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

Explanation

Given,

$$\Delta H=20\ \text{kJ mol}^{-1}$$

$$E_{a(\text{uncat,fwd})}=300\ \text{kJ mol}^{-1}$$

The catalysed reaction at $$27^\circ C$$ and the uncatalysed reaction at $$327^\circ C$$ have the same rate.

Converting temperatures into Kelvin,

$$T_1=27+273=300\ K$$

$$T_2=327+273=600\ K$$

Using the Arrhenius equation,

$$k=Ae^{-E_a/RT}$$

Since the rates are equal, the corresponding rate constants are equal.

Therefore,

$$Ae^{-E_{a(\text{cat,fwd})}/RT_1}=Ae^{-E_{a(\text{uncat,fwd})}/RT_2}$$

Cancelling $$A$$ and taking logarithms,

$$\frac{E_{a(\text{cat,fwd})}}{T_1}=\frac{E_{a(\text{uncat,fwd})}}{T_2}$$

Substituting the given values,

$$\frac{E_{a(\text{cat,fwd})}}{300}=\frac{300}{600}$$

$$E_{a(\text{cat,fwd})}=300\times\frac{300}{600}$$

$$E_{a(\text{cat,fwd})}=150\ \text{kJ mol}^{-1}$$

For a reaction,

$$\Delta H=E_{a(\text{fwd})}-E_{a(\text{bwd})}$$

A catalyst lowers the activation energies of both forward and backward reactions by the same amount, so the value of $$\Delta H$$ remains unchanged.

Using the catalysed pathway,

$$20=150-E_{a(\text{cat,bwd})}$$

$$E_{a(\text{cat,bwd})}=150-20$$

$$E_{a(\text{cat,bwd})}=130\ \text{kJ mol}^{-1}$$

Hence, the activation energy of the catalysed backward reaction is

$$130\ \text{kJ mol}^{-1}$$