When 1.0 mole of H₂(g) and 3.0 moles of I₂ vapours are allowed to react, 'x' moles of HI(g) is forme — Chemical Equilibrium Chemistry Question
Question
When 1.0 mole of H₂(g) and 3.0 moles of I₂ vapours are allowed to react, 'x' moles of HI(g) is formed at equilibrium. Addition of a further 2.0 moles of H₂(g) gave an additional 'x' moles of HI(g) at new equilibrium. The value of equilibrium constant for the reaction: H₂(g) + I₂(g) ⇌ 2HI(g) is
💡 Solution & Explanation
Eq 1: H₂ = 1 - x/2, I₂ = 3 - x/2, HI = x. Eq 2 (after adding 2 mol H₂ and forming additional x mol HI): Total HI = 2x. Total reacted H₂ and I₂ = x. H₂ = 1 + 2 - x = 3 - x. I₂ = 3 - x. K_c = x² / ((1 - x/2)(3 - x/2)) = (2x)² / ((3 - x)(3 - x)). x² / (3 - 2x + x²/4) = 4x² / (9 - 6x + x²). 1 / (3 - 2x + x²/4) = 4 / (9 - 6x + x²). 9 - 6x + x² = 12 - 8x + x². 2x = 3 => x = 1.5. Substituting x into K_c: K_c = 4(1.5)² / (3 - 1.5)² = 4(2.25) / 2.25 = 4. Therefore, correct answer is 4.