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

Options

A. `10^(-1)`
B. `10^(-2)`
C. `2xx10^(-3)`
D. `2xx10^(-2)`

Answer

A

Explanation

`2.303logk=2.303logA-(E_(a))/(RT)`
`E_(a)=2.303RT" (given)"`
Putting the values of `E_(a)` and dividing by `2.303` on both sides, we get,
`logk=logA-1," "logk-logA=-1`
`"log"((k)/(A))=-1`
Taking antilog `((k)/(A))=10^(-1)`.