The rate constants of a certain reaction are $1.6 \times 10^{-3}$ and $1.625 \times 10^{-2}(\mathrm{~s})^{-1}$ at $10^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. Calculate the activation energy.

Solution :

Given :

$ \mathrm k_{1}=1.6 \times 10^{-2}$ at $\mathrm T_{1}=10^{°} \mathrm{C}=283 \mathrm{~K}$

$\mathrm k_{2}=1.625 \times 10^{-2}$ at $\mathrm T_{2} =30^{\circ} \mathrm{C}=308 \mathrm{~K} $

Activation energy can be calculated by using Arrhenius law

The value of ideal gas constant is , $\mathrm{R} =1.987 (~ \mathrm{cal} / \mathrm{mol.} \mathrm{K}) $

$$ \begin{aligned} \ln \left(\mathrm {\frac{k_{2}}{k_{1}}}\right) &=\frac{-\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm T_{2}}-\frac{1}{\mathrm{~T}_{1}}\right] \\ \ln \left(\mathrm{k}_{2} / \mathrm{k}_{1}\right) &=\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right] \\ \ln \left(\frac{1.625 \times 10^{-2}}{1.6 \times 10^{-3}}\right) &=\frac{\mathrm{E}}{1.987}\left[\frac{1}{283}-\frac{1}{303}\right] \\ \mathrm{E} &=19500 ~\mathrm{cal/mol} \end{aligned} $$