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).

A vessel of 10L was filled with 6 mole of Sb_(2)S_(3) and 6 mole of H_

1.	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).
Answer» `0.08` `0.8` `0.4` `0.04` Solution :MOLE of `PbS=708//236="mole of" H_(2)S` `Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)` `{:("Initial",6,6,0,0),("at EQ.",5,3,2,3):}` `K_(c)=((3//10)^(3)xx(2//10)^(2))/((5//10)xx(3//10)^(3))=(4)/(50)=0.08`