The activation energy for a reaction at temperature T K was found to be or `2.303 "RT J mol"^(-1)`. The ratio of the rate constant to Arrhenius factor is

Options

A. `10^(-1)`
B. `10^(-2)`
C. `2 xx 10^(-3)`
D. `2 xx 10^(-2)`

Answer

A

Explanation

`2.303 log k = 2.303 log A-E_(a)//RT`
Putting `E_(a) = 2.303 RT` (Given)
2.303 log k = 2.303 log `A - (2.303 RT)/(RT)`
or log k = log A -1
or log `(k)/(A)=-1 or (k) (A)=10^(-1)`.