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

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Solution

Activation energy, $E a = 209.5 K J / m o l = 209500 J / m o l$
Temperature $= 581 K, R = 8.314 J K^{- 1} m o l^{- 1}$
We know that according to Arrhenius equation, $K = A e^{- \frac{E a}{R T}}$
Here, $e^{- \frac{E a}{R T}}$ represents the number of molecules which have energy equal or more than activation energy,
So, $K = A e^{- \frac{E a}{R T}}$
$e^{- \frac{E a}{R T}} = 1.47 \times 10^{- 19}$

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