Question

The energy of activation for an uncatalysed reaction is $100 \,kJ\,mol ^{-1}$. 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, $K_{1}=A e^{-100 / R T}$
In presence of catalyst ; $K _{2}= Ae ^{-25 / RT }$
$\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 } $
$\Rightarrow \log _{10} \frac{ K _{2}}{ K _{1}}=\frac{75}{2.303 \times 8.314 \times 10^{-3} \times 300}=13.057$