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=5.10^(-4)s^(-1),T_1`=300K
`k_2=2.0xx10^(-3)s^(-1),T_2`=320K
R=8.314 `JK^(-1)mol^(-1),E_a` = ?
`log(k_2)/(k_1)=(E_a)/(2.303 R)[(T_2-T_1)/(T_1T_2)]`
`log(2.0xx10^(-3))/(5.0xx10^(-4))=(E_a)/(2.303xx8.314 JK^(-1)mol^(-1))`
`=(E_a)/(2.303xx8.314 JK^(-1)mol^(-1))`
`log4=(E_a)/(19.15JK^(-1)mol^(-1))xx(20)/(320xx300K)`
`E_a-0.6021xx19.15xx4800" K mol"^(-1)`
=55345.0 J `mol^(-1)`
=55.345 kJ `mol^(-1)`