Principle used: Le Chatelier’s principle

Increasing the volume of the container decreases the pressure. The equilibrium shifts in the direction having greater number of moles of गैसीय species to oppose this decrease in pressure.

So, for each reaction, compare the total gaseous moles on reactant and product sides.

Option A

$2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$

• Reactant side gaseous moles $= 2$
• Product side gaseous moles $= 2+1=3$

Since products have more gaseous moles, increasing volume will shift equilibrium to the right.

A is correct.

Option B

$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

• Reactant side gaseous moles $= 1+1=2$
• Product side gaseous moles $= 2$

Number of gaseous moles is same on both sides, so change in volume has no effect on equilibrium position.

B is incorrect.

Option C

$4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(l)$

Only gaseous species are counted; $H_2O(l)$ is a liquid, so it is not counted.

• Reactant side gaseous moles $= 4+5=9$
• Product side gaseous moles $= 4$

Products have fewer gaseous moles, so increasing volume will shift equilibrium to the left, not toward products.

C is incorrect.

Option D

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

• Reactant side gaseous moles $= 3$
• Product side gaseous moles $= 2$

Products have fewer gaseous moles, so increasing volume shifts equilibrium to the left.

D is incorrect.

Final conclusion

The reaction in which increase in volume favours product formation is:

$\boxed{A}$

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.