A Level Chemistry: Lattice Energy, Entropy and Gibbs Free Energy — Revision Notes

Condensed for the final weeks. For the full explanation, use the Lattice Energy, Entropy and Gibbs Free Energy study guide.

Definitions that must be exact

Term	Definition
Lattice energy ΔH_latt	Enthalpy change when one mole of an ionic compound is formed from its gaseous ions
Enthalpy of atomisation	Enthalpy change when one mole of gaseous atoms is formed from the element in its standard state
First electron affinity	Enthalpy change when one mole of gaseous atoms each gain an electron to form 1− ions
Enthalpy of hydration	Enthalpy change when one mole of gaseous ions is dissolved in an excess of water
Enthalpy of solution	Enthalpy change when one mole of solute dissolves to infinite dilution

Two details carry marks: “one mole” in every definition, and “gaseous” in the lattice, atomisation, affinity and hydration definitions.

Lattice energy defined this way is always exothermic (negative) — bonds are being formed. First electron affinity is exothermic; second and subsequent electron affinities are endothermic, because an electron is being forced onto an already negative ion.

What controls lattice energy magnitude

lattice energy  proportional to  (charge+ x charge-) / (r+ + r-)

• Greater ionic charge → more exothermic. This dominates.
• Smaller ionic radius → more exothermic.

So MgO (2+/2−, small ions) has a far more exothermic lattice energy than NaCl (1+/1−). Charge matters more than size — compare MgO with NaCl and the charge product quadruples.

Born–Haber cycles

Apply Hess’s law around the cycle. The reliable method:

1. Write ΔH_f at the bottom, elements in standard states.
2. Go up via atomisation, then ionisation energies (cation) and electron affinities (anion) to reach gaseous ions.
3. Lattice energy takes you from gaseous ions down to the solid.
4. Set the two routes equal.

delta-H_f  =  sum(atomisation)  +  sum(IE)  +  sum(EA)  +  delta-H_latt

Watch the multipliers. Cl₂ → 2Cl needs 2 × atomisation; Mg²⁺ needs both first and second ionisation energy.

(Background, beyond the 9701 syllabus — included for interest/extension only, not examinable.) Comparing the experimental lattice energy with the theoretical value from a purely ionic model reveals covalent character — a large discrepancy means significant polarisation of the anion by the cation.

Enthalpies of solution and hydration

Enthalpy of hydration ΔH_hyd — enthalpy change when one mole of a gaseous ion dissolves in water to form aqueous ions. Always exothermic, dominated by the attraction between the ion and surrounding polar water molecules.

Enthalpy of solution ΔH_sol — enthalpy change when one mole of an ionic solid dissolves to form an infinitely dilute solution. Can be either sign, depending on the balance between the lattice-breaking energy and the hydration energy released.

The cycle: dissolving is broken into two steps — first break the lattice apart into gaseous ions (the reverse of lattice energy, so +|ΔH_latt|), then hydrate those gaseous ions:

delta-H_sol = -delta-H_latt + delta-H_hyd(cation) + delta-H_hyd(anion)

Worked example. ΔH_latt(NaCl) = −787 kJ mol⁻¹, ΔH_hyd(Na⁺) = −406 kJ mol⁻¹, ΔH_hyd(Cl⁻) = −364 kJ mol⁻¹:

delta-H_sol = -(-787) + (-406) + (-364) = 787 - 770 = +17 kJ/mol

A small positive value is typical for simple 1:1 salts like NaCl — the two large terms nearly cancel, so the small residual can fall on either side of zero.

Charge and radius: hydration enthalpy becomes more exothermic as ionic charge increases and ionic radius decreases — the same size/charge logic as lattice energy, which is why the two often move together and partly cancel in the solution cycle.

Entropy

Entropy S is a measure of the disorder, or the number of ways energy and particles can be arranged.

delta-S = S(products) - S(reactants)      units J K^-1 mol^-1

Note the units: J, not kJ. Gibbs calculations mix them, and this is where most arithmetic marks vanish.

Order of entropy: gas ≫ liquid > solid.

Predict the sign by counting moles of gas:

Change	ΔS
More moles of gas produced	Positive
Fewer moles of gas	Negative
Solid → liquid → gas	Positive
Dissolving a solid	Usually positive

If the number of gas moles is unchanged, ΔS is small and its sign needs closer thought.

Gibbs free energy

delta-G = delta-H - T delta-S       (T in KELVIN, delta-S converted to kJ)

ΔG negative → reaction is feasible. ΔG = 0 gives the temperature at which feasibility changes:

T = delta-H / delta-S

ΔH	ΔS	Feasible when
−	+	Always
+	−	Never
−	−	Low temperature
+	+	High temperature

Those four rows answer most Gibbs questions on sight.

As always: feasible ≠ fast. A reaction with a very negative ΔG may have an activation energy so high that no observable change occurs.

Exam traps

• Omitting “one mole” or “gaseous” from a definition.
• Mixing J and kJ in ΔG = ΔH − TΔS.
• Using °C instead of K.
• Forgetting the second ionisation energy for a 2+ ion, or the ×2 for a diatomic atomisation.
• Saying second electron affinity is exothermic.
• Concluding that a feasible reaction will be observed.

Self-test

1. Define lattice energy precisely.
2. Why is the second electron affinity endothermic?
3. Which has the more exothermic lattice energy, NaCl or MgO, and why?
4. State the Gibbs equation and the condition for feasibility.
5. A reaction has ΔH positive and ΔS positive. When is it feasible?

Answers: 1. The enthalpy change when one mole of an ionic compound is formed from its gaseous ions. 2. An electron is being added to an already negatively charged ion, so energy must be supplied to overcome the repulsion. 3. MgO — the ions carry double the charge and are smaller, and lattice energy is proportional to the product of the charges divided by the sum of the radii. 4. ΔG = ΔH − TΔS; the reaction is feasible when ΔG is negative. 5. At high temperature, where TΔS exceeds ΔH.