$$\mathrm{\Delta }H=\mathrm{\Sigma }{D}_{bondsbroken}-\mathrm{\Sigma }{D}_{bondsformed}$$In this expression, the symbol Ʃ means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 7.3) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.

Consider the following reaction:

or

To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds are formed (bond energy = 432 kJ/mol), releasing 2 × 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:

[latex]\Delta H = \Sigma D_{bonds \; broken} - \Sigma D_{bonds \; formed}[/latex]

[latex]= \left[ D_{H-H} + D_{Cl-Cl} \right] - 2 \, D_{H-Cl}[/latex]

[latex]=  \left[ 436 \: kJ + 243\:  kJ \right] - 2 \left( 432 \: kJ \right) = -185 \: kJ[/latex]

This excess energy is released as heat, so the reaction is exothermic. Appendix G gives a value for the standard molar enthalpy of formation of HCl(g), ΔH_f°​, of –92.307 kJ/mol. Twice that value is –184.6 kJ, which agrees well with our answer for the formation of two moles of HCl.

 

Ionic Bond Strength and 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 (ΔH_lattice) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:

Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔH_lattice = 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.

The lattice energy ΔH_lattice of an ionic crystal can be expressed by the following equation (derived from Coulomb’s law, governing the forces between electric charges):

in which C is a constant that depends on the type of crystal structure; Z⁺ and Z^– are the charges on the ions; and R_o is the interionic distance (the sum of the radii of the positive and negative ions). Thus, 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 (R_o is nearly the same—about 200 pm for both compounds).

Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF₂ (2957 kJ/mol) to that of MgI₂ (2327 kJ/mol) to observe the effect on lattice energy of the smaller ionic size of F^– as compared to I^–.

 

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:

• ΔH_f​°      the standard enthalpy of formation of the compound
• IE         the ionization energy of the metal
• EA           the electron affinity of the nonmetal
• ΔH_s​°       the enthalpy of sublimation of the metal
• D              the bond dissociation energy of the nonmetal
• ΔH_lattice  the lattice energy of the compound

Figure 7.13 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.

.

We begin with the elements in their most common states, Cs(s) and F₂(g). ΔH_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, ΔH_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.

Table 7.4  Individual stepts for enthalpy of formation of CsF.

Reaction	Chemical equation	Reaction enthalpy
Enthalpy of sublimation of Cs(s)	Cs(s) → Cs(g)	ΔH = ΔHs​° = 76.5kJ/mol
One-half of the bond energy of F2	1/2 F2(g) → F(g)	ΔH = 1/2 D = 79.4kJ/mol
Ionization energy of Cs(g)	Cs(g) → Cs+(g) + e−	ΔH = IE = 375.7kJ/mol
Electron affinity of F	F(g) + e− → F−(g)	ΔH = EA = −328.2kJ/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	Cs(s) + 1/2 F2(g) → CsF(s)	ΔH = ΔHf° = ΔHs​° + 1/2 D + IE + EA + (−ΔHlattice) = −553.5 kJ/mol

By rearranging the final expression in Table 7.4 for the enthalpy of formation (ΔH_f°), we can solve for the lattice enthalpy in terms of other, known quantities:

ΔH_lattice = ΔH_s​° + 1/2 D + IE + EA – ΔH_f°

For cesium fluoride, using the data provided in Table 7.4, the lattice energy is:

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 ΔH_s​°, ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔH_lattice, and standard enthalpy of formation ΔH_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.