At 500 K the rate of a bimolecular reaction is ten times the rate at 400 K. Find the activation energy for this reaction

(a) from Arrhenius law,

(b) from collision theory,

(c) what is the percentage difference in rate of reaction at 600 K predicted by these two methods ?

Solution :

Given :

$\mathrm {r_{2}=10 ~r_{1}}$

$\mathrm T_{2}=400 \mathrm{~K}, \mathrm T_{2}=500 \mathrm{~K}, \mathrm T_3=600~\mathrm K$

$\mathrm{R} =1.987 \mathrm{~cal} /(\mathrm{mol}.\mathrm{K})$

(a) From Arrhenius law

$$ \begin{aligned} \ln (\mathrm{r}) &=-\frac{\mathrm{E}}{\mathrm{RT}}+\ln \mathrm{k}_{0} \\ \ln \left(\mathrm{r}_{2}/\mathrm{r}_{\mathrm{1}}\right)&=-\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{2}}-\frac{1}{\mathrm{~T}_{1}}\right] \\ \ln \left(\mathrm{r}_{2} /\mathrm{r}_{1}\right) &=\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]\\ \ln \left(10 r_{1}/r_{1}\right)&=\frac{\mathrm{E}}{1.987}\left[\frac{1}{400}-\frac{1}{500}\right]\\ \mathrm E &= 9150 \mathrm {~cal/mol} \end{aligned} …