Lattice
energy (∆Hlattice)
Lattice energy is defined as the amount of energy required
to completely remove the constituent ions from its crystal lattice to an
infinite distance. It is also referred as lattice enthalpy.
NaCl(s) → Na+(g) + Cl–(g)
∆Hlattice = + 788 kJ mol-1
From the above equation it is clear that 788 kJ of energy
is required to separate Na+ and Cl- ions from 1 mole of
NaCl.
The Born–Haber cycle is an approach to analyse reaction
energies. It was named after two German scientists Max Born and Fritz Haber who
developed this cycle. The cycle is concerned with the formation of an ionic
compound from the reaction of a metal with a halogen or other non-metallic
element such as oxygen.
Born–Haber cycle is primarily used in calculating lattice
energy, which cannot be measured directly. The Born–Haber cycle applies Hess's
law to calculate the lattice enthalpy. For example consider the formation of a
simple ionic solid such as an alkali metal halide MX, the following steps are
considered.
∆H1 - enthalpy change for the sublimation M(s)
to M(g)
ΔH2- enthalpy change for the dissociation of ½ X2
(g) to X(g)
∆H3- Ionisation energy for M(g) to M+(g)
∆H4 - electron affinity for the conversion of X(g) to X-(g)
U - the lattice enthalpy for the formation of solid MX
∆Hf - enthalpy change for the formation of
solid MX directly form elements According to Hess's law of heat summation
∆Hf
= ∆H1 + ∆H 2
+ ∆H 3 + ∆H 4 + U
Let us use the Born - Haber cycle for determining the
lattice enthalpy of NaCl as follows:
Since the reaction is carried out with reactants in
elemental forms and products in their standard states, at 1 bar, the overall
enthalpy change of the reaction is also the enthalpy of formation for NaCl.
Also, the formation of NaCl can be considered in 5 steps. The sum of the
enthalpy changes of these steps is equal to the enthalpy change for the overall
reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride
using Born-Haber cycle
ΔHf = heat of formation of sodium chloride = –
411.3 kJ mol–1
ΔH1 = heat of sublimation of Na(S) = 108.7 kJ
mol–1
ΔH2 = ionisation energy of Na(S) = 495.0 kJ mol–1
ΔH3 = dissociation energy of Cl2(S)
= 244 kJ mol–1
ΔH4 = Electron affinity of Cl(S)= – 349.0 kJ
mol–1
U = lattice energy of NaCl
∆Hf = ∆H1 + ∆H2 + 1/2∆H3
+ ∆H4 +U
U = (–411.3) – (108.7 + 495.0 + 122 – 349)
U = (–411.3) – (376.7)
U = – 788 kJ mol–1
This negative sign in lattice energy indicates that the
energy is released when sodium is formed from its constituent gaseous ions Na+
and Cl-
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