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In inorganic chemicals the lattice enthalpy fulfills the same role as a bond energy does in organic chemistry. It gives a measure of how strong the attractive force is between the negative and positive ions in the ionic lattice.
The lattice enthalpy, D Hlatt is the energy released by the reaction :
A+(g) + B-(g) AB(s)
That is, the formation of a solid ionic compound from its gaseous constituent ions.
A Born-Haber cycle is a very special type of Hess' Law cycle. In AS, Hess' Law was used to calculate enthalpy changes of reaction using formation or combustion data for compounds.
With the calculation of lattice enthalpies the data required are atomisation energies for the elements, ionisation energies for the metal component, D Hi, electron affinities for the non-metal component, DHea, and the enthalpy change of formation, DHf, for the compound.
These data are put together in the following manner :
where, M is the metal and A the non-metal (both univalent in this example, e.g. NaCl).
Thus the lattice enthalpy for a compound is found by the following calculation :
DHlatt = DHf(MA(s)) - DHat(M(s)) - DHat(1/2A2(g)) - DHi1(M(g)) - DHea(A(g))
This formula works for compounds from univalent positive and negative ions such as group I metal halides.
With higher valency ions 2nd ionisation energies (or higher) are needed for the metal components and 2nd electron affinity energies are needed for non-metal components. The need for more than one atom of an ion also multiplies the relevant data by that number ->
The following diagram works for compounds from univalent positive ions, such as group I metals, combined with divalent negative ions, such as oxide ions.
DHlatt = DHf(M2A(s)) - 2DHat(M(s)) - DHat(1/2A2(g)) - 2DHi1(M(g)) - DHea(A(g)) - DHea(A-(g))
The following diagram works for compounds from divalent positive ions, such as group II metals, combined with univalent negative ions, such as halide ions.
DHlatt = DHf(MA2(s)) - DHat(M(s)) - 2DHat(1/2A2(g)) - DHi1(M(g)) - DHi2(M(g)) - 2DHea(A(g))
The following diagram works for compounds from divalent positive ions, such as group II metals, combined with divalent negative ions, such as oxide ions.
DHlatt = DHf(MA(s)) - DHat(M(s)) - DHat(1/2A2(g)) - DHi1(M(g)) - DHi2(M(g)) - DHea(A(g)) - DHea(A-(g))
Compound ions, such as NH4+, NO3-, CO32-, OH- and SO42-, cannot be used in these calculations as they don't have atomisation energies, ionisation energies or electron affinities.
The stronger the charge on an ion the stronger the attractive force that will result in an ionic lattice. Therefore, 1+/- ions form compounds with lower lattice enthalpies than 2+/- ions.
The following table of data illustrates that point :
|Ionic compound :||Lattice enthalpy :|
Another variable that can be changed with ionic compounds is the ionic radius. In general, the larger the ionic radius the lower the lattice enthalpy as the amount of interaction between the ions is smaller and the packing of the ions is less efficient :
|NaCl||-780 kJmol-1||MgO||-3791 kJmol-1||NaF||-918 kJmol-1|
|KCl||-711 kJmol-1||CaO||-3401 kJmol-1||NaCl||-780 kJmol-1|
|RbCl||-685 kJmol-1||SrO||-3223 kJmol-1||NaBr||-742 kJmol-1|
|CsCl||-661 kJmol-1||BaO||-3054 kJmol-1||NaI||-705 kJmol-1|
The ease decomposition of Groups II carbonates is determined mainly by the size of the metal cation.
The smaller the size of the cation the more polarizing the cation is. That is, the metal cation interacts with the electrons in the carbonate ion, destabilizing the bonds leading to the formation of the metal oxide and carbon dioxide gas.
Therefore, the carbonates become less easy to decompose as the group is descended :
for the reaction : MCO3(s) MO(s) + CO2(g)
the trend in decomposition temperature is
MgCO3 < CaCO3 < SrCO3 < BaCO3
Also, the lattice enthalpy of the metal oxide formed decreases down the group (see above). Therefore the oxide product is not formed as readily, making the carbonate more stable.
written by Dr Richard Clarkson : © Saturday, 1 November 1997
Updated : Saturday, 14th January, 2012
Created with the aid of ACD/ChemSketch Freeware 12