Chemistry - Chemical Kinetics Question with Solution | TestHub

$\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right];$ If $\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=2$ $$\begin{array}{l}\log 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{1}{300}-\frac{1}{310}\right]\\ \begin{array}{l}{\mathrm{E}}_{\mathrm{a}}=0.3010\times 2.303\times 8.314\left(\frac{300\times 310}{10}\right)\\ =53598.59{\mathrm{Jmol}}^{-1}=54\mathrm{kJ}\end{array}\end{array}$$