An equilibrium mixture for the reaction
`2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)`
had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` is

Options

A. 0.004
B. 0.08
C. 0.016
D. 0.16

Answer

C

Explanation

`K_(c) =[[H_(2)]^(2)[S_(2)]]/[[H_(2)S]^(2)]`
`[H_(2)] =(2xx0.2)/(2) =0.2 " mole"//L`
`[S_(2)]=(0.8)/(2) =0.4 "mole"//L ,[H_(2)S]=1 "mole"//L`
`:. K_(c) =[[0.2]^(2)[0.4]]/(1) =0.016 "mole"//L`