Question

Two moles each of the gases P$_2$, Q$_2$ and PQ are present in a vessel at equilibrium. If 1 mole each of P$_2$ and Q$_2$ is added at equilibrium, then determine the composition (in mole) of each species at the new equilibrium.

• A. $n_{P_2}=0.5,\ n_{Q_2}=0.5,\ n_{PQ}=1$
• B. $n_{P_2}=1.33,\ n_{Q_2}=1.33,\ n_{PQ}=1.67$
• C. $n_{P_2}=2.67,\ n_{Q_2}=2.67,\ n_{PQ}=2.33$
• D. $n_{P_2}=2.67,\ n_{Q_2}=2.67,\ n_{PQ}=2.67$

Hint:

When equilibrium constant is 1, products and reactants tend to equalize at equilibrium.

Answer 1

The Correct Option is D

Step 1: Write the given equilibrium reaction

P₂(g) + Q₂(g) ⇌ 2PQ(g)

At equilibrium, initially:
n(P₂) = 2,   n(Q₂) = 2,   n(PQ) = 2

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Step 2: Calculate the equilibrium constant K_c

K_c = [PQ]² / ([P₂][Q₂])

K_c = 2² / (2 × 2) = 1

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Step 3: Add reactants and assume shift in equilibrium

After adding 1 mole each of P₂ and Q₂:

n(P₂) = 3,   n(Q₂) = 3,   n(PQ) = 2

Let x moles of P₂ and Q₂ react further ‘‘towards right’’.

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Step 4: Write equilibrium mole expressions

At new equilibrium:

n(P₂) = 3 − x
n(Q₂) = 3 − x
n(PQ) = 2 + 2x

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Step 5: Apply equilibrium constant condition

K_c = 1 = (2 + 2x)² / (3 − x)²

(2 + 2x)² = (3 − x)²

4 + 8x + 4x² = 9 − 6x + x²

3x² + 14x − 5 = 0

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Step 6: Solve the quadratic equation

x = [−14 ± √(14² + 60)] / 6

x = [−14 + 14] / 6 = 0.33

(Only the positive root is physically acceptable.)

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Step 7: Find the new equilibrium composition

n(P₂) = 3 − 0.33 = 2.67

n(Q₂) = 3 − 0.33 = 2.67

n(PQ) = 2 + 2(0.33) = 2.67

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Final Answer:

The new equilibrium composition is:
P₂ = 2.67,   Q₂ = 2.67,   PQ = 2.67