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 ?

Options

A. `1.82xx10^(-18)`
B. `1.47xx10^(19)`
C. `2.67xx10^(16)`
D. `3.89xx10^(19)`

Answer

B

Explanation

Fraction of molecules (x) having energy equal to or greater than activation energy may be calculated as follows:
`x=n//N=e^(-Ea)"//RT"`
In `x=-E_(a)/(RT)or logx=-E_(a)/(2.303RT)`
or `logx=-(209xx10^(3)J" mol"^(-1))/(2.303xx(8.314Jk^(-1)" mol"^(-1))xx581k)`
`=-18.8323`
`x=Antilog(-18.8323)`
`=1.471xx10^(-19)`
Fraction of molecules `=1.47xx10^(-19`