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For the gaseous phase reaction $2 A + B ⇌ 2 C + D$ , initially there are 2 mole each of A&B. If 0.4 mol of D is present at equilibrium at a given T&P, in-correct relationship is

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$P_{A} < P_{B} & P_{D} < P_{C}$

$P_{A} = P_{C} & P_{B} = P_{D}$

$P_{C} = 2 P_{D} & P_{A} = 3 P_{B} / 4$

$P_{A} > P_{D} & P_{B} > P_{C}$

answer is B.

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Detailed Solution

Equilibrium for 2A + B ⇌ 2C + D

Equilibrium Calculation for Reaction

Reaction: 2A + B ⇌ 2C + D

Initial moles: A = 2, B = 2

At equilibrium: D = 0.4 mole

Species	A	B	C	D
Initial	2	2	0	0
Change	-0.8	-0.4	+0.8	+0.4
Equilibrium	1.2	1.6	0.8	0.4

Any relationship that claims A < 1.2, B < 1.6, C ≠ 0.8, or D ≠ 0.4 is incorrect.

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