The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate activation energy for the reaction.

Explanation

`k_2//k_1=2,T_1=300K,T_2=400K`
`E_a=?,R=8.314Jk^(-1)mol^(-1)`
`log=(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
log2
`=(E_a)/(2.303xx8.314JK^(-1)mol^(-1))[(400-300)/(400xx300)]`
`0.3010=(E_a)/(19.15 JK^(-1)mol^(-1))[(100)/(400xx300k)]`
`E_a=0.3010xx19.15 JK^(-1)mol^(-1)xx1200K`
`=6916.98" J mol"^(-1)=6.92kJmol^(-1)`