The following values for the first order rate constant were obtained for a certain reaction.

Calculate the activation energy, `R=8.31 J K^(-1) mol^(-1)`.

Explanation

According to the available data,
`k_(1) = 3.46 xx 10^(-5)s^(-1), k_(2)=13.5 xx 10^(-5) s^(-1), T_(1)=298 K, T_(2)=308 K`
`log k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`log (13.5 xx 10^(-5)s^(-1))/(3.46 xx 10^(-5)s^(-1)) = (E_(a))/(2.303 xx 8.31 J K^(-1) mol^(-1))[1/298 K-1/308K]`
`log 3.9017 = (E_(a))/(2.303 xx 8.31 J mol^(-1))[(308-298)/(298 xx 308)]`
`0.59125 = (E_(a))/(19.138 J mol^(-1)) [(10)/(298 xx 308)]`
`E_(a) = (0.59125 xx 19.138 xx 298 xx 308)/(10) (J mol^(-1))`
`=103857.42 J mol^(-1)=103.857 = 103.9 kJ mol^(-1)`.