A-Level Chemistry Prep: Born-Haber Cycles

Understanding A-Level Chemistry, particularly intricate subjects like Born-Haber cycles, can be challenging. These cycles play a crucial role in comprehending the creation of ionic compounds and the associated energy changes. This post will analyse a question from the OCR A 2019 Paper 1, explaining each stage in detail to aid in mastering the fundamental principles and improving analytical abilities. Whether you aim to enhance your comprehension or require efficient exam tactics, this manual offers beneficial tips to elevate your confidence and academic performance.

The task at hand involves completing the Born-Haber cycle for the creation of Potassium Oxide.

We will now analyse each step of the cycle starting from the base.

Initially, the formation of Potassium Oxide (K₂O) necessitates two K+ ions to counterbalance the -2 charge on 1/2 O2.

Enthalpy of formation: The enthalpy change when one mole of a substance is formed in its standard state from the pure elements in their standard states under standard conditions of 298K and 1 bar pressure.

Next, we observe the atomisation of Oxygen, where instead of stating 1/2 O2, we can represent it as O. It's important to note that Oxygen is diatomic and gaseous under standard conditions.

Subsequently, we witness the atomisation of Potassium, transitioning its state from solid to gas.

Enthalpy of Atomisation: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state, under standard conditions.

Following this, we encounter the first ionization energy of Potassium, which applies to EVERY Potassium ion. This explains the presence of 2e- in the equation.

First Ionisation Energy: the energy required to remove one mole of electrons from one mole of gaseous atoms from its outermost shell.

This only needs to occur once in order to balance the charges.

Subsequently, we observe the first electron affinity of Oxygen, succeeded by the second electron affinity. This process happens twice to balance the charges. As a result, Potassium ions carry a +2 charge, while the Oxygen ion holds a -2 charge: 2K⁺ and O²⁻.

Electron affinities: the enthalpy changes when one mole of electrons is added to one mole of gaseous atoms/ions.

Lattice enthalpy is defined as the enthalpy change when one mole of an ionic solid is formed from its constituent ions in their gaseous states.

This value is always exothermic (negative) because it represents the energy released when the oppositely charged ions attract each other to form a solid lattice. The Born-Haber cycle combines various enthalpy changes, such as ionisation energy, electron affinity, and enthalpy of formation, to determine this lattice enthalpy. This comprehensive approach helps in understanding the stability and properties of ionic compounds.

In this next question, we have been asked to work out the lattice enthalpy of Potassium Oxide, K₂O.

A trick I used, during my A-levels, was:

ΔH(f) = Everything else + ΔH(LE)

Remember: there will an atomisation and ionisation of EACH Potassium ion, therefore (89 x 2) and (419 x 2)

Enthalpy of Formation = -363 kJ mol-1

Everything else = (89 x 2) + 249 + (419 x 2) + 790 + (-141)

This equals = +1914 kJ mol-1

Using the following formula:

ΔH(f) = Everything else + ΔH(LE) We can work out Lattice Enthalpy:

-363 = 1914 + LE

-363 - 1914 = LE

Therefore Lattice Enthalpy is -2277 kJ mol-1

Trends and Enthalpy Changes

Part ei)

In this part of the question, we need to explain why the first Ionisation Energy of Sodium is more endothermic than that of Potassium.

To begin with, let's break down the concept of endothermic reactions.

Endothermic reactions involve the absorption of heat energy from the surroundings. Exothermic reactions involve the release of energy from the chemicals into the surroundings.

Essentially, Sodium requires a higher energy input for its first Ionisation Energy compared to Potassium. But what is the reason for this?

Even though both Sodium and Potassium are part of Group 1, they differ in their Periods. Sodium is found in Period 3, while Potassium is located in Period 4. Atomic radius increases across periods and down groups, resulting in Sodium having a smaller atomic radius.

This difference is due to the varying electron counts. Sodium has 11 electrons, whereas Potassium has 19. Therefore, Potassium has more electron shells to accommodate all 19 electrons, unlike Sodium.

The first Ionisation Energy for Sodium involves removing its outermost electron from the 3s orbital. In contrast, the first Ionisation Energy for Potassium removes the electron from the 4s orbital. Consequently, the 3s orbital in Sodium experiences a stronger nuclear attraction due to its proximity to the nucleus, leading to a higher electron-nucleus attraction. On the other hand, Potassium experiences a weaker nuclear attraction due to its distance from the nucleus. The increased nuclear attraction in Sodium means that more energy is required to extract the electron from the orbital, which in this case is the 3s orbital in Sodium.

A perfect answer to this question would be:

There are fewer shells in Sodium and therefore the outermost electron experiences a greater nuclear attraction.

Part eii)

In the final part of this question, we are looking into comparing the lattice enthalpy of Sodium Oxide compared to Potassium Oxide and why the former is more exothermic than the latter.

The lattice enthalpy of sodium oxide (Na₂O) is more exothermic than that of potassium oxide (K₂O) due to differences in the ionic sizes of sodium and potassium, which affect the strength of the ionic bonds formed in the lattice.

Sodium (Na⁺): Sodium ions have a smaller ionic radius compared to potassium ions, which are larger. Sodium ions, being smaller, have a higher charge density than potassium ions. A more exothermic lattice enthalpy indicates a more stable ionic lattice. Due to the higher charge density and the smaller ionic distance in Sodium Oxide, the electrostatic forces between the Na⁺ and O²⁻ ions are stronger than those between K⁺ and O²⁻ ions in potassium oxide.

Stronger electrostatic forces lead to a more stable ionic lattice and thus a more exothermic lattice enthalpy.

A perfect answer to this question would be:

The ionic radius for Na is smaller and so Na⁺ has a stronger attraction to O²⁻.

Understanding the intricacies of lattice enthalpy and the various steps involved in constructing a Born-Haber cycle provides a solid foundation for predicting and explaining the stability of ionic compounds. By familiarising yourself with these concepts, you can approach such questions with confidence, ensuring a thorough and accurate analysis. Keep practicing, and remember that a clear grasp of the underlying principles will not only help you ace your exams but also deepen your appreciation for the fascinating world of chemistry.