AQA A-Level Chemistry Paper 1, 2021: Question 1

11 marks · Medium difficulty · State/Explain/Numerical

Define enthalpy change, complete a Born–Haber cycle for calcium chloride, calculate lattice and solution enthalpies, and explain the trend in hydration enthalpy.

Practise this question

Question

Question text

01 This question is about enthalpy changes for calcium chloride and

magnesium chloride.

01.1 State the meaning of the term enthalpy change.

[1 mark]

Figure 1 shows an incomplete Born–Haber cycle for the formation of calcium chloride.

Figure 1

01.2 Complete Figure 1 by writing the formulas, including state symbols, of the

appropriate species on each of the three blank lines.

[3 marks]

01.3 Table 1 shows some enthalpy data.

Table 1

Enthalpy change

/ kJ mol–1

Enthalpy of formation of calcium chloride –795

Enthalpy of atomisation of calcium +193

First ionisation energy of calcium +590

Second ionisation energy of calcium +1150

Enthalpy of atomisation of chlorine +121

*02* Electron affinity of chlorine –364

Use Figure 1 and the data in Table 1 to calculate a value for the

enthalpy of lattice dissociation of calcium chloride.

[2 marks]

4 –1

Enthalpy of lattice dissociation kJ mol

01.4 Magnesium chloride dissolves in water.

Give an equation, including state symbols, to represent the process that occurs when

the enthalpy of solution of magnesium chloride is measured.

[1 mark]

01.5 Table 2 shows some enthalpy data.

*03* Table 2

Enthalpy change

/ kJ mol–1

Enthalpy of lattice dissociation of MgCl2 +2493

Enthalpy of hydration of Mg2+(g) –1920

Enthalpy of hydration of Cl–(g) –364

Use your answer to Question 01.4 and the data in Table 2 to calculate a value for the

enthalpy of solution of magnesium chloride.

[2 marks]

Enthalpy of solution kJ mol–1

01.6 The enthalpy of hydration of Ca2+(g) is –1650 kJ mol–1

Suggest why this value is less exothermic than that of Mg2+(g)

[2 marks]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

Ignore conditions even if wrong 1

01.1 Heat (energy) change at constant pressure

Ignore energy change

M2 Ca2+(g) + 2 e– + Cl (g) Alternative M2 Ca+(g) + e‒ + 2 Cl(g) 1

01.2 M3 Ca2+(g) + 2 Cl –(g) 1

M1 Ca(s) + Cl 2(g) 1

M1 –795 + LE = 193 + 590 +1150 + ( 2 x 121) + (2 x –364) Numbers and factors used correctly from cycle 1

Rearrangement to calculate LE 1

01.3 –1

M2 LE = (+) 2242 (kJ mol ) If one or both factors of 2 missing award 1 mark

for (+) 2485, (+)2121 or (+)2606 (kJ mol–1)

Allow 1 mark for – 2242 (kJ mol–1)

Allow MgCl (s) ⇌ Mg2+(aq) + 2 Cl–(aq) 1

01.4 MgCl (s) → Mg2+(aq) + 2 Cl‒(aq) 2+ –

2 Allow MgCl2(s) + aq ⇌ Mg (aq) + 2 Cl (aq)

– A-LEVEL CHEMISTRY – –

M1 for expression with or without numbers 1

M1 ΔH soln MgCl = ΔH latt diss+ ΔH hyd Mg2+ + 2ΔH hyd Cl–

01.5 OR 2493 –1920 + (2 x –364) M2 for answer 1

If factor of 2 missing for ΔH hyd Cl– , allow 1 mark

M2 = – 155 (kJ mol–1)

for 209

Allow converse answers 1

M1 Ca2+ (ion) bigger/lower charge to size ratio (than Mg2+)

M1 Do not accept Ca2+ is a bigger atom/molecule

M1 Allow Ca2+ has more shells/ more distance of 1

01.6 outer e to nucleus

Ignore more shielding

M2 weaker attraction/bond to (Oδ- in) water

How to answer it

Energetics: Born–Haber Cycles & Solution Enthalpies

📌 What this question tests

This question assesses fundamental thermodynamic definitions, constructing and navigating Born–Haber cycles for Group 2 halides, calculating lattice dissociation enthalpies, writing balanced thermochemical equations for dissolution, computing enthalpies of solution via Hess's Law cycles, and explaining trends in hydration enthalpies using ionic charge and ionic radius.

Question 01.1: Definition of Enthalpy Change

Part (a) • 1 Mark

✅ Correct Answer

Heat (energy) change at constant pressure

🧠 Exam Technique

This is a standard recall definition from physical chemistry. You must include both components: heat/energy change AND constant pressure. Simply writing "energy change" is insufficient and will not score.

Mark Scheme: [1 mark] Heat (energy) change at constant pressure. (Ignore standard conditions if added, ignore standalone "energy change").

Question 01.2: Completing the Born–Haber Cycle

Part (b) • 3 Marks

✅ Correct Species for the Blank Lines

Bottom line (elements in standard states): Ca(s) + Cl₂(g) [1 mark]
Upper middle line (between 1st & 2nd IE of Ca): Ca²⁺(g) + 2 e⁻ + Cl₂(g) (or Ca⁺(g) + e⁻ + 2 Cl(g)) [1 mark]
Right-hand lower line (after electron affinity): Ca²⁺(g) + 2 Cl⁻(g) [1 mark]

❌ Common Errors

Missing state symbols: Forgetting (s) or (g) loses the marks immediately.
Incorrect stoichiometry: Writing Cl⁻(g) instead of 2 Cl⁻(g) .
Missing electrons: Forgetting to account for released electrons (e.g. leaving off 2 e⁻ ).

💡 Key Knowledge: Reading the Cycle Sequence

Follow the arrows step-by-step from bottom to top:

Ca(s) + Cl₂(g) (elements) → Ca(g) + Cl₂(g) (atomisation of Ca)
Ca(g) + Cl₂(g) → Ca⁺(g) + e⁻ + Cl₂(g) (first IE of Ca)
Ca⁺(g) + e⁻ + Cl₂(g) → Ca²⁺(g) + 2 e⁻ + Cl₂(g) (second IE of Ca)
Ca²⁺(g) + 2 e⁻ + Cl₂(g) → Ca²⁺(g) + 2 e⁻ + 2 Cl(g) (atomisation of Cl₂ / bond dissociation)
Ca²⁺(g) + 2 e⁻ + 2 Cl(g) → Ca²⁺(g) + 2 Cl⁻(g) (electron affinity of 2 Cl atoms)
Ca²⁺(g) + 2 Cl⁻(g) → CaCl₂(s) (lattice formation enthalpy, downwards arrow)

Question 01.3: Calculating Lattice Dissociation Enthalpy

Part (c) • 2 Marks

📐 Step-by-Step Calculation

Lattice dissociation involves turning solid CaCl₂(s) into gaseous ions Ca²⁺(g) + 2 Cl⁻(g) (an endothermic process, so the value must be positive).

Step 1: Write the energy cycle relationship
Going via standard route: formation + lattice dissociation = atomisation & ionisation steps.
Δ_fH(CaCl₂) + Δ_{L diss}H = Δ_atH(Ca) + IE₁(Ca) + IE₂(Ca) + 2(Δ_atH(Cl)) + 2(EA(Cl))

Step 2: Substitute the data with stoichiometric multipliers
-795 + Δ_{L diss}H = +193 + 590 + 1150 + 2(+121) + 2(-364)
-795 + Δ_{L diss}H = 193 + 590 + 1150 + 242 - 728
-795 + Δ_{L diss}H = +1447

Step 3: Solve for Δ_{L diss}H
Δ_{L diss}H = +1447 - (-795) = +2242 kJ mol⁻¹

✅ Final Answer

+2242 kJ mol⁻¹

❌ Common Errors & Mark Penalties

Forgetting factor of 2: There are two moles of chlorine atoms! Forgetting to double the atomisation or electron affinity gave partial credit values of +2485, +2121, or +2606 (1 mark max).
Sign errors: Giving -2242 kJ mol⁻¹ scored only 1 mark because dissociation is strictly endothermic (+).

Mark Scheme:
• M1: Correct cycle expression using numbers and correct factors: -795 + LE = 193 + 590 + 1150 + (2 × 121) + (2 × -364)
• M2: LE = (+) 2242 (kJ mol⁻¹)

Question 01.4: Equation for Enthalpy of Solution

Part (d) • 1 Mark

✅ Correct Equation

MgCl₂(s) → Mg²⁺(aq) + 2 Cl⁻(aq)

(Also acceptable: MgCl₂(s) + aq → Mg²⁺(aq) + 2 Cl⁻(aq) or reversible arrows ⇌ )

🧠 Exam Technique

Enthalpy of solution represents 1 mole of an ionic solid dissolving in water to form aqueous ions at infinite dilution. Always ensure:

Solid state symbol on the reactant: (s)
Aqueous state symbols on the products: (aq)
Stoichiometry is balanced: 2 Cl⁻ , not Cl₂²⁻ or Cl⁻

Question 01.5: Enthalpy of Solution Calculation

Part (e) • 2 Marks

📐 Step-by-Step Calculation

Step 1: Recall the relationship
Δ_solnH = Δ_{L diss}H + ΣΔ_hydH(ions)
Δ_solnH = Δ_{L diss}H(MgCl₂) + Δ_hydH(Mg²⁺) + 2 × Δ_hydH(Cl⁻)

Step 2: Substitute values from Table 2
Δ_solnH = 2493 + (-1920) + 2 × (-364)
Δ_solnH = 2493 - 1920 - 728

Step 3: Calculate the final value
Δ_solnH = -155 kJ mol⁻¹

✅ Final Answer

-155 kJ mol⁻¹

❌ Common Errors

Students frequently forget the factor of 2 for Cl⁻ ions:
2493 - 1920 - 364 = +209 kJ mol⁻¹ (Awarded only 1 mark).

Mark Scheme:
• M1: Expression with or without values: 2493 - 1920 + (2 × -364)
• M2: -155 (kJ mol⁻¹)

Question 01.6: Comparing Hydration Enthalpies

Part (f) • 2 Marks

✅ Key Points to Score 2 Marks

Mark 1 (Size/Charge Density): Ca²⁺ is bigger than Mg²⁺ (or has a lower charge density / lower charge-to-size ratio).
Mark 2 (Attraction to Water): Ca²⁺ has a weaker attraction to the partially negative oxygen atom ( δ⁻ O ) in water molecules.

💡 Scientific Principle

Both Ca²⁺ and Mg²⁺ carry a 2+ charge. However, Ca²⁺ has more electron shells, resulting in a larger ionic radius. The positive charge is spread over a larger volume, resulting in a lower charge density and thus weaker ion-dipole attractions with water dipoles.

❌ Examiner Trap Warnings

Never say "calcium atom": Mentioning calcium atoms or calcium molecules forfeits M1. You must specifically refer to the Ca²⁺ ion.
Don't just mention "weaker bonds": Be precise about what is being attracted. Specify that the ion attracts water or the lone pair / δ⁻ on oxygen.
Shielding alone is ignored: Explaining solely in terms of shielding will not gain credit for the size difference.

Mark Scheme:
• M1: Ca²⁺ (ion) bigger / lower charge to size ratio (than Mg²⁺ ). [Allow converse for Mg²⁺. Do NOT accept Ca²⁺ atom/molecule. Allow more shells.]
• M2: Weaker attraction/bond to ( O^δ- in) water.

Topics

Physical Chemistry · 3.1.4 Energetics · 3.1.8 Thermodynamics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.