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)`

Explanation

According to available data:
`k_(2)/k_(1)=2, T_(1)=27+ 273=300 K, T_(2)=37 + 273=310 K, R=8.314 JK^(-1)mol^(-1)`
`log k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`log 2=(E_(a))/(2.303 xx 8.314(JK^(-1) mol^(-1))[1/300K - 1/310K]`
`0.3010 = (E_(a))/(2.303 xx (8.314 J mol^(-1)) xx (310-300)/(300 xx 310)`
`0.3010 = (E_(a))/(2.303 xx 8.314 J mol^(-1)) xx (10)/(300 xx 310)`
`E_(a) = 0.3010 xx 2.303 xx 8.314(J mol^(-1) xx 300 xx 310)/(10)`
`=53598.5 J mol^(-1)=53.59 kJ mol^(-1)`