Energetics

See Working Through Chemistry Section 5

Syllabus requirements
Subject Focus: Energetics

Learning Outcome 5: I can show an understanding of enthalpy changes, entropy and Gibbs
free energy.

Topic Sub-Topic Assessment Criteria

5.1.1 1. Recognise the Joule (J) as the unit of energy.
5.1 Exothermic 2. Describe endothermic and exothermic energy changes for
Changes and phase changes and chemical changes.
in energy endothermic
processes
3. Sketch energy level (enthalpy) diagrams for exothermic
5.1.2 Energy and endothermic processes.
level
(enthalpy)
diagrams
1. Define the following enthalpy changes: standard enthalpy
change of reaction, formation, atomisation, ionisation,
combustion, solution (or dissolution), neutralisation, solvation
including hydration, electron affinity, bond enthalpy terms,
bond dissociation enthalpies and lattice enthalpies.
5.2 2. Relate the enthalpy of solution to the lattice enthalpy and
Standard enthalpies of hydration.
enthalpy
3. Distinguish between the first and second electron affinity
changes
for di-negative ions.
4. Explain the difference between bond enthalpy terms and
bond dissociation enthalpies.
5. Estimate the enthalpy change of a reaction using bond
enthalpies.
6. Explain why enthalpy values obtained from bond enthalpy
calculations are approximate.
5.3 Hess’s 1. State Hess’s Law.
5.3.1 Hess’s
law and
law and its 2. Calculate enthalpy changes indirectly by the construction
the Born-
use in simple of simple enthalpy cycles or energy level diagrams from
Haber
calculations given thermochemical data.
cycle.
3. Construct Born-Haber cycles.
5.3.2 Born-
4. Calculate enthalpy changes indirectly by the construction
Haber cycle
of Born-Haber cycles from given thermochemical data.

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 5.4 1. Explain the deviation between theoretical and
Theoretical experimental lattice enthalpies of silver halides.
and
experimental 2. Interpret the difference in terms of a degree of covalency
values of in the bonding of this formally ionic structure.
enthalpy 3. Explain the difference between expected and experimental
changes values of enthalpies of formation or combustion of benzene
and similar simple hydrocarbons as evidence of delocalised
structures.

Topic Sub-Topic Assessment Criteria

1. Plot temperature-time curves from the data obtained
from calorimetry experiments.
2. Interpret temperature-time curves to determine
5.5.1 temperature change.
Determination
3. Describe the relationship ΔH = –mcΔT.
of enthalpy
5.5
change of 4. Calculate enthalpy changes of neutralization, solution
Calorimetry
neutralization, and reaction from temperature changes and/or
solution and temperature-time curves using ΔH = –mcΔT
reaction 5. Calculate the enthalpy change of an acid-base reaction
using ΔH = –mcΔT.

6. Describe a simple thermometric acid-base titration.

5.5.2 Simple 7. Plot temperature-volume curves from the data obtained
thermometric from simple thermometric acid-base titrations.
acid-base
titrations 8. Determine the end-point from the temperature-volume
curve.
1. Distinguish between the concept of a system and its
5.6 Entropy 5.6.1 The surroundings.
and Gibbs system and its
2. State that ΔH alone is not able to fully predict
free energy surroundings
spontaneous change.
3. Describe entropy as a measure of disorder of a system.
4. Predict whether ΔS is positive, negative or nearly zero
5.6.2 Entropy in simple changes exemplified by state change,
combustion, dissolution, and dimerisation.
5. Calculate ΔS from absolute entropy values.
6. Explain that processes are spontaneous when the free
energy change is negative.
5.6.3 Gibbs
free energy 7. Describe the relationship between free energy, enthalpy
and entropy, namely, ΔG = ΔH – TΔS.

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 8. Calculate the temperature at which processes start or
cease to be spontaneous by use of the equation ΔG = ΔH –
TΔS.
5.6.4 Kinetic
versus 9. Explain the difference between the kinetic and
thermodynamic thermodynamic stability of chemical processes.
stability

Changes in energy
• A reaction can be endothermic or exothermic.
• Exothermic reaction: A reaction in which heat energy is given off.
• Endothermic reaction: A reaction in which heat energy is absorbed.

Exothermic reaction Endothermic reaction

Notice that in an exothermic change, the In an endothermic reaction, the products have
products have a lower energy than the a higher energy than the reactants. The system
reactants. The energy that the system loses absorbs this extra energy as heat from the
is given out as heat. The surroundings warm surroundings.
up.

ΔH = Energy products – Energy reactants

ΔH = Energy products – Energy reactants
ΔH = +ve
ΔH = -ve
ΔH will be positive (the value will have a
ΔH will be negative (the value will have a
positive sign)
negative sign)

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Standard enthalpy changes
Definitions
a) Standard enthalpy change of reaction, ΔH°: the heat given off or absorbed at 298 K
and 101 300 Pa when reagents react according to the stoichiometric equation.

b) Formation: energy given off or absorbed when one mole of a compound is formed
form its elements, all substances being in their standard states.
Example:

½ N2 (g) + 1½ H2 (g) → NH3 (g)

c) Combustion: energy given off when one mole of substance reacts completely with
oxygen under standard conditions.
Example:

CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (l)

d) Neutralisation: energy given off when one mole of H2O is formed from an acid and a
base under standard conditions.

HCl (aq) + NaOH (aq) → H2O (l)

Note:
• The value of the enthalpy change of neutralisation for a strong acid and a strong
base in aqueous solution is -57.3 kJ mol-1 (no need to remember value).
• Values are usually lower (i.e. less exothermic) when one of the acid or base (or
both) is weak.
• Some of the energy released is used up to ionize the weak acid or weak base.

e) Atomisation: energy required to form one mole of gaseous atoms from a substance in
its standard state under standard conditions.

½ Br2 (l) → Br (g)

C (graphite) → C (g)

f) Sublimation: energy required to convert one mole of a substance in the solid state to
one mole of the substance in the gaseous state under standard conditions.

I2 (s) → I2 (g)

C (graphite) → C (g)
Note: In some cases, such as in the case of graphite, the enthalpy of sublimation may also
be the enthalpy of sublimation.

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 g) Solvation: energy given off when one mole of a substance in the gaseous state is
surrounded by solvent molecules under standard conditions.
Note: when the solvent is water, this may be referred to as hydration.

K+ (g) + aq → K+ (aq)

h) Solution (or dissolution): energy given off or absorbed when one mole of solute
dissolves in excess solvent under standard conditions.

KBr (s) + aq → K+(aq) + Br-(aq)

i) Lattice enthalpy: energy given off when one mole of ionic solid in its standard state is
formed from its ions in the gaseous state.

Mg2+ (g) + 2 Br- (g) → MgBr2 (s)

Note: The value of the lattice enthalpy increases as the charge of the ions involved
increases and as the size of the ions involves decreases.

j) Ionisation: The first ionisation energy is the energy required to remove one mole of
electrons from one mole of gaseous atoms to form one mole of unipositive ions in the
gaseous state.

Na (g) → Na+ (g) + e-

The second ionisation energy is the energy required to remove one mole of electrons
from one mole of unipositive ions to form one mole of dipositive ions in the gaseous
state.

Na+(g) → Na2+ (g) + e-

k) Electron affinity: The first electron affinity is the energy given off or absorbed when
one mole of electrons is added to one mole of gaseous atoms to form one mole of
gaseous uninegative ions.

O (g) + e- → O- (g)
The second electron affinity is the energy required to add one mole of electrons to one
mole of gaseous uninegative ions to form one mole of gaseous dinegative ions.

O- (g) + e → O2- (g)

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Hess’s law
The energy change for any process is the same, independent of the pathway.

Hess cycle examples

Example 1: Determine the standard enthalpy change of formation of H2O2 (l), given the
following information.

H2 (g) + ½ O2 (g) → H2O (l) ΔH° = -285.8 kJ mol-1

H2O2 (l) → H2O (l) + ½ O2 (g) ΔH° = -98.2 kJ mol-1

Answer = -187.7 kJ mol-1

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Example 2: Consider the formation of dinitrogen tetroxide from nitrogen dioxide.

2NO2 (g) N2O4 (g) ΔHө = -57.2 kJ mol–1

Calculate the enthalpy change of formation of N2O4 (g) if that of NO2 (g) is +33.2 kJ mol-1.
(2 marks)

Taken from Matsec May 2005 Paper 1, Answer = 9.2 kJ mol-1

Example 3: Consider the dissociation of gaseous phosphorus pentachloride according to the

following equation:

PCl5 (g) PCl3 (g) + Cl2 (g) ΔHө = +88 kJ mol–1

Calculate the standard enthalpy of formation of PCl3 (g) if that of PCl5 (g) is -375 kJ mol-1.
(2 marks)
Taken from May 2007 Paper 1, Answer = -287 kJ mol-1

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Example 4: Calculate the enthalpy change of formation of benzene, formula C6H6, given the
following data.
C6H6 (l) ΔH°combustion = -3267 kJ mol-1
C (s) ΔH°combustion = -394 kJ mol-1
H2 (g) ΔH°combustion = -286 kJ mol-1

Answer = +45 kJ mol-1

Example 5: Calculate the enthalpy change for the reaction of ethene with HCl, given the
information shown below.

C2H4 (g) + HCl (g) → C2H5Cl (g)

C2H4 (g) ΔH°formation = +52.2 kJ mol-1
HCl (g) ΔH°formation = -92.3 kJ mol-1
C2H5Cl ΔH°formation = -109 kJ mol-1

Answer: -68.9 kJ mol-1

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Born-Haber cycle
• The Born-Haber cycle is concerned with the formation of an ionic compound from a
metal and a non-metallic element.
• Born-Haber cycles are used primarily as a means of calculating lattice energy which
cannot otherwise be measured directly.
• In practice, any unknown enthalpy change involved in the cycle can be determined by
applying Hess’s law to the cycle.
• The cycle centres on the formation of an ionic compound. Two enthalpies related to
this are the enthalpy of formation and the lattice enthalpy.
The enthalpies involved in the Born-Haber cycle are:
a) Enthalpy of atomisation of the metal (or enthalpy of sublimation)
b) Ionisation energy (or energies) of the metal
c) Enthalpy of atomisation of the non-metal
d) Electron affinity (or affinities) of the non-metal
e) Lattice enthalpy.
An example of a Born-Haber cycle is shown below.

By Hess’s law:
-411 = 107 + 496 +122 -349 + L.E.
L.E. = -787 kJ mol-1
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A schematic illustration of the Born-Haber cycle is shown below.

Example 1: Construct a Born-Haber cycle for NaBr and using the values shown below
calculate the lattice enthalpy of NaBr.
ΔH°sublimation Na = 108.4 kJ mol-1 ΔH°ionisation Na = 495.8 kJ mol-1
ΔH°atomisation Br2 = 111.7 kJ mol-1 ΔH°electron affinity Br = -324 kJ mol-1
ΔH°formation NaBr = -360 kJ mol-1

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Example 2: Consider the following data:
Enthalpy change Value (kJ mol-1)
Lattice enthalpy of strontium chloride -2150
First ionisation enthalpy for strontium +549
Second ionisation enthalpy for strontium +1064
The first electron affinity of chlorine -349
The enthalpy of atomisation of chlorine +122
The enthalpy of sublimation of strontium +164

i) Construct a Born-Haber cycle for strontium chloride. (4 marks)

ii) Calculate the enthalpy of formation of strontium chloride. (3 marks)

Taken from Matsec May 2023 Paper 1

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Example 3: (a) Match each of the energy changes in the table below with one of the processes
shown as (i) – (vi). The first answer is given for guidance. (2 marks)
Process ΔHө / kJ mol-1
(i) O2 (g) 2 O (g) 496
(ii) O (g) + e- O- (g) -148
(iii) O- (g) + e- O2- (g) 850
(iv) Ca (s) Ca (g) 193
(v) Ca (g) Ca+ (g) + e- 590
+
(vi) Ca (g) Ca2+ (g) + e- 1150
(vii) Ca (s) + ½ O2 (g) CaO (s) -635

Name of energy change Process

Enthalpy of sublimation (iv)
First ionisation enthalpy
Bond dissociation enthalpy
First electron affinity
Standard enthalpy change of formation

(b) (i) How will the enthalpy change for the process Mg (g) Mg+ (g) + e- compare
with that for process (v) above? Explain your answer.
(ii) How will the enthalpy change for the process K (g) K+(g) + e- compare with
that for process (v) above? Explain your answer. (part (i) and (ii) 4 marks)
(c) Using a Born-Haber cycle, calculate the lattice enthalpy of calcium oxide. (4 marks)
(d) How would you expect the lattice enthalpy MgO to compare with that of CaO? Explain
your answer. (2 marks)
Total 12 marks. Taken from Matsec May 2004 Paper 1

Gambin 12
Comparing Theoretical (Calculated) and Experimental lattice enthalpies

The lattice enthalpy can be determined in 2 ways:

1) An experimental value is obtained from a Born-Haber cycle. Knowing all the other
enthalpy changes, the lattice enthalpy is determined as the only unknown.

2) The lattice enthalpy can be calculated using a theoretical model. This model is based
on 2 charges separated by a certain distance. The theoretical model assumes 100 %
ionic character.

If the value calculated theoretically is significantly different from that obtained from the Born-
Haber cycle, this means that the compound has a significant degree of covalent character, and
it is not made up of 2 oppositely charged ions but has only partial charges with electron pairs
being shared.

Difference between Theoretical (Calculated) and Experimental lattice enthalpies

• The theoretical model (calculated) assumes 100 % ionic character. The greater the
covalent character of a compound, the greater the disagreement between theoretical and
experimental values.
• The degree of ionic character in a compound may be predicted by looking at the
difference in electronegativity between the bonded atoms.
• The greater the difference in electronegativity between the bonded atoms, the greater
the degree of ionic bonding.

Difference in electronegativity between bonded atoms

0 1.6 3.2
100 % ≈50 % covalent >90% ionic

• Where the difference in electronegativity between the bonded atoms is great, (i.e.
greater degree of ionic bonding) example in NaCl, the ionic model gives good
agreement with experimental values.
• Where the difference in electronegativity is smaller (i.e. greater degree of covalent
character), there is a disagreement between theoretical and experimental values.

N.B. The covalent character of a compound may also be explained on the basis of the polarising
power of the cation (positive ion) and polarizability of the anion (see bonding notes).

Gambin 1
 Comparing Theoretical (Calculated) and Experimental lattice enthalpies for NaCl and
AgCl

Less ionic
than NaCl
i.e. has a
higher High percentage
degree of difference
covalent suggests
bonding compound is not
purely ionic

Mostly
ionic

• The theoretical model (calculated) assumes 100 % ionic character. The greater the
covalent character of a compound, the greater the disagreement between theoretical and
experimental values.

• Sodium chloride has a very small percentage difference between its experimental and
theoretical lattice enthalpy. This suggests that NaCl is almost completely ionic.

• The larger difference between experimental and theoretical lattice enthalpy for AgCl
means that AgCl is not as close to the theoretical mathematical model as NaCl.

• This suggests that AgCl is less ionic in nature than NaCl and that AgCl has a degree of
covalent bonding.

Comparing the experimental and theoretical lattice enthalpies for silver halides.
The table below compares the theoretical (ΔHcal shown in the first column) and experimental
lattice (ΔHexp shown in the second column) enthalpies for silver halides. The third column
shows the difference between experimental and theoretical values. (Note that the values may
vary slightly depending on the source. We are only interested in the trend and not the specific
values.)

Most ionic

Gambin 2
Least ionic
 • The theoretical model (calculated) assumes 100 % ionic character. The greater the
covalent character of a compound, the greater the disagreement between theoretical and
experimental values.

• As you go from F to I, the electronegativity of the halogen is decreasing.

• That means that the electronegativity difference between the silver and the halogen will
be greatest with AgF, and least with AgI.

• The greater the difference in electronegativity between the bonded atoms, the more
purely ionic the bonding will be.

• Thus, AgF has the greatest degree of ionic bonding whilst AgI has the greatest degree
of covalent bonding.

• This explains why AgF has the smallest difference between experimental and
theoretical lattice enthalpies while AgI has the largest difference.

• The fluorine is the most electronegative atom in the group and so will have the greatest
tendency to pull bonding electrons to its end of the bond, and so form ions.

• Iodine is the least electronegative, and so doesn't have as strong a tendency to pull the
bonding electrons, and so the bonding isn't so purely ionic i.e. has a greater degree of
covalent character.

Calorimetry

This relates the enthalpy change of a process to a temperature change.

In a calorimetric experiment, the heat released (or absorbed) by the reaction is equal to the
amount of heat absorbed (or released) by the calorimeter.

Calorimetry is often used to calculate enthalpies of:

i) Solution
ii) Neutralisation
iii) Reaction
iv) Combustion

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The equation used is:
ΔH = -mcΔT
Where:
ΔH = Enthalpy change
m = mass of substance whose temperature is changing
(typically mass of the solvent used in the reaction)
c = Specific heat capacity of substance whose
temperature is changing
ΔT = Change in temperature

ΔT = T2 – T1

Examples:

1) 2.00 g of anhydrous CaCl2 were dissolved in 100 cm3 of water at 20 °C. On mixing, the
temperature of the mixture rose to 23.5 °C. Assuming that the solution formed has a density of
1.00 g cm-3 and a specific heat capacity of 4.20 J g-1 K-1, calculate the enthalpy of solution of
CaCl2. Required data: Molar mass (g mol-1): Cl = 35.5, Ca = 40 (Ans = -81.7 kJ mol-1)

Note: Enthalpy of solution: energy given off or absorbed when one mole of solute dissolves in
excess solvent under standard conditions.

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2) 100 cm3 of 0.500 mol dm-3 NaOH were mixed with 100 cm3 of 0.500 mol dm-3 HCl.
The initial temperature of both solutions was 22.00 °C. The temperature, after suitable
correction, reached a maximum of 25.43 °C. Given that the specific heat capacity of the
resulting mixture is 4.18 J cm-3 K-1 calculate the enthalpy of neutralization.
(Ans = -57.3 kJ mol-1)

Note: Enthalpy of neutralization: the energy given off when one mole of H2O is formed form
an acid and a base under standard conditions.

Gambin 5
3) An experiment was carried out to determine the enthalpy change of the reaction of
CuSO4 (aq) with Mg. These react in a redox reaction as shown below.

CuSO4 (aq) + Mg (s) → MgSO4 (aq) + Cu (s)

By means of a burette 100 cm3 of CuSO4 was transferred into a polystyrene cup which was
placed inside an empty beaker. The concentration of the CuSO4 is 0.0756 mol dm-3. A
thermometer was placed inside the CuSO4 solution and the temperature was recorded at 0.5
minute intervals. At 2 minutes, an excess of Mg was poured all at once into the CuSO4 solution.
The mixture was stirred and the temperature readings taken at 0.5 minute intervals. The results
obtained are shown in the table below.

Time (minutes) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Temperature (°C) 19.5 19.5 19.5 X 28.8 28.5 28.2 27.9 27.6 27.3

a) On the graph paper provided, plot a graph of temperature against time for the experimental
results shown above.

Gambin 6
b) Determine the corrected maximum temperature change of the reaction.

T1 = 19.5 °C T2 = 29.1 °C

ΔT = T2 – T1 = 29.1 – 19.5 = 9.6 °C

c) Assuming that the solution has a specific heat capacity of 4.18 kJ kg-1 K-1, calculate the
standard enthalpy change for the reaction of CuSO4 with Mg (reaction shown above) in
kJ mol-1. Note that the density of the CuSO4 solution may be assumed to be equal to
1.00 kg dm-3.

ΔH = –mc ΔT

100 cm3 ÷ 103 = 0.1 dm3

1 dm3 → 1 kg
0.1 dm3 → ?

m = 0.1 kg

ΔH = –mcΔT

ΔH = –(0.1)(4.18)(9.6)

ΔH = –4.013 kJ

Moles CuSO4 used:

1000 cm3 → 0.0756 mol CuSO4

100 cm3 → ?

100 × 0.0756/ 1000 = 7.560 ×10-3 mol CuSO4

Finding energy given off for 1 mol of CuSO4:

7.560 ×10-3 mol CuSO4 → –4.013 kJ

1 mol CuSO4 → ?

1 × –4.013/7.560 ×10-3 = -531 kJ mol-1

Gambin 7
 Bond Energies

Bond dissociation enthalpy: This is the energy required to break one mole of a given bond in
a specific molecule by homolytic fission (i.e. homolysis), all species being in the gaseous state.

Example:

CH4 (g) → CH3. (g) + H. (g) ΔH° = 444 kJ mol-1

Note:
• Homolytic fission (also known as homolysis) is the symmetrical breaking of a
covalent bond.

• The bond enthalpy of a particular type of bond will vary depending on what is around
it in the molecule. Examples of different values of the bond dissociation enthalpy of
the C—H bond are shown below (no need to remember values):

Bond Bond dissociation enthalpy (kJ mol-1)

H3C—H 439
CH3CH2—H 423
(CH3)2CH—H 414

Bond enthalpy term (also known as mean bond enthalpy): This is the average value of
the bond dissociation enthalpies for all the bonds of the same type averaged over a number of
species that contain that bond.

Note:

• Data tables use average values which will work well enough in most cases.

• That means that if you use the C—H value (as an example) in some calculation, you
can't be sure that it exactly fits the molecule you are working with. As a result
calculations using mean bond enthalpies will give less reliable data compared to other
methods (for example, finding ΔH of reaction using standard enthalpies of formation).

Gambin 1
Bond breaking vs bond formation

• Bond breaking is endothermic (requires energy), therefore the enthalpy value will be
positive.
• Bond formation is exothermic (gives off energy), therefore the enthalpy value will be
negative.

Examples of bond energies:

Bond Bond Energy (kJ mol-1) Bond Bond Energy (kJ mol-1)
C—H 414 N−−N 941.4
C—C 347 N—H 393
C=C 620 H—H 436
O—H 460 O=O 498.7

Example 1: Calculate the standard enthalpy change of the following reaction using the
standard bond energies provided.

CH2=CH2 (g) + H2 (g) → CH3CH3 (g)

Bonds broken (endothermic) Bonds formed (exothermic)

1 C=C = 620 1 C—C = 347
4 C—H = 414 × 4 = 1656 6 C—H = 414 × 6 = 2484
1 H—H = 436
Total = 2712 Total = – 2831
(negative sign is added since bond
formation is exothermic)

ΔH = 2713 – 2831 = – 118 kJ mol-1

Gambin 2
Example 2: Using the bond energies provided, calculate the enthalpy change for the reaction
between nitrogen and hydrogen to give ammonia, equation shown below, and hence determine
the enthalpy change of formation of ammonia.

N2 (g) + 3H2 (g) → 2NH3 (g)

Example 3: Taken from Working through Chemistry 2nd edition 5.26

a) Use bond energies to estimate the enthalpy change for the reaction of ethene with
oxygen.
CH2=CH2 (g) + 3O2 (g) → 2CO2(g) + 2H2O (g)

Gambin 3
b) Use the enthalpies of formation given in the table below to obtain a second value for the
enthalpy of reaction of ethene with oxygen.

Process ΔH° (kJ mol-1)

Enthalpy of formation of CH2=CH2 (g) 55.3
Enthalpy of formation of CO2 (g) -393.5
Enthalpy of formation of H2O (g) -241.8

c) Comment on any difference in the answers to (a) and (b).

The answer to (b) is more accurate. The answer to (a) uses mean bond enthalpies which are not
exactly the same as the bond energies in the compounds being considered. In (b) the actual
enthalpies of formation of the compounds being considered are used (and not mean values).

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Example 4: Taken from Working through Chemistry 2nd edition 5.28

The enthalpy of combustion of hydrazine, NH2NH2, is given below.

NH2NH2 + O2 → N2 + 2H2O ΔH° = -622 kJ mol-1

Use this value and the bond enthalpy values of the other bonds to calculate the N-N bond
enthalpy.

Gambin 5
 Enthalpy change, entropy change and change in Gibbs free energy

Entropy
• Symbol is S
• Entropy is a measure of the disorder of a system.
• Unit of S = J K-1 mol-1

Note: It is possible to work out the entropy of 1 mole of any substance. Calculations of
entropy are not required but we will need to calculate the entropy change, ΔS°.

• The standard entropy change of a reaction is a measure of the change in disorder

brought about by the process when measured under standard conditions.
• The entropy change is the difference between the entropy of the products and the
entropy of the reactants.

ΔS° = S°products - S°reactants

• The entropy of a substance depends on its state.

• Solids are most ordered so have the lowest entropy, while gases are most disordered
so have the highest entropy.

Entropy of a system increases, ΔS° is positive, if:

i) There is a change in state from solid → liquid → gas.

ii) The number of moles of particles increases during a reaction.

Examples for point (i) include:

H2O (l) → H2O (g)

CO2 (s) → CO2 (g)

Gambin 1
Examples for point (ii) include:

2O3 (g) ⇌ 3O2 (g)

N2O4 (g) ⇌ 2 NO2 (g)

Note: ΔS° is positive for the forward reaction. Reversing any of these processes will reverse
the sign of ΔS°, from positive to negative.

Examples which involve both (i) and (ii) include:

2H2O2 (l) → 2H2O (l) + O2 (g)

2Ca(NO3)2 (s) → 2CaO (s) + 4NO2 (g) + O2 (g)

Example 1: Calculate the change in entropy for the following reaction.

H2 (g) + Br2 (l) → 2HBr (g)

Substance S° (J mol-1 K-1)

H2 (g) 131.0
Br2 (l) 152.3
HBr (g) 198.5

ΔS° = S°products - S°reactants

ΔS° = 2(198.5) – (131+152.3)

ΔS° = +113.7 J K-1 mol-1

Note: You should be able to qualitatively predict if the sign for ΔS° will be positive or negative.

In this reaction, 1 mole of gas and 1 mole of liquid formed 2 moles of gas. Therefore, there was
an increase in disorder on going from reactants to products, thus the entropy change is positive.

Gambin 2
Example 2: Calculate the change in entropy for the following reaction.

NH3 (g) + HCl (g) → NH4Cl (s)

Substance S° (J mol-1 K-1)

NH3 (g) 193.0
HCl (g) 187.0
NH4Cl (s) 94.6

ΔS° = S°products - S°reactants

ΔS° = 94.6 – (187+193)

ΔS° = -285.4 J K-1 mol-1

Note: In this reaction, 2 moles of gas have formed 1 mole of solid. Both the decrease in the
number of moles and the formation of a solid from a gas will decrease the disorder. This
explains why the entropy change is negative.

Example 3: For the following reactions, predict if the entropy change is positive, negative
or approximately zero.

i) CaCO3 (s) → CaO (s) + CO2 (g)

• The entropy change is positive, meaning there is an increase in entropy, hence an

increase in disorder.
• 1 mole of reactant forms 2 moles of product. The increase in the number of moles results
in an increase in entropy.
• Also, 1 mole of solid forms 1 mole of gas. The formation of a gas from a solid will
result in increase in entropy.

ii) N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

• The entropy change is negative, meaning there is a decrease in entropy, hence a

decrease in disorder.
• In the forward reaction, 4 moles of gas react to form 2 moles of gas. The decrease in
the number of moles of gas on going from reactants to products, will result in a decrease
in disorder, hence a decrease in entropy.

Gambin 3
 iii) NaCl (s) + aq → NaCl (aq)

• The entropy change is positive, meaning there is an increase in entropy, hence an

increase in disorder.
• In this reaction, a solid is going into solution. The solid is highly ordered while the
solution is disordered. The entropy of the solution is higher than the entropy of the solid.

iv) CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

• The entropy change is negative, meaning there is a decrease in entropy, hence a

decrease in disorder.
• 3 moles of gas are reacting to form 1 mole of gas and 2 moles of liquid. The entropy of
a gas is higher than that of a liquid. The decrease in the number of moles of gas on
going from reactants to products, will result in a decrease in disorder, hence a decrease
in entropy.

Notice that if the water had been formed as steam, you couldn't easily predict whether there
was an increase or a decrease in entropy, because there would be three moles of gas on each
side. In that case, the entropy change would be approximately zero.

Change in Gibbs Free Energy, ΔG

ΔG° = ΔH° - TΔS°

Unit of ΔG° is J mol-1

For a reaction or process to be energetically feasible, the change in Gibbs free energy, ΔG°
must be negative.

Note: the value of ΔS° is typically given in J K-1 mol-1 while the value of ΔH° is typically given
in kJ mol-1. The values used need to be consistent, i.e. the value ΔH° needs to be converted in
J mol-1.

Gambin 4
Example

i) Consider the following reaction:

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

Given that ΔH° = -92.38 kJ mol-1 and ΔS° = -198.3 J mol-1 K-1, calculate the Gibbs free
energy change at 25 °C and 500 °C.

The Gibbs free energy change is calculated from the equation ΔGo = ΔHo – TΔSo.
At 25 oC:
ΔGo = ΔHo – TΔSo
= -92.38 × 103 - (298 × (-198.3))
= -33.3 × 103 J mol-1

At 500 oC:
ΔGo = ΔHo – TΔSo
= -92.38 × 103 - (773 × (-198.3))
= +60.9 × 103 J mol-1

Care must be taken to ensure that the units used are consistent. In the
above question the change in enthalpy is given in kJ mol-1 while the
change in entropy is given in J mol-1K-1. So, in the above example the
enthalpy is converted to J mol-1 by multiplying by 1000. The Gibbs free
energy is then given in the units of J mol-1. Alternatively, the entropy
units could have been converted to kJ mol-1 K-1 and the Gibbs free
energy would then have been given in the units of kJ mol-1.

The temperature must be converted to K by adding 273 to the temperature in

oC.

ii) At which of the two temperatures is the forward reaction spontaneous? Explain.

The forward reaction is spontaneous at 25 oC. The

forward reaction will be spontaneous when the value of The positive value of ΔGo at
the Gibbs free energy change, ΔGo, is negative. 500 oC implies not only that the
forward reaction is not
spontaneous, but that the
reverse process is.

Question taken from Matsec 2020 1st session P1 no 2 (part question).

Solution taken from Model answers Plus+ To Matriculation Advanced Level Chemistry
2017-2020 by J.P. Tabone Adami and D. Gambin

Gambin 5
Sign of ΔG° for different combinations of ΔH° and ΔS°

• The spontaneity of a process is favoured by two factors:

o a negative ΔH° (decreasing the energy of the system)
o a positive ΔS° (increasing disorder of the system)

• Each of these is an individual ‘thumbs up’  for a process.

• Reactions for which ΔH° is negative and ΔS° is positive are always spontaneous since
ΔG° is negative for all values of T.
• Conversely, if ΔH° is positive and ΔS° is negative, the value of ΔG° is always positive
and the reaction is never spontaneous.
• If ΔH° and ΔS° are both positive or both negative, then the spontaneity of the reaction
depends on the temperature.

ΔH° ΔS° ΔG°

-ve  +ve  -v, always spontaneous
-ve  -ve  -ve at low temperature
+ve  +ve  -ve at high temperature
+ve  -ve  +ve, never spontaneous

Spontaneous endothermic reaction

Although the enthalpy of solution of sodium chloride is endothermic, the compound is highly
soluble. Explain this observation. (3 marks)

NaCl (s) → Na+ (aq) + Cl- (aq)

The process increases disorder as one mole of crystalline solid forms two moles of aqueous
ions. The increase in entropy makes for a positive value of ΔS. A process is spontaneous if
ΔG is negative.

ΔG = ΔH – TΔS

ΔH° of the process is positive but if TΔS is larger than ΔH, then the value of ΔG is negative
and the process of solution of the solid is spontaneous.

Question taken from Matsec May 2019 Paper 2 no 1d

Gambin 6
ΔG° = 0

• If ΔG° is negative, the reaction is spontaneous.

• If ΔG° = 0, the system is at equilibrium. The reaction/process just becomes
spontaneous.
• If ΔG° is positive, the reaction/process is not spontaneous, but occurs spontaneously
in the reverse direction.

At the temperature at which a process becomes just spontaneous, ΔG° = 0. Therefore, by

knowing ΔH° and ΔS°, the temperature at which the process just becomes spontaneous can
be determined by using ΔG° = ΔH° - TΔS° and setting ΔG° = 0.

ΔG° = ΔH° - TΔS°

0 = ΔH° - TΔS°
TΔS° = ΔH°
T = ΔH°/ ΔS°

Example:

Reaction A is spontaneous at room temperature, whilst reaction B is not.

A NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)

B NH4+ (aq) + NO3-(aq) → NH4NO3 (s)

a) What is understood by the term spontaneous ? (1 mark)

A spontaneous process is one in which the change in Gibbs free energy, ΔG, is negative
(process, involves a decrease in free energy).

b) At what temperature does the reaction B become just spontaneous, knowing that
ΔH° = -28.05 kJ mol-1 and ΔS° = -108.7 J K-1 mol-1 ? (3 marks)

At the temperature at which a process becomes just spontaneous, ΔG° = 0.

ΔG° = ΔH° - TΔS°

0 = -28.05 ×103 – (T × -108.7)
T = 28.05 ×103/108.7
T = 258 K

Gambin 7
Note 1: the value of ΔS° is given in J K-1 mol-1 while the value of ΔH° is given in kJ mol-1. The
values used need to be consistent, i.e. the value ΔH° needs to be converted in J mol-1.

Note 2: The value for temperature obtained is an approximation because the values of ΔH°
and ΔS° are used and these are only correct at 298 K. Actually, the values of ΔH° and ΔS°
change slightly with temperature and as a result the actual temperature will be slightly off
from the calculated one.
Question taken from Matsec May 2016 P1 no 6 (part question)

Kinetic stability vs thermodynamic stability

• If G for a process/reaction is negative, we can state that the reactants are

thermodynamically unstable.

• When a reaction does not appear to take place due to a high activation energy we say
that the reactants are kinetically stable.

• C(diamond) is thermodynamically unstable but kinetically stable under standard

conditions.

• C(graphite) is the thermodynamically stable form of carbon under standard conditions.

Example: Allotropes of carbon

• The change C(diamond) → C(graphite) has a negative value of ΔG under ambient

conditions so the reaction is energetically feasible.

• The fact that the reaction does not appear to occur is linked to the high activation energy
of the reaction since strong C-C bonds need to be broken before the atoms in diamond
can be rearranged to form graphite.

• This will make the reaction rate so slow as to be unobservable under ambient
conditions.

• If the temperature is raised sufficiently, the

activation energy barrier is overcome and the
reaction takes place at a measurable rate.

Gambin 8
Example: Allotropes of sulfur

• The slow rate of conversion from diamond to graphite can be contrasted by the faster
rate of change from one sulfur allotrope to another1.

• Below 95.3°C, monoclinic sulfur changes to the rhombic form. Both forms are made
up of cyclic S8 molecules differing only in the way in which the S8 molecules are packed
within the crystal structure.

• Intermolecular induced dipole-induced dipole forces need to be broken as the packing

of the S8 molecules changes.

• These intermolecular forces are weaker than the covalent bonds which need to be
broken to change from diamond to graphite.

• Thus, the conversion from monoclinic sulfur to rhombic sulfur requires a much smaller
activation energy than the conversion of diamond to graphite.

• Thus, the conversion between sulfur allotropes is faster than the conversion between
the allotropes of carbon.

Note that:

• S(monoclinic) is both thermodynamically unstable and kinetically unstable under

standard conditions.

• S(rhombic) is the thermodynamically stable form of S under standard conditions.

1
The conversion from one allotrope to another may take up to 2 days. This is faster than the conversion of
diamond to graphite.
Gambin 9