The energies of activation for forward and reverse reactions for $$A_2 + B_2 \rightleftharpoons 2AB$$ are 180 kJ $$mol^{-1}$$ and 200 kJ $$mol^{-1}$$ respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ $$mol^{-1}$$. The enthalpy change of the reaction $$(A_2 + B_2 \longrightarrow 2AB)$$ in the presence of catalyst will be (in kJ $$mol^{-1}$$)

For any reaction, the energy profile looks like:

Reactants $$\xrightarrow{\;E_{a(fwd)}\;}$$ Transition state $$\xrightarrow{\;E_{a(rev)}\;}$$ Products

Hence, the difference between the two activation energies equals the enthalpy change of reaction:

$$\Delta H = E_{a(fwd)} - E_{a(rev)} \; -(1)$$

Given (without catalyst):

$$E_{a(fwd)} = 180\;{\rm kJ\,mol^{-1}}, \qquad E_{a(rev)} = 200\;{\rm kJ\,mol^{-1}}$$

Substituting in (1):

$$\Delta H = 180 - 200 = -20\;{\rm kJ\,mol^{-1}}$$

The negative sign means the forward reaction $$A_2 + B_2 \rightarrow 2AB$$ is exothermic by 20 kJ mol−1.

In the presence of a catalyst, the activation energy of each direction is reduced by 100 kJ mol−1:

$$E_{a(fwd,\,cat)} = 180 - 100 = 80\;{\rm kJ\,mol^{-1}}$$

$$E_{a(rev,\,cat)} = 200 - 100 = 100\;{\rm kJ\,mol^{-1}}$$

Using equation (1) again:

$$\Delta H_{\rm(cat)} = 80 - 100 = -20\;{\rm kJ\,mol^{-1}}$$

Thus the catalyst lowers both activation energies equally, but the difference—and therefore the enthalpy change—remains the same. Only its magnitude is usually reported in such problems.

Therefore, the enthalpy change of the reaction in the presence of the catalyst is $$20\;{\rm kJ\,mol^{-1}}$$.

Option D which is: 20