Calculations Using Born-Haber Cycles
- Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice energy (ΔHlattꝋ) by applying Hess’s law and rearranging:
ΔHfꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔHlattꝋ
- If we simplify this into three terms, this makes the equation easier to see:
- ΔHlattꝋ
- ΔHfꝋ
- ΔH1ꝋ (the sum of all of the various enthalpy changes necessary to convert the elements in their standard states to gaseous ions)
- The simplified equation becomes
ΔHfꝋ = ΔH1ꝋ + ΔHlattꝋ
So, if we rearrange to calculate the lattice energy, the equation becomesΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
- When calculating the ΔHlattꝋ, all other necessary values will be given in the question
- A Born-Haber cycle could be used to calculate any stage in the cycle
- For example, you could be given the lattice energy and asked to calculate the enthalpy change of formation of the ionic compound
- The principle would be exactly the same
- Work out the direct and indirect route of the cycle (the stage that you are being asked to calculate will always be the direct route)
- Write out the equation in terms of enthalpy changes and rearrange if necessary to calculate the required value
- Remember: sometimes a value may need to be doubled or halved, depending on the ionic solid involved
- For example, with MgCl2 the value for the first electron affinity of chlorine would need to be doubled in the calculation, because there are two moles of chlorine atoms
- Therefore, you are adding 2 moles of electrons to 2 moles of chlorine atoms, to form 2 moles of Cl- ions
Worked example: Calculating the lattice energy of KCl
Answer
- Step 1: The corresponding Born-Haber cycle is:
- Step 2: Applying Hess’ law, the lattice energy of KCl is:
ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ K) + (ΔHatꝋ Cl) + (IE1 K) + (EA1 Cl)]
- Step 3: Substitute in the numbers:
ΔHlattꝋ = (-437) - [(+90) + (+122) + (+418) + (-349)] = -718 kJ mol-1
Worked example: Calculating the lattice energy of MgO
Answer
- Step 1: The corresponding Born-Haber cycle is:
- Step 2: Applying Hess’ law, the lattice energy of MgO is:
ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ Mg) + (ΔHatꝋ O) + (IE1 Mg) + (IE2 Mg) + (EA1 O) + (EA2 O)]- Step 3: Substitute in the numbers:
ΔHlattꝋ = (-602) - [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]
= -3812 kJ mol-1
Examiner Tip
Make sure you use brackets when carrying out calculations using Born-Haber cycles as you may forget a +/- sign which will affect your final answer!