Q25 / 75 · JEE Main 2025 (4 Apr Shift 1) · Chemistry

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

  • A
    $\mathrm{T}=\mathrm{Te}$
  • B
    $\mathrm{Te} \gt \mathrm{T}$
  • Correct answer
    C
    $\mathrm{T} \gt \mathrm{Te}$
  • D
    $\mathrm{Te}=5 \mathrm{~T}$

Solution

For reaction to be spontaneous according to $2^{\text {nd }}$ law:
$\begin{aligned}<br />& \Delta \mathrm{G} \lt 0 \\<br>& \Rightarrow \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S} \lt 0 \\<br>& \Rightarrow \mathrm{~T} \gt \left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\<br>& \Rightarrow \mathrm{~T} \gt \mathrm{T}_{\mathrm{e}}<br />\end{aligned}$

Hence, the correct answer is option: 'C'.

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