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The activation energy for the reaction ...

The activation energy for the reaction `:` ltbr. `2Hl(g) rarr H_(2)(g)+I_(2)(g)`
is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

Text Solution

Verified by Experts

The correct Answer is:
`1.47xx10^(19)`
Fraction of molecules having energy equal to or greater than activation energy is given by `:`
`(n)/(N)=(x)=e^(-Ea//RT)`
`:.ln (x)=(-Ea)/(RT)`
`or log (x)=(-E_(a))/(2.303RT)`
`=(209.5xxJ mol^(-1))/(2.303xx8.314J K^(-1) mol ^(-1)xx581K)`
`=-18.8323`
`:. x=Antilog(-18.8323)`
`=Antilog(-18.8323+1-1)`
`=Antilogbar(19).1677=1.471xx10^(-19)`
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