5.2.1 (a,b) Lattice enthalpy, Born–Haber and related enthalpy cycles
Syllabus
(a) explanation of the term lattice enthalpy (formation of 1 mol of ionic lattice from gaseous ions, ∆HLE) and use as a measure of the strength of ionic bonding in a giant ionic lattice
{Definition required.}
(b) use of the lattice enthalpy of a simple ionic solid (i.e. NaCl, MgCl2) and relevant energy terms for:
(i) the construction of Born–Haber cycles
(ii) related calculations
{Relevant energy terms: enthalpy change of formation, ionisation energy, enthalpy change of atomisation and electron affinity.}
{Definition required for first ionisation energy and enthalpy change of formation only}
What does this mean?
You'll have learned definitions for the enthalpy change of formation and combustion in Year 12
The lattice enthalpy follows the same pattern:
"The enthalpy change when 1 mole of an ionic compound forms from its gaseous ions under standard conditions"
More specifically, this definition is Lattice Formation enthalpy
And so should be exothermic.
But, were you given Lattice Dissociation enthalpy - it would be the same but endothermic.
The lattice energy depends on the strength of the ionic bonds
So, comparing LiF to LiCl we would expect the Lattice enthalpy to be lower because the charges of the fluoride and chloride ions are both -1 but Fluoride ions are smaller and so can pack more closely to the Lithium ions
It would be harder to compare NaF to MgF2 or AlF3 - because it is not only the sizes of the ions that are changing but also the charges of the Metal ions and therefore the formulae and structure.
Constructing a Born-Haber cycle
Sodium Chloride
Step 1
Step one is forming one mole of the required ionic substance from its elements, in their standard states and in their usual formula.
In this case that means we must use Chlorine as Cl2 - even though it would be more convenient not to - because Chlorine doesn't exist naturally as individual Cl atoms.
That means that we only need half a mole of Chlorine.
Remember that we're not allowed to set up the reaction as 2Na + Cl2--> 2 NaCl - because the definition says we must form one mole of substance.
We can assume that all formation enthalpies are exothermic because they always are in A level questions even if not in reality!
Step 2
Next we start atomising the reactant elements.
It doesn't matter at all which one you choose to atomise first because, in reality, if you heated the mixture the processes would both start happening at the same time. Doing them sequentially is theoretical
In this case we have turned the one mole of Sodium metal (as a solid) into one mole of sodium ions in the gaseous state.
The A level examiner will want to see the state symbols.
Step 3
The next step is either to atomise the Chlorine or to ionise the Sodium by removing an electron.
The example below is for ionising Sodium
This energy change is the first ionisation energy (Δ 1st IEH) that you learned about in Year 12
Step 4
Having dealt with the Sodium, this example now atomises the Chlorine
This is the same process as breaking the Cl-Cl bond
But the Atomisation Energy (Δ AH) of Chlorine makes 1 mole of Cl atoms
While the bond dissociation energy of Chlorine breaks one mole of Cl-Cl bonds and so makes 2 moles of Cl atoms.
As a result - Bond dissociation energy of Chlorine = 2 x Atomisation energy of chlorine.
Examiners sometimes give you the bond dissociation, sometimes the Atomisation energy.
So, be careful!
Step 5
Now the Chloride ion must be made by gaining the electron that we removed from the Sodium
This is not the ionisation energy - that's for losing electrons.
This is the Electron Affinity Δ EAH
Step 6
Now that we have Sodium and Chlorine ions they can combine to make Sodium Chloride
This is Δ LEH - the Lattice Enthalpy
Calculation
Hess' Law states that enthalpy change is independent of route taken.
Δ fH (NaCl) = Δ atH (Na) + Δ ieH (Na) + Δ atH (Cl) + Δ EAH (Cl) + Δ LEH (NaCl)
You'll be given all but one of these values and asked to calculate the missing one.