InterviewSolution
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InterviewSolution
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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. `0.08`B. `0.8`C. `0.4`D. `0.04` |
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Answer» Correct Answer - A::B::D 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` |
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