AQA A-Level Chemistry Paper 1, 2023: Question 5

📐 Step-by-Step Calculation

1. State the thermodynamic relationship:
   ΔH_solution = ΔH_{lattice dissociation} + ΣΔH_hydration
2. Convert lattice formation to dissociation:
   Table 2 gives lattice formation: Ca²⁺(g) + 2Cl⁻(g) → CaCl₂(s) = -2237 kJ mol⁻¹.
   Therefore, ΔH_dissociation = +2237 kJ mol⁻¹.
3. Sum the hydration enthalpies (account for 2 Cl⁻ ions!):
   ΣΔH_hyd = ΔH_hyd(Ca²⁺) + 2 × ΔH_hyd(Cl⁻)
   ΣΔH_hyd = -1650 + [2 × (-364)] = -1650 - 728 = -2378 kJ mol⁻¹
4. Calculate ΔH_solution:
   ΔH_solution = +2237 + (-2378) = -141 kJ mol⁻¹

✅ Correct Answer

The temperature increases / goes up.

🧠 Exam Technique

ΔH_solution is negative (-141 kJ mol⁻¹), which means the dissolution process is exothermic. Exothermic reactions release thermal energy to the surroundings, raising the temperature of the water.

✅ Mark Scheme Requirements

• [M1]: Fluoride ions (F⁻) are smaller than chloride ions (Cl⁻)
  OR F⁻ has a higher charge density.
• [M2]: Resulting in a stronger attraction between the fluoride ion and the δ+ H (partially positive / electron-deficient hydrogen) atom in water.

❌ Examiner Pitfalls & Rejections

• Do NOT say "fluorine atoms": The comparison must strictly be between fluoride ions (F⁻) and chloride ions (Cl⁻).
• Never mention "ionic bonds" breaking: Hydration is an ion-dipole attraction forming between ions and water molecules, not ionic bonding.
• Do NOT write attraction to H⁺: Water contains partially charged δ+ H atoms, not free H⁺ ions.

✅ Correct Species on Diagram

• Lower blank line: Ca(g) + Cl₂(g)
  (Enthalpy of atomisation of calcium converts Ca(s) to Ca(g))
• Top blank line: Ca²⁺(g) + 2e⁻ + 2Cl(g)
  (Chlorine molecules atomised into gaseous chlorine atoms before electron affinity occurs)

💡 Key Knowledge: Cycle Logic

1. Ca(s) + Cl₂(g) → Ca(g) + Cl₂(g) [Ca atomisation]
2. Ca(g) + Cl₂(g) → Ca⁺(g) + e⁻ + Cl₂(g) [1st IE of Ca]
3. Ca⁺(g) + e⁻ + Cl₂(g) → Ca²⁺(g) + 2e⁻ + Cl₂(g) [2nd IE of Ca]
4. Ca²⁺(g) + 2e⁻ + Cl₂(g) → Ca²⁺(g) + 2e⁻ + 2Cl(g) [Bond dissociation / atomisation of Cl₂]
5. Ca²⁺(g) + 2e⁻ + 2Cl(g) → Ca²⁺(g) + 2Cl⁻(g) [2 × Electron affinity of Cl]

📐 Calculation Walkthrough

By Hess’s Law, the direct route equals the indirect route:

1. Insert values from Table 3:
   -795 = +193 + 590 + 2nd IE + 2(+121) + 2(-364) + (-2237)
2. Simplify known terms:
   -795 = 193 + 590 + 2nd IE + 242 - 728 - 2237
   -795 = -1940 + 2nd IE
3. Rearrange for 2nd IE:
   2nd IE = -795 + 1940 = +1145 kJ mol⁻¹

✅ Any One of the Following Points

• The electron is being removed from a positive ion (Ca⁺) rather than a neutral atom.
• The electron is being removed from a smaller ion (closer to the nucleus).
• There is a stronger electrostatic attraction between the same number of protons (20) and fewer electrons (19).

❌ Common Misconception

Do not claim the electron is being removed from a different principal quantum shell. For calcium, both the 1st and 2nd electrons are removed from the 4s subshell (Ca is [Ar]4s²; Ca⁺ is [Ar]4s¹). The jump to the 3p shell happens at the 3rd ionisation energy!

Stage 1: Comparing Theoretical Ionic Values

• 1a: The theoretical lattice dissociation enthalpy for CaCl₂ (2223 kJ mol⁻¹) is significantly larger than KCl (690 kJ mol⁻¹) and AgCl (770 kJ mol⁻¹). Values for KCl and AgCl are similar.
• 1b: Ca²⁺ has a higher charge (+2 vs +1) and is a smaller ion (higher charge density) compared to K⁺ and Ag⁺.
• 1c: Therefore, CaCl₂ has much stronger electrostatic attraction between oppositely charged ions (stronger ionic bonding).

Stage 2: Similarities (CaCl₂ and KCl)

• 2a & 2b: Both CaCl₂ and KCl have theoretical values that are almost identical / very close to their experimental Born–Haber values:
  • CaCl₂: 2223 vs 2237 kJ mol⁻¹
  • KCl: 690 vs 701 kJ mol⁻¹
• 2c: This indicates that CaCl₂ and KCl are almost purely (100%) ionic; their ions act as perfectly spherical point charges with negligible polarisation.

Stage 3: Discrepancy in AgCl

• 3a: AgCl shows a significant difference between the theoretical model (770 kJ mol⁻¹) and experimental value (905 kJ mol⁻¹) — a difference of 135 kJ mol⁻¹.
• 3b: This indicates that AgCl has substantial covalent character in addition to ionic bonding.
• 3c: The Ag⁺ cation polarises the chloride ion (distorting its electron cloud) because Ag⁺ has transition metal d-electrons that poorly shield its nuclear charge.