Born-Haber Cycles
- A Born-Haber cycle is a specific application of Hess's Law for ionic compounds and enables us to calculate lattice enthalpy, 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
Start with your ionic solid
- 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) ΔHꝋat = +108 kJ mol-1
- The enthalpy of atomisation of chlorine is
½Cl2 (g) → Cl (g) ΔHꝋat = +121 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
Create the gaseous atoms
- Now that 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– ΔHꝋIE = +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) ΔHꝋEA = -364 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
Create 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) ΔHꝋf = -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 enthalpy, which is shown as the change from solid to gaseous ions. This means the arrow must point upwards. For sodium chloride, the equation is
NaCl (s) → Na+(g) + Cl–(g) ΔHꝋlatt
Drawing a Born-Haber cycle- Step 4
Complete the cycle
- The cycle is now complete
- The cycle is usually used to calculate the lattice enthalpy of an ionic solid, but can be used to find other enthalpy changes if you are given the lattice enthalpy
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) | ΔHꝋat |
| 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) | ΔHꝋat |
| Convert Cl (g) atoms into Cl- (g) ions | Cl (g) + e- → Cl-(g) | EA1 |
| Add up all the values to get ΔHꝋ1 |
ΔHꝋ1 |
|
| Apply to Hess's law to get ΔHꝋlat |
ΔHꝋlat |
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) | ΔHꝋat |
| 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+ + e- | IE2 |
| Convert O2 (g) molecules into O (g) atoms | ½O2 (g) → O(g) | ΔHꝋat |
| Convert O(g) atoms into O-(g) ions | O (g) + e- → O- (g) | EA1 |
| Convert O-(g) ions into O2-(g) ion | O-(g) + e- →O2- (g) | EA2 |
| Add up all the values to get ΔHꝋ1 |
ΔHꝋ1 |
|
| Apply to Hess's law to get ΔHꝋlat |
ΔHꝋlat |
Answer:
Exam Tip
You will not be asked to drawn an entire Born-Haber cycle from scratch but could be asked to complete a partially drawn one. When constructing Born-Haber cycles, the direction of the changes is important, but the relative size of the steps does not matter so don’t worry if the steps don’t correspond to the magnitude of the energy changes.
You don’t need to show the energy axis in a Born-Haber cycle, but you do need to show the electron(s) in the ionisation step otherwise you might lose marks in an exam.
