AQA A-Level Chemistry Paper 3, June 2025: Question 1

✅ Correct Answer

KOH(s) → K⁺(aq) + OH⁻(aq)

Also allowed: KOH(s) + aq → K⁺(aq) + OH⁻(aq) or KOH(s) → KOH(aq)

💡 Key Knowledge

Enthalpy of solution is the enthalpy change when 1 mole of an ionic solid dissolves completely in water to give infinitely dilute aqueous ions.

• LHS must be 1 mole of solid: KOH(s).
• RHS must be separated aqueous ions: K⁺(aq) and OH⁻(aq).

❌ Common Errors

• Writing H₂O on the LHS as a reactant (e.g. KOH + H₂O → ... ) — 0 marks. H₂O acts as the solvent, not a stoichiometric reactant.
• Using a reversible arrow (⇌) instead of an irreversible arrow (→) — 0 marks.
• Omitting state symbols — the question explicitly states: "including state symbols".

🧠 Exam Technique

Always inspect the wording: if it asks for "enthalpy of solution", start from the solid ionic lattice and finish with hydrated ions.

✅ Final Answer

Volume of water = 53.4 (or 53)

Units = cm³ (or 0.0534 dm³)

❌ Common Calculation Traps

• Unit clash in q: Substituting ΔH in kJ directly into q = mcΔT where c is in J K⁻¹ g⁻¹. You must multiply kJ by 1000 to get J!
• Missing unit on the dotted line: Students often leave the separate "Units" line blank, forfeiting M5.
• Temperature in Kelvin: Attempting to convert 12.3 °C by adding 273. The difference ΔT is 12.3 whether in °C or K!

📐 Calculation & Cycle

By Hess's Law:

ΔH_soln = ΣΔH_hyd − ΔH_{latt formation}

or:

ΔH_soln = ΔH_{latt dissociation} + ΣΔH_hyd

Substituting values:

ΔH_soln = (−322) + (−528) − (−789)

ΔH_soln = −850 + 789 = −61 kJ mol⁻¹

💡 Visualising the Hess Cycle

Consider the two routes from KOH(s) to K⁺(aq) + OH⁻(aq):

• Direct route: ΔH_soln
• Indirect route:
  1. Break lattice into gaseous ions: −ΔH_{latt form} = −(−789) = +789 kJ mol⁻¹
  2. Hydrate gaseous ions: ΔH_hyd(K⁺) + ΔH_hyd(OH⁻) = (−322) + (−528) = −850 kJ mol⁻¹

Direct = Indirect: +789 + (−850) = −61 kJ mol⁻¹

✅ Correct Answer

Enthalpy of solution = −61 kJ mol⁻¹

❌ Common Sign Traps

• Confusing formation vs. dissociation: The table gives lattice formation (−789). Dissociating the solid into ions requires +789 kJ mol⁻¹.
• Writing +61 (gives 1/2 marks) — sign error.
• Writing −1639 (gives 1/2 marks) — added all three numbers directly without reversing lattice formation.

✅ Accepted Answers

• Heat losses (or thermal energy lost to surroundings/cup).
• Incomplete dissolving of the KOH solid.
• Also allowed: Incomplete dissociation; solution was not at infinite dilution.

🧠 Why does this explain the difference?

The calculated theoretical value was −61 kJ mol⁻¹, whereas the experimental value was only −51.3 kJ mol⁻¹ (less exothermic).

If heat is lost to the surroundings or if some solid remains undissolved, the observed temperature rise will be lower than expected, resulting in a less negative calculated experimental enthalpy.