At 25°C, 300 g of deuterium oxide, D 2 O (d = 1.10 g/mL) and 180 g of H 2 O (d = 0.997 g/mL) are mixed. The volumes are additive. Fifty per cent of the H 2 O reacts to form HDO. The value of K c at 25

Question

At 25°C, 300 g of deuterium oxide, D 2 O (d = 1.10 g/mL) and 180 g of H 2 O (d = 0.997 g/mL) are mixed. The volumes are additive. Fifty per cent of the H 2 O reacts to form HDO. The value of K c at 25°C for the reaction: H 2 O + D 2 O ⇌ 2HDO is

TestHub · Accepted Answer

Moles of D 2 O = 300/20 = 15 mol Moles of H 2 O = 180/18 = 10 mol H 2 O reacted = 0.5 × 10 = 5 mol, so HDO formed = 10 mol At equilibrium: H 2 O = 5 mol, D 2 O = 15 - 5 = 10 mol, HDO = 10 mol K c = (10) 2 /[(5)(10)] as Δn = 0 and the volume cancels K c = 2

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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