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TOPIC 15: ENERGETICS
15. 1. Energy cycles
E. I.: The concept of the energy change in a single step reaction being equivalent to the summation
of smaller steps can be applied to changes involving ionic compounds.
Nature of science: Making quantitative measurements with replicates to ensure reliability—energy cycles
allow for the calculation of values that cannot be determined directly. (3.2)
Understandings
• 15.1 U1 Representative equations (e.g. M+(g)  M+(aq)) can be used for enthalpy/energy of hydration,
ionization, atomization, electron affinity, lattice, covalent bond and solution.
• 15.1 U2 Enthalpy of solution, hydration enthalpy and lattice enthalpy are related in an energy cycle.
Application and skills
• 15.1 AS1 Construction of Born-Haber cycles for group 1 and 2 oxides and chlorides.
• 15.1 AS1 Construction of energy cycles from hydration, lattice and solution enthalpy. For example
dissolution of solid NaOH or NH4Cl in water.
• 15.1 AS3 Calculation of enthalpy changes from Born-Haber or dissolution energy cycles.
• 15.1 AS4 Relate size and charge of ions to lattice and hydration enthalpies.
• 15.1 AS5 Perform lab experiments which could include single replacement reactions in aqueous solutions.
Using Born-Haber cycles to calculate lattice enthalpy
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 that 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.
Lattice enthalpy cannot be measured directly but can be calculated using a Born-Haber cycle which is an
energy cycle based on Hess’s Law and that relates lattice energy to ionization energy, electron affinity and
atomization. See table 18 in IB data booklet for lattice enthalpies.
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:




atomization of both metal (from solid to gas = sublimation) and non-metal
ionization of the metal
electron affinity of the non-metal
lattice energy
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Standard ionization energy:
First ionization energy is the amount of energy required to remove one mole of electrons from one mole of
gaseous atoms or gaseous ions in standard conditions.
Standard 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 in standard conditions; most common first enthalpies are negative. See table 8 in data booklet.
 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.
Standard enthalpy of atomization 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 of enthalpies of atomization:
 carbon: enthalpy of atomisation = Hsublimation (carbon) =
 Na(s)
 Na (g) = + 108.4 kJ mol-1
C (s)  C (g) = 715 kJ mol-1
Exercises: The Born-Haber cycle can be used to calculate for instance the lattice enthalpy of an ionic
compound as well as its enthalpy of formation.
1. Draw a Born-Haber cycle and use it to calculate the lattice enthalpy for KCl (s)  K+(g) + Cl-(g).
In the table below name each enthalpy change.
Change
Enthalpy change
K(s) + ½ Cl2 (g)  KCl (s)
Value in kJ
mol-1
-562.6
K(s)  K (g)
+89.6
K(g)  K+ (g)
+419.0
½ Cl2 (g)  Cl (g)
+122.0
Cl (g)  Cl- (g)
-349.0
2. Calculate the standard enthalpy of formation of NaCl (s). Name each enthalpy change.
Change
Enthalpy change
NaCl (s)  Na+(g) + Cl-(g)
Value in kJ
mol-1
+790.0
Na(s)  Na (g)
+108.4
Na(g)  Na+ (g)
+495.8
½ Cl2 (g)  Cl (g)
+122.0
Cl (g)  Cl- (g)
-349.0
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3. Draw a Born-Haber cycle and use it to calculate the lattice enthalpy for MgO (s). Write an equation to
represent the lattice enthalpy for magnesium oxide. Name each enthalpy change.
Change
Mg (s) + ½ O2 (g)  MgO (s)
Enthalpy change
Value in kJ mol-1
-552
Mg (s)  Mg (g)
+148
Mg (g)  Mg+ (g)
+738
Mg+(g)  Mg2+ (g)
+1451
½ O2 (g)  O (g)
+249
O (g)  O- (g)
-141
O- (g)  O2- (g)
+798
4. Calculate the standard enthalpy of atomization of bromine using the standard enthalpy values below.
Change
K(s) + ½ Br2 (g)  KBr (s)
Value in kJ mol-1
-392
K(s)  K (g)
+89
K(g)  K+ (g)
+419
Br(g)  Br- (g)
-325
K+ (g) + Br -(g)  KBr (s)
-672
5. Draw a Born-Haber cycle and calculate the enthalpy of formation for CaO using the values below.
Change
Enthalpy change
Ca2+ (g) + O2- (g)  CaO (s)
Value in
kJ mol-1
-3401
Ca (s)  Ca (g)
+177
Ca (g)  Ca+ (g)
+590
Ca+ (g)  Ca2+ (g)
+1100
½ O2 (g)  O (g)
+249
O (g)  O- (g)
-141
O- (g)  O2- (g)
+790
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Factors affecting magnitude of lattice enthalpy: ionic charge and ionic radius
Study the chart below or you could also use table 18 in your IB data booklet.
larger anions
larger
cations
F-
Cl-
Br-
I-
Li+
+1031
+848
+803
+759
Na+
+918
+780
+742
K+
+817
+711
Rb+
+783
Cs+
+747
larger anions
O2-
S2-
Be2+
+4443
+3832
+705
Mg2+
+3791
+3299
+679
+651
Ca2+
+3401
+3013
+685
+656
+628
Sr2+
+3223
+2843
+661
+635
+613
Ba2+
+3054
+2725
larger
cations
The lattice enthalpy depends on:

size of ion: the smaller the ion , the larger the electrostatic attraction and the more energy is needed to
sublime the ions.
 charge of the ion: the larger the charge, the greater the electrostatic attraction.
The smaller the ionic radius and the greater the charge of the ion the greater the charge density, the greater
the electrostatic attraction, the more energy is needed.
Using energy cycles to calculate enthalpy of solution
The enthalpy of solution is the enthalpy change when one mole of a solute dissolves in a solute to make an
infinitely dilute solution under standard conditions. As the enthalpy of solution depends on the concentration of
the solution, an infinitely dilute solution is considered to ensure all ions in the ionic compound have broken
away from the lattice (sublimed) and are hydrated by the maximum number of water molecules. Table 19 has
enthalpy of solution values for a number of aqueous solutions.
The enthalpy of solution can have a positive or negative value as it is the sum of the enthalpy values of two
connected processes:


The ions need to break away from the lattice; each ion needs to sublime. This is an endothermic process
and its value of course is known as the lattice enthalpy.
Once the ions have separated from the ionic lattice each ion is hydrated by water molecules. This is an
exothermic process as the water molecules interact with the ions as there is an electrostatic attraction
between the polar water molecules and the positive and negative ions. Table 20 in the data booklet has
enthalpies of hydration.
The enthalpy of hydration of an ion is the enthalpy change when one mole of gaseous ions is dissolved to form
an infinitely dilute solution of one mole of aqueous solutions under standard conditions. As hydration is an
exothermic process all values are negative.
Equations defining the enthalpy of hydration of a sodium ion and a chloride ion:
Na+ (g)  Na+ (aq) Hθhydr = - 424 kJ mol-1
and
Cl- (g)  Cl- (aq) Hθhydr = - 359 kJ mol-1
Using the values in table 20 in the data booklet it can easily be observed that the enthalpies of hydration
become less negative or less exothermic as you go down the groups. This is because as you go down a group
ionic radii increase lowering the charge density of the ions and lowering the strength of the electrostatic forces
between the polar water molecule and the ion and as a result lower (less negative) the enthalpy of hydration.
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The greater the charge of the ion the greater the charge density and therefore the electrostatic attraction and
as a result a more negative/exothermic enthalpy of hydration.
The enthalpy of solution can be calculated using an energy cycle from which a formula can be derived. The
energy cycle relates the lattice enthalpy and enthalpies of hydration to the enthalpy of solution as shown
below:
Na+ (g) + Cl- (g)
Enthalpy
Hθlattice (NaCl)
Hθhydr (Na+) + Hθhydr (Cl-)
= - 424 + (-359) = -783 kJ mol-1
+790 kJ mol-1
Na+ (aq) + Cl- (aq)
Hθsolution (NaCl) = + 7 kJ mol-1
NaCl(s)
Hθsolution (NaCl) = Hθlattice (NaCl) + Hθhydr (Na+) + Hθhydr (Cl-) = +790 -424 – 359 = +7
Exercises
1. Using the appropriate data from the IB data booklet calculate to calculate the enthalpy of solution of
calcium chloride.
2. Using the data in the IB data booklet to calculate the lattice enthalpy of potassium bromide.
3. Using the data below calculate the enthalpy of solution of sodium hydroxide.
NaOH (s)  Na+ (g) + OH- (g)
Na+ (g)  Na+ (aq)
OH- (g)  OH- (aq)
Hθlattice enthalpy= +900 kJ mol-1
Hθhydr = - 424 kJ mol-1
Hθhydr = - 519 kJ mol-1
4. Use the data in the table below to calculate the enthalpy of solution of ammonium
nitrate and ammonium chloride. Draw the Born-Haber cycles for both of the calculations.
Lattice energy of NH4Cl (s)
Enthalpy change / kJ mol
+705
Lattice energy of NH4NO3 (s)
+646
Enthalpy of hydration NH4+ (g)
–307
-
–381
Enthalpy of hydration Cl (g)
Enthalpy of hydration NO3- (g)
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15. 2. ENTROPY
E.I.: A reaction is spontaneous if the overall transformation leads to an increase in total entropy
(system plus surroundings). The direction of spontaneous change always increases the total entropy
of the universe at the expense of energy available to do useful work. This is known as the second law
of thermodynamics.
Nature of science: Theories can be superseded—the idea of entropy has evolved through the years as a
result of developments in statistics and probability. (2.2)
Understandings
 15.2 U1 Entropy (S) refers to the distribution of available energy among the particles. The more ways the
energy can be distributed the higher the entropy.
 15.2 U2 Gibbs free energy (G) relates the energy that can be obtained from a chemical reaction to the
change in enthalpy (ΔH), change in entropy (ΔS), and absolute temperature (T).
 15.1 U3 Entropy of gas>liquid>solid under same conditions.
Application and skills
 15.2 AS1 Prediction of whether a change will result in an increase or decrease in entropy by considering
the states of the reactants and products.
 15.2 AS2 Calculation of entropy changes (ΔS) from given standard entropy values (Sº).
 15.2 AS3 Application of Δ�G�° = Δ�H�° − ��TΔS��° in predicting spontaneity and calculation of
various conditions of enthalpy and temperature that will affect this.
 15.2 AS4 Relation of ΔG to position of equilibrium.
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.
Entropy is a measure of the different ways in which energy can be distributed in a system. The more ways the
energy can be distributed the higher the entropy.
It is sometimes also referred to as a measure of the randomness or disorder of a system; the less regularly
arranged, the greater the entropy or disorder.
Entropy increase
We are interested in the entropy change, S, of a reaction system.
an increase in entropy: S = +
a decrease in entropy: S = -
The entropy of any system is increased as a result of the following changes:
 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);
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 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;
All of the above increase the different ways in which energy can be distributed as either the number of
particles in the system increases or the amount of energy in the system increases.
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 energy of system;
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
giving a reason 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
The standard entropy of a system is the entropy of the system at 1 atm and 298 K is 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 K-1
mol-1 because it has the most perfect order (this is also known as the 3rd Law of Thermodynamics).
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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)
Sθ 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
Sθ 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)
Sθ 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:
Sθ system = a Sθ (products)
-
b Sθ (reactants)
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-1 for 1 mole of reactant as written in the
equation.
(a) CaCO3 (s) 
CaO (s)
+
CO2 (g)
Using the above expression:
Sθsystem = [Sθ (CaO) + Sθ (CO2 )] – [Sθ (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)
Sθsystem = [2Sθ (NH3)] – [ Sθ (N2) + 3Sθ (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)
(c) Cl2(g) +
H2 (g) 
2HCl (g)
Sθsystem = [2Sθ (HCl)] – [ Sθ (Cl2) + Sθ (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)
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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:
O2 (g) 
(a) N2(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)
3O2 (g)  2CO2 (g) + 4 H2O (g)
(g) 2CH3OH (l) +
O2 (g)  2Fe2O3 (s)
(h) 4FeO (s) +
(i) CH4 (g) +
2O2 (g)  CO2 (g) + 2H2O (g)
Spontaneity of a reaction
According to the Second Law of Thermodynamics, when a spontaneous process occurs, there must be an
increase in the total 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.
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
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S sur
+
-
spontaneous?
S uni
+
+/7 hours
S universe = S surroundings+ S system
Yes
No
Yes if S sys  S surr
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-
if
+
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 that S surroundings depends on both 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
We can now substitute S surroundings
- Hsystem
T
in the expression below.
S universe
=
=
-H system/T
+
S system
Rearranging the above expression produces a new expression and quantity, G, called Gibbs free energy
which is the energy available to do useful work.
At constant temperature and pressure, a chemical reaction is spontaneous if there is a decrease in the total
Gibbs free energy of the system i.e. Gibbs free energy change must have a negative value:
Gθ  0.
Gθ reaction = H θ system - TS θ system
Gibbs free energy relates the free energy that can be obtained from a chemical reaction to the changes in
enthalpy, and entropy of the system during the reaction and the absolute temperature in Kelvin at which the
reaction takes place. Gibbs free energy is measured in kJ mol-1.
It is important to appreciate that in the above expression the temperature must be the absolute temperature
and be expressed in Kelvin.
We can use the above expression to predict the spontaneity of a reaction as shown in the table below
if
Hsystem
(KJmol-1)
positive
S universe
(JK-1mol-1)
positive
G (KJmol-1)
if
positive
negative
depends on
temperature
positive
if
negative
positive
negative
if
negative
negative
depends on
temperature
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
Gibbs free energy enables to predict the spontaneity of a reaction using the enthalpy and entropy change of
the reaction as well as the temperature at which the reaction occurs. It describes the temperature dependence
of most reactions.
Calculations of Gθ to find out if a reaction is spontaneous or not
There are 3 different ways in which we can calculate the Gθ of a reaction.
Using standard Gibbs free energies of formation

Standard Gibbs free energy of formation is the free energy change when 1 mole of the compound is
formed from its elements in their standard states;
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
Standard Gibbs free energy of formation of elements in their standard states is
formed spontaneously as they already exist in that state.
Gθreaction = a Gθf (products)
-
zero as they are not
b Gθf (reactants)
Worked examples: Calculate the standard free-energy changes for the following:
(a)
2O2 (g)  CO2 (g) + 2H2O (l)
CH4 (g) +
Gθreaction = [Gθf (CO2) + 2Gθf (H2O)] - [Gθf (CH4) + 2Gθf (O2)]
= [(1 mol) (-394.4 kJ mol-1) + (2 mol) (-237.2 kJ mol-1)] - [(1 mol) (-50.8kJ mol-1) +(2 mol) (0kJ mol1
)]
= - 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)
Gθreaction = [2Gθf (Mg) + Gθf (O2)] - [2Gθf (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 Gθ = Hθsystem - TSθsystem
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 Gθ for the following reaction is:
3H2 (g) + N2 (g) 
2NH3(g) at 500 C
Gθ = Hθsystem - TSθsystem
= - 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.
Exercises
1. Calculate the free energy for the following reaction using the free energies of formation in the table below:
substance
NH3 (g)
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Gθf (in kJ mol-1)
- 26.5
substance
N2H4 (g)
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Gθf (in kJ mol-1)
159.4
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HCl (g)
H2O (l)
MgCl2(s)
MgO (s)
N2H4 (g)
H2O (g)
-95.27
-236.81
-592.1
-569.6
159.4
-105.98
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)
-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 Hθ298 and Sθ298.
(i) 2SO2 (g) + O2 (g)  2SO3 (g) at 400 K.
Hθ298 = -196.6 kJ
(ii) H2O2 (g)  H2O (g) + ½O2 (g) at 298K
Hθ298 = -106 kJ
(iii) N2(g) +
O2 (g) 
Sθ = - 189.6 J K-1.
Sθ298 = + 58 J K-1.
Hθ298 = +180 kJ mol-1 Sθ298 = + 58 J K-1mol-
2NO (g) at 298K
1
.
(iv) Ca2+ (aq) + CO32- (aq)  CaCO3 (s) at 298K
Hθ298 = +13 kJ mol-1
Sθ298 = -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, Gθ, for the reaction
Fe3 O4 (s) + CO (g)  Fe (s) + Fe2 O3 (s) + CO2 (g)
Fe2 O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)
Gθ = - 29.4 kJ
3Fe2 O3 (s) + CO (g)  2 Fe3 O4 (s) + CO2 (g)
Gθ = - 61.6 kJ
Relating G to equilibrium
Gibbs free energy change allows us to have a better idea of the extent to which a reaction goes to completion
as opposed to reaching an equilibrium between reactants and products.
The table below shows the link between Gθ and the position of the equilibrium. More on this in topic 17.
Gθ reaction
Gθ reaction > + 30 kJ mol-1
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Position of equilibrium
No reaction possible – position of equilibrium towards reactants.
7 hours
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0 kJ mol < G reaction < + 30 kJ mol
Gθ reaction = 0 kJ mol-1
0 kJ mol-1 > Gθ reaction >- 30 kJ mol-1
Gθ reaction < - 30 kJ mol-1
-1
Topic15
θ
-1
Kc  1
Partial reaction producing equilibrium mixture. Kc  1
Partial reaction producing equilibrium mixture. K = 1
Partial reaction producing equilibrium mixture. Kc 1
Complete reaction. Kc  1
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