Gibbs Free Energy and Equilibrium

This is an extension topic and while it is not explicitly outlined in the HSC Chemistry syllabus, it has been examined in the past e.g. HSC 2023 Q35(b).

HSC Chemistry Syllabus

deduce the equilibrium expression for homogeneous reactions occurring in solution (ACSCH079, ACSCH096)
perform calculations to find the value of K_eq and concentrations of substances within an equilibrium system, and use these values to make predictions on the direction in which a reaction may proceed (ACSCH096)

Gibbs Free Energy and Equilibrium Explained

Key Points

Change in Gibbs free energy ( $\Delta G$ ) determines the spontaneity of reaction, whether a system is at equilibrium, and which direction it will proceed to reach an equilibrium.

Change in Standard Gibbs free energy ( $\Delta G^\circ$ ) is related to the equilibrium constant of a system, that is the relative concentration of products and reactants at equilibrium.

What is Gibbs Free Energy (G)?

Gibbs free energy represents the maximum amount of work a thermodynamic system can perform at constant temperature and pressure. It's given by the equation:

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

The change in Gibbs free energy $\Delta G$ during a process is what indicates spontaneity:

If $\Delta G < 0$ , the reaction is spontaneous, it will proceed in the forward direction until it reaches equilibrium.

If $\Delta G > 0$ , the reaction is non-spontaneous. This means the reverse reaction is spontaneous and the system will proceed in the reverse direction until it reaches equilibrium.

If $\Delta G = 0$ , the system is in a state of equilibrium.

Standard Gibbs Free Energy (∆Gº)

Standard Gibbs free energy change is the change in Gibbs free energy when a reaction occurs under standard conditions (concentrations of reactant and products are 1 mol/K, 1 bar of pressure and 298.15 K).

It's calculated using:

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

Standard Gibbs free energy is related to Gibbs free energy according to the following equation:

$\Delta G = \Delta G^\circ + RT \ln Q$

where:

$R$ is the gas constant (8.314 J mol^–1 K^–1)
$T$ is the temperature (in Kelvin)
$Q$ is the reaction quotient

For conditions not at equilibrium, we use the reaction quotient $Q$ , which has the same expression as $K$ but uses the current concentrations.

This equation helps predict the direction of a reaction under any given set of conditions. For a reversible reaction, its $\Delta G$ value will always approach zero as it progresses towards an equilibrium state.

When $Q < K$ , the equilibrium position will continue to shift to the product side until it reaches equilibrium. This means $\Delta G < 0$ , and will increase in value until it equals 0 as an equilibrium is established.

When $Q > K$ , the equilibrium position will continue to shift to the reactant side until it reaches equilibrium. This means $Delta G > 0$ , and will decrease in value until it equals 0 as an equilibrium is established.

When $Q = K$ , the system is at equilibrium, and $\Delta G = 0$ .

Standard Gibbs Free Energy and Equilibrium

At equilibrium, the free energy change is zero ( $\Delta G = 0$ ), which leads us to an important equation that relates $\Delta G^\circ$ to the equilibrium constant:

$0 = \Delta G^\circ + RT \ln Q$

$\Delta G^\circ = -RT \ln K$

where:

$R$ is the gas constant (8.314 J mol^–1 K^–1)
$K K$ is the equilibrium constant

This equation implies that the standard Gibbs free energy change is a measure of the direction a reaction will proceed in order to reach equilibrium from the standard states of its reactants and products.

Under standard conditions, the reaction quotient, $Q$ , equals 1 (concentrations of all substances equal 1 mol/L).

Therefore,

When $\Delta G^\circ = 0$ , the reaction is already at an equilibrium so the value of $K = 1$ .

When $\Delta G^\circ < 0$ , the reaction is not at an equilibrium yet. This suggests that the reaction will proceed in the forward reaction from standard states to reach an equilibrium, and the equilibrium concentration of products will be greater than the concentration of reactants. Thus, $K > 1$ .

When $\Delta G^\circ > 0$ , the reaction is not at an equilibrium yet. This suggests the reaction will proceed in the reverse reaction from standard states to reach an equilibrium, and the equilibrium concentration of products will be less than the concentration of reactants. Thus, $K < 1$ .

Example 1

The standard Gibbs free energy change ( $\Delta G^\circ$ ) for the following reaction is -20 kJ/mol.

$2NO_2(g) \rightleftharpoons N_2O_4(g)$

(a) Calculate the Gibbs free energy change ( $Delta G$ ) at 298 K when $[NO_2] = 0.010 \ mol \ L^{-1}$ and $[N_2O_4] = 0.10 \ mol \ L^{-1}$ .

(b) Calculate the value of the equilibrium constant for the reaction.

Solution for part (a):

$\Delta G = \Delta G^\circ + RT \ln Q$

$\Delta G = -20 000 + (8.314)(298) \ln \frac{(0.10)}{(0.010)^2}$

$\Delta G = -2.9 \times 10^3 \, J \, mol^{-1}$

Since $\Delta G < 0$ , the forward reaction is spontaneous, and more products will form until the system reaches an equilibrium.

Solution for part (b):

$\Delta G^\circ = -RT \ln K$

$-20 000 = -(8.314)(298) \ln K$

$\ln K = \frac{20 000}{(8.314)(298)}$

$K = e^{\frac{20000}{(8.314)(298)}} = 3205$

Previous section: Using Colourimetry to Calculate Equilibrium Constant

Next section: Dissolution of Ionic Compounds

Gibbs Free Energy and Equilibrium

HSC Chemistry Syllabus