Chemical Kinetics

The graph obtained between $\ln k$ ( $k=$ rate constant) on $y$-axis and $1 / T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \mathrm{~K}$. The activation energy of the reaction (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\left(R=831 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$