Gibbs Free Energy And Spontaneity

Question

For a reaction at 500 K, $\Delta H = -40$ kJ/mol and $\Delta S = -100$ J/mol·K. Which statement correctly describes the spontaneity of this reaction?

RankGuru · Accepted Answer

The Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ determines spontaneity: if $\Delta G < 0$, the reaction is spontaneous. Here $\Delta H = -40{,}000$ J/mol and $\Delta S = -100$ J/mol·K. For spontaneity: $-40{,}000 - T(-100) < 0 \Rightarrow -40{,}000 + 100T < 0 \Rightarrow T < 400$ K. At temperatures below 400 K, the exothermic enthalpy term dominates, making $\Delta G < 0$. Above 400 K, the negative entropy penalty $(-T\Delta S)$ becomes larger, making $\Delta G > 0$. Option 'Spontaneous at all temperatures' is wrong because with $\Delta S < 0$, increasing temperature always penalizes $\Delta G$. Option 'Non-spontaneous at all temperatures' is incorrect — at low T the reaction is indeed spontaneous. Option 'Spontaneous only above 400 K' contradicts the calculation since high temperatures amplify the unfavorable entropy term. This classic JEE pattern tests understanding of the temperature crossover point. Plausibility check: the crossover temperature $T = \Delta H / \Delta S = 40{,}000/100 = 400$ K, which is dimensionally consistent.

Gibbs Free Energy And Spontaneity

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