Enthalpy and Bond Strength

Lattice Energy

Lattice energies can be calculated by adding the energy changes involved in breaking and forming bonds as well as the energy changes that occur during changes of state.

Ionic compounds do not have covalent bonds or covalently bonded molecules. An ionic compound forms a lattice structure instead. The energy needed to form a lattice structure (which always has a negative value) is called lattice energy. Often, it is not possible to experimentally measure the energy change involved in the formation of the lattice directly.

German scientists Max Born and Fritz Haber developed a method, called the Born-Haber cycle, to calculate the enthalpy of formation of a lattice structure. The Born-Haber cycle relies on Hess's law and formation enthalpies. Consider how the lattice energy of sodium chloride (NaCl) can be calculated using the Born-Haber cycle.

Recall that bond enthalpy is defined as the energy needed to break a bond. Similarly, lattice energy is the energy needed to form a lattice, and the energy needed to break the lattice will have the opposite sign but will be equal in magnitude. Determining the lattice energy of sodium chloride (NaCl) begins with determining the enthalpy change associated with breaking the bonds between sodium ions and chloride ions.
NaCl(s)Na+(g)+Cl(g){\rm{NaCl}}(s)\rightarrow{\rm{Na}}^{+}(g)+{\rm{Cl}}^{-}(g)
ΔH\Delta H for this change is ΔHlattice\Delta H_{\rm{lattice}}. The first step of the cycle involves the standard enthalpy of formation (ΔHf\Delta H_{f}^\circ) for NaCl. The enthalpy of formation for NaCl is –411.2 kJ/mol.
Na(s)+12Cl2(g)NaCl(s)ΔHf=411.2kJ{\rm{Na}}(s)+{\textstyle\frac12}{\rm{Cl}}_{2}(g)\rightarrow{\rm{NaCl}}(s)\;\;\;\;\;\Delta {H_{f}^\circ}=-411.2\;{\rm{kJ}}
The reverse reaction is required:
NaCl(s)Na(s)+12Cl2(g)ΔH=411.2kJ{\rm{NaCl}}(s)\rightarrow{\rm{Na}}(s)+{\textstyle\frac12}{\rm{Cl}}_{2}(g)\;\;\;\;\;\;\Delta H=411.2\;{\rm{kJ}}
Sodium in this reaction is a solid metal, so it must be converted to gaseous sodium. Sodium gas has an enthalpy of formation of 107.3 kJ/mol.
Na(s)Na(g)ΔHf=107.3kJ{\rm{Na}}(s)\rightarrow{\rm{Na}}(g)\;\;\;\;\;\Delta {H_{f}^\circ}=107.3\;{\rm{kJ}}
Chlorine is a diatomic molecule. It is already a gas, so it does not have to be converted. However, the diatomic molecule must be broken into chlorine atoms. This involves another enthalpy change:
12Cl2(g)Cl(g)ΔHf=121.7kJ{\textstyle\frac12}{\rm{Cl}}_{2}(g)\rightarrow{\rm{Cl}}(g)\;\;\;\;\;\Delta {H_{f}^\circ}=121.7\;{\rm{kJ}}
Breaking a lattice produces ions, not atoms. Therefore, the next step involves producing ions from the atoms. For sodium this is the first ionization energy of the element:
Na(g)Na+(g)+eΔH=496kJ{\rm{Na}}(g)\rightarrow{\rm{Na}}^{+}(g)+{\rm{e}}^{-}\;\;\;\;\;\Delta H=496\;{\rm{kJ}}
For chlorine, this change involves the electron affinity of chlorine:
Cl(g)+eCl(g)ΔH=349kJ{\rm{Cl}}(g)+{\rm{e}}^{-}\rightarrow{\rm{Cl}}^{-}(g)\;\;\;\;\;\Delta H=-349\;{\rm{kJ}}
Combine the results of each reaction:
NaCl(s)Na(s)+12Cl2(g)ΔH=411.2kJNa(s)Na(g)ΔHf=107.3kJ12Cl2(g)Cl(g)ΔHf=121.7kJNa(g)Na+(g)+eΔH=496kJCl(g)+eCl(g)ΔH=349kJ\begin{aligned}&{\rm{NaCl}}(s)\rightarrow{\rm{Na}}(s)+{\textstyle\frac12}{\rm{Cl}}_{2}(g) \;\;\;\;\;&&{\Delta H=411.2\;{\rm{kJ}}}\\&{\rm{Na}}(s)\rightarrow{\rm{Na}}(g)\;\;\;\;\;&&{\Delta {H_{f}^\circ}=107.3\;{\rm{kJ}}}\\&{\textstyle\frac12}{\rm{Cl}}_{2}(g)\rightarrow{\rm{Cl}}(g)\;\;\;\;\;&&{\Delta {H_{f}^\circ}=121.7\;{\rm{kJ}}}\\&{\rm{Na}}(g)\rightarrow{\rm{Na}}^{+}(g)+{\rm{e}}^{-}\;\;\;\;\;&&{\Delta H=496\;{\rm{kJ}}}\\&{\rm{Cl}}(g)+{\rm{e}}^{-}\rightarrow{\rm{Cl}}^{-}(g)\;\;\;\;\;&&{\Delta H={-}349\;{\rm{kJ}}}\end{aligned}
NaCl(s)+Na(s)+12Cl2(g)+Na(g)+Cl(g)+eNa(s)+12Cl2(g)+Na(g)+Cl(g)+Na+(g)+e+Cl(g)ΔHlattice=411.2kJ+107.3kJ+121.7kJ+496kJ349=787.2kJ\begin{aligned}&\overline{{\rm{NaCl}}(s)+{\rm{Na}}(s)+{\textstyle\frac12}{\rm{Cl}}_{2}(g)+{\rm{Na}}(g)+{\rm{Cl}}(g)+{\rm{e}}^{-}\rightarrow{\rm{Na}}(s)+{\textstyle\frac12}{\rm{Cl}}_{2}(g)+{\rm{Na}}(g)+{\rm{Cl}}(g)+{\rm{Na}}^{+}(g)+{\rm{e}}^{-}+{\rm{Cl}}^{-}(g)}\\&{\Delta H_{\rm{lattice}}=411.2\;{\rm{kJ}}+107.3\;{\rm{kJ}}+121.7\;{\rm{kJ}}+496\;{\rm{kJ}}-349=787.2\;{\rm{kJ}}}\end{aligned}
All terms except three cancel:
NaCl(s)Na+(g)+Cl(g)ΔHlattice=787.2kJ{\rm{NaCl}}(s)\rightarrow{\rm{Na}}^{+}(g)+{\rm{Cl}}^{-}(g)\;\;\;\;\;\Delta H_{\rm{lattice}}=787.2\;{\rm{kJ}}
Note that the Born-Haber cycle is an approximation, similar to bond enthalpies. It does not account for all the forces involved in formation of a lattice. The results obtained from the Born-Haber cycle are different from the actual values.