At 25°C, 560 g of deuterium oxide, D₂O (d = 1.10 g/ml) and 504 g H₂O (d = 0.997 g/ml) are mixed. The volumes are additive. Fifty per cent of the H₂O reacts to form HDO. The value of K_c at 25°C for th

Question

At 25°C, 560 g of deuterium oxide, D₂O (d = 1.10 g/ml) and 504 g H₂O (d = 0.997 g/ml) are mixed. The volumes are additive. Fifty per cent of the H₂O reacts to form HDO. The value of K_c at 25°C for the reaction: H₂O + D₂O ⇌ 2HDO is

Prashant Kotian · Accepted Answer

Correct answer: 4. Moles of D₂O = 560 / 20 = 28 mol. Moles of H₂O = 504 / 18 = 28 mol. 50% of H₂O reacts, which is 14 mol. Moles reacted: D₂O = 14 mol. Moles formed: HDO = 2 × 14 = 28 mol. Equilibrium moles: H₂O = 28 - 14 = 14 mol, D₂O = 28 - 14 = 14 mol, HDO = 28 mol. Volume cancels out in the K_c expression since Δn = 0. K_c = [HDO]² / ([H₂O][D₂O]) = (28)² / (14 × 14) = 4. Therefore, correct answer is 4.

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