Chemical Thermodynamics · Lesson 7 of 10
Lattice, Solution, and Dilution Enthalpies
“Use energy cycles to explain how ionic crystals form, dissolve, and respond to added solvent.”
• State the lattice-dissociation enthalpy convention and construct a Born–Haber cycle. • Combine atomisation, ionisation, bond dissociation, and electron-gain steps. • Explain how hydration competes with lattice separation during dissolution. • Distinguish solution enthalpy at a specified composition from infinite dilution. • Calculate dilution enthalpy by subtracting consistent solution processes.
Separating a crystal into gaseous ions
An ionic crystal is an organised arrangement of oppositely charged ions. Pulling its ions apart to infinite separation in the gas phase requires energy. Lattice enthalpy makes that separation a definite thermodynamic process. Because different books sometimes use the reverse convention, always begin by writing the equation associated with the value.
The enthalpy change when one mole of an ionic solid separates into its gaseous ions. For sodium chloride, has positive lattice dissociation enthalpy, about in the dataset used here.
The reverse process, gaseous ions forming the crystal, has the opposite value, about −788 kJ/mol. It is difficult to measure either route directly as a simple experiment. Hess’s law provides another route involving quantities that can be obtained separately: the Born–Haber cycle.
Constructing a Born–Haber cycle
Begin with sodium metal and chlorine molecules in their reference states, then imagine producing the gaseous ions step by step. Each step must conserve atoms and charge, and all values must refer to the same reaction amount. Completing the route to sodium chloride gives the same endpoint as direct formation from the elements.
| Step for one mole of NaCl | Enthalpy / kJ mol−1 |
|---|---|
| : sublimation | +108.4 |
| : half the dissociation | +121.0 |
| : ionisation | +496.0 |
| : electron gain | −348.6 |
| : lattice formation | |
| : direct formation | −411.2 |
Problem
Use the steps above to find the positive lattice dissociation enthalpy of NaCl.
- 1.First add the changes that produce gaseous ions from the reference elements: .
- 2.Let L be the positive dissociation value. Forming the crystal from those ions contributes .
- 3.Hess’s law gives . Therefore .
- 4.For the dissociation process there are two moles of gaseous products and no gaseous reactants, so . At 298 K this is approximately .
Ionisation enthalpy and electron-gain enthalpy are thermal reaction quantities. Their relation to zero-temperature ionisation energy and electron affinity includes heat-capacity corrections. This is why “electron affinity” should not be used as an unsigned synonym for electron-gain enthalpy: gaining an electron can have a negative enthalpy change while the corresponding affinity is quoted as a positive energy release.
In that model, , whereas . Here is positive for energy released by electron attachment. These are model-based thermal relations, not a reason to change the sign of a tabulated electron-gain step.
Dissolving balances separation and hydration
When sodium chloride dissolves, ions leave the crystal and become surrounded by water molecules. Separation costs enthalpy, but interactions with water release enthalpy. Hydration means solvation by water; solvation is the more general term for surrounding a solute with solvent particles. A useful Hess route separates these two effects even though actual dissolution does not release a container of bare gaseous ions first.
Problem
Use lattice dissociation enthalpy +788 kJ/mol and total hydration enthalpy −784 kJ/mol to find solution enthalpy.
- 1.Separate the lattice contribution from the hydration contribution. Their signs represent opposite energy transfers.
- 2.Add them: .
- 3.The near cancellation gives a small positive net enthalpy change: dissolving absorbs a small amount of heat under these specified conditions.
- 4.An endothermic dissolution can still proceed. Direction also depends on entropy and Gibbs energy, developed in the next lessons.
Lattice enthalpy influences solubility but cannot determine it alone. Hydration and entropy also matter. Some fluorides have strong lattices and lower solubility than related chlorides, but this is not a universal rule for every fluoride. Similarly, many endothermically dissolving salts become more soluble as temperature rises, yet actual solubility trends require the complete balance.
How much solvent is present?
An enthalpy of solution refers to dissolving one mole of solute in a specified amount of solvent at stated temperature and pressure. Changing that amount changes interactions in the final solution. At infinite dilution, adding further solvent changes the molar enthalpy negligibly: solute particles are so far apart that their mutual interactions become negligible, though their interactions with solvent remain.
| Final solution for 1 mol HCl gas | Solution enthalpy / kJ mol−1 |
|---|---|
| With 10 mol water | −69.01 |
| With 25 mol water | −72.03 |
| With 40 mol water | −72.79 |
| At infinite dilution | −74.85 |
The notation HCl with a stated number of moles of water describes a solution composition, not a distinct stoichiometric hydrate molecule. The enthalpy of dilution is the change when extra solvent is added to a solution of known initial composition. A more negative solution value means more energy has been released in reaching that final solution from the same starting gas and solvent reference.
Problem
Find the enthalpy change when the solution formed from 1 mol HCl and 25 mol water is diluted by another 15 mol water at the same temperature.
- 1.The initial composition has solution enthalpy relative to HCl gas plus water. The final composition has value .
- 2.Subtract the first formation route from the second: .
- 3.The common HCl gas reference cancels, leaving a change from the 25-water solution plus 15 water to the 40-water solution.
- 4.The negative result means this dilution releases heat. “Dilution” does not automatically mean cooling or zero heat transfer.
A sign convention must follow its written process: lattice formation is negative while lattice dissociation is positive. Hydration does not mean the ions stop interacting; it means their surroundings now include strongly interacting solvent molecules.
Quiz
Which process defines positive lattice dissociation enthalpy?
What principle makes a Born–Haber cycle work?
Why is only half the chlorine bond-dissociation enthalpy needed for one mole of NaCl formation?
A salt has lattice dissociation enthalpy +900 kJ/mol and hydration enthalpy −930 kJ/mol. What is its solution enthalpy?
What becomes negligible at infinite dilution?
The initial and final solution enthalpies are −60 and −65 kJ/mol relative to the same references. What is the dilution change?
Practice Problems
- Write the lattice dissociation and lattice formation equations for one mole of sodium chloride, including gaseous-ion charges and enthalpy signs.
- Use formation enthalpy −400, metal sublimation +100, metal ionisation +500, half halogen bond dissociation +120, and electron-gain enthalpy −350 kJ/mol to find a model salt’s lattice dissociation enthalpy.
- Explain why a very large lattice dissociation enthalpy does not alone prove that a salt is insoluble.
- A salt has solution enthalpy +12 kJ/mol and lattice dissociation enthalpy +760 kJ/mol. Find the hydration sum.
- Calculate the enthalpy of diluting the 10-water HCl solution to the 40-water solution using the table.
- Distinguish infinite dilution from absence of solute–solvent interactions, and explain why the amount of solvent must be stated in a solution-enthalpy measurement.
Key Takeaways
A Born–Haber cycle combines gaseous-atom and gaseous-ion steps with crystal formation. Lattice dissociation and formation have opposite signs. Solution enthalpy balances lattice separation against solvation or hydration. Solution and dilution enthalpies depend on the initial and final compositions, not simply the solute name.