Let us consider a reversible reaction at temperature T. In this reaction, both H and S were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at:

T Te. 🧠 When H 0 and S 0, temperature must exceed Te = H / S for spontaneity G = H - T S < 0 only when the entropy term T S exceeds H. The equilibrium temperature Te is where G = 0. For T Te: T S Te S = H, so G < 0 (spontaneous). Answer: (c) T Te

JEE Main · 2025 · Shift-IeasyTHERMO-147

Let us consider a reversible reaction at temperature T. In this reaction, both H and S were observed to have positive…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Let us consider a reversible reaction at temperature $T$ . In this reaction, both $\Delta H$ and $\Delta S$ were observed to have positive values. If the equilibrium temperature is $T_e$ , then the reaction becomes spontaneous at:

Options

a
$T = T_e$
b
$T_e > T$
c
$T > T_e$
✓
d
$T_e = 5T$

Correct Answerc

$T > T_e$

Detailed Solution

🧠 When $\Delta H > 0$ and $\Delta S > 0$ , temperature must exceed $T_e = \Delta H / \Delta S$ for spontaneity $\Delta G = \Delta H - T\Delta S < 0$ only when the entropy term $T\Delta S$ exceeds $\Delta H$ . The equilibrium temperature $T_e$ is where $\Delta G = 0$ .

For $T > T_e$ : $T\Delta S > T_e \Delta S = \Delta H$ , so $\Delta G < 0$ (spontaneous).

$\boxed{\text{Answer: (c) } T > T_e}$

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