Calculating lattice enthalpies with Born-Haber cycles

Hess' law states that the overall enthalpy change for a reaction is independent of the route taken.

As lattice enthalpies cannot be measured directly, Born-Haber cycles are used to determine the enthalpy change for an alternative, indirect pathway.

Here's an example of a Born-Haber cycle for calculating the lattice enthalpy of sodium chloride (NaCl).

Worked example 1 - Calculating lattice enthalpy of sodium chloride

Given the following enthalpy changes, calculate the lattice enthalpy of sodium chloride (NaCl) using a hypothetical Born-Haber cycle:

Step 1: Apply Hess's Law for the Born-Haber cycle

The overall enthalpy change for the formation of NaCl can be calculated by summing the enthalpy changes for each step of the Born-Haber cycle.

Step 2: Substitution and correct evaluation

ΔH_latt(NaCl) = $-$ΔH_ea(Cl) $-$ ΔH_IE1(Na) $-$ ΔH_at(Cl) $-$ ΔH_at(Na) + ΔH_f

= $-$($-$349) $-$ 496 $-$ 122 $-$ 107 + ($-$411)

= 349 $-$ 496 $-$ 122 $-$ 107 $-$ 411

ΔH_latt(NaCl) = $-$787 kJ mol^-1

Therefore, the lattice enthalpy of sodium chloride is $-$787 kJ mol^-1.

Adapting Born-Haber cycles for different compounds

The Born-Haber cycle can be modified for various compounds.

For example, when calculating the lattice enthalpy of a group 2 compound like magnesium chloride (MgCl₂), extra steps are needed:

• Magnesium forms a 2+ ion, so the second ionisation energy of magnesium is included.
• There are 2 moles of chlorine per mole of MgCl₂, so the atomisation enthalpy of chlorine is doubled.

Similarly, when constructing a Born-Haber cycle for sodium oxide (Na₂O), additional steps are required:

• There are 2 moles of sodium per mole of Na₂O, so the atomisation enthalpy and first ionisation energy of sodium are doubled.
• Oxygen forms a 2- ion, so the second electron affinity of oxygen is included.

For compounds with ions that have charges greater than 1, additional ionisation energies and electron affinities are incorporated into the Born-Haber cycle, as demonstrated in the following examples.

Worked example 2 - Calculating lattice enthalpy of magnesium chloride

Given the following enthalpy changes, calculate the lattice enthalpy of magnesium chloride (MgCl₂) using a hypothetical Born-Haber cycle:

Step 1: Apply Hess's law for the Born-Haber cycle

The overall enthalpy change for the formation of MgCl₂ can be calculated by summing the enthalpy changes for each step of the Born-Haber cycle.

Step 2: Substitution and correct evaluation

$\mathrm{\Delta }{\text{H}}_{\text{latt}}({\text{MgCl}}_2)=-2\mathrm{\Delta }{\text{H}}_{\text{ea}}(\text{Cl})-\mathrm{\Delta }{\text{H}}_{\text{IE1}}(\text{Mg})-\mathrm{\Delta }{\text{H}}_{\text{IE2}}(\text{Mg})-2\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{Cl})-\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{Mg})+\mathrm{\Delta }{\text{H}}_{\text{f}}$

$=-(-698)-738-\mathrm{1,451}-244-148+(-641)$

$=698-738-\mathrm{1,451}-244-148-641$

$\mathrm{\Delta }{\text{H}}_{\text{latt}}({\text{MgCl}}_2)=-\mathrm{2,524}{\text{ kJ mol}}^{-1}$

Therefore, the lattice enthalpy of magnesium chloride is $-$2,524 kJ mol^-1.

Worked example 3 - Calculating lattice enthalpy of sodium oxide

Given the following enthalpy changes, calculate the lattice enthalpy of sodium oxide (Na₂O) using a hypothetical Born-Haber cycle:

Step 1: Apply Hess's law for the Born-Haber cycle

The overall enthalpy change for the formation of Na₂O can be calculated by summing the enthalpy changes for each step of the Born-Haber cycle.

Step 2: Substitution and correct evaluation

$\mathrm{\Delta }{\text{H}}_{\text{latt}}({\text{Na}}_2\text{O})=-\mathrm{\Delta }{\text{H}}_{\text{ea1}}(\text{O})-\mathrm{\Delta }{\text{H}}_{\text{ea2}}(\text{O})-2\mathrm{\Delta }{\text{H}}_{\text{IE1}}(\text{Na})-2\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{Na})-\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{O})+\mathrm{\Delta }{\text{H}}_{\text{f}}$

= $-(-$141) $-790-992-214-249+(-416)$

= $141-790-992-214-249-416$

ΔH_latt(Na₂O) = $-$2,520 kJ mol^-1

Therefore, the lattice enthalpy of sodium oxide is $-$2,520 kJ mol^-1.

Born-Haber cycles can calculate various enthalpy and energy values

Born-Haber cycles are versatile and can be used to determine any unknown enthalpy or energy value within the cycle. The process is similar to calculating lattice enthalpy.

For instance, here's an example of a Born-Haber cycle for calculating the enthalpy of formation of magnesium oxide (MgO).

Worked example 4 - Calculating the enthalpy of formation of magnesium oxide

Given the following enthalpy changes, calculate the enthalpy of formation (ΔH_f) of magnesium oxide (MgO) using a hypothetical Born-Haber cycle:

Step 1: Applying Hess's law for the Born-Haber cycle

The overall enthalpy change for the formation of MgO can be calculated by summing the enthalpy changes for each step of the Born-Haber cycle.

Step 2: Substitution and correct evaluation

$\mathrm{\Delta }{\text{H}}_{\text{f}}=\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{Mg})+\mathrm{\Delta }{\text{H}}_{\text{at}}(\text{O})+\mathrm{\Delta }{\text{H}}_{\text{IE1}}(\text{Mg})+\mathrm{\Delta }{\text{H}}_{\text{IE2}}(\text{Mg})+\mathrm{\Delta }{\text{H}}_{\text{ea1}}(\text{O})+\mathrm{\Delta }{\text{H}}_{\text{ea2}}(\text{O})+\mathrm{\Delta }{\text{H}}_{\text{latt}}$

= 148 + 249 + 738 + 1,451 + ($-141)+790+(-$3,791)

= 148 + 249 + 738 + 1,451 $-141+790-\mathrm{3,791}$

ΔH_f = $-$556 kJ mol^-1

Therefore, the enthalpy of formation of magnesium oxide is $-$556 kJ mol-¹.