AQA A-Level Chemistry Paper 1, 2018: Question 1

6 marks · Medium difficulty · State/Explain/Numerical

Complete the Born–Haber cycle for magnesium oxide and calculate its enthalpy of lattice formation using given thermodynamic data.

Practise this question

Question

Figure 1 shows an incomplete Born–Haber cycle for the formation of MgO(s) from Mg(s) and 1/2 O2(g), with levels for atomisation labelled, but three intermediate energy levels empty before reaching Mg2+(g) + O2-(g). Table 1 provides enthalpy values in kJ mol-1: enthalpy of formation of MgO (-602), atomisation of Mg (+150), first IE of Mg (+736), second IE of Mg (+1450), bond dissociation of oxygen (+496), first EA of oxygen (-142), and second EA of oxygen (+844).
Question text

01 This question is about lattice enthalpies.

01.1 Figure 1 shows a Born–Haber cycle for the formation of magnesium oxide.

Complete Figure 1 by writing the missing symbols on the appropriate energy levels.

[3 marks]

Figure 1

01.2 Table 1 contains some thermodynamic data.

Table 1

Enthalpy change

/ kJ mol–1

Enthalpy of formation for magnesium oxide –602

Enthalpy of atomisation for magnesium +150

First ionisation energy for magnesium +736

Second ionisation energy for magnesium +1450

Bond dissociation enthalpy for oxygen +496

First electron affinity for oxygen –142

Second electron affinity for oxygen +844

Calculate a value for the enthalpy of lattice formation for magnesium oxide.

[3 marks]

Enthalpy of lattice formation kJ mol–1

Mark scheme

Show the mark scheme Mark scheme for 01.1 shows the three missing energy levels: Mg+(g) + e- + O(g), Mg2+(g) + 2e- + O(g), and Mg2+(g) + e- + O-(g), with 1 mark per correct level including state symbols. For 01.2, it shows the expression setting Delta f H equal to the sum of all cycle steps, leading to an enthalpy of lattice formation of -3888 or -3890 kJ mol-1.

Question Answers Additional Comments/Guidance Mark

One mark for each level with correct state symbols

Mg2+(g) + 2e– + O(g)

Mg2+(g) + e– + O–(g)

01.1

Mg+(g) + e – + O (g)

ΔfH = ΔaH (Mg) + ½ ΔBDH (O2) + Δ1st IEH (Mg) + Δ2nd IEH (Mg) +

Δ1st EAH (O) + Δ2nd EAH(O) + ΔLEH (MgO)

01.2 - 602 = 150 + (½ x 496) + 736 +1450 – 142 + 844 + Δ H (MgO)

LE 1

-1 Allow answers to 2sf or more

ΔLEH (MgO) = -3888 / -3890 (kJ mol ) 1 mark for +3888 or +3890 1

1 mark for -4136 or -4140 (not 496 x 1/2)

Total 6

How to answer it

Born–Haber Cycle & Lattice Formation Enthalpy of MgO

📌 What this question tests

This question assesses your ability to complete an energy level diagram for a Born–Haber cycle involving group 2 oxides, correctly account for successive ionisation energies and electron affinities, track electrons and state symbols, and use Hess's Law to calculate a value for lattice enthalpy of formation.

Question 01.1 • 3 Marks

Completing the Born–Haber Energy Levels for Magnesium Oxide

Identifying the intermediate species and electron transitions

✅ Correct Labels (from lowest to highest missing level)

  1. First missing level (going up from Mg(g) + O(g)):
    Mg⁺(g) + e⁻ + O(g) [1 Mark]
  2. Top missing level (going up again):
    Mg²⁺(g) + 2e⁻ + O(g) [1 Mark]
  3. Level after arrow pointing downwards:
    Mg²⁺(g) + e⁻ + O⁻(g) [1 Mark]
State symbols (g) are strictly required for every single species on all three levels. Missing or incorrect state symbols forfeit the mark for that level.

💡 Key Knowledge

  • 1st Ionisation Energy of Mg: Endothermic (arrow points up) → forms Mg⁺(g) + e⁻ . Oxygen remains unreacted as O(g) .
  • 2nd Ionisation Energy of Mg: Endothermic (arrow points up) → forms Mg²⁺(g) + 2e⁻ .
  • 1st Electron Affinity of Oxygen: Exothermic (arrow points down) because an electron is attracted to a neutral atom → forms O⁻(g) , leaving 1e⁻ unreacted.
  • 2nd Electron Affinity of Oxygen: Endothermic (arrow points up to Mg²⁺(g) + O²⁻(g) ) due to strong electrostatic repulsion between incoming electron and negative O⁻ ion.
Visual Walkthrough of Figure 1: Starting from Mg(g) + O(g) :
↑ Arrow up (1st IE of Mg) → write Mg⁺(g) + e⁻ + O(g)
↑ Arrow up (2nd IE of Mg) → write Mg²⁺(g) + 2e⁻ + O(g)
↓ Arrow down (1st EA of O, exothermic) → write Mg²⁺(g) + e⁻ + O⁻(g)
↑ Arrow up (2nd EA of O, endothermic) → leads directly to the given level Mg²⁺(g) + O²⁻(g)

🧠 Exam Technique

  • Account for electrons: Always keep track of lost/gained electrons. When Mg loses one electron, write + e⁻ . When it loses a second, write + 2e⁻ . When O takes one, the balance reduces back to + e⁻ .
  • Check direction of arrows: An arrow pointing downwards represents an exothermic enthalpy change (1st EA). Arrows pointing upwards represent endothermic changes.

❌ Common Errors

  • Forgetting the free electrons (e.g. writing just Mg⁺(g) + O(g) without the e⁻ ).
  • Omitting state symbols or writing (s) instead of (g) . All atomisation, ionisation, and electron affinity species must be gaseous.
  • Writing molecular oxygen O₂(g) instead of atomic oxygen O(g) in these upper steps.
Question 01.2 • 3 Marks

Calculation of Enthalpy of Lattice Formation for MgO

Applying Hess's Law using Born–Haber cycle data

📐 Step-by-Step Calculation

Step 1: Write down the Hess's Law relationship
Direct formation = sum of indirect pathway steps:
ΔfH = ΔaH(Mg) + ½ ΔdissH(O₂) + 1st IE(Mg) + 2nd IE(Mg) + 1st EA(O) + 2nd EA(O) + ΔLEH
[Mark 1]

Step 2: Substitute the numerical values into the equation
Watch the halogen/oxygen dissociation: only ½ mole of O=O bonds is broken to produce 1 mole of O(g):
½ × (+496) = +248 kJ mol⁻¹.
-602 = +150 + (½ × 496) + 736 + 1450 + (-142) + (+844) + ΔLEH
[Mark 2]

Step 3: Simplify and solve for ΔLEH
-602 = 150 + 248 + 736 + 1450 - 142 + 844 + ΔLEH
-602 = +3286 + ΔLEH
ΔLEH = -602 - 3286
ΔLEH = -3888 kJ mol⁻¹ (or -3890 kJ mol⁻¹)
[Mark 3]

✅ Final Answer

-3888 kJ mol⁻¹

(Also accepts -3890 kJ mol⁻¹ to 3 significant figures).

❌ Calculation Traps & Common Losses

  • Missing the ½ factor on Bond Dissociation Enthalpy: Using +496 instead of (½ × 496 = 248) gives -4136 kJ mol⁻¹ (scores max 1/3). Remember the formula of magnesium oxide is MgO, which needs only 1 atom of O from ½ O₂!
  • Sign error (+3888 instead of -3888): Lattice enthalpy of formation is strongly exothermic (always negative). An answer of +3888 scores only 1/3.
  • Sign slip on 1st EA: Note that 1st electron affinity is negative (-142) while 2nd electron affinity is positive (+844).

🧠 Reality Check

Lattice enthalpy of formation represents the formation of a giant ionic lattice from gaseous ions:
Mg²⁺(g) + O²⁻(g) → MgO(s)
Because bond-making is exothermic, this value must be negative. Furthermore, because both ions have a 2+/2- charge, the magnitude should be very large (roughly 4× the magnitude of a 1+/1- lattice like NaCl ~ -787 kJ mol⁻¹).

Topics

Physical Chemistry · 3.1.8 Thermodynamics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.