For a reaction, the graph of $\ln k$ (on $y$-axis) and $\frac{1}{T}$ (on $x$-axis) is a straight line with a slope $-2 \times 10^4 \mathrm{~K}$. The activation energy of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is ( $R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

The following graph is obtained for a first order reaction $(A \rightarrow P)$. The activation energy ( $E_a$ in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ and heat of reaction $\left(|\Delta H|\right.$ in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for this reaction are respectively

$\left(x=\right.$ reaction coordinate; $y=E$ in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$

For a first order reaction, the ratio between the time taken to complete $\frac{3}{4}$ th of the reaction and time taken to complete half of the reaction is

The following equation is obtained for a first order reaction at 300 K

$$ \log _{10} \frac{k}{A}=0.00174 $$

What is the activation energy (in $\mathrm{J} \mathrm{mol}^{-1}$ ) of the reaction?

$$ \left(R=8314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) $$