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

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:

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

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

$⇒ \dfrac{{10}^{3}}{2} = \dfrac{Ea}{300}$

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

$⇒ {E}_{a} = 150kJ{mol}^{– 1}$

$∴$ Activation energy of catalysed backward reaction

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

$= 150 - 20 = 130kJ{mol}^{– 1}$