AQA Chem5: Enthalpy Change

Revision notes on enthalpy change including Born-Haber cycles from AQA Chem5

  • Created by: anna
  • Created on: 09-06-13 17:17

Born Haber Cycles

  • Born-Haber cycles are used to determine enthalpy changes associated with ionic compounds
  • Hess' Law is used to determine enthalpy changes associated with covalent compounds but can still be applied to Born-Haber cycles
  • For an ionic compound the lattice enthalpy is the heat energy released when one mole of solid in its standard state is formed from its ions in the gaseous state
  • This value cannot be directly determined so we use an enthalpy cycle involving  the enthalpy of formation
  • For example, there are two routes from the elements Na and Cl to the ionic compound NaCl, one is through the enthalpy of formation the other involves various enthalpy changes including: enthalpy of atomisation (Na) -> enthalpy of atomisation (Cl) -> enthalpy of 1st ionisation energy (Na) -> enthalpy of 1st electron affinity (Cl) -> lattice association enthalpy
  • Once a Born-Haber cycle is created you can apply Hess' Law to calculate the unknown enthalpy change e.g. lattice association enthalpy
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Additional Steps in Born-Haber Cycles


Some Born-Haber Cycles require additional steps:

  • For dipositive metal ions there will be a 2nd ionisation energy, M+ (g) -> M2+(g) + e-
  • For oxides there will be an additional electron affinity, the 2nd electron affinity, this 2nd one will be endothermic as the O- will repel the e-,  O- (g) + e-  -> O2- (g)
  • In the case of bromine which is a liquid before it can be atomised it has to be vapourised using the enthalpy of vapourisation, Br2(l) -> Br2(g), it is an endothermic reaction as energy has to be taken in to boil something
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Comparing Lattice Enthalpies

  • Theoretical values for lattice enthalpies are determined assuming that the crystal lattice is 100% ionic
  • Experimental values often differ from theoretical values if the bonding in the crystal is much less than 100% ionic
  • Formation of calcium flouride: as you look at Ca+F- CaF2 and CaF3 the lattice enthalpies become more negative -820kJmol-1, -2630kJmol-1, -3246kJmol-1 this is due to the progressively stronger electrostatic attraction, when you consider their enthalpies of formation CaF2 is the most feasible flouride as it has the most negative value (-1220kJmol-1), CaF3 has a positive enthalpy of formation (+700kJmol-1) as the third electron requires more energy to be removed
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Ions in Solution

  • when ionic compounds dissolve in water the process may be exothermic or endothermic
  • both the enthalpy of solution and the enthalpy of hydration are involved
  • when ions dissolve in water ion-dipole forces are formed between the ions and the polar water molecules
  • the formation of ion-dipole bonds results in the reaction being highly exothermic (the enthalpy of hydration is negative)
  • to calculate the enthalpy of solution you use either a Born-Haber cycle or a Hess cycle both of which must include enthalpy of hydration
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