The rate constants for a reaction at $400 \mathrm{K}$ and $500 \mathrm{K}$ are $2.60\times {10}^{-5} {\mathrm{s}}^{-1}$ and $2.60\times {10}^{-3} {\mathrm{s}}^{-1}$,respectively. The activation energy of the reaction in ${\mathrm{kJmol}}^{-1}$ is ______

The relation between rate constant at two different temperatures and activation energies is given by :

$\log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$

So,

 $$\begin{gathered} \log \left(\frac{2.60\times {10}^{-3}}{2.60\times {10}^{-5}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{1}{400}-\frac{1}{500}\right] \\ {\mathrm{E}}_{\mathrm{a}}= \log \left({10}^2\right)\times 2.303\times 8.314\left[\frac{400\times 500}{100}\right] \\ {\mathrm{E}}_{\mathrm{a}}=2\times 2.303\times 8.314\times \frac{200000}{100} \\ {\mathrm{E}}_{\mathrm{a}}=76.6 \mathrm{KJ}/\mathrm{mol} \end{gathered}$$