The rate constants of a reaction at 300 and 320 K are `0.0231 s^-1` and `0.0693 s^-1` respectively. Calculate the value of activation energy of the reaction. [R=8.314J `K^-1 mol^-1`, `log3=0.4771`]

Explanation

`k_1=0.023 s^(-1),T_1=300K`
`k_2=0.0693s^(-1),T_2=320K`
R=8.314 J `K^(-1)mol^(-1),E_a=?`
log`(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
log`(0.0693)/(0.0231)=(E_a)/(2.303xx8.31JK^(-1)mol^(-1) [(320-300)/(320xx300K)]`
log 3= `(E_a)/(19.15JK^(-1)mol^(-1))xx(20)/(320xx300K)`
`0.4771=(E_a)/(19.15xx4800)J" mol"^(-1)`
`E_a=0.4771xx19.15xx4800" J mol"^(-1)`
=43855 J `mol^(-1)`=43.855 kJ `mol^(-1)`