Consider the following gaseous equilibrium in a closed container of volume ‘V’ at T (K). P₂(g) + Q₂(g) ⇌ 2PQ(g) 2 moles each of P₂(g), Q₂(g) and PQ(g) are present at equilibrium. Now one mole each of ‘P₂’ and ‘Q₂’ are added to the equilibrium keeping the temperature at T (K). The number of moles of P₂, Q₂ and PQ at the new equilibrium, respectively, are

Consider the following gaseous equilibrium in a closed container of volume V at T (K). P2 + Q2 ⇌ 2PQ. Find the moles at new equilibrium.

Q. Consider the following gaseous equilibrium in a closed container of volume ‘V’ at T (K).

P₂(g) + Q₂(g) ⇌ 2PQ(g)

2 moles each of P₂(g), Q₂(g) and PQ(g) are present at equilibrium. Now one mole each of P₂ and Q₂ are added to the equilibrium keeping the temperature at T (K). The number of moles of P₂, Q₂ and PQ at the new equilibrium, respectively, are

(A) 2.56, 1.62, 2.24

(B) 2.67, 2.67, 2.67

(D) 1.66, 1.66, 1.66

Correct Answer: 2.67, 2.67, 2.67

Explanation

For the equilibrium:

P₂ + Q₂ ⇌ 2PQ

At initial equilibrium:

n(P₂) = 2, n(Q₂) = 2, n(PQ) = 2

So equilibrium constant:

K = (2)² / (2 × 2) = 1

After adding 1 mole each of P₂ and Q₂:

P₂ = 3, Q₂ = 3, PQ = 2

Let x moles react forward to re-establish equilibrium.

P₂ = 3 − x Q₂ = 3 − x PQ = 2 + 2x

Apply equilibrium constant:

K = (2 + 2x)² / [(3 − x)(3 − x)] = 1

Solving:

2 + 2x = 3 − x 3x = 1 x = 1/3

Final moles:

P₂ = 3 − 1/3 = 2.67 Q₂ = 2.67 PQ = 2 + 2/3 = 2.67

Hence, the correct answer is option (B).

Explanation

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