J}$ |\n| British thermal unit (BTU) | $=1054.804 \\mathrm{~J}^{\\underline{1}}$ |"} {"id": "textbook_7707", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C4"} {"id": "textbook_7708", "title": "2260. Units and Conversion Factors - ", "content": "| Units of Pressure | |\n| :--- | :--- |\n| torr | $=1 \\mathrm{~mm} \\mathrm{Hg}$ (exact, definition) |"} {"id": "textbook_7709", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C5"} {"id": "textbook_7710", "title": "2260. Units and Conversion Factors - ", "content": "1 BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. $59^{\\circ} \\mathrm{F}\\left(15^{\\circ} \\mathrm{C}\\right)$ is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table."} {"id": "textbook_7711", "title": "2260. Units and Conversion Factors - ", "content": "| pascal (Pa) | $=\\mathrm{N} \\mathrm{m}^{-2}$ (exact, definition)
$=\\mathrm{kg} \\mathrm{m}^{-1} \\mathrm{~s}^{-2}$ (exact, definition) |\n| :--- | :--- |\n| atmosphere (atm) $=760 \\mathrm{~mm} \\mathrm{Hg}$ (exact, definition)
$=760$ torr (exact, definition)
$=101,325 \\mathrm{~N} \\mathrm{~m}$
-2
$=101,325 \\mathrm{~Pa}$ (exact, definition) | |\n| bar | $=10^{5} \\mathrm{~Pa}$ (exact, definition)
$=10^{5} \\mathrm{~kg} \\mathrm{~m}^{-1} \\mathrm{~s}^{-2}$ (exact, definition) |"} {"id": "textbook_7712", "title": "2260. Units and Conversion Factors - ", "content": "TABLE C5"} {"id": "textbook_7713", "title": "2260. Units and Conversion Factors - ", "content": "1094 C\u2022 Units and Conversion Factors\n"} {"id": "textbook_7714", "title": "2261. APPENDIX D - ", "content": ""} {"id": "textbook_7715", "title": "2262. Fundamental Physical Constants - ", "content": "\nFundamental Physical Constants"} {"id": "textbook_7716", "title": "2262. Fundamental Physical Constants - ", "content": "| Name and Symbol | Value |\n| :---: | :---: |\n| atomic mass unit (amu) | $1.6605402 \\times 10^{-27} \\mathrm{~kg}$ |\n| Avogadro's number | $6.02214076 \\times 10^{23} \\mathrm{~mol}^{-1}$ |\n| Boltzmann's constant (k) | $1.380649 \\times 10^{-23} \\mathrm{~J} \\mathrm{~K}^{-1}$ |\n| charge-to-mass ratio for electron $\\left(e / m_{e}\\right)$ | $1.75881962 \\times 10^{11} \\mathrm{C} \\mathrm{kg}^{-1}$ |\n| fundamental unit of charge (e) | $1.602176634 \\times 10^{-19} \\mathrm{C}$ |\n| electron rest mass ( $m_{e}$ ) | $9.1093897 \\times 10^{-31} \\mathrm{~kg}$ |\n| Faraday's constant (F) | $9.6485309 \\times 10^{4} \\mathrm{C} \\mathrm{mol}^{-1}$ |\n| gas constant ( $R$ ) | $8.205784 \\times 10^{-2} \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=8.314510 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$ |\n| molar volume of an ideal gas, $1 \\mathrm{~atm}, 0^{\\circ} \\mathrm{C}$ | $22.41409 \\mathrm{~L} \\mathrm{~mol}^{-1}$ |\n| molar volume of an ideal gas, $1 \\mathrm{bar}, 0^{\\circ} \\mathrm{C}$ | $22.71108 \\mathrm{~L} \\mathrm{~mol}^{-1}$ |\n| neutron rest mass ( $m_{n}$ ) | $1.6749274 \\times 10^{-27} \\mathrm{~kg}$ |\n| Planck's constant (h) | $6.62607015 \\times 10^{-34} \\mathrm{~J} \\mathrm{~S}$ |\n| proton rest mass ( $m_{p}$ ) | $1.6726231 \\times 10^{-27} \\mathrm{~kg}$ |\n| Rydberg constant (R) | $1.0973731534 \\times 10^{7} \\mathrm{~m}^{-1}=2.1798736 \\times 10^{-18} \\mathrm{~J}$ |\n| speed of light (in vacuum) (c) | $2.99792458 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$ |"} {"id": "textbook_7717", "title": "2262. Fundamental Physical Constants - ", "content": "TABLE D1\n"} {"id": "textbook_7718", "title": "2263. APPENDIX E - ", "content": ""} {"id": "textbook_7719", "title": "2264. Water Properties - ", "content": "\n| Temperature | Density ( $\\mathrm{g} / \\mathrm{mL}$ ) |\n| :---: | :---: |\n| 0 | 0.9998395 |\n| 4 | 0.9999720 (density maximum) |\n| 10 | 0.9997026 |\n| 15 | 0.9991026 |\n| 20 | 0.9982071 |\n| 22 | 0.9977735 |\n| 25 | 0.9970479 |\n| 30 | 0.9956502 |\n| 40 | 0.9922 |\n| 60 | 0.9832 |\n| 80 | 0.9718 |\n| 100 | 0.9584 |"} {"id": "textbook_7720", "title": "2264. Water Properties - ", "content": "TABLE E1"} {"id": "textbook_7721", "title": "2264. Water Properties - ", "content": "Density of Water as a Function of Temperature\n"} {"id": "textbook_7722", "title": "2264. Water Properties - ", "content": "| Temperature | Vapor Pressure (torr) | Vapor Pressure (Pa) |\n| :---: | :---: | :---: |\n| 0 | 4.6 | 613.2812 |\n| 4 | 6.1 | 813.2642 |\n| 10 | 9.2 | 1226.562 |\n| 15 | 12.8 | 1706.522 |\n| 20 | 17.5 | 2333.135 |\n| 22 | 19.8 | 2639.776 |\n| 25 | 23.8 | 3173.064 |\n| 30 | 31.8 | 4239.64 |\n| 35 | 42.2 | 5626.188 |\n| 40 | 55.3 | 7372.707 |\n| 45 | 71.9 | 9585.852 |\n| 50 | 92.5 | 12332.29 |\n| 55 | 118.0 | 15732 |"} {"id": "textbook_7723", "title": "2264. Water Properties - ", "content": "TABLE E2"} {"id": "textbook_7724", "title": "2264. Water Properties - ", "content": "| Temperature | | Vapor Pressure (torr) | | Vapor Pressure (Pa) |\n| :--- | :--- | :--- | :---: | :---: |\n| 60 | 149.4 | 19918.31 | | |\n| 65 | 187.5 | 24997.88 | | |\n| 70 | 233.7 | 31157.35 | | |\n| 75 | 289.1 | 38543.39 | | |\n| 80 | 435.1 | 47342.64 | | |\n| 85 | 525.8 | 57808.42 | | |\n| 90 | 760.0 | 70100.71 | | |\n| 95 | | 84512.82 | | |\n| 100 | | | | |\n| TABLE E2 | | | | |"} {"id": "textbook_7725", "title": "2264. Water Properties - ", "content": ""} {"id": "textbook_7726", "title": "2264. Water Properties - ", "content": "Water $\\mathrm{K}_{\\mathrm{w}}$ and $\\mathrm{pK}_{\\mathrm{W}}$ at Different Temperatures $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n| Temperature | $\\mathrm{K}_{\\mathrm{w}} \\mathbf{1 0}^{\\mathbf{- 1 4}}$ | $\\mathrm{pK}_{\\mathrm{w}}{ }^{\\mathbf{1}}$ |\n| :--- | :--- | :--- |\n| 0 | 0.112 | 14.95 |"} {"id": "textbook_7727", "title": "2264. Water Properties - ", "content": "TABLE E3"} {"id": "textbook_7728", "title": "2264. Water Properties - ", "content": "| Temperature | $\\mathrm{K}_{\\mathrm{w}} 10^{-14}$ | $\\mathrm{pK}_{\\mathrm{w}}{ }^{1}$ |\n| :---: | :---: | :---: |\n| 5 | 0.182 | 14.74 |\n| 10 | 0.288 | 14.54 |\n| 15 | 0.465 | 14.33 |\n| 20 | 0.671 | 14.17 |\n| 25 | 0.991 | 14.00 |\n| 30 | 1.432 | 13.84 |\n| 35 | 2.042 | 13.69 |\n| 40 | 2.851 | 13.55 |\n| 45 | 3.917 | 13.41 |\n| 50 | 5.297 | 13.28 |\n| 55 | 7.080 | 13.15 |\n| 60 | 9.311 | 13.03 |\n| 75 | 19.95 | 12.70 |\n| 100 | 56.23 | 12.25 |"} {"id": "textbook_7729", "title": "2264. Water Properties - ", "content": "TABLE E3\n"} {"id": "textbook_7730", "title": "2264. Water Properties - ", "content": "Standard Water Melting and Boiling Temperatures and Enthalpies of the Transitions"} {"id": "textbook_7731", "title": "2264. Water Properties - ", "content": "| | Temperature (K) | $\\Delta H(\\mathrm{~kJ} / \\mathrm{mol})$ |\n| :--- | :--- | :--- |\n| melting | 273.15 | 6.088 |\n| boiling | 373.15 | $40.656(44.016$ at 298 K$)$ |"} {"id": "textbook_7732", "title": "2264. Water Properties - ", "content": "TABLE E5"} {"id": "textbook_7733", "title": "2264. Water Properties - ", "content": "Water Cryoscopic (Freezing Point Depression) and Ebullioscopic (Boiling Point Elevation) Constants\n$\\mathrm{K}_{\\mathrm{f}}=1.86^{\\circ} \\mathrm{C} \\cdot \\mathrm{kg} \\cdot \\mathrm{mol}^{-1}($ cryoscopic constant $)$\n$\\mathrm{K}_{\\mathrm{b}}=0.51^{\\circ} \\mathrm{C} \\cdot \\mathrm{kg} \\cdot \\mathrm{mol}^{-1}($ ebullioscopic constant)"} {"id": "textbook_7734", "title": "2264. Water Properties - ", "content": "TABLE E6\n"} {"id": "textbook_7735", "title": "2264. Water Properties - ", "content": "FIGURE E1 The plot shows the extent of light absorption versus wavelength for water. Absorption is reported in reciprocal meters and corresponds to the inverse of the distance light may travel through water before its intensity is diminished by $1 / e$ ( $\\sim 37 \\%$ ).\n"} {"id": "textbook_7736", "title": "2265. APPENDIX F - ", "content": ""} {"id": "textbook_7737", "title": "2266. Composition of Commercial Acids and Bases - ", "content": "\nComposition of Commercial Acids and Bases"} {"id": "textbook_7738", "title": "2266. Composition of Commercial Acids and Bases - ", "content": "| Acid or Base $^{\\mathbf{1}}$ | Density (g/mL) | Percentage by Mass | Molarity |\n| :--- | :--- | :--- | :--- |\n| acetic acid, glacial | 1.05 | $99.5 \\%$ | 17.4 |\n| aqueous ammonia ${ }^{\\frac{3}{2}}$ | 0.90 | $28 \\%$ | 14.8 |\n| hydrochloric acid | 1.18 | $36 \\%$ | 11.6 |\n| nitric acid | 1.42 | $71 \\%$ | 16.0 |\n| perchloric acid | 1.67 | 1.70 | $50 \\%$ |\n| phosphoric acid | 1.53 | $96 \\%$ | 11.65 |\n| sodium hydroxide | 1.84 | 19.7 | |\n| sulfuric acid | | | 18.0 |"} {"id": "textbook_7739", "title": "2266. Composition of Commercial Acids and Bases - ", "content": "TABLE F1"} {"id": "textbook_7740", "title": "2266. Composition of Commercial Acids and Bases - ", "content": "[^17]"} {"id": "textbook_7741", "title": "2267. APPENDIX G - ", "content": ""} {"id": "textbook_7742", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "\nStandard Thermodynamic Properties for Selected Substances"} {"id": "textbook_7743", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| aluminum | | | |\n| $\\mathrm{Al}(\\mathrm{s})$ | 0 | 0 | 28.3 |\n| $\\mathrm{Al}(\\mathrm{g})$ | 324.4 | 285.7 | 164.54 |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})$ | -531 | -485 | -321.7 |\n| $\\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ | -1676 | -1582 | 50.92 |\n| $\\mathrm{AlF}_{3}(s)$ | -1510.4 | -1425 | 66.5 |\n| $\\mathrm{AlCl}_{3}(S)$ | -704.2 | -628.8 | 110.67 |\n| $\\mathrm{AlCl}_{3} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s)$ | -2691.57 | -2269.40 | 376.56 |\n| $\\mathrm{Al}_{2} \\mathrm{~S}_{3}(\\mathrm{~S})$ | -724.0 | -492.4 | 116.9 |\n| $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}(\\mathrm{~s})$ | -3445.06 | -3506.61 | 239.32 |\n| antimony | | | |\n| $\\mathrm{Sb}(\\mathrm{s})$ | 0 | 0 | 45.69 |\n| $\\mathrm{Sb}(\\mathrm{g})$ | 262.34 | 222.17 | 180.16 |\n| $\\mathrm{Sb}_{4} \\mathrm{O}_{6}(s)$ | -1440.55 | -1268.17 | 220.92 |\n| $\\mathrm{SbCl}_{3}(\\mathrm{~g})$ | -313.8 | -301.2 | 337.80 |\n| $\\mathrm{SbCl}_{5}(\\mathrm{~g})$ | -394.34 | -334.29 | 401.94 |\n| $\\mathrm{Sb}_{2} \\mathrm{~S}_{3}(\\mathrm{~s})$ | -174.89 | -173.64 | 182.00 |\n| $\\mathrm{SbCl}_{3}(\\mathrm{~s})$ | -382.17 | -323.72 | 184.10 |\n| SbOCl(s) | -374.0 | - | - |"} {"id": "textbook_7744", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7745", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7746", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "arsenic"} {"id": "textbook_7747", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{As}(\\mathrm{s})$ | 0 | 0 | 35.1 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{As}(\\mathrm{g})$ | 302.5 | 261.0 | 174.21 |\n| $\\mathrm{As}_{4}(\\mathrm{~g})$ | 143.9 | 92.4 | 314 |\n| $\\mathrm{As}_{4} \\mathrm{O}_{6}(s)$ | -1313.94 | -1152.52 | 214.22 |\n| $\\mathrm{As}_{2} \\mathrm{O}_{5}(\\mathrm{~s})$ | -924.87 | -782.41 | 105.44 |\n| $\\mathrm{AsCl}_{3}(\\mathrm{~g})$ | -261.50 | -248.95 | 327.06 |\n| $\\mathrm{As}_{2} \\mathrm{~S}_{3}(\\mathrm{~s})$ | -169.03 | -168.62 | 163.59 |\n| $\\mathrm{AsH}_{3}(\\mathrm{~g})$ | 66.44 | 68.93 | 222.78 |\n| $\\mathrm{H}_{3} \\mathrm{AsO}_{4}(s)$ | -906.3 | - | - |"} {"id": "textbook_7748", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "barium"} {"id": "textbook_7749", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{Ba}(s)$ | 0 | 0 | 62.5 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Ba}(g)$ | 180 | 146 | 170.24 |\n| $\\mathrm{Ba}^{2+}(a q)$ | -537.6 | -560.8 | 9.6 |\n| $\\mathrm{BaO}(s)$ | -548.0 | -520.3 | 72.1 |\n| $\\mathrm{BaCl}_{2}(s)$ | -855.0 | -806.7 | 123.7 |\n| $\\mathrm{BaSO}_{4}(s)$ | -1473.2 | -1362.3 | 132.2 |"} {"id": "textbook_7750", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "beryllium"} {"id": "textbook_7751", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\operatorname{Be}(s)$ | 0 | 0 | 9.50 |\n| :--- | :--- | :--- | :--- |\n| $\\operatorname{Be}(g)$ | 324.3 | 286.6 | 136.27 |\n| $\\operatorname{BeO}(s)$ | -609.4 | -580.1 | 13.8 |"} {"id": "textbook_7752", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "bismuth"} {"id": "textbook_7753", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\operatorname{Bi}(s)$ | 0 | 0 | 56.74 |\n| :--- | :--- | :--- | :--- |\n| $\\operatorname{Bi}(g)$ | 207.1 | 168.2 | 187.00 |"} {"id": "textbook_7754", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7755", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Bi}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ | -573.88 | -493.7 | 151.5 |\n| $\\mathrm{BiCl}_{3}(s)$ | -379.07 | -315.06 | 176.98 |\n| $\\mathrm{Bi}_{2} \\mathrm{~S}_{3}(S)$ | -143.1 | -140.6 | 200.4 |\n| boron | | | |\n| B(s) | 0 | 0 | 5.86 |\n| $\\mathrm{B}(\\mathrm{g})$ | 565.0 | 521.0 | 153.4 |\n| $\\mathrm{B}_{2} \\mathrm{O}_{3}(S)$ | -1273.5 | -1194.3 | 53.97 |\n| $\\mathrm{B}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ | 36.4 | 87.6 | 232.1 |\n| $\\mathrm{H}_{3} \\mathrm{BO}_{3}(\\mathrm{~s})$ | -1094.33 | -968.92 | 88.83 |\n| $\\mathrm{BF}_{3}(\\mathrm{~g})$ | -1136.0 | -1119.4 | 254.4 |\n| $\\mathrm{BCl}_{3}(\\mathrm{~g})$ | -403.8 | -388.7 | 290.1 |\n| $\\mathrm{B}_{3} \\mathrm{~N}_{3} \\mathrm{H}_{6}(\\mathrm{l})$ | -540.99 | -392.79 | 199.58 |\n| $\\mathrm{HBO}_{2}(s)$ | -794.25 | -723.41 | 37.66 |\n| bromine | | | |\n| $\\mathrm{Br}_{2}(1)$ | 0 | 0 | 152.23 |\n| $\\mathrm{Br}_{2}(\\mathrm{~g})$ | 30.91 | 3.142 | 245.5 |\n| $\\operatorname{Br}(\\mathrm{g})$ | 111.88 | 82.429 | 175.0 |\n| $\\mathrm{Br}^{-}(\\mathrm{aq})$ | -120.9 | -102.82 | 80.71 |\n| $\\mathrm{BrF}_{3}(\\mathrm{~g})$ | -255.60 | -229.45 | 292.42 |\n| $\\operatorname{HBr}(\\mathrm{g})$ | -36.3 | -53.43 | 198.7 |\n| cadmium | | | |\n| $\\mathrm{Cd}(s)$ | 0 | 0 | 51.76 |\n| $\\mathrm{Cd}(\\mathrm{g})$ | 112.01 | 77.41 | 167.75 |\n| $\\mathrm{Cd}^{2+}(\\mathrm{aq})$ | -75.90 | -77.61 | -73.2 |"} {"id": "textbook_7756", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7757", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CdO}(s)$ | -258.2 | -228.4 | 54.8 |\n| $\\mathrm{CdCl}_{2}(s)$ | -391.5 | -343.9 | 115.3 |\n| $\\mathrm{CdSO}_{4}(\\mathrm{~S})$ | -933.3 | -822.7 | 123.0 |\n| $\\operatorname{CdS}(s)$ | -161.9 | -156.5 | 64.9 |\n| calcium | | | |\n| $\\mathrm{Ca}(s)$ | 0 | 0 | 41.6 |\n| $\\mathrm{Ca}(\\mathrm{g})$ | 178.2 | 144.3 | 154.88 |\n| $\\mathrm{Ca}^{2+}(\\mathrm{aq})$ | -542.96 | -553.04 | -55.2 |\n| $\\mathrm{CaO}(s)$ | -634.9 | -603.3 | 38.1 |\n| $\\mathrm{Ca}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -985.2 | -897.5 | 83.4 |\n| $\\mathrm{CaSO}_{4}(\\mathrm{~s})$ | -1434.5 | -1322.0 | 106.5 |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(s)$ | -2022.63 | -1797.45 | 194.14 |\n| $\\mathrm{CaCO}_{3}(s)$ (calcite) | -1220.0 | -1081.4 | 110.0 |\n| $\\mathrm{CaSO}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}(s)$ | -1752.68 | -1555.19 | 184.10 |\n| carbon | | | |\n| C(s) (graphite) | 0 | 0 | 5.740 |\n| $\\mathrm{C}(s)$ (diamond) | 1.89 | 2.90 | 2.38 |\n| $\\mathrm{C}(\\mathrm{g})$ | 716.681 | 671.2 | 158.1 |\n| $\\mathrm{CO}(\\mathrm{g})$ | -110.52 | -137.15 | 197.7 |\n| $\\mathrm{CO}_{2}(\\mathrm{~g})$ | -393.51 | -394.36 | 213.8 |\n| $\\mathrm{CO}_{3}{ }^{2-}(\\mathrm{aq})$ | -677.1 | -527.8 | -56.9 |\n| $\\mathrm{CH}_{4}(\\mathrm{~g})$ | -74.6 | -50.5 | 186.3 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{I})$ | -239.2 | -166.6 | 126.8 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$ | -201.0 | -162.3 | 239.9 |"} {"id": "textbook_7758", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7759", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CCl}_{4}(\\mathrm{I})$ | -128.2 | -62.5 | 214.4 |\n| $\\mathrm{CCl}_{4}(\\mathrm{~g})$ | -95.7 | -58.2 | 309.7 |\n| $\\mathrm{CHCl}_{3}(\\mathrm{I})$ | -134.1 | -73.7 | 201.7 |\n| $\\mathrm{CHCl}_{3}(\\mathrm{~g})$ | -103.14 | -70.34 | 295.71 |\n| $\\mathrm{CS}_{2}($ I) | 89.70 | 65.27 | 151.34 |\n| $\\mathrm{CS}_{2}(\\mathrm{~g})$ | 116.9 | 66.8 | 238.0 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{2}(\\mathrm{~g})$ | 227.4 | 209.2 | 200.9 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ | 52.4 | 68.4 | 219.3 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ | -84.0 | -32.0 | 229.2 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{I})$ | -484.3 | -389.9 | 159.8 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{g})$ | -434.84 | -376.69 | 282.50 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I})$ | -277.6 | -174.8 | 160.7 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{g})$ | -234.8 | -167.9 | 281.6 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -691.11 | -587.06 | 95 |\n| $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$ | -103.8 | -23.4 | 270.3 |\n| $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{~g})$ | 82.927 | 129.66 | 269.2 |\n| $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{I})$ | 49.1 | 124.50 | 173.4 |\n| $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}($ I | -124.2 | -63.2 | 177.8 |\n| $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -95.4 | -65.90 | 270.2 |\n| $\\mathrm{CH}_{3} \\mathrm{Cl}(\\mathrm{g})$ | -81.9 | -60.2 | 234.6 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}(\\mathrm{I})$ | -136.52 | -59.31 | 190.79 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}(\\mathrm{g})$ | -112.17 | -60.39 | 276.00 |\n| $\\mathrm{C}_{2} \\mathrm{~N}_{2}(\\mathrm{~g})$ | 308.98 | 297.36 | 241.90 |\n| HCN( $)$ | 108.9 | 125.0 | 112.8 |"} {"id": "textbook_7760", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7761", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HCN}(\\mathrm{g})$ | 135.5 | 124.7 | 201.8 |\n| cesium | | | |\n| $\\mathrm{Cs}^{+}(\\mathrm{aq})$ | -248 | -282.0 | 133 |\n| chlorine | | | |\n| $\\mathrm{Cl}_{2}(\\mathrm{~g})$ | 0 | 0 | 223.1 |\n| $\\mathrm{Cl}(\\mathrm{g})$ | 121.3 | 105.70 | 165.2 |\n| $\\mathrm{Cl}^{-}(a q)$ | -167.2 | -131.2 | 56.5 |\n| $\\mathrm{ClF}(\\mathrm{g})$ | -54.48 | -55.94 | 217.78 |\n| $\\mathrm{ClF}_{3}(\\mathrm{~g})$ | -158.99 | -118.83 | 281.50 |\n| $\\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})$ | 80.3 | 97.9 | 266.2 |\n| $\\mathrm{Cl}_{2} \\mathrm{O}_{7}(\\mathrm{I})$ | 238.1 | - | - |\n| $\\mathrm{Cl}_{2} \\mathrm{O}_{7}(\\mathrm{~g})$ | 272.0 | - | - |\n| $\\mathrm{HCl}(\\mathrm{g})$ | -92.307 | -95.299 | 186.9 |\n| $\\mathrm{HClO}_{4}($ l | -40.58 | - | - |\n| chromium | | | |\n| $\\mathrm{Cr}(\\mathrm{s})$ | 0 | 0 | 23.77 |\n| $\\mathrm{Cr}(\\mathrm{g})$ | 396.6 | 351.8 | 174.50 |\n| $\\mathrm{CrO}_{4}{ }^{2-}(\\mathrm{aq})$ | -881.2 | -727.8 | 50.21 |\n| $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}(\\mathrm{aq})$ | -1490.3 | -1301.1 | 261.9 |\n| $\\mathrm{Cr}_{2} \\mathrm{O}_{3}(S)$ | -1139.7 | -1058.1 | 81.2 |\n| $\\mathrm{CrO}_{3}(S)$ | -589.5 | - | - |\n| $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(\\mathrm{~S})$ | -1806.7 | - | - |\n| cobalt | | | |\n| $\\mathrm{Co}(\\mathrm{S})$ | 0 | 0 | 30.0 |"} {"id": "textbook_7762", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7763", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Co}^{2+}(\\mathrm{aq})$ | -67.4 | -51.5 | -155 |\n| $\\mathrm{Co}^{3+}(\\mathrm{aq})$ | 92 | 134 | -305.0 |\n| $\\mathrm{CoO}(s)$ | -237.9 | -214.2 | 52.97 |\n| $\\mathrm{Co}_{3} \\mathrm{O}_{4}(s)$ | -910.02 | -794.98 | 114.22 |\n| $\\mathrm{Co}_{\\left(\\mathrm{NO}_{3}\\right)_{2}(s)}$ | -420.5 | - | - |"} {"id": "textbook_7764", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "copper"} {"id": "textbook_7765", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{Cu}(s)$ | 0 | 0 | 33.15 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Cu}(g)$ | 338.32 | 298.58 | 166.38 |\n| $\\mathrm{Cu}^{+}(a q)$ | 51.9 | 50.2 | -26 |\n| $\\mathrm{Cu}^{2+}(a q)$ | 64.77 | 65.49 | -99.6 |\n| $\\mathrm{CuO}(s)$ | -157.3 | -129.7 | 42.63 |\n| $\\mathrm{Cu}_{2} \\mathrm{O}(s)$ | -168.6 | -146.0 | 93.14 |\n| $\\mathrm{CuS}^{(s)}$ | -53.1 | -53.6 | 66.5 |\n| $\\mathrm{Cu}_{2} \\mathrm{~S}(s)$ | -79.5 | -86.2 | 120.9 |\n| $\\mathrm{CuSO}_{4}(s)$ | -771.36 | -662.2 | 109.2 |\n| $\\mathrm{Cu}^{\\left(\\mathrm{NO}_{3}\\right)_{2}(s)}$ | -302.9 | - | - |"} {"id": "textbook_7766", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "fluorine"} {"id": "textbook_7767", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{F}_{2}(g)$ | 0 | 0 | 202.8 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~F}(g)$ | 79.4 | 62.3 | 158.8 |\n| $\\mathrm{~F}^{-}(\\mathrm{aq})$ | -332.6 | -278.8 | -13.8 |\n| $\\mathrm{~F}_{2} \\mathrm{O}(g)$ | 24.7 | 41.9 | 247.43 |\n| $\\mathrm{HF}(g)$ | -273.3 | -275.4 | 173.8 |"} {"id": "textbook_7768", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "hydrogen"} {"id": "textbook_7769", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{H}_{2}(g)$ | 0 | 0 | 130.7 |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7770", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7771", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{H}(\\mathrm{g})$ | 217.97 | 203.26 | 114.7 |\n| $\\mathrm{H}^{+}(a q)$ | 0 | 0 | 0 |\n| $\\mathrm{OH}^{-}(a q)$ | -230.0 | -157.2 | -10.75 |\n| $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$ | -285.8 | | 69.91 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -285.83 | -237.1 | 70.0 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.82 | -228.59 | 188.8 |\n| $\\mathrm{H}_{2} \\mathrm{O}_{2}($ I) | -187.78 | -120.35 | 109.6 |\n| $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{~g})$ | -136.3 | -105.6 | 232.7 |\n| $\\mathrm{HF}(\\mathrm{g})$ | -273.3 | -275.4 | 173.8 |\n| $\\mathrm{HCl}(\\mathrm{g})$ | -92.307 | -95.299 | 186.9 |\n| $\\operatorname{HBr}(\\mathrm{g})$ | -36.3 | -53.43 | 198.7 |\n| $\\mathrm{HI}(\\mathrm{g})$ | 26.48 | 1.70 | 206.59 |\n| $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g})$ | -20.6 | -33.4 | 205.8 |\n| $\\mathrm{H}_{2} \\mathrm{Se}(\\mathrm{g})$ | 29.7 | 15.9 | 219.0 |\n| $\\mathrm{HNO}_{3}$ | -206.64 | - | - |\n| iodine | | | |\n| $\\mathrm{I}_{2}(\\mathrm{~s})$ | 0 | 0 | 116.14 |\n| $\\mathrm{I}_{2}(\\mathrm{~g})$ | 62.438 | 19.3 | 260.7 |\n| $\\mathrm{I}(\\mathrm{g})$ | 106.84 | 70.2 | 180.8 |\n| $\\mathrm{I}^{-}(a q)$ | -55.19 | -51.57 | 11.13 |\n| $\\mathrm{IF}(\\mathrm{g})$ | 95.65 | -118.49 | 236.06 |\n| $\\mathrm{ICl}(\\mathrm{g})$ | 17.78 | -5.44 | 247.44 |\n| $\\operatorname{IBr}(\\mathrm{g})$ | 40.84 | 3.72 | 258.66 |\n| $\\mathrm{IF}_{7}(\\mathrm{~g})$ | -943.91 | -818.39 | 346.44 |"} {"id": "textbook_7772", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7773", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HI}(\\mathrm{g})$ | 26.48 | 1.70 | 206.59 |\n| iron | | | |\n| $\\mathrm{Fe}(s)$ | 0 | 0 | 27.3 |\n| $\\mathrm{Fe}(\\mathrm{g})$ | 416.3 | 370.7 | 180.5 |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})$ | -89.1 | -78.90 | -137.7 |\n| $\\mathrm{Fe}^{3+}(\\mathrm{aq})$ | -48.5 | -4.7 | -315.9 |\n| $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(S)$ | -824.2 | -742.2 | 87.40 |\n| $\\mathrm{Fe}_{3} \\mathrm{O}_{4}(S)$ | -1118.4 | -1015.4 | 146.4 |\n| $\\mathrm{Fe}(\\mathrm{CO})_{5}(\\mathrm{l})$ | -774.04 | -705.42 | 338.07 |\n| $\\mathrm{Fe}(\\mathrm{CO})_{5}(\\mathrm{~g})$ | -733.87 | -697.26 | 445.18 |\n| FeCl $2(S)$ | -341.79 | -302.30 | 117.95 |\n| $\\mathrm{FeCl}_{3}(S)$ | -399.49 | -334.00 | 142.3 |\n| $\\mathrm{FeO}(s)$ | -272.0 | -255.2 | 60.75 |\n| $\\mathrm{Fe}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -569.0 | -486.5 | 88. |\n| $\\mathrm{Fe}(\\mathrm{OH})_{3}(\\mathrm{~S})$ | -823.0 | -696.5 | 106.7 |\n| $\\mathrm{FeS}(s)$ | -100.0 | -100.4 | 60.29 |\n| $\\mathrm{Fe}_{3} \\mathrm{C}(s)$ | 25.10 | 20.08 | 104.60 |\n| lead | | | |\n| $\\mathrm{Pb}(\\mathrm{s})$ | 0 | 0 | 64.81 |\n| $\\mathrm{Pb}(\\mathrm{g})$ | 195.2 | 162. | 175.4 |\n| $\\mathrm{Pb}^{2+}(\\mathrm{aq})$ | -1.7 | -24.43 | 10.5 |\n| $\\mathrm{PbO}(s)$ (yellow) | -217.32 | -187.89 | 68.70 |\n| $\\mathrm{PbO}(s)$ (red) | -218.99 | -188.93 | 66.5 |\n| $\\mathrm{Pb}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -515.9 | - | - |"} {"id": "textbook_7774", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7775", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{PbS}(s)$ | -100.4 | -98.7 | 91.2 |\n| $\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{~s})$ | -451.9 | - | - |\n| $\\mathrm{PbO}_{2}(S)$ | -277.4 | -217.3 | 68.6 |\n| $\\mathrm{PbCl}_{2}(S)$ | -359.4 | -314.1 | 136.0 |\n| lithium | | | |\n| Li(s) | 0 | 0 | 29.1 |\n| $\\mathrm{Li}(\\mathrm{g})$ | 159.3 | 126.6 | 138.8 |\n| $\\mathrm{Li}^{+}(a q)$ | -278.5 | -293.3 | 13.4 |\n| $\\mathrm{LiH}(s)$ | -90.5 | -68.3 | 20.0 |\n| Li(OH)(S) | -487.5 | -441.5 | 42.8 |\n| $\\operatorname{LiF}(s)$ | -616.0 | -587.5 | 35.7 |\n| $\\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)$ | -1216.04 | -1132.19 | 90.17 |"} {"id": "textbook_7776", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "magnesium"} {"id": "textbook_7777", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\operatorname{Mg}^{2+}(a q)$ | -466.9 | -454.8 | -138.1 |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7778", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "manganese"} {"id": "textbook_7779", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{Mn}(s)$ | 0 | 0 | 32.0 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Mn}(g)$ | 280.7 | 238.5 | 173.7 |\n| $\\mathrm{Mn}^{2+}(a q)$ | -220.8 | -228.1 | -73.6 |\n| $\\mathrm{MnO}(s)$ | -385.2 | -362.9 | 59.71 |\n| $\\mathrm{MnO}_{2}(s)$ | -520.03 | -465.1 | 53.05 |\n| $\\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)$ | -958.97 | -881.15 | 110.46 |\n| $\\mathrm{Mn}_{3} \\mathrm{O}_{4}(s)$ | -541.4 | -1283.23 | 155.64 |\n| $\\mathrm{MnO}_{4}{ }^{-}(\\mathrm{aq})$ | -653.0 | -500.7 | 59 |\n| $\\mathrm{MnO}_{4}{ }^{2-}(\\mathrm{aq})$ | | 191.2 | |"} {"id": "textbook_7780", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7781", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7782", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "\n| $\\mathrm{Hg}(1)$ | 0 | 0 | 75.9 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Hg}(\\mathrm{g})$ | 61.4 | 31.8 | 175.0 |\n| $\\mathrm{Hg}^{2+}(\\mathrm{aq})$ | | 164.8 | |\n| $\\mathrm{Hg}^{2+}(\\mathrm{aq})$ | 172.4 | 153.9 | 84.5 |\n| $\\mathrm{HgO}(\\mathrm{s})(\\mathrm{red})$ | -90.83 | -58.5 | 70.29 |\n| $\\mathrm{HgO}(s)$ (yellow) | -90.46 | -58.43 | 71.13 |\n| $\\mathrm{HgCl}_{2}(\\mathrm{~s})$ | -224.3 | -178.6 | 146.0 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s)$ | -265.4 | -210.7 | 191.6 |\n| $\\mathrm{HgS}(\\mathrm{s})(\\mathrm{red})$ | -58.16 | -50.6 | 82.4 |\n| $\\operatorname{HgS}(s)$ (black) | -53.56 | -47.70 | 88.28 |\n| $\\mathrm{HgSO}_{4}(\\mathrm{~S})$ | -707.51 | -594.13 | 0.00 |"} {"id": "textbook_7783", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "nickel"} {"id": "textbook_7784", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{Ni}^{2+}(a q)$ | -64.0 | -46.4 | -159 |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7785", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "nitrogen"} {"id": "textbook_7786", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{N}_{2}(g)$ | 0 | 0 | 191.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~N}(g)$ | 472.704 | 455.5 | 153.3 |\n| $\\mathrm{NO}(g)$ | 90.25 | 87.6 | 210.8 |\n| $\\mathrm{NO}_{2}(g)$ | 33.2 | 51.30 | 240.1 |\n| $\\mathrm{~N}_{2} \\mathrm{O}(g)$ | 81.6 | 103.7 | 220.0 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{3}(g)$ | 83.72 | 139.41 | 312.17 |\n| $\\mathrm{NO}_{3}-(a q)$ | -205.0 | -108.7 | 146.4 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{4}(g)$ | 11.1 | 99.8 | 304.4 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{5}(g)$ | 11.3 | 115.1 | 355.7 |"} {"id": "textbook_7787", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7788", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{NH}_{3}(\\mathrm{~g})$ | -45.9 | -16.5 | 192.8 |\n| $\\mathrm{NH}_{4}{ }^{+}(\\mathrm{aq})$ | -132.5 | -79.31 | 113.4 |\n| $\\mathrm{N}_{2} \\mathrm{H}_{4}(\\mathrm{l})$ | 50.63 | 149.43 | 121.21 |\n| $\\mathrm{N}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ | 95.4 | 159.4 | 238.5 |\n| $\\mathrm{NH}_{4} \\mathrm{NO}_{3}(S)$ | -365.56 | -183.87 | 151.08 |\n| $\\mathrm{NH}_{4} \\mathrm{Cl}(\\mathrm{s})$ | -314.43 | -202.87 | 94.6 |\n| $\\mathrm{NH}_{4} \\mathrm{Br}(\\mathrm{s})$ | -270.8 | -175.2 | 113.0 |\n| $\\mathrm{NH}_{4} \\mathrm{I}(\\mathrm{s})$ | -201.4 | -112.5 | 117.0 |\n| $\\mathrm{NH}_{4} \\mathrm{NO}_{2}(s)$ | -256.5 | - | - |\n| $\\mathrm{HNO}_{3}(\\mathrm{I})$ | -174.1 | -80.7 | 155.6 |\n| $\\mathrm{HNO}_{3}(\\mathrm{~g})$ | -133.9 | -73.5 | 266.9 |\n| $\\mathrm{HNO}_{3}(\\mathrm{aq})$ | -207.4 | -110.5 | 146 |"} {"id": "textbook_7789", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "oxygen"} {"id": "textbook_7790", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{O}_{2}(g)$ | 0 | 0 | 205.2 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{O}(g)$ | 249.17 | 231.7 | 161.1 |\n| $\\mathrm{O}_{3}(g)$ | 142.7 | 163.2 | 238.9 |"} {"id": "textbook_7791", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "phosphorus"} {"id": "textbook_7792", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{P}_{4}(s)$ | 0 | 0 | 164.4 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{P}_{4}(g)$ | 58.91 | 24.4 | 280.0 |\n| $\\mathrm{P}(g)$ | 314.64 | 278.25 | 163.19 |\n| $\\mathrm{PH}_{3}(g)$ | 5.4 | 13.5 | 210.2 |\n| $\\mathrm{PCl}_{3}(g)$ | -287.0 | -267.8 | 311.78 |\n| $\\mathrm{PCl}_{5}(g)$ | -374.9 | -305.0 | 364.4 |\n| $\\mathrm{P}_{4} \\mathrm{O}_{6}(s)$ | -1640.1 | - | - |"} {"id": "textbook_7793", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7794", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{P}_{4} \\mathrm{O}_{10}(s)$ | -2984.0 | -2697.0 | 228.86 |\n| $\\mathrm{PO}_{4}{ }^{3-}(\\mathrm{aq})$ | -1277 | -1019 | -222 |\n| $\\mathrm{HPO}_{3}(\\mathrm{~s})$ | -948.5 | - | - |\n| $\\mathrm{HPO}_{4}{ }^{2-}(\\mathrm{aq})$ | -1292.1 | -1089.3 | -33 |\n| $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{2-}(\\mathrm{aq})$ | -1296.3 | -1130.4 | 90.4 |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{2}(s)$ | -604.6 | - | - |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{3}(s)$ | -964.4 | - | - |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}(S)$ | -1279.0 | -1119.1 | 110.50 |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}(\\mathrm{l})$ | -1266.9 | -1124.3 | 110.5 |\n| $\\mathrm{H}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7}(S)$ | -2241.0 | - | - |\n| $\\mathrm{POCl}_{3}(\\mathrm{l})$ | -597.1 | -520.8 | 222.5 |\n| $\\mathrm{POCl}_{3}(\\mathrm{~g})$ | -558.5 | -512.9 | 325.5 |\n| potassium | | | |\n| K(s) | 0 | 0 | 64.7 |\n| K(g) | 89.0 | 60.5 | 160.3 |\n| $\\mathrm{K}^{+}(a q)$ | -252.4 | -283.3 | 102.5 |\n| KF( $s$ ) | -576.27 | -537.75 | 66.57 |\n| $\\mathrm{KCl}(\\mathrm{s})$ | -436.5 | -408.5 | 82.6 |\n| rubidium | | | |\n| $\\mathrm{Rb}^{+}(\\mathrm{aq})$ | -246 | -282.2 | 124 |\n| silicon | | | |\n| Si(s) | 0 | 0 | 18.8 |\n| $\\mathrm{Si}(\\mathrm{g})$ | 450.0 | 405.5 | 168.0 |\n| $\\mathrm{SiO}_{2}(\\mathrm{~S})$ | -910.7 | -856.3 | 41.5 |"} {"id": "textbook_7795", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7796", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{SiH}_{4}(\\mathrm{~g})$ | 34.3 | 56.9 | 204.6 |\n| $\\mathrm{H}_{2} \\mathrm{SiO}_{3}(S)$ | -1188.67 | -1092.44 | 133.89 |\n| $\\mathrm{H}_{4} \\mathrm{SiO}_{4}(\\mathrm{~S})$ | -1481.14 | -1333.02 | 192.46 |\n| $\\mathrm{SiF}_{4}(\\mathrm{~g})$ | -1615.0 | -1572.8 | 282.8 |\n| $\\mathrm{SiCl}_{4}(I)$ | -687.0 | -619.8 | 239.7 |\n| $\\mathrm{SiCl}_{4}(\\mathrm{~g})$ | -662.75 | -622.58 | 330.62 |\n| SiC(s, beta cubic) | -73.22 | -70.71 | 16.61 |\n| SiC(s, alpha hexagonal) | -71.55 | -69.04 | 16.48 |\n| silver | | | |\n| $\\mathrm{Ag}(s)$ | 0 | 0 | 42.55 |\n| $\\mathrm{Ag}(\\mathrm{g})$ | 284.9 | 246.0 | 172.89 |\n| $\\mathrm{Ag}^{+}(\\mathrm{aq})$ | 105.6 | 77.11 | 72.68 |\n| $\\mathrm{Ag}_{2} \\mathrm{O}(s)$ | -31.05 | -11.20 | 121.3 |\n| $\\mathrm{AgCl}(\\mathrm{s})$ | -127.0 | -109.8 | 96.3 |\n| $\\mathrm{Ag}_{2} \\mathrm{~S}(\\mathrm{~S})$ | -32.6 | -40.7 | 144.0 |\n| sodium | | | |\n| $\\mathrm{Na}(S)$ | 0 | 0 | 51.3 |\n| $\\mathrm{Na}(\\mathrm{g})$ | 107.5 | 77.0 | 153.7 |\n| $\\mathrm{Na}^{+}(\\mathrm{aq})$ | -240.1 | -261.9 | 59 |\n| $\\mathrm{Na}_{2} \\mathrm{O}(s)$ | -414.2 | -375.5 | 75.1 |\n| $\\mathrm{NaCl}(\\mathrm{s})$ | -411.2 | -384.1 | 72.1 |\n| strontium | | | |\n| $\\mathrm{Sr}^{2+}(\\mathrm{aq})$ | -545.8 | -557.3 | -32.6 |\n| sulfur | | | |"} {"id": "textbook_7797", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7798", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{S}_{8}(s)$ (rhombic) | 0 | 0 | 256.8 |\n| $\\mathrm{S}(\\mathrm{g})$ | 278.81 | 238.25 | 167.82 |\n| $S^{2-}(a q)$ | 41.8 | 83.7 | 22 |\n| $\\mathrm{SO}_{2}(\\mathrm{~g})$ | -296.83 | -300.1 | 248.2 |\n| $\\mathrm{SO}_{3}(\\mathrm{~g})$ | -395.72 | -371.06 | 256.76 |\n| $\\mathrm{SO}_{4}{ }^{2-}(\\mathrm{aq})$ | -909.3 | -744.5 | 20.1 |\n| $\\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}(\\mathrm{aq})$ | -648.5 | -522.5 | 67 |\n| $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g})$ | -20.6 | -33.4 | 205.8 |\n| $\\mathrm{HS}^{-}(\\mathrm{aq})$ | -17.7 | 12.6 | 61.1 |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{I})$ | -813.989 | -690.00 | 156.90 |\n| $\\mathrm{HSO}_{4}{ }^{2-}(\\mathrm{aq})$ | -885.75 | -752.87 | 126.9 |\n| $\\mathrm{H}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{7}(\\mathrm{~S})$ | -1273.6 | - | - |\n| $\\mathrm{SF}_{4}(\\mathrm{~g})$ | -728.43 | -684.84 | 291.12 |\n| $\\mathrm{SF}_{6}(\\mathrm{~g})$ | -1220.5 | -1116.5 | 291.5 |\n| $\\mathrm{SCl}_{2}($ I) | -50 | - | - |\n| $\\mathrm{SCl}_{2}(\\mathrm{~g})$ | -19.7 | - | - |\n| $\\mathrm{S}_{2} \\mathrm{Cl}_{2}(\\mathrm{I})$ | -59.4 | - | - |\n| $\\mathrm{S}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -19.50 | -29.25 | 319.45 |\n| $\\mathrm{SOCl}_{2}(\\mathrm{~g})$ | -212.55 | -198.32 | 309.66 |\n| $\\mathrm{SOCl}_{2}(\\mathrm{I})$ | -245.6 | - | - |\n| $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}($ I | -394.1 | - | - |\n| $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -354.80 | -310.45 | 311.83 |\n| tin | | | |\n| $\\mathrm{Sn}(\\mathrm{s})$ | 0 | 0 | 51.2 |"} {"id": "textbook_7799", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7800", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\operatorname{Sn}(\\mathrm{g})$ | 301.2 | 266.2 | 168.5 |\n| $\\mathrm{SnO}(\\mathrm{s})$ | -285.8 | -256.9 | 56.5 |\n| $\\mathrm{SnO}_{2}(\\mathrm{~S})$ | -577.6 | -515.8 | 49.0 |\n| $\\mathrm{SnCl}_{4}(\\mathrm{I})$ | -511.3 | -440.1 | 258.6 |\n| $\\mathrm{SnCl}_{4}(\\mathrm{~g})$ | -471.5 | -432.2 | 365.8 |\n| titanium | | | |\n| $\\mathrm{Ti}(s)$ | 0 | 0 | 30.7 |\n| $\\mathrm{Ti}(\\mathrm{g})$ | 473.0 | 428.4 | 180.3 |\n| $\\mathrm{TiO}_{2}(s)$ | -944.0 | -888.8 | 50.6 |\n| $\\mathrm{TiCl}_{4}($ ) | -804.2 | -737.2 | 252.4 |\n| $\\mathrm{TiCl}_{4}(\\mathrm{~g})$ | -763.2 | -726.3 | 353.2 |"} {"id": "textbook_7801", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "tungsten"} {"id": "textbook_7802", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{W}(s)$ | 0 | 0 | 32.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~W}(g)$ | 849.4 | 807.1 | 174.0 |\n| $\\mathrm{WO}_{3}(s)$ | -842.9 | -764.0 | 75.9 |"} {"id": "textbook_7803", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "zinc"} {"id": "textbook_7804", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\mathrm{Zn}(s)$ | 0 | 0 | 41.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Zn}(g)$ | 130.73 | 95.14 | 160.98 |\n| $\\mathrm{Zn}^{2+}(a q)$ | -153.9 | -147.1 | -112.1 |\n| $\\mathrm{ZnO}(s)$ | -350.5 | -320.5 | 43.7 |\n| $\\mathrm{ZnCl}_{2}(s)$ | -415.1 | -369.43 | 111.5 |\n| $\\mathrm{ZnS}^{(s)}$ | -206.0 | -201.3 | 57.7 |\n| $\\mathrm{ZnSO}_{4}(s)$ | -982.8 | -871.5 | 110.5 |\n| $\\mathrm{ZnCO}_{3}(s)$ | -812.78 | -731.57 | 82.42 |"} {"id": "textbook_7805", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1"} {"id": "textbook_7806", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |"} {"id": "textbook_7807", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "complexes"} {"id": "textbook_7808", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}, \\mathrm{cis}$ | -898.7 | - | - |\n| :---: | :---: | :---: | :---: |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}$, trans | -896.2 | - | - |\n| $\\mathrm{NH}_{4}\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{2}\\left(\\mathrm{NO}_{2}\\right)_{4}\\right]$ | -837.6 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{2}\\left(\\mathrm{NO}_{2}\\right)_{4}\\right]_{3}$ | -2733.0 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl}$, cis | -874.9 | - | - |\n| [ $\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2} \\mathrm{Cl}$, trans | -877.4 | - | - |\n| $\\left[\\mathrm{Co}(\\mathrm{en})_{2}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}$, cis | -689.5 | - | - |\n| [Co(en) $2_{2} \\mathrm{Cl}_{2}$ ]Cl, cis | -681.2 | - | - |\n| [Co(en) $)_{2} \\mathrm{Cl}_{2}$ ]Cl, trans | -677.4 | - | - |\n| $\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]\\left(\\mathrm{ClO}_{4}\\right)_{3}$ | -762.7 | - | - |\n| [Co(en) ${ }_{3} \\mathrm{Br}_{2}$ | -595.8 | - | - |\n| [Co(en) ${ }_{3} \\mathrm{I}_{2}$ | -475.3 | - | - |\n| [Co(en) $3_{3} \\mathrm{I}_{3}$ | -519.2 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{ClO}_{4}\\right)_{3}$ | -1034.7 | -221.1 | 615 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]\\left(\\mathrm{NO}_{3}\\right)_{2}$ | -1088.7 | -412.9 | 331 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{NO}_{3}\\right)_{3}$ | -1282.0 | -524.5 | 448 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl]Cl}{ }_{2}\\right.$ | -1017.1 | -582.5 | 366.1 |\n| $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right] \\mathrm{Cl}_{2}$ | -725.5 | - | - |\n| $\\left[\\mathrm{Ni}^{\\left.\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{2}}\\right.$ | -994.1 | - | - |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Br}_{2}$ | -923.8 | - | - |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6} \\mathrm{II}_{2}\\right.$ | -808.3 | - | - |"} {"id": "textbook_7809", "title": "2268. Standard Thermodynamic Properties for Selected Substances - ", "content": "TABLE G1\n"} {"id": "textbook_7810", "title": "2269. APPENDIX H - ", "content": ""} {"id": "textbook_7811", "title": "2270. Ionization Constants of Weak Acids - ", "content": "\nIonization Constants of Weak Acids"} {"id": "textbook_7812", "title": "2270. Ionization Constants of Weak Acids - ", "content": "| Acid | Formula | $K_{a}$ at $25^{\\circ} \\mathrm{C}$ | Lewis Structure |\n| :---: | :---: | :---: | :---: |\n| acetic | $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathbf{H}$ | $1.8 \\times 10^{-5}$ |