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86 Gibbs Energy and Equilibrium

Learning Objectives

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

  • Understand how free energy change is a driving force, related to reaction quotients, Q.
  • 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 ΔG 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 ΔG 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 chapter 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 less 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:

ΔG = ΔGo + RT lnQ

R is the gas constant (8.314 J K-1 mol-1), T is the absolute temperature (in Kelvin), 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 the example below.

Example:

Calculating ΔG under Nonstandard Conditions

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

2 NH3(g) → 3 H2(g) + N2(g);       ΔGo = 33.0 kJ mol-1 

T = 25 oC, PN2 = 0.870 atm, PH2 = 0.250 atm, and PNH3 = 12.9 atm.

Solution

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

\begin{array}{c}\Delta}G=\Delta}G\text{°}+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}}×\phantom{\rule{0.2em}{0ex}}\text{298 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.5em}{0ex}}\frac{\left({0.250}^{3}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.870}{{12.9}^{2}}\right)\\ \\ =9680\phantom{\rule{0.2em}{0ex}}\frac{\text{J}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}\text{or 9.68 kJ/mol}\end{array}

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.

Check Your Learning

Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Answer: ΔG = −47 kJ; yes

Since, for a system at equilibrium, Q = K and ΔG = 0, the previous equation may be written as

0=\text{Δ}G\text{°}+RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}K\phantom{\rule{5em}{0ex}}\left(\text{at equilibrium}\right)
\text{Δ}G\text{°}=\text{-}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{Δ}G\text{°}}{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 86-1.

Table 86-1:  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

Example:

Calculating an Equilibrium Constant using Standard Free Energy Change

Given that the standard free energies of formation of Ag+(aq), Cl−(aq), and AgCl(s) are 77.1 kJ mol-1, −131.2 kJ mol-1, and −109.8 kJ mol-1, respectively, calculate the solubility product, Ksp, for AgCl.

Solution

The reaction of interest is the following:

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

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

\begin{array}{}\\ \Delta}G\text{°}=\left[\Delta{G}_{\text{f}}^{°}\left({\text{Ag}}^{\text{+}}\left(aq\right)\right)\phantom{\rule{0.2em}{0ex}}+\Delta}{G}_{\text{f}}^{°}\left({\text{Cl}}^{\text{−}}\left(aq\right)\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left[\text{Δ}{G}_{\text{f}}^{°}\left(\text{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}

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

ΔGo = – RT lnK
55.7 kJ mol-1 = -(8.314 x 10-3 kJ mol-1 K-1) (298.15K) ln K
ln K = 55.7 kJ mol-1 / (-2.479 kJ mol-1) = -22.47
K = e22.47 = 1.74 x 10-10

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

Check Your Learning

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

{\text{2NO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}{\text{O}}_{4}\left(g\right)
Answer: K = 6.9

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 86.1). 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.

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 86-1:  These plots show the free energy versus reaction progress for systems whose standard free energy 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.

 

End-of-Chapter Exercises

1. Calculate ΔGo for each of the following reactions from the equilibrium constant at the temperature given.

(a) N2(g) + O2(g) → 2NO(g) T=2000 oC Kp = 4.1 x 10-4
(b) H2(g) + I2(g) → 2HI(g) T=400 oC Kp = 50.0
(c) CO2(g) + H2(g) → CO(g) + H2O(g) T = 980 oC Kp = 1.67
(d) CaCO3(s) → CaO(s) + CO2(g) T = 900 oC Kp = 1.04
(e) HF(aq) + H2O(l) → H3O+(aq) + F–(aq) T = 25 oC Kp = 7.2 x 10-4
(f) AgBr(s) → Ag+(aq) + Br–(aq) T = 25 oC Kp = 3.3 x 10-13

 

2. Calculate ΔGo for each of the following reactions from the equilibrium constant at the temperature given.

(a) Cl2(g) + Br2(g) → 2BrCl(g) T=25 oC Kp = 4.7 x 10-2
(b) 2SO2(g) + O2(g) → 2SO3(g) T=500 oC Kp = 48.2
(c) H2O(l)  →  H2O(g) T = 60 oC Kp = 0.196
(d) CoO(s) + CO(g) → Co(s) + CO2(g) T = 550 oC Kp = 4.9 x 102
(e) CH3NH2(aq) + H2O(l) → CH3NH3+(aq) + OH–(aq) T = 25 oC Kp = 4.4 x 10-4
(f) PbI2(s) → Pb2+(aq) + 2I–(aq) T = 25 oC Kp = 8.7 x 10-9

 

3. Calculate the equilibrium constant at 25 oC for each of the following reactions from the value of the ΔGo provided.

(a) O2(g) + 2F2(g) → 2OF2(g) ΔGo = -9.2 kJ
(b) I2(s) + Br2(l) → 2IBr(g) ΔGo = 7.3 kJ
(c) 2 LiOH(s) + CO2(g) → Li2CO2(s) + H2O(g) ΔGo = -79 kJ
(d) N2O3(g) → NO(g) + NO2(g) ΔGo = -1.6 kJ

 

4. What happens to ΔG (i.e., does it become more positive or more negative) for the following chemical reactions when the partial pressure of oxygen is increased?

(a)  S(s) + O2(g) → SO2(g)

(b)  2 SO2(g) + O2(g) → 2 SO3(g)

(c)  2 HgO(s) → 2 Hg(l) + O2(g)

 

5.  One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P): G6P \rightleftharpoons F6P, which has a standard Gibbs Energy change of 1.7 kJ (ΔGo = 1.7 kJ. 

(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 ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.

 

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Introductory Chemistry Copyright © by D. Latimer; D. Vanderwel; and J. Hollett. All Rights Reserved.

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