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

Answer

A

Explanation

Given : `t_(1//2)(1)=20s `
`t_(1//2)(2)=2s " "[because t_(1//2)prop(1)/(k)]`
`T_(1)=300 K T_(2)=300 K T_(2)=350 K`
`because " " log ""(K_(2))/(K_(1))=-(E_(a))/(2.303R)[(1)/(T_(2))-(1)/(T_(1))]`
`:." " log ""((0.693)/(2))/((0.693)/(20))=-(E_(a))/(2.303xx8.314)[(1)/(350)-(1)/(300)]`
`implies log ""(20)/(2)=- (E_(a))/(2.303xx8.314)[(-50)/(350xx300)]`
`:." " E_(a)= (2.303xx8.314xx350xx300)/(50)`
`E_(a)=40.209 `kj/mol