Welcome to Lattice Enthalpy
Have you ever wondered why table salt (\(\text{NaCl}\)) forms hard, brittle crystals that require temperatures over \(800^\circ\text{C}\) to melt? The secret lies in lattice enthalpy — the tremendous electrostatic "glue" that binds oppositely charged ions together into an orderly 3D crystal lattice.
In this chapter of A2 1: Further Physical and Organic Chemistry, we will explore how ionic lattices are built, how to calculate the energy holding them together using Born-Haber cycles, why some ionic compounds have unexpected covalent character, and what happens energetically when a solid dissolves in water.
Don't worry if thermodynamic cycles seem tricky at first! By breaking down each step into simple, logical pieces, you will master these calculations with confidence.
1. Key Enthalpy Definitions
Before building cycles, you must know your key definitions and standard states. In CCEA examinations, writing precise definitions (including standard conditions of \(298\text{ K}\) and \(100\text{ kPa}\)) is worth easy marks.
Lattice Enthalpy (\(\Delta_{\text{latt}}H^\ominus\))
There are two ways to describe lattice enthalpy, but in CCEA A Level, standard lattice formation enthalpy is most commonly defined as:
The enthalpy change when one mole of an ionic compound is formed from its constituent gaseous ions under standard conditions (\(298\text{ K}\), \(100\text{ kPa}\)).
Equation example:
\(\text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)} \rightarrow \text{NaCl(s)}\) \(\Delta_{\text{latt}}H^\ominus = -787\text{ kJ mol}^{-1}\)
• Sign: Because strong bonds are being formed between gaseous ions, lattice formation enthalpy is always exothermic (a negative value, \(\Delta H < 0\)).
• Note: Lattice dissociation enthalpy is the exact opposite process (\(\text{NaCl(s)} \rightarrow \text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)}\)), which is always endothermic (\(+787\text{ kJ mol}^{-1}\)). Always check which direction your question specifies!
Enthalpy of Formation (\(\Delta_{\text{f}}H^\ominus\))
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
Equation example:
\(\text{Na(s)} + \frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{NaCl(s)}\)
Enthalpy of Atomisation (\(\Delta_{\text{at}}H^\ominus\))
The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.
Equation examples:
\(\text{Na(s)} \rightarrow \text{Na(g)}\) (for a solid metal)
\(\frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{Cl(g)}\) (for a diatomic gas, this is equal to half the bond dissociation enthalpy)
Ionisation Enthalpy (\(\Delta_{\text{ie}}H^\ominus\))
• First Ionisation Enthalpy (\(\Delta_{\text{ie1}}H^\ominus\)): The enthalpy change required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous \(1+\) ions.
\(\text{Na(g)} \rightarrow \text{Na}^+\text{(g)} + \text{e}^-\)
• Second Ionisation Enthalpy (\(\Delta_{\text{ie2}}H^\ominus\)): The enthalpy change to remove one mole of electrons from one mole of gaseous \(1+\) ions to form one mole of gaseous \(2+\) ions.
\(\text{Mg}^+\text{(g)} \rightarrow \text{Mg}^{2+}\text{(g)} + \text{e}^-\)
Memory Tip: Ionisation enthalpy is always endothermic (\(\Delta H > 0\)) because energy is needed to pull a negatively charged electron away from the positive pull of the nucleus.
Electron Affinity (\(\Delta_{\text{ea}}H^\ominus\))
• First Electron Affinity (\(\Delta_{\text{ea1}}H^\ominus\)): The enthalpy change when one mole of gaseous atoms each gain one electron to form one mole of gaseous \(1-\) ions.
\(\text{Cl(g)} + \text{e}^- \rightarrow \text{Cl}^-\text{(g)}\) (Exothermic, \(\Delta H < 0\), because the incoming electron is attracted to the nucleus).
• Second Electron Affinity (\(\Delta_{\text{ea2}}H^\ominus\)): The enthalpy change when one mole of gaseous \(1-\) ions each gain an electron to form one mole of gaseous \(2-\) ions.
\(\text{O}^-\text{(g)} + \text{e}^- \rightarrow \text{O}^{2-}\text{(g)}\) (Endothermic, \(\Delta H > 0\), because energy must be supplied to overcome the electrostatic repulsion between the negative ion and the incoming negative electron!).
Key Takeaway
Every enthalpy term in lattice energy calculations must have balanced state symbols. Gaseous ions, gaseous atoms, and standard states are crucial for picking up full marks!
2. The Born-Haber Cycle
Lattice enthalpy cannot be measured directly in a lab experiment because you cannot simply take isolated gaseous ions and easily compress them into a crystal. Instead, we use an indirect route based on Hess's Law called a Born-Haber cycle.
Step-by-Step Construction of a Born-Haber Cycle
Think of a Born-Haber cycle like an energy ladder. Going up means adding energy (endothermic steps, \(\Delta H > 0\)), and going down means energy is released (exothermic steps, \(\Delta H < 0\)).
Let's trace the steps to make \(\text{NaCl(s)}\) starting from elements in standard states: \(\text{Na(s)} + \frac{1}{2}\text{Cl}_2\text{(g)}\):
1. Atomise the metal: Convert solid sodium into gaseous atoms.
\(\text{Na(s)} \rightarrow \text{Na(g)}\) [\(\Delta_{\text{at}}H^\ominus(\text{Na})\)]
2. Atomise the non-metal: Break chlorine gas into gaseous atoms.
\(\frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{Cl(g)}\) [\(\Delta_{\text{at}}H^\ominus(\text{Cl})\)]
3. Ionise the metal: Remove electrons to form gaseous cations.
\(\text{Na(g)} \rightarrow \text{Na}^+\text{(g)} + \text{e}^-\) [\(\Delta_{\text{ie1}}H^\ominus(\text{Na})\)]
4. Add electrons to the non-metal: Form gaseous anions.
\(\text{Cl(g)} + \text{e}^- \rightarrow \text{Cl}^-\text{(g)}\) [\(\Delta_{\text{ea1}}H^\ominus(\text{Cl})\)]
5. Form the crystal lattice: Combine gaseous ions to form the solid ionic compound.
\(\text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)} \rightarrow \text{NaCl(s)}\) [\(\Delta_{\text{latt}}H^\ominus\)]
6. Direct Route: Direct formation of the solid compound from standard elements.
\(\text{Na(s)} + \frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{NaCl(s)}\) [\(\Delta_{\text{f}}H^\ominus(\text{NaCl})\)]
The Mathematical Formula
By Hess's Law, the total energy of the direct route equals the total energy of the indirect route:
\(\Delta_{\text{f}}H^\ominus = \Delta_{\text{at}}H^\ominus(\text{metal}) + \Delta_{\text{ie}}H^\ominus(\text{metal}) + \Delta_{\text{at}}H^\ominus(\text{non-metal}) + \Delta_{\text{ea}}H^\ominus(\text{non-metal}) + \Delta_{\text{latt}}H^\ominus\)
Rearranging to find lattice formation enthalpy (\(\Delta_{\text{latt}}H^\ominus\)):
\(\Delta_{\text{latt}}H^\ominus = \Delta_{\text{f}}H^\ominus - \left[ \Delta_{\text{at}}H^\ominus(\text{metal}) + \Delta_{\text{ie}}H^\ominus(\text{metal}) + \Delta_{\text{at}}H^\ominus(\text{non-metal}) + \Delta_{\text{ea}}H^\ominus(\text{non-metal}) \right]\)
Worked Example: Calculating \(\Delta_{\text{latt}}H^\ominus\) for Magnesium Chloride (\(\text{MgCl}_2\))
Data:
• \(\Delta_{\text{f}}H^\ominus(\text{MgCl}_2) = -641\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{at}}H^\ominus(\text{Mg}) = +148\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{ie1}}H^\ominus(\text{Mg}) = +738\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{ie2}}H^\ominus(\text{Mg}) = +1451\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{at}}H^\ominus(\text{Cl}) = +122\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{ea1}}H^\ominus(\text{Cl}) = -349\text{ kJ mol}^{-1}\)
Step-by-Step Solution:
Watch out for stoichiometry! Magnesium forms a \(2+\) ion (needs 1st and 2nd IE), and there are two moles of chlorine atoms formed and ionised (multiply by 2!).
• Total energy to gaseous ions \(= \Delta_{\text{at}}H(\text{Mg}) + \Delta_{\text{ie1}}H(\text{Mg}) + \Delta_{\text{ie2}}H(\text{Mg}) + 2 \times \Delta_{\text{at}}H(\text{Cl}) + 2 \times \Delta_{\text{ea1}}H(\text{Cl})\)
• Total \(= (+148) + (+738) + (+1451) + 2(+122) + 2(-349)\)
• Total \(= 148 + 738 + 1451 + 244 - 698 = +1883\text{ kJ mol}^{-1}\)
Now use: \(\Delta_{\text{f}}H^\ominus = \text{Total} + \Delta_{\text{latt}}H^\ominus\)
\(-641 = +1883 + \Delta_{\text{latt}}H^\ominus\)
\(\Delta_{\text{latt}}H^\ominus = -641 - 1883 = \mathbf{-2524\text{ kJ mol}^{-1}}\)
Common Mistakes to Avoid
• Forgetting multipliers: If the formula is \(\text{MgCl}_2\), remember you need \(2 \times \Delta_{\text{at}}H(\text{Cl})\) and \(2 \times \Delta_{\text{ea}}H(\text{Cl})\).
• Missing the 2nd Ionisation Energy: Group 2 metals (\(\text{Mg}\), \(\text{Ca}\), \(\text{Ba}\)) require both \(\Delta_{\text{ie1}}H\) and \(\Delta_{\text{ie2}}H\).
• Sign errors: Be careful when subtracting negative values like electron affinities.
3. Factors Affecting Lattice Enthalpy
The magnitude of lattice enthalpy depends directly on the electrostatic force of attraction between oppositely charged ions, governed by two main factors:
1. Ionic Charge: Greater charge on the ions means stronger electrostatic attraction between them. For example, \(\text{MgO}\) (\(\text{Mg}^{2+}\) and \(\text{O}^{2-}\)) has a much higher magnitude of lattice enthalpy than \(\text{NaCl}\) (\(\text{Na}^+\) and \(\text{Cl}^-\)).
2. Ionic Radius: Smaller ions can pack closer together, placing the positive nucleus and negative electrons closer together. Smaller inter-ionic distance leads to stronger attraction and a more exothermic lattice enthalpy.
Example Comparison:
• \(\text{NaF}\) has a more exothermic lattice enthalpy than \(\text{NaCl}\) because the fluoride ion (\(\text{F}^-\)) is smaller than the chloride ion (\(\text{Cl}^-\)).
• \(\text{CaO}\) has a significantly more exothermic lattice enthalpy (\(-3401\text{ kJ mol}^{-1}\)) than \(\text{NaCl}\) (\(-787\text{ kJ mol}^{-1}\)) because both ions carry \(2+\) and \(2-\) charges compared to \(1+\) and \(1-\).
Key Takeaway
Greatest lattice enthalpy (most exothermic) \(\rightarrow\) High ionic charge + Small ionic radius (high charge density).
4. Theoretical vs. Experimental Lattice Enthalpy (Covalent Character)
Scientists can calculate theoretical lattice enthalpies using purely physical electrostatic calculations (assuming the purely ionic model). In this model, ions are assumed to be perfect, hard spheres with point charges and purely electrostatic attractions.
Comparing Values
• For compounds like \(\text{NaCl}\), the theoretical value (\(-770\text{ kJ mol}^{-1}\)) and experimental Born-Haber value (\(-787\text{ kJ mol}^{-1}\)) match almost identically. This confirms that \(\text{NaCl}\) is nearly 100% ionic.
• For compounds like \(\text{AgI}\), the theoretical value is \(-758\text{ kJ mol}^{-1}\), but the experimental Born-Haber value is \(-890\text{ kJ mol}^{-1}\). The experimental lattice enthalpy is significantly more exothermic than predicted!
Why the Discrepancy? (Polarisation and Fajans' Rules)
When there is a large difference between experimental and theoretical values, it means the bonding is not purely ionic — it has significant covalent character.
This happens due to polarisation of the anion by the cation:
• Cation Polarising Power: A small cation with a high positive charge (high charge density, e.g., \(\text{Al}^{3+}\), \(\text{Mg}^{2+}\), \(\text{Ag}^+\)) strongly attracts the electron cloud of a neighbouring anion.
• Anion Polarisability: A large anion with a high negative charge (e.g., \(\text{I}^-\)) has loosely held outer electrons that are easily distorted/polarised by the cation.
• Result: The electron cloud is pulled into the region between the nuclei, creating orbital overlap and additional covalent character. This makes the lattice stronger and more exothermic than the purely ionic model predicts.
Did you know? Fajans' Rules explain why \(\text{AlCl}_3\) is covalent and sublimes at low temperatures, while \(\text{AlF}_3\) remains predominantly ionic!
Key Takeaway
A large difference between theoretical and Born-Haber lattice enthalpy indicates polarisation and covalent character in the ionic lattice.
5. Enthalpy of Solution and Hydration
What happens when you dissolve an ionic solid in water? The process involves two competing energetic steps: breaking down the crystal lattice and hydrating the isolated ions.
Definitions
• Standard Enthalpy of Solution (\(\Delta_{\text{sol}}H^\ominus\)): The enthalpy change when one mole of an ionic solid dissolves completely in an amount of water large enough that the ions are separated and do not interact with each other under standard conditions.
\(\text{NaCl(s)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(g)}\)
• Standard Enthalpy of Hydration (\(\Delta_{\text{hyd}}H^\ominus\)): The enthalpy change when one mole of specified gaseous ions dissolves in water to form one mole of aqueous ions under standard conditions.
\(\text{Mg}^{2+}\text{(g)} + \text{aq} \rightarrow \text{Mg}^{2+}\text{(aq)}\) (Always exothermic, \(\Delta H < 0\), due to ion-dipole attractions between ions and polar \(\text{H}_2\text{O}\) molecules!).
The Solution Cycle Formula
We can link these enthalpy changes in an energy cycle:
\(\Delta_{\text{sol}}H^\ominus = \Delta_{\text{latt dissociation}}H^\ominus + \sum \Delta_{\text{hyd}}H^\ominus\)
Or using lattice formation enthalpy (\(\Delta_{\text{latt formation}}H^\ominus\)):
\(\Delta_{\text{sol}}H^\ominus = -\Delta_{\text{latt formation}}H^\ominus + \sum \Delta_{\text{hyd}}H^\ominus\)
Factors Affecting Enthalpy of Hydration
Just like lattice enthalpy, hydration enthalpy depends on charge density:
1. Higher ionic charge \(\rightarrow\) stronger attraction to water dipoles \(\rightarrow\) more exothermic \(\Delta_{\text{hyd}}H^\ominus\).
2. Smaller ionic radius \(\rightarrow\) water molecules can get closer to the ion's charge center \(\rightarrow\) more exothermic \(\Delta_{\text{hyd}}H^\ominus\).
Example: \(\Delta_{\text{hyd}}H^\ominus\) of \(\text{Mg}^{2+}\) (\(-1920\text{ kJ mol}^{-1}\)) is much more exothermic than that of \(\text{Na}^+\) (\(-406\text{ kJ mol}^{-1}\)).
Worked Example: Enthalpy of Solution Calculation
Data:
• \(\Delta_{\text{latt formation}}H^\ominus(\text{CaCl}_2) = -2258\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{hyd}}H^\ominus(\text{Ca}^{2+}) = -1650\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{hyd}}H^\ominus(\text{Cl}^-) = -364\text{ kJ mol}^{-1}\)
Calculation:
• \(\Delta_{\text{latt dissociation}}H^\ominus = +2258\text{ kJ mol}^{-1}\)
• Total hydration enthalpy \(= \Delta_{\text{hyd}}H(\text{Ca}^{2+}) + 2 \times \Delta_{\text{hyd}}H(\text{Cl}^-)\)
• Total hydration enthalpy \(= (-1650) + 2(-364) = -1650 - 728 = -2378\text{ kJ mol}^{-1}\)
• \(\Delta_{\text{sol}}H^\ominus = (+2258) + (-2378) = \mathbf{-120\text{ kJ mol}^{-1}}\)
Since \(\Delta_{\text{sol}}H^\ominus\) is exothermic (\(-120\text{ kJ mol}^{-1}\)), \(\text{CaCl}_2\) dissolves readily and releases heat (which is why calcium chloride is used in commercial self-heating cans and de-icers!).
Key Takeaway
Dissolving is an energy balance between breaking the lattice (endothermic) and hydrating the ions (exothermic). If the enthalpy of solution is slightly endothermic or exothermic, the substance is likely to dissolve soluble in water (supported by increases in entropy).
Quick Chapter Summary Checklist
Before sitting your exam, ensure you can:
• Write standard definitions including state symbols for \(\Delta_{\text{latt}}H^\ominus\), \(\Delta_{\text{at}}H^\ominus\), \(\Delta_{\text{ie}}H^\ominus\), \(\Delta_{\text{ea}}H^\ominus\), \(\Delta_{\text{sol}}H^\ominus\), and \(\Delta_{\text{hyd}}H^\ominus\).
• Construct and calculate unknown values from a Born-Haber cycle (taking care of stoichiometric factors).
• Explain the difference between theoretical and experimental lattice enthalpy using polarisation and covalent character.
• Explain trends in lattice and hydration enthalpies based on ionic radius and ionic charge.
• Set up and calculate enthalpy of solution using hydration and lattice enthalpies.