2.2a The Born Haber Cycle
This section looks
at the application of Hess's Law
to the theoretical formation of an ionic lattice from its constituent
elements in their standard states i.e. most stable state at 298K (25oC),
101kP (1 atm).
 Brown
arrows = exothermic
Blue arrows
= endothermic
Fig
1. The simplest way, and the easiest way to learn to construct
and present a Born–Haber cycle – in this case
for sodium chloride NaCl. |
 |
 Right Fig
2. The enthalpy level diagram for the Born–Haber cycle for
sodium chloride.
Left Fig 3. A way of thinking to solve the cycle
for one unknown enthalpy component in a Born–Haber cycle
presented in the style of Fig 2.
All the delta H values
are defined and their use explained below. You need to be able
to recognise any of ΔH 1–6 and be familiar with the 'styles' of
figs 1–2 and be able to complete/construct a cycle and solve a
problem to obtain an unknown value. |
- In any enthalpy value with a
superscript theta, (θ), i.e. ΔHθx =
??, the θ denotes a standard enthalpy value,
which usually means 1 atm. pressure (101 kPa) and a 298 K
temperature (25oC). Where relevant, any such criteria need to be added to the
definitions below if quoted as standard values.
-
ΔH1 = ΔHθf(NaCl)
the enthalpy of formation of sodium chloride:
- Na(s)
+ 1/2Cl2(g) ==> NaCl(s)
ΔHf(NaCl) = –411 kJ mol–1
- The standard enthalpy of
formation is defined as the energy released or
absorbed when 1 mole of a compound is formed from its
constituent elements
in their normal stable states at 298K and 1atm.
-
ΔH2 = ΔHθatom(Na)
the enthalpy of atomisation of sodium:
- Na(s)
==> Na(g) ΔHatom(Na) = +107 kJ mol–1
- The standard enthalpy of
atomisation is defined as the energy absorbed when
1 mole of gaseous atoms is formed from the element in its normal
stable state at 298K and 1atm.
- other examples, but not
needed here are ...
- 1/2O2(g)
==> O(g) ΔHatom(O) = +249 kJ mol–1
- 1/2Br2(l)
==> Br(g) ΔHatom(Br) = +112 kJ mol–1
- 1/2I2(l)
==> I(g) ΔHatom(I) = +107 kJ mol–1
- Note that the
enthalpy of atomisation for gaseous species is related to
the 'gaseous' bond enthalpy value
- e.g. the enthalpy of
atomisation of oxygen is half the bond enthalpy of the O=O
double bond,
-
ΔH3 = ΔHθatom(Cl2)
the enthalpy of atomisation of chlorine:
- 1/2Cl2(g)
==> Cl(g) ΔHatom(Cl) = +121 kJ mol–1
- Already defined above and is
also half the bond enthalpy for chlorine molecules
- Cl2(g)
==> 2Cl(g) ΔHBE(Cl2) =
+242 kJ mol–1
- so take care with how
the data is presented.
-
ΔH4 = ΔHθel.affin.(Cl) the 1st electron affinity of chlorine:
- Cl(g)
+ e– ==> Cl–(g)
ΔHelec.affin.(Cl) = –355 kJ mol–1
- The standard enthalpy
change for the first electron affinity is defined as the energy released or
absorbed when one mole of gaseous neutral atoms gain one
electron each to form one mole of singly charged negative gaseous
ions at 298K and 1atm.
- I wouldn't worry about
the 2nd of chlorine, the Cl2– ion will be too endothermic
to be form in a chemical change and is
electronically very unstable.
- The 2nd electron
affinity would be defined as the energy absorbed or released
when 1 mole of singly charged negative ions gain one
electron each to form 1 mole of doubly charged negative
ions.
- However in a Born–Haber Cycle for the
formation of an ionic metal oxide e.g. MgO (shown below) you
need two electron affinities
- 1st electron
affinity of oxygen is exothermic
- O(g) ==>
O–(g) ΔH1st elec.affin.(O) =
–142 kJ mol–1
- but the 2nd is very
endothermic
- O–(g)
+ e– ==>
O2–(g) ΔH2nd elec.affin.(O)
= +844 kJ mol–1
- because
of the O–... e– repulsion, but must be
considered in the cycle for MgO etc. (see further down the page
- The 2nd electron
affinity (not needed here) would be defined as:
- The energy released or
absorbed when one mole of gaseous singly charged negative ions gain one
electron each to form one mole of doubly charged negative gaseous
ions at 298K and 1atm.
- This would be needed for
the Born–Haber cycle for magnesium oxide (Mg2+O2–).
-
ΔH5 = ΔHθ1st IE(Na)
the 1st ionization enthalpy of sodium:
- Na(g)
==> Na+(g) + e–
ΔH1st IE(Na) = +502 kJ mol–1
- The standard first
ionisation energy is defined as the energy absorbed when
the most loosely bond electron is removed from one mole of
neutral gaseous atoms to form one mole of singly positively
charged gaseous ions at 298K and 1atm.
- Again, in a B–H cycle
for MgO etc. the 2nd ionisation energy must be taken into
consideration and would be defined as:
- The energy absorbed when
the most loosely bond electron is removed from one mole of
singly positively charged gaseous ions to form one mole of
doubly positively
charged gaseous ions at 298K and 1atm.
-
ΔH6 = ΔHθLE(NaCl)
the lattice enthalpy of sodium chloride:
- Na+(g)
+ Cl–(g) ==> NaCl(s)
ΔHLE(NaCl) = –786 kJ mol–1
- The standard enthalpy
change, known as the lattice enthalpy (lattice energy) is defined as
the energy released when
1 mole of an ionic compound is formed from its constituent gaseous
positive and negative ions at 298K and 1atm.
- Factors affecting the
value of lattice enthalpy – lattice enthalpy
previously discussed – brief summary
below.
- The greater the force of
attraction between the ions, the greater the energy release, in
coming together to the point of minimum potential energy in forming
the most stable ionic crystal lattice.
- From the laws of
electrostatics, the force of attraction is proportional to
...
- +ve charge x –ve
charge/(distance between the centres of the charges)2,
- translating this into the
ionic lattice situation, the force of attraction is proportional to
the ..
- charge on cation x charge
on anion/(radius of cation + radius of anion)2,
- so quite simply, the smaller
the radius of an ion or the greater the charge on an ion, the
greater will be the lattice enthalpy because the electric field
intensity is increases, hence the force of attraction is stronger as
the ions are closer together, hence the
greater amount of energy released when the ions come together.
- the sum of the cation and
anion radius = the distance between the positive and
negative centres of attraction
- From this simple trends can
be noted e.g. in terms of ∆LE (in kJmol–1)
...
- LiF (1022) > LiCl
(846) > LiBr (800) > LiI (744), from L to R, the anion
radius becomes larger,
- Al2O3
(15916) > MgO (3889) > Na2O
(2478), from L to
R, the cation charge decreases (and the cation radius increases),
- MgO (3889)
is over
4 x NaCl (780), charges 2+ x 2– versus 1+ x 1–, ignoring ionic radii
differences.
TOP OF PAGE
2.2b Problem solving using a
Born–Haber Cycle as presented above
-
In
Fig. 1 above for sodium
chloride ΔH1
= ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
because all the arrows in the cycle point from the start to the
finish
- This shows how to
present and solve a Hess's Law cycle for the formation of an
ionic metal chloride and is an example of the so–called
Born–Haber Cycle.
- ΔH1
= ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
- ΔHf(NaCl)
= ΔHatom(Na)
+ ΔHatom(Cl)
+ ΔHelecaffin(Cl)
+ ΔH1stIE(Na)
+ ΔHLE(NaCl)
- –411 = (+107) +
(+121) + (–355) + (+502) + (–786)
- I often work in ()
at first retaining the original delta H sign, it reduces the
risk of 'sign error'!
- –411 = +107 +121 –355
+502 –786 = –411
- If you know five of
the six values you can theoretically calculate the 6th enthalpy
value which is the most important application of this particular Hess's Law
cycle.
-
Fig 2. shows one how to
present a Born–Haber cycle diagram for the formation of an
ionic metal chloride by using an enthalpy level diagram.
- However, in solving
problems via enthalpy diagrams you need to define your start and
end points and watch the signs!
- Fig. 3 shows one
possible start and end point.
- The sum of the 'upper loop'
enthalpy changes must equal the sum of the 'lower loop' changes,
but the signs of the numerical enthalpy values must be
appropriate,
- so ΔH2 + ΔH3 + ΔH5 + ΔH4
equals the upper loop, and no change in signs required, the
arrows point towards the end products,
- but the sum of the lower
loop enthalpies = ΔH1 + (–ΔH6)
because the arrow must point in the opposite direction to Fig. 2
and the sign of ΔH6 must therefore be
reversed,
- so ΔH2 + ΔH3 + ΔH5 + ΔH4
= ΔH1 –ΔH6 to solve a
problem.
- +107 +121 + 502 –355 =
–411 – –786
- so, +375 = +375, so, if 5
values known, the 6th can be theoretically calculated.
2.2c Problem
solving using other cycles (not using enthalpy level diagrams –
see 2.2d)
-
Δ abbreviations used
(all are defined further up the page)
-
f = enthalpy
of formation
-
atom =
atomisation energy
-
BE = bond
enthalpy
-
IE =
ionisation energy (1st or 2nd etc.)
-
LE = lattice
enthalpy expressed endothermically i.e. from ionic crystals to
free gaseous ions.
-
elec.affin = electron
affinity (1st or 2nd)
-
Each cycle involves 6–8
enthalpy values, of which you must know all of them except one!
-
You can then calculate
the unknown enthalpy value by substitution and simple algebraic
rearrangement.
-
No numerical values
are shown on these Born–Haber cycle diagrams, but they are in
section 2.2d enthalpy level diagrams.
-
(a) ΔHθf(KI)
= ΔHθatom(K)
+ ΔHθatom(I2)
+ ΔHθ1st IE(K) + ΔHθel.affin.(I)
= + ΔHθLE(KI)
-
(b) ΔHθatom(M)
+ {2 x ΔHθatom(X2)}
+ ΔHθ1st IE(M) + ΔHθ2nd IE(M)
+ {2 x ΔHθel.affin.(X)} = ΔHθf(MX2)
+ ΔHθLE(MX2)
TOP OF PAGE
2.2d
– more Born–Haber Cycle Enthalpy Level Diagrams
-
Δ abbreviations used
(all are defined further up the page)
-
f = enthalpy
of formation
-
at =
atomisation energy
-
IE =
ionisation energy (1st or 2nd etc.)
-
LE = lattice
enthalpy expressed endothermically i.e. from ionic crystals to
free gaseous ions.
-
ea = electron
affinity (1st or 2nd)
-
Each cycle involves 6–8
enthalpy values, of which you must know all of them except one!
-
You can then calculate
the unknown enthalpy value by substitution and simple algebraic
rearrangement.
-
All the numerical values
on the Born–Haber cycle diagrams are in kJ mol–1.
Born–Haber Cycle for
Sodium Chloride
-
route A = route B =
+377 kJ mol–1 (but watch the signs!)
-
ΔHθatom(Na)
+ ΔHθatom(Cl2)
+ ΔHθ1st IE(Na) + ΔHθel.affin.(Cl)
= ΔHθf(NaCl) + ΔHθLE(NaCl)
-
This Born–Haber cycle
can be adapted for any Group 1 Alkali Metal Halide MX eg LiF, NaBr,
KCl, KBr etc.
Born–Haber Cycle for
Magnesium Chloride
-
route A = route B =
+1883 kJ mol–1 (but
watch the signs!)
-
ΔHθatom(Mg)
+ {2 x ΔHθatom(Cl2)}
+ ΔHθ1st IE(Mg) + ΔHθ2nd IE(Mg)
+ {2 x ΔHθel.affin.(Cl)} = ΔHθf(MgCl2)
+ ΔHθLE(MgCl2)
-
This Born–Haber cycle
can be adapted for any Group 2 Alkaline Earth Halide MX2
eg MgF2, CaCl2, CaBr2 etc.
-
Why not MgCl?
-
-
route A = route B =
+659 kJ mol–1 (but watch the signs!)
-
ΔHθatom(Mg)
+ ΔHθatom(Cl2)
+ ΔHθ1st IE(Mg) + ΔHθel.affin.(Cl)
= ΔHθf(MgCl) + ΔHθLE(MgCl)
-
The lattice enthalpy for
MgCl can be theoretically calculated (its similar to NaCl, and so
with known values for the rest of the cycle, you can then calculate
the enthalpy of formation for MgCl. This turns out to –94 kJ mol–1,
which is much less exothermic than the energy released when the
favoured MgCl2 is formed.
-
Why not MgCl3?
-
I've found one quoted
value of +3949 kJ mol–1 for the enthalpy of formation of
MgCl3.
-
I think one could
reasonably deduce that this highly endothermic compound is hardly
likely to exist!
-
Although the more highly
charged Mg3+ ion would increase the lattice enthalpy,
favouring MgCl3 formation, the formation of Mg3+
requires far more energy because the 3rd ionisation requires the
removal of an electron from an inner less shielded shell.
TOP OF PAGE
Born–Haber Cycle for
Sodium Oxide
-
route A = route B =
+2114 kJ mol–1 (but watch the signs!)
-
{2 x ΔHθatom(Na)}
+ ΔHθatom(O2)
+ {2 x ΔHθ1st IE(Na)} + ΔHθ1st
elec.affin.(O) + ΔHθ2nd elec.affin.(O)
= ΔHθf(Na2O) + ΔHθLE(Na2O)
-
2nd electron affinity of
oxygen is the same as the 1st electron affinity of the O–
ion, but it seems a lot safer to think of these as the 1st and 2nd
electron affinities of oxygen atoms themselves, in the same way that you
refer to the 1st and 2nd ionisation energies of gaseous sodium
atoms.
-
This Born–Haber cycle
can be adapted for any Group 1 Alkali Metal oxide M2O eg
Li2O, K2O etc.
Born–Haber Cycle for
Magnesium Oxide
-
route A = route B =
+3243 kJ mol–1 (but
watch the signs!)
-
ΔHθatom(Mg)
+ ΔHθatom(O2)
+ ΔHθ1st IE(Mg) + ΔHθ2nd IE(Mg) + ΔHθ1st
elec.affin.(O) + ΔHθ2nd elec.affin.(O)
= ΔHθf(MgO) + ΔHθLE(MgO)
-
Using this data gives a
lattice energy of kJ mol–1, but one of my data books
quotes 3889.
-
This Born–Haber cycle
can be adapted for any Group 2 Alkaline Earth oxide MO eg CaO, BaO
etc.
 Comparison
of experimental and theoretical lattice enthalpies
The factors affecting the values of
lattice enthalpies were briefly discussed in section 2.1c.
BUT, here I want to quote discuss a
comparison between experimental lattice enthalpies and theoretically
calculated lattice enthalpies.
By comparing experimental lattice
enthalpies calculated from Born–Haber cycles with those from
theoretical calculations based on a perfect ionic model, the
difference in values provide evidence for some covalent character in
ionic compounds.
Despite the evidence of a 'little'
covalent character, most 'ionic compounds' have sufficient ionic
character to produce a crystal lattice where the atoms/ions are
essentially in the same positions in the lattice as they would be if it
was perfect (as in the diagram of sodium chloride on the right).
I've collected some data from my
textbook library and come up with some good sets of data to make the
point. I've quoted the difference in ΔHθLE
and then quoted the theoretically calculated value as a percentage of
the experimentally derived value from a Born - Haber cycle (like those
on this page). I'm afraid data varies from book to book, so I've tried
to judge a consensus!
Considering differences in
electronegativity also fit in with
| Ionic compound |
formula |
ΔHθLE(Born
- Haber experimental) |
ΔHθLE(theoretical
calculation) |
ΔHθLE
Difference |
Electronegativity difference,
non-metal - metal, Pauling scale |
Comments on whether, or not, the ΔHθLE
data indicates covalent character in the crystal |
| lithium
fluoride |
LiF |
1037 |
1033 |
4 (99.6%) |
4.0-1.0 =
3.0 |
Little
evidence of covalent character until LiBr and LiI with the
larger anion. BUT clear trend of increasing covalent character,
however small, in covalent character down the group 7 halides,
note the decrease in electronegativity difference too. |
| lithium
chloride |
LiCl |
852 |
845 |
7 (99.2%) |
3.0 - 1.0 =
2.0 |
| lithium
bromide |
LiBr |
815 |
798 |
17 (97.9%) |
2.8 - 1.0 =
1.8 |
| lithium
iodide |
LII |
761 |
740 |
21 (97.2%) |
2.5 - 1.0 =
1.5 |
| caesium
fluoride |
CsF |
750 |
748 |
2 (99.7%) |
4.0 - 0.7 =
3.3 |
Not a clear
trend, CsF highly ionic, the rest showing a trace of covalent
character with a significantly smaller electronegativity
difference. |
| caesium
chloride |
CsCl |
676 |
652 |
24 (96.4%) |
3.0 - 0.7 =
2.3 |
| caesium
bromide |
CsBr |
654 |
632 |
22 (96.6%) |
2.8 - 0.7 =
2.1 |
| caesium
iodide |
CsI |
620 |
601 |
19 (96.9%) |
2.5 - 0.7 =
1.8 |
| silver
fluoride |
AgF |
953 |
920 |
33 (96.5%) |
4.0 - 1.9 =
2.1 |
All the
silver halides show some covalent character. A clear trend of
increasing covalent character down group 7 as the halide ion
radius increases and can be more easily polarised by the Ag+
ion drawing the electron cloud of the anion towards itself, thus
creating the covalent character. The decreasing
electronegativity difference fits the trend too |
| silver
chloride |
AgCl |
903 |
832 |
71 (92.1%) |
3.0 - 1.9 =
1.1 |
| silver
bromide |
AgBr |
895 |
815 |
80 (91.1%) |
2.8 - 1.9 =
0.9 |
| silver
iodide |
AgI |
882 |
777 |
105 (88.1%) |
2.5 - 1.9 =
0.6 |
| sodium chloride |
NaCl |
781 |
777 |
4 (99.5%) |
3.0 - 0.9 = 2.1 |
Very highly ionic, big difference in
electronegativity. |
| potassium bromide |
KBr |
679 |
667 |
12 (98.2%) |
2.8 - 0.8 = 2.0 |
A trace of covalent character, but pretty
ionic, big difference in electronegativity. |
| calcium oxide |
CaO |
3607 |
3519 |
88 (97.6%) |
3.5 - 1.0 = 2.5 |
Little covalent character, perhaps more
than sodium chloride because of more highly charged cation?, but
pretty ionic with a large electronegativity difference |
| calcium fluoride |
CaF2 |
2611 |
2586 |
25 (99.0%) |
4.0 - 1.0 = 3.0 |
A big electronegativity difference, very
ionic, little evidence of covalent character |
| cadmium iodide |
CdI2 |
2435 |
1986 |
449 (81.6%) |
2.5 - 1.7 = 0.8 |
Large discrepancy suggesting a fair
degree of covalent character in the bonding and note the small
difference in electronegativity, from which you should predict
some covalent character anyway. |
| |
|
|
|
|
|
|
Energetics-Thermochemistry-Thermodynamics Notes INDEX
TOP OF PAGE
Keywords:
Born–Haber cycle for sodium chloride NaCl, Born–Haber cycle for sodium oxide
Na2O, Born–Haber cycle for magnesium chloride MgCl2 MgCl MgCl3, Born–Haber cycle
for magnesium oxide, Born–Haber cycle for potassium iodide KI, Born–Haber cycle
for Group 1 halides MX eg LiF NaF NaBr NaI KF KCl KBr, Born–Haber cycle for
Group 1 oxides M2O Li2O K2O, Born–Haber cycle for Group 2 halides MX2, MgF2,
MgBr2 CaF2 CaBr2 Born–Haber cycle for Group 2 oxides MO CaO BaO Born Haber Cycle
calculations for AQA AS chemistry Born Haber Cycle calculations for Edexcel A
level AS chemistry Born Haber Cycle calculations for A level OCR AS chemistry A
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chemistry A Born Haber Cycle calculations for A level OCR Salters A level
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draw diagrams of a Born Haber cycle, how to do calculations with a Born Haber
cycle, how to apply Hess's Law to the Born Haber cycle, how to use a Born Haber
cycle to solve enthalpy change problems, explaining the Born Haber cycle for
sodium chloride NaCl, how to construct an enthalpy level diagram to represent
the Born Haber cycle for sodium chloride, using enthalpy of formation, enthalpy
of ionisation, lattice enthalpy, enthalpy of atomisation, electron affinity
enthalpy to set up a Born Haber cycle for NaCl, how to solve problems using a
Born Haber cycle, constructing the Born Haber cycle for potassium iodide KI,
diagram of the Born Haber cycle for a group 2 halide salt MX2 e.g. MgCl2, how to
work out the Born Haber cycle for a group 2 oxide MO, using a Born Haber cycle
to show why MgCl cannot exist (isn't stable), constructing the Born Haber cycle
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Energetics-Thermochemistry-Thermodynamics Notes INDEX |
ΔHθf(KI)
ΔHθatom(I)
ΔHθ1st IE(K)
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