AQA A-Level Chemistry Paper 1, 2017: Question 1

✅ Correct answer (what to write)

The enthalpy change (heat energy change) when 1 mol of an ionic solid (1 mol of ionic lattice) is formed from its gaseous ions.

💡 Key knowledge

• “Lattice formation” is always gaseous ions → solid lattice.
• Standard units: kJ mol⁻¹.
• For most ionic solids, lattice formation enthalpy is negative (exothermic), but the definition is about the process, not the sign.

🧠 Exam technique (how to secure both marks)

• Include both marking points explicitly: “1 mol” and “from gaseous ions”.
• State the product as a solid ionic lattice (or “solid ionic compound”).
• If you use an equation, make sure it matches formation, not dissociation.

❌ Common errors (seen by examiners)

• Describing lattice dissociation (solid → gaseous ions) instead of formation.
• Using atoms or molecules instead of ions (examiner note: CE=0/2 if wrong process/particles).
• Missing “1 mol”.
• Talking only about “energy released” without defining the process (mark scheme: ignore “heat energy released”).

💡 What the data mean

• AgI(s) → Ag⁺(aq) + I⁻(aq) is the enthalpy of solution of AgI: +112 kJ mol⁻¹.
• Ag⁺(g) → Ag⁺(aq) and I⁻(g) → I⁻(aq) are hydration enthalpies (both negative here).
• To go from solid → aqueous ions, you can go via solid → gaseous ions (lattice dissociation) then gaseous → aqueous (hydration).

📐 Calculations (step-by-step)

1. Write the Hess relationship for dissolving AgI(s):
   AgI(s) → Ag⁺(aq) + I⁻(aq) ΔH = +112
2. Express this as:
   ΔH(solution) = ΔH(lattice dissociation) + ΔH(hydration of Ag⁺) + ΔH(hydration of I⁻)
3. Substitute values (note: hydration enthalpies are given for ion(g) → ion(aq) ):
   +112 = ΔH(diss) + (−464) + (−293)
4. Rearrange:
   ΔH(diss) = 112 + 464 + 293 = +869 kJ mol⁻¹
5. Convert to lattice formation (reverse process, so change sign):
   ΔH(lattice formation) = −869 kJ mol⁻¹

🧠 Exam technique (how to avoid the sign trap)

• Always decide first: are you finding dissociation (positive) or formation (negative)? The question asks formation.
• Circle the state symbols: formation must be Ag⁺(g) + I⁻(g) → AgI(s) .
• Put units on the final answer: kJ mol⁻¹.

❌ Common calculation errors

• Not reversing at the end (giving +869 instead of −869).
• Subtracting hydration enthalpies incorrectly (they are already negative).
• Mixing up “lattice formation” with “lattice dissociation”.
• Missing units or writing mol⁻¹ incorrectly (keep it as kJ mol⁻¹).

✅ Full-mark explanation (2 clear points)

• AgI contains covalent character (it is not completely ionic).
• Therefore the forces/bonds holding the lattice together are stronger than predicted by a perfect ionic model.

Link to the data: stronger real interactions make the lattice formation enthalpy more negative (larger magnitude) than the purely ionic prediction.

💡 Key knowledge (what “perfect ionic model” assumes)

• Assumes ions are perfect point charges with purely electrostatic attraction.
• Assumes no electron density sharing (i.e., no covalency).
• Ag⁺ is relatively polarising and I⁻ is polarisable, increasing covalent character (you don’t need to mention this detail to score marks, but it can help your explanation sound precise).

🧠 Exam technique (how marks are awarded here)

• Make two distinct statements: (1) covalent character, (2) stronger lattice forces.
• Keep it about the model vs reality. The examiner is rewarding recognition of non-ideal ionic bonding.
• Be careful with wording: “smaller numerical value” means less negative in magnitude.

❌ Common errors (from the mark scheme notes)

• Talking about electronegativity instead of covalent character (ignored).
• Vague “stronger bonds” without stating covalent character (often loses a mark).
• Referring to atoms/molecules instead of ions (can lose credit).
• Just saying “more energy” without explaining why the model differs (mark scheme: ignore references to energy alone).

✅ Correct answers (either route earns full marks)

Option 1: Silver nitrate test (halide precipitation)

• Reagent: AgNO₃(aq)
• Observation: yellow precipitate (AgI)

Option 2: Halogen displacement

• Reagent: Cl₂(aq) or Br₂(aq)
• Observation: brown solution and/or black precipitate (iodine formed)

💡 Key knowledge

• With AgNO₃: iodide forms AgI(s) which is yellow.
• Halogen displacement: a more reactive halogen oxidises I⁻ to I₂.
• Observations must be specific colour/appearance, not just “a precipitate forms”.

🧠 Exam technique

• For 2 marks you need both: a valid reagent and the correct observation.
• If you choose AgNO₃, the key scoring phrase is “yellow precipitate”.
• If you choose Cl₂/Br₂, state the visual change to iodine: brown solution and/or black solid.

❌ Common errors / examiner notes

• Adding unnecessary conditions like acidified reagents or ammonia and thinking they are required (mark scheme: “Ignore ammonia/acidified…”).
• Using AgNO₃ but giving the wrong precipitate colour (e.g., “white” is for Cl⁻).
• For displacement tests, not naming an observation linked to iodine (must mention brown/black).

💡 Before you move on, can you do these?

• State lattice formation definition with 1 mol + gaseous ions.
• Use +112, −464, −293 to reach −869 kJ mol⁻¹ with correct sign logic.
• Explain model vs real using covalent character + stronger lattice forces.
• Pick a test for I⁻ and give a specific observation.

🧠 What distinguishes top answers

• Definitions are tight and mark-scheme precise (no missing “gaseous ions”).
• Calculations show clear sign handling and a final statement: ΔH(latt, form) = −869 kJ mol⁻¹.
• Explanations explicitly name covalent character (not just “not perfect”).