The energies of activation for forward and reverse reactions for $A_2+B_2 \rightleftharpoons 2 A B$ are $180 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy change of the reaction $\left(\mathrm{A}_2+\mathrm{B}_2 \longrightarrow 2 \mathrm{AB}\right)$ in the presence of catalyst will be (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ )

• A

  $300$
• B

  $120$
• C

  $280$
• Correct answer
  D

  $20$

Solution

So, $\Delta H_{\text {Reaction }}=E_f-E_b$ $=80-100=-20$ Hence, (D) is correct.

Hence, the correct answer is option: 'D'.