Cambridge AS & A Level9701

Lattice energy and Born-Haber cycles

Chemistry 9701 Chapter Notes

What this chapter covers

Lattice energy and Born-Haber cyclesEnthalpies of solution and hydrationEntropy change, ΔSGibbs free energy change, ΔG
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In this chapter

  1. 1Defining Key Enthalpy Changes
  2. 2The Born-Haber Cycle
  3. 3Calculations using Born-Haber Cycles
  4. 4Factors Affecting Lattice Enthalpy

1. Defining Key Enthalpy Changes

To understand Born-Haber cycles, you first need to be fluent in the language of enthalpy changes. These definitions describe the specific energy changes that occur when we transform elements in their standard states into the gaseous ions needed to form an ionic lattice. Each definition is for one mole of substance under standard conditions (298 K and 100 kPa). Getting the state symbols right is crucial.

Lattice Enthalpy (formation): M⁺(g) + X⁻(g) → MX(s)

Enthalpy of Atomisation (metal): M(s) → M(g)

Enthalpy of Atomisation (non-metal): ½X₂(g) → X(g)

First Ionisation Energy: M(g) → M⁺(g) + e⁻

First Electron Affinity: X(g) + e⁻ → X⁻(g)

Key term

Lattice Enthalpy (ΔH_latt): The standard enthalpy change when one mole of a solid ionic compound is formed from its constituent gaseous ions.

Examiner insight

Examiners frequently test definitions. A perfect definition includes 'one mole', 'gaseous' (for IE, EA, LE), standard states, and correct state symbols in any accompanying equation.

Common pitfall

Forgetting that ionisation energies and electron affinities apply only to substances in the gaseous state. You must 'pay' the energy cost of atomisation first.

Worked example 13 marks

Write balanced chemical equations, including state symbols, for the following enthalpy changes:(a) The first ionisation energy of potassium.(b) The enthalpy of atomisation of bromine.(c) The second electron affinity of sulfur.

  1. 1

    Step 1 (a): First ionisation energy is the removal of one electron from each atom in one mole of gaseous atoms. For potassium (K), this is: K(g) → K⁺(g) + e⁻

  2. 2

    Step 2 (b): Enthalpy of atomisation is the formation of one mole of gaseous atoms from the element in its standard state. Bromine is Br₂(l) at standard state. So, the equation is: ½Br₂(l) → Br(g). Note: If starting from Br₂(g), it would be ½Br₂(g) → Br(g), which is half the bond enthalpy.

  3. 3

    Step 3 (c): Second electron affinity is adding an electron to each ion in one mole of gaseous 1- ions. For sulfur, this means starting with S⁻(g). The equation is: S⁻(g) + e⁻ → S²⁻(g)

Recap

  • Lattice enthalpy is the formation of a solid lattice from gaseous ions and is always exothermic.
  • Enthalpy of atomisation forms 1 mole of gaseous atoms from an element in its standard state and is always endothermic.
  • Ionisation energy is the removal of electrons from gaseous atoms and is always endothermic.
  • First electron affinity is often exothermic, but second and subsequent electron affinities are always endothermic.
  • State symbols (s), (l), (g) are not optional; they are essential for correct definitions.

Quick check

  1. Why is the lattice enthalpy of any ionic compound always an exothermic value?1 mark
  2. Write the equation for the enthalpy of atomisation of solid iodine, I₂(s).1 mark

2. The Born-Haber Cycle

Lattice enthalpy cannot be measured directly. We can't just get a flask of gaseous sodium and chloride ions and measure the heat released as they form a salt. Instead, we use an energy cycle called a Born-Haber cycle. It's a clever application of Hess's Law, which states that the total enthalpy change for a reaction is independent of the route taken. The cycle connects the standard enthalpy of formation (the 'direct' route) with a multi-step 'indirect' route that involves atomising elements, ionising them, and then combining the gaseous ions (lattice enthalpy). By knowing all other values, we can find the one we don't know – usually the lattice enthalpy.

ΔH_f = ΔH_at(metal) + ΣIE(metal) + ΔH_at(non-metal) + ΣEA(non-metal) + ΔH_latt

ΔH_latt = ΔH_f - (ΔH_at(metal) + ΣIE(metal) + ΔH_at(non-metal) + ΣEA(non-metal))

Key term

Born-Haber Cycle: An energy cycle relating the lattice enthalpy of an ionic compound to its standard enthalpy of formation and other enthalpy changes.

Examiner insight

Clarity is key. Always draw Born-Haber cycles as energy level diagrams with clear labels for each step, including the species present at each level and the name of the enthalpy change for each arrow.

Fun fact

The cycle is named after German scientists Max Born and Fritz Haber, who developed it in 1919. Haber is also famous for the Haber-Bosch process for making ammonia.

Worked example 14 marks

Draw a fully labelled Born-Haber cycle for the formation of sodium chloride (NaCl).

  1. 1

    Step 1: Start with the elements in their standard states at the zero-energy level: Na(s) + ½Cl₂(g).

  2. 2

    Step 2: Draw an arrow down to the ionic lattice, NaCl(s). Label this arrow as the enthalpy of formation, ΔH_f.

  3. 3

    Step 3: From the elements line, draw an arrow up for the atomisation of sodium: Na(s) → Na(g). The line is now Na(g) + ½Cl₂(g). Label the arrow ΔH_at[Na].

  4. 4

    Step 4: Draw another arrow up for the first ionisation energy of sodium: Na(g) → Na⁺(g). The line is now Na⁺(g) + ½Cl₂(g) + e⁻. Label the arrow IE₁[Na].

  5. 5

    Step 5: Draw another arrow up for the atomisation of chlorine: ½Cl₂(g) → Cl(g). The line is now Na⁺(g) + Cl(g) + e⁻. Label the arrow ΔH_at[Cl].

  6. 6

    Step 6: Now, draw an arrow down for the electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g). The line is now Na⁺(g) + Cl⁻(g). Label the arrow EA₁[Cl].

  7. 7

    Step 7: Finally, draw a large arrow down from the gaseous ions, Na⁺(g) + Cl⁻(g), to the ionic lattice, NaCl(s). This completes the cycle and represents the lattice enthalpy, ΔH_latt.

Recap

  • A Born-Haber cycle is a practical application of Hess's Law.
  • The cycle calculates an unknown enthalpy change, usually lattice enthalpy.
  • Upward arrows on the cycle represent endothermic changes (positive ΔH).
  • Downward arrows represent exothermic changes (negative ΔH).
  • The cycle starts with elements in standard states and ends with the ionic lattice.
  • The 'indirect route' must end up at the same energy level as the 'direct route'.

Quick check

  1. Which single enthalpy change in a Born-Haber cycle for NaCl corresponds to the 'direct route'?1 mark

3. Calculations using Born-Haber Cycles

The main purpose of a Born-Haber cycle is to calculate a value. The most common calculation is finding the lattice enthalpy. Using the principle of Hess's Law, the sum of enthalpy changes going 'up' the cycle must equal the sum of enthalpy changes going 'down'. A more formal way is to state that the direct route (formation) is equal to the indirect route (all other steps). Be very careful with signs (+/-) and stoichiometry, especially for compounds like MgCl₂ or K₂O where you need to multiply certain values by two.

Clockwise energies = Anticlockwise energies

ΔH_f = Σ(indirect route steps) + ΔH_latt

Key term

Stoichiometry: The quantitative relationship between reactants and products in a chemical reaction, which in this context means accounting for the number of moles of each ion in the formula.

Examiner insight

Show your full working. A correct final answer with no working may not get full marks. Lay out the calculation clearly, showing the formula used, the substitution of values, and the final result.

Common pitfall

Forgetting to multiply enthalpy values by the stoichiometric coefficients. For K₂O, students often forget to use 2 x ΔH_at[K] and 2 x IE₁[K].

Worked example 14 marks

Use the data below to calculate the lattice energy of potassium oxide, K₂O. Enthalpy of formation of K₂O: –361 kJ mol⁻¹ Enthalpy of atomisation of K: +89 kJ mol⁻¹ First ionisation energy of K: +418 kJ mol⁻¹ Enthalpy of atomisation of O: +249 kJ mol⁻¹ First electron affinity of O: –141 kJ mol⁻¹ Second electron affinity of O: +798 kJ mol⁻¹

  1. 1

    Step 1: Set up the Hess's Law equation for the cycle. ΔH_latt = ΔH_f - (all other steps).

  2. 2

    Step 2: Identify all the 'other steps' and account for stoichiometry. For K₂O, we need 2 moles of K atoms and 1 mole of O atoms. Steps are: 2x atomisation of K, 2x ionisation of K, 1x atomisation of O, 1x first EA of O, 1x second EA of O.

  3. 3

    Step 3: Substitute the values into the equation. Be careful with signs. Sum of other steps = (2 * ΔH_at[K]) + (2 * IE₁[K]) + (ΔH_at[O]) + (EA₁[O]) + (EA₂[O])

  4. 4

    Step 4: Calculate the sum of the other steps. Sum = (2 * 89) + (2 * 418) + (249) + (-141) + (798) Sum = 178 + 836 + 249 - 141 + 798 = +1920 kJ mol⁻¹

  5. 5

    Step 5: Calculate the lattice energy. ΔH_latt = ΔH_f - (Sum of other steps) ΔH_latt = -361 - (+1920)

  6. 6

    Step 6: Final answer. ΔH_latt = -2281 kJ mol⁻¹

Worked example 22 marks

The second electron affinity of oxygen is a large positive value (+798 kJ mol⁻¹). Explain why.

  1. 1

    Step 1: The second electron affinity is the process of adding an electron to a negative ion, in this case O⁻(g) to form O²⁻(g).

  2. 2

    Step 2: The incoming electron is being added to an already negatively charged O⁻ ion.

  3. 3

    Step 3: There is a strong electrostatic repulsion between the negative ion and the negative electron. Energy must be supplied to overcome this repulsion, making the process highly endothermic (positive enthalpy change).

Recap

  • Always start by writing the main equation: ΔH_latt = ΔH_f - (sum of other steps).
  • Check the chemical formula (e.g., MgCl₂) to determine stoichiometry.
  • Multiply atomisation and ionisation/electron affinity values by the correct stoichiometric coefficient.
  • Pay close attention to the signs of each enthalpy value you substitute.
  • The final lattice enthalpy value must be a large negative number.
  • You can also rearrange the formula to find any other unknown value, like the enthalpy of formation.

Quick check

  1. For the formation of MgCl₂, which two enthalpy values must be multiplied by 2?2 marks

4. Factors Affecting Lattice Enthalpy

The magnitude of the lattice enthalpy tells us how strong the electrostatic attraction is in an ionic lattice. A more negative (more exothermic) value means stronger bonding. Two key factors determine this strength: ionic charge and ionic radius. The attraction between ions is described by Coulomb's Law, which states that the force is proportional to the product of the charges and inversely proportional to the distance between them (related to their radii).

Key term

Charge Density: The ratio of an ion's charge to its volume, which is a key factor in determining the strength of ionic and intermolecular forces.

Examiner insight

When comparing lattice enthalpies, always state and compare both the ionic charges and the ionic radii of the ions involved. A complete answer addresses both factors, even if one is constant.

Common pitfall

Stating that one ion is smaller but failing to explain that this leads to a smaller internuclear distance and thus a stronger electrostatic force of attraction.

Worked example 13 marks

Explain why the lattice enthalpy of magnesium oxide (MgO, -3791 kJ mol⁻¹) is significantly more exothermic than that of sodium chloride (NaCl, -787 kJ mol⁻¹).

  1. 1

    Step 1: Identify the ions and their charges in each compound. MgO consists of Mg²⁺ and O²⁻ ions. NaCl consists of Na⁺ and Cl⁻ ions.

  2. 2

    Step 2: Compare the ionic charges. The ions in MgO have charges of +2 and -2, while the ions in NaCl have charges of +1 and -1.

  3. 3

    Step 3: Relate charge to electrostatic attraction. The force of attraction between ions is proportional to the product of their charges. For MgO, the product is (+2) * (-2) = -4. For NaCl, it is (+1) * (-1) = -1.

  4. 4

    Step 4: Conclude. The much greater product of charges in MgO leads to a far stronger electrostatic attraction between the ions in the lattice. This results in a much more exothermic (larger negative) lattice enthalpy.

Worked example 23 marks

Explain why the lattice enthalpy of lithium fluoride (LiF, -1030 kJ mol⁻¹) is more exothermic than that of potassium fluoride (KF, -821 kJ mol⁻¹).

  1. 1

    Step 1: Identify the ions and their charges. Both compounds contain the F⁻ anion and a Group 1 cation (Li⁺ and K⁺). The ionic charges are the same (+1 and -1 in both).

  2. 2

    Step 2: Compare the ionic radii of the cations. Li⁺ is in Period 2 and K⁺ is in Period 4. The Li⁺ ion is smaller than the K⁺ ion as it has fewer electron shells.

  3. 3

    Step 3: Relate ionic size to attraction. Because the Li⁺ ion is smaller, the distance between the centers of the positive and negative ions (the internuclear distance) is smaller in the LiF lattice.

  4. 4

    Step 4: Conclude. The smaller distance between ions in LiF results in a stronger electrostatic attraction, and therefore a more exothermic lattice enthalpy compared to KF.

Recap

  • Lattice enthalpy becomes more exothermic (stronger ionic bond) as ionic charge increases.
  • Lattice enthalpy becomes more exothermic (stronger ionic bond) as ionic radius decreases.
  • Ionic charge has a greater effect on lattice enthalpy than ionic radius.
  • Charge density is a useful concept combining both charge and size.
  • Higher charge density leads to stronger ionic bonding and more exothermic lattice enthalpy.

Quick check

  1. Which compound would you expect to have a more exothermic lattice enthalpy: MgCl₂ or CaCl₂? Explain your choice.2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define the term 'first ionisation energy' and write an equation for the first ionisation energy of calcium.3 marks
  2. The lattice enthalpy of caesium chloride (CsCl) is -657 kJ mol⁻¹. That of sodium chloride (NaCl) is -787 kJ mol⁻¹. Explain this difference in terms of the ions involved.3 marks
  3. Construct a fully labelled Born-Haber cycle for the formation of magnesium oxide, MgO.5 marks
  4. Given the following data, calculate the standard enthalpy of formation of silver chloride, AgCl. Enthalpy of atomisation of Ag(s): +285 kJ mol⁻¹ First ionisation energy of Ag(g): +731 kJ mol⁻¹ Enthalpy of atomisation of Cl₂(g): +122 kJ mol⁻¹ Electron affinity of Cl(g): -349 kJ mol⁻¹ Lattice enthalpy of AgCl(s): -905 kJ mol⁻¹3 marks
  5. Explain why the third ionisation energy of magnesium is significantly larger than its second ionisation energy.3 marks
  6. Predict, with reasoning, whether the lattice enthalpy of calcium sulfide (CaS) is more or less exothermic than that of potassium chloride (KCl).4 marks
  7. The experimental lattice enthalpy for sodium chloride is -787 kJ mol⁻¹. A theoretical value can also be calculated assuming a perfectly ionic model. The experimental value is often slightly different. Suggest a reason for this difference.2 marks
  8. Use the data to calculate the second electron affinity of oxygen. Lattice enthalpy of MgO: -3791 kJ mol⁻¹ Enthalpy of formation of MgO: -602 kJ mol⁻¹ Enthalpy of atomisation of Mg: +148 kJ mol⁻¹ First IE of Mg: +738 kJ mol⁻¹ Second IE of Mg: +1451 kJ mol⁻¹ Enthalpy of atomisation of O: +249 kJ mol⁻¹ First EA of O: -141 kJ mol⁻¹5 marks
  9. A student incorrectly draws a Born-Haber cycle for Al₂O₃. They use 2 x IE₁ for aluminium and 3 x EA₁ for oxygen. Identify and explain the two main errors in their choice of ionisation and electron affinity steps.4 marks
  10. Explain why it is necessary to supply energy to form a Mg²⁺ ion from a Mg atom, yet magnesium oxide, containing Mg²⁺ ions, is a stable compound that forms exothermically.3 marks

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