Question

The energy of activation for an uncatalysed reaction is $100kJmo{l}^{-1}$. Presence of a catalyst lowers the energy of activation by $75\%$. The ${\log }_{10}\frac{K_2}{K_1}$ of the ratio of rate constant of catalysed and uncatalysed reactions at $27^{\circ }C$ is?
Assume the frequency factor is same for both reactions.
(Given $2.303\times 8.314=19.147$ )

• A. 13.05
• B. 26.10
• C. 6.52
• D. None of these

Answer 1

Answer: (A)

In absence of catalyst, $K_1=A{e}^{-100/RT}$
In presence of catalyst ; $K_2=A{e}^{-25/RT}$
$\therefore \frac{K_2}{K_1}=e^{75/RT}$
$\Rightarrow {\log }_e\frac{K_2}{K_1}=\frac{75}{RT}$
$\Rightarrow {\log }_{10}\frac{K_2}{K_1}=\frac{75}{2.303\times R\times T}$
$\Rightarrow {\log }_{10}\frac{K_2}{K_1}=\frac{75}{2.303\times 8.314\times 10^{-3}\times 300}=13.057$