Calculations Using Born-Haber Cycles

A Born-Haber cycle is an application of Hess's law: it breaks the formation of an ionic solid from its elements into a series of separate, measurable enthalpy steps. Because lattice enthalpy cannot be measured directly, the cycle lets you calculate it indirectly, or use it to check whether a compound behaves as a purely ionic model predicts. Before working through calculations, it helps to be confident with the individual terms involved, covered in lattice energy, atomisation and electron affinity.

The key idea is that enthalpy is a state function. If you go from elements in their standard states all the way to the solid ionic compound by two different routes, the total enthalpy change must be the same for both routes. One route is the single-step enthalpy of formation, \(\Delta H_f\). The other route is the sum of every individual step in the Born-Haber cycle. Setting the two routes equal to each other gives you an equation you can solve for whichever term is unknown.

For a simple compound like sodium chloride, the cycle usually needs these terms:

• Enthalpy of formation, \(\Delta H_f\): elements to compound.
• Enthalpy of atomisation, \(\Delta H_{at}\): converting an element into gaseous atoms (for a diatomic gas, this is half the bond enthalpy).
• Ionisation energy, \(IE\): removing electrons from a gaseous atom or ion (always endothermic).
• Electron affinity, \(EA\): adding an electron to a gaseous atom or ion (usually exothermic for a first electron, but a second electron affinity is endothermic because it is added to a negatively charged ion).
• Lattice enthalpy of formation, \(\Delta H_{latt}\): gaseous ions coming together to form the solid lattice (strongly exothermic).

The diagram below shows the route from the elements \(Na(s)\) and \(Cl_2(g)\) to solid \(NaCl(s)\), broken into individual steps, alongside the single direct step of formation.

Using the values shown in the diagram, the indirect route through the five separate steps must have the same total enthalpy change as the direct route, \(\Delta H_f\). Written as an equation:

\(\Delta H_f(NaCl) = \Delta H_{at}(Na) + IE_1(Na) + \Delta H_{at}(Cl) + EA(Cl) + \Delta H_{latt}(NaCl)\)

Step	Process	Value (kJ/mol)
1	Atomisation of Na(s)	+107
2	First ionisation energy of Na(g)	+496
3	Atomisation of Cl2(g) (half bond enthalpy)	+122
4	Electron affinity of Cl(g)	−349
5	Enthalpy of formation of NaCl(s)	−411

Substituting the known values and solving for the unknown lattice enthalpy:

\(-411 = 107 + 496 + 122 + (-349) + \Delta H_{latt}\)

\(-411 = 376 + \Delta H_{latt}\)

\(\Delta H_{latt}(NaCl) = -411 - 376 = -787 \) kJ mol\(^{-1}\)

The large negative value confirms strong electrostatic attraction between \(Na^+\) and \(Cl^-\) ions in the solid lattice.

The same Hess's law approach applies to any ionic compound, but the cycle needs extra steps whenever an ion carries more than one charge:

• For \(MgCl_2\), you must include both the first and second ionisation energies of magnesium, since forming \(Mg^{2+}\) removes two electrons, and you need two lots of chlorine atomisation and electron affinity because there are two chloride ions per formula unit.
• For \(MgO\), you need the first and second ionisation energies of magnesium and both the first electron affinity of oxygen (exothermic) and the second electron affinity (endothermic, since an electron is being added to an already negative \(O^-\) ion).

Lattice enthalpies calculated from Born-Haber cycles for compounds like \(MgCl_2\) and \(MgO\) are far more negative than for \(NaCl\), because higher ionic charges attract each other much more strongly. When the experimental (Born-Haber) lattice enthalpy differs noticeably from a value calculated purely from an ionic model, this points to some covalent character in the bonding, which is explored further under polarisability.

• Forgetting to halve the bond enthalpy when atomising a diatomic element such as \(Cl_2\) or \(O_2\).
• Mixing up the sign of an ionisation energy (always positive) with an electron affinity (usually negative, but positive for a second electron affinity).
• Not doubling steps that involve two of the same ion, such as the two chloride ions in \(MgCl_2\).
• Losing track of which direction each arrow points when rearranging the Hess's law equation to isolate the unknown term.

Related lessons

• Uk A Level Chemistry