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An equilibrium mixture of the reaction 2...

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

A

`0.004`

B

`0.008`

C

`0.016`

D

`0.160`

Text Solution

Verified by Experts

The correct Answer is:
C
`K=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))=([0.10]^(2)[0.4])/([0.5]^(2))=0.016`
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