The rate constant `(K')` of one reaction is double of the rate constant (K") of another reaction. Then the relationship between the corresponding activation energies of the two reactions `(E_(a)^(') "and" E_(a)^(''))` will be

Options

A. `E'_(a) gt E''_(a)`
B. `E'_(a) = E''_(a)`
C. `E'_(a) lt E''_(a)`
D. `E'_(a) = 4E''_(a)`

Answer

C

Explanation

As `K' gt K'', E'_(a) lt E''_(a)` (Greater the rate constant, less is the activation energy).
`K=2K'', K'=Ae^(-E'_(a)//RT)` (1)
`K''=Ae^((E''_(a))/(RT))`...(2)
Equation (1)`//` Equation (2)
`(2K'')/(K'')=e^(-(E'_(a))/(RT)+(E''_(a))/(RT))`
`In (2) =(1)/(RT)[E''_(a)-E'_(a)]`
`RT In (2)+E'_(a)=E'_(a)`
`implies E''_(a) gt E_(a)`