19

In this chapter, we learn to relate the reaction quotient (Q) to Gibb's free energy mathematically through the expression ΔG = ΔG° + RT ln Q. We also learn about the relationship between the equilibrium constant (K) and ΔG°, which can be expressed through the equation ΔG° = –RTlnKeq. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium.

By the end of this section, you will be able to:

- Relate standard free energy changes to equilibrium constants

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

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

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

$$\text{\Delta}G=\text{\Delta}G\text{\xb0}+RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}Q$$

*R* is the gas constant (8.314 J/K mol), *T* is the kelvin or absolute temperature, and *Q* is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 19.1.

*T* = 25 °C, ${P}_{{\text{N}}_{2}}=\text{0.870 atm},$ ${P}_{{\text{H}}_{2}}=\text{0.250 atm},$ and ${P}_{{\text{NH}}_{3}}=\text{12.9 atm}$

$${\text{2NH}}_{3}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3H}}_{2}(g)+{\text{N}}_{2}(g)\phantom{\rule{5em}{0ex}}\text{\Delta}G\text{\xb0}=\text{33.0 kJ/mol}$$

$$\text{\Delta}G=\text{\Delta}G\text{\xb0}+RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}Q=33.0\phantom{\rule{0.2em}{0ex}}\frac{\text{kJ}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\left(8.314\phantom{\rule{0.2em}{0ex}}\frac{\text{J}}{\text{mol K}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{298 K}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.5em}{0ex}}\frac{\left({0.250}^{3}\right)\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}0.870}{{12.9}^{2}}\right)\phantom{\rule{0.2em}{0ex}}=9680\phantom{\rule{0.2em}{0ex}}\frac{\text{J}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}\text{or 9.68 kJ/mol}$$

Since the computed value for Δ*G* is positive, the reaction is nonspontaneous under these conditions.

Δ*G* = −47 kJ/mol; yes

For a system at equilibrium, *Q* = *K* and Δ*G* = 0, and the previous equation may be written as

$$0=\text{\Delta}G\text{\xb0}+RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}K\phantom{\rule{5em}{0ex}}(\text{at equilibrium})$$

$$\text{\Delta}G\text{\xb0}=\text{\u2212}RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}K\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}K={e}^{-\phantom{\rule{0.1em}{0ex}}\frac{\text{\Delta}G\text{\xb0}}{RT}}$$

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 19.1.

Relations between Standard Free Energy Changes and Equilibrium Constants

K | ΔG° | Composition of an Equilibrium Mixture |
---|---|---|

> 1 | < 0 | Products are more abundant |

< 1 | > 0 | Reactants are more abundant |

= 1 | = 0 | Reactants and products are comparably abundant |

$$\text{AgCl}(s)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons \phantom{\rule{0.2em}{0ex}}{\text{Ag}}^{\text{+}}(aq)+{\text{Cl}}^{\text{\u2212}}(aq)\phantom{\rule{5em}{0ex}}{K}_{\text{sp}}=[{\text{Ag}}^{\text{+}}][{\text{Cl}}^{\text{\u2212}}]$$

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

$$\begin{array}{}\\ \text{\Delta}G\text{\xb0}=\left[\text{\Delta}{G}_{\text{f}}^{\xb0}\left({\text{Ag}}^{\text{+}}(aq)\right)\phantom{\rule{0.2em}{0ex}}+\text{\Delta}{G}_{\text{f}}^{\xb0}\left({\text{Cl}}^{\text{\u2212}}(aq)\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left[\text{\Delta}{G}_{\text{f}}^{\xb0}\left(\text{AgCl}(s)\right)\right]\hfill \\ =\left[\text{77.1 kJ/mol}-\text{131.2 kJ/mol}\right]-\left[-\text{109.8 kJ/mol}\right]=\text{55.7 kJ/mol}\hfill \end{array}$$

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

$${K}_{\text{sp}}={e}^{-\frac{\text{\Delta}G\text{\xb0}}{RT}}=\text{exp}\left(-\phantom{\rule{0.1em}{0ex}}\frac{\text{\Delta}G\text{\xb0}}{RT}\right)\phantom{\rule{0.2em}{0ex}}=\text{exp}\left(-\phantom{\rule{0.1em}{0ex}}\frac{55.7\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J/mol}}{8.314\phantom{\rule{0.2em}{0ex}}\text{J/mol\xb7K}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}298.15\phantom{\rule{0.2em}{0ex}}\text{K}}\right)=\text{exp}\left(-22.470\right)={e}^{-22.470}=1.74\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u221210}}$$

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

$${\text{N}}_{2}{\text{O}}_{4}(g)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons \phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}(g)$$

*K* = 0.32

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of *Q*), equilibrium is established when the system’s free energy is minimized (Figure 19.1). If a system consists of reactants and products in nonequilibrium amounts (*Q* ≠ *K*), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Previous Citation(s)

Flowers, P., et al. (2019). Chemistry: Atoms First 2e. https://openstax.org/details/books/chemistry-atoms-first-2e (13.4)

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