Question

The energies of activation for forward and reverse reactions for ${A2 + B2<=>2AB}$ 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 (${A2 + B2->2AB}$) in the presence of catalyst will be (in $kJ\, mol^{-1}$)

• A. $300$
• B. $120$
• C. $280$
• D. $-20$

Answer 1

Answer: (D)

$-20$

Answer 2

So, $\Delta H_{Reaction}=E_{f}-E_{b}$ $= 80 - 100 = -20$ Hence, $\left(4\right)$ is correct.