JEE MAIN - Chemistry (2025 - 4th April Morning Shift - No. 16)
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:
$\mathrm{Te}>\mathrm{T}$
$\mathrm{T}>\mathrm{Te}$
$\mathrm{T}=\mathrm{Te}$
$\mathrm{Te}=5 \mathrm{~T}$
Explanation
For reaction to be spontaneous according to $2^{\text {nd }}$ law:
$$\begin{aligned} & \Delta \mathrm{G}<0 \\ & \Rightarrow \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S}<0 \\ & \Rightarrow \mathrm{~T}>\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\ & \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}} \end{aligned}$$
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