Chemical Kinetics And Nuclear Chemistry

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$