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 :

Question

Aicharya · Accepted Answer

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

Let us consider a reversible reaction at temperature, T. In this reaction, both $\Delta \mathrm{H}$ and…

Solution