Born-Haber Cycles Calculations

Are you wondering what to do with the Born-Haber cycle you just drew? You can actually use it to figure out the lattice enthalpy for an ionic solid or any other unknown enthalpy change shown in the cycle. It's easy! All you have to remember is Hess' Law, which states that the enthalpy change of a reaction remains the same no matter what the route taken is. As long as you start and end at the same places, you can calculate the enthalpy change for any part of the cycle. So, if you rearrange the equation, you can use it to calculate lattice enthalpy. Born-Haber cycles calculations just got simpler!

Born Haber cycle for potassium chloride
Born Haber cycle for potassium chloride

 

In this article, we will explore the world of lattice enthalpies and learn how to calculate them using Born-Haber calculations. We'll dive into how to compare the lattice enthalpies of different substances, and examine the factors that affect them. Along the way, we'll go through some example calculations to help you get the hang of it. Finally, we'll look at how physicists calculate lattice energy theoretically, and explain why these values sometimes differ from their corresponding experimental values.

Let's jump right into an example calculation! Using the information provided in the table below, we can calculate the lattice formation enthalpy of potassium chloride (KCl). 

Step 1

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

Born-Haber KCl
Born-Haber KCl

Step 2

Apply Hess’ Law to the diagram.

Step 3

Fill in the values from the table above.

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).

Step 1

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

Born-Haber cycle MgO
Born-Haber cycle MgO

Step 2

Apply Hess’ Law to the diagram.

Step 3

Fill in the values from the table above

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

The lattice enthalpies of sodium chloride and magnesium oxide are different because the charges on the ions and the ionic radii are different. Magnesium ions and oxide ions are smaller than sodium and chloride ions, which 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.

The ions of sodium chloride and magnesium oxide
The ions of sodium chloride and magnesium oxide

That's a great explanation! You're absolutely right that larger ionic charges lead to stronger electrostatic attraction between ions, resulting in a larger lattice dissociation enthalpy. And you're correct that smaller ions are closer together, leading to stronger ionic attraction and a larger lattice enthalpy.

As you go down a group on the periodic table, the size of the positive ions increases, leading to larger radii and weaker electrostatic attraction between the ions. This weaker attraction results in smaller lattice enthalpies for their chloride salts. This trend is observed in Group 1, as you mentioned.

Overall, the size and charge of ions are important factors to consider when comparing lattice enthalpies of different substances. Thanks for explaining this so clearly!

Lattice enthalpies get smaller as you go down a group
Lattice enthalpies get smaller as you go down a group

Theoretical values for lattice enthalpy

That's a great explanation of the theoretical method used by physicists to calculate lattice enthalpies! It's important to note that while this method assumes that the substance is highly ionic with only electrostatic attraction between the ions, in reality, some compounds may have a covalent character that leads to a larger difference between theoretical and experimental lattice enthalpies.

When a compound has a covalent character, the electrons in the ionic bond behave a little like those in a covalent bond. This happens when there is not enough electronegativity difference between the atoms for a complete electron transfer. The anion becomes polarized, meaning that a cation attracts the electrons on an anion, causing the electrons to overlap with the electron cloud of the cation and distorting the anion's electron density.

Overall, it's important to consider both theoretical and experimental methods when calculating lattice enthalpies, and to recognize when a compound may have a covalent character that affects its lattice enthalpy. Thanks for explaining this so clearly!

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. 

Covalent character
Covalent character

Have a look at the following image: 

Covalent character and polarisation
Covalent character and polarisation

As you can see, not all cations have the same polarising power. Smaller cations with a high positive charge, like  and , 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  and , 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.

Trends in lattice enthalpy
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.

Additionally, a trend in the periodic table shows that as we move from left to right, lattices become less ionic and more covalent. This is due to the smaller size and higher positive charge of cations on the left side of the table, which have a greater polarizing power and can cause greater distortion of electron density in anions. Larger anions are also more easily polarized due to weaker attraction between the nucleus and electrons. This trend leads to larger differences between theoretical and experimental lattice enthalpies for compounds on the right side of the periodic table. Overall, understanding the factors that affect lattice enthalpy is important in predicting and explaining the properties of ionic compounds.

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 elementThe first ionisation energy of the metalSubsequent ionisation enthalpies if appropriateFirst electron affinity of the non-metalSubsequent electron affinities if appropriate

Quiz questions showing the correct answer and a leaderboard with friends.

Create chemistry notes and questions for free

96% of learners report doubling their learning speed with Shiken

Join Shiken for free

Try Shiken Premium for free

Start creating interactive learning content in minutes with Shiken. 96% of learners report 2x faster learning.
Try Shiken for free
Free 14 day trial
Cancel anytime
20k+ learners globally
Shiken UI showing questions and overall results.