IB DP Chemistry Reactivity 1.2: Energy cycles in reactions

This guide teaches Reactivity 1.2, “Energy cycles in reactions”, from scratch. It follows the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023) and covers understandings Reactivity 1.2.1 to 1.2.5. The first two are for SL and HL; the last three sit in the additional higher level (AHL) part and are HL only. It applies to exams from 2025 onwards.

After this guide, condense the ideas with the energy cycles revision notes and test yourself with the energy cycles practice questions. Every other unit sits on the DP Chemistry hub, the progress checklist tracks your progress, the syllabus guide shows how the course fits together, and a free diagnostic quiz shows where to start.

What this unit covers

Understanding	Skills you must show	Level
Reactivity 1.2.1	Calculate ΔH from average bond enthalpies; explain why they are averages and why answers differ from measured values	SL and HL
Reactivity 1.2.2	Use Hess’s law to find ΔH for a reaction from several other steps	SL and HL
Reactivity 1.2.3	Deduce equations for ΔHc⦵ and ΔHf⦵ and solve problems with these terms	HL only
Reactivity 1.2.4	Calculate ΔH⦵ of a reaction from ΔHf⦵ data or from ΔHc⦵ data	HL only
Reactivity 1.2.5	Interpret and find values in a Born–Haber cycle for compounds of univalent and divalent ions	HL only

Every method below applies one idea: conservation of energy. The guide states that average bond enthalpies, ΔHc⦵ and ΔHf⦵ values, and the two summation equations for 1.2.4 are given in the data booklet. In this guide, all data are supplied with each example.

Reactivity 1.2.1 Bond enthalpies

To break a covalent bond you must put energy in, so bond breaking is endothermic. When a bond forms, the same amount of energy is given out, so bond forming is exothermic.

The bond enthalpy is the energy required to split one mole of a particular bond, with every species in the gas phase. Bond enthalpies are always positive.

A C–H bond in methane is not quite the same as a C–H bond in ethanol or in chloroform, because the neighbouring atoms change its electron distribution. Tables therefore give an average bond enthalpy, taken over many different compounds. That is why a calculated ΔH is an estimate.

The method:

1. Draw or picture every bond in every reactant and product.
2. Add up the bond enthalpies of all bonds broken (energy in).
3. Add up the bond enthalpies of all bonds formed (energy out).
4. ΔH = Σ(bonds broken) − Σ(bonds formed).

If the bonds formed are stronger in total than those broken, ΔH is negative and the reaction is exothermic.

Worked example 1: hydrogenating ethyne

Estimate ΔH for C₂H₂(g) + 2H₂(g) → C₂H₆(g). Use these values (kJ mol⁻¹): C≡C 839, C–C 346, C–H 414, H–H 436.

• Bonds broken: one C≡C, two C–H, two H–H = 839 + 2(414) + 2(436) = 2539 kJ
• Bonds formed: one C–C, six C–H = 346 + 6(414) = 2830 kJ
• ΔH = 2539 − 2830 = −291 kJ mol⁻¹

A shortcut gives the same answer: two C–H bonds survive unchanged, so cancel them and count only the four new C–H bonds formed.

A value calculated from measured enthalpies of formation is about −312 kJ mol⁻¹. The estimate differs for two reasons:

• The bond enthalpies are averages, not the actual values in ethyne and ethane.
• Bond enthalpies apply only to gases. If any reactant or product is a liquid or solid, the energy of the intermolecular forces is left out. (Here all species are gases, so the averages are the main cause.)

Reactivity 1.2.2 Hess’s law

Hess’s law says that ΔH for a reaction does not depend on the route taken from the initial state to the final state. This follows from conservation of energy. If two routes gave different energy changes, you could go out by one and back by the other and create energy from nothing.

Rules for combining equations:

• Reverse an equation: change the sign of ΔH.
• Multiply an equation by a number: multiply ΔH by the same number.
• Add equations: add their ΔH values. Species on both sides cancel.

Worked example 2: nitrogen dioxide oxidizing sulfur dioxide

Find ΔH for SO₂(g) + NO₂(g) → SO₃(g) + NO(g) from:

• (1) 2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = −197.8 kJ mol⁻¹
• (2) N₂(g) + 2O₂(g) → 2NO₂(g) ΔH = +66.4 kJ mol⁻¹
• (3) N₂(g) + O₂(g) → 2NO(g) ΔH = +180.6 kJ mol⁻¹

Work from the target, one species at a time:

• One SO₂ on the left and one SO₃ on the right: use (1) halved, ½ × (−197.8) = −98.9.
• One NO₂ on the left, but NO₂ is a product in (2): reverse (2) and halve it, −½ × (+66.4) = −33.2.
• One NO on the right: use (3) halved, ½ × (+180.6) = +90.3.

Check the cancelling: ½N₂ appears on the right of reversed (2) and the left of halved (3). The O₂ balances too: ½O₂ + ½O₂ used on the left, one O₂ released on the right.

ΔH = −98.9 − 33.2 + 90.3 = −41.8 kJ mol⁻¹

Reactivity 1.2.3 Formation and combustion equations (HL only)

Standard conditions mean a pressure of 100 kPa, with data usually quoted at 298 K. Each substance is in its standard state, its pure form at that pressure.

• Standard enthalpy change of formation, ΔHf⦵: the enthalpy change for making one mole of a compound out of its constituent elements, each in its standard state.
• Standard enthalpy change of combustion, ΔHc⦵: the enthalpy change when one mole of a substance burns completely in oxygen.

Both definitions fix “one mole” of a particular substance, so fractions in the equation are normal. An element in its standard state has ΔHf⦵ = 0, because forming it from itself involves no change.

Worked example 3: writing the equations

• ΔHf⦵ of sodium carbonate: 2Na(s) + C(s, graphite) + 1½O₂(g) → Na₂CO₃(s)
• ΔHf⦵ of ethane-1,2-diol: 2C(s, graphite) + 3H₂(g) + O₂(g) → HOCH₂CH₂OH(l)
• ΔHc⦵ of pentane: C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(l)

Check that only elements appear on the left of a formation equation, that each element is in its standard state (O₂ gas, not O atoms), and that the substance named has a coefficient of 1.

Some values do two jobs. Burning graphite, C(s) + O₂(g) → CO₂(g), is both ΔHc⦵ of carbon and ΔHf⦵ of carbon dioxide.

Allotropes. Graphite is the standard state of carbon, so ΔHf⦵ of graphite is zero. Diamond is a different form, so its ΔHf⦵ is not zero. Using ΔHc⦵ values of −393.5 kJ mol⁻¹ (graphite) and −395.4 kJ mol⁻¹ (diamond), the route graphite → CO₂ ← diamond gives ΔH(graphite → diamond) = −393.5 − (−395.4) = +1.9 kJ mol⁻¹, which is ΔHf⦵ of diamond.

Reactivity 1.2.4 Calculating ΔH⦵ from formation or combustion data (HL only)

Two equations come from Hess cycles:

using formation data:   ΔH⦵ = Σ ΔHf⦵(products) − Σ ΔHf⦵(reactants)
using combustion data:  ΔH⦵ = Σ ΔHc⦵(reactants) − Σ ΔHc⦵(products)

The order is reversed for combustion. With formation data, the cycle goes down from the elements to both sides. With combustion data, it goes down from both sides to the combustion products. Multiply each value by its coefficient in the balanced equation.

Worked example 4: ΔHf⦵ of ethyne from combustion data

Use these ΔHc⦵ values (kJ mol⁻¹): C(s, graphite) −393.5; H₂(g) −285.8; C₂H₂(g) −1301.1.

The formation reaction is 2C(s, graphite) + H₂(g) → C₂H₂(g). The “reactants” here are the elements.

ΔHf⦵ = [2(−393.5) + (−285.8)] − (−1301.1) = −1072.8 + 1301.1 = +228.3 kJ mol⁻¹

Ethyne has a positive ΔHf⦵: it is less stable than its elements.

Worked example 5: ethyne from calcium carbide

Find ΔH⦵ for CaC₂(s) + 2H₂O(l) → C₂H₂(g) + Ca(OH)₂(s). Use these ΔHf⦵ values (kJ mol⁻¹): CaC₂(s) −59.8; H₂O(l) −285.8; C₂H₂(g) +228.3; Ca(OH)₂(s) −986.1.

• Σ ΔHf⦵(products) = 228.3 + (−986.1) = −757.8
• Σ ΔHf⦵(reactants) = −59.8 + 2(−285.8) = −631.4
• ΔH⦵ = −757.8 − (−631.4) = −126.4 kJ mol⁻¹

Reactivity 1.2.5 Born–Haber cycles (HL only)

You cannot measure the energy holding an ionic lattice together directly. A Born–Haber cycle uses Hess’s law to find it from steps that can be measured.

The terms, each per mole:

• Enthalpy of atomization: forming gaseous atoms from the element in its standard state. For a metal this is sublimation. For a diatomic gas it is half the bond enthalpy per atom formed (so 2 mol of Cl atoms from Cl₂ costs one full bond enthalpy).
• Ionization energy (IE): removing electrons from gaseous atoms or ions. A divalent cation such as Mg²⁺ needs IE₁ + IE₂.
• Electron affinity: the enthalpy change when gaseous atoms gain electrons. The first electron affinity of most non-metals is negative. Adding a second electron to an anion, as in forming O²⁻, is positive because the electron is repelled.
• Lattice enthalpy: here taken as the enthalpy change to separate one mole of a solid ionic compound into its gaseous ions, MX(s) → M⁺(g) + X⁻(g). This is positive. Some books write the reverse process, with the same size and a negative sign, so read the direction in each question.
• Enthalpy of formation: as in 1.2.3.

Going round the cycle from the elements, two routes reach the solid: directly (ΔHf⦵), or through gaseous atoms and ions and then the lattice forming. By Hess’s law:

ΔHf⦵ = Σ(atomization) + Σ(IE) + Σ(electron affinity) − lattice enthalpy

Building a complete cycle from nothing is not assessed, but you must read one and calculate any missing value.

Worked example 6: potassium iodide (univalent ions)

Use these values (kJ mol⁻¹): atomization of K +89; IE₁(K) +419; atomization of iodine (½I₂(s) → I(g)) +107; first electron affinity of I −295; ΔHf⦵(KI) −328.

• Sum of steps from elements to gaseous ions: 89 + 419 + 107 + (−295) = +320
• −328 = 320 − lattice enthalpy
• Lattice enthalpy = 320 + 328 = +648 kJ mol⁻¹

Worked example 7: magnesium iodide (divalent cation)

Use these values (kJ mol⁻¹): atomization of Mg +148; IE₁(Mg) +738; IE₂(Mg) +1451; atomization of iodine +107; first electron affinity of I −295; ΔHf⦵(MgI₂) −364.

Two iodine atoms are needed, so double both iodine terms.

• Steps: 148 + 738 + 1451 + 2(107) + 2(−295) = +1961
• Lattice enthalpy = 1961 − (−364) = +2325 kJ mol⁻¹

Interpreting the result: the MgI₂ lattice enthalpy is over three times that of KI. Mg²⁺ carries twice the charge of K⁺ and is smaller, so the attraction to the iodide ions is much stronger (Structure 2.1). Notice also that IE₂ is the biggest endothermic step. It is paid back by the large lattice enthalpy, which is why MgI₂ still forms exothermically.

Common errors

• Writing ΔH = Σ(bonds formed) − Σ(bonds broken), which flips the sign.
• Missing bonds that look hidden, such as the C–H bonds in a condensed formula like CH₃OCH₃.
• Using bond enthalpies for a reaction with liquid water without allowing for vaporization.
• Forgetting to reverse the sign of ΔH when reversing an equation, or to multiply it when scaling.
• Swapping the two summation equations: formation is products minus reactants, combustion is reactants minus products.
• Assigning a ΔHf⦵ value other than zero to graphite, O₂(g) or another element in its standard state, or giving O₂ a ΔHc⦵.
• Writing a formation equation that makes two moles of product.
• In Born–Haber cycles, forgetting IE₂ for a 2+ ion or failing to double the anion terms for MgX₂.

Next steps

Read the one-sitting summary notes next, then try the practice set with full worked answers. Use the checklist to tick off each understanding.

Official syllabus

Source document: International Baccalaureate Organization, Diploma Programme Chemistry guide, first assessment 2025 (published February 2023), sub-topic Reactivity 1.2.