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

Answer

A

Explanation

From 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)=310k,T_(1)=300k`
On substituting the values,
`"log"2=(-E_(a))/(2.303xx8.314)((1)/(310)-(1)/(300))` `orE_(a)=(log2xx2.303xx8.314xx310xx300)/(10)`
`E_(a)=53598.6J//"mol"=53.6kJ//"mol"`