JEE Chemistry 2026 Question with Solution

Given: The equilibrium is $P_2(\text{g}) + Q_2(\text{g}) \rightleftharpoons 2PQ(\text{g})$. Initially, $2$ moles each of $P_2$, $Q_2$ and $PQ$ are at equilibrium. Then $1$ mole each of $P_2$ and $Q_2$ is added.

Find: The new equilibrium moles of $P_2$, $Q_2$ and $PQ$ and the correct option.

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, the solution gives

$$K = 1$$

Step 3: After addition of reactants. New moles become $P_2 = 3$, $Q_2 = 3$, $PQ = 2$. Let $x$ moles react forward.

So at the new equilibrium,

$$P_2 = 3 - x, \quad Q_2 = 3 - x, \quad PQ = 2 + 2x$$

Step 4: Solving using $K = 1$.

$$\frac{(2+2x)^2}{(3-x)(3-x)} = 1 \Rightarrow x = \frac{2}{3}$$

Step 5: Final moles.

$$P_2 = Q_2 = PQ = 2.67$$

Therefore, the new equilibrium moles are $2.67, 2.67, 2.67$ and the correct option is B.