Consider the following hypothetical reactions
`Aoverset(k_(1))underset(k_(2))hArrB " , "Boverset(k_(3))underset("slow")rarrC`
The overall rate constant `k_(theta)` of the reaction is

Options

A. `(k_(3)k_(1))/(k_(2))`
B. `(k_(1)k_(2))/(k_(3))`
C. `(k_(1))/(k_(2)k_(3))`
D. `(k_(1)-k_(2))/(k_(3))`

Answer

A

Explanation

For the reaction
`A overset(k_(1))underset(k_(2))hArr B`
at equilibrium
`k_(1)[A]=k_(2)[B]`
`[B]=(k_(1)[A])/(k_(2))`
From the slowest step overall rate of reaction
`=k_(3)[B]=(k_(3)k_(1))/(k_(2))[A]=k_(0)[A]`
where `k_(0)=(k_(3)k_(1))/(k_(2))`