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5.4.3Physical Chemistry and Transition Elements

Calculations Using Born-Haber Cycles — OCR A Level Chemistry A H432 Revision Notes & Topical Questions

5 min read 5 parts Lesson 22 of 40

📚 OCR A Level Chemistry A H432 Hub / Topical Revision
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Part 1 of 5The Calculation Method

5.4.3 Calculations Using Born-Haber Cycles

Because the cycle is closed, the two routes must give the same total.

Equation
  • ΔHf = ΔHat(metal) + ΣIE + ΔHat(non-metal) + ΣEA + ΔH(LE), where each term includes any multiples needed.
Rearranged
  • ΔH(LE) = ΔHf − (ΔHat(metal) + ΣIE + ΔHat(non-metal) + ΣEA), which is the usual form when finding the lattice enthalpy.
Multipliers
  • Atomisation and electron affinity terms must be multiplied by the number of atoms or ions of the non-metal in the formula.
Signs
  • Take care to include the sign of each term, because several are negative and the final answer should be strongly exothermic.

Part 2 of 5Step-by-Step Approach

Always check the cycle before substituting values.

1
Sketch the cycle or list all stages so that no step, such as a second ionisation, is missed.
2
Identify the unknown quantity, which is often the lattice enthalpy but may be an electron affinity or the enthalpy of formation.
3
Write Hess's law as an equation with the unknown on one side and all known terms on the other.
4
Substitute values with signs and multipliers, keeping units in kJ mol⁻¹.
5
Check the answer is sensible: lattice enthalpies are large and negative, and Mg²⁺ compounds have bigger values than Na⁺ compounds.

Part 3 of 5Experimental versus Theoretical

Experimental versus Theoretical
Experimental value
  • The Born–Haber value uses measured enthalpies and so includes any covalent character in the bonding.
Theoretical value
  • A calculation assuming purely ionic bonding with perfect spherical ions gives a value that may differ from the experimental one.
Agreement
  • Close agreement suggests the compound is almost purely ionic, as in NaCl.
Disagreement
  • A larger difference points to covalent character, which arises when a small cation distorts a large, polarisable anion.

Part 4 of 5Worked Example

Question

  • Calculate the lattice enthalpy of NaCl using: ΔHf(NaCl) = −411, ΔHat(Na) = +107, IE₁(Na) = +496, ΔHat(½Cl₂) = +122, EA₁(Cl) = −349, all in kJ mol⁻¹.
  • Then calculate the lattice enthalpy of MgCl₂ using ΔHf = −641, ΔHat(Mg) = +148, IE₁ = +738, IE₂ = +1451, bond enthalpy Cl–Cl = +244, EA₁(Cl) = −349.

  1. NaClsum of other terms = 107 + 496 + 122 − 349 = +376.
  2. ΔH(LE) = −411 − 376 = −787 kJ mol⁻¹.
  3. MgCl₂148 + 738 + 1451 + 244 + 2(−349) = 148 + 738 + 1451 + 244 − 698 = +1883.
  4. ΔH(LE) = −641 − 1883 = −2524 kJ mol⁻¹.

Answer

  • NaCl: −787 kJ mol⁻¹;
    • MgCl₂: −2524 kJ mol⁻¹.

Part 5 of 5Exam Checklist & Tips

5.4.3 – Key Points for the Exam

  • Hess's law The overall enthalpy change is independent of the route taken.
  • Lattice enthalpy The enthalpy change when one mole of an ionic lattice forms from its gaseous ions.
  • Covalent character Partial sharing of electrons in a bond that is mainly ionic.
  • Polarisation Distortion of the electron cloud of an anion by a neighbouring cation.

Common Mistakes

  • Forgetting to multiply the electron affinity by two when the formula contains two anions.
  • Using the bond enthalpy of Cl₂ in the cycle for NaCl without halving it.
  • Leaving out the second ionisation energy for a 2+ cation.
  • Giving the lattice enthalpy a positive sign, when the formation of a lattice is exothermic.

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