Chemistry » Chemical Bonding » Strengths of Ionic and Covalent Bonds

# The Born-Haber Cycle

## The Born-Haber Cycle

It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess’s law that breaks down the formation of an ionic solid into a series of individual steps:

• $$\text{Δ}{H}_{\text{f}}^{°},$$ the standard enthalpy of formation of the compound
• IE, the ionization energy of the metal
• EA, the electron affinity of the nonmetal
• $$\text{Δ}{H}_{s}^{°},$$ the enthalpy of sublimation of the metal
• D, the bond dissociation energy of the nonmetal
• ΔHlattice, the lattice energy of the compound

The figure below diagrams the Born-Haber cycle for the formation of solid cesium fluoride.

The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.

We begin with the elements in their most common states, Cs(s) and F2(g). The $$\text{Δ}{H}_{s}^{°}$$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F–F bond to produce fluorine atoms.

Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the y-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride.

The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $$\text{Δ}{H}_{\text{f}}^{°},$$ of the compound from its elements. In this case, the overall change is exothermic.

Hess’s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. The table below shows this for fluoride, CsF.

 Enthalpy of sublimation of Cs(s) $$\text{Cs}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cs}\left(g\right)$$ $$\text{Δ}H=\text{Δ}{H}_{s}^{°}=76.5\text{kJ/mol}$$ One-half of the bond energy of F2 $$\frac{1}{2}\phantom{\rule{0.2em}{0ex}}{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{F}\left(g\right)$$ $$\text{Δ}H=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}D=79.4\text{kJ/mol}$$ Ionization energy of Cs(g) $$\text{Cs}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cs}}^{\text{+}}\left(g\right)+{\text{e}}^{\text{−}}$$ $$\text{Δ}H=IE=375.7\text{kJ/mol}$$ Negative of the electron affinity of F $$\text{F}\left(g\right)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{F}}^{\text{−}}\left(g\right)$$ $$\text{Δ}H=\text{−}EA=-328.2\text{kJ/mol}$$ Negative of the lattice energy of CsF(s) $${\text{Cs}}^{\text{+}}\left(g\right)+{\text{F}}^{\text{−}}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CsF}\left(s\right)$$ $$\text{Δ}H=\text{−Δ}{H}_{\text{lattice}}=?$$ Enthalpy of formation of CsF(s), add steps 1–5 $$\begin{array}{l}\text{Δ}H=\text{Δ}{H}_{f}^{°}=\text{Δ}{H}_{s}^{°}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}D+IE+\left(-EA\right)+\left(-\text{Δ}{H}_{\text{lattice}}\right)\\ \text{Cs}\left(s\right)+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CsF}\left(s\right)\end{array}$$ $$\text{Δ}H=-553.5\text{kJ/mol}$$

Thus, the lattice energy can be calculated from other values. For cesium fluoride, using this data, the lattice energy is:

$$\text{Δ}{H}_{\text{lattice}}=\left(553.5+76.5+79.4+375.7+328.2\right)\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=1413.3\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$$

The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $$\text{Δ}{H}_{s}^{°},$$ ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation $$\text{Δ}{H}_{\text{f}}^{°}$$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.

Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds.

Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.

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