Question

Consider the following reaction with its thermodynamic data: \(2 \mathrm{~A}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(g) \Delta H^{\circ}<0 ; \Delta S^{\circ}<0 ; \Delta G^{\circ}\) at \(60^{\circ} \mathrm{C}=+10 \mathrm{~kJ}\) Which statements about the reaction are true? (a) When \(\Delta G=1,\) the reaction is at equilibrium. (b) When \(Q=1, \Delta G=\Delta G^{\circ}\). (c) At \(75^{\circ} \mathrm{C}\), the reaction is definitely nonspontaneous. (d) At \(100^{\circ} \mathrm{C}\), the reaction has a positive entropy change. (e) If \(\mathrm{A}\) and \(\mathrm{B}_{2}\) are elements in their stable states, \(S^{\circ}\) for \(\mathrm{A}\) and \(\mathrm{B}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(0 .\) (f) \(K\) for the reaction at \(60^{\circ} \mathrm{C}\) is less than \(1 .\)

Short answer

a) If ΔG = 1 kJ, the reaction is at equilibrium. b) If Q = 1, then ΔG = ΔG°. c) If the reaction is spontaneous at 60°C and ΔH° < 0 and ΔS° < 0, then the reaction will be nonspontaneous at 75°C. d) If ΔS° < 0 at 60°C, then ΔS° < 0 at 100°C. e) Elements in their standard states have 0 entropy at 25°C. f) If ΔG° > 0, then K < 1. Answer: The true statements are (b) and (f).

Step-by-step solution

Statement (a) Analysis

We know that when a reaction is at equilibrium, ΔG = 0. Given that when ΔG = 1, the statement claims the reaction is at equilibrium. This is incorrect, so the statement is false.

Statement (b) Analysis

The relationship between ΔG, ΔG°, and Q is given as: \(\Delta G = \Delta G^{\circ} + RT \ln{Q}\) When Q equals 1, the equation becomes: \(\Delta G = \Delta G^{\circ} + RT \ln{1}\) Since ln(1) = 0, this simplifies to: \(\Delta G = \Delta G^{\circ}\) This statement is true.

Statement (c) Analysis

We are given that ΔH° < 0 and ΔS° < 0. Usually, reactions become more spontaneous at higher temperatures if ∆H < 0 and ΔS > 0, but in this case ΔS < 0. Therefore, increasing the temperature to 75°C does not necessarily mean it will be nonspontaneous. Although a qualitative answer might suggest that with ΔH < 0 and ΔS < 0, a conclusion can be made, the given data is not sufficient to determine this without calculation, so this statement is false.

Statement (d) Analysis

The data given is at 60°C, and we are given that ΔS° < 0. The temperature change would not change the sign of ΔS°. Therefore, even at 100°C, the reaction would still have a negative entropy change. This statement is false.

Statement (e) Analysis

Elements in their stable states, i.e., standard states, have an entropy of 0 at 0 K (absolute zero) according to the Third Law of thermodynamics. At 25°C, elements in their standard states would have nonzero entropy values due to the temperature difference. Thus, this statement is false.

Statement (f) Analysis

When a reaction has a ΔG° > 0, the equilibrium constant K < 1. In this case, ΔG° at 60°C is given as +10 kJ, so K < 1. Therefore, this statement is true. In conclusion, the true statements are (b) and (f).