Topic 12 · page 12 of 17

Born-Haber calculations

Rearrange Hess's law to get lattice energy = formation enthalpy minus the sum of the other steps. Watch the multiples for MgCl₂, the second ionisation energy, and the endothermic second electron affinity of oxygen.

14 min read12.13

Read first: Building a Born-Haber cycle

Lattice energy from the cycle 12.13

Page 11 gave the rule: ΔfH\Delta_\text{f}H equals the sum of the steps on the staircase, including the lattice energy. Rearranged for the unknown:

lattice energy=ΔfH−(ΣΔatH+ΣIE+ΣEA) \text{lattice energy} = \Delta_\text{f}H - (\Sigma\Delta_\text{at}H + \Sigma\text{IE} + \Sigma\text{EA})

The bracket holds every step except the lattice energy, each with its own sign and each multiplied by the number of moles in the formula.

Worked exampleSodium chloride

ΔfH\Delta_\text{f}H = −411, ΔatH\Delta_\text{at}H(Na) = +107, ΔatH\Delta_\text{at}H(Cl) = +122, first IE(Na) = +496, first EA(Cl) = −349 (all kJ mol⁻¹).

  1. Sum of the steps = 107 + 122 + 496 + (−349) = +376
  2. Lattice energy = −411 − 376 = −787 kJ mol⁻¹

Magnesium chloride: multiples and a second step 12.13

MgCl₂ has two chlorine atoms and a 2+ metal ion. Compared with NaCl:

An enthalpy level diagram for magnesium chloride, drawn to scale. From zero at Mg solid plus Cl2 gas the arrows go up 148, up 244, up 738 and up 1451, reaching 2581 at Mg 2 plus gas plus two electrons plus 2 Cl gas, then down 698 to Mg 2 plus gas plus 2 Cl minus gas at 1883. A downward arrow of minus 2524 labelled lattice energy reaches the solid MgCl2 at minus 641. A long arrow from the elements to the solid is labelled delta H f, minus 641.
Born-Haber cycle for MgCl₂, drawn to scale. The staircase reaches 2581 kJ mol⁻¹ above the elements before the lattice energy brings it down.

Worked exampleMagnesium chloride

ΔfH\Delta_\text{f}H = −641; ΔatH\Delta_\text{at}H(Mg) = +148; ΔatH\Delta_\text{at}H(Cl) = +122; IE(Mg) = +738 and +1451; first EA(Cl) = −349 (kJ mol⁻¹).

  1. Sum of the steps = 148 + (2 × 122) + 738 + 1451 + (2 × (−349))
  2. = 148 + 244 + 738 + 1451 − 698 = +1883
  3. Lattice energy = −641 − 1883 = −2524 kJ mol⁻¹

Magnesium oxide: an endothermic second electron affinity 12.13

Oxygen needs two electrons. The first electron affinity is exothermic (−141), but the second is endothermic (+798), because the electron is forced onto an ion that is already negative.

Worked exampleMagnesium oxide

ΔfH\Delta_\text{f}H = −602; ΔatH\Delta_\text{at}H(Mg) = +148; ΔatH\Delta_\text{at}H(O) = +249 per mole of O atoms; IE(Mg) = +738 and +1451; EA(O) = −141 and +798 (kJ mol⁻¹).

  1. Sum of the steps = 148 + 249 + 738 + 1451 + (−141) + 798
  2. = +3243
  3. Lattice energy = −602 − 3243 = −3845 kJ mol⁻¹

The lattice energy of MgO is much more exothermic than that of NaCl (−787) or MgCl₂ (−2524). Page 15 explains why: higher charges and smaller ions.

Working backwards 12.13

The cycle is one equation, so any single unknown can be found, not just the lattice energy.

Worked exampleAn unknown formation enthalpy

Find ΔfH\Delta_\text{f}H of KCl from: lattice energy −718; ΔatH\Delta_\text{at}H(K) = +89; ΔatH\Delta_\text{at}H(Cl) = +122; first IE(K) = +419; first EA(Cl) = −349 (kJ mol⁻¹).

  1. ΔfH\Delta_\text{f}H = 89 + 122 + 419 + (−349) + (−718)
  2. = 281 − 718 = −437 kJ mol⁻¹

Worked exampleAn unknown electron affinity

Find the first electron affinity of chlorine from the NaCl cycle: ΔfH\Delta_\text{f}H = −411; lattice energy = −787; ΔatH\Delta_\text{at}H(Na) = +107; ΔatH\Delta_\text{at}H(Cl) = +122; first IE(Na) = +496 (kJ mol⁻¹).

  1. ΔfH\Delta_\text{f}H = 107 + 122 + 496 + EA + (−787)
  2. −411 = −62 + EA, so EA = −411 + 62 = −349 kJ mol⁻¹

Check yourself

All values in kJ mol⁻¹.

  1. Calculate the lattice energy of KCl: ΔfH\Delta_\text{f}H = −437; ΔatH\Delta_\text{at}H(K) = +89; ΔatH\Delta_\text{at}H(Cl) = +122; first IE(K) = +419; first EA(Cl) = −349.
  2. Find the first ionisation energy of sodium from the NaCl cycle, given ΔfH\Delta_\text{f}H = −411; lattice energy = −787; ΔatH\Delta_\text{at}H(Na) = +107; ΔatH\Delta_\text{at}H(Cl) = +122; first EA(Cl) = −349.
  3. Calculate ΔfH\Delta_\text{f}H of MgCl₂ from lattice energy = −2524 and the steps on this page.
  4. Explain why the second electron affinity of oxygen is endothermic.
  5. Both MgO and MgCl₂ contain Mg²⁺. Suggest why the lattice energy of MgO (−3845) is much more exothermic than that of MgCl₂ (−2524).
  6. A student leaves out the "× 2" for chlorine in the MgCl₂ cycle. What value does the student get for the lattice energy, and by how much is it wrong?

Answers cover these until you have tried the questions

  1. Sum = 89 + 122 + 419 − 349 = +281. Lattice energy = −437 − 281 = −718 kJ mol⁻¹.
  2. ΔfH\Delta_\text{f}H = 107 + 122 + IE + (−349) + (−787), so −411 = −907 + IE. IE = −411 + 907 = +496 kJ mol⁻¹.
  3. ΔfH\Delta_\text{f}H = 148 + 244 + 738 + 1451 − 698 + (−2524) = 1883 − 2524 = −641 kJ mol⁻¹.
  4. The second electron is added to O⁻, which is already negatively charged. It is repelled, so energy has to be supplied, making the process endothermic.
  5. In MgO the anion is O²⁻, with charge 2− and a smaller radius than Cl⁻. The attraction between 2+ and 2− ions that are close together is far stronger than between 2+ and 1− ions, so more energy is released when the lattice forms.
  6. Without the doubling, the sum = 148 + 122 + 738 + 1451 − 349 = 2110, so the lattice energy = −641 − 2110 = −2751 kJ mol⁻¹. This is 227 kJ mol⁻¹ too exothermic (−2751 compared with the correct −2524).