In areversible reaction, study of its mechanism says that both the forward and reverse reaction follows first-order kinetics. If the halflife of forward reaction `(t_(1//2))_(f)` is `400 s` and that of reverse reaction `(t_(1//2))_(b)` is `250 s`, the equilibrium of the reaction is

Options

A. `1.6`
B. `0.433`
C. `0.625`
D. `1.109`

Answer

C

Explanation

`t_(1//2)=0.693/K` (for first-order kinetics)
`K_(f)=0.693/400 s^(-1), K_(b)=0.693/250 s^(-1)`
`K=K_(f)/K_(b)=250/400=0.625`