The equilibrium constant for decomposition of H2O(g) H2O(g) H2(g) + 1 O2(g) (Gº = 92.34 kJ mol–1) is 8.0 × 10–3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition the degree of

Question

The equilibrium constant for decomposition of H2O(g) H2O(g) H2(g) + 1 O2(g) (Gº = 92.34 kJ mol–1) is 8.0 × 10–3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition the degree of dissociation () of water is _________ × 10–2. (nearest integer value) (Assume is negligible with respect to 1)

Prashant Kotian · Accepted Answer

Correct answer: .. # Solution: Degree of Dissociation of H₂O(g) **Step 1: Relate ΔG° to Kp** Use: ΔG° = –RT ln(Kp) 92,340 = –8.314 × 2300 × ln(Kp) ln(Kp) = –4.82, so Kp = 8.0 × 10⁻³ ✓ (confirms given value) **Step 2: Set up ICE table for dissociation** H₂O(g) ⇌ H₂(g) + ½O₂(g) Initial: 1 mol, 0, 0 Change: –α, +α, +α/2 Equilibrium: (1–α), α, α/2 (Since α << 1, we have 1–α ≈ 1) **Step 3: Calculate partial pressures at equilibrium** Total moles at equilibrium = (1–α) + α + α/2 = 1 + α/2 At Ptotal = 1 bar: - P(H₂O) = (1–α)/(1+α/2) × 1 ≈ 1–α - P(H₂) = α/(1+α/2) × 1 ≈ α - P(O₂) = (α/2)/(1+α/2) × 1 ≈ α/2 **Step 4: Apply Kp expression** Kp = [P(H₂) × P(O₂)^0.5] / P(H₂O) 8.0 × 10⁻³ = [α × (α/2)^0.5] / 1 8.0 × 10⁻³ = α × √(α/2) 8.0 × 10⁻³ = α^1.5 / √2 **Step 5: Solve for α** α^1.5 = 8.0 ×

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