Born-Haber Cycles (Edexcel A Level Chemistry): Revision Note
Exam code: 9CHO
Born-Haber Cycles
A Born-Haber cycle is a specific application of Hess’ Law for ionic compounds and enable us to calculate lattice energy which cannot be found by experiment
The basic principle of drawing the cycle is to construct a diagram in which energy increases going up the diagram
The basic principle of a Born-Haber cycle
The cycle shows all the steps needed to turn atoms into gaseous ions and from gaseous ions into the ionic lattice
The alternative route to the ionic lattice begins from the enthalpy of formation of the elements in their standard states
Drawing the cycle for sodium chloride
A good starting point is to draw the elements with their state symbols about a third of the way up the diagram
This is shown as the left hand side of the equation for the process indicated
The location is marked by drawing a horizontal bar or line which represents the starting energy level
Drawing a Born-Haber cycle step 1
Next, we need to create the gaseous ions
This is a two step process of first creating the gaseous atoms and then turning them into ions
Creating gaseous atoms is a bond breaking process, so arrows must be drawn upwards
It doesn't matter whether you start with sodium or chlorine
The enthalpy of atomisation of sodium is
Na (s) → Na (g) ΔHatꝋ = +108 kJ mol -1
The enthalpy of atomisation of chlorine is
½Cl2 (g) → Cl (g) ΔHatꝋ = +122 kJ mol -1
We can show the products of the process on the horizontal lines and the energy value against a vertical arrow connecting the energy levels
Drawing a Born-Haber cycle step 2 - creating the gaseous atoms
Now the ions are created
The sodium ion loses an electron, so this energy change is the first ionisation energy for sodium
Na (g) → Na+ (g) + e– ΔHieꝋ = +500 kJ mol-1
The change is endothermic so the direction continues upwards
The chlorine atom gains an electron, so this is electron affinity
Cl (g) + e– → Cl- (g) ΔHeaꝋ = -349 kJ mol-1
The exothermic change means this is downwards
The change is displaced to the right to make the diagram easier to read
Drawing a Born-Haber cycle step 3 - creating the gaseous ions
The two remaining parts of the cycle can now be completed
The enthalpy of formation of sodium chloride is added at the bottom of the diagram
Na (s) + ½Cl2 (g) → NaCl (s) ΔHfꝋ = -411 kJ mol -1
This is an exothermic change for sodium chloride so the arrow points downwards
Enthalpy of formation can be exothermic or endothermic, so you may need to show it above the elements ( and displaced to the right) for a endothermic change
The final change is lattice energy, which is usually shown as a formation. For sodium chloride the equation is
Na+(g) + Cl-(g) → NaCl (s) ΔHlattꝋ
Drawing a Born-Haber cycle step 4 - completing the cycle
The cycle is now complete
The cycle is usually used to calculate the lattice energy of an ionic solid, but can be used to find other enthalpy changes if you are given the lattice energy
Worked Example
Constructing a Born-Haber cycle for KCl
Construct a Born-Haber Cycle which can be used to calculate the lattice energy of potassium chloride
Step | Equation | Enthalpy change |
|---|---|---|
Convert K(s) atoms into K(g) atoms | K (s) → K (g) | ΔHat |
Convert K(g) atoms into K+(g) ions | K (g) → K+ (g) + e- | IE1 |
Convert Cl2(g) molecules into Cl(g) atoms | ½Cl2 (g) → Cl (g) | ΔHat |
Convert Cl(g) atoms into Cl-(g) ions | Cl (g) + e- → Cl- (g) | EA1 |
Add up all values to get ΔH1 | ΔH1 | |
Apply Hess's law to find ΔHlatt | ΔHlatt |
Answer
Worked Example
Constructing a Born-Haber cycle for MgO
Construct a Born-Haber Cycle which can be used to calculate the lattice energy of magnesium oxide
Step | Equation | Enthalpy change |
|---|---|---|
Convert Mg(s) atoms into Mg(g) atoms | Mg (s) → Mg (g) | ΔHat |
Convert Mg(g) atoms into Mg+(g) ions | Mg (g) → Mg+ (g) + e- | IE1 |
Convert Mg+(g) ions into Mg2+(g) ions | Mg+ (g) → Mg2+ (g) + e- | IE2 |
Convert O2(g) molecules into O(g) atoms | ½O2 (g) → O (g) | ΔHat |
Convert O(g) atoms into O-(g) ions | O (g) + e- → O- (g) | EA1 |
Convert O-(g) ions into O2-(g) ions | O- (g) + e- → O2- (g) | EA2 |
Add up all values to get ΔH1 | ΔH1 | |
Apply Hess's law to find ΔHlatt | ΔHlatt |
Answer
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 becomes
ΔHlattꝋ = 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 both the enthalpy of atomisation and 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 making 2 moles of Cl (g) atoms, and then adding 2 moles of electrons to them, to form 2 moles of Cl- ions
Doubling the electron affinity but forgetting to double the atomisation is the most common error in Born-Haber calculations
Worked Example
Calculating the lattice energy of KCl
Given the data below, calculate the ΔHlattꝋ of potassium chloride (KCl)
ΔHat / kJ mol-1 | IE / EA / kJ mol-1 | |
|---|---|---|
K | +90 | +418 |
Cl | +122 | -349 |
ΔHf [KCl] = -437 kJ mol-1
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
Given the data below, calculate the of ΔHlattꝋ magnesium oxide of magnesium oxide (MgO)
ΔHatꝋ / kJ mol-1 | IE1 or EA1 / kJ mol-1 | IE2 or EA2 / kJ mol-1 | |
|---|---|---|---|
Mg | +148 | +738 | +1451 |
O | +249.2 | -141.1 | +798 |
ΔHfꝋ (MgO) = -602 kJ mol-1
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) + (+249.2) + (+738) + (+1451) + (-141.1) + (+798)]
= -3845 kJ mol-1
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