Calculations using Born-Haber cycles

In this lesson, we will learn:

• To use Born-Haber cycles to calculate the lattice energies of ionic compounds.
• To use Born-Haber cycles to calculate the solution enthalpies of ionic compounds.

• In the last two lessons we saw several new enthalpy terms for the individual steps that take place during a chemical process, such as the forming of an ionic compound or the dissolving of an ionic solid to give an aqueous solution.
  These terms can be combined when drawing a Born-Haber cycle, which applies Hess’s law that an enthalpy change from reactants to products is the same regardless of the route taken. This means we can calculate a few things:
• Even though we can’t calculate the lattice energy of an ionic solid directly, we can indirectly work it out by using enthalpy of formation, atomisation enthalpy, electron affinity and ionization energies of the ionic compound.
• If we do not know any one of the lattice energy, enthalpy change of hydration or enthalpy change of solution, we can use two of them to find the other out.
This lesson will show some worked examples of this.


• Worked example: Calculating lattice energy.
  The table below^1,2 shows some enthalpy terms for the ionic compound CaCl₂ in kJ mol^-1.



Enthalpy term	Value (kJ mol-1)
$\triangle H$at (Ca)	178
$\triangle H$at (Cl)	121
1st ionization energy, Ca	589.7
2nd ionization energy, Ca	1145
1st electron affinity, Cl	-348.7
$\triangle H$f (CaCl2)	-795.8


Use this data to construct a Born-Haber cycle and find the lattice energy of the compound CaCl₂.

Step 1: Begin your Born-Haber cycle with CaCl_{2 (s)}. As the complete, solid ionic lattice, it will be the lowest energy state in this chemical process. This state is connected to enthalpy of formation, $\triangle H$_f, and the lattice enthalpy $\triangle H$_lat so it is usually drawn as a large ‘platform’ at the bottom of the Born-Haber cycle diagram.

Step 2: Above the ionic lattice are the elements in their standard states: solid calcium and chlorine, a diatomic gas. This will be written Ca _(s) + Cl_{2 (g)}. The gap between the lattice and the standard state elements is $\triangle H$_f or the standard enthalpy of formation.
The direction of the arrow is important – the definition is to form one mole of product from the elements, so the arrow points downwards.

Step 3: Atomise the calcium metal. This costs energy, so it will be represented by an upwards pointing arrow from the elements to Ca _(s) and Cl_{2 (g)}.

Step 4: Form Ca²⁺ ions using the 1^st and 2^nd ionisation energies. These are both endothermic processes costing energy, so an upward arrow from Ca _(g) to Ca²⁺ _(aq) should be drawn. Your current “state” should be Ca²⁺ _(g) and Cl_{2 (g)}, which hasn’t been changed yet.

Step 5: Atomise the Cl₂ molecule. This is an endothermic process costing energy. Remember the definition of atomisation enthalpy is to form one mole of gaseous atoms. By atomizing one mole of Cl₂ we have formed two moles of Cl atoms. The value we have for $\triangle H$_at then must be multiplied by two. The current state of our reactants should contain Ca²⁺ _(g) and 2Cl _(g). Because the electron affinity of chlorine is negative, this is the highest energy state, or ‘platform’ in your Born-Haber diagram.

Step 6: Form 2Cl^- using the 1^st electron affinity. This is an exothermic step that releases energy, so there is a downward arrow from Ca²⁺ _(g) + 2Cl _(g) to Ca²⁺ _(g) + 2Cl^- _(g).

Step 7: From the previous step we now have the two oppositely charged ions in their gaseous state, ready to form the ionic solid. This step from Ca²⁺ _(g) and 2Cl^- _(g) to CaCl_{2 (s)} is the lattice enthalpy we are trying to work out.
To calculate lattice enthalpy, sum all the enthalpy terms together by following the route in these steps from CaCl_{2 (s)} to the gaseous ions. Going in the opposite direction of an enthalpy arrow means the sign must be reversed. In short, the enthalpy steps should be:
• - $\triangle H$_f (CaCl₂) for the change: CaCl_{2 (s)} $\,$→$\,$ Ca _(s) + Cl_{2 (g)}
• $\triangle H$_at (Ca) for the change: Ca _(s) + Cl_{2 (g)} $\,$→$\,$ Ca _(g) + Cl_{2 (g)}
• 1^st and 2^nd ionization of Ca for the change: Ca _(g) + Cl_{2 (g)} $\,$→$\,$ Ca²⁺ _(g) + Cl_{2 (g)}
• 2 x $\triangle H$_at (Cl) for the change: Ca²⁺ _(g) + Cl_{2 (g)} $\,$→$\,$ Ca²⁺ _(g) + 2Cl _(g)
• 2 x 1^st electron affinity of Cl for the change: Ca²⁺ _(g) + 2Cl _(g) $\,$→$\,$ Ca²⁺ _(g) + 2Cl^- _(g)
The calculation should be:
(-(-795.8) + 178 + (589.7 + 1145) + (2*121) + (2*-348.7)) = 2253.1 kJ mol^-1. See the image below:


• Worked example: Calculating enthalpy of solution

The tables below¹ shows the lattice energies and enthalpy of hydration of some common ions and compounds.



Compound	Lattice energy (kJ mol-1)
NaCl	787
NaBr	752
KCl	717
KBr	689




Cation	$\triangle H$hyd (kJ mol-1)
Li+	-520
Na+	-405
K+	-321
Anion	$\triangle H$hyd (kJ mol-1)
F-	-506
Cl-	-364
Br-	-337



Use this data to construct a Born-Haber cycle to find the enthalpy of solution, $\triangle H$_sol , for:
1. NaBr
2. KCl
Which is the more exothermic process?

First, we need to construct a Born-Haber cycle for the processes, walking through with NaBr.

Remember that $\triangle H$_sol is the sum of the lattice dissociation energy (the positive value) and the hydration enthalpy of the individual ions (which are always negative values). In steps:
• Start your Born-Haber cycle with the ionic solid, NaBr (s)
• Draw an up arrow to represent lattice dissociation of NaBr _(s). This will have a positive value of +752 kJ mol^-1.
• The next step in the energy cycle will be the dispersed gaseous ions, in the form of Na⁺ _(g) and Br^- _(g)
• These gaseous ions now make stabilising bonds with the solvent water molecules. This is represented by a downward arrow with the hydration enthalpies of Na⁺ and Br^- added together. As the compound is a 1:1 molar ratio, you will sum -405 kJ mol^-1 (Na⁺) and -337 kJ mol^-1 (Br^-) which results in -742 kJ mol^-1.
• This final step is the dissolved aqueous ions, Na⁺ _(aq) and Br^- _(aq). The gap between this step and the step with the solid ionic compound is the enthalpy of solution, $\triangle H$_sol. Sum the two terms you have already drawn by following the direction of the arrows to find this enthalpy change, which is:
  +752 – 742 = +10 kJ mol^-1.
The direction of the arrows is important. This +10 kJ mol^-1 value means that going from solid NaBr to dissolved aqueous Na⁺ and Br^- ions is slightly endothermic.
See below for a diagram:

For the KCl compound, following these same steps will give you an overall $\triangle H$_sol for KCl of +32 kJ mol^-1.

From this, we can say that dissolving one mole of NaBr is more exothermic (a more negative $\triangle H$_sol value) than dissolving one mole of KCl as shown by their solution enthalpies here.
• This finding that dissolving NaBr is more exothermic should be backed up by the solubility data. As a more spontaneous, exothermic process, NaBr should dissolve more readily at a given temperature than KCl, and it does.


¹ Source for enthalpy terms used on this page: ATKINS, P. W., & DE PAULA, J. (2006).?Atkins' Physical chemistry. Oxford, Oxford University Press.

² Source for some enthalpy of atomisation data: https://www.webelements.com/calcium/thermochemistry.html