WJEC A-Level Chemistry Unit 3, June 2025: Question 7

✅ Mark Scheme Answer

Gases are produced and they have higher entropy (more disorder) than solids.

🧠 Exam Technique

• Look at the stoichiometry: 2 moles of solid produce 2 moles of solid + 5 moles of gas (4NO₂ + O₂).
• Always mention both the production of gas and that gases have significantly higher entropy than solids.

📐 Step-by-Step Calculation

1. Convert temperature to Kelvin:
   T = 1500 + 273 = 1773 K
2. Apply Gibbs Feasibility Condition:
   At the minimum decomposition temperature, ΔG = 0:
   ΔG = ΔH − TΔS = 0  ⇒  ΔH = TΔS
3. Convert ΔS to kJ K⁻¹ mol⁻¹ and calculate:
   ΔS = 379 J K⁻¹ mol⁻¹ = 0.379 kJ K⁻¹ mol⁻¹
   ΔH = 1773 × 0.379 = +672 kJ mol⁻¹ (or 671.97 kJ mol⁻¹)

❌ Common Errors

• Unit Clash: Leaving ΔS in J K⁻¹ mol⁻¹ while calculating ΔH in kJ mol⁻¹ leads to an answer off by a factor of 1000.
• Celsius vs Kelvin: Using 1500 °C directly instead of converting to absolute temperature (1773 K).
• Missing Sign: Decomposition is endothermic; omitting the positive sign or writing a negative value loses the mark.

✅ Correct Answers

I. Ionic Equation:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

II. Enthalpy Cycle Calculation:
General relationship:
Δ_solH = Δ_hydH(M²⁺) + Δ_hydH(SO₄²⁻) + Δ_lattH(breaking)

• For MgSO₄: Δ_solH = (−1920) + (−1059) + 2833 = −146 kJ mol⁻¹
• For BaSO₄: Δ_solH = (−1360) + (−1059) + 2474 = +55 kJ mol⁻¹

Explanation:
Dissolving MgSO₄ is exothermic (Δ_solH < 0), favouring solubility. Dissolving BaSO₄ is endothermic (Δ_solH > 0), meaning it is insoluble.

💡 Key Knowledge

• Lattice Breaking vs Lattice Formation: The question gives lattice breaking enthalpy, which is positive (endothermic). Do not invert its sign.
• Hydration Enthalpy: Always negative (exothermic), as gaseous ions form bonds with polar water molecules.
• Solubility Trend: Down Group 2, sulfate solubility decreases because cation hydration enthalpy drops significantly faster (due to larger ionic radius) than the lattice breaking enthalpy.

✅ Mark Scheme Answer

pH value: Any value between 4.0 and 6.8 (i.e. weakly acidic).

Reason: The ammonium ion (NH₄⁺) is a conjugate acid and partially dissociates to release H⁺ ions:
NH₄⁺ ⇌ NH₃ + H⁺
or NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

❌ Common Errors

• Stating pH = 7 under the assumption that all salts are neutral.
• Confusing NH₄⁺ with a base because it is related to ammonia.
• Writing a fully irreversible reaction rather than indicating partial dissociation / reversible hydrolysis.

✅ Observations

🧠 Exam Technique

Be specific with precipitate colours. Writing just "yellow" often loses marks in WJEC schemes for lead iodide—use bright yellow or canary yellow to ensure full credit.

📐 Step-by-Step Calculation

1. Deduce phase behaviour from boiling points:
   • Boiling points: O₂ = 90 K, CO₂ = 195 K.
   • At 100 K: Only O₂ is gaseous (CO₂ has condensed/frozen).
   • At 200 K: Both O₂ and CO₂ are gases.
2. Calculate moles of O₂ at 100 K:
   V = 1.028 × 10³ cm³ = 1.028 × 10⁻³ m³
   p = 1 atm = 1.01 × 10⁵ Pa, T = 100 K, R = 8.31 J K⁻¹ mol⁻¹
   n(O₂) = pV / RT = (1.01 × 10⁵ × 1.028 × 10⁻³) / (8.31 × 100) = 0.125 mol
3. Calculate total moles of gas at 200 K:
   V = 1.193 × 10⁴ cm³ = 1.193 × 10⁻² m³
   n(total) = (1.01 × 10⁵ × 1.193 × 10⁻²) / (8.31 × 200) = 0.725 mol
4. Find moles of CO₂ and CaCO₃:
   n(CO₂) = n(total) − n(O₂) = 0.725 − 0.125 = 0.600 mol
   From equation: CaCO₃(s) → CaO(s) + CO₂(g) (1:1 ratio)
   Therefore, n(CaCO₃) = 0.600 mol
5. Calculate percentage mass of CaCO₃:
   Mᵣ(CaCO₃) = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹
   Mass of CaCO₃ = 0.600 mol × 100.1 g mol⁻¹ = 60.06 g
   % mass = (60.06 / 81.3) × 100 = 73.9%

❌ Common Errors

• Volume Unit Conversion: Forgetting that 1 cm³ = 10⁻⁶ m³. Multiplying incorrectly ruins the ideal gas substitution.
• Ignoring Boiling Points: Failing to realise that CO₂ is a solid/liquid at 100 K. Students who assumed both gases were present at 100 K could not solve the problem.
• Stoichiometry Mix-up: The question asks for percentage of CaCO₃, not NaNO₃. Remember the CaCO₃ : CO₂ ratio is 1:1.