The rate constant of a first order reaction increases from `2xx10^(-2)` to `8xx10^(-2)` when the temperature changes from 300 K to 320 K. Calculate the energy of activation.

Explanation

We shall use the Arrhenius equation
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given : `k_(1)=2xx10^(-2), k_(2)=8xx10^(-2), T_(1)=300K, T_(2)=320K and R=8.314JK^(-1)"mol"^(-1)`
Substituting these values in the above equation, we get
`"log"(8xx10^(-2))/(2xx10^(-2))=(E_(a))/(2.303xx8.314)[(320-300)/(320xx300)]`
`or log 4=(E_(a))/(19.147)xx(20)/(96000) or 0.6021=(E_(a))/(19.147)xx(1)/(4800)`
`or E_(a)=0.6021xx19.147xx4800 or E_(a)=55.34"kJ mol"^(-1)`