The degree of dissociation of NH₃ at T K, in the vessel of volume V₁, is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

For 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), let α be the degree of dissociation. Total moles = 1 + α/2. Partial pressures: P_SO₃ = (1-α)P / (1+α/2), P_SO₂ = αP / (1+α/2), P_O₂ = (α/2)P / (1+α/2). K_p = P_SO₂² P_O₂ / P_SO₃² = (α² (α/2) P) / ( (1-α)² (1+α/2) ). We are given that P = K_p. Substituting this gives: 1 = (α³/2) / [ (1-α)² (1+α/2) ] => 1 = α³ / [ (1-α)² (2+α) ]. Expanding: (2+α)(1 - 2α + α²) = α³ => 2 - 4α + 2α² + α - 2α² + α³ = α³ => 2 - 3α = 0 => α = 2/3 ≈ 0.67. The degree of dissociation for NH₃ is the same. Therefore, correct answer is C.