Born-Haber Cycle Calculations
- Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice enthalpy (ΔHlatꝋ) by applying Hess’s law and rearranging:
ΔHꝋf = ΔHꝋat + ΔHꝋat + IE + EA - ΔHꝋlat
- If we simplify this into three terms, this makes the equation easier to see:
- ΔHꝋlat
- ΔHꝋf
- ΔHꝋ1 (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:
ΔHꝋf = ΔHꝋ1 - ΔHꝋlat
- So, if we rearrange to calculate the lattice enthalpy, the equation becomes
ΔHꝋlat = - ΔHꝋf + ΔHꝋ1
- When calculating the ΔHlatꝋ, 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 enthalpy 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 chloride ions, i.e. 2Cl-
Worked example
Calculating the lattice enthalpy of KCl
Given the data below, calculate the ΔHꝋlat of potassium chloride (KCl).
ΔHꝋat (kJmol-1) |
IE / EA (kJmol-1) | |
K | +90 | +418 |
Cl | +122 | -349 |
ΔHꝋf (kJmol-1) |
||
KCl | -437 |
Answer:
- Step 1: Construct the Born-Haber cycle
- Step 2: Applying Hess’ law, the lattice enthalpy of KCl is:
ΔHꝋlat = -ΔHꝋf + ΔHꝋ1
ΔHꝋlat = -ΔHꝋf + [(ΔHꝋat K) + (ΔHꝋat Cl) + (IE1 K) + (EA1 Cl)]
- Step 3: Substitute in the numbers:
ΔHꝋlat = -(-437) + [(+90) + (+122) + (+418) + (-349)] = 718 kJ mol-1
Worked example
Calculating the lattice enthalpy of MgO
Given the data below, calculate the of ΔHꝋlat magnesium oxide of magnesium oxide (MgO)
ΔHꝋat (kJmol-1) |
IE1 / EA1(kJmol-1) | IE1 / EA1(kJmol-1) | |
Mg | +148 | +736 | +1450 |
O | +248 | -142 | +770 |
ΔHꝋf (kJmol-1) |
|||
MgO | -602 |
Answer:
- Step 1: Construct the Born-Haber cycle
- Step 2: Applying Hess’ law, the lattice enthalpy of MgO is:
ΔHꝋlat = -ΔHꝋf + ΔHꝋ1
ΔHꝋlat = -ΔHꝋf + [(ΔHꝋat Mg) + (ΔHꝋat O) + (IE1 Mg) + (IE2 Mg) + (EA1 O) + (EA2 O)]
- Step 3: Substitute in the numbers:
ΔHꝋlat = -(-602) + [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]
= 3812 kJ mol-1