A vessel of `2.50 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 dissolving it in water and treating with excess of `Pb^(2+)` to give `1.029g` of `PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`? (At weight of `Pb=206`)

A vessel of `2.50 litre` was filled with `0.01` mole of `Sb_(2)S_(3)`

1.	A vessel of `2.50 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 dissolving it in water and treating with excess of `Pb^(2+)` to give `1.029g` of `PbS` as precipitate. What is value of `K_(c)` of the reaction at `440^(@)C`? (At weight of `Pb=206`)
Answer» Correct Answer - `4.3xx10^(-1)` ,

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