9.4:
Der Born-Haber-Kreisprozess
9.4:
Der Born-Haber-Kreisprozess
Lattice Energy
An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy (ΔHlattice) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions.
Here, the convention is used where the ionic solid is separated into ions, meaning the lattice energies will be endothermic (positive values). Another way is to use an equivalent, but opposite convention, wherein the lattice energy is exothermic (negative values) and described as the energy released when ions combine to form a lattice. Thus, make sure to confirm which definition is used when looking up lattice energies in another reference.
In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔHlattice = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl– ions. When one mole each of gaseous Na+ and Cl– ions form solid NaCl, 769 kJ of heat is released.
Determination of Lattice Energy of an Ionic Compound
It is not possible to measure lattice energies directly. However, the lattice energy can be calculated 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:
| Enthalpy of sublimation of Cs (s) | Cs (s) → Cs (g) | ΔH = ΔHs° = 76.5 kJ/mol | |
| One-half of the bond energy of F2 | ½ F2 (g) → F (g) | ΔH = ½ D = 79.4 kJ/mol | |
| Ionization energy of Cs (g) | Cs (g) → Cs+ (g) + e− | ΔH = IE = 375.7 kJ/mol | |
| Electron affinity of F | F (g) + e− → F− (g) | ΔH = EA = −328.2 kJ/mol | |
| Negative of the lattice energy of CsF (s) | Cs+ (g) + F− (g) → CsF (s) | ΔH = −ΔHlattice = ? | |
| Enthalpy of formation of CsF (s), add steps 1–5 | ΔH = ΔHf° = ΔHs°+ ½ D + IE + (EA) + (−ΔHlattice) Cs (s) + ½ F2 (g) → CsF (s) |
ΔH = −553.5 kJ/mol | |
- Consider the elements in their most common states, Cs (s) and F2 (g).
- The ΔHs° represents the conversion of solid cesium into a gas (sublimation), and then the ionization energy converts the gaseous cesium atoms into cations.
- In the next step, the energy required to break the F–F bond to produce fluorine atoms needs to be accounted for.
- Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity)
- Now, one mole of Cs cations and one mole of F anions is present. 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, ΔHf°, of the compound from its elements. In this case, the overall change is exothermic.
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 related to the interaction of just two atoms.
Lattice Energy as a Function of Ion Radius and Charge
The lattice energy of an ionic crystal increases rapidly as the charges of the ions increase, and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z+ and Z– = 1) is 1023 kJ/mol, whereas that of MgO (Z+ and Z– = 2) is 3900 kJ/mol (Ro = the interionic distance defined as the sum of the radii of the positive and negative ions, is nearly the same — about 200 pm for both compounds).
Different interatomic distances produce different lattice energies. For example, compare the lattice energy of MgF2 (2957 kJ/mol) to that of MgI2 (2327 kJ/mol), which demonstrates the effect on lattice energy of the smaller ionic size of F– as compared to I–.
Other Applications of the Born-Haber Cycle
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 ΔHs°, ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation ΔHf° are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.
This text is adapted from Openstax, Chemistry 2e, Section 7.5: Strengths of Ionic and Covalent Bonds.