Chemistry - Chemical Kinetics Question with Solution | TestHub

The Arrhenius equation is
$\mathrm{k}=\text{A}{\text{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\text{RT}}$
In absence of catalyst,${\mathrm{k}}_1=\text{A}{\text{e}}^{-100/\text{RT}}$
In presence of catalyst,${\mathrm{k}}_2=\text{A}{\text{e}}^{-25/\text{RT}}$
So$\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}={\mathrm{e}}^{-75/\text{RT}}$or 2.303$\text{lo}{\text{g}}^{\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}}=\frac{75}{\text{RT}}$
or$2.303 \text{log}\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{75}{8.314\times 10^{-3}\times 293}$
or$\text{log}\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{75}{8.314\times 10^{-3}\times 293\times 2.303}$
or$\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=2.34\times 10^{13}$
As the things being equal in presence or absence of a catalyst,
${\text{k}}_{\text{2}}=$rate in presence of catalyst
${\text{k}}_{\text{1}}\text{=}$rate in absence of catalyst
i.e.,$\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=2.34\times 10^{13}$