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. `2x10^(-2)`

Answer

A

Explanation

(A) According to Arrhenius equation
`k = Ae^(-E_(a)//RT) `
or log k = log `A-(Ea)/(2.303RT)`
`E_(a)= 2.303 RT`
`:. ` log k = log A `-(2.303RT)/(2.303RT)`
log k = log A-1
or log k - log A=-1
log `(k)/(A) =-1`
`(k)/(A) = 10^(-1)`