Question

Consider the following gaseous equilibrium in a closed container of volume $V$ at temperature $T$:
P$_2$(g) + Q$_2$(g) $\rightleftharpoons$ 2PQ(g)
Initially, 2 moles each of P$_2$(g), Q$_2$(g) and PQ(g) are present at equilibrium. One mole each of P$_2$ and Q$_2$ are added. The number of moles of P$_2$, Q$_2$ and PQ at the new equilibrium respectively are

• A. 1.21, 2.24, 1.56
• B. 2.67, 2.67, 2.67
• C. 1.66, 1.66, 1.66
• D. 2.56, 1.62, 2.24

Hint:

When equilibrium constant equals 1, equilibrium tends to equalize concentrations.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the change in moles of each component when additional reactants are added to the system at equilibrium.

Initially, we have the equilibrium system:

P₂(g) + Q₂(g) ⇌ 2PQ(g)

with 2 moles each of P₂, Q₂, and PQ. On adding 1 mole each of P₂ and Q₂, the initial moles become:

• P₂: 3 moles
• Q₂: 3 moles
• PQ: 2 moles

Since the system will try to achieve a new equilibrium, we assume that at new equilibrium, 'x' moles of P₂ and Q₂ have reacted to form '2x' moles of PQ. Therefore, at equilibrium, we have:

• P₂ = 3 - x
• Q₂ = 3 - x
• PQ = 2 + 2x

Since the reaction is symmetric in terms of stoichiometry, when the reaction reaches equilibrium, the changes in moles must satisfy the following condition:

The equilibrium constant K should remain unchanged:

K = \(\frac{(PQ)^2}{(P_2)(Q_2)}\)

On substituting the expressions for equilibrium concentrations, we have:

K = \(\frac{(2 + 2x)^2}{(3-x)(3-x)}\)

For simplification and solving, assume the system returns almost to its initial concentration values due to the completeness of reaction being achieved similarly on both sides by the Le Chatelier's principle. The practicality of exam-styled problems allows us to proceed directly using intelligent guesswork by checking given options.

Reach based on the simplest balance — let's check given each moles approximates back to a balanced state when the total addition leads to a nearly equal redistribution.

• Check for 2.67: With P₂ = 2.67, Q₂ = 2.67, PQ = 2.67, we intuitively realize, involving argument continuity and practical exam strategies, that maintaining approximated central values satisfies stoichiometric adjustments made.

Hence, the answer based on option guesswork, aligned indirectly with reaction symmetry tests or comparisons, concludes:

2.67, 2.67, 2.67.