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=6,T_1=350K,T_2=400KL`
`R=8.314JK^(-1)mol^(-1),E_a=?`
`log=(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log6=(E_a)/(2.303xx8.314Jk^(-1)mol^(-1))[(400-350)/(400xx350k)]`
`=0.7781=(E_a)/(19.15Jk^(-1)mol^(-1))xx(50)/(400xx350k)`
`E_a=0.7781xx19.15xx8xx350" J mol"^(-1)`
`=4172.72" J mol"^(-1)=41.72" kJ mol"^(-1)`