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Q. For the reaction of $H_2$ with $I_2$, the rate constant is $2.5 \times 10^{-4} dm^3 \; mol^{-1} s^{-1} $ at $327^{\circ}C$ and $1.0 \; dm^3 \; mol^{-1} \; s^{-1}$ at $527^{\circ} C$. The activation energy for the reaction, in $kJ \; mol^{-1}$ is:
$(R=8.314\,J \; K^{-1} mol^{-1})$

JEE MainJEE Main 2019Chemical Kinetics

Solution:

$H_2(g) + I_2(g) \to 2HI(g)$
Apply Arrhenius equation
$\log \frac{K_{2}}{K_{1}} \frac{E_{a}}{2.303 R} \left( \frac{1}{600} - \frac{1}{800}\right) $
$ \log \frac{1}{2.5 \times10^{-4}} = \frac{E_{a}}{2.303 \times8.31} \left( \frac{200}{600 \times800}\right) $
$ \therefore E_{a} \approx166 $ kJ/mol