Chemistry - Chemical Equilibrium Question with Solution | TestHub
ChemistryChemical EquilibriumDegree of Dissociation/Vapour Density/Molar MassMedium2 minQB
ChemistryMediumnumerical
The per cent dissociation of H2S(g) if 0.25 mole of H2S is kept in a 0.5 L vessel at 1100 K for the reaction: 2H2S(g) ⇌ 2H2(g) + S2(g); Kc = 6.75 × 10-6. (Answer as the nearest integer.)
Answer:
3.00
Solution:
Initial [H2S] = 0.25/0.5 = 0.5 M Let the degree of dissociation be α
[H2S] = 0.5(1-α), [H2] = 0.5α, [S2] = 0.25α
Kc = (0.5α)2(0.25α)/[0.5(1-α)]2 = 0.25α3/(1-α)2
For small α, 1-α ≈ 1, so 0.25α3 = 6.75 × 10-6
α3 = 2.7 × 10-5, so α = 0.03
Per cent dissociation = 3
Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Degree of Dissociation/Vapour Density/Molar Mass
⏱ 2mℹ️ Source: QB
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