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2.2.2 Constructing Born-Haber Cycles
A Born-Haber cycle is similar to a Hess’ law energy cycle and allows the calculation of an
enthalpy change which cannot be measured directly, providing all the other energy
changes are known.
In the Born-Haber cycle:
 All ∆H values pointing upwards are endothermic values.
 All ∆H values pointing downwards are exothermic values.
Use a Born-Haber cycle to calculate the lattice enthalpy of NaCl(s)
Step 1
Start the cycle with the elements in their standard states at zero energy. The elements
always sit on the zero (datum) line.
Step 2
Add the formation of the ionic solid from its elements in their standard states. This is the
enthalpy change of formation of the ionic solid, which is exothermic (arrow downwards, ∆H
= -ve). This is drawn immediately below the elements.
Step 3
Add the lattice enthalpy (formation of the ionic solid from its ions in the gas state). This
step is very exothermic and the gaseous ions will have much more energy than the ionic
solid.
Step 4
Complete the cycle by converting the elements into the gaseous ions in a number of steps
using equations and data usually provided.
For example:
Atomisation of sodium
Atomisation of chlorine
Ionisation of sodium
Electron affinity of chlorine
Enthalpy of Hydration, (
)
Definition: This is the standard enthalpy change when …
Examples:
Note: Hydration enthalpies are always negative (exothermic)
Reason:
Enthalpy of Solution (
)
Definition: This is the standard enthalpy change when …
Example:
Example: Use a Born-Haber cycle to calculate the lattice enthalpy for MgO(s).
Calculating Enthalpy of Solution
This is the enthalpy change that accompanies the dissolving of 1 mol of ionic compound:
The dissolving of an ionic compound can be considered a two step process, one involving
bond breaking (endothermic), the other involving bond making (exothermic).
1. Breaking up the ionic lattice to give separate (gaseous) ions. This requires an
amount of energy equivalent to the _________________________________.
2. Hydration of the separate gaseous ions to give the separate aqueous ions. This
releases an amount of energy equivalent to the
_____________________________________________.
Example: Calculate the enthalpy of solution of sodium chloride NaCl(s)
Solution: Write out the relevant steps with thermochemical values and add together OR
draw thermochemical cycle.
Note: Enthalpies of Solution have negative (exothermic) or small positive (endothermic)
values. If lattice enthalpy is large when compared to the hydration enthalpies then the
ionic solid will be insoluble.
Mean Bond Enthalpies
Lattice dissociation enthalpy (
compound.
) is a measure of the strength of ionic bonds in an ionic
Bond dissociation enthalpy (
) is a measure of the strength of specific covalent bonds
within specific gaseous species.
Definition: The mean bond enthalpy is the …
Mean bond enthalpies provide a simple way of calculating approximate values (usually
within 10% of true value) of overall enthalpy changes for molecular reactions, particularly
where other data are not available.
Example:
Calculate the enthalpy change for the complete combustion of propane using mean bond
enthalpy data.
Solution:
Note: It is perfectly legitimate and possibly quicker to count only the actual bonds broken
and formed rather than to completely atomise the reactant molecules.
Example: CH3CH2CH2OH + O2  CH3CH2COOH + H2O
The only bonds broken and formed are –
Entropy Change, ∆S, and Free Energy Change, ∆G.
Spontaneous Change
Def: A spontaneous change is …
Note 1: If a change in one direction is spontaneous, the reverse process will require an
input of external energy and will not be spontaneous.
Note 2: The word “spontaneous” in thermodynamics does not have exactly the same
meaning as it does in everyday speech where it means there is virtual certainty in
something happening. In thermodynamics it means that it can happen, not that it will
happen. A better word is ___________
There are two factors that drive spontaneous changes:
1. The enthalpy factor
Most spontaneous changes are ___________________.
The natural direction of a spontaneous change is from higher to lower enthalpy with
a release of the difference in energy (usually as heat) between the two.
(From an enthalpy point of view, the products are ____________ stable than the
reactants.)
Example:
However, some changes are spontaneous even though they are endothermic.
Example:
Clearly, there must be an additional factor that causes reactions to occur, over and above
a favourable decrease in enthalpy.
2. The entropy factor
The additional factor that helps to drive spontaneous change is _____________, which is
given the symbol _____.
Entropy can be thought of as the degree of
____________________________________________
The natural direction for a spontaneous change is from a lower to higher entropy.
Sθ298 / J mol-1 K-1
Diamond
2.4
He(g)
126
Graphite
5.7
Ar(g)
155
NaCl(s)
72
O2(g)
205
NaHCO3(s)
102
CO(g)
198
SiO2(s)
42
CO2(g)
214
H2O(l)
70
H2O(g)
189
CH3OH(l)
127
CH3OH(g)
240
From the standard entropy values given in the table it can be seen that:

Simple molecules generally seem to have _____________ entropies than more
complicated molecules.
Example:

For substances of similar complexity entropy ______________ on going from solid
to liquid to gas.
Example:
Variation of entropy with temperature
Heating leads to an increase in disorder which fits with the increase in entropy as a
substance changes from solid to liquid to gas mentioned previously.
It can be seen that:

Entropy ________________ slowly with temperature in solids, liquids and gases.

A change of _______________ causes a sudden increase in entropy.

Boiling causes a _____________ increase in entropy than melting.
Absolute Entropy, Sθ
Since entropy is closely linked with disorder and it _______________ when the
temperature is lowered, it is reasonable to assume that, at 0 K (absolute zero), all disorder
disappears and all substances will be perfectly ordered and have __________ entropy.
This assumption turns out to be true for most substances, and especially for perfectly
ordered crystals.
The idea of zero entropy has an important consequence. Unlike enthalpy, where only
enthalpy changes can be considered, there is a well defined starting point to the entropy
scale (zero entropy at zero Kelvin). This means that absolute standard entropy values,
Sθ, exist for substances and from these, standard entropy changes, ∆Sθ for reactions can
be calculated.
Entropy Change in Chemical Reactions, ∆Sθ
Entropy calculations using standard entropy values use the following equation:
Example:
Calculate the entropy change that accompanies the combustion of graphite:
C(graphite) + O2(g)  CO2(g)
Answer:
Note: In this example the entropy change is small. Carbon dioxide and oxygen are both
gases and have similar entropies. Graphite is a low-entropy solid. So even though two
moles form one, the number of gas moles does not change and there is not much change
in entropy.
Example:
Calculate the entropy change that accompanies the thermal decomposition of solid sodium
hydrogencarbonate. (Sθ (Na2CO3(s)) = 136 J mol-1 K-1).
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
Answer:
Gibbs Free Energy Change in Chemical Reactions, ∆Gθ
A combination of two factors, enthalpy and entropy, govern the feasibility of reactions.
Exothermic processes tend to happen quite often but, equally endothermic processes
which lead to an increase in entropy also happen quite often.
The sometimes conflicting demands of enthalpy and entropy are brought together in the
relationship:
This combines the influence of both enthalpy change and entropy change.
With ∆H negative (exothermic) and ∆S positive (increase in entropy) then ∆G will be
___________
If ∆Gθ for a reaction is negative (or zero) then the reaction is feasible; if it is positive
then the reaction is not feasible.
Note 1: If ∆H is positive (endothermic) then, providing T∆S is __________ positive then
the reaction will be feasible.
Note 2: The term T∆S is temperature dependent meaning that some reactions may be
feasible at one temperature but not at another.
Note 3: A reaction with ∆G negative means it can go. It does not necessarily mean that it
will go. There is the kinetics of a reaction to take into account as well as the
thermodynamics. For example, a feasible reaction (thermodynamically unstable) may
have a very high _________ preventing it from proceeding to any significant extent
(kinetically stable).
Example:
Calculate the standard free-energy change for the combustion of graphite at 298K.
C(s) + O2(g)  CO2(g)
(∆H = -394 kJ mol-1 ∆S = +3.3 J K-1 mol-1)
Answer:
Note: Because ∆S is so small, ∆G will not vary much with temperature, though in practice
the reaction is extremely slow at 298K due to high activation energy.
Example:
Calculate the standard free-energy change for the rusting of iron at 298K
2Fe(s) + 1.5O2(g)  Fe2O3(s)
(∆H = -825 kJ mol-1 ∆S = -272 J K-1 mol-1)
Answer:
Note: ∆G is very negative so the rusting of iron is highly feasible at room temperature
even though ∆S is negative. The reaction would become less feasible at higher
temperatures as T∆S gets closer in value to ∆H.
Example:
Calculate the standard free-energy change for the decomposition of 1 mol sodium
hydrogencarbonate at 298K.
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
(∆H = +130 kJ mol-1 ∆S = +335 J K-1 mol-1)
Answer:
Note: At 298K ∆G is positive so the reaction is not feasible. However, ∆G is not very
positive so raising the temperature will make T∆S more positive and at a certain
temperature the reaction will become feasible.
Effect of Temperature on Feasibility
Reaction
∆S positive (entropy
gain)
∆H negative
(exothermic)
Feasible at all T
∆S negative (entropy loss)
∆H positive
(endothermic)
Not feasible at any T.
Calculating the Temperature at which a reaction becomes Feasible
A reaction that is not feasible at one temperature may become feasible if the temperature
is changed. The temperature at which feasibility is just achieved is the temperature at
which there is no tendency for the reaction to go one way or the other. At this temperature
∆G = 0 so:
Example:
Calculate the temperature at which the thermal decomposition of sodium
hydrogencarbonate becomes feasible.
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
(∆H = +130 kJ mol-1 ∆S = +335 J K-1 mol-1)
Answer:
Note: The feasibility temperature gives the “break-even” point for the reaction. That is the
temperature at which the concentrations of reactants and products become roughly equal.
In this example, the reaction becomes more feasible as the temperature is increased
above this point and product formation will be increasingly favoured.
Example:
Calculate the temperature above which limestone will spontaneously decompose to
quicklime and carbon dioxide.
CaCO3(s)  CaO(s) + CO2(g)
(∆H = +178 kJ mol-1 ∆S = +161 J K-1 mol-1)
Answer:
Entropy in Physical Changes
In many physical changes there is an increase or decrease in disorder (and hence
entropy).
1. Melting (Fusion)
At 00C (273K) and 100 kPa a mixture of ice and water has no spontaneous tendency to
either solidify or liquefy. The mixture is at equilibrium and ∆G = 0
H2O(s) = H2O(l) ∆H = +6 kJ mol-1
The entropy change ∆S for melting of ice at 273K can be determined as follows:
2..Freezing (Solidification)
H2O(l) = H2O(s) ∆H = -6 kJ mol-1
The entropy change ∆S for freezing of water at 273K can be determined as follows:
3. Boiling (Vaporisation)
H2O(l) = H2O(g) ∆H = +44 kJ mol-1
The entropy change ∆S for boiling of water at 373K can be determined as follows: