What is the equilibrium constant of the reaction: 1/2H 2 (g) + 1/2I 2 (g) ⇌ HI(g), at 450°C?

Take 1 mol of HI, α = 0.25 At equilibrium: HI = 0.75 mol, H 2 = I 2 = 0.125 mol For 2HI(g) ⇌ H 2 (g) + I 2 (g), Δn_g = 0 K = (0.125)(0.125)/(0.75) 2 = 1/36 For H 2 (g) + I 2 (g) ⇌ 2HI(g), K = 36 For the half-stoichiometry reaction, K = √(36) = 6 This is option (d).

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What is the equilibrium constant of the reaction: 1/2H₂(g) + 1/2I₂(g) ⇌ HI(g), at 450°C?

Options:

Answer:

Solution:

Take 1 mol of HI, α = 0.25 At equilibrium: HI = 0.75 mol, H₂ = I₂ = 0.125 mol

For 2HI(g) ⇌ H₂(g) + I₂(g), Δn_g = 0

K = (0.125)(0.125)/(0.75)² = 1/36

For H₂(g) + I₂(g) ⇌ 2HI(g), K = 36

For the half-stoichiometry reaction, K = √(36) = 6

This is option (d).

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kc

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