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. `48.6kJ`
B. `58.5 kJ mol^(-1)`
C. `60.5 kJ`
D. `53.6 kJ mol^(-1)`

Answer

D

Explanation

According to Arrhenius equation
log `(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Substituting the data, we have
log `2=(E_(a))/(2.303(8.314JK^(-1)mol^(-1)))((310K-300K)/(300Kxx310K))`
`E_(a)=53.6KJ//mole`