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23. Chemical energetics

Syllabus
9701–2028–2029
Section
23
Level
A2

23.1 Lattice energy and Born-Haber cycles

Syllabus
9701–2028–2029
Topic
23.1
Level
A2

Define atomisation and lattice energy by their exact particles and direction

Term Defining process for one mole Typical sign
enthalpy change of atomisation, ΔHₐₜ element in its standard state → 1 mol gaseous atoms positive
lattice energy, ΔHₗₐₜₜ stoichiometric gaseous ions → 1 mol solid ionic lattice negative

Na(s)→Na(g)\ce{Na(s) -> Na(g)}

12 ClX2(g)→Cl(g)\ce{1/2Cl2(g) -> Cl(g)}

NaX+(g)+ClX−(g)→NaCl(s)\ce{Na+(g) + Cl-(g) -> NaCl(s)}

Write the species, coefficients and states before inserting a value. Atomising a molecular element may require a fraction of its standard-state molecule; reversing lattice formation to separate the solid into gaseous ions changes the sign.

The official ΔHₗₐₜₜ convention here is gas-phase ions to solid lattice. Do not mix it with a positive lattice-dissociation value or call any solid-to-gas change atomisation without forming gaseous atoms.

First electron affinity balances nuclear attraction and electron repulsion

X(g)+eX−→XX−(g)EA1=ΔH for this process\ce{X(g) + e- -> X-(g)}\qquad EA_1=\Delta H\text{ for this process}

Factor Effect on attraction of the incoming electron
greater nuclear charge with similar shielding makes EA₁ more exothermic
larger atomic radius / greater electron distance makes EA₁ less exothermic
more inner-shell shielding makes EA₁ less exothermic
strong repulsion in a compact or already occupied orbital makes EA₁ less exothermic

Group 17 atoms gain an electron to complete the p subshell, so their first electron affinities are strongly exothermic. Chlorine is more exothermic than fluorine because the incoming electron experiences greater repulsion in fluorine's very compact 2p orbital; from Cl down to I, increasing radius and shielding make EA₁ less exothermic.

Group 16 shows the parallel anomaly: sulfur has a more exothermic EA₁ than oxygen because oxygen's compact 2p orbital gives greater electron repulsion. From S down the group, increasing distance and shielding make EA₁ less exothermic. Group 17 values are generally more exothermic than the corresponding Group 16 values because the stronger nuclear attraction and p-subshell completion favour electron gain.

Electron affinity is not ionisation energy in reverse. A second electron affinity adds an electron to X⁻(g), so repulsion from the negative ion makes that separate process endothermic.

Construct a Born–Haber cycle by making the same gaseous ions

A Born–Haber cycle applies Hess's law between the elements in their standard states, the ionic solid, and one common set of gaseous ions. Every alternative path must end at exactly the same stoichiometric gaseous ions before lattice formation.

MgCl₂ cycle step Process / coefficient Enthalpy term
formation Mg(s) + Cl₂(g) → MgCl₂(s) ΔH°f
atomise Mg Mg(s) → Mg(g) ΔHₐₜ(Mg)
atomise chlorine Cl₂(g) → 2Cl(g) 2ΔHₐₜ(Cl)
form Mg²⁺ Mg(g) → Mg²⁺(g) + 2e⁻ IE₁ + IE₂
form 2Cl⁻ 2Cl(g) + 2e⁻ → 2Cl⁻(g) 2EA₁(Cl)
form lattice Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s) ΔHₗₐₜₜ

For a +2 cation include both successive ionisation energies. For a –2 anion include EA₁ and the endothermic EA₂. Multiply every atomisation or electron-affinity term by the number of atoms/ions in one formula unit; the allowed charge range is ±1 and ±2.

Do not use bond dissociation and atomisation for the same non-metal atoms twice. A labelled cycle is valid only when atoms, electrons, charges, states and stoichiometric coefficients are conserved on every route.

Calculate a missing Born–Haber term with one signed Hess equation

Write the formation path first, then sum the alternative gas-ion path in the same direction. Insert tabulated values with their given signs and coefficients; only then rearrange for the unknown.

ΔHf∘=ΔHat(Na)+ΔHat(Cl)+IE1(Na)+EA1(Cl)+ΔHlatt\Delta H_f^\circ=\Delta H_{at}(Na)+\Delta H_{at}(Cl)+IE_1(Na)+EA_1(Cl)+\Delta H_{latt}

NaCl term Value / kJ mol⁻¹
ΔH°f[NaCl(s)] −411
ΔHₐₜ[Na(s) → Na(g)] +108
ΔHₐₜ[½Cl₂(g) → Cl(g)] +121
IE₁(Na) +496
EA₁(Cl) −349

ΔHlatt=−411−[108+121+496−349]=−787 kJ mol−1\Delta H_{latt}=-411-[108+121+496-349]=-787\ \mathrm{kJ\ mol^{-1}}

The negative result matches the official lattice-formation direction: attraction releases energy when gaseous Na⁺ and Cl⁻ form NaCl(s). Check that a second IE/EA or a factor of two has not been omitted for multivalent ions.

Higher ionic charge and smaller radius increase lattice-energy magnitude

∣ΔHlatt∣ increases roughly with ∣q+q−∣r++r−|\Delta H_{latt}|\ \text{increases roughly with}\ \frac{|q_+q_-|}{r_++r_-}

Change while other factors are comparable Electrostatic consequence Formation ΔHₗₐₜₜ
larger charge magnitude larger charge product and stronger attraction more negative; larger magnitude
smaller ionic radius charge centres are closer more negative; larger magnitude
larger ionic radius charge centres are farther apart less negative; smaller magnitude

LiF has a larger lattice-energy magnitude than LiI because F⁻ is smaller than I⁻ at the same charges. MgO has a much larger magnitude than NaCl because the 2+/2− charge product is four times the 1+/1− product, alongside radius differences.

Say larger magnitude or more negative for the formation convention; “larger” alone is ambiguous. Charge and radius both matter, so compare one factor at a time where possible and acknowledge structural differences when compounds are not otherwise similar.

23.2 Enthalpies of solution and hydration

Syllabus
9701–2028–2029
Topic
23.2
Level
A2

Define hydration and solution enthalpy from exact state changes

Term Process for one mole Sign
hydration enthalpy, ΔHₕyd one mole of a specified gaseous ion → that aqueous ion always exothermic / negative
solution enthalpy, ΔHₛₒₗ one mole of ionic solid → its stoichiometric aqueous ions may be positive or negative

NaX+(g)→NaX+(aq)\ce{Na+(g) -> Na+(aq)}

ClX−(g)→ClX−(aq)\ce{Cl-(g) -> Cl-(aq)}

NaCl(s)→NaX+(aq)+ClX−(aq)\ce{NaCl(s) -> Na+(aq) + Cl-(aq)}

Hydration releases energy when ion–dipole attractions form between ions and water. Dissolution also requires lattice separation, so the balance between endothermic separation and exothermic hydration determines the sign of ΔHₛₒₗ.

A hydration value belongs to one ion, not a whole salt. State charge and phase: Na⁺(g) → Na⁺(aq) is hydration, whereas NaCl(s) → aqueous ions is solution.

Construct the solution cycle through the same gaseous ions

Connect the solid directly to its aqueous ions by ΔHₛₒₗ, and indirectly through the stoichiometric gaseous ions. The indirect route reverses lattice formation, then hydrates each gaseous ion.

MgCl₂ path State equation Enthalpy contribution
direct dissolution MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq) ΔHₛₒₗ
lattice separation MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g) −ΔHₗₐₜₜ (formation)
hydrate cation Mg²⁺(g) → Mg²⁺(aq) ΔHₕyd(Mg²⁺)
hydrate anions 2Cl⁻(g) → 2Cl⁻(aq) 2ΔHₕyd(Cl⁻)

ΔHsol=−ΔHlatt+∑ΔHhyd\Delta H_{sol}=-\Delta H_{latt}+\sum \Delta H_{hyd}

The summation includes stoichiometric coefficients. Do not use one hydration value for MgCl₂ as a whole, omit the factor 2 for chloride, or insert negative lattice formation without reversing its direction.

Calculate a missing cycle term without changing its defined sign

Use one convention throughout. With lattice formation, the gaseous-ion-to-aqueous route gives ΣΔHₕyd = ΔHₗₐₜₜ + ΔHₛₒₗ. Apply coefficients first, substitute signed data second, and rearrange last.

KCl datum Value / kJ mol⁻¹
ΔHₗₐₜₜ[KCl] (formation) −711
ΔHₛₒₗ[KCl] +26
ΔHₕyd[K⁺] −322

ΔHhyd(K+)+ΔHhyd(Cl−)=ΔHlatt+ΔHsol\Delta H_{hyd}(K^+)+\Delta H_{hyd}(Cl^-)=\Delta H_{latt}+\Delta H_{sol}

ΔHhyd(Cl−)=−711+26−(−322)=−363 kJ mol−1\Delta H_{hyd}(Cl^-)=-711+26-(-322)=-363\ \mathrm{kJ\ mol^{-1}}

Hydration must be exothermic, so the negative answer is physically consistent. A positive result here signals a likely direction, coefficient or rearrangement error; do not flip a sign merely to force the expectation.

Higher charge density makes hydration more exothermic

∣ΔHhyd∣ increases as ionic charge density ∣q∣r increases|\Delta H_{hyd}|\ \text{increases as ionic charge density}\ \frac{|q|}{r}\ \text{increases}

Ion change Ion–dipole attraction to water ΔHₕyd
higher charge at similar radius stronger more negative; larger magnitude
smaller radius at same charge charge is closer to water dipoles; stronger more negative; larger magnitude
larger radius at same charge weaker less negative; smaller magnitude

Water's Oδ− end points towards a cation, while its Hδ+ ends point towards an anion. Stronger attraction releases more energy as the hydration shell forms.

Mg²⁺ has a much more negative hydration enthalpy than Na⁺ because it has higher charge and high charge density. Within Group 1, Na⁺ is more negative than K⁺ because Na⁺ is smaller at the same +1 charge.

Say more negative or larger magnitude; “larger hydration enthalpy” is ambiguous. Compare charge first and radius among equal-charge ions rather than using radius alone across unlike charges.

23.3 Entropy change, ΔS

Syllabus
9701–2028–2029
Topic
23.3
Level
A2

Entropy counts possible arrangements of particles and energy

Entropy, S, is the number of possible arrangements of the particles and their energy in a given system. More possible arrangements mean higher entropy.

The arrangements include where particles can be and how the system's energy can be distributed. A gas therefore usually has higher entropy than the same amount of the same substance as a liquid or solid because its particles and energy have many more possible arrangements.

S(gas)>S(liquid)>S(solid)for the same substance under comparable conditionsS(\mathrm{gas})>S(\mathrm{liquid})>S(\mathrm{solid})\quad\text{for the same substance under comparable conditions}

Use the syllabus definition rather than treating entropy as everyday 'messiness'. Entropy is a property of the stated system, so identify its particles, state and conditions before comparing arrangements.

Predict the sign of ΔS from the change in accessible arrangements

Change in the system Sign of ΔS Particle-and-energy reason
solid → liquid; liquid → gas + particles gain freedom and more arrangements become possible
gas → liquid; liquid → solid − particles become more constrained
solid solute → dilute solution + solute particles become dispersed through the solvent
crystallisation from solution − dispersed particles enter an ordered lattice
temperature increases without a phase change + energy can be distributed among more accessible energy states
temperature decreases − fewer energy arrangements are accessible
reaction forms more gaseous molecules + gas particles have more positional and energy arrangements
reaction forms fewer gaseous molecules − gas-particle arrangements decrease

Count gaseous molecules using the balanced-equation coefficients and ignore solid, liquid and aqueous coefficients for this specific gas-count test. For N₂(g) + 3H₂(g) → 2NH₃(g), four gaseous molecules become two, so ΔS is predicted to be negative.

Melting and boiling have positive ΔS; freezing and condensing have negative ΔS. CaCO₃(s) → CaO(s) + CO₂(g) has positive ΔS because gas is produced from solids.

If the number of gaseous molecules is unchanged, gas count alone gives no sign. A qualitative prediction is not a numerical calculation, and a positive ΔS alone does not establish feasibility; Gibbs free energy is treated in 23.4.

Calculate ΔS° with coefficients and the correct physical states

ΔS∘=∑νS∘(products)−∑νS∘(reactants)\Delta S^\circ=\sum \nu S^\circ(\mathrm{products})-\sum \nu S^\circ(\mathrm{reactants})

Use the standard molar entropy for each substance in its stated physical state. Multiply every S° value by its stoichiometric coefficient, total the product side and reactant side separately, then subtract reactants from products.

For 2Mg(s) + O₂(g) → 2MgO(s):

Substance S° / J K⁻¹ mol⁻¹ Coefficient Contribution
Mg(s) 32.60 2 65.20
O₂(g) 205.0 1 205.0
MgO(s) 38.20 2 76.40

ΔS∘=(2×38.20)−[(2×32.60)+205.0]=−193.8 J K−1 mol−1\Delta S^\circ=(2\times38.20)-[(2\times32.60)+205.0]=-193.8\ \mathrm{J\ K^{-1}\ mol^{-1}}

The negative sign agrees with the qualitative change: one mole of gas is consumed and only solids remain. Keep entropy in J K⁻¹ mol⁻¹; do not convert it to kJ until a later equation explicitly requires matching kJ units.

The tabulated quantities are S°, not ΔS° values for individual substances. Do not reverse products minus reactants, omit coefficients, use the wrong physical state, or introduce ΔSsurr; the latter is explicitly not required here.

23.4 Gibbs free energy change, ΔG

Syllabus
9701–2028–2029
Topic
23.4
Level
A2

Use the Gibbs equation with kelvin and consistent energy units

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ

Symbol Meaning Typical unit before substitution
ΔG° standard Gibbs free energy change kJ mol⁻¹
ΔH° standard enthalpy change kJ mol⁻¹
T absolute temperature K
ΔS° standard entropy change of the system J K⁻¹ mol⁻¹

Before substitution, convert ΔS° from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000 when ΔH° is in kJ mol⁻¹. Then TΔS° has the same kJ mol⁻¹ unit as ΔH°.

The equation combines the enthalpy contribution with the temperature-weighted entropy contribution. Use the signed values exactly as given: a negative ΔS° makes the term −TΔS° positive.

T must be in kelvin, not °C. Do not combine kJ and J, discard a negative sign, or treat T as carrying a stoichiometric coefficient.

Calculate ΔG° from a complete, unit-safe data route

For CH₃OH(l) + HBr(g) → CH₃Br(g) + H₂O(l) at 298 K, ΔH° = −47 kJ mol⁻¹. The standard molar entropies are 240, 99.0, 246 and 70.0 J K⁻¹ mol⁻¹ respectively in reaction order.

ΔS∘=(246+70.0)−(240+99.0)=−23.0 J K−1 mol−1\Delta S^\circ=(246+70.0)-(240+99.0)=-23.0\ \mathrm{J\ K^{-1}\ mol^{-1}}

−23.0 J K−1 mol−1=−0.0230 kJ K−1 mol−1-23.0\ \mathrm{J\ K^{-1}\ mol^{-1}}=-0.0230\ \mathrm{kJ\ K^{-1}\ mol^{-1}}

ΔG∘=−47−[298×(−0.0230)]=−40.146≈−40.1 kJ mol−1\Delta G^\circ=-47-[298\times(-0.0230)]=-40.146\approx-40.1\ \mathrm{kJ\ mol^{-1}}

Calculate ΔS° with products minus reactants if it is not supplied, convert its energy unit, substitute signed quantities, and round only the final answer. The negative result is then interpreted separately as feasible under the stated standard conditions.

A negative entropy does not make TΔS° negative after the Gibbs subtraction is applied twice: −T(negative ΔS°) is positive. Keep brackets around the signed entropy term.

Use the sign of ΔG to state thermodynamic feasibility

Gibbs free energy change Conclusion for the forward process
ΔG < 0 thermodynamically feasible under the stated conditions
ΔG = 0 equilibrium
ΔG > 0 not thermodynamically feasible under the stated conditions

Feasibility is a thermodynamic conclusion, not a rate prediction. A feasible reaction may be imperceptibly slow if its activation energy is high; combustion can have negative ΔG yet still need ignition.

State the direction and the conditions attached to the value. A standard ΔG° conclusion applies to standard states at the specified temperature; changing temperature or composition can change the actual Gibbs free energy change.

Do not translate negative ΔG into 'instantaneous', 'safe', 'goes to completion' or 'high yield'. Those claims require kinetic or equilibrium evidence not supplied by the sign alone.

Predict how temperature changes feasibility from the signs of ΔH° and ΔS°

ΔH° ΔS° Temperature effect on ΔG° Feasibility in the constant-ΔH°/ΔS° model
− + both ΔH° and −TΔS° are negative feasible at all temperatures
+ − both ΔH° and −TΔS° are positive not feasible at any temperature
+ + raising T makes −TΔS° more negative feasible above the crossover temperature
− − raising T makes −TΔS° more positive feasible below the crossover temperature

ΔG∘=0⇒Tcrossover=ΔH∘ΔS∘with consistent units\Delta G^\circ=0\quad\Rightarrow\quad T_{crossover}=\frac{\Delta H^\circ}{\Delta S^\circ}\quad\text{with consistent units}

If ΔH° = +50 kJ mol⁻¹ and ΔS° = +150 J K⁻¹ mol⁻¹, convert ΔS° to +0.150 kJ K⁻¹ mol⁻¹. The crossover is 50/0.150 ≈ 333 K, so the reaction is feasible above 333 K and not feasible below it in this model.

Always explain the change through −TΔS°. Raising T favours a positive ΔS° because the negative entropy term grows in magnitude, but opposes a negative ΔS° because −T(negative ΔS°) becomes increasingly positive.

Do not say that higher temperature always increases feasibility. Inspect both signs, convert units before finding a crossover, and recognise that the four-case table assumes ΔH° and ΔS° do not change appreciably with temperature.