Chemistry - Chemical Kinetics Question with Solution | TestHub

In absence of catalyst,${\text{K}}_1={\text{Ae}}^{-100/\text{RT}}$

In presence of catalyst ;${\text{K}}_2={\text{Ae}}^{-25/\text{RT}}$

$\therefore \frac{{\text{K}}_2}{{\text{K}}_1}={\text{e}}^{75/\text{RT}}\Rightarrow {\text{log}}_{\text{e}}\frac{{\text{K}}_2}{{\text{K}}_1}=\frac{75}{\text{RT}}$

$\Rightarrow {\text{log}}_{10}\frac{{\text{K}}_2}{{\text{K}}_1}=\frac{75}{\text{2.303}\times \text{R}\times \text{T}}$

$\Rightarrow {\text{log}}_{10}\frac{{\text{K}}_2}{{\text{K}}_1}=\frac{75}{\text{2.303}\times \text{8.314}\times 10^{-3}\times 300}\mathrm{=\; 13.057}$