Which of the following conditions will most enhance the spontaneity of a reaction?

Analyze the question: We need the condition that makes a reaction most spontaneous. Spontaneity is decided by the sign of the Gibbs free energy change, ΔG. The more negative ΔG is, the more spontaneous the reaction.

Concept:

ΔG = ΔH − TΔS

• ΔG < 0: spontaneous
• ΔG = 0: equilibrium
• ΔG > 0: non-spontaneous

Check each option:

• A. ΔH is negative and greater than TΔS. This is ambiguous. If "greater" means greater in magnitude, then |ΔH| > |TΔS|, and ΔG = ΔH − TΔS is negative whatever the sign of ΔS. Take ΔH = −100 and TΔS = +20: ΔG = −120, strongly spontaneous. Take TΔS = −20: ΔG = −80, still spontaneous.
• B. ΔH negative and ΔS = 0. ΔG = ΔH, which is negative, so the reaction is spontaneous. But it gets no extra help from entropy.
• C. ΔH positive and less than TΔS. Here ΔG = ΔH − TΔS is negative, so the reaction is spontaneous. But this only works at high temperature, and the process is endothermic.
• D. ΔH positive and equal to TΔS. ΔG = 0, so the system is at equilibrium. It is not spontaneous.

Which is "most enhanced"? Options B, C and A can all give negative ΔG. Under the intended reading, A describes a favourable enthalpy change (exothermic) that clearly dominates the TΔS term, so ΔG is definitely negative. Option B is exothermic but has no entropy contribution. Option C depends on temperature and on the entropy term overcoming an unfavourable enthalpy.

Answer: A

Tip: For ΔG = ΔH − TΔS, remember "negative ΔH helps, positive ΔS helps". Exothermic reactions with a dominant enthalpy term are reliably spontaneous. If ΔG = 0, the system is at equilibrium, not spontaneous.