If the half-lives of the first order reaction at 350 K and 300 K are 2 and 20 seconds respectively, the activation energy of the reaction in kJ `"mol"^(-1)` is :

Options

A. `40.2`
B. `20.1`
C. `60.3`
D. `30.2`

Explanation

`log((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`log.((t_(1//2))_(t))/((t_(1//2))_(2))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`log. ((20)/(20))=(E_(a))/(2.303xx8.314)[(1)/(300)-(1)/(350)]`
`E_(a)=(1xx2.303xx8.314xx300xx350)/(50)J "mol"^(-1)`
`=40208J "mol"^(-1)`
`=40.2kJ"mol"^(-1)`