Thermodynamics

In a nutshell

Thermodynamics explains why reactions happen. It uses energy cycles to find lattice enthalpies that cannot be measured directly, and it combines enthalpy change with entropy (disorder) to decide whether a reaction is feasible.

The key result is the Gibbs equation ΔG = ΔH − TΔS: a reaction is feasible only when ΔG is zero or negative.

Assumed knowledge: Energetics, Bonding.

Core content

The enthalpy changes you must define

A Born-Haber cycle is built from several standard enthalpy changes. AQA credits precise wording, so learn each with its equation and state symbols.

Two points examiners hammer:

• Bond enthalpy is twice atomisation for a diatomic molecule. Cl₂(g) ⟶ 2Cl(g) is the bond enthalpy; ½Cl₂(g) ⟶ Cl(g) is the atomisation enthalpy, which is half of it.
• The second electron affinity is endothermic (positive). For O⁻(g) + e⁻ ⟶ O²⁻(g) energy is needed to force an electron onto an already negative ion against the repulsion.

Born-Haber cycles

A lattice enthalpy cannot be measured directly, so we find it indirectly. A Born-Haber cycle links the lattice enthalpy to changes that can be measured, then applies Hess's law: the enthalpy change by the direct route equals the sum of the steps round the cycle.

Reading the cycle up from the elements and back down to the solid gives:

Δ_fH^⦵ = Δ_atH^⦵(Na) + IE₁ + Δ_atH^⦵(Cl) + EA₁ + Δ_LH^⦵

Rearranged, this is how you find the lattice enthalpy (worked in full below).

For compounds with more than one ion of a kind, scale the steps:

• MgCl₂, CaF₂ need 2 × atomisation and 2 × electron affinity of the halogen (two halide ions form).
• A Group 2 metal needs both the 1st and 2nd ionisation energies (it forms a 2+ ion).

Lattice enthalpy, the perfect ionic model and covalent character

The size of a lattice enthalpy depends on the ions:

• Charge: higher ionic charge gives a more negative (stronger) lattice enthalpy.
• Radius: smaller ions sit closer together, giving a more negative lattice enthalpy.

So MgO (2+ and 2−, both small) has a far more exothermic lattice enthalpy than NaCl (1+ and 1−).

A theoretical lattice enthalpy can be calculated from the perfect ionic model, which treats the ions as point charges (perfect spheres) with purely electrostatic attraction.

Compare it with the experimental value from a Born-Haber cycle:

• Close agreement means the compound is well described as purely ionic (for example NaCl).
• A large difference is evidence for covalent character: the lattice is held more strongly than a perfect ionic model predicts, so the experimental value is more exothermic than the theoretical one.

Still don't get it? · covalent character from lattice enthalpies

Imagine two magnets wrapped in soft foam. The "perfect ionic" prediction treats each ion as a hard, perfectly round ball that only feels the pull of the opposite charge, nothing more. That gives one predicted strength of attraction, the theoretical lattice enthalpy.

Now the real picture: a small, strongly charged positive ion tugs on the soft electron cloud of a big negative ion and pulls some of that cloud into the space between them. That shared electron density is the start of a covalent bond. It is an extra bit of bonding on top of the pure electrostatic pull, so the real lattice is harder to pull apart than the foam-ball model said.

The exam version: the experimental (Born-Haber) lattice enthalpy is more exothermic than the theoretical (perfect ionic model) value. The gap is caused by covalent character, from the cation polarising the anion, and it is largest when the cation is small and highly charged and the anion is large. A fluoride ion is small and hard to polarise, so fluorides agree well with the ionic model.

Enthalpies of hydration and solution

When an ionic solid dissolves, the lattice breaks apart into gaseous ions and those ions are then surrounded by water.

Enthalpy of hydration, Δ_hydH^⦵ is the enthalpy change when 1 mol of gaseous ions dissolves in water to form 1 mol of aqueous ions: X⁺(g) + aq ⟶ X⁺(aq). It is always exothermic because bonds form between the ions and the polar water molecules.

• Positive ions are attracted to the δ− oxygen of water.
• Negative ions are attracted to the δ+ hydrogen of water.
• Hydration is more exothermic for ions of higher charge density (smaller ions, or higher charge).

Enthalpy of solution, Δ_solH^⦵ is the enthalpy change when 1 mol of a solid ionic compound dissolves in water. A Hess cycle links it to the lattice and hydration enthalpies:

Δ_solH^⦵ = Δ_LH^⦵(dissociation) + ΣΔ_hydH^⦵

(The lattice dissociation enthalpy is positive; if a question gives the lattice formation value instead, change its sign.)

Entropy, ΔS

Entropy (S) measures the disorder of a system: the number of ways its particles and their energy can be arranged. More disorder means higher entropy. Its unit is J K⁻¹ mol⁻¹.

Key facts:

• Entropy rises solid < liquid < gas, with a much bigger jump on boiling than on melting.
• A perfect crystal at 0 K has zero entropy (no disorder).
• Making more gas or more molecules increases entropy.

The small step is melting; the large step is boiling, where disorder increases far more.

Predict the sign of ΔS from the equation. A change from solid or liquid to gas, or an increase in the number of gas molecules, gives ΔS positive:

• NH₄Cl(s) ⟶ NH₃(g) + HCl(g): ΔS positive (solid to gas, more molecules).
• N₂(g) + 3H₂(g) ⟶ 2NH₃(g): ΔS negative (4 mol gas to 2 mol gas).

Calculate it from absolute entropies:

$\Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants})$

Gibbs free-energy change, ΔG, and feasibility

Enthalpy alone does not decide whether a reaction happens: some feasible reactions are endothermic. The balance of enthalpy and entropy is captured by the Gibbs free-energy change:

$\Delta G = \Delta H - T\Delta S$

(derivation not required). A reaction is feasible when ΔG ≤ 0 (negative or zero).

Watch the units: ΔH is in kJ mol⁻¹ but ΔS is in J K⁻¹ mol⁻¹, so divide TΔS by 1000, and use T in kelvin.

Because the −TΔS term grows with temperature, the sign of ΔS decides how feasibility changes with T:

Still don't get it? · how temperature flips feasibility

Think of a tug-of-war. On one side is enthalpy (ΔH): exothermic reactions pull towards "go". On the other side is entropy, but entropy's pull is scaled by temperature in the term −TΔS, so it is like a player whose strength grows the hotter it gets.

If ΔS is positive, the entropy player pulls towards "go" and gets stronger as T rises, so heating makes the reaction feasible (an endothermic decomposition that only works when hot). If ΔS is negative, the entropy player pulls towards "stop" and also gets stronger with T, so heating can drag an otherwise-feasible exothermic reaction back to "stop".

The exam version: feasibility is set by the sign of ΔG in ΔG = ΔH − TΔS. The temperature where it switches is found by setting ΔG = 0, which gives T = ΔH/ΔS.

Feasible does not mean fast. If ΔG is negative but the activation energy is high, the reaction can still be immeasurably slow.

Finding ΔS and ΔH from a ΔG against T graph (skills opportunity MS 3.3)

Rewriting ΔG = ΔH − TΔS as $y = mx + c$ with ΔG as $y$ and T as $x$:

$\Delta G = (-\Delta S)\,T + \Delta H$

So a plot of ΔG against T is a straight line with gradient = −ΔS and intercept (at T = 0) = ΔH. The reaction is feasible wherever the line is below ΔG = 0.

The positive gradient shows ΔS is negative; the line crosses ΔG = 0 at about 462 K, so the reaction is feasible below that temperature.

Worked examples

1. Born-Haber cycle: the lattice enthalpy of NaCl

By Hess's law, formation by the direct route equals the sum round the cycle, so rearrange for the lattice enthalpy:

$\Delta_{L}H^{\ominus} = \Delta_{f}H^{\ominus} - [\,\Delta_{at}H^{\ominus}(\text{Na}) + \text{IE}_1 + \Delta_{at}H^{\ominus}(\text{Cl}) + \text{EA}_1\,]$

$\Delta_{L}H^{\ominus} = -411 - [\,(+107) + (+496) + (+122) + (-349)\,]$

$\Delta_{L}H^{\ominus} = -411 - (+376)$

$\Delta_{L}H^{\ominus} = -787\ \text{kJ mol}^{-1}$

The answer is negative (lattice formation is exothermic). The marks here are the correct use of the atomisation value for Cl, not the bond enthalpy, and keeping the sign.

2. Enthalpy of solution of NaCl

Data: lattice dissociation enthalpy = +787 kJ mol⁻¹; Δ_hydH^⦵(Na⁺) = −406; Δ_hydH^⦵(Cl⁻) = −364 kJ mol⁻¹.

$\Delta_{sol}H^{\ominus} = \Delta_{L}H^{\ominus}(\text{dissociation}) + \Sigma\,\Delta_{hyd}H^{\ominus}$

$\Delta_{sol}H^{\ominus} = (+787) + [\,(-406) + (-364)\,]$

$\Delta_{sol}H^{\ominus} = +787 - 770 = +17\ \text{kJ mol}^{-1}$

Slightly endothermic: NaCl still dissolves because the entropy increase makes ΔG negative. For a compound like MgCl₂ you would add 2 × the chloride hydration enthalpy.

3. Entropy change from absolute entropies

For N₂(g) + 3H₂(g) ⟶ 2NH₃(g), with S^⦵ values 192 (N₂), 131 (H₂) and 193 (NH₃) J K⁻¹ mol⁻¹:

$\Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants})$

$\Delta S^{\ominus} = (2 \times 193) - (192 + 3 \times 131)$

$\Delta S^{\ominus} = 386 - 585 = -199\ \text{J K}^{-1}\,\text{mol}^{-1}$

Negative, because 4 mol of gas become 2 mol of gas, so disorder falls. Keep the sign and the units J K⁻¹ mol⁻¹.

4. Gibbs free energy and feasibility

For the same reaction at 298 K, ΔH = −92 kJ mol⁻¹ and ΔS = −199 J K⁻¹ mol⁻¹ = −0.199 kJ K⁻¹ mol⁻¹:

$\Delta G = \Delta H - T\Delta S$

$\Delta G = -92 - (298 \times -0.199)$

$\Delta G = -92 - (-59.3) = -92 + 59.3$

$\Delta G = -32.7\ \text{kJ mol}^{-1}$

ΔG is negative, so the reaction is feasible at 298 K. The commonest lost mark is forgetting to convert ΔS to kJ (÷1000) before subtracting.

5. The temperature at which a reaction becomes feasible

For CaCO₃(s) ⟶ CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹ = +0.161 kJ K⁻¹ mol⁻¹.

The reaction just becomes feasible when ΔG = 0:

$0 = \Delta H - T\Delta S \quad\Rightarrow\quad T = \frac{\Delta H}{\Delta S}$

$T = \frac{178}{0.161} = 1106\ \text{K}$

Both ΔH and ΔS are positive, so the reaction is feasible above about 1110 K (3 sig figs). Always check the two values are in the same energy unit before dividing.

Common exam mistakes

• Using the bond enthalpy instead of the atomisation enthalpy (or not halving it) for a diatomic gas in a Born-Haber cycle. For Cl₂, the atomisation enthalpy (½Cl₂ ⟶ Cl) is half the bond enthalpy; you need one mole of gaseous atoms.
• Forgetting to scale for the number of moles: MgCl₂ and CaF₂ need 2 × atomisation and 2 × electron affinity of the halogen, and a Group 2 metal needs both ionisation energies.
• Mixing up lattice formation and lattice dissociation: they are equal and opposite, so using the wrong one flips the sign of the whole answer.
• Confusing the terms: lattice enthalpy mistaken for enthalpy of formation or solution, and electron affinity mistaken for ionisation energy.
• Making the second electron affinity exothermic. Adding an electron to an already negative ion (O⁻ ⟶ O²⁻) is endothermic because the electron is repelled.
• Leaving conditions out of definitions: not saying enthalpy change (not "energy"), 1 mole, gaseous ions/atoms, or standard states for formation.
• Describing hydration loosely. The mark needs the named ion (the chloride ion, not "chlorine") attracted to the δ+ hydrogen of water, and cations to the δ− oxygen. "Intermolecular forces", "hydrogen bonding" and "electronegativity" are rejected.
• Getting signs wrong in the solution cycle, or forgetting 2 × a hydration enthalpy when two of that ion form. The shortcut ΔH = ΣΔ_fH(products) − ΣΔ_fH(reactants) works only with formation data, not here.
• Treating entropy as the energy of particles rather than their disorder, and writing the units as kJ instead of J K⁻¹ mol⁻¹.
• In a ΔG calculation, mixing energy units (not converting TΔS to kJ), not converting T to kelvin, or dropping the units and sign of the final answer.
• Arguing feasibility in terms of rate, or rate in terms of thermodynamics. A negative ΔG makes a reaction feasible, but a high activation energy can still make it too slow to observe.

Key definitions

• Enthalpy of formation (Δ_fH^⦵): the enthalpy change when 1 mole of a compound is formed from its elements, with all reactants and products in their standard states, under standard conditions.
• Enthalpy of atomisation (Δ_atH^⦵): the enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state.
• Bond dissociation enthalpy: the enthalpy change when 1 mole of a covalent bond is broken into gaseous atoms.
• First ionisation energy: the enthalpy change when 1 mole of electrons is removed from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions.
• First electron affinity: the enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons to form 1 mole of gaseous 1− ions.
• Lattice enthalpy of formation (Δ_LH^⦵): the enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions under standard conditions.
• Lattice enthalpy of dissociation: the enthalpy change when 1 mole of a solid ionic compound is separated into its gaseous ions under standard conditions.
• Enthalpy of hydration (Δ_hydH^⦵): the enthalpy change when 1 mole of gaseous ions is dissolved in water to form 1 mole of aqueous ions.
• Enthalpy of solution (Δ_solH^⦵): the enthalpy change when 1 mole of a solid ionic compound is dissolved in water (the ions ending up far enough apart not to interact).
• Entropy (S): a measure of the disorder of a system (the number of ways its particles and energy can be arranged); unit J K⁻¹ mol⁻¹.
• Standard entropy change: ΔS^⦵ = ΣS^⦵(products) − ΣS^⦵(reactants).
• Gibbs free-energy change: ΔG = ΔH − TΔS; a reaction is feasible when ΔG ≤ 0.
• Perfect ionic model: a model in which ions are treated as point charges (perfect spheres) with purely electrostatic attraction.

Specification

• I can define lattice enthalpy (as either lattice formation or lattice dissociation).
• I can define enthalpy of formation, ionisation energy, enthalpy of atomisation, bond enthalpy and electron affinity.
• I can construct a Born-Haber cycle and use it to calculate a lattice enthalpy, or one of the other enthalpy changes.
• I can compare a Born-Haber lattice enthalpy with one from the perfect ionic model as evidence for covalent character.
• I can define enthalpy of hydration.
• I can use a cycle to calculate an enthalpy of solution from lattice enthalpy and enthalpies of hydration.
• I can calculate an entropy change from absolute entropy values.
• I can use ΔG = ΔH − TΔS to find ΔG and decide whether a reaction is feasible.
• I can use ΔG = ΔH − TΔS to find how ΔG varies with temperature and the temperature at which a reaction becomes feasible.

Related notes

• Energetics
• Bonding
• Rate equations