Lattice Energy — Born-Haber Cycle & Examples

We have been learning about ionic bonding. When two elements have large electronegativity differences they will form ionic bonds. The less electronegative element will donate its electron to form a cation. The more electronegative element will take the electron to form an anion. The opposite electrostatic charges hold the ions together. We also know that ionic compounds form a crystal lattice.

Which Crystal Lattice is More Stable?

Have you ever wondered which ionic compounds are more stable? For example, is sodium chloride ($NaCl$) or potassium chloride ($KCl$) stronger. We would expect the strength of the crystal lattice to depend on two factors. The first is the strength of the charges. The second is the distance between the ions. We can answer our question based on the periodic trends.

Let's look at an example where the distance between atoms is important. In sodium chloride and potassium chloride, the charges are the same. Sodium and potassium ions both have a positive charge. The chloride ions are the same in both compounds. So, the strength of the ionic bond would depend on the distance between the atoms. We know that atomic radii increase as you move down a group of the periodic table. Potassium has a larger radius than sodium. The distance between the atoms is greater in potassium chloride than it is in sodium chloride. Sodium chloride is more stable than potassium chloride. We could draw the same conclusions if we changed the halogen. For example, fluorine has a smaller atomic radius than chlorine. So, sodium fluoride is stronger than sodium chloride.

Let's look at an example where the difference in charge is the determining factor. Let's compare sodium chloride with magnesium chloride. Magnesium has a positive two charge. The magnesium is going to interact more with the chloride ions than the sodium. So, magnesium chloride is more stable than sodium chloride. We are seeing that as the ionic charge increases, so does the stability of the crystal lattice. Magnesium also has a smaller atomic radius than sodium. As we move across a period, the change in atomic radius is small. So, the difference in the charges is more prominent.

Lattice Energy

Can we determine which crystal lattice is more stable without periodic trends? Yes! Thankfully, physical chemists have calculated the lattice energy for almost every ionic compound. Lattice energy is also called lattice enthalpy. The lattice energy is the energy required to separate one mole of a solid ionic compound into its gaseous ions. The larger the lattice energy the more stable the crystal lattice. The best part is that you can look up the lattice energies! Here are the lattice energies for $NaCl$, $KCl$, and $MgCl_{2}$

$\Delta$$H_{lat}$[$NaCl$] = 788 kJ/mol

$\Delta$$H_{lat}$[$KCl$] = 715 kJ/mol

$\Delta$$H_{lat}$[$MgCl_{2}$] = 2326 kJ/mol

The lattice energies are positive values. The positive energy value means that the process is endothermic. You have to put energy into the system. The difference between $NaCl$ and $KCl$ is not that different. That's because the determining factor was the differences in atomic radii. The difference between $MgCl_{2}$ and $NaCl$ is very large. The determining difference is the charges.

Calculating Lattice Energies

Unfortunately, you will have to calculate the lattice energy. We calculate the lattice energy by using the Born-Haber cycle. The Born-Haber cycle is an application of Hess's Law. The Born-Haber cycle looks at two paths for the formation of an ionic compound. The direct path is the enthalpy of formation. The indirect path looks at the energy of atomization, ionization energy, electron affinity, and lattice energy. The direct path is equal to the indirect path.

Let's calculate the $\Delta$$H_{lat}$[NaCl]. The reaction is $NaCl$(s)$\rightarrow$$Na^{+}$(g)+$Cl^{-}$(g).

The direct path is the standard enthalpy of formation.

$Na$(s) + 1/2 $Cl_2$(g) $\rightarrow$ NaCl(s). $\Delta$$H_{f}$[NaCl] = -411 kJ.

The indirect path has five steps:

First, we need sodium metal to be gaseous sodium atoms.

$Na$(s) $\rightarrow$ $Na$(g). $\Delta$$H_{f}$[Na(g)]=108 kJ.

We need diatomic chlorine ($Cl_{2}$) to be two chlorine atoms.

1/2 $Cl_2$(g) $\rightarrow$ $Cl$(g). $\Delta$$H_{f}$[Cl(g)]=122 kJ.

Now, we can ionize the gaseous sodium atoms. This is the ionization energy.

$Na$(g) $\rightarrow$ $Na^{+}$(g)+ $e^{-}$. $I_{E}$[$Na^{+}$(g)]=496 kJ.

Next, the chlorine atoms each gain an electron. This is the electron affinity.

$Cl$(g) + $e^{-}$ $\rightarrow$ $Cl^{-}$(g). $E_{A}$[$Cl^{-}$(g)]=-349 kJ.

Finally, the ions react to form sodium chloride. This reaction is the -$\Delta$$H_{lat}$[NaCl]. It is negative because it is the reverse reaction of the lattice energy.

$Na^{+}$(g) + $Cl^{-}$(g) $\rightarrow$ $NaCl$(s).

Now, we use Hess's law to calculate the lattice energy.

$\Sigma_{Direct}=\Sigma_{indirect}$

$\Delta$$H_{f}$[NaCl]= $\Delta$$H_{f}$[Na(g)]+$\Delta$$H_{f}$[Cl(g)]+$I_{E}$[$Na^{+}$(g)]+$E_{A}$[$Cl^{-}$]+-$\Delta$$H_{lat}$[NaCl]

-411kJ=108kJ+122kJ+496kJ+-349kJ+-$\Delta$$H_{lat}$[NaCl]

$\Delta H_{lat}$ [NaCl] = 788kJ

(Video) Lattice Energy of Ionic Compounds, Basic Introduction, Charge vs Ionic Radius

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In this video, The Organic Chemistry Tutor walks you through lattice energy. You'll learn how ionic charge and atomic radius affect it. He'll also teach you to compare different compounds qualitatively. He works through two unique example problems. Both could easily be test questions! If you want to master lattice energy, this video can help!

(Video) How to Draw Born Haber Cycle of Ionic Compounds - Energetics

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