For first order reaction, if rate constant at ${17}^{°}\text{C }$ is$2.8\times {10}^{-5}{\text{s}}^{-1}$ and at ${27}^{°}\text{C}$ is$2.8\times {10}^{-4}{\text{s}}^{-1}$ . Then activation energy is

The Arrhenius Equation is given below.

$\log \frac{k_{\text{2}}}{k_{\text{1}}}=\frac{{\text{E}}_{\text{a}}}{2.303R}\left[\frac{{\text{T}}_{\text{2}}-{\text{T}}_1}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right]$

Put the given values in this equation to get the activation energy.

$\log \left[\frac{2.8\times {10}^{-4}}{2.8\times {10}^{-5}}\right]=\frac{{\text{E}}_{\text{a}}}{2.303\times 8.314}\left[\frac{1}{290}-\frac{1}{300}\right]$

     ${\text{E}}_{\text{a}}=\frac{2.303\times 8.314\times 290\times 300\times \log 10}{10}$

          $=166299.6{\text{ J mol}}^{-1}$

 $\therefore {\text{E}}_{\text{a}}=166.3{\text{ KJ mol}}^{-1}$