The rate constants of a reaction at `500 K and 700 K ` are `0.02s^(-1)` resp-ectively. Calculate the value of activation energy , `E_(a)^(@)` (`R=8.314J K^(-1) mol^(-1)`).

Explanation

`log.(k_(2))/(k_(1))=(E_(a))/(2,303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log.(0.07)/(0.02)=((E_(a))/(2.203xx8.314JK^(-1)mol^(-1)))[(700-500)/(700xx500)]`
`0.544=E_(a)xx5.714xx10^(-4)//19.15`
`E_(a)=(0.544xx19.5)/(5.714xx10^(-4))= 18230.8J`
Since
`k=Ae^(Ea//RT)`
`log k=logA-(E_(a))/(2.303RT)`
`logA=logk+(E_(a))/(2.303RT)`
`logA=log(0.02)+(18230.8Jmol^(-1))/(2.303xx8.314JK^(-1)mol^(-1)xx500K)`
`= -1.699+1.90= 0.201~~0.2`
`A="Antilog"(0.2)~~1.585~~1.6`