The degree of dissociation of HI at a particular temperature of 0.8. Calculate the volume (in litre) — Chemical Equilibrium Chemistry Question
Question
The degree of dissociation of HI at a particular temperature of 0.8. Calculate the volume (in litre) of 1.6 M-Na₂S₂O₃ solution required to neutralize the iodine present in an equilibrium mixture of a reaction when 2 moles each of H₂ and I₂ are heated in a closed vessel of 2 l capacity.
💡 Solution & Explanation
For 2HI ⇌ H₂ + I₂, α = 0.8. K_c(dissociation) = (0.4)² / (1 - 0.8)² = 0.16 / 0.04 = 4. The formation reaction H₂ + I₂ ⇌ 2HI has K_c = 1/4 = 0.25. Initial H₂ = 2, I₂ = 2. Let x moles react: eq amounts are H₂ = 2-x, I₂ = 2-x, HI = 2x. K_c = (2x)² / (2-x)² = 0.25. Taking the square root: 2x / (2-x) = 0.5 => 4x = 2 - x => 5x = 2 => x = 0.4. Moles of I₂ at equilibrium = 2 - 0.4 = 1.6 mol. Neutralization reaction: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Moles of S₂O₃²⁻ needed = 2 × 1.6 = 3.2 mol. Volume of 1.6 M Na₂S₂O₃ required = 3.2 mol / 1.6 M = 2.0 L. Therefore, correct answer is 2.