For the equilibrium $\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right),\mathrm{\Delta }H\text{ is }-40\mathrm{kJ}/\mathrm{mol}$, If the ratio of activation energies of the forward $\left(E_f\right)$ and reverse $\left(E_{\mathrm{b}}\right)$ is 2/3, then

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The given reaction is,

$\begin{array}{l}\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)\\ \mathrm{△}\mathrm{H}=-40\mathrm{KJ}/\mathrm{mol}\\ \mathrm{\&}\frac{{\mathrm{E}}_{\mathrm{f}}}{{\mathrm{E}}_{\mathrm{b}}}=\frac{2}{3}\Rightarrow {\mathrm{E}}_{\mathrm{f}}=\frac{2}{3}{\mathrm{E}}_{\mathrm{b}}\end{array}$

From graph we have,

$\begin{array}{l}\mathrm{△}\mathrm{H}={\mathrm{E}}_{\mathrm{b}}-{\mathrm{E}}_{\mathrm{f}}\\ \Rightarrow -40={\mathrm{E}}_{\mathrm{b}}-\frac{2}{3}{\mathrm{E}}_{\mathrm{b}}\\ \Rightarrow -40=\frac{1}{3}{\mathrm{E}}_{\mathrm{b}}\Rightarrow {\mathrm{E}}_{\mathrm{b}}=-120\mathrm{KJ}/\mathrm{mol}\\ \mathrm{\&}{\mathrm{E}}_{\mathrm{f}}={\mathrm{E}}_{\mathrm{b}}-\mathrm{\Delta H}\\ =-120-(-40)\\ -80\mathrm{KJ}/\mathrm{mol}\end{array}$