1. Use Arrhenius equation

For a reaction, $k = A e^{-E_a/(RT)}$ where:

• $k$ = rate constant
• $A$ = pre-exponential factor
• $E_a$ = activation energy
• $R$ = gas constant
• $T$ = temperature

Since the rate remains unchanged, the rate constants are equal for the uncatalysed and catalysed reactions.

Also, it is given that the pre-exponential factor is the same.

So, $A e^{-E_{a1}/(R\cdot 700)} = A e^{-E_{a2}/(R\cdot 500)}$

Cancelling $A$: $e^{-E_{a1}/(700R)} = e^{-E_{a2}/(500R)}$

Therefore, $\frac{E_{a1}}{700} = \frac{E_{a2}}{500}$

2. Use the decrease in activation energy

Catalyst decreases activation energy by $30\ \text{kJ mol}^{-1}$, so $E_{a1} - E_{a2} = 30$

Let the catalysed activation energy be $E_{a2}=x$. Then $E_{a1}=x+30$

Substitute into $\frac{E_{a1}}{700} = \frac{E_{a2}}{500}$

So, $\frac{x+30}{700} = \frac{x}{500}$

3. Solve for $x$

Cross-multiplying: $500(x+30)=700x$ $500x+15000=700x$ $15000=200x$ $x=75$

Thus, $E_{a2}=75\ \text{kJ mol}^{-1}$

4. Check options

• A: $198$ kJ/mol ❌
• B: $135$ kJ/mol ❌
• C: $105$ kJ/mol ❌
• D: $75$ kJ/mol ✅

Hence, the activation energy for the catalysed reaction is: $\boxed{75\ \text{kJ mol}^{-1}}$