What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? `(R = 8.314 J "mol K"^(-))`

Options

A. `342" kJ mol"^(-1)`
B. `269" kJ mol"^(-1)`
C. `34.7" kJ mol"^(-1)`
D. `15.1" kJ mol"^(-1)`

Answer

C

Explanation

`"log" (k_(2))/(k_(1))=(Ea)/(2.303xxR)((1)/(293)-(1)/(308))`
`log 2=(Ea)/(2.303R)((308-293)/(293xx308))`
`E_(a)=34673 J=34.67 kJ`