The rate temperature changes from 300K to 310K. Activation energy of such a reaction will be `(R = 8.314 JK^(-1) mol^(-1) " and " log 2 = 0.3010)`

Options

A. `53.6kJmol^(-1)`
B. `48.6kJmol^(-1)`
C. `58.5kJmol^(-1)`
D. `60.5kJmol^(-1)`

Answer

A

Explanation

`log.(K_(2))/(K_(1))=(-E_(a))/(2.030R)((1)/(T_(2))-(1)/(T_(1)))`
`(K_(2))/(K_(1))=2`, `T_(2)=310K` `T_(1)=300K`
`implieslog2=(-E_(a))/(2.303xx8.134)((1)/(310)-(1)/(300))`
`impliesE_(a)=53598.6J//mol=53.6KJ//mol`