The rate of reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation `(E _(a))` of the reacton assuming that it does not change with temperature. `[ R = 8.314 J K ^(-1) mol ^(-1), log 4 = 0. 6021]`

Explanation

Given `( k _(2))/(k _(1)) = 4. T _(1) = 292 K, T _(2) = 313 K, R = 8. 314 J mol ^(-1) K ^(-1).`
According to Arrhenius equation `log "" ( k _(2))/( k _(1)) = ( E _(a))/( 2. 303 R) [ ( T _(2) - T _(1))/( T _(1)T _(2))]`
`log 4 = ( E _(a ))/( 2. 303 xx 8.3.4) [ ( 313 - 293)/( 293 xx 313) ] , 0. 6020 = ( E _(a))/( 19. 14) [ ( 20)/( 91709) ]`
`E _(a) = ( 0. 6020 xx 19. 14 xx 91709)/( 20) = 52834.84 J = 52.834 kJ mol ^(-1).`