15.2: Examples and Applications
Example \(15.9\)
Using the Sylow Theorems, we can determine that \(A_5\) has subgroups of orders \(2\text{,}\) \(3\text{,}\) \(4\text{,}\) and \(5\text{.}\) The Sylow \(p\)-subgroups of \(A_5\) have orders \(3\text{,}\) \(4\text{,}\) and \(5\text{.}\) The Third Sylow Theorem tells us exactly how many Sylow \(p\)-subgroups \(A_5\) has.
Solution
Since the number of Sylow \(5\)-subgroups must divide \(60\) and also be congruent to \(1 \pmod{5}\text{,}\) there are either one or six Sylow \(5\)-subgroups in \(A_5\text{.}\) All Sylow \(5\)-subgroups are conjugate. If there were only a single Sylow \(5\)-subgroup, it would be conjugate to itself; that is, it would be a normal subgroup of \(A_5\text{.}\) Since \(A_5\) has no normal subgroups, this is impossible; hence, we have determined that there are exactly six distinct Sylow \(5\)-subgroups of \(A_5\text{.}\)
The Sylow Theorems allow us to prove many useful results about finite groups. By using them, we can often conclude a great deal about groups of a particular order if certain hypotheses are satisfied.
Theorem \(15.10\)
If \(p\) and \(q\) are distinct primes with \(p \lt q \text{,}\) then every group \(G\) of order \(pq\) has a single subgroup of order \(q\) and this subgroup is normal in \(G\text{.}\) Hence, \(G\) cannot be simple. Furthermore, if \(q \not\equiv 1 \pmod{p}\text{,}\) then \(G\) is cyclic.
- Proof
-
We know that \(G\) contains a subgroup \(H\) of order \(q\text{.}\) The number of conjugates of \(H\) divides \(pq\) and is equal to \(1 + kq\) for \(k = 0, 1, \ldots\text{.}\) However, \(1 + q\) is already too large to divide the order of the group; hence, \(H\) can only be conjugate to itself. That is, \(H\) must be normal in \(G\text{.}\)
The group \(G\) also has a Sylow \(p\)-subgroup, say \(K\text{.}\) The number of conjugates of \(K\) must divide \(q\) and be equal to \(1 + kp\) for \(k = 0, 1, \ldots\text{.}\) Since \(q\) is prime, either \(1 + kp = q\) or \(1 + kp = 1\text{.}\) If \(1 + kp = 1\text{,}\) then \(K\) is normal in \(G\text{.}\) In this case, we can easily show that \(G\) satisfies the criteria, given in Chapter 9, for the internal direct product of \(H\) and \(K\text{.}\) Since \(H\) is isomorphic to \({\mathbb Z}_q\) and \(K\) is isomorphic to \({\mathbb Z}_p\text{,}\) \(G \cong {\mathbb Z}_p \times {\mathbb Z}_q \cong {\mathbb Z}_{pq}\) by Theorem \(9.21\).
Example \(15.11\)
Every group of order \(15\) is cyclic.
Solution
This is true because \(15 = 5 \cdot 3\) and \(5 \not\equiv 1 \pmod{3}\text{.}\)
Example \(15.12\)
Let us classify all of the groups of order \(99 = 3^2 \cdot 11\) up to isomorphism.
Solution
First we will show that every group \(G\) of order \(99\) is abelian. By the Third Sylow Theorem, there are \(1 + 3k\) Sylow \(3\)-subgroups, each of order \(9\text{,}\) for some \(k = 0, 1, 2, \ldots\text{.}\) Also, \(1 + 3k\) must divide \(11\text{;}\) hence, there can only be a single normal Sylow \(3\)-subgroup \(H\) in \(G\text{.}\) Similarly, there are \(1 +11k\) Sylow \(11\)-subgroups and \(1 +11k\) must divide \(9\text{.}\) Consequently, there is only one Sylow \(11\)-subgroup \(K\) in \(G\text{.}\) By Corollary 14.16, any group of order \(p^2\) is abelian for \(p\) prime; hence, \(H\) is isomorphic either to \({\mathbb Z}_3 \times {\mathbb Z}_3\) or to \({\mathbb Z}_9\text{.}\) Since \(K\) has order \(11\text{,}\) it must be isomorphic to \({\mathbb Z}_{11}\text{.}\) Therefore, the only possible groups of order \(99\) are \({\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_{11}\) or \({\mathbb Z}_9 \times {\mathbb Z}_{11}\) up to isomorphism.
To determine all of the groups of order \(5 \cdot 7 \cdot 47 = 1645\text{,}\) we need the following theorem.
Theorem \(15.13\)
Let \(G' = \langle a b a^{-1} b^{-1} : a, b \in G \rangle\) be the subgroup consisting of all finite products of elements of the form \(aba^{-1}b^{-1}\) in a group \(G\text{.}\) Then \(G'\) is a normal subgroup of \(G\) and \(G/G'\) is abelian
The subgroup \(G'\) of \(G\) is called the commutator subgroup of \(G\text{.}\) We leave the proof of this theorem as an exercise (Exercise \(10.4.14\) in Chapter 10).
Example \(15.14\)
We will now show that every group of order \(5 \cdot 7 \cdot 47 = 1645\) is abelian, and cyclic by Theorem \(9.21\).
Solution
By the Third Sylow Theorem, \(G\) has only one subgroup \(H_1\) of order \(47\text{.}\) So \(G/H_1\) has order 35 and must be abelian by Theorem \(15.10\). Hence, the commutator subgroup of \(G\) is contained in \(H\) which tells us that \(|G'|\) is either \(1\) or \(47\text{.}\) If \(|G'|=1\text{,}\) we are done. Suppose that \(|G'|=47\text{.}\) The Third Sylow Theorem tells us that \(G\) has only one subgroup of order \(5\) and one subgroup of order \(7\text{.}\) So there exist normal subgroups \(H_2\) and \(H_3\) in \(G\text{,}\) where \(|H_2| = 5\) and \(|H_3| = 7\text{.}\) In either case the quotient group is abelian; hence, \(G'\) must be a subgroup of \(H_i\text{,}\) \(i= 1, 2\text{.}\) Therefore, the order of \(G'\) is \(1\text{,}\) \(5\text{,}\) or \(7\text{.}\) However, we already have determined that \(|G'| =1\) or \(47\text{.}\) So the commutator subgroup of \(G\) is trivial, and consequently \(G\) is abelian.
Finite Simple Groups
Given a finite group, one can ask whether or not that group has any normal subgroups. Recall that a simple group is one with no proper nontrivial normal subgroups. As in the case of \(A_5\text{,}\) proving a group to be simple can be a very difficult task; however, the Sylow Theorems are useful tools for proving that a group is not simple. Usually, some sort of counting argument is involved.
Example \(15.15\)
Let us show that no group \(G\) of order \(20\) can be simple.
Solution
By the Third Sylow Theorem, \(G\) contains one or more Sylow \(5\)-subgroups. The number of such subgroups is congruent to \(1 \pmod{5}\) and must also divide \(20\text{.}\) The only possible such number is \(1\text{.}\) Since there is only a single Sylow \(5\)-subgroup and all Sylow \(5\)-subgroups are conjugate, this subgroup must be normal.
Example \(15.16\)
Let \(G\) be a finite group of order \(p^n\text{,}\) \(n \gt 1\) and \(p\) prime. By Theorem \(14.15\), \(G\) has a nontrivial center.
Solution
Since the center of any group \(G\) is a normal subgroup, \(G\) cannot be a simple group. Therefore, groups of orders \(4\text{,}\) \(8\text{,}\) \(9\text{,}\) \(16\text{,}\) \(25\text{,}\) \(27\text{,}\) \(32\text{,}\) \(49\text{,}\) \(64\text{,}\) and \(81\) are not simple. In fact, the groups of order \(4\text{,}\) \(9\text{,}\) \(25\text{,}\) and \(49\) are abelian by Corollary \(14.16\).
Example \(15.17\)
No group of order \(56= 2^3 \cdot 7\) is simple. We have seen that if we can show that there is only one Sylow \(p\)-subgroup for some prime \(p\) dividing 56, then this must be a normal subgroup and we are done.
Solution
By the Third Sylow Theorem, there are either one or eight Sylow \(7\)-subgroups. If there is only a single Sylow \(7\)-subgroup, then it must be normal.
On the other hand, suppose that there are eight Sylow \(7\)-subgroups. Then each of these subgroups must be cyclic; hence, the intersection of any two of these subgroups contains only the identity of the group. This leaves \(8 \cdot 6 = 48\) distinct elements in the group, each of order \(7\text{.}\) Now let us count Sylow \(2\)-subgroups. There are either one or seven Sylow \(2\)-subgroups. Any element of a Sylow \(2\)-subgroup other than the identity must have as its order a power of \(2\text{;}\) and therefore cannot be one of the \(48\) elements of order \(7\) in the Sylow \(7\)-subgroups. Since a Sylow \(2\)-subgroup has order \(8\text{,}\) there is only enough room for a single Sylow \(2\)-subgroup in a group of order \(56\text{.}\) If there is only one Sylow \(2\)-subgroup, it must be normal.
For other groups \(G\text{,}\) it is more difficult to prove that \(G\) is not simple. Suppose \(G\) has order \(48\text{.}\) In this case the technique that we employed in the last example will not work. We need the following lemma to prove that no group of order \(48\) is simple.
Theorem \(15.18\)
Let \(H\) and \(K\) be finite subgroups of a group \(G\text{.}\) Then
- Proof
-
Recall that
\[ HK = \{ hk : h \in H, k \in K \}\text{.} \nonumber \]Certainly, \(|HK| \leq |H| \cdot |K|\) since some element in \(HK\) could be written as the product of different elements in \(H\) and \(K\text{.}\) It is quite possible that \(h_1 k_1 = h_2 k_2\) for \(h_1, h_2 \in H\) and \(k_1, k_2 \in K\text{.}\) If this is the case, let
\[ a = (h_1)^{-1} h_2 = k_1 (k_2)^{-1}\text{.} \nonumber \]Notice that \(a \in H \cap K\text{,}\) since \((h_1)^{-1} h_2\) is in \(H\) and \(k_2 (k_1)^{-1}\) is in \(K\text{;}\) consequently,
\begin{align*} h_2 & = h_1 a^{-1}\\ k_2 & = a k_1\text{.} \end{align*}Conversely, let \(h = h_1 b^{-1}\) and \(k = b k_1\) for \(b \in H \cap K\text{.}\) Then \(h k = h_1 k_1\text{,}\) where \(h \in H\) and \(k \in K\text{.}\) Hence, any element \(hk \in HK\) can be written in the form \(h_i k_i\) for \(h_i \in H\) and \(k_i \in K\text{,}\) as many times as there are elements in \(H \cap K\text{;}\) that is, \(|H \cap K|\) times. Therefore, \(|HK| = (|H| \cdot |K|)/|H \cap K|\text{.}\)
Example \(15.19\)
To demonstrate that a group \(G\) of order \(48\) is not simple, we will show that \(G\) contains either a normal subgroup of order \(8\) or a normal subgroup of order \(16\text{.}\)
Solution
By the Third Sylow Theorem, \(G\) has either one or three Sylow \(2\)-subgroups of order \(16\text{.}\) If there is only one subgroup, then it must be a normal subgroup.
Suppose that the other case is true, and two of the three Sylow \(2\)-subgroups are \(H\) and \(K\text{.}\) We claim that \(|H \cap K| = 8\text{.}\) If \(|H \cap K| \leq 4\text{,}\) then by Lemma 15.18,
which is impossible. Notice that \(H \cap K\) has index two in both of \(H\) and \(K\text{,}\) so is normal in both, and thus \(H\) and \(K\) are each in the normalizer of \(H \cap K\text{.}\) Because \(H\) is a subgroup of \(N(H \cap K)\) and because \(N(H \cap K)\) has strictly more than \(16\) elements, \(|N(H \cap K)|\) must be a multiple of \(16\) greater than \(1\text{,}\) as well as dividing \(48\text{.}\) The only possibility is that \(|N(H \cap K)|= 48\text{.}\) Hence, \(N(H \cap K) = G\text{.}\)
The following famous conjecture of Burnside was proved in a long and difficult paper by Feit and Thompson [2].
Theorem \(15.20\). Odd Order Theorem
Every finite simple group of nonprime order must be of even order
The proof of this theorem laid the groundwork for a program in the 1960s and 1970s that classified all finite simple groups. The success of this program is one of the outstanding achievements of modern mathematics.


6.6: RC Circuits
By the end of the section, you will be able to:
- Describe the charging process of a capacitor
- Describe the discharging process of a capacitor
- List some applications of RC circuits
When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and several other phenomena that involve charging and discharging capacitors are discussed in this module.
Circuits with Resistance and Capacitance
An RC circuit is a circuit containing resistance and capacitance. As presented in Capacitance , the capacitor is an electrical component that stores electric charge, storing energy in an electric field.
Figure \(\PageIndex{1a}\) shows a simple RC circuit that employs a dc (direct current) voltage source \(ε\), a resistor \(R\), a capacitor \(C\), and a two-position switch. The circuit allows the capacitor to be charged or discharged, depending on the position of the switch. When the switch is moved to position \( A\) , the capacitor charges, resulting in the circuit in Figure \(\PageIndex{1b}\). When the switch is moved to position B , the capacitor discharges through the resistor.
Charging a Capacitor
We can use Kirchhoff’s loop rule to understand the charging of the capacitor. This results in the equation \(\epsilon - V_R - V_C = 0\). This equation can be used to model the charge as a function of time as the capacitor charges. Capacitance is defined as \(C = q/V\), so the voltage across the capacitor is \(V_C = \frac{q}{C}\). Using Ohm’s law, the potential drop across the resistor is \(V_R = IR\), and the current is defined as \(I = dq/dt\).
\[\epsilon - V_R - V_C =0,\]
\[\epsilon - IR - \frac{q}{C} = 0,\]
\[\epsilon - R\frac{dq}{dt} - \frac{q}{C} = 0.\]
This differential equation can be integrated to find an equation for the charge on the capacitor as a function of time.
\[\epsilon - R\frac{dq}{dt} - \frac{q}{C} = 0.\]
\[\frac{dq}{dt} = \frac{\epsilon C - q}{RC},\]
\[\int_0^q \frac{dq}{\epsilon C - q} = \frac{1}{RC} \int_0^t dt.\]
Let \(u = \epsilon C - q\), then \(du = -dq\). The result is
\[-\int_0^q \frac{du}{u} = \frac{1}{RC} \int_0^t dt,\]
\[\ln \left(\frac{\epsilon C - q}{\epsilon C}\right) = - \frac{1}{RC} t.\]
\[\frac{\epsilon C - q}{\epsilon C} = e^{-t/RC}.\]
Simplifying results in an equation for the charge on the charging capacitor as a function of time:
\[q(t) = C\epsilon \left(1 - e^{-\frac{t}{RC}}\right) = Q\left(1 - e^{-\frac{t}{\tau}}\right).\]
A graph of the charge on the capacitor versus time is shown in Figure \(\PageIndex{2a}\). First note that as time approaches infinity, the exponential goes to zero, so the charge approaches the maximum charge \(Q = C\epsilon\) and has units of coulombs. The units of RC are seconds, units of time. This quantity is known as the time constant :
\[\tau = RC.\]
At time \(t = \tau = RC\), the charge equal to \(1 - e^{-1} = 1 - 0.368 = 0.632\) of the maximum charge \(Q = C\epsilon\). Notice that the time rate change of the charge is the slope at a point of the charge versus time plot. The slope of the graph is large at time \(t - 0.0 \, s\) and approaches zero as time increases.
As the charge on the capacitor increases, the current through the resistor decreases, as shown in Figure \(\PageIndex{2b}\). The current through the resistor can be found by taking the time derivative of the charge.
\[ \begin{align*} I(t) &= \frac{dq}{dt} \\[4pt] &= \frac{d}{dt}\left[C\epsilon \left( 1 - e^{-\frac{t}{RC}} \right) \right], \\[4pt] &= C\epsilon \left(\frac{1}{RC}\right) e^{-\frac{t}{RC}} \\[4pt] &= \frac{\epsilon}{R} e^{-\frac{t}{TC}} \\[4pt] &= I_0 e^{-\frac{t}{RC}}, \end{align*}\]
\[I(t) = I_0 e^{-t/\tau}.\]
At time \(t = 0.0 \, s\), the current through the resistor is \(I_0 = \frac{\epsilon}{R}\). As time approaches infinity, the current approaches zero. At time \(t = \tau\), the current through the resistor is \(I (t = \tau) = I_0e^{-1} = 0.368 I_0\).
Figures \(\PageIndex{2c}\) and Figure \(\PageIndex{2d}\) show the voltage differences across the capacitor and the resistor, respectively. As the charge on the capacitor increases, the current decreases, as does the voltage difference across the resistor \(V_R(t) = (I_0R)e^{-t/\tau} = \epsilon e^{-t/\tau}\). The voltage difference across the capacitor increases as \(V_C (t) = \epsilon (1 - e^{-t/\tau} )\).
Discharging a Capacitor
When the switch in Figure \(\PageIndex{3a}\) is moved to position B , the circuit reduces to the circuit in part (c), and the charged capacitor is allowed to discharge through the resistor. A graph of the charge on the capacitor as a function of time is shown in Figure \(\PageIndex{3a}\). Using Kirchhoff’s loop rule to analyze the circuit as the capacitor discharges results in the equation \(-V_R -V_C = 0\), which simplifies to \(IR + \frac{q}{C} = 0\). Using the definition of current \(\frac{dq}{dt}R = - \frac{q}{C}\) and integrating the loop equation yields an equation for the charge on the capacitor as a function of time:
\[q(t) = Qe^{-t/\tau}.\]
Here, Q is the initial charge on the capacitor and \(\tau = RC\) is the time constant of the circuit. As shown in the graph, the charge decreases exponentially from the initial charge, approaching zero as time approaches infinity.
The current as a function of time can be found by taking the time derivative of the charge:
\[I(t) = - \frac{Q}{RC}e^{-t/\tau}.\]
The negative sign shows that the current flows in the opposite direction of the current found when the capacitor is charging. Figure \(\PageIndex{3b}\) shows an example of a plot of charge versus time and current versus time. A plot of the voltage difference across the capacitor and the voltage difference across the resistor as a function of time are shown in Figures \(\PageIndex{3c}\) and \(\PageIndex{3d}\). Note that the magnitudes of the charge, current, and voltage all decrease exponentially, approaching zero as time increases.
Now we can explain why the flash camera mentioned at the beginning of this section takes so much longer to charge than discharge: The resistance while charging is significantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower.
One application of an RC circuit is the relaxation oscillator, as shown below. The relaxation oscillator consists of a voltage source, a resistor, a capacitor, and a neon lamp. The neon lamp acts like an open circuit (infinite resistance) until the potential difference across the neon lamp reaches a specific voltage. At that voltage, the lamp acts like a short circuit (zero resistance), and the capacitor discharges through the neon lamp and produces light. In the relaxation oscillator shown, the voltage source charges the capacitor until the voltage across the capacitor is 80 V. When this happens, the neon in the lamp breaks down and allows the capacitor to discharge through the lamp, producing a bright flash. After the capacitor fully discharges through the neon lamp, it begins to charge again, and the process repeats. Assuming that the time it takes the capacitor to discharge is negligible, what is the time interval between flashes?
Strategy
The time period can be found from considering the equation \(V_C(t) = \epsilon (1 - e^{=t/\tau})\). where \(\tau = (R + r)C.\)
Solution
The neon lamp flashes when the voltage across the capacitor reaches 80 V. The RC time constant is equal to \(\tau = (R + r) = (101 \, \Omega) (50 \times 10^{-3} F) = 5.05 \, s\). We can solve the voltage equation for the time it takes the capacitor to reach 80 V:
\[V_C (t) = \epsilon (1 - e^{-t/\tau} ),\]
\[e^{-t/\tau} = 1 - \frac{V_C(t)}{\epsilon},\]
\[ln (e^{-t/\tau}) = ln \left(1 - \frac{V_C(t)}{\epsilon} \right),\]
\[t = -\tau ln \left(1 - \frac{V_C(t)}{\epsilon} \right) = -5.05 \, s \cdot ln \left(1 - \frac{80 \, V}{100 \, V} \right) = 8.13 \, s.\]
Significance
One application of the relaxation oscillator is for controlling indicator lights that flash at a frequency determined by the values for R and C . In this example, the neon lamp will flash every 8.13 seconds, a frequency of \( f = \frac{1}{T} = \frac{1}{8.13 \, s} = 0.55 \, Hz\). The relaxation oscillator has many other practical uses. It is often used in electronic circuits, where the neon lamp is replaced by a transistor or a device known as a tunnel diode. The description of the transistor and tunnel diode is beyond the scope of this chapter, but you can think of them as voltage controlled switches. They are normally open switches, but when the right voltage is applied, the switch closes and conducts. The “switch” can be used to turn on another circuit, turn on a light, or run a small motor. A relaxation oscillator can be used to make the turn signals of your car blink or your cell phone to vibrate.
RC circuits have many applications. They can be used effectively as timers for applications such as intermittent windshield wipers, pace makers, and strobe lights. Some models of intermittent windshield wipers use a variable resistor to adjust the interval between sweeps of the wiper. Increasing the resistance increases the RC time constant, which increases the time between the operation of the wipers.
Another application is the pacemaker . The heart rate is normally controlled by electrical signals, which cause the muscles of the heart to contract and pump blood. When the heart rhythm is abnormal (the heartbeat is too high or too low), pace makers can be used to correct this abnormality. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during physical activities, thus meeting the increased need for blood and oxygen, and an RC timing circuit can be used to control the time between voltage signals to the heart.
Looking ahead to the study of ac circuits ( Alternating-Current Circuits ), ac voltages vary as sine functions with specific frequencies. Periodic variations in voltage, or electric signals, are often recorded by scientists. These voltage signals could come from music recorded by a microphone or atmospheric data collected by radar. Occasionally, these signals can contain unwanted frequencies known as “noise.” RC filters can be used to filter out the unwanted frequencies.
In the study of electronics, a popular device known as a 555 timer provides timed voltage pulses. The time between pulses is controlled by an RC circuit. These are just a few of the countless applications of RC circuits.
A relaxation oscillator is used to control a pair of windshield wipers. The relaxation oscillator consists of a 10.00-mF capacitor and a \(10.00 \, k\Omega\) variable resistor known as a rheostat. A knob connected to the variable resistor allows the resistance to be adjusted from \(0.00 \, \Omega\) to \(10.00 \, k\Omega\). The output of the capacitor is used to control a voltage-controlled switch. The switch is normally open, but when the output voltage reaches 10.00 V, the switch closes, energizing an electric motor and discharging the capacitor. The motor causes the windshield wipers to sweep once across the windshield and the capacitor begins to charge again. To what resistance should the rheostat be adjusted for the period of the wiper blades be 10.00 seconds?
Strategy
The resistance considers the equation \(V_{out}(t) = V(1 - e^{-t/\tau})\), where \(\tau = RC\). The capacitance, output voltage, and voltage of the battery are given. We need to solve this equation for the resistance.
Solution
The output voltage will be 10.00 V and the voltage of the battery is 12.00 V. The capacitance is given as 10.00 mF. Solving for the resistance yields
\[V_{out}(t) = V(1 - e^{-t/\tau})\]
\[e^{-t/RC} = 1 - \frac{V_{out}(t)}{V},\]
\[ln (e^{-t/RC}) = ln \left(1 - \frac{V_{out}(t)}{V} \right),\]
\[-\frac{t}{RC} = ln \left(1 - \frac{V_{out}(t)}{V} \right),\]
\[R = \frac{-t}{C \, ln\left( 1 - \frac{V_C(t)}{V}\right)} = \frac{-10.00 \, s}{10 \times 10^{-3} F \, ln \left(1 - \frac{10 \, V}{12 \, V}\right)} = 558.11 \, \Omega.\]
Significance
Increasing the resistance increases the time delay between operations of the windshield wipers. When the resistance is zero, the windshield wipers run continuously. At the maximum resistance, the period of the operation of the wipers is:
\[t = -RC ln\left(1 - \frac{V_{out}(t)}{V}\right) = - (10 \times 10^{-3} F)(10 \times 10^3 \, \Omega) ln \left(1 - \frac{10 \, V}{12 \, V}\right) = 179.18 \, s = 2.98 \, min.\]
The RC circuit has thousands of uses and is a very important circuit to study. Not only can it be used to time circuits, it can also be used to filter out unwanted frequencies in a circuit and used in power supplies, like the one for your computer, to help turn ac voltage to dc voltage.


29.1: The Kennedy Promise
In the 1950s, President Dwight D. Eisenhower presided over a United States that prized conformity over change. Although change naturally occurred, as it does in every era, it was slow and greeted warily. By the 1960s, however, the pace of change had quickened and its scope broadened, as restive and energetic waves of World War II veterans and baby boomers of both sexes and all ethnicities began to make their influence felt politically, economically, and culturally. No one symbolized the hopes and energies of the new decade more than John Fitzgerald Kennedy, the nation’s new, young, and seemingly healthful, president. Kennedy had emphasized the country’s aspirations and challenges as a “new frontier” when accepting his party’s nomination at the Democratic National Convention in Los Angeles, California.
THE NEW FRONTIER
The son of Joseph P. Kennedy, a wealthy Boston business owner and former ambassador to Great Britain, John F. Kennedy graduated from Harvard University and went on to serve in the U.S. House of Representatives in 1946. Even though he was young and inexperienced, his reputation as a war hero who had saved the crew of his PT boat after it was destroyed by the Japanese helped him to win election over more seasoned candidates, as did his father’s fortune. In 1952, he was elected to the U.S. Senate for the first of two terms. For many, including Arthur M. Schlesinger, Jr., a historian and member of Kennedy’s administration, Kennedy represented a bright, shining future in which the United States would lead the way in solving the most daunting problems facing the world.
Kennedy’s popular reputation as a great politician undoubtedly owes much to the style and attitude he personified. He and his wife Jacqueline conveyed a sense of optimism and youthfulness. “Jackie” was an elegant first lady who wore designer dresses, served French food in the White House, and invited classical musicians to entertain at state functions. “Jack” Kennedy, or JFK, went sailing off the coast of his family’s Cape Cod estate and socialized with celebrities (Figure 29.1.2). Few knew that behind Kennedy’s healthful and sporty image was a gravely ill man whose wartime injuries caused him daily agony.
Nowhere was Kennedy’s style more evident than in the first televised presidential debate held on September 23, 1960, between him and his Republican opponent Vice President Richard M. Nixon. Seventy million viewers watched the debate on television; millions more heard it on the radio. Radio listeners judged Nixon the winner, whereas those who watched the debate on television believed the more telegenic Kennedy made the better showing.
View television footage of the first Kennedy-Nixon debate at the JFK Presidential Library and Museum.
Kennedy did not appeal to all voters, however. Many feared that because he was Roman Catholic, his decisions would be influenced by the Pope. Even traditional Democratic supporters, like the head of the United Auto Workers, Walter Reuther, feared that a Catholic candidate would lose the support of Protestants. Many southern Democrats also disliked Kennedy because of his liberal position on civil rights. To shore up support for Kennedy in the South, Lyndon B. Johnson, the Protestant Texan who was Senate majority leader, was added to the Democratic ticket as the vice presidential candidate. In the end, Kennedy won the election by the closest margin since 1888, defeating Nixon with only 0.01 percent more of the record sixty-seven million votes cast. His victory in the Electoral College was greater: 303 electoral votes to Nixon’s 219. Kennedy’s win made him both the youngest man elected to the presidency and the first U.S. president born in the twentieth century.
Kennedy dedicated his inaugural address to the theme of a new future for the United States. “Ask not what your country can do for you; ask what you can do for your country,” he challenged his fellow Americans. His lofty goals ranged from fighting poverty to winning the space race against the Soviet Union with a moon landing. He assembled an administration of energetic people assured of their ability to shape the future. Dean Rusk was named secretary of state. Robert McNamara, the former president of Ford Motor Company, became secretary of defense. Kennedy appointed his younger brother Robert as attorney general, much to the chagrin of many who viewed the appointment as a blatant example of nepotism.
Kennedy’s domestic reform plans remained hampered, however, by his narrow victory and lack of support from members of his own party, especially southern Democrats. As a result, he remained hesitant to propose new civil rights legislation. His achievements came primarily in poverty relief and care for the disabled. Unemployment benefits were expanded, the food stamps program was piloted, and the school lunch program was extended to more students. In October 1963, the passage of the Mental Retardation Facilities and Community Mental Health Centers Construction Act increased support for public mental health services.
KENNEDY THE COLD WARRIOR
Kennedy focused most of his energies on foreign policy, an arena in which he had been interested since his college years and in which, like all presidents, he was less constrained by the dictates of Congress. Kennedy, who had promised in his inaugural address to protect the interests of the “free world,” engaged in Cold War politics on a variety of fronts. For example, in response to the lead that the Soviets had taken in the space race when Yuri Gagarin became the first human to successfully orbit the earth, Kennedy urged Congress to not only put a man into space (Figure 29.1.3) but also land an American on the moon, a goal finally accomplished in 1969. This investment advanced a variety of military technologies, especially the nation’s long-range missile capability, resulting in numerous profitable spin-offs for the aviation and communication industries. It also funded a growing middle class of government workers, engineers, and defense contractors in states ranging from California to Texas to Florida—a region that would come to be known as the Sun Belt—becoming a symbol of American technological superiority. At the same time, however, the use of massive federal resources for space technologies did not change the economic outlook for low-income communities and underprivileged regions.
To counter Soviet influence in the developing world, Kennedy supported a variety of measures. One of these was the Alliance for Progress, which collaborated with the governments of Latin American countries to promote economic growth and social stability in nations whose populations might find themselves drawn to communism. Kennedy also established the Agency for International Development to oversee the distribution of foreign aid, and he founded the Peace Corps, which recruited idealistic young people to undertake humanitarian projects in Asia, Africa, and Latin America. He hoped that by augmenting the food supply and improving healthcare and education, the U.S. government could encourage developing nations to align themselves with the United States and reject Soviet or Chinese overtures. The first group of Peace Corps volunteers departed for the four corners of the globe in 1961, serving as an instrument of “soft power” in the Cold War.
Kennedy’s various aid projects, like the Peace Corps, fit closely with his administration’s flexible response, which Robert McNamara advocated as a better alternative to the all-or-nothing defensive strategy of mutually assured destruction favored during Eisenhower’s presidency. The plan was to develop different strategies, tactics, and even military capabilities to respond more appropriately to small or medium-sized insurgencies, and political or diplomatic crises. One component of flexible response was the Green Berets, a U.S. Army Special Forces unit trained in counterinsurgency—the military suppression of rebel and nationalist groups in foreign nations. Much of the Kennedy administration’s new approach to defense, however, remained focused on the ability and willingness of the United States to wage both conventional and nuclear warfare, and Kennedy continued to call for increases in the American nuclear arsenal.
Cuba
Kennedy’s multifaceted approach to national defense is exemplified by his careful handling of the Communist government of Fidel Castro in Cuba. In January 1959, following the overthrow of the corrupt and dictatorial regime of Fulgencio Batista, Castro assumed leadership of the new Cuban government. The progressive reforms he began indicated that he favored Communism, and his pro-Soviet foreign policy frightened the Eisenhower administration, which asked the Central Intelligence Agency (CIA) to find a way to remove him from power. Rather than have the U.S. military invade the small island nation, less than one hundred miles from Florida, and risk the world’s criticism, the CIA instead trained a small force of Cuban exiles for the job. After landing at the Bay of Pigs on the Cuban coast, these insurgents, the CIA believed, would inspire their countrymen to rise up and topple Castro’s regime. The United States also promised air support for the invasion.
Kennedy agreed to support the previous administration’s plans, and on April 17, 1961, approximately fourteen hundred Cuban exiles stormed ashore at the designated spot. However, Kennedy feared domestic criticism and worried about Soviet retaliation elsewhere in the world, such as Berlin. He cancelled the anticipated air support, which enabled the Cuban army to easily defeat the insurgents. The hoped-for uprising of the Cuban people also failed to occur. The surviving members of the exile army were taken into custody.
The Bay of Pigs invasion was a major foreign policy disaster for President Kennedy and highlighted Cuba’s military vulnerability to the Castro administration. The following year, the Soviet Union sent troops and technicians to Cuba to strengthen its new ally against further U.S. military plots. Then, on October 14, U.S. spy planes took aerial photographs that confirmed the presence of long-range ballistic missile sites in Cuba. The United States was now within easy reach of Soviet nuclear warheads (Figure 29.1.4).
On October 22, Kennedy demanded that Soviet premier Nikita Khrushchev remove the missiles. He also ordered a naval quarantine placed around Cuba to prevent Soviet ships from approaching. Despite his use of the word “quarantine” instead of “blockade,” for a blockade was considered an act of war, a potential war with the Soviet Union was nevertheless on the president’s mind. As U.S. ships headed for Cuba, the army was told to prepare for war, and Kennedy appeared on national television to declare his intention to defend the Western Hemisphere from Soviet aggression.
The world held its breath awaiting the Soviet reply. Realizing how serious the United States was, Khrushchev sought a peaceful solution to the crisis, overruling those in his government who urged a harder stance. Behind the scenes, Robert Kennedy and Soviet ambassador Anatoly Dobrynin worked toward a compromise that would allow both superpowers to back down without either side’s seeming intimidated by the other. On October 26, Khrushchev agreed to remove the Russian missiles in exchange for Kennedy’s promise not to invade Cuba. On October 27, Kennedy’s agreement was made public, and the crisis ended. Not made public, but nevertheless part of the agreement, was Kennedy’s promise to remove U.S. warheads from Turkey, as close to Soviet targets as the Cuban missiles had been to American ones.
The showdown between the United States and the Soviet Union over Cuba’s missiles had put the world on the brink of a nuclear war. Both sides already had long-range bombers with nuclear weapons airborne or ready for launch, and were only hours away from the first strike. In the long run, this nearly catastrophic example of nuclear brinksmanship ended up making the world safer. A telephone “hot line” was installed, linking Washington and Moscow to avert future crises, and in 1963, Kennedy and Khrushchev signed the Limited Test Ban Treaty, prohibiting tests of nuclear weapons in Earth’s atmosphere.
Vietnam
Cuba was not the only arena in which the United States sought to contain the advance of Communism. In Indochina, nationalist independence movements, most notably Vietnam’s Viet Minh under the leadership of Ho Chi Minh, had strong Communist sympathies. President Harry S. Truman had no love for France’s colonial regime in Southeast Asia but did not want to risk the loyalty of its Western European ally against the Soviet Union. In 1950, the Truman administration sent a small military advisory group to Vietnam and provided financial aid to help France defeat the Viet Minh.
In 1954, Vietnamese forces finally defeated the French, and the country was temporarily divided at the seventeenth parallel. Ho Chi Minh and the Viet Minh controlled the North. In the South, the last Vietnamese emperor and ally to France, Bao Dai, named the French-educated, anti-Communist Ngo Dinh Diem as his prime minister. But Diem refused to abide by the Geneva Accords, the treaty ending the conflict that called for countrywide national elections in 1956, with the victor to rule a reunified nation. After a fraudulent election in the South in 1955, he ousted Bao Dai and proclaimed himself president of the Republic of Vietnam. He cancelled the 1956 elections in the South and began to round up Communists and supporters of Ho Chi Minh.
Realizing that Diem would never agree to the reunification of the country under Ho Chi Minh’s leadership, the North Vietnamese began efforts to overthrow the government of the South by encouraging insurgents to attack South Vietnamese officials. By 1960, North Vietnam had also created the National Liberation Front (NLF) to resist Diem and carry out an insurgency in the South. The United States, fearing the spread of Communism under Ho Chi Minh, supported Diem, assuming he would create a democratic, pro-Western government in South Vietnam. However, Diem’s oppressive and corrupt government made him a very unpopular ruler, particularly with farmers, students, and Buddhists, and many in the South actively assisted the NLF and North Vietnam in trying to overthrow his government.
When Kennedy took office, Diem’s government was faltering. Continuing the policies of the Eisenhower administration, Kennedy supplied Diem with money and military advisors to prop up his government (Figure 29.1.5). By November 1963, there were sixteen thousand U.S. troops in Vietnam, training members of that country’s special forces and flying air missions that dumped defoliant chemicals on the countryside to expose North Vietnamese and NLF forces and supply routes. A few weeks before Kennedy’s own death, Diem and his brother Nhu were assassinated by South Vietnamese military officers after U.S. officials had indicated their support for a new regime.
TENTATIVE STEPS TOWARD CIVIL RIGHTS
Cold War concerns, which guided U.S. policy in Cuba and Vietnam, also motivated the Kennedy administration’s steps toward racial equality. Realizing that legal segregation and widespread discrimination hurt the country’s chances of gaining allies in Africa, Asia, and Latin America, the federal government increased efforts to secure the civil rights of African Americans in the 1960s. During his presidential campaign, Kennedy had intimated his support for civil rights, and his efforts to secure the release of civil rights leader Martin Luther King, Jr., who was arrested following a demonstration, won him the African American vote. Lacking widespread backing in Congress, however, and anxious not to offend white southerners, Kennedy was cautious in assisting African Americans in their fight for full citizenship rights.
His strongest focus was on securing the voting rights of African Americans. Kennedy feared the loss of support from southern white Democrats and the impact a struggle over civil rights could have on his foreign policy agenda as well as on his reelection in 1964. But he thought voter registration drives far preferable to the boycotts, sit-ins, and integration marches that had generated such intense global media coverage in previous years. Encouraged by Congress’s passage of the Civil Rights Act of 1960, which permitted federal courts to appoint referees to guarantee that qualified persons would be registered to vote, Kennedy focused on the passage of a constitutional amendment outlawing poll taxes, a tactic that southern states used to disenfranchise African American voters. Originally proposed by President Truman’s Committee on Civil Rights, the idea had been largely forgotten during Eisenhower’s time in office. Kennedy, however, revived it and convinced Spessard Holland, a conservative Florida senator, to introduce the proposed amendment in Congress. It passed both houses of Congress and was sent to the states for ratification in September 1962.
Kennedy also reacted to the demands of the civil rights movement for equality in education. For example, when African American student James Meredith, encouraged by Kennedy’s speeches, attempted to enroll at the segregated University of Mississippi in 1962, riots broke out on campus (Figure 29.1.6). The president responded by sending the U.S. Army and National Guard to Oxford, Mississippi, to support the U.S. Marshals that his brother Robert, the attorney general, had dispatched.
Following similar violence at the University of Alabama when two African American students, Vivian Malone and James Hood, attempted to enroll in 1963, Kennedy responded with a bill that would give the federal government greater power to enforce school desegregation, prohibit segregation in public accommodations, and outlaw discrimination in employment. Kennedy would not live to see his bill enacted; it would become law during Lyndon Johnson’s administration as the 1964 Civil Rights Act.
TRAGEDY IN DALLAS
Although his stance on civil rights had won him support in the African American community and his steely performance during the Cuban Missile Crisis had led his overall popularity to surge, Kennedy understood that he had to solidify his base in the South to secure his reelection. On November 21, 1963, he accompanied Lyndon Johnson to Texas to rally his supporters. The next day, shots rang out as Kennedy’s motorcade made its way through the streets of Dallas. Seriously injured, Kennedy was rushed to Parkland Hospital and pronounced dead.
The gunfire that killed Kennedy appeared to come from the upper stories of the Texas School Book Depository building; later that day, Lee Harvey Oswald, an employee at the depository and a trained sniper, was arrested (Figure 29.1.7). Two days later, while being transferred from Dallas police headquarters to the county jail, Oswald was shot and killed by Jack Ruby, a local nightclub owner who claimed he acted to avenge the president.
Almost immediately, rumors began to circulate regarding the Kennedy assassination, and conspiracy theorists, pointing to the unlikely coincidence of Oswald’s murder a few days after Kennedy’s, began to propose alternate theories about the events. To quiet the rumors and allay fears that the government was hiding evidence, Lyndon Johnson, Kennedy’s successor, appointed a fact-finding commission headed by Earl Warren, chief justice of the U.S. Supreme Court, to examine all the evidence and render a verdict. The Warren Commission concluded that Lee Harvey Oswald had acted alone and there had been no conspiracy. The commission’s ruling failed to satisfy many, and multiple theories have sprung up over time. No credible evidence has ever been uncovered, however, to prove either that someone other than Oswald murdered Kennedy or that Oswald acted with co-conspirators.
Section Summary
The arrival of the Kennedys in the White House seemed to signal a new age of youth, optimism, and confidence. Kennedy spoke of a “new frontier” and promoted the expansion of programs to aid the poor, protect African Americans’ right to vote, and improve African Americans’ employment and education opportunities. For the most part, however, Kennedy focused on foreign policy and countering the threat of Communism—especially in Cuba, where he successfully defused the Cuban Missile Crisis, and in Vietnam, to which he sent advisors and troops to support the South Vietnamese government. The tragedy of Kennedy’s assassination in Dallas brought an early end to the era, leaving Americans to wonder whether his vice president and successor, Lyndon Johnson, would bring Kennedy’s vision for the nation to fruition.
Review Questions
The term Kennedy chose to describe his sealing off of Cuba to prevent Soviet shipments of weapons or supplies was ________.
- interdiction
- quarantine
- isolation
- blockade
B
Kennedy proposed a constitutional amendment that would ________.
- provide healthcare for all Americans
- outlaw poll taxes
- make English the official language of the United States
- require all American men to register for the draft
B
What steps did Kennedy take to combat Communism?
Kennedy’s economic development programs, supported by the Peace Corps, were intended to reduce poverty in developing nations so their citizens would be less attracted to Communism. After the Bay of Pigs invasion failed to overthrow the government of Fidel Castro, Kennedy demanded that the Soviet Union remove intermediate-range missiles from Cuba. He also increased support for the anti-Communist government in South Vietnam and sent advisors and troops to train the South Vietnamese army.
Glossary
- counterinsurgency
- a new military strategy under the Kennedy administration to suppress nationalist independence movements and rebel groups in the developing world
- flexible response
- a military strategy that allows for the possibility of responding to threats in a variety of ways, including counterinsurgency, conventional war, and nuclear strikes
- naval quarantine
- Kennedy’s use of ships to prevent Soviet access to Cuba during the Cuban Missile Crisis


2.10: Finding indexes
The library provides a list of Article Indexes & Databases that you have access to as an ISU student. You can access these from the library homepage .
You can browse the indexes by subject using the All Subjects drop-down menu. Once you choose your subject you’ll get a list of indexes and databases that cover that area, with Best Bets featured at the top.
The Subjects dropdown on the Article Indexes & Databases page brings you to a list of article indexes hand-picked by the librarian for that subject. If you need help with your research, that librarian’s contact information is listed there so you can contact them with questions.
If you know the name of the article index you want, you can find it by using the default A-Z list or by using the search box. Then click on the name of the index you want to use. Note: this search box does not find articles within the index, but instead it searches names and descriptions of indexes or databases. When using the search box, it is best to use very general terms, such as “chemistry” or “design.” If you’re not sure which index to use, you can also use a general article database like Academic Search Ultimate.
General purpose indexes
General purpose article indexes cover many different subject areas and often include articles from magazines and newspapers as well as from scholarly journals. One good example of a general purpose index is Academic Search Ultimate . Because it covers many different subject areas, a general purpose index like this can be a good place to start.
In Academic Search Ultimate, options for filtering your search results are located on the left side of your screen in the Refine Results section. You can use these to focus your search results in various ways.
Academic Search Ultimate has some interesting and unique features, including text to speech technology that can read articles to you. Other features include saving your search results, filtering articles to peer-reviewed journals only, and generating citations for your bibliography.
Indexes like Academic Search Ultimate are great for finding information when you’re not sure which subject area your topic falls under. However, if you want to find in-depth research articles in a specific subject area, subject-focused indexes are a better choice.


4.1: Introduction
Introduction
The 1855 Colton's Map of the World on Mercator's Projection (4.1.1) was positioned to focus on the Americas while incorporating the entire known world. Originally hand-colored in pastels and bordered with Colton's characteristic spiral motif, the map was a re-engraving. Colton also marked the routes of a few explorers, including Captain Cook. Antarctica is only partially defined, although a note on the bottom noted the recent discovery by Captain Ross of Mt. Terror and Mt. Erebus. Much of Africa was still designated as unexplored regions. The Mercator Projection became the standard methodology for map definitions after the late 1500s, when the concept was defined in 1569 by Gerardus Mercator. The 1500s was the age of exploration and traversing the oceans on the seas; the Mercator style of map embodied any straight line between destinations. However, the projection distorts the shape of the continents and their size.
By the 19 th century, the continents were generally known, trade had expanded across the oceans, the balance of power had changed, settlers moved freely from one continent to another, and the industrial age fueled massive economic changes. Power also moved from royalty and the church to wealthy merchants and a burgeoning middle class. The century also endured multiple revolutions and wars, uprisings from slavery, or faraway rulers. Changes in art were evident in this period; technologies invented new products for artists, multiple movements developed and spread across the world, and artists traveled and learned new methodologies. In some contemporary art movements, the general art world became more accessible for female artists and artists of color.
Europe
In Europe at the beginning of the 19 th century, France and Napoleon controlled most of the continent of Europe; his defeat brought significant changes to the countries. After the defeat, the Congress of Vienna, composed of representatives from each region met to redefine the borders of the continent's countries by changing the size of significant areas and balancing power in the desire to maintain peace. Prussia and Austria added territory, and Russia absorbed some of Poland, the neutrality of Switzerland was established, the Papal States were restored, among other realignments. Although there were different local wars, the basic structure of the agreements generally remained in place until World War One. With relative peace, the science and technology of the industrial revolution developed across Europe, and essential concepts developed; Darwin's natural selection, Pasteur's vaccine, theories of heat, energy, atoms, electrical transmission, the induction motor, and other significant contributions.
Europe changed significantly in the 19 th century with the collapse of previous empires and kingdoms and the establishment of new power centers. The Russian Empire was geographically immense, extending from parts of Asia, Europe, and North America, third in the world in population with diverse languages, religions, and economic status. The government was based on a monarchy with primarily an agricultural economy and lagged behind Europe in the growth of industrial changes. The empire was also engaged in continual warfare in some parts of its territory, starting with the invasion of Napoleon, a disaster for Napoleon, uprisings within its borders, and wars with other European countries. In 1861, Tsar Alexander II did emancipate the serfs, ending the aristocracy's hold on land and power; however, the serfs now owned a small amount of land and had to pay excessive taxes, keeping them in poverty. By the end of the century, railroads extended trade, and agricultural growth helped improve the life of the peasantry while continuing to enhance the aristocracy. Social and political changes developed with multiple factions and views demanding reforms, sowing the seeds of revolution in the next century.
Although America won its independence from England earlier, in the 19 th century, the British Empire was the world's powerhouse, overseeing and controlling extensive territory in Canada, South Africa, parts of Africa, different islands, Egypt, India, and ports in China. When the century ended, the empire exerted its control over a fifth of the world, including twenty-five percent of its population. After Napoleon was defeated in 1815 at the Battle of Waterloo, the British did not have major rivals. The Royal Navy was able to exercise its control of the oceans for trade and supremacy. The navy was so powerful it stopped the slave trade and blocked most piracy. They integrated the entire empire into a global power based on a Pax Britannica (British Peace), establishing the empire as the world's arbitrator and controller of the territories. The British ideals of free markets helped spur the industrial revolution and economic growth, developing shipping lines and trading policies with steamships and linking the world with telegraph cables. However, by the end of the century, the old Ottoman Empire was crumbling, generating the Crimean War. The development and industrialization of Germany and the United States started the decline of British dominance.
As the economies in Europe transformed in the early 19 th century to industrialization and revolutions altered the political status, poverty and desperation prevailed. Art moved from romantic portrayals to Realism, depicting the poverty of the people and their hardships. Industrialization also brought new inventions and methodologies, even benefiting the art community, making the materials of the artist portable, and improving printing procedures. Artists moved outdoors, and the naturalistic ideas of the Realistic transformed into the approaches of Impressionism using light, color, and brushstrokes in new and revolutionary methods. The Post-Impressionists expanded the concept of color and lines. The movements brought new ideas about how nudes were portrayed, what subject matter was painted and how emotions were expressed.
America
America was a fledgling and inexperienced nation at the onset of the 19 th century. Expanding westward with the Louisiana Purchase started the movement of people out of the original coastal states and into the Midwestern regions. In 1820, approximately ten million people lived in the twenty-two states and territories; 1.5 million were slaves. There were eleven free states and ten slave states when the Missouri Compromise became law, bringing Missouri in as a slave state; however, no new slave states could be established west of the Mississippi River. Other new eastern states were admitted in pairs, one free and one slave, until 1850, when California became a state under the Compromise of 1850. California won its independence from Mexico in 1848; the gold rush in 1849 brought a surge in population, and the need for an organized government as the state was never a territory. As part of the agreement, the regions of Utah and New Mexico had no slave restrictions, and the Fugitive Slave Law was enacted; now, slaves who escaped to the free states had to be returned. The official institution of slavery remained solidly supported in the south from its beginnings until the end of the Civil War in 1865. With the addition of California, the United States existed from coast to coast.
Major industries developed and grew in the eastern half of the country before the civil war. The large rivers were a primary method of transportation aided by the invention of the steamboat. The telegraph opened up communication across the states, and the sewing machine created the tools for the mass manufacture of clothing. Most of the significant industrial factories were in the northern states. In contrast, the southern states focused on large agricultural plantations supported by slave labor, growing cotton for the mills of the north and tobacco for export. The railroads were one of the integral elements in the expansion of the United States. The railroads also helped accelerate the government's move to confiscate Native American territory. Native people were removed from their natural regions when the original colonists came. As America became an independent country, they moved large numbers of the tribes from eastern and Midwestern states further west. With the opening of the country's western half, the Native populations were again forced onto reservations in undesirable lands.
After the civil war, the transcontinental railroad fueled another industrial expansion, including steel manufacturing, the telephone, and the light bulb. The western states held vast mineral resources transported by the railroads. After the civil war, fourteen million people immigrated to the United States to work in factories by the end of the century. People's lives also changed; craft industries and small shops disappeared as men, women, and children went to work in the factories, a transformation from a rural country to urban centers. Factories were often dangerous and unsanitary places; people were forced into long hours at low pay, and the labor unions began to organize and grow. The industrial revolution helped propel America's growth and economic status, the internal destruction of Native Americans, ending formal slavery, and opening the lands to the world's immigrants, all within one century.
Early in the 1800s, artists were still influenced by European styles, painting portraits, and everyday life until the new ideas of American landscape painting were established and instilled with concepts of the new country and natural wonders, creating a new genre. Artists still went to Europe for training; American Realism painters embodied themes of democracy, heroic American life, and westward expansion instead of religious art or paintings of royalty. After the civil war and the opportunities to move throughout the country, photography, prints, and magazine illustrations became popular art expressions.
Color of Impressionism
Before the Impressionists, color was meant to be literal, the authentic color of the actual item in ordinary light. Color might be used from light to dark to develop shadows, folds, and light sources, as the object remains identifiable by its color; the sky is blue, and a cape is red. The Impressionists changed the concept of how color was used based on lighting or elements of nature as the wind blowing the grasses or the sun's position and effect, or weather conditions. The Impressionists sought out the conditions of nature and used multiple colors to create the right feeling and impact of the wide range of colors they found, painting en plein air .
Earlier artists usually painted a medium brown or rust color base on the canvas as a base or undercoat, building up to the actual colors. The Impressionists preferred a light-colored background, helping to strengthen the brighter paint colors to develop their light-saturated images. The favorite color of the Impressionists was white, a color mixed with others to create the bright light-reflective colors. They seldom used the darker earthy colors of previous painters, instead preferring the lighter side of the spectrum. The Impressionists also uniquely used complementary colors. Mixing complementary colors produces a dull mud color; however, they placed strokes of complementary colors next to each other, giving each color more intensity. The eye sees a single or local color and knows it is red or blue; however, when the local color is placed next to another color, the perception of the local color changes. That intensity was essential to achieve the luminous quality of their plein air paintings, capturing the influence of light from the sun, reflections on the water, or the movement of leaves at a point in time. They discovered the two-color were detectable up-close yet farther away; the combination looked as brilliant as the original color.
Color of Post-Impressionism
The Post-Impressionists moved from the luminescent quality of paint and light to a color palette of artificial colors and dynamic presentations. Their palette was based on saturated hues of a broad range of imaginative or unreal colors, stressing artificiality. They used intense and seemingly random colors as a symbolic definition of emotions. The vibrant colors were frequently expressed with geometric planes, distorted forms, and more diffused light. The Post-Impressionists were motivated by Asian art, especially the unique woodblock prints, and captured their essence by flattening the geometric planes in their work.
Although the Impressionists used visible brushstrokes to accentuate color changes and movement, the Post-Impressionist went further using broad strokes, sometimes flowing across the scene and other times thick and bold. The type of brushstroke helped create the emotion and feel of the image. They used color and line to depict the image and emotion by symbolic expressions, and although the result may appear distorted, the object had an expressive truth.
Landscapes in North America
Although landscapes had long been a subject of art in Asia, it was not an acceptable genre in the European or American tradition until the 1800s. The genre was ignored because landscape painting was not part of any classical or mythological stories and did not include any human anatomy. John Constable, the English painter, started painting the region around his house in the early 1800s, developing the standard for landscapes and an influence on the American landscape painters. As America began to develop westward, artists traveled with the explorers. The breathtaking vistas inspired them they found as they moved further westward, generating a new-style free from European influence.
The composition of the landscape painting was essential; the depth and distance of the horizon, the hazy mountains in the far background, the foreground including a detailed tree or large boulders with deep cracks. Artists also changed some of the elements and might add a small person to demonstrate a person's relationship to the grand, vast landscape. Their encounters led the artists to paint monumental images, light, and luminosity, an essential element of how the sun reflected off the water or created shadows in the trees. Instead, they could not make light; they camped in a location waiting to capture the same light. Whether the sun overhead, a storm on the horizon, or the deep shadows of the sun setting, the light was an important quality and required perseverance to obtain the perfect composition. Color has also developed relationships throughout the paintings. The color of the mountains-- a hazy purple or snowcapped gray-white, how the canyons presented erosion-- yellow-brown cliffs or green and rolling, colors of the sky, shadows, or trees; the color of the elements in the painting created the mood and unlimited vastness of the work. Landscape paintings were an important part of American history, the struggle to destroy nature, and the future of modern environmentalism.


5.2A: Tight Junctions
Tight junctions serve as selectively permeable seals in our body’s internal and external surfaces.
LEARNING OBJECTIVE
Describe the characteristics of tight junctions
Key Takeaways
Key Points
- Tight junctions are the closely associated areas of two cells whose membranes join together to form a virtually impermeable barrier to fluid.
- Tight junctions perform vital functions—such as holding cells together—and form protective and functional barriers.
- Tight junctions are composed of a branching network of sealing strands with each strand acting independently from the others.
- The major types of proteins in junctions are the claudins and the occludins.
- Each strand is formed from a row of transmembrane proteins embedded in both plasma membranes, with extracellular domains joining one another directly.
Key Terms
- blood-brain barrier : A structure in the central nervous system (CNS) that keeps various substances found in the bloodstream out of the brain while allowing in the substances essential to metabolic function, e.g., oxygen.
- Claudins : Proteins that form the backbone of the tight junction strands.
- cell adhesion molecule : Molecules that help cells stick to each other and to their surroundings. The proteins located on the cell surface bind with other cells or with the extracellular matrix (ECM).
- cytoskeleton : A cellular structure like a skeleton, contained within the cytoplasm.
- epithelia : The covering of internal and external body surfaces, where tight junctions are found.
- zonula occludens : Another name for tight junctions.
Tight Junction: An electron micrograph showing a tight junction in rat kidney tissue. The three dark lines of density correspond to the tight junction and the light lines in between correspond to the paracellular space.
Imagine a largely waterproof zipper connecting the sides of two different jackets. That zipper is like a tight junction (TJ), also called an occluding junction. A TJ creates a small zone that occludes the extracellular space (the space between cells).
This is why tight junctions are also called zonula occludens. The word zonula comes from words that mean small zone or encircling belt, while occludens comes from the Latin word occludere, which means to close up.
Location and Function
Tight junctions are virtually (but also partly selectively) impermeable seals that encircle cells and bind them together into leakproof sheets. In other words, the plasma membranes of adjacent cells essentially fuse together tightly in order to limit the leakage of various substances between the two cells.
What can and cannot go through all depends on the substance’s size, charge, as well as the location and precise composition of the tight junctions in the part of the body in question.
Tight junctions are located within our body’s epithelia. Epithelia is the plural of epithelium. Epithelium is a word that refers to the covering of the body’s internal and external surfaces. This includes organs (such as skin), blood vessels, and cavities.
Occludin: Model of the protein structure of the coiled-coil domain of human occludin.
Thus, these tight junctions serve various functions, depending on what epithelium is in question. In the skin, they keep us somewhat watertight and help keep allergens out of our body. In the digestive system, they help prevent the leakage of digestive enzymes into our bloodstream.
Tight junctions also serve as a structural support mechanism that help keep the epithelium together.
Composition
A tight junction—a kind of symmetrical cell junction—is composed of numerous important proteins that are either directly involved in its composition or intimately involved with connecting the tight junction to and between the cells in one way or another. These proteins include:
- Occludins, which maintain the barrier between adjacent cells.
- Claudins, which form the backbone of tight junction strands.
- Junctional adhesion molecules (JAMs) are immunoglobulin (antibody) proteins that help seal the intercellular space between two cells.
- Zonula occludens (ZO) are proteins that help link the tight junction to each cell’s internal skeleton (cytoskeleton).
The occludins and claudins are the major components of tight junction strands. When fully formed, a tight junction is not one, long, continuous seal. Instead, it looks like a series of local seals joined together in a maze-like fashion.
Tight junction: Diagram of tight junction components.


16.1: Introduction (2015)
Prince Hall was among the most noteworthy Citizens of Boston during the revolutionary period, where little was known of him beforehand. He was the founder of the African Lodge of Honorable Society of Free and Accepted Masons of Boston, the worlds first black Freemasonry and the first society in America dedicated on social, political and economic improvement.
Born in 1735 (although other sources say 1748) in Barbados, he was eventually brought to Boston as the Slave of William Hall. In 1756 his son Primus was born to another servant of a different household, Delia. In 1762 Hall joined the Congregational church and soon after married Sarah Ritchie, another enslaved woman. Following her death, which occurred eight years later, he married Flora Gibbs of Gloucester.
A month after the Boston Massacre William Hall freed Prince. He was no longer a slave but accounted as a free man where he made his living as a peddler, caterer and leather dresser. He was listed as a voter and taxpayer. He owned a house and a leather workshop in Boston.
He was thought to have conceivably taken part in the revolution fighting at Bunker Hill being one of the six Prince Hall’s listed on military records.
In 1775 Hall joined a British army lodge of Masons stationed in Boston. Accompanying him were 14 other free black men, and when the British departed they formed their own lodge, African Lodge Number 1. Twelve years later they became a permanent charter, Hall being its first Grand Master.
Hall used his position as “Worshipful Master” of the black Masons to speak out against slavery and the denial of black rights. He protested the lack of schools for black children until he finally established one of his own. His protest later became a basis for abolition of slavery and fighting for black rights.
Prince Hall died in 1807 at 72. A year after his death his lodge honored him by changing its name to Prince Hall Grand Lodge.
Works Cited
“People & Events Prince Hall.” PBS . PBS, n.d. Web. 02 Nov. 2015.
“Summary of Prince Hall, the Pioneer of Negro Masonry. Proofs of the Legitimacy of Prince Hall Masonry.” Summary of Prince Hall, the Pioneer of Negro Masonry. Proofs of the Legitimacy of Prince Hall Masonry . N.p., n.d. Web. 02 Nov. 2015.


8.E: Entropy and Free Energy (Exercises)
8.1: Spontaneity
What is a spontaneous reaction?
- Answer
-
A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.
Indicate whether the following processes are spontaneous or nonspontaneous.
- Liquid water freezing at a temperature below its freezing point
- Liquid water freezing at a temperature above its freezing point
- The combustion of gasoline
- A ball thrown into the air
- A raindrop falling to the ground
- Iron rusting in a moist atmosphere
- Answer
-
a. spontaneous
b. nonspontaneous
c. spontaneous
d. nonspontaneous
e. spontaneous
f. spontaneous
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
- Answer
-
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
8.2: Entropy
In the figure below all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, Δ S , for the system when it is converted from distribution to distribution (d).
- Answer
-
There are four initial microstates and four final microstates.
\[ΔS=k\ln\dfrac{W_\ce{f}}{W_\ce{i}}=\mathrm{1.38×10^{−23}\:J/K×\ln\dfrac{4}{4}=0}\]
Consider a system similar to the one below, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to \(\dfrac{1}{8}\). What does this comparison tell us about even larger systems?
- Answer
-
The probability for all the particles to be on one side is \(\dfrac{1}{32}\). This probability is noticeably lower than the \(\dfrac{1}{8}\) result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
At room temperature, the entropy of the halogens increases from I 2 to Br 2 to Cl 2 . Explain.
- Answer
-
The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I 2 is a solid, Br 2 is a liquid, and Cl 2 is a gas.
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
- C 2 H 5 OH( l ) or C 3 H 7 OH( l )
- C 2 H 5 OH( l ) or C 2 H 5 OH( g )
- 2Hor H( g )
- Answer
-
C 3 H 7 OH( l ) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. C 2 H 5 OH( g ) as it is in the gaseous state. 2H( g ), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.
- \(\ce{Pb^2+}(aq)+\ce{S^2-}(aq)⟶\ce{PbS}(s)\)
- \(\ce{2Fe}(s)+\ce{3O2}(g)⟶\ce{Fe2O3}(s)\)
- \(\ce{2C6H14}(l)+\ce{19O2}(g)⟶\ce{14H2O}(g)+\ce{12CO2}(g)\)
- Answer
-
Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. Negative. There is a net loss of three moles of gas from reactants to products. Positive. There is a net increase of seven moles of gas from reactants to products.
Write the balanced chemical equation for the combustion of benzene, C 6 H 6 ( l ), to give carbon dioxide and water vapor. Would you expect Δ S to be positive or negative in this process?
- Answer
-
\[\ce{C6H6}(l)+7.5\ce{O2}(g)⟶\ce{3H2O}(g)+\ce{6CO2}(g)\]
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and Δ S is positive.
8.3: The Second and Third law
Calculate \(ΔS^\circ_{298}\) for the following changes.
- \(\ce{SnCl4}(l)⟶\ce{SnCl4}(g)\)
- \(\ce{CS2}(g)⟶\ce{CS2}(l)\)
- \(\ce{Cu}(s)⟶\ce{Cu}(g)\)
- \(\ce{H2O}(l)⟶\ce{H2O}(g)\)
- \(\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(l)\)
- \(\ce{2HCl}(g)+\ce{Pb}(s)⟶\ce{PbCl2}(s)+\ce{H2}(g)\)
- \(\ce{Zn}(s)+\ce{CuSO4}(s)⟶\ce{Cu}(s)+\ce{ZnSO4}(s)\)
- Answer
-
107 J/K; −86.4 J/K; 133.2 J/K; 118.8 J/K; −326.6 J/K; −171.9 J/K; (g) −7.2 J/K
Determine the entropy change for the combustion of gaseous propane, C 3 H 8 , under standard state conditions to give gaseous carbon dioxide and water.
- Answer
-
100.6 J/K
Using the relevant \(S^\circ_{298}\) values listed in Resource Table 2, calculate \(S^\circ_{298}\) for the following changes:
- \(\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)\)
- \(\ce{N2}(g)+\dfrac{5}{2}\ce{O2}(g)⟶\ce{N2O5}(g)\)
- Answer
-
−198.1 J/K; −348.9 J/K
By calculating Δ S univ at each temperature, determine if the melting of 1 mole of NaCl( s ) is spontaneous at 500 °C and at 700 °C.
\[S^\circ_{\ce{NaCl}(s)}=\mathrm{72.11\:\dfrac{J}{mol⋅K}}\hspace{40px} S^\circ_{\ce{NaCl}(l)}=\mathrm{95.06\:\dfrac{J}{mol⋅K}}\hspace{40px ΔH^\circ_\ce{fusion}=\mathrm{27.95\: kJ/mol}\]
\[S_{\mathrm{NaCl}(s)}^{\circ}=72.11 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} \quad S_{\mathrm{NaCl}(l)}^{\circ}=95.06 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} \quad \Delta H_{\text {fusion }}^{\circ}=27.95 \mathrm{~kJ} / \mathrm{mol}\]
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
- Answer
-
As Δ S univ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
Use the standard entropy data in Resource Table T1 to determine the change in entropy for each of the reactions listed. All are run under standard state conditions and 25 °C.
a. \[\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)\]
b. \[\mathrm{O}_2(g)+\mathrm{N}_2(g) \longrightarrow 2 \mathrm{NO}(g)\]
c. \[2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_2 \mathrm{~S}(s)\]
d. \[\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_2(s)\]
e. \[\mathrm{Fe}_2 \mathrm{O}_3(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_2(g)\]
f. \[\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}(s) \longrightarrow \mathrm{CaSO}_4(s)+2 \mathrm{H}_2 \mathrm{O}(g)\]
- Answer
-
(a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K
Q16.3.0
What is the difference between Δ S , Δ S °, and \(ΔS^\circ_{298}\) for a chemical change?
Q16.3.1
Calculate \(ΔS^\circ_{298}\) for the following changes.
- \(\ce{SnCl4}(l)⟶\ce{SnCl4}(g)\)
- \(\ce{CS2}(g)⟶\ce{CS2}(l)\)
- \(\ce{Cu}(s)⟶\ce{Cu}(g)\)
- \(\ce{H2O}(l)⟶\ce{H2O}(g)\)
- \(\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(l)\)
- \(\ce{2HCl}(g)+\ce{Pb}(s)⟶\ce{PbCl2}(s)+\ce{H2}(g)\)
- \(\ce{Zn}(s)+\ce{CuSO4}(s)⟶\ce{Cu}(s)+\ce{ZnSO4}(s)\)
S16.3.1
107 J/K; −86.4 J/K; 133.2 J/K; 118.8 J/K; −326.6 J/K; −171.9 J/K; (g) −7.2 J/K
Q16.3.2
Determine the entropy change for the combustion of liquid ethanol, C 2 H 5 OH, under standard state conditions to give gaseous carbon dioxide and liquid water.
Q16.3.3
Determine the entropy change for the combustion of gaseous propane, C 3 H 8 , under standard state conditions to give gaseous carbon dioxide and water.
S16.3.3
100.6 J/K
Q16.3.4
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is:
\[\ce{Fe2O3}(s)+\ce{2Al}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)\]
Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
Q16.3.5
Using the relevant \(S^\circ_{298}\) values listed in Appendix G , calculate \(S^\circ_{298}\) for the following changes:
- \(\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)\)
- \(\ce{N2}(g)+\dfrac{5}{2}\ce{O2}(g)⟶\ce{N2O5}(g)\)
S16.3.5
−198.1 J/K; −348.9 J/K
Q16.3.6
From the following information, determine \(ΔS^\circ_{298}\) for the following:
- \(\ce{N}(g)+\ce{O}(g)⟶\ce{NO}(g) \hspace{20px} ΔS^\circ_{298}=\,?\)
- \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{24.8\: J/K}\)
- \(\ce{N2}(g)⟶\ce{2N}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{115.0\: J/K}\)
- \(\ce{O2}(g)⟶\ce{2O}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{117.0\: J/K}\)
Q16.3.7
By calculating Δ S univ at each temperature, determine if the melting of 1 mole of NaCl( s ) is spontaneous at 500 °C and at 700 °C.
\[S^\circ_{\ce{NaCl}(s)}=\mathrm{72.11\:\dfrac{J}{mol⋅K}}\hspace{40px} S^\circ_{\ce{NaCl}(l)}=\mathrm{95.06\:\dfrac{J}{mol⋅K}}\hspace{40px ΔH^\circ_\ce{fusion}=\mathrm{27.95\: kJ/mol}\]
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
S16.3.7
As Δ S univ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
Q16.3.8
Use the standard entropy data in Appendix G to determine the change in entropy for each of the reactions listed in [link] . All are run under standard state conditions and 25 °C.
Q16.3.8
2.86 J/K; 24.8 J/K; −113.2 J/K; −24.7 J/K; 15.5 J/K; 290.0 J/K
16.4: Free Energy
Q16.4.1
What is the difference between Δ G , Δ G °, and \(ΔG^\circ_{298}\) for a chemical change?
Q16.4.2
A reactions has \(ΔH^\circ_{298}\) = 100 kJ/mol and \(ΔS^\circ_{298}=\textrm{250 J/mol⋅K}\). Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
S16.4.2
The reaction is nonspontaneous at room temperature. Above 400 K, Δ G will become negative, and the reaction will become spontaneous.
Q16.4.3
Explain what happens as a reaction starts with Δ G < 0 (negative) and reaches the point where Δ G = 0.
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
- \(\ce{MnO2}(s)⟶\ce{Mn}(s)+\ce{O2}(g)\)
- \(\ce{H2}(g)+\ce{Br2}(l)⟶\ce{2HBr}(g)\)
- \(\ce{Cu}(s)+\ce{S}(g)⟶\ce{CuS}(s)\)
- \(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)\)
- \(\ce{CH4}(g)+\ce{O2}(g)⟶\ce{C}(s,\,\ce{graphite})+\ce{2H2O}(g)\)
- \(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g)\)
S16.4.3
465.1 kJ nonspontaneous; −106.86 kJ spontaneous; −53.6 kJ spontaneous; −83.4 kJ spontaneous; −406.7 kJ spontaneous; −30.0 kJ spontaneous
Q16.4.4
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
- \(\ce{C}(s,\, \ce{graphite})+\ce{O2}(g)⟶\ce{CO2}(g)\)
- \(\ce{O2}(g)+\ce{N2}(g)⟶\ce{2NO}(g)\)
- \(\ce{2Cu}(s)+\ce{S}(g)⟶\ce{Cu2S}(s)\)
- \(\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)\)
- \(\ce{Fe2O3}(s)+\ce{3CO}(g)⟶\ce{2Fe}(s)+\ce{3CO2}(g)\)
- \(\ce{CaSO4⋅2H2O}(s)⟶\ce{CaSO4}(s)+\ce{2H2O}(g)\)
Given:
\[\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s) \hspace{20px} ΔG^\circ_{298}=\mathrm{−2697.0\: kJ/mol}\]
\[\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{−457.18\: kJ/mol}\]
\[\ce{6H2O}(g)+\ce{P4O10}(g)⟶\ce{4H3PO4}(l) \hspace{20px} ΔG^\circ_{298}=\mathrm{−428.66\: kJ/mol}\]
Q16.4.5
- Determine the standard free energy of formation, \(ΔG^\circ_\ce{f}\), for phosphoric acid.
- How does your calculated result compare to the value in Appendix G ? Explain.
S16.4.5
−1124.3 kJ/mol for the standard free energy change. The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.
Q16.4.6
Is the formation of ozone (O 3 ( g )) from oxygen (O 2 ( g )) spontaneous at room temperature under standard state conditions?
Q16.4.7
Consider the decomposition of red mercury(II) oxide under standard state conditions.
\[\ce{2HgO}(s,\,\ce{red})⟶\ce{2Hg}(l)+\ce{O2}(g)\]
- Is the decomposition spontaneous under standard state conditions?
- Above what temperature does the reaction become spontaneous?
S16.4.7
The reaction is nonspontaneous; Above 566 °C the process is spontaneous.
Q16.4.8
Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
- Ammonia: \(\ce{2NH3}(g)⟶\ce{N2}(g)+\ce{3H2}(g)\)
- Diborane: \(\ce{B2H6}(g)⟶\ce{2B}(g)+\ce{3H2}(g)\)
- Hydrazine: \(\ce{N2H4}(g)⟶\ce{N2}(g)+\ce{2H2}(g)\)
- Hydrogen peroxide: \(\ce{H2O2}(l)⟶\ce{H2O}(g)+\dfrac{1}{2}\ce{O2}(g)\)
Q16.4.9
Calculate Δ G ° for each of the following reactions from the equilibrium constant at the temperature given.
- \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} \mathrm{T=2000\:°C} \hspace{20px} K_p=4.1×10^{−4}\)
- \(\ce{H2}(g)+\ce{I2}(g)⟶\ce{2HI}(g) \hspace{20px} \mathrm{T=400\:°C} \hspace{20px} K_p=50.0\)
- \(\ce{CO2}(g)+\ce{H2}(g)⟶\ce{CO}(g)+\ce{H2O}(g) \hspace{20px} \mathrm{T=980\:°C} \hspace{20px} K_p=1.67\)
- \(\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g) \hspace{20px} \mathrm{T=900\:°C} \hspace{20px} K_p=1.04\)
- \(\ce{HF}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{F-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=7.2×10^{−4}\)
- \(\ce{AgBr}(s)⟶\ce{Ag+}(aq)+\ce{Br-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=3.3×10^{−13}\)
S16.4.9
1.5 × 10 2 kJ; −21.9 kJ; −5.34 kJ; −0.383 kJ; 18 kJ; 71 kJ
Q16.4.10
Calculate Δ G ° for each of the following reactions from the equilibrium constant at the temperature given.
- \(\ce{Cl2}(g)+\ce{Br2}(g)⟶\ce{2BrCl}(g) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.7×10^{−2}\)
- \(\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} \mathrm{T=500\:°C} \hspace{20px} K_p=48.2\)
- \(\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} \mathrm{T=60\:°C} \hspace{20px} K_p=\mathrm{0.196\: atm}\)
- \(\ce{CoO}(s)+\ce{CO}(g)⇌\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} \mathrm{T=550\:°C} \hspace{20px} K_p=4.90×10^2\)
- \(\ce{CH3NH2}(aq)+\ce{H2O}(l)⟶\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.4×10^{−4}\)
- \(\ce{PbI2}(s)⟶\ce{Pb^2+}(aq)+\ce{2I-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=8.7×10^{−9}\)
Q16.4.11
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ G ° given.
- \(\ce{O2}(g)+\ce{2F2}(g)⟶\ce{2OF2}(g) \hspace{20px} ΔG°=\mathrm{−9.2\: kJ}\)
- \(\ce{I2}(s)+\ce{Br2}(l)⟶\ce{2IBr}(g) \hspace{20px} ΔG°=\mathrm{7.3\: kJ}\)
- \(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} ΔG°=\mathrm{−79\: kJ}\)
- \(\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g) \hspace{20px} ΔG°=\mathrm{−1.6\: kJ}\)
- \(\ce{SnCl4}(l)⟶\ce{SnCl4}(l) \hspace{20px} ΔG°=\mathrm{8.0\: kJ}\)
S16.4.11
K = 41; K = 0.053; K = 6.9 × 10 13 ; K = 1.9; K = 0.04
Q16.4.2
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ G ° given.
- \(\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} ΔG°=\mathrm{−10.88\: kJ}\)
- \(\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} ΔG°=\mathrm{3.4\: kJ}\)
- \(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} ΔG°=\mathrm{−39\: kJ}\)
- \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} ΔG°=\mathrm{−141.82\: kJ}\)
- \(\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} ΔG°=\mathrm{−1.88\: kJ}\)
Q16.4.13
Calculate the equilibrium constant at the temperature given.
- (a) \(\ce{O2}(g)+\ce{2F2}(g)⟶\ce{2F2O}(g) \hspace{20px} \mathrm{(T=100\:°C)}\)
- \(\ce{I2}(s)+\ce{Br2}(l)⟶\ce{2IBr}(g) \hspace{20px} \mathrm{(T=0.0\:°C)}\)
- \(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} \mathrm{(T=575\:°C)}\)
- \(\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g) \hspace{20px} \mathrm{(T=−10.0\:°C)}\)
- \(\ce{SnCl4}(l)⟶\ce{SnCl4}(g) \hspace{20px} \mathrm{(T=200\:°C)}\)
S16.4.13
In each of the following, the value of Δ G is not given at the temperature of the reaction. Therefore, we must calculate Δ G ° from the values Δ H ° and Δ S ° and then calculate Δ G from the relation ΔG° = ΔH° − TΔS°.
- K = 1.29
- K = 2.51 × 10 −3
- K = 4.83 × 10 3
- K = 0.219
- K = 16.1
Q16.4.14
Calculate the equilibrium constant at the temperature given.
- (a) \(\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} \mathrm{(T=100\:°C)}\)
- \(\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} \mathrm{(T=0.0\:°C)}\)
- \(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} \mathrm{(T=125\:°C)}\)
- \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} \mathrm{(T=675\:°C)}\)
- \(\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} \mathrm{(T=90\:°C)}\)
Q16.4.15
Consider the following reaction at 298 K:
\[\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_P=0.142\]
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
S16.4.16
The standard free energy change is \(ΔG^\circ_{298}=−RT\ln K=\mathrm{4.84\: kJ/mol}\). When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and \(ΔG_{298}\) becomes less positive as it approaches zero. At equilibrium, Q = K , and Δ G = 0.
Q16.4.17
Determine the normal boiling point (in kelvin) of dichloroethane, CH 2 Cl 2 . Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
Q16.4.18
Under what conditions is \(\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g)\) spontaneous?
S16.4.18
The reaction will be spontaneous at temperatures greater than 287 K.
Q16.4.19
At room temperature, the equilibrium constant ( K w ) for the self-ionization of water is 1.00 × 10 −14 . Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
Q16.4.20
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction \(\ce{2H2S}(g)+\ce{SO2}(g)⇌\dfrac{3}{8}\ce{S8}(s,\,\ce{rhombic})+\ce{2H2O}(l)\). What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
S16.4.20
K = 5.35 × 10 15
The process is exothermic.
Q16.4.21
Consider the decomposition of CaCO 3 ( s ) into CaO( s ) and CO 2 ( g ). What is the equilibrium partial pressure of CO 2 at room temperature?
Q16.4.22
In the laboratory, hydrogen chloride (HCl( g )) and ammonia (NH 3 ( g )) often escape from bottles of their solutions and react to form the ammonium chloride (NH 4 Cl( s )), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH 3 in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
S16.4.22
1.0 × 10 −8 atm. This is the maximum pressure of the gases under the stated conditions.
Q16.4.23
Benzene can be prepared from acetylene. \(\ce{3C2H2}(g)⇌\ce{C6H6}(g)\). Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
Q16.4.24
Carbon dioxide decomposes into CO and O 2 at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO 2 was 1.15 atm?
\[x=\mathrm{1.29×10^{−5}\:atm}=P_{\ce{O2}}\]
Q16.4.25
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
\[\ce{CH4}(g)+\ce{4Cl2}(g)⟶\ce{CCl4}(g)+\ce{4HCl}(g)\]
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
Q16.4.25B
Acetic acid, CH 3 CO 2 H, can form a dimer, (CH 3 CO 2 H) 2 , in the gas phase.
\[\ce{2CH3CO2H}(g)⟶\ce{(CH3CO2H)2}(g)\]
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
At 25 °C, the equilibrium constant for the dimerization is 1.3 × 10 3 (pressure in atm). What is Δ S ° for the reaction?
S16.4.25B
−0.16 kJ
Q16.4.26
Nitric acid, HNO 3 , can be prepared by the following sequence of reactions:
\[\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g)\]
\[\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\]
\[\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(l)+\ce{NO}(g)\]
How much heat is evolved when 1 mol of NH 3 ( g ) is converted to HNO 3 ( l )? Assume standard states at 25 °C.
Q16.4.27A
Determine Δ G for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
\[\ce{SbCl5}(g)⟶\ce{SbCl3}(g)+\ce{Cl2}(g)\]
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl 5 , 9.14 g of SbCl 3 , and 2.84 g of Cl 2 .
Chlorine molecules dissociate according to this reaction:
\[\ce{Cl2}(g)⟶\ce{2Cl}(g)\]
1.00% of Cl 2 molecules dissociate at 975 K and a pressure of 1.00 atm.
S16.4.27A
- (a) −22.1 kJ;
- 61.6 kJ/mol
Q16.4.27
Given that the \(ΔG^\circ_\ce{f}\) for Pb 2 + ( aq ) and Cl − ( aq ) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K sp , for PbCl 2 ( s ).
Q16.4.28
Determine the standard free energy change, \(ΔG^\circ_\ce{f}\), for the formation of S 2− ( aq ) given that the \(ΔG^\circ_\ce{f}\) for Ag + ( aq ) and Ag 2 S( s ) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag 2 S( s ) is 8 × 10 −51 .
S16.4.28
90 kJ/mol
Q16.4.29
Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
Q16.4.30
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
\[\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{8.58\: kJ}\]
- (a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
- Determine the equilibrium constant, K P , for this physical process.
- By calculating ∆ G , determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, \(P_{\ce{H2O}}\), is 0.011 atm.
- If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of \(P_{\ce{H2O}}\) in the air?
S16.4.30
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; K p = 0.031; The evaporation of water is spontaneous; \(P_{\ce{H2O}}\) must always be less than K p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
Q16.4.31
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
\[\mathrm{Glu + ATP ⟶ G6P + ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−17\: kJ}\]
In this process, ATP becomes ADP summarized by the following equation:
\[\mathrm{ATP⟶ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−30\: kJ}\]
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
\[\mathrm{Glu⟶G6P} \hspace{20px} ΔG^\circ_{298}=\:?\]
Q16.4.32
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
\[\mathrm{G6P⇌F6P} \hspace{20px} ΔG^\circ_{298}=\mathrm{1.7\: kJ}\]
- (a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
- Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M , however, in a typical cell, they are not even close to these values. Calculate Δ G when the concentrations of G6P and F6P are 120 μ M and 28 μ M respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
S16.4.32
(a) Nonspontaneous as \(ΔG^\circ_{298}>0\); \(ΔG^\circ_{298}=−RT\ln K,\) \(ΔG = 1.7×10^3 + \left(8.314 × 335 × \ln\dfrac{28}{128}\right) = \mathrm{−2.5\: kJ}\). The forward reaction to produce F6P is spontaneous under these conditions.
Q16.4.33
Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
- (a) \(\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)\)
- \(\ce{HCl}(g)+\ce{NH3}(g)⟶\ce{NH4Cl}(s)\)
- \(\ce{(NH4)2Cr2O7}(s)⟶\ce{Cr2O3}(s)+\ce{4H2O}(g)+\ce{N2}(g)\)
- \(\ce{2Fe}(s)+\ce{3O2}(g)⟶\ce{Fe2O3}(s)\)
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of Δ G , Δ H , and Δ S for this process, and justify your choices.
S16.4.33
Δ G is negative as the process is spontaneous. Δ H is positive as with the solution becoming cold, the dissolving must be endothermic. Δ S must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
Q16.4.34
An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu 2 S decomposes to form copper and sulfur described by the following equation:
\[\ce{Cu2S}(s)⟶\ce{Cu}(s)+\ce{S}(s)\]
- (a) Determine \(ΔG^\circ_{298}\) for the decomposition of Cu 2 S( s ).
- The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine \(ΔG^\circ_{298}\) for the process.
- The production of copper from chalcocite is performed by roasting the Cu 2 S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
Q16.4.35
What happens to \(ΔG^\circ_{298}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
- (a) \(\ce{S}(s)+\ce{O2}(g)⟶\ce{SO2}(g)\)
- \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{SO3}(g)\)
- \(\ce{HgO}(s)⟶\ce{Hg}(l)+\ce{O2}(g)\)
S16.4.35
- (a) Increasing \(P_{\ce{O2}}\) will shift the equilibrium toward the products, which increases the value of K . \(ΔG^\circ_{298}\) therefore becomes more negative.
- Increasing \(P_{\ce{O2}}\) will shift the equilibrium toward the products, which increases the value of K . \(ΔG^\circ_{298}\) therefore becomes more negative.
- Increasing \(P_{\ce{O2}}\) will shift the equilibrium the reactants, which decreases the value of K . \(ΔG^\circ_{298}\) therefore becomes more positive.