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

Answer

B

Explanation

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