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}) $ …