The activation energy for the reaction ${2 H I}_{(g)} \to H_{2} I_{2 (g)}$ is 209.5 $209.5 {m o l}^{- 1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

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Solution

The Arrenhius Equation is given by,
$k = A e^{\frac{- E_{a}}{R T}}$
here in this equation $E_{a}$ is the activation energy and the term $e^{\frac{- E_{a}}{R T}}$ is the fraction of molecules of reactants having energy equal to or greater than activation energy.
Therefore, $e^{\frac{_{E_{a}}}{R T}} = e^{\frac{- 209500}{8.314 \times 581}}$
= $e^{- 43.4}$
= $\frac{1}{e^{43.4}}$
finding the value of $\frac{1}{a n t i l o g (43.4)}$ we get
$e^{\frac{_{E_{a}}}{R T}} = 1.47 \times 10^{- 19}$

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