Summary - A level

BORN HABER CYCLES

The Born Haber cycles is an adaption of Hess’s law to calculate lattice enthalpy from other data
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

Generally Lattice Enthalpy refers to the enthalpy change when a lattice is made from gaseous ions.
Some exam boards refer to the opposite process at times. Hence the two definitions below
Enthalpy of lattice formation Enthalpy of lattice dissociation
The Enthalpy of lattice formation is the standard enthalpy The Enthalpy of lattice dissociation is the standard enthalpy
change when 1 mole of an ionic crystal lattice is formed change when 1 mole of an ionic crystal lattice form is
from its constituent ions in gaseous form. separated into its constituent ions in gaseous form.
Na+ (g) + Cl- (g)  NaCl (s) [Latt H = -787 kJ mol-1] NaCl (s)  Na+ (g) + Cl- (g) [Latt H = +787 kJ mol-1]




+
+
+




+
lattice formation from gaseous ions

This process cannot be measured experimentally directly. The following data can be used to work out the
lattice enthalpy

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]


Enthalpy of atomisation The enthalpy change for a solid metal turning to
The enthalpy of atomisation of an element is the enthalpy change when gaseous atoms can also be called the Enthalpy of
1 mole of gaseous atoms is formed from the element in its sublimation and will numerically be the same as the
standard state enthalpy of atomisation
Na (s)  Na(g) [atH = +148 kJ mol-1] Na (s)  Na(g) [subH = +148 kJ mol-1]
½ O2 (g)  O (g) [atH = +249 kJ mol-1]

Bond dissociation enthalpy (bond energy) For diatomic molecules the dissH of the molcule is
The bond dissociation enthalpy is the standard molar enthalpy change the same as 2x atH of the element
when one mole of a covalent bond is broken into two gaseous Cl2 2Cl (g) Hdiss = +242 kJ mol-1
(g) 
atoms (or free radicals)
Cl2 (g)  2Cl (g) dissH = +242 kJ mol-1 ½ Cl2 (g)  Cl (g) atH = +121 kJ mol-1
Or
CH4 (g)  CH3 (g) + H(g) dissH = +435 kJ mol-1


First Ionisation enthalpy Second Ionisation enthalpy
The first ionisation enthalpy is the enthalpy change required to The second ionisation enthalpy is the enthalpy change to
remove 1 mole of electrons from 1 mole of gaseous atoms to remove 1 mole of electrons from one mole of gaseous 1+ ions
form 1 mole of gaseous ions with a +1 charge to produces one mole of gaseous 2+ ions.
Mg (g)  Mg+ (g) + e- [ IE 1H] Mg+ (g)  Mg 2+ (g) + e- [ IE 2H]

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, First Electron affinity second 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 The second electron affinity is the enthalpy change when one
of gaseous ions with a –1 charge mole of gaseous 1- ions gains one electron per ion to produce
gaseous 2- ions.
O (g) + e-  O- (g) [ea1H] = -141.1 kJ mol-1]
O – (g) + e-  O2- (g) [ea2H = +798 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 The second electron affinity for oxygen is endothermic
and there is an attraction between the nucleus and the electron because it take energy to overcome the repulsive force
between the negative ion and the electron

The Born Haber cycle links all these enthalpy changes in an enlarged version of a Hess’s law cycle.

H lattice
Born-Haber cycles
ions in gaseous state ionic lattice calculate a measure of
This arrow is a combination of ionic bond strength based
several enthalpy changes that f H on experimental data.
turn elements into gaseous
ions elements in standard state



Born Haber cycle: sodium Chloride
Pay attention to state
Na+ (g) + e- Cl (g) symbols and direction of
+
arrows.
eaH (Cl)
atH (Cl)
Na+ (g) + Cl- (g) Usually all pieces of data are given
Na+(g) + e- + ½ Cl2(g) except the one that needs to be
calculated
IE 1H(Na)
 Latt H (NaCl) Careful
This Born Haber cycle has been
Na (g) + ½ Cl2(g) constructed using a lattice enthalpy
of formation. Sometimes questions
atH (Na) will give the enthalpy of lattice
dissociation which has the
Na (s) + ½ Cl2(g) opposite sign and the arrow points
in the opposite direction. This
changes the calculation
fH (NaCl)
NaCl (s)


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) + LattH

Rearrange to give LattH = fH - (atH (Na) + IEH(Na)+ atH (Cl) eaH(Cl) )


LattH =-411 – (+107 + 496 + 122 + -349) = -787 kJmol-1




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