JEE Advance - Chemistry (2018 - Paper 2 Offline - No. 12)
Consider the following reversible reaction, $$A\left( g \right) + B\left( g \right) \to AB\left( g \right).$$
The activation energy of the backward reaction exceeds that of the forward reaction by $$2RT$$ (in $$J\,mo{l^{ - 1}}$$). If the pre-exponential factor of the forward reaction is $$4$$ times that of the reverse reaction, the absolute value of $$\Delta {G^ \circ }$$ (in $$J\,mo{l^{ - 1}}$$ ) for the reaction at $$300$$ $$K$$ is ____________.
(Given; $$\ln \left( 2 \right) = 0.7,RT = 2500$$ $$J\,mo{l^{ - 1}}$$ at $$300$$ $$K$$ and $$G$$ is the Gibbs energy)
The activation energy of the backward reaction exceeds that of the forward reaction by $$2RT$$ (in $$J\,mo{l^{ - 1}}$$). If the pre-exponential factor of the forward reaction is $$4$$ times that of the reverse reaction, the absolute value of $$\Delta {G^ \circ }$$ (in $$J\,mo{l^{ - 1}}$$ ) for the reaction at $$300$$ $$K$$ is ____________.
(Given; $$\ln \left( 2 \right) = 0.7,RT = 2500$$ $$J\,mo{l^{ - 1}}$$ at $$300$$ $$K$$ and $$G$$ is the Gibbs energy)
Answer
8500
Explanation
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 $8500 \mathrm{~J} \mathrm{~mol}^{-1}$.
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 $8500 \mathrm{~J} \mathrm{~mol}^{-1}$.
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