AQA A-Level Chemistry Paper 1, June 2024: Question 8
14 marks · Medium difficulty · State/Explain/Numerical
Answer questions on thermodynamics, Born–Haber cycles, lattice enthalpy calculations, and feasibility of reactions involving sodium compounds.
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Question text
08 This question is about enthalpy changes.
08.1 Theoretical values for enthalpies of lattice dissociation can be calculated using a
perfect ionic model.
State the meaning of the term perfect ionic model.
[1 mark]
08.2 Enthalpies of lattice dissociation can also be obtained from Born–Haber cycles.
Figure 3 shows an incomplete Born–Haber cycle for the formation of sodium oxide.
Figure 3
Complete Figure 3 by writing formulas, including state symbols, of the appropriate
species on each of the two blank lines.
[2 marks]
08.3 Table 4 shows some enthalpy changes.
Table 4
Enthalpy change ΔH / kJ mol–1
Enthalpy of atomisation of oxygen +248
Enthalpy of atomisation of sodium +109
Enthalpy of formation of sodium oxide –416
First ionisation energy of sodium +494
First electron affinity of oxygen –142
Second electron affinity of oxygen +844
Use the data in Table 4 to calculate the enthalpy of lattice dissociation of
sodium oxide.
[2 marks]
Enthalpy of lattice dissociation kJ mol–1
08.4 Explain why the second electron affinity of oxygen has a positive value.
[1 mark]
08.5 Explain why the enthalpy of lattice dissociation for sodium oxide is greater than the
enthalpy of lattice dissociation for sodium chloride.
[2 marks]
08.6 Sodium chloride dissolves in water.
Table 5 shows some more enthalpy changes.
Table 5
Enthalpy change ΔH / kJ mol–1
Enthalpy of hydration for Cl– ions –364
Enthalpy of hydration for Na+ ions –406
Enthalpy of lattice dissociation for NaCl +771
Use the data in Table 5 to calculate the enthalpy of solution for sodium chloride.
[2 marks]
Enthalpy of solution27 kJ mol–1
08.7 Give a reason why data books do not contain a value for the enthalpy of solution of
sodium oxide.
[1 mark]
08.8 Calculate the temperature, in °C, at which this reaction becomes feasible.
1 –1
*26* NaCl(s) → Na(s) + Cl2(g) ΔH = +411 kJ mol
ΔS = +90.1 J K–1 mol–1
[3 marks]
Temperature °C
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark
(Ions are) point charges Do not accept atoms or molecules in answer
Or Allow no polarisation of ions
(Ions are) perfect spheres 1
08.1
Or (1 x AO1)
No covalent character
2 Na+ (g) + 2 e– + O(g)
08.2 1
2 Na(s) + O2(g) (2 x AO2)
–416 + x = 248 + (2 × 109) + (2 × 494) – 142 + 844
08.3 –1
enthalpy of lattice dissociation = (+) 2572 (kJ mol ) –2572 (kJ mol–1) scores 1 mark (2 x AO2)
O– repels the electron (being added) Allow negative ion repels electron 1
08.4
(1 x AO1)
Oxide ions
Ignore electronegativity
M1 have higher (negative) charge
Or
smaller size 2
08.5
Or (2 x AO3)
higher charge density/higher charge:size ratio
(than chloride ions)
M2 stronger attraction between (O2– and Na+/oppositely charged)
ions
enthalpy of solution = 771 – 406 – 364 2
08.6 -1
–1 Allow 1 mark for -1 (kJmol ) (2 x AO2)
= (+)1 (kJ mol )
It reacts with water Do not accept – It dissolves in water
08.7 Or
(1 x AO3)
It reacts to form (a solution of) NaOH
M1 T = ΔH/ΔS
M2 T = –3 = 4562 (K)
90.1×10 3
08.8 M3 = M2 – 273 (1 x AO1,
M3 T = 4562 – 273 = 4289 (°C) M3: Allow 4290 (°C)
2 x AO2)
How to answer it
Enthalpy Changes, Born–Haber Cycles & Feasibility
📋 What this question tests
This question comprehensively assesses AQA Year 2 Physical Chemistry (Thermodynamics):
- The perfect ionic model and its physical assumptions.
- Completing and applying Born–Haber cycles to calculate lattice dissociation enthalpy.
- Explaining trends in electron affinities and factors influencing lattice enthalpy (ionic charge & ionic radius).
- Calculating enthalpy of solution using hydration and lattice dissociation enthalpies.
- Chemical reactivity of metal oxides with water versus dissolution.
- Applying Gibbs Free Energy (ΔG = ΔH − TΔS) to determine the temperature of feasibility (including unit conversion and temperature scale conversion to °C).
Theoretical Lattice Enthalpy: The Perfect Ionic Model
Definition & Core Assumptions [1 mark]
✅ Acceptable Answers
- Ions are point charges
- Ions are perfect spheres
- There is no covalent character (or purely/100% ionic)
- Allow: No polarisation of ions
❌ Common Errors & Traps
- Writing "atoms" or "molecules" instead of ions (immediately disqualifies the mark).
- Vague phrasing such as "the bonds are strong" or "it has ionic bonds".
Completing the Born–Haber Cycle for Na₂O
Filling Missing Species and State Symbols [2 marks]
✅ Correct Missing Levels
Upper missing line:
2 Na⁺(g) + 2 e⁻ + O(g)
(Forms after the first ionisation of both Na atoms; O is still neutral gas)
Lower missing line (baseline elements in standard states):
2 Na(s) + ½ O₂(g)
(The elements in their standard states at 298 K, 100 kPa)
🧠 Exam Technique: Reading the Arrows
- Stoichiometry: Sodium oxide is Na₂O. You need 2 Na and 1 O. Every species prior to lattice dissociation must account for 2 sodiums!
- State Symbols: Must be included and correct. Omitting (s) or (g) loses the mark.
- Notice the arrow going down from the upper line: it represents adding 1 electron to O(g) to make O⁻(g), which leaves 1 free electron ( e⁻ ). Therefore, the level above must have 2 e⁻ !
Calculation: Lattice Dissociation Enthalpy of Na₂O
Born–Haber Cycle Energy Balance [2 marks]
📐 Step-by-Step Calculation
By Hess's Law, going clockwise equals going anticlockwise, or:
ΔfH(Na₂O) + ΔLH = 2×ΔatH(Na) + 2×IE₁(Na) + ΔatH(O) + EA₁(O) + EA₂(O)
- Identify required stoichiometry:
• Atomisation of Na: 2 × (+109) = +218 kJ mol⁻¹
• 1st Ionisation Energy of Na: 2 × (+494) = +988 kJ mol⁻¹
• Atomisation of oxygen: +248 kJ mol⁻¹ (gives 1 mol of O(g))
• 1st Electron Affinity of O: −142 kJ mol⁻¹
• 2nd Electron Affinity of O: +844 kJ mol⁻¹ - Set up the equation:
−416 + ΔLH = 248 + (2 × 109) + (2 × 494) + (−142) + 844
−416 + ΔLH = 248 + 218 + 988 − 142 + 844
−416 + ΔLH = +2156 - Solve for lattice dissociation enthalpy (ΔLH):
ΔLH = +2156 − (−416) = +2572 kJ mol⁻¹
❌ Common Traps
- Forgetting to multiply by 2: Neglecting to double the atomisation or ionisation energy of Na.
- Sign Error: Lattice dissociation enthalpy is endothermic ( +2572 ). If you give −2572 (formation), you lose 1 mark.
✅ Final Answer
+2572 kJ mol⁻¹
Note: −2572 scores 1 mark out of 2.
Second Electron Affinity of Oxygen
Explaining Why ΔH is Endothermic (+844 kJ mol⁻¹) [1 mark]
💡 The Chemical Principle
O⁻(g) + e⁻ → O²⁻(g)
The first electron affinity forms a negative oxide ion ( O⁻ ). Adding a second negatively charged electron requires overcoming electrostatic repulsion between two negative species.
✅ Model Answer
The negative O⁻ ion repels the incoming electron (being added).
(Also allowed: "negative ion repels the electron")
Comparing Lattice Dissociation Enthalpies: Na₂O vs NaCl
Ionic Size and Charge Factors [2 marks]
✅ Required Points (2 Marks)
- Mark 1 (Anion comparison): Oxide ion (O²⁻) has a higher negative charge OR is smaller than the chloride ion (Cl⁻) (or oxide has a greater charge density).
- Mark 2 (Attraction): Results in stronger electrostatic attraction between O²⁻ and Na⁺ ions (compared to Na⁺ and Cl⁻).
❌ Common Mistakes
- Comparing cations: Both compounds contain Na⁺; talking about sodium loses credit.
- Using "electronegativity": The mark scheme explicitly states "Ignore electronegativity". Lattice enthalpy is governed by ionic radii and charges, not electronegativity differences!
- Failing to mention comparative terms: say "stronger attraction", not just "attraction".
Enthalpy of Solution of NaCl
Enthalpy Cycle Calculation [2 marks]
📐 Step-by-Step Calculation
The enthalpy of solution links lattice dissociation enthalpy and enthalpies of hydration:
ΔsolH = ΔLH (dissociation) + ΣΔhydH
- Identify the values from Table 5:
• Lattice dissociation enthalpy: +771 kJ mol⁻¹
• Hydration of Na⁺: −406 kJ mol⁻¹
• Hydration of Cl⁻: −364 kJ mol⁻¹ - Substitute into formula:
ΔsolH = +771 + (−406) + (−364)
ΔsolH = 771 − 770 = +1 kJ mol⁻¹
✅ Final Answer
+1 kJ mol⁻¹
(Sign is required. If sign omitted or −1 given, max 1 mark awarded)
❌ Sign Warning
Lattice dissociation is breaking the lattice ( +771 ). Hydration is forming ion-dipole bonds (exothermic, negative ). Do not flip the sign of lattice dissociation if it is already given as dissociation!
Why Data Books Lack ΔsolH for Na₂O
Dissolution vs Chemical Reaction [1 mark]
✅ Correct Reason
- Sodium oxide reacts with water (to form NaOH / hydroxide ions).
- Na₂O(s) + H₂O(l) → 2 NaOH(aq)
❌ What NOT to Write
- "It dissolves in water" — specifically rejected in the mark scheme! Dissolution is a physical process; Na₂O chemically reacts.
- "It is insoluble" — incorrect, it reacts vigorously.
Calculation: Feasibility Temperature
Gibbs Free Energy & Unit Conversions [3 marks]
📐 Step-by-Step Calculation
NaCl(s) → Na(s) + ½ Cl₂(g) ΔH = +411 kJ mol⁻¹, ΔS = +90.1 J K⁻¹ mol⁻¹
- Condition for feasibility:
A reaction becomes feasible when ΔG ≤ 0 . At the transition point:
ΔG = ΔH − TΔS = 0 &implies; T = ΔH / ΔS - Convert units of ΔS to kJ K⁻¹ mol⁻¹ (or ΔH to J mol⁻¹):
ΔS = 90.1 / 1000 = 0.0901 kJ K⁻¹ mol⁻¹
(or ΔH = 411 × 1000 = 411,000 J mol⁻¹) - Calculate Temperature in Kelvin (T in K):
T = 411 / (90.1 × 10⁻³) = 4561.598... K ≈ 4562 K - Convert Kelvin to Celsius (°C):
Temperature in °C = T(K) − 273 = 4561.6 − 273 = 4288.6 ≈ 4289 °C
(Allow 4290 °C)
✅ Final Answer
4289 °C (or 4290 °C)
❌ Critical Traps to Avoid
- Forgetting to convert to °C: Stopping at 4562 K is the most frequent blunder. The question specifies "in °C"!
- Unit mismatch between ΔH and ΔS: ΔH is in kJ whereas ΔS is in J. Forgetting the factor of 1000 gives 4.56 K.
Topics
Physical Chemistry · Inorganic Chemistry · 3.1.8 Thermodynamics · 3.1.3 Bonding · 3.2.4 Properties of Period 3 Elements and Their Oxides
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.