The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be:
`(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`

Options

A. `53.6 JK^(-1) "mol"^(-1)`
B. `48.6 JK^(-1) "mol"^(-1)`
C. `58.5 JK^(-1) "mol"^(-1)`
D. `60.5 JK^(-1) "mol"^(-1)`

Answer

A

Explanation

Form Arrhenius equation,
`log` `(k_(2))/(k_(1)) = (-E_(a))/(2.303R)((1)/(T_(2)) - (1)/(T_(1)))`
Given, `(k_(2))/(k_(1)) = 2, T_(2) = 310 K`
`T_(1) = 300 K`
On putting values,
`rArr log 2 = (-E_(a))/(2.303 xx 8.314) ((1)/(310) - (1)/(310))`
`rArr E_(a) = 53603.93 J//mol = 53.6 kJ//mol`