OxfordAQA A-Level Chemistry: Thermodynamics (9620)

This study guide covers section 3.1.8 Thermodynamics of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It teaches both subsections: 3.1.8.1 Born–Haber cycles and 3.1.8.2 Gibbs free-energy change, ΔG, and entropy change, ΔS. The specification places 3.1.8 in International A2 physical chemistry, so all of it is International A-level only. It is assessed in Unit 3 (Inorganic 2 and Physical 2), and Unit 5 (Practical and synoptic) can draw on all content.

Use it with the Thermodynamics revision notes and the Thermodynamics practice questions. See also the OxfordAQA A-level Chemistry hub, the printable checklist and the free 10-minute diagnostics.

This topic builds on three International AS pages: Energetics (enthalpy of formation, Hess’s law, bond enthalpy), Atomic structure (ionisation energies) and Bonding and structure (ionic bonding).

What this topic covers

Spec section	What you must be able to do	Stage
3.1.8.1 Born–Haber cycles	Define the five data terms and lattice enthalpy; construct cycles to calculate a lattice enthalpy or one other term; compare with perfect ionic model values as evidence for covalent character	International A-level only
3.1.8.1 Born–Haber cycles	Define enthalpy of hydration; calculate enthalpies of solution and other changes from lattice and hydration enthalpies	International A-level only
3.1.8.2 ΔG and ΔS	Explain why ΔH is not enough to decide feasibility; calculate ΔS from absolute entropies; use ΔG = ΔH – TΔS to see how ΔG varies with T and to find the temperature at which a reaction becomes feasible	International A-level only

No required practical activity is listed under 3.1.8.

3.1.8.1 The enthalpy terms

A Born–Haber cycle is a Hess’s law cycle linking an ionic solid to its elements through gaseous atoms and ions. Each term refers to 1 mole; learn its equation with state symbols.

Term	Definition	Example equation
Enthalpy of formation, ΔfH⦵	Enthalpy change when 1 mole of a compound forms from its elements in their standard states	Cs(s) + ½F₂(g) → CsF(s)
Enthalpy of atomisation, ΔatH⦵	Enthalpy change when 1 mole of gaseous atoms forms from the element in its standard state	Ba(s) → Ba(g)
Bond enthalpy	Enthalpy change to break 1 mole of a covalent bond, with all species gaseous	F₂(g) → 2F(g)
First ionisation energy	Enthalpy change to remove 1 mole of electrons from 1 mole of gaseous atoms, forming 1 mole of gaseous 1+ ions	Cs(g) → Cs⁺(g) + e⁻
Second ionisation energy	Enthalpy change to remove 1 mole of electrons from 1 mole of gaseous 1+ ions, forming gaseous 2+ ions	Ba⁺(g) → Ba²⁺(g) + e⁻
First electron affinity	Enthalpy change when each atom in 1 mole of gaseous atoms gains one electron, forming 1 mole of gaseous 1– ions	F(g) + e⁻ → F⁻(g)
Lattice enthalpy of dissociation	Enthalpy change when 1 mole of an ionic solid separates into its gaseous ions	CsF(s) → Cs⁺(g) + F⁻(g)
Lattice enthalpy of formation	Enthalpy change when 1 mole of an ionic solid forms from its gaseous ions	Cs⁺(g) + F⁻(g) → CsF(s)

Key points:

• The two lattice enthalpies have the same size and opposite signs. Always name the one you use.
• Atomisation is per mole of atoms; bond enthalpy is per mole of bonds. For a gaseous diatomic element, ΔatH⦵ = ½ × bond enthalpy. Iodine is a solid, so ½I₂(s) → I(g) also includes turning the solid into a gas; its value is given separately.
• Ionisation energies are always endothermic. A 2+ ion needs the first and second values.
• First electron affinities of halogens and oxygen are exothermic. The second electron affinity of oxygen, O⁻(g) + e⁻ → O²⁻(g), is endothermic: the negative ion repels the incoming electron.

Constructing a Born–Haber cycle

Draw an energy level diagram. The elements in their standard states sit at zero. Endothermic steps are arrows pointing up; exothermic steps point down. Hess’s law says both routes from the elements to the solid have the same total:

ΔfH = ΔatH(metal) + ΔatH(non-metal) + IE(s) + EA(s) + lattice enthalpy of formation

Multiply every step by the number of times it happens in the formula.

Worked example 1: lattice enthalpy of caesium fluoride

Data (kJ mol⁻¹): ΔfH⦵(CsF) –554; ΔatH⦵(Cs) +76; first IE(Cs) +376; F–F bond enthalpy +158; first EA(F) –328. ΔatH(F) = ½ × 158 = +79.

Build the diagram level by level (energies relative to the elements):

Level	Step to reach it	Energy (kJ mol⁻¹)
Cs(s) + ½F₂(g)	start	0
Cs(g) + ½F₂(g)	ΔatH(Cs), up	+76
Cs(g) + F(g)	ΔatH(F), up	+155
Cs⁺(g) + e⁻ + F(g)	first IE(Cs), up	+531
Cs⁺(g) + F⁻(g)	first EA(F), down	+203
CsF(s)	lattice enthalpy of formation, down	–554

Lattice enthalpy of formation = ΔfH – (level of gaseous ions)
                              = –554 – 203 = –757 kJ mol⁻¹

So the lattice enthalpy of dissociation of CsF is +757 kJ mol⁻¹.

Worked example 2: finding a different term (atomisation of barium)

The same cycle can be solved for any one unknown. For BaCl₂, barium loses two electrons and two chlorine atoms each gain one.

Data (kJ mol⁻¹): ΔfH⦵(BaCl₂) –859; first IE(Ba) +503; second IE(Ba) +965; Cl–Cl bond enthalpy +242; first EA(Cl) –349; lattice enthalpy of formation of BaCl₂ –2051.

Two Cl atoms need one whole Cl–Cl bond broken: +242
Two electron affinities: 2 × (–349) = –698
–859 = ΔatH(Ba) + 503 + 965 + 242 – 698 – 2051
ΔatH(Ba) = –859 – 503 – 965 – 242 + 698 + 2051 = +180 kJ mol⁻¹

The answer is positive, as every atomisation must be; a negative answer signals a sign or multiplier error.

Born–Haber values and the perfect ionic model

Theory gives a calculated value from a perfect ionic model: perfectly spherical ions with evenly spread charge, held only by electrostatic attraction, with no covalent character. A Born–Haber value uses measured data, so it includes the actual bonding.

• Values almost equal: the bonding is very close to purely ionic.
• Born–Haber value clearly more exothermic: the real lattice is held more strongly than pure electrostatics predicts: evidence for covalent character.

Covalent character comes from polarisation: the cation distorts the anion’s electron cloud, pulling electron density between the ions. A small or highly charged cation and a large anion give the most polarisation.

Worked example 3: silver iodide

Data (kJ mol⁻¹): ΔfH⦵(AgI) –62; ΔatH⦵(Ag) +285; first IE(Ag) +731; ΔatH⦵(I) +107; first EA(I) –295. A perfect ionic model value of –808 kJ mol⁻¹ is used for this example.

Steps to gaseous ions = 285 + 731 + 107 + (–295) = +828
Born–Haber lattice enthalpy of formation = –62 – 828 = –890 kJ mol⁻¹
Difference from model = –890 – (–808) = –82 kJ mol⁻¹ (about 10% more exothermic)

The experimental value is 82 kJ mol⁻¹ more exothermic than the model predicts, so AgI has significant covalent character. The large, easily distorted iodide ion is polarised by Ag⁺.

Enthalpy of hydration and enthalpy of solution

Enthalpy of hydration, ΔhydH⦵: the enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions, for example F⁻(g) → F⁻(aq). It is always exothermic: the δ– oxygen atoms of water are attracted to a cation and the δ+ hydrogen atoms to an anion. Smaller or more highly charged ions have more exothermic values.

Enthalpy of solution, ΔsolH⦵: the enthalpy change when 1 mole of an ionic solid dissolves in water, for example RbF(s) → Rb⁺(aq) + F⁻(aq).

The cycle has two routes from solid to aqueous ions: dissolve directly, or break the lattice into gaseous ions and then hydrate them:

ΔsolH = lattice enthalpy of dissociation + Σ ΔhydH(each ion)

Worked example 4: enthalpy of solution of rubidium fluoride

Data (kJ mol⁻¹): lattice enthalpy of dissociation of RbF +785; ΔhydH(Rb⁺) –296; ΔhydH(F⁻) –506.

ΔsolH = +785 + (–296) + (–506) = –17 kJ mol⁻¹

Hydration releases slightly more energy than breaking the lattice needs, so dissolving is exothermic.

Worked example 5: lattice enthalpy from an enthalpy of solution

Data (kJ mol⁻¹): ΔsolH(SrCl₂) –52; ΔhydH(Sr²⁺) –1445; ΔhydH(Cl⁻) –381.

–52 = lattice dissociation + (–1445) + 2 × (–381)
lattice dissociation = –52 + 1445 + 762 = +2155 kJ mol⁻¹

ΔhydH(Cl⁻) counts twice because SrCl₂ contains two Cl⁻ ions.

3.1.8.2 Entropy and feasibility

Why ΔH is not enough

Ice melts on a warm day although melting is endothermic, so ΔH alone cannot decide feasibility. The missing factor is entropy, S, a measure of disorder, in J K⁻¹ mol⁻¹.

Entropy increases (ΔS positive) in both kinds of change:

• Physical changes: solid → liquid → gas. Gas particles move freely and are the most disordered.
• Chemical changes: a reaction that makes more moles of gas than it uses, or turns a solid into a gas.

Calculating ΔS from absolute entropies

ΔS = ΣS(products) – ΣS(reactants)

Scale each entropy by its equation coefficient; elements have non-zero absolute entropies.

The Gibbs free-energy change

The specification gives the relationship

ΔG = ΔH – TΔS        (derivation not required)

with T in kelvin. A reaction is feasible when ΔG is zero or negative. Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ (÷ 1000) to match ΔH.

Melting ice shows the balance. For H₂O(s) → H₂O(l), take ΔH = +6.01 kJ mol⁻¹ and ΔS = +22.0 J K⁻¹ mol⁻¹:

At 263 K: ΔG = 6.01 – 263 × 0.0220 = +0.22 kJ mol⁻¹   (not feasible)
At 283 K: ΔG = 6.01 – 283 × 0.0220 = –0.22 kJ mol⁻¹   (feasible)
ΔG = 0 at T = 6.01 / 0.0220 = 273 K

How ΔG varies with temperature

Treat ΔH and ΔS as fixed. ΔG then changes in a straight line with T, and –TΔS sets the direction.

ΔH	ΔS	As T rises	Feasible
–	+	ΔG stays negative	At all temperatures
+	–	ΔG stays positive	At no temperature
+	+	ΔG becomes more negative	Above T = ΔH/ΔS
–	–	ΔG becomes less negative	Below T = ΔH/ΔS

To find the temperature at which a reaction becomes feasible, set ΔG = 0:

T = ΔH / ΔS        (ΔS in kJ K⁻¹ mol⁻¹)

“Feasible” says nothing about rate: a feasible reaction can be very slow if its activation energy is high.

Worked example 6: ΔH and ΔS both positive

N₂O₄(g) → 2NO₂(g), ΔH⦵ = +57 kJ mol⁻¹. Absolute entropies (J K⁻¹ mol⁻¹): N₂O₄(g) 304; NO₂(g) 240.

ΔS = (2 × 240) – 304 = +176 J K⁻¹ mol⁻¹ = +0.176 kJ K⁻¹ mol⁻¹
ΔG at 298 K = 57 – (298 × 0.176) = +4.6 kJ mol⁻¹   → not feasible
T = 57 / 0.176 = 324 K

ΔS is positive because 1 mole of gas becomes 2. The reaction is feasible at 324 K and above.

Worked example 7: ΔH and ΔS both negative

PCl₃(g) + Cl₂(g) → PCl₅(g), ΔH⦵ = –88 kJ mol⁻¹. Absolute entropies (J K⁻¹ mol⁻¹): PCl₃(g) 312; Cl₂(g) 223; PCl₅(g) 365.

ΔS = 365 – (312 + 223) = –170 J K⁻¹ mol⁻¹ = –0.170 kJ K⁻¹ mol⁻¹
ΔG = –88 – T(–0.170) = –88 + 0.170T
At 300 K: ΔG = –37 kJ mol⁻¹   At 500 K: ΔG = –3 kJ mol⁻¹   At 700 K: ΔG = +31 kJ mol⁻¹
T = –88 / –0.170 = 518 K

ΔG rises as T rises, so the reaction is feasible below 518 K and not above it.

Common errors

• Using one ionisation energy for a 2+ ion, or one hydration enthalpy for an ion that appears twice.
• Writing Cl₂(g) → 2Cl(g) as an atomisation: it makes 2 moles of atoms.
• Halving a bond enthalpy for iodine or bromine, whose standard states are not gases.
• Treating a less exothermic Born–Haber value as evidence for covalent character.
• Mixing J and kJ in ΔG = ΔH – TΔS, using T in °C, or not saying “above” or “below” with T = ΔH/ΔS.

Next steps

Revise from the revision notes, then test yourself with the practice questions.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.8 Thermodynamics (International A2).