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Calculating Lattice Energies

The lattice energy is the energy associated with the chemical reaction of gaseous ions combining to form 1 mole of a solid ionic compound. This value cannot be determined directly. However, related equations with energy values that can be determined directly can be summed to get the value for lattice energy. The key to doing these types of problems is to be able to write the chemical reactions correctly, which must include physical states. The products and reactants can be algebraically added, multiplied, canceled, and so on, to obtain the correct reaction. Whatever is done to the reaction is also done to the associated energy value.

>> Example 1

Write the chemical reactions associated with the following energy values. (You may want to refer to Chapter 3 for some of them and maybe review the naming rules.)

1. the lattice energy of calcium chloride
2. the ionization energy of potassium
3. the second ionization energy of strontium
4. the electron affinity of bromine atom
5. the sublimation of lithium
6. the lattice energy of silver iodide

Solution:

1. Lattice energy results from the gaseous ions forming a solid. The formula for calcium chloride is CaCl₂. This substance is made of the ions Ca²⁺ and Cl^–. Therefore the reaction is

   Ca²⁺(g) + 2 Cl^–(g) CaCl₂(s)

2. Ionization energy is the energy required to remove an ion from an atom in the gaseous state, so

   K(g) K⁺(g) + e^–

3. The second ionization energy is the energy required to remove a second electron from the gaseous cation.

   Sr⁺(g) Sr²⁺(g) + e^–

4. Electron affinity is the energy associated with a gaseous atom gaining an electron.

   Br(g) + e^– Br^–

5. Sublimation is a substance going from the solid state to the gaseous state.

   Li(s) Li(g)

6. Lattice energy is the result of gaseous ions creating an ionic solid. The formula for silver iodide is AgI, made of Ag⁺ and I^–.

   Ag⁺(g) + I^–(g) AgI(s)

When adding equations together, keep the reactants as reactants and the products as products. Any substance that is exactly the same (including both charge and physical state) on both the product and reactant side can be canceled. Remember that stoichiometric coefficients are not part of the formula, but denote the amount of the substance. Therefore if there are two atoms on the reactant side and one on the product, one atom on each side will cancel, leaving one atom on the reactant side.

>> Example 2

When the following reactions are all added together, what is the resulting reaction?

Li(s) Li(g)

Li(g) Li⁺(g) + e^–

Br(g) + e^– Br^–(g)

Li⁺(g) + Br^–(g) LiBr(s)

Br₂(l) 2 Br(g)

Solution:

Leaving all the reactants together and all the products together, the resulting equation is

Li(g) + Br(g) + e^– + Li⁺(g) + Br^–(g) + Li(s) + Br₂(l)
                     2 Br(g) + LiBr(s) + Br^–(g) + Li⁺(g) + e^– + Li(g)

Canceling all substances that are exactly the same (including stoichiometric coefficients), you get

Br(g) + Li(s) + Br₂(l) 2 Br(g) + LiBr(s)

Canceling the the same substances with different coefficients, you get

Li(s) + Br₂(l) Br(g) + LiBr(s)

If you change an equation (to get the desired results), change the associated energies as follows:

If you multiply by a factor, multiply the energy by the same factor.

If you reverse the equation, change the sign of the energy.

If you add equations, add the energies.

>> Example 3

What is the lattice energy of LiBr? The ionization energy (IE) of lithium is 520 kJ/mol, the vaporization energy (VE) of lithium is 134.7 kJ/mol, the VE for Br₂(l) is 15.46 kJ/mol, the energy of atomization (AE) of gaseous bromine (Br₂2 Br) is 111.7 kJ/mol, the electron affinity (EA) of bromine is –324 kJ/mol, the formation energy (FE) (Li(s) + 1/2 Br₂(l)  LiBr(s)) is –351.2 kJ/mol.

Solution:

Writing the equation for each energy value given,

IE = Li (g) Li⁺(g) + e^–

VE = Li(s) Li(g)

VE = Br₂(l) 2 Br₂(g)

AE = Br₂(g) 2 Br(g)

EA = Br(g) + e^– Br^–(g)

FE = Li(s) + 1/2 Br₂(l) LiBr(s)

The equations need to be rearranged to get Li⁺(g) + Br^–(g) LiBr(s).

A good way to begin is to move the substance to the position needed. This requires reversing the equations for ionization energy and electron affinity and keeping the formation energy the same.

IE:	Li+(g) + e– Li (g)	E = –520 kJ
EA:	Br–(g) Br(g) + e–	E = –324 kJ
FE:	Li(s) + 1/2 Br2(l) LiBr(s)	E = –351.2 kJ

To get rid of the products in the formation energy equation, reverse the vaporization energies. In addition, the vaporization energy for bromine should be halved.

VE:	Li(g) Li(s)	E = –134.7 kJ
VE:	1/2 Br2(g) 1/2 Br2(l)	E = –7.73 kJ

Adding these equations gives

Li⁺(g) + Br^–(g) + 1/2 Br₂(g) LiBr(s) + Br(g)

E = sum of values above = –1337 kJ

To get rid of the remaining bromine, you need to add in the reverse of one-half of the atomization equation.

Br(g) 1/2 Br₂(g)       E = –55.85 kJ

Adding that in gives the lattice energy of –1395 kJ/mol.

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