Temperature Dependence of Equilibrium Constant and Entropy Changes

To understand the influence of temperature variations on the equilibrium constant $K$ in terms of entropy principles, we use the fundamental thermodynamic relationships between free energy, entropy, and the equilibrium constant.

1. Thermodynamic Framework

The standard Gibbs free energy change of a reaction, $\Delta G^{\circ}$, is related to the equilibrium constant $K$ by: $\Delta G^{\circ} = -RT \ln K$

Rearranging for $\ln K$: $\ln K = -\frac{\Delta G^{\circ}}{RT}$

We also know that $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S_{\text{sys}}^{\circ}$. Substituting this gives: $\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S_{\text{sys}}^{\circ}}{R}$

The entropy change of the surroundings, $\Delta S_{\text{surr}}^{\circ}$, at constant pressure and temperature $T$ is given by the heat transferred to the surroundings: $\Delta S_{\text{surr}}^{\circ} = -\frac{\Delta H^{\circ}}{T}$

Therefore, the total entropy change of the universe, $\Delta S_{\text{total}}^{\circ}$, is: $\Delta S_{\text{total}}^{\circ} = \Delta S_{\text{sys}}^{\circ} + \Delta S_{\text{surr}}^{\circ} = \Delta S_{\text{sys}}^{\circ} - \frac{\Delta H^{\circ}}{T}$

Substituting this back into the expression for $\ln K$: $\ln K = \frac{\Delta S_{\text{total}}^{\circ}}{R}$

Thus, the equilibrium constant $K$ is directly related to the total entropy change: $K = \exp\left(\frac{\Delta S_{\text{total}}^{\circ}}{R}\right) = \exp\left(\frac{\Delta S_{\text{sys}}^{\circ} + \Delta S_{\text{surr}}^{\circ}}{R}\right)$

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2. Analysis of Endothermic Reactions ($\Delta H^{\circ} > 0$)

• For an endothermic reaction, heat is absorbed from the surroundings, so: $\Delta S_{\text{surr}}^{\circ} = -\frac{\Delta H^{\circ}}{T} < 0$ This negative value represents an unfavourable entropy change of the surroundings.

• When the temperature $T$ is increased: $\left| \Delta S_{\text{surr}}^{\circ} \right| = \frac{\Delta H^{\circ}}{T} \quad \text{decreases}$ This means $\Delta S_{\text{surr}}^{\circ}$ becomes less negative (i.e., approaches zero from the negative side).

• Therefore, as $T$ increases, the unfavourable change in entropy of the surroundings diminishes (decreases in magnitude). As a result, $\Delta S_{\text{total}}^{\circ}$ increases, which leads to an increase in the value of $K$.

This confirms that Option B is correct.

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3. Analysis of Exothermic Reactions ($\Delta H^{\circ} < 0$)

• For an exothermic reaction, heat is released to the surroundings, so: $\Delta S_{\text{surr}}^{\circ} = -\frac{\Delta H^{\circ}}{T} = \frac{|\Delta H^{\circ}|}{T} > 0$ This positive value represents a favourable entropy change of the surroundings.

• When the temperature $T$ is increased: $\Delta S_{\text{surr}}^{\circ} = \frac{|\Delta H^{\circ}|}{T} \quad \text{decreases}$

• Therefore, as $T$ increases, the favourable change in entropy of the surroundings diminishes. As a result, $\Delta S_{\text{total}}^{\circ}$ decreases, which leads to a decrease in the value of $K$.

This confirms that Option D is correct.

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Conclusion

• Option A is incorrect because the decrease in $K$ for an exothermic reaction is due to the temperature dependence of $\Delta S_{\text{surr}}^{\circ}$, not merely a positive $\Delta S_{\text{sys}}^{\circ}$.
• Option B is correct because for an endothermic reaction, $\Delta S_{\text{surr}}^{\circ} < 0$ (unfavourable), and increasing $T$ makes it less negative (diminishes the unfavourable effect), thereby increasing $K$.
• Option C is incorrect because the sign of $\Delta S_{\text{sys}}^{\circ}$ is not inherently negative for all endothermic reactions, nor is it the primary cause of the temperature dependence of $K$.
• Option D is correct because for an exothermic reaction, $\Delta S_{\text{surr}}^{\circ} > 0$ (favourable), and increasing $T$ decreases $\Delta S_{\text{surr}}^{\circ}$ (diminishes the favourable effect), thereby decreasing $K$.

Correct Options: B and D