For a reversible reaction A ⇌ B , the ∆ H forward reaction = 20   kJ   mol – 1 .…

As per the question 

To determine the activation energy of the catalyzed backward reaction, we need to use the Arrhenius equation and the given information.

The Arrhenius equation is given by: 

$\mathrm{k}={\mathrm{Ae}}^{(-\mathrm{Ea}/\mathrm{RT})}$

${\mathrm{Ae}}^{-\frac{300\times {10}^3}{600\times \mathrm{R}}}={\mathrm{Ae}}^{\frac{-\mathrm{Ea}}{300\times \mathrm{R}}}$

$\Rightarrow \frac{{10}^3}{2}=\frac{\mathrm{Ea}}{300}$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}}=150\times {10}^3\mathrm{J} {\mathrm{mol}}^{–1}$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}}=150 \mathrm{kJ} {\mathrm{mol}}^{–1}$

$\therefore$ Activation energy of catalysed backward reaction

Energy of activation for backward reaction${\mathrm{E}}_{\mathrm{b}}={\mathrm{E}}_{\mathrm{a}}-∆\mathrm{H}$

$=150-20=130 \mathrm{kJ} {\mathrm{mol}}^{–1}$