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)D(g)` After equilibrium, the `H_(2)S` formed was analysed was analysed by dissloved it in water and treating with execedd of `Pb^(20+)` to give `1.19` g of PbS as precipitate. What is the value of `K_(c)` at `440^(@)C` ?