Therefore, the activation energy can be calculated by using the expression shown below.

$\log \frac{K_{2}}{K_{1}}=\frac{E_{ a }}{2.303 \times R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$

Substitute all the values.

$\log \frac{2 K_{1}}{K_{1}}=\frac{E_{ a }}{2.303 \times 8.314}\left(\frac{1}{300}-\frac{1}{400}\right)$

$\log 2=\frac{E_{ a }}{2.303 \times 8.314} \times \frac{1}{1200}$

$E_{ a }=0.3 \times 2.303 \times 8.314 \times 1200$

$=6.88 \,kJ$