Lattice enthalpy is a direct measure of the strength of the ionic bonding in a giant ionic lattice. Because it is impossible to measure this change directly by experiment, we construct thermodynamic cycles known as Born-Haber cycles. These cycles apply Hess's law to link lattice enthalpy with other enthalpy changes that can be measured directly.
Key Principle
A Born-Haber cycle is essentially a closed loop of enthalpy changes. By summing the steps in the loop, we can calculate any missing value, typically the lattice enthalpy, because the total enthalpy change around a complete cycle is zero.
Important Enthalpy Definitions
To construct and use Born-Haber cycles, you must learn the exact definitions of the constituent enthalpy changes. Pay close attention to state symbols in the definitions.
The enthalpy change when 1 mole of a solid ionic compound is formed from its constituent gaseous ions under standard conditions. This process is highly exothermic:
\( \text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s) \quad \Delta H_L^\theta < 0 \)
The enthalpy change when 1 mole of a solid ionic compound is completely separated into its constituent gaseous ions under standard conditions. This process is highly endothermic:
\( \text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g) \quad \Delta H_{\text{L-diss}}^\theta > 0 \)
The enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. This is an endothermic process:
\( \text{Na}(s) \rightarrow \text{Na}(g) \quad \Delta H_{at}^\theta = +107\text{ kJ mol}^{-1} \)
\( \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad \Delta H_{at}^\theta = +122\text{ kJ mol}^{-1} \)
The enthalpy change when 1 mole of gaseous 1- ions is formed from 1 mole of gaseous atoms by gaining 1 mole of electrons. This is usually exothermic because the incoming electron is attracted to the positive nucleus:
\( \text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g) \quad \Delta H_{EA1}^\theta = -348\text{ kJ mol}^{-1} \)
Second Electron Affinity (\( \Delta H_{EA2}^\theta \))
In contrast to the first, the second electron affinity is always endothermic. This is because energy is required to overcome the electrostatic repulsion between the negative ion and the incoming negative electron:
\( \text{O}^-(g) + e^- \rightarrow \text{O}^{2-}(g) \quad \Delta H_{EA2}^\theta = +798\text{ kJ mol}^{-1} \)
Make sure you understand the difference between bond enthalpy and atomisation enthalpy for diatomic molecules like chlorine. The bond enthalpy of Cl-Cl is +244 kJ mol⁻¹, which represents \( \text{Cl}_2(g) \rightarrow 2\text{Cl}(g) \). The atomisation enthalpy represents forming only 1 mole of atoms, \( \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \), which is exactly half of the bond enthalpy (+122 kJ mol⁻¹).
Born-Haber cycle: NaCl
Explore the energy journey
From elements to an ionic crystal
Follow one route from sodium and chlorine to sodium chloride. Watch what happens to the particles, predict the energy change, then connect each step to the cycle.
This is one valid route through the cycle. The atomisation steps can be arranged in a different order if the states and enthalpy changes remain consistent. Particle sketches illustrate each change; they do not represent mole counts or actual particle sizes.
Atomise the sodium
Na(s) → Na(g)
Predict the energy change:
ΔatH = +107 kJ mol⁻¹Relative enthalpy: 0 → +107 kJ mol⁻¹
Energy rises. Energy is absorbed to separate sodium atoms from the metal into the gas phase.
Atomise the chlorine
½Cl₂(g) → Cl(g)
Predict the energy change:
ΔatH = +122 kJ mol⁻¹Relative enthalpy: +107 → +229 kJ mol⁻¹
Energy rises. Breaking the Cl-Cl bond requires energy. Atomisation forms one mole of Cl atoms, so the value is half the Cl₂ bond dissociation enthalpy: +244 ÷ 2 = +122 kJ mol⁻¹.
Remove an electron
Na(g) → Na⁺(g) + e⁻
Predict the energy change:
ΔieH = +496 kJ mol⁻¹Relative enthalpy: +229 → +725 kJ mol⁻¹
Energy rises. The first ionisation energy is positive: energy is needed to overcome the attraction between the electron and the sodium nucleus.
Add an electron to chlorine
Cl(g) + e⁻ → Cl⁻(g)
Predict the energy change:
ΔeaH = −348 kJ mol⁻¹Relative enthalpy: +725 → +377 kJ mol⁻¹
Energy falls. The first electron affinity of chlorine is negative because attraction between the incoming electron and the nucleus releases energy. The gaseous ions still lie above the neutral gaseous atoms.
Build the ionic lattice
Na⁺(g) + Cl⁻(g) → NaCl(s)
Predict the energy change:
ΔLH = −788 kJ mol⁻¹Relative enthalpy: +377 → −411 kJ mol⁻¹
Energy falls. Oppositely charged ions attract as the extended ionic lattice forms, releasing energy. Lattice enthalpy of formation is −788 kJ mol⁻¹; reversing this step gives a dissociation enthalpy of +788 kJ mol⁻¹.
The complete energy cycle
Match the numbered arrows to the steps above.
Arrow heights follow the enthalpy changes. All values are in kJ mol⁻¹. Scroll across the diagram on a small screen.
- ΔfH: enthalpy of formation
- Na(s) + ½Cl₂(g) → NaCl(s)
−411 kJ mol⁻¹ - 1 · ΔatH: sodium atomisation
- Na(s) → Na(g)
+107 kJ mol⁻¹ - 2 · ΔatH: chlorine atomisation
- ½Cl₂(g) → Cl(g)
+122 kJ mol⁻¹ - 3 · ΔieH: first ionisation energy of sodium
- Na(g) → Na⁺(g) + e⁻
+496 kJ mol⁻¹ - 4 · ΔeaH: first electron affinity of chlorine
- Cl(g) + e⁻ → Cl⁻(g)
−348 kJ mol⁻¹ - 5 · ΔLH: lattice enthalpy of formation
- Na⁺(g) + Cl⁻(g) → NaCl(s)
−788 kJ mol⁻¹
The numbers 1 to 5 follow the indirect route in order. The worked calculation below writes the same quantities as ΔHfθ and ΔHLθ.
Step 1: Apply Hess's Law. According to Hess's law, the enthalpy change of the direct route (formation) equals the sum of the enthalpy changes of the indirect route:
\( \Delta H_f^\theta = \Delta H_{at}^\theta(\text{Na}) + \Delta H_{at}^\theta(\text{Cl}) + \text{IE}_1(\text{Na}) + \Delta H_{EA1}^\theta(\text{Cl}) + \Delta H_L^\theta \)
Step 2: Substitute the known values:
\( -411 = +107 + 122 + 496 + (-348) + \Delta H_L^\theta \)
\( -411 = +377 + \Delta H_L^\theta \)
Step 3: Solve for \( \Delta H_L^\theta \):
\( \Delta H_L^\theta = -411 - 377 = -788\text{ kJ mol}^{-1} \)
The standard lattice enthalpy of formation of NaCl is −788 kJ mol⁻¹.
Theoretical vs Experimental Lattice Enthalpy
We can calculate a theoretical lattice enthalpy by assuming the ionic compound is a purely ionic lattice. This is called the ionic model.
The Purely Ionic Model Assumptions
- The ions are perfectly spherical.
- The charge is distributed evenly across each sphere.
- There is only electrostatic attraction between oppositely charged ions: no covalent character.
If we compare this theoretical value with the experimental value calculated from the Born-Haber cycle, we find two possible scenarios:
- Close Agreement: For compounds like NaCl or KF, the theoretical and experimental values are very similar (within 1 to 2%). This shows that their bonding is almost 100% ionic, matching the ionic model.
- Significant Discrepancy: For compounds like AgCl or MgI₂, the experimental value is significantly more exothermic than the theoretical value. This indicates the presence of covalent character.
The sharing of electron density that occurs when a cation distorts (polarises) the electron cloud of a neighbouring anion, drawing electron density between the two nuclei.
The polarisation of an anion depends on two factors:
- Cation Polarising Power: High charge density (small ionic radius and high charge, such as Li⁺ or Mg²⁺) polarises anions strongly.
- Anion Polarisability: Large ionic radius and high charge (such as I⁻ or S²⁻) mean the outer electron cloud is easily distorted.
The silver ion (Ag⁺) is relatively small and has a polarising effect on the large chloride ion (Cl⁻). This distorts the electron cloud of the chloride ion, drawing electron density between the nuclei.
This polarisation gives AgCl significant covalent character. The covalent interactions make the bonds in the lattice stronger than predicted by purely electrostatic forces. Consequently, the experimental value is more exothermic than the theoretical value.
In contrast, the sodium ion (Na⁺) is a simple spherical ion with low polarising power, meaning NaCl has negligible covalent character and fits the purely ionic model well.
Study this topic on the go
Get active recall flashcards, notes, and topic quizzes in ChemEasy, or build your revision schedule with ChemPlan IB.