AQA A-Level Chemistry Paper 1, 2019: Question 1

9 marks · Medium difficulty · State/Explain/Numerical

Complete a Born–Haber cycle for caesium iodide, calculate the enthalpy of atomisation of iodine, evaluate the bonding character, and determine the feasibility of decomposition using Gibbs free energy.

Practise this question

Question

The question presents an incomplete Born-Haber cycle for caesium iodide with two blank energy levels to fill in. Table 1 lists standard enthalpy changes: atomisation of Cs (+79 kJ/mol), first ionisation energy of Cs (+376 kJ/mol), electron affinity of I (-314 kJ/mol), lattice formation enthalpy of CsI (-585 kJ/mol), and enthalpy of formation of CsI (-337 kJ/mol). Questions 01.1 to 01.3 ask to fill the missing species, calculate the enthalpy of atomisation of iodine, and deduce bonding type compared to the perfect ionic model (-582 kJ/mol). Question 01.4 provides Table 2 with standard entropies for CsI(s) (130 J K⁻¹ mol⁻¹), Cs(s) (82.8 J K⁻¹ mol⁻¹), and I₂(s) (117 J K⁻¹ mol⁻¹) to show the decomposition of CsI is not feasible at 298 K.
Question text

01 Figure 1 shows an incomplete Born–Haber cycle for the formation of caesium iodide.

The diagram is not to scale.

Figure 1

Table 1 gives values of some standard enthalpy changes.

Table 1

Name of enthalpy change ∆Ho/ kJ mol–1

Enthalpy of atomisation of caesium +79

First ionisation energy of caesium +376

Electron affinity of iodine –314

Enthalpy of lattice formation of caesium iodide –585

Enthalpy of formation of caesium iodide –337

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

species on each of the two blank lines.

[2 marks]

01.2 Use Figure 1 and the data in Table 1 to calculate the standard enthalpy of

atomisation of iodine.

[2 marks]

Standard enthalpy of atomisation of iodine kJ mol-1

01.3 The enthalpy of lattice formation for caesium iodide in Table 1 is a value obtained

by experiment.

The value obtained by calculation using the perfect ionic model is –582 kJ mol–1

Deduce what these values indicate about the bonding in caesium iodide.

[1 mark]

01.4 Use data from Table 2 to show that this reaction is not feasible at 298 K

1 o –1

CsI(s) → Cs(s) + I2(s) ∆H = +337 kJ mol

Table 2

CsI(s) Cs(s) I2(s)

So / J K–1 mol–1 130 82.8 117

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for question 01: 01.1 awards 1 mark for Cs+(g) + e- + I(g) on the top line and 1 mark for Cs(s) + 1/2 I2(s) on the lower line. 01.2 awards 2 marks for setting up the Born-Haber expression and calculating +107 kJ mol⁻¹. 01.3 gives 1 mark for stating the bonding is almost purely/perfectly ionic. 01.4 awards 4 marks: M1 for calculating ΔS = +11.3 J K⁻¹ mol⁻¹, M2 for stating ΔG = ΔH - TΔS, M3 for unit conversion, and M4 for obtaining ΔG = +334 kJ mol⁻¹ (or +334000 J mol⁻¹) showing non-feasibility.

Question Answers Additional Comments/Guidelines Mark

Top line Cs+(g) + e– + I(g)

01.1 1 1

Lower line Cs(s) + 2I2(s)

79 + x + 376 –314 = –337 +585 1

01.2 -1 Allow I mark for -107 (kJmol-1)

So enthalpy change = 107 (kJmol ) 1

Allow answer to 2sf or more

(Almost/Mostly) purely/ perfectly ionic If ionic not mentioned, allow no/little covalent

bonding/character

01.3 1

Penalise references to atoms/molecules

Ignore electronegativity

1 –1 –1 M1 Correct entropy change value 1

M1 ΔS = [(82.8 + 2 x 117) –130 ] = 11.3 (J K mol )

M2 equation or equation with numbers 1

M2 ΔG = ΔH – TΔS

01.4 –3 M3 for converting units:

M3 ΔG = 337 - 298× 11.3 x 10 OR 337000 – 298 x 11.3 -1 -1 -1 1

∆S into kJK mol or ∆H into Jmol

–1 –1 M4 answer with correct units

M4 ΔG = (+)334 kJ mol or 334000 J mol Any negative answer loses M4 1

How to answer it

Thermodynamics & Born–Haber Cycles: Caesium Iodide

📌 What This Question Tests

This question assesses mastery of Year 2 Physical Chemistry thermodynamics:

  • Born–Haber cycle construction: Identifying intermediate species, ionization, atomisation, and state symbols.
  • Enthalpy calculations via Hess's Law: Calculating an unknown enthalpy (atomisation of iodine) using cycle data.
  • Ionic model comparison: Comparing theoretical (electrostatic/point-charge) vs. experimental lattice enthalpies to deduce bonding nature.
  • Gibbs Free Energy (ΔG = ΔH − TΔS): Calculating standard entropy change (ΔS), executing unit conversions (J to kJ), and evaluating reaction feasibility.
Part 01.1 • 2 Marks

Completing the Born–Haber Cycle Diagram

Writing the formulas, including state symbols, of the appropriate species on each blank line

✅ Correct Answers

Top blank line:

Cs⁺(g) + e⁻ + I(g)

Lower blank line:

Cs(s) + ½I₂(s)

💡 Key Knowledge

  • Lower Line: Represents the elements in their standard states at 298 K, 100 kPa. Iodine exists naturally as diatomic solid crystals: ½I₂(s) .
  • Top Line: Reached after atomisation of both elements AND the 1st ionisation of caesium. Caesium loses an electron to form a gaseous cation: Cs⁺(g) + e⁻ , while iodine remains atomic gas I(g) awaiting electron affinity.

🧠 Exam Technique

  • Always follow the single change per arrow:
    • From Cs(g) + I(g) up to top line involves 1st IE of Cs: caesium forms Cs⁺(g) + e⁻ .
    • The arrow coming down from the top line shows electron affinity of I, consuming the free electron to make I⁻(g) .
  • Double-check state symbols on every single entity: (s) vs (g) .

❌ Common Errors

  • Omitting the free electron ( + e⁻ ) on the top line.
  • Writing iodine as liquid or gas in standard state (iodine is I₂(s) ).
  • Writing I(s) instead of ½I₂(s) on the elements line.
Mark Scheme: 1 mark for Cs⁺(g) + e⁻ + I(g) ; 1 mark for Cs(s) + ½I₂(s) . State symbols essential.
Part 01.2 • 2 Marks

Calculating the Standard Enthalpy of Atomisation of Iodine

Using the Born–Haber cycle energy balance to find ΔH°at(I)

📐 Step-by-Step Calculation

Apply Hess's Law across the cycle: Clockwise route = Anticlockwise route (or equate Formation to the sum of all other steps).

1 State the cycle relationship:
ΔH°f(CsI) = ΔH°at(Cs) + ΔH°at(I) + 1st IE(Cs) + EA(I) + ΔH°latt form(CsI)

2 Substitute known values into the equation:
−337 = +79 + ΔH°at(I) + 376 + (−314) + (−585)

3 Collect and simplify numerical terms:
79 + 376 − 314 − 585 = −444
−337 = ΔH°at(I) − 444

4 Solve for the unknown enthalpy:
ΔH°at(I) = −337 + 444 = +107 kJ mol⁻¹

🧠 Exam Technique

Sense-check your value: Atomisation involves breaking bonds (covalent bonds in ½I₂(s) plus solid lattice dispersion forces). Atomisation is strictly endothermic, so the value must be positive.

❌ Common Errors

  • Sign reversal: Concluding with −107 kJ mol⁻¹ loses the final accuracy mark (only 1/2 awarded).
  • Multiplying by 2: This question asks for atomisation producing 1 mole of I(g), not bond dissociation enthalpy of I₂. No multiplication by 2 is needed.
Mark Scheme:
• M1: Correct cycle expression: 79 + x + 376 − 314 = −337 + 585 (or equivalent rearrangement)
• M2: +107 (kJ mol⁻¹) [Allow −107 for 1 mark; accept 2 or more sig figs].
Part 01.3 • 1 Mark

Deducing Bonding from Experimental vs Theoretical Lattice Enthalpy

Comparing Experimental (−585 kJ mol⁻¹) and Perfect Ionic Model (−582 kJ mol⁻¹)

✅ Correct Answer

The bonding is almost completely purely ionic (or "has little to no covalent character").

💡 Key Knowledge

  • The perfect ionic model assumes ions are point charges and completely spherical with no distortion of electron clouds.
  • When the experimental value is very close to the theoretical value (here, a tiny difference of 3 kJ mol⁻¹), the assumption holds true: no significant polarisation occurs.
  • Both Cs⁺ and I⁻ are large ions with low charge densities, so polarisation is negligible.

❌ Common Misconceptions & Examiner Traps

  • Referring to atoms or molecules: Penalised immediately. CsI consists of ions in a giant lattice, not molecules.
  • Saying "it is covalent": Incorrect. A large discrepancy (e.g. in AgCl or Al₂O₃) indicates covalent character; a tiny difference indicates purely ionic character.
  • Discussing electronegativity: The question specifically directs you to use the enthalpy values provided. Mentioning electronegativity alone gains no credit.
Mark Scheme: (Almost/mostly) purely/perfectly ionic. [If ionic is not explicitly stated, allow "no / little covalent bonding/character". Penalise references to "atoms" or "molecules"].
Part 01.4 • 4 Marks

Feasibility Calculation: Gibbs Free Energy (ΔG)

Reaction: CsI(s) → Cs(s) + ½I₂(s)   |   ΔH° = +337 kJ mol⁻¹ at 298 K

📐 Step-by-Step Calculation

1 Calculate entropy change (ΔS):
ΔS = ΣS°(products) − ΣS°(reactants)
Take careful account of the stoichiometric coefficient ½ for I₂(s):
ΔS = [S°(Cs(s)) + ½ × S°(I₂(s))] − S°(CsI(s))
ΔS = [82.8 + (½ × 117)] − 130
ΔS = [82.8 + 58.5] − 130 = 141.3 − 130 = +11.3 J K⁻¹ mol⁻¹

2 Recall Gibbs Free Energy equation:
ΔG = ΔH − TΔS

3 Convert units (Crucial Step):
Convert ΔS to kJ K⁻¹ mol⁻¹: 11.3 / 1000 = +0.0113 kJ K⁻¹ mol⁻¹
(Alternatively, convert ΔH to J mol⁻¹: 337 000 J mol⁻¹)

4 Calculate ΔG at 298 K and conclude:
ΔG = 337 − (298 × 0.0113) = 337 − 3.3674 = +333.6 kJ mol⁻¹ (or +334 kJ mol⁻¹)
Since ΔG > 0 (positive), the reaction is not feasible at 298 K.

🧠 Exam Technique

  • Write the formula first: Writing ΔG = ΔH − TΔS guarantees Mark 2 even if arithmetic errors follow.
  • Always state units with the final ΔG: Either kJ mol⁻¹ or J mol⁻¹ . Leaving units blank loses Mark 4.

❌ Common Errors

  • Stoichiometry blunder: Forgetting the ½ for I₂(s) in ΔS calculation gives ΔS = +69.8, losing M1.
  • The 1000× Unit Trap: Adding or subtracting J directly with kJ (e.g. 337 − 298 × 11.3). This is the single most common error in A-Level thermodynamics.
  • Negative result: Any calculation yielding a negative ΔG cannot score M4 because the question explicitly asks you to show the reaction is not feasible.
Mark Scheme:
• M1: ΔS = [(82.8 + ½ × 117) − 130] = +11.3 (J K⁻¹ mol⁻¹)
• M2: ΔG = ΔH − TΔS (stated in words or symbols)
• M3: Converting units: ΔG = 337 − 298 × (11.3 × 10⁻³) OR 337000 − 298 × 11.3
• M4: ΔG = +334 kJ mol⁻¹ (or +334000 J mol⁻¹ ) with units. Any negative answer loses M4.

Topics

Physical Chemistry · 3.1.8 Thermodynamics · 3.1.3 Bonding

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.