Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

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Solution

In the given case:
$E_{a} = 209.5 k J$ $m o l^{- 1}$ = $209500 J$ $m o l^{- 1}$
$T = 581 K$
$R = 8.314 J$ $K^{- 1}$ $m o l^{- 1}$
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
$x = e^{- E a / R T}$
$l n^{x} = - E_{a} / R T$
$l o g x = - \frac{E_{a}}{2.303 R T}$
$l o g x = - \frac{209500 J m o l^{- 1}}{2.303 \times 8.314 J K^{- 1} m o l^{- 1} \times 581}$ $= 18.8323$
Now, x= Anti log $(18.8323)$
$= A n t i l o g ¯ 19 .1677$

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