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. `2xx10^(-3)`
D. `2xx10^(-2)`

Answer

A

Explanation

`2.303logk=2.303 logA. -(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)`