Consider the following reversible reaction,\ $A_{(g)}+B_{(g)} ightleftharpoons A B_{(g)}$\ The a

Question

Consider the following reversible reaction,\ $A_{(g)}+B_{(g)} ightleftharpoons A B_{(g)}$\ The activation energy of the backward reaction exceeds that of the forward reaction by $2 R T$ (in $\mathrm{J} \mathrm{mol}^{-1}$ ). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta G^{\circ}\left(ight.$ in $\left.\mathrm{mol}^{-1}ight)$ for the reaction at 300 K is _______ .\ (Given : $\ln (2)=0.7, R T=2500 \mathrm{~J} \mathrm{~mol}^{-1}$ at 300 K and $G$ is the Gibbs energy)\

Professor PC Thomas Classes · Answer

${(8500)}: A_{(g)}+B_{(g)} ightleftharpoons A B_{(g)}$ Given, $E_{t_{i b}}-\mathbb{E}_{a_{j j}}=22 R X I$ and $\frac{A_{f}}{A_{b}}=4$\ $\Rightarrow K_{e q}=\frac{K_{f}}{K_{b}}$\ Now, $\frac{K_{f}}{K_{b}}=\frac{A_{f}}{A_{b}} e^{\left(\left(E_{a_{b}}-\Psi_{a_{j}}ight) /, R Tight.}$\ or $\quad K_{e q}=4 e^{2 R T / R T} ; \quad K_{e q}=4 e^{2}$\ $\Delta G^{\circ}=-R T \ln K_{e q}=-R T \ln \left(4 e^{2}ight)$ $$ \begin{aligned} & =-R T(2+\ln 4) \ & =-2500(2+2 	imes 0.7) \ & =-8500 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $$ $	herefore \quad$ Absolute value of $\Delta G^{\circ}$ is 8500 .\