Lattice Enthalpy (OCR A-Level Chemistry A): Revision Notes

Lattice Enthalpy

Introduction to lattice enthalpy

Ionic compounds in their solid state exhibit remarkable stability, characterized by their strong ionic bonds and typically high melting points. This stability originates from the powerful electrostatic forces of attraction between oppositely charged ions arranged in the ionic lattice structure. Understanding and quantifying this stability is essential in physical chemistry.

Lattice enthalpy provides a quantitative measure of the strength of ionic bonding within a giant ionic lattice. Specifically, it is defined as the enthalpy change that occurs when one mole of an ionic compound forms from its constituent gaseous ions under standard conditions.

For example, the formation of solid potassium chloride from gaseous ions can be represented as:

$\text{K}^+(g) + \text{Cl}^-(g) \rightarrow \text{KCl}(s) \qquad \Delta_{\text{LE}}H^\circ = -711 \text{ kJ mol}^{-1}$

This process involves bringing separate gaseous ions together to form the ionic lattice structure. Since ionic bond formation releases energy, lattice enthalpy is always exothermic, meaning the enthalpy change will always be negative.

The Born-Haber cycle

Lattice enthalpy cannot be measured directly through experiment. Instead, we must calculate it indirectly using an energy cycle approach. This indirect determination employs a special type of thermochemical cycle called a Born-Haber cycle.

The Born-Haber cycle demonstrates two different pathways (routes) for converting elements in their standard states into an ionic lattice:

• Route 1 (indirect pathway): Elements → Gaseous atoms → Gaseous ions → Ionic lattice
• Route 2 (direct pathway): Elements → Ionic lattice

According to Hess's Law, the total enthalpy change is independent of the route taken. Since Route 2 involves just one step (the enthalpy change of formation), and Route 1 involves multiple steps with known enthalpy values, we can calculate the unknown lattice enthalpy.

Route 1: The indirect pathway

Route 1 requires three distinct stages:

Formation of gaseous atoms:

• Converting elements from their standard states into gaseous atoms
• This change is endothermic because it involves breaking bonds

Formation of gaseous ions:

• Converting gaseous atoms into positive and negative gaseous ions
• Overall, this change is endothermic (though electron affinity contributes an exothermic component)

Lattice formation:

• Bringing gaseous ions together to form the solid ionic lattice
• This is the lattice enthalpy and is exothermic

Route 2: The direct pathway

Route 2 converts elements in their standard states directly to the ionic lattice through a single step - the enthalpy change of formation ($\Delta_f H^\circ$). This process is typically exothermic.

Key enthalpy changes in Born-Haber cycles

Several specific enthalpy changes appear in Born-Haber cycles. Understanding each definition is crucial for constructing accurate cycles and performing calculations.

Standard enthalpy change of formation ($\Delta_f H^\circ$)

This represents the enthalpy change occurring when one mole of a compound forms from its constituent elements under standard conditions, with all reactants and products in their standard states.

For sodium chloride formation:

$\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{NaCl}(s) \qquad \Delta_f H^\circ = -411 \text{ kJ mol}^{-1}$

Standard enthalpy change of atomization ($\Delta_{at} H^\circ$)

This is the enthalpy change for forming one mole of gaseous atoms from the element in its standard state under standard conditions.

Examples include:

$\text{Na}(s) \rightarrow \text{Na}(g) \qquad \Delta_{at}H^\circ = +107 \text{ kJ mol}^{-1}$

$\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \qquad \Delta_{at}H^\circ = +121 \text{ kJ mol}^{-1}$

The atomization process is always endothermic because bonds must be broken to produce gaseous atoms. When the element exists as a gas in its standard state (like $\text{Cl}_2$), the enthalpy of atomization relates to the bond enthalpy of the bond being broken.

First ionization energy ($\Delta_{IE1} H^\circ$)

The first ionization energy is the enthalpy change required to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.

For example:

$\text{Na}(g) \rightarrow \text{Na}^+(g) + e^- \qquad \Delta_{IE}H^\circ = +496 \text{ kJ mol}^{-1}$

Ionization energies are endothermic because energy must be supplied to overcome the electrostatic attraction between the negatively charged electron and the positively charged nucleus.

First electron affinity ($\Delta_{EA} H^\circ$)

Electron affinity represents the opposite of ionization energy. The first electron affinity is the enthalpy change when one electron is added to each atom in one mole of gaseous atoms, forming one mole of gaseous 1− ions.

For example:

$\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g) \qquad \Delta_{EA}H^\circ = -346 \text{ kJ mol}^{-1}$

First electron affinities are exothermic because the electron being added is attracted towards the positive nucleus.

Determining lattice enthalpies: worked examples

By constructing a complete Born-Haber cycle, you can calculate lattice enthalpies using Hess's Law. The following examples demonstrate the systematic approach needed for different types of ionic compounds.

Successive electron affinities

The magnesium chloride example included two ionization energies to account for the formation of $\text{Mg}^{2+}(g)$. Similarly, when an anion has a charge greater than 1−, such as $\text{O}^{2-}$, successive electron affinities are required. These are handled analogously to successive ionization energies.

The first two electron affinities for oxygen are:

First EA: $\text{O}(g) + e^- \rightarrow \text{O}^-(g) \qquad \Delta_{EA1}H^\circ = -141 \text{ kJ mol}^{-1}$

Second EA: $\text{O}^-(g) + e^- \rightarrow \text{O}^{2-}(g) \qquad \Delta_{EA2}H^\circ = +790 \text{ kJ mol}^{-1}$

A Born-Haber cycle involving $\text{O}^{2-}(g)$ would contain both electron affinity energy changes.

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