1343. Solution -
$$
\begin{gathered}
=\left[1 \mathrm{~mol}(75.9 \mathrm{~J} / \mathrm{mol} \mathrm{~K})+\frac{1}{2} \operatorname{mol}(205.2 \mathrm{~J} / \mathrm{mol} \mathrm{~K})\right]-1 \mathrm{~mol}(71.13 \mathrm{~J} / \mathrm{mol} \mathrm{~K})=107.4 \mathrm{~J} / \mathrm{mol} \mathrm{~K} \\
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=90.46 \mathrm{~kJ}-298.15 \mathrm{~K} \times 107.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{~mol} \times \frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}} \\
\Delta G^{\circ}=(90.46-32.01) \mathrm{kJ} / \mathrm{mol}=58.45 \mathrm{~kJ} / \mathrm{mol}
\end{gathered}
$$

Both ways to calculate the standard free energy change at $25^{\circ} \mathrm{C}$ give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.


1344. Check Your Learning -
Calculate $\Delta G^{\circ}$ using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at $25^{\circ} \mathrm{C}$ ?

$$
\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g)
$$


1345. Answer: -
(a) $140.8 \mathrm{~kJ} / \mathrm{mol}$, nonspontaneous
(b) $141.5 \mathrm{~kJ} / \mathrm{mol}$, nonspontaneous


1346. Free Energy Changes for Coupled Reactions -
The use of free energies of formation to compute free energy changes for reactions as described above is possible because $\Delta \mathrm{G}$ is a state function, and the approach is analogous to the use of Hess' Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:

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

An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:

$$
\begin{array}{ll}
\mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta G_{\mathrm{f}, \mathrm{gas}}^{\circ} \\
\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) & -\Delta G_{\mathrm{f}, \text { liquid }}^{\circ} \\
\hline \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta G=\Delta G_{\mathrm{f}, \mathrm{gas}}^{\circ}-\Delta G_{\mathrm{f}, \text { liquid }}^{\circ}
\end{array}
$$

This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for $\Delta G^{\circ}$ :

$$
\mathrm{ZnS}(s) \rightarrow \mathrm{Zn}(s)+\mathrm{S}(s) \quad \Delta G_{1}^{\circ}=201.3 \mathrm{~kJ}
$$

The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:


1346. Free Energy Changes for Coupled Reactions -
The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:

$$
\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) \quad \Delta G_{2}^{\circ}=-300.1 \mathrm{~kJ}
$$

The coupled reaction exhibits a negative free energy change and is spontaneous:

$$
\mathrm{ZnS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Zn}(s)+\mathrm{SO}(g) \quad \Delta G^{\circ}=201.3 \mathrm{~kJ}+-300.1 \mathrm{~kJ}=-98.8 \mathrm{~kJ}
$$

This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.


1348. Calculating Free Energy Change for a Coupled Reaction -
Is a reaction coupling the decomposition of ZnS to the formation of $\mathrm{H}_{2} \mathrm{~S}$ expected to be spontaneous under standard conditions?


1349. Solution -
Following the approach outlined above and using free energy values from Appendix G:

$$
\begin{array}{lll}
\text { Decomposition of zinc sulfide: } & \mathrm{Zn}(s) \rightarrow \mathrm{Zn}(s)+\mathrm{S}(s) & \Delta G_{1}^{\circ}=201.3 \mathrm{~kJ} \\
\text { Formation of hydrogen sulfide: } & \mathrm{S}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{~S}(g) & \Delta G_{2}^{\circ}=-33.4 \mathrm{~kJ} \\
\text { Coupled reaction: } & \mathrm{ZnS}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{~S}(g) & \Delta G^{\circ}=201.3 \mathrm{~kJ}+-33.4 \mathrm{~kJ}=167.9 \mathrm{~kJ}
\end{array}
$$

The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.


1350. Check Your Learning -
What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?

$$
\mathrm{FeS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{SO}_{2}(g)
$$


1351. Answer: -
-199.7 kJ; spontaneous


1352. Key Terms -
entropy ( $\boldsymbol{S}$ ) state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates; often described as a measure of the disorder of the system
Gibbs free energy change ( $\boldsymbol{G}$ ) thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in $G$
microstate possible configuration or arrangement of matter and energy within a system
nonspontaneous process process that requires continual input of energy from an external source
reversible process process that takes place so slowly as to be capable of reversing direction in response to an infinitesimally small change in conditions; hypothetical construct that can only be approximated by real processes
second law of thermodynamics all spontaneous processes involve an increase in the entropy of the universe


1353. Key Equations -
$\Delta S=\frac{q_{\mathrm{rev}}}{T}$
$S=k \ln W$
$\Delta S=k \ln \frac{W_{\mathrm{f}}}{W_{\mathrm{i}}}$
$\Delta S^{\circ}=\sum \nu S^{\circ}($ products $)-\sum \nu S^{\circ}($ reactants $)$
$\Delta S=\frac{q_{\mathrm{rev}}}{T}$
$\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}$
$\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T}$
$\Delta G=\Delta H-T \Delta S$


1354. Summary - 1354.1. Spontaneity
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system.


1354. Summary - 1354.2. Entropy
Entropy ( $S$ ) is a state function that can be related to
spontaneous change process that takes place without a continuous input of energy from an external source
standard entropy ( $\boldsymbol{S}^{\circ}$ ) entropy for one mole of a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K
standard entropy change ( $\boldsymbol{\Delta} \boldsymbol{S}^{\circ}$ ) change in entropy for a reaction calculated using the standard entropies
standard free energy change ( $\boldsymbol{\Delta} \boldsymbol{G}^{\mathbf{0}}$ ) change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)
standard free energy of formation ( $\Delta \boldsymbol{G}_{\mathrm{f}}^{\circ}$ ) change in free energy accompanying the formation of one mole of substance from its elements in their standard states
third law of thermodynamics entropy of a perfect crystal at absolute zero ( 0 K ) is zero
the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the "disorder" of the system.

For a given substance, entropy depends on phase with $S_{\text {solid }}<S_{\text {liquid }}<S_{\text {gas }}$. For different substances in the same physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for
some chemical reactions and physical changes may be reliably predicted.


1354. Summary - 1354.3. The Second and Third Laws of Thermodynamics
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, $S_{\text {univ }}>0$. If $\Delta S_{\text {univ }}<0$, the process is nonspontaneous, and if $\Delta S_{\text {univ }}=0$, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K . With only one possible
microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.


1354. Summary - 1354.4. Free Energy
Gibbs free energy $(G)$ is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A number of approaches to the computation of free energy changes are possible.


1355. Exercises - 1355.1. Spontaneity
1. What is a spontaneous reaction?
2. What is a nonspontaneous reaction?
3. Indicate whether the following processes are spontaneous or nonspontaneous.
(a) Liquid water freezing at a temperature below its freezing point
(b) Liquid water freezing at a temperature above its freezing point
(c) The combustion of gasoline
(d) A ball thrown into the air
(e) A raindrop falling to the ground
(f) Iron rusting in a moist atmosphere
4. A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
5. Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.


1355. Exercises - 1355.2. Entropy
6. In Figure 12.8 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, $\Delta S$, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
7. In Figure 12.8 all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, $\Delta S$, for the system when it is converted from distribution (b) to distribution (d).
8. How does the process described in the previous item relate to the system shown in Figure 12.4?
9. Consider a system similar to the one in Figure 12.8, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to $\frac{1}{8}$. What does this comparison tell us about even larger systems?
10. Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
11. Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
12. Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
(a) $\mathrm{H}_{2}(g), \mathrm{HBrO}_{4}(g), \mathrm{HBr}(g)$
(b) $\mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{s})$
(c) $\mathrm{He}(g), \mathrm{Cl}_{2}(g), \mathrm{P}_{4}(g)$


1355. Exercises - 1355.2. Entropy
(c) $\mathrm{He}(g), \mathrm{Cl}_{2}(g), \mathrm{P}_{4}(g)$
13. At room temperature, the entropy of the halogens increases from $I_{2}$ to $B r_{2}$ to $\mathrm{Cl}_{2}$. Explain.
14. Consider two processes: sublimation of $I_{2}(s)$ and melting of $I_{2}(S)$ (Note: the latter process can occur at the same temperature but somewhat higher pressure).
$\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)$
$\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)$
Is $\Delta S$ positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
15. Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
(a) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)$ or $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(I)$
(b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{I})$ or $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)$
(c) $2 \mathrm{H}(\mathrm{g})$ or $\mathrm{H}(\mathrm{g})$
16. Predict the sign of the entropy change for the following processes.
(a) An ice cube is warmed to near its melting point.
(b) Exhaled breath forms fog on a cold morning.
(c) Snow melts.
17. Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
(a) $\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{NaCl}(s)$
(b) $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{2}(s)$


1355. Exercises - 1355.2. Entropy
(b) $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{2}(s)$
(c) $2 \mathrm{C}_{6} \mathrm{H}_{14}(l)+19 \mathrm{O}_{2}(g) \longrightarrow 14 \mathrm{H}_{2} \mathrm{O}(g)+12 \mathrm{CO}_{2}(g)$
18. Write the balanced chemical equation for the combustion of methane, $\mathrm{CH}_{4}(g)$, to give carbon dioxide and water vapor. Explain why it is difficult to predict whether $\Delta S$ is positive or negative for this chemical reaction.
19. Write the balanced chemical equation for the combustion of benzene, $\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})$, to give carbon dioxide and water vapor. Would you expect $\Delta S$ to be positive or negative in this process?


1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics
20. What is the difference between $\Delta S$ and $\Delta S^{\circ}$ for a chemical change?
21. Calculate $\Delta S^{\circ}$ for the following changes.
(a) $\mathrm{SnCl}_{4}(l) \longrightarrow \mathrm{SnCl}_{4}(g)$
(b) $\mathrm{CS}_{2}(g) \longrightarrow \mathrm{CS}_{2}(l)$
(c) $\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}(g)$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)$
(e) $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)$
(f) $2 \mathrm{HCl}(g)+\mathrm{Pb}(s) \longrightarrow \mathrm{PbCl}_{2}(s)+\mathrm{H}_{2}(g)$
$(\mathrm{g}) \mathrm{Zn}(s)+\mathrm{CuSO}_{4}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{ZnSO}_{4}(s)$
22. Determine the entropy change for the combustion of liquid ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, under the standard conditions to give gaseous carbon dioxide and liquid water.
23. Determine the entropy change for the combustion of gaseous propane, $\mathrm{C}_{3} \mathrm{H}_{8}$, under the standard conditions to give gaseous carbon dioxide and water.
24. "Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s)$. Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb $851.8 \mathrm{~kJ} / \mathrm{mol}$ of heat.


1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics
25. Using the relevant $S^{\circ}$ values listed in Appendix G, calculate $\Delta S^{\circ}{ }_{298}$ for the following changes:
(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$
(b) $\mathrm{N}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)$
26. From the following information, determine $\Delta S^{\circ}$ for the following:
$\mathrm{N}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g) \quad \Delta S^{\circ}=$ ?
$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta S^{\circ}=24.8 \mathrm{~J} / \mathrm{K}$
$\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{~N}(g) \quad \Delta S^{\circ}=115.0 \mathrm{~J} / \mathrm{K}$
$\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \quad \Delta S^{\circ}=117.0 \mathrm{~J} / \mathrm{K}$
27. By calculating $\Delta S_{\text {univ }}$ at each temperature, determine if the melting of 1 mole of $\mathrm{NaCl}(s)$ is spontaneous at $500^{\circ} \mathrm{C}$ and at $700^{\circ} \mathrm{C}$.
$S_{\mathrm{NaCl}(s)}^{\circ}=72.11 \frac{\mathrm{~J}}{\mathrm{~mol} \cdot \mathrm{~K}} \quad S_{\mathrm{NaCl}(l)}^{\circ}=95.06 \frac{\mathrm{~J}}{\mathrm{~mol} \cdot \mathrm{~K}} \quad \Delta H_{\text {fusion }}^{\circ}=27.95 \mathrm{~kJ} / \mathrm{mol}$


1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
28. Use the standard entropy data in Appendix $G$ to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and $25^{\circ} \mathrm{C}$.
(a) $\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{O}_{2}(g)$
(b) $\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(l) \longrightarrow 2 \mathrm{HBr}(g)$
(c) $\mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{CuS}(s)$
(d) $2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)$
(e) $\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{C}(s$, graphite $)+2 \mathrm{H}_{2} \mathrm{O}(g)$
(f) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)$
29. Use the standard entropy data in Appendix $G$ to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and $25^{\circ} \mathrm{C}$.
(a) $\mathrm{C}(s$, graphite $)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$
(b) $\mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$
(c) $2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)$


1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics
(c) $2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)$
(d) $\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)$
(e) $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$
(f) $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$


1355. Exercises - 1355.4. Free Energy
30. What is the difference between $\Delta G$ and $\Delta G^{\circ}$ for a chemical change?
31. A reaction has $\Delta H^{\circ}=100 \mathrm{~kJ} / \mathrm{mol}$ and $\Delta S^{\circ}=250 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
32. Explain what happens as a reaction starts with $\Delta G<0$ (negative) and reaches the point where $\Delta G=0$.
33. Use the standard free energy of formation data in Appendix $G$ to determine the free energy change for each of the following reactions, which are run under standard state conditions and $25^{\circ} \mathrm{C}$. Identify each as either spontaneous or nonspontaneous at these conditions.
(a) $\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{O}_{2}(g)$
(b) $\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(l) \longrightarrow 2 \mathrm{HBr}(g)$
(c) $\mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{CuS}(s)$
(d) $2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)$
(e) $\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{C}(s$, graphite $)+2 \mathrm{H}_{2} \mathrm{O}(g)$
(f) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)$
34. Use the standard free energy data in Appendix $G$ to determine the free energy change for each of the following reactions, which are run under standard state conditions and $25^{\circ} \mathrm{C}$. Identify each as either spontaneous or nonspontaneous at these conditions.


1355. Exercises - 1355.4. Free Energy
(a) $\mathrm{C}(s$, graphite $)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$
(b) $\mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$
(c) $2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)$
(d) $\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)$
(e) $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$
(f) $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$
35. Given:
$\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$
$\Delta G^{\circ}=-2697.0 \mathrm{~kJ} / \mathrm{mol}$
$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$
$\Delta G^{\circ}=-457.18 \mathrm{~kJ} / \mathrm{mol}$
$6 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(l)$
$\Delta G^{\circ}=-428.66 \mathrm{~kJ} / \mathrm{mol}$
(a) Determine the standard free energy of formation, $\Delta G_{\mathrm{f}}^{\circ}$, for phosphoric acid.


1355. Exercises - 1355.4. Free Energy
(a) Determine the standard free energy of formation, $\Delta G_{\mathrm{f}}^{\circ}$, for phosphoric acid.
(b) How does your calculated result compare to the value in Appendix G? Explain.
36. Is the formation of ozone $\left(\mathrm{O}_{3}(\mathrm{~g})\right)$ from oxygen $\left(\mathrm{O}_{2}(\mathrm{~g})\right)$ spontaneous at room temperature under standard state conditions?
37. Consider the decomposition of red mercury(II) oxide under standard state conditions.
$2 \mathrm{HgO}(s$, red $) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
38. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
(a) Ammonia: $2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
(b) Diborane: $\mathrm{B}_{2} \mathrm{H}_{6}(g) \longrightarrow 2 \mathrm{~B}(g)+3 \mathrm{H}_{2}(g)$
(c) Hydrazine: $\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)$
(d) Hydrogen peroxide: $\mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$
39. Calculate $\Delta G^{\circ}$ for each of the following reactions from the equilibrium constant at the temperature given.
(a) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$


1355. Exercises - 1355.4. Free Energy
(a) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$
$\mathrm{T}=2000^{\circ} \mathrm{C} \quad K_{p}=4.1 \times 10^{-4}$
(b) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)$
$\mathrm{T}=400^{\circ} \mathrm{C} \quad K_{p}=50.0$
(c) $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ $\mathrm{T}=980^{\circ} \mathrm{C} \quad K_{p}=1.67$
$(\mathrm{d}) \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
$\mathrm{T}=900^{\circ} \mathrm{C} \quad K_{p}=1.04$
(e) $\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)$
$\mathrm{T}=25^{\circ} \mathrm{C}$
$K_{p}=7.2 \times 10^{-4}$
(f) $\mathrm{AgBr}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Br}^{-}(a q)$ $\mathrm{T}=25^{\circ} \mathrm{C} \quad K_{p}=3.3 \times 10^{-13}$
40. Calculate $\Delta G^{\circ}$ for each of the following reactions from the equilibrium constant at the temperature given.
(a) $\mathrm{Cl}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{BrCl}(g)$
$\mathrm{T}=25^{\circ} \mathrm{C} \quad K_{p}=4.7 \times 10^{-2}$


1355. Exercises - 1355.4. Free Energy
$\mathrm{T}=25^{\circ} \mathrm{C} \quad K_{p}=4.7 \times 10^{-2}$
(b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
$\mathrm{T}=500^{\circ} \mathrm{C}$
$K_{p}=48.2$
(c) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)$
$\mathrm{T}=60^{\circ} \mathrm{C}$
$K_{p}=0.196$
(d) $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(\mathrm{s})+\mathrm{CO}_{2}(g)$
$\mathrm{T}=550^{\circ} \mathrm{C} \quad K_{p}=4.90 \times 10^{2}$
(e) $\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}(a q)+\mathrm{OH}^{-}(a q)$
$\mathrm{T}=25^{\circ} \mathrm{C} \quad K_{p}=4.4 \times 10^{-4}$
(f) $\mathrm{PbI}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)$
$\mathrm{T}=25^{\circ} \mathrm{C}$
$K_{p}=8.7 \times 10^{-9}$
41. Calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for each of the following reactions from the value of $\Delta G^{\circ}$ given.
(a) $\mathrm{O}_{2}(g)+2 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{OF}_{2}(g)$
$\Delta G^{\circ}=-9.2 \mathrm{~kJ}$


1355. Exercises - 1355.4. Free Energy
$\Delta G^{\circ}=-9.2 \mathrm{~kJ}$
(b) $\mathrm{I}_{2}(s)+\mathrm{Br}_{2}(l) \longrightarrow 2 \operatorname{IBr}(g)$
$\Delta G^{\circ}=7.3 \mathrm{~kJ}$
(c) $2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)$
$\Delta G^{\circ}=-79 \mathrm{~kJ}$
$(d) \mathrm{N}_{2} \mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)$
$\Delta G^{\circ}=-1.6 \mathrm{~kJ}$
(e) $\mathrm{SnCl}_{4}(l) \longrightarrow \mathrm{SnCl}_{4}(l)$
$\Delta G^{\circ}=8.0 \mathrm{~kJ}$
42. Determine $\Delta G^{\circ}$ for the following reactions.
(a) Antimony pentachloride decomposes at $448^{\circ} \mathrm{C}$. The reaction is:
$\mathrm{SbCl}_{5}(\mathrm{~g}) \longrightarrow \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$
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 $\mathrm{Cl}_{2}$.
(b) Chlorine molecules dissociate according to this reaction:
$\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)$
$1.00 \%$ of $\mathrm{Cl}_{2}$ molecules dissociate at 975 K and a pressure of 1.00 atm .


1355. Exercises - 1355.4. Free Energy
$1.00 \%$ of $\mathrm{Cl}_{2}$ molecules dissociate at 975 K and a pressure of 1.00 atm .
43. Given that the $\Delta G_{\mathrm{f}}^{\circ}$ for $\mathrm{Pb}^{2+}(\mathrm{aq})$ and $\mathrm{Cl}^{-}(\mathrm{aq})$ is $-24.3 \mathrm{~kJ} / \mathrm{mole}$ and $-131.2 \mathrm{~kJ} /$ mole respectively, determine the solubility product, $K_{\mathrm{sp}}$, for $\mathrm{PbCl}_{2}(s)$.
44. Determine the standard free energy change, $\Delta \boldsymbol{G}_{\mathrm{f}}^{\circ}$, for the formation of $S^{2-}(a q)$ given that the $\Delta \boldsymbol{G}_{\mathrm{f}}^{\circ}$ for $\mathrm{Ag}^{+}(\mathrm{aq})$ and $\mathrm{Ag}_{2} \mathrm{~S}(s)$ are $77.1 \mathrm{~kJ} /$ mole and $-39.5 \mathrm{~kJ} /$ mole respectively, and the solubility product for $\mathrm{Ag}_{2} \mathrm{~S}(s)$ is $8 \times 10^{-51}$.
45. Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
46. The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ .
$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta G^{\circ}=8.58 \mathrm{~kJ}$
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, $K_{P}$, for this physical process.


1355. Exercises - 1355.4. Free Energy
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, $K_{P}$, for this physical process.
(c) By calculating $\Delta G$, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, $P_{\mathrm{H}_{2} \mathrm{O}}$, is 0.011 atm .
(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of $P_{\mathrm{H}_{2} \mathrm{O}}$ in the air?
47. In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
$\mathrm{Glu}+\mathrm{ATP} \longrightarrow \mathrm{G} 6 \mathrm{P}+\mathrm{ADP} \quad \Delta G^{\circ}=-17 \mathrm{~kJ}$
In this process, ATP becomes ADP summarized by the following equation:
ATP $\longrightarrow$ ADP $\Delta G^{\circ}=-30 \mathrm{~kJ}$
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
Glu $\longrightarrow$ G6P $\quad \Delta G^{\circ}=$ ?
48. One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
$\mathrm{G6P} \rightleftharpoons \mathrm{~F} 6 \mathrm{P} \quad \Delta G^{\circ}=1.7 \mathrm{~kJ}$
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?


1355. Exercises - 1355.4. Free Energy
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be $1 M$, however, in a typical cell, they are not even close to these values. Calculate $\Delta G$ when the concentrations of G6P and F6P are $120 \mu M$ and $28 \mu M$ respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is $37^{\circ} \mathrm{C}$.
49. Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$
(b) $\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$
(c) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+4 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)$
(d) $2 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$
50. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of $\Delta G, \Delta H$, and $\Delta S$ for this process, and justify your choices.
51. An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the $\mathrm{Cu}_{2} \mathrm{~S}$ decomposes to form copper and sulfur described by the following equation:
$\mathrm{Cu}_{2} \mathrm{~S}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{S}(s)$


1355. Exercises - 1355.4. Free Energy
$\mathrm{Cu}_{2} \mathrm{~S}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{S}(s)$
(a) Determine $\Delta \boldsymbol{G}^{\circ}$ for the decomposition of $\mathrm{Cu}_{2} \mathrm{~S}(S)$.
(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine $\Delta G^{\circ}$ for the process.
(c) The production of copper from chalcocite is performed by roasting the $\mathrm{Cu}_{2} \mathrm{~S}$ in air to produce the Cu . By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
52. What happens to $\Delta G^{\circ}$ (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
(a) $\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g})$
(b) $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{3}(\mathrm{~g})$
(c) $\mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$


1355. Exercises - 1355.4. Free Energy
626 12•Exercises


1356. CHAPTER 13 <br> Fundamental Equilibrium Concepts -
Figure 13.1 Transport of carbon dioxide in the body involves several reversible chemical reactions, including hydrolysis and acid ionization (among others).


1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria
13.2 Equilibrium Constants
13.3 Shifting Equilibria: Le Châtelier's Principle
13.4 Equilibrium Calculations

INTRODUCTION Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure 13.1). This chapter provides a thorough introduction to the essential aspects of chemical equilibria.

We now have a good understanding of chemical and physical change that allow us to determine, for any given process:

1. Whether the process is endothermic or exothermic
2. Whether the process is accompanied by an increase of decrease in entropy
3. Whether a process will be spontaneous, non-spontaneous, or what we have called an equilibrium process

Recall that when the value $\Delta G$ for a reaction is zero, we consider there to be no free energy change-that is, no
free energy available to do useful work. Does this mean a reaction where $\Delta G=0$ comes to a complete halt? No, it does not. Just as a liquid exists in equilibrium with its vapor in a closed container, where the rates of evaporation and condensation are equal, there is a connection to the state of equilibrium for a phase change or a chemical reaction. That is, at equilibrium, the forward and reverse rates of reaction are equal. We will develop that concept and extend it to a relationship between equilibrium and free energy later in this chapter.

In the explanation that follows, we will use the term $Q$ to refer to any reactant or product concentration or pressure. When the concentrations or pressure of reactants and products are at equilibrium, the term $K$ will be used. This will be more clearly explained as we go along in this chapter.


1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria
Now we will consider the connection between the free energy change and the equilibrium constant. The fundamental relationship is:
$\Delta G^{\circ}=-\mathrm{R} T \ln K$-this can be for $K_{c}$ or $K_{p}$ (and we will see later, any equilibrium constant we encounter).
We also know that the form of $K$ can be used in non-equilibrium conditions as the reaction quotient, $Q$. The defining relationship here is

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

Without the superscript, the value of $\Delta G$ can be calculated for any set of concentrations.
Note that since $Q$ is a mass-action reaction of productions/reactants, as a reaction proceeds from left to right, product concentrations increase as reactant concentrations decrease, until $Q=K$, and at which time $\Delta G$ becomes zero:
$0=\Delta G^{\circ}+R T \ln K$, a relationship that reduces to our defining connection between $Q$ and $K$.
Thus, we can see clearly that as a reaction moves toward equilibrium, the value of $\Delta G$ goes to zero.
Now, think back to the connection between the signs of $\Delta G^{\circ}$ and $\Delta H^{\circ}$
$\Delta H^{\circ} \quad \Delta S^{\circ} \quad$ Result
Negative Positive Always spontaneous
Positive Negative Never spontaneous
Positive Positive Spontaneous at high temperatures
Negative Negative Spontaneous at low temperatures
Only in the last two cases is there a point at which the process swings from spontaneous to non-spontaneous (or the reverse); in these cases, the process must pass through equilibrium when the change occurs. The concept of the connection between the free energy change and the equilibrium constant is an important one that we will expand upon in future sections. The fact that the change in free energy for an equilibrium process is zero, and that displacement of a process from that zero point results in a drive to re-establish equilibrium is fundamental to understanding the behavior of chemical reactions and phase changes.


1358. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the nature of equilibrium systems
- Explain the dynamic nature of a chemical equilibrium

The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of "reactant" and "product," a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.

Figure 13.2 illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:

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

Note that a special double arrow is used to emphasize the reversible nature of the reaction.

FIGURE 13.2 (a) A sealed tube containing colorless $\mathrm{N}_{2} \mathrm{O}_{4}$ darkens as it decomposes to yield brown $\mathrm{NO}_{2}$. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal.

For this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:

$$
\begin{aligned}
\operatorname{rate}_{f} & =k_{f}\left[\mathrm{~N}_{-2} \mathrm{O}_{4}\right] \\
\operatorname{rate}_{r} & =k_{r}\left[\mathrm{NO}_{2}\right]^{2}
\end{aligned}
$$


1358. LEARNING OBJECTIVES -
As the reaction begins ( $t=0$ ), the concentration of the $\mathrm{N}_{2} \mathrm{O}_{4}$ reactant is finite and that of the $\mathrm{NO}_{2}$ product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N.
${ }_{2} \mathrm{O}_{4}$ is consumed and its concentration falls, while $\mathrm{NO}_{2}$ is produced and its concentration increases (Figure $13.2 \mathbf{b})$. The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure 13.2c). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure $13.2 \mathbf{b}$ and Figure 13.2c). It's important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not "stopped," but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.

FIGURE 13.3 A two-person juggling act illustrates the dynamic aspect of chemical equilibria. Each person is throwing and catching clubs at the same rate, and each holds a (approximately) constant number of clubs.

Physical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:

$$
\operatorname{Br}_{2}(l) \rightleftharpoons \mathrm{Br}_{2}(g)
$$

When liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in Figure 13.4.


1358. LEARNING OBJECTIVES -
FIGURE 13.4 A sealed tube containing an equilibrium mixture of liquid and gaseous bromine. (credit: http://images-of-elements.com/bromine.php)


1359. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
- Relate the magnitude of an equilibrium constant to properties of the chemical system

The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( $\boldsymbol{Q}$ ). For a reversible reaction described by

$$
m \mathrm{~A}+n \mathrm{~B} \rightleftharpoons x \mathrm{C}+y \mathrm{D}
$$

the reaction quotient is derived directly from the stoichiometry of the balanced equation as

$$
Q_{c}=\frac{[\mathrm{C}]^{x}[\mathrm{D}]^{y}}{[\mathrm{~A}]^{m}[\mathrm{~B}]^{n}}
$$

where the subscript $c$ denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:

$$
Q_{p}=\frac{P_{\mathrm{C}}^{x} P_{\mathrm{D}}^{y}}{P_{\mathrm{A}}^{m} P_{\mathrm{B}}^{n}}
$$

Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing $Q$. In most cases, this will introduce only modest errors in calculations involving reaction quotients.


1361. Writing Reaction Quotient Expressions -
Write the concentration-based reaction quotient expression for each of the following reactions:
(a) $3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$
(b) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
(c) $4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$


1362. Solution -
(a) $Q_{c}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]^{3}}$
(b) $Q_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}$
(c) $Q_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}}{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}$


1363. Check Your Learning -
Write the concentration-based reaction quotient expression for each of the following reactions:
(a) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
(b) $\mathrm{C}_{4} \mathrm{H}_{8}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{4}(g)$
(c) $2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \rightleftharpoons 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)$


1364. Answer: -
(a) $Q_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}$; (b) $Q_{c}=\frac{\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]^{2}}{\left[\mathrm{C}_{4} \mathrm{H}_{8}\right]} ;$ (c) $Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]^{8}\left[\mathrm{H}_{2} \mathrm{O}\right]{ }^{10}}{\left[\mathrm{C}_{4} \mathrm{H}_{10}\right]^{2}\left[\mathrm{O}_{2}\right]^{13}}$

FIGURE 13.5 Changes in concentrations and $Q_{C}$ for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.

The numerical value of $Q$ varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction's status. To illustrate this point, consider the oxidation of sulfur dioxide:

$$
2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)
$$

Two different experimental scenarios are depicted in Figure 13.5, one in which this reaction is initiated with a mixture of reactants only, $\mathrm{SO}_{2}$ and $\mathrm{O}_{2}$, and another that begins with only product, $\mathrm{SO}_{3}$. For the reaction that begins with a mixture of reactants only, $Q$ is initially equal to zero:

$$
Q_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{0^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=0
$$


1364. Answer: -
As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of $Q_{c}$ ), product concentration increases (as does the numerator of $Q_{c}$ ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of $Q_{c}$.

If the reaction begins with only product present, the value of $Q_{c}$ is initially undefined (immeasurably large, or infinite):

$$
Q_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{0} \rightarrow \infty
$$

In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of $Q_{c}$ decrease with time, the reactant concentrations and the denominator of $Q_{c}$ increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.

The constant value of $Q$ exhibited by a system at equilibrium is called the equilibrium constant, $\boldsymbol{K}$ :

$$
K \equiv Q \text { at equilibrium }
$$

Comparison of the data plots in Figure 13.5 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant.


1366. Evaluating a Reaction Quotient -
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

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

When $0.10 \mathrm{~mol} \mathrm{NO}_{2}$ is added to a 1.0-L flask at $25^{\circ} \mathrm{C}$, the concentration changes so that at equilibrium, $\left[\mathrm{NO}_{2}\right]=$ 0.016 M and $\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.042 \mathrm{M}$.
(a) What is the value of the reaction quotient before any reaction occurs?
(b) What is the value of the equilibrium constant for the reaction?


1367. Solution -
As for all equilibrium calculations in this text, use the simplified equations for $Q$ and $K$ and disregard any concentration or pressure units, as noted previously in this section.
(a) Before any product is formed, $\left[\mathrm{NO}_{2}\right]=\frac{0.10 \mathrm{~mol}}{1.0 \mathrm{~L}}=0.10 M$, and $\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0 \mathrm{M}$. Thus,

$$
Q_{c}=\frac{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\left[\mathrm{NO}_{2}\right]^{2}}=\frac{0}{0.10^{2}}=0
$$

(b) At equilibrium, $K_{c}=Q_{c}=\frac{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\left[\mathrm{NO}_{2}\right]^{2}}=\frac{0.042}{0.016^{2}}=1.6 \times 10^{2}$. The equilibrium constant is $1.6 \times 10^{2}$.


1368. Check Your Learning -
For the reaction $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$, the concentrations at equilibrium are $\left[\mathrm{SO}_{2}\right]=0.90 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.35$ $M$, and $\left[\mathrm{SO}_{3}\right]=1.1 \mathrm{M}$. What is the value of the equilibrium constant, $K_{c}$ ?


1369. Answer: -
$K_{c}=4.3$

By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large $K$ will reach equilibrium when most of the reactant has been converted to product, whereas a small $K$ indicates the reaction achieves equilibrium after very little reactant has been converted. It's important to keep in mind that the magnitude of $K$ does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.

The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing $Q$ to $K$ for
an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below:

$$
\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \quad K_{c}=0.640 \quad \mathrm{~T}=800^{\circ} \mathrm{C}
$$

The bar charts in Figure 13.6 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction's equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.


1369. Answer: -
FIGURE 13.6 Compositions of three mixtures before $\left(Q_{c} \neq K_{c}\right)$ and after $\left(Q_{c}=K_{c}\right)$ equilibrium is established for the reaction $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$.


1371. Predicting the Direction of Reaction -
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:

$$
\begin{gathered}
\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \\
K_{c}=0.64
\end{gathered}
$$

Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.

| Reactants/Products |  | Experiment 1 | Experiment 2 |
| :--- | :--- | :--- | :--- |
| $[\mathrm{CO}]_{\mathrm{i}}$ | 0.020 M | 0.011 M | 0.0094 M |
| $\left[\mathrm{H}_{2} \mathrm{O}\right]_{\mathrm{i}}$ | 0.020 M | 0.0011 M | 0.0025 M |
| $\left[\mathrm{CO}_{2}\right]_{\mathrm{i}}$ | 0.0040 M | 0.037 M | 0.0015 M |
| $\left[\mathrm{H}_{2}\right]_{\mathrm{i}}$ | 0.0040 M | 0.046 M | 0.0076 M |


1372. Solution -
Experiment 1:

$$
Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040
$$

$Q_{c}<K_{C}(0.040<0.64)$
The reaction will proceed in the forward direction.
Experiment 2:

$$
Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^{2}
$$

$Q_{c}>K_{c}(140>0.64)$
The reaction will proceed in the reverse direction.
Experiment 3:

$$
Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48
$$

$Q_{c}<K_{c}(0.48<0.64)$
The reaction will proceed in the forward direction.


1373. Check Your Learning -
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
(a) A 1.00-L flask containing 0.0500 mol of $\mathrm{NO}(\mathrm{g}), 0.0155 \mathrm{~mol}^{2} \mathrm{Cl}_{2}(\mathrm{~g})$, and 0.500 mol of NOCl :

$$
2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) \quad K_{c}=4.6 \times 10^{4}
$$

(b) A 5.0-L flask containing 17 g of $\mathrm{NH}_{3}, 14 \mathrm{~g}$ of $\mathrm{N}_{2}$, and 12 g of $\mathrm{H}_{2}$ :

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \quad K_{c}=0.060
$$

(c) A 2.00-L flask containing 230 g of $\mathrm{SO}_{3}(\mathrm{~g})$ :

$$
2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=0.230
$$


1374. Answer: -
(a) $Q_{c}=6.45 \times 10^{3}$, forward. (b) $Q_{c}=0.23$, reverse. (c) $Q_{c}=0$, forward.


1375. Homogeneous Equilibria -
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

$$
\begin{aligned}
\mathrm{C}_{2} \mathrm{H}_{2}(a q)+2 \mathrm{Br}_{2}(a q) & \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Br}_{4}(a q) & K_{c} & =\frac{\left[\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Br}_{4}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{2}} \\
\mathrm{I}_{2}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{I}_{3}^{-}(a q) & K_{c} & =\frac{\left[\mathrm{I}_{3}^{-}\right]}{\left[\mathrm{I}_{2}\right]\left[\mathrm{I}^{-}\right]} \\
\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) & K_{c} & =\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]} \\
\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) & K_{c} & =\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}
\end{aligned}
$$


1375. Homogeneous Equilibria -
These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two
examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the $Q$ (or $K$ ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn't be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

The equilibria below all involve gas-phase solutions:


1375. Homogeneous Equilibria -
The equilibria below all involve gas-phase solutions:

$$
\begin{aligned}
\mathrm{C}_{2} \mathrm{H}_{6}(g) & \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) & K_{c} & =\frac{\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]} \\
3 \mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{O}_{3}(g) & K_{c} & =\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]^{3}} \\
\mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g) & \rightleftharpoons 2 \mathrm{NH}_{3}(g) & K_{c} & =\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}} \\
\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) & \rightleftharpoons 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) & K_{c} & =\frac{\left[\mathrm{CO}_{2}\right]^{3}\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}{\left[\mathrm{C}_{3} \mathrm{H}_{8}\right]\left[\mathrm{O}_{2}\right]^{5}}
\end{aligned}
$$

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ( $K_{c}$ ) or partial pressures $\left(K_{p}\right)$ of the reactants and products. A relation between these two $K$ values may be simply derived from the ideal gas equation and the definition of molarity:


1375. Homogeneous Equilibria -
$$
\begin{gathered}
P V=n R T \\
\begin{array}{c}
P=\left(\frac{n}{V}\right) R T \\
=M R T
\end{array}
\end{gathered}
$$

where $P$ is partial pressure, $V$ is volume, $n$ is molar amount, $R$ is the gas constant, $T$ is temperature, and $M$ is molar concentration.

For the gas-phase reaction $m \mathrm{~A}+n \mathrm{~B} \rightleftharpoons x \mathrm{C}+y \mathrm{D}$ :

$$
\begin{gathered}
K_{P}=\frac{\left(P_{C}\right)^{x}\left(P_{D}\right)^{y}}{\left(P_{A}\right)^{m}\left(P_{B}\right)^{n}} \\
=\frac{([\mathrm{C}] \times R T)^{x}([\mathrm{D}] \times R T)^{y}}{([\mathrm{~A}] \times R T)^{m}([\mathrm{~B}] \times R T)^{n}} \\
=\frac{[\mathrm{C}]^{x}[\mathrm{D}]^{y}}{[\mathrm{~A}]^{m}[\mathrm{~B}]^{n}} \times \frac{(R T)^{x+y}}{(R T)^{m+n}} \\
=K_{c}(R T)^{(x+y)-(m+n)} \\
=K_{c}(R T)^{\Delta n}
\end{gathered}
$$

And so, the relationship between $K_{c}$ and $K_{P}$ is

$$
K_{P}=K_{c}(R T)^{\Delta n}
$$

where $\Delta n$ is the difference in the molar amounts of product and reactant gases, in this case:

$$
\Delta n=(x+y)-(m+n)
$$


1377. Calculation of $\boldsymbol{K}_{\boldsymbol{P}}$ -
Write the equations relating $K_{c}$ to $K_{P}$ for each of the following reactions:
(a) $\mathrm{C}_{2} \mathrm{H}_{6}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$
(b) $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$
(c) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
(d) $K_{c}$ is equal to 0.28 for the following reaction at $900^{\circ} \mathrm{C}$ :

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

What is $K_{P}$ at this temperature?


1378. Solution -
(a) $\Delta n=(2)-(1)=1$
$K_{P}=K_{c}(R T)^{\Delta n}=K_{c}(R T)^{1}=K_{c}(R T)$
(b) $\Delta n=(2)-(2)=0$
$K_{P}=K_{c}(R T)^{\Delta n}=K_{c}(R T)^{0}=K_{c}$
(c) $\Delta n=(2)-(1+3)=-2$
$K_{P}=K_{c}(R T)^{\Delta n}=K_{c}(R T)^{-2}=\frac{K_{c}}{(R T)^{2}}$
(d) $K_{P}=K_{c}(\mathrm{RT})^{\Delta n}=(0.28)[(0.0821)(1173)]^{-2}=3.0 \times 10^{-5}$


1379. Check Your Learning -
Write the equations relating $K_{c}$ to $K_{P}$ for each of the following reactions:
(a) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
(b) $\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$
(c) $\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$
(d) At $227^{\circ} \mathrm{C}$, the following reaction has $K_{c}=0.0952$ :

$$
\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)
$$

What would be the value of $K_{P}$ at this temperature?
Answer:
(a) $K_{P}=K_{c}(R T)^{-1}$; (b) $K_{P}=K_{c}(R T)$; (c) $K_{P}=K_{c}(R T)$; (d) 160 or $1.6 \times 10^{2}$


1380. Heterogeneous Equilibria -
A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:

$$
\begin{aligned}
\mathrm{PbCl}_{2}(s) & \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) & K_{c} & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2} \\
\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CaCO}_{3}(s) & K_{c} & =\frac{1}{\left[\mathrm{CO}_{2}\right]} \\
\mathrm{C}(s)+2 \mathrm{~S}(g) & \rightleftharpoons \mathrm{CS}_{2}(g) & K_{c} & =\frac{\left[\mathrm{CS}_{2}\right]}{[\mathrm{S}]^{2}} \\
\mathrm{Br}_{2}(l) & \rightleftharpoons \mathrm{Br}_{2}(g) & K_{c} & =\left[\mathrm{Br}_{2}(g)\right]
\end{aligned}
$$

Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.
Two of the above examples include terms for gaseous species only in their equilibrium constants, and so $K_{p}$ expressions may also be written:

$$
\begin{aligned}
\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CaCO}_{3}(s) & K_{P} & =\frac{1}{P_{\mathrm{CO}_{2}}} \\
\mathrm{C}(s)+2 \mathrm{~S}(g) & \rightleftharpoons \mathrm{CS}_{2}(g) & K_{P} & =\frac{P_{\mathrm{CS}_{2}}}{\left(P_{\mathrm{S}}\right)^{2}}
\end{aligned}
$$


1381. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the ways in which an equilibrium system can be stressed
- Predict the response of a stressed equilibrium using Le Châtelier's principle

A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier's principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.

Reaction rates are affected primarily by concentrations, as described by the reaction's rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.


1382. Effect of a Change in Concentration -
If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction

$$
\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad K_{c}=50.0 \text { at } 400^{\circ} \mathrm{C}
$$

The rate laws for the forward and reverse reactions are

$$
\begin{array}{lll}
\text { forward } & \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g) & \operatorname{rate}_{f}=k_{f}\left[\mathrm{H}_{2}\right]^{m}\left[\mathrm{I}_{2}\right]^{n} \\
\text { reverse } & 2 \mathrm{HI}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) & \operatorname{rate}_{r}=k_{r}[\mathrm{HI}]^{x}
\end{array}
$$

When this system is at equilibrium, the forward and reverse reaction rates are equal.

$$
\operatorname{rate}_{f}=\operatorname{rate}_{r}
$$

If the system is stressed by adding reactant, either $\mathrm{H}_{2}$ or $\mathrm{I}_{2}$, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:

$$
\operatorname{rate}_{f}>\operatorname{rate}_{r}
$$

The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.

The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction

$$
\operatorname{rate}_{f}<\operatorname{rate}_{r}
$$


1382. Effect of a Change in Concentration -
$$
\operatorname{rate}_{f}<\operatorname{rate}_{r}
$$

and a temporary net reaction in the reverse direction to re-establish equilibrium.
As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,

$$
Q_{c}=\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=K_{c}
$$

If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then $Q_{c}<K_{c}$ and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If $\mathrm{H}_{2}$ is added, the right shift will consume $\mathrm{I}_{2}$ and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of $\mathrm{H}_{2}$ and HI and a lesser concentration of $\mathrm{I}_{2}$ than was present before. If $\mathrm{I}_{2}$ is added, the new equilibrium mixture will have greater concentrations of $\mathrm{I}_{2}$ and HI and a lesser concentration of $\mathrm{H}_{2}$. Finally, if HI
is removed, the new equilibrium mixture will have greater concentrations of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case $Q_{c}>K_{c}$ and the equilibrium will shift left.

For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure $P$ of an ideal gas is proportional to its molar concentration $M$,

$$
M=\frac{n}{V}=\frac{P}{R T}
$$


1382. Effect of a Change in Concentration -
$$
M=\frac{n}{V}=\frac{P}{R T}
$$

and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.

At equilibrium, the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$ is described by the reaction quotient

$$
Q_{p}=\frac{P_{\mathrm{HI}^{2}}}{P_{\mathrm{H}^{2}} P_{\mathrm{I}^{2}}}=K_{p}
$$

If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3 , the partial pressures of all three species will be increased by a factor of 3:

$$
\begin{aligned}
& Q_{p}^{\prime}=\frac{\left(3 P_{\mathrm{HI}^{2}}\right)^{2}}{3 P_{\mathrm{H}^{2}}{ }^{3 \mathrm{I}^{2}}}=\frac{9 P_{\mathrm{HI}^{2}}}{9 P_{\mathrm{H}^{2}} P_{\mathrm{I}^{2}}}=\frac{P_{\mathrm{HI}^{2}}}{P_{\mathrm{H}^{2}} P_{\mathrm{I}^{2}}}=Q_{p}=K_{p} \\
& Q_{p}{ }^{\prime}=Q_{p}=K_{p}
\end{aligned}
$$

And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.


1382. Effect of a Change in Concentration -
And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.

A similar treatment of a different system, $2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$, however, yields a different result:

$$
\begin{aligned}
& Q_{p}=\frac{P_{\mathrm{NO}^{2}} P_{\mathrm{O}^{2}}}{P}=K_{p} \\
& \left(\mathrm{NO}^{2}\right)^{2} \\
& Q_{p}{ }^{\prime}=\frac{\left(3 P_{\mathrm{NO}^{2}}\right)^{2} 3 P_{\mathrm{O}^{2}}}{\left(3 P_{\mathrm{NO}^{2}}\right)^{2}}=\frac{9 P_{\mathrm{NO}^{2}}{ }^{3 P} \mathrm{O}^{2}}{9 P}\left(\mathrm{NO}^{2}\right)^{2}
\end{aligned} \frac{27 P_{\mathrm{NO}^{2} P_{\mathrm{O}^{2}}}^{P}\left(\mathrm{NO}^{2}\right)^{2}}{}=3 Q_{p}>K_{p} .
$$

In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.


1382. Effect of a Change in Concentration -
In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.

These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better "accommodates" the volume change. In the second example, two moles of reactant $\left(\mathrm{NO}_{2}\right)$ yield three moles of product ( $2 \mathrm{NO}+\mathrm{O}_{2}$ ), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas ( 2 mol ) than the forward reaction ( 3 mol ). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.


1383. LINK TO LEARNING -
Check out this link (http://openstax.org/l/16equichange) to see a dramatic visual demonstration of how
equilibrium changes with pressure changes.


1385. Equilibrium and Soft Drinks -
The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733-1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestly's approach involved production of carbon dioxidey reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).

The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:

$$
\begin{array}{ll}
\text { dissolution } & \mathrm{CO}_{2}(g)=\mathrm{CO}_{2}(a q) \\
\text { hydrolysis } & \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)=\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \\
\text { ionization } & \mathrm{H}_{2} \mathrm{CO}_{3}(a q)=\mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q)
\end{array}
$$

These same equilibrium reactions are the basis of today's soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.


1385. Equilibrium and Soft Drinks -
When a carbonated beverage container is opened, a hissing sound is heard as pressurized $\mathrm{CO}_{2}$ escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved $\mathrm{CO}_{2}$ and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the $\mathrm{CO}_{2}(\mathrm{aq})$ concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as "flat."

FIGURE 13.7 Opening a soft-drink bottle lowers the $\mathrm{CO}_{2}$ pressure above the beverage, shifting the dissolution equilibrium and releasing dissolved $\mathrm{CO}_{2}$ from the beverage. (credit: modification of work by "D Coetzee"/Flickr)


1386. Effect of a Change in Temperature -
Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to reestablish equilibrium without any change in the value of the equilibrium constant, $K$. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.

To understand this phenomenon, consider the elementary reaction

$$
\mathrm{A} \rightleftharpoons \mathrm{~B}
$$

Since this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation's stoichiometry:

$$
\begin{aligned}
& \operatorname{rate}_{f}=k_{f}[\mathrm{~A}] \\
& \operatorname{rate}_{r}=k_{r}[\mathrm{~B}]
\end{aligned}
$$

When the system is at equilibrium,

$$
\operatorname{rate}_{r}=\operatorname{rate}_{f}
$$

Substituting the rate laws into this equality and rearranging gives

$$
\begin{aligned}
k_{f}[\mathrm{~A}] & =k_{r}[\mathrm{~B}] \\
\frac{[\mathrm{B}]}{[\mathrm{A}]} & =\frac{k_{f}}{k_{r}}=K_{c}
\end{aligned}
$$

The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.

Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:


1386. Effect of a Change in Temperature -
$$
\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H=57.20 \mathrm{~kJ}
$$

For purposes of applying Le Chatelier's principle, heat $(q)$ may be viewed as a reactant:

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

Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.


1387. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
- Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
- Explain how temperature affects the spontaneity of some proceses
- Relate standard free energy changes to equilibrium constants

Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology-for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.

Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:

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

As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either $\mathrm{NH}_{3}$ only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount $x$ :

$$
\Delta\left[\mathrm{N}_{2}\right]=+x
$$

the corresponding changes in the other species concentrations are


1387. LEARNING OBJECTIVES -
$$
\Delta\left[\mathrm{N}_{2}\right]=+x
$$

the corresponding changes in the other species concentrations are

$$
\begin{aligned}
\Delta\left[\mathrm{H}_{2}\right] & =\Delta\left[\mathrm{N}_{2}\right]\left(\frac{3 \mathrm{~mol} \mathrm{H}_{2}}{1 \mathrm{~mol} \mathrm{~N}_{2}}\right)=+3 x \\
\Delta\left[\mathrm{NH}_{3}\right] & =-\Delta\left[\mathrm{N}_{2}\right]\left(\frac{2 \mathrm{~mol} \mathrm{NH}_{3}}{1 \mathrm{~mol} \mathrm{~N}_{2}}\right)=-2 x
\end{aligned}
$$

where the negative sign indicates a decrease in concentration.


1389. Determining Relative Changes in Concentration -
Derive the missing terms representing concentration changes for each of the following reactions.
(a)

(b)
$\mathrm{I}_{2}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{3}^{-}(a q)$
$-\quad x$
(c)
$\begin{array}{ll}\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ x & -\end{array}$


1390. Solution -
(a) $\begin{array}{rl}\mathrm{C}_{2} \mathrm{H}_{2}(g)+\underset{2}{2} & 2 \mathrm{Br}_{2}(g)\end{array} \underset{2 x}{ } \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Br}_{4}(g)$
(b) $\begin{array}{lll}\mathrm{I}_{2}(a q)+ & \mathrm{I}^{-}(a q) & \rightleftharpoons \\ -x & -x & \mathrm{I}_{3}^{-}(a q) \\ x\end{array}$
(c)


1391. Check Your Learning -
Complete the changes in concentrations for each of the following reactions:
(a)

(b)

$4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
(c)


1392. Answer: -
(a) $2 x, x,-2 x$; (b) $x,-2 x$; (c) $4 x, 7 x,-4 x,-6 x$ or $-4 x,-7 x, 4 x, 6 x$


1393. Calculation of an Equilibrium Constant -
The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the $K$ expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.


1395. Calculation of an Equilibrium Constant -
Iodine molecules react reversibly with iodide ions to produce triiodide ions.

$$
\mathrm{I}_{2}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{3}^{-}(a q)
$$

If a solution with the concentrations of $\mathrm{I}_{2}$ and $\mathrm{I}^{-}$both equal to $1.000 \times 10^{-3} \mathrm{M}$ before reaction gives an equilibrium concentration of $\mathrm{I}_{2}$ of $6.61 \times 10^{-4} \mathrm{M}$, what is the equilibrium constant for the reaction?


1396. Solution -
To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:

$$
\begin{aligned}
K_{C}= & {\left[\mathrm{I}_{3}{ }^{-}\right] } \\
& {\left[\mathrm{I}_{2}\right]\left[\mathrm{I}^{-}\right] }
\end{aligned}
$$

Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.

|  |  |  | $\mathrm{I}_{3}{ }^{-}$ |
| :---: | :---: | :---: | :---: |
| Initial concentration (M) | $1.000 \times 10^{-3}$ | $1.000 \times 10^{-3}$ | 0 |
| Change ( $M$ ) | -x | -x | $+x$ |
| Equilibrium concentration ( $M$ ) | $1.000 \times 10^{-3}-x$ | $1.000 \times 10^{-3}-x$ | $x$ |

At equilibrium the concentration of $\mathrm{I}_{2}$ is $6.61 \times 10^{-4} \mathrm{M}$ so that

$$
\begin{gathered}
1.000 \times 10^{-3}-x=6.61 \times 10^{-4} \\
x=1.000 \times 10^{-3}-6.61 \times 10^{-4} \\
=3.39 \times 10^{-4} M
\end{gathered}
$$

The ICE table may now be updated with numerical values for all its concentrations:


1396. Solution -
The ICE table may now be updated with numerical values for all its concentrations:

|  |  | r | $\mathrm{I}_{3}{ }^{-}$ |
| :---: | :---: | :---: | :---: |
| Initial concentration (M) | $1.000 \times 10^{-3}$ | $1.000 \times 10^{-3}$ | 0 |
| Change ( $M$ ) | $-3.39 \times 10^{-4}$ | $-3.39 \times 10^{-4}$ | $+3.39 \times 10^{-4}$ |
| Equilibrium concentration ( $M$ ) | $6.61 \times 10^{-4}$ | $6.61 \times 10^{-4}$ | $3.39 \times 10^{-4}$ |

Finally, substitute the equilibrium concentrations into the $K$ expression and solve:

$$
\begin{gathered}
K_{c}=\frac{\left[\mathrm{I}_{3}-\right]}{\left[\mathrm{I}_{2}\right]\left[\mathrm{I}^{-}\right]} \\
=\frac{3.39 \times 10^{-4} M}{\left(6.61 \times 10^{-4} M\right)\left(6.61 \times 10^{-4} M\right)}=776
\end{gathered}
$$


1397. Check Your Learning -
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some
nail polish removers.

$$
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}
$$

When 1 mol each of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ and $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\frac{1}{3} \mathrm{~mol}$ of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)


1398. Answer: -
$K_{C}=4$


1399. Calculation of a Missing Equilibrium Concentration -
When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.


1401. Calculation of a Missing Equilibrium Concentration -
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At $2000^{\circ} \mathrm{C}$, the value of the $K_{c}$ for the reaction, $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$, is $4.1 \times 10^{-4}$. Calculate the equilibrium concentration of $\mathrm{NO}(\mathrm{g})$ in air at 1 atm pressure and $2000^{\circ} \mathrm{C}$. The equilibrium concentrations of $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ at this pressure and temperature are 0.036 M and 0.0089 M , respectively.


1402. Solution -
Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

$$
\begin{gathered}
K_{c}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
{[\mathrm{NO}]^{2}=K_{c}\left[\mathrm{~N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
{[\mathrm{NO}]=\sqrt{K_{c}\left[\mathrm{~N}_{2}\right]\left[\mathrm{O}_{2}\right]}} \\
=\sqrt{\left(4.1 \times 10^{-4}\right)(0.036)(0.0089)} \\
=\sqrt{1.31 \times 10^{-7}} \\
=3.6 \times 10^{-4}
\end{gathered}
$$

Thus [NO] is $3.6 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$ at equilibrium under these conditions.
To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for $K$ :

$$
\begin{aligned}
K_{c} & =\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
& =\frac{\left(3.6 \times 10^{-4}\right)^{2}}{(0.036)(0.0089)} \\
& =4.0 \times 10^{-4}
\end{aligned}
$$

This result is consistent with the provided value for $K$ within nominal uncertainty, differing by just 1 in the least significant digit's place.


1403. Check Your Learning -
The equilibrium constant $K_{c}$ for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is $6.00 \times 10^{-2}$. Calculate the equilibrium concentration of ammonia if the equilibrium
concentrations of nitrogen and hydrogen are $4.26 M$ and $2.09 M$, respectively.


1404. Answer: -
$1.53 \mathrm{~mol} / \mathrm{L}$


1405. Calculation of Equilibrium Concentrations from Initial Concentrations -
Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Develop an ICE table.
3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
4. Confirm the calculated equilibrium concentrations.

The last two example exercises of this chapter demonstrate the application of this strategy.


1407. Calculation of Equilibrium Concentrations -
Under certain conditions, the equilibrium constant $K_{c}$ for the decomposition of $\mathrm{PCl}_{5}(g)$ into $\mathrm{PCl}_{3}(g)$ and $\mathrm{Cl}_{2}(g)$ is 0.0211. What are the equilibrium concentrations of $\mathrm{PCl}_{5}, \mathrm{PCl}_{3}$, and $\mathrm{Cl}_{2}$ in a mixture that initially contained only $\mathrm{PCl}_{5}$ at a concentration of 1.00 M ?


1408. Solution -
Use the stepwise process described earlier.
Step 1.
Determine the direction the reaction proceeds.
The balanced equation for the decomposition of $\mathrm{PCl}_{5}$ is

$$
\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)
$$

Because only the reactant is present initially $Q_{c}=0$ and the reaction will proceed to the right.
Step 2.
Develop an ICE table.

|  | $\mathbf{P C l}_{\mathbf{5}}$ |  |  |
| :---: | :---: | :---: | :---: |
|  | $\mathrm{PCl}_{3}$ | + | $\mathbf{C l}_{\mathbf{2}}$ |
| Initial concentration $(M)$ | 1.00 | 0 | 0 |
| Change $(M)$ | $-x$ | $+x$ | $+x$ |
| Equilibrium concentration $(M)$ | $1.00-x$ | $x$ | $x$ |


1409. Step 3. -
Solve for the change and the equilibrium concentrations.
Substituting the equilibrium concentrations into the equilibrium constant equation gives

$$
\begin{gathered}
K_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=0.0211 \\
=\frac{(x)(x)}{(1.00-x)}
\end{gathered}
$$

$$
\begin{gathered}
0.0211=\frac{(x)(x)}{(1.00-x)} \\
0.0211(1.00-x)=x^{2} \\
x^{2}+0.0211 x-0.0211=0
\end{gathered}
$$

Appendix B shows an equation of the form $a x^{2}+b x+c=0$ can be rearranged to solve for $x$ :

$$
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}
$$

In this case, $a=1, b=0.0211$, and $c=-0.0211$. Substituting the appropriate values for $a, b$, and $c$ yields:

$$
\begin{gathered}
x=\frac{-0.0211 \pm \sqrt{(0.0211)^{2}-4(1)(-0.0211)}}{2(1)} \\
=\frac{-0.0211 \pm \sqrt{\left(4.45 \times 10^{-4}\right)+\left(8.44 \times 10^{-2}\right)}}{2} \\
=\frac{-0.0211 \pm 0.291}{2}
\end{gathered}
$$

The two roots of the quadratic are, therefore,

$$
x=\frac{-0.0211+0.291}{2}=0.135
$$

and

$$
x=\frac{-0.0211-0.291}{2}=-0.156
$$

For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so $x=0.135 \mathrm{M}$.

The equilibrium concentrations are


1409. Step 3. -
For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so $x=0.135 \mathrm{M}$.

The equilibrium concentrations are

$$
\begin{gathered}
{\left[\mathrm{PCl}_{5}\right]=1.00-0.135=0.87 \mathrm{M}} \\
{\left[\mathrm{PCl}_{3}\right]=x=0.135 \mathrm{M}} \\
{\left[\mathrm{Cl}_{2}\right]=x=0.135 \mathrm{M}}
\end{gathered}
$$

Step 4.
Confirm the calculated equilibrium concentrations.
Substitution into the expression for $K_{c}$ (to check the calculation) gives

$$
K_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{(0.135)(0.135)}{0.87}=0.021
$$

The equilibrium constant calculated from the equilibrium concentrations is equal to the value of $K_{c}$ given in the problem (when rounded to the proper number of significant figures).


1410. Check Your Learning -
Acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, reacts with ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, to form water and ethyl acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}$.

$$
\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}
$$

The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, 0.15 \mathrm{M}$ in $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, 0.40 \mathrm{M}$ in $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}$, and $0.40 \mathrm{Min}_{2} \mathrm{O}$ ?


1411. Answer: -
$\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.18 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.18 \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.37 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.37 \mathrm{M}$
Check Your Learning
A 1.00-L flask is filled with 1.00 moles of $\mathrm{H}_{2}$ and 2.00 moles of $\mathrm{I}_{2}$. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{I}_{2}$, and HI in moles/L?

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


1412. Answer: -
$\left[\mathrm{H}_{2}\right]=0.06 \mathrm{M},\left[\mathrm{I}_{2}\right]=1.06 \mathrm{M},[\mathrm{HI}]=1.88 \mathrm{M}$


1414. Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption -
What are the concentrations at equilibrium of a 0.15 M solution of HCN?

$$
\mathrm{HCN}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CN}^{-}(a q) \quad K_{c}=4.9 \times 10^{-10}
$$


1415. Solution -
Using " $x$ " to represent the concentration of each product at equilibrium gives this ICE table.

|  | $\mathrm{HCN}(\mathrm{aq}) \rightleftharpoons$ |  | $\mathrm{H}^{+}(\mathrm{aq})$ |
| :---: | :---: | :---: | :---: |
|  | + | $\mathrm{CN}^{-}(\mathrm{aq})$ |  |
| Initial concentration $(M)$ | 0.15 | 0 | 0 |
| Change $(M)$ | $-x$ | $+x$ | $+x$ |
| Equilibrium concentration $(M)$ | $0.15-x$ | $x$ | $x$ |

Substitute the equilibrium concentration terms into the $K_{c}$ expression

$$
K_{c}=\frac{(x)(x)}{0.15-x}
$$

Rearrange to the quadratic form and solve for $x$

$$
\begin{gathered}
x^{2}+4.9 \times 10^{-10}-7.35 \times 10^{-11}=0 \\
x=8.56 \times 10^{-6} M(3 \text { sig. figs. })=8.6 \times 10^{-6} M(2 \text { sig. figs. })
\end{gathered}
$$

Thus $\left[\mathrm{H}^{+}\right]=\left[\mathrm{CN}^{-}\right]=x=8.6 \times 10^{-6} \mathrm{M}$ and $[\mathrm{HCN}]=0.15-x=0.15 \mathrm{M}$.
Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small $K$ ), and so the initial concentration experiences a negligible change:

$$
\text { if } x \ll 0.15 \mathrm{M} \text {, then }(0.15-x) \approx 0.15
$$

This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:


1415. Solution -
This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

$$
\begin{gathered}
K_{c}=\frac{(x)(x)}{0.15-x} \approx \frac{x^{2}}{0.15} \\
4.9 \times 10^{-10}=\frac{x^{2}}{0.15} \\
x^{2}=(0.15)\left(4.9 \times 10^{-10}\right)=7.4 \times 10^{-11}
\end{gathered}
$$

$$
x=\sqrt{7.4 \times 10^{-11}}=8.6 \times 10^{-6} M
$$

The value of $x$ calculated is, indeed, much less than the initial concentration

$$
8.6 \times 10^{-6} \ll 0.15
$$

and so the approximation was justified. If this simplified approach were to yield a value for $x$ that did not justify the approximation, the calculation would need to be repeated without making the approximation.


1416. Check Your Learning -
What are the equilibrium concentrations in a $0.25 \mathrm{MNH}_{3}$ solution?

$$
\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{c}}=1.8 \times 10^{-5}
$$


1417. Answer: -
$\left[\mathrm{OH}^{-}\right]=\left[\mathrm{NH}_{4}{ }^{+}\right]=0.0021 \mathrm{M} ;\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M}$


1418. Temperature Dependence of Spontaneity -
As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

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

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since $T$ is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:


1418. Temperature Dependence of Spontaneity -
1. Both $\boldsymbol{\Delta H}$ and $\boldsymbol{\Delta S}$ are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, $\Delta G$ will be negative if the magnitude of the $T \Delta S$ term is greater than $\Delta H$. If the $T \Delta S$ term is less than $\Delta H$, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both $\boldsymbol{\Delta H}$ and $\boldsymbol{\Delta S}$ are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, $\Delta G$ will be negative if the magnitude of the $T \Delta S$ term is less than $\Delta H$. If the $T \Delta S$ term's magnitude is greater than $\Delta H$, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. $\boldsymbol{\Delta H}$ is positive and $\boldsymbol{\Delta} \boldsymbol{S}$ is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, $\Delta G$ will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. $\boldsymbol{\Delta} \boldsymbol{H}$ is negative and $\boldsymbol{\Delta} \boldsymbol{S}$ is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, $\Delta G$ will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 13.8.


1418. Temperature Dependence of Spontaneity -
These four scenarios are summarized in Figure 13.8.

|  | $\begin{gathered} \Delta \mathrm{H}>0 \\ \text { (endothermic) } \end{gathered}$ | $\begin{gathered} \Delta \mathrm{H}<0 \\ \text { (exothermic) } \end{gathered}$ |
| :---: | :---: | :---: |
| $\Delta S>0$
 (increase in entropy) | $\Delta \mathrm{G}<0$ at high temperature $\Delta G>0$ at low temperature Process is spontaneous at high temperature | $\Delta G<0$ at any temperature
 Process is spontaneous at any temperature |
| $\Delta \mathrm{S}<0$ (decrease in entropy) | $\Delta \mathrm{G}>0$ at any temperature
 Process is nonspontaneous at any temperature | $\Delta \mathrm{G}<0$ at low temperature $\Delta G>0$ at high temperature Process is spontaneous at low temperature |

FIGURE 13.8 There are four possibilities regarding the signs of enthalpy and entropy changes.


1420. Predicting the Temperature Dependence of Spontaneity -
The incomplete combustion of carbon is described by the following equation:

$$
2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)
$$

How does the spontaneity of this process depend upon temperature?