JEE Chemistry 2025 Question with Solution

Given: $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2\mathrm{AB}$, activation energy for the forward reaction is $180 \, \text{kJ mol}^{-1}$ and for the backward reaction is $200 \, \text{kJ mol}^{-1}$. The catalyst lowers both activation energies by $100 \, \text{kJ mol}^{-1}$.

Find: Which statement is correct.

For a reaction profile,

$$\Delta H = E_{a,\text{forward}} - E_{a,\text{backward}}$$

So,

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

Thus, the reaction enthalpy is $-20 \, \text{kJ mol}^{-1}$, not $+20 \, \text{kJ mol}^{-1}$.

A catalyst lowers the activation energy of both forward and backward reactions by the same amount, but it does not change thermodynamic state functions such as $\Delta H$ or $\Delta G$. Therefore, it cannot make a non-spontaneous reaction spontaneous.

So:

1. Option A is correct because catalyst does not alter Gibbs energy change.
2. Option B is incorrect because spontaneity depends on $\Delta G$, which remains unchanged.
3. Option C is incorrect because $\Delta H = -20 \, \text{kJ mol}^{-1}$.
4. Option D is incorrect because catalysis does not change enthalpy change.

Therefore, the correct option is A.

Given: Forward activation energy $= 180 \, \text{kJ mol}^{-1}$, backward activation energy $= 200 \, \text{kJ mol}^{-1}$.

Find: The correct statement about the effect of catalyst.

First compute the enthalpy change:

$$\Delta H = E_f - E_b$$ $$\Delta H = 180 - 200 = -20 \, \text{kJ mol}^{-1}$$

Now after adding catalyst:

$$E_f' = 180 - 100 = 80 \, \text{kJ mol}^{-1}$$ $$E_b' = 200 - 100 = 100 \, \text{kJ mol}^{-1}$$

Then,

$$\Delta H' = E_f' - E_b' = 80 - 100 = -20 \, \text{kJ mol}^{-1}$$

So the enthalpy change remains unchanged even after catalysis.

Also, catalyst changes only the rate by providing an alternative pathway of lower activation energy. It does not change $\Delta G$, so it does not change spontaneity.

Therefore, the correct option is A.