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

Answer

B

Explanation

`log (k_(2))/(k_(1))=(E_(a))/(2.303 R)((1)/(T_(1))-(1)/(T_(2)))`
`log 2=(E_(a))/(2.303 xx 8.314)[(1)/(293)-(1)/(308)]`
`0.3=(E_(a))/(2.303 xx 8.314) xx (15)/(293 xx 308)`
`E_(a)=(0.3 xx 2.303 xx 8.314 xx 293 xx 308)/(15)`
`=34673 mol^(-1)=34.7 kJ mol^(-1)`