QJEEAdv18Chef8d: JEE Advanced 2018 Chemistry - Class 11 JEEAdv18Chef8d Solution: For a reaction, the plots of and with time at temperature and…

For the reaction at equilibrium

$\mathrm{A} \rightleftharpoons \mathrm{P}$

At any temperature:

Concentration of $A$ at equilibrium [A]_{e q}<5 Concentration of $\mathrm{B}$ at equilibrium [\mathrm{P}]_{e q}>5

Equilibrium constant $\left(\mathrm{K}_{e q}\right)=\frac{[\mathrm{P}]_{e q}}{[\mathrm{~A}]_{e q}}$

\mathrm{K}_{e q}>1

\text{Also,}~~ \frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}}>\frac{\mathrm{T}_2}{\mathrm{~T}_1}

$\begin{aligned}& \text{If}~\mathrm{T}_2 >\mathrm{T}_1 \\\\& \text{Hence}~\frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}} >\frac{\mathrm{T}_2}{\mathrm{~T}_1}>1 \\\\& \frac{\mathrm{K}_{\mathrm{T}_1}}{\mathrm{~K}_{\mathrm{T}_2}} >1 \\\\& \mathrm{~K}_{\mathrm{T}_1} >\mathrm{K}_{\mathrm{T}_2}\end{aligned}$

It can be clearly seen from diagram that concentration of reactant is lesser and that of product is higher at $T_1$ compared to $T_2$ . Hence,

equilibrium constant is higher at $\mathrm{T}_1$ than $\mathrm{T}_2$ . The reaction is exothermic.

i.e., \Delta \mathrm{H}^{\circ}<0 for forward reaction

The expression for the spontaneity and Gibbs free energy of reaction:

$\begin{aligned}\Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\\\\Delta \mathrm{S}^{\circ} & =\frac{\Delta \mathrm{H}^{\circ}-\Delta \mathrm{G}^{\circ}}{\mathrm{T}}\end{aligned}$

Since, both $\Delta \mathrm{G}^{\circ}$ and $\Delta \mathrm{H}^{\circ}$ are less than zero \left\{\Delta \mathrm{H}^{\circ}<0\right. and \left.\Delta \mathrm{G}^{\circ}<0\right\}

If \Delta \mathrm{H}^{\circ}>\Delta \mathrm{G}^{\circ} , then \Delta \mathrm{S}^{\circ}>0

Otherwise,

If \Delta \mathrm{H}^{\circ}<\Delta \mathrm{G}^{\circ} , then \Delta \mathrm{S}^{\circ}<0

we know,

$\begin{aligned}\Delta \mathrm{G}^{\circ} & =-\mathrm{RT} \ln \mathrm{K} . \\\\\Delta \mathrm{G}_1^{\circ} & =-\mathrm{RT}_1 \ln \mathrm{K}_{\mathrm{T}_1} \\\\\text { and } \Delta \mathrm{G}_2^{\circ} & =-\mathrm{RT} \ln \mathrm{K}_{\mathrm{T}_2}\end{aligned}$

Its given, \ln \mathrm{K}_{\mathrm{T}_1}>\ln \mathrm{K}_{\mathrm{T}_2}

$\begin{gathered}\Delta \mathrm{G}_1^{\circ}<\Delta \mathrm{G}_2^{\circ} \\\\\Delta \mathrm{H}_{\mathrm{T}_1}^{\circ}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_1}^{\mathrm{o}}<\Delta \mathrm{H}_{\mathrm{T}_2}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_2}^{\mathrm{o}}\end{gathered}$

This is possible only if \Delta S^{\circ}<0

Hence,

$\begin{array}{r}\Delta \mathrm{G}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0 \\\\\Delta \mathrm{H}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0\end{array}$