An equilibrium mixture of the reaction `2H_(2)S(g)hArr2H_(2)(g) + S_(2)(g)` had `0.5` mole `H_(2)S`, `0.10` mole `H_(2)` and `0.4` mole `S_(2)` in one litre vessel. The value of equilibrium constants (K) in mole `litre^(-1)` is

Options

A. `0.004`
B. `0.008`
C. `0.016`
D. `0.160`

Answer

C

Explanation

`K = ([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2)) = ([0.10]^(2)[0.4])/([0.5]^(2)) = 0.016`