Consider the following reversible reaction wherein the moles of species at equilibrium are given

P₂ + Q₂ ⇌ 2PQ

	P2	Q2	PQ
At equilibrium	2 mol	2 mol	2 mol

If one mole of P₂ and one mole of Q₂ are added at equilibrium, find the number of moles of P₂, Q₂ and PQ respectively at new equilibrium.

The correct option is 8/3 , 8/3 , 8/3.

Let us solve the problem step-by-step.
We are given the following reversible reaction:
$P_2+Q_2\rightleftharpoons 2PQ$

Initially, at the first equilibrium, the moles of each species in the container of volume V are:
Moles of $P_2=2\text{mol}$
Moles of $Q_2=2\text{mol}$
Moles of $PQ=2\text{mol}$

The equilibrium constant $K_c$ for the reaction is given by:
$K_c=\frac{{\left[PQ\right]}^2}{\left[P_2\right]\left[Q_2\right]}=\frac{{\left(2/V\right)}^2}{\left(2/V\right)·\left(2/V\right)}=\frac{4}{4}=1$

Now, 1 mole of $P_2$ and 1 mole of $Q_2$ are added to the system.
Immediately after addition, the moles of the species become:
Moles of $P_2=2+1=3\text{mol}$
Moles of $Q_2=2+1=3\text{mol}$
Moles of $PQ=2\text{mol}$

Let $x$ moles of $P_2$ react with $x$ moles of $Q_2$ to establish a new equilibrium.
At the new equilibrium, the moles of each species will be:
Moles of $P_2=3-x$
Moles of $Q_2=3-x$
Moles of $PQ=2+2x$

Since the temperature remains constant, the value of $K_c$ remains unchanged:
$K_c=\frac{{\left(2+2x\right)}^2}{\left(3-x\right)\left(3-x\right)}=1$

Taking the square root on both sides:
$\frac{2+2x}{3-x}=1$
$2+2x=3-x$
$3x=1$
$x=\frac{1}{3}$

Substituting the value of $x$ to find the moles at the new equilibrium:
Moles of $P_2=3-\frac{1}{3}=\frac{8}{3}\text{mol}$
Moles of $Q_2=3-\frac{1}{3}=\frac{8}{3}\text{mol}$
Moles of $PQ=2+2\left(\frac{1}{3}\right)=2+\frac{2}{3}=\frac{8}{3}\text{mol}$

Thus, the number of moles of $P_2$, $Q_2$, and $PQ$ at the new equilibrium are 8/3, 8/3, and 8/3, respectively.