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

Step 1: Writing equilibrium constant expression.
\[ K = \frac{(PQ)^2}{P_2 Q_2} \]
Step 2: Initial equilibrium condition.
Since moles of all species are equal, $K = 1$.
Step 3: After addition of reactants.
New moles become P$_2$ = 3, Q$_2$ = 3, PQ = 2. Let $x$ moles react forward.
Step 4: Solving using $K=1$.
\[ \frac{(2+2x)^2}{(3-x)(3-x)} = 1 \Rightarrow x = \frac{2}{3} \]
Step 5: Final moles.
P$_2$ = Q$_2$ = PQ = 2.67