A reactant `(A)` forms two products
`A overset (k_(1))rarr B`, Activation energy `E_(a1)`
`A overset (k_(2))rarr C`, Activation energy `E_(a2)`
If `E_(a_(2)) = 2E_(a_(1))` then `k_(1)` and `k_(2)` are related as

Options

A. `k_(1)=2k_(2) e^(E_(a_(2))//RT)`
B. `k_(1)=k_(2) e^(E_(a_(1))//RT)`
C. `k_(2) =k_(1) e^(E_(a_(2))//RT)`
D. `k_(1)=Ak_(2)e^(E_(a_(1))//RT)`

Answer

B

Explanation

`A overset(k_(1))rarrB, A overset(k_(2))rarr C`
By Arrhenius equation,
`k_(1) =A'e^(-E_(a_(1))//RT) and k_(2) =A'e^(-E_(a_(2))//RT)` [A' is Arrhenius constant]
`because E_(a_(2)) =2E_(a_(1))`
`therefore k_(2) =A'e^(-2E_(a_(1))//RT) therefore (k_(1))/(k_(2))=(A'e^(-E_(a_(1))//RT))/(A'e^(-2E_(a_(1))//RT))=e^(E_(a_(1))//RT) therefore k_(1) =k_(2) e^(E_(a_(1))//RT)`