Chemistry - Chemical Equilibrium Question with Solution | TestHub
ChemistryChemical EquilibriumProblems involving KpHard2 minQB
ChemistryHardnumerical
A vessel of 4.0 L was filled with 0.02 mole of Sb2S3 and 0.03 mole of H2 to attain equilibrium at 440°C as:
Sb2S3(s) + 3H2(g) ⇌ 2Sb(s) + 3H2S(g)
After equilibrium, the H2S formed was analysed by dissolving it in water and treating with excess of Pb2+ to give 4.78 g of PbS as precipitate. The value of Kc of the reaction at 440°C is (Pb = 207, S = 32)
Answer:
8.00
Solution:
Molar mass of PbS = 207 + 32 = 239 g/mol Moles of PbS = 4.78/239 = 0.02 mol, so H2S at equilibrium = 0.02 mol
H2 used = 0.02 mol, so H2 left = 0.03 - 0.02 = 0.01 mol
Kc = [H2S]3/[H2]3 = (0.02/4.0)3/(0.01/4.0)3 = (2)3
Kc = 8
Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp
⏱ 2mℹ️ Source: QB
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