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

Explanation

Given, `T_1 = 293K, T_2 = 313 K, k_2 // k_1 =4`
`R= 8.314" JK"^(-1)" mol"^(-1), log 4 = 0.6021`
We know, `log"" (k_2)/(k_1) = (E_a)/( 2.303 R) [(T_2 - T_1)/( T_1 T_2) ]`
`therefore log 4= (E_a)/( 2.303 xx 8.314) [(313 - 293)/(313 xx 293) ]`
`rArr 0.6021 = (E_a)/( 19.147) [ (20)/( 91709) ]`
`rArr E_a = (0.6021 xx 19.147 xx 91709)/( 20)`
`rArr E_a = 52862" J mol"^(-1)`