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:
Solution
<p>For reaction to be spontaneous according to $2^{\text {nd }}$ law:</p>
<p>$$\begin{aligned}
& \Delta \mathrm{G}<0 \\
& \Rightarrow \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S}<0 \\
& \Rightarrow \mathrm{~T}>\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\
& \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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