The rate constant of a first order reaction are 0.58 `s^(-1)` at 313 K and `0.045 s^(-1)` at 293 K. what is the energy of activation for the reaction?

Explanation

Given : Rate constant `K_(1) = 0.58 s^(-1)`
Rate constant `K_(2) = 0.045 s^(-1)`
`T_(1) = 313 K`
`T_(2) = 293 K`
Assume R gas constant = 8.314
To find : Activation energy `[E_(A)]`
Formula : `log (k_(2))/(k_(1)) = (E_(a))/(2.303R) [(T_(2) - T_(1))/(T_(1)T_(2))]`
`log (0.045)/(0.58) = (E_(a))/(2.303 xx 8.314) xx [(293 - 313)/(313 xx 293)]`
`log (0.045)/(0.48) = (E_(a))/(19.1471) xx [(-20)/(91709)]`
`log 0.045 - log 0.58 = (E_(a))/(19.1471) xx [(-20)/(91709)]`
`-2.556 = - 0.00001138977 E_(a)`
`E_(a) = 224411.90`
`E_(a) = 224.411 kJ mol^(-1)`