For a reversible reaction $\mathrm{A} \rightleftharpoons \mathrm{B}$, the $\Delta \mathrm{H}_{\text{forward reaction}} = 20 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The activation energy of the uncatalysed forward reaction is $300 \mathrm{~kJ} \mathrm{~mol}^{-1}$. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at $27^{\circ} \mathrm{C}$ is found to be same as that of the uncatalysed reaction at $327^{\circ} \mathrm{C}$.

The activation energy of the catalysed backward reaction is ___________ $\mathrm{kJ}~ \mathrm{mol}^{-1}$.

Using the Arrhenius equation: <br/><br/> $k = Ae^{-E_a/RT}$ <br/><br/> where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin. <br/><br/> The equation is used for both the uncatalyzed forward reaction and the catalyzed backward reaction. By setting the two equations equal to each other and cancelling out the pre-exponential factor, we get: <br/><br/> $e^{\frac{300 \times 10^3}{600 \times R}} = e^{\frac{-E_a}{300 \times R}}$ <br/><br/> Simplifying the equation, we get: <br/><br/> $\frac{300 \times 10^3}{600 \times R} = \frac{E_a}{300 \times R}$ <br/><br/> Solving for $E_a$, we get: <br/><br/> $$E_a = \frac{10^3}{2}\times 300 = 150 \times 10^3\ \mathrm{J~mol^{-1}} = 150\ \mathrm{kJ~mol^{-1}}$$ <br/><br/> This gives us the activation energy for the uncatalyzed forward reaction. <br/><br/> To find the activation energy for the catalyzed backward reaction, we use the relationship: <br/><br/> $$E_{\text{rev,catalysed}} = E_{\text{fwd,uncat}} - \Delta H_{\text{forward reaction}}$$ <br/><br/> Substituting the given values, we get: <br/><br/> $$E_{\text{rev,catalysed}} = 150\ \mathrm{kJ~mol^{-1}} - 20\ \mathrm{kJ~mol^{-1}} = 130\ \mathrm{kJ~mol^{-1}}$$ <br/><br/> Therefore, the activation energy for the catalyzed backward reaction is $\boxed{130\ \mathrm{kJ~mol^{-1}}}$.