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For the reaction 2AB2(g) ⇌ 2AB(g) + B2(g), initially 2 moles of AB2 are present. At equilibrium, 2(1-x) moles of AB2 and 2x moles of AB are present. Calculate the partial pressure of AB2 and AB at equilibrium.

1. P_AB2 = 2(1-x)P / (2+x)
2. P_AB = xP / (2+x)
3. P_AB2 = (1-x)P
4. P_AB = xP

Correct answer: P_AB2 = 2(1-x)P / (2+x)

Solution

The partial pressure of a gas is proportional to its mole fraction. For AB2, the mole fraction is (2(1-x))/(2+x), and multiplying by total pressure P gives the partial pressure as 2(1-x)P/(2+x).

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