In a reaction carried out at `500` K, `0.001%` of the total number of collisions are effective. The energy of activation of the reaction is approximately

Options

A. `15.8 kcal mol^(-1)`
B. `11.5 kcal mol^(-1)`
C. `12.8 kcal mol^(-1)`
D. zero

Answer

B

Explanation

The fraction of molecules having energy equal to or greater than `E_(a)` is :
`x = (n)/(N) = e^(-E_(a)//RT) (x = (n)/(N) = 0.001% = (0.001)/(100) = 10^(-5))`
`:. log x = (-E_(a))/(2.3 xx RT) = (R = 2 cal mol^(-1) K^(-1), T = 500 K)`
`log 10^(-5) = (-E_(a))/(2.3 xx 2 xx 500)`
`E_(a) = 11.5 xx 103 cal mol^(-1) = 11.5 kcal mol^(-1)`