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

Answer

C

Explanation

`log_(10)((k_(2))/(k_(1)))=(E_(a))/(2.303R)xx((1)/(T_(1))-(1)/(T_(2)))`
`log2=(E_(a))/(2.303xx8.314)xx((1)/(300)-(1)/(310))`
`E_(a)=(0.301xx2.303xx8.314xx300xx310)/(10)`
`=53598 J "mol"^(-1)`
`~~53.6kJ "mol"^(-1)`