In which of the following reactions, an increase in the volume of the container will favour the formation of products?

According to Le Chatelier’s principle, whenever a change is imposed on an equilibrium system, the system tries to oppose that change. If the volume of the container is increased, the total pressure of the gaseous mixture decreases. The equilibrium therefore shifts in the direction in which the number of moles of gaseous species is larger, because that shift will increase the pressure again.

We now examine each reaction one by one, counting only the gaseous moles on each side.

For option A we have

$$4\text{NH}_3(g)+5\text O_2(g)\rightleftharpoons 4\text{NO}(g)+6\text H_2\text O(l).$$

Number of gaseous moles on the left = $$4+5=9,$$ while on the right only the gaseous $$4\text{NO}(g)$$ counts (the water is liquid), so gaseous moles on the right = $$4.$$ Because $$9\gt4,$$ an increase in volume will shift the equilibrium toward the left, not toward the products. So option A is not favoured.

For option B we have

$$2\text{NO}_2(g)\rightleftharpoons 2\text{NO}(g)+\text O_2(g).$$

Gaseous moles on the left = $$2.$$ Gaseous moles on the right = $$2+1=3.$$ Now $$3\gt2,$$ so the product side contains more gas molecules. When the volume is increased, the equilibrium will shift to the side with more gaseous moles, i.e. toward the products. Hence the formation of products is favoured in option B.

For option C we have

$$3\text O_2(g)\rightleftharpoons 2\text O_3(g).$$

Left side moles = $$3,$$ right side moles = $$2.$$ Because the left has more gaseous moles, an increase in volume will shift the equilibrium to the left, so product formation is not favoured.

For option D we have

$$\text H_2(g)+\text I_2(g)\rightleftharpoons 2\text{HI}(g).$$

Left side moles = $$1+1=2,$$ right side moles = $$2.$$ The number of gaseous moles is the same on both sides, so a change in volume produces no shift in equilibrium; product formation is neither favoured nor opposed.

Among the four reactions, only option B shows an increase in the number of gaseous moles from reactants to products, so only that reaction is driven toward products when the volume is increased.

Hence, the correct answer is Option B.