In which of the following equations can Le-chartelier’s principle be applied?

Analyze the question: Le Chatelier's principle says that when a system in dynamic equilibrium is disturbed (by a change in concentration, pressure, or temperature), it shifts in the direction that opposes the change. So we are looking for a reversible reaction, one that can reach equilibrium.

Explain the concept:

• Reversible reactions are shown with the double arrow ⇌.
• One-way reactions (→) go to completion, so there is no equilibrium to shift.
• Gases are especially useful here, because pressure changes affect them.

Check each option:

• A. H₂(g) + S(g) → H₂(g) is written with a single arrow and is not even balanced or sensible (the product should be H₂S). It is not an equilibrium.
• B. 2H₂S + O₂ → 2H₂O + 2SO₂ is a one-way combustion reaction, so no equilibrium.
• C. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is the Haber process, a reversible reaction. It reaches dynamic equilibrium, so Le Chatelier's principle applies.
• D. NaOH + CO₂ → Na₂CO₃ + H₂O is a one-way neutralisation, and it is also unbalanced as written.
• E. S + O₂ → SO₂ is a one-way combustion that goes essentially to completion.

Application to option C (to confirm):

• Increasing pressure moves the equilibrium toward fewer gas moles (4 moles → 2 moles), so more NH₃ forms.
• The forward reaction is exothermic, so lowering the temperature favours NH₃, though in industry a moderate temperature (~450°C) is used for a reasonable rate.
• Adding more N₂ or H₂ shifts the equilibrium right, and removing NH₃ also shifts it right.

Answer: C, because it is the only reversible reaction, and Le Chatelier's principle only applies to systems at equilibrium.

Exam tip: Look for the ⇌ sign. If you see only a single arrow (→), Le Chatelier's principle usually does not apply. Also remember the Haber process (N₂ + H₂ ⇌ NH₃) and the Contact process (2SO₂ + O₂ ⇌ 2SO₃), the two favourite examples in NECO, WASSCE and JAMB.