`K_f` and `K_b` are the velocity constants of forward and backward reactions. The equilibrium constant `K_(eq)` of the reversible reaction will be

Options

A. `K_b//K_f`
B. `K_fxxK_b`
C. `K_f//K_b`
D. `K_f-K_b`

Answer

C

Explanation

Suppose a reversible reaction occurs by a one-step mechanism:
`2A+B=A_2B`
The rate of the forward reaction is
`Rate_f=K_f[A]^2[B]`
The rate of the backward reaction is
`Rate_b=K_b[A_2B]`
At equilibrium,
`Rate_f=Rate_b`
`K_f[A]^2[B]=K_b[A_2B]`
Dividing both sides of this equation by `K_b` and by `[A]^2[B]` gives
`K_(eq)=K_f/K_b=([A_2B])/([A]^2[B])`