This means:

$\mathbf{\u2206}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{K}\mathbf{\left)}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{Cl}\mathbf{\left)}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\left(}{\mathbf{IE}}_{\mathbf{1}}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\left(}{\mathbf{EA}}_{\mathbf{1}}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\u2206}{}_{\mathbf{LE}}\mathbf{H}^{\mathbf{\Theta}}$

If we rearrange this equation we can use it to calculate lattice enthalpy:

$\mathbf{\u2206}{}_{\mathbf{LE}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{\u2206}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{)}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{[}}{\mathbf{}}\mathbf{(}\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{\left(}\mathbf{K}\mathbf{\right)}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{\left(}\mathbf{Cl}\mathbf{\right)}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}{\mathbf{IE}}_{\mathbf{1}}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}{\mathbf{EA}}_{\mathbf{1}}\mathbf{)}{\mathbf{}}{\mathbf{]}}$

- In this article, you will practice calculating lattice enthalpies using Born-Haber calculations.
- We will look at how you can compare the lattice enthalpies of different substances and the factors that affect lattice enthalpy.
- At the end, we will see how physicists calculate lattice energy theoretically, and why these values sometimes differ from their corresponding experimental values.

## Examples of Born-Haber calculations

You need to practice to get the hang of lattice enthalpy calculations, so let us try a few example calculations together!

Use the information on the table below to calculate the lattice formation enthalpy of potassium chloride (KCl).

$\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$ | ${\mathbf{IE}}_{\mathbf{1}}\mathbf{/}{\mathbf{EA}}_{\mathbf{1}}$ | |

K | +90 | +418 |

Cl | +122 | -349 |

$\mathbf{\u2206}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$ | ||

KCl | -437 |

**Step 1**

Draw the Born-Haber cycle for KCl as shown below.

**Step 2**

Apply Hess’ Law to the diagram.

$\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=(\u2206{}_{\mathrm{f}}\mathrm{H}^{\mathrm{\Theta}})-[(\u2206{}_{\mathrm{at}}\mathrm{H}^{\mathrm{\Theta}}{\left(}{\mathrm{K}}{\right)})+(\u2206{}_{\mathrm{at}}\mathrm{H}^{\mathrm{\Theta}}{\left(}{\mathrm{Cl}}{\right)})+({\mathrm{IE}}_{1})+({\mathrm{EA}}_{1})]$

**Step 3**

Fill in the values from the table above.

$\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=-437-[(90)+(122)+(418)+(-349)]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=-718\mathrm{kJ}{\mathrm{mol}}^{-1}$

Make sure to use brackets when calculating the lattice enthalpy since the signs will affect your answer.

Use the information in the table below to calculate the lattice formation enthalpy of magnesium oxide (MgO).

$\mathbf{\u2206}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$ | ${\mathbf{IE}}_{\mathbf{1}}\mathbf{/}{\mathbf{EA}}_{\mathbf{1}}$ | ${\mathbf{IE}}_{\mathbf{2}}\mathbf{/}{\mathbf{EA}}_{\mathbf{2}}$ | |

Mg | +148 | +736 $\left({\mathrm{IE}}_{1}\right)$ | +1450 $\left({\mathrm{IE}}_{2}\right)$ |

O | +248 | -142 $\left({\mathrm{EA}}_{1}\right)$ | +770 $\left({\mathrm{EA}}_{2}\right)$ |

$\mathbf{\u2206}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta}}\mathbf{}\mathbf{(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$ | |||

Mg | -602 |

**Step 1**

Draw the Born-Haber cycle for MgO as shown below.

**Step 2**

Apply Hess’ Law to the diagram.

$\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=\u2206{}_{\mathrm{f}}\mathrm{H}^{\mathrm{\Theta}}-[\u2206{}_{\mathrm{at}}\mathrm{H}^{\mathrm{\Theta}}(\mathrm{Mg})+\u2206{}_{\mathrm{at}}\mathrm{H}^{\mathrm{\Theta}}(\mathrm{O})+{\mathrm{IE}}_{1}+{\mathrm{IE}}_{2}+{\mathrm{EA}}_{1}+{\mathrm{EA}}_{2}]$

**Step 3**

Fill in the values from the table above

$\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=-602-[(148)+(248)+(736)+(1450)+(-142)+(770)]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\u2206{}_{\mathrm{LE}}\mathrm{H}^{\mathrm{\Theta}}=-3812\mathrm{kJ}{\mathrm{mol}}^{-1}$

## Comparing lattice enthalpies

For your chemistry exams, you must be able to explain the differences between lattice enthalpies of different substances. You will also be expected to give a reason for the differences between the theoretical and experimental values of lattice enthalpy. Not to worry- we’ve got you covered!

### Factors that affect lattice enthalpy

Two factors that affect lattice enthalpy are:

- The charges on the ions
- The radius of the ions

#### The charge on the ion

Compare the lattice enthalpies of sodium chloride and magnesium oxide below. Their crystal lattices have the same geometry, so why are their enthalpies different?

Have a look now at their ions. Can you see a reason for the difference in their lattice enthalpies?

Magnesium and oxygen ions have larger ionic charges than sodium and chloride ions. Ions with larger ionic charges have **stronger electrostatic attraction** between the ions. More energy is needed to overcome the attraction between the ions and break up the lattice. So the lattice dissociation enthalpy of magnesium oxide (MgO) is **larger** than that of sodium chloride (NaCl).

#### The radius of the ion

Magnesium ions and oxide ions are smaller than sodium and chloride ions. This means the ions in the magnesium oxide lattice are closer together. The strength of ionic attraction depends on the **closeness of the centres** of the attracted ions, so there is a stronger attraction between the magnesium and oxygen ions.

Ions with larger radii tend to have **smaller lattice enthalpies**, because the electrostatic **attraction** between larger ions is **weaker**.

You can observe this effect as you go down a group on the periodic table. As an example look at Group 1. As their positive ions get bigger, the lattice enthalpies of their chloride salts decrease.

### Theoretical values for lattice enthalpy

Physicists use a theoretical method to calculate lattice enthalpies. This is based on the assumption that the substance is **highly ionic** with only **electrostatic attraction** between the ions. In other words, they consider the ions as point charges that bond together to make a lattice. They also take into account the geometry of the lattice and the distance between the atoms.

What physicists actually calculate is the theoretical **lattice energy **and not lattice enthalpy. It is possible to do a calculation that converts one to the other, but you won’t need to know how to do this for your exams! However, you will need to talk about the differences in their values.

The table below shows the difference between lattice enthalpies calculated using this theoretical method and enthalpies calculated using Born-Haber cycles. Can you see much of a difference between the values?

Compound | Experimental Lattice Energy/ ${{\mathbf{kJmol}}}^{\mathbf{-}\mathbf{1}}$ | Theoretical Lattice Energy/ ${{\mathbf{kJmol}}}^{\mathbf{-}\mathbf{1}}$ |

NaF | -918 | -912 |

NaCl | -780 | -770 |

NaBr | -742 | -735 |

AgF | -958 | -920 |

AgCl | -905 | -833 |

AgBr | -891 | -816 |

- A small difference, as in the case of NaCl,
**suggests that the assumption the compound is ionic is fairly accurate**.

- A large difference, as in the case of AgCl, suggests that the compound is not purely ionic. Instead, the bonds show a
**covalent character.**

We say Born-Haber cycles are experimental because they use enthalpy changes that can be measured to calculate lattice enthalpy. On the other hand, the physics-style calculations are based on the assumption* *that an ionic compound is completely ionic. So the lattice enthalpy values that you get with the second method are theoretical.

A **covalent character** means that the electrons in an ionic bond behave a little like those in a covalent bond. This happens when there is not enough **electronegativity **between the atoms for a complete electron transfer. The anion (negative ion) then becomes **polarised**.

Essentially, a cation (positive ion) attracts the electrons on an anion (negative ion). This attraction makes the electrons on the anion overlap the electron cloud of the cation and distorts the anion’s electron density (shown below).

The anion becomes **polarised** because its electrons are no longer evenly distributed in the orbitals. Instead, some electrons become clustered between the cation and the anion- a little bit like a shared pair of electrons in a covalent bond.

Have a look at the following image:

As you can see, not all cations have the same polarising power. **Smaller cations** with a **high positive charge**, like ${\mathrm{Mg}}^{2+}$ and ${\mathrm{Al}}^{3+}$, have a greater polarising power. This means the smaller a cation is, the more likely it is to distort the electron density of an anion.

**Larger anions** with a **high negative charge**, like ${\mathrm{Cl}}^{-}$ and ${\mathrm{Br}}^{-}$, are more easily polarised, since their outer shell electrons are further from the nucleus. In other words, the attraction between the nucleus and the electrons is weaker so they are more easily distorted.

We can observe **greater covalent character** in an ionic bond when there is **more polarisation of the anion**. This leads us to a trend in the periodic table: as we go from left to right the lattices become less ionic and more covalent. The differences between the theoretical and experimental values for lattice energy are greater. This trend supports an **ionic model** for compounds like sodium chloride.

Polarisation as a result of the distortion of electron density shows us that bonding is neither purely ionic or covalent, but somewhere between the two.

Getting the hang of Born-Haber calculations takes practice! Review this article as many times as you need and use the flashcards in this section to strengthen your skills.

## Born-Haber Cycles Calculations - Key takeaways

- Lattice enthalpies calculated using Born-Haber cycles are
**experimental**, because they use enthalpy changes that can be measured. - Two factors that affect lattice enthalpy are the
**charge**and the**radius**of the ions. - Ions with
**large ionic charges**have**larger lattice dissociation enthalpies**, because of stronger**electrostatic attraction**between the ions. - More energy is needed to break up the lattice. Ions with
**larger radii**tend to have**smaller lattice enthalpies**, because the**electrostatic attraction**between larger ions is weaker. - Theoretical physics-style calculations for lattice enthalpy are based on the assumption that the substance is
**highly ionic.** - Small differences between the theoretical and experimental values for a lattice enthalpy suggest that the substance is
**highly ionic**. - Large differences between theoretical and experimental values of lattice enthalpy suggest there is a
**covalent character**between the bonds of the ions. - A covalent character means that the electrons in an ionic bond behave a little like those in a covalent bond. There is not enough
**electronegativity**between the atoms for a complete electron transfer. - An anion becomes polarised when its
**electron density is distorted**by the attraction of a small highly charged cation.

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##### Frequently Asked Questions about Born-Haber Cycles Calculations

What is the formula for calculating lattice enthalpy?

We can calculate lattice enthalpy by drawing a Born-Haber cycle with enthalpy changes we can measure. When we draw Born-Haber cycles we must show these enthalpy changes in the following order:

- The Enthalpy of formation of the compound.
- Enthalpy of atomisation of each element
- The first ionisation energy of the metal
- Subsequent ionisation enthalpies if appropriate
- First electron affinity of the non-metal
- Subsequent electron affinities if appropriate

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