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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 $KJmol^{-1}$ is :

$\left ( R=8.314\; J\; K^{-1} mol^{-1}\right )$
Option 1)
$166$
Option 2)
$150$
Option 3)
$72$
Option 4)
$59$

Answers (1)

$ln\left ( \frac{P_{1}}{P_{2}} \right )=\frac{Ea}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$

Given $T_{1}=327^{\circ}C=600 K$

$T_{2}=527^{\circ}C=800 K$

$P_{1}=2.5\times 10^{-4}$

$P_{2}=1\; atm$

$ln\left ( \frac{1}{2.5\times 10^{-4}} \right )=\frac{Ea}{8.314}\times \left ( \frac{1}{600}-\frac{1}{800} \right )$

$E_{a}=166\; kJ$

Option 1)
$166$
Option 2)

$150$

Option 3)

$72$

Option 4)

$59$

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