Question

Activation energy of a reaction is 100 kJ/mol. By using a catalyst $E_a$ decreases by 75%. What will be the effect on rate of reaction at ${20}^{0}C$?

Consider all other things are equal.

Answer 1

According to Arrhenius equation,

$K=A{e}^{-Ea/RT}$

In the absence of catalyst $K_1=A{e}^{-100/RT}$

in the presence of catalyst, $K_2=A{e}^{-25/RT}$

$\therefore \frac{K_1}{K_2}=\frac{e^{-100/RT}}{e^{-25/RT}}=e^{-75/RT}$

or $2303\log \frac{K_1}{K_2}=\frac{75}{RT}=\frac{75}{8.314\times {10}^{-3}\times 298}=30.27$

$\frac{K_1}{K_2}=1.38\times {10}^{13}$