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R1.2 Energy cycles in reactionsIB Chemistry HL: Revision notes

Section 1

Bond enthalpies

Bond breaking absorbs energy; bond forming releases energy. Using average bond enthalpies for gaseous species:

ΔH = Σ(bonds broken) − Σ(bonds formed)

Example, N₂ + 3H₂ → 2NH₃: broken 945 + 3(436) = 2253; formed 6(391) = 2346; ΔH = −93 kJ. Values are approximate because average bond enthalpies are means over many compounds and apply only to gases.

Key termsaverage bond enthalpy

Section 2

Hess's law

Hess's law: the enthalpy change for a reaction is independent of the pathway between the initial and final states. Combine given equations to reach the target: reverse an equation → change the sign of ΔH; multiply an equation → multiply ΔH. Every calculation below (formation data, combustion data, Born–Haber) is an application of Hess's law.

Key termsHess's law

Section 3

Standard enthalpies of formation and combustion (HL)

Standard enthalpy of formation, ΔHf⦵: the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, usually 298 K). ΔHf⦵ of an element in its standard state is zero. Example: ½N₂(g) + 1½H₂(g) → NH₃(g).

Standard enthalpy of combustion, ΔHc⦵: the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. Example: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).

Key termsstandard enthalpy of formationstandard enthalpy of combustion
Common mistake

Writing a formation equation that makes two moles, or starts from gaseous atoms, or gives the product in the wrong state.

Section 4

Calculations with ΔHf⦵ and ΔHc⦵ data (HL)

Both equations are in the data booklet:

  • ΔH⦵ = Σ ΔHf⦵(products) − Σ ΔHf⦵(reactants)
  • ΔH⦵ = Σ ΔHc⦵(reactants) − Σ ΔHc⦵(products)

Multiply each value by its coefficient. Combustion data are ideal for formation enthalpies that cannot be measured directly: for propane, 3C + 4H₂ → C₃H₈, ΔHf⦵ = [3(−394) + 4(−286)] − (−2219) = −107 kJ mol⁻¹.

Key termsproducts minus reactantsreactants minus products
Exam tip

Formation: products minus reactants. Combustion: reactants minus products. The order flips because combustion arrows point away from the substances.

Section 5

Born–Haber cycles (HL)

A Born–Haber cycle applies Hess's law to the formation of an ionic compound. Route from elements to gaseous ions:

  1. atomisation of the metal and of the non-metal (per atom formed);
  2. ionisation energies of the metal (IE₁ + IE₂ for M²⁺);
  3. electron affinities of the non-metal;
  4. ions come together to form the lattice (the reverse of the lattice enthalpy).

The sum equals ΔHf⦵. For KCl: lattice enthalpy = 437 + 89 + 419 + 121 − 349 = +717 kJ mol⁻¹.

Key termsBorn–Haber cyclelattice enthalpyenthalpy of atomisationelectron affinity

Section 6

Interpreting Born–Haber values (HL)

First electron affinities are usually exothermic (the electron is attracted by the nucleus); second electron affinities (O⁻ → O²⁻) are endothermic because the electron is repelled by a negative ion.

For divalent ions, count everything: MgCl₂ needs IE₁ + IE₂ and two chlorine atomisations and electron affinities. Larger ionic charge and smaller ionic radius give a larger lattice enthalpy, which is why MgCl₂ (lattice 2526) forms rather than MgCl (about 753) despite the cost of IE₂.

Key termsdivalent ion
Common mistake

Forgetting to double the atomisation and electron affinity of chlorine for MgCl₂.

Must know

  • ΔH = Σ bonds broken − Σ bonds formed.
  • Hess's law: ΔH is independent of the route.
  • (HL) ΔHf⦵ of an element in its standard state = 0.
  • (HL) Formation: products − reactants; combustion: reactants − products.
  • (HL) Born–Haber: atomisation + IE + EA − lattice enthalpy = ΔHf⦵.
  • (HL) Higher charge, smaller ions → larger lattice enthalpy.

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