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The relationship between the equilibrium constant, $K_{eq}$, and $\Delta G^\circ$ for any reaction is shown in the graph, for a limited range of $\Delta G^\circ$ values - HSC - SSCE Chemistry - Question 37 - 2024 - Paper 1

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Question 37

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The relationship between the equilibrium constant, $K_{eq}$, and $\Delta G^\circ$ for any reaction is shown in the graph, for a limited range of $\Delta G^\circ$ val... show full transcript

Worked Solution & Example Answer:The relationship between the equilibrium constant, $K_{eq}$, and $\Delta G^\circ$ for any reaction is shown in the graph, for a limited range of $\Delta G^\circ$ values - HSC - SSCE Chemistry - Question 37 - 2024 - Paper 1

Step 1

Explain, with reference to the information provided, why this reaction proceeds to completion rather than coming to equilibrium.

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Answer

The graph indicates that reactions with large negative ΔG\Delta G^\circ values correspond to extremely large equilibrium constants, implying that these reactions will proceed effectively to completion.

For the given reaction, we calculate:

ΔG=ΔHTΔS=217 kJ mol1(3 kJ mol1)=214 kJ mol1\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -217 \text{ kJ mol}^{-1} - (-3 \text{ kJ mol}^{-1}) = -214 \text{ kJ mol}^{-1}

This value of ΔG\Delta G^\circ is large and negative, which signifies that the reaction has a strong tendency to proceed to completion. The substantial negative value of ΔH\Delta H^\circ indicates that the reaction is highly exothermic, while the small TΔST\Delta S^\circ term contributes little to the overall Gibbs free energy change.

Thus, due to the significant negative ΔH\Delta H^\circ relative to the TΔST\Delta S^\circ, this reaction predominantly moves toward completion, rather than achieving a state of equilibrium.

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