A vessel of `250` litre 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 dissolved it in water and treating with execess of `Pb^(2+)` to give `1.19` g of PbS as precipitate. What is the value of `K_(c)` at `440^(@)C` ?

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`