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TOPIC 15: ENERGETICS (HL)
In this section we will study why some chemical reactions are spontaneous (or feasible) in terms of energetics)
and other reactions are not. The spontaneity or feasibility of a change (physical or chemical) depends on the
interaction of two concepts, one of which was introduced in the SL course:
 Enthalpy: during a spontaneous change most systems show a tendency to release energy to take on a
more stable lower energy state; however, this is not always the case e.g. dissolving of ammonia chloride in
water is endothermic. This shows the limitations of using enthalpy change only to explain spontaneity and
proves that reducing the enthalpy of a system is not the only factor that determines spontaneity.
 Entropy which refers to the state of disorder; most spontaneous changes involve an increase in disorder but
not all e.g. condensation; such an example shows the limitations of using entropy to explain spontaneity of
changes.
To indicate the combined effect of these two concepts (or how they interact with each other) a third concept
has been introduced which is called “free energy”. This concept relates the enthalpy and entropy changes to
the system itself (as opposed to also considering the surroundings) as that gives us the advantage that we can
measure both the enthalpy and entropy changes.
15. 1 STANDARD ENTHALPY CHANGES OF REACTION
15.1.1 Define and apply the terms standard state, standard enthalpy change of formation and standard enthalpy
change of combustion
15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

Standard enthalpy changes, indicated by the symbol H , are enthalpy changes measured or calculated with
both reactants and products in standard
conditions i.e. 101 kPa (=1 atm) and 298 K and in
the case of solutions a concentration of 1.0 mol dm-3.
The state of substances in which they exist in these standard conditions is called the standard (physical) state
of these substances.

Therefore in an equation defining a standard enthalpy change, H , the reactants and products should be in
their standard physical states.
When we refer to an enthalpy of reaction we are referring to the energy change when the reaction is carried
according to the given stoichiometric equation (i.e. the balanced equation as given) and not necessarily for 1
mole of reactant or product. It is therefore often expressed in kJ; if the equation is for 1 mole of a particular
reactant or product e.g. 1 mole of octane, than enthalpy change can be expressed in kJ mol-1.
Standard enthalpy change of formation= Hf 
The standard enthalpy of formation = the enthalpy change of the reaction in which one mole of a substance is
formed from its elements with all chemicals in their standard states. As it is molar it is in kJ mol-1.
Examples:
H2 (g)
¼P4 (s)
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+
+
½ O2 (g) 
3/2 Cl2 (g) 
H2O (l)
PCl3 (l)
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Hf  of water = - 285.8 kJ mol -1
Hf  of PCl3 = - 319.7 kJ mol -1
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The above 2 equations tell us that in both reactions the products, i.e. the compounds, are more stable than the
reactants, i.e. the elements, as they are at a lower enthalpy level.
Reference point:
As the absolute enthalpies of chemicals cannot be measured and a reference point on the vertical axis
of a energy level diagram is needed, it has been decided to give the absolute enthalpy content of an
element in its standard state (which is their most stable state) the relative value of zero enthalpy.
It is the standard against which the other enthalpies and energetic stability are measured.
Standard enthalpy changes of formation of elements = zero!!
The standard enthalpy change of formation of elements in their standard states is zero as no formation
reaction is needed. The formation of an element from its elements is not considered a reaction as no new
substance is formed so therefore it cannot have an enthalpy change.
Examples of standard enthalpy changes of formation for some compounds
reaction
Ca (s) + ½O2 (g) CaO (s)
Hf  in kJ mol -1
-635.5
Na (s) + ½Cl2 (g) NaCl (s)
-411.0
C (s) + O2 (g)  CO2 (g)
-393.5
H2 (g) + ½O2 (g) H2O(l)
-285.9
H2 (g) + ½O2 (g) H2O(g)
-241.8
C (s) + 3H2 (g) + ½O2 (g)  C2H5OH (l)
-277.7
½H2 (g) + ½F2 (g) HF (g)
-271.1
8C (s) + 4H2 (g)  C8H8 (l)
-224.4
2H2 (g) + N2 (g)  N2H4 (l)
+ 50.6
3/2 H2 (g) + ½N2 (g)  NH3 (l)
- 46.0
Using standard enthalpy changes of formation to find enthalpy change of reaction
In any reaction, the reactants and the products are made form the same elements in their standard states.
Hreaction = a Hf (products)
-
b Hf (reactants)
The above mathematical expression means that to get the enthalpy change of a reaction simply add up the
enthalpies of formation of the products (multiplied by their respective coefficients symbolised by a) and
subtract from this the sum of the enthalpies of formation of the reactants (multiplied by their respective
coefficients, b).
The sum of the enthalpies changes of formation of the reactants is the energy released or absorbed when the
reactants are decomposed; by subtracting this from the sum of the products we are reversing the sign of the
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individual enthalpies of formation of the reactants. This makes sense as during the decomposition of the
reactants the formation reaction is reversed.
Worked example 1:
Using enthalpies changes of formation, calculate the enthalpy change of combustion of benzene as shown by
the equation below. (find the values in table 11 in your IB data booklet or in the table on the previous page in
this handout)
C6H6 (l)
Hreaction  = a Hf (products)
+
-
7½O2 (g)  6CO2 (g)
+
3H2O (l)
b Hf (reactants)
= [6 Hf (CO2 (g)) + 3Hf (H2O (l))] - [7½ Hf (O2 (g)) + 1Hf (C6H6 (l)) ]
= [6 mol (-393.5 kJ mol-1) + 3 mol (-285.8 kJ mol-1)] - [7½mol (+0 kJ mol-1) +1mol (+49 kJ mol-1)]
= - 3267 kJ (it could be expressed as kJ mol-1 as it is for one mole of C6H6.
Worked example 2:
Calculate the standard enthalpy for the combustion of 1 mole of propane.
+ 5O2 (g)  3 CO2 (g)
C3H8 (l)
+
4H2O (l)
Hr  = [3 Hf (CO2 (g)) + 4 Hf (H2O (l))] - [5 Hf (O2 (g)) + 1 Hf (C3H8 (l)) ]
= [3 mol (-393.5 kJ mol-1) + 4 mol (-285.9 kJ mol-1)] - [ 5 mol (+0 kJ mol-1) + 1 mol (-105 kJ mol-1)]
= - 2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ (mol-1)
We can also consider the above example 2 from the perspective of an enthalpy level diagram.
Enthalpy diagram
(energy in kJ mol-1)
3C (s) + 4H 2 (g) + 5 O2 (g)
0 Hf (reactants) = - 105 kJ = H1
C3 H8 (g) + 5 O2 (g)
Hf (products) = -2324.1 kJ = H2
Hreaction = - 2219.1 kJ
3CO2 (g) +
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4H2O (g)
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Hreaction =
H2
H1 = Hf (products) - Hf (reactants) =
-
- 2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ
We can also represent the above calculations using an enthalpy cycle
Hreaction
C3 H8 (g) + 5O2 (g)
3CO2 (g) +
H2
4H2O (l)
H1
3C (s) + 4H 2 (g) + 5O2 (g)
Hreaction =
H1
H2 =
-
-2324.1 kJ - (- 105 kJ )
= - 2219.1 kJ
General enthalpy cycle
Hreaction
reactants
H2 = enthalpies formation reactants
products
H1 = enthalpies of formation
products
elements in standard states of both reactants and products
Hreaction =
H1
-
H2 = products - reactants
Exercises
Use enthalpies changes of formation values in your IB data booklet table 11 and the table below to complete
the following calculations using the expression on page 2. No need to draw cycles and profile diagrams.
CH4 (g)
O3 (g)
CO2 (g)
H2O (l)
C6H6 (l
CaBr2 (s)
Br- (g)
Hf in kJ mol-1 for
-74.8
NaHCO3 (s)
+143
Na2CO3 (s)
-394
N2O (g)
-286
SO2 (g)
+ 49
H2O (g)
-682.8
Ca2+ (g)
-233.9
CuO (s)
-948
-1131
+81.6
-296.9
- 242
+1925.9
-157.3
1. Using standards heats or enthalpies of formation, calculate H in kJ for the reaction below
3CH4 (g) + 4O3 (g) 
3CO2 (g)
+ 6H2O (l)
2. The combustion of benzene may be written as
2C6H6 (l) + 15O2 (g) 
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12CO2 (g)
+ 6H2O (l)
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Using Hf calculate the standard enthalpy change of combustion for benzene, kJ mol-1.
3. Using the enthalpies of formation and the equation below, calculate Hf of hydrazine, N2H4 (l).
N2O (g) + 3H2 (g)  N2H4 (l)
4. Given the following information
H = -317.0 kJ
+ H2O (l)
2CH3OH (l) + 3O2 (g)  2CO2 (g) + 4H2O (l)
H = -727.0 kJ mol-1
calculate the standard enthalpy change of formation of methanol, CH3OH(l).
5. Given the enthalpies of formation, what is the value of H in kJ for the reaction below?
CaBr2 (s)  Ca2+ (g) + 2Br- (g)
6. Given the thermochemical equation:
2Cu2O (s) + O2 (g) 
4CuO (s)
H = -292.0 kJ
calculate Hf in kJ mol-1 for Cu2O (s)
7. Nitroglycerine (Hf (C3H5 (NO3 )3 ) = – 364 kJ mol-1 ) decomposes violently when it is detonated
according to the equation:
2C3H5 (NO3 )3 (l)
 ½O2 (g) + 6CO2 (g) + 3N2 (g) + 5H2O (l)
What is the enthalpy change for the decomposition of 2 moles of nitroglycerine?
8. Use the standard enthalpies of formation, calculate the enthalpy for the reaction
2NaHCO3 (s)
 Na2CO3 (s) + CO2 (g) + H2O (l)
Enthalpies changes of formation also allow us to comment on the stability of a compound relative to its
constituent elements.
Example:
H2 (g)
+
½ O2 (g) 
H2O (l)
Hf  = - 285.8 kJ mol –1
Using the equation above, 1 mole of water molecules has an absolute enthalpy which is 285.8 kJ lower than
the absolute enthalpy of oxygen and hydrogen combined and is therefore more stable than these two
elements.
Using standard enthalpy of combustion to find enthalpy change of a reaction
Hc =
standard enthalpy change of combustion for 1 mole of the substance.
This is the amount of energy released when 1 mole of a substance is combusted in sufficient supply of oxygen
in standard conditions.
This method only works if all substances in the reaction can be combusted.
The equation below is the equation describing the standard enthalpy of formation of water but it is also the
standard enthalpy of combustion of hydrogen!!
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H2 (g)
+
½ O2 (g) 

Hc (H2(g)) = - 285.8 kJ mol –1
H2O (l)
The same relationship applies to carbon and carbon dioxide i.e. standard enthalpy of formation of carbon
dioxide is equal to the standard enthalpy of combustion of carbon.
General enthalpy cycle
Hreaction
reactants
products
H2 = enthalpies combustion reactants
H1 = enthalpies of combustion
products
combustion products of both reactants and products
Hreaction =
a H2 (reactants) - b H1 (products)
Worked example: Calculate the enthalpy change for hydrogenation of propene using enthalpy changes of
combustion only. You will find enthalpy of combustions in your data booklet and propene = -2058 kJ mol-1. This
only works if all species are flammable.
C3H6 (g) + H 2 (g)  C3 H8 (g)
We can also represent the above calculations using an enthalpy cycle
Hreaction
C3H6 (g) + H 2 (g)
C3 H8 (g)
H2
Enthalpies combustion C and H
+5
O2(g)
H1 enthalpy combustion product
+ 5 O2 (g)
3CO2 (g) +
Hhydrogenation = Hc (reactants) - Hc (product) =
4H2O (l)
(-2058 + -286 kJ) - (- 2219 kJ )
= - 125 kJ
Exercise
Calculate the enthalpy of hydrogenation of ethane first using enthalpies of formation and then using enthalpy
of combustion. In each case draw enthalpy cycles and add equations.
Using enthalpy changes of combustion and enthalpy changes of formation in Hess Law
Worked example 1

Construct a simple enthalpy cycle and calculate the value of Hf (C2 H5OH(l)) given the following data
compound
H2O (l)
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Hf in kJ mol-1
-286
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Hcombustion in kJ mol-1
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CO2 (g)
C2 H5OH (l)
-394
-1371
Answer: The equation for the enthalpy of formation for ethanol is:
2C (s) + 4H 2 (g) + ½ O2 (g)  C2 H5OH (l)
Hf
2C (s) + 4H 2 (g) + ½ O2
C2 H5OH (l)
(g)
Hf (CO2 and H20)
Hcombustion
2CO2 (g) +
Hf (C2H5OH ) = Hf (products) - Hcombustion=
4H2O (l)
- 1645 kJ - (- 1371 kJ )
= - 274 kJ
15. 2. Born-Haber cycle (FOR IONIC SUBSTANCES ONLY!)
15.2.1 Define and apply the terms lattice enthalpy and electron affinity.
15. 2.2. Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.
15.2.3 Construct a Born–Haber cycle for group 1 and 2 oxides and chlorides, and use it to calculate an
enthalpy change.
15.2.4 Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds
in terms of their covalent character.
Lattice enthalpy is the ionic equivalent to bond enthalpies. Bond enthalpies in molecular substances indicate
the strength of covalent bonds. Lattice enthalpy is an indicator of the strength of an ionic bond within an ionic
lattice.
Lattice enthalpy = is the enthalpy change when one mole of a solid ionic crystal is separated into each of its
component ions in the gaseous state (this is to indicate that the ions should be completely separated and free
to move independently), at standard temperature and pressure. This an endothermic process and therefore it
has a positive sign. A negative lattice enthalpy value refers to the formation of an ionic lattice which is
shown by reversing the equation shown below.
separation:
NaCl (s)
formation:
Na+ (g) +

Na+ (g) +
Cl- (g)
Cl- (g)  NaCl (s)
Hlattice  = + 776 kJ mol -1
Hlattice  = - 776 kJ mol -1
The lattice energy is an also indicator of the stability of an ionic compound: the larger the value the more
stable the compound, the more tightly held are the ions within the lattice.
Calculation of lattice energy using Born-Haber cycles
Lattice enthalpies cannot be measured directly but can be calculated using Born-Haber cycles which are
thermodynamic cycles based on Hess’s Law.
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Born-Haber cycles do this by relating lattice energy to ionisation energies, electron affinities and other atomic
and molecular properties.
Each Born-Haber cycle involves, starting from the elements in their standard state, all the changes which
occur to form an ionic compound from the elements:




atomisation of both metal (from solid to gas = sublimation) and non-metal
ionisation of metal
electron affinity of non-metal
lattice energy
Electron affinity
 the first electron affinity of an element is the enthalpy change that occurs when one electron is gained by
each atom in a mole of gaseous atoms of the element to give one mole of ions, each with a single negative
charge (at standard temperature and pressure); most common first enthalpies are negative.
 the second electron affinity is the enthalpy change when a second electron is gained by one mole of
gaseous ions with one negative charge to give one mole of gaseous ions with a double negative charge;
most second electron enthalpies are positive as it takes energy to add an electron to an already negative
ion; however, in most cases that is compensated by the gain in lattice energy as a result of the stronger
attraction caused by an ion with a negative charge of –2 as opposed to –1.
Standard enthalpy of atomisation of an element
Standard enthalpy of atomisation of an element is the enthalpy increase that takes place when one mole of
gaseous atoms is made from the element in its standard physical state; in the case of hydrogen this is equal to
half the value of the H-H bond enthalpy so the same value is known under two different terms. In the case of
solid elements it is the same value as the heat of sublimation.
Examples:
 carbon: enthalpy of atomisation = Hsublimation (carbon) =
 Na (s)
 Na (g) = + 108.4 kJ mol-1
C (s)  C (g) = 715 kJ mol-1
Factors affecting lattice enthalpy
Study the chart below.
larger anions
larger
cations
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larger anions
F-
Cl-
Br-
I-
O2-
S2-
Li+
+1031
+848
+803
+759
Be2+
+4443
+3832
Na+
+918
+780
+742
+705
Mg2+
+3791
+3299
K+
+817
+711
+679
+651
Ca2+
+3401
+3013
Rb+
+783
+685
+656
+628
Sr2+
+3223
+2843
Cs+
+747
+661
+635
+613
Ba2+
+3054
+2725
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larger
cations
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The lattice enthalpy depends on:


size of ion or inter-ionic distance: the smaller the ion or the inter-ionic distance, the larger the attraction
charge of the ion: the larger the charge, the greater the attraction
Uses of energy cycles: to test ionic and bonding models
See table 13 in your IB data booklet which compares experimental values (obtained using Born-Haber) and
theoretical values. If the experimental value is greater than the theoretical value calculated using electrostatic
principles, this means that the actual bonding is stronger than what the electrostatic model suggest i.e.
bonding has a stronger covalent character as opposed to an ionic one.
Conclusions from the data in the table:
 Na-compounds: theoretical (predicted from physical calculations) and experimental (from Born-Haber
which uses experimental values) enthalpies show little discrepancy which means that the ionic model (ions
are discrete spherical ions) is correct.
 Ag-compounds: large discrepancy: actual lattice energy are higher than predicted by ionic model. Silver is a
transition metal in which the 3d electrons are less effective at shielding the attractive effect of the nucleus
which can attract the electron cloud from the anion. This polarises the anion and making it return partly the
electron that silver had donated to it. This gives the bond a more covalent character which at the same
time makes it stronger than an ionic bond.
15. 3. ENTROPY
15. 3. 1 State and explain the factors that increase the entropy in a system.
15. 3. 2 Predict whether the entropy change (ΔS) for a given reaction or process is positive or negative.
15. 3. 3 Calculate the standard entropy change for a reaction ( S°) using standard entropy values (S°) .
Entropy
Energetics or Thermodynamics allows us to determine the feasibility or spontaneity of a reaction; it establishes
whether a reaction is possible in principle (kinetics establishes whether it is possible in practice). To determine
the spontaneity we need to find out what is the driving force of spontaneous changes.
Reactions which are highly exothermic are generally spontaneous but lowering of the enthalpy is not always
the driving force e.g. evaporation and the dissolving of ammonium nitrate both of which occur spontaneously
although they are endothermic. There are other factors which also affect spontaneity and they are included in
the concept of entropy which is =
 a measure of the randomness or disorder of a system; the less regularly arranged, the greater the entropy
or disorder
 a measure of the random dispersal of energy of a system.
Entropy increase
We are interested in the entropy change, S, of a reaction system.
an increase in entropy: S = +
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a decrease in entropy: S = -
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A proper workable definition of entropy is that it is a measure of the different ways in which molecules can be
arranged and their quanta of energy be distributed; as such the entropy of any system can be increased in the
following ways:
 number of molecules/moles increased i.e. when there are more molecules/moles on the product side than
on the reactant side of a reaction (e.g. decomposition of a metal carbonate);
 the production of a greater number of gaseous particles (greatest increase in entropy);
 when molecules are more spread out (greater distance between them) e.g. formation of a gas, increase in
volume of a gas, evaporation, melting (more disorder over a greater area);
 increase in complexity of molecule:
 larger number of atoms/electrons; more ways of distributing energy/more energy levels;
 non-linear molecules instead of linear which fit together nicely, more orderly
 increase in temperature (more units of energy or quanta available for distribution amongst the particles)
also resulting in greater movement of particles;
Simple rules for recognising entropy increases:






change in state: entropy always increases when a substance changes from solid to liquid to gas;
greatest increase = increase in number of particles in the gaseous state;
when a pure solid/liquid dissolves in a solvent;
when heating a substance; increased movement of particles;
when the number of moles is greater on the product side;
when a gas is produced during a reaction.
Predicting the sign of entropy S
Predict whether the entropy change is positive, negative or close to zero for each of the following reactions,
give reasons for your answer:
(a) S (g) +
O2 (g)  SO2 (g)
(b) O (g) +
O (g)  O2 (g)
(c) NH4NO3 (s)  2H2O(g) + N2O (g)
(d) Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
(e) 2H2O2 (l)  2H2O (l) + O2 (g)
(f) Ca2+ (aq) + CO32- (aq)  CaCO3 (s)
(g) 2LiOH (aq) + CO2 (g)  Li2CO3 (aq) + H2O (l)
(h) N2(g)

(i)
N2(g) +
(j)
HCl (g) +
2N (g)
O2 (g) 
NH3 (g)  NH4Cl (s)
O2 (g)  2SO3 (g)
(k) 2SO2 (g) +
(l) 2HBr (g) +
F2 (g)  2HF (g) + Br2 (g)
3O2 (g)  2CO2 (g) + 4 H2O (g)
(m) 2CH3OH (l) +
(n) 4FeO (s) +
2NO (g)
O2 (g)  2Fe2O3 (s)
Standard entropy
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The standard entropy of a system is the entropy of the system at 1 atm and 298 K which is the temperature
and pressure at which most reactions occur and is therefore measured in J K-1 mol –l . As entropy is a much
smaller quantity than enthalpy it is measured in joules and not kilojoules which means either enthalpy or
entropy will need to be converted when both quantities are involved in the same calculations.
Unlike enthalpy, absolute entropy can be measured and is expressed on a positive scale.
The scale starts with zero enthalpy which is the entropy of a crystal at O K; it has been given a value of 0 J K1
mol-1 because it has the most perfect order (this is also known as the 3rd Law of Thermodynamics).
As this is now an absolute value – it is the most orderly system and therefore has the lowest entropy possible
– standard entropies of any other substance, elements and compounds, are therefore always positive.
Examples of some values of standard entropies: these are absolute values as entropy can be measured.
Substance
H2 (g)
N2 (g)
O2 (g)
Cl2 (g)
I2 (s)
H2O(l)
H2O(g)
NH3 (g)
SO2 (g)
SO3 (g)
CO (g)
Br2(g)
F2 (g)
S J K-1 mol-1
+ 130.6
+ 191.4
+ 204.9
+ 223.0
+ 116.1
+ 70.0
+ 188.7
+ 193
+ 248
+ 257
+ 198
+ 245.3
+202.7
S J K-1 mol-1
+ 213.6
+ 210.5
+ 172.8
+ 186.2
+ 219.5
+ 160.7
+ 201.8
+ 187
+ 199
+ 127
+ 173.51
+94.6
+240.45
substance
CO2 (g)
NO (g)
C6H6 (l)
CH4 (g)
C2H4 (g)
C2H5OH(l)
CHCl3 (l)
HCl (g)
HBr (g)
CH3OH (l)
HF (g)
NH4Cl (s)
NO2 (g)
substance
C (s) (graphite)
S (s)
Na (s)
Zn (s)
Cu (s)
CaCO3 (s)
CuSO4.H2O (s)
CaO (s)
FeO (s)
Fe2O3 (s)
S J K-1 mol-1
+ 5.7
+ 31.9
+ 51.0
+ 41.4
+ 33.3
+ 92.9
+ 305.4
+ 39.8
+ 61
+ 90
Calculating the standard entropy change for a reaction
The entropy change can be calculated using the following equation:
Ssystem = a S (products)
-
b S (reactants)
You need to multiply by the coefficients because the greater the number of particles the greater the entropy.
Worked examples:
Using absolute entropy values given, calculate the standard entropy change for the following reactions at
25C. The product is in J K-1 but could also be left in J K-1 mol-1for 1 mole of reactant as written in the
equation.
(a) CaCO3 (s) 
CaO (s)
+
CO2 (g)
Using the above expression:
Ssystem = [S (CaO) + S (CO2 )] – [S (CaCO3)]
= [ (1 mol) (39.8 J K-1 mol-1) + (1 mol)(213.6 J K-1 mol-1)] - [(1 mol)(92.9 J K-1 mol-1)]
= 160.5 J K-1
(increase in entropy for this reaction)
(b) N2(g) +
3H2 (g) 
2NH3 (g)
(unit of result left in J K-1 mol-1)
Ssystem = [2S (NH3)] – [ S (N2) + 3S (H2)]
= [( 2 x (193 J K-1 mol-1) - (192 J K-1 mol-1)] + 2 x (131 J K-1 mol-1)
= -199 J K-1
(decrease in entropy)
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(c) Cl2(g) +
H2 (g) 
2HCl (g)
Ssystem = [2S (HCl)] – [ S (Cl2) + S (H2)]
= [ (2 mol) (187 J K-1 mol-1)] - [(1 mol)(223 J K-1 mol-1) + [(1 mol)(131 J K-1 mol-1)]
= 20 J K-1
(increase in entropy)
Exercises
For the following reactions, first predict whether the entropy of the system will increase (= + sign), decrease
(= - sign) or be close to zero and then calculate the standard entropy change:
(a) N2(g) +
O2 (g) 
2NO (g)
(b) HCl (g) +
NH3 (g)  NH4Cl (s)
(c) 2SO2 (g) +
O2 (g)  2SO3 (g)
(d) 2CO (g) +
O2 (g)  2CO2 (g)
(e) 2HBr (g) +
F2 (g)  2HF (g) + Br2 (g)
(f) 2NO (g) +
O2 (g)  2NO2 (g)
(g) 2CH3OH (l) +
(h) 4FeO (s) +
(i) CH4 (g) +
3O2 (g)  2CO2 (g) + 4 H2O (g)
O2 (g)  2Fe2O3 (s)
2O2 (g)  CO2 (g) + 2H2O (g)
15. 4. SPONTANEITY OF A REACTION
15.4.1 Predict whether a reaction or process will be spontaneous by using the sign of G°.
15.4.2 Calculate G° for a reaction using the equation G° = H° − T S° and by using values of the standard free
energy change of formation, ΔGf°
15.4.3 Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and
enthalpy changes and the equation G°= H°− T S°.
According to the second law of thermodynamics, when a spontaneous process occurs, there must be an
increase in the entropy of the universe.
The entropy of the universe can be divided into the entropy of the system and the entropy of the surroundings.
For a reaction to be spontaneous, the entropy change of the system and the entropy change of the
surroundings that it caused by the reaction in the system when added together must have a positive value.
S universe
=
S surroundings
+
S system
During a reaction, the entropy change of the system causes a change in entropy of the surroundings in either
of the following two ways:
1.
In an exothermic reaction heat energy flows from the system to the surroundings; this causes a
decrease in the entropy of the system (S system = negative ) but causes the particles of the
surroundings to move faster and so they become more disordered. This entropy change of the
surroundings, S surroundings , will be positive.
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2.
In an endothermic reaction, the surroundings lose heat to the system. This has the opposite effect: the
entropy of the surroundings will decrease as the particles become more ordered and S surroundings will
be negative. The entropy of the system however increases so S system = positive.
The spontaneity of a reaction depends on how these entropy changes interact as shown by the table below.
Predicting whether a reaction will be spontaneous or not using, S uni , S surr and S sys
S sys
+
+
-
if
if
if
if
S sur
spontaneous?
S uni
+
+
S universe = S surroundings+ S system
Yes
+
+/+/-
No
Yes if S sys  S surr
Yes if S system   S surr
If we can calculate S universe we can have an idea of the spontaneity of a reaction. The problem is we can not
measure S surroundings.
But we know S surroundings depends on the enthalpy change of the system and the temperature of the
surroundings at the time of the reaction which are two quantities we can measure.
S surr
- Hsystem
T
=
We can now substitute S surroundings in the expression below.
S universe
=
-H system/T
+
S system
Rearranging the above expression produces a new expression and quantity called free energy. In our
calculations we are interested in the change in free energy in a reaction. Change in free energy is expressed
in kJ mol-1.
G = Hsystem - TSsystem
For a reaction to be spontaneous its free energy change must have a negative value: G  0.
We can use the above expression to predict the spontaneity of a reaction as shown in the table below
H (KJmol-1)
if
positive
S universe
(JK-1mol-1)
positive
if
positive
negative
depends on
temperature
positive
if
negative
positive
negative
if
negative
negative
depends on
temperature
Topic15
G (KJmol-1)
Spontaneity
Spontaneous but only at high temperatures when
T S system is greater than H
Reaction will not occur (reaction will occur in the
opposite direction)
Spontaneous at all temperatures
Spontaneous but only at low temperatures when
TS system is smaller than H
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Calculations of G to find out if a reaction is spontaneous or not
There are 3 different ways in which we cam calculate the G of a reaction.
Using standard free energies of formation

Standard free energy of formation is the free energy change when 1 mole of the compound is formed from
its elements in their standard states;

Standard free energy of formation of elements in their standard states is
spontaneously as they already exist in that state.
Greaction = a Gf (products)
-
zero as they are not formed
b Gf (reactants)
Worked examples: Calculate the standard free-energy changes for the following:
(a)
2O2 (g)  CO2 (g) + 2H2O (l)
CH4 (g) +
Greaction = [Gf (CO2) + 2Gf (H2O)] - [Gf (CH4) + 2Gf (O2)]
= [(1 mol) (-394.4 kJ mol-1) + (2 mol) (-237.2 kJ mol-1)]
-
[(1 mol) (-50.8 kJ mol-1) +(2 mol) (0 kJ mol-1)]
= - 818.0 kJ this means reaction is very spontaneous!! (the answer could also be stated in kJ mol-1)
(b) 2MgO (s)
 O2 (g) + 2Mg (s)
(unit left in kJ mol-1)
Greaction = [2Gf (Mg) + Gf (O2)] - [2Gf (MgO)]
= [2 x (0 kJ mol-1) + (0 kJ mol-1)]
= 1139 kJ mol-1
-
[2 x (-569.6 kJ mol-1)]
which means reaction is not spontaneous at all.
Using Hess Law
We can also add up equations of which we know the free energy values to find the free energy value of
another equation; again the same rules apply.
Using G = Hsystem - TSsystem
The main problem in this type of calculation is that entropy and enthalpy are measured in different units; either
you change the entropy into kJ or do it the other way around.
Worked example:
Calculate G for the following reaction is:
3H2 (g) + N2 (g) 
2NH3(g) at 500 C
G = Hsystem - TSsystem
= - 92.38 kJ - (773 K) (0.198.3 kJ K-1) = - 92.38 kJ + 153.29 kJ = 60.91 kJ
This positive value means that at 500C the forward reaction is not spontaneous.
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Exercises
1. Calculate the free energy for the following reaction using the free energies of formation in the table below:
Gf (in kJ mol-1)
- 26.5
-95.27
-236.81
-592.1
-569.6
159.4
-105.98
substance
NH3 (g)
HCl (g)
H2O (l)
MgCl2(s)
MgO (s)
N2H4 (g)
H2O (g)
substance
N2H4 (g)
CH3OH (l)
CO2 (g)
NO (g)
SO2 (g)
SO3 (g)
(a) N2(g) +
3H2 (g) 
(b) H2(g) +
Cl2 (g) 
(c) H2O (l) +
MgCl2 (s)  MgO (s) + 2HCl (g)
Gf (in kJ mol-1)
159.4
-166.23
-394.4
86.71
-300.4
-370.4
2NH3 (g)
2HCl (g)
(d) 2NH3(g)  N2H4 (g) + H2 (g)
(e) 2CH3OH (l) +
(f)
N2(g) +
3O2 (g)  2CO2 (g) + 4 H2O (g)
O2 (g) 
(g) 2SO2 (g) +
2NO (g)
O2 (g)  2SO3 (g)
2. (a) Calculate the free energy of the following reactions when given H298 and S298.
(i) 2SO2 (g) + O2 (g)  2SO3 (g) at 400 K.
H298 = -196.6 kJ
(ii) H2O2 (g)  H2O (g) + ½O2 (g) at 298K
H298 = -106 kJ
(iii) N2(g) +
O2 (g) 
2NO (g) at 298K
H298 = +180 kJ mol-1
S = - 189.6 J K-1.
S298 = + 58 J K-1.
S298 = + 58 J K-1mol-
1
.
(iv) Ca2+ (aq) + CO32- (aq)  CaCO3 (s) at 298K
H298 = +13 kJ mol-1
S298 = -205 J K-1mol-1.
(b) For the reactions above that are not spontaneous
(i) Decide if you could make them spontaneous by manipulating the temperature.
(ii) If that is the case, calculate the temperature at which the reaction would become spontaneous.
3. From the information below, calculate the standard free energy change, G , for the reaction
Fe3 O4 (s) + CO (g)  Fe (s) + Fe2 O3 (s) + CO2 (g)
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Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)
G = - 29.4 kJ
3Fe2 O3 (s) + CO (g)  2 Fe3 O4 (s) + CO2 (g)
G = - 61.6 kJ
4. Repeat question 2 (b) for question 1.
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