For a decomposition reaction, the values of rate constants, k at two different temperatures are given below :
`k_1=2.15xx10^-7 L mol^-1 s^-1` at 650K
`k_2=2.39xx10^-7 L mol^-1 s^-1` at 700K
calculate activation energy for the reaction.

Explanation

`k_1=2.15xx10^(-8)" L mol"^(-1)s^(-1)`
`T_1=650 K`
`k_2=2.39xx10^(-7)" L mol"^(-1)s^(-1)`
`T_2=700" "K`
`R=8.314" JK"^(-1)" mol"^(-1)s^(-1)`,
`R=8.314" "JK^(-1)mol^(-1),E_a=?`
`(r_2)/(r_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log(2.39xx10^(-7))/(2.15xx10^(-8))=(E_a)/(2.303xx8.314 JK^(-1)mol^(-1))[(700-650)/(700xx650K)]`
`log(23.9)/(2.15)=(E_a)/(19.15 JK^(-1)mol^(-1)xx(50)/(700xx650K)`
`log11.12=(E_a)/(19.15xx9100J mol^(-1))`
`1.0461=(E_a)/(19.15xx9100)J" mol"^(-1)`
`E_a=19.15xx9100xx1.0461" J mol"^(-1)`
`=182298.6" J mol"^(-1)`
`=182.3" kJ mol"^(-1)`