1421. Solution -
Combustion processes are exothermic ( $\Delta H<0$ ). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, $\Delta S>0$ ). The reaction is therefore spontaneous $(\Delta G<0)$ at all temperatures.


1422. Check Your Learning -
Popular chemical hand warmers generate heat by the air-oxidation of iron:

$$
4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)
$$

How does the spontaneity of this process depend upon temperature?


1423. Answer: -
$\Delta H$ and $\Delta S$ are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms "high" and "low" mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in "spontaneity" (as reflected by its $\Delta G$ ) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which $\Delta G$ is plotted on the $y$ axis versus $T$ on the $x$ axis:

$$
\begin{gathered}
\Delta G=\Delta H-T \Delta S \\
y=b+m x
\end{gathered}
$$

Such a plot is shown in Figure 13.9. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative $\Delta G$ ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the $x$-intercept of the line, that is, the value of $T$ for which $\Delta G$ is zero:

$$
\Delta G=0=\Delta H-T \Delta S
$$

$$
T=\frac{\Delta H}{\Delta S}
$$

So, saying a process is spontaneous at "high" or "low" temperatures means the temperature is above or below, respectively, that temperature at which $\Delta G$ for the process is zero. As noted earlier, the condition of $\Delta \mathrm{G}=0$ describes a system at equilibrium.

FIGURE 13.9 These plots show the variation in $\Delta G$ with temperature for the four possible combinations of arithmetic sign for $\Delta H$ and $\Delta S$.


1425. Equilibrium Temperature for a Phase Transition -
As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.


1426. Solution -
The process of interest is the following phase change:

$$
\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)
$$

When this process is at equilibrium, $\Delta G=0$, so the following is true:

$$
0=\Delta H^{\circ}-T \Delta S^{\circ} \quad \text { or } \quad T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}
$$

Using the standard thermodynamic data from Appendix G,


1426. Solution -
Using the standard thermodynamic data from Appendix G,

$$
\begin{aligned}
\Delta H^{\circ} & =1 \mathrm{~mol} \times \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(g)\right)-1 \mathrm{~mol} \times \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(l)\right) \\
& =(1 \mathrm{~mol})-241.82 \mathrm{~kJ} / \mathrm{mol}-(1 \mathrm{~mol})(-241.82 \mathrm{~kJ} / \mathrm{mol})=44.01 \mathrm{~kJ} \\
\Delta S^{\circ} & =1 \mathrm{~mol} \times \Delta S^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(g)\right)-1 \mathrm{~mol} \times \Delta S^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(l)\right) \\
& =(1 \mathrm{~mol}) 188.8 \mathrm{~J} / \mathrm{K} \cdot \mathrm{~mol}-(1 \mathrm{~mol}) 70.0 \mathrm{~J} / \mathrm{K} \cdot \mathrm{~mol}=118.8 \mathrm{~J} / \mathrm{K} \\
T & =\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{44.01 \times 10^{3} \mathrm{~J}}{118.8 \mathrm{~J} / \mathrm{K}}=370.5 \mathrm{~K}=97.3^{\circ} \mathrm{C}
\end{aligned}
$$


1426. Solution -
The accepted value for water's normal boiling point is $373.2 \mathrm{~K}\left(100.0^{\circ} \mathrm{C}\right)$, and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.


1427. Check Your Learning -
Use the information in Appendix G to estimate the boiling point of $\mathrm{CS}_{2}$.


1428. Answer: -
313 K (accepted value 319 K )


1429. Free Energy and Equilibrium -
The free energy change for a process may be viewed as a measure of its driving force. A negative value for $\Delta G$ represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When $\Delta G$ is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).

In the section on equilibrium, the reaction quotient, $Q$, was introduced as a convenient measure of the status of an equilibrium system. Recall that $Q$ is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When $Q$ is lesser than the equilibrium constant, $K$, the reaction will proceed in the forward direction until equilibrium is reached and $Q=K$. Conversely, if $Q>K$, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M ) is related to the standard free energy change, according to this equation:

$$
\Delta G=\Delta G^{\circ}+R T \ln Q
$$

$R$ is the gas constant ( $8.314 \mathrm{~J} / \mathrm{K} \mathrm{mol}$ ), $T$ is the kelvin or absolute temperature, and $Q$ is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.12.


1431. Calculating $\boldsymbol{\Delta} \boldsymbol{G}$ under Nonstandard Conditions -
What is the free energy change for the process shown here under the specified conditions?

$$
\begin{aligned}
T=25^{\circ} \mathrm{C}, P_{\mathrm{N}_{2}}=0.870 \mathrm{~atm}, P_{\mathrm{H}_{2}} & =0.250 \mathrm{~atm}, \text { and } P_{\mathrm{NH}_{3}}=12.9 \mathrm{~atm} \\
2 \mathrm{NH}_{3}(g) & \longrightarrow 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \quad \Delta G^{\circ}=33.0 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$


1432. Solution -
The equation relating free energy change to standard free energy change and reaction quotient may be used directly:

$$
\Delta G=\Delta G^{\circ}+R T \ln Q=33.0 \frac{\mathrm{~kJ}}{\mathrm{~mol}}+\left(8.314 \frac{\mathrm{~J}}{\mathrm{~mol} \mathrm{~K}} \times 298 \mathrm{~K} \times \ln \frac{\left(0.250^{3}\right) \times 0.870}{12.9^{2}}\right)=9680 \frac{\mathrm{~J}}{\mathrm{~mol}} \text { or } 9.68 \mathrm{~kJ} / \mathrm{mol}
$$

Since the computed value for $\Delta G$ is positive, the reaction is nonspontaneous under these conditions.


1433. Check Your Learning -
Calculate the free energy change for this same reaction at $875^{\circ} \mathrm{C}$ in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?


1434. Answer: -
$\Delta G=-47 \mathrm{~kJ} / \mathrm{mol}$; yes

For a system at equilibrium, $Q=K$ and $\Delta G=0$, and the previous equation may be written as

$$
\begin{array}{rlrr}
0 & =\Delta G^{\circ}+R T \ln K & & \text { (at equilibrium) } \\
\Delta G^{\circ} & =-R T \ln K & \text { or } & K=e^{-\frac{\Delta G^{\circ}}{R T}}
\end{array}
$$

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1.

Relations between Standard Free Energy Changes and Equilibrium Constants

| $\boldsymbol{K}$ | $\Delta \boldsymbol{G}^{\circ}$ | Composition of an Equilibrium Mixture |
| :--- | :--- | :--- |
| $>1$ | $<0$ | Products are more abundant |
| $<1$ | $>0$ | Reactants are more abundant |
| $=1$ | $=0$ | Reactants and products are comparably abundant |

TABLE 13.1


1436. Calculating an Equilibrium Constant using Standard Free Energy Change -
Given that the standard free energies of formation of $\mathrm{Ag}^{+}(\mathrm{aq}), \mathrm{Cl}^{-}(\mathrm{aq})$, and $\mathrm{AgCl}(s)$ are $77.1 \mathrm{~kJ} / \mathrm{mol},-131.2 \mathrm{~kJ} /$ mol , and $-109.8 \mathrm{~kJ} / \mathrm{mol}$, respectively, calculate the solubility product, $K_{\mathrm{sp}}$, for AgCl .


1437. Solution -
The reaction of interest is the following:

$$
\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]
$$

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

$$
\begin{aligned}
& \Delta G^{\circ}=\left[\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}(a q)\right)+\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}(a q)\right)\right]-\left[\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl}(s))\right] \\
& =[77.1 \mathrm{~kJ} / \mathrm{mol}-131.2 \mathrm{~kJ} / \mathrm{mol}]-[-109.8 \mathrm{~kJ} / \mathrm{mol}]=55.7 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

The equilibrium constant for the reaction may then be derived from its standard free energy change:

$$
K_{\mathrm{sp}}=e^{-\frac{\Delta G^{\circ}}{R T}}=\exp \left(-\frac{\Delta G^{\circ}}{R T}\right)=\exp \left(-\frac{55.7 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K} \times 298.15 \mathrm{~K}}\right)=\exp (-22.470)=e^{-22.470}=1.74 \times 10^{-10}
$$

This result is in reasonable agreement with the value provided in Appendix J.


1438. Check Your Learning -
Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at $25^{\circ} \mathrm{C}$.

$$
\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)
$$


1439. Answer: -
$K=0.32$

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of $Q$ ), equilibrium is established when the system's free energy is minimized (Figure 13.10). If a system consists of reactants and products in nonequilibrium amounts $(Q \neq K)$, the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

FIGURE 13.10 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.


1440. Key Terms -
equilibrium state of a reversible reaction in which the forward and reverse processes occur at equal rates
equilibrium constant ( $\boldsymbol{K}$ ) value of the reaction quotient for a system at equilibrium; may be expressed using concentrations ( $K_{c}$ ) or partial pressures ( $K_{p}$ )
heterogeneous equilibria equilibria in which reactants and products occupy two or more different phases
homogeneous equilibria equilibria in which all reactants and products occupy the same phase
law of mass action when a reversible reaction has
attained equilibrium at a given temperature, the reaction quotient remains constant
Le Châtelier's principle an equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium
reaction quotient $(Q)$ mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations $\left(Q_{c}\right)$ or pressures $\left(Q_{p}\right)$
reversible reaction chemical reaction that can proceed in both the forward and reverse directions under given conditions


1441. Key Equations -
$$
\begin{array}{lr}
Q_{c}=\frac{[\mathrm{C}]^{x}[\mathrm{D}]^{y}}{[\mathrm{~A}]^{m}[\mathrm{~B}]^{n}} & \text { for the reaction } m \mathrm{~A}+n \mathrm{~B} \rightleftharpoons x \mathrm{C}+y \mathrm{D} \\
Q_{P}=\frac{\left(P_{C}\right)^{x}\left(P_{D}\right)^{y}}{\left(P_{A}\right)^{m}\left(P_{B}\right)^{n}} & \text { for the reaction } m \mathrm{~A}+n \mathrm{~B} \rightleftharpoons x \mathrm{C}+y \mathrm{D} \\
P=M R T & \\
K_{c}=Q_{c} \text { at equilibrium } & \\
K_{p}=Q_{p} \text { at equilibrium } & \\
K_{P}=K_{c}(R T)^{\Delta n} &
\end{array}
$$


1442. Summary - 1442.1. Chemical Equilibria
A reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species.


1442. Summary - 1442.2. Equilibrium Constants
The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, $Q$. For a reaction at equilibrium, the composition is constant, and $Q$ is called the equilibrium constant, $K$.

A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.


1442. Summary - 1442.3. Shifting Equilibria: Le Châtelier's Principle
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system's response to
these disturbances is described by Le Châtelier's principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture's composition ( $K$ does not change).


1442. Summary - 1442.4. Equilibrium Calculations
Calculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. We also learned that a negative value for $\Delta G$ indicates a spontaneous process; a positive $\Delta G$ indicates a nonspontaneous process; and a $\Delta G$ of zero indicates that the system is at equilibrium. We also saw how free energy, spontaneity, and equilibrium relate.


1443. Exercises - 1443.1. Chemical Equilibria
1. What does it mean to describe a reaction as "reversible"?
2. When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
3. If a reaction is reversible, when can it be said to have reached equilibrium?
4. Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
5. If the concentrations of products and reactants are equal, is the system at equilibrium?


1443. Exercises - 1443.2. Equilibrium Constants
6. Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
7. Explain why an equilibrium between $\mathrm{Br}_{2}(I)$ and $\mathrm{Br}_{2}(g)$ would not be established if the container were not a closed vessel shown in Figure 13.4.
8. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure $\mathrm{NO}_{2}$ or with pure $\mathrm{N}_{2} \mathrm{O}_{4}$ ? $2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$
9. Among the solubility rules previously discussed is the statement: All chlorides are soluble except $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$, $\mathrm{AgCl}, \mathrm{PbCl}_{2}$, and CuCl .
(a) Write the expression for the equilibrium constant for the reaction represented by the equation
$\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)$. Is $K_{c}>1,<1$, or $\approx 1$ ? Explain your answer.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $\mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{PbCl}_{2}(s)$. Is $K_{c}>1$, <1, or $\approx 1$ ? Explain your answer.
10. Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates-except those of the ammonium ion and the alkali metals-are insoluble.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation


1443. Exercises - 1443.2. Equilibrium Constants
(a) Write the expression for the equilibrium constant for the reaction represented by the equation
$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$. Is $K_{c}>1,<1$, or $\approx 1$ ? Explain your answer.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}{ }^{3-}(a q) \rightleftharpoons \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)$. Is $K_{c}>1,<1$, or $\approx 1$ ? Explain your answer.
11. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$. Which value of $K_{c}$ would make this reaction most useful commercially? $K_{c} \approx 0.01, K_{c} \approx 1$, or $K_{c} \approx 10$. Explain your answer.
12. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $\mathrm{KI}(a q)+\mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{KI}_{3}(a q)$ give the same expression for the reaction quotient. $\mathrm{KI}_{3}$ is composed of the ions $\mathrm{K}^{+}$and $\mathrm{I}_{3}{ }^{-}$.
13. For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be $100 \%$. Is $K_{c}>1,<1$, or $\approx 1$ for a titration reaction?


1443. Exercises - 1443.2. Equilibrium Constants
14. For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is $K_{c}>1,<1$, or $\approx 1$ for a useful precipitation reaction?
15. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:
(a) $\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)$
(b) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
(c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
(d) $\mathrm{BaSO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{SO}_{2}(g)$
(e) $\mathrm{P}_{4}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(s)$
(f) $\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$
(g) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$
(h) $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)$
16. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:
(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$


1443. Exercises - 1443.2. Equilibrium Constants
(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
(b) $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
(c) $\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$
(d) $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
(e) $\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$
(f) $2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{PbO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$
(g) $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)$
(h) $\mathrm{S}_{8}(g) \rightleftharpoons 8 \mathrm{~S}(g)$
17. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
(a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
$K_{c}=17 ;\left[\mathrm{NH}_{3}\right]=0.20 M,\left[\mathrm{~N}_{2}\right]=1.00 M,\left[\mathrm{H}_{2}\right]=1.00 M$


1443. Exercises - 1443.2. Equilibrium Constants
(b) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
$K_{P}=6.8 \times 10^{4} ; \mathrm{NH}_{3}=3.0 \mathrm{~atm}, \mathrm{~N}_{2}=2.0 \mathrm{~atm}, \mathrm{H}_{2}=1.0 \mathrm{~atm}$
(c) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$
$K_{c}=0.230 ;\left[\mathrm{SO}_{3}\right]=0.00 \mathrm{M},\left[\mathrm{SO}_{2}\right]=1.00 \mathrm{M},\left[\mathrm{O}_{2}\right]=1.00 \mathrm{M}$
(d) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$ $K_{P}=16.5 ; \mathrm{SO}_{3}=1.00 \mathrm{~atm}, \mathrm{SO}_{2}=1.00 \mathrm{~atm}, \mathrm{O}_{2}=1.00 \mathrm{~atm}$
(e) $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$ $K_{c}=4.6 \times 10^{4} ;[\mathrm{NO}]=1.00 M,\left[\mathrm{Cl}_{2}\right]=1.00 M,[\mathrm{NOCl}]=0 M$
(f) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
$K_{P}=0.050 ; \mathrm{NO}=10.0 \mathrm{~atm}, \mathrm{~N}_{2}=\mathrm{O}_{2}=5 \mathrm{~atm}$


1443. Exercises - 1443.2. Equilibrium Constants
$K_{P}=0.050 ; \mathrm{NO}=10.0 \mathrm{~atm}, \mathrm{~N}_{2}=\mathrm{O}_{2}=5 \mathrm{~atm}$
18. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
(a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \quad K_{c}=17 ;\left[\mathrm{NH}_{3}\right]=0.50 \mathrm{M},\left[\mathrm{N}_{2}\right]=0.15 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.12 \mathrm{M}$
(b) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
$K_{P}=6.8 \times 10^{4} ; \mathrm{NH}_{3}=2.00 \mathrm{~atm}, \mathrm{~N}_{2}=10.00 \mathrm{~atm}, \mathrm{H}_{2}=10.00 \mathrm{~atm}$
(c) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$
$K_{c}=0.230 ;\left[\mathrm{SO}_{3}\right]=2.00 \mathrm{M},\left[\mathrm{SO}_{2}\right]=2.00 \mathrm{M},\left[\mathrm{O}_{2}\right]=2.00 \mathrm{M}$
(d) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$


1443. Exercises - 1443.2. Equilibrium Constants
(d) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$
$K_{P}=6.5 \mathrm{~atm} ; \mathrm{SO}_{2}=1.00 \mathrm{~atm}, \mathrm{O}_{2}=1.130 \mathrm{~atm}, \mathrm{SO}_{3}=0 \mathrm{~atm}$
(e) $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$
$K_{P}=2.5 \times 10^{3} ; \mathrm{NO}=1.00 \mathrm{~atm}, \mathrm{Cl}_{2}=1.00 \mathrm{~atm}, \mathrm{NOCl}=0 \mathrm{~atm}$
(f) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
$K_{c}=0.050 ;\left[\mathrm{N}_{2}\right]=0.100 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.200 \mathrm{M},[\mathrm{NO}]=1.00 \mathrm{M}$
19. The following reaction has $K_{P}=4.50 \times 10^{-5}$ at 720 K .
$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? $P\left(\mathrm{NH}_{3}\right)=93 \mathrm{~atm}, P\left(\mathrm{~N}_{2}\right)=48 \mathrm{~atm}$, and $P\left(\mathrm{H}_{2}\right)=52$ atm
20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?


1443. Exercises - 1443.2. Equilibrium Constants
20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$
$\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.12 \mathrm{M},\left[\mathrm{Cl}_{2}\right]=0.16 \mathrm{M}$ and $\left[\mathrm{SO}_{2}\right]=0.050 \mathrm{M} . K_{c}$ for the reaction is 0.078 .
21. Which of the systems described in Exercise 13.15 are homogeneous equilibria? Which are heterogeneous equilibria?
22. Which of the systems described in Exercise 13.16 are homogeneous equilibria? Which are heterogeneous equilibria?
23. For which of the reactions in Exercise 13.15 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?
24. For which of the reactions in Exercise 13.16 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?
25. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.
(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
$K_{c}=0.50$ at $400^{\circ} \mathrm{C}$
(b) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$
$K_{c}=50.2$ at $448^{\circ} \mathrm{C}$


1443. Exercises - 1443.2. Equilibrium Constants
$K_{c}=50.2$ at $448^{\circ} \mathrm{C}$
(c) $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g)$
$K_{P}=4.08 \times 10^{-25}$ at $25^{\circ} \mathrm{C}$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)$
$K_{P}=0.122$ at $50^{\circ} \mathrm{C}$
26. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.
(a) $\mathrm{Cl}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g)$
$K_{c}=4.7 \times 10^{-2}$ at $25^{\circ} \mathrm{C}$
(b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
$K_{P}=48.2$ at $500^{\circ} \mathrm{C}$
(c) $\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CaCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g)$
$K_{P}=5.09 \times 10^{-44}$ at $25^{\circ} \mathrm{C}$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)$
$K_{P}=0.196$ at $60^{\circ} \mathrm{C}$


1443. Exercises - 1443.2. Equilibrium Constants
$K_{P}=0.196$ at $60^{\circ} \mathrm{C}$
27. What is the value of the equilibrium constant expression for the change $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)$ at $30^{\circ} \mathrm{C}$ ? (See Appendix E.)
28. Write the expression of the reaction quotient for the ionization of HOCN in water.
29. Write the reaction quotient expression for the ionization of $\mathrm{NH}_{3}$ in water.
30. What is the approximate value of the equilibrium constant $K_{P}$ for the change $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}(l) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}(g)$ at $25^{\circ} \mathrm{C}$. (The equilibrium vapor pressure for this substance is 570 torr at $25^{\circ} \mathrm{C}$.)


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
31. The following equation represents a reversible decomposition:
$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
Under what conditions will decomposition in a closed container proceed to completion so that no $\mathrm{CaCO}_{3}$ remains?
32. Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium.
33. What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
34. The following reaction occurs when a burner on a gas stove is lit:
$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
Is an equilibrium among $\mathrm{CH}_{4}, \mathrm{O}_{2}, \mathrm{CO}_{2}$, and $\mathrm{H}_{2} \mathrm{O}$ established under these conditions? Explain your answer.
35. A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, $\mathrm{SO}_{3}$, from sulfur dioxide, $\mathrm{SO}_{2}$, and oxygen, $\mathrm{O}_{2}$, shown here. At high temperatures, the rate of formation of $\mathrm{SO}_{3}$ is higher, but the equilibrium amount (concentration or partial pressure) of $\mathrm{SO}_{3}$ is lower than it would be at lower temperatures.
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
(b) Is the reaction endothermic or exothermic?
36. Suggest four ways in which the concentration of hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$, could be increased in an equilibrium described by the following equation:


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
36. Suggest four ways in which the concentration of hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$, could be increased in an equilibrium described by the following equation:
$\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g) \quad \Delta H=95 \mathrm{~kJ}$
37. Suggest four ways in which the concentration of $\mathrm{PH}_{3}$ could be increased in an equilibrium described by the following equation:


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
$$
\mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{PH}_{3}(g) \quad \Delta H=110.5 \mathrm{~kJ}
$$

38. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
$\Delta H=92 \mathrm{~kJ}$
(b) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$
$\Delta H=181 \mathrm{~kJ}$
(c) $2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g}) \quad \Delta H=-285 \mathrm{~kJ}$
(d) $\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CaCO}_{3}(s)$
$\Delta H=-176 \mathrm{~kJ}$
39. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) $2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$
$\Delta H=484 \mathrm{~kJ}$
(b) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
$\Delta H=-92.2 \mathrm{~kJ}$
(c) $2 \mathrm{Br}(g) \rightleftharpoons \mathrm{Br}_{2}(g)$
(d) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \rightleftharpoons 2 \mathrm{HI}(g)$


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
$$
\begin{aligned}
\Delta H=-224 \mathrm{~kJ} \\
\Delta H=53 \mathrm{~kJ}
\end{aligned}
$$


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
40. Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
(a) Write the expression for the equilibrium constant $\left(K_{c}\right)$ for the reversible reaction
$2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \quad \Delta H=-90.2 \mathrm{~kJ}$
(b) What will happen to the concentrations of $\mathrm{H}_{2}, \mathrm{CO}$, and $\mathrm{CH}_{3} \mathrm{OH}$ at equilibrium if more $\mathrm{H}_{2}$ is added?
(c) What will happen to the concentrations of $\mathrm{H}_{2}, \mathrm{CO}$, and $\mathrm{CH}_{3} \mathrm{OH}$ at equilibrium if CO is removed?
(d) What will happen to the concentrations of $\mathrm{H}_{2}, \mathrm{CO}$, and $\mathrm{CH}_{3} \mathrm{OH}$ at equilibrium if $\mathrm{CH}_{3} \mathrm{OH}$ is added?
(e) What will happen to the concentrations of $\mathrm{H}_{2}, \mathrm{CO}$, and $\mathrm{CH}_{3} \mathrm{OH}$ at equilibrium if the temperature of the system is increased?
41. Nitrogen and oxygen react at high temperatures.
(a) Write the expression for the equilibrium constant $\left(K_{c}\right)$ for the reversible reaction
$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{~kJ}$
(b) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if more $\mathrm{O}_{2}$ is added?
(c) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if $\mathrm{N}_{2}$ is removed?
(d) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if NO is added?


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
(d) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if NO is added?
(e) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentrations of $\mathrm{N}_{2}, \mathrm{O}_{2}$, and NO at equilibrium if the temperature of the system is increased?
42. Water gas, a mixture of $\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
(a) Write the expression for the equilibrium constant for the reversible reaction
$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g) \quad \Delta H=131.30 \mathrm{~kJ}$
(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2} \mathrm{O}$ is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
43. Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
(a) Write the expression for the equilibrium constant $\left(K_{c}\right)$ for the reversible reaction
$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=98.7 \mathrm{~kJ}$
(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2} \mathrm{O}$ is removed?


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
(c) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2} \mathrm{O}$ is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if $\mathrm{H}_{2}$ is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
44. Ammonia is a weak base that reacts with water according to this equation:
$\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q)$
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
(a) Addition of NaOH
(b) Addition of HCl
(c) Addition of $\mathrm{NH}_{4} \mathrm{Cl}$
45. Acetic acid is a weak acid that reacts with water according to this equation:
$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)$
Will any of the following increase the percent of acetic acid that reacts and produces $\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}$ion?
(a) Addition of HCl
(b) Addition of NaOH
(c) Addition of $\mathrm{NaCH}_{3} \mathrm{CO}_{2}$
46. Suggest two ways in which the equilibrium concentration of $\mathrm{Ag}^{+}$can be reduced in a solution of $\mathrm{Na}^{+}, \mathrm{Cl}^{-}$, $\mathrm{Ag}^{+}$, and $\mathrm{NO}_{3}{ }^{-}$, in contact with solid AgCl .


1443. Exercises - 1443.3. Shifting Equilibria: Le Châtelier's Principle
$\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{Ag}^{+}(a q)+\mathrm{NO}_{3}{ }^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s)+\mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}{ }^{-}(a q)$
$\Delta H=-65.9 \mathrm{~kJ}$
47. How can the pressure of water vapor be increased in the following equilibrium?
$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=41 \mathrm{~kJ}$
48. A solution is saturated with silver sulfate and contains excess solid silver sulfate:
$\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$
A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive $\mathrm{Ag}^{+}$ ions. Explain this observation.
49. When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?
(a) HCl
(b) $\mathrm{H}^{+}+\mathrm{Cl}^{-}$
(c) HOCl
(d) $\mathrm{H}^{+}+\mathrm{OCl}^{-}$


1443. Exercises - 1443.4. Equilibrium Calculations
50. A reaction is represented by this equation: $\mathrm{A}(a q)+2 \mathrm{~B}(a q) \rightleftharpoons 2 \mathrm{C}(a q) \quad K_{c}=1 \times 10^{3}$
(a) Write the mathematical expression for the equilibrium constant.
(b) Using concentrations $\leq 1 M$, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
51. A reaction is represented by this equation: $2 \mathrm{~W}(a q) \rightleftharpoons \mathrm{X}(a q)+2 \mathrm{Y}(a q)$

$$
K_{c}=5 \times 10^{-4}
$$


1443. Exercises - 1443.4. Equilibrium Calculations
(a) Write the mathematical expression for the equilibrium constant.
(b) Using concentrations of $\leq 1 M$, identify two sets of concentrations that describe a mixture of $\mathrm{W}, \mathrm{X}$, and Y at equilibrium.
52. What is the value of the equilibrium constant at $500^{\circ} \mathrm{C}$ for the formation of $\mathrm{NH}_{3}$ according to the following equation?
$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
An equilibrium mixture of $\mathrm{NH}_{3}(g), \mathrm{H}_{2}(g)$, and $\mathrm{N}_{2}(g)$ at $500^{\circ} \mathrm{C}$ was found to contain $1.35 \mathrm{MH}_{2}, 1.15 \mathrm{M}_{2}$, and $4.12 \times 10^{-1} \mathrm{MNH}_{3}$.
53. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. $\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 3 \mathrm{H}_{2}(g)+\mathrm{CO}(g)$
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: $\mathrm{CH}_{4}, 0.126 \mathrm{M} ; \mathrm{H}_{2} \mathrm{O}, 0.242 \mathrm{M} ; \mathrm{CO}, 0.126 \mathrm{M} ; \mathrm{H}_{2} 1.15 \mathrm{M}$, at a temperature of $760^{\circ} \mathrm{C}$ ?


1443. Exercises - 1443.4. Equilibrium Calculations
54. A $0.72-\mathrm{mol} \mathrm{sample}$ of $\mathrm{PCl}_{5}$ is put into a $1.00-\mathrm{L}$ vessel and heated. At equilibrium, the vessel contains 0.40 mol of $\mathrm{PCl}_{3}(g)$ and 0.40 mol of $\mathrm{Cl}_{2}(g)$. Calculate the value of the equilibrium constant for the decomposition of $\mathrm{PCl}_{5}$ to $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$ at this temperature.
55. At 1 atm and $25^{\circ} \mathrm{C}, \mathrm{NO}_{2}$ with an initial concentration of 1.00 M is $0.0033 \%$ decomposed into NO and $\mathrm{O}_{2}$. Calculate the value of the equilibrium constant for the reaction.
$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$
56. Calculate the value of the equilibrium constant $K_{P}$ for the reaction $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$ from these equilibrium pressures: NO, $0.050 \mathrm{~atm} ; \mathrm{Cl}_{2}, 0.30 \mathrm{~atm} ; \mathrm{NOCl}, 1.2 \mathrm{~atm}$.
57. When heated, iodine vapor dissociates according to this equation:
$\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})$
At 1274 K , a sample exhibits a partial pressure of $\mathrm{I}_{2}$ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm . Determine the value of the equilibrium constant, $K_{P}$, for the decomposition at 1274 K .
58. A sample of ammonium chloride was heated in a closed container.


1443. Exercises - 1443.4. Equilibrium Calculations
58. A sample of ammonium chloride was heated in a closed container.
$\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$
At equilibrium, the pressure of $\mathrm{NH}_{3}(g)$ was found to be 1.75 atm . What is the value of the equilibrium constant $K_{P}$ for the decomposition at this temperature?
59. At a temperature of $60^{\circ} \mathrm{C}$, the vapor pressure of water is 0.196 atm . What is the value of the equilibrium constant $K_{P}$ for the vaporization equilibrium at $60^{\circ} \mathrm{C}$ ?
$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)$
60. Complete the following partial ICE tables.
(a)


1443. Exercises - 1443.4. Equilibrium Calculations
$$
2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)
$$

change $-\quad+x$
(b)

$$
4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)
$$

change $\quad+x$
(c)

$$
2 \mathrm{CH}_{4}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{2}(g)+3 \mathrm{H}_{2}(g)
$$

change
(d)

$$
\begin{array}{ll}
\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\
\text { change_ } & \rightleftharpoons-
\end{array}
$$

(e)

$$
\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)
$$

change $+x$
(f)

$$
\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)
$$

change $+x$
61. Complete the following partial ICE tables.
(a)

$$
2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)
$$

change

$$
\begin{array}{llll}
& - & +x
\end{array}
$$

(b)

$$
\mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g)
$$


1443. Exercises - 1443.4. Equilibrium Calculations
change $+x$
(c)
$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$
change $+x$
(d)
$2 \mathrm{NH}_{3}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)$
change $\qquad$
$\qquad$ $+x$
(e)

$$
\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)
$$

change

$$
+x
$$

(f)

$$
\mathrm{Fe}(s)+5 \mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_{5}(g)
$$


1443. Exercises - 1443.4. Equilibrium Calculations
change $+x$
62. Why are there no changes specified for Ni in Exercise 13.60, part (f)? What property of Ni does change?
63. Why are there no changes specified for $\mathrm{NH}_{4} \mathrm{HS}$ in Exercise 13.61, part (e)? What property of $\mathrm{NH}_{4} \mathrm{HS}$ does change?
64. Analysis of the gases in a sealed reaction vessel containing $\mathrm{NH}_{3}, \mathrm{~N}_{2}$, and $\mathrm{H}_{2}$ at equilibrium at $400^{\circ} \mathrm{C}$ established the concentration of $\mathrm{N}_{2}$ to be 1.2 M and the concentration of $\mathrm{H}_{2}$ to be 0.24 M .
$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \quad K_{c}=0.50$ at $400^{\circ} \mathrm{C}$
Calculate the equilibrium molar concentration of $\mathrm{NH}_{3}$.
65. Calculate the number of moles of HI that are at equilibrium with 1.25 mol of $\mathrm{H}_{2}$ and $1.25 \mathrm{~mol}^{\text {of }} \mathrm{I}_{2}$ in a 5.00-L flask at $448^{\circ} \mathrm{C}$.
$\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \quad K_{c}=50.2$ at $448^{\circ} \mathrm{C}$
66. What is the pressure of BrCl in an equilibrium mixture of $\mathrm{Cl}_{2}, \mathrm{Br}_{2}$, and BrCl if the pressure of $\mathrm{Cl}_{2}$ in the mixture is 0.115 atm and the pressure of $\mathrm{Br}_{2}$ in the mixture is 0.450 atm ?
$\mathrm{Cl}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g) \quad K_{P}=4.7 \times 10^{-2}$


1443. Exercises - 1443.4. Equilibrium Calculations
67. What is the pressure of $\mathrm{CO}_{2}$ in a mixture at equilibrium that contains 0.50 atm $\mathrm{H}_{2}, 2.0$ atm of $\mathrm{H}_{2} \mathrm{O}$, and 1.0 atm of CO at $990^{\circ} \mathrm{C}$ ?
$\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \quad K_{P}=1.6$ at $990^{\circ} \mathrm{C}$
68. Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
$\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \quad K_{c}=4.90 \times 10^{2}$ at $550^{\circ} \mathrm{C}$
What concentration of CO remains in an equilibrium mixture with $\left[\mathrm{CO}_{2}\right]=0.100 \mathrm{M}$ ?
69. Carbon reacts with water vapor at elevated temperatures.
$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g) \quad K_{c}=0.2$ at $1000^{\circ} \mathrm{C}$
Assuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with $\left[\mathrm{H}_{2} \mathrm{O}\right]=0.500 \mathrm{M}$ at $1000^{\circ} \mathrm{C}$ ?
70. Sodium sulfate 10 -hydrate, $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$, dehydrates according to the equation


1443. Exercises - 1443.4. Equilibrium Calculations
70. Sodium sulfate 10 -hydrate, $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$, dehydrates according to the equation
$\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=4.08 \times 10^{-25}$ at $25^{\circ} \mathrm{C}$
What is the pressure of water vapor at equilibrium with a mixture of $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$ and $\mathrm{NaSO}_{4}$ ?
71. Calcium chloride 6-hydrate, $\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, dehydrates according to the equation
$\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CaCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=5.09 \times 10^{-44}$ at $25^{\circ} \mathrm{C}$
What is the pressure of water vapor at equilibrium with a mixture of $\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ and $\mathrm{CaCl}_{2}$ at $25{ }^{\circ} \mathrm{C}$ ?


1443. Exercises - 1443.4. Equilibrium Calculations
72. A student solved the following problem and found the equilibrium concentrations to be $\left[\mathrm{SO}_{2}\right]=0.590 \mathrm{M}$, $\left[\mathrm{O}_{2}\right]=0.0450 \mathrm{M}$, and $\left[\mathrm{SO}_{3}\right]=0.260 \mathrm{M}$. How could this student check the work without reworking the problem? The problem was: For the following reaction at $600^{\circ} \mathrm{C}$ :
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \quad K_{c}=4.32$
What are the equilibrium concentrations of all species in a mixture that was prepared with $\left[\mathrm{SO}_{3}\right]=0.500$ $\mathrm{M},\left[\mathrm{SO}_{2}\right]=0 \mathrm{M}$, and $\left[\mathrm{O}_{2}\right]=0.350 \mathrm{M}$ ?
73. A student solved the following problem and found $\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.16 \mathrm{M}$ at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of $\mathrm{N}_{2} \mathrm{O}_{4}$ in a mixture formed from a sample of $\mathrm{NO}_{2}$ with a concentration of 0.10
$M$ ?
$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) \quad K_{c}=160$
74. Assume that the change in concentration of $\mathrm{N}_{2} \mathrm{O}_{4}$ is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\mathrm{N}_{2} \mathrm{O}_{4}$ with chloroform as the solvent.


1443. Exercises - 1443.4. Equilibrium Calculations
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\mathrm{N}_{2} \mathrm{O}_{4}$ with chloroform as the solvent.
$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{c}=1.07 \times 10^{-5}$ in chloroform
(b) Confirm that the change is small enough to be neglected.
75. Assume that the change in concentration of $\mathrm{COCl}_{2}$ is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of $\mathrm{COCl}_{2}$ with an initial concentration of 0.3166 M .
$\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \quad K_{c}=2.2 \times 10^{-10}$
(b) Confirm that the change is small enough to be neglected.
76. Assume that the change in pressure of $\mathrm{H}_{2} \mathrm{~S}$ is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of $\mathrm{H}_{2} \mathrm{~S}$ with an initial pressure of 0.824 atm .


1443. Exercises - 1443.4. Equilibrium Calculations
$$
2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \quad K_{P}=2.2 \times 10^{-6}
$$

(b) Confirm that the change is small enough to be neglected.
77. What are all concentrations after a mixture that contains $\left[\mathrm{H}_{2} \mathrm{O}\right]=1.00 \mathrm{M}$ and $\left[\mathrm{Cl}_{2} \mathrm{O}\right]=1.00 \mathrm{M}$ comes to equilibrium at $25^{\circ} \mathrm{C}$ ?
$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{c}=0.0900$
78. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of $\mathrm{H}_{2}$ and 63.5 g of iodine at 448 ${ }^{\circ} \mathrm{C}$.
$\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \quad K_{c}=50.2$ at $448^{\circ} \mathrm{C}$
79. Butane exists as two isomers, $n$-butane and isobutane.


1443. Exercises - 1443.4. Equilibrium Calculations
$K_{P}=2.5$ at $25^{\circ} \mathrm{C}$
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
80. What is the minimum mass of $\mathrm{CaCO}_{3}$ required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant $\left(K_{c}\right)$ is 0.50 for the decomposition reaction of $\mathrm{CaCO}_{3}$ at that temperature?
$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
81. The equilibrium constant $\left(K_{c}\right)$ for this reaction is 1.60 at $990^{\circ} \mathrm{C}$ :
$\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of $\mathrm{H}_{2}, 2.00 \mathrm{~mol}$ of $\mathrm{CO}_{2}, 0.750 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}$, and 1.00 mol of CO to a $5.00-\mathrm{L}$ container at $990^{\circ} \mathrm{C}$.
82. In a 3.0-L vessel, the following equilibrium partial pressures are measured: $\mathrm{N}_{2}, 190$ torr; $\mathrm{H}_{2}, 317$ torr; $\mathrm{NH}_{3}, 1.00 \times 10^{3}$ torr.
$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$


1443. Exercises - 1443.4. Equilibrium Calculations
$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
(a) How will the partial pressures of $\mathrm{H}_{2}, \mathrm{~N}_{2}$, and $\mathrm{NH}_{3}$ change if $\mathrm{H}_{2}$ is removed from the system? Will they increase, decrease, or remain the same?
(b) Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
83. The equilibrium constant $\left(K_{c}\right)$ for this reaction is 5.0 at a given temperature.
$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$
(a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of $\mathrm{CO}, 0.30 \mathrm{~mol}$ of water vapor, and 0.90 mol of $\mathrm{H}_{2}$ in a liter. How many moles of $\mathrm{CO}_{2}$ were there in the equilibrium mixture?
(b) Maintaining the same temperature, additional $\mathrm{H}_{2}$ was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of $\mathrm{CO}, 0.30$ mol of water vapor, and 1.2 mol of $\mathrm{H}_{2}$ in a liter. How many moles of $\mathrm{CO}_{2}$ were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
84. Antimony pentachloride decomposes according to this equation:
$\mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)$


1443. Exercises - 1443.4. Equilibrium Calculations
$\mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)$
An equilibrium mixture in a 5.00 -L flask at $448{ }^{\circ} \mathrm{C}$ contains 3.85 g of $\mathrm{SbCl}_{5}, 9.14 \mathrm{~g}$ of $\mathrm{SbCl}_{3}$, and 2.84 g of $\mathrm{Cl}_{2}$. How many grams of each will be found if the mixture is transferred into a $2.00-\mathrm{L}$ flask at the same temperature?
85. Consider the equilibrium
$4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g)$
(a) What is the expression for the equilibrium constant $\left(K_{c}\right)$ of the reaction?
(b) How must the concentration of $\mathrm{NH}_{3}$ change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
(c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of $\mathrm{NO}_{2}$ ?
(d) If the change in the pressure of $\mathrm{NO}_{2}$ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of $\mathrm{O}_{2}$ change?
86. The binding of oxygen by hemoglobin $(\mathrm{Hb})$, giving oxyhemoglobin $\left(\mathrm{HbO}_{2}\right)$, is partially regulated by the concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$and dissolved $\mathrm{CO}_{2}$ in the blood. Although the equilibrium is complicated, it can be summarized as


1443. Exercises - 1443.4. Equilibrium Calculations
$\mathrm{HbO}_{2}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}-\mathrm{Hb}-\mathrm{H}^{+}+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
(a) Write the equilibrium constant expression for this reaction.
(b) Explain why the production of lactic acid and $\mathrm{CO}_{2}$ in a muscle during exertion stimulates release of $\mathrm{O}_{2}$ from the oxyhemoglobin in the blood passing through the muscle.
87. Liquid $\mathrm{N}_{2} \mathrm{O}_{3}$ is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and $\mathrm{NO}_{2}$. At $25^{\circ} \mathrm{C}$, a value of $K_{P}=1.91$ has been established for this decomposition. If 0.236 moles of $\mathrm{N}_{2} \mathrm{O}_{3}$ are placed in a 1.52 - L vessel at $25^{\circ} \mathrm{C}$, calculate the equilibrium partial pressures of $\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}), \mathrm{NO}_{2}(\mathrm{~g})$, and $\mathrm{NO}(\mathrm{g})$.
88. A 1.00-L vessel at $400^{\circ} \mathrm{C}$ contains the following equilibrium concentrations: $\mathrm{N}_{2}, 1.00 \mathrm{M} ; \mathrm{H}_{2}, 0.50 \mathrm{M}$; and $\mathrm{NH}_{3}, 0.25 \mathrm{M}$. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 $M$ ? The equilibrium reaction is $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$


1443. Exercises - 1443.4. Equilibrium Calculations
89. Calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for each of the following reactions from the value of $\Delta G^{\circ}$ given.
(a) $\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g)$
$\Delta G^{\circ}=-10.88 \mathrm{~kJ}$
(b) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)$
$\Delta G^{\circ}=3.4 \mathrm{~kJ}$
(c) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g) \quad \Delta G^{\circ}=-39 \mathrm{~kJ}$
(d) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta G^{\circ}=-141.82 \mathrm{~kJ}$
(e) $\mathrm{CS}_{2}(g) \longrightarrow \mathrm{CS}_{2}(l) \quad \Delta G^{\circ}=-1.88 \mathrm{~kJ}$
90. Calculate the equilibrium constant at the temperature given.
(a) $\mathrm{O}_{2}(g)+2 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{~F}_{2} \mathrm{O}(g) \quad\left(\mathrm{T}=100^{\circ} \mathrm{C}\right)$
(b) $\mathrm{I}_{2}(s)+\mathrm{Br}_{2}(l) \longrightarrow 2 \mathrm{IBr}(g) \quad\left(\mathrm{T}=0.0^{\circ} \mathrm{C}\right)$


1443. Exercises - 1443.4. Equilibrium Calculations
(c) $2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad\left(\mathrm{T}=575^{\circ} \mathrm{C}\right)$
$(d) \mathrm{N}_{2} \mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g) \quad\left(\mathrm{T}=-10.0^{\circ} \mathrm{C}\right)$
(e) $\mathrm{SnCl}_{4}(l) \longrightarrow \mathrm{SnCl}_{4}(g) \quad\left(\mathrm{T}=200^{\circ} \mathrm{C}\right)$
91. Calculate the equilibrium constant at the temperature given.
(a) $\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g) \quad\left(\mathrm{T}=100^{\circ} \mathrm{C}\right)$
(b) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) \quad\left(\mathrm{T}=0.0^{\circ} \mathrm{C}\right)$
(c) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g) \quad\left(\mathrm{T}=125^{\circ} \mathrm{C}\right)$
(d) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad\left(\mathrm{T}=675^{\circ} \mathrm{C}\right)$
(e) $\mathrm{CS}_{2}(g) \longrightarrow \mathrm{CS}_{2}(l) \quad\left(\mathrm{T}=90^{\circ} \mathrm{C}\right)$
92. Consider the following reaction at 298 K :


1443. Exercises - 1443.4. Equilibrium Calculations
92. Consider the following reaction at 298 K :
$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{P}=0.142$
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.
93. Determine the normal boiling point (in kelvin) of dichloroethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
94. Under what conditions is $\mathrm{N}_{2} \mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)$ spontaneous?
95. At room temperature, the equilibrium constant $\left(K_{W}\right)$ for the self-ionization of water is $1.00 \times 10^{-14}$. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
96. Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction $2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons \frac{3}{8} \mathrm{~S}_{8}(s$, rhombic $)+2 \mathrm{H}_{2} \mathrm{O}(l)$. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
97. Consider the decomposition of $\mathrm{CaCO}_{3}(s)$ into $\mathrm{CaO}(s)$ and $\mathrm{CO}_{2}(g)$. What is the equilibrium partial pressure of $\mathrm{CO}_{2}$ at room temperature?


1443. Exercises - 1443.4. Equilibrium Calculations
98. In the laboratory, hydrogen chloride $(\mathrm{HCl}(g))$ and ammonia $\left(\mathrm{NH}_{3}(g)\right)$ often escape from bottles of their solutions and react to form the ammonium chloride $\left(\mathrm{NH}_{4} \mathrm{Cl}(s)\right)$, the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and $\mathrm{NH}_{3}$ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
99. Benzene can be prepared from acetylene. $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$. Determine the equilibrium constant at $25^{\circ} \mathrm{C}$ and at $850^{\circ} \mathrm{C}$. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
100. Carbon dioxide decomposes into CO and $\mathrm{O}_{2}$ at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at $1000{ }^{\circ} \mathrm{C}$ for which the initial pressure of $\mathrm{CO}_{2}$ was 1.15 atm?
101. Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
$\mathrm{CH}_{4}(g)+4 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+4 \mathrm{HCl}(g)$
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?


1443. Exercises - 1443.4. Equilibrium Calculations
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?
102. Acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, can form a dimer, $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)_{2}$, in the gas phase. $2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{g}) \longrightarrow\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)_{2}(\mathrm{~g})$
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.


1443. Exercises - 1443.4. Equilibrium Calculations
At $25^{\circ} \mathrm{C}$, the equilibrium constant for the dimerization is $1.3 \times 10^{3}$ (pressure in atm). What is $\Delta S^{\circ}$ for the reaction?


1444. CHAPTER 14 <br> Acid-Base Equilibria -
Figure 14.1 Sinkholes such as this are the result of reactions between acidic groundwaters and basic rock formations, like limestone. (credit: modification of work by Emil Kehnel)


1446. 2 pH and pOH -
14.3 Relative Strengths of Acids and Bases
14.4 Hydrolysis of Salts
14.5 Polyprotic Acids
14.6 Buffers
14.7 Acid-Base Titrations

INTRODUCTION Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$, is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases). These H+ transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation (Figure 14.1) to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class.


1447. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
- Write equations for acid and base ionization reactions
- Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
- Describe the acid-base behavior of amphiprotic substances

The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be $\mathrm{CO}_{2}$ ), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.

Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, $\mathrm{H}^{+}$. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, ${ }^{1} \mathrm{H}$.) A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).


1447. LEARNING OBJECTIVES -
The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates $\mathrm{H}^{+}$, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts $\mathrm{H}^{+}$, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, $\mathrm{OH}^{-}$, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, $\mathrm{NH}_{4}{ }^{+}$, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.

The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:

Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, $\mathrm{C}_{5} \mathrm{NH}_{5}$, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:

The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acidbase reactions for two amphiprotic species, bicarbonate ion and water:


1447. LEARNING OBJECTIVES -
$$
\begin{aligned}
\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \\
\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}
$$

The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.

In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:

The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at $25^{\circ} \mathrm{C}$, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, $\boldsymbol{K}_{\mathbf{w}}$ :

$$
\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]
$$


1447. LEARNING OBJECTIVES -
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at $25^{\circ} \mathrm{C}, K_{\mathrm{w}}$ has a value of $1.0 \times 10^{-14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at $100^{\circ} \mathrm{C}$, the value for $K_{\mathrm{w}}$ is about $5.6 \times 10^{-13}$, roughly 50 times larger than the value at $25^{\circ} \mathrm{C}$.


1449. Ion Concentrations in Pure Water -
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at $25^{\circ} \mathrm{C}$ ?


1450. Solution -
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=x$. At $25^{\circ} \mathrm{C}$ :

$$
K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=(x)(x)=x^{2}=1.0 \times 10^{-14}
$$

So:

$$
x=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{1.0 \times 10^{-14}}=1.0 \times 10^{-7} \mathrm{M}
$$

The hydronium ion concentration and the hydroxide ion concentration are the same, $1.0 \times 10^{-7} \mathrm{M}$.


1451. Check Your Learning -
The ion product of water at $80^{\circ} \mathrm{C}$ is $2.4 \times 10^{-13}$. What are the concentrations of hydronium and hydroxide ions in pure water at $80^{\circ} \mathrm{C}$ ?


1452. Answer: -
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=4.9 \times 10^{-7} \mathrm{M}$


1454. The Inverse Relation between $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right.$] and $\left[\mathrm{OH}^{-}\right.$] -
A solution of an acid in water has a hydronium ion concentration of $2.0 \times 10^{-6} \mathrm{M}$. What is the concentration of hydroxide ion at $25^{\circ} \mathrm{C}$ ?


1455. Solution -
Use the value of the ion-product constant for water at $25^{\circ} \mathrm{C}$

$$
2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}
$$

to calculate the missing equilibrium concentration.
Rearrangement of the $K_{\mathrm{w}}$ expression shows that $\left[\mathrm{OH}^{-}\right]$is inversely proportional to $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$:

$$
\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{1.0 \times 10^{-14}}{2.0 \times 10^{-6}}=5.0 \times 10^{-9}
$$

Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Châtelier's principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.

Substituting the ion concentrations into the $K_{\mathrm{w}}$ expression confirms this calculation, resulting in the expected value:

$$
K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left(2.0 \times 10^{-6}\right)\left(5.0 \times 10^{-9}\right)=1.0 \times 10^{-14}
$$


1456. Check Your Learning -
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 $M$ at $25^{\circ} \mathrm{C}$ ?

Answer:
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1 \times 10^{-11} \mathrm{M}$


1458. Representing the Acid-Base Behavior of an Amphoteric Substance -
Write separate equations representing the reaction of $\mathrm{HSO}_{3}{ }^{-}$
(a) as an acid with $\mathrm{OH}^{-}$
(b) as a base with HI


1459. Solution -
(a) $\mathrm{HSO}_{3}{ }^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{SO}_{3}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
(b) $\mathrm{HSO}_{3}{ }^{-}(a q)+\mathrm{HI}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{I}^{-}(a q)$


1460. Check Your Learning -
Write separate equations representing the reaction of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$
(a) as a base with HBr
(b) as an acid with $\mathrm{OH}^{-}$


1461. Answer: -
(a) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{HBr}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Br}^{-}(a q)$; (b)
$\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$


1463. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the characterization of aqueous solutions as acidic, basic, or neutral
- Express hydronium and hydroxide ion concentrations on the pH and pOH scales
- Perform calculations relating pH and pOH

As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ( $K_{\mathrm{w}}$ ). The concentrations of these ions in a solution are often critical determinants of the solution's properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.

A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where " X " is the quantity of interest and "log" is the base-10 logarithm:

$$
\mathrm{pX}=-\log \mathrm{X}
$$

The $\mathbf{p H}$ of a solution is therefore defined as shown here, where $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$is the molar concentration of hydronium ion in the solution:

$$
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]
$$

Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:

$$
\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-\mathrm{pH}}
$$

Likewise, the hydroxide ion molarity may be expressed as a p-function, or $\mathbf{p O H}$ :

$$
\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]
$$

or


1463. LEARNING OBJECTIVES -
$$
\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]
$$

or

$$
\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}
$$

Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_{\mathrm{w}}$ expression:

$$
K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]
$$

$$
\begin{gathered}
-\log K_{\mathrm{w}}=-\log \left(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\right)=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+-\log \left[\mathrm{OH}^{-}\right] \\
\mathrm{p} K_{\mathrm{w}}=\mathrm{pH}+\mathrm{pOH}
\end{gathered}
$$

At $25^{\circ} \mathrm{C}$, the value of $K_{\mathrm{W}}$ is $1.0 \times 10^{-14}$, and so:

$$
14.00=\mathrm{pH}+\mathrm{pOH}
$$

As was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times$ $10^{-7} \mathrm{M}$ at $25^{\circ} \mathrm{C}$. The pH and pOH of a neutral solution at this temperature are therefore:

$$
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \\
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00
\end{aligned}
$$


1463. LEARNING OBJECTIVES -
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{-7} \mathrm{M}$ and hydroxide ion molarities less than $1.0 \times 10^{-7} \mathrm{M}$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{-7} \mathrm{M}$ and hydroxide ion molarities greater than $1.0 \times 10^{-7} M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).

Since the autoionization constant $K_{\mathrm{w}}$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than $25^{\circ} \mathrm{C}$. For example, the "Check Your Learning" exercise accompanying Example 14.1 showed the hydronium molarity of pure water at $80^{\circ} \mathrm{C}$ is $4.9 \times 10^{-7} \mathrm{M}$, which corresponds to pH and pOH values of:

$$
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \\
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31
\end{aligned}
$$


1463. LEARNING OBJECTIVES -
At this temperature, then, neutral solutions exhibit $\mathrm{pH}=\mathrm{pOH}=6.31$, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31 . This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around $36-40^{\circ} \mathrm{C}$. Unless otherwise noted, references to pH values are presumed to be those at $25^{\circ} \mathrm{C}$ (Table 14.1).

| Summary of Relations for Acidic, Basic and Neutral Solutions |  |  |
| :---: | :---: | :---: |
| Classification | Relative lon Concentrations | pH at $\mathbf{2 5}^{\circ} \mathrm{C}$ |
| acidic | $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]>\left[\mathrm{OH}^{-}\right]$ | $\mathrm{pH}<7$ |
| neutral | $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]$ | $\mathrm{pH}=7$ |
| basic | $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]<\left[\mathrm{OH}^{-}\right]$ | $\mathrm{pH}>7$ |

TABLE 14.1

Figure 14.2 shows the relationships between $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH}$, and pOH for solutions classified as acidic, basic, and neutral.

FIGURE 14.2 The pH and pOH scales represent concentrations of $\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{OH}^{-}$, respectively. The pH and pOH values of some common substances at $25^{\circ} \mathrm{C}$ are shown in this chart.


1465. Calculation of pH from $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ -
What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \times 10^{-3} \mathrm{M}$ ?


1466. Solution -
$$
\begin{gathered}
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\
=-\log \left(1.2 \times 10^{-3}\right) \\
=-(-2.92)=2.92
\end{gathered}
$$

(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)


1467. Check Your Learning -
Water exposed to air contains carbonic acid, $\mathrm{H}_{2} \mathrm{CO}_{3}$, due to the reaction between carbon dioxide and water:

$$
\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)
$$

Air-saturated water has a hydronium ion concentration caused by the dissolved $\mathrm{CO}_{2}$ of $2.0 \times 10^{-6} \mathrm{M}$, about 20-times larger than that of pure water. Calculate the pH of the solution at $25^{\circ} \mathrm{C}$.


1470. Calculation of Hydronium Ion Concentration from pH -
Calculate the hydronium ion concentration of blood, the pH of which is 7.3.


1471. Solution -
$$
\begin{gathered}
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.3 \\
\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-7.3 \\
{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-7.3} \text { or }\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\text {antilog of }-7.3} \\
{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-8} \mathrm{M}}
\end{gathered}
$$

(On a calculator take the antilog, or the "inverse" log, of -7.3 , or calculate $10^{-7.3}$.)


1472. Check Your Learning -
Calculate the hydronium ion concentration of a solution with a pH of -1.07 .


1473. Answer: -
12 M


1475. Environmental Science -
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved $\mathrm{CO}_{2}$ which forms carbonic acid:

$$
\begin{gathered}
\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \\
\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)
\end{gathered}
$$

Acid rain is rainwater that has a pH of less than 5 , due to a variety of nonmetal oxides, including $\mathrm{CO}_{2}, \mathrm{SO}_{2}, \mathrm{SO}_{3}$, NO , and $\mathrm{NO}_{2}$ being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:

$$
\begin{gathered}
\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{3}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \\
\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{HSO}_{4}^{-}(a q)
\end{gathered}
$$

Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of "roasting" ores of metal sulfides in metalrefining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.


1475. Environmental Science -
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until
the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.

For further information on acid rain, visit this website (http://openstax.org/l/16EPA) hosted by the US Environmental Protection Agency.

FIGURE 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by "Eden, Janine and Jim"/Flickr)


1477. Calculation of pOH -
What are the pOH and the pH of a $0.0125-M$ solution of potassium hydroxide, KOH ?


1478. Solution -
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding $\left[\mathrm{OH}^{-}\right]=0.0125 \mathrm{M}$ :

$$
\begin{aligned}
\mathrm{pOH}= & =-\log \left[\mathrm{OH}^{-}\right]=-\log 0.0125 \\
& =-(-1.903)=1.903
\end{aligned}
$$

The pH can be found from the pOH :

$$
\begin{gathered}
\mathrm{pH}+\mathrm{pOH}=14.00 \\
\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.903=12.10
\end{gathered}
$$


1479. Check Your Learning -
The hydronium ion concentration of vinegar is approximately $4 \times 10^{-3} \mathrm{M}$. What are the corresponding values of pOH and pH ?


1480. Answer: -
$\mathrm{pOH}=11.6, \mathrm{pH}=2.4$

The acidity of a solution is typically assessed experimentally by measurement of its pH . The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4).

FIGURE 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of $\pm 0.002 \mathrm{pH}$ units, and may cost in excess of $\$ 1000$. (b) A portable pH meter has lower resolution ( 0.01 pH units), lower accuracy ( $\pm 0.2 \mathrm{pH}$ units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)

The pH of a solution may also be visually estimated using colored indicators (Figure 14.5). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.

FIGURE 14.5 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH . (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa)


1481. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Assess the relative strengths of acids and bases according to their ionization constants
- Rationalize trends in acid-base strength in relation to molecular structure
- Carry out equilibrium calculations for weak acid-base systems


1482. Acid and Base Ionization Constants -
The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure 14.6.

| 6 Strong Acids |  | 6 Strong Bases |  |
| :--- | :--- | :--- | :--- |
| $\mathrm{HClO}_{4}$ | perchloric acid | LiOH | lithium hydroxide |
| HCl | hydrochloric acid | NaOH | sodium hydroxide |
| HBr | hydrobromic acid | KOH | potassium hydroxide |
| HI | hydroiodic acid | $\mathrm{Ca}(\mathrm{OH})_{2}$ | calcium hydroxide |
| $\mathrm{HNO}_{3}$ | nitric acid | $\mathrm{Sr}(\mathrm{OH})_{2}$ | strontium hydroxide |
| $\mathrm{H}_{2} \mathrm{SO}_{4}$ | sulfuric acid | $\mathrm{Ba}(\mathrm{OH})_{2}$ | barium hydroxide |

FIGURE 14.6 Some of the common strong acids and bases are listed here.
The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, $\boldsymbol{K}_{\mathbf{a}}$. For the reaction of an acid HA:

$$
\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q)
$$

the acid ionization constant is written


1482. Acid and Base Ionization Constants -
the acid ionization constant is written

$$
K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
$$

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include $\left[\mathrm{H}_{2} \mathrm{O}\right]$ in the equation. The larger the $K_{\mathrm{a}}$ of an acid, the larger the concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{A}^{-}$relative to the concentration of the nonionized acid, HA , in an equilibrium mixture, and the stronger the acid. An acid is classified as "strong" when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large ( $K_{\mathrm{a}} \approx \infty$ ). Acids that are partially ionized are called "weak," and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.

To illustrate this idea, three acid ionization equations and $K_{\mathrm{a}}$ values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}<\mathrm{HNO}_{2}<\mathrm{HSO}_{4}{ }^{-}$:


1482. Acid and Base Ionization Constants -
$$
\begin{array}{ccc}
\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}(a q) & K_{\mathrm{a}}=1.8 \times 10^{-5} \\
\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{NO}_{2}{ }^{-}(a q) & K_{\mathrm{a}}=4.6 \times 10^{-4} \\
\mathrm{HSO}_{4}{ }^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) & \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) & K_{\mathrm{a}}=1.2 \times 10^{-2}
\end{array}
$$

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:

$$
\% \text { ionization }=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]_{\mathrm{eq}}}{[\mathrm{HA}]_{0}} \times 100
$$

where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, $\left[\mathrm{A}^{-}\right]=$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$). Unlike the $K_{\mathrm{a}}$ value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.


1484. Calculation of Percent Ionization from pH -
Calculate the percent ionization of a $0.125-M$ solution of nitrous acid (a weak acid), with a pH of 2.09 .


1485. Solution -
The percent ionization for an acid is:

$$
\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]_{\mathrm{eq}}}{\left[\mathrm{HNO}_{2}\right]_{0}} \times 100
$$

Converting the provided pH to hydronium ion molarity yields

$$
\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=10^{-2.09}=0.0081 \mathrm{M}
$$

Substituting this value and the provided initial acid concentration into the percent ionization equation gives

$$
\frac{8.1 \times 10^{-3}}{0.125} \times 100=6.5 \%
$$

(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)


1486. Check Your Learning -
Calculate the percent ionization of a $0.10-M$ solution of acetic acid with a pH of 2.89 .


1487. Answer: -
1.3\% ionized


1488. LINK TO LEARNING -
View the simulation (http://openstax.org/l/16AcidBase) of strong and weak acids and bases at the molecular level.

Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant $\left(K_{\mathbf{b}}\right)$ in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:

$$
\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q)
$$

the ionization constant is written as

$$
K_{\mathrm{b}}=\frac{\left[\mathrm{HB}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}
$$

Inspection of the data for three weak bases presented below shows the base strength increases in the order $\mathrm{NO}_{2}{ }^{-}<\mathrm{CH}_{2} \mathrm{CO}_{2}^{-}<\mathrm{NH}_{3}$.


1488. LINK TO LEARNING -
$$
\begin{array}{ll}
\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b}}=2.17 \times 10^{-11} \\
\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b}}=5.6 \times 10^{-10} \\
\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b}}=1.8 \times 10^{-5}
\end{array}
$$

A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a
base is also reflected in its percent ionization, computed as

$$
\% \text { ionization }=\left[\mathrm{OH}^{-}\right]_{e q} /[\mathrm{B}]_{0} \times 100 \%
$$

but will vary depending on the base ionization constant and the initial concentration of the solution.


1489. Relative Strengths of Conjugate Acid-Base Pairs -
Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, $K_{\mathrm{a}}$ or $K_{\mathrm{b}}$, which represents the extent of the acid or base ionization reaction. For the conjugate acidbase pair HA / $\mathrm{A}^{-}$, ionization equilibrium equations and ionization constant expressions are

$$
\begin{aligned}
\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q) & K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \\
\mathrm{A}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{HA}(a q) & K_{\mathrm{b}}=\frac{[\mathrm{HA}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{A}^{-}\right]}
\end{aligned}
$$

Adding these two chemical equations yields the equation for the autoionization for water:

$$
\begin{gathered}
\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{A}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{HA}(a q) \\
2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)
\end{gathered}
$$


1489. Relative Strengths of Conjugate Acid-Base Pairs -
As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so

$$
K_{\mathrm{a}} \times K_{\mathrm{b}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \times \frac{[\mathrm{HA}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{A}^{-}\right]}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=K_{\mathrm{w}}
$$

This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, $K_{\mathrm{w}}$. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:

$$
K_{\mathrm{a}}=K_{\mathrm{w}} / K_{\mathrm{b}} \text { or } K_{\mathrm{b}}=K_{\mathrm{w}} / K_{\mathrm{a}}
$$

The inverse proportional relation between $K_{\mathrm{a}}$ and $K_{\mathrm{b}}$ means the stronger the acid or base, the weaker its conjugate partner. Figure 14.7 illustrates this relation for several conjugate acid-base pairs.

Relative conjugate base strength

FIGURE 14.7 Relative strengths of several conjugate acid-base pairs are shown.

FIGURE 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.