The rate constant of a reaction is `1.5 xx 10^(7)s^(-1)` at `50^(@)`C and `4.5 xx 10^(7)s^(-1)` at `100^(@)`C. Calculate the value of activation energy for the reaction `(R=8.314 J K^(-1)mol^(-1))`

Explanation

`log.k_2/k_1=E_a/(2.303R)(1/T_1-1/T_2)`
`log.(4.5xx10^7)/(1.5xx10^7)=E_a/(2.303xx8.314)(1/323-1/373)`
`log3=E_a/(2.303xx8.314)xx50/(323xx373)`
`E_a=(0.471xx19.147xx323xx373)/50=(1100579.725)/50=22011.595"J mol"^(-1)`
`E_a=22.01"kJ mol"^(-1)`