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Chemical Equilibrium: Degree of Dissociation of H₂O(g)

Question 74:
The equilibrium constant for decomposition of H₂O(g) is given as:

H₂O(g) ⇌ H₂(g) + ½O₂(g)    (ΔG° = 92.34 kJ mol⁻¹)

At 2300 K and total pressure of 1 bar, the equilibrium constant (K_p) is 8.0 × 10⁻³. Under these conditions, the degree of dissociation (α) of water is _____ × 10⁻² (nearest integer value).

Assume α is negligible with respect to 1.

• (1) 2
• (2) 3
• (3) 4
• (4) 6
• (5) 5

Correct Answer: (5) → α = 5 × 10⁻²

Solution:
For the reaction: H₂O(g) ⇌ H₂(g) + ½O₂(g)

At t = 0 → H₂O = 1 mol
At equilibrium → H₂O = (1−α), H₂ = α, O₂ = α/2

Total moles: n_T = 1 + α/2 ≈ 1 (since α ≪ 1)

Expression for equilibrium constant: K_p = (P_H₂ × P_O₂^1/2) / P_H₂O
K_p = (αP) × (α/2 × P)^1/2 / ((1−α)P)

For P = 1 bar, simplifying gives: 8 × 10⁻³ = α^3/2 / √2

Therefore: α^3/2 = 8√2 × 10⁻³
α³ = 128 × 10⁻⁶
α = (128 × 10⁻⁶)^1/3 = 5.03 × 10⁻²

Hence, the degree of dissociation α ≈ 5 × 10⁻².

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