Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

4.2.7Additional lessons

Born-Haber Cycle Calculations (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 79 of 172

📚 IB Chemistry HL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 5Setting Up the Calculation

4.2.7 Born-Haber Cycle Calculations (HL)
Sum rule
  • The enthalpy of formation equals the sum of atomisation, ionisation, electron affinity and the negative of the lattice enthalpy.
Lattice sign
  • Because IB lattice enthalpy is endothermic, it appears with a minus sign when written in the formation equation.
Rearranged form
  • ΔHlatt = ΔHatom(metal) + ΔHatom(non-metal) + IE + EA − ΔHf⦵ for a compound with 1+ and 1− ions.
Multiple ions
  • For MgCl₂ the cycle uses both ionisation energies of magnesium and two chlorine atoms, each atomised and given an electron.

Part 2 of 5Method for Any Unknown Term

1
Draw the cycle with all species and arrows so each term is placed in its correct direction.
2
Mark upward arrows as positive enthalpy changes and downward arrows as negative ones.
3
Write an equation stating that the direct route equals the indirect route going round the cycle.
4
Substitute the known values, multiplying by coefficients where more than one atom or ion is involved.
5
Solve for the unknown and check that its sign and size make chemical sense.

Part 3 of 5Comparing Experimental and Theoretical Values

Comparing Experimental and Theoretical Values
Theoretical model
  • Assumes perfectly spherical ions with purely electrostatic attraction, so it predicts lattice enthalpy from charges and radii.
Experimental value
  • From the Born-Haber cycle, includes any covalent contribution to the bonding.
Close agreement
  • Suggests bonding is almost purely ionic, as in NaCl.
Larger experimental value
  • Indicates extra covalent character, as expected when a small cation polarises a large anion.

Part 4 of 5Worked Example

Question

  • Calculate the lattice enthalpy of NaCl.
  • Data (kJ mol⁻¹): ΔHf⦵(NaCl) = −411;
    • atomisation Na = +107;
    • atomisation ½Cl₂ = +122;
    • first ionisation energy Na = +496;
    • electron affinity Cl = −349.

  1. Form-by-route equationΔHf⦵ = ΔHatom(Na) + ΔHatom(Cl) + IE + EA − ΔHlatt.
  2. Add the steps107 + 122 + 496 + (−349) = +376 kJ mol⁻¹.
  3. RearrangeΔHlatt = 376 − (−411) = 376 + 411 = +787 kJ mol⁻¹.

Answer Lattice enthalpy of NaCl = +787 kJ mol⁻¹

Part 5 of 5Exam Checklist & Tips

4.2.7 – Key Points for the Exam

  • Polarisation Distortion of the electron cloud of an anion by a nearby cation, adding covalent character.
  • Ionisation energy
    • Energy needed to remove electrons from gaseous atoms;
      • always positive.
  • Experimental lattice enthalpy Value obtained from a Born-Haber cycle using measured data.
  • Theoretical lattice enthalpy Value calculated from the pure ionic model using charges and radii.

Common Mistakes

  • Using the bond enthalpy of Cl₂ rather than half of it when only one chlorine atom is needed.
  • Giving the lattice enthalpy a negative sign while using the IB definition.
  • Leaving out the second ionisation energy for a 2+ cation.
  • Claiming that agreement between values proves 100 percent ionic bonding rather than very high ionic character.

Last updated:

\n