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# Born-Haber Cycles Calculations

Now you know how to draw a Born-Haber cycle like the one shown below, what's next? Well, you can use it to calculate the lattice enthalpy of an ionic solid, or any other unknown enthalpy change shown in the cycle. How? Remember, Hess’ Law states that the enthalpy change of a reaction remains the same regardless of the route taken. This allows us to calculate the enthalpy change for any part of the cycle as long as we start and end at the same places.

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This means:

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

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

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

Fig. 1 - Born-Haber cycle for potassium chloride

• 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).

 $\mathbf{∆}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta }}\mathbf{}\mathbf{\left(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}$ ${\mathbf{IE}}_{\mathbf{1}}\mathbf{/}{\mathbf{EA}}_{\mathbf{1}}$ K +90 +418 Cl +122 -349 $\mathbf{∆}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta }}\mathbf{}\mathbf{\left(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}$ KCl -437

Step 1

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

Fig. 2 - Born-Haber KCl

Step 2

Apply Hess’ Law to the diagram.

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

Step 3

Fill in the values from the table above.

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

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).

 $\mathbf{∆}{}_{\mathbf{at}}\mathbf{H}^{\mathbf{\Theta }}\mathbf{}\mathbf{\left(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}$ ${\mathbf{IE}}_{\mathbf{1}}\mathbf{/}{\mathbf{EA}}_{\mathbf{1}}$ ${\mathbf{IE}}_{\mathbf{2}}\mathbf{/}{\mathbf{EA}}_{\mathbf{2}}$ Mg +148 +736 $\left({\mathrm{IE}}_{1}\right)$ +1450 $\left({\mathrm{IE}}_{2}\right)$ O +248 -142 $\left({\mathrm{EA}}_{1}\right)$ +770 $\left({\mathrm{EA}}_{2}\right)$ $\mathbf{∆}{}_{\mathbf{f}}\mathbf{H}^{\mathbf{\Theta }}\mathbf{}\mathbf{\left(}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}$ Mg -602

Step 1

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

Fig. 3 - Born-Haber cycle MgO

Step 2

Apply Hess’ Law to the diagram.

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

Step 3

Fill in the values from the table above

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

## 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?

Fig. 4 - Lattice enthalpies of magnesium oxide and sodium chloride

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

Fig. 5 - The ions of sodium chloride and magnesium oxide

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.

Fig. 6 - Lattice enthalpies get smaller as you go down a group

### 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).

Fig. 7 - Covalent character

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.

Fig. 8 - Covalent character

Have a look at the following image:

Fig. 9 - Covalent character and polarisation

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.

Fig. 10 - Trends in lattice enthalpy

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:

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

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