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

Using the Arrhenius equation:

k = A e^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

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:

e^(300 × 10^3 / 600 × R) = e^(-Ea / 300 × R)

Simplifying the equation, we get:

300 × 10^3 / 600 × R = Ea / 300 × R

Solving for Ea, we get:

Ea = (10^3 / 2) × 300 = 150 × 10^3 J mol^−1 = 150 kJ mol^−1

This gives us the activation energy for the uncatalyzed forward reaction.

To find the activation energy for the catalyzed backward reaction, we use the relationship:

E_rev,catalysed = E_fwd,uncat – ΔH_forward reaction

Substituting the given values, we get:

E_rev,catalysed = 150 kJ mol^−1 – 20 kJ mol^−1 = 130 kJ mol^−1

Therefore, the activation energy for the catalyzed backward reaction is 130 kJ mol^−1.