Chemical Energetics (A Level)

0. What Examiners Want

This guide covers Topic 23 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught.

The seven energy terms in this topic on one page. Learn the equation and the sign of each: they are the pieces of every cycle in the sections below.

1. Atomisation, Lattice Energy and the Terms

An ionic solid such as sodium chloride is held together by the strong electrostatic attraction between its ions in a giant lattice. We cannot measure the energy of forming this lattice directly from gaseous ions, so we calculate it with an energy cycle. First we need two new definitions (23.1.1) and one you met in outline at AS.

Standard enthalpy change of atomisation, ΔH_at. The enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.

The result is always endothermic, because bonds between atoms must be broken. For a diatomic element the equation shows half a molecule, because the definition is per mole of atoms: ΔH_at(Cl) = ½ × 244 = +122 kJ mol⁻¹, where 244 is the Cl–Cl bond energy in the Data Booklet.

Lattice energy, ΔH_latt. The enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions (under standard conditions). The syllabus puts it as ‘the change from gas phase ions to solid lattice’.

Na⁺(g) + Cl⁻(g) → NaCl(s)    ΔH_latt ≈ −785 kJ mol⁻¹
Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)    ΔH_latt ≈ −2520 kJ mol⁻¹

Because the definition describes ions coming together, ΔH_latt is always exothermic (negative). The more negative it is, the stronger the ionic bonding. Standard conditions are 298 K and 100 kPa.

2. Electron Affinity

For a non-metal the step that forms the negative ion is not an ionisation energy but an electron affinity.

First electron affinity, EA₁. The enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous ions with a single negative charge.

Cl(g) + e⁻ → Cl⁻(g)    EA₁ = −349 kJ mol⁻¹
O(g) + e⁻ → O⁻(g)    EA₁ = −141 kJ mol⁻¹

EA₁ is usually exothermic, because the incoming electron is attracted by the nucleus. The second electron affinity is always endothermic: the electron is being added to an ion that is already negative, so it is repelled.

O⁻(g) + e⁻ → O²⁻(g)    EA₂ = +798 kJ mol⁻¹
O(g) + 2e⁻ → O²⁻(g)    EA₁ + EA₂ = (−141) + (+798) = +657 kJ mol⁻¹

The overall change to O²⁻(g) is endothermic, yet oxides form easily, because the very large lattice energy of an oxide more than pays for it. For sulfur, EA₁ = −200 and EA₂ = +640, so S(g) + 2e⁻ → S²⁻(g) is (−200) + (+640) = +440 kJ mol⁻¹.

Factors affecting electron affinity (23.1.2b). The value depends on how strongly the incoming electron is attracted by the nucleus:

1. Nuclear charge: a greater nuclear charge attracts the electron more strongly, so EA₁ is more exothermic. Chlorine (17+) is more exothermic than sulfur (16+) in the same period.
2. Distance from the nucleus (atomic radius): the further the new electron enters from the nucleus, the weaker the attraction.
3. Shielding: more inner shells shield the nucleus, so the attraction is weaker.
4. Repulsion between electrons: in a very small atom the incoming electron is repelled by the electrons already in the compact outer shell, which reduces the energy released.

Trends in Groups 17 and 16 (23.1.2c).

EA₁ becomes less exothermic down each group from the second element onwards. The first element (F, O) is out of line.

• Down the group (Cl to I, and S to Te): EA₁ becomes less exothermic. Each atom has one more shell, so the added electron is further from the nucleus and is more shielded. The attraction is weaker, so less energy is released.
• The first element is the exception: fluorine is less exothermic than chlorine, and oxygen is less exothermic than sulfur. The F and O atoms are so small that the added electron is repelled strongly by the dense cloud of electrons already in the 2p subshell. This repulsion offsets the strong attraction of the nucleus.

3. Born–Haber Cycles

A Born–Haber cycle is Hess’s law applied to the formation of an ionic solid. There are two routes from the elements in their standard states to the solid:

• Route 1 (direct): form the solid from its elements, ΔH_f.
• Route 2 (indirect): turn the elements into gaseous atoms, then into gaseous ions, then let the ions form the lattice.

Both routes have the same overall enthalpy change, so for a compound of one metal M and one non-metal X:

ΔH_f = ΔH_at(M) + IE(M) + ΔH_at(X) + EA(X) + ΔH_latt

so that ΔH_latt = ΔH_f − (the sum of all the other steps). The cycle is limited to cations with a charge of 1+ or 2+ and anions with a charge of 1− or 2−.

Which steps are needed for each ion.

How to draw the cycle as an energy level diagram.

1. Put the elements in their standard states (for example Na(s) + ½Cl₂(g)) as the first level, near the bottom left.
2. Draw the atomisation of the metal (arrow up), then its ionisation(s) (arrows up). Show the electrons you have removed on the next level.
3. Draw the atomisation of the non-metal (arrow up).
4. Draw the electron affinity step(s) (EA₁ arrow down, EA₂ arrow up) to reach the gaseous ions.
5. Draw the lattice energy as an arrow down from the gaseous ions to the solid, and the enthalpy change of formation as an arrow down from the elements to the same solid.
6. Give every level its state symbols, and put the right multiplier in front of each species and each arrow (2Na, 2Cl).

Arrows pointing up are endothermic (+). Arrows pointing down are exothermic (−). The diagram is not drawn to scale.

Born–Haber cycle for sodium chloride (1+ cation, 1− anion). Read it from the bottom left: atomise Na, ionise Na, atomise Cl, add an electron, then form the lattice. ΔH_f is the direct route.

Worked example 1: NaCl (1+ and 1−). Calculate the lattice energy of sodium chloride from: ΔH_f[NaCl] = −411; ΔH_at[Na] = +107; IE₁[Na] = +494; ΔH_at[Cl] = +122; EA₁[Cl] = −349 (all kJ mol⁻¹).

Sum of the steps to the gaseous ions = (+107) + (+494) + (+122) + (−349) = +374 kJ mol⁻¹.
ΔH_latt = ΔH_f − (sum of steps) = (−411) − (+374) = −785 kJ mol⁻¹.

Born–Haber cycle for magnesium chloride (2+ cation, 1− anions). There are two chlorine atoms, so the atomisation and the electron affinity are each doubled, and the metal needs IE₁ and IE₂.

Worked example 2: MgCl₂ (2+ and 1−). ΔH_f[MgCl₂] = −641; ΔH_at[Mg] = +148; IE₁[Mg] = +736; IE₂[Mg] = +1450 (Data Booklet); ΔH_at[Cl] = +122; EA₁[Cl] = −349.

Sum of the steps = (+148) + (+736) + (+1450) + 2(+122) + 2(−349) = 148 + 736 + 1450 + 244 − 698 = +1880 kJ mol⁻¹.
ΔH_latt = (−641) − (+1880) = −2521 kJ mol⁻¹.

Born–Haber cycle for sodium oxide (1+ cation, 2− anion). Two sodium atoms, one oxygen atom, and two electron affinities. EA₂ is an upward (endothermic) step.

Worked example 3: Na₂O (1+ and 2−). ΔH_f[Na₂O] = −414; ΔH_at[Na] = +107; IE₁[Na] = +494; ΔH_at[O] = +249; EA₁[O] = −141; EA₂[O] = +798.

Sum of the steps = 2(+107) + 2(+494) + (+249) + (−141) + (+798) = 214 + 988 + 249 − 141 + 798 = +2108 kJ mol⁻¹.
ΔH_latt = (−414) − (+2108) = −2522 kJ mol⁻¹.

Born–Haber cycle for calcium chloride, with all the values filled in. Check each arrow against the data table: IE₁ and IE₂ for Ca, and 2 × for chlorine.

Worked example 3b: CaCl₂ (2+ and 1−). ΔH_f[CaCl₂] = −796; ΔH_at[Ca] = +178; IE₁ = +590; IE₂ = +1150; Cl–Cl bond energy 244; EA₁[Cl] = −349.

Sum of the steps = 178 + 590 + 1150 + 244 − 698 = +1464 kJ mol⁻¹. ΔH_latt = (−796) − (+1464) = −2260 kJ mol⁻¹. This is less exothermic than MgCl₂ (−2521) because Ca²⁺ is larger than Mg²⁺.

Worked example 4: finding a different unknown. The cycle works for any one missing value. The lattice energy of KCl is −717 kJ mol⁻¹; ΔH_f[KCl] = −437; ΔH_at[K] = +89; IE₁[K] = +419; ΔH_at[Cl] = +122. Find EA₁[Cl].

ΔH_f = ΔH_at(K) + IE₁(K) + ΔH_at(Cl) + EA₁(Cl) + ΔH_latt
−437 = 89 + 419 + 122 + EA₁ + (−717)
EA₁ = −437 − (89 + 419 + 122 − 717) = −437 − (−87) = −350 kJ mol⁻¹ (the usual value, −349, to rounding).

4. What Controls Lattice Energy

The lattice energy comes from the electrostatic attraction between oppositely charged ions. The force between ions of charge q⁺ and q⁻ whose centres are a distance d apart increases with q⁺ × q⁻ and decreases as d increases, where d is the sum of the two ionic radii (r⁺ + r⁻). So the numerical size of the lattice energy increases with ionic charge and decreases with ionic radius.

• Ionic charge. The greater the charge on the ions, the stronger the attraction and the more exothermic the lattice energy. MgO (2+, 2−) has a far more exothermic lattice energy than NaCl (1+, 1−), partly because the product of the charges is four times greater and partly because Mg²⁺ and O²⁻ are small.
• Ionic radius. The larger the ions, the further apart their centres and the weaker the attraction, so the lattice energy is less exothermic. It becomes less exothermic from LiF to NaF to KF (larger cation) and from LiF to LiCl to LiBr (larger anion). Another way to say it: smaller ions of the same charge have a higher charge density.

Ions drawn to scale. The bar chart shows q⁺q⁻/d for four compounds. The order (MgO > LiF > NaCl > KBr) is the order of the lattice energies, from most to least exothermic.

Worked example. Place NaCl, MgO, KBr and CaCl₂ in order, from the most to the least exothermic lattice energy.
MgO > CaCl₂ > NaCl > KBr. MgO has 2+ and 2− ions that are both small, so it has the greatest attraction. CaCl₂ has a 2+ ion, so its lattice energy is greater than that of the compounds with only 1+ ions. NaCl has smaller ions than KBr (Na⁺ is smaller than K⁺ and Cl⁻ is smaller than Br⁻), so the ions are closer, the attraction is stronger and its lattice energy is more exothermic.

5. Hydration and Solution

When an ionic solid dissolves in water, the lattice breaks up and each ion becomes surrounded by water molecules. Water is polar: the δ− oxygen is attracted to cations and the δ+ hydrogens are attracted to anions. These attractions are ion–dipole bonds.

Hydrated ions. Oxygen (δ−) points to the cation, hydrogen (δ+) points to the anion. Forming these ion–dipole attractions releases energy.

Enthalpy change of hydration, ΔH_hyd. The enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution.

Na⁺(g) + aq → Na⁺(aq)    ΔH_hyd = −406 kJ mol⁻¹
Cl⁻(g) + aq → Cl⁻(aq)    ΔH_hyd = −364 kJ mol⁻¹

ΔH_hyd is always exothermic because ion–dipole bonds are formed. It is defined for a specified ion, so each ion has its own value.

Enthalpy change of solution, ΔH_sol. The enthalpy change when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution.

NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)    ΔH_sol = +3.9 kJ mol⁻¹
MgCl₂(s) + aq → Mg²⁺(aq) + 2Cl⁻(aq)    ΔH_sol = −55 kJ mol⁻¹

ΔH_sol may be positive or negative. It is the small difference between two large numbers: the energy released by hydrating the ions and the energy needed to break the lattice.

Effect of charge and radius on hydration enthalpy (23.2.4). The size of ΔH_hyd depends on the charge density of the ion, which is its charge divided by its size. The greater the charge density, the stronger the attraction to the water dipoles and the more exothermic ΔH_hyd.

ΔH_hyd for some ions, with the ionic radius shown below each bar. For ions of the same charge the smaller ion is more exothermic; for ions of similar size the more highly charged ion is far more exothermic.

6. Energy Cycles for Dissolving

The three quantities are linked by Hess’s law. Start from the gaseous ions. They can reach the aqueous ions directly by being hydrated, or by first forming the solid (lattice energy) and then dissolving it.

The cycle for an ionic solid MX. Route 1: gaseous ions → solid → aqueous ions (ΔH_latt then ΔH_sol). Route 2: gaseous ions → aqueous ions (the two hydration enthalpies). The routes give the same total.

ΔH_latt + ΔH_sol = ΔH_hyd(cation) + ΔH_hyd(anion)

So ΔH_sol = ΣΔH_hyd − ΔH_latt. Remember to multiply a hydration enthalpy by the number of that ion in the formula (for MgCl₂: ΔH_hyd[Mg²⁺] + 2 × ΔH_hyd[Cl⁻]).

Worked example 1 (find ΔH_sol). ΔH_latt[LiCl] = −846; ΔH_hyd[Li⁺] = −519; ΔH_hyd[Cl⁻] = −364 (kJ mol⁻¹).
ΔH_sol = (−519) + (−364) − (−846) = −883 + 846 = −37 kJ mol⁻¹. Exothermic.

Worked example 2 (find ΔH_sol, endothermic). ΔH_latt[AgCl] = −905; ΔH_hyd[Ag⁺] = −475; ΔH_hyd[Cl⁻] = −364.
ΔH_sol = (−475) + (−364) − (−905) = −839 + 905 = +66 kJ mol⁻¹. Endothermic, and AgCl is almost insoluble in water.

The same cycle as an energy level diagram. When the energy released on hydration is greater than the lattice energy, ΔH_sol is negative (LiCl). When the lattice is held more strongly, ΔH_sol is positive (AgCl).

Worked example 3 (find a lattice energy). ΔH_sol[NaBr] = −0.6; ΔH_hyd[Na⁺] = −406; ΔH_hyd[Br⁻] = −335.
ΔH_latt = ΣΔH_hyd − ΔH_sol = (−406 − 335) − (−0.6) = −741 + 0.6 = −740 kJ mol⁻¹.

Worked example 4 (find a hydration enthalpy). ΔH_latt[NaF] = −913; ΔH_sol[NaF] = +1; ΔH_hyd[Na⁺] = −406.
ΔH_hyd[F⁻] = ΔH_latt + ΔH_sol − ΔH_hyd[Na⁺] = (−913) + (+1) − (−406) = −506 kJ mol⁻¹.

Application: the solubility of the Group 2 sulfates. Solubility decreases down Group 2 from MgSO₄ (very soluble) to BaSO₄ (insoluble). Both energies become less exothermic down the group, but not by the same amount:

• ΔH_hyd of the cation depends strongly on its size, so it becomes much less exothermic from Mg²⁺ to Ba²⁺.
• The sulfate ion is large, so the lattice energy depends on the radius of the sulfate as well as on the cation. The lattice energy therefore changes less than the hydration enthalpy.
• So ΔH_sol = ΣΔH_hyd − ΔH_latt becomes less exothermic (more positive) down the group, and the sulfates become less soluble.

A schematic (not to scale). The energy released on hydration falls faster than the energy needed to break the lattice, so the gap, −ΔH_sol, closes and then reverses.

7. Entropy

Energetically favourable (exothermic) changes are not the only ones that happen: ammonium nitrate dissolves in water and the solution gets cold, and ice melts at room temperature while absorbing energy. A second quantity, entropy, is needed.

Entropy, S. Entropy is a measure of the number of possible arrangements of the particles and their energy in a given system. The more ways of arranging the particles and their energy, the greater the entropy. It is often described as the disorder of a system, but the definition in terms of arrangements is the one to use. The units are J K⁻¹ mol⁻¹.

Three molecules and two flasks. Once the partition is removed there are 8 equally likely arrangements, and only 2 of them have all the molecules on one side. With a mole of gas the spread-out arrangements overwhelm all others, so gases mix spontaneously.

Standard molar entropy, S°. The entropy of 1 mole of a substance in its standard state (298 K, 100 kPa), in J K⁻¹ mol⁻¹. Unlike enthalpy of formation, the S° of an element is not zero: every substance has a positive S°. (A perfect crystal at 0 K has zero entropy.)

Standard molar entropies. Gases have much higher S° than liquids, and liquids higher than solids. More atoms per formula unit, and a softer or less tightly bonded solid, also give a higher S°.

Predicting the sign of ΔS (23.3.2).

Entropy increases gradually as a substance is heated and jumps at the melting point and the boiling point. The jump on boiling is much larger because a gas has far more arrangements than a liquid.

Examples. CaCO₃(s) → CaO(s) + CO₂(g): one gas molecule is produced from a solid, so ΔS is positive. N₂(g) + 3H₂(g) → 2NH₃(g): 4 mol of gas become 2 mol, so ΔS is negative. 2NO₂(g) → N₂O₄(g): 2 mol become 1 mol, so ΔS is negative. When the number of gas molecules does not change (for example H₂(g) + I₂(g) → 2HI(g)), predict only a small ΔS and say that it is difficult to decide.

Calculating the entropy change of a reaction (23.3.3).

ΔS° = ΣS°(products) − ΣS°(reactants)

Multiply each S° by its coefficient in the equation, use the value for the correct state, and give the answer in J K⁻¹ mol⁻¹ with its sign.

Worked example 1. CaCO₃(s) → CaO(s) + CO₂(g). S°: CaCO₃ 92.9; CaO 39.7; CO₂ 213.6.
ΔS° = (39.7 + 213.6) − 92.9 = +160.4 J K⁻¹ mol⁻¹. Positive, as predicted.

Worked example 2. N₂(g) + 3H₂(g) → 2NH₃(g). S°: N₂ 191.6; H₂ 130.6; NH₃ 192.3.
ΔS° = 2(192.3) − [191.6 + 3(130.6)] = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹. Negative, as predicted.

Worked example 3. 2H₂(g) + O₂(g) → 2H₂O(l). S°: H₂ 130.6; O₂ 205.0; H₂O(l) 69.9.
ΔS° = 2(69.9) − [2(130.6) + 205.0] = 139.8 − 466.2 = −326.4 J K⁻¹ mol⁻¹. Three moles of gas become a liquid, so the large negative value is expected.

8. Gibbs Free Energy

A reaction is feasible when the total entropy of the system and its surroundings increases. The Gibbs free energy change, ΔG, takes both into account in one quantity that refers to the reaction itself:

ΔG° = ΔH° − TΔS°

ΔH° is the enthalpy change of the reaction, T is the temperature in kelvin, and ΔS° is the entropy change of the reaction. (You do not need to use ΔS_total = ΔS_sys + ΔS_surr.)

• If ΔG is negative, the reaction is feasible (spontaneous) at that temperature.
• If ΔG is positive, the reaction is not feasible at that temperature.
• If ΔG = 0, the reaction is just on the borderline: the temperature at which feasibility changes is T = ΔH° / ΔS°.

Calculating ΔG. Check the units first: ΔH is in kJ mol⁻¹, ΔS is in J K⁻¹ mol⁻¹. Convert ΔH to J (×1000) or ΔS to kJ (÷1000).

Worked example 1. ZnCO₃(s) → ZnO(s) + CO₂(g); ΔH° = +71.0 kJ mol⁻¹. S°: ZnCO₃ 82.4; ZnO 43.6; CO₂ 213.6 J K⁻¹ mol⁻¹.
ΔS° = (43.6 + 213.6) − 82.4 = +174.8 J K⁻¹ mol⁻¹.
At 298 K: ΔG° = 71 000 − 298 × 174.8 = +18 910 J = +18.9 kJ mol⁻¹. Positive: not feasible at 298 K.
At 1200 K: ΔG° = 71 000 − 1200 × 174.8 = −138 760 J = −139 kJ mol⁻¹. Negative: feasible at 1200 K.

The effect of temperature (23.4.4). Because ΔG° = ΔH° − TΔS°, a graph of ΔG against T is a straight line with gradient −ΔS and intercept ΔH (at T = 0). The signs of ΔH and ΔS decide what happens:

ΔG against T for the four combinations of signs. Where ΔG is below zero the reaction is feasible. The lines that change sign cross ΔG = 0 at T = ΔH / ΔS.

The zinc carbonate example drawn as a graph. The intercept is ΔH, the gradient is −ΔS, and the reaction becomes feasible above 406 K.

Worked example 2: the minimum temperature (endothermic, ΔS positive). CaCO₃(s) → CaO(s) + CO₂(g); ΔH° = +178.0 kJ mol⁻¹; ΔS° = +160.4 J K⁻¹ mol⁻¹ (worked out in section 7).
At 298 K: ΔG° = 178 000 − 298 × 160.4 = +130 201 J = +130 kJ mol⁻¹, so limestone does not decompose at room temperature.
ΔG° = 0 when T = ΔH° / ΔS° = 178 000 / 160.4 = 1110 K. Above about 1110 K the reaction is feasible.

Worked example 3: the maximum temperature (exothermic, ΔS negative). N₂(g) + 3H₂(g) → 2NH₃(g); ΔH° = −92.2 kJ mol⁻¹; ΔS° = −198.8 J K⁻¹ mol⁻¹ (section 7).
At 298 K: ΔG° = −92 200 − 298 × (−198.8) = −92 200 + 59 242 = −33.0 kJ mol⁻¹: feasible.
ΔG° = 0 when T = −92 200 / −198.8 = 464 K. The reaction is feasible only below about 464 K. (This is why a high temperature favours the reactants in the Haber process, and why a compromise temperature is used industrially to get a reasonable rate.)

Worked example 4: a salt that dissolves endothermically. Ammonium nitrate dissolves with ΔH_sol = +25.7 kJ mol⁻¹ and ΔS = +108 J K⁻¹ mol⁻¹. At 298 K: ΔG = 25 700 − 298 × 108 = −6484 J = −6.5 kJ mol⁻¹. The process is feasible because the entropy gain (ions spreading through the solution) outweighs the positive enthalpy change.

9. Quick Sheet and Checklist

Equations to learn.

• ΔH_f = ΔH_at(M) + IE(M) + ΔH_at(X) + EA(X) + ΔH_latt, with the multipliers and all the steps for each ion.
• ΔH_latt + ΔH_sol = ΣΔH_hyd.
• ΔS° = ΣS°(products) − ΣS°(reactants).
• ΔG° = ΔH° − TΔS°; feasible if ΔG is negative; T = ΔH / ΔS when ΔG = 0.

Checklist. Before the exam you should be able to:

• [ ] write each definition from memory with its starting and finishing species, and give an equation with state symbols for each
• [ ] explain the trend in EA₁ down Groups 16 and 17, name the exception and give its reason
• [ ] explain why EA₂ is endothermic
• [ ] draw a Born–Haber cycle for a compound with 1+ or 2+ cations and 1− or 2− anions, with the multipliers and state symbols
• [ ] pick only the relevant data and calculate a lattice energy, a formation enthalpy or an electron affinity from a cycle
• [ ] explain how ionic charge and ionic radius change a lattice energy and a hydration enthalpy
• [ ] build the lattice–solution–hydration cycle and calculate any one of the three quantities
• [ ] explain why the Group 2 sulfates become less soluble down the group, using both energies
• [ ] predict the sign of ΔS for a change of state, a temperature change and a change in the number of gas molecules
• [ ] calculate ΔS° from standard entropies and ΔG° from ΔH° and ΔS° with consistent units
• [ ] work out the temperature at which a reaction becomes feasible, and say whether it is a minimum or a maximum