The rate of reaction triples when temperature changes form `20^(@)C` to `50^(@)C`. Calculate the energy of activation for the reaction `(R= 8.314JK^(-1)mol^(-1))`.

Explanation

According to the available data:
`k_(2)/k_(1) = 3, T_(1)=20+273=293 K, T_(2)=50+273=323K, R=8.314 J K6(-1) mol^(-1)`
`log k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`log3 = (E_(a))/(2.303 xx 8.314 JK^(-1)mol^(-1)) [1/293K-1/323K]`
`0.4771 = (E_(a))/(2.303 xx 8.314 (J mol^(-1))[(323-293)/(293 xx 323)]`
`E_(a) = (0.4771 xx 2.303 xx 8.314 (J mol^(-1)) xx 293 xx 323)/(30)`
`=28817.8 J mol^(-1)=28.82 kJ mol^(-1)`