The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation `(E_(a))` of the reaction assuming that it does not change with temperature. `" "[R=8.314JK^(-1)"mol"^(-1), log 4=0.6021]`

Explanation

`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Substituting the values, we have
`log 4=(E_(a))/(2.303xx8.314JK^(-1)"mol"^(-1))[(313K-293K)/(293K xx 313K)]`
or `0.6021=(E_(a))/(19.147JK^(-1)"mol"^(-1))xx(20)/(91709K)`
or `E_(a)=(0.6021xx19.147xx91709)/(20)"J mol"^(-1)=52863"J mol"^(-1)=52.86"kJ mol"^(-1)`