IB DP Chemistry Reactivity 1.2: Energy cycles in reactions – Revision Notes

These notes condense Reactivity 1.2. The energy cycles study guide has the full teaching and more worked examples, so read that first if anything here is unfamiliar.

They follow the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023) and cover understandings Reactivity 1.2.1 to 1.2.5. Bond enthalpies (1.2.1) and Hess’s law (1.2.2) are for SL and HL. Formation and combustion data (1.2.3, 1.2.4) and Born–Haber cycles (1.2.5) are additional higher level material, so they are HL only. Use them for the course examined from 2025 onwards.

Once you can recall everything here, try the practice set for this unit. Mark off each understanding on the course checklist, find other units on the course hub, and take a free diagnostic to see which topics need the most work.

Key definitions

Term	Meaning	Level
Bond enthalpy	Energy absorbed when one mole of a particular covalent bond is broken in the gas phase; always positive	SL and HL
Average bond enthalpy	Mean of that bond’s enthalpy over a range of compounds	SL and HL
Hess’s law	ΔH for a reaction is the same whatever pathway links the initial and final states	SL and HL
ΔHf⦵	ΔH for making one mole of a substance from its elements, every species in its standard state	HL only
ΔHc⦵	ΔH for burning one mole of a substance completely in excess oxygen, every species in its standard state	HL only
Enthalpy of atomization	ΔH to turn the standard-state element into one mole of gas-phase atoms	HL only
Ionization energy	Energy taken in when each atom (or ion) in a mole of gas-phase particles loses an electron	HL only
Electron affinity	ΔH when one mole of gaseous atoms (or ions) gains one mole of electrons	HL only
Lattice enthalpy	ΔH to separate one mole of an ionic solid into its gaseous ions; positive in this direction	HL only

Reactivity 1.2.1 Bond enthalpies

Energy in to break bonds, energy out when bonds form. The balance decides the sign of ΔH.

Method in steps

1. Write out the displayed formula of every species, so no bond is hidden.
2. List bonds broken and add their enthalpies.
3. List bonds formed and add their enthalpies.
4. ΔH = Σ(bonds broken) − Σ(bonds formed).
5. Bonds that survive the reaction unchanged may be left out of both lists.

Worked reminder. Propene adds chlorine: CH₂=CHCH₃(g) + Cl₂(g) → CH₂ClCHClCH₃(g). Data (kJ mol⁻¹): C=C 614, Cl–Cl 242, C–C 346, C–Cl 324.

• Broken: C=C + Cl–Cl = 614 + 242 = 856
• Formed: C–C + 2 C–Cl = 346 + 2(324) = 994
• ΔH = 856 − 994 = −138 kJ mol⁻¹

Only the double bond and the chlorine molecule change. The six C–H bonds and the original C–C bond are unaffected, so they cancel.

Why the answer is only an estimate

• Each value is a mean over many molecules. The real strength of a bond depends on what else is attached to the atoms.
• The values hold for gaseous species. With liquids or solids, the enthalpy of the intermolecular forces (vaporization, for example) is ignored.

Reactivity 1.2.2 Hess cycles

Hess’s law is conservation of energy applied to a reaction. Any route between the same start and end states has the same ΔH.

Method in steps

1. Write the target equation and keep it in view.
2. For each species in the target, pick a given equation that contains it.
3. Reverse an equation if the species is on the wrong side (flip the sign of ΔH).
4. Scale an equation so the coefficient matches (scale ΔH by the same factor).
5. Add the adjusted equations. Cancel species that appear on both sides, and check the sum gives the target.
6. Add the adjusted ΔH values.

Worked reminder. Sulfur recovery from natural gas uses 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l). Given:

• (a) H₂S(g) + 1½O₂(g) → SO₂(g) + H₂O(l) ΔH = −562.0 kJ mol⁻¹
• (b) S(s) + O₂(g) → SO₂(g) ΔH = −296.8 kJ mol⁻¹

Target = 2 × (a) − 3 × (b): ΔH = 2(−562.0) + 3(+296.8) = −1124.0 + 890.4 = −233.6 kJ mol⁻¹

Check: 3O₂ cancels, and two of the three SO₂ from reversed (b) are cancelled by the two SO₂ made in (a), leaving one SO₂ on the left.

Reactivity 1.2.3 Writing ΔHf⦵ and ΔHc⦵ equations (HL only)

Checklist for a ΔHf⦵ equation:

• exactly one mole of the compound on the right;
• only elements on the left, each in its standard state (C as graphite, O₂(g), Br₂(l), Na(s));
• fractional coefficients allowed on the left.

Checklist for a ΔHc⦵ equation:

• exactly one mole of the fuel on the left, with oxygen;
• complete combustion products, for example CO₂(g) and H₂O(l) under standard conditions.

Key facts:

• ΔHf⦵ of an element in its standard state is zero. A different allotrope (diamond, O₃) has a non-zero value.
• One equation can serve both definitions: H₂(g) + ½O₂(g) → H₂O(l) is ΔHc⦵ of hydrogen and ΔHf⦵ of water.
• Combustion products such as CO₂ and H₂O have no ΔHc⦵, and neither does oxygen.

Reactivity 1.2.4 ΔH⦵ from ΔHf⦵ or ΔHc⦵ data (HL only)

Data given	Equation	Memory hook
ΔHf⦵	ΔH⦵ = Σ ΔHf⦵(products) − Σ ΔHf⦵(reactants)	Formation: “products minus reactants”
ΔHc⦵	ΔH⦵ = Σ ΔHc⦵(reactants) − Σ ΔHc⦵(products)	Combustion: “reactants minus products”
Bond enthalpies	ΔH = Σ(broken) − Σ(formed)	Bonds: same order as combustion

Worked reminder: combustion data. Find ΔHf⦵ of ethyl ethanoate from ΔHc⦵ values (kJ mol⁻¹): graphite −393.5, hydrogen −285.8, ethyl ethanoate −2238.1.

Formation equation: 4C(s, graphite) + 4H₂(g) + O₂(g) → CH₃COOCH₂CH₃(l). Oxygen has no combustion value.

• Σ reactants = 4(−393.5) + 4(−285.8) = −2717.2
• ΔHf⦵ = −2717.2 − (−2238.1) = −479.1 kJ mol⁻¹

Worked reminder: formation data. Roasting zinc blende: 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g). ΔHf⦵ (kJ mol⁻¹): ZnS(s) −206.0; ZnO(s) −350.5; SO₂(g) −296.8.

• Σ products = 2(−350.5) + 2(−296.8) = −1294.6
• Σ reactants = 2(−206.0) + 0 = −412.0
• ΔH⦵ = −1294.6 − (−412.0) = −882.6 kJ mol⁻¹

Reactivity 1.2.5 Reading Born–Haber cycles (HL only)

Step	Typical sign	Watch for
Atomization of the metal	+	Once per metal atom in the formula
Atomization of the non-metal	+	Half the bond enthalpy per atom from a diatomic gas; doubled for MX₂
First ionization energy	+	Always needed
Second ionization energy	+	Only for a 2+ cation, and larger than the first
First electron affinity	usually −	Doubled for MX₂
Second electron affinity	+	Only for a 2− anion such as O²⁻ (repulsion)
Lattice enthalpy (separating)	+	Read the direction the question uses
ΔHf⦵ of the compound	usually −	Direct route from elements to solid

Hess’s law around the cycle:

ΔHf⦵ = Σ(atomization) + Σ(ionization energies) + Σ(electron affinities) − lattice enthalpy

Rearrange for whichever term is missing. You will not be asked to build a full cycle from nothing, but you must read one and extract any value from it.

Worked reminder. Find ΔHf⦵ of lithium iodide. Data (kJ mol⁻¹): Li(s) → Li(g) +159; IE₁ of lithium +520; ½I₂(s) → I(g) +107; I(g) + e⁻ → I⁻(g) −295; lattice enthalpy +759.

• Elements to gaseous ions: 159 + 520 + 107 + (−295) = +491
• ΔHf⦵ = 491 − 759 = −268 kJ mol⁻¹

The large lattice term more than pays for the endothermic steps.

Must-know distinctions

• Bond enthalpy vs ΔHf⦵. Bond enthalpies are averages and refer to gases. ΔHf⦵ values come from measurements on specific substances in their standard states.
• Atomization vs bond enthalpy. Atomizing ½Cl₂ to one Cl atom costs half the Cl–Cl bond enthalpy.
• Ionization energy vs electron affinity. One removes electrons from the metal; the other adds electrons to the non-metal.
• Lattice enthalpy direction. Separating the solid is endothermic; forming it from gaseous ions has the same size and the opposite sign.
• Formation vs combustion sums. Formation: products minus reactants. Combustion: reactants minus products.

Quick self-test

1. Is breaking a bond exothermic or endothermic?
2. Give two reasons why ΔH calculated from average bond enthalpies may differ from a measured value.
3. A → B, ΔH = +40 kJ mol⁻¹; C → B, ΔH = −25 kJ mol⁻¹. Find ΔH for A → C.
4. X + Y → Z, ΔH = −60 kJ mol⁻¹. State ΔH for 2Z → 2X + 2Y.
5. (HL only) Write the equation for ΔHf⦵ of ethanoic acid, CH₃COOH(l).
6. (HL only) Calculate ΔH⦵ for TiCl₄(l) + 2Mg(s) → Ti(s) + 2MgCl₂(s). ΔHf⦵ values: TiCl₄(l) −804.2; MgCl₂(s) −641.3 kJ mol⁻¹.
7. (HL only) Calculate ΔH⦵ for C₂H₄(g) + H₂O(l) → C₂H₅OH(l). ΔHc⦵ values: C₂H₄(g) −1411.0; C₂H₅OH(l) −1366.8 kJ mol⁻¹.
8. (HL only) Why is ΔHf⦵ of Br₂(l) zero but ΔHf⦵ of Br₂(g) not?
9. (HL only) Find the first electron affinity of bromine. Data for NaBr (kJ mol⁻¹): atomization of Na +107; IE₁(Na) +496; atomization of bromine +112; lattice enthalpy +751; ΔHf⦵ −361.
10. (HL only) Which ionization energy terms appear in a Born–Haber cycle for MgBr₂?

Answers

1. Endothermic: energy must be absorbed to overcome the attraction.
2. The values are averages over many compounds; and they apply to gases, so the intermolecular forces in any liquid or solid are not counted.
3. A → B then B → C (reverse of C → B): +40 + 25 = +65 kJ mol⁻¹.
4. Reversed and doubled: +120 kJ mol⁻¹.
5. 2C(s, graphite) + 2H₂(g) + O₂(g) → CH₃COOH(l).
6. 2(−641.3) − (−804.2) = −478.4 kJ mol⁻¹ (Mg and Ti are zero).
7. −1411.0 − (−1366.8) = −44.2 kJ mol⁻¹ (water has no ΔHc⦵).
8. The standard state of bromine is the liquid. Forming the gas involves vaporization, so its ΔHf⦵ is positive.
9. −361 = 107 + 496 + 112 + EA − 751, so EA = −325 kJ mol⁻¹.
10. Both IE₁ and IE₂ of magnesium, because Mg²⁺ has lost two electrons.

Where marks are usually lost

• Reversing the bond enthalpy formula, which gives the right size with the wrong sign.
• Counting bonds from a molecular formula and missing some, especially in esters, ethers and alcohols.
• Not multiplying a bond enthalpy by the number of that bond present (six C–H in ethane, not one).
• Forgetting to change the sign when a Hess equation is reversed, or to scale ΔH with the equation.
• Using the ΔHf⦵ formula with ΔHc⦵ data, or the other way round.
• Including a combustion term for oxygen, or a formation term for an element such as Cl₂(g) or Fe(s).
• Writing state symbols that do not match the standard state, such as H₂O(g) in a ΔHc⦵ equation or I₂(g) in a ΔHf⦵ equation.
• In Born–Haber cycles, leaving out IE₂ or failing to double the anion terms in MX₂ compounds.
• Mixing up the sign of the lattice enthalpy when the question defines it in the forming direction.

Official syllabus

International Baccalaureate Organization, Diploma Programme Chemistry guide, first assessment 2025 (published February 2023): Reactivity 1.2, “Energy cycles in reactions”.