WAEC 2013 Chemistry · Question 35

Consider the reaction represented by the equation: N 2 O 4 (g) ⇌ 2NO 2 (g) ; ∆H = + x kJ mol^ -1 What happens…

Question

Consider the reaction represented by the equation: N $_2$ O $_4$ (g) ⇌ 2NO $_2$ (g) ; ∆H = + x kJ $mol^{-1}$ What happens when the temperature is reduced at equilibrium?

A.Concentration of N 2O4(g) decreases
B.Concentration of N 2O4(g) increases.
C.Pressure exerted by the gases increases
D.Pressure exerted by the gases remains constant

Answer: B. Concentration of N 2O4(g) increases.

Explanation written with AI and checked against the answer key

Analyze the question

The reaction N₂O₄(g) ⇌ 2NO₂(g) has ΔH = +x kJ/mol, so the forward reaction is endothermic (absorbs heat). We need to find what happens when the temperature is lowered at equilibrium.

Concept: Le Chatelier's Principle

When a system at equilibrium is disturbed, it shifts in the direction that opposes the change.

Increasing temperature favours the endothermic direction.
Decreasing temperature favours the exothermic direction.

Solution

Forward reaction (N₂O₄ → 2NO₂): endothermic (ΔH positive).
Backward reaction (2NO₂ → N₂O₄): exothermic (releases heat).
When the temperature is reduced, the system tries to produce heat to oppose the cooling. It does this by favouring the exothermic reaction, which is the backward reaction.
The equilibrium shifts to the left, so: • [N₂O₄] increases • [NO₂] decreases

Checking the other options

A: N₂O₄ decreases. Wrong, it increases.
C and D: The shift to the left means 2 moles of gas become 1 mole, so the number of gas moles decreases. At constant volume, the pressure therefore decreases. So pressure neither increases (C) nor stays constant (D).

Exam tip: Remember "Cool down → favour exothermic; heat up → favour endothermic." With a positive ΔH, cooling pushes the equilibrium toward the reactant side.

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