✏️ HSC Analyse Question (6 marks)

📝 Task (6 marks) — Suggested: 200–250 words

Qualitatively analyse the effect of temperature on the value of Kₑq. In your response, compare the effect for both exothermic and endothermic forward reactions, explain WHY temperature is the only factor that changes Kₑq, and use a specific chemical example to illustrate your analysis.

Word count: 0 / 200–250 suggested

🌟 Band 6 Model Answer

Temperature is the only factor that changes the value of Kₑq because it is the only change that affects the forward and reverse reaction rates unequally on a fundamental thermodynamic level. When concentration or pressure is changed, the system shifts to restore the original product-to-reactant ratio (Kₑq remains constant). Temperature, however, changes the proportion of particles exceeding each activation energy differently — because the forward and reverse reactions have different activation energies — producing a genuinely new equilibrium ratio and therefore a new Kₑq value.

For an endothermic forward reaction (ΔH > 0), increasing temperature favours the endothermic direction (forward), shifting equilibrium to the right. More products and fewer reactants are present at the new equilibrium, so Kₑq increases. Conversely, decreasing temperature shifts equilibrium to the left, producing fewer products, and Kₑq decreases.

For an exothermic forward reaction (ΔH < 0), the opposite occurs. Increasing temperature favours the endothermic reverse direction, shifting equilibrium to the left. Fewer products and more reactants result, so Kₑq decreases. Decreasing temperature shifts right, increasing Kₑq.

For example, in the equilibrium N₂O₄(g) ⇌ 2NO₂(g) (ΔH = +57 kJ/mol), heating produces a darker brown colour as more NO₂ forms and Kₑq increases. Cooling produces a paler colour as Kₑq decreases. This has industrial significance: for the exothermic Haber process, high temperatures decrease Kₑq (lower yield), requiring a compromise temperature of 450°C.

💡 Why This Scores 6/6

1. Addresses WHY temperature is unique (different activation energies → unequal rate changes → new thermodynamic ratio). Most students only state WHAT happens.

2. Covers BOTH endothermic and exothermic with clear direction of Kₑq change for each.

3. Uses a specific chemical example (N₂O₄/NO₂) with observable evidence (colour change).

4. Links to real-world application (Haber process compromise temperature).

5. Uses precise terminology: Kₑq, ΔH, endothermic, exothermic, equilibrium position.

6. Analyses rather than just describes: explains the relationship (WHY Kₑq changes) and draws implications (industrial compromise).

Memory framework: For any temperature-Kₑq question, cover: (1) The rule (endo favoured by heating), (2) Both directions (endo fwd → Kₑq up, exo fwd → Kₑq down), (3) WHY (different Eₐ = unequal rate changes), (4) Example with evidence, (5) Industrial implication.