Question

A catalyst lowers the activation energy for a certain reaction from $75kJ$ to $25kJ$ $mo{l}^{-1}$. The effect on the rate of reaction at ${25}^{\circ }$C, other things being equal is :

• A. rate of reaction decreases $5.81\times {10}^8$ times
• B. rate of reaction decreases $6.34\times {10}^8$ times
• C. rate of reaction increases $6.34\times {10}^8$ times
• D. rate of reaction increases $5.81\times {10}^8$ times

Answer 1

Answer: (D)

The relationship between the rate of the reaction and the activation energy is $logk=-\frac{E_a}{2.303RT}+logA$

This equation in presence and in absence of catalyst can be written as $log\left(\frac{k_2}{k_1}\right)=-\frac{1}{2.303RT}(E_{a2}-E_{a1})$

Here, $E_{a_2}$ and $E_{a_1}$ represents the activation energies in presence and in absence of catalyst and $k_2$ and $k_1$ represents the rate constants in presence and in absence of catalyst.

Hence, $log\left(\frac{k_2}{k_1}\right)=-\frac{1}{2.303\times 8.314\times 298}(25000-75000)=8.763$

$\left(\frac{k_2}{k_1}\right)=5.81\times {10}^8$

Hence, the rate of the reaction increase $5.81\times {10}^8$ times.