The rate constants ofa reaction at 500 K and 700 K are `0.025 sec^-1` and `0.075 sec^-1` respectively. Calculate the energy ofactivation of the reaction. (`R=8.314JK^-1` and log 3 = 0.447)

Explanation

Given, `T_1=500K.T_2=700K`
Rate constant `(K_1)=0.025sec^(-1)`
Rate constant `(k_2)=0.075sec^(-1)`
`R=8.3.14JK^(-1)mol^(-1)`
`thereforelog""K_2/K_1=E_a/(2.303R)[(T_2-T_1)/(T_1.T_2)]`
or `E_a=(2.303xxRxxT_1xxT_2xxlog_3)/(T_2-T_1)`
`[therefore""K_2/K_1=log""(0.075)/(0.025)=log3]`
`E_a=(2.303xx8.314xx500xx700xx0.477)/(700-500)`
`E_a=15986.43Jmol^(-1)=15.99kJmol^(-1)`