AOT - Gibbs Free Energy and the Equilibrium Constant (Lesson)

Gibbs Free Energy and the Equilibrium Constant

In a previous lesson, the concept of Le Chatelier's principle was used to explain how a chemical reaction at equilibrium can be influenced by changes imposed on the system. Depending on the nature of the stress that is imposed on a system at equilibrium, the reaction will shift in one direction or another. When stated in more thermodynamic terms, the reaction was either made to be spontaneous in one direction or another. By analyzing a system at equilibrium in this manner, Le Chatelier's principle allows for little more than a qualitative observation regarding this phenomenon. In this lesson, the influence of concentration on the Gibbs free energy change is examined, as that allows for a more quantitative view.

Concentration and Gibbs Free Energy

In order to relate Gibbs free energy to the equilibrium constant (K_eq), it must first be recognized that up to this point, only reactions under standard conditions have been considered. By way of reminder, standard conditions for thermodynamics include 1 atm of pressure for gases, 1 M concentrations for solutions, and a temperature of 298 K. What about systems that are not at standard conditions? How can the Gibbs free energy change be determined in these situations? The expression below allows for $LaTeX: \Delta$ $\Delta$ G to be calculated under any set of conditions.

$LaTeX: \Delta$ $\Delta$ G = $LaTeX: \Delta$ $\Delta$ G^{$^\circ$} + RT ln (Q)

Notice that this equation essentially applies a correctional factor to the standard Gibbs free energy change that allows for a nonstandard temperature to be applied. Also, the equation contains the reaction quotient, Q which was discussed in the Equilibrium module. The gas constant R is also included with a value of 8.314 and units of J/ mol $LaTeX: \cdot$ $\cdot$ K.

Consider the following reaction as an example:

2 NO (g) + O₂ (g) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ 2 NO₂ (g)

At 298 K, the standard Gibbs free energy change for this reaction is -70.52 kJ/mol. The 298 K represents standard temperature; however, what if the partial pressures of each substance are as follows?

$LaTeX: P_{NO}=4.0\times10^{-2}atm$ $P_{NO}=4.0\times10^{-2}atm$

$LaTeX: P_{O_2}=4.2\times10^{-3}atm$ $P_{O_2}=4.2\times10^{-3}atm$

$LaTeX: P_{NO_2}=0.40atm$ $P_{NO_2}=0.40atm$

First, the Q expression should be written:

$LaTeX: Q=\frac{\left(P_{NO_{2}^{}{}}\right)^2}{(P_{NO})^2(P_{O_{2}}){}}$ $Q=\frac{\left(P_{NO_{2}^{}{}}\right)^2}{(P_{NO})^2(P_{O_{2}}){}}$

After substituting all variables, the equation is shown below. Notice that the value for R has been changed from 8.314 to 0.008314. The reasoning behind this is so that the units for the standard Gibbs free energy change (kJ/mol) will now cancel with the units for R (expressed in J/molK).

$LaTeX: \Delta$ $\Delta$ G = $LaTeX: \Delta$ $\Delta$ G^{$^\circ$} + RT ln (Q)

$LaTeX: \Delta$ $\Delta$ G = -70.52 + (0.008314)(298) ln $LaTeX: \frac{}{}\frac{\left(0.40\right)^2}{\left(4.0\times10^{-2}\right)^2\left(4.2\times10^{-3}\right)}$ $\frac{}{}\frac{\left(0.40\right)^2}{\left(4.0\times10^{-2}\right)^2\left(4.2\times10^{-3}\right)}$

$LaTeX: \Delta$ $\Delta$ G = -34.1 kJ/mol

This value indicates that the reaction remains spontaneous under these conditions although it is less spontaneous than it would be under standard conditions. Keep in mind that it is entirely possible for the spontaneity to change such that the reverse reaction would now be spontaneous.

You Try It!

In the following self-assessment activity, determine the Gibbs free energy change. Click on the plus sign to check your answer!

Gibbs Free Energy and the Equilibrium Constant

As noted in the previous lesson, when a system is at equilibrium, $LaTeX: \Delta$ $\Delta$ G = 0 due to the fact that both the forward and reverse reactions are equally spontaneous. With this in mind, the Gibbs free energy relationship for nonstandard conditions can be simplified for a system at equilibrium as follows:

$LaTeX: \Delta$ $\Delta$ G = $LaTeX: \Delta$ $\Delta$ G^{$^\circ$} + RT ln (Q)

At equilibrium $LaTeX: \Delta$ $\Delta$ G = 0 and Q = K

0 = $LaTeX: \Delta$ $\Delta$ G^{$^\circ$} + RT ln (K)

$LaTeX: \Delta$ $\Delta$ G^{$^\circ$} = -RT ln (K)

Using this relationship it is now possible to determine the equilibrium constant for any reaction as long as the standard Gibbs free energy change can be calculated. For example, given the chemical reaction shown below and the $LaTeX: \Delta G^{^{\circ}}_f$ $\Delta G^{^{\circ}}_f$ values, the equilibrium constant for this precipitation reaction at 298 K can be determined.

AgBr (s) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ Ag⁺ (aq) + Br^- (aq)

Table
Substance	$LaTeX: \Delta G^{^{\circ}}_f$ $\Delta G^{^{\circ}}_f$ (kJ/mol)
AgBr (s)	-96.9
Ag⁺ (aq)	77.11
Br^- (aq)	-102.8

The first step in this instance would be to calculate the standard Gibbs free energy change using Hess's Law:

$LaTeX: \Delta G^{^{\circ}}=\Sigma\Delta G_f^{^{\circ}}\left(products\right)-\Sigma\Delta G^{^{\circ}}_f\left(reactants\right)$ $\Delta G^{^{\circ}}=\Sigma\Delta G_f^{^{\circ}}\left(products\right)-\Sigma\Delta G^{^{\circ}}_f\left(reactants\right)$

$LaTeX: \Delta G^{^{\circ}}=\left\lbrace\Delta G_f^{^{\circ}}\left\lbrack Ag^+\left(aq\right)\right\rbrack+\Delta G_f^{^{\circ}}\left\lbrack Cl^-\left(aq\right)\right\rbrack\right\rbrace-\lbrace\Delta G_f^{^{\circ}}\lbrack AgBr\left(s\right)]\}$ $\Delta G^{^{\circ}}=\left\lbrace\Delta G_f^{^{\circ}}\left\lbrack Ag^+\left(aq\right)\right\rbrack+\Delta G_f^{^{\circ}}\left\lbrack Cl^-\left(aq\right)\right\rbrack\right\rbrace-\lbrace\Delta G_f^{^{\circ}}\lbrack AgBr\left(s\right)]\}$

$LaTeX: \Delta G_{}^{^{\circ}}$ $\Delta G_{}^{^{\circ}}$ =[1(77.11) + 1(-102.8)] - [1(-96.9)]

$LaTeX: \Delta G_{}^{^{\circ}}$ $\Delta G_{}^{^{\circ}}$ = 71.2 kJ/mol

From this value, it can be clearly seen that the reaction is non-spontaneous under these conditions due to the positive Gibbs free energy change value. At this point, it is possible to make some qualitative predictions regarding the value of the equilibrium constant. Because the reaction is non-spontaneous in the direction that it is written it should be expected that the equilibrium constant, K should have a relatively low value due to a limited amount of products being formed. This is verified with the calculation below:

$LaTeX: \Delta$ $\Delta$ G^{$^\circ$} = -RT ln (K)

71.2 = (0.008314)(298) ln(K)

K = 3.31 x 10^-13

Alternatively, the equation can be rearranged in advance to calculate K:

$LaTeX: K=e^{\frac{-\Delta G^{^{\circ}}}{RT}}$ $K=e^{\frac{-\Delta G^{^{\circ}}}{RT}}$

$LaTeX: K=e^{\frac{-71.2}{\left(0.008314\right)\left(298\right)}}$ $K=e^{\frac{-71.2}{\left(0.008314\right)\left(298\right)}}$

$LaTeX: K=3.31\times10^{-13}$ $K=3.31\times10^{-13}$

In fact, the equilibrium constant for this system is extremely low under these conditions which further demonstrates the non-spontaneous nature of this equilibrium.

You Try It!

In the following self-assessment activity, calculate the equilibrium constant. Click on the plus sign to check your answer!

[CC BY 4.0] UNLESS OTHERWISE NOTED | IMAGES: LICENSED AND USED ACCORDING TO TERMS OF SUBSCRIPTION