The activation energy for a reaction at the temperature T K was found to be `2.303RT 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

a) According to Arrhenius equation,
log k = log A`-E/(2.303R) xx 1/T`
`2.303 log k = 2.303 log A-(E_(a))/(RT)`
Substituting the value of `E_(a)` as 2.303 RT and dividing by 2.303 on both sides, we get.
log k`-log A=-1`
`log(k/A)=log 10^(-1)`
Taking anti log.
`(k/A)=10^(-1)`