For following reactions
A $\overset{700 K}{\to}$ Product. A $\frac{500 K}{catalyst} \to$ Product
It was found that the $E_{a}$ is decreased by $30 kj /$ mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

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$198 kJ / mol$

$75 kJ / mol$

$135 kJ / mol$

answer is D.

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Detailed Solution

$K_{catalyst} = K A \cdot e^{- \frac{E_{a_{1}}}{{RT}_{1}}} = A \cdot E^{- \frac{E_{2}}{{RT}_{2}}} A \cdot e^{- \frac{E_{a_{1}}}{{RT}_{1}}} = A \cdot E^{- \frac{E_{2}}{{RT}_{2}}} \begin{aligned} \frac{{Ea}_{1}}{T_{1}} = \frac{{Ea}_{2}}{T_{2}} \\ ∵ {Ea}_{1} = {Ea}_{2} - 30 \\ \frac{{Ea}_{2} - 30}{500} = \frac{{Ea}_{2}}{700} \\ 5 {Ea}_{2} = 7 {Ea}_{2} - 210 \\ {Ea}_{2} = \frac{210}{2} = 105 kJ / mol \\ Activation energy of catalysed reaction = 105 - 30 \\ = 75 kJ / mol \end{aligned}$