QJEEAdv18Ch1e74: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Ch1e74 Solution: Consider the following reversible reaction, The activation…

At $300 \mathrm{~K}$ ,

For the reversible reaction,

$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{AB}(g)$

(i) Pre-exponential factor for forward reaction $\left(A_f\right)=4 \times$ pre-exponential factor for backward reverse reaction $\left(A_b\right)$

$A_f=4 \times A_b$

(ii) Activation energy for backward reaction $\left(E_b\right)_f-$ Activation energy $=2 \mathrm{RT}$ for forward reaction $\left(\mathrm{E}_a\right)_{f^{-}}$

$\mathrm{A}_b-\mathrm{A}_f=2 \mathrm{RT}$

$\begin{aligned}& \text {(iii) Equilibrium constant }\left(\mathrm{K}_{e q}\right)=\frac{\mathrm{K}_f}{\mathrm{~K}_b} \\\\& \mathrm{~K}_{e q}=\frac{\mathrm{A}_f \times e^{-\left(\mathrm{E}_a\right)_f} / \mathrm{RT}}{\mathrm{A}_b \times e^{-\left(\mathrm{E}_a\right)_b} / \mathrm{RT}}\end{aligned}$

$\begin{aligned}\mathrm{K}_{e q} & =\frac{\mathrm{A}_f}{\mathrm{~A}_b} \times e^{\left[-\left(\mathrm{E}_a\right)_f-\left(\mathrm{E}_a\right)_b\right]} / \mathrm{RT} \\\\\mathrm{K}_{e q} & =\frac{4 \mathrm{~A}_b}{\mathrm{~A}_b} \times e^{-[2 \mathrm{RT} / \mathrm{RT}]} \\\\\mathrm{K}_{e q} & =4 \times e^{-2} \\\\\ln \mathrm{K}_{e q} & =\ln 4+2 \ln \\\\\ln \mathrm{K}_{e q} & =2 \ln 2+2=2 \times 0.7+2 \\\\& =(1.4+2)=3.4\end{aligned}$

The expression for the change in Gibbs free energy of reaction $(\Delta \mathrm{G})$ is given as:

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln k$

At equilibrium, $\Delta \mathrm{G}=0$

$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{e q}$

Substituting the value of RT and $\ln \mathrm{K}_{e q}$

$\begin{aligned}\Delta \mathrm{G}^{\circ} & =-2500 \mathrm{~J} \mathrm{~mol}^{-1} \times 3.4 \\\\\Delta \mathrm{G}^{\circ} & =-8500 \mathrm{~J} \mathrm{~mol}^{-1}\end{aligned}$

The absolute value of Gibbs free energy $\left(\Delta \mathrm{G}^{\circ}\right)$ for the reaction at $300 \mathrm{~K}$ is $8500 \mathrm{~J} \mathrm{~mol}^{-1}$ .