Tutorials

# What is Lattice Energy?

## Core Concepts

In this tutorial about lattice energy, we will cover its definition, relevant periodic table trends, factors that influence it, and how to calculate it.

## What is Lattice Energy?

During the formation of solid ionic compounds, electropositive metals react with electronegative nonmetals. Both the generation and dissolution of such compounds involve the concept of lattice energy, a type of potential energy expressed in units of kJ/mol. Lattice energy maintains the fixed positions of cations and anions within ionic compounds. We can further investigate this term in two different ways, depending on our perspective.

The key to understanding this concept lies in the crystalline structure of ionic compounds. Their strong, rigid composition enables interactions between each charged ion and its oppositely charged counterparts. These interactions involve large amounts of energy, explaining the high melting and boiling points characteristic of ionic compounds.

Lattice energy can be described as a certain quantity of energy is released when gaseous ions react during the formation of one mole of a solid ionic compound; however, it also describes the energy that facilitates the dissociation of one mole of a solid ionic compound into its constituent gaseous ions. Depending on our chosen definition, the lattice energy of a given ionic compound may either be a positive or negative value.

### Exothermic versus Endothermic

We can view lattice energies as either endothermic or exothermic processes depending on which definition we focus on. A process is exothermic when it releases energy. Our first definition, the formation of an ionic compound, involves exothermic lattice energy, corresponding to a negative value.

On the other hand, if we use the alternate definition, the dissolution of an ionic compound, the nature of the lattice energy value changes. Because this process requires energy, it falls into the endothermic category, corresponding to a positive value.

## Factors that affect Lattice Energy

#### 1. The charges held by the constituent ions (represented by the variables Q1 and Q2)

As we increase the ion charge variable, lattice energy increases. This means that ions with larger charge values will produce ionic compounds with greater lattice energies. In turn, ions possessing weaker charges decrease the lattice energies of their compounds.

#### 2. The distance between the constituent ions (represented by the variable R)

As we increase the distance variable, lattice energy decreases. Essentially, larger ions compose ionic compounds with smaller lattice energies due to the increased distance between them. Smaller ions produce larger lattice energies in their ionic compounds.

## Lattice Energy Trends

To summarize, lattice energy increases as we increase ion charge and decrease the distance. More specifically, it increases from left to right across periods and from bottom to top up groups.

We can summarize lattice energy periodic table trends in the following image:

## Finding Lattice Energies

When presented with multiple ionic compounds, chemists must often determine which exhibits the highest lattice energy. To do so, they consider both the ion charge variable and the distance variable.

### Calculating Lattice Energies

Although calculating exact lattice energies can prove complicated, we often simplify the process using Coulomb’s Law. This law provides the following equation describing the lattice energy of a given ionic compound:

Q1 & Q2 = the relative charges of the constituent ions in an ionic compound
R = the distance between charges
K = 2.31 x 10^-19 J-nm.

The final answer should be written in units of Joules (J).

#### Steps to Solve:

1. Solve the equation for each ionic compound, inputting the charge and distance values specific to it.
2. Compare the results; the largest quantity denotes the ionic compound with the largest lattice energy.

### Approximating Lattice Energies

During comparisons, we can also use the charge and distance variables to estimate relative lattice energies.

#### Steps to Solve:

1. First look at the relative charges displayed by each ion in a given compound—if one compound has much higher ionic charges, then it will likely have the higher lattice energy.

2. If the charge discrepancies between compounds do not seem clear, calculate Q1 x Q2 for each compound and compare those values. For example, a calculated charge of -3 has 3 times the magnitude of a calculated charge of -1; this would denote that the ionic compound with the charge of -3 exhibits a much higher lattice energy than the ionic compound with the -1 charge (roughly 3 times as high).

3. If the charges of multiple compounds are the same or too similar in value to produce distinctions, consider the sizes of the ions. Juxtapose atomic size, comparing cation to cation and anion to anion between compounds. If you see a significant discrepancy in size either between cations or between anions, the component with the larger atomic radius will lower the lattice energy of its respective compound. Conversely, the ion with a smaller atomic radius will increase the energy value of its respective compound.

## Applications of Lattice Energy

### The Born-Haber Cycle

Lattice energy is implicated in the Born-Haber Cycle, which helps chemists analyze reaction energies. This cycle typically informs investigations of ionic compound formation from different elements. It clarifies the overall reaction process by breaking it down into a series of steps. This approach to chemical reaction analysis stems from Hess’s Law, which states that overarching energy changes can be determined by exploring individual steps, then combining their effects.

As lattice energy forms part of the Born-Haber Cycle equation, we can solve for it when the other factors are plugged in. The equation reads as follows:

Heat of formation= lattice energy + heat of atomization + dissociation energy + (sum of Ionization energies) + (sum of electron affinities)

The Born-Haber Cycle applies Hess’s law to calculate lattice energies by juxtaposing a given ionic compound’s enthalpy change of formation to the enthalpy required to form gaseous ions from its components.

### Other Applications of Lattice Energies

Scientists use lattice energies more broadly to evaluate electron relationships and fluoride relationships. The factors, in turn, inform investigations about the relative strengths of different ionic solids as well as predictions about ionic compound identities, components, and properties.

## Further Examples of Calculating Lattice Energies

### Examples: Using Approximation Techniques

First, we will practice solving for the charge variable.

Problem 1: Given the compound MgO, determine its combined charge.

#### Steps to Solve:

1. Write out the charges of its ions: Mg+2  and O-2

2. Multiply these charges: (2) x (-2) = -4

Problem 2: Given the compound KCl, determine its combined charge.

1. Write out the charges of its ions: K+1  and Cl-1

2. Multiply these charges: (1) x (-1) = -1

We can compare the -4 charge of MgO to the -1 charge of KCl as discussed. As the former is 4 times the quantity of the latter, we can predict that its lattice energy would be approximately 4 times greater as well.

Now, we will practice solving for the size variable.

Problem 3: Given the compound CaO, determine the sizes of its ions.

#### Steps to Solve:

1. Determine the ionic radii of its cation: Ca+2 has an ionic radius of 0.100 nm.

2. Determine the ionic radii of its anion: O-2 has an ionic radius of 0.140nm.

We can compare these values to those of another ionic compound as discussed. This provides insight into which exhibits larger lattice energy.

### Born-Haber Cycle Examples

Now, we will practice solving for exact lattice energy using the Born-Haber Cycle.

Problem 1: Given the compound NaCl, determine its lattice energy.

#### Steps to Solve:

1. Write the reaction describing the formation of NaCl under normal conditions:

Na(s)+12Cl2(g)→NaCl(s)

1. Change the reactants into their ionic gas components.

Na(s)→Na(g)
Na(g)→Na+(g)+e−

12Cl2(g)→Cl(g)
Cl(g)+e−→Cl−(g)

Na+(g)+Cl−(g)→NaCl(s)

This final transformation shows the creation of the “lattice” compound itself.

3.  In accordance with Hess’s law, separate the parts of the reaction and consider them in isolation.

NaCl(s)→Na(s)+12Cl2
−ΔHf,NaCl(s)=+411 kJ

Na, Δ=107 kJ

Na(g)→Na+(g)+e−
IE1,Na(g)=502 kJ

12Cl2(g)→Cl(g)
12ΔHbond,Cl2(g)= 12×242 kJ

Cl(g)+e−→Cl−(g)
EA1,Cl(g)=−355 kJ

Na+(g)+Cl−(g)→NaCl(s), ΔHlattice=?

4. Re-combine these parts to give the final Born-Habor Cycle equation:

0=ΔHcycle=ΔHf,NaCl(s)+ΔHsub,Na+IE1,Na(g)+12ΔHbond,Cl2(g)−EA1,Cl(g)−ΔHlattice

Then, solve for lattice variable: ΔHlattice,NaCl(s)

=−[411+107+502+12(242)−355]kJ

Lattice energy of NaCl=−786 kJ