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Given that the enthalpy change of a reac...

Given that the enthalpy change of a reaction at 325 K is 0.12 kcal and energy of activation for the backward reaction is 0.02 kcal, calculate the percentage of the reactant molecules, crossing over the energy barrier.

Text Solution

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Activation energy for forward reaction `=0.12+0.02" kcal"=0.14" kcal"=140" cal"`
Fraction of molecules crossing the energy barrier `=e^(-E_(a)//RT)`
`:." "%" of molecules crossing the energy barrier "=e^(-E_(a)/RT)xx100`
or `x=e^(-140//(2xx325))xx100" or "x=100xxe^(-140//650)`
In `x=" In "100-(140)/(650)`
`logx=log100-(1)/(2.303)xx(14)/(65)`
`=2-0.0935=1.9065`
or `x="Antilog "1.9065=80.63%`
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