Consider the following reactions at 300K

 $\text{X\hspace{0.17em}}\to \text{\hspace{0.17em}Y\hspace{0.17em}}\left(\text{uncatalysed\hspace{0.17em}reaction}\right)$

 $\text{X\hspace{0.17em}}\stackrel{\text{catalyst}}{\to }\text{\hspace{0.17em}Y\hspace{0.17em}}\left(\text{catalysed\hspace{0.17em}reaction}\right)$

 The energy of activation is lowered by $0.314{\text{\hspace{0.17em}KJ\hspace{0.17em}mol}}^{-1}$for the catalyzed reaction. The rate of   reaction is

${\text{k}}_{\text{uncat}}{\text{\hspace{0.17em}=\hspace{0.17em}Ae}}^{{\text{-E}}_{\text{uncat}}\text{/RT}}\mathrm{.....}\left(\text{i}\right)$

${\text{k}}_{\text{cat}}{\text{\hspace{0.17em}=\hspace{0.17em}Ae}}^{{\text{-E}}_{\text{cat}}\text{/RT}}\mathrm{.....}\left(\text{ii}\right)$

Let the $E_a$for u ncatalysed reaction be A then for catalyzed reaction it will be 

.$(A-8.314\times {10}^3)\text{\hspace{0.17em}cal/mole}$Dividing (ii) by (i),

$\frac{{\text{k}}_{\text{cat}}}{{\text{k}}_{\text{uncat}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{e}}^{\text{-}\left({\text{A\hspace{0.17em}-\hspace{0.17em}8.314\hspace{0.17em}×\hspace{0.17em}10}}^{\text{3}}\right)\text{/RT}}}{{\text{e}}^{\text{-A/RT}}}{\text{\hspace{0.17em}=e}}^{{\text{8.314\hspace{0.17em}×\hspace{0.17em}10}}^{\text{3}}\text{/RT}}$

$\frac{{\text{k}}_{\text{cat}}}{{\text{k}}_{\text{uncat}}}{\text{\hspace{0.17em}=\hspace{0.17em}e}}^{\text{3.33}}\text{\hspace{0.17em}=\hspace{0.17em}28\hspace{0.17em}}\left(\text{take\hspace{0.17em}log\hspace{0.17em}then\hspace{0.17em}anti\hspace{0.17em}log}\right)$