🔬 Welcome to Chemical Energetics: Lattice Energy and Born-Haber Cycles ⚛️
Hello! This chapter takes the concept of energy changes (enthalpy) you learned in AS Level and applies it to one of the most fundamental structures in chemistry: ionic solids. You might think ionic compounds are simple, but the energy holding them together is complex!
We are going to learn how to measure, or rather, *calculate* the immense stability of salts like NaCl, using a brilliant application of Hess's Law called the Born-Haber cycle. This is a crucial A-Level topic that helps explain why certain ionic compounds form easily and how they interact with water. Don't worry if this seems tricky at first; we will break down the energy steps one by one!
Part 1: The Essential Enthalpy Terms (Revisited and New)
Before we build the Born-Haber cycle, we need to define the fundamental energy changes involved in turning elements into gaseous ions, which is the starting point for forming an ionic lattice.
1. Enthalpy Change of Atomisation (\(\Delta H_{\text{at}}\))
Definition: The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state, under standard conditions (298 K, 101 kPa).
- \(\Delta H_{\text{at}}\) is always endothermic (positive $\Delta H$) because you always need energy to break the bonds holding the atoms together (e.g., metallic bonds in a metal or covalent bonds in a non-metal).
- Example for Sodium: \(Na(s) \rightarrow Na(g)\)
- Example for Chlorine: \(\frac{1}{2}Cl_2(g) \rightarrow Cl(g)\) (Note: We only need one mole of atoms, not molecules).
2. First Ionisation Energy (IE)
Definition: The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous $1+$ ions.
- IE is always endothermic. It takes energy to pull an electron away from the nucleus.
- Example: \(Na(g) \rightarrow Na^+(g) + e^-\)
3. Electron Affinity (EA) - A Closer Look
Electron affinity is the energy change when an atom gains an electron. This is crucial for forming the negative ion (anion).
Definition of First Electron Affinity (\(EA_1\)) (Syllabus 23.1.2a)
Definition: The enthalpy change when one electron is added to each atom in one mole of gaseous atoms to form one mole of gaseous $1-$ ions.
- $EA_1$ is usually exothermic (negative $\Delta H$) because the incoming electron is attracted to the positive nucleus, releasing energy.
- Example: \(Cl(g) + e^- \rightarrow Cl^-(g)\)
Second Electron Affinity (\(EA_2\))
What happens if you try to add a second electron? For example, forming $O^{2-}$ from $O^{-}$?
- The second electron affinity is always highly endothermic (positive $\Delta H$).
- This is because you are forcing a negative electron into an already negative ion ($O^{-}$), leading to very strong electrostatic repulsion. This process requires a significant input of energy.
- Example: \(O^-(g) + e^- \rightarrow O^{2-}(g)\) (\(\Delta H\) is large and positive)
Trends in Electron Affinity (Groups 16 and 17) (Syllabus 23.1.2c)
Halogens (Group 17) and Group 16 elements tend to have the most negative (most exothermic) first electron affinities.
- Why? These elements are close to achieving a stable noble gas configuration. Gaining just one (Group 17) or two (Group 16) electrons results in a highly stable electron arrangement, making the process energetically favourable.
- Trends within a group: $EA_1$ generally becomes less negative (less exothermic) as you move down the group. Why? The atomic radius increases, meaning the nucleus has less attraction for the incoming electron, releasing less energy.
🔑 Quick Review: Energy Signs
- Endothermic (\(\Delta H\) is +): Atomisation (\(\Delta H_{\text{at}}\)), Ionisation Energy (IE), Second Electron Affinity (\(EA_2\)). Energy goes IN.
- Exothermic (\(\Delta H\) is -): First Electron Affinity (\(EA_1\)) (usually), Lattice Energy (\(\Delta H_{\text{latt}}\)), Enthalpy of Formation (\(\Delta H_{\text{f}}\)) (usually). Energy comes OUT.
Part 2: Lattice Energy (\(\Delta H_{\text{latt}}\))
This is the energy released when the gaseous ions come together to form the solid ionic structure.
1. Definition and Convention (Syllabus 23.1.1b)
The Cambridge syllabus uses a specific definition for lattice energy, defining it as the formation of the solid lattice.
Definition: The enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions.
- Example for Sodium Chloride: \(Na^+(g) + Cl^-(g) \rightarrow NaCl(s)\)
- Lattice energy is a measure of the strength of the electrostatic forces in the ionic solid. Since forming strong bonds releases energy, $\Delta H_{\text{latt}}$ is always exothermic (negative $\Delta H$).
2. Factors Affecting the Magnitude of Lattice Energy (Syllabus 23.1.5)
We are interested in the numerical magnitude of the lattice energy (how large the number is, ignoring the negative sign). A larger magnitude means stronger attractive forces in the solid lattice.
The strength of the electrostatic force (and thus the magnitude of $\Delta H_{\text{latt}}$) is determined by two main factors:
(a) Ionic Charge (Q)
- The force of attraction is proportional to the product of the charges of the ions (\(Q_1 \times Q_2\)).
- Rule: Higher charge leads to stronger attraction.
- Example: MgS ($+2$ and $-2$ ions) has a much larger lattice energy magnitude than NaCl ($+1$ and $-1$ ions), because the forces are four times as strong (2x2 vs 1x1).
(b) Ionic Radius (r)
- The force of attraction is inversely proportional to the square of the distance between the nuclei (approximately the sum of the ionic radii, $r_{cation} + r_{anion}$).
- Rule: Smaller ionic radius leads to stronger attraction (ions are closer).
- Example: LiF (small ions) has a larger lattice energy magnitude than CsI (large ions).
💡 Friendly Tip: Explaining Magnitude
When asked to explain the difference in $\Delta H_{\text{latt}}$ magnitude between two salts, always follow this order:
- Check Charge first: If charges are different (e.g., NaCl vs MgO), charge dominates. State which product of charges is larger.
- Check Radius second: If charges are the same (e.g., NaCl vs KCl), discuss the radii. State which ion is smaller, resulting in a smaller internuclear distance.
- Conclusion: Stronger electrostatic attraction means a more negative $\Delta H_{\text{latt}}$ (larger numerical magnitude).
Part 3: The Born-Haber Cycle (Syllabus 23.1.3 & 23.1.4)
The Born-Haber cycle uses Hess's Law to calculate a specific enthalpy change ($\Delta H_{\text{latt}}$ or $\Delta H_{\text{EA}}$) that cannot be measured directly. It links the overall formation of an ionic solid ($\Delta H_{\text{f}}$) to a sequence of steps involving gaseous atoms and ions.
1. Principle: Hess's Law Applied
Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken. The Born-Haber cycle compares two pathways:
- The Direct Route: Enthalpy of formation ($\Delta H_{\text{f}}$).
- The Indirect Route: The sum of all the individual steps (Atomisation, Ionisation, Electron Affinity, Lattice Formation).
According to Hess's Law:
$$ \Delta H_{\text{f}} = \Delta H_{\text{at, metal}} + \Delta H_{\text{at, non-metal}} + \text{IE} + \text{EA} + \Delta H_{\text{latt}} $$By knowing all but one value, we can calculate the unknown value (usually $\Delta H_{\text{latt}}$ or the $\Delta H_{\text{EA}}$). This is a great way to measure $\Delta H_{\text{latt}}$ indirectly!
2. Constructing the Cycle (Step-by-Step Example: KCl)
The cycle starts with the elements in their standard states at the bottom, and progresses upwards to the gaseous ions, before dropping back down to the ionic solid.
- Start with the Target: Write the equation for the standard enthalpy of formation ($\Delta H_{\text{f}}$).
\(K(s) + \frac{1}{2}Cl_2(g) \rightarrow KCl(s)\) (This is the bottom step in the diagram) - Atomise the Metal: Turn the solid metal into gaseous atoms (endothermic, $\Delta H_{\text{at}}$).
\(K(s) \rightarrow K(g)\) - Atomise the Non-Metal: Turn the non-metal molecule into gaseous atoms (endothermic, $\Delta H_{\text{at}}$).
\(\frac{1}{2}Cl_2(g) \rightarrow Cl(g)\) - Ionise the Metal: Turn the gaseous metal atoms into gaseous cations (endothermic, IE).
\(K(g) \rightarrow K^+(g) + e^-\) (Use multiple IE values if forming a $+2$ ion, e.g., $Mg^{2+}$) - Electron Affinity for Non-Metal: Turn the gaseous non-metal atoms into gaseous anions (exothermic for $EA_1$, endothermic for $EA_2$).
\(Cl(g) + e^- \rightarrow Cl^-(g)\) (If forming $O^{2-}$, use $EA_1 + EA_2$) - Lattice Formation: Bring the gaseous ions together to form the solid lattice (highly exothermic, $\Delta H_{\text{latt}}$).
\(K^+(g) + Cl^-(g) \rightarrow KCl(s)\) (This closes the cycle)
Note: The syllabus limits Born-Haber cycle calculations to solids containing $+1$ and $+2$ cations, and $-1$ and $-2$ anions (e.g., NaCl, MgCl\(_2\), MgO, Na\(_2\)O).
3. Calculation Example (Using the Cycle)
If we want to calculate the lattice energy of KCl, we rearrange the Hess's Law equation:
$$ \Delta H_{\text{latt}} = \Delta H_{\text{f}} - (\Delta H_{\text{at, K}} + \Delta H_{\text{at, Cl}} + \text{IE}_1 + \text{EA}_1) $$When performing the calculation, remember the signs!
🛑 Common Mistake Alert: Sign Errors
Always pay close attention to the definition of the term given in the question versus the direction in your cycle. If a process is reversed (e.g., using $\Delta H_{\text{dissociation}}$ instead of $\Delta H_{\text{formation}}$), you must flip the sign.
- IE and $\Delta H_{\text{at}}$ are always positive when defined correctly.
- $\Delta H_{\text{latt}}$ (formation from gaseous ions) is always negative.
Part 4: Enthalpies of Solution and Hydration
When an ionic solid dissolves in water, two major energy changes occur. We can link these changes using an energy cycle analogous to the Born-Haber cycle (Syllabus 23.2.2).
1. Defining Solution and Hydration (Syllabus 23.2.1)
(a) Enthalpy Change of Solution (\(\Delta H_{\text{sol}}\))
Definition: The enthalpy change when one mole of an ionic solid dissolves completely in a large excess of water to form an infinitely dilute solution.
- Example: \(NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)\)
- $\Delta H_{\text{sol}}$ can be positive (endothermic, feels cold) or negative (exothermic, feels hot).
(b) Enthalpy Change of Hydration (\(\Delta H_{\text{hyd}}\))
Definition: The enthalpy change when one mole of specified gaseous ions dissolves in a large excess of water to form an infinitely dilute solution of aqueous ions.
- Example: \(Na^+(g) \rightarrow Na^+(aq)\)
- The process of hydration involves the attractive forces between the polar water molecules and the ions. This attraction releases energy, so $\Delta H_{\text{hyd}}$ is always exothermic (negative $\Delta H$).
- Analogy: Imagine a gaseous ion floating alone. As soon as it enters the water, it gets surrounded by water molecules (the solvent shell, or hydration sphere). This "wrapping" releases energy.
2. The Energy Cycle for Solution
Dissolving an ionic solid can be broken down into two theoretical steps:
- Break the solid lattice into gaseous ions (The opposite of $\Delta H_{\text{latt}}$).
- Hydrate the gaseous ions (Add the $\Delta H_{\text{hyd}}$ for all ions).
This gives us the critical relationship (derived from Hess's Law):
$$ \Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \sum \Delta H_{\text{hyd}} $$(Or, more commonly written, if $\Delta H_{\text{latt}}$ is defined as $\text{gas ions} \rightarrow \text{solid}$, then: $\Delta H_{\text{sol}} = \Delta H_{\text{dissociation}} + \sum \Delta H_{\text{hyd}}$)
Let's use the syllabus definitions (where $\Delta H_{\text{latt}}$ is formation: $G \to S$):
$$ \Delta H_{\text{sol}} = \text{Energy to break the lattice (endothermic)} + \text{Energy released by hydration (exothermic)} $$Since breaking the lattice is the reverse of $\Delta H_{\text{latt}}$ formation, the equation is:
$$ \mathbf{\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \sum \Delta H_{\text{hyd}}} $$3. Factors Affecting the Magnitude of Hydration Enthalpy (Syllabus 23.2.4)
Just like lattice energy, the magnitude of $\Delta H_{\text{hyd}}$ (how negative it is) depends on the strength of the attraction between the ion and the water molecules. This is governed by the charge density of the ion.
(a) Ionic Charge (Q)
- Higher charge leads to a more negative $\Delta H_{\text{hyd}}$. A $2+$ ion attracts the water dipoles much more strongly than a $1+$ ion.
- Example: $\Delta H_{\text{hyd}}$ for $Mg^{2+}$ is much more negative than for $Na^+$.
(b) Ionic Radius (r)
- Smaller radius leads to a more negative $\Delta H_{\text{hyd}}$. A smaller ion concentrates the charge over a smaller volume (high charge density), creating a stronger electric field to attract water molecules.
- Example: $\Delta H_{\text{hyd}}$ for $Li^+$ is more negative than for $Cs^+$.
In summary, both $\Delta H_{\text{latt}}$ and $\Delta H_{\text{hyd}}$ become more negative (larger magnitude) when ions have higher charges and smaller radii.
✅ Key Takeaway and Connection
The stability of an ionic compound in its solid state is measured by its Lattice Energy ($\Delta H_{\text{latt}}$). Its ability to dissolve in water is determined by the balance between the energy required to break the lattice (the cost, $-\Delta H_{\text{latt}}$) and the energy gained from surrounding the ions with water (the profit, $\sum \Delta H_{\text{hyd}}$).
For a salt to be soluble, the energy gained from hydration must be large enough to compensate for the energy required to break the strong ionic bonds. High charge density ions tend to have large (negative) $\Delta H_{\text{latt}}$ *and* large (negative) $\Delta H_{\text{hyd}}$, often leading to high solubility trends (e.g., Group 2 metal compounds).