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 aassuming that it does not change with temperature.

Explanation

Given if rate at 293 K is R thus at 313 K rate becomes = 4R
`log.(k_(2))/(k_(1)) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]`
`log.(4R)/(R ) = (E_(a))/(2.303 xx 8.314)[(313 - 298)/(293xx 313)]`
`log4 = (E_(a))/(19.1471)[(20)/(91709)]`
`0.6021 = (E_(a))/(19.1471)[(20)/(91709)]`
`(0.6021 xx 19.1471 xx 91709)/(20) = E_(a)`
`E_(a) = 52863.2177 J mol^(-1)`
or `= 52.863 kJ mol^(-1)`