Arrhenius equation for the rate constant of a reaction in terms of energy of activation at two different temperatures is given by

$\log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.3\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$

Rise in temperature $=9 \mathrm{K}$

Initial temperature $=300 \mathrm{K}$

$\log \left(\frac{{\mathrm{K}}_{309}}{{\mathrm{K}}_{300}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.3\mathrm{R}}\left[\frac{1}{300}-\frac{1}{309}\right]$

$\log 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.3\mathrm{R}}\left[\frac{9}{300\times 309}\right]$

${\mathrm{E}}_{\mathrm{a}}=\frac{2.3\times 0.30\times 300\times 309}{9}$

${\mathrm{E}}_{\mathrm{a}}=58.988\times {10}^3 \mathrm{J}=58.988\mathrm{kJ}=59\mathrm{kJ}$