A vessel of 250 L was filled with 0.01 mole of `Sb_(2)S_(3)` and 0.01 mole of `H_(2)` to attain the equilibrium at `440^(@)C` as
`Sb_(2)S_(3(s))+3H_(2(g))hArr2Sb_(s)+3H_(2)S_(g)`
After equilibrium the `H_(2)S` formed was analysed by dissolving it in water and treating with excess of `Pb^(2+)` to give 1.19 g of PbS as precipitate. The value of `K_(C)` at `440^(@)C` is `(Pb=20S)`

Answer

1

Explanation

`underset(0.01-x, 0.01-3x)(Sb_(2)S_(3(s))+3H_(2(g)))hArrunderset(2x)(2Sb_(s))+underset(3x)(3H_(2)S_(g))`
where, `3x=0.005`
(`H_(2)S+Pb^(2+)toPbS+2H^(+)`
number of moles of PbS formed `=(1.19)/(239)=0.005` mol)
At equilibrium `[H_(2)]=((0.005)/(250))`
`K_(C)=((0.005)/(00.5))=1`