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

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

(b) From collision theory

$$\begin{aligned} \mathrm k &=\mathrm T^{\mathrm{1/2}}.\mathrm{{e}^{-E/RT}} \\ \mathrm r &=\mathrm k'_{o} \mathrm T^{1/2}.\mathrm{{e}^{-E/RT}}\\ \ln \mathrm r &=-\frac{\mathrm{E}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}}\right)+\ln \mathrm{T}^{1/2}+\ln \mathrm{k}_{o}^{\prime}\\ \ln \mathrm r_{1} &=-\frac{E}{R}\left(\frac{1}{\mathrm T_{1}}\right)+\ln \mathrm T_{1}^{1/2}+\ln \mathrm k_{o}^{\prime}\\ \ln \mathrm r_{2} &=-\mathrm {\frac{E}{R}}\left(\frac{1}{\mathrm T_{2}}\right)+\ln \mathrm T_{2}^{1/2}+\ln \mathrm k_{o}^{\prime}\\ \ln \left(\mathrm{r}_{2} /\mathrm r_{1}\right) &=\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]+\ln \left(\frac{\mathrm{T}_{2}^{1 / 2}}{\mathrm{~T}_{1}^{1/2}}\right)\\ \ln \left(10\mathrm r_{1} / \mathrm r_{1} \right) &=\frac{ \mathrm{E}}{1.987}\left[\frac{1}{400}-\frac{1}{500}\right]+\ln \left[\frac{(500)^{1 / 2}}{(400)^{1/2}}\right]\\ \mathrm{E} &=8707 \mathrm{~cal} / \mathrm{mol} \end{aligned}$$

(c) Difference in two methods

From Arrhenius law

$$\begin {aligned}\ln \left(\mathrm{r}_{3} / \mathrm{r}_{2}\right) &=\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{2}}-\frac{1}{\mathrm{~T}_{3}}\right]\\ \ln \mathrm {\left(r_3 / r_{2}\right) }&=\frac{9150}{1.987}\left[\frac{1}{500}-\frac{1}{600}\right]\\ \mathrm {r_{3} / r_{2}} &=4.64~\therefore \mathrm r_{3}=4.64 ~\mathrm r_{2} \end{aligned}$$

From Collision theory :

$$ \begin{aligned} \ln \left(\mathrm r_{3} / \mathrm r_{2}\right) &=\mathrm {\frac{E}{\mathrm{R}}\left[\frac{1}{T_{2}}-\frac{1}{T_{2}}\right]}+\ln \mathrm {\left(\frac{T_{3}^{1 / 2}}{T_{2}^{1 / 2}}\right)} \\ \ln \mathrm {\left(r_{3} / r_{2}\right) }&=\frac{8707}{1.987}\left[\frac{1}{500}-\frac{1}{600}\right]+\ln \left[\frac{(600)^{1/2}}{{(500)^{1 / 2}}}\right]\\ \mathrm r_{3} / \mathrm r_{2} &=4.72 \\ \mathrm r_{3} &=4.72 \mathrm r_{2} \end{aligned} $$

From Arrhenius law , $\mathrm r_{3} =4.64 \mathrm{r}_{2} $

From collision theory , $\mathrm r_{3} =4.72 \mathrm r_{2}$

The % difference in rates,

$$ =\frac{(4.72-4.64)\mathrm r_2}{4.72}×100$$

The rate of reaction from Collision theory is $1.7 \%$ more than it is given by Arrhenius law.