7.1: Balancing Chemical Equations
18
Harper College Chemistry Department
Summary
By the end of this section, you will be able to:
- Derive chemical equations from narrative descriptions of chemical reactions.
- Write and balance chemical equations in molecular, total ionic, and net ionic formats.
A preceding chapter 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 combine 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 (CH 4 ) and two diatomic oxygen molecules (O 2 ) to produce one carbon dioxide molecule (CO 2 ) and two water molecules (H 2 O). The chemical equation representing this process is provided in the upper half of Figure 1 , with space-filling molecular models shown in the lower half of the figure.
This example illustrates the fundamental aspects of any chemical equation:
- The substances undergoing reaction are called reactants and their formulas are placed on the left side of the equation.
- The substances generated by the reaction are called products and their formulas are placed on the right side of the equation.
- Plus signs (+) separate individual reactant and product formulas and an arrow (⟶) separates the reactant and product (left and right) sides of the equation.
- 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.
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 2 ). 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:
- One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
- One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
- One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules. The mole concept will be discussed in a different chapter.
Balancing Equations
The chemical equation described in the previous section 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, CO 2 and H 2 O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is
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:
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| C | 1 × 1 = 1 | 1 × 1 = 1 | 1 = 1, yes |
| H | 4 × 1 = 4 | 2 × 2 = 4 | 4 = 4, yes |
| O | 2 × 2 = 4 | (1 × 2) + (2 × 1) = 4 | 4 = 4, yes |
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:
Comparing the number of H and O atoms on either side of this equation confirms its imbalance:
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| H | 1 × 2 = 2 | 1 × 2 = 2 | 2 = 2, yes |
| O | 1 × 1 = 1 | 1 × 2 = 2 | 1 ≠ 2, no |
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 H 2 O to H 2 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 H 2 O to 2.
$\ce{2 H2O -> H2 + O2} \hspace{4em} \textrm{(unbalanced)}$
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| H | 2 × 2 = 4 | 1 × 2 = 2 | 4 ≠ 2, no |
| O | 2 × 1 = 2 | 1 × 2 = 2 | 2 = 2, yes |
The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H 2 product to 2.
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| H | 2 × 2 = 4 | 2 × 2 = 4 | 4 = 4, yes |
| O | 2 × 1 = 2 | 1 × 2 = 2 | 2 = 2, yes |
These coefficients yield equal numbers of both H and O atoms on the reactant and product sides and the balanced equation is, therefore:
Balancing Chemical Equations
Write a balanced equation for the reaction of molecular nitrogen (N 2 ) and oxygen (O 2 ) to form dinitrogen pentoxide.
Solution:
First, write the unbalanced equation.
Next, count the number of each type of atom present in the unbalanced equation.
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| N | 1 × 2 = 2 | 1 × 2 = 2 | 2 = 2, yes |
| O | 1 × 2 = 2 | 1 × 5 = 5 | 2 ≠ 5, no |
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 O 2 and N 2 O 5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| N | 1 × 2 = 2 | 2 × 2 = 4 | 2 ≠ 4, no |
| O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |
The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N 2 to 2.
$\ce{2 N2 + 5 O2 -> 2 N2O5} \hspace{4em} \textrm{(unbalanced)}$
| Element | Reactants | Products | Balanced? |
|---|---|---|---|
| N | 2 × 2 = 4 | 2 × 2 = 4 | 4 = 4, yes |
| O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |
The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.
Check Your Learning
Write 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.)
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 (C 2 H 6 ) with oxygen to yield H 2 O and CO 2 , represented by the unbalanced equation:
$\ce{C2H6 + O2 -> H2O + CO2}\hspace{4em}\textrm{unbalanced}$
Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:
$\ce{C2H6 + O2 -> 3H2O + 2CO2}\hspace{4em}\textrm{unbalanced}$
This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O 2 reactant to yield an odd number, so a fractional coefficient, $\frac{7}{2}$ , is used instead to yield a provisional balanced equation:
$\ce{C2H6 + \frac{7}{2}O2 -> 3H2O + 2CO2}\hspace{4em}$
A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:
$\ce{2C2H6 + 7O2 -> 6H2O + 4CO2}\hspace{4em}$
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,
$\ce{3N2 + 9H2 -> 6NH3}$
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:
$\ce{N2 + 3H2 -> 2NH3}$
Use this interactive tutorial for additional practice balancing equations.
Additional Information in Chemical Equations
The 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 ). These notations are illustrated in the example equation here:
$\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}$
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).
Key Concepts and Summary
Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products.
End of Chapter Exercises
1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
2. Balance the following equations:
(a) $\ce{PCl5(s) + H2O(l) -> POCl3(l) + HCl(aq)}$
(b) $\ce{Cu(s) + HNO3(aq) -> Cu(NO3)2(aq) + H2O(l) + NO(g)}$
(c) $\ce{H2(g) + I2(s) -> HI(s)}$
(d) $\ce{Fe(s) + O2(g) -> Fe2O3(s)}$
(e) $\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g)}$
(f) $\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + H2O(g)}$
(g) $\ce{P4(s) + Cl2(g) -> PCl3(l)}$
(h) $\ce{PtCl4(s) -> Pt(s) + Cl2(g)}$
3. Balance the following equations:
(a) Ag(s) +H 2 S(g) + O 2 (g) → Ag 2 S(s) + H 2 O(l)
(b) P 4 (s) + O 2 (g) → P 4 O 10 (s)
(c) Pb(s) + H 2 O(l) + O 2 (g) → Pb(OH) 2 (s)
(d) Fe(s) + H 2 O(l) → Fe 3 O 4 + H 2 (g)
(e) Sc 2 O 3 + SO 3 → Sc 2 (SO 4 ) 3
(f) Ca 3 (PO 4 ) 2 (aq) + H 3 PO 4 (aq) → Ca(H 2 PO 4 ) 2 (aq)
(g) Al(s) + H 2 SO 4 (aq) → Al 2 SO 4 (aq) + H 2 (g)
(h) TiCl 4 (s) + H 2 O(l) → TiO 2 (s) + HCl
4. Write a balanced molecular equation describing each of the following chemical reactions.
(a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
(b) Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
(c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
(d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
5. Write a balanced equation describing each of the following chemical reactions.
(a) Solid potassium chlorate, KClO 3 , decomposes to form solid potassium chloride and diatomic oxygen gas.
(b) Solid aluminum metal reacts with solid diatomic iodine to form solid Al 2 I 6 .
(c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
(d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.
6. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.
(a) Write the formulas of barium nitrate and potassium chlorate.
(b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
(c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
7. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.
(a) The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
(b) The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
(c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
(d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
(e) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.
Glossary
- balanced equation
- chemical equation with equal numbers of atoms for each element in the reactant and product
- chemical equation
- symbolic representation of a chemical reaction
- coefficient
- number placed in front of symbols or formulas in a chemical equation to indicate their relative amount
- molecular equation
- chemical equation in which all reactants and products are represented as neutral substances
- product
- substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation
- reactant
- substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation


6.9: Hypothetical Case Study- Foodborne Illness Outbreak
Attack rates are also used to help identify the source of an outbreak from a specific source, such as in a foodborne illness investigation. Let’s pretend you are a public health officer investigating an outbreak of foodborne illness associated with a single meal (point source) at a U.S. Army basic training location. Several recruits experienced vomiting and diarrhea within a few hours of breakfast. Although samples have been sent to the lab for testing, you won’t get the results for 3-5 days. In order to prevent further illness among the recruits, you can use epidemiological methods (like calculating attack rates) to determine which food item was likely the source of contamination. After interviewing all of the recruits, you come up with the following data on two suspected food items:
Breakfast food items: scrambled eggs and sausage.
Number of recruits who ate scrambled eggs and became ill: 186/400 attack rate = 46.5%
Number of recruits who ate sausage and became ill: 140/400 attack rate = 35%
Number of recruits who didn’t eat scrambled eggs and became ill: 4/400 attack rate = 1%
Number of recruits who didn’t eat sausage and became ill: 54/400 attack rate = 13.5%
Next, you’ll calculate a new statistic: the risk ratio. This is the mathematical probability of becoming ill after eating each food vs. becoming ill after not eating each food. Your results are as follows:
Scrambled eggs risk ratio: 46.5/1 = 46.5
Sausage risk ratio: 35/13.5 =2.59
At this point you could draw a reasonable conclusion that the most likely source of the food poisoning was the scrambled eggs, since the risk ratio of becoming ill after eating scrambled eggs is 46.5, compared to the risk ratio of eating the sausage which was only 2.59. You would probably still do further investigation by taking a sample of the scrambled eggs and testing it for a microorganism, interviewing the chefs and cooking staff on their preparation methods, and finding out the source of the eggs used and perhaps testing that source as well. But you don’t even need to know the pathogenic cause of the outbreak to advise the kitchen staff to discontinue serving the scrambled eggs until the exact cause is identified and removed.


1.5: Raman Spectroscopy
Raman spectroscopy is a chemical instrumentation technique that exploits molecular vibrations.
-
- 1.5.1: Raman- Application
- If one can extract all of the vibrational information corresponds a molecule, its molecular structure can then be determined. In the field of spectroscopy, two main techniques are applied in order to detect molecular vibrational motions: Infrared spectroscopy (IR) and Raman spectroscopy. Raman Spectroscopy has its unique properties which have been used very commonly and widely.
-
- 1.5.2: Raman- Theory
- The phenomenon of Raman scattering of light was first postulated by Smekai in 1923 and first observed experimentally in 1928 by Raman and Krishnan. Raman scattering is most easily seen as the change in frequency for a small percentage of the intensity in a monochromatic beam as the result of coupling between the incident radiation and vibrational energy levels of molecules. A vibrational mode will be Raman active only when it changes the polariazbility of the moleculeat.


12.2: Normal Microbiota of the Body
-
- Last updated
- Save as PDF
Learning Objectives
- Compare and contrast the microbiomes of various body sites
- Identify features of the body sites that might affect with microbes reside in each space
PART 1
Michael, a 10-year-old boy in generally good health, went to a birthday party on Sunday with his family. He ate many different foods but was the only one in the family to eat the undercooked hot dogs served by the hosts. Monday morning, he woke up feeling achy and nauseous, and he was running a fever of 38 °C (100.4 °F). His parents, assuming Michael had caught the flu, made him stay home from school and limited his activities. But after 4 days, Michael began to experience severe headaches, and his fever spiked to 40 °C (104 °F). Growing worried, his parents finally decide to take Michael to a nearby clinic.
Exercise \(\PageIndex{1}\)
- What signs and symptoms is Michael experiencing?
- What do these signs and symptoms tell us about the stage of Michael’s disease?
Normal Flora
Microbes are ubiquitous is a phrase that has been repeated often, but many people do not realize how close to home it is.  Many are living with us side-by-side in a mutualistic relationship that benefits everyone. In this section we explore those that live in harmony with us, at least temporarily.
Normal Microbiota of the Skin
The skin is home to a wide variety of normal microbiota, consisting of commensal organisms that derive nutrition from skin cells and secretions such as sweat and sebum. The normal microbiota of skin tends to inhibit transient-microbe colonization by producing antimicrobial substances and outcompeting other microbes that land on the surface of the skin. This helps to protect the skin from pathogenic infection.
The skin’s properties differ from one region of the body to another, as does the composition of the skin’s microbiota. The availability of nutrients and moisture partly dictates which microorganisms will thrive in a particular region of the skin. Relatively moist skin, such as that of the nares (nostrils) and underarms, has a much different microbiota than the dryer skin on the arms, legs, hands, and top of the feet. Some areas of the skin have higher densities of sebaceous glands. These sebum-rich areas, which include the back, the folds at the side of the nose, and the back of the neck, harbor distinct microbial communities that are less diverse than those found on other parts of the body.
Different types of bacteria dominate the dry, moist, and sebum-rich regions of the skin. The most abundant microbes typically found in the dry and sebaceous regions are Betaproteobacteria and Propionibacteria, respectively. In the moist regions, Corynebacterium and Staphylococcus are most commonly found (Figure \(\PageIndex{1}\)). Viruses and fungi are also found on the skin, with Malassezia being the most common type of fungus found as part of the normal microbiota. The role and populations of viruses in the microbiota, known as viromes, are still not well understood, and there are limitations to the techniques used to identify them. However, Circoviridae, Papillomaviridae, and Polyomaviridaeappear to be the most common residents in the healthy skin virome. 1 2 3
Exercise \(\PageIndex{2}\)
What are the four most common bacteria that are part of the normal skin microbiota?
Microbiota of the Eye
The surfaces of the eyeball and inner eyelid are mucous membranes called conjunctiva. The normal conjunctival microbiota has not been well characterized, but does exist. One small study (part of the Ocular Microbiome project) found twelve genera that were consistently present in the conjunctiva. 4 These microbes are thought to help defend the membranes against pathogens. However, it is still unclear which microbes may be transient and which may form a stable microbiota. 5
Use of contact lenses can cause changes in the normal microbiota of the conjunctiva by introducing another surface into the natural anatomy of the eye. Research is currently underway to better understand how contact lenses may impact the normal microbiota and contribute to eye disease. The watery material inside of the eyeball is called the vitreous humor. Unlike the conjunctiva, it is protected from contact with the environment and is almost always sterile, with no normal microbiota (Figure \(\PageIndex{2}\)).
Mouth and GI tract
Microbes such as bacteria and archaea are abundant in the mouth and coat all of the surfaces of the oral cavity. The most well known genus is Streptococcus , but there are many others that have not been well characterized. Even within the mouth, different structures, such as the teeth or cheeks, host unique communities of both aerobic and anaerobic microbes. The number of microbes goes down and is essentially zero by the stomach. But after the stomach, microorganisms present in the small intestine can include lactobacilli, diptherioids and the fungus Candida . On the other hand, the large intestine (colon) contains a diverse and abundant microbiota that is important for normal function. These microbes include Bacteriodetes (especially the genera Bacteroides and Prevotella ) and Firmicutes (especially members of the genus Clostridium ). Methanogenic archaea and some fungi are also present, among many other species of bacteria. These microbes all aid in digestion and contribute to the production of feces, the waste excreted from the digestive tract, and flatus, the gas produced from microbial fermentation of undigested food. They can also produce valuable nutrients. For example, lactic acid bacteria such as bifidobacteria can synthesize vitamins, such as vitamin B12, folate, and riboflavin, that humans cannot synthesize themselves. E. coli found in the intestine can also break down food and help the body produce vitamin K, which is important for blood coagulation.
Normal Microbiota of the Respiratory System
The upper respiratory tract contains an abundant and diverse microbiota. The nasal passages and sinuses are primarily colonized by members of the Firmicutes, Actinobacteria, and Proteobacteria. The most common bacteria identified include Staphylococcus epidermidis , viridans group streptococci (VGS), Corynebacterium spp. (diphtheroids), Propionibacterium spp., and Haemophilus spp. The oropharynx includes many of the same isolates as the nose and sinuses, with the addition of variable numbers of bacteria like species of Prevotella , Fusobacterium , Moraxella , and Eikenella , as well as some Candida fungal isolates. In addition, many healthy humans asymptomatically carry potential pathogens in the upper respiratory tract. As much as 20% of the population carry Staphylococcus aureus in their nostrils. 6 The pharynx, too, can be colonized with pathogenic strains of Streptococcus , Haemophilus , and Neisseria .
The lower respiratory tract, by contrast, is scantily populated with microbes. Of the organisms identified in the lower respiratory tract, species of Pseudomonas , Streptococcus , Prevotella , Fusobacterium , and Veillonella are the most common. It is not clear at this time if these small populations of bacteria constitute a normal microbiota or if they are transients.
Normal Microbiota of the Urogenital System
Below the bladder, the normal microbiota of the male urogenital system is found primarily within the distal urethra and includes bacterial species that are commonly associated with the skin microbiota- think Staphylococcus . In women, the normal microbiota is found within the distal one third of the urethra and the vagina. The normal microbiota of the vagina becomes established shortly after birth and is a complex and dynamic population of bacteria that fluctuates in response to environmental changes. Members of the vaginal microbiota play an important role in the nonspecific defense against vaginal infections and sexually transmitted infections by occupying cellular binding sites and competing for nutrients. In addition, the production of lactic acid by members of the microbiota provides an acidic environment within the vagina that also serves as a defense against infections. For the majority of women, the lactic-acid–producing bacteria in the vagina are dominated by a variety of species of Lactobacillus . For women who lack sufficient lactobacilli in their vagina, lactic acid production comes primarily from other species of bacteria such as Leptotrichia spp., Megasphaera spp., and Atopobium vaginae . Lactobacillus spp. use glycogen from vaginal epithelial cells for metabolism and production of lactic acid. This process is tightly regulated by the hormone estrogen. Increased levels of estrogen correlate with increased levels of vaginal glycogen, increased production of lactic acid, and a lower vaginal pH. Therefore, decreases in estrogen during the menstrual cycle and with menopause are associated with decreased levels of vaginal glycogen and lactic acid, and a higher pH. In addition to producing lactic acid, Lactobacillus spp. also contribute to the defenses against infectious disease through their production of hydrogen peroxide and bacteriocins (antibacterial peptides).
Key Concepts and Summary
- Normal flora lives in a mutualistic relationship with humans. Humans benefit from microbes being there and microbes benefit too.
- What genera of microbes are represented in each area are highly diverse and dependent on defenses and resources are available in each area.
Footnotes
- 1 Belkaid, Y., and J.A. Segre. “Dialogue Between Skin Microbiota and Immunity,” Science 346 (2014) 6212:954–959.
- 2 Foulongne, Vincent, et al. “Human Skin Microbiota: High Diversity of DNA Viruses Identified on the Human Skin by High Throughput Sequencing.” PLoS ONE (2012) 7(6): e38499. doi: 10.1371/journal.pone.0038499.
- 3 Robinson, C.M., and J.K. Pfeiffer. “Viruses and the Microbiota.” Annual Review of Virology (2014) 1:55–59. doi: 10.1146/annurev-virology-031413-085550.
- 4 Abelson, M.B., Lane, K., and Slocum, C.. “The Secrets of Ocular Microbiomes.” Review of Ophthalmology June 8, 2015. www.reviewofophthalmology.com...isease/c/55178. Accessed Sept 14, 2016.
- 5 Shaikh-Lesko, R. “Visualizing the Ocular Microbiome.” The Scientist May 12, 2014. http://www.the-scientist.com/?articl...lar-Microbiome . Accessed Sept 14, 2016.
- 6 J. Kluytmans et al. “Nasal Carriage of Staphylococcus aureus : Epidemiology, Underlying Mechanisms, and Associated Risks.” Clinical Microbiology Reviews 10 no. 3 (1997):505–520.
Contributor
-
Nina Parker, (Shenandoah University), Mark Schneegurt (Wichita State University), Anh-Hue Thi Tu (Georgia Southwestern State University), Philip Lister (Central New Mexico Community College), and Brian M. Forster (Saint Joseph’s University) with many contributing authors. Original content via Openstax (CC BY 4.0; Access for free at https://openstax.org/books/microbiology/pages/1-introduction )


1.9: I Glossary
Air embolism: The presence of air in the vascular system that occurs when air is introduced into the venous system and travels to the right ventricle and/or pulmonary circulation.
Arterial blood sampling: Blood is obtained via venipuncture into an artery.
Basal infusion: Continuous rate of medication administration, regardless of demand attempts.
Blown vein: A ruptured vein that is leaking blood.
Breakthrough bolus dose: A dose of opioid or non-opioid medication administered by the nurse for breakthrough pain when a patient is receiving patient-controlled analgesia.

Catheter embolism: An embolism that occurs when a small part of the cannula breaks off and flows into the vascular system.
Catheter-related bloodstream infection (CR-BSI): An infection caused by microorganisms introduced into the bloodstream through the puncture site, the hub, or contaminated IV tubing or IV solution, leading to bacteremia or sepsis. A CR-BSI is a hospital-acquired preventable infection and considered an adverse event.
Central venous access device (CVAD): A type of vascular access that involves the insertion of a tube into a vein in the neck, chest, or groin and threaded into a central vein (most commonly the internal jugular, subclavian, or femoral) and advanced until the terminal lumen resides within the inferior vena cava, superior vena cava, or right atrium.
Demand dose: Medication dose given on activation of demand (pressing the demand button).
Epidural: Administration of analgesics and anesthetics into the spinal fluid via an epidural catheter for severe pain management associated with surgical procedures or during labor and delivery.
Extravasation: A condition that occurs when vesicant solution (medication) is administered and inadvertently leaks into surrounding tissue, causing damage to surrounding tissue. It is characterized by the same signs and symptoms as infiltration but also includes burning, stinging, redness, blistering, or necrosis of the tissue.
Hypertonic solutions: IV fluids with a higher concentration of dissolved particles than blood.
Hypotonic solutions: IV fluids with a lower concentration of dissolved solutes than blood.
Infiltration: A condition that occurs when a nonvesicant solution (IV solution) is inadvertently administered into surrounding tissue. Signs and symptoms include pain, swelling, redness, the skin surrounding the insertion site is cool to touch, there is a change in the quality or flow of IV, the skin is tight around the IV site, IV fluid is leaking from IV site, or there are frequent alarms on the IV pump.
Intravenous therapy (IV therapy): Administration of a substance directly into a person’s vein.
Isotonic solutions: IV fluids with a similar concentration of dissolved particles as blood.
Loading dose: Ordered amount of medication administered at the time of PCA initiation.
Lockout interval: Time period in which no follow-up demand dose may be administered (even if demand button is activated).
Lockout maximum: The maximum dose of medication that can be administered within a certain period, commonly prescribed to 1 hour limit.
Midline peripheral catheters: Larger peripheral catheters (i.e., 16-18 gauge) that allow for rapid infusions and blood sampling and can be used for longer duration that traditional peripheral catheters. They are ultrasound-guided and can be inserted by RNs with additional training or other trained professionals.
Patient-controlled analgesia (PCA): A method of pain management that allows hospitalized patients with severe pain to safely self-administer opioid medications using a programmed pump according to their level of discomfort.
Peripheral IV (PIV): A short intravenous catheter inserted by percutaneous venipuncture into a peripheral vein and held in place with a sterile transparent dressing.
Phlebitis: The inflammation of the vein’s inner lining, the tunica intima. Clinical indications are localized redness, pain, heat, and swelling that can track up the vein leading to a palpable venous cord.
Pulmonary edema: A condition caused by excess fluid accumulation in the lungs due to excessive fluid in the circulatory system. It is characterized by decreased oxygen saturation; increased respiratory rate; fine or coarse crackles in the lung bases; restlessness; breathlessness; dyspnea; and coughing up pink, frothy sputum. Pulmonary edema requires prompt medical attention and treatment.
Saline locks: A short extension set that allows intermittent IV access without ongoing infusion.
Venipuncture: The process of introducing a needle into a patient’s vein to collect a blood sample or insert an IV catheter.


13.3: Equilibrium Constants
-
- Last updated
- Save as PDF
Learning Objectives
By the end of this section, you will be able to:
- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
- Relate the magnitude of an equilibrium constant to properties of the chemical system
The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( Q ) . For a reversible reaction described by
\[m A +n B +\rightleftharpoons x C +y D \nonumber \]
the reaction quotient is derived directly from the stoichiometry of the balanced equation as
\[Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber \]
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:
\[Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber \]
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.  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.
Example \(\PageIndex{1}\): Writing Reaction Quotient Expressions
Write the concentration-based reaction quotient expression for each of the following reactions:
- \(\ce{3 O2(g) \rightleftharpoons 2 O3(g)}\)
- \(\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}\)
- \(\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}\)
Solution
- \[Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber \]
- \[Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber \]
- \[Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber \]
Exercise \(\PageIndex{1}\)
Write the concentration-based reaction quotient expression for each of the following reactions:
- \(\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}\)
- \(\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}\)
- \(\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}\)
- Answer
-
- \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber \]
- \[Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber \]
- \[Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber \]
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:
\[\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber \]
Two different experimental scenarios are depicted in Figure \(\PageIndex{1}\), one in which this reaction is initiated with a mixture of reactants only, SO 2 and O 2 , and another that begins with only product, SO 3 . For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:
\[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber \]
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 .
If the reaction begins with only product present, the value of Q c is initially undefined (immeasurably large, or infinite):
\[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber \]
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.
The constant value of Q exhibited by a system at equilibrium is called the equilibrium con stant, \(K\):
\[K \equiv Q \text { at equilibrium } \nonumber \]
Comparison of the data plots in Figure \(\PageIndex{1}\) 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 .
Definition: Law of Mass Action
At a given temperature, the reaction quotient for a system at equilibrium is constant.
Example \(\PageIndex{2}\): Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
\[\ce{2 NO2(g) <=> N2O4(g)} \nonumber \]
When 0.10 mol NO 2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO 2 ] = 0.016 M and [N 2 O 4 ] = 0.042 M .
- What is the value of the reaction quotient before any reaction occurs?
- What is the value of the equilibrium constant for the reaction?
Solution
As 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.
(a) Before any product is formed
\[\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber \]
\[[\ce{N2O4}] = 0~\text{M} \nonumber \]
Thus
\[Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber \]
(b) At equilibrium,
\[K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber \]
The equilibrium constant is \(1.6 \times 10^{2}\).
Exercise \(\PageIndex{2}\)
For the reaction
\[\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber \]
the equilibrium concentrations are [SO 2 ] = 0.90 M , [O 2 ] = 0.35 M , and [SO 3 ] = 1.1 M . What is the value of the equilibrium constant, K c ?
- Answer
-
K c = 4.3
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.
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 an 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:
\[\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber \]
The bar charts in Figure \(\PageIndex{2}\) 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.
Example \(\PageIndex{3}\): Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
\[\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber \]
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
| Reactants/Products | Experiment 1 | Experiment 2 | Experiment 3 |
|---|---|---|---|
| [CO] i | 0.020 M | 0.011 M | 0.0094 M |
| [H 2 O] i | 0.020 M | 0.0011 M | 0.0025 M |
| [CO 2 ] i | 0.0040 M | 0.037 M | 0.0015 M |
| [H 2 ] i | 0.0040 M | 0.046 M | 0.0076 M |
Solution
Experiment 1:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber \]
Q c < K c (0.040 < 0.64)
The reaction will proceed in the forward direction.
Experiment 2:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber \]
Q c > K c (140 > 0.64)
The reaction will proceed in the reverse direction.
Experiment 3:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber \]
Q c < K c (0.48 < 0.64)
The reaction will proceed in the forward direction.
Exercise \(\PageIndex{3}\)
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
- A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl 2 (g), and 0.500 mol of NOCl: \[\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber \]
- A 5.0-L flask containing 17 g of NH 3 , 14 g of N 2 , and 12 g of H 2 : \[\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber \]
- A 2.00-L flask containing 230 g of SO 3 (g): \[\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber \]
- Answer
-
(a) Q c = 6.45 \times 10^{3} DELMAR, forward. (b) Q c = 0.23, reverse. (c) Q c = 0, forward.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions . These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:
\[\begin{aligned}
\ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \\[4pt]
\ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \\[4pt]
\ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \\[4pt]
\ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]}
\end{aligned} \nonumber \]
These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included . Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K a = K eq [H 2 O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K a = K eq (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K a = K eq (1), or K a = K eq .
The equilibria below all involve gas-phase solutions:
\[\begin{aligned}
\ce{C2H6(g) & \rightleftharpoons C2H4(g) + H2(g)} & K_c & =\frac{\left[ C_2 H_4\right]\left[ H_2\right]}{\left[ C_2 H_6\right]} \\[4pt]
\ce{3 O2(g) & \rightleftharpoons 2 O3(g)} & K_c & =\frac{\left[ O_3\right]^2}{\left[ O_2\right]^3} \\[4pt]
\ce{N2(g) + 3 H2(g) & \rightleftharpoons 2 NH3(g)} & K_c & =\frac{\left[ NH_3\right]^2}{\left[ N_2\right]\left[ H_2\right]^3} \\[4pt]
\ce{C3H8(g) + 5 O2(g) & \rightleftharpoons 3 CO2(g) + 4 H2O(g)} & K_c & =\frac{\left[ CO_2\right]^3\left[ H_2 O \right]^4}{\left[ C_3 H_8\right]\left[ O_2\right]^5}
\end{aligned} \nonumber \]
For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ( K c ) or partial pressures ( K p ) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:
\[\begin{align*}
P V &=n R T \\[4pt]
P & =\left(\frac{n}{V}\right) R T \\[4pt]
& =M R T
\end{align*} \nonumber \]
where \(P\) is partial pressure, \(V\) is volume, \(n\) is molar amount, \(R\) is the gas constant, \(T\) is temperature, and \(M\) is molar concentration.
For the gas-phase reaction
\[a A +b B \rightleftharpoons c C +d D \nonumber \]
\[\begin{align*}
K_P &=\frac{\left(P_C\right)^c\left(P_D\right)^d}{\left(P_A\right)^a\left(P_B\right)^b} \\[4pt]
&= \dfrac{([ C ] \times R T)^c([ D ] \times R T)^d}{([ A ] \times R T)^a([ B ] \times R T)^b} \\[4pt]
&= \dfrac{[ C ]^c[ D ]^d}{[ A ]^a[ B ]^b} \times \frac{(R T)^{c+d}}{(R T)^{a+b}} \\[4pt]
&= K_c(R T)^{(c+d)-(a+b)} \\[4pt]
&= K_c(R T)^{\Delta n}
\end{align*} \nonumber \]
And so, the relationship between K c and K P is
\[K_P=K_c(R T)^{\Delta n} \nonumber \]
where \(Δn\) is the difference in the molar amounts of product and reactant gases, in this case:
\[\Delta n=(c+d)-(a+b) \nonumber \]
Example \(\PageIndex{4}\): Calculation of K P
Write the equations relating K c to K P for each of the following reactions:
- \(\ce{C2H6(g) <=> C2H4(g) + H2(g)}\)
- \(\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\)
- \(\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\)
- K c is equal to 0.28 for the following reaction at 900 °C: \[\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g)} \nonumber \] What is K P at this temperature?
Solution
- \(Δn = (2) − (1) = 1\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^1 = K_c (RT) \nonumber \]
- \(Δn = (2) − (2) = 0\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^0 = K_c \nonumber \]
- \(Δn = (2) − (1 + 3) = −2\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^{−2} = \dfrac{K_c}{(R T)^2} \nonumber \]
- \[K_P = K_c (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 \times 10^{−5} \nonumber \]
Exercise \(\PageIndex{4}\)
Write the equations relating K c to K P for each of the following reactions:
- \(\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}\)
- \(\ce{N2O4(g) <=> 2 NO2(g)}\)
- \(\ce{C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O (g)}\)
- At 227 °C, the following reaction has K c = 0.0952: \[\ce{CH3OH(g) <=> CO(g) + 2 H2(g)} \nonumber \] What would be the value of K P at this temperature?
- Answer
-
(a) K P = K c ( RT ) −1 ; (b) K P = K c ( RT ); (c) K P = K c ( RT ); (d) 160 or 1.6 \times 10^{2} DELMAR
Heterogeneous Equilibria
A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:
\[\begin{align*}
\ce{PbCl2(s) & \rightleftharpoons Pb^{2+}(aq) + 2 Cl^{-}(aq)} & K_c & =\left[ \ce{Pb^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \\[4pt]
\ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_c & =\frac{1}{\left[ \ce{CO_2}\right]} \\[4pt]
\ce{C(s) + 2S(g) & \rightleftharpoons CS2(g)} & K_c & =\frac{\left[ CS_2\right]}{\left[ \ce{S^2} \right.} \\[4pt]
\ce{Br2(l) & \rightleftharpoons Br2(g)} & K_c & =\left[ \ce{Br_2(g)} \right]
\end{align*} \nonumber \]
Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.
Two of the above examples include terms for gaseous species only in their equilibrium constants, and so K p expressions may also be written:
\[\begin{align*}
\ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_P & =\frac{1}{P_{ \ce{CO2}}} \\[4pt]
\ce{C(s) + 2 S(g) & \rightleftharpoons CS2(g)} & K_P & =\frac{P_{ \ce{CS2}}}{\left(P_{ \ce{S} }\right)^2}
\end{align*} \nonumber \]
Coupled Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]}
\end{array} \nonumber \]
\[K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber \]
2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }}
\end{array} \nonumber \]
\[K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber \]
3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]}
\end{array} \nonumber \]
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:
\[\begin{aligned}
& A + B \rightleftharpoons B + C \\[4pt]
& A + B \rightleftharpoons B + C \\[4pt]
& A \rightleftharpoons C
\end{aligned} \nonumber \]
\[K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber \]
Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:
\[K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber \]
\[K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber \]
Example \(\PageIndex{5}\) demonstrates the use of this strategy in describing coupled equilibrium processes.
Example \(\PageIndex{5}\): Equilibrium Constants for Coupled Reactions
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
\[\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber \]
Use the information below to calculate Kc for this reaction.
\[\begin{align*}
\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \\[4pt]
\ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C
\end{align*} \nonumber \]
Solution
The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.
Reverse the first coupled reaction equation:
\[2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber \]
Multiply the second coupled reaction by 3:
\[3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber \]
Finally, add the two revised equations:
\[\begin{align*}
\ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \\[4pt]
\ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)}
\end{align*} \nonumber \]
\[K_{ c }= K_{ c 1}, K_{ c 2},=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber \]
Exercise \(\PageIndex{5}\)
Use the provided information to calculate Kc for the following reaction at 550 °C:
\[\begin{align*}
\ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \\[4pt]
\ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \\[4pt]
\ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67
\end{align*} \nonumber \]
- Answer
-
K c = 0.14


7.9: Chapter 7 Activity
selected template will load here
This action is not available.
Go to Career Services on hacc.edu (hacc.edu > Current Students > Career Services) and click on "Career Resources" in the left menu bar then look for “What Can I Do With This Major?” near the center of the page. Select “View All Majors” then select a major you might find interesting.
What are three “Areas” listed under your chosen major?
Under each of those three “Areas,” what are two types of employers that would hire graduates with this degree?
What are three occupations listed under “Occupational Outlook” near the bottom of the entry?
What are two professional organizations related to this occupation?
What are two actual job openings listed under “Employment Opportunities?”


11.1: Organs and Structures of the Respiratory System
The major organs of the respiratory system function primarily to provide oxygen to body tissues for cellular respiration, remove the waste product carbon dioxide, and help to maintain acid-base balance. Portions of the respiratory system are also used for non-vital functions, such as sensing odors, speech production, and for straining, such as during childbirth or coughing.
Figure 11.1: Major Respiratory Structures
Functionally, the respiratory system can be divided into a conducting zone and a respiratory zone. The conducting zone of the respiratory system includes the organs and structures not directly involved in gas exchange. The gas exchange occurs in the respiratory zone.
Conducting Zone
The major functions of the conducting zone are to provide a route for incoming and outgoing air, remove debris and pathogens from the incoming air, and warm and humidify the incoming air. Several structures within the conducting zone perform other functions as well. The epithelium of the nasal passages, for example, is essential to sensing odors, and the bronchial epithelium that lines the lungs can metabolize some airborne carcinogens.
The Nose and its Adjacent Structures
The major entrance and exit for the respiratory system is through the nose. When discussing the nose, it is helpful to divide it into two major sections: the external nose, and the nasal cavity or internal nose.
The external nose consists of the surface and skeletal structures that result in the outward appearance of the nose and contribute to its numerous functions. The root is the region of the nose located between the eyebrows. The bridge is the part of the nose that connects the root to the rest of the nose. The dorsum nasi is the length of the nose. The apex is the tip of the nose. On either side of the apex, the nostrils are formed by the alae (singular = ala). An ala is a cartilaginous structure that forms the lateral side of each naris (plural = nares), or nostril opening. The philtrum is the concave surface that connects the apex of the nose to the upper lip.
Figure 11.2: Nose
Underneath the thin skin of the nose are its skeletal features. While the root and bridge of the nose consist of bone, the protruding portion of the nose is composed of cartilage. As a result, when looking at a skull, the nose is missing. The nasal bone is one of a pair of bones that lies under the root and bridge of the nose. The nasal bone articulates superiorly with the frontal bone and laterally with the maxillary bones. Septal cartilage is flexible hyaline cartilage connected to the nasal bone, forming the dorsum nasi. The alar cartilage consists of the apex of the nose; it surrounds the naris.
The nares open into the nasal cavity, which is separated into left and right sections by the nasal septum. The nasal septum is formed anteriorly by a portion of the septal cartilage (the flexible portion you can touch with your fingers) and posteriorly by the perpendicular plate of the ethmoid bone (a cranial bone located just posterior to the nasal bones) and the thin vomer bones (whose name refers to its plough shape). Each lateral wall of the nasal cavity has three bony projections, called the superior, middle, and inferior nasal conchae. The inferior conchae are separate bones, whereas the superior and middle conchae are portions of the ethmoid bone. Conchae serve to increase the surface area of the nasal cavity and to disrupt the flow of air as it enters the nose, causing air to bounce along the epithelium, where it is cleaned and warmed. The conchae and meatuses also conserve water and prevent dehydration of the nasal epithelium by trapping water during exhalation. The floor of the nasal cavity is composed of the palate. The hard palate at the anterior region of the nasal cavity is composed of bone. The soft palate at the posterior portion of the nasal cavity consists of muscle tissue. Air exits the nasal cavities via the internal nares and moves into the pharynx.
Figure 11.3: Upper Airway
Several bones that help form the walls of the nasal cavity have air-containing spaces called the paranasal sinuses, which serve to warm and humidify incoming air. Sinuses are lined with a mucosa. Each paranasal sinus is named for its associated bone: frontal sinus, maxillary sinus, sphenoidal sinus, and ethmoidal sinus. The sinuses produce mucus and lighten the weight of the skull.
The nares and anterior portion of the nasal cavities are lined with mucous membranes, containing sebaceous glands and hair follicles that serve to prevent the passage of large debris, such as dirt, through the nasal cavity. An olfactory epithelium used to detect odors is found deeper in the nasal cavity.
The conchae, meatuses, and paranasal sinuses are lined by respiratory epithelium composed of pseudostratified ciliated columnar epithelium. The epithelium contains goblet cells, one of the specialized, columnar epithelial cells that produce mucus to trap debris. The cilia of the respiratory epithelium help remove the mucus and debris from the nasal cavity with a constant beating motion, sweeping materials towards the throat to be swallowed. Interestingly, cold air slows the movement of the cilia, resulting in accumulation of mucus that may in turn lead to a runny nose during cold weather. This moist epithelium functions to warm and humidify incoming air. Capillaries located just beneath the nasal epithelium warm the air by convection. Serous and mucus-producing cells also secrete the lysozyme enzyme and proteins called defensins, which have antibacterial properties. Immune cells that patrol the connective tissue deep to the respiratory epithelium provide additional protection.
Figure 11.4: Pseudostratified Ciliated Columnar Epithelium
Pharynx
The pharynx is a tube formed by skeletal muscle and lined by mucous membrane that is continuous with that of the nasal cavities. The pharynx is divided into three major regions: the nasopharynx, the oropharynx, and the laryngopharynx.
Figure 11.5: Divisions of the Pharynx
The nasopharynx is flanked by the conchae of the nasal cavity, and it serves only as an airway. At the top of the nasopharynx are the pharyngeal tonsils. A pharyngeal tonsil, also called an adenoid, is an aggregate of lymphoid reticular tissue similar to a lymph node that lies at the superior portion of the nasopharynx. The function of the pharyngeal tonsil is not well understood, but it contains a rich supply of lymphocytes and is covered with ciliated epithelium that traps and destroys invading pathogens that enter during inhalation. The pharyngeal tonsils are large in children, but interestingly, tend to regress with age and may even disappear. The uvula is a small bulbous, teardrop-shaped structure located at the apex of the soft palate. Both the uvula and soft palate move like a pendulum during swallowing, swinging upward to close off the nasopharynx to prevent ingested materials from entering the nasal cavity. In addition, auditory (Eustachian) tubes that connect to each middle ear cavity open into the nasopharynx. This connection is why colds often lead to ear infections.
The oropharynx is a passageway for both air and food. The oropharynx is bordered superiorly by the nasopharynx and anteriorly by the oral cavity. The fauces is the opening at the connection between the oral cavity and the oropharynx. As the nasopharynx becomes the oropharynx, the epithelium changes from pseudostratified ciliated columnar epithelium to stratified squamous epithelium. The oropharynx contains two distinct sets of tonsils, the palatine and lingual tonsils. A palatine tonsil is one of a pair of structures located laterally in the oropharynx in the area of the fauces. The lingual tonsil is located at the base of the tongue. Similar to the pharyngeal tonsil, the palatine and lingual tonsils are composed of lymphoid tissue, and trap and destroy pathogens entering the body through the oral or nasal cavities.
The laryngopharynx is inferior to the oropharynx and posterior to the larynx. It continues the route for ingested material and air until its inferior end, where the digestive and respiratory systems diverge. The stratified squamous epithelium of the oropharynx is continuous with the laryngopharynx. Anteriorly, the laryngopharynx opens into the larynx, whereas posteriorly, it enters the esophagus.
Larynx
The larynx is a cartilaginous structure inferior to the laryngopharynx that connects the pharynx to the trachea and helps regulate the volume of air that enters and leaves the lungs. The structure of the larynx is formed by several pieces of cartilage. Three large cartilage pieces—the thyroid cartilage (anterior), epiglottis (superior), and cricoid cartilage (inferior)—form the major structure of the larynx. The thyroid cartilage is the largest piece of cartilage that makes up the larynx. The thyroid cartilage consists of the laryngeal prominence, or “Adam’s apple,” which tends to be more prominent in males. The thick cricoid cartilage forms a ring, with a wide posterior region and a thinner anterior region. Three smaller, paired cartilages—the arytenoids, corniculates, and cuneiforms—attach to the epiglottis and the vocal cords and muscle that help move the vocal cords to produce speech.
Figure 11.6: Larynx
The epiglottis, attached to the thyroid cartilage, is a very flexible piece of elastic cartilage that covers the opening of the trachea. When in the “closed” position, the unattached end of the epiglottis rests on the glottis. The glottis is composed of the vestibular folds, the true vocal cords, and the space between these folds. A vestibular fold, or false vocal cord, is one of a pair of folded sections of mucous membrane. A true vocal cord is one of the white, membranous folds attached by muscle to the thyroid and arytenoid cartilages of the larynx on their outer edges. The inner edges of the true vocal cords are free, allowing oscillation to produce sound. The size of the membranous folds of the true vocal cords differs between individuals, producing voices with different pitch ranges. Folds in males tend to be larger than those in females, which create a deeper voice. The act of swallowing causes the pharynx and larynx to lift upward, allowing the pharynx to expand and the epiglottis of the larynx to swing downward, closing the opening to the trachea. These movements produce a larger area for food to pass through, while preventing food and beverages from entering the trachea.
Figure 11.7: Vocal Cords
Continuous with the laryngopharynx, the superior portion of the larynx is lined with stratified squamous epithelium, transitioning into pseudostratified ciliated columnar epithelium that contains goblet cells. Similar to the nasal cavity and nasopharynx, this specialized epithelium produces mucus to trap debris and pathogens as they enter the trachea. The cilia beat the mucus upward towards the laryngopharynx, where it can be swallowed down the esophagus.
Trachea
The trachea (windpipe) extends from the larynx toward the lungs. The trachea is formed by 16 to 20 stacked, C-shaped pieces of hyaline cartilage that are connected by dense connective tissue. The trachealis muscle and elastic connective tissue together form the fibroelastic membrane, a flexible membrane that closes the posterior surface of the trachea, connecting the C-shaped cartilages. The fibroelastic membrane allows the trachea to stretch and expand slightly during inhalation and exhalation, whereas the rings of cartilage provide structural support and prevent the trachea from collapsing. In addition, the trachealis muscle can be contracted to force air through the trachea during exhalation. The trachea is lined with pseudostratified ciliated columnar epithelium, which is continuous with the larynx. The esophagus borders the trachea posteriorly.
Figure 11.8: Trachea
Bronchial Tree
The trachea branches into the right and left primary bronchi at the carina. These bronchi are also lined by pseudostratified ciliated columnar epithelium containing mucus-producing goblet cells. The carina is a raised structure that contains specialized nervous tissue that induces violent coughing if a foreign body, such as food, is present. Rings of cartilage, similar to those of the trachea, support the structure of the bronchi and prevent their collapse. The primary bronchi enter the lungs at the hilum, a concave region where blood vessels, lymphatic vessels, and nerves also enter the lungs. The bronchi continue to branch into bronchial a tree. A bronchial tree (or respiratory tree) is the collective term used for these multiple-branched bronchi. The main function of the bronchi, like other conducting zone structures, is to provide a passageway for air to move into and out of each lung. In addition, the mucous membrane traps debris and pathogens.
A bronchiole branches from the tertiary bronchi. Bronchioles, which are about 1 mm in diameter, further branch until they become the tiny terminal bronchioles, which lead to the structures of gas exchange. There are more than 1000 terminal bronchioles in each lung. The muscular walls of the bronchioles do not contain cartilage like those of the bronchi. This muscular wall can change the size of the tubing to increase or decrease airflow through the tube.
Respiratory Zone
In contrast to the conducting zone, the respiratory zone includes structures that are directly involved in gas exchange. The respiratory zone begins where the terminal bronchioles join a respiratory bronchiole, the smallest type of bronchiole, which then leads to an alveolar duct, opening into a cluster of alveoli.
Figure 11.9: Respiratory Zone
Alveoli
An alveolar duct is a tube composed of smooth muscle and connective tissue, which opens into a cluster of alveoli. An alveolus is one of the many small, grape-like sacs that are attached to the alveolar ducts.
An alveolar sac is a cluster of many individual alveoli that are responsible for gas exchange. An alveolus is approximately 200 mm in diameter with elastic walls that allow the alveolus to stretch during air intake, which greatly increases the surface area available for gas exchange. Alveoli are connected to their neighbors by alveolar pores, which help maintain equal air pressure throughout the alveoli and lung.
Figure 11.10: Structures of the Respiratory Zone
The alveolar wall consists of three major cell types: type I alveolar cells, type II alveolar cells, and alveolar macrophages. A type I alveolar cell is a squamous epithelial cell of the alveoli, which constitute up to 97 percent of the alveolar surface area. These cells are about 25 nm thick and are highly permeable to gases. A type II alveolar cell is interspersed among the type I cells and secretes pulmonary surfactant, a substance composed of phospholipids and proteins that reduces the surface tension of the alveoli. Roaming around the alveolar wall is the alveolar macrophage, a phagocytic cell of the immune system that removes debris and pathogens that have reached the alveoli.
The simple squamous epithelium formed by type I alveolar cells is attached to a thin, elastic basement membrane. This epithelium is extremely thin and borders the endothelial membrane of capillaries. Taken together, the alveoli and capillary membranes form a respiratory membrane that is approximately 0.5 mm thick. The respiratory membrane allows gases to cross by simple diffusion, allowing oxygen to be picked up by the blood for transport and CO 2 to be released into the air of the alveoli.
References
Bizzintino J, Lee WM, Laing IA, Vang F, Pappas T, Zhang G, Martin AC, Khoo SK, Cox DW, Geelhoed GC, et al. Association between human rhinovirus C and severity of acute asthma in children. Eur Respir J [Internet]. 2010 [cited 2013 Mar 22]; 37(5):1037–1042. Available from: erj.ersjournals.com/gca?submit=Go&gca=erj%3B37%2F5%2F1037&allch=
Kumar V, Ramzi S, Robbins SL. Robbins Basic Pathology. 7th ed. Philadelphia (PA): Elsevier Ltd; 2005.
Martin RJ, Kraft M, Chu HW, Berns, EA, Cassell GH. A link between chronic asthma and chronic infection. J Allergy Clin Immunol [Internet]. 2001 [cited 2013 Mar 22]; 107(4):595-601. Available from: http://erj.ersjournals.com/gca?submit=Go&gca=erj%3B37%2F5%2F1037&allch=
Contributors and Attributions
- How Asthma Causes a shortness in breath.A look inside your body. Authored by : UrhealthIsUrWealth. Located at : https://youtu.be/Zlnsx8zKLvQ . License : All Rights Reserved . License Terms : Standard YouTube LIcense


2.3: Patterns of Global Biodiversity
Biological diversity is not evenly distributed across the Earth; certain latitudes, ecosystems, and regions contain more diversity than others. There are several factors that influence how many species a given ecosystem supports and these factors lead to predictable patterns of diversity across the globe.
Some regions are deemed biodiversity hotspots (Fig 2.3.1); these regions contain more than 1,500 endemic (found only in that region) plant species and have lost at least 70% of their original area to anthropogenic uses ( Myers et al 2000 ). Consequently, hotspots contain a high number of species that are unique to that ecosystem and which face serious conservation threats. The United States contains two biodiversity hotspots. The California Floristic Province stretches from southern Oregon to northern Baja Mexico, covers the majority of the state of California, contains ~2,125 endemic plant species and has lost over 75% of its area to agriculture and urban expansion. Nearly 40% of the Province is under federal protection to preserve the remaining biodiversity. The other biodiversity hotspot in the United States is the state of Hawai’i, which is part of the Polynesia-Micronesia hotspot incorporating the majority of the southern Pacific Ocean island systems.


1.9: Divide Whole Numbers (Part 1)
- Use division notation
- Model division of whole numbers
- Divide whole numbers
- Translate word phrases to math notation
- Divide whole numbers in applications
Before you get started, take this readiness quiz.
- Multiply: \(27 • 3\). If you missed this problem, review Example 1.4.6 .
- Subtract: \(43 − 26\). If you missed this problem, review Example 1.3.4 .
- Multiply: \(62(87)\). If you missed this problem, review Example 1.4.8 .
Use Division Notation
So far we have explored addition, subtraction, and multiplication. Now let’s consider division. Suppose you have the \(12\) cookies in Figure \(\PageIndex{1}\) and want to package them in bags with \(4\) cookies in each bag. How many bags would we need?
Figure \(\PageIndex{1}\)
You might put \(4\) cookies in first bag, \(4\) in the second bag, and so on until you run out of cookies. Doing it this way, you would fill \(3\) bags.
Figure \(\PageIndex{2}\)
In other words, starting with the \(12\) cookies, you would take away, or subtract, \(4\) cookies at a time. Division is a way to represent repeated subtraction just as multiplication represents repeated addition. Instead of subtracting \(4\) repeatedly, we can write
\[12 \div 4 \nonumber \]
We read this as twelve divided by four and the result is the quotient of \(12\) and \(4\). The quotient is \(3\) because we can subtract \(4\) from \(12\) exactly \(3\) times. We call the number being divided the dividend and the number dividing it the divisor . In this case, the dividend is \(12\) and the divisor is \(4\). In the past you may have used the notation \(4 \overline{\smash{)}12}\), but this division also can be written as \(12 ÷ 4\), \(12/4\), \(\dfrac{12}{4}\). In each case the \(12\) is the dividend and the \(4\) is the divisor.
To represent and describe division, we can use symbols and words.
| Operation | Notation | Expression | Read as | Result |
|---|---|---|---|---|
| Division | ÷ | 12 ÷ 4 | Twelve divided by four | the quotient of 12 and 4 |
| \(\dfrac{a}{b}\) | \(\dfrac{12}{4}\) | |||
| \(b \overline{)a}\) | \(4 \overline{\smash{)}12}\) | |||
| a/b | 12/4 |
Division is performed on two numbers at a time. When translating from math notation to English words, or English words to math notation, look for the words of and and to identify the numbers.
Translate from math notation to words.
- \(64 ÷ 8\)
- \(\dfrac{42}{7}\)
- \(4 \overline{\smash{)}28}\)
Solution
- We read this as sixty-four divided by eight and the result is the quotient of sixty-four and eight .
- We read this as forty-two divided by seven and the result is the quotient of forty-two and seven .
- We read this as twenty-eight divided by four and the result is the quotient of twenty-eight and four .
Translate from math notation to words:
- \(84 ÷ 7\)
- \(\dfrac{18}{6}\)
- \(8 \overline{\smash{)}24}\)
- Answer a
-
eighty-four divided by seven; the quotient of eighty-four and seven
- Answer b
-
eighteen divided by six; the quotient of eighteen and six.
- Answer c
-
twenty-four divided by eight; the quotient of twenty-four and eight
Translate from math notation to words:
- \(72÷ 9\)
- \(\dfrac{21}{3}\)
- \(6 \overline{\smash{)}54}\)
- Answer a
-
seventy-two divided by nine; the quotient of seventy-two and nine
- Answer b
-
twenty-one divided by three; the quotient of twenty-one and three
- Answer c
-
fifty-four divided by six; the quotient of fifty-four and six
Model Division of Whole Numbers
As we did with multiplication, we will model division using counters. The operation of division helps us organize items into equal groups as we start with the number of items in the dividend and subtract the number in the divisor repeatedly.
Model the division: \(24 ÷ 8\).
Solution
To find the quotient \(24 ÷ 8\), we want to know how many groups of \(8\) are in \(24\).
Model the dividend. Start with \(24\) counters.
The divisor tell us the number of counters we want in each group. Form groups of \(8\) counters.
Count the number of groups. There are \(3\) groups.
\[24 \div 8 = 3 \nonumber\]
Model: \(24 ÷ 6\).
- Answer
-
Model: \(42 ÷ 7\).
- Answer
-
Divide Whole Numbers
We said that addition and subtraction are inverse operations because one undoes the other. Similarly, division is the inverse operation of multiplication. We know \(12 ÷ 4 = 3\) because \(3 • 4 = 12\). Knowing all the multiplication number facts is very important when doing division.
We check our answer to division by multiplying the quotient by the divisor to determine if it equals the dividend. In Example \(\PageIndex{2}\), we know \(24 ÷ 8 = 3\) is correct because \(3 • 8 = 24\).
Divide. Then check by multiplying.
- \(42 ÷ 6\)
- \(\dfrac{72}{9}\)
- \(7 \overline{\smash{)}63}\)
Solution
| Divide 42 by 6. | 42 ÷ 6 = 7 |
| Check by multiplying. | 7 • 6 = 42 ✓ |
| Divide 72 by 9. | \(\dfrac{72}{9}\) |
| Check by multiplying. | 8 • 9 = 72 ✓ |
| Divide 63 by 7. | \(7 \overline{\smash{)}63}\) |
| Check by multiplying. | 9 • 7 = 63✓ |
Divide. Then check by multiplying:
- \(54 ÷ 6\)
- \(\dfrac{27}{9}\)
- Answer a
-
\(9\)
- Answer b
-
\(3\)
Divide. Then check by multiplying:
- \(\dfrac{36}{9}\)
- \(8 \overline{\smash{)}40}\)
- Answer a
-
\(4\)
- Answer b
-
\(5\)
What is the quotient when you divide a number by itself?
\[\dfrac{15}{15} = 1 \quad \text{because} \quad 1 \cdot 15 = 15 \nonumber \]
Dividing any number (except \(0\)) by itself produces a quotient of \(1\). Also, any number divided by \(1\) produces a quotient of the number. These two ideas are stated in the Division Properties of One.
| Any number (except 0) divided by itself is one. | a ÷ a = 1 |
| Any number divided by one is the same number. | a ÷ 1 = a |
Divide. Then check by multiplying:
- \(11 ÷ 11\)
- \(\dfrac{19}{1}\)
- \(1 \overline{\smash{)}7}\)
Solution
| A number divided by itself is 1. | 11 ÷ 11 = 1 |
| Check by multiplying. | 1 • 11 = 11 ✓ |
| A number divided by 1 equals itself. | \(\dfrac{19}{1} = 19\) |
| Check by multiplying. | 19 • 1 = 19 ✓ |
| A number divided by 1 equals itself. | \(1 \overline{\smash{)}7} = 7\) |
| Check by multiplying. | 7 • 1 = 7✓ |
Divide. Then check by multiplying:
- \(14 ÷ 14\)
- \(\dfrac{27}{1}\)
- Answer a
-
\(1\)
- Answer b
-
\(27\)
Divide. Then check by multiplying:
- \(\dfrac{16}{1}\)
- \(1 \overline{\smash{)}4}\)
- Answer a
-
\(16\)
- Answer b
-
\(4\)
Suppose we have \($0\), and want to divide it among \(3\) people. How much would each person get? Each person would get \($0\). Zero divided by any number is \(0\).
Now suppose that we want to divide \($10\) by \(0\). That means we would want to find a number that we multiply by \(0\) to get \(10\). This cannot happen because \(0\) times any number is \(0\). Division by zero is said to be undefined .
These two ideas make up the Division Properties of Zero.
| Zero divided by any number is 0. | 0 ÷ a = 0 |
| Dividing a number by zero is undefined. | a ÷ 0 = undefined |
Another way to explain why division by zero is undefined is to remember that division is really repeated subtraction. How many times can we take away \(0\) from \(10\)? Because subtracting \(0\) will never change the total, we will never get an answer. So we cannot divide a number by \(0\).
Divide. Check by multiplying:
- \(0 ÷ 3\)
- \(10 / 0\)
Solution
| Zero divided by any number is zero. | 0 ÷ 3 = 0 |
| Check by multiplying. | 0 • 3 = 0 ✓ |
| Dividing a number by zero is undefined. | 10/0 = undefined |
Divide. Then check by multiplying:
- \(0 ÷ 2\)
- \(17 / 0\)
- Answer a
-
\(0\)
- Answer b
-
undefined
Divide. Then check by multiplying:
- \(0 ÷ 6\)
- \(13 / 0\)
- Answer a
-
\(0\)
- Answer b
-
undefined
When the divisor or the dividend has more than one digit, it is usually easier to use the \(4 \overline{\smash{)}12}\) notation. This process is called long division. Let’s work through the process by dividing \(78\) by \(3\).
| Divide the first digit of the dividend, 7, by the divisor, 3. | |
| The divisor 3 can go into 7 two times since 2 × 3 = 6. Write the 2 above the 7 in the quotient. | |
| Multiply the 2 in the quotient by 3 and write the product, 6, under the 7. | |
| Subtract that product from the first digit in the dividend. Subtract 7 − 6. Write the difference, 1, under the first digit in the dividend. | |
| Bring down the next digit of the dividend. Bring down the 8. | |
| Divide 18 by the divisor, 3. The divisor 3 goes into 18 six times. | |
| Write 6 in the quotient above the 8. | |
| Multiply the 6 in the quotient by the divisor and write the product, 18, under the dividend. Subtract 18 from 18. |
We would repeat the process until there are no more digits in the dividend to bring down. In this problem, there are no more digits to bring down, so the division is finished. So \(78 ÷ 3 = 26\).
Check by multiplying the quotient times the divisor to get the dividend. Multiply \(26 × 3\) to make sure that product equals the dividend, \(78\).
It does, so our answer is correct.
Step 1. Divide the first digit of the dividend by the divisor. If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on.
Step 2. Write the quotient above the dividend.
Step 3. Multiply the quotient by the divisor and write the product under the dividend.
Step 4. Subtract that product from the dividend.
Step 5. Bring down the next digit of the dividend.
Step 6. Repeat from Step 1 until there are no more digits in the dividend to bring down.
Step 7. Check by multiplying the quotient times the divisor.
Divide \(2,596 ÷ 4\). Check by multiplying.
Solution
| Let's rewrite the problem to set it up for long division. | |
| Divide the first digit of the dividend, 2, by the divisor, 4. | |
| Since 4 does not go into 2, we use the first two digits of the dividend and divide 25 by 4. The divisor 4 goes into 25 six times. | |
| We write the 6 in the quotient above the 5. | |
| Multiply the 6 in the quotient by the divisor 4 and write the product, 24, under the first two digits in the dividend. | |
| Subtract that product from the first two digits in the dividend. Subtract 25 − 24 . Write the difference, 1, under the second digit in the dividend. | |
| Now bring down the 9 and repeat these steps. There are 4 fours in 19. Write the 4 over the 9. Multiply the 4 by 4 and subtract this product from 19. | |
| Bring down the 6 and repeat these steps. There are 9 fours in 36. Write the 9 over the 6. Multiply the 9 by 4 and subtract this product from 36. | |
| Check by multiplying. |
It equals the dividend, so our answer is correct. So \(2,596 ÷ 4 = 649\).
Divide. Then check by multiplying: \(2,636 ÷ 4\).
- Answer
-
\(659\)
Divide. Then check by multiplying: \(2,716 ÷ 4\).
- Answer
-
\(679\)
Divide \(4,506 ÷ 6\). Check by multiplying.
Solution
| Let's rewrite the problem to set it up for long division. | |
| First we try to divide 6 into 4. | |
| Since that won't work, we try 6 into 45. There are 7 sixes in 45. We write the 7 over the 5 | |
| Multiply the 7 by 6 and subtract this product from 45. | |
| Now bring down the 0 and repeat these steps. There are 5 sixes in 30. Write the 5 over the 0. Multiply the 5 by 6 and subtract this product from 30. | |
| Now bring down the 6 and repeat these steps. There is 1 six in 6. Write the 1 over the 6. Multiply 1 by 6 and subtract this product from 6. | |
| Check by multiplying. |
It equals the dividend, so our answer is correct.
Divide. Then check by multiplying: \(4,305 ÷ 5\).
- Answer
-
\(861\)
Divide. Then check by multiplying: \(3,906 ÷ 6\).
- Answer
-
\(651\)
Divide \(7,263 ÷ 9\). Check by multiplying.
Solution
| Let's rewrite the problem to set it up for long division. | |
| First we try to divide 9 into 7. | |
| Since that won't work, we try 9 into 72. There are 8 nines in 72. We write the 8 over the 2. | |
| Multiply the 8 by 9 and subtract this product from 72. | |
| Now bring down the 6 and repeat these steps. There are 0 nines in 6. Write the 0 over the 6. Multiply the 0 by 9 and subtract this product from 6. | |
| Now bring down the 3 and repeat these steps. There are 7 nines in 63. Write the 7 over the 3. Multiply the 7 by 9 and subtract this product from 63. | |
| Check by multiplying. |
It equals the dividend, so our answer is correct.
Divide. Then check by multiplying: \(4,928 ÷ 7\).
- Answer
-
\(704\)
Divide. Then check by multiplying: \(5,663 ÷ 7\).
- Answer
-
\(809\)
So far all the division problems have worked out evenly. For example, if we had \(24\) cookies and wanted to make bags of \(8\) cookies, we would have \(3\) bags. But what if there were \(28\) cookies and we wanted to make bags of \(8\)? Start with the \(28\) cookies as shown in Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\)
Try to put the cookies in groups of eight as in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\)
There are \(3\) groups of eight cookies, and \(4\) cookies left over. We call the \(4\) cookies that are left over the remainder and show it by writing \(R4\) next to the \(3\). (The \(R\) stands for remainder.)
To check this division we multiply \(3\) times \(8\) to get \(24\), and then add the remainder of \(4\).
Divide \(1,439 ÷ 4\). Check by multiplying.
Solution
| Let's rewrite the problem to set it up for long division. | |
| First we try to divide 4 into 1. Since that won't work, we try 4 into 14. There are 3 fours in 14. We write the 3 over the 4. | |
| Multiply the 3 by 4 and subtract this product from 14. | |
| Now bring down the 3 and repeat these steps. There are 5 fours in 23. Write the 5 over the 3. Multiply the 5 by 4 and subtract this product from 23. | |
| Now bring down the 9 and repeat these steps. There are 9 fours in 39. Write the 9 over the 9. Multiply the 9 by 4 and subtract this product from 39. There are no more numbers to bring down, so we are done. The remainder is 3. | |
| Check by multiplying. |
So \(1,439 ÷ 4\) is \(359\) with a remainder of \(3\). Our answer is correct.
Divide. Then check by multiplying: \(3,812 ÷ 8\).
- Answer
-
\(476\) with a remainder of \(4\)
Divide. Then check by multiplying: \(4,319 ÷ 8\).
- Answer
-
\(539\) with a remainder of \(7\)
Divide and then check by multiplying: \(1,461 ÷ 13\).
Solution
| Let's rewrite the problem to set it up for long division. | |
| First we try to divide 13 into 1. Since that won't work, we try 13 into 14. There is 1 thirteen in 14. We write the 1 over the 4. | |
| Multiply the 1 by 13 and subtract this product from 14. | |
| Now bring down the 6 and repeat these steps. There is 1 thirteen in 16. Write the 1 over the 6. Multiply the 1 by 13 and subtract this product from 16. | |
| Now bring down the 1 and repeat these steps. There are 2 thirteens in 31. Write the 2 over the 1. Multiply the 2 by 13 and subtract this product from 31. There are no more numbers to bring down, so we are done. The remainder is 5. 1,462 ÷ 13 is 112 with a remainder of 5. | |
| Check by multiplying. |
Our answer is correct.
Divide. Then check by multiplying: \(1,493 ÷ 13\).
- Answer
-
\(114\) R \(11\)
Divide. Then check by multiplying: \(1,461 ÷ 12\).
- Answer
-
\(121\) R \(9\)
Divide and check by multiplying: \(74,521 ÷ 241\).
Solution
| Let's rewrite the problem to set it up for long division. | |
| First we try to divide 241 into 7. Since that won’t work, we try 241 into 74. That still won’t work, so we try 241 into 745. Since 2 divides into 7 three times, we try 3. Since 3 × 241 = 723, we write the 3 over the 5 in 745. Note that 4 would be too large because 4 × 241 = 964, which is greater than 745. | |
| Multiply the 3 by 241 and subtract this product from 745. | |
| Now bring down the 2 and repeat these steps. 241 does not divide into 222.  Try 9. Since 9 × 241 = 2,169, we write the 9 over the 1. Multiply the 9 by 241 and subtract this product from 2,221. | |
| There are no more numbers to bring down, so we are finished. The remainder is 52. So 74,521 ÷ 241 is 309 with a remainder of 52. | |
| Check by multiplying. |
Sometimes it might not be obvious how many times the divisor goes into digits of the dividend. We will have to guess and check numbers to find the greatest number that goes into the digits without exceeding them.
Divide. Then check by multiplying: \(78,641 ÷ 256\).
- Answer
-
\(307\) R \(49\)
Divide. Then check by multiplying: \(76,461 ÷ 248\).
- Answer
-
\(308\) R \(77\)
Contributors and Attributions
- Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.


10.8: Major Solar Features
The Photosphere is the visible portion of the sun. It consists of the area in which gaseous layers change from completely opaque to being transparent. The photosphere is also the lowest observable layer of the solar atmosphere.
Faculae are bright areas visible on the Photosphere. The term faculae is Latin for “little torch. ”
Granulation is vast gas bubbles, with rising centers and sinking edges. This bubbling pattern is due to solar convection, as noted in the sun’s regions. Each granulation cell is about 1,000 miles across; compare that to the diameter of Earth at a little under 8,000 miles. Granulation is also a photosphere feature.
We observe limb darkening because the sun is a sphere, near what is seen as the edge of the solar disk, the light must travel farther through the solar atmosphere. This causes the limb to be dimmer than the rest of the disk.
Sunspots are temporary disturbed magnetic areas in the Photosphere. Sunspots come in many shapes and sizes, according to the status of the sun’s magnetic field. This magnetic field traps gas, slowing its motion, and making it cooler than the surrounding area. Sunspots look dark because it is cooler than the surrounding area. Usually, sunspots have two parts: a darker central region called the umbra surrounded by a lighter region called the penumbra . Astronomers have noted a sunspot cycle. This cycle averages in duration of slightly more than 11 years, in which sunspot frequencies varies from a maximum to a minimum and back to a maximum again.
The Chromosphere layer of the sun’s atmosphere located above the Photosphere / beneath the Corona. The term ‘chromosphere’ means the sphere of color. This is due to the fact that hydrogen atoms emit energy called hydrogen-alpha (H-α) radiation. The chromosphere is reddish in appearance, therefore it is hard to see.
Prominences are bright gas clouds ejected from the sun and shaped by the sun’s magnetic field. They appear as spikes, loops, “trees,” detached regions, and more when they occur at the sun’s edge as seen from Earth. When seen straight on, however, prominences look like dark lines, known as filaments , silhouetted against the solar disk because they are slightly cooler than the surface beneath.
Flares occur when the sun’s atmosphere suddenly releases built-up magnetic energy. These emit radiation storms and are by far the Solar System’s largest explosions.
The Corona is the sun’s outer atmosphere; charged particles and dust at low density and very high temperature (10 million K).
Solar Wind streams off of the sun in all directions at speeds of about 1 million miles/hour); the source of the solar wind is the sun’s Corona.
A Coronal Mass Ejection or CME occurs when huge gas bubbles threaded with magnetic field lines are ejected from the sun over several hours. If a CME directly hits Earth, the results can be catastrophic; from disruption of orbiting satellites to destroying electrical power stations. It takes 3 to 72 hours for a CME to strike Earth.


7.10: Translating English to Math
- Translate English phrases to mathematical expressions
Words Are Important!
When working with probability and statistics, words such as “more than” or “less than” can drastically change the answer. Table \(\PageIndex{1}\) shows some of the common phrases you may run into while reading a problem. It will be essential later in the course that you can correctly match these phrases with their correct symbol.
| = | ≤ | ≥ |
|---|---|---|
| is the same as | is less than or equal to | is greater than or equal to |
| is equal to | is at most | is at least |
| is exactly the same as | is not greater than | is not less than |
| has not changed from | within |
| ≠ | > | < |
|---|---|---|
| is not | is more than | is less than |
| is not equal to | is greater than | is below |
| is different from | is above | is lower than |
| has changed from | is higher than | is shorter than |
| is not the same as | is longer than | is smaller than |
| is bigger than | has decreased | |
| has increased | is reduced |
Table \(\PageIndex{1}\)
Translate the following to mathematical expressions when rolling 2 dice:
- Rolling a sum of less than 5.
- Rolling a sum that is above 5.
- Rolling a sum of no less than 8.
- Rolling a sum of at most 8.
Solution
Let \(x\) be the sum of the dice.
- The mathematical expression for rolling a sum of less than 5 would be:
\[x < 5 \nonumber\]
- The mathematical expression for rolling a sum that is above 5 would be:
\[x > 5 \nonumber\]
- The mathematical expression for rolling a sum of no less than 8 would be:
\[x ≥ 8 \nonumber\]
- The mathematical expression for rolling a sum of at most 8 would be:
\[x ≤ 8 \nonumber\]