for a given reaction at temeperature T, the velocity constant k, is expressed as :
`k=ae^(-27000K'//T)` (K' = Boltzmann constant)
Given , `R=2 cal K^(-1) "mol"^(-1)`. Calculate the value of energy of activation. Comment on the results.

Answer

`E=18xx10^(-20)"cal"`

Explanation

`k=Ae^(-E_(a)//RT) " ".....(i)`
`k=Ae^(-27000k//T)" ".....(ii)`
`(E_(a))/(RT)=(27000K')/(T)`
`E_(a)=27000K'`
`=27000Rxx(R)/(N)=(27000xx(2)^(2))/(6.023xx10^(23))`
`=18xx10^(-20)"cal"`
Activation energy is low, therefore reaction will be fast