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A vessel of 250 L was filled with 0.01 m...

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)`

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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 ?

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 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 ?

A vessel of 10L was filled with 6 mole of Sb_(2)S_(3) and 6 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 708 g "of" PbS as precipitate. What is value of K_(c) of the reaction at 440^(@)C ?(At. weight of Pb=206) .

Sb_(2)S_(3) is -

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