The rate of a reaction doubles when its temperature changes from `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. `48.6 kJ mol^(-1)`
B. `58.5 kJ mol^(-1)`
C. `60.5 kJ mol^(-1)`
D. `53.6 kJ mol^(-1)`

Answer

D

Explanation

As per Arrhenius equation:
`log. (K_(2))/(K_(1)) = (E_(a))/(2.3R)[(T_(2)-T_(1))/(T_(2)T_(1))]`
`2.3 log2 = (E_(a))/(8.314)[(10)/(300 xx 310)]`
`:. E_(a) = 53.6 kJ mol^(-1)`