For the reaction of $\mathrm{H}_{2}$ with $\mathrm{I}_{2}$, the rate constant is $2.5 \times 10^{-4} \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at $327^{\circ} \mathrm{C}$ and $1.0 \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at $527^{\circ} \mathrm{C}$. The activation energy for the reaction, in $\mathrm{kJ} \mathrm{mol}^{-1}$ is : $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
Correct Option: 1
$K=e^{-\frac{E_{a}}{R T}}$ or $\log K=\frac{-E_{a}}{2.303 R T}$
So, $\log \frac{K_{2}}{K_{1}}=\frac{E_{a}}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$
$\log \frac{1}{2.5 \times 10^{-4}}=\frac{E_{a}}{8.314 \times 2.303}\left(\frac{1}{600}-\frac{1}{800}\right)$
$3.6=\frac{E_{a}}{8.314 \times 2.303} \times \frac{200}{600 \times 800}$
$E_{a}=165.4 \mathrm{~kJ} / \mathrm{mol} \approx 166 \mathrm{~kJ} / \mathrm{mol}$
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