For a reversible reaction at temperature T , both Δ H and Δ S are positive. If the equilibrium temperature is T e , then the reaction becomes spontaneous at:

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 4 April, Morning Shift — Question 12

For a reversible reaction at temperature $T$ , both $\Delta H$ and $\Delta S$ are positive. If the equilibrium temperature is $T_e$ , then the reaction becomes spontaneous at:

Option A:
$T_e = 5T$
Option B:
$T_e > T$
Option C:
$T > T_e$
Correct
Option D:
$T = T_e$

Answer: C

Step-by-step solution

For spontaneity, $\Delta G = \Delta H - T\Delta S < 0$ . Since $\Delta H > 0$ and $\Delta S > 0$ , we require $T\Delta S > \Delta H \Rightarrow T > \frac{\Delta H}{\Delta S}$

The equilibrium temperature is given by $T_e = \frac{\Delta H}{\Delta S}$

Therefore, the reaction becomes spontaneous when $T > T_e$

Thus, the correct answer is Option C.

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry

One mole of an ideal gas expands isothermally and reversibly from 10 dm^3 to 20 dm^3 at 300 K . Delta U, q and work done in the process…

Answer: C

Step-by-step solution

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