Lattice Enthalpy

Enthalpy of atomisation The enthalpy of atomisation of an element is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state Na (s)
Na(g) [∆atH = +148 kJ mol-1] ½ O2 (g)
O (g) [∆atH = +249 kJ mol-1] The enthalpy change for a solid metal turning to gaseous atoms can also be called the Enthalpy of sublimation and will numerically be the same as the enthalpy of atomisation Na (s)
Na(g) [∆subH = +148 kJ mol-1] First Ionisation enthalpy The first ionisation enthalpy is the enthalpy change required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous ions with a +1 charge Mg (g)
Mg+ (g) + e- [∆ IE 1 H] Second Ionisation enthalpy The second ionisation enthalpy is the enthalpy change to remove 1 mole of electrons from one mole of gaseous 1+ ions to produces one mole of gaseous 2+ ions. Mg+ (g)
Mg 2+ (g) + e- [∆ IE 2 H] First Electron affinity The first electron affinity is the enthalpy change that occurs when 1 mole of gaseous atoms gain 1 mole of electrons to form 1 mole of gaseous ions with a –1 charge O (g) + e-
O- (g) [∆eaH] = -141.1 kJ mol-1] The first electron affinity is exothermic for atoms that normally form negative ions because the ion is more stable than the atom and there is an attraction between the nucleus and the electron second electron affinity The second electron affinity is the enthalpy change when one mole of gaseous 1- ions gains one electron per ion to produce gaseous 2- ions. O – (g) + e-
O2- (g) [∆eaH = +798 kJ mol-1] The second electron affinity for oxygen is endothermic because it take energy to overcome the repulsive force between the negative ion and the electron Lattice Enthalpy The Lattice Enthalpy is the standard enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent ions in gaseous form. Na+ (g) + Cl- (g)
NaCl (s) [∆ LE H = -787 kJ mol-1] Enthalpy of Hydration ∆Hhyd Enthalpy change when one mole of gaseous ions become aqueous ions . X+ (g) + aq
X+ (aq) For Li+ ∆hydH = -519 kJ mol-1 or X- (g) + aq
X- (aq) For F- ∆hydH = -506 kJ mol-1 This always gives out energy (exothermic, -ve) because bonds are made between the ions and the water molecules Enthalpy of solution The enthalpy of solution is the standard enthalpy change when one mole of an ionic solid dissolves in an large enough amount of water to ensure that the dissolved ions are well separated and do not interact with one another NaCl (s) + aq
Na+ (aq) + Cl- (aq) ∆solH Enthalpy change of formation The standard enthalpy change of formation of a compound is the energy transferred when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states Na (s) + ½Cl2 (g)
NaCl (s) [∆fH = – 411.2 kJ mol-1] N Goalby chemrevise.org 1 5. The lattice enthalpy can be used as a measure of ionic bond strength. Definitions of enthalpy changes

BORN HABER CYCLES The lattice enthalpy cannot be determined directly. We calculate it indirectly by making use of changes for which data are available and link them together in an enthalpy cycle the Born Haber cycle Born Haber cycle: sodium Chloride By applying Hess’s law the heat of formation equals to the sum of everything else ∆fH =∆atH (Na) + ∆IEH(Na)+ ∆atH(Cl) + ∆EaH(Cl) + ∆LEH Rearrange to give ∆LEH = ∆fH – (∆atH (Na) + ∆IEH(Na)+ ∆atH (Cl) ∆EaH(Cl) ) Pay attention to state symbols and direction of arrows. Usually all pieces of data are given except the one that needs to be calculated ∆LEH =-411 – (+107 + 496 + 122 + -349) = -787 kJmol-1 Born Haber cycle: magnesium Chloride The data for the ∆at H (Cl) could also be given as the bond energy for E(Cl-Cl ) bond. Remember : E(Cl-Cl ) = 2 x ∆at H (Cl) Note in this example the first and second ionisation energies of magnesium are needed as Mg is a +2 ion. Born Haber cycle: calcium oxide CaO (s) Ca (s) + ½ O2 (g) Ca (g) + ½ O2 (g) ∆Ea1H (O) ∆atH(Ca) ∆atH(O) ∆LEH ∆fH (CaO) + e- + O- Ca (g) 2+ (g) Ca + O (g) 2+ (g) + 2eCa2+ (g) + 2e- + ½ O2 (g) Ca + (g) + e- + ½ O2 (g)) ∆IE1H(Ca) ∆IE2H (Ca) Ca2+ (g) + O2- (g) ∆Ea1H(O) Notice the second electron affinity for oxygen is endothermic because it take energy to overcome the repulsive force between the negative ion and the electron

Trends in Lattice Enthalpies The strength of a enthalpy of lattice formation depends on the following factors 1. The sizes of the ions: The larger the ions, the less negative the enthalpies of lattice formation (i.e. a weaker lattice). As the ions are larger the charges become further apart and so have a weaker attractive force between them. 2. The charges on the ion: The bigger the charge of the ion, the greater the attraction between the ions so the stronger the lattice enthalpy (more negative values). The lattice enthalpies become less negative down any group. e.g. LiCl, NaCl, KCl, RbCl e.g group 1 halides (eg NaF KI) have lattice enthalpies of around – 700 to -1000 group 2 halides (eg MgCl2 ) have lattice enthalpies of around –2000 to –3500 group 2 oxides eg MgO have lattice enthalpies of around –3000 to – 4500 kJmol-1