Chapter 14: Fundamental Equilibrium Concepts

14.5 Equilibrium and Thermodynamics

Learning Outcomes

  • Relate standard free energy changes to equilibrium constants

Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for [latex]\Delta[/latex]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 [latex]\Delta[/latex]G is zero, the forward and reverse driving forces are equal, and so 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, [latex]\Delta[/latex]G, is related to the standard free energy change, [latex]\Delta[/latex]G°, according to this equation:

[latex]\Delta G=\Delta G^{\circ }+RT\text{ln}Q[/latex]

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example 6.

Example 14.5.1: Calculating [latex]\Delta[/latex]G under Nonstandard Conditions

What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, [latex]{P}_{{\ce{N}}_{2}}=\text{0.870 atm}[/latex], [latex]{P}_{{\ce{H}}_{2}}=\text{0.250 atm}[/latex], and [latex]{P}_{{\ce{NH}}_{3}}=\text{12.9 atm}[/latex]

[latex]{\ce{2NH}}_{3}\left(g\right)\longrightarrow {\ce{3H}}_{2}\left(g\right)+{\ce{N}}_{2}\left(g\right)\qquad\Delta G^{\circ }=\text{33.0 kJ/mol}[/latex]

Show Solution

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

[latex]\Delta G=\Delta G^{\circ }+RT\text{ln}Q=33.0\frac{\text{kJ}}{\text{mol}}+\left(8.314\frac{\text{J}}{\text{mol K}}\times \text{298 K}\times \text{ln}\dfrac{\left({0.250}^{3}\right)\times 0.870}{{12.9}^{2}}\right)=9680\frac{\text{J}}{\text{mol}}\text{or 9.68 kJ/mol}[/latex]

Since the computed value for [latex]\Delta[/latex]G is positive, the reaction is nonspontaneous under these conditions.

Check Your Learning

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

[latex]0=\Delta G^{\circ }+RT\text{ln}K\qquad\left(\text{at equilibrium}\right)[/latex]
[latex]\Delta G^{\circ }=-RT\text{ln}K\qquad\text{ or }\qquad{K}={e}^{-\frac{\Delta G^{\circ }}{RT}}[/latex]

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 14.5.1.

Table 14.5.1. Relations between Standard Free Energy Changes and Equilibrium Constants
K [latex]\Delta[/latex]G° Comments
> 1 < 0 Products are more abundant at equilibrium.
< 1 > 0 Reactants are more abundant at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.

Example 14.5.2: Calculating an Equilibrium Constant using Standard Free Energy Change

Given that the standard free energies of formation of [latex]\ce{Ag+}[/latex](aq), [latex]\ce{Cl-}[/latex](aq), and [latex]\ce{AgCl}[/latex](s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for [latex]\ce{AgCl}[/latex].

Show Solution

The reaction of interest is the following:

[latex]\ce{AgCl}\left(s\right)\rightleftharpoons {\ce{Ag}}^{\text{+}}\left(aq\right)+{ce{Cl}}^{-}\left(aq\right){K}_{\text{sp}}=\left[{\ce{Ag}}^{\text{+}}\right]\left[{\ce{Cl}}^{-}\right][/latex]

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

[latex]\begin{array}{rll}\Delta G^{\circ }&=&\Delta {G}_{298}^{\circ }=\left[\Delta {G}_{\text{f}}^{\circ }\left({\text{Ag}}^{\text{+}}\left(aq\right)\right)+\Delta {G}_{\text{f}}^{\circ }\left({\ce{Cl}}^{-}\left(aq\right)\right)\right]-\left[\Delta {G}_{\text{f}}^{\circ }\left(\ce{AgCl}\left(s\right)\right)\right]\\{}&=&\left[\text{77.1 kJ/mol}-\text{131.2 kJ/mol}\right]-\left[-\text{109.8 kJ/mol}\right]=\text{55.7 kJ/mol}\end{array}[/latex]

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

[latex]{K}_{\text{sp}}={e}^{-\dfrac{\Delta G^{\circ }}{RT}}=\text{exp}\left(-\frac{\Delta G^{\circ }}{RT}\right)=\text{exp}\left(-\dfrac{55.7\times {10}^{3}\text{J/mol}}{8.314\text{J/mol}\cdot\text{K}\times 298.15\text{K}}\right)=\text{exp}\left(-22.470\right)={e}^{-22.470}=1.74\times {10}^{-\text{10}}[/latex]

This result is in reasonable agreement with the value provided in Solubility Products.

Check Your Learning

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 14.5.1). If a system is present with reactants and products present in nonequilibrium amounts (Q ≠ K), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Three graphs, labeled, “a,” “b,” and “c” are shown where the y-axis is labeled, “Gibbs free energy ( G ),” and, “G superscript degree sign ( reactants ),” while the x-axis is labeled, “Reaction progress,” and “Reactants,” on the left and, “Products,” on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G less than 0,” while the lowest point on the graph is labeled, “Q equals K greater than 1.” In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G greater than 0,” while the lowest point on the graph is labeled, “Q equals K less than 1.” In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is equal to the starting point on the y-axis which is labeled, “G superscript degree sign ( reactants ).” The lowest point on the graph is labeled, “Q equals K equals 1.” At the top of the graph is the label, “Delta G superscript degree sign equals 0.”
Figure 14.5.1. These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.

Key Concepts and Summary

Reactions proceed in the direction that establishes equilibrium, where Q = K and free energy is minimized

Key Equations

  • [latex]\Delta{G}=\Delta G^{\circ }+RT\text{ln}K[/latex]
  • [latex]\Delta G^{\circ }=-RT\text{ln}K[/latex]

Try It

  1. Calculate [latex]\Delta[/latex]G° for each of the following reactions from the equilibrium constant at the temperature given.
    1. [latex]{\ce{Cl}}_{2}\left(g\right)+{\ce{Br}}_{2}\left(g\right)\longrightarrow \ce{2BrCl}\left(g\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=4.7\times {10}^{-\text{2}}[/latex]
    2. [latex]{\ce{2SO}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightleftharpoons {\ce{2SO}}_{3}\left(g\right)\,\,\,{;}\,\,\,\text{T}=500^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=48.2[/latex]
  2. Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of [latex]\Delta[/latex]G° given.
    1. [latex]{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\longrightarrow {\ce{2OF}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{9.2 kJ}[/latex]
Show Selected Solutions

2. Equilibrium constants are calculated from [latex]\text{ln}K=\frac{-\Delta G^{\circ }}{RT}[/latex]. Note that K is a function of T and thus changes as T changes.

  1. [latex]\text{ln}K=-\left[\frac{-9.2\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times \text{298.15 K}\right)}\right]=3.71[/latex], K = 41

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