TOPIC 5
ENERGETICS
5.1 Exothermic and Endothermic Reactions (SL/HL)
 There are often energy changes that take place during a chemical
reaction.
 These may be in the form of light, sound, but much more commonly
heat energy.
 Reactions that release heat energy are called exothermic reactions.
These cause a rise in temperature.
 When heat energy is taken in from the surroundings by the chemicals,
causing a temperature drop, this is called an endothermic reaction.
 The internal energy stored within the chemical is called the enthalpy
(H). This cannot be measured by itself, only H
 The absolute enthalpy stored is very difficult to measure. Usually, the
enthalpy change (H) of the reactants compared to the products is
measured.
 If this is carried out in the lab at atmospheric pressure (101 kPa) and
298 K (250C) then this is called the standard enthalpy change of
reaction. (H0)
Enthalpy Level Diagrams
 In an exothermic reaction, the enthalpy of the reactants is
greater than that of the products.
 This leads to a release of energy so that H is negative and the
products are more stable than the reactants.
Exothermic Reaction
reactants
Enthalpy
products
reaction pathway
5.1.2
Combustion and Neutralisation are Exothermic reactions
CH4 +
NaOH +
2O2
CO2
HCl
NaCl
+ 2H2O
H = -ve
+
H = -ve
H2O
(Anotate your diagram with the reactants and products)
 In an endothermic reaction, the reactants have less enthalpy than
the products.
 There is an increase in energy taken from the surroundings so that H
is positive and the products are less stable than the reactants.
Endothermic Reaction
products
Enthalpy
reactants
Reaction pathway
5.2 Calculations of Enthalpy Changes (SL/HL)
 The enthalpy change during a chemical reaction depends on 3
factors. (H or q)
1. The temperature change.
2. The mass that changes temperature.
3. The specific heat capacity of the material that changes temperature.
H = m C T
H in kJ, m in Kg (in aqueous solutions it will be the mass of water
which is the same as it’s volume in dm3)
C is the specific heat capacity and in an aqueous solution this will be
4.18 kJ kg-1 K-1
 Usually we want the enthalpy change per mole H mol-1 of the
limiting reagent.
 In order to measure the enthalpy change in an aqueous solution a
simple calorimeter can be used.
 Errors due to heat loss to the surroundings can be minimized by
plotting the temperature rise against time and then extrapolating the
graph back to estimate the temperature rise for an instantaneous
reaction.
 Use a polystyrene cup and lid to prevent heat loss
TOK: What criteria do we use in judging whether discrepancies between
experimental and theoretical values are due to experimental limitations or
theoretical assumptions?
E.g. The temperature rise during the neutralization of 50 cm3 of 0.5 mol
dm-3 HCl with 50 cm3 of 0.5 mol dm-3 NaOH is 3.50C. Calculate the
enthalpy change.
HCl +
NaOH
NaCl
+
H2O
Moles of HCl = CV = 0.5 x 0.05 = 0.025
Moles of NaOH = “ “ “ “ = 0.025
Both are limiting reagents.
H = mCT
=
0.1 x 4.18 x 3.5
=
1.46 kJ
H mol-1 =
1.46 kJ for 0.025 moles of reactant.
=
1.46 / 0.025
=
-58.5 kJ mol-1
Since there is a temp. rise we must insert a negative sign to indicate this is
exothermic.
E.g. 2 If 50 cm3 of 0.2 mol dm-3 CuSO4 reacts with 1.2 g of Zn, then
there is a temperature rise of 10.40C. Calculate the enthalpy change.
Zn
+
CuSO4
ZnSO4
+
Cu
Moles of CuSO4 = CV= 0.2 x 0.05 = 0.01 moles
Moles of Zn = m/Mr = 1.2 / 65 = 0.0184 moles
CuSO4 is the limiting reagent.
H = mCT
= 0.05 x 4.18 x 10.4
= 2.17 kJ for 0.01 moles of reactants.
H mol-1 = 2.17 / 0.01
=
-217 kJmol-1
E.g. 3 When 0.535g of NH4Cl is dissolved in 25cm3 water, the
temperature drops by 3.70C. Calculate the enthalpy change per mole of
ammonium chloride for this endothermic process.
Answer = + 38.7 kJ mol-1
5.3 Hess’s Law (SL/HL)
The enthalpy for a reaction depends only on the difference between the
enthalpy of the products and the enthalpy of the reactants. It is
independent on the route by which the reaction may occur.
X
Z
Y
Route 1 = Route 2
The enthalpy change for a reaction is the sum of the individual
enthalpy changes for each step.
H1 = H2 + H3
 Hess’s law is particularly useful for determining the enthalpy
change for a reaction that is difficult to measure directly.
Note: Units for Hess law calculations are usually given as kJ rather than kJ mol-1. This is because
the enthalpy change for a specific reaction has been calculated, rather than per mole of limiting
reactant.
E.g. Calculate the enthalpy change in the following reaction:
C
+
½ O2
CO
Given the following information:
C
+
CO +
O2
H = -393.5 kJ
CO2
½ O2
H = -283 kJ
CO2
Construct enthalpy level diagram:
C
+
½ O2
CO
CO2
H1 = H2
-
H3
H1 =-393.5
-
(-283)
H2 = -110.5 kJ
E.g. 2 Calculate the enthalpy change in the following reaction:
C
+
2H2
CH4
Given the following information:
C
+
O2
CO2
H = -393 kJ
H2
+
½ O2
H2O
H = -286 kJ
CH4 +
2O2
CO2 +
2H2O
H = -890 kJ
Construct enthalpy cycle:
C
CO2
+
+
2H2
CH4
2H2O
Route 1 = Route 2
H1 = H2
+
H1 = (-393) +
2 xH3
(2 x -286) -
H1 = -75 kJ
-
H4
(-890)
3.
Calculate the enthalpy change of the following:
Fe +
Cl2
FeCl2
Given the following information.
4.
2 Fe +
3Cl2
2FeCl2
+
H = -800 kJ
2FeCl3
Cl2
H = -120 kJ
2FeCl3
Calculate the enthalpy change for the following:
NaHCO3 +
HCl
NaCl + CO2 +
H2O
Given the following:
2NaHCO3
Na2CO3 + CO2
Na2CO3 + 2HCl
H2O
H = -214 kJ
+ H2O
H = -175 kJ
+
2NaCl + CO2
5. Calculate the following enthalpy change.
2C
+
3H2 +
½ O2
C2H5OH
Given the following enthalpies of combustion. (Tip: React with
oxygen giving carbon dioxide and water).
Carbon = -393.5, Hydrogen = -285.8, Ethanol = -1371 kJmol-1
Answers = 3) –340 kJ
4) –194.5 kJ
5) –273.4 kJmol-1
5.4 Exothermic and endothermic reactions
 Energy is required to break bonds.
 Energy is released when bonds form.
 In an exothermic reaction, the amount of energy required to break the
bonds of the reactants is less then the amount of energy released when
the bonds form in the products.
Enthalpy
reactants
products
 This suggests that the bonds in the reactant are weaker than those in
the product ( and that the product is therefore more stable).
 In an endothermic reaction the reverse is true. The reactants have
stronger bonds than the products.
Enthalpy
products
reactants
5.4 Bond Enthalpies (SL/HL)
 This is the enthalpy change to break a bond in the gaseous state.
E.g. CH4(g)
CH3(g) + H(g)
 The reverse of this will give the enthalpy change when this bond forms.
 Since the enthalpy change on breaking this C-H bond will not be the
same for all C-H bonds (due to different chemical environments in
different compounds), the bond enthalpy is given as an average.
 Average bond enthalpies are given in the data booklet.
E.g. Use bond enthalpies to calculate the enthalpy change in the
following:
CH4 +
Cl2
CH3Cl
+
HCl
Bonds breaking = (1 x C-H) + (1 x Cl-Cl)
= 412
+
242
-1
= + 654 kJmol
Bonds forming = (1 x C-Cl) + (1 x H-Cl)
= 338
+
341
-1
= 679 kJmol
Overall = ENERGY INPUT – OUTPUT
= +654
(+679)
= -25 kJmol-1
E.g. 2 Calculate the enthalpy changes in the following:
a)
CH4 +
b)
2O2
CH2=CH2 +
CO2
H2
+
2H2O
CH3CH3
c)
NH2-NH2
Answers = a) - 698
+
b)-128,
O2
N2 +
2H2O
c)-585
 There are 2 major errors in the answers given when using bond
enthalpies.
1. They are averages and so the actual strength of the bond in the
molecule may be different to the one used.
2. The values are in the gaseous state, so that with if the reactants were
liquids extra energy would be required to vaporize them.
15.1 Standard Enthalpy Change of Combustion (HL ONLY)
 The standard enthalpy change of combustion is the enthalpy change
when 1 mole of compound is burnt completely in excess oxygen
under standard conditions.
H2(g) +
½ O2(g)
H2O(l)
H = -285.8 kJmol-1
15.1 Standard Enthalpy Change of Formation
 The enthalpy of formation of a compound is
1.
2.
3.
The enthalpy change when 1 mole of a compound
is formed
from its elements in their standard states.
under standard conditions of temperature and
pressure.
 The enthalpy of formation of an element in its
standard (normal) state is zero.
E.g. The enthalpy of formation for sodium chloride can be
represented by the following equation.
Na(s) +
½ Cl2(g)
NaCl(s)
E.g. 2 Write down equations to show the enthalpy of formation of a)
Magnesium bromide b) Aluminium iodide c) methane d) Methanol
(CH3OH).
 To find the enthalpy of formation of a compound, Hess’s law can be
used.
E.g. Give the equation which shows the enthalpy of formation of ethanol
(C2H5OH) and then determine its values if the enthalpy of combustion of
carbon, hydrogen and ethanol are –393.5, -285.8 and –1371 kJmol-1
2C(s) +
3H2(g)
+
½ O2(g)
C2H5OH(l)
2CO2 + 3H2O
Route 1 = Route 2
H(f) = 2 xH1
H(f) = (2 x –393.5)
+
+
3 x H2
(3 x –285.8)
= -273.4 kJmol-1
H3
-
-
(-1371)
 If enthalpies of formation (Hf ) are known then they can be used
to find the overall enthalpy change in a reaction using the equation.
H =  Hf (products) -  Hf (reactants)
Where  is the sum of all the enthalpies.
E.g. Calculate the enthalpy change in the following reaction
3CO + Fe2O3
2Fe + 3CO2
Given the enthalpies of formation of carbon monoxide, iron oxide and
carbon dioxide are –111, -822 and –394 kJmol-1.
H = (products) – (reactants)
= [(3 x CO2)+ (2 x Fe)] – [(3 x CO) + Fe2O3]
= [(3 x –394)] – [(3 x –111) + -822]
= -27 kJmol-1
E.g. 2 Calculate the enthalpy change in the following reaction.
2NaHCO3
Na2CO3 + CO2 + H2O
Given the following enthalpies of formation for sodium hydrogen
carbonate, sodium carbonate, carbon dioxide and water are –948, -1131, 395, -286 kJmol-1.
E.g. 3 Calculate the enthalpy of reaction of the following.
B2H6 + 3O2
B2O3 + 3H2O
If the enthalpies of formation for boron hydride, boron oxide and water
are +32, -1225 and –286 kJmol-1.
E.g. 4 Calculate H formation of CH3N2H3
4CH3N2H3 + 5N2O4
4CO2 + 12H2O + 9N2
Enthalpy of reaction is –5116 kJmol-1
Enthalpies of formation for nitrogen oxide, carbon dioxide and water are
–20, -393 and –286 kJmol-1.
Answers =
+84 kJ mol-1
-2115 kJ mol-1
+53 kJ mol-1
15.2 Born Haber Cycle (HL ONLY)
The Born Haber cycle shows the individual steps involved in the enthalpy of
formation of a compound.
Na+(g) +
Cl(g)
ΔHelectron affinity
ΔHionisation
Na(g) +
Na+(g) + Cl-(g)
Cl(g)
ΔHatomization
Na(s) +
ΔHlattice
½ Cl2(g)
ΔHformation
NaCl(s)
Enthalpy of atomization is the energy required to form one mole of gaseous
atoms. It is an endothermic process. (+103 KJ/mol for Na and +121 KJ/mol for Cl)
Ionisation Energy is the energy required to make one mole of gaseous metal
ions. This is also endothermic. (+500 KJ/mol for Na)
Electron Affinity is the energy released when one mole of gaseous non-metal
ions are produced. This is exothermic. (-364 KJ/mol for Cl)
Lattice Enthalpy is the amount of energy released when one mole of gaseous
ionic metal and non-metal ions produces a compound. This is also exothermic. (-771
KJ/mol for NaCl).
Enthalpy of Formation is the sum of all these individual values.
E.g. +103 + 121+ 500 –364 – 771 = -441 KJ/mol (the formation of NaCl is
therefore an exothermic process)
Using the Born Haber Cycle
 The cycle can be used to determine the enthalpy of formation of an ionic
compound, or it can be used to work out the lattice enthalpy (as this is a
difficult value to measure directly.)
 If there is a large difference between the lattice enthalpy obtained by
experimental use of the Born-Haber cycle and between theoretical
determination then this leads to the indication that the compound under
discussion is not ionic.
E.g. NaCl has good agreement, but AlCl3 does not since it is more covalent in
character.
15.2 Lattice Enthalpy (HL ONLY)
 This is the energy released when gaseous ions form 1 mole of a crystalline
compound.
E.g. Na+(g) +
Cl-(g)
NaCl(s)
 The size of the lattice enthalpy is dependant on 2 factors.
1. The size of the ions.
The smaller the ions the greater the attractive force between the protons
of one ion and the electrons of another.
E.g. LiCl = -846 KJ/mol compared with KCl = -701 KJ/mol
2. The size of the charge on the ions.
The greater the charge, the greater the attractive force between the ions.
E.g. NaCl = -771 KJ/mol compared with MgCl2 = -2493 KJ/mol.
Note: See data booklet for more comparisons.
15.3 Entropy (S) (HL)
 The universe tends towards a natural state of disorder.

Entropy (S) is a measure of the energy of disorder.
 If an entropy change (S) is positive then the reaction has
increased the disorder of the chemicals.
 If an entropy change is negative, then the reaction has made the
chemicals more ordered.
Examples of Entropy Changes.
1.
Changes of state: If a solid changes into a liquid or more
importantly a gas, then there is an increase in the disorder of the
chemicals and a positive entropy change.
2.
Formation of gaseous products. If one of the products
of a reaction is a gas, whilst none of the reactants were gases,
there will be a large positive entropy change due to the increase
in disorder of gases compared to solids and liquids.
Zn(s) + 2HCl(aq)
3.
An increase in the number of particles. If some of the
reactant particles split up into two or three product particles
then this is an increase in disorder.
NH4Cl
4.
ZnCl2(aq) + H2(g)
NH3 + HCl
Dissolving. Mixing or dissolving two types of particles
together causes an increase in disorder.
15.4 Spontaneity (HL)
 When considering if a reaction will take place spontaneously or not,
3 factors must be considered.
1.
2.
3.
The enthalpy change (H). The internal energy
stored in the chemical bonds.
The entropy change (S). The disorder of the
chemicals.
The Temperature.
 Spontaneity (G) is measured as the Gibbs free energy change.
G = H - TS
 If G is negative, a reaction will take place spontaneously at
that temperature (in theory).
 If G is positive it will not.
Effects of Enthalpy and Entropy on Spontaneity (HL)
H
S
Endo
More disordered
(+)
(+)
Endo
More ordered
(+)
Exo
(-)
More disordered
(-)
Exo
(+)
More ordered
(-)
(-)
G
Spontaneity
Depends on T
Spontaneous
only at high
temps when
TS is greater
than H
Always (+)
Never
spontaneous at
any temperature.
Always (-)
Always
spontaneous at
all temperatures
Depends on T
Spontaneous
only at low
temps when
TS is less than
H
 Even if a reaction is predicted to be spontaneous at a particular
temperature, it must also have sufficient collision energy (activation
energy) to react.
E.g. Carbon should react spontaneously with oxygen at all
temperatures, but it requires heating to give it’s particles enough
energy so that when they collide they will react.
15.3 Calculating Entropy Changes (HL ONLY)
 Unlike enthalpy (H), the absolute entropy (S) of a chemical can be
measured directly.
 So that the entropy change can be calculated as:
S = S (products) - S (reactants)
E.g. Calculate the entropy change in the following reaction.
3H2(g) + N2(g)
2NH3(g)
The absolute entropies of hydrogen, nitrogen and ammonia are 131, 192
and 192 JK-1mol-1.
S = S (products) - S (reactants)
= (2 x 192) - [ (3 x 131) + 192) ]
= -201 JK-1mol-1
Decrease in disorder
E.g. 2 Calculate the entropy change for:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The absolute entropies for methane, oxygen, carbon dioxide and water are
186, 205, 214, and 70 JK-1mol-1
Answer = -242 JK-1mol-1 (decrease in disorder)
Calculating Gibbs Free Energy Changes (G) (HL ONLY)
 This can be determined in 2 ways:
1.
2.
Using entropy and enthalpy values.
Using Gibbs Free energy of formation values.
From Enthalpy and Entropy Values
E.g. Is the following reaction spontaneous at 500K?
If not, at what temperature does it become spontaneous?
CaCO3(s)
CaO(s) + CO2(g)
Given the following enthalpy of formation values for calcium carbonate,
calcium oxide and carbon dioxide of –1207, -636 and –394 kJmol-1.
Also the following entropy values for these chemicals are 93, 40 and 214
JK-1mol-1.
Enthalpy change H = H (products) - H (reactants)
= [-636 + (-394)]
-
[-1207]
= + 177 kJmol-1
This is an endothermic reaction in kilojoules.
Entropy ChangeS = S (products)
= (40 + 214)
-
-
S (reactants)
(93)
= +161 JK-1mol-1
Increase in entropy, more disordered, measured in Joules.
Gibbs free energy change (at 500K) G = H - TS
= (+177 000) - ( 500 x 161)
= 96 500 J mol-1
= 95.5 kJ mol-1
Since G is positive, this reaction will be non spontaneous at 500K
For this reaction to be spontaneous, G must be negative (less than
zero).
TS must be greater than H
( T x 161) > +177 000
T > 177 000 / 161
T > 1 099 K
(8060 C)
If the temperature rises above 8060 then this reaction will be spontaneous
(providing the particles also have enough activation energy to react when
they collide).
If the temperature was at exactly 8060 then the system is said to be at
equilibrium in that G is zero and the system is neither spontaneous nor
non spontaneous.
2. From Gibbs Free energy of formation changes
G (reaction) = G (products) - G (reactants)
E.g. Calculate the Gibbs Free Energy change in the following reaction,
CH4(g) + O2(g)
CO2(g) + H2O(l)
Given the following free energy changes of formation for methane,
carbon dioxide and water are –50, -394 and –237 kJmol-1 (Oxygen as an
element has a G formation value of zero).
G = G (products) - G (reactants)
= [ -394 + ( 2 x –237)] - (-50)
= -818 kJmol-1
Questions
Calculate G for the following reactions and which are spontaneous at
298 K.
Reaction
S (J K-1mol-1) H (kJmol-1)
C + O2
CO2
+3.36
-393.5
H2O(g)
H2O(l)
-119.1
-44.1
Mg + ½ O2
MgO
-218.8
-1204
H2O(l)
H2O(g)
-22.2
-6.01
First 3 are spontaneous, 4th is non spontaneous
©MAC/ JMH Shatin College, Hong Kong.