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. `0.01`
B. `0.1`
C. `0.02`
D. `0.001`

Answer

B

Explanation

We know that,
`log(K)/(A)=(-E_(a))/(2.303RT)=(-2.303RT)/(2.303RT)=-1`
`log(K)/(A)=-1implies(K)/(A)="antilog"(-1)`=0.1