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1.8Physical Chemistry

Thermodynamics — Oxford AQA International A Level Chemistry Revision Notes & Topical Questions

10 min read 25 parts Lesson 8 of 13

📚 Oxford AQA International A Level Chemistry Hub / Topical Revision
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Complete PDF with figures and original page layoutPage 1 Oxford AQA International Your notes A Level (IAL) Chemistry Thermodynamics Contents Born–Haber Cycles Constructing Born–Haber Cycles Born-Haber Calculations Enthalpy of Hydration Entropy Change,…

Part 1 of 9Born–Haber cycles

1Definition
  • Definition a Born–Haber cycle is a specific application of Hess’s law to ionic compounds
2Purpose
  • Purpose it allows the lattice enthalpy, which cannot be measured directly, to be calculated
3Requirement
  • Requirement it needs several different enthalpy changes

Part 2 of 9Enthalpy of formation and combustion

Standard enthalpy of formation, ΔHf = the enthalpy change when one mole of a substance forms from its elements under standard conditions with all in standard states,
  • such as H2(g) + ½O2(g) → H2O(l), −286 kJ mol-1
Sign
  • Sign it can be positive or negative, and it is 0 for elements
Standard enthalpy of combustion, ΔHc = the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions,
  • such as CH4 + 2O2 → CO2 + 2H2O(l), −890 kJ mol-1
Sign
  • Sign it is always negative because combustion is exothermic
Overlap
  • Overlap the ΔHc of carbon is the ΔHf of carbon dioxide

Part 3 of 9Enthalpy of atomisation and ionisation energies

1
Enthalpy of atomisation, ΔHat = the enthalpy change when one mole of gaseous atoms forms from the element in its standard state,
  • such as Mg(s) → Mg(g), +148 kJ mol-1,
  • and it is always positive
2
First ionisation energy = the enthalpy change when one mole of gaseous atoms forms one mole of gaseous 1+ ions,
  • such as Mg(g) → Mg+(g) + e-, +738 kJ mol-1, always positive
3
Second ionisation energy = the enthalpy change when one mole of gaseous 1+ ions forms one mole of gaseous 2+ ions,
  • such as Mg+(g) → Mg2+(g) + e-, +1451 kJ mol-1
4Second against first
  • Second against first it is larger because the electron is more strongly attracted to the nucleus of a 1+ ion

Part 4 of 9Electron affinity

1
First electron affinity = the enthalpy change when one mole of gaseous atoms forms one mole of gaseous 1- ions,
  • such as O(g) + e- → O-(g), −141 kJ mol-1
2First value
  • First value it can be exothermic, endothermic or neutral
3
Second electron affinity = the enthalpy change when one mole of gaseous 1- ions forms one mole of gaseous 2- ions,
  • such as O-(g) + e- → O2-(g), +798 kJ mol-1
4Second value
  • Second value it is always positive because energy is needed to overcome repulsion between the 1- ion and the electron

Part 5 of 9Lattice enthalpy

Enthalpy of lattice formation = the enthalpy change when one mole of solid ionic compound forms from its gaseous ions,
  • such as Na+(g) + Cl-(g) → NaCl(s), −788 kJ mol-1
Sign
  • Sign it is always negative because new bonds form and energy is released
Enthalpy of lattice dissociation = the enthalpy change when one mole of solid ionic compound dissociates into gaseous ions,
  • such as NaCl(s) → Na+(g) + Cl-(g), +788 kJ mol-1
Sign
  • Sign it is always positive, and is the opposite of lattice formation
Measurement
  • Measurement neither can be measured directly

Part 6 of 9Constructing a Born–Haber cycle: elements and atoms

1Principle
  • Principle energy increases going up the diagram,
    • and the cycle shows the steps from atoms to gaseous ions and then to the lattice, with an alternative route by the enthalpy of formation
2
Step 1: starting elements
  • Place the elements on the left, about a third of the way up.
  • Give each its own energy level with state symbols.
3Step 2
Step 2 gaseous atoms
  • bond breaking, so arrows point upwards
  • Na(s) → Na(g), +108 kJ mol-1.
  • ½Cl2(g) → Cl(g), +121 kJ mol-1.

Part 7 of 9Constructing a Born–Haber cycle: ions and lattice

1
Step 3: gaseous ions
  • Sodium loses an electronNa(g) → Na+(g) + e-, +500 kJ mol-1, an upward arrow.
  • Chlorine gains an electronCl(g) + e- → Cl-(g), −364 kJ mol-1, a downward arrow, displaced to the right.
2
Step 4: lattice
  • Enthalpy of formation, Na(s) + ½Cl2(g) → NaCl(s), −411 kJ mol-1, on the left at the bottom with a downward arrow.
  • Lattice formation, Na+(g) + Cl-(g) → NaCl(s), on the right at the bottom with a downward arrow.
3Use
  • Use the completed cycle gives the lattice enthalpy, or any other enthalpy if the lattice enthalpy is given

Part 8 of 9Born–Haber cycle for MgCl<sub>2</sub>

  • Arrow requirements
    • each enthalpy change needs its own arrow, a label, an equation with state symbols and electrons, and a value
Enthalpy changeValue / kJ mol-1
Atomisation of Mg+148
Atomisation of Cl+122
First ionisation of Mg+738
Second ionisation of Mg+1451
Electron affinity of Cl-349
Formation of MgCl2-641
Lattice formation-2524
  • Watch-outs
    • Include both ionisation energies of magnesium because Mg2+ forms.
    • Double the atomisation and electron affinity of chlorine because there are two chlorine atoms.
    • Exothermic (negative) arrows point down, and endothermic (positive) arrows point up.

Part 9 of 9Lattice enthalpy calculation

1Equation
  • ΔHf = ΔHat + ΔHie + ΔHea + ΔHlatt
    • (of both elements)
2Simplified
  • ΔHf = ΔH1 + ΔHlatt, where ΔH1 is the sum of all the changes from the elements to gaseous ions
3Rearranged
  • ΔHlatt = ΔHf − ΔH1
4Any stage
  • Any stage the stage being found is always the direct route, so write the equation in terms of enthalpy changes and rearrange
5Remember
  • Remember include all required enthalpies, doubling or halving values where the formula needs it,
    • such as two ionisation energies for Mg2+ and doubling the chlorine electron affinity in MgCl2

Key terms

TermMeaning
Born–Haber cycleA Hess’s law cycle that links lattice enthalpy to other enthalpy changes of an ionic compound.
Enthalpy of atomisationThe enthalpy change to form one mole of gaseous atoms from an element.
Electron affinityThe enthalpy change when gaseous atoms or ions gain electrons.
Lattice formation enthalpyThe enthalpy change when one mole of ionic solid forms from gaseous ions.
Ionisation energyThe enthalpy change to remove electrons from gaseous atoms or ions.

Avoid these mistakes

1
Drawing an arrow in the wrong direction — endothermic steps go up and exothermic steps go down.
2
Forgetting to double values for two atoms of an element — MgCl2 needs two chlorine atoms.
3
Leaving out the second ionisation energy for a 2+ ion — both are needed.
4
Mixing up lattice formation and dissociation — they are equal and opposite in sign.

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