The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy `(E_(a))` of the reaction ? (R = gas constant)

Options

A. `1842.4R`
B. `921.2R`
C. `460.6R`
D. `230.3R`

Answer

B

Explanation

`log((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log.(10)/(1)=(E_(a))/(2.303)[(400-200)/(400xx200)]`
`1=(E_(a))/(2.303R)[(200)/(400xx200)]`
`E_(a)=2.303Rxx400=921.2R`