Section
Section 3
3
Energetics
Page
1
Energetics
Conservation of energy
There are many forms of energies: heat, light, chemical energy, nuclear energy, etc. Energy can
be converted from one form into another but it is never created and never destroyed
during the conversion. This is the important concept of the First Law of Thermodynamic:
Energy can neither created nor destroyed, but can be transformed or exchanged between a
system and its surrounding.
Exothermic and Endothermic Reactions
During the course of a chemical reaction, chemical bonds in the reactants are broken and new
ones formed in the products. Bond-breaking is energy absorbing, i.e. endothermic, whereas
bond-forming is energy releasing, i.e. exothermic. Therefore, during the course of a chemical
reaction, heat is either given out or taken in from the surroundings. Therefore, the total energy
absorbed or released during the reaction at constant pressure is called the enthalpy change of
a reaction (H).
If the enthalpy of the product is higher than the enthalpy of the reactants, the reaction is said
to be endothermic and H has a positive value. The temperature of the reaction mixture
decrease because energy is absorbed during the reaction.

heat
content
heat
content
Products
Reactants

Reactants
Endothermic Reaction
Products
Exothermic Reaction
If the enthalpy of the product is lower than that of the reactants, the reaction is said to be
exothermic and H has a negative value. The energy that is lost during an exothermic reaction
is first transferred to the reaction mixture. The temperature of the reaction mixture therefore
increases.
.
1.
(a)
Classify the following changes as exothermic or endothermic:
(i)
Sodium hydroxide dissolves in water and the temperature of the solution
rises.
(ii)
Ammonium chloride dissolves in water and the temperature of the
surroundings drops.
(iii)
Hydrogen and oxygen combine explosively to form water.
(iv)
Liquid water condenses to ice at
(v)
Liquid nitrogen (boiling point = 77 K ) boils spontaneously at room
temperature.
0 C.
(b)
Explain why  H
has a negative value for an exothermic reaction and a
positive value for an endothermic reaction.
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
(c)
A student argues : “If heat is taken in during an endothermic reaction,
surely the temperature of the chemicals should go up, not down.” Can
you help the student to understand what is happening ?
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__________________________________________________________________
__________________________________________________________________
Section
3
Energetics
Page
2
Standard Enthalpy of Changes
The enthalpy changes that occur during a chemical reaction vary depending on temperature,
pressure, the physical state and the amount of the substances involved.
The standard molar enthalpy change of a reaction is the enthalpy change under standard
conditions per mole of the reaction as specified by a balanced chemical equation. The standard
conditions chosen are 298 K and one atmospheric pressure.
Some standard enthalpy changes are defined below:
1.
Standard Enthalpy (change) of Formation
It refers to the formation of one mole of the substance from its elements (in standard
states) under the standard conditions. For example:
e.g.
N.B. The Standard Enthalpy of Formation of all elements is assigned to be O.
2.
Standard Enthalpy (change) of Combustion
It refers to the complete combustion of one mole of the substance in oxygen under
standard conditions.
e.g.
3.
Standard Enthalpy (change) of Solution
It refers to the dissolution of one mole of substance in infinite dilution (volume of
solvent is so large that on further dilution there is no further heat
exchange) under standard conditions.
e.g.
4.
Standard Enthalpy (change) of Neutralization
It refers to the formation of one mole of water by neutralization of an appropriate amount of
acid under standard conditions.
e.g.
5.
Standard Enthalpy (change) of reaction
It is a general name for the enthalpy change of any reaction.
always better to give an equation.
e.g.
.
2.
To avoid confusion, it is
Write complete thermochemical equations showing
(a)
the standard heat of combustion of methane,
CH4
.
(b)
the standard heat of formation of calcium oxide,
CaO
.
(c)
the standard heat of neutralization of H2SO4 (aq)
and
KOH (s)
(d)
the standard heat of solution of potassium hydroxide
.
.
Section
3
Energetics
Page
3
Determination of the Standard Enthalpy of Reaction
Experimental Determination of the Standard Enthalpy of Reaction
Example 1
To
determine
the
enthalpy
of
combustion
of
ethanol
A metal calorimeter has a mass of 200 g and a specific heat capacity of 0.42 J g -1 K-1. Into it
are put 500 cm3 of water. Combustion of 1.84 g of ethanol rises the temperature of the
water by 25 oC. Calculate the standard enthalpy of combustion of ethanol.
Given: Density of water = 1 g cm-3 ; specific heat capacity of water = 4.2 J g-1 K-1
screen to
prevent
heat loss
Heat absorbed by water
= m c 
=
Heat absorbed by calorimeter
= m c 
=
thermometer
 Total heat evolved by ethanol
metal
calorimeter
No. of mole of ethanol burnt
 Heat evolved by 1 mole ethanol
spirit
burner
contains
ethanol
=
=
=
 Standard enthalpy of combustion
=
of ethanol , Hc [C2H5OH]
This value is lower than the listed value (Hc [C2H5OH] = -1368 kJ mol-1) because of the following
sources of error :
heat lost from the top and sides of the calorimeter ;
-
heat which goes into the material of the beaker rather than the water ;
-
incomplete combustion :
burn completely.
Example 2
To
determine
the
and
in a limited supply of oxygen, some of the ethanol will not
enthalpy
of
neutralization
A glass calorimeter has a mass of 200 g and a specific heat capacity of 0.42 J g 1K-1 . Into it
are put 50 cm3 of 1.25 M hydrochloric acid and 50 cm3 of 1.25 M potassium
hydroxide solution at the same temperature. The temperature of the calorimeter and contents
rises by 7.0 oC. Calculate the standard enthalpy of neutralization. Assuming that the
specific heat capacity of all the solution is 4200 J kg-1 K-1.
-
Equation :
Heat absorbed by acid-alkali mixture
=
Heat absorbed by glass calorimeter
=
 Total heat evolved by neutralization
=
No. of mole of water formed
 Standard enthalpy of neutralization
Hneu
=
=
This value is lower than the listed value (Hneu = - 56.7 kJ mol-1) because of the following sources of
error :
heat lost due to evaporation, convection and conduction ;
-
specific heats of the solutions cannot be exactly 4.2Jg-1K-1 as that of water.
____________________________________________________________________________________
.
3.
The enthalpy of neutralization of ethanoic acid with aqueous sodium hydroxide is 55.2 kJ mol-1 while that of hydrochloric acid is - 57.3 kJ mol-1.
Account for the
difference in these two values.
(2 marks) (91 IIA 1(b)(ii))
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
Indirect
Determination of
Section 3
Energetics
Standard Enthalpy of Reaction
the
Page
4
Many reaction enthalpies cannot be determined experimentally because the reaction cannot be
brought about in the laboratory, therefore these standard enthalpy of reactions must be found
indirectly. Such reaction enthalpies can be calculated from the enthalpies of other reactions by
applying the Hess‘s Law.
Hess‘s Law states that the total energy change accompanying a chemical change is
independent of the route by which the chemical change takes place.
For example:
The great value of Hess' Law is that it can be used to calculate enthalpy change that cannot be
determined directly by experiments.
Example 1:
Determination of enthalpy of hydration of copper(II) sulphate
CuSO4 (s)
Example 2:
+

5 H2 O
CuSO4. 5 H2O (s)
Determination of enthalpy change of formation (Hf) of CaCO3
Consider the following enthalpy level diagram:
Ca(s)
+
C (graphite)
+
1 21 O2(g)
+
2 HCl(aq)
+
+
C (graphite)
+
1 21 O2(g)
H1
CaCl2(aq)
H2(g)
Hf
CaCO3(s)
+
2HCl(aq)
H2
H3
CaCl2(aq)
+
CO2(g)
+
H2O(l)
H1 and H3 are both determined experimentally. For example, H1 can be determined
experimentally by measuring the temperature change when a known mass of calcium metal
reacts with a known volume of excess HCl.
is known from literature. Hf
of CaCO3
can be calculated from these known
H2
information by applying Hess’s law.
Application of Standard Enthalpies of Reaction
By making use of these values, enthalpy changes of some other reactions can also be calculated.
For example, fuel engineers and dieticians can made use of these data to choose the correct fuel
or food for a particular purpose.
For example, during the Apollo 11 project which landed the first man on the moon on 21 July
1969, the engines of the lunar module used methylhydrazine and dinitrogen tetraoxide. These
liquids were carefully chosen since they ignite spontaneously and very exothermically on
contact.
Example:
Calculate the enthalpy change for the following reaction.
+
5 N2O4 (l) 
Hf (CH3NHNH2(l))
Hf (CO2 (g))
= + 53 kJ mol-1
4 CH3NHNH2 (l)
Given:
=
- 393 kJ mol-1
4 CO2 (g)
+
12 H2O (l)
+
9 N2 (g)
Hf (N2O4(l)) = - 20 kJ mol-1
Hf (H2O (l)) = - 286 kJ mol-1
Section
.
4.
3
Energetics
Page
5
When one mole of graphite is burned completely in oxygen to produce carbon dioxide,
394 kJ of heat is evolved whereas, when diamond is similarly burned, 396 kJ is
evolved.
(a)
Write thermochemical equations summarizing this information.
(b)
Draw an energy level diagram comparing the combustion of diamond and
graphite.
(c)
What is the enthalpy change for the conversion of graphite into diamond?
H as a measure of Energetic Stability
For those exothermic reaction, because the products are of lower energy content than the
reactants, therefore they are energetically more stable. And for those endothermic reactions, the
products are therefore energetically unstable.
N.B. 'Stability' is in a relative sense. Because the feasibility of a reaction does not depend on the
enthalpy change alone since the reaction may be kinetically controlled.
For example:
Both diamond and graphite are energetically unstable and are liable to react with O2 to form
carbon dioxide. However, graphite and diamond can be stored for long periods at room
conditions.
It is because the reactions are so slow that both diamond and graphite are kinetically stable (
high activation energy is required to break the giant covalent network of diamond or graphite)
____________________________________________________________________________________
.
5.
The enthalpy change of combustion of diamond and graphite are as follows :
C(diamond) +
C(graphite) +
O2(g)
O2(g)
 CO2(g)
 CO2(g)
 =
 =
- 395.4 kJ mol-1
- 393.4 kJ mol-1
Which one is the more stable allotrope, diamond or graphite ?
Comment on the accuracy of the jeweller‘s advertisement “Diamonds last forever”.
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
.
6.
Draw energy level diagrams for the following reactions:
Section
.
7.
(a)
C (s)
+
2 S (s)
(b)
Zn(s)
+
2 H+ (aq)

CS2 (l)

Zn2+ (aq)
Comment on the thermodynamic stability of
equations:
N2 (g) +
NO (g) +
1
2
O2 (g) 
1

2 O2 (g)
1
2
+
3
H
Energetics
Page
-1
= + 88 kJ mol
H2 (g)
H
=
6
- 152 kJ mol-1
NO (g). Base your answer on the two
NO(g)
H =
NO2 (g)
H =
kJ mol-1
- 74 kJ mol-1
+ 90
Energetic of Formation of Covalent Compounds
Bond
Enthalpies
The Standard Bond Dissociation Enthalpy is the amount of energy (per mole) required
when a particular bond in a specific compound is broken with the molecules and the resulting
fragments being in their standard state. This value is always positive.
Bond Energy Term is the average value of energy associated with a bond. This is the
average of the dissociation energies of a particular type of bond.
In the methane molecule there are four C-H
enthalpies:
CH4 (g)
 CH3 (g) + H (g)
bonds and thus four bond dissociation
435 kJ mol-1
453 kJ mol-1
Hd1 (298 K)
=
+
Hd2 (298 K)
=
+
CH2 (g)
 CH2 (g) + H (g)
 CH (g)
+ H (g)
Hd3 (298 K)
=
+
CH (g)
 C (g)
Hd4 (298 K)
=
+
Hd (298 K)
=
+ 1652 kJ mol -1
CH3 (g)
+
H (g)
In the complete dissociation of methane:
CH4 (g)
 C (g)
+ 4 H (g)
425 kJ mol-1
339 kJ mol-1
Dividing the standard enthalpy change equally among the four bonds gives an average value for
the C-H bond of 413 kJ mol-1. This value is the average standard bond enthalpy (bond
energy term) of the CH bond.
.
8.
(a)
(b)
Explain the difference between ' bond energy term ' and ' bond dissociation
energy ' .
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
Name a molecule for which the bond dissociation energy is the same as the bond
energy term.
__________________________________________________________________
Evidence for the Consistency of Bond Enthalpies
Section 3
Energetics
Page 7
By burning a series of alcohols and measure the enthalpy change per mole of each, it was
found that the difference in heat of combustion between successive alcohols in a homologous
series is about the same:
Alcohol
Enthalpy
of
Combustion
propan-1-ol
- 2017 kJ mol-1
butan-1-ol
- 2675 kJ mol-1
pentan-1-ol
- 3323 kJ mol-1
hexan-1-ol
- 3976 kJ mol-1
heptan-1-ol
- 4623 kJ mol-1
Difference
-------------
The more or less constant increase in the standard enthalpy of combustion can be explained in
that by passing up the homologous series, one more - CH2 group was exist and thus on
combustion it need to break one more CC and 2 CH bonds (and to breakdown 1.5
oxygen molecules for complete reaction) as passing up the series . On the other hand, one more
carbon dioxide ( 2 C=O bonds) and one water ( 2 OH bonds) will be formed on passing up
the series.
The increase in standard enthalpy of combustion could therefore be explained by the increase
in energy released as the number of new bonds formed are increased.
N.B. We have assumed that each
amount of energy.
CO,
OH,
CH
and
C = O
has a definite
Determination of Bond Enthalpies by Additivity Rule
Example:
In case of methane, the CH
bond energy can be calculated from the total
amount of energy required to break the molecule into constituent atoms, i.e. to atomize the
molecule.
CH4(g)

C (g)
+
Hd (298 K)
4 H (g)
=
+ 1652 kJ mol -1
Because it involve the breaking of four CH bonds and we have assumed that
bonds are of the same energy. Therefore
The bond energy
(bond energy term) of each CH bond
=
_____________
all
the
C- H
kJ mol-1
N. B. The bond energy of a specific type of bond may change when the environment changed.
For example:
CCl4
E(CCl) = + 327 kJ mol-1
CH3Cl
E(CCl) = + 335 kJ mol-1
C2H5Cl
E(CCl) =
+ 342
kJ mol-1
The bond energy depends upon the environment of the bond, e.g. to what atom or group of
atoms do the bond attached.
.
7.
Calculate the bond energy of
C4H10 (g)
C5H12(g)
C-C
and
C-H
from the following information:

4 C (g)
+
10 H (g)
H
=

5 C (g)
+
12 H (g)
H
=
+ 5165 kJ mol-1
+ 6340 kJ mol-1
Section
3
Energetics
Page
8
Application of Bond Enthalpies
Bond enthalpies can be used in the calculation of enthalpies of reactions by assuming that the
bond enthalpy of a particular type of bond is constant. And because of this assumption, the
calculation is not as accurate as that derived from experiments.
Example:
Calculate the enthalpy of reaction for the reaction:
C2H4

Given that:
+
E (C=C)
E (CH)

H2
=
=

C2H6
+ 619 kJ mol-1 ;
+ 413 kJ mol-1 ;
E (CC)
E (HH)
=
=
+ 347 kJ mol-1;
+ 435 kJ mol-1 .
Sometimes, the additivity rule may break down. For example:
+
H2 (g)
H = -120 kJ mol-1

cyclohexene
cyclohexane
+ 3 H2 (g) 
benzenecyclohexane
(one of the proposed structures)
Benzene is
___________ kJ mol -1
H(expected)
=
H(actual)
=
_______
kJ mol-1
- 208 kJ mol-1
more stable than the expected structure.
Conclusion: This proposed structure of benzene is incorrect.
Energetics of Formation of Ionic Compounds
Lattice energy
Lattice energy of an ionic crystal is the standard enthalpy of formation of the crystal lattice
from its constituent ions in the gaseous state. For example, the lattice energy of sodium
chloride is the enthalpy change of the following reaction:
Electron Affinity
Electron affinity (E.A.) is the energy change when one mole of electrons is
to one mole of neutral atoms or ions of the element.
X (g) + e - 
X - (g)
H = 1 st E.A.
2X (g) + e

X (g)
H = 2 nd E.A.
____________________________________________________________________________________
.
7.
(i)
added
Write an equation to represent the change related to the second electron
affinity of oxygen.
(ii)
The first and second electron affinities of oxygen are -142 kJ mol-1 and
79 kJ mol-1 respectively. Explain why they have opposite signs.
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
+
(iii) Explain why all the inert gases have positive first electron affinities.
__________________________________________________________________
__________________________________________________________________
(4 marks) (94 IA 1(a))
Section
3
Energetics
Page
9
The Born-Haber Cycle
Direct determination of lattice energy is very difficult because it is very difficult to get isolated
sodium and chloride ions. Therefore the values are usually calculated from other experimentally
determined data by applying the Hess Law. The Born-Haber Cycle is a technique of applying
Hess‘s Law to the standard enthalpy changes which occur when an ionic compound is formed.
Take the reaction between sodium and chlorine to give sodium chloride as an example. The
reaction can be considered to occur by means of the following steps:
In order for the above compound to form, Hf must be negative. In the above equation, the most
sufficient value is ionization energy. Therefore if the lattice energy can compensate for the
ionization energy, the compound will most probably be formed.
.
8.
Draw a Born-Haber Cycle for the formation of
[i] magnesium chloride
[ii]
aluminium oxide
Energy Change in Dissolving Ionic Solid
When an ionic solid dissolves in a solvent, two enthalpy terms are involved.
1.
The ions must be separated from the ionic lattice. The energy required is the lattice
dissociation energy.
2.
The separate ions interact with the molecules of water. If the solvent is polar, a charged ion
can be attracted to one end of a polar solvent molecule. The energy released as these
attractive forces come into play is compensation for the energy required to dissociate the
lattice.
The attachment of polar molecules to ions is known as solvation. (If water is the solvent, the
process is called hydration.) The energy released when one mole of a substance in the form of
gaseous ion is solvated is called solvation energy (hydration energy).
The enthalpy of solution is the difference between the lattice energy and the hydration energy.
The heat of solution is usually small as compared with the lattice energy. It may be positive or
negative.
Hsolution
=
Hhydration
-
Hlattice
Section
3
Page 10
Energetics
Past Paper
-
1
Hf
for CuSO4 . 5 H2O
Hsoln [CuSO4 . 5 H2O(s)]
can be determined using the following data:
= + 8 kJ mol -1
Hf
[CuSO4(s)]
Hsoln [CuSO4(s)]
=
=
- 773 kJ mol -1
- 66 kJ mol -1
Hf
=
- 286 kJ mol -1
Show how
[H2O(l)]
(3 marks)
(90IA2(b))

2
(i)
Calculate the enthalpy of formation of NaCl(s) from the following data :
Reaction
NaOH (aq) + HCl (aq)  NaCl (aq) + H20 (l)
H2 (g)
+ 21 O2 (g)  H2O (l)
H /kJ mol -1
- 57.3
H2 (g) +
Cl2 (g)  HCl (g)
HCl (g)
+
aq
 HCl (aq)
Na(s) + 21 02 (g) + 21 H2 (g) + aq  NaOH (aq)
NaCl (s) +
aq
 NaCl (aq)
- 92.3
1
2
3
- 285.9
1
2
- 71.9
- 425.6
+ 3.9
(ii)
When solid sodium chloride is dissolved in water, the process is endothermic.
Explain this observation.
(8 marks) (912A1(b))
(i)
Define the standard enthalpy of formation of a compound, using CH3OH (l) as
an illustration.
(ii)
Given the following thermochemical data at 298 K :
Standard enthalpy of formation of CO 2(g)
- 726.6 kJ mol -1
- 393.5 kJ mol -1
Standard enthalpy of formation of H2O(l)
- 285.8 kJ mol -1
Calculate the standard enthalpy of formation of CH3OH(l)
at 298 K.
(4 marks) (922A1(b))
Standard enthalpy of combustion of CH 3OH(l)
4
Given the following thermochemical data at
Compound
cyclopropane(g)
propene (g)
propane (g)
water (l)
Hcombustion/ kJ mol -1
- 2091
- 2058
- 2220

298 K:
H
formation
/ kJ mol -1



-285.8
(i)
Calculate the enthalpy change involved in the hydrogenation of cyclopropane to
propane.
(ii)
Calculate the enthalpy change involved in the conversion of cyclopropane to
propene. Comment on the relative stability of cyclopropane and propene.
(8 marks) (93IIA3(b))
Section
3
Page 11
Energetics
Marking Scheme
1.
90IA2(b)
Cu(s)
+
S(s)
5 (-286)
Cu(s)
f
+
+
2O2(g)
S(s)
+
+
5H2(g)
2O2(g)
+
+
5
2
O2(g)
5H2O(l)
-773
CuSO4(s)
+
5H2O(l)
CuSO4(aq)
+
5H2O(l)
- 66
+8
CuSO4  5H2O(s)
2
f [CuSO4  5H2O(s)] =
2.
5 (-286) - 773 - 66 - (+ 8) =
-2277 kJ mol-1
1
91IIA1(b)
(i)
Na(s)
+
1
2
Cl2(g)
+
O2(g)
1
2
+
H2(g)
+
aq
f
-425.6
NaOH(aq)
+
1
2
Cl2(g)
NaOH(aq)
+
HCl(g)
NaOH(aq)
+
HCl(aq)
NaCl(aq)
+
H2O(l)
NaCl(s)
+ H2O(l)
+
1
2
NaCl(s)
+
H2(g)
+
1
2
O2(g)
H2(g)
- 92.3
+
aq
- 71.9
- 285.9
- 57.3
+ 3.9
f [NaCl(s)]
- 365.1 kJ mol-1
1
(ii) The dissolving of a salt in water is a case of an enthalpy
change
accompanying
a
chemical
process.
The
first
process
requires an input of energy to break the lattice.
The second
process involves the release of energy when new bonds are made
between ions and water.
The enthalpy of solution is a measure
of the difference between these two processes.
In case of NaCl,
the lattice energy is larger than the enthalpy of hydration.
2
3.
=
3
- 425.6 - 92.3 - 71.9 - 57.3 - (+ 3.9) - (- 285.9)
=
92IIA1(b)
(i)
The standard enthalpy of formation of a compound is the standard
enthalpy change that occurs when one mole of the compound is
made from its constituent elements under standard conditions (298K
1
and 1 atmospheric pressure). e.g. C(s) + 2H 2(g) +

2 O2(g)
CH3OH(l)
(ii)
C(s)
1
+
2H2(g)
+
1
2
O2(g)
- 393.5
1.5O2
+
2(-285.8)
CO2(g)
f [CH3OH]
=
f
1.5O2
+
CH3OH(l)
- 726.6
2H2O(l)
- 393.5 + 2 (- 285.8) - (- 726.6)
2
=
- 238.5 kJ mol-1
1
Section
4.
(i)
+
- 2091
- 285.8
3H2O(l)
=
+
3CO2(g)
3H2O(l)
=
+
H2O(l)
- 2091 - 285.8 - (-2220)
- 2091

Page 12
-2220

(ii)
Energetics
hydrogenation CH3CH2CH3
H2(g)
hydrogenation
3
=
2
- 156.8 kJ mol-1
1+1
CH3CH=CH2
-2220
+
3CO2(g)
- 2091 - (-2058)
=
1
- 33 kJ mol-1
1
The conversion from cyclopropane to propene is exothermic, hence
propene is more stable because in the structure of cyclopropane
there is squeezing of bond angles / the ring is highly strained.
2