A TEXT BOOK OF
ENGINEERING
CHEMISTRY - I
FOR
SEMESTER - I
F.E. DEGREE COURSES IN ENGINEERING
(COMMON FOR ALL BRANCHES)
As Per Revised Syllabus of UNIVERSITY OF MUMBAI, MUMBAI
Choice Based Credit and Grading System w.e.f. Academic Year 2019-2020
Dr. (Mrs.) Trupti S. Paradkar
Associate Professor
Dwarkadas J. Sanghvi College of Engineering
Vile-Parle (W), Mumbai - 56
N6001
ENGINEERING CHEMISTRY - I
Second Edition
©
:
:
ISBN 978-81-942538-2-2
September 2019
Author
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DEDICATED
TO
MY
HUSBAND
SANJAY
&
LOVING
DAUGHTER
OWEE
For their unconditional love and support
PREFACE
It gives me immense pleasure to present the first edition of the book on Engineering
Chemistry - I to the students of first year engineering. This book is prepared according to
the new syllabus (2019) of University of Mumbai; which comes into effect from the
academic year 2019-2020.
The syllabus has been meticulously followed throughout the book with emphasis on
minute details. The topics have been structured in order to develop the subject in a
logical sequence. The coverage of this book will definitely help students to understand
and grasp the subject with ease.
Numerous solved problems including the ones from university examinations of past
years on different topics have been incorporated wherever applicable. This would help
the students in solving a wide variety of problems.
At the end of every chapter there are ample unsolved problems for practice along
with answer keys. Review questions on theory are added along with references to the
appropriate sections from the book. University problems and theory questions of recent
past are listed year wise under every chapter.
Any constructive suggestions for enhancing the utility of this book and for removing
errors that might have gone unnoticed will be gratefully acknowledged.
Dr. (Mrs.) Trupti S. Paradkar
ACKNOWLEDGEMENT
I honestly acknowledge Shri Vile Parle Kelavani Mandal for providing the
infrastructural facilities in the college.
My thanks are also due to Principal Dr. Hari Vasudevan, Dwarkadas J. Sanghavi
College of Engineering for his continued support and encouragement. I sincerely thank
my friends for their suggestions, and well-wishers for their moral support while writing
this book.
I would like to express my deepest gratitude to Mr. Dinesh K. Furia and Mr. Pradeep
K. Furia of M/s Nirali Prakashan.
I am indeed thankful to Mr. Shashikant Patel for facilitating the coordination required
for publishing this edition. I am also thankful to Paras Rambhia, Riddhi Hardik Vira of
Decent Typesetters for their excellent computer typesetting work.
I am also grateful to my husband Mr. Sanjay Paradkar and loving daughter Owee for
bearing with me throughout this exercise.
Last but not the least; special thanks to my students for their overwhelming response
to all my earlier books on Applied Chemistry I & II.
Dr. (Mrs.) Trupti S. Paradkar
SYLLABUS
Module 1 : Atomic and Molecular Structure

Atomic orbitals (s, p, d, f) orbital shapes, Electronic Configuration, Molecular orbital theory
(MOT), bonding and anti-bonding orbitals, Molecular orbital diagrams of Homonuclear and
Heteronuclear diatomic molecules-Be2, O2, CO, NO their bond order and magnetic
properties.
Module 2 : Aromatic Systems and their Molecular Structure

05 Hours
Introduction: Definition-Polymer, polymerization, Properties of Polymers-Molecular weight
(Number average and Weight average), Numerical problems on molecular weight, effect of
heat on polymers (glass transition temperature), Viscoelasticity, Conducting Polymers,
Classification-Thermoplastic and Thermosetting polymers; Compounding of plastic,
Fabrication of plastic by Compression, Injection, Transfer and Extrusion moulding,
Preparation, properties and uses of PMMA and Kevlar.
Module 6 : Water

05 Hours
Statement of Gibbs’ Phase Rule, Terms involved with examples, One Component System
(Water), Reduced Phase Rule, Two Component System (Pb-Ag), Advantages and Limitations
of Phase Rule. Numerical problems on Phase Rule.
Module 5 : Polymers

03 Hours
Ionic, dipolar and Vander Waal’s interactions, Equations of state of real gases and critical
phenomena.
Module 4 : Phase Rule-Gibb’s Phase Rule

02 Hours
Define Aromaticity, Huckel’s rule, Structure and bonding of benzene and pyrrole.
Module 3 : Intermolecular Forces and Critical Phenomena

04 Hours
05 Hours
Introduction: Impurities in water, hardness of water-units (no conversions), types and
numerical problems, determination of hardness of water by EDTA method and numerical
problems. Softening of water by Ion Exchange process and numerical problems, BOD, CODdefinition, significance and Numerical problems. Water purification-membrane technologyElectrodialysis, Reverse osmosis, and Ultra filtration.
CONTENTS
Module 1 : Atomic and Molecular Structure
1.1 - 1.20
1.1
Concept of Atomic Orbital .............................................................................................................................. 1.1
1.2
Electronic Configuration .................................................................................................................................. 1.3
1.3
Molecular Orbital Theory (MOT) .................................................................................................................. 1.4
1.4
Linear Combination of Atomic Orbitals (LCAO) Method .................................................................... 1.6
1.5
Shapes of Molecular Orbitals ........................................................................................................................ 1.8
1.6
Molecular Orbital Treatment for Homonuclear Diatomic Molecules .......................................... 1.12
1.7
Molecular Orbital Treatment for Heteronuclear Diatomic Molecules ......................................... 1.16
1.8
Review Questions ............................................................................................................................................. 1.18
Module 2 : Aromatic Systems and their Molecular Structure
2.1 - 2.9
2.1
Introduction .......................................................................................................................................................... 2.1
2.2
Aromaticity ............................................................................................................................................................ 2.1
2.3
Huckel’s Rule ........................................................................................................................................................ 2.2
2.4
Structure and Bonding of Benzene ............................................................................................................. 2.4
2.5
Structure and Bonding of Pyrrole ................................................................................................................ 2.7
2.6
Review Questions ............................................................................................................................................... 2.8
Module 3 : Intermolecular Forces and Critical Phenomena
3.1 - 3.10
3.1
Introduction .......................................................................................................................................................... 3.1
3.2
Ionic Interactions ................................................................................................................................................ 3.1
3.3
Van Der Waal’s Interactions ........................................................................................................................... 3.2
3.4
Dipolar Interactions ........................................................................................................................................... 3.3
3.5
Hydrogen Bonding ............................................................................................................................................ 3.3
3.6
Ion-Dipole Force ................................................................................................................................................ 3.6
3.7
Equations of State of Real Gases and Critical Phenomena ................................................................ 3.6
3.8
Review Questions ............................................................................................................................................... 3.9
Module 4 : Phase Rule-Gibb’s Phase Rule
4.1 - 4.24
4.1
Introduction .......................................................................................................................................................... 4.1
4.2
Gibb’s Phase Rule ............................................................................................................................................... 4.1
4.3
4.2.1
Phase ...................................................................................................................................................... 4.2
4.2.2
Component .......................................................................................................................................... 4.3
4.2.3
Degree of Freedom of Variance ................................................................................................... 4.4
Application of Phase Rule to One Component System ...................................................................... 4.5
4.4
Application of the Phase Rule to Two Component Systems ............................................................. 4.9
4.4.1
General Characteristics of Two Component Systems ........................................................ 4.9
4.4.2
Thermal Analysis .............................................................................................................................. 4.10
4.4.3
Eutectic System ................................................................................................................................. 4.12
4.4.4
The Lead Silver System .................................................................................................................. 4.12
4.5
Advantages of Phase Rule ............................................................................................................................ 4.16
4.6
Limitations of Phase Rule .............................................................................................................................. 4.16
4.7
Review Questions ............................................................................................................................................. 4.17
4.8
University Questions ....................................................................................................................................... 4.19
Module 5 : Polymers
5.1 - 5.40
5.1
Introduction .......................................................................................................................................................... 5.1
5.2
What are Polymers? ........................................................................................................................................... 5.2
5.3
Polymerisation, Degree of Polymerisation and Functionality ........................................................... 5.3
5.4
Classification of Polymers ............................................................................................................................... 5.5
5.4.1
Natural and Synthetic Polymers .................................................................................................. 5.5
5.4.2
Organic and Inorganic Polymers ................................................................................................. 5.5
5.4.3
Homopolymers and Co-Polymers ............................................................................................... 5.5
5.4.4
Linear, Branched and Cross Linked Polymers ......................................................................... 5.8
5.5
Molecular Weight ............................................................................................................................................. 5.10
5.6
Melting and Glass Transition Phenomena .............................................................................................. 5.13
5.7
Viscoelasticity ..................................................................................................................................................... 5.14
5.8
Plastics .................................................................................................................................................................. 5.15
5.8.1
5.9
Classification of Plastics ................................................................................................................ 5.15
Compounding of Plastics .............................................................................................................................. 5.17
5.10 Moulding or Fabrication of Plastics .......................................................................................................... 5.19
5.11 Some Individual Plastics ................................................................................................................................ 5.23
5.12 Conducting Polymers ..................................................................................................................................... 5.28
5.13 Polymers in Medicine and Surgery ............................................................................................................ 5.29
5.14 Review Questions ............................................................................................................................................. 5.30
5.15 University Questions ....................................................................................................................................... 5.32
Module 6 : Water
6.1 - 6.60
6.1
Introduction .......................................................................................................................................................... 6.1
6.2
Sources of Water ................................................................................................................................................ 6.1
6.3
Hardness of Water ............................................................................................................................................. 6.2
6.3.1
Types of Hardness ............................................................................................................................. 6.3
6.3.2
Causes of Hardness ........................................................................................................................... 6.4
6.4
Units of Hardness ............................................................................................................................................... 6.5
6.5
Estimation of Hardness .................................................................................................................................. 6.14
6.6
Undesirable Effect of Hard Water .............................................................................................................. 6.27
6.7
Water Treatment (Softening) ....................................................................................................................... 6.28
6.7.1
Ion Exchange or Deionization or Demineralization Process ............................................ 6.28
6.8
Desalination of Brackish Water ................................................................................................................... 6.32
6.9
Water Pollution ................................................................................................................................................. 6.37
6.9.1
BOD - Biological Oxygen Demand ........................................................................................... 6.38
6.9.2
COD - Chemical Oxygen Demand ............................................................................................ 6.40
6.10 Sewage Treatment by Activated Sludge Process ................................................................................ 6.45
6.11 Review Questions ............................................................................................................................................. 6.47
6.12 Unsolved Problems .......................................................................................................................................... 6.48
6.13 University Questions ....................................................................................................................................... 6.51
References
R.1 - R.1
CHAPTER
1
ATOMIC AND MOLECULAR STRUCTURE
SYLLABUS
x
x
x
Atomic orbitals (s,p,d,f) orbital shapes, Electronic Configuration
Molecular orbital theory (MOT), bonding and anti-bonding orbitals
Molecular orbital diagrams of Homonuclear and Heteronuclear diatomic molecules- Be2, O2,
CO, NO their bond order and magnetic properties
1.1 CONCEPT OF ATOMIC ORBITAL
According to the wave-mechanical theory, electrons patrol in three-dimensional space
around the nucleus; sometimes near the nucleus, and sometimes farther away. Thus,
electrons effectively occupy a relatively vast space around the nucleus. It has been found
that the probability of finding an electron around the nucleus is maximum in a space,
called atomic orbital. Thus, atomic orbital is the region in the space around the nucleus
within which there is high probability of finding the electron.
An atom can have many possible number of orbitals. These orbitals can be
categorized on the basis of their size, shape or orientation. A smaller sized orbital means
there is a greater chance of getting an electron near the nucleus. The orbital wave
function or ψ is a mathematical function used for representing the coordinates of an
electron. The square of the orbital wave function represents the probability of finding an
electron. This wave function also helps us in drawing boundary surface diagrams.
Boundary surface diagrams of the constant probability density for different orbitals help
us understand the shape of orbitals. Let us represent the shapes of orbitals with the help
of boundary surface diagrams.
Determination of shapes of orbitals:
s-orbital: Boundary surface diagram for s orbital
looks like a sphere having the nucleus as its center
which in two dimensions can be seen as a circle.
Fig. 1.1 : s-orbital
1.1
F.E. Sem.-I Engineering Chemistry-I
1.2
Atomic and Molecular Structure
Hence, we can say that s-orbitals are spherically symmetric having the probability of
finding the electron at a given distance, equal in all the directions. The size of the s
orbital is also found to increase with the increase in the value of principal quantum
number (n), thus, 4s > 3s > 2s > 1s.
p-orbitals:
Each
p
orbital consists of two
sections known as lobes
which lie on either side of
the plane passing through the
nucleus.
Fig. 1.2 : p-orbital
The three p orbitals differ in the way the lobes are oriented; whereas they are identical
in terms of size, shape and energy.
As the lobes lie along one of the x, y or z-axis, these three orbitals are given the
designations 2px, 2py, and 2pz. Thus, we can say that there are three p orbitals whose axes
are mutually perpendicular. Similar to s orbitals, size, and energy of p orbitals increases
with an increase in the principal quantum number (4p > 3p > 2p).
d-orbital: Magnetic orbital quantum number for d orbitals is given as (-2, -1, 0, 1, 2).
Hence, we can say that there are five d-orbitals. These orbitals are designated as dxy, dyz,
dxz, dx2–y2 and dz2. Out of these five d orbitals, shapes of the first four d-orbitals are
similar to each other, which is different from the dz2 orbital. However, the energy of all
the five d orbitals is the same.
Fig. 1.3 : d-orbital
Comparison: Orbit and Orbital:
Orbit
Orbital
1. Orbit is a fixed circular path, 1. Orbital is a three-dimensional region
described by a moving electron around
in space around the nucleus, where
a nucleus.
there is maximum probability of
finding an electron.
2
2. Maximum capacity of any orbit is 2n 2. Maximum capacity of any orbital is
two elecrons.
electrons, where n is the number of
orbit.
F.E. Sem.-I Engineering Chemistry-I
1.3
Atomic and Molecular Structure
3. It has two dimensional representation.
3. It
has
three-dimensional
representation.
4. The position and velocity of a moving 4. The position and velocity of moving
electron, in an orbit can be calculated
electron, in an orbital can be
with great accuracy.
determined
within
limits
of
Heisenberg’s uncertainty principle.
5. The distance of the orbit from the 5. It is impossible to know the exact
nucleus for a given electron is fixed.
trajectory of an electron in an orbital
of an atom.
1.2 ELECTRONIC CONFIGURATION
The distribution of electrons in various orbitals is known as electronic configuration
of the atom. The atom is built up by filling electrons in various orbitals one at a time, by
placing it in the lowest energy orbital. The atom is said to be in the ground state when it
is in its lowest energy state. This is the most stable state for the atom. The filling of
orbitals by electrons in the ground state is determined by following rules.
1. Aufbau principle: The Aufbau principle states that in the ground state of an
atom, the orbital with lower energy is filled first, before the filling of the orbital with
higher energy. In other words, the electrons enter the orbitals in order of their increasing
energies. The order in which the energies of the orbitals increase, and hence the order in
which the orbitals are filled is as follows.
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s,…..
The order may be remembered by using the graphical representation shown in the
Fig. 1.4.
Fig. 1.4 : Order of filling the Orbitals
F.E. Sem.-I Engineering Chemistry-I
1.4
Atomic and Molecular Structure
Starting from the top, the direction of the arrows gives the order of filling of orbitals.
2. Pauli’s exclusion principle: According to this principle, no two electrons in an
atom can have the same set of four quantum numbers. It can also be stated as “Only two
electrons may exist in the same orbital and they must have opposite spin”. This means
that the two electrons can have the same value of three quantum numbers n, l and m, but
must have the opposite spin quantum number.
3. Hund’s Rule of maximum multiplicity: This rule deals with the filling of
electrons into the orbitals belonging to the same subshell i.e. orbitals of equal energy. It
states “pairing of electrons in the orbitals belonging to the same subshell (p, d, f) does not
take place until each orbital belonging to that subshell has got one electron each i.e. it is
singly occupied”.
E.g. There are three p orbitals (px, py, pz). According to Hund’s rule, each of the three
p orbitals must get one electron each of parallel spin before any one of them gets the
second electron of opposite spin.
1.3 MOLECULAR ORBITAL THEORY (MOT)
We know that the electrons have a dual character and they are considered either as
particles or as waves. An electron in an atom may therefore be described as occupying an
atomic orbital, or described by a wave function Ψ (an abstract mathematical concept). By
itself, Ψ has no physical significance except that it represents the amplitude of the wave,
and it is obtained as a solution of the Schrodinger wave equation, where Ψ2 stands for the
probability of finding an electron.
The fundamental assumption of MOT is that if two nuclei are placed at equilibrium
distance and electrons are added, they will go into molecular orbitals (which are in many
respects analogues to the atomic orbitals).
The basic assumption of MOT pinpoints that the valance electrons are essentially
associated with all the nuclei in the molecule. To visualize this, atomic orbitals from
different atoms have to be combined so that the molecular orbitals are produced.
Hence, we are going to discuss only one approximation which is rather simple,
qualitative and easy to understand, known as the linear combination of atomic orbitals,
abbreviated as LCAO method.
Qualitative molecular orbital (MO) theory was introduced by Robert S Mulliken and
Friedrich Hund, and a mathematical description was provided by Douglas Hartree and
Vladimir Fock in 1930. Basically, molecular orbital theory (MOT) was developed to
explain,
F.E. Sem.-I Engineering Chemistry-I
1.5
Atomic and Molecular Structure
1. The formation of chemical bond
2. Relative bond strength
3. Paramagnetic or diamagnetic nature of molecules.
The salient features of MOT are as follows :
1. When two atoms approach each other, their atomic orbitals lose their identity and
mutually overlap to form new orbitals called molecular orbitals.
2. The number of MO formed is equal to the number of overlapping atomic orbitals.
3. Maximum capacity of a MO is two electrons with opposite spins. MO is a
polycentric region in space defined by its size and shape, associated with two or
more atoms in a molecule and each has a capacity of two electrons with opposite
spins.
4. Only atomic orbitals having comparable energies as well as proper orientations
interact significantly. For example, 1s atomic orbital can overlap with 1s atomic
orbital but not with 2s atomic orbital or 2s atomic orbital can overlap with 2px
atomic orbital.
5. When two atomic orbital overlap, they interact to form two molecular orbitals, in
the following two ways.
x
When the atomic orbitals overlap in-phase it leads to an increase in the
intensity of the negative charge in the region of overlap. The molecular orbital
thus formed has lower potential energy than the separate atomic orbitals and is
called bonding molecular orbital. The difference in energy between the
combining atomic orbitals and the bonding molecular orbital formed is called
the stabilization energy. Thus bonding molecular orbital stabilizes the
molecule.
Characteristics of bonding molecular orbital:
(a) It possesses lower energy than that of the combining atomic orbitals.
(b) It imparts stability to the molecules.
(c) Every electron in it contributes to the attraction of two combining atoms.
(d) It possesses high electron density between the two nuclei.
(e) It is formed when the lobes of combining atomic orbitals have the same
signs.
x
When the atomic orbitals overlap out-of-phase, it leads to a decrease in the
intensity of the negative charge between the nuclei and leads to higher
potential energy. The molecular orbital of this type is called antibonding
molecular orbital. The difference in energy between the antibonding
F.E. Sem.-I Engineering Chemistry-I
1.6
Atomic and Molecular Structure
molecular orbital and combining atomic orbitals is called destabilization
energy. Thus antibonding molecular orbital destabilizes the molecule.
Characteristics of antibonding molecular orbital:
(a) It possesses higher energy than that of the combining atomic orbitals.
(b) It imparts instability to the molecules.
(c) Every electron in it contributes to the repulsion of two combining atoms.
(d) It possesses low electron density between the two nuclei.
(e) It is formed when the lobes of combining atomic orbitals have the opposite
signs.
6. The shape of MO formed depends on the type of combining atomic orbitals.
7. The bonding MO are represented by V, S, G etc., whereas antibonding MO are
represented by V*, S*, G* etc.
8. Inner orbital MO which do not take part in bond formation are called non bonding
MO.
1.4 LINEAR COMBINATION OF
ATOMIC ORBITALS (LCAO) METHOD
Molecular orbitals of a molecule, i.e. solutions to the molecular Schrodinger equation
can be obtained by a method known as linear combination of atomic orbitals (LCAO). As
per this method, linear combination is brought about by taking the summation or
difference of atomic orbital wave functions. Consider two atoms A and B which have
atomic orbitals described by the wave functions ΨA and ΨB, then the wave function for
the molecule AB can be obtained by a linear combination of the atomic orbitals ψA and
ψB.
Ψ = N(C1ΨA ± C2ΨB)
Where Ψ is the wave function of MO, ΨA and ΨB are wave functions of AOs
belonging to atom A and B respectively, C1 and C2 are constants chosen to give minimum
energy for Ψ, N is a normalizing constant chosen to ensure that the probability of finding
an electron in the whole of the space is unity.
The equation clearly shows that linear combination gives rise to formation of two
MOs. One is obtained by taking the sum of two AO wave functions i.e. Ψb = N(C1 ΨA +
C2ΨB). This results in an increased electron density between the two nuclei. This is the
bonding MO having a lower energy than the combining AOs. The other MO is obtained
by taking the difference of two AO wave functions i.e. Ψa = N(C1 ΨA - C2ΨB). It has a
node with zero electron density between the nuclei due to cancellation of two wave
F.E. Sem.-I Engineering Chemistry-I
1.7
Atomic and Molecular Structure
functions. This is the antibonding MO having a higher energy than the combining AOs.
The following energy level diagram (Fig. 1.5) illustrates the process.
Fig. 1.5 : Formation of bonding and antibonding MOs
Significance of bonding and antibonding MOs :
Taking squares of the bonding MO wave function (Ψb), we get
(Ψb)2 = ΨA2 + ΨB2 + 2ΨAΨB (for identical atoms C1 = C2 = 1).
The terms ΨA2 and ΨB2 indicate electronic charge densities of wave functions ΨA and
ΨB belonging to isolated atoms A and B respectively. (Ψb)2 is the electronic charge
density of the wave function Ψb, i.e., the bonding MO. Thus, the more is the value of
(Ψb)2, the more is the charge density between the two combining nuclei. It is evident from
the equation that (Ψb)2 > ΨA2 + ΨB2 by a term 2ΨAΨB. This term 2ΨAΨB results from the
overlap of AOs and is called overlap integral. Larger is the overlap integral more is the
charge density between the nuclei and stable is the bond formed. The plot of electron
charge density as a function of inter-nuclear distance for bonding MO is shown in the
Fig. 1.6.
Fig. 1.6 : Plot of Electron Charge Density for bonding MO
Taking square of the antibonding MO wave function (Ψa), we get
(Ψa)2 = ΨA2 + ΨB2 - 2ΨAΨB
It is evident from the above equation that (Ψa)2 ΨA2 + ΨB2 by a term 2ΨAΨB. That’s
how, in the antibonding MO, there is less charge density between the nuclei. Therefore,
the energy of the antibonding MO is more than the sum of the energies of the two
F.E. Sem.-I Engineering Chemistry-I
1.8
Atomic and Molecular Structure
interacting atoms A and B. The charge density touches the axis at mid-point between the
nuclei. This mid-point is called a node and at this point, the electron charge density is
zero. The plot of electron charge density as a function of inter-nuclear distance for
antibonding MO is shown in the Fig. 1.7.
Fig. 1.7 : Plot of Electron Charge Density for antibonding MO
1.5 SHAPES OF MOLECULAR ORBITALS
Conditions for the formation of molecular orbitals:
There are certain limitations to the combination of atomic orbitals which are as
follows :
1. The energies of combining atomic orbitals should be of similar magnitude.
Formation of a homonuclear diatomic molecule is not possible if 1s-orbital of one atom
overlaps with 2s-orbital of another atom.
2. According to rule of maximum overlap, combination of atomic orbitals takes
place only if overlapping is to a considerable extent. Greater is the overlapping of atomic
orbitals, the greater is the buildup of the charge between the nuclei.
3. The combining atomic orbitals should possess the symmetry about the molecular
axis. A px-orbital will combine with an s-orbital as they have same symmetry. But a pzorbital of an atom will not overlap with an s-orbital of another atom as they don’t have
same symmetries.
When two atomic orbitals overlap along the internuclear axis, then the resulting
molecular orbital is called sigma (V) molecular orbital. When two atomic orbitals overlap
sideways, then the resulting molecular orbital is called pi (S) molecular orbital. Some
simple cases of combinations of atomic orbitals to form molecular orbitals are given.
(a) s-s combination of orbitals (1s with 1s or 2s with 2s) : The two molecular
orbitals formed may be designated as bonding V (1s) or V (2s) and antibonding
F.E. Sem.-I Engineering Chemistry-I
1.9
Atomic and Molecular Structure
V* (1s) or V* (2s). This indicates that the overlap is along internuclear axis. V (1s)
is formed by constructive overlapping and V* (1s) is formed by destructive
overlapping of the two s-orbitals as shown in the Fig. 1.8.
Fig. 1.8 : s-s overlap of orbitals
(b) coaxial s-p combination of orbitals (s-orbital with px orbital) : An s-orbital
may combine with a p-orbital provided that the lobes of the p-orbital are pointing
along the axis joining the nuclei. When the lobes which overlap have the same
sign, results in a bonding MO with an increased electron density between the
nuclei. When the overlapping lobes have opposite signs, it gives an antibonding
MO with reduced electron density in between the nuclei as shown in the Fig. 1.9.
Fig. 1.9: Coaxial s-p overlap of orbitals
(c) axial p-p combination : Consider the combination of two p-orbitals which have
lobes pointing along the axis joining the nuclei. In this case, both a bonding MO
and an antibonding MOs are produced as shown in the Fig. 1.10.
Fig. 1.10 : Axial p-p overlap of orbitals
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1.10
Atomic and Molecular Structure
(d) Side-to-side p-p combination: Consider the combination of two orbitals which
both have lobes perpendicular to the axis joining the nucleus. Lateral overlap of
orbitals results in the formation of S bonding MO and S antibonding MO as
shown in the Fig. 1.11.
Fig. 1.11 : Side-to-side p-p overlap of orbitals
Molecular orbitals give the probability of distribution of electron clouds around a
group of nuclei. Shape and size of a molecular orbital depends upon the shape, size,
number and orientation of the combining atomic orbitals. Each molecular orbital can
accommodate a maximum of two electrons with opposite spins.
The filling of molecular orbital takes place as:
(a) The aufbau principle: Molecular orbitals are filled in order of increasing energy.
(b) Pauli Exclusion Principle: Two electrons in the same orbital cannot have set of
all the four quantum numbers identical. It means maximum number of electrons
in an orbital is two, and they must have opposite spins.
(c) Hund’s rule of maximum multiplicity: pairing of electrons in the degenerate
molecular orbitals does not occur until each of them has one electron each.
The relative energies of molecular orbitals in increasing order have been found to be
as follows:
(a) For H2 to N2: V (1s) V* (1s) V (2s) V* (2s) [S (2px) = S (2py)] V (2pz) [S* (2px) = S* (2py)] V* (2pz)
(b) For O2 to Ne2: V (1s) V* (1s) V (2s) V* (2s) V (2pz) [S (2px) = S (2py)]
[S* (2px) = S* (2py) V* (2pz)]
The basis of the above orders is:
x
Within the same pair of molecular orbitals like (V and V*) or (S and S*), the
antibonding molecular orbitals is less stable and has higher energy than the
corresponding bonding molecular orbitals.
F.E. Sem.-I Engineering Chemistry-I
1.11
Atomic and Molecular Structure
x
Pair of molecular orbitals (V1s and V*1s) possesses lower energies than pair of
molecular orbitals (V2s and V*2s).
x
Degenerate molecular orbitals have similar energies. e.g.,
[S (2px) = S (2py)] and [S* (2px) = S* (2py)]
x
The extent of stabilization and destabilization during the formation of bonding and
antibonding molecular orbitals is equal.
x
V molecular orbital is more stable compared to S molecular orbital.
This explains, from O2 to Ne2, the order of molecular orbital energy level is:
V2pz [S (2px) = S (2py)] [S* (2px) = S* (2py)] V*(2pz)
Bond order of a molecule according to MOT:
Bond order of a molecule is a measure of strength of a bond between two atoms.
It is defined mathematically as:
1
Bond order = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number of
electrons in antibonding molecular orbitals (Na)]
Nb – Na
=
2
(a) When the bond order is zero, there is no net force of attraction due to equal and
opposite influence of identical number of bonding and antibonding molecular
orbitals. Hence, the molecule is unstable and it does not exist.
(b) A positive value of bond order reveals that the molecule exists and is stable.
(c) A bond order of +1, +2, +3 indicates that the molecule is formed by joining of
two atoms by a single, double or triple bond respectively.
(d) Greater the value of bond order, greater is the stability of the molecule. Thus
higher the bond order, higher is the dissociation energy.
(e) Bond length is inversely proportional to bond order.
(f) Higher the bond order higher is the dissociation energy.
Magnetic nature of a molecule according to MOT:
A molecule is paramagnetic in nature, if it contains one or more unpaired electrons in
its molecular orbitals. Greater the number of unpaired electrons in the molecular orbitals
of a substance, higher is its paramagnetic character. But if the molecule does not have
any unpaired electron, then it is diamagnetic in nature.
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1.12
Atomic and Molecular Structure
Comparisons:
Bonding Molecular Orbitals and Anti-Bonding Molecular Orbitals:
Bonding Molecular Orbitals
Anti-Bonding Molecular Orbitals
1. Formed by addition overlap of atomic 1. Formed by subtraction overlap of
orbitals.
atomic orbitals.
2. Possesses lower energy than that of 2. Possesses higher energy than that of
the atomic orbitals.
the atomic orbitals.
3. Ψb = ψA + ψB
3. Ψa = ψA - ψB
4. Imparts stability to the molecule.
4. Imparts instability to the molecule.
5. Possesses high electron-density in the 5. Possesses low electron-density in the
region between the two nuclei.
region between the two nuclei.
6. It is formed when lobes of the 6. It is formed when lobes of the
combining atomic orbitals possess
combining atomic orbitals possess
same sign.
opposite sign.
V and S Molecular orbitals:
1.
2.
3.
4.
Sigma Molecular Orbital
The sigma molecular orbital is
obtained when the atomic orbitals
overlap collinearly to a large extent.
For sigma overlap, the lobes of atomic
orbitals point along the internuclear
axis.
Sigma molecular orbital is denoted by
V.
For sigma molecular orbital, ψ is
along the internuclear axis and the
electron density ψ2 is distributed
uniformly along the axis.
Pi Molecular Orbital
1. The pi molecular orbital is obtained
when atomic orbitals overlap along the
side to a lesser extent.
2. For pi overlap, the lobes of the atomic
orbitals are perpendicular to the axis.
3. Pi molecular orbital is denoted by S.
4. For pi molecular orbital, ψ is zero
along the internuclear axis and the
electron density ψ2 on the axis is zero.
1.6 MOLECULAR ORBITAL TREATMENT FOR
HOMONUCLEAR DIATOMIC MOLECULES
(1) Hydrogen Molecule, H2 :
Hydrogen molecule (H2) is formed by the overlap of 1s atomic orbitals of two
hydrogen atoms having one electron each. Two molecular orbitals viz. bonding and
antibonding are generated from the overlap of these atomic orbitals. In a hydrogen
F.E. Sem.-I Engineering Chemistry-I
1.13
Atomic and Molecular Structure
molecule, there are two electrons with opposite spins which are accommodated in lower
energy bonding (V1S) molecular orbital as shown in the Fig. 1.12.
Fig. 1.12 : MO energy level diagram for H2
Hence, molecular orbital electronic configuration of H2 molecule = V1S2.
1
Bond order (H2) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
2–0
Nb – Na
=
=
2
2 = 1
As the bond order is greater than zero therefore H2 molecule is stable and it exists.
Magnetic Behaviour of H2: H2 molecule is diamagnetic as both the electrons are paired.
(2) Lithium molecule, Li2 :
Lithium molecule is formed by the
overlap of two lithium atoms each having
the electronic configuration of 1s2 2s1. So,
there are total six electrons which have to
be accommodated in 4 molecular orbitals
viz. V1s, V*1s, V2s and V*2s.
The six electrons are filled in these 4
molecular orbitals according to aufbau
principle and Pauli’s exclusion principle as
shown in the Fig. 1.13.
Hence, Molecular orbital electronic
configuration of Li2 molecule = V1s2 V*1s2
V2s2 = KKV2s2.
Fig. 1.13 : MO energy level
diagram for Li2
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1.14
Atomic and Molecular Structure
Since, the inner shell of filled V1s and V*1s molecular orbitals do not contribute to the
bonding, and is written as KK which means K-shell is completely filled.
1
Bond Order (Li2) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
1
= 2 (Nb – Na)
1
= 2 (4 – 2) = 1
Positive bond order for Li2 suggests that it is stable molecule and it exists.
In fact, Li2 molecules do exist in the vapour state. However, it is energetically more
favourable for lithium to form a metallic structure in the solid.
Magnetic behaviour of Li2: Li2 molecule is diamagnetic since both the electrons are
paired.
(3) Beryllium molecule, Be2 :
Beryllium molecule may be formed by the overlap of two beryllium atoms each
having the electronic configuration of 1s2 2s2. So, there are total of eight electrons which
have to be accommodated in 4 molecular orbitals viz. V1s, V*1s, V2s and V*2s.
The eight electrons are filled in these 4 molecular orbitals according to auf bau
principle and Pauli’s exclusion principle as shown in the Fig. 1.14.
Hence, Molecular orbital electronic configuration of Be2 molecule = V1s2 V*1s2 V2s2
* 2
V 2s = KKV2s2V*2s2.
Fig. 1.14 : MO energy level diagram for Be2
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1.15
Atomic and Molecular Structure
Since, the inner shell of filled V1s and V*1s molecular orbitals do not contribute to the
bonding and is sometimes written as KK which means K-shell is completely filled.
1
Bond Order (Be2) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
1
= 2 (Nb – Na)
1
= 2 (4 – 4)
=0
The bonding effect is cancelled by antibonding effect where equal number of bonding
and antibonding electrons is present in Be2. Since the bond order is zero, beryllium is an
imaginary nonexistent molecule.
(4) Oxygen molecule, O2 :
Oxygen molecule is formed by the overlap of two oxygen atoms each having
electronic configuration as 1s2 2s2 2px2, 2py1, 2pz1. The total of 16 electrons in oxygen
molecules can be filled according to auf bau principle, pauli’s exclusion principle and
Hund’s rule of maximum multiplicity. The filled molecular orbitals of oxygen molecule
are shown in the Fig. 1.15.
Hence, the molecular orbital electronic
configuration of oxygen molecule is
O2 [KK (V2s)2 (V*2s)2 (V2px)2 (S2py
S2pz)4 (S*2py S*2pz)2]
In oxygen, the inner shell does not
participate in bonding. The bonding and
antibonding 2s orbitals cancel each other.
A V bond results from the filling of V2px2.
Since S*2py1 is half filled and therefore
cancels half the effect of the completely
filled S2py2 orbital, half of a S bond results.
Similarly, another half of a S bond arises
from S2pz2 and S*2pz1 giving a total of 1 +
1
1
+
2
2 = 2 bonds. Thus, there are two
covalent bonds in oxygen one sigma and
one pi.
Fig. 1.15 : MO energy level
diagram for O2
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1.16
Atomic and Molecular Structure
1
Bond Order (O2) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
1
= 2 (Nb – Na)
1
= 2 (8 – 4)
=2
Positive bond order for O2 suggests that it is stable molecule and it should exist.
Magnetic behaviour of O2: The antibonding S*2py and S*2pz orbitals are singly
occupied in accordance with Hund’s rule. Since there are two unpaired electrons with
parallel spins, oxygen molecule shows paramagnetism. This was the success of MOT as
valance bond theory failed to explain the paramagnetism of oxygen.
1.7 MOLECULAR ORBITAL TREATMENT FOR
HETERONUCLEAR DIATOMIC MOLECULES
1. NO Molecule:
NO molecule is formed by the overlap of nitrogen and oxygen atom. The nitrogen
atom has seven electrons and the oxygen atom has eight electrons. Therefore, there are
fifteen electrons in the NO molecule.
Fig. 1.16 : MO energy level diagram for NO
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1.17
Atomic and Molecular Structure
Hence, the molecular orbital electronic configuration of NO molecule is
[KK (V2s)2 (V*2s)2 (V2px)2 (S2py S2pz)4 (S*2py1 S*2pz0)]
The filled molecular orbitals of NO molecule are shown in the Fig. 1.16.
The inner shell is nonbonding. The bonding and antibonding 2s orbitals cancel each
other. A V bond is formed by the filled V2px2 orbital. A S bond is formed by the filled
S2pz2 orbital. The half-filled S*2py1 half cancels the filled S2py2 orbital, giving half a
bond.
1
Bond Order (NO) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
1
= 2 (Nb – Na)
1
= 2 (8 – 3)
= 2.5
The bond order for NO molecule is 2.5 i.e. in between a double and triple bond.
Magnetic behaviour of NO: The NO molecule is paramagnetic since it contains an
unpaired electron.
2. CO Molecule:
CO molecule is formed by the overlap of carbon and oxygen atom. The carbon atom
has six electrons and oxygen atom has eight electrons. Therefore, there are fourteen
electrons in the CO molecule.
Hence, the molecular orbital electronic configuration of CO molecule is
[KK (V2s)2 (V*2s)2 (V2px)2 (S2py S2pz)4]
The filled molecular orbitals of CO molecule are shown in the Fig. 1.17.
The inner shell is nonbonding. The bonding and antibonding 2s orbitals cancel each
other. A V bond is formed by the filled V2px2 orbital. Two S bonds are formed by the
filled S2py2 and S2pz2 orbitals.
F.E. Sem.-I Engineering Chemistry-I
1.18
Atomic and Molecular Structure
Fig. 1.17 : MO energy level diagram for CO
1
Bond Order (CO) = 2 [Number of electrons in bonding molecular orbitals (Nb) – Number
of electrons in antibonding molecular orbitals (Na)]
1
= 2 (Nb – Na)
1
= 2 (8 – 2)
=3
The bond order for CO molecule is 3. Thus in CO molecule, there is a triple bond out
of which one is a sigma bond and the other two are pi bonds and therefore CO is a very
stable molecule.
Magnetic behaviour of CO: As in the CO molecule all the electrons are paired, it is
diamagnetic molecule.
1.8 REVIEW QUESTIONS
1. Write a note on atomic orbital.
Ans. Refer Section 1.1.
F.E. Sem.-I Engineering Chemistry-I
1.19
Atomic and Molecular Structure
2. What is the electronic configuration?
Ans. Refer Section 1.2.
3. Explain the rules for filling of atomic orbitals.
Ans. Refer Section 1.2.
4. What are the main features of MOT?
Ans. Refer Section 1.3.
5. Explain the characteristics of bonding MO.
Ans. Refer Section 1.3.
6. Explain the characteristics of antibonding MO.
Ans. Refer Section 1.3.
7. What are the conditions for the formation of MO?
Ans. Refer Section 1.5.
8. Distinguish between orbit and orbital.
Ans. Refer Section 1.1.
9. Distinguish between bonding MO and antibonding MO.
Ans. Refer Section 1.5.
10. Explain s-s overlap of orbitals.
Ans. Refer Section 1.5.
11. Explain coaxial s-p overlap of orbitals.
Ans. Refer Section 1.5.
12. Explain axial p-p overlap of orbitals.
Ans. Refer Section 1.5.
13. Write a note on bond order.
Ans. Refer Section 1.5.
14. With the help of MO energy level diagram explain formation of H2.
Ans. Refer Section 1.6.
15. With the help of MO energy level diagram explain formation of Li2.
Ans. Refer Section 1.6.
16. With the help of MO energy level diagram explain formation of Be2.
Ans. Refer Section 1.6.
17. With the help of MO energy level diagram explain formation of O2.
Ans. Refer Section 1.6.
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1.20
Atomic and Molecular Structure
18. Discuss the formation of NO on the basis of MOT.
Ans. Refer Section 1.7.
19. Discuss the formation of CO on the basis of MOT.
Ans. Refer Section 1.7.
20. Differentiate between V and S MO.
Ans. Refer Section 1.5.
21. Explain why O2 is paramagnetic.
Ans. Refer Section 1.6.
22. Explain why Be2 doesn’t exist.
Ans. Refer Section 1.6.
23. Calculate the bond order and magnetic behaviour for NO molecule.
Ans. Refer Section 1.7.
24. Calculate the bond order and magnetic behaviour for CO molecule.
Ans. Refer Section 1.7.
25. Calculate the bond order and magnetic behaviour for Li2 molecule.
Ans. Refer Section 1.6.
CHAPTER
2
AROMATIC SYSTEMS AND THEIR
MOLECULAR STRUCTURE
SYLLABUS
x
x
Define Aromaticity, Huckel’s rule
Structure and bonding of benzene and pyrrole
2.1 INTRODUCTION
In earlier times, during the study of organic chemistry some sweet smelling
compounds were obtained from natural sources. These compounds showed different
properties compared to aliphatic compounds. They were called aromatic (Greek, aroma
= pleasant smell) compounds. Further studies revealed that these compounds contain
benzene rings involving six carbon atoms in a ring. Later on, a large number of aromatic
compounds were discovered which lacked the sweet smell. Thus this word aroma lost it’s
significance.
2.2 AROMATICITY
Aromatic compounds form the most important group of organic compounds and
benzene is the simplest member of this group. The determination of the structural formula
of benzene was not as simple as its molecular formula C6H6. The molecular formula
shows that benzene contains eight hydrogen atoms less than a saturated hydrocarbon
containing six carbon atoms i.e. Hexane (C6H14). This leads to expectation that it should
be a highly unsaturated compound containing double or triple bonds between its carbon
atoms. In such a case, benzene was supposed to give addition reactions which are typical
of alkenes or alkynes. Further, it should be readily oxidisable. Surprisingly, benzene was
found to behave in an almost exactly opposite manner which is clear from the following
Table 2.1.
2.1
F.E. Sem.-I Engineering Chemistry-I
2.2
Aromatic Systems and their Molecular Structure
Table 2.1 : Comparison of properties of benzene with alkenes
Reagent
1.
KMnO4
Reaction with alkenes
Reaction with benzene
Rapid oxidation
No reaction
(cold, dilute, aqueous)
2.
Br2 in CCl4 (in the dark)
Rapid addition
No reaction
3.
HBr
Rapid addition
No reaction
4.
H2 in presence of Ni
Rapid hydrogenation at Slow hydrogenation at
25oC and under a 100-200oC and under a
pressure of about 1 atm
pressure of about 100 atm
Organic compounds which resemble benzene in their chemical behaviour are called
aromatic compounds. They exhibit certain characteristic properties which are different
from those of aliphatic and alicyclic compounds. These characteristic properties are
collectively referred to as aromaticity or aromatic character. Aromaticity may
therefore be defined as the ability of many cyclic compounds containing conjugation in
their rings to undergo substitution reactions rather than the addition reactions, to resist
oxidation and to exhibit a marked stability of their rings.
Characteristics of aromatic compounds:
1. They are highly unsaturated as shown by the lesser number of hydrogen atoms in
their molecular formulae.
2. They are cyclic compounds with five, six or seven membered rings.
3. Their molecules are flat or nearly flat as shown by physical methods such as x-ray
and electron diffraction methods.
4. They undergo readily certain electrophilic substitution reactions such as nitration,
halogenation, sulphonation, Friedel-Crafts alkylation and acylation etc.
5. Although their molecular formulae suggest a high degree of unsaturation, yet they
do not respond to tests characteristics of unsaturated compounds. They fail to decolourize
an aqueous solution of potassium permanganate (Baeyer’s test).
6. They are associated with high thermodynamic stability as is indicated by their low
heats of combustion and hydrogenation.
2.3 HUCKEL’S RULE
This rule is based upon molecular orbital treatment and is employed for predicting
aromaticity in organic compounds. The main theoretical requirements for a substance to
possess aromaticity are :
F.E. Sem.-I Engineering Chemistry-I
2.3
Aromatic Systems and their Molecular Structure
1. The molecule or ion must be flat or nearly flat.
2. It must have cyclic clouds of delocalized S-electrons above and below the plane
of the molecule. The S-electron clouds should encompass all the carbon atoms of
the cyclic system.
3. The S-clouds in the molecule or ion must contain a total of (4n + 2) S electrons
where n = 0, 1, 2, 3, … etc.
The above requirements are collectively known as Huckel rule or (4n + 2) rule.
Huckel rule for aromaticity states that a cyclic system containing (4n + 2) S electrons,
where n is an integer, such as 0, 1, 2, 3 etc, would have special stability. This special
stability is called aromaticity.
This rule can be applied successfully to cyclic polyenes, polycyclic compounds and
non-benzenoid compounds to predict armaticity in them.
The rule is based on quantum mechanics and is strongly supported by facts. The
values of n are called Huckel numbers.
Thus,
n=0
4n + 2 = 2
n=1
4n + 2 = 6
n=2
4n + 2 = 10
n=3
4n + 2 = 14 etc.
Therefore, for a molecule to be aromatic, the number of S-electrons on the ring
should be either 2 or 6 or 10 or 14, etc.
Thus, benzene, naphthalene and anthracene exhibit aromaticity as they contain 6, 10
and 14, S-electrons respectively as shown in the Fig. 2.1.
Benzene
Naphthalene
Anthracene
Fig. 2.1 : Aromatic compounds containing 6S, 10S and 14S electrons
Theoretical justification of Huckel rule: On quantum mechanical ground, an
aromatic system possesses a close shell of S-electrons corresponding to inert gas
configuration. Huckel pointed out that energies of molecular orbitals of aromatic systems
have a pattern where there is always one orbital of lowest energy followed by degenerate
(having same energy) pair of orbitals in order of increasing energy. Filling of orbitals
takes place as per Hund’s rule. For a conjugated planer, monocyclic system having 2, 6,
10, 14, etc. S-electrons, the lowest energy MO and all other degenerate pairs of MOs
F.E. Sem.-I Engineering Chemistry-I
2.4
Aromatic Systems and their Molecular Structure
would be occupied by two electrons each. Such systems are stable due to their closed
shell filling of orbitals. For a monocyclic planer conjugated system with 4, 8, 12, etc Selectrons, there will be always two singly occupied degenerate orbitals. Such systems are
highly unstable and are anti-aromatic, e.g. cyclobutadiene (4 S-electrons) and cyclooctatetraene (8 S-electrons).
Study of some compounds for their aromaticity:
In spite of conjugation in the ring and the planarity of
the ring, if the number of S-electrons on the ring is not in
accordance with Hackel’s rule, the compound would not
exhibit aromaticity, e.g., cyclopropenyl anoin. It satisfies
both the conditions of conjugation and planarity of the
Fig. 2.2 :
ring. However, the number of S-electrons on the ring
Cyclopropenyl anion
being four, which is not a Huckel number, it does not
show aromaticity. It is an anti-aromatic compound.
The cycloheptatriene has three conjugated double
bonds. Thus there are six S-electrons. Even though six is
a Huckel number, cycloheptatriene is not an aromatic
compound since for a cyclic compound to be aromatic, all
the double bonds must be continuously conjugated. Only
Fig. 2.3 : Cycloheptatriene
in that case, there will be a circular cloud of S-electrons.
Cyclopentadiene contains only four S-electrons which is not a Huckel number. As it
does not satisfy Huckel’s rule, it is anti-aromatic. The corresponding Cyclopentadienyl
anion will have six electrons (i.e. four S-electrons and a pair of electrons) for resonance.
That satisfies Huckel’s rule for n = 1. Therefore, cyclopentadienyl anion is an aromatic
compound.
Cyclopentadienyl anion
Cyclopentadiene
Fig. 2.4
2.4 STRUCTURE AND BONDING OF BENZENE
According to Kekule, six carbon atoms of benzene are linked to each other by
alternate single and double covalent bonds to form a hexagonal ring as shown in the
Fig. 2.5.
F.E. Sem.-I Engineering Chemistry-I
2.5
Aromatic Systems and their Molecular Structure
Fig. 2.5 : Kekule structure of benzene
Each carbon atom is linked to one hydrogen atom thus conforming to its molecular
formula.
Kekule structure does not explain the following:
1. Benzene does not give addition reactions and fails to decolorize Baeyer’s reagent.
Instead, it readily undergoes electrophilic substitution reactions in which the benzene ring
is retained.
2. On the basis of Kekule structure, the heat of combustion of benzene is expected to
be 3449.0 kJ/mol, however the experimental value is 3298.5 kJ/mol. Benzene is therefore
more stable than expected from Kekule structure.
3. On the basis of Kekule structure, the heat of hydrogenation of benzene is expected
to be 358.0 kJ/mol, however the experimental value is 208.5 kJ/mol. This shows that
benzene is more stable than expected from Kekule structure.
4. X-ray diffraction studies show that all the carbon-carbon bond lengths are
identical and lie in between that of single and double bonds. This is not in accordance
with Kekule structure which contains two kinds of carbon-carbon bonds.
Molecular orbital structure of benzene:
Structure of benzene can be best described by
using the orbital concept. The orbital picture of
benzene shows that each carbon atom in benzene
is sp2 hybridised. The C-H bonds in benzene are
sp2-s, V bond. The C-C bonds are sp2-sp2, V
bonds. The sp2 hybridisation of the carbon atoms
indicates that all the carbon atoms of the ring are
in the same plane with their bonds separated by
angles of 120o as shown in the Fig. 2.6.
Fig. 2.6
F.E. Sem.-I Engineering Chemistry-I
2.6
Aromatic Systems and their Molecular Structure
The six hydrogen atoms radiate from the carbon atom in the same plane like the
spokes of a wheel. The whole molecule is planer. Each carbon atom has a pure p orbital
with one electron. It overlaps laterally with p-orbitals of the adjacent carbon atoms on
either side. It results in the formation of three S bonds. The 6S electrons of benzene are
enough to fill all the bonding S- molecular orbitals. However, the p orbitals may overlap
simultaneously with both adjacent p orbitals as in the Fig. 2.7.
Fig. 2.7
This results in two continuous doughnut-shaped S-electron clouds, one lying above
and the other lying below the plane of the ring as shown in the Fig. 2.8. This participation
of the p orbital electrons is called delocalization and is responsible for the enhanced
stability of the benzene ring.
Fig. 2.8
Thus in terms of molecular orbital theory, resonance is delocalization of electrons in
number of adjacent molecular orbitals.
Thus, for a compound to be aromatic it is not just enough to have Huckel number of
electrons on the planer ring. Therefore, there must be a circular delocalized cloud of S
electrons. Thus the Huckel rule is modified as follows in terms of molecular orbital
theory. The cyclic unsaturated compounds will be aromatic if they contain (4n + 2)
delocalized S electrons in the form of a circular cloud above and below the plane of the
ring.
F.E. Sem.-I Engineering Chemistry-I
2.7
Aromatic Systems and their Molecular Structure
Evidence in support of orbital structure of benzene:
1. Unusual stability: Benzene molecule exhibits unusual stability and resists the
formation of addition products. This can easily be understood in terms of delocalization
of S-electrons which is responsible for aromaticity.
2. Isomer number: According to orbital concept, all the six carbons in benzene are
completely equivalent. Similarly, all the six hydrogen atoms also occupy identical
positions. Thus, benzene should form only one monosubstituted and three disubstituted
products. This has been found to be in actual practice.
3. Electrophilic substitution reactions: There are two continuous ring like Selectron clouds one above and the other below the plane of carbon atoms. The S-electrons
are easily attacked by electrophiles. Hence, benzene undergoes electrophilic substitution
reactions.
2.5 STRUCTURE AND BONDING OF PYRROLE
Heterocyclic compounds are cyclic compounds containing hetero atoms such as
nitrogen, sulphur, oxygen as a part of the ring system. The heterocyclic compounds may
be classified on the basis of the ring size, number of hetero atoms in the ring, number of
rings, and nature of the hetero atom. Pyrrole is five membered ring made up of four
carbon atoms and the hetero atom nitrogen.
Fig. 2.9 : Pyrrole
Since pyrrole has two conjugated double bonds, it is expected to exhibit the properties
of conjugated dienes. Being unsaturated compound, it should undergo addition reaction.
Also, pyrrole is expected to behave as a secondary amine. However, pyrrole does not
show any of these expected properties. It does not give addition reactions, with few
exceptions. It readily gives substitution reactions like nitration, sulphonation, FriedelCrafts reactions, Riemer-Tiemann reaction etc. Instead of showing the properties of a
secondary amine, pyrrole is found to be an extremely weak base. Pyrrole has lower heats
of hydrogenation and combustion than expected. This indicates that it is more stable than
expected. Pyrrole has planar ring structure i.e. all the atoms of the ring including the
nitrogen atom are in the same plane. There are two double bonds i.e. 4 S-electrons. These
F.E. Sem.-I Engineering Chemistry-I
2.8
Aromatic Systems and their Molecular Structure
double bonds are conjugated with the lone pair of electrons. Thus there are total six
electrons. Therefore, pyrrole satisfies Huckel’s rule for n = 1 and hence it is aromatic in
nature.
Molecular orbital structure of pyrrole:
All the atoms of pyrrole heterocyclic rings are in the same plane, and angles between
the bonds in the ring are 120o. Further, all the atoms including the nitrogen atom are sp2
hybridised. Two sp2 orbitals of each atom are utilized to form V bond between the carbon
atoms and also the nitrogen atom. The third sp2 orbital forms V bond between carbon and
hydrogen atoms. Then each of the carbon atoms contains a p orbital with one electron and
the nitrogen atom contains a p orbital with a pair of electrons. These p orbitals overlap
with adjacent p orbitals on either side to produce circular clouds of S-electron. There will
be delocalized circular cloud of six S-electrons above and below the plane of the ring.
Thus, Huckel’s rule is fully satisfied and hence pyrrole ring (Fig. 2.10) shows
aromaticity.
Fig. 2.10
Due to the presence of the S-electron cloud, the molecule is reactive towards
electrophilic reagents. If this S-electron cloud is disturbed, the aromaticity is lost.
Therefore, pyrrole generally gives substitution reactions. In the case of pyrrole, the lone
pair of electrons on the nitrogen atom is also involved in the S-electron cloud. Therefore,
it is not available to be shared with a proton. Hence, pyrrole is an extremely weak base.
2.6 REVIEW QUESTIONS
1. What is aromaticity?
Ans. Refer Section 2.2.
2. Describe the aromatic character of benzene.
Ans. Refer Section 2.4.
F.E. Sem.-I Engineering Chemistry-I
2.9
Aromatic Systems and their Molecular Structure
3. State and explain Huckel’s rule.
Ans. Refer Section 2.3.
4. What are the conditions which must be satisfied for a compound to show
aromaticity?
Ans. Refer Section 2.3.
5. Explain why cycloheptatriene is anti-aromatic compound.
Ans. Refer Section 2.3.
6. Comment on aromatic character of cyclopentadiene and cyclopentadiene anion.
Ans. Refer Section 2.3.
7. Write a note on Huckel’s rule and aromaticity.
Ans. Refer Section 2.2 and 2.3.
8. Explain the aromaticity of pyrrole.
Ans. Refer Section 2.5.
9. Which of the following molecules are aromatic and why?
(i) Cyclo-octatetraene, (ii) Cyclobutadiene, (iii) Cyclopropenyl anion.
Ans. Refer Section 2.3.
10. What are limitations of Kekule structure of benzene?
Ans. Refer Section 2.4.
CHAPTER
3
INTERMOLECULAR FORCES AND
CRITICAL PHENOMENA
SYLLABUS
x
x
Ionic, dipolar and Van der Waal’s interactions
Equations of state of real gases and critical phenomena
3.1 INTRODUCTION
Intermolecular forces are the forces which mediate interaction between the molecules,
including forces of attraction or repulsion which act between the molecules and other
types of neighboring particles, e.g. atoms or ions. Intermolecular forces are weak relative
to intramolecular forces (the forces which hold a molecule together) e.g. the covalent
bond, involving sharing electron pairs between atoms, is much stronger than the forces
present between neighboring molecules.
The strength of the intermolecular forces between the molecules or atoms that
compose a substance determines the states (solid, liquid or gas) of the substance at the
room temperature. Strong intermolecular forces tend to result in liquids and solids with
high meting and boiling points. Weak intermolecular forces tend to result in gases with
low melting and boiling points. Here, we will focus on the fundamental types of
intermolecular forces.
3.2 IONIC INTERACTIONS
Ionic bonds are formed due to transfer of one or more electrons from one atom to the
other between a metal and non-metal atom. The metallic atom loses its electron present in
its valence shell and converts into a cation. The non-metallic atom gains electrons and
converts into an anion. The electrostatic force of attraction holds the oppositely charged
ions together.
The number of electrons that an atom gains or loses while forming an ionic bond is
called its electrovalency. The atom which loses electrons is called electropositive and the
one which gains electrons is called electronegative atom.
3.1
F.E. Sem.-I Engineering Chemistry-I
3.2
Intermolecular Forces and Critical Phenomena
Since ionic bond is the strongest bond, it takes a lot of energy to break it. The greater
the charge difference, the stronger the attraction.
3.3 VAN DER WAAL’S INTERACTIONS
This
is
the
weak
attractive
intermolecular force present in all
molecules and atoms. It is known as Van
der Waals forces or dispersion forces or
London forces. They occur between the
molecules.
Fig. 3.1
Due to constant movement of electrons, even in nonpolar molecule ,at any instant
more electrons may be present in one region than in another. It causes an unequal charge
distribution which causes temporary dipole. This temporary dipole can induce a similar
temporary dipole on a nearby molecule as shown in the Fig. 3.1.
This temporary dipole can extend over large number of molecules in a synchronized
way. It holds them together in a lattice. Collectively many such temporary dipoles and
induced dipoles result in the weak electrostatic force of attraction, that is, van der Waals
forces.
They exist in all types of molecules along with dipole or any other intermolecular
forces. These forces are weaker than the covalent bonds present within the molecules.
The exact value of these forces cannot be predicted as they vary with size, shape and
polarisability of the molecule. Molecules containing large, diffuse electron cloud like
polyatomic ions or molecules with multiple bonds, they are as strong as covalent bonding
forces.
x
Van der Waals forces explain the condensation of gases and freezing of liquids on
cooling.
x
High molecular weight indicates more electrons and more powerful attractive
forces than Van der Waals forces. This is the explanation for why high molecular
weight compounds tend to be solids or liquids and low molecular weight
compounds tend to be gases.
x
At the boiling point of a liquid, the amount of molecular agitation is enough to
overcome Van der Waals force of attraction. Hence, boiling point is the measure
of these forces. This explains the periodic trend in boiling points for the noble
gases e.g. Ne boils at much lower temperature than Xe. Since Ne is having lesser
electrons than Xe, its Van der Waals forces are more easily overcome by thermal
motion.
F.E. Sem.-I Engineering Chemistry-I
3.3
Intermolecular Forces and Critical Phenomena
3.4 DIPOLAR INTERACTIONS
The dipole-dipole force exists in all polar molecules.
Polar molecules have permanent dipoles that interact with
the permanent dipoles of neighboring molecules. The
positive end of one permanent dipole is attracted to the
negative end of another permanent dipole. This attraction is
the dipole-dipole force as shown in the Fig. 3.2.
Fig. 3.2
The distance between the two dipoles and their orientation determines the dipoledipole interaction. All molecules including the polar ones have dispersion forces. In
addition to dispersion forces, polar molecules have dipole-dipole forces. These additional
forces raise the melting and boiling points of the polar molecules. Non-polar molecules
having same molecular weight and shape but lacking the presence of dipole movement
has relatively low melting and boiling point. The same is illustrated by comparing
Formaldehyde and Ethane in the Table 3.1.
Table 3.1: Comparison of melting & boiling points of Formaldehyde & Ethane
Molar mass Boiling Point
Melting Point
Name
Formula
o
(g/mol)
( C)
(oC)
Formaldehyde
30
–19.5
–92
CH2O
(Polar)
Ethane
C2H6
30.1
–88
–172
(Non-polar)
Formaldehyde being polar, it shows higher melting point and boiling point than nonpolar ethane despite their molecular weights being almost the same.
The polarity of molecules also helps in determining the liquids miscibility i.e. its
ability to mix without separating into two phases. Polar liquids are miscible with other
polar liquids, but are not miscible with non-polar liquids e.g. water, a polar liquid does
not mix with the oil, a non-polar liquid.
3.5 HYDROGEN BONDING
Polar molecules containing hydrogen atoms bonded directly to fluorine, oxygen or
nitrogen exhibit an additional intermolecular force called a hydrogen bond e.g. HF, NH3
and H2O show hydrogen bonding. The large electronegativity difference between
hydrogen and these electronegative elements, as well as the small size of these atoms
gives rise to a strong attraction between the hydrogen in each of these molecules and the
F, O or N on neighboring molecules. This attraction between a hydrogen atom and an
electronegative atom is the hydrogen bond. The example of the same is shown in the
Fig. 3.3.
F.E. Sem.-I Engineering Chemistry-I
3.4
G– G+
NH
Intermolecular Forces and Critical Phenomena
G– G+
OH
G– G+
FH
Fig. 3.3
Similarly, the hydrogen atom in each
water molecule is hydrogen bonded to the
oxygen in four other water molecules as
shown in the Fig. 3.4.
As shown in the figure, a normal
covalent bond length is 0.96 Ao and the
hydrogen bond length is 1.97 Ao. As the
bond length indicates the bond strength,
hydrogen bond is weaker than the covalent
and the ionic bonds.
Fig. 3.4: Hydrogen bonding in
water molecule
There are two types of hydrogen bonds namely, intramolecular and intermolecular.
Intramolecular Hydrogen Bond:
The hydrogen bond formed between
hydrogen and an electro negative atom
(F, O & N) within the same molecule is
intramolecular hydrogen bond. It
results in the cyclization of the
molecules
and
prevents
their
association. It does not affect physical
properties of the compound. The
intramolecular hydrogen bonds present
in different molecules is shown in the
Fig. 3.5.
Fig. 3.5: Intramolecular Hydrogen Bonds
The intramolecular hydrogen bonding is important in biological molecules as the
shapes of proteins and nucleic acids are largely influenced by it e.g. the two strands of the
double helix in DNA are held together by hydrogen bond between hydrogen atoms of one
strand with the lone pairs on the nitrogen or oxygen on the other strand.
Intermolecular Hydrogen Bond: The hydrogen bond formed between the hydrogen
atom of one molecule and an electronegative atom of the other molecule is called
Intermolecular hydrogen bond e.g. Hydrogen Chloride, Water, Ammonia, Alcohol etc.
F.E. Sem.-I Engineering Chemistry-I
3.5
Intermolecular Forces and Critical Phenomena
Effect of intermolecular hydrogen bonding is reflected in many properties of the
compounds such as increase in the melting point, boiling point, solubility etc.
Effects of hydrogen bonding:
1. Solubility : Solubility of substances in certain solvents is influenced by hydrogen
bonding. Covalent compounds generally do not dissolve in water. But those which form a
hydrogen bond with water readily dissolve in it e.g. Ammonia, Amines, Ethanol, Lower
Aldehydes and Ketones are soluble in water due to the formation of hydrogen bonds
between hydrogen atom of water molecule and the electronegative atom of these
molecules.
2. Physical State : Intermolecular hydrogen bonding causes two or more molecules
of a compound to exist as associated molecules. This results in an increase in the size and
molecular mass of the compound which gets reflected in the physical state of the
substance e.g. H2O is liquid and H2S is a gas, although oxygen and sulphur belong to the
same group. In water, oxygen is highly electronegative and forms intermolecular
hydrogen bonds. It results in getting water molecules associated due to which molecular
mass is increased. Hence, water exists as liquid at room temperature. On the other hand,
electronegativity difference of hydrogen and sulphur is less and there is negligible
hydrogen bonding in H2S. They are not associated and hence H2S exists as a gas at room
temperature. The same applies for existence of HF as liquid and HCL as a gas at room
temperature.
3. Melting and Boiling Point : Hydrogen bonds are very strong intermolecular
forces. Due to hydrogen bonding and a consequent association of molecules, larger
energy is required to separate these molecules before they can melt or boil. Therefore
these compounds usually show elevated melting and boiling points e.g. methanol and
ethane as shown in the Table 3.2.
Table 3.2: Comparison of melting & boiling points of Methanol & Ethane
Name
Formula
Molar mass
(g/mol)
Boiling Point
(oC)
Melting Point
(oC)
Methanol
CH3OH
32
64.7
–97.8
Ethane
C2H6
30.1
–88
–172
Since Methanol has hydrogen directly bonded to oxygen, its molecule shows
hydrogen bonding. This hydrogen is strongly attracted to the oxygen on neighboring
molecules. Due to this strong attraction, the boiling point of methanol is higher than
ethane.
F.E. Sem.-I Engineering Chemistry-I
3.6
Intermolecular Forces and Critical Phenomena
The melting and boiling points of the hydrides of group 14 increase with increase in
molecular weight. However, in case of the hydrides of elements of Groups 15, 16, 17 the
melting and boiling points of H2O, NH3 and HF are exceptionally higher than the
hydrides of other members of the group because of hydrogen bonding.
3.6 ION-DIPOLE FORCE
The ion-dipole force occurs in the mixtures of ionic compounds and polar
compounds. It is very important in aqueous solutions of ionic compounds. e.g. When
NaCl is mixed with water, the sodium and chloride ions interact with water molecules
via. ion-dipole forces. Ion-dipole forces are the strongest of all types of intermolecular
forces. They are responsible for the ability of ionic substances to form solutions with
water.
Comparison of different types of intermolecular forces is shown in the Table 3.3.
Table 3.3: About Intermolecular forces:
Type of Force
Relative Strength
Present In
Van der Waals force weak, but increases with all atoms and molecules
(dispersion force)
increasing molar mass
dipole-dipole force
moderate
only polar molecules
hydrogen bond
strong
molecules containing H bonded
directly to F, O or N
ion-dipole
very strong
mixtures of ionic compounds
and polar compounds
3.7 EQUATIONS OF STATE OF REAL GASES AND
CRITICAL PHENOMENA
The gases which follow the gas laws at a low pressure and a high temperature are
called real gases. But their behaviour deviates significantly from that of ideal gas at high
pressure and low temperature. The force of attraction between the molecules of real gases
cannot be ignored at low pressure and high temperature.
The combination of all the gas laws viz. Boyle’s Law, Charles Law and Avogadro’s
Law gives the ideal gas equation which relates the four parameters – pressure (P), volume
(V), absolute temperature (T) and number of moles (n) of ideal gas. This equation is
called ideal gas equation. It is expressed as,
PV = nRT
F.E. Sem.-I Engineering Chemistry-I
3.7
Intermolecular Forces and Critical Phenomena
Van der Waals equation of State for Real Gases:
The Van der Waals equation of state is the equation that modifies the ideal gas laws
by disapproving the following points.
1. The force of attraction between gaseous molecules is negligible. The ideal gas law
treats gas molecules as point particles that interact with their containers but not
with each other.
2. The volume of gaseous molecules is negligible as compared to the total volume of
the gas.
The two corrections explained by Van der Waal are as follows.
Volume Correction: Since the real gas molecules take up some volume, the actual
space available for the movement of gas molecules inside the vessel is not the real
volume of the gas. The real volume of the gas is calculated by the following equation:
V ideal gas = V – b
Where,
V = Volume of the container
b = Volume occupied by gas molecules
So the ideal gas equation becomes P (V – b) = RT
Pressure Correction: The force with which the molecules of the gas inside a
container collide with the walls of the container is called the pressure of the gas.
According to Van der Waal, the pressure of real gas is the combination of pressure
developed due to collision of the molecules of gas with the walls of the container. In
addition, in case of real gas, there is pressure loss due to inward pull of adjacent
molecules. This inward pull makes the gas molecules striking the walls of the container
experience some backward drag. Therefore, the observed pressure is smaller compared to
the real gas pressure.
P ideal gas = P + p
Where,
P = Observed pressure of the gas.
p = pressure lost by the gas molecules due to molecular attractions
n2
= a V2
a = Van der Waals constant
The value of a depends on the nature of gas
n = Number of moles of real gas
V = Volume of gas
F.E. Sem.-I Engineering Chemistry-I
3.8
Intermolecular Forces and Critical Phenomena
So the ideal gas equation after correction is rewritten as follows:
ªP + a º
«
V2»¼ (V – b) = RT (For 1 mol)
¬
2
ªP + a n º (V – nb) = nRT (For n number of moles)
«
V2»¼
¬
Critical Phenomena: The essential condition for the liquefaction of the gas is
described by the study of critical temperature, critical pressure and critical volume and
their inter relationships.
When a gaseous system is transformed to its liquid state, there is a tremendous
decrease in the volume. This decrease in volume can be effectively brought about by
lowering of temperature, or by increasing pressure (or) by both. In both these effects, the
gaseous molecules come closer to each other and experience an increase in force of
attraction which results in liquefaction of gases. At any constant temperature when
pressure is increased volume is decreased and vice versa. Such P-V curves at constant
temperature are known as isotherms.
The continuous decrease in pressure with increase in volume is seen in both ideal and
real gases. There is a definite deviation exhibited by the real gas from ideal gas behaviour
at high pressure and low volumes.
Critical Temperature (Tc): It is defined as the characteristic temperature of a gas at
which increase in pressure results in liquefaction of gas. But above this temperature, there
will be no liquefaction even by increasing the pressure. For instance, Tc of CO2 is 31.1 o
C. This means that it is not possible to liquefy CO2 by applying pressure when its
temperature is above 31.1 o C.
Critical Pressure (Pc): It is defined as the minimum pressure at which a gas can be
liquefied at its critical temperature.
Critical volume (Vc): The volume occupied by 1 mole of a gas at its critical pressure
and at critical temperature is the critical volume of the gas.
A gas is said to be at its critical state when its pressure, volume and temperature are
Pc, Vc and Tc.
Let us derive the values of critical constants Tc (critical temperature), Vc (critical
volume) and Pc (critical pressure) in terms of the Van der Waal's constants `a' and `b'.
The Van der Waal's equation is given by
8a
Tc = 27Rb
F.E. Sem.-I Engineering Chemistry-I
3.9
Intermolecular Forces and Critical Phenomena
Hence the critical constants can be calculated using the values of Van der Waal's
constants of a gas and vice versa. Since Pc and Tc can often be determined
experimentally, these values may be employed to calculate the constants a and b.
a = 3Vc2Pc
Vc
b= 3
Based on the critical temperature values, gases are categorised as "permanent" and
"temporary" gases. H2, N2, He gases having very low critical temperature belong to the
permanent type. Gases like NH3, CO2, SO2, HCl etc. having critical temperature in the
ordinary range of temperatures belong to the temporary gases type.
3.8 REVIEW QUESTIONS
1. Write a note on Van der Waal’s forces.
Ans. Refer Section 3.3.
2. What is hydrogen bonding?
Ans. Refer Section 3.5.
3. Discuss the two types of hydrogen bonding.
Ans. Refer Section 3.5.
4. What are the consequences of hydrogen bonding?
Ans. Refer Section 3.5.
5. Justify the existence of HF as liquid and HCl as gas.
Ans. Refer Section 3.5.
6. Write a note on ion dipole force.
Ans. Refer Section 3.6.
7. What are intermolecular forces?
Ans. Refer Section 3.1.
8. Compare the intermolecular forces.
Ans. Refer Section 3.6.
9. Explain the effect of hydrogen bonding on physical state of the substance.
Ans. Refer Section 3.5.
10. Explain dipolar interactions.
Ans. Refer Section 3.4.
F.E. Sem.-I Engineering Chemistry-I
3.10
Intermolecular Forces and Critical Phenomena
11. Discuss the effect of hydrogen bonding on melting point and boiling point of
substances.
Ans. Refer Section 3.5.
12. How hydrogen bonding influences the solubility of substances?
Ans. Refer Section 3.5.
13. Explain formaldehyde shows higher melting and boiling point than ethane.
Ans. Refer Section 3.4.
14. Define critical temperature, critical volume and critical pressure.
Ans. Refer Section 3.7.
15. Write a note on critical phenomena.
Ans. Refer Section 3.7.
CHAPTER
4
PHASE RULE – GIBB’S PHASE RULE
SYLLABUS
x
x
x
x
x
x
Statement of Gibbs’ Phase Rule, Terms involved with examples
One Component System (Water)
Reduced Phase Rule
Two Component System (Pb- Ag)
Advantages and Limitations of Phase Rule
Numerical problems on Phase Rule
4.1 INTRODUCTION
Phase equilibrium deals with the study of equilibrium conditions of heterogeneous
systems. Substances are present in different phases in heterogeneous systems. Such a
system can be conveniently studied with the help of a generalization called Phase Rule.
This rule was deduced on the basis of Thermodynamic principles by Willard Gibbs
(1876) and was later developed by Ostwald, Van’t Hoff, H.W.B. Roozeboom and many
more. Phase rule is an important tool in the study of heterogeneous equilibrium. It relates
the conditions which must be specified to describe the state of a system at equilibrium.
The plot indicating the relationships between various phases under different temperature,
pressure and concentration is known as phase diagram. Phase rule, with the help of a
phase diagram, is useful in predicting the effects of temperature, pressure and
concentration on the equilibrium of heterogeneous systems.
A substance can exist in various states depending on the external conditions. For
example, water can exist in the form of solid, liquid and vapour. A change in the state of
a substance is called phase transition. Equilibrium in a system in which phase transitions
occur in addition to a chemical process is called phase equilibrium.
4.2 GIBB’S PHASE RULE
Gibb’s phase rule states that in every heterogeneous system in equilibrium, the sum of
the number of phases and degree of freedom is greater than the number of components
by 2.
4.1
F.E. Sem.-I Engineering Chemistry-I
4.2
Phase Rule - Gibb’s Phase Rule
This is mathematically expressed as
P+F=C+2
or
F=C–P+2
Where P is the number of phases present in equilibrium, C is the number of
components for the system and F is the number of degrees of freedom for the
equilibrium. This rule is valid for any system at equilibrium at definite temperature and
pressure provided the equilibrium between any number of phases is not influenced by
gravity, by electrical or magnetic forces or by surface action; and is only influenced by
temperature, pressure and concentration.
For an accurate and effective interpretation of the phase rule, it is important to
understand clearly the meaning of the terms involved in the phase rule.
4.2.1 Phase
A phase is defined as any homogeneous, physically distinct, mechanically separable
portion of the system which is separated from other parts of the system by definite
boundary surfaces.
We all know that a homogeneous system is the one in which both the reactants and
products are of one phase (i.e. solid, liquid, gas) and a heterogeneous system is one in
which they are in different phases.
A homogeneous system is one which is uniform throughout in physical and chemical
properties e.g. a solution of salt in water. A heterogeneous system comprises of two or
more different parts, each of which is homogeneous in itself and is separated from others.
These homogeneous, physically distinct and mechanically separable parts of a
heterogeneous system existing in equilibrium are called phases.
Common example that describe phase are,
(1) In a freezing water system, ice, liquid water and water vapour constitute three
phases. Each of the phases is distinct, homogeneous and mechanically separable.
Ice (s) œ water (l) œ water vapour (g).
(2) A gaseous mixture, being thoroughly miscible in all proportions, constitutes one
phase only. Thus, a mixture of N2 and H2 forms one phase only.
(3) Two miscible liquids form one liquid phase only.
For example: ethyl alcohol and water.
(4) Two immiscible liquids form two different phases and can be separated by
boundary surfaces.
For example: kerosene and water or benzene and water
F.E. Sem.-I Engineering Chemistry-I
4.3
Phase Rule - Gibb’s Phase Rule
(5) A solution of a substance in a solvent consists of one phase.
For example: glucose solution in water.
(6) A system containing CaCO3, CaO and CO2 has three phases – two solid phases
and one gaseous phase.
CaCO3 œ CaO (s) + CO2 (g)
(7) A homogeneous solid solution of salt constitutes a single phase.
For example: Mohr’s salt [FeSO4. (NH4)2SO4.6H2O] solution constitutes a single
phase although it consists of FeSO4, (NH4)2SO4, H2O.
(8) Each solid makes up a separate phase except in the case of a solid solution. For
example, many forms of sulphur can exist together and all are separate phases.
4.2.2 Component
The components of the system do not represent the number of the constituents or
chemical individuals present in the system. As given in the statement of phase rule, the
number of components of a system at equilibrium is the smallest number of
independently variable constituents by means of which the composition of each phase
present can be expressed, either directly or in the form of chemical equation.
This concept of component can be explained in connection with phase rule with the
help of the following examples,
(1) In the water system,
Ice (s) œ water (l) œ water vapour (g)
The composition of each one of the solid, liquid and vapour can be expressed
using one constituent, namely H2O. Hence, it is one component system.
(2) The sulphur system consists of four phases rhombic, monoclinic, liquid and
vapour, the chemical composition of all the phase is sulphur. Hence, it is one
component.
(3) In the thermal decomposition of CaCO3,
CaCO3 (s) œ CaO (s) + CO2 (g)
The composition of each of the three phases can be expressed in terms of at least
any two of the independently variable constituents, CaCO3, CaO and CO2.
Suppose CaCO3 and CaO are chosen as the two components, then the
composition of different phases is represented as follows:
Phase CaCO3 = CaCO3 + OCaO
Phase CaO = OCaCO3 + CaO
Phase CO2 = CaCO3 – CaO
Thus it is a two component system.
F.E. Sem.-I Engineering Chemistry-I
4.4
Phase Rule - Gibb’s Phase Rule
(4) In the dissociation of NH4Cl in a closed vessel
NH4Cl (s) œ NH4Cl (g) œ NH3 (g) + HCl (g)
The proportions of NH3 and HCl are equivalent and hence, the composition of
both phases (solid and gaseous) can be expressed in terms of NH4Cl alone. Hence,
the number of component is one. However, if NH3 or HCl is in excess, the system
becomes a two component system.
(5) The acetic acid system is a one component system because the composition of
each phase can be expressed in terms of each phase of CH3COOH, although it
exists in the form of double molecules entirely in the solid state, to a great extent
in the liquid state and only to a smaller extent in the vapour state.
There may be equilibrium such as:
2 CH3COOH œ (CH3COOH)2
in the solid and liquid phases due to association, but the only independent species
in all the three phases is CH3COOH.
(6) In the equilibrium,
Fe (s) + H2O (g) œ FeO (s) + H2 (g)
The minimum components requires to express the composition of each phase is
three. So it is three component system.
(7) A salt solution is a two component system.
4.2.3 Degree of Freedom of Variance
The ‘Degree of Freedom’ means the minimum number of independently variable
factors, such as temperature, pressure and composition of the phases, which must be
arbitrarily specified in order to describe completely the state of the system.
A system having one, two, three or zero degrees of freedom are usually called
univariant, bivariant, trivariant and invariant respectively.
Common examples include,
(1) In case of water system,
Ice (s) œ Water (l) œ Water Vapour (g)
no specification of conditions is necessary because the three phases can occur in
equilibrium only at a particular temperature and pressure. This situation exists
only at triple point; hence this system does not have any degree of freedom
(invariant). If condition (e.g. pressure or temperature) is altered, three phases will
not remain in equilibrium and one of the phases disappears.
F.E. Sem.-I Engineering Chemistry-I
4.5
Phase Rule - Gibb’s Phase Rule
(2) Consider a system containing water in contact with its vapour.
Water (l) œ Water Vapour (g)
The vapour will exert a certain pressure at a particular temperature. Hence, it is
enough if we fix the temperature of the system. Then, the pressure that would be
exerted by the vapour is automatically fixed. Hence of the two variable factors of
temperature and pressure, it is enough if we fix one. Therefore this system has
only one degree of freedom or it is univariant.
(3) A system of a pure gas should satisfy the gas equation PV = RT. If we assign
values arbitrarily for any two of the three variables, the value of the third gets
automatically fixed and the system is completely defined. This system has two
degrees of freedom or it is bivariant.
(4) For a system consisting of
NaCl (s) œ NaCl Water (aq) œ Water Vapour (g)
is completely defined if the temperature or pressure is specified. The saturation
solubility is fixed at a particular temperature or pressure. Hence the system is
univariant.
4.3 APPLICATION OF PHASE RULE TO
ONE COMPONENT SYSTEM
The Water System:
Water is the most common example of one component system.
The water system consists of three phases, namely, ice, water and water vapour.
Ice (s) œ Water (l) œ Water Vapour (g)
The three phases may occur in four possible combinations in equilibrium as follows :
(i) Liquid l Vapour
(ii) Liquid l Solid
(iii) Solid l Vapour
(iv) Solid l Liquid l Gas.
As water (H2O) is the only compound involved in the system, therefore, it is single or
one component system. From the phase rule, when C = 1,
F=C–P+2=1–P+2=3–P
The degree of freedom depends on the number of phases present at equilibrium.
Three different cases are possible.
(i) P = 1;
F = 2 (bivariant system)
(ii) P = 2;
F = 1 (univariant system)
(iii) P = 3;
F = 0 (invariant system)
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4.6
Phase Rule - Gibb’s Phase Rule
It is clear that for any one component- system the maximum number of degrees of
freedom is two. Therefore, such a system can be represented completely by a two –
dimensional diagram.
On the basis of experimental data obtained for the water system, a plot of
relationships between the various phases (ice, water and vapour) under different
conditions of temperature and pressure. (Temperature on x-axis and pressure on y-axis) is
drawn and resulting phase diagram is shown in Fig 4.1.
Fig. 4.1 : The Phase Diagram of Water System
(A) Area :
As can be seen from the diagram, there are three areas namely BOC (ice), COA
(water) and AOB (vapour). In any area, only one of the phases is present. To define the
system completely at any point in an area, both temperature and pressure should be fixed.
So the system in an area has two degrees of freedom (bivariant). The same conclusion is
drawn by substituting the values of C = 1 and P = 1 in the phase rule equation
F=C–P+2
F=1–1+2
F=2
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4.7
Phase Rule - Gibb’s Phase Rule
(B) Boundary Lines :
Separating the areas are lines OA (vaporization curve), OB and OC, which connect
the point at which two phases can co-exist in equilibrium. In order to locate any point on
the particular line, either temperature or pressure co-ordinate should be known. In other
words, if the value of either the temperature or the pressure is fixed, the value of the other
is automatically fixed and the system is completely defined. Any point on the boundary
lines has one degree of freedom or is univariant.
F=C–P+2
F=1–2+2
F=1
(1) The curve OA (Vaporization Curve) : This curve is known as the vaporization
curve as it represents the equilibrium between liquid water and vapour at different
temperatures.
Liquid water l Vapour
Along the curve OA, water and vapour are in equilibrium with each other. This is
known as vaporization or vapour pressure curve. This curve describes the effect of
temperature on the vaporization of water. At any given temperature, there is one and only
one pressure at which water vapour is in equilibrium with liquid water. Similarly at any
given pressure, there is one temperature at which water vapour is in equilibrium with
liquid water. i.e. the system has one degree of freedom (univariant).
For example, at atmospheric pressure, water and vapour can exist in equilibrium only
at one temperature i.e. the boiling point of water.
The curve OA ends at point A corresponding to a temperature of 374qC (critical
temperature) and pressure of 218.5 atmospheric pressure, beyond which there is only one
phase (liquid and vapour phases merge into each other). At the lower end, the curve
terminates at the point O, where water freezes to form ice. But by careful elimination of
solid particles which induce crystallization, water may be cooled far below its freezing
point, without the crystallization of ice. Dotted curve OA’ shows the vapour pressure of
super cooled water. It represents a meta-stable equilibrium and is situated above the
sublimation curve, OB. The vapour pressure of the meta-stable phase at any point is
greater than that of the stable phase. The super cooled water will at once change into
solid ice even with the slightest disturbance.
(2) The curve OB (Sublimation Curve) : Along the curve OB, ice and vapour are
in equilibrium with each other. It is known as sublimation curve. It is seen that the curve
OB terminates at the lower end at absolute zero (−273qC) where no vapour can be present
and only ice exists.
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4.8
Phase Rule - Gibb’s Phase Rule
(3) The curve OC (Melting Point Curve) : Along the curve OC, two phases,
namely ice and water, are in equilibrium with each other. This is known as melting point
or freezing point curve. This curve describes the effect of pressure on the melting point of
the ice. The slope of the curve shows that
(a) Increase of pressure decreases the melting point of ice.
(b) Ice melts with decrease of volume.
At any point on this curve, if either the pressure or the temperature is changes one of
the phases disappears.
(4) The curve OA′ (Metastable Curve) : The dotted curve OA′ is a continuation of
the OA curve and represents the vapour pressure of supercooled water. With due care, it
is possible to cool water (any liquid) below its freezing point without separation of ice
and the water is then said to be super cooled. The liquid l vapour system along the
curve OA′ is said to be in metastable equilibrium as even a slight disturbance in this
phase changes the super cool water into ice and the curve merges into OB. This curve
OA′ lies above the curve OB and shows that the metastable has a higher vapour pressure
than the stable one at the same temperature.
(C) Triple Point :
It has been found experimentally that the curves OA, OB and OC meet in a point ‘O’.
This is called the ‘triple point’ where all the three phases, viz., ice, water and vapour coexist. Only at a particular value of temperature and pressure represented by the triple
point, all the three phases can co-exist. So, the system has no degrees of freedom
(invariant) at this point. The triple point corresponds to a temperature of 0.0075qC and a
pressure of 4.58 mm. At this point, if we change either temperature or pressure, one of
the phases will disappear.
The salient features of the phase diagram of the water system are summarized in
Table 4.1.
Table 4.1 : Water System
Number of components = 1
Name of the
represented in
diagram
system as Phase in equilibrium
the phase
Degree of freedom or
variance
F=C–P+2
Areas
(1) BOC (ice)
(2) COA (water)
(3) AOB (vapour)
Ice
Water
vapour
Two
(Bi-variant)
F=1–1+2=2
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4.9
Phase Rule - Gibb’s Phase Rule
Curve
(1) OC (Melting point curve)
(2) OA (Vaporization curve)
(3) OB (Sublimation curve)
Ice and water
One
(univariant)
Water and vapour
F=1–1+2=1
Ice and vapour
Point
Ice, water and vapour
(1) O (Triple point)
Zero (Invariant)
F=1–3+2=0
4.4 APPLICATION OF THE PHASE RULE TO
TWO COMPONENT SYSTEMS
4.4.1 General Characteristics of Two Component Systems
(1) The two component systems have three variables namely pressure, temperature
and concentration.
(2) The composition of all the individual phases of the system can be expressed by
means of not less than two components.
(3) The maximum number of phases in a two-component system will be four.
P=C–F+2
P=2–0+2
P=4
(Maximum number of phases exist when degrees of freedom = 0. Negative degree
of freedom can not exist.)
(4) The maximum number of degrees of freedom in a two component system will be
three.
F=C–P+2
F=2–1+2
F=3
(i.e. when the system exists as a single phase)
(5) For constructing a phase diagram of a two component system, a three dimensional
space model is required using the three variables (viz, temperature, pressure and
concentration) as its co-ordinates.
4.10
Phase Rule - Gibb’s Phase Rule
Pressure
F.E. Sem.-I Engineering Chemistry-I
Concentration
Temperature
Fig. 4.2 : Three dimensional space model of a two component system
To simplify the diagram, one of the variables is kept constant. If the pressure is kept
constant, the diagram is called isobaric; if temperature is kept constant it is called
isothermal and if composition is kept constant, the diagram is called isoplethal.
Phase rule for two component system : In a two-component system, when P = 2,
degree of freedom (F) has the highest value i.e. 3. Three variables temperature, pressure
and concentration of one of the two components must be specified in order to describe
the system completely. Since the maximum number of degrees of freedom in a two
component system is 3, the phase behavior of a binary system may be represented by a
three dimensional diagram of pressure, temperature and composition. Solid-liquid
equilibrium of an alloy has practically no gas phase and the effect of pressure is small on
this type of equilibrium. Usually the experiments are conducted under atmospheric
pressure. Thus keeping the pressure constant of a system, in which vapour phase is not
considered, is known as condensed system. It will reduce the degrees of freedom of the
system by one. For such a system the phase rule becomes F = C – P + 1. This is known as
the reduced or condensed phase rule, having only two variables, temperature and
concentration (composition) of the constituents.
4.4.2 Thermal Analysis
Thermal analysis is the study of the cooling curves of various compositions of a
system during solidification. Basically, the form of the cooling curve depicts the
composition of the solid. It can be understood from the following considerations:
(1) When a pure substance in the fused or liquid state is allowed to cool slowly and
the temperature noted at definite times, the graphic representation of the rate of cooling
will be continuous curve (see Fig. 4.3 (a)). When the freezing point is reached and the
solid makes its appearances, it is indicated by a break in the continuity of the cooling
curve and the temperature will remain constant, until the liquid is completely solidified.
Thereafter, the fall in temperature will again become continuous.
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4.11
Phase Rule - Gibb’s Phase Rule
(2) If a mixture of two solids in the fused state be cooled slowly and the cooling
curve is obtained in a similar manner. We likewise obtain a continuous cooling curve, so
long as the mixture (or solution) is in the liquid state. When a solid phase begins to form,
the rate of cooling abruptly alters and the cooling curve exhibits a break. However, the
temperature does not remain constant, as in the previous case of cooling of a pure
substance. The temperature decreases continuously, but at a different rate and if the
mixture forms a eutectic, the fall of temperature continues, till the eutectic point is
reached. The system now becomes invariant from the point of view of the phase rule and
the temperature remains constant, until solidification is complete. (See Fig 4.3 (b)).
Thereafter, the fall of temperature becomes uniform, but the rate of fall is quite different
from the previous one.
a
Beginning of
freezing
c
End of
freezing
b
Freezing point
d
Temperature
Temperature
a
Time
Freezing point
Beginning of
freezing
b
End of
freezing
c
Eutectic point
d
e
Time
(a)
(b)
Fig. 4.3 : Cooling Curves
Using the cooling curve for any mixture of a definite composition, it is possible to
obtain its: (i) freezing point, and (ii) eutectic temperature.
(a) The freezing point varies with the composition of the system, but the eutectic
point remains constant for a given system.
(b) The nearer the composition of the system to the eutectic, the shorter is the portion
bc and the more prolonged is the halt cd.
(c) If the mixture coincides with the eutectic compositions, the curve shows no break
corresponding to bc, but the break appears only at the eutectic point c.
(d) If the cooling curves of a series of alloys of known compositions are worked out
and their freezing points are noted, by plotting freezing point against composition.
It is known as the T – C curve for the alloy system. However, in order to complete
the diagram, it is necessary to know the freezing points of the pure components
also.
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4.12
Phase Rule - Gibb’s Phase Rule
(e) Now the cooling curve of an alloy of the same metals, but of unknown
composition is determined and its freezing point located in the T – C diagram.
The composition corresponding to this freezing point yields the composition of
the alloy.
(f) The thermal analysis procedure can also be used to derive the phase diagram of
any two component system.
4.4.3 Eutectic System
A binary system consisting of two substances, which are miscible in all proportions is
the liquid phase, but which do not react chemically, is known as the “eutectic system”,
for example, a mixture of lead and silver comprises of such a system.
Eutectic mixture is a ‘solid solution’ of two or more substances having the lowest
freezing point of all the possible mixture of the components. This is taken advantage of in
“alloying of low melting point”, which are generally eutectic mixtures.
Eutectic point: Two or more solid substances capable of forming solid solutions with
each other have the property of lowering each other’s freezing point; and the minimum
freezing point attainable corresponding to the eutectic mixture, is termed the eutectic
point (means lowest melting point).
Application of eutectics: Low melting alloy are used in safety devices (for example,
as plugs in automobiles), fire-sprinklers and as ‘fail safe’ device in boilers). By suitable
choice of metals, very low melting alloys can be obtained, for example, wood’s metal
(alloy containing 50% Bi, 25% Cd and 12.5% Cd) melts at 65qC only.
4.4.4 The Lead Silver System
It is a two-component system with four possible phases- solid Ag, solid Pb, solution
of Ag + Pb and vapour. Since the pressure has nearly no effect on equilibrium, so the
system can be conveniently represented by a temperature- concentration diagram
(see Fig. 2.4) at constant atmospheric pressure. As the gaseous phase is practically absent
and one variable pressure is neglected, the condensed phase rule: F = C – P + 1 = 3 – P
will be applicable.
(1) Curve AO (freezing point curve of Ag) shows the effect on freezing point of Ag
on addition of lead in small quantities. The curve starts from A (961qC), the melting point
of Ag, where pure Ag co-exists as solid and liquid (vapour being neglected). The curve
indicates that the melting point of Ag falls gradually on adding Pb, along AO, till the
lowest point O (303qC) is reached, where the solution gets saturated with respect to lead.
At O, no more lead can go in solution and consequently, melting point of Ag does not fall
any further; and if any lead is added, it separates as the solid phase. Along this curve,
F.E. Sem.-I Engineering Chemistry-I
4.13
Phase Rule - Gibb’s Phase Rule
solid Ag and solution (vapour being negligible) co-exists and hence, according to reduced
phase rule equation: F = 3 – P = 3 – 2 = 1 i.e., the system is univariant. The point O
(303qC) corresponds to a fixed composition of 2.6% Ag and 97% Pb and is known as
eutectic composition. On cooling, the whole mass crystallizes out as such.
(2) Curve BO (freezing point curve of Pb). It represents the effect on freezing point
of Pb on gradual addition of small amounts of Ag to it. Point B is the melting point of
pure lead (327qC). Along BO, the melting point gradually falls on the addition of Ag, till
lowest point O is reached, where the solution gets saturated with respect to Ag and the
melting point of lead does not fall any more. On cooling, the whole mass (having eutectic
composition) crystallizes out. The system is univariant like AO.
(3) Point O (eutectic point). The two curves AO and BO meet at O, where three
phases (solid Ag, solid Pb and their solution) co-exists and according to condensed phase
rule, the system will be invariant (F = 3 – P = 3 – 3 = 0). The point O (303qC) represents
a fixed composition (Ag = 2.6%, Pb = 97.4%) and is called eutectic composition. No
mixture of lead and silver has a melting point lower than the eutectic temperature. At this
point, the temperature remains constant, until the whole of the melt solidified en-block to
become solid of eutectic composition. However, further cooling results in the
simultaneous crystallization of a mixture of Ag and Pb in relative amounts corresponding
to eutectic point O. Below the temperature line of eutectic temperature, we have two
regions in the diagram, viz.,
(i) The region marked eutectic + solid Ag, in which crystalline silver and solid
eutectic are stable and
(ii) The region marked eutectic + solid Pb, in which crystalline lead and solid eutectic
are stable.
Fig. 4.4 : T – C diagram of Pb-Ag system
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4.14
Phase Rule - Gibb’s Phase Rule
(4) Area AOB represents solution of Pb-Ag. If a sample of lead containing less than
2.6% Ag is taken, say P. On allowing the mass to cool, the temperature gradually falls
without any change in composition, till point P′ is reached on the curve BO. On lowering
the temperature, lead begins to separate out and the composition varies along P′O, till
point O is reached. On further cooling, the whole mass solidifies en-block to the eutectic
composition (2.6% Ag; 97.4% Pb.)
Application to Pattinson’s Process : The above principle is utilized in the
Pattinson’s process of desilverization of lead. If a sample of argentiferrous lead,
containing less than 2.6% Ag, is allowed to cool gradually, lead will separate out and the
solution will become progressively richer in Ag, till the percentage 2.6 of Ag is reached;
and on further cooling, the whole mass will solidify as such. On the other hand, if leadsilver alloy containing Ag greater than 2.6% is allowed to cool, then pure silver separate
along the curve AO, till the eutectic composition at O is reached.
NUMERICALS BASED ON PHASE RULE
Problem 1: An alloy of tin and lead contain 73% tin. Find the mass of eutectic in
1Kg of solid alloy if the eutectic contains 64% of tin.
Solution: 1Kg of alloy contains 730 g tin and 270 g lead. In the eutectic composition, tin
is 64% and lead is 36%. Therefore, corresponding to 270 g of lead, the mass of tin,
270 u 64
=
= 480 g
36
Total mass of eutectic in alloy = 270 g + 480 g = 750 g
Total mass of eutectic in alloy = 750 g
Problem 2: An alloy AB of 10 g weight contains 25% of A. The molten AB on
cooling gave out B and a eutectic alloy with A and B at equal percentage. What is
the amount of B that has formed?
Solution: Weight of A in 10 g of alloy = 2.5 g and weight of B in 10 g of alloy = 7.5 g
On cooling eutectic containing equal amount of A and B separate i.e. 2.5 g each of A and
B forms eutectic.
Therefore, the amount of B formed = original amount – amount in eutectic.
= 7.5 g – 2.5 g = 5 g
Amount of B formed = 5 g
Problem 3: An alloy of Cd and Bi contains 25% of Cd. Find the mass of eutectic in 1
Kg of alloy, if the eutectic system contains 40% Cd.
Solution: 1 Kg of alloy contains 250 g of Cd and 750 g Bi. In the eutectic system, Cd is
40% and Bi is 60%.
F.E. Sem.-I Engineering Chemistry-I
4.15
Phase Rule - Gibb’s Phase Rule
250 u 60
= 375 g
40
Total mass of eutectic in 1 Kg alloy = 250 g + 375 g = 625 g
Therefore, corresponding to 250 g of Cd, mass of Bi =
Total mass of eutectic in alloy = 625 g
Problem 4: 1000 Kg of a sample of argentiferrous lead containing 0.1 % silver is
melted and then allowed to cool. If eutectic contains, 2.6% Ag, what mass of:
(i) mass of eutectic will be formed, and (ii) mass of lead will separate out?
Solution:
(i) Mass of Ag present in 1000 Kg argentiferrous lead = 1 Kg
1 u 100
Mass of eutectic = 2.6 = 38.46 Kg
(ii) Mass of lead = 1000 - 38.46 = 961.54 Kg
(i)Mass of eutectic = 38.46 Kg, (ii) Mass of lead = 961.54 Kg
Problem 5: An alloy of tin and lead contain 80% tin. Find the mass of eutectic in
1Kg of solid alloy if the eutectic contains 60% of tin.
Solution: 1Kg of alloy contains 800 g tin and 200 g lead. In the eutectic composition, tin
is 60% and lead is 40%. Therefore, corresponding to 200 g of lead, the mass of tin,
200 u 60
= 300 g
=
40
Total mass of eutectic in alloy = 200 g + 300 g = 500 g
Total mass of eutectic in alloy = 500 g
Problem 6: An alloy AB of 20 g weight contains 20% of A. The molten AB on
cooling gave out B and a eutectic alloy with A and B at equal percentage. What is
the amount of B that has formed?
Solution: Weight of A in 20 g of alloy = 4 g and weight of B in 20 g of alloy = 16 g
On cooling eutectic containing equal amount of A and B separate i.e. 4 g each of A and B
forms eutectic.
Therefore, the amount of B formed = original amount – amount in eutectic.
= 16 g – 4 g = 12 g
Amount of B formed = 12 g
Problem 7: An alloy of Cd and Bi contains 20% of Cd. Find the mass of eutectic in 2
Kg of alloy, if the eutectic system contains 50% Cd.
Solution: 2 Kg of alloy contains 400 g of Cd and 1600 g Bi. In the eutectic system, Cd is
50% and Bi is 50%.
400 u 50
Therefore, corresponding to 400 g of Cd, mass of Bi =
= 400 g
50
F.E. Sem.-I Engineering Chemistry-I
4.16
Phase Rule - Gibb’s Phase Rule
Total mass of eutectic in 2 Kg alloy = 400 g + 400 g = 800 g
Total mass of eutectic in alloy = 800 g
4.5 ADVANTAGES OF PHASE RULE
Applications of Phase Rule are as follows:
(1) It applies to physical as well as chemical phase reactions.
(2) It provides a convenient basis for classification of equilibrium states of systems
with the help of phases, components and degree of freedom.
(3) It applies to macroscopic systems and hence information about molecular
structures is not essential.
(4) Phase rule does not take any cognizance of the nature or the amounts of
substances present in the system.
(5) It indicates that different systems having the same degree of freedom behave in a
similar fashion. Further, it helps in predicting the behaviour of a system under different
conditions of the governing variables.
(6) It helps in deciding whether the given number of substances together would exist
in equilibrium under a given set of conditions or whether some of them will have to be
inter-converted or eliminated.
4.6 LIMITATIONS OF PHASE RULE
Limitations of the Phase Rule are as follows:
(1) Phase rule can be applied only for systems in equilibrium. It is not of much help
in case of systems which attain the equilibrium state very slowly.
(2) All the phases of the system must be present under the same conditions of
temperature, pressure and gravitational force.
(3) It applies only to a single equilibrium state. It does not indicate the other possible
equilibria in the system.
(4) Phase rule considers only the number of phases but not their quantities. Even a
minute quantity of the phase, when present, accounts towards the number of phases.
Hence much care has to be taken in deciding the number of phases existing in the
equilibrium state.
(5) The solid and liquid phases should not be so finely subdivided as to bring about
deviation from their normal values of vapour pressure.
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4.17
Phase Rule - Gibb’s Phase Rule
4.7 REVIEW QUESTIONS
1. State phase rule.
Ans. Refer Section 4.2.
2. What is invariant system? Give any one example of it.
Ans. Refer Section 4.2.3.
3. What is triple point?
Ans. Refer Section 4.2.3 (C).
4. Define phase, component and degree of freedom.
Ans. Refer Section 4.2.1, 4.2.2, 4.2.3.
5. What is metastable state?
Ans. Refer Section 4.3 (B).
6. Calculate the number of phases in the following systems :
(i) MgCO3 (s) œ MgO (s) + CO2 (s)
(ii) I2 (s) œ I2 (g)
(iii) Ice (s) œ Water (l) œ Water Vapour (g)
(iv) Emulsion of oil in water
(v) Rhombic sulphur (s) œ monoclinic sulphur (s).
Ans. (i) 3, (ii) 2, (iii) 3, (iv) 2, (v) 2.
7. Discuss the significance of triple point.
Ans. Refer Section 4.3 (C).
8. Calculate the number of component present in the following systems:
(i) CaCO3 (s) œ CaO (s) + CO2 (g)
(ii) H2O (l) œ 1/2 H2 (g) + O2 (g)
(iii) NH4Cl (s) œ NH3 (g) + HCl (g)
(iv) NaCl (s) œ NaCl (aq)
(v) H2O (s) œ H2 (g) + H2O (g).
Ans. (i) 2, (ii) 1, (iii) 1, (iv) 2, (v) 2.
9. What is the number of phases in a system consisting of benzene and water in a
closed beaker?
Ans. 3.
10. What is the number of phases in a gaseous mixture consisting of N2, H2 and NH3
in equilibrium?
Ans. 1
11. Apply phase rule to calculate the degree of freedom at the triple point of a one
component system.
Ans. Refer Section 4.3.
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4.18
Phase Rule - Gibb’s Phase Rule
12. Mention the applications of phase rule.
Ans. Refer Section 4.5.
13. Mention the limitations of phase rule.
Ans. Refer Section 4.6.
14. State Gibb’s phase rule and explain the terms involved in it by giving two
examples.
Ans. Refer Section 4.2.
15. Explain the following :
(i) Triple point,
(ii) Condensed phase rule with explanation of the terms involved.
Ans. Refer Section (i) 4.3 (C), (ii) 4.4.1.
16. Water system is a representation system for explaining phase rule and phase
equilibria. Explain.
Ans. Refer Section 4.3.
17. Draw and explain the phase diagram of ice-water-water vapour system. How does
the melting point of ice change with variation of pressure?
Ans. Refer Section 4.3.
18. In a phase diagram for water system, explain
(1) a bivariant system, (2) an univariant system, (3) an invariant system.
Ans. Refer Section 4.3.
19. Write notes on :
(i) Reduced phase rule,
(ii) Application of phase rule.
Ans. Refer Section (i) 4.4.1, (ii) 4.5.
20. What is triple point? Explain the triple point with reference to water system.
Ans. Refer Section 4.3 (C).
21. How many degrees of freedom are present in the following systems?
(i) A gas in equilibrium with its solution in a liquid.
(ii) A solution of a solid in a liquid in equilibrium with solvent vapour.
(iii) Two partially miscible liquids in the absence of vapour.
(iv) I2 (s) œ I2 (g)
(v) NH3 (g).
Ans. (i) 2, (ii) 2, (iii) 2, (iv) 1, (v) 2.
F.E. Sem.-I Engineering Chemistry-I
4.19
Phase Rule - Gibb’s Phase Rule
4.8 UNIVERSITY QUESTIONS
December 2007
1. Give the demerits of phase rule.
(3 M)
Ans. Refer Section 4.6.
2. State and explain phase rule. Discuss the application of phase rule to one
component water system.
(7 M)
Ans. Refer Section 4.2 and 4.3.
May 2008
1. What is triple point in phase diagram? Explain it with reference to one component
water system phase diagram.
(3 M)
Ans. Refer Section 4.3.
2. Explain any two of the following terms:
(6 M)
(i) Phases, (ii) Components, (iii) Degrees of freedom.
Ans. Refer Section 4.2.1, 4.2.2 and 4.2.3.
3. State and explain condensed phase rule.
(3 M)
Ans. Refer Section 4.4.1.
December 2008
1. Define Gibb’s phase rule equation. Explain the application to one component
system.
(5 M)
Ans. Refer Section 4.2 and 4.3.
2. State and explain condensed phase rule.
(3 M)
Ans. Refer Section 4.4.1.
May 2009
1. What is condensed phase rule equation? Explain its application with the help of
phase diagram to two component lead-silver system.
(6 M)
Ans. Refer Section 4.4.1 and 4.4.4.
2. State phase rule equation. Mention any three applications.
(3 M)
Ans. Refer Section 4.2 and 4.5.
3. What is triple point in phase diagram? Explain it with reference to one component
water system phase diagram.
(3 M)
Ans. Refer Section 4.3.
December 2009
1. State the limitations of phase rule.
(3 M)
Ans. Refer Section 4.6.
2. What is triple point in phase diagram? Explain it with reference to one component
water system phase diagram.
(3 M)
Ans. Refer Section 4.3.
F.E. Sem.-I Engineering Chemistry-I
4.20
Phase Rule - Gibb’s Phase Rule
May 2010
1. What is phase rule? Discuss in brief lead-silver equilibrium with diagram. (5 M)
Ans. Refer Section 4.4.4.
2. Explain any two of the following terms:
(3 M)
(i) Component, (ii) Degrees of freedom.
Ans. Refer Section 4.2.2 and 4.2.3.
3. Write short note on one component system-water.
(5 M)
Ans. Refer Section 4.3.
December 2010
1. What are the merits of phase rule?
(3 M)
Ans. Refer Section 4.5.
2. State and explain phase rule. Discuss the application of phase rule to one
component water system.
(5 M)
Ans. Refer Section 4.3.
May 2011
1. What is condensed phase rule equation? Explain its application with the help of
phase diagram to two component lead-silver system.
(5 M)
Ans. Refer Section 4.4.4.
2. State the limitations of phase rule.
(3 M)
Ans. Refer Section 4.6.
3. Write short note on one component system water.
(5 M)
Ans. Refer Section 4.3.
December 2011
1. Explain the application of phase rule to one component system.
(5 M)
Ans. Refer Section 4.3.
June 2012
1. Explain application of Gibbs Phase Rule to one component system-water system.
Ans. Refer Section 4.3.
(5 M)
2. Using phase rule, find the number of degrees of freedom in the following systems
at equilibrium.
(5 M)
(i) In the water system when
Ice (s) l Water (l) l Water Vapour (g)
(ii) A gaseous mixture of Nitrogen and Hydrogen.
Ans. (i) 0, (ii) 3.
December 2012
1. State limitations of phase rule.
(3 M)
Ans. Refer Section 4.6.
F.E. Sem.-I Engineering Chemistry-I
4.21
Phase Rule - Gibb’s Phase Rule
2. Define phase rule. Explain the terms phase, component and degree of freedom.
Ans. Refer Section 2.2, 2.2.1, 2.2.2 and 2.2.3.
(5 M)
3. Draw neat and labelled phase diagram for water system.
(4 M)
Ans. Refer Section 4.3.
June 2013
1. Define phase, component and degree of freedom.
(3 M)
Ans. Refer Section 4.2.1, 4.2.2 and 4.2.3.
2. Draw and explain phase diagram of water system.
(5 M)
Ans. Refer Section 4.3.
3. Give limitations of phase rule.
(4 M)
Ans. Refer Section 4.6.
December 2013
1. Define Gibbs phase rule. State the number of phases, components and degree of
freedom for the following equations : CaCO3(s) l CaO(s) + CO2(g).
(3 M)
Ans. Refer Section 4.2, 4.2.1, 4.2.2 and 4.2.3.
2. Draw and explain phase diagram of water system with reference to curves and
tripple points.
(5 M)
Ans. Refer Section 4.3.
3. Give limitations of phase rule.
(4 M)
Ans. Refer Section 4.6.
May 2014
1. What is a condensed system? State the condensed phase rule equation.
(3 M)
Ans. Refer Section 4.4.1.
2. Define phase, components and degrees of freedom. State the number of phases,
components and degrees of freedom for the following equilibrium.
(5 M)
H2O(s) l H2O(l) l H2O(vapour).
Ans. Refer Section 4.2.1, 4.2.2, 4.2.3 and 4.3.
3. Draw the phase diagram of water system. Explain tripple point.
(4 M)
Ans. Refer Section 4.3.
December 2014
1. Explain the reduced phase rule.
(3 M)
Ans. Refer Section 4.4.1.
2. Explain the two component, Pb-Ag system with an appropriate phase diagram.
Ans. Refer Section 4.4.4.
(5 M)
3. Explain the following terms giving two examples :
(4 M)
(i) Phase, (ii) Component.
Ans. Refer Section 4.2.1 and 4.2.2.
F.E. Sem.-I Engineering Chemistry-I
4.22
Phase Rule - Gibb’s Phase Rule
May 2015
1. Define Phase, Component and Degree of freedom.
(3 M)
Ans. Refer Section 4.2.1, 4.2.2 and 4.2.3.
2. State Gibbs phase rule. Give its applications to one component system.
(5 M)
Ans. Refer Section 4.2 and 4.3.
3. What is a condensed phase system? Draw the phase diagram of lead silver system
with proper labeling.
(4 M)
Ans. Refer Section 4.4.1 and 4.4.4.
December 2015
1. What is Triple point? Explain it with reference to water system.
(3 M)
Ans. Refer Section 4.3.
2. What is the phase rule? Draw a neat labeled diagram for water system. Using
phase rule find the number of degrees of freedom for the following systems. (5 M)
(a) Ice(s) l Water (l) l Water Vapour (g).
(b) Water(l) l Water Vapour (g).
Ans. Refer Section 4.2 and 4.3.
3. Discuss the limitations of phase rule.
(4 M)
Ans. Refer Section 4.6.
May 2016
1. Give the number of phases in the following systems.
(a) Saturated solution of NaCl
(b) Mixture of rhombic and mono-clinic sulphur
(c) Mixture Oxygen and Nitrogen
(d) Ice l Water.
Ans. (a) 1, (b) 2, (c) 1 and (d) 2.
2. Explain the following terms :
(a) Condensed phase rule, (b) Tripple point.
Ans. (a) Refer Section 4.4.1.
(b) Refer Section 4.3.
3. Discuss the limitations of phase rule.
(3 M)
(5 M)
(4 M)
Ans. Refer Section 4.6.
December 2016
1. State the limitations of phase rule.
Ans. Refer Section 4.6.
2. State the condensed phase rule.
Ans. Refer Section 4.4.1.
(3 M)
(3 M)
F.E. Sem.-I Engineering Chemistry-I
4.23
Phase Rule - Gibb’s Phase Rule
3. What is the application of phase rule to one component water system? Explain
with the help of phase diagram.
(4 M)
Ans. Refer Section 4.3.
4. Define phase, component and degree of freedom.
(3 M)
Ans. Refer Section 4.2.1, 4.2.2 and 4.2.3.
June 2017
1. What is reduced phase rule?
(3 M)
Ans. Refer Section 4.4.1.
2. Draw and explain the important features of phase diagram of water system . (3 M)
Ans. Refer Section 4.3.
3. What is Gibbs phase rule equation? Explain the meaning of each term involved in
it.
(4 M)
Ans. Refer Section 4.2.
4. What is meant by Triple point? Find the degree of freedom for triple point. (3 M)
Ans. Refer Section 4.3.
December 2017
1. State the number of phases and components for the following equilibrium. (3 M)
(i) H2O(s) l H2O(l) l H2O(g)
(ii) Mixture of Rhombic and Monoclinic sulphur.
Ans. (i) Phase = 3, Component = 1, (ii) Phases = 2, Component = 1.
2. What are the limitations of phase rule?
(3 M)
Ans. Refer Section 4.6.
3. What is reduced phase rule? Draw the phase diagram of Ag-Pb system.
(4 M)
Ans. Refer Section 4.4.1 and 4.4.2.
4. Discuss the triple point in one component system.
(3 M)
Ans. Refer Section 4.3.
May 2018
1. What are the limitations of phase rule?
(3 M)
Ans. Refer Section 4.6.
2. Explain the term ‘Phase’ with appropriate example.
(3 M)
Ans. Refer Section 4.2.1.
3. Discus the application of phase rule to one component system based on diagram
and triple point.
(4 M)
Ans. Refer Section 4.3.
4. What is reduced phase rule?
(3 M)
Ans. Refer Section 4.4.1.
F.E. Sem.-I Engineering Chemistry-I
4.24
Phase Rule - Gibb’s Phase Rule
December 2018
1. Explain the reduced phase rule.
(3 M)
Ans. Refer Section 4.4.1.
2. Discuss the advantages and limitations of phase rule.
(3 M)
Ans. Refer Section 4.5 and 4.6.
3. Draw the phase diagram of one component system and find out the number of
degree of freedom along the curves and areas.
(4 M)
Ans. Refer Section 4.3.
4. Define phase, component and degree of freedom.
(3 M)
Ans. Refer Section 4.2.1, 4.2.2 and 4.2.3.
May 2019
1. Discuss the advantages and limitations of phase rule.
(3 M)
Ans. Refer Section 4.5 and 4.6.
2. What is Triple Point? Write the condition at which Triple Point exists for water
system.
(3 M)
Ans. Refer Section 4.3.
3. Draw and explain phase diagram of Pb-Ag system
(4 M)
Ans. Refer Section 4.4.4.
4. Write Gibbs equation of phase rule and define the terms involved in it.
(3 M)
Ans. Refer Section 4.2, 4.2.1, 4.2.2 and 4.2.3.
CHAPTER
5
POLYMERS
SYLLABUS
x
x
x
x
x
x
Introduction: Definition - Polymer, polymerization
Properties of Polymers - Molecular weight (Number average and Weight average), Numerical
problems on molecular weight, effect of heat on polymers (glass transition temperature),
Viscoelasticity, Conducting Polymers
Classification - Thermoplastic and Thermosetting polymers
Compounding of plastic
Fabrication of plastic by Compression, Injection, Transfer and Extrusion moulding
Preparation, properties and uses of PMMA and Kevlar
5.1 INTRODUCTION
Natural occurring polymers – those derived from plants and animal have been used
for many centuries, these materials include wood, rubber, cotton, wool, leather and silk.
Other natural polymers such as proteins, enzymes, starches and cellulose are important in
biological and physiological processes in plants and animals. Modern scientific research
tools have made possible the determination of the molecular structures of this group of
materials and the development of numerous polymers which are synthesized from small
organic molecules. Many of our useful plastics, rubber and fiber materials are synthetic
polymers. In fact, since the conclusion of World War II, the field of materials has been
virtually revolutionized by the advent of synthetic polymers.
Synthetic polymers like plastics, nylon etc. have been developed to serve as
substitutes for natural polymers. This has led to a tremendous growth of the polymer
industry. They can be used in fact anything from clothing to powering a space vehicle to
even replacing a human organ. This explains why this age is called as ‘polymer age’.
The disposal of used polymeric materials has also become a serious environmental
problem as they are non-biodegradable. Therefore, there are attempts to produce
biodegradable polymers.
5.1
F.E. Sem.-I Engineering Chemistry-I
5.2
Polymers
5.2 WHAT ARE POLYMERS?
For one thing, they are complex and giant molecules and are different from low
molecular weight compounds. The small molecules, which combine to form a big
molecule, can be of one or more chemical compounds. To illustrate, imagine a set of
rings of the same size and made of the same material. When these rings are interlinked,
the chain formed can be considered as representing a polymer from molecules of the
same compound. Alternatively, individual rings could be of different sizes and materials
and interlinked to represent a polymer from molecules of different compounds.
The two situations are shown in the Fig. 5.1.
Chains made of interlinked rings (representing polymer molecules)
Fig. 5.1 : Conceptual representation of polymer formation
This interlinking of many units has given this name ‘polymer’.
Thus the word polymer is derived from two Greek words, Poly (many) and Meros
(parts or units). A polymer is a large molecule which is formed by repeated linking of
small molecules.
For example, polyethene is a polymer formed by linking together a large number of
ethene (C2H4) molecule.
ethene (monomer)
polyethene, PE (polymer)
Polystyrene is formed by linking together a large number of styrene molecules.
F.E. Sem.-I Engineering Chemistry-I
5.3
(monomer)
Polymers
Polystyrene, PS (polymer)
Thus, polymers are the giant complex high molecular weight compounds made up of
a large number of small molecules of small molecular weight. The small molecules which
combine with each other to form polymer molecules are termed as monomers. They are
the “building blocks” of polymers.
5.3 POLYMERISATION, DEGREE OF POLYMERISATION
AND FUNCTIONALITY
Polymerization: The process of joining a large number of molecules of monomer or
monomers to form large molecule of polymer is known as polymerization. The polymer
molecule formed contains a structural identity, repeating itself several times. These
repeating entities are called the repeat units or the monomeric units of the polymer
molecule. The size of the polymer molecule is decided by the number of repeat units
present in it. This number denotes the degree of polymerization.
For example, 5 molecules of ethylene monomer can add on to each other to form a
single molecule of polyethylene. Here, the repeat unit is –CH2 –CH2– and the polymer
molecule contains 5 such repeat units. Hence, the degree of polymerization is 5.
5 CH2 = CH2 (Ethylene monomer)
Polymerization
 CH2  CH2  CH2  CH2  CH2  CH2  CH2  CH2  CH2  CH2 
1
2
3
4
5
OR
-- CH2 – CH2-5
DP = 5
(Polyethlene molecule containing 5 repeat units of  CH2  CH2 )
F.E. Sem.-I Engineering Chemistry-I
5.4
Polymers
A PVC polymer may be formed by joining 250 to 400 molecules of vinyl chloride
monomer. Natural rubber may consist of 5,000-10,000 monomer molecules joined
together. A gaseous compound called butadiene with a molecular weight of 54, combines
nearly 4,000 times and gives a polymer known as polybutadine with about 2,00,000
molecular weight. It is essentially the ‘giantness’ of the size of polymer molecule that
makes its behaviour different from that of a chemical compound.
Functionality:
Simple molecules combine together to form macromolecules with or without
elimination of water, ammonia etc. During this process, new C-C linkages result. For a
substance to serve as a monomer, there must be at least two reactive (or bonding) sites.
The number of such reactive sites present in a monomer is referred to as its functionality.
Some monomer molecules have two functional groups but some can have more than two.
A bifunctional monomer has two reactive groups and forms linear polymer. Some of
the examples of bifunctional monomers are given below:
CH2 = CH2
CH2 = CH
ethene
H2N  (CH2)  COOH
CH2 = CH  Cl
glycine
vinyl chloride
C6H5
styrene
A trifunctional monomer has three reactive sites or three functional groups and it
forms a highly branched polymer. Some of its examples of trifunctional monomers are
given bellow:
OH
HO ― CH2  CH  CH2  HO
│
OH
glycerol
Phenol
Ortho, meta and para position provide functionality
In a tetrafunctional monomer, there are four reactive groups and it forms cross linked
polymers. Some of its examples of tetrafunctional monomers are given below:
HC { CH
H2C = CH  CH = CH2
acetylene
(2S bonds make 4 reactive sites)
butadiene
(2 double bonds make 4 reactive sites)
F.E. Sem.-I Engineering Chemistry-I
5.5
Polymers
5.4 CLASSIFICATION OF POLYMERS
Polymers can have different chemical structure, physical properties, mechanical
behaviour, thermal characteristics etc. and can be classified as follows:
5.4.1 Natural and Synthetic Polymers
Depending on their origin, polymers can be grouped as natural or synthetic. Polymers
isolated from natural material are called natural polymers. Typical examples are cotton,
silk, wool and rubber. Celloluse rayon, leather are chemical modifications of natural
polymers.
Polymers synthesized from low molecular weight compounds are called synthetic
polymers. Typical examples are polyethylene, PVC, nylon and terylene.
5.4.2 Organic and Inorganic Polymers
This classification is based on chemical composition of polymer chain.
[A] Organic polymers:
A polymer whose backbone chain is essentially made of carbon atoms is termed an
organic polymer. The atoms attached to the side valencies of the backbone carbon atoms
are usually those of hydrogen, oxygen, nitrogen etc. The majority of synthetic polymers
is organic. Examples of organic polymers are polyethylene, polypropelene, PVC etc.
The body of livings contains organic polymer in the form of proteins, DNA, RNA,
polysaccharides etc.
[B] Inorganic polymers:
If the chain backbone of polymer contains atoms other than carbon, then it is called
inorganic polymer. Some of the examples of inorganic polymers are given below:
H H H H H
│ │ │ │ │
―Si―Si―Si―Si―Si―
│ │ │ │ │
H H H H H
R
R
R
│
│
│
O―Si―O―Si―O―Si―
│
│
│
R
R
R
O
O
║
║
―P―O―P―O
│
│
OR
OR
Polysilane
Silicone polymer
Polyphosphate
5.4.3 Homopolymers and Co-Polymers
Many polymers are better known by their trivial or trade names. A polymer may be
prepared from identical monomers or from monomers of different chemical structures;
accordingly, they are called homopolymers or copolymers.
F.E. Sem.-I Engineering Chemistry-I
5.6
Polymers
(a) Homopolymer : When the polymer is obtained by polymerization of a large
number of identical monomers, it is called Homopolymer.
Consider ‘M’ as the repeat unit.
MMMMMM
Linear Polymer
M
|
MMMMMM
|
|
M
M
Branched Polymer
|
|
M
M
For example, Polyethylene is made from ethylene monomer,
n CH2 = CH2
o
( CH2 – CH2 ) n
Ethene
Polyethylene
Polystyrene is made from styrene
o
( CH2 – CH2 )
n CH2 = CH2
¸
¸
C6H5
C6H5
Styrene
n
Polyethylene
(b) Co-polymer: When the polymer is obtained by polymerization of two or more
different monomers, it is called as co-polymer.
Consider the repeat units as ‘M’ and ‘N’
(i) When two different units in co-polymers are distributed randomly through the
chain, it is called Random co-polymer.
For example,
―M―M—N—MN—N—N―M―
(ii) When two different monomers are distributed alternately through the chain, it is
called alternating co-polymer.
For example,
MNMNMN
(iii)When the sequence of one monomer is followed by sequence of the other
monomer and thus the chain continues, then the polymer is known as block co-polymer.
F.E. Sem.-I Engineering Chemistry-I
5.7
Polymers
For example,
—MMMNNMMMNN—
(block of 3 monomers)
(block of 2 monomer)
(iv) When in a branched polymer, the monomer segments on the branches and the
backbone are different, then it is known as Graft co-polymer.
For example,
— M1 — M1 — M1 — M1 — M1 — M1 —
│
│
M2
M2
│
│
M2
M2
│
│
M2
M2
│
│
or example, Styrene butadiene rubber or SBR is made from styrene and butadiene
monomers.
mx CH2 = CH – CH = CH2 + nx CH2 = CH
Butadiene
|
C6H5
Styrene
(-- CH2 – CH = CH – CH2)m-----(--CH2 – CH)n
|
C6H5 x
Styrene – butadiene rubber or SBR
(a copolymer)
Homochain and Heterochain :
Homochain: If the main chain is made up of same species of atoms, the polymer is
called ‘homochain polymer’.
For example,
¸
¸
¸
¸
¸
¸
CCCCC C
¸
¸
¸
¸
¸
¸
F.E. Sem.-I Engineering Chemistry-I
5.8
Polymers
Heterochain: If the main chain is made up of different atoms, then it is called
‘heterochain polymer’.
For example,  C  C  O  C  C  O  C  C  O  C  C  O 
5.4.4 Linear, Branched and Cross Linked Polymers
The polymers are classified as linear, branched and cross linked polymers on the basis
of their structures.
(i) Linear polymers: The linear polymers are the polymer molecules in which each
repeating structural part is joined to two other such monomers. The bifunctional
monomers generally form the linear polymers.
Consider M and N as the two different monomers.
MMMMMM
or
NNNNNN
Linear homopolymer
MNMNMN
Linear Co-polymer
(ii) Branched polymers: The branched polymers are the polymer molecules that
have short branches growing out from the main chain or branch. These short branches
comprise of the same repeat units as the main chain. The branches result from side
reactions during polymerization. A polymer chain having branch points that connect three
or more chain segments is termed as a branched polymer. Usually the trifunctional
monomers form the branched polymer. For example, Low-density Polyethylene (LDPE).
Slightly branched
polymer
More branched
polymer
F.E. Sem.-I Engineering Chemistry-I
5.9
Polymers
│
M
│
M
│
--M – M – M – M – M – M – M – M – M – M –
│
│
M
M
│
│
M
M
Branched homo-polymer
│
│
M
M
│
│
│
│
N
M
│
│
M
N
│
│
– M – N – M – N—N – M – N – M – N –
Branched Co-polymer
│
N
│
M
│
(iii) Cross- linked polymers: When the monomer molecules are connected to each
other by covalent bonds to form a three dimensional network like polymer molecule, it is
called as cross-linked polymer.
For example, Bakelite (phenol formaldehyde resin)
H2C
F.E. Sem.-I Engineering Chemistry-I
5.10
Polymers
This structure formed is huge, strong and rigid. These polymers are hard, brittle,
mechanically strong and do not soften on heating. Usually polyfunctional monomers give
such type of structures. Most of the thermosetting polymer have such structures.
5.5 MOLECULAR WEIGHT
Molecular weight is an extremely important variable in polymers because it relates
directly to the physical properties of macro molecules. In general, higher the molecular
weight, tougher the polymer; however, too high molecular weight leads to processing
difficulties. The molecular weight and its distribution determine the viscous and elastic
properties of the molten polymer. A variety of methods are available for molecular
weight determination and they are applicable in different ranges of molecular weight.
−
−
(1) Number-average molecular weight (Mn): Mn is defined as the total mass (w) of
all the molecules in a polymer sample divided by the total number of molecules present.
−
Mn is determined by methods such as end group analysis and the measurements of
colligative properties such as osmometry (osmotic pressure measurement), ebullimetry
(elevation in boiling point measurement) and cryoscopy (freezing point depression
measurement).
−
¦ NiMi
Thus the number average molecular weight is Mn =
¦ Ni
Where, Ni is the number of molecules of mass Mi.
The number-average molecular mass is a good index of physical properties such as
impact and tensile strength, but is not a good index of other properties such as flow.
−
(2) Weight- average molecular mass (Mw) : When the molecular weight is averaged
−
according to the weight of molecules of each type, Mw is obtained. It is determined by
light scattering measurement and sedimentation-equilibrium methods. Both the methods
depend on the weights of the molecules present.
NiMi 2
−
¦
The weight-average molecular weight Mw =
¦ NiMi
Where, Ni is the number of molecules of molecular weight Mi
Polydispersity Index :
−
It is defined as the ratio of weight-average molecular mass (Mw) and Number average
molecular mass.
F.E. Sem.-I Engineering Chemistry-I
5.11
Polymers
−
Mw
It can be represented as −
Mn
If value of Polydispersity index is one, polymer is identified as mono-dispersed where
all constituting molecules have identical molecular mass. When Polydispersity index is
more than unity, polymer is identified as polydispersed where constituting molecules
have different molecular mass.
NUMERICALS BASED ON MOLECULAR WEIGHT
Problem 1: A polymer has the following composition: 100 molecules of molecular
mass 1000, 200 molecules of molecular mass 2000 and 500 molecules of molecular
mass 5000 . Calculate the number and weight average of molecular weight and the
polydispersity index.
Solution: Given M1 = 1000 g/mol, N1 = 100; M2 = 2000/ g/mol, N2 = 200; M3 = 5000
g/mol, N3 = 500.
The number average molecular weight:
−
Mn =
¦NiMi
¦Ni
=
100 u 1000 + 200 u 2000 + 500 u 5000
100 + 200 + 500
1 u 105 + 4 u 105 + 25 u 105
=
800
= 3.75 × 103
The weight average molecular weight:
¦NiMi 100 u (1000)2 + 200 u (2000)2 + 500 u (5000)2
−
=
Mw =
100 u 1000 + 200 u 2000 + 500 u 5000
¦NiMi
2
1 u 108 + 8 u 108 + 125 u 108
=
30 u 105
= 4.46 × 103
The polydispersity index (PDI) is
−
4.46 u 103
Mw
=
= 1.19
−
3.75 u 103
Mn
Thus,
−
−
Mn = 3.75 × 103, Mw = 4.46 × 103, PDI = 1.19
F.E. Sem.-I Engineering Chemistry-I
5.12
Polymers
Problem 2: If a polymer sample has population as:
10 molecules of molecular mass = 5000, 20 molecules of molecular mass = 7500
20 molecules of molecular mass = 10000, 25 molecules of molecular mass = 15000
20 molecules of molecular mass = 20000, 5 molecules of molecular mass = 25000
Calculate its number-average and weight –average molecular mass of the polymer.
Solution: The number average molecular weight:
−
Mn =
¦NiMi
¦Ni
10 u 5000 + 20 u 7500 + 20 u 10000 + 25 u 15000 + 20 u 20000 + 5 u 25000
10 + 20 + 20 + 25 + 20 + 5
6
1.3 u 10
= 1.3 × 104
=
100
=
The weight average molecular weight:
−
Mw =
¦NiM2i
¦NiMi
=
10 u (5000)2 + 20 u (7500)2 + 20 u (10000)2 + 25 u (15000)2 + 20 u (20000)2 + 5 u (25000)2
1.3 u 106
=
20125 u 106
= 1.548× 104
1.3 u 106
Thus,
−
−
Mn = 1.3× 104, Mw = 1.548× 104
Problem 3: In a polymer, there are 100 molecules of molecular weight 100, 200
molecules of molecular weight 1000 and 300 molecules of molecular weight 10000.
− −
Find Mn, Mw and PDI.
Solution: The number average molecular weight:
−
Mn =
=
¦NiMi
¦Ni
100 u 100 + 200 u 1000 + 300 u 10000
100 + 200 + 300
3.21 u 106
=
600
= 5.35 × 103
F.E. Sem.-I Engineering Chemistry-I
5.13
Polymers
The weight average molecular weight:
¦NiMi
−
Mw =
¦NiMi
2
=
100 u (100)2 + 200 u (1000)2 + 300 u (10000)2
100 u 100 + 200 u 1000 + 300 u 10000
= 9.4 × 103
The polydispersity index (PDI) is
−
9.4 u 103
Mw
− = 5.35 u 103 = 1.757
Mn
Thus,
−
−
Mn = 5.35 × 103, Mw = 9.4 × 103, PDI = 1.757
5.6 MELTING AND GLASS TRANSITION PHENOMENA
Amorphous polymers do not have melting point but softening point. At low
temperature, polymers exits as glassy material. As the temperature of polymer is
increased, it eventually softens and becomes more flexible. The temperature at which it
becomes soft and rubbery is the Glass transition temperature (Tg). After this crystalline
and amorphous thermoplastic polymer behaves differently.
(a) Amorphous polymers
(b) Crystalline polymers
Fig. 5.2: Effect of temperature changes on amorphous and crystalline polymers
Glass transition temperature (Tg) can be defined as the temperature at which the
internal energy of the molecules of the polymer increases to such an extent that the chain
F.E. Sem.-I Engineering Chemistry-I
5.14
Polymers
segments of the polymer molecules just starts leaving their lattice sites. The properties of
the polymer depend on whether the temperature is above or below Tg .It decides the
characteristics of a polymer Polystyrene (PS) (Tg = 100qC) and polymethyl methacrylate
(PMMA) (Tg > 100qC) are hard and stiff at room temperature. Polyethyl acrylate
(Tg = −70qC) and rubbers (Tg < −50qC) are soft and rubbery. The value of Tg can be
modified by adding plasticizers or through co-polymerization. The factor affecting free
rotation, like introduction of aryl rings, causes stiffening of the chain. In such case, Tg is
raised as more thermal energy would be required to bring mobility of the chains. For
example Tg for poly(oxyethylene) is 206 K and for (polyphenylene oxide) is 356 K.
Facts about Tg:
(1) The value of Tg depends upon:
(a) chain-length, (b) extent of cross linking, (c) the barrier which hinders the
internal rotation around the chain links.
(2) Tg of a given polymer varies with the rate of heating or cooling.
(3) Tg of a linear polymer is fairly sharp, as movements of chain segments from one
side to another does not involve any exchange of bonds.
(4) Tg of partially cross linked polymer is blurred, as movement of chain segments
involves both the exchange and breaking of bonds.
(5) Tg of a linear polymer is lower than that of partially cross linked polymer.
(6) A cross linked polymer does not possess any Tg as polymer does not soften on
heating and gets destroyed at high temperature.
(7) Below Tg, the polymer is hard and brittle.
5.7 VISCOELASTICITY
The physical state of polymer is dependent on their chain length and molecular
weight. They are either elastic solids or viscous liquids depending on their physical state.
The change of the physical properties of the polymer is due to the effect of temperature
on them. The rate of cooling also affects the properties of polymers. If polymer is cooled
slowly it transforms to hard and brittle or vice versa.
All polymers are hard rigid solids at low temperatures and change from the solid to
liquid state either by melting (crystalline polymers) or by softening (amorphous
polymers) at high temperature. Amorphous polymers with random arrangements of their
molecular chains soften and become rubber like at a particular temperature. The rubber
like polymer softens to viscous liquid as the temperature is increased further. The
changes from solid to rubbery polymer to viscous liquid are reversible.
F.E. Sem.-I Engineering Chemistry-I
5.15
Polymers
Thus this behaviour of the polymer to exhibit different range of viscosities with
variation in temperature and the rate of change of temperatures is known as
viscoelasticity.
5.8 PLASTICS
The word plastic originated from Greek, meaning a material which can be moulded or
formed into any shape of one’s own choice. In a broad sense, a plastic refers to a material
which exhibits plasticity i.e. the ability to get deformed or to undergo change of shape
under pressure. Plastic may be defined as organic materials of high molecular weight,
which can be moulded into any desired form, when subjected to heat and pressure in the
presence of a catalyst. They are relatively new materials of construction which rapidly
found extensive industrial application. In recent years, they have attained great
importance in every walk of our life, due to their certain unique properties as follows:
(1) Lightness in weight
(2) Good thermal and electrical insulation.
(3) Good corrosion resistivity.
(4) Easy workability and adhesiveness
(5) Low fabrication cost and decorative surface effect.
(6) Insect resistivity, good strength and toughness
(7) Chemically inert to action of light, oils, acids and dampness
(8) Low maintenance cost and high refractive index.
Plastics must be differentiated from resins. In technical sense, resins are the basic
binding materials which form a major part of plastics and which actually has undergone
polymerization and condensation reactions, during their preparation. Thus there is a little
difference between plastic and resin. These two terms are currently used interchangeably.
The resins may be liquid or viscous or solid but it can be polymerized further to give
required shape in the mould. During the moulding, compounding agents like fillers,
colourants, and catalysts may be added in them. These ingredients provide sturdiness and
other desired characteristics to polymers.
5.8.1 Classification of Plastics
On the basis of setting manner in final shape stage of manufacture on application of
heat and pressure, plastics may be categorized as
(i) Thermoplastics (TP)
(ii) Thermosetting plastics (TS).
F.E. Sem.-I Engineering Chemistry-I
5.16
Polymers
(i) Thermoplastics: They may be defined as those which can be softened on heating
and harden on cooling reversibly. They soften on heating and remain in the same state till
hot. On cooling, they acquire rigidity and hardness. The changes involved in heating and
cooling process are purely of the physical nature. The chemical nature of the material is
not affected. These plastics have linear polymer chains or negligibly branched chains.
They are obtained by addition polymerization.
Mechanically these plastics are soft. They have lesser tensile strength, better
flexibility and poor abrasion resistance. Their molecular weight is relatively less and are
soluble in certain organic solvents. Some of the examples are polyethene, PVC,
polystyrene, polyvinyl acetate, polyacrylonitrile, styrene acrylonitrile co-polymer, nylons,
polyesters etc.
(ii) Thermosets or Thermosetting plastics: Thermosetting plastics are those which
set up on heating and can not be reformed once they are set. In general, they are the
plastics which are formed by condensation. They have three dimensional network
structures and have high molecular weights. They have predominant covalent cross links
between the long chains which are responsible for the three dimensional network
structure.
When they are moulded, additional cross-linking are formed between the long chains
leading to increase in molecular weight. Due to the cross-linkings formed, they acquire
properties as hardness, toughness, brittleness, non-swelling and non-softening etc. The
strength of bonds are retained on heating and they can be softened or reclaimed once they
are cured.
Some of the examples of thermosetting plastics are phenol formaldehyde,
polyurethanes, melamine formaldehyde, urea formaldehyde etc.
Comparison between Thermoplastics and Thermosetting:
Thermoplastics (TP)
Thermosetting (TS)
(1) Softens on heating and stiffen on (1) On heating, they are converted into
cooling, reversibly, by the action of
an infusible mass and once set,
heat.
can’t be reshaped.
(2) They are formed
polymerization.
by
addition (2) They are formed by condensation
polymerization.
(3) Contain long chains with negligible (3) Contains three dimensional array of
cross links.
network.
(4) They can be remoulded, reshaped and (4) They can not be remoulded and
can be used.
reused.
F.E. Sem.-I Engineering Chemistry-I
5.17
Polymers
(5) They are usually soluble in organic (5) They are usually insoluble in any
solvents; some swell up in solvent.
solvents.
(6) They can be reclaimed from the wastes. (6) They can not be reclaimed.
(7) They are usually soft, weak and less (7) They are usually hard, strong and
brittle.
more brittle.
(8) They have low molecular weight as (8) They have high molecular weight as
compared to thermosetting.
compared to thermoplastics
(9) Neighbouring polymeric chains are (9) Neighbouring polymeric chains are
held together by weak Vander Waal’s
held together by strong covalent
forces.
bonds in the form of crosslinks.
(10) Heating and cooling do not alter (10) Chemical changes occur on heating.
chemical nature. Only physical
changes are involved on heating.
(11) For
example,
polyethylene, (11) For example, polyurethane, phenol
polystyrene, PVC, Teflon, nylon etc.
formaldehyde, urea formaldehyde.
5.9 COMPOUNDING OF PLASTICS
In order to impart certain definite properties to finished products, plastic is
compounded with other ingredients. These ingredients either discharge a useful function
during moulding or imparts some useful property to the finished product. This is called a
mixture. The various constituents added are −
(1) Binders, (2) Fillers, (3) Plasticizers, (4) Dyes and pigment, (5) Lubricants,
(6) Catalyst, (7) Stabilizers.
(1) Binders: Usually plastic is classified depending on the type of binder used for its
manufacture. As the name implies, the main purpose of binder is to hold the other
constituents of the plastic together. It also determines the type of treatment needed to
mould the articles from the plastic material. A binder may compose of 30 to 100% of the
plastics.
The binders used may be natural or synthetic resins or cellulose derivatives. In
presence of catalyst they get converted into infusible cross link form. They also influence
the properties of the plastic. e.g. If low molecular weight binder is used, the plastic gets
moulded very easily.
(2) Fillers: Fillers are the substance added to plastic to serve the following purposes:
(a) They reduce the cost of the plastic.
(b) They reduce shrinkage on setting and brittleness
F.E. Sem.-I Engineering Chemistry-I
5.18
Polymers
(c) They impart better tensile strength hardness of the plastic, opacity, finish and
workability. Fillers are added to impart special characters to the finished products.
For example, (i) Barium salts make plastic impervious to X-rays. (ii) Asbestos
provide corrosion and heat resistances. (iii) Carborundum, quartze, mica provide
extra hardness. (iv) Addition of carbon black increases tensile strength.
(v) Shredded textiles increase tensile impact strength.
In addition to above, the other materials in common use as fillers are cotton, corn
husks, graphite, clay, paper pulp, wood flour, pumice, metallic oxides (e.g. ZnO, PbO),
saw dust and metal powders (such as Fe, Cu, Pb and Al). The proportion of the filler
added can reach as high as 50% of the plastic.
(3) Plasticizers: The important function of plasticizer is to improve plasticity and
flexibility, so as to reduce the temperature and pressure required for moulding. They play
very important role in determining the properties of finished product and hence are
chosen accordingly. Their plasticizing properties are believed to be mainly due to the
partial neutralization of the intermolecular forces of attraction in the resin molecules.
They impart greater freedom of movement between the polymeric macromolecules of the
resin. (thus increases flexibility and plasticity). At the same time, it reduces its strength
and decreases its chemical resistance.
Fig. 5.3 : Action of plasticizer in plastics
The proportion of plasticizer can be up to 60% of the plastics. They are often used
with thermosetting plastics. For instance, cellulose derivatives has the tendency to
discolour when moulded unless moulding temperature is reduced by the addition of
suitable plasticizer.
Most commonly used plasticizers are vegetable oils, camphor, esters of oleic, stearic
or phthalic acids, tributyl phosphate, triphenyl phosphate, triacetin etc. When used with
cellulose acetate, camphor increases surface hardness; tributyl phosphate and triphenyl
phosphate impart flame proofness; triacetin and tributyl phosphate improves toughness.
(4) Dyes and pigment: Colour appeal is often of prime importance in high polymer
artifacts. Thus they are meant for providing decorative colours to the plastic. The main
colouring materials are organic dyestuffs and opaque inorganic pigments.
F.E. Sem.-I Engineering Chemistry-I
5.19
Polymers
Table 5.1 : Materials used as colourants
Colorants
Imparted colour
Ultramarine
Blue
Carbon black
Black
Calcium carbonate
White
Zinc oxide
White
Chromium trioxide
Green
Ferric oxide
Red
Antimony sulphide
Crimson red
(5) Lubricants: Theses additives promote a good finish to the plastic materials, in
addition to improving the flow characteristics and friction reduction in the processing
machines. Waxes, oils, stearates, oleates and soaps are employed to make the moulding
of the plastics easier and help in providing a glossy finish to the products.
(6) Catalyst: Accelerators are added in case of thermosetting plastics to accelerate
the polymerization of fusible resin during moulding process, into crosslinked infusible
form. Catalysts which are used for this purpose are hydrogen peroxide, benzoyl peroxide,
acetyl sulphuric acid, ammonia and its salts, metals like Ag, Cu and Pb; metallic oxides
such as zinc oxide etc.
(7) Stabilizers: Alkaline earth oxides, organometallic salts, epoxy compounds and
amine type compounds serve as stabilizers which are added to polymers to prevent their
degradation. Their purpose is to improve the thermal stability during processing. For
example, during the moulding of vinyl chloride and vinylidine chloride polymers, heat
stabilizers are used, as these polymers show a tendency to undergo decomposition and
discolouration at the moulding temperature. Lead salts like white lead, lead chromate,
litharge, red lead serve as opaque moulding compounds. Stearates of lead, cadmium and
barium are transparent moulding compounds.
5.10 MOULDING OR FABRICATION OF PLASTICS
Fabrication of plastic or moulding is the technique of giving desired shape to the
plastic with the help of a mould. This technique involves fabrication of plastic under
severe heat and pressure and is applicable to both the types of resins i.e. thermosetting
and thermoplastic. Before moulding, it is essential to dry the resins in order to achieve
optimum performance of finished product. If moisture is present during moulding, it
generally lowers the density and impairs mechanical and optical properties. The
moulding process involves different techniques which are described below:
F.E. Sem.-I Engineering Chemistry-I
5.20
Polymers
(A) Compression Moulding
(B) Injection Moulding
(C) Transfer Moulding
(D) Extrusion Moulding.
(A) Compression Moulding: This method is applicable to both thermoplastics and
thermosetting resins.
Fig. 5.4 : Compression moulding of plastics
In this method, a synthetic plastic material that has to be moulded is mixed with filler
and the other ingredients in proper proportions and then placed in the mould. The mould
is closed under low pressure. Then it is heated with simultaneous application of the
pressure according to specifications (high pressure of the order of 100 to 500 Kg/cm2 and
temperature of the order of 100 to 200qC). The cavities get filled with fluidized plastic.
Once the moulding is over, the material is withdrawn after cooling. Finally curing is done
either by heating (in case of thermosetting) or cooling (in case of thermoplastics). After
this process is over, the moulded article is taken out by opening the mould parts.
Currently fully automatic moulding presses are available to quicken the process. Door
handles, handles of electrical iron, bottle caps, screw caps are obtained by this method.
(B) Injection Moulding: This method is applicable to thermoplastic resins. The
plastic in the form of powder is fed into a hot cylinder through hopper. The hot softened
F.E. Sem.-I Engineering Chemistry-I
5.21
Polymers
plastic is forced at a controlled rate into a tightly locked mould by means of screw
arrangement or piston. The temperature at the nozzle is increased (between 130q-260qC)
which makes plastic fluidized and is injected in the mould. The mould is kept cooled to
enable the hot plastic to be cured and become rigid. Moulded object is ejected
mechanically without any deformation. Telephones, buckets, dustbins are made by this
technique.
Advantages:
(1) It is widely used for moulding of thermoplastics.
(2) It has high speed production.
(3) It has low finished cost.
(4) It has low loss of material.
Limitation: As a large number of cavities can not be filled simultaneously, there is
limitation of design of articles to be moulded.
Fig. 5.5 : Injection Moulding
(C) Transfer Moulding: This method is used for thermosetting materials. In this
process, the principle of injection moulding is operative. In this the moulding powder is
placed in a heated chamber, maintained at the minimum temperature at which it begins to
become plastic. This plastic material is injected through an orifice, into the mould by a
plunger, working at a high pressure. Due to the friction developed at the orifice, the
temperature of the material rises to such an extent that the moulded material becomes a
liquid and flows into the mould. Curing takes place under the influence of heat and
pressure. The moulded article is then ejected mechanically.
F.E. Sem.-I Engineering Chemistry-I
5.22
Polymers
Fig. 5.6: Transfer Moulding
Advantages:
(1) Due to ejection of the material from orifice to the mould at high speed and in
highly plasticized condition, delicate articles may be handled without distortion or
displacement.
(2) Intricate shapes not attainable by compression moulding can be produced.
(3) Blistering is almost eliminated.
(4) Article free from flow marks are produced.
(5) Thick pieces cure completely and uniformly.
(6) Mechanical strength and density of fabricated piece is higher.
(7) Finished cost of fabrication article is eliminated.
(D) Extrusion Moulding : Extrusion moulding is applicable to thermoplastic resins.
It is mainly used for continuous moulding of thermoplastic materials into articles of
uniform cross section. The thermoplastic ingredients are heated to plastic condition and
then pushed by means of a screw conveyor into a die, having the required outer shape of
the article to be manufactured. The finished product extruding out is cooled by
atmospheric exposure or by blowing air or by spraying water. A long conveyor carries
away continuously the cooled product.
The articles of uniform cross section like tubes, rods, strips insulated electric cables
are manufactured by this technique.
As per the requirement, extrusion moulding can be carried out in two ways:
(a) Vertical extrusion moulding.
(b) Horizontal extrusion moulding.
F.E. Sem.-I Engineering Chemistry-I
5.23
Polymers
Fig. 5.7 (a) Vertical extrusion moulding
Fig. 5.7 (b) Horizontal extrusion moulding
5.11 SOME INDIVIDUAL PLASTICS
1. Polymethyl methacrylate (PMMA) or Lucite or Plexiglass:
Polymethyl methacrylate is the best known plastic from the acrylates because of its
outstanding optical properties. This polymer has the following structure.
CH3
│
 CH2  C 
│
COOCH3
n
Polymethyl methacrylate (PMMA)
F.E. Sem.-I Engineering Chemistry-I
5.24
Polymers
Preparation: The monomer used is methyl methacrylate (an ester of methyl acrylic
acid is produced from acetone). Polymethyl methacrylate is obtained by polymerization
of methyl methacrylate in presence of acetyl peroxide or hydrogen peroxide. It is an
acrylic polymer.
H COOCH3
│ │
nC = C
│ │
H CH3
Polymerization
Methyl methacrylate
(MMA)
H
COOCH3
│
│
―― C ---- C ――――
│
│
H
CH3
n
Polymethyl methacrylate
(PMMA)
Properties:
(1) It is colourless, transparent, hard and fairly rigid material.
(2) It is amorphous because of the presence of bulky side groups in the molecules.
(3) It has high softening point of about 130qC to 140qC and becomes rubber like at a
temperature above 65qC.
(4) It has high optical transparency.
(5) It has high resistance to sunlight and ability of transmitting light accurately, even
in curved sections.
(6) It has low chemical resistance to hot acids and alkalies.
(7) It has low scratch resistance.
Uses:
(1) It is used for automotive applications such as in tail and signal light lenses, dials,
medallions etc.
(2) For making aircraft light fixtures, bomber noses, gun turrets, cockpit canopies,
transparent models of complicated mechanisms.
(3) For making bone splints, artificial eyes, dentures, emulsions, paints, adhesives,
wind screens, jewellery, T.V. screens, guards etc.
(4) It’s sheets are used for signs, glazing skylights and decorative purposes.
2. Kevlar:
It is an aromatic polyamide with benzene rings linked to the amide groups –CONH–.
Preparation: It is prepared by poly condensation between aromatic dichloride and
aromatic diamines.
F.E. Sem.-I Engineering Chemistry-I
n ClOC
COCl
5.25
+
Terephthalic acid dichloride
n H2N
Polymers
NH2
p-Aminoaniline (1, 3-diamobenzene)
- 2 n HCl
O
║
―C
O
H
║
│
C――N
H
│
N―
n
Kevlar
Properties:
(i) It is exceptionally strong (five times stronger than steel).
(ii) It has high heat stability and flexibility.
(iii) It is more rigid than nylon.
Uses:
(i) It is used in the aerospace and aircraft industries.
(ii) It is used in car parts (tyres, brakes etc.)
(iii) It is used in ropes, cables, bullet proof vests, motor cycle helmets etc.
3. Urea formaldehyde resin:
Amino resins are condensation products, obtained by the reaction of urea or
melamine with formaldehyde. Commercially important amino resin is ureaformaldehyde, which is prepared by the reaction between 2 parts of urea and 1 part of
formaldehyde, in basic medium, in a stainless steel vessel at about 50qC. Primary
products are mono and dimethylol ureas.
NHCH2OH
NH2
│
│
OC + HCOH
OC
│
│
NH2
NH2
Urea
Monomethylol urea
NH2
│
CO + 2 HCOH
│
NH2
Urea
NHCH2OH
│
OC
│
NHCH2OH
Dimethylol urea
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5.26
Polymers
For moulding, the methylol derivatives are compounded (with fillers, platicizers,
pigments, and catalyst) and then cured (by applying heat and pressure). During curing,
long – C – N – C – N – chains are formed.
Properties:
1. Urea formaldehyde resins give clear, water-white products of good tensile
strength.
2. It also gives good electrical insulation and good chemical resistance.
3. It gives great hardness, great light-stability and good abrasion resistance.
Uses:
1. It is used for bonding grinding wheels.
2. It is used as binder of glass fibres, rockwool etc., which are used for filtration and
insulation purposes.
3. It is used for foundary cores.
4. It is used as electrical insulator.
It is used as decorative articles like plates, drinking glasses, dishes etc.
4. Phenol formaldehyde resin:
It is prepared by condensing phenol with formaldehyde in the presence of
acidic/alkaline catalyst. Initially, o- and p- hydroxy methyl phenol are formed which react
to form linear polymer novolac.
Phenol
o-Hydroxy methyl
phenol
p-Hydroxy methyl
phenol
F.E. Sem.-I Engineering Chemistry-I
Monomethyl
phenol
5.27
Polymers
Phenol
Novolac
Hexamethylene tetramine is added to novolac which provides formaldehyde. This
converts the soluble and fusible novolac into a hard infusible and insoluble solid of crosslinked bakelite.
Cross-linked polymer bakelite
Properties:
1. It is rigid, hard, infusible, insoluble solid.
2. It is scratch resistant, water resistant, and resistant to non-oxidizing acids and
salts.
3. Possesses excellent electrical insulating character.
Uses:
1. It is used for making telephone parts, cabinets for radio and television.
2. It is used for making bearings used in propeller shaft in paper industry.
3. It is used for making switches, plugs, switch boards etc.
F.E. Sem.-I Engineering Chemistry-I
5.28
Polymers
4. It is used in paints and varnishes.
5. It is used as adhesive for grinding wheels.
5.12 CONDUCTING POLYMERS
Most of the polymeric materials are poor conductors of electricity due to
unavailability of large number of free electrons. In past several years, polymeric materials
have been synthesized possessing electrical conductivities on par with metallic
conductors. Organic polymers with highly delocalized π-electron system having electrical
conductance of the order of conductors are called conducting polymers. The basic
requirement for them is the formation of continuous conjugation through the polymer
chain.
Different types of conducting polymers are as follows.
1. Intrinsically conducting polymers: It is a polymer whose backbone or
associated groups consists of delocalized electron pair or residual charge. In an
electric field, conjugated π-electron of the polymer get excited, and hence can be
transported through the solid polymeric materials. Overlapping of orbitals over
the entire backbone results in the formation of valance bands as well as
conduction bands. Some common examples are polyacetylene polymers,
polyaniline, polyanthrylene, polypyrrole, polythiophene, polyazomethine etc.
2. Extrinsically conducting polymers: It is a polymer whose conductivity is due to
the presence of externally added ingredient in them. They are of two types:
(a) Conductive elements-filled polymers: It is a resin filled with conducting
elements such as carbon black or metal oxides, in which the polymer acts as a
binder to hold the conducting elements. Along with conductivity these
polymers are low in cost, light in weight, mechanically durable and strong.
(b) Blended conducting polymer: It is a polymer obtained by blending a
conventional polymer with conducting polymer. Such polymers possess better
physical, chemical and mechanical properties.
3. Doped conducting polymers: It is obtained by exposing a polymer to a charge
transfer agent either in gas phase or in solution phase. The conductivity of
intrinsically conducting polymers (ICP) can be increased by creating either
positive or negative charges on the polymer backbone by oxidation or reduction
known as doping. It is of two types:
(a) P-doping: In this an intrinsically conducting polymer (ICP) is treated with a
Lewis acid where oxidation process takes place and positive charges on the
polymer backbone are created. Commonly used p-dopants are I2, Br2, AsF5,
PF6, naphthylamine, etc.
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5.29
Polymers
(C2H2)n + 2FeCl3 → (C2H2)n+ FeCl4‾ + FeCl2
2(C2H2)n + 3I2 → 2[(C2H2)n+ I3‾
(b) N-doping: In this an intrinsically conducting polymer (ICP) is treated with a
Lewis base where reduction takes place and negative charges are created on
the polymer backbone. Commonly used n-dopants are Lithium (Li),
Sodium (Na), Calcium (Ca) etc.
Re duction
…–CH = CH – CH = CH – … + B 
o … – CH = CH – CH = CH –…
│
B+
(Lewis Base)
4. Co-ordination conducting polymer: It is a charge transfer complex containing
polymer obtained by combining a metal atom with a poly dentate ligand.
Applications:
(1) In rechargeable light weight batteries based on perchlorate doped polyacetylenelithium system. These are about 10 times lighter than conventional lead storage
batteries.
(2) In wiring in aircrafts and aerospace components.
(3) In electronic devices such as transistors and diodes and in telecommunication
systems.
(4) In antistatic coating for clothing.
(5) In electromagnetic screening materials.
(6) In photovoltaic devices.
(7) In molecular wires and molecular switches.
5.13 POLYMERS IN MEDICINE AND SURGERY
Biomaterials are materials that can be implanted in the body or used in diagnostic,
surgical and therapeutic applications without causing adverse effect on blood and other
tissues. They are developed from metals, ceramics and polymers. Use of biomaterials
made from polymers is increasing day-by-day. Their appeal and acceptability is mainly
due to versatility and the fact that they can be modified at will to suit specific body
functions.
Characteristics of biomedical polymers:
(1) It should be bio compatible.
(2) It should be fabricated into the desired shape or form without being degraded.
(3) It should have purity and reproducibility.
(4) They can be easily sterilized with no alteration in properties.
F.E. Sem.-I Engineering Chemistry-I
5.30
Polymers
(5) They should have optimum physical and chemical properties.
(6) They should not destroy cellular elements of blood, enzyme or produce toxic or
allergic reactions.
(7) They should not deplete electrolytes in the body.
Table 5.2: Applications of biomedical polymers:
Polymer
Applications
1. Polyethylene
Disposable syringe
2. Polypropylene
Heart walls, blood filters
3. Polyvinylchloride
Disposable syringe
4. Acrylic Hydrogels
Grafting
5. PMMA
Contact lenses
6. Polyalkyl sulphone
Membrane Oxygenator
7. Silicone Rubber
Heart walls, drain tubes
8. Polyurethane
Heart wall, blood filters, artificial heart
5.14 REVIEW QUESTIONS
1. Define: (i) monomer, (ii) functionality, (iii) polymer.
Ans. Refer Section 5.2 and 5.3.
2. What are polymers? How they are classified?
Ans. Refer Section 5.2 and 5.4.
3. What is meant by homopolymer or copolymer? Give an example of each.
Ans. Refer Section 5.4.3.
4. What are organic and inorganic polymers?
Ans. Refer Section 5.4.2.
5. How polymers are classified on the basis of their structure?
Ans. Refer Section 5.4.4.
6. Define degree of polymerization?
Ans. Refer Section 5.3.
7. What are plasticizers? What are their functions?
Ans. Refer Section 5.9.
8. Write a note on fillers.
Ans. Refer Section 5.9.
F.E. Sem.-I Engineering Chemistry-I
5.31
Polymers
9. What is meant by compounding of plastics? What are the different constituents of
compounding? Write their uses.
Ans. Refer Section 5.9.
10. What are the advantages of transfer moulding?
Ans. Refer Section 5.10.
11. Describe extrusion moulding.
Ans. Refer Section 5.10.
12. Give the classification of plastics
Ans. Refer Section 5.8.1.
13. What are the methods used for fabrication of plastic? Explain any one of these
methods in brief.
Ans. Refer Section 5.10.
14. Write a note on compression moulding.
Ans. Refer Section 5.10.
15. Differentiate between thermosetting plastics and thermosoftening plastics.
Ans. Refer Section 5.8.1.
16. Describe the method of preparation, properties and application of polymethyl
methacryalate (PMMA).
Ans. Refer Section 5.11.
17. Give the preparation, properties and uses of Kevlar.
Ans. Refer Section 5.11.
18. Write a note on conducting polymer.
Ans. Refer Section 5.12.
19. Explain in brief the role of polymers in medicine and surgery.
Ans. Refer Section 5.13.
20. What is glass transition temperature? Explain its significance.
Ans. Refer Section 5.6.
21. Write a note on viscoelasticity.
Ans. Refer Section 5.7.
22. Give preparation, properties and uses of urea formaldehyde resin.
Ans. Refer Section 5.11.
23. Give preparation, properties and uses of phenol formaldehyde resin.
Ans. Refer Section 5.11.
24. Write a note on injection moulding.
Ans. Refer Section 5.10.
F.E. Sem.-I Engineering Chemistry-I
5.32
Polymers
25. Write a note on molecular weight of polymer.
Ans. Refer Section 5.5.
5.15 UNIVERSITY QUESTIONS
December 2007
1. Explain the term ‘glass transition temperature’. What is its significance?
(3 M)
Ans. Refer Section 5.6.
2. What is polymerization?
(2 M)
Ans. Refer Section 5.3.
3. What is meant by fabrication of plastics? Name different methods of fabrications.
Explain transfer moulding with the help of a neat diagram.
(6 M)
Ans. Refer Section 5.10.
May 2008
1. What is polymerization? What are the conditions of polymerization?
(3 M)
Ans. Refer Section 5.3.
2. Distinguish between thermoplastics and thermosetting plastics.
(3 M)
Ans. Refer Section 5.8.1.
3. What do you mean by fabrication of plastics? Explain transfer moulding with the
help of a neat diagram.
(6 M)
Ans. Refer Section 5.10.
4. Explain Glass transition temperature.
Ans. Refer Section 5.6.
5. Write short note on conducting polymers.
Ans. Refer Section 5.12.
6. Write the preparation, properties and uses of urea formaldehyde.
(2 M)
(5 M)
(3 M)
Ans. Refer Section 5.11.
December 2008
1. State applications of conducting polymers.
(3 M)
Ans. Refer Section 5.12.
2. Explain preparation, method and uses of PMMA and urea formaldehyde. (7 M)
Ans. Refer Section 5.11.
3. With reference to polymers explain glass transition temperature.
(3.5 M)
Ans. Refer Section 5.6.
F.E. Sem.-I Engineering Chemistry-I
5.33
Polymers
May 2009
1. Explain preparation, properties and uses of PMMA.
(3 M)
Ans. Refer Section 5.11.
2. What are the main constituents of plastics? Write the functions and examples of
each constituent.
(5 M)
Ans. Refer Section 5.9.
3. Write short note on injection moulding process.
(5 M)
Ans. Refer Section 5.10.
4. Write the preparation, properties and uses of urea formaldehyde.
(3 M)
Ans. Refer Section 5.11.
December 2009
1. Distinguish between thermoplastics and thermosetting plastics.
(3 M)
Ans. Refer Section 5.8.1.
2. Name the methods for fabrication of plastics. With the help of labelled diagram,
describe transfer moulding and injection moulding.
(7 M)
Ans. Refer Section 5.10.
3. Write short note on conducting polymers.
(5 M)
Ans. Refer Section 5.12.
4. Write the preparation, properties and uses of phenol formaldehyde.
(3 M)
Ans. Refer Section 5.11.
May 2010
1. Give synthesis properties and uses of PMMA and urea formaldehyde.
(5 M)
Ans. Refer Section 5.11.
2. Define conducting polymers. Explain how polymers are made conductors with
suitable examples.
(2.5 M)
Ans. Refer Section 5.12.
3. Write a note on glass transition phenomenon.
(2.5 M)
Ans. Refer Section 5.6.
4. What is fabrication? Explain any two methods of fabrication of plastics in detail.
Ans. Refer Section 5.10.
(5 M)
December 2010
1. Explain the following additives used for compounding of plastics:
(i) plasticizers, (ii) resins, (iii) catalysts.
Ans. Refer Section 5.9.
(6 M)
F.E. Sem.-I Engineering Chemistry-I
5.34
2. Explain in detail the injection moulding for fabrication of plastics.
Ans. Refer Section 5.10.
Polymers
(4 M)
May 2011
1. Distinguish between thermoplastics and thermosetting plastics.
(3 M)
Ans. Refer Section 5.8.1.
2. What is meant by fabrication of plastics? Explain extrusion moulding with the
help of neat diagram.
(5 M)
Ans. Refer Section 5.10.
3. What are the main constituents of plastics? Write the functions and examples of
each constituent.
(5 M)
Ans. Refer Section 5.9.
4. Write short note on conducting polymers.
(5 M)
Ans. Refer Section 5.12.
5. Write the preparation, properties and uses of phenol formaldehyde.
(3 M)
Ans. Refer Section 5.11.
December 2011
1. What is degree of polymerization? Give its significance.
(2 M)
Ans. Refer Section 5.3.
2. Distinguish between thermoplastics and thermosetting plastics.
(5 M)
Ans. Refer Section 5.8.1.
3. Explain the term ‘glass transition temperature’. What is its significance?
(5 M)
Ans. Refer Section 5.6.
4. What is fabrication? What are the various types? With a neat diagram explain any
one methods of fabrication of plastics in detail.
(6 M)
Ans. Refer Section 5.10.
5. Advanced polymeric materials like conducting polymers have gained increasing
importance in the recent years. Explain, what are their structural features with one
example are.
(4 M)
Ans. Refer Section 5.12.
June 2012
1. What are the functions of fillers and plasticizers in the compounding of plastics?
Ans. Refer Section 5.9.
(4 M)
2. Write preparation, properties and uses of Polymethyl methacrylate (PMMA) and
urea formaldehyde.
(6 M)
Ans. Refer Section 5.11.
F.E. Sem.-I Engineering Chemistry-I
5.35
Polymers
3. What is meant by fabrication of plastics? Explain compression moulding with the
help of diagram.
(6 M)
Ans. Refer Section 5.10.
4. Define glass transition temperature of polymer. What factors influence its value?
Ans. Refer Section 5.6.
(5 M)
December 2012
1. Write synthesis, properties and applications of Kevlar.
(3 M)
Ans. Refer Section 5.11.
2. What is fabrication technology? Mention various moulding techniques and
explain injection moulding with the help of diagram.
(6 M)
Ans. Refer Section 5.10.
3. Write a note on compounding of plastic.
(5 M)
Ans. Refer Section 5.9.
4. Write the preparation, properties and uses of phenol formaldehyde.
(3 M)
Ans. Refer Section 5.11.
June 2013
1. What are plasticizers? Give its function.
Ans. Refer Section 5.9.
2. Explain fabrication of plastic with example of injection moulding.
Ans. Refer Section 5.10.
3. Write preparation, properties and uses of Kevlar.
Ans. Refer Section 5.11.
4. Explain effects of heat on polymers and factors affecting it.
Ans. Refer Section 5.6.
5. Write short notes on.
(i) Conducting Polymers,
(ii) Polymers in medicine and surgery.
(3 M)
(5 M)
(3 M)
(5 M)
(5 M)
Ans. Refer Section 5.12 and 5.13.
December 2013
1. What is the function of plasticiser in the compounding of plastic? Give two
examples.
(3 M)
Ans. Refer Section 5.9.
2. Distinguish between thrmoplastic and thermosetting resins :
Ans. Refer Section 5.8.1.
(3 M)
F.E. Sem.-I Engineering Chemistry-I
5.36
Polymers
3. Define moulding and discuss the injection moulding method of fabrication of
plastic.
(5 M)
Ans. Refer Section 5.10.
4. Give the preparation, properties and uses of PMMA.
(3 M)
Ans. Refer Section 5.11.
5. What is glass transition temperature? What are the factors that affect it, and what
is the significance of it?
(5 M)
Ans. Refer Section 5.6.
6. Define conducting polymers. Explain intrinsic and doped polymers with
appropriate examples.
(5 M)
Ans. Refer Section 5.12.
May 2014
1. Name the various ingredients in the compounding of plastics and give two
examples of it.
(3 M)
Ans. Refer Section 5.9.
2. What are thermoplastic polymers? Give two examples. Give the preparation,
properties and uses of any one of it.
(5 M)
Ans. Refer Section 5.8.1 and 5.11.
3. Write short notes on any two :
(6 M)
(a) Glass transition temperature,
(b) Conducting Polymers,
(c) Polymers used in medicine and surgery.
Ans. Refer Section 5.6, 5.12 and 5.13.
4. What is moulding? Explain with the help of diagram extrusion moulding. (5 M)
Ans. Refer Section 5.10.
5. Give the preparation and uses of Kevlar.
(2 M)
Ans. Refer Section 5.11.
December 2014
1. Explain the role of plasticizers and lubricants in the compounding of plastics.
Ans. Refer Section 5.9.
2. Distinguish between thermoplastic and thermosetting resins.
Ans. Refer Section 5.8.1.
3. Give the preparation, properties and uses of :
(a) PMMA, (b) Kevlar.
Ans. Refer Section 5.11.
(3 M)
(6 M)
F.E. Sem.-I Engineering Chemistry-I
5.37
Polymers
4. What is fabrication of plastics? Explain the injection moulding method with the
help of a neat diagram.
(5 M)
Ans. Refer Section 5.10.
5. Discuss any two of the following :
(5 M)
(a) Glass transition temperature,
(b) Polymers in medicine and surgery,
(c) Conducting Polymers.
Ans. Refer Section 5.6, 5.13 and 5.12.
May 2015
1. Explain Glass transition temperature of Polymer and its significance.
Ans. Refer Section 5.6.
2. Explain compounding of plastics. (Five ingredients)
Ans. Refer Section 5.9.
3. Define fabrication. Explain compression moulding with labelled diagram.
Ans. Refer Section 5.10.
4. Give the preparation and applications of the following :
(a) PMMA, (b) Kevlar, (c) Bakelite.
(3 M)
(5 M)
(5 M)
(5 M)
Ans. Refer Section 5.11.
December 2015
1. Distinguish between thermoplastic and thermosetting resins.
(3 M)
Ans. Refer Section 5.8.1.
2. Explain the functions of the following constituents in the compounding of plastics
(any two).
(5 M)
(a) Fillers, (b) Plasticizers, (c) Lubricants.
Ans. Refer Section 5.9.
3. Give the preparation, properties and uses of :
(a) PMMA, (b) Kevlar.
(6 M)
Ans. Refer Section 5.11.
4. What is fabrication of plastics? Explain the injection moulding method with the
help of a neat diagram.
(5 M)
Ans. Refer Section 5.10.
May 2016
1. Thermosetting polymers cannot be reshaped and reused. Give reasons.
Ans. Refer Section 5.8.1.
(3 M)
F.E. Sem.-I Engineering Chemistry-I
5.38
Polymers
2. Describe a moulding method suitable for thermoplastic resins.
(5 M)
Ans. Refer Section 5.10.
3. Give the preparation, properties and uses of PMMA.
(3 M)
Ans. Refer Section 5.11.
4. What is glass transition temperature? What are the factors affecting it? What is its
significance?
(5 M)
Ans. Refer Section 5.6.
5. Explain the functions of the following constituents in the compounding of
plastics. (any two)
(5 M)
(a) Plasticizers, (b) Lubricants, (c) Stabilizers
Ans. Refer Section 5.9.
December 2016
1. Explain the role of polymers in medicine and surgery with the help of three
examples.
(3 M)
Ans. Refer Section 5.13.
2. Distinguish between thrmoplastic and thermosetting resins.
(3 M)
Ans. Refer Section 5.8.1.
3. What is meant by fabrication of plastics? Explain injection moulding with the
help of diagram.
(6 M)
Ans. Refer Section 5.10.
4. What are functions of fillers and plasticizers in the compounding of plastics?
Ans. Refer Section 5.9.
(4 M)
5. Write the properties, preparations and uses of PMMA.
(3 M)
Ans. Refer Section 5.11.
June 2017
1. Give the preparation, properties and uses of PMMA.
(3 M)
Ans. Refer Section 5.11.
2. Write a note on conducting polymers.
(3 M)
Ans. Refer Section 5.12.
3. Write short notes on :
(6 M)
(a) Injection Moulding method for plastics.
(b) Polymers in medicine and surgery.
Ans. Refer Section 5.10.
4. What are the additives mixed with plastics for its compounding? Explain their
functions.
(4 M)
Ans. Refer Section 5.9.
F.E. Sem.-I Engineering Chemistry-I
5.39
5. Distinguish between thermoplastic and thermo setting plastic.
Ans. Refer Section 5.8.1.
6. Write a note on viscoelastic state.
Ans. Refer Section 5.7.
December 2017
1. Give the preparation, properties and uses of Kevlar.
Ans. Refer Section 5.11.
2. What are plasticizers? Give it’s uses and examples.
Ans. Refer Section 5.9.
3. Distinguish between thermoplastic and thermosetting plastics.
Ans. Refer Section 5.8.1.
4. Define fabrication. Discuss extrusion moulding in detail.
Ans. Refer Section 5.10.
5. List the uses of polymers in medicine and surgery.
Ans. Refer Section 5.13.
6. Write notes on :
(a) Glass Transition Temperature,
(b) Conducing polymers.
Polymers
(3 M)
(3 M)
(3 M)
(3 M)
(3 M)
(6 M)
(4 M)
(6 M)
Ans. Refer Section 5.6 and 5.12.
May 2018
1. Discuss the role of polymers in medicines and surgery.
(3 M)
Ans. Refer Section 5.13.
2. Give the preparation, properties and uses of Kevlar.
(3 M)
Ans. Refer Section 5.11.
3. Define moulding. Explain injection moulding with the help of neat diagram. (6 M)
Ans. Refer Section 5.10.
4. Discuss the following additives in compounding of plastics.
(a) Fillers, (b) Plasticizers.
Ans. Refer Section 5.9.
5. Write short notes on Glass Transition Temperature
(4 M)
(3 M)
Ans. Refer Section 5.6.
December 2018
1. Give the preparation, properties and uses of Kevlar.
Ans. Refer Section 5.11.
(3 M)
F.E. Sem.-I Engineering Chemistry-I
5.40
2. What is Glass Transition Temperature? Write its significance.
Ans. Refer Section 5.6.
3. Distinguish between thermoplastic and thermosetting plastics.
Ans. Refer Section 5.8.1.
4. What is fabrication of plastics? Explain injection moulding in detail.
Ans. Refer Section 5.10.
Polymers
(3 M)
(3 M)
(6 M)
May 2019
1. Discuss the effect of temperature on polymers.
(3 M)
Ans. Refer Section 5.6.
2. What is compounding of plastics? Explain the role played by various constituents
used during compounding of plastics.
(6 M)
Ans. Refer Section 5.9.
3. Write short notes on Glass Transition Temperature.
(3 M)
Ans. Refer Section 5.6.
CHAPTER
6
WATER
SYLLABUS
x
x
x
x
x
Introduction - Impurities in water, hardness of water- units (no conversions), types and
numerical problems
Determination of hardness of water by EDTA method and numerical problems
Softening of water by Ion Exchange process and numerical problems
BOD, COD- definition, significance and numerical problems
Water purification-membrane technology- Electrodialysis, Reverse osmosis, and Ultra
filtration
6.1 INTRODUCTION
Water is nature’s most wonderful, abundant and useful compound. It is essential for
the lives of all living kingdom including man, animal and plants. It has occupied a unique
position in industries. It acts as a coolant in power and chemical plants. It is widely used
in other fields such as rayon, paper, chemicals, ice, textiles, and production of steel. In
addition to that, it is used for drinking, bathing, sanitary, washings, irrigation, fire
fighting etc. It’s most important use as an engineering material is in ‘Steam Generation’.
Thus, the science of water has become most important aspect for the mankind.
Earth, our planet on which we live is called blue planet because 4/5th of the earth’s
crust is covered with water. But the hard fact of life is that about 97% of it is locked in
ocean which is too saline to drink and for direct use for agricultural and industrial
purposes. Out of what is left, about 80% is trapped in polar ice caps. Another 10% of it is
locked in rock crevices and minerals lying as deep as 800 meters below the Earth’s crust.
Only about 0.3% of the world’s water resources is available for domestic, agriculture and
industrial purposes. Still we do not understand the importance of water and continuously
misuse it.
6.2 SOURCES OF WATER
While studying the various sources of water, our focus remains on the difference in
the chemical composition of water with respect to its source.
6.1
F.E. Sem.-I Engineering Chemistry-I
6.2
Water
The available water sources are:
(i) Rain water: It is purest form of naturally occurring water since it is obtained
from evaporation of surface water which is nature’s own wonderful distillation
unit. By the time, rain water reaches the Earth’s surface, it gets contaminated due
to the several gases like Carbon dioxide, Sulphur dioxide, nitrogen dioxide,
ammonia etc. and also the suspended solid particles present in the atmosphere.
(ii) River water: Rivers are fed by rain and spring waters. Water from these sources
flow over the surface of land, dissolves the soluble minerals of the soil and falls
into the river. River water thus contains dissolved minerals of the soil such as
chlorides, sulphates, bicarbonates of sodium, calcium, magnesium and iron. It
also contains organic matter derived from decomposition of plants and small
particles of sand and rock in suspension.
(iii) Lake water: It has more constant chemical composition with high amount of
organic matter.
(iv) Sea water: It is the most impure form of natural water. It contains 3.5% of
dissolved salts of which sodium chloride constitutes about 2.6%. The other salts
present are sulphates of sodium, bicarbonates of potassium, magnesium and
calcium, bromides of potassium and magnesium etc.
(v) Spring and well water: Rain water percolates underground to form springs and
wells which contain more dissolved salts and high hardness.
6.3 HARDNESS OF WATER
Hardness in water is that characteristic which prevents the lathering of soap. Water is
said to be hard when it does not produce lather readily with soap in contrast to soft water.
Hardness in water is attributed to the presence of dissolved salts of calcium, magnesium
and to a minor extent certain heavy metals such as iron. These metal ions form insoluble
precipitate with the fatty acid components of soap.
Soap generally consists of sodium salts of long chain fatty acid such as oleic acid,
palmetic acid and stearic acid. Hard water when treated with soap, sodium or potassium
salt of higher fatty acid does not produce lather but forms precipitate. Thus, it does not
possess any detergent value. Typical reactions of soap (sodium stearate) with calcium
chloride and magnesium sulphates can be written as follows:
+ 2NaCl
2C17H35COONa + CaCl2 o (C17H35COO)2 Ca p
Sodium stearate Hardness
Calcium stearate (Insoluble)
F.E. Sem.-I Engineering Chemistry-I
6.3
Water
2C17H35COONa + MgSO4 o (C17H35COO)2 Mg p + Na2SO4
Sodium stearate Hardness
Magnesium stearate (Insoluble)
Hard and Soft Water : Water which does not produce lather with soap solution
readily but forms a white curd is called hard water. On the other hand, water which
lathers easily on shaking with soap solution is called soft water. Such water does not
contain dissolved calcium and magnesium salts in it. Hard water consumes much more
amount of soap to produce the same amount of lather as compared to soft water because
initially the soap is utilized to precipitate the hardness causing metal ions. The anions
usually associated with these metal ions include chloride, sulphate, and bicarbonate but
these do not contribute to the hardness of water.
6.3.1 Types of Hardness
There are two types of hardness :
(i) Temporary or Carbonate or Alkaline Hardness :
It is caused by the presence of dissolved bicarbonates of calcium, magnesium and
other heavy metals and the carbonate of iron.
The temporary hardness can be easily destroyed by mere boiling of water. On boiling,
bicarbonates are decomposed, yielding insoluble carbonates or hydroxides which are
deposited as a crust at the bottom of the vessel and carbon dioxide sets free.
heat
Ca (HCO3 ) 2 
o CaCO3 p H 2O CO 2 n
(inso lub le)
Calcium
bicarbonate
Calcium
carbonate
(ii) Permanent or Non-carbonate or Non-alkaline Hardness :
It is caused due to the presence of chlorides, nitrates, sulphates etc. of calcium,
magnesium and other metals. This type of hardness cannot be removed by adapting easy
means like boiling. It requires special chemical treatment for removal of hardness causing
salts, such as internal conditioning or external treatment which involves the softening
methods.
Permanent hardness is obtained by subtracting the value of temporary hardness from
total hardness.
Thus, Total Hardness = [Temporary Hardness + Permanent Hardness].
F.E. Sem.-I Engineering Chemistry-I
6.4
Water
6.3.2 Causes of Hardness
When water flows over or percolates through the ground rocks or solids, there are
various physical and chemical changes that take place. Due to these changes, it gets
contaminated with dissolved salts of calcium and magnesium. We call it as hardness of
water.
The various changes causing hardness of water are as follows:
(i) Dissolution: Mineral constituents of rocks like sodium chloride, gypsum
(CaSO4.2H2O) etc. dissolved in it.
(ii) Hydration: Some minerals like anhydrite (CaSO4), Olivine (Mg2SiO4) etc
undergo hydration leading to the formation of products of increased volume, due
to which the disintegration of mineral bearing rocks take place.
Hydration
CaSO 4 2H 2O 
o CaSO 4 ˜ 2H 2O
(Anhydrite)
(Gypsum)
Hydration
Mg 2SiO 4 xH 2O 
o Mg 2SiO 4 ˜ xH 2O
(Olivine)
(Serpentine)
(iii) Action of dissolved oxygen: Dissolved oxygen brings about oxidation and
hydration reactions.
(iv) Action of dissolved carbon dioxide:
(a) It converts insoluble carbonates of calcium, magnesium and iron into soluble
bicarbonates.
For example,
CaCO3 CO 2 H 2O 
o Ca(HCO3 ) 2
Inso lub le
So lub le
MgCO3 CO 2 H 2O 
o Mg(HCO3 ) 2
Inso lub le
So lub le
(b) It converts rock forming silicates and alumino silicates of sodium, potassium,
calcium and iron into soluble carbonates, bicarbonates and silica.
For example,
K2O.Al2O3.6SiO2 + CO2 + 2 H2O o Al2O3.2Si.O2.2H2O + K2CO3 + 4SiO2
F.E. Sem.-I Engineering Chemistry-I
6.5
Water
(c) Rocks containing felspar disintegrate and charge nearby river water with
dissolved salts, fine clay and silica in suspension.
Comparisons:
Temporary Hardness
Permanent Hardness
1. It is due to bicarbonates and carbonates 1. It is due to chlorides, sulphates,
nitrates of Ca2+, Fe2+, Mg2+etc. other
of Ca2+, Fe2+, Mg2+etc.
than carbonates and bicarbonates.
2. It is known as carbonate or alkaline 2. It is known as non-carbonate or nonhardness.
alkaline hardness.
3. Temporary hardness leads to formation 3. Permanent
hardness
leads
of loose deposits of carbonates and
formation of adherent scales.
2+
2+
hydroxides of Ca , Mg respectively,
if used in boilers.
to
4. Temporary hardness can be removed by 4. Permanent hardness cannot be
simple techniques such as boiling and
removed by simple techniques such
filtering.
as boiling and filtering.
Hard Water
Soft Water
1. Hard water is one which does not 1. Soft water gives lather easily on
shaking it with soap solution.
produce lather with soap solution
readily but forms a white curd.
2. Hard water contains dissolved calcium 2. Soft water does not contain dissolved
and magnesium salts.
calcium and magnesium salts in it.
3. Cleansing quality of soap is depressed 3. Cleansing quality of soap is not
and a lot of soap is wasted during
depressed and soap is not wasted
washing and bathing.
during washing and bathing.
4. Due to presence of dissolved hardness 4. Less fuel and time are required for
cooking in the soft water.
producing salt, boiling point of water is
elevated. Consequently more fuel and
time are required for cooking.
6.4 UNITS OF HARDNESS
Although hardness of water is never present in the form of calcium carbonate as it is
insoluble in water, hardness of water is conveniently expressed in terms of equivalent
F.E. Sem.-I Engineering Chemistry-I
6.6
Water
amount of CaCO3. Calcium carbonate is chosen particularly for reporting the hardness of
water since it simplifies the calculations as its molecular weight is 100. Moreover, it is
the most insoluble salt that can be precipitated in water treatment.
The equivalents of CaCO3
[Mass of hardness producing substance] u [Chemical equivalent of CaCO3]
=
Chemical equivalent of hardness producing substance
Mass of hardness producing substance u 50
= Chemical equivalent of hardness producing substance
Table 6.1 : Calculation of Equivalents of Calcium Carbonate
Molar Mass
Chemical
Equivalent
Multiplication Factor for
converting into
Equivalents of CaCO3
Ca(HCO3)2
162
81
100/162
Mg(HCO3)2
146
73
100/146
CaCO3
100
50
100/100
MgCO3
84
42
100/84
CaCl2
111
55.5
100/111
MgCl2
95
47.5
100/95
CaSO4
136
68
100/136
MgSO4
120
60
100/120
Mg(NO3)2
148
74
100/148
CO2
44
22
100/44
61
61
100/61 u 2
CO3
60
30
100/60
OH–
17
17
100/34
NaAlO2
82
82
100/82 u 2
Al2(SO4)3
342
57
100/114
FeSO4.7H2O
278
139
100/278
H+
1
1
100/2
HCl
36.5
1
100/73
Dissolved Salt/Ion
HCO3
–
2–
F.E. Sem.-I Engineering Chemistry-I
6.7
Water
Following are the units for expressing hardness of water in terms of CaCO3 equivalent:
(1) Parts per million (ppm): It is the part of calcium carbonate equivalent hardness
present per 106 parts of water.
1 ppm = 1 part of CaCO3 equivalent hardness in 106 parts of water
(2) Milligrams per litre (mg/L): It is the number of milligrams of calcium carbonate
equivalent hardness present per litre of water.
1 mg/L = 1 mg of CaCO3 equivalent hardness of 1litre water
But 1 L weighs 1 Kg.
?
1 L = 1 Kg = 1000g = 1000 u 1000 mg = 106 mg
1 mg/L = 1 mg of CaCO3 equivalent per 106 mg of water
= 1 parts of CaCO3 equivalent per 106 mg of water
= 1 ppm
(3) Clarke’s Degree (qCl): It is number of grains (1/7000 lb) of CaCO3 equivalent
hardness per gallon (10 lb) of water or it is part of CaCO3 equivalent hardness per
70,000 parts of water.
1 qClarke =1 grain of CaCO3 equivalent hardness per gallon of water.
Or 1 qClarke =1 part of CaCO3 equivalent hardness per 70,000 parts of water.
(4) Degree French (qFr): It is the part of CaCO3 equivalent hardness per 105 parts of
water.
1 qFr = 1 part of CaCO3 equivalent hardness per 105 parts of water.
Relationship between various units:
1 ppm = 1 mg/L = 0.1q Fr = 0.07q Cl
1 mg/L = 1 ppm = 0.1q Fr = 0.07q Cl
1q Cl = 1.43q Fr = 14.3 ppm = 14.3 mg/L
1q Fr = 10 ppm = 10 mg/L = 0.7q Cl
Degree of hardness is total quantity of hardness causing salts present in water and it is
expressed in terms of CaCO3 equivalent quantity. It may be called total hardness of
water. The degree of hardness in a water sample can be found by methods like EDTA
titration, soap solution titration. The water sample may contain temporary hardness
causing salts or permanent hardness causing salts. Thus, degree of hardness or total
hardness can be represented as
Degree of hardness = Permanent hardness + Temporary hardness
The degree of hardness in natural water varies from sample to sample.
F.E. Sem.-I Engineering Chemistry-I
6.8
Water
NUMERICALS BASED ON HARDNESS OF WATER
Problem 1: Calculate temporary hardness and permanent hardness of water sample
from the following data:
Mg(HCO3)2 = 16.8 mg/L, MgCl2 = 19 mg/L, MgSO4 = 24 mg/L,
Mg(NO3)2 = 29.6 mg/L, CaCO3 = 4 mg/L, MgCO3 = 10 mg/L
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in mg/L
Multiplication factor
CaCO3 equivalent
16.8
100/146
11.5
MgCl2
19
100/95
20
MgSO4
24
100/120
20
29.6
100/148
20
Ca CO3
04
100/100
4
MgCO3
10
100/84
11.9
Mg(HCO3)2
Mg(NO3)2
Temporary Hardness =
=
=
Permanent Hardness =
=
=
Hardness due to [Mg(HCO3)2 + CaCO3 + MgCO3]
11.5 + 04 + 11.9
27.4 mg CaCO3 equivalent / litre
Hardness due to [MgCl2 + MgSO4 + Mg(NO3)2]
20 + 20 +20
60 mg CaCO3 equivalent / litre
Thus,
Temporary hardness = 27.4 mg CaCO3 equivalent / litre
Permanent hardness = 60 mg CaCO3 equivalent / litre
Problem 2: Calculate carbonate and non-carbonate hardness of water containing
following impurities in mg/L:
Mg(HCO3)2 = 7.1, Ca(HCO3)2 = 8.1, MgCO3 = 4.2, CaCO3 = 10, MgSO4 = 24.
Solution: Conversion into CaCO3 equivalent:
Quantity in mg/L
Multiplication factor
CaCO3 equivalent
Mg(HCO3)2
7.1
100/146
05
Ca(HCO3)2
8.1
100/162
05
MgCO3
4.2
100/84
05
CaCO3
10
100/100
10
MgSO4
24
100/120
20
Constituents
F.E. Sem.-I Engineering Chemistry-I
6.9
Water
Carbonate Hardness = Hardness due to
[Mg(HCO3)2 + Ca(HCO3)2 + CaCO3 + MgCO3]
= 05 + 05 + 05 + 10
= 25 ppm
Non-carbonate Hardness = Hardness due to MgSO4
= 20 ppm
Thus,
Carbonate hardness = 25 ppm
Non-carbonate hardness = 20 ppm
Problem 3: A sample of ground water has 150 mg/L of Ca2+ and 60 mg/L of Mg2+.
Find the total hardness of water sample.
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in mg/L
Multiplication factor
CaCO3 equivalent
Ca
150
100/40
375
Mg2+
60
100/24
250
2+
Total Hardness = Hardness due to [Ca2+ + Mg2+]
= 375 + 250
= 625 mg/L
Thus,
Total hardness = 625 mg/L
Problem 4: How many grams of FeSO4 dissolved per litre gives 210.5 ppm of
hardness? (Fe = 56, S = 32, O = 16, Ca = 40, C = 12).
Solution:
FeSO4 = CaCO3
136 g = 100 g
?
100 ppm of hardness = 136 ppm of FeSO4
210.5
?
210.5 ppm of hardness = 100 u 136 = 286.3 ppm of FeSO4
Thus,
286.3 ppm of FeSO4 is required
Problem 5: A water sample contains:
Ca(HCO3)2 = 32.4 mg/l, Mg(HCO3)2 = 29.2 mg/l, CaSO4 -13.6 mg/l.
F.E. Sem.-I Engineering Chemistry-I
6.10
Calculate temporary, permanent and total hardness.
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in mg/l
Multiplication factor
Ca(HCO3)2
32.4
100/162
Mg(HCO3)2
29.2
100/146
CaSO4
13.6
100/136
Temporary Hardness =
=
=
Permanent Hardness =
=
Total Hardness =
=
=
Water
[May 2008]
CaCO3 equivalent
20
20
10
Hardness due to [Mg(HCO3)2 + Ca(HCO3)2]
20 + 20
40 ppm
Hardness due to [CaSO4]
10 ppm
Temporary Hardness + Permanent Hardness
40 + 10
50 ppm
Thus,
Temporary hardness = 40 ppm, Permanent hardness = 10 ppm, Total hardness = 50 ppm
Problem 6: A water sample on analysis has been found to contain following
impurities in ppm:
Mg(HCO3)2 = 14.6, Mg(NO3)2 = 29.6, Ca(HCO3)2 = 8.1, MgCl2 = 19, MgSO4 = 24.
Calculate the temporary and permanent hardness of water sample. [May 2009]
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
Mg(HCO3)2
14.6
100/146
10
Ca(HCO3)2
8.1
100/162
05
Mg(NO3)2
29.6
100/148
20
MgCl2
19
100/95
20
MgSO4
24
100/120
20
Temporary Hardness =
=
=
Permanent Hardness =
=
=
Hardness due to [Mg (HCO3)2 + Ca (HCO3)2]
10 + 5
15 ppm
Hardness due to [Mg (NO3)2 + MgCl2 + MgSO4]
20 + 20 +20
60 ppm
F.E. Sem.-I Engineering Chemistry-I
6.11
Water
Thus,
Temporary hardness = 15 ppm
Permanent hardness = 60 ppm
Problem 7: The water sample contains the following impurities in mg/L.
Mg(HCO3)2 = 7.3, MgCl2 = 9.5, Ca(HCO3)2 = 16.2, CaSO4 = 13.6
Calculate the temporary permanent and total hardness.
[Dec 2009]
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in mg/L
Multiplication factor
CaCO3 equivalent
Mg(HCO3)2
7.3
100/146
05
MgCl2
9.5
100/95
10
Ca(HCO3)2
16.2
100/162
10
CaSO4
13.6
100/136
10
Temporary Hardness = Hardness due to [Mg(HCO3)2 + Ca(HCO3)2]
= 5 + 10
= 15 ppm
Permanent Hardness = Hardness due to [MgCl2 + CaSO4]
= 10 +10
=
Total Hardness =
=
=
20 ppm
Temporary Hardness + Permanent Hardness
15 + 20
35 ppm
Thus,
Temporary hardness = 15 ppm, Permanent hardness = 20 ppm, Total hardness = 35 ppm
Problem 8: A water sample on analysis has been found to contain
MgCl2 = 19 ppm, CaCO3 = 5 ppm, Ca(HCO3)2 = 29.5 ppm, CaSO4 = 13 ppm.
Calculate temporary, permanent and total hardness.
[May 2011]
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
MgCl2
19
100/95
20
CaCO3
5
100/100
5
29.5
100/162
18.21
13
100/136
9.56
Ca(HCO3)2
CaSO4
F.E. Sem.-I Engineering Chemistry-I
6.12
Water
Temporary Hardness = Hardness due to [CaCO3 + Ca(HCO3)2]
= 5 + 18.21
= 23.21 ppm
Permanent Hardness = Hardness due to [MgCl2 + CaSO4]
= 20 +9.56
= 29.56 ppm
Total Hardness = Temporary Hardness + Permanent Hardness
= 23.21 + 29.56
= 52.77 ppm
Thus,
Temporary hardness = 23.21ppm, Permanent hardness= 29.56ppm, Total hardness = 52.77ppm
Problem 9: Calculate temporary, permanent and total hardness of water sample
containing Mg(HCO3)2 = 7.3 ppm, Ca(HCO3)2 = 16.2 ppm, MgCl2 = 9.5 ppm,
CaSO4 = 13.6 ppm.
[Dec 2012]
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
MgCl2
9.5
100/95
10
Mg(HCO3)2
7.3
100/146
5
Ca(HCO3)2
16.2
100/162
10
CaSO4
13.6
100/136
10
Temporary Hardness = Hardness due to [Mg(HCO3)2 + Ca(HCO3)2]
= 5 + 10
= 15 ppm
Permanent Hardness = Hardness due to [MgCl2 + CaSO4]
= 10 + 10
= 20 ppm
Total Hardness = Temporary Hardness + Permanent Hardness
= 15 + 20
= 35 ppm
Thus,
Temporary hardness = 15ppm, Permanent hardness= 20ppm, Total hardness = 35ppm
F.E. Sem.-I Engineering Chemistry-I
6.13
Water
Problem 10: What is the total hardness of water sample which has following
impurities in mg/l? Ca(HCO3)2 = 162, CaCl2 = 22.2, Mg Cl2 = 95, NaCl = 20.
[June 2013]
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
MgCl2
95
100/95
100
CaCl2
22.2
100/111
20
Ca(HCO3)2
162
100/162
100
NaCl
20
-
-
Total Hardness = Hardness due to [Ca(HCO3)2 + MgCl2 + CaCl2]
= 100 + 100 + 20 = 220 ppm
Thus,
Total hardness = 220ppm
Problem 11: Two samples of water A and B were analysed for their salt content:
(i) Sample A was found to contain 168 mg MgCO3 per litre.
(ii) Sample B was found to contain 820 mg Ca (NO3)2 per litre and 2mg SiO2 per
litre.
Calculate the total hardness of each sample and state which sample is more hard.
[Dec 2013]
Solution: Hardness of sample A = CaCO3 equivalent of MgCO3
§168·
= ¨ 84 ¸ u 100 = 200 ppm
© ¹
Hardness of sample B = CaCO3 equivalent of Ca(NO3)2
§820·
= ¨164¸ u 100 = 500 ppm
© ¹
Thus, the sample B is more hard than sample A.
Problem 12: Calculate all types of hardness of water sample containing:
Ca (HCO3)2 – 81 ppm, MgSO4 – 60 ppm, Mg CO3 – 42 ppm, Ca(NO3)2 – 82 ppm.
Solution: Conversion into CaCO3 equivalent:
[May 2015]
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
Ca(HCO3)2
81
100/162
50
MgSO4
60
100/120
50
MgCO3
42
100/84
50
Ca(NO3)2
82
100/164
50
F.E. Sem.-I Engineering Chemistry-I
6.14
Water
Temporary Hardness = Hardness due to [MgCO3 + Ca(HCO3)2]
=
Permanent Hardness =
=
Total Hardness =
=
50 + 50 = 100 ppm
Hardness due to [MgSO4 + Ca(NO3)2]
50 + 50 = 100 ppm
Temporary Hardness + Permanent Hardness
100 + 100 = 200 ppm
Thus,
Temporary hardness = 100ppm, Permanent hardness= 100ppm, Total hardness = 200ppm
Problem 13: A hard water sample contains following impurities in ppm. Mg(HCO3)2
= 150, NaCl =77, CaCl2 = 135, MgSO4 = 85. Calculate temporary, permanent and
total hardness of the given water sample.
(May 2019)
Solution: Conversion into CaCO3 equivalent:
Constituents
Quantity in ppm
Multiplication factor
CaCO3 equivalent
Mg(HCO3)2
150
100/146
102.74
MgSO4
85
100/120
70.83
CaCl2
135
100/111
121.62
NaCl
77
-
-
Temporary Hardness = Hardness due to [ Mg(HCO3)2]
=
Permanent Hardness =
=
Total Hardness =
=
102.74 ppm
Hardness due to [MgSO4 + CaCl2]
70.83 + 121.62 = 192.45 ppm
Temporary Hardness + Permanent Hardness
102.74 + 192.45 = 295.19 ppm
Thus,
Temporary hardness = 102.74ppm, Permanent hardness= 192.45ppm,
Total hardness = 295.19 ppm
6.5 ESTIMATION OF HARDNESS
Hardness of water can be determined by two methods:
(1) Soap Solution Method: Total hardness of water can be determined by titrating a
fixed volume of water sample (100 ml) against standard alcoholic soap solution.
Appearance of stable lather which persists for two minutes is the end point of titration. In
the beginning, sodium soap will precipitate all the hardness causing metal ions in the
F.E. Sem.-I Engineering Chemistry-I
6.15
Water
form of soap and then it will form-free lather. If same water sample is boiled for 30
minutes and then titrated against same soap solution, the titration reading corresponds to
permanent hardness. The difference between two measurements corresponds to the
temporary hardness of water.
(2) EDTA Method: Hardness of water can be determined more accurately by EDTA
method. This is a complexometric method, as EDTA forms complex ions with hardness
causing Ca2+ and Mg2+ ions present in water. It is an accurate, convenient, rapid method
and hence is popular.
The structure of EDTA (Ethylene diamine tetra acetic acid) is
HOOC – H2C
CH2 − COOH
N--- CH2 ---CH2 ---N
HOOC – H2C
CH2 − COOH
In the form of its sodium salt yields the anion
‾OOC – H2C
CH2 − COO‾
N--- CH2 ---CH2 ---N
‾OOC – H2C
CH2 − COO‾
EDTA is used in the form of its disodium salt. It reacts quickly with hardness causing
metal ions to form cyclic coordination complex. (Chelate) Thus,
Complexation of EDTA with Ca2+ and Mg2+ ions:
O–C=O
O=C–O
M
H2C – N
CH2
--
OO.C.CH2
Where,
M = Ca or Mg
N – CH2
CH2
CH2 ˜ OC ˜ O--
F.E. Sem.-I Engineering Chemistry-I
6.16
Water
EDTA is usually employed as its disodium salt.
NaO.OC – H2C
CH2.CO.ONa
N--- CH2 ---CH2 ---N
HOOC – H2C
CH2 − COOH
In order to determine the equivalence point, alcoholic solution of blue coloured dye
Erichrome Black-T is used which forms unstable wine red complex with Ca2+ and Mg2+
ions. pH of about 10 is required for this reaction otherwise it will not go to completion.
M 2 EBT
(Ca 2 or Mg 2 of hard water)
pH
10

o [M EBT] complex
Wine red
During the course of titration against EDTA solution, EDTA combines with M2+ ions
to form stable complex M-EDTA releasing EBT which combines with M2+ ions present
in the solution and wine red colour is retained.
[M – EBT] complex + EDTA
titration
o
[M-EDTA] complex + EBT
Wine red
M2+ + EBT
(Blue)
[M-EBT] complex
Wine red
When all M2+ ions formed [M-EDTA] complex, the next drop of EDTA added
displaces the EBT indicator from [M-EBT] complex and the wine red colour changes to
blue. This is called the equivalence point. Thus, at equivalence point,
[M-EBT] complex + EDTA o [M-EDTA] complex + EBT (blue)
Thus, change of wine red colour to distinct blue marks the end point of titration.
Various steps involved in EDTA titration:
(1) Preparation of standard hard water : Dissolve 1g of pure, dry CaCO3 in
minimum quantity of dilute HCl. Boil it to dryness to expel excess of acid and
CO2. Dissolve the residue in distilled water to make 1 litre solution. Each ml of
the solution contains 1 mg of CaCO3 equivalent hardness.
(2) Preparation of EDTA solution: Dissolve 4g of pure EDTA and 0.1g MgCl2 in
1 litre distilled water.
(3) Preparation of indicator: Dissolve 0.5g of Eriochrome Black-T in 100 ml of
alcohol.
(4) Preparation of Buffer solution: Add 67.5g of NH4Cl to 570 ml of concentrated
ammonia solution and then dilute with distilled water to 1 litre.
F.E. Sem.-I Engineering Chemistry-I
6.17
Water
(5) Titration 1: Standardization of EDTA: Rinse and fill the burette with EDTA.
Pipette out 50 ml of standard hard water in a conical flask. Add 10-15 ml buffer
and 4-5 drops of EBT indicator. Titrate with EDTA solution till wine red colour
changes to blue. Let volume used be V1 ml.
(6) Titration 2: Titration of unknown hard water: Titrate 50ml of hard water
sample, mixed with 15 ml buffer solution and 4-5 drops of EBT with EDTA
solution till wine red colour changes to blue. Let volume used be V2 ml.
(7) Titration 3: Titration of permanent hardness : Take 250 ml of the water
sample in a large beaker. Boil it, till the volume is reduced to about 50 ml. Filter,
wash the precipitate with distilled water. Collect the filtrate and washings in a
250 ml measuring flask. Make up the volume to 250 ml with distilled water.
Titrate 50 ml of it with EDTA as in Titration (1). Let the volume be V3 ml.
Calculations:
50 ml of standard hard water = V1 ml of EDTA.
?
50 u 1 mg CaCO3 = V1 ml of EDTA.
?
1 ml of EDTA = 50 / V1 mg of CaCO3 equivalent
Now, 50 ml of given hard water = V2 ml of EDTA
V2 u 50
= V
mg of CaCO3 equivalent
1
? 1 L (1000 ml) of given hard water = 1000 V2/V1 mg of CaCO3 equivalent
? Total hardness of water = 1000 V2/V1 mg / L
= 1000 V2/V1 ppm
} (i)
Now 50 ml of boiled water = V3 ml of EDTA
V3 u 50
= V
mg of CaCO3 equivalent
1
? 1000 ml of boiled water = 1000 V3/V1, mg of CaCO3 equivalent
?
Permanent hardness = 1000 V3/V1 ppm
And Temporary hardness = [Total hardness – Permanent hardness]
= 1000 [V2/V2 − V3/V1] ppm
ªV2 − V3º
= 1000 « V » ppm
¬
¼
1
} (ii)
} (iii)
Numericals Based on Calculation of Hardness by EDTA Method :
Problem 1: 50 ml of standard hard water (1.1 mg of CaCO3 per ml) requires 38 ml
of disodium EDTA. 100 ml of water sample consumes 21 ml of EDTA during
titration. Find degree of hardness of the water sample.
F.E. Sem.-I Engineering Chemistry-I
6.18
Water
Solution:
Strength of standard hard water = 1.1 mg CaCO3 per ml
50 ml standard hard water = 1.1 u 50 = 55 mg CaCO3
50 ml standard hard water = 38 ml of EDTA equivalent
= 55 mg CaCO3
?
1 ml EDTA = 55/38 mg CaCO3
100 ml of water sample = 21 ml of EDTA
1000 u 21
1000 ml of water sample =
100
?
=
1 ml EDTA =
210 ml EDTA =
=
Q
?
210 ml EDTA
55/38 mg CaCO3
210 u 55/38
303.9 mg CaCO3
Thus,
Degree of hardness of water sample = 303.9 mg/L
Problem 2: 0.28 g of CaCO3 was dissolved in HCl and the solution made up to 1 litre
with distilled water. 100 ml of the above solution required 28 ml of EDTA solution.
100 ml of hard water sample required 33 ml of EDTA solution. After boiling of this
water, cooling and filtering, 100 ml of this solution on titration required 10 ml of
EDTA solution. Calculate each type of hardness of water.
Solution: Step I:
1000 ml contains 0.28 g or 280 mg
280
?
1 ml will contain = 1000
?
Strength of Hard water = 0.28 mg of CaCO3 per ml
Step II: Now, 100 ml of standard hard water = 28 ml of EDTA
And 100 ml of standard hard water = (100 u 0.28) mg of CaCO3
?
28ml of EDTA = (100 u 0.28) mg of CaCO3
1 ml of EDTA =
100 u 0.28
28
1 ml of EDTA = 1mg of CaCO3 equivalent
Step III:
100 ml of unknown hard water = 33 ml of EDTA solution
= 33 u 1 mg of CaCO3
F.E. Sem.-I Engineering Chemistry-I
6.19
? 1000 ml of unknown hard water =
?
Water
1000 u 33
100
= 330 ppm
Total Hardness = 330 ppm
Step IV:
100 ml of boiled filtered water = 10 ml of EDTA
= 10 u 1 mg CaCO3 equivalent / litre.
1000 u 10
? 1000 ml of boiled filtered water =
100
= 100 mg of CaCO3 / litre
Permanent Hardness = 100 ppm.
Temporary Hardness = Total Hardness – Permanent Hardness
= 330 – 100
= 230 ppm
Thus,
Total Hardness = 330 ppm, Permanent Hardness = 100 ppm, Temporary Hardness = 230 ppm
Problem 3: 50 ml of standard hard water containing 1 mg pure CaCO3 per ml
consumed 20 ml of EDTA. 50 ml of water sample consumed 25 ml of EDTA solution
using Eriochrome Black-T indicator. 50 ml of water sample after boiling, filtering
consumed 18 ml of EDTA. Calculate various types of hardness of sample.
Solution: Step I:
Strength of standard hard water = 1 mg/ml
50 ml of standard hard water = 50 mg of CaCO3
50 ml of standard hard water = 20 ml of EDTA
?
1 ml of EDTA = 50/20 mg of CaCO3 equivalent
Step II:
?
Q
?
= 2.5 mg of CaCO3 equivalent
50 ml of water sample = 25 ml of EDTA
1000 u 25
1000 ml of Water Sample =
ml of EDTA
50
= 500 ml of EDTA
1 ml EDTA = 50/20 mg of CaCO3 equivalent
50
500 ml of EDTA = 500 u 20 of CaCO3 equivalent
Total Hardness = 1250 mg/L
F.E. Sem.-I Engineering Chemistry-I
Step III:
?
Step IV:
6.20
Water
50 ml of boiled water = 18 ml of EDTA
50
= 18 u 20 mg of CaCO3
1000 u 18 u 50
1000 ml of boiled water =
50 u 20
= 900 mg/L
Permanent Hardness = 900 ppm
Temporary Hardness = Total Hardness – Permanent Hardness
= 1250 – 900
= 350 ppm
Thus,
Total Hardness = 1250 ppm, Permanent Hardness = 900 ppm, Temporary Hardness = 350 ppm
Problem 4: A standard hard water contains 15 g/l calcium carbonate, 20 ml of this
water required 25 ml of EDTA solution. 100 ml of sample water required 18 ml of
EDTA solution. The same sample after boiling required 12 ml of EDTA solution.
Calculate temporary hardness of water.
[May 2008]
Solution: Step I:
Strength of standard hard water = 15 g/l
= 15000 mg/ 1000 ml
= 15 mg/ml
Q 20 ml standard hard water ≡ 25 ml of EDTA solution
ª20
º
? 1 ml EDTA ≡ «25 u 15» mgs of CaCO3 Equivalent hardness
¬
¼
300
≡ 25 mgs of CaCO3 Equivalent hardness
≡ 12 mgs of CaCO3 Equivalent hardness
Step II:
100 ml water sample ≡ 18 ml EDTA solution
≡ [18 × 12] mgs of CaCO3 equivalent hardness
≡ 216 mgs of CaCO3 equivalent hardness
? Total hardness
= 2160 ppm
Step III:
100 ml water sample (after boiling) ≡ 12 ml EDTA solution
≡ [12 × 12] mg of CaCO3 equivalent
≡ 144 mgs of CaCO3 equivalent
F.E. Sem.-I Engineering Chemistry-I
?
?
6.21
Water
Permanent hardness ≡ 1440 ppm
Temporary hardness ≡ 2160 – 1440 ppm = 720 ppm
Total Hardness = 2160ppm, Permanent Hardness = 720ppm, Temporary Hardness = 1440ppm
Problem 5: 0.5 gm of CaCO3 was dissolved in HCl and the solution made up to
500 ml with distilled water. 50 ml of the solution required 48 ml of EDTA solution
for titration. 50 ml of hard water sample required 15 ml of EDTA and after boiling
and filtering required 10 ml of EDTA solution. Calculate temporary hardness of
water.
[May 2009]
Solution: Step I:
Strength of standard hard water = 0.5 gm CaCO3/ 500 ml of distilled water.
= 500 mg in 500 ml water
= 1mg/ml
Step II: 50 ml of standard hard water required ≡ 48 ml EDTA solution
i.e. 48 ml EDTA solution ≡ 50 mg CaCO3 equivalent hardness
§50·
? 1 ml EDTA solution ≡ ¨48¸ mg CaCO3 equivalent hardness
© ¹
Now, 50 ml water sample ≡ 15 ml EDTA solution
50º
ª
? Hardness of sample ≡ «15 u 48» mg CaCO3 equivalent for 50 ml hardness
¬
¼
50º
1000
ª
? Hardness per litre of sample ≡ «15 u 48» u 50 mg/l
¬
¼
? Total hardness ≡ 312.5 ppm
Step III: 50 ml water sample after boiling ≡ 10 ml EDTA solution
50·
§
? Permanent hardness of sample ≡ ¨10 u 48¸ mg CaCO3 equivalent for 50 ml
©
¹
50· 1000
§
? Permanent hardness of 1 litre sample ≡ ¨10 u 48¸ u 50 mg/l
©
¹
? Permanent hardness ≡ 208.33 ppm
? Temporary hardness ≡ Total hardness – Permanent hardness
≡ 312.5 – 208.33
? Temporary hardness ≡ 104.17 ppm
Total Hardness = 312.5ppm, Permanent Hardness = 208.33ppm, Temporary Hardness = 104.17ppm
Problem 6: 1 g of CaCO3 was dissolved in 1 lit of distilled water. 50 ml of solution
required 45 ml EDTA for titration. 50 ml of hard water required 25 ml of EDTA for
titration. The same sample of water after boiling consumed 15 ml of EDTA for
titration. Calculate the hardness of water.
[Dec 2010]
F.E. Sem.-I Engineering Chemistry-I
6.22
Water
Solution: Step I: Strength of standard hard water = 1 mg/ml
50 ml of Standard hard water = 50 mg of CaCO3
50 ml of standard hard water = 45 ml EDTA
§50·
? 1 ml EDTA ≡ ¨45¸ mg CaCO3 equivalent
© ¹
Step II: 50 ml of hard water sample = 25 ml of EDTA
1000
?
1000 ml of hard water sample = 50 u 25 ml EDTA
= 500 ml EDTA
50
500 ml EDTA = 500 u 45 of CaCO3 equivalent
= 555.55 ppm
?
Total hardness = 555.55 ppm
Step III:
50 ml of boiled water = 15 ml EDTA
1000
50
?
1000 ml of boiled water = 50 u 15 u 45
= 333.33 ppm
Permanent hardness = 333.33 ppm
?
Temporary hardness = Total hardness – Permanent hardness
?
Temporary hardness = 555.55 ppm – 333.33 ppm
= 222.22 ppm
Total Hardness = 555.55ppm, Permanent Hardness = 333.33ppm, Temporary Hardness = 222.22ppm
Problem 7: 50 ml sample of water required 7.2 ml of N/20 disodium EDTA for
titration. After boiling and filtration, the same volume required 4 ml of EDTA.
Calculate both the types of hardness in it.
[May 2011]
Solution:
N/20 EDTA = 0.05 N EDTA.
Step I:
1000 ml 1N EDTA = 50 g of CaCO3
1 ml of 1N EDTA = 50 mg of CaCO3
50 ml water sample = 7.2 ml of 0.05 N EDTA
1000
1000 ml of water sample = 50 u 7.2 ml of 0.05 N EDTA
?
= 144 ml of 0.05 N EDTA
As 1ml of 1N EDTA = 50 mg of CaCO3
?
144 ml of 0.05 N EDTA = 144 u 0.05 u 50 mg of CaCO3
= 360 ppm
Total Hardness of water = 360 ppm
F.E. Sem.-I Engineering Chemistry-I
Step II:
?
6.23
50 ml of boiled water = 4 ml of 0.05 N EDTA
1000
1000 ml of boiled water = 50 u 4 ml of 0.05 N EDTA
As, 1 ml of 1 N EDTA
80 ml of 0.05 N EDTA
?
Step III:
Water
Permanent Hardness
Temporary Hardness
=
=
=
=
=
=
=
=
80 ml of 0.05 N EDTA
50 mg of CaCO3
80 u 0.05 u 50
200 mg of CaCO3 equivalent
200 ppm
Total Hardness – Permanent Hardness
360 ppm – 200 ppm
160 ppm
Thus,
Total Hardness = 360 ppm, Permanent Hardness = 200 ppm, Temporary Hardness = 160 ppm
Problem 8: 20 ml of standard hard water (containing 1.2 g CaCO3 per litre)
required 35 ml of EDTA. 50 ml of hard water sample required 30 ml of same EDTA.
100 ml of hard water sample after boiling required 25 ml of same EDTA. Calculate
various hardnesses.
[Dec 2011]
Solution: Step I:
Strength of hard water = 1200/1000 mg of CaCO3 per ml
= 1.2 mg of CaCO3 per ml
Step II: 20 ml of standard hard water = 35 ml of EDTA
20 ml of standard hard water = (20 × 1.2) mg of CaCO3
(20 u 1.2)
= 0.685 mg of CaCO3 equivalent
1 ml of EDTA =
35
Step III: 50 ml of hard water sample = 30 ml of EDTA
§1000
·
? 1000 ml of hard water sample = ¨ 50 u 30¸ u 0.685
©
¹
= 411 ppm
?
Total hardness = 411 ppm
Step IV:
100 ml boiled hard water = 25 ml of EDTA
1000 u 25
?
1000 ml of boiled hard water =
u 0.685
100
= 171.25 ppm
?
Permanent hardness = 171.25 ppm
F.E. Sem.-I Engineering Chemistry-I
6.24
Water
Temporary hardness = Total hardness – Permanent hardness
= 411 – 171.25
= 239.75 ppm
Thus,
Total Hardness = 411 ppm, Permanent Hardness = 171.25 ppm,
Temporary Hardness = 239.75 ppm
Problem 9: 0.5 gm of CaCO3 was dissolved in HCl and the solution made up to
500 ml with distilled water. 50 ml of the solution required 45 ml of EDTA solution
for titration. 50 ml of hard water sample required 15 ml of EDTA and after boiling
and filtering required 10 ml of EDTA solution. Calculate temporary hardness and
total hardness of water.
[June 2013]
Solution: Step I:
Strength of standard hard water = 0.5 gm CaCO3/ 500 ml of distilled water.
= 500 mg in 500 ml water = 1mg/ml
Step II: 50 ml of standard hard water required ≡ 45 ml EDTA solution
i.e. 45 ml EDTA solution ≡ 50 mg CaCO3 equivalent hardness
50
mg CaCO3 equivalent hardness
? 1 ml EDTA solution ≡
45
Now, 50 ml water sample ≡ 15 ml EDTA solution
50º
ª
? Hardness of sample ≡ «15 u 45» mg CaCO3 equivalent for 50 ml hardness
¬
¼
50º
1000
ª
? Hardness per litre of sample ≡ «15 u 45» u 50 mg/l
¬
¼
? Total hardness ≡ 333.33 ppm
Step III: 50 ml water sample after boiling ≡ 10 ml EDTA solution
50º
ª
? Permanent hardness of sample ≡ «10 u 45» mg CaCO3 equivalent for 50 ml
¬
¼
50º 1000
ª
? Permanent hardness of 1 litre sample ≡ «10 u 45» u 50 mg/l
¬
¼
? Permanent hardness ≡ 222.22 ppm
? Temporary hardness ≡ Total hardness – Permanent hardness
≡ 333.33 – 222.22
? Temporary hardness ≡ 111.11 ppm
Total Hardness = 333.33ppm, Permanent Hardness = 222.22ppm,
Temporary Hardness = 111.11ppm
F.E. Sem.-I Engineering Chemistry-I
6.25
Water
Problem 10: 50ml of standard hard water containing 1 mg of pure CaCO3 per ml
consumed 20 ml of EDTA. 50 ml of the water sample consumed 30 ml of same
EDTA solution. After boiling and filtering, 50 ml of the water sample required 10 ml
of the same EDTA for titration. Calculate the total and permanent hardness of
water sample.
[Dec 2013]
Solution: Step I:
Strength of standard hard water = 1 mg/ml
50 ml of standard hard water = 50 mg of CaCO3
50 ml of standard hard water = 20 ml of EDTA
50
?
1 ml of EDTA = 20 mg of CaCO3 equivalent
Step II:
?
Q
?
Step III:
?
Step IV:
= 2.5 mg of CaCO3 equivalent
50 ml of water sample = 30 ml of EDTA
1000 u 30
1000 ml of Water Sample =
ml of EDTA
50
= 600 ml of EDTA
1 ml EDTA = 50/20 mg of CaCO3 equivalent
50
600 ml of EDTA = 600 u 20 of CaCO3 equivalent
Total Hardness = 1500 ppm
50 ml of boiled water = 10 ml of EDTA
50
= 10 u 20 mg of CaCO3
1000 u 10 u 50
1000 ml of boiled water =
50 u 20
= 500 ppm
Permanent Hardness = 500 ppm
Temporary Hardness = Total Hardness – Permanent Hardness
= 1500 – 500
= 1000 ppm
Thus,
Total Hardness = 1500 ppm, Permanent Hardness = 500 ppm, Temporary Hardness = 1000 ppm
Problem 11: Calculate total hardness in ppm in given water sample. (i) 50ml
standard hard water containing 1 mg pure CaCO3 per ml, consumed 20ml EDTA
solution (ii) 50ml water sample consumed 30ml EDTA solution using Erio-Black T
indicator.
[Dec. 2017, May 2018]
F.E. Sem.-I Engineering Chemistry-I
Solution: Step I:
Strength of standard hard water
50 ml of standard hard water
50 ml of standard hard water
?
1 ml of EDTA
Step II:
=
=
=
=
=
50 ml of water sample =
?
1000 ml of Water Sample =
Q
=
1 ml EDTA =
?
600 ml of EDTA =
Total Hardness =
6.26
Water
1 mg/ml
50 mg of CaCO3
20 ml of EDTA
50/20 mg of CaCO3 equivalent
2.5 mg of CaCO3 equivalent
30 ml of EDTA
1000 u 30
ml of EDTA
50
600 ml of EDTA
50/20 mg of CaCO3 equivalent
50
600 u 20 of CaCO3 equivalent
1500 mg/L
Thus,
Hardness of water = 1500 ppm
Problem 12: In the process of determination of hardness, standard hard water
sample was prepared by dissolving 2.5g CaCO3 and making solution up to 1 liter.
50ml of above hard water required 45ml of EDTA. 50ml of unknown hard water
sample consumed 30ml EDTA solution using Erio-Black T indicator. The unknown
hard water sample was boiled and filtered. 50ml of this boiled solution required
20ml of EDTA. Calculate hardness of all types of unknown hard water sample.
[May 2019]
Solution: Step I:
Strength of hard water = 2500/1000 mg of CaCO3 per ml
= 2.5 mg of CaCO3 per ml
Step II: 50 ml of standard hard water = 45 ml of EDTA
50 ml of standard hard water = (50 × 2.5) mg of CaCO3
(50 u 2.5)
1 ml of EDTA =
45
= 2.777 mg of CaCO3 equivalent
Step III: 50 ml of hard water sample = 30 ml of EDTA
§1000
·
? 1000 ml of hard water sample = ¨ 50 u 30¸ u 2.777
©
¹
= 1666.2 ppm
?
Total hardness = 1666.2 ppm
F.E. Sem.-I Engineering Chemistry-I
Step IV:
?
?
6.27
Water
100 ml boiled hard water = 20 ml of EDTA
1000
1000 ml of boiled hard water = 50 u 20 u 2.777
= 1110.8 ppm
Permanent hardness = 1110.8 ppm
Temporary hardness = Total hardness – Permanent hardness
= 1666.2 – 1110.8
= 555.4 ppm
Thus,
Total Hardness = 1666.2 ppm, Permanent Hardness = 1110.8 ppm,
Temporary Hardness = 555.4 ppm
6.6 UNDESIRABLE EFFECT OF HARD WATER
The use of hard water is not possible for domestic purpose, industrial purpose or even
in steam generation in boilers as it has a lot of disadvantages for various purposes.
In Domestic Use :
(1) Washing: Hard water does not lather freely with soap. It produces sticky
precipitates of calcium and magnesium soaps which adhere on the fabric giving spots and
streaks. This causes wastage of soap. Iron salts present in the water cause staining of
cloth.
(2) Bathing: Hard water does not lather freely decreasing cleansing quality of soap
and wastages of soap.
(3) Drinking: Hard water has adverse effect on our digestive system. There is an
increased possibility of forming calcium oxalate crystals in urinary tracks.
(4) Cooking: Because of the presence of dissolved salts, the boiling point of water is
increased. It causes wastage of fuel and time. It also gives unpleasant taste to tea and
coffee. The dissolved salts are deposited as carbonates on the inner walls of the water
heating utensils.
In Industrial Use :
(1) Sugar industry: Water containing sulphates, nitrates, alkali, carbonates etc. if
used in sugar refining causes difficulties in the crystallization of sugar. The sugar so
produced may be deliquescent.
(2) Textile industry: Hard water causes soap to go waste as it cannot produce good
quality of lather. Precipitates of calcium and magnesium soaps adhere to the fabric giving
improper shades on dyeing. Iron and manganese cause coloured spots on fabric.
F.E. Sem.-I Engineering Chemistry-I
6.28
Water
(3) Paper industry: Calcium and Magnesium salts tend to react with chemicals to
provide a smooth and glossy finish to paper. Iron salts affect the colour of the paper.
(4) Laundry: There is wastage of soap and colouration of the clothes due to iron.
(5) Concrete making: Water containing chlorides and sulphates affect the hydration
of cement and final strength of hardened concrete.
(6) Pharmaceutical industry: Hard water, if used for preparing pharmaceutical
products like drugs, injection, ointments etc., may produce certain undesirable products
in them.
(7) Dyeing industry: The dissolved calcium, magnesium and iron salts react with
costly dyes forming undesirable precipitates which give impure shades and spots on the
fabric.
6.7 WATER TREATMENT (SOFTENING)
The process whereby the hardness of water can be reduced or removed irrespective of
whether it is temporary or permanent is called ‘softening’ of water. It is very essential
process since hard water is unsuitable for domestic as well as industrial use. The hardness
causing salts can be removed from water by following two types:
(1) External treatment
(2) Internal treatment.
External Treatment:
The external treatment of water is carried out before its entry into the boiler.
In industry, the important methods generally employed for softening are :
(A) Lime-Soda Process,
(B) Zeolite Permutit Process,
(C) Ion exchange Process.
6.7.1 Ion Exchange or Deionization or
Demineralization Process
Now-a-days synthetic ion exchange resins are used in the industry. Ion exchange
resins are insoluble, cross-linked; long chain organic polymers with a microporous
structure and the functional groups attached to the chains which are responsible for the
ion exchanging properties. They may be classified as follows.
(i) Cation Exchange Resins (RH+): Resins containing acidic functional groups
(−COOH, −SO3H etc) are capable of exchanging their H+ ions with other cations. They
are mainly styrene divinyl benzene copolymers, which on sulphonation or carboxylation
F.E. Sem.-I Engineering Chemistry-I
6.29
Water
acquire capability to exchange their hydrogen ions with the cations in the water. They can
be simply represented as RH+ where R represents the insoluble polymeric matrix. Their
exchange reaction with other cation is shown below.
2RH+ + M+2 o R2M + 2H+
Amberlite IR -120 and Dowex -50 are some of the examples of commercially
available cation excahnge resins.
For example,
Fig. 6.1 : Cation Exchange Resin
(ii) Anion Exchange Resins (R′OH--): Resins containing basic functional groups
−NH2, = NH etc. as hydrochloride are capable of exchanging their anions with other
anions. They are styrene divinyl benzene or amine formaldehyde copolymers, which
contain amino or quaternary ammonium or quaternary phosphonium or tertiary
sulphonium groups as an integral part of resin matrix. This after treatment with dilute
NaOH solution becomes capable to exchange their OH-- ions with anions of water.
They can be represented as R′OH-- where R′ represents insoluble organic matrix.
Their exchange reaction with other anions is shown below.
R′OH-- + X-- o R′ X-- + OH-Amberlite – 400 & Dowex – 3 are some of the examples of commercially available
anion exchange resins.
F.E. Sem.-I Engineering Chemistry-I
6.30
Water
For example,
Fig. 6.2 : Anion Exchange Resin
Principle:
When raw water passes through cation exchanger resin, all cations from water are
absorbed in exchange of H+ and when it passes through anion exchanger resins, all anions
are absorbed in exchange of OH--. Thus, the water obtained is of distilled quality.
Process:
The hard water is passed first through cation exchange column, which removes all the
cations like Ca2+, Mg2+ etc. from it and equivalent amount of H+ ions are released from
this column to water.
Reaction:
2RH+ + Ca2+ o R2Ca2+ + 2H+
2RH+ + Mg2+ o R2Mg2+ + 2H+
Then the hard water is passed through anion exchange column, which removes all the
anions like SO42−, Cl-- etc. present in the water and equivalent amount of OH-- ions are
released from this column to water.
Reaction:
RcOH− + Cl− o RcCl− + OH−
2−
2−
2RcOH− + SO4 o Rc SO4 + 2OH−
2
2−
3
2RcOH− + CO
o R2c CO3 + 2OH−
2−
H+ ions released from cation exchange and OH-- ions released from anion exchange
columns combine to produce water molecule.
H+ + OH− o H2O
F.E. Sem.-I Engineering Chemistry-I
6.31
Water
As the water coming out is free from all ions (deionised water), the process is called
as deionisation or demineralisation process.
Fig. 6.3 : Demineralization of Water
Regeneration:
When cation exchangers and anion exchangers are exhausted completely (i.e. they
lose their capacity to exchange H+ and OH- ions), regeneration process is carried out. The
exhausted cation exchangers is regenerated by passing a solution of dilute HCl or dilute
H2SO4. The regeneration can be represented as:
R2Ca2+ + 2H+ o 2 RH+ + Ca2+
The column is washed with deionised water and washings are sent to sink. The
exhausted anion exchanger is regenerated by passing a solution of dilute NaOH. The
regeneration can be represented as
R′2SO42- + 2OH– o 2 R′OH + SO42Then the column is washed with deionised water and washings are passed to sink.
Advantages of Ion Exchange Process:
(1) It produces water of very low hardness (2 ppm).
(2) Treated water contains negligible amount of total dissolved solids.
(3) Water obtained can be used for high pressure boilers.
(4) Highly acidic and alkaline water can be softened.
Disadvantages of Ion Exchange Process:
(1) The equipment is costly.
(2) More costly chemicals are needed.
(3) The turbidity of water should be below 10 ppm as efficiency of the process gets
affected.
F.E. Sem.-I Engineering Chemistry-I
6.32
Water
Numerical Based on Ion Exchange Process
Problem 1: After treating 104 Litre of water by ion exchanger, the cationic resin
required 200 Litre of 0.1 N HCL and anionic resin required 200 Litre of 0.1 N
NaOH solutions. Find the hardness of the above sample of water.
Solution: In an ion exchanger all hardness causing cations are removed by cation
exchanger. Hence the amount of acid used for regeneration of cation resin refers to
hardness.
Hardness in 104 litres of water ≡ 200 L of 0.1 N HCl
= 200 L of 0.1 N CaCO3 eq
= 200 × 0.1 L of 1 N CaCO3 eq
= 20 L of 1N CaCO3 eq
= 20 × 50 g of CaCO3 eq
?
= 1000 g of CaCO3 eq
1000
Hardness in 1L of water = 104 g of CaCO3 eq
= 0.1 u 1000 mg of CaCO3 eq
= 100 mg of CaCO3 eq
Thus,
Hardness of water sample = 100 mg/L
6.8 DESALINATION OF BRACKISH WATER
Water containing high concentration of dissolved solids with a peculiar salty taste is
called brackish water. Sea water is a good example of brackish water as it contains 3.5%
of dissolved salts. It is totally unfit for most of the domestic and industrial applications.
The process of removing common salt (NaCl) from such water is known as desalination.
Commonly used techniques for the desalination of brackish water are as follows.
1. Electrodialysis:
Electrodialysis (ED) is an electro membrane process in which ions are transported
through ion permeable membranes from one solution to another under the influence of
potential gradient. It depends on the following general princples.
Most salts dissolved in water are ionic, being positively (cationic) or negatively
(anionic) charged. These ions are attracted to electrodes with an opposite charge.
Membranes can be constructed to permit selective passage of either anions or cations.
F.E. Sem.-I Engineering Chemistry-I
6.33
Water
It is an electrically driven membrane separation process that is capable of separating,
concentrating, and purifying selected ions from aqueous solutions. The process is based
on the property of ion exchange membranes to selectively reject anions or cations.
Electrodialysis is based on the fact that the ions present in the saline water migrate
towards respective electrodes through ion-selective membranes under the influence of
applied emf.
Fig. 6.4 illustrates the method of electro dialysis. The unit consists of electrodes and
ion selective membranes. These membranes are thin and rigid, which are also permeable
to either cation or anion. The anode is placed near anion selective membrane while the
cathode is placed near the cation selective membrane.
Fig. 6.4 : Electro Dialysis of Brackish Water
Under the influence of an applied emf across the electrodes cations move toward the
cathode and anions move towards the anode through respective membranes. There is
depletion of ions in the central compartment while it increases in the two side
compartments. Desalinated water is taken out from the central compartment and brackish
water is replaced by fresh samples.
An electro dialysis cell based on the same principle is used for practical purposes. It
consists of a large number of paired sets of rigid plastic membranes. Under the pressure
of about 5-6 kg/m2 saline water is passed between membrane pairs. An electric field is
applied perpendicular to the direction of water flow. Fixed positive charges inside the
membrane repel positively charged ions (Na+) and permit negatively charged ions (Cl-) to
pass through. Similarly, the fixed negative charges inside the membrane repel chloride
ions but permit sodium ions. Thus water in one compartment is deprived of salts while
the salt concentration in the adjacent compartment is increased. Thus alternate streams of
pure water and brackish water are obtained.
F.E. Sem.-I Engineering Chemistry-I
6.34
Water
Advantages:
(1) Easy operation and variability of ED equipment.
(2) The unit is compact.
(3) Demineralization of biological solutions without affecting the quality.
(4) Separation of salts and ions without changing phase and adding chemicals.
(5) Higher feed recovery in many applications.
(6) The process is economical as the cost of installation of the plant and its
operational expenses are less.
Disadvantages:
(1) Sometimes pretreatment is necessary before the electrodialysis.
(2) Suspended solids with a diameter that exceeds 10 µm need to be removes else
they may plug the membrane pours.
(3) Substances such as large organic anions, colloids, ion oxides and manganese
oxide can disturb the selective effect of the membrane.
2. Reverse Osmosis:
There are various membrane techniques available for the separation of solutes on the
basis of pore size which include reverse osmosis, ultra filtration etc.
Osmosis:
When two solutions of unequal concentrations are separated by semi-permeable
membrane (which does not allow movement of solute particles) flow of solvent takes
place from low concentration solution to high concentration solution side due to osmosis.
High concentration
solution compartment
ooooooooooooo
ooooooooooooo
ooooooooooooo
ooooooooooooo
---------------------------------- ---
Low concentration
solution compartment
Semi-permeable membrane
Fig. 6.5: Osmosis
This passage of solvent from low concentration solution side compartment to high
concentration solution side compartment is due to difference in vapour pressure of the
two compartments. The flow continues till the concentration is equal on both the sides.
The driving force for osmosis is called osmotic pressure.
Reverse Osmosis:
If a hydrostatic pressure in excess of osmotic pressure is applied on higher
concentration solution side, solvent starts moving from higher concentration to lower
F.E. Sem.-I Engineering Chemistry-I
6.35
Water
concentration side compartment through semi-permeable membrane; this is the principle
of reverse osmosis.
Thus, in the process of reverse osmosis pure solvent is separated from its
contaminants, rather than removing contaminants from water. This membrane filtration is
also called ‘super filtration’ or ‘hyper filtration’.
Advantages:
(1) Reverse osmosis can be used to remove ionic as well as non-ionic, colloidal and
high molecular weight organic matter.
(2) It removes colloidal silica which is not removed by deminerization.
(3) The process is economical, simple, highly reliable and has low capital and
operating cost.
(4) The life of the semipermeable membrane is about two years and can be replaced
with in few minutes there by nearly uninterrupted water supply can be provided.
Disadvantages:
(1) RO units use a lot of water but have production efficiency close to 48% (for large
scale industrial and municipal system).
(2) It is not practical for household systems to have low back pressure as it recovers
only 5-15% of water that enters the system. This adds to the load on the
household septic system.
Industrial Applications:
(i) Due to low capital cost, simplicity, low operating cost and high reliability, reverse
osmosis is used for desalination and purification of brackish and sea water for drinking
and industrial use (for high pressure boilers).
Fig. 6.6 : Reverse Osmosis Cell
F.E. Sem.-I Engineering Chemistry-I
6.36
Water
The process of removing common salt from sea water is known as desalination. The
principle of reverse osmosis is applied to treat sea water. The membrane consists of thin
films of cellulose acetate. Recently membranes made of polymethacrylate and polyamide
polymers are used. In this process, pressure (15 to 40 kg cm−2) is applied to sea water and
pure water is forced through semi-permeable membrane. [Ionic and non-ionic, dissolved
salts are left behind].
(ii) Water recycle and recovery system: Reverse osmosis can be used for a variety
of specialized membrane application for chemical recovery and waste water reclamation.
For example, Recovery of Nickel and Chromium solutions, purification of pickling acids,
water from recycle in textile, electroplating, paper and pulp industries.
3. Ultrafilteration:
Ultra filtration is a separation process using membranes with pore sizes in the range
of 0.1 to 0.001 micron. It is membrane filtration in which hydrostatic pressure forces a
liquid against a semi-permeable membrane. Suspended solids and solute of high molecule
weight are retained, while water and low molecular weight solutes pass through the
membrane. Ultra filtration does this by pressuring the solution flow, which is tangential
to the surface of supported membrane. The solvent and other dissolved components that
pass through the membrane are known as permeate and the components that do not pass
through membrane are known as retenate. It is fundamentally not different from reverse
osmosis. Generally low applied pressures are sufficient to achieve high flux rates from an
ultra filtration membrane. Flux of a membrane is defined as the amount of permeate
produced per unit area of membrane surface per unit time which is expressed as gallons
per square foot per day or as cubic meters per square meters per day.
Advantages:
(1) It removes high molecular weight substances, colloidal materials, organic and
inorganic polymeric molecules.
(2) As only high molecular weight species are removed, the osmotic pressure
differential across the membrane surface is negligible.
(3) Low applied pressures are sufficient to achieve high flux rates from an
ultrafiltration membrane.
Disadvantages:
(1) Low molecular weight organics and ions such as sodium, calcium, magnesium,
chloride and sulfate are not removed.
Applications:
(1) This separation is used in industry and research for purifying and concentrating
macromolecular solution especially protein solutions.
F.E. Sem.-I Engineering Chemistry-I
(2)
(3)
(4)
(5)
6.37
Water
It is used in industry to separate suspended solids from solution.
It is used in paint recovery in automotive industry.
It is used in the fractionation of milk and whey.
It is used in removal of colloids.
6.9 WATER POLLUTION
Water pollution is contamination of water by foreign matter that deteriorates the
quality of the water. Water pollution covers pollutions in liquid forms like ocean
pollution and river pollution. As the term applies, liquid pollution occurs in the oceans,
lakes, streams, rivers, underground water and bays, in short liquid-containing areas. It
involves the release of toxic substances, pathogenic germs, substances that require much
oxygen to decompose, easy soluble substances, radioactivity, etc. that becomes deposited
upon the bottom and their accumulations will interfere with the condition of aquatic
ecosystems. For example, the eutrophication: lack of oxygen in a water body caused by
excessive algae growths because of enrichment of pollutants. Polluted water means water
that does not meet even the minimum standards for any function and purposes for which
it would be suitable in its natural state. The water pollution may be defined as any
alteration in the physical, chemical and biological properties of water, as well as
contamination with any foreign substance which would constitute a health hazard or
decrease the utility of water.
Sewage:
Sewage is the liquid wastes which includes human and household waste waters,
industrial waste, ground waste, and street and storm waters.
Constituents of Sewage:
(1) Domestic sewage (human excreta, discharges from bath, kitchen, lavatories etc.)
from public and private buildings.
(2) Industrial and trade wastes from manufacturing processes (tanneries, slaughter
houses, distilleries, textile mills, chemical plants etc.)
(3) Ground water entering through severs.
(4) Storm water i.e. rain water from houses and roads.
If the water polluted by domestic effluent, industrial effluent and agricultural waste is
consumed by human beings, it causes various types of diseases. 5-7 ppm of dissolved
oxygen is present in unpolluted water and is essential for supporting aquatic life. In
presence of good amount of dissolved oxygen (> 8 ppm) aerobic bacteria lead to
oxidation of organic compound present in water (which is called aerobic oxidation).
Oxidation products are inoffensive smelling, non putrefying nitrites, nitrates, sulphates,
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6.38
Water
phosphates etc. But if the dissolved oxygen is less (< 5 ppm) the sewage is called stale
and anaerobic bacteria bring about putrefaction producing methane, hydrogen sulphide
and phosphine which give offensive odour (which called anaerobic oxidation). In an
anaerobic oxidation the bacteria extract combined oxygen contain inorganic matter,
nitrates, nitrites and sulphates of sewage. When the anaerobic decomposition is
continuing the sewage is known as septic sewage.
6.9.1 BOD – Biological Oxygen Demand
The water when gets polluted with large amount of organic matter, a lot of dissolved
oxygen is consumed in the biological aerobic decay. The decreased DO affects the
aquatic lives. The biological oxygen demand (BOD) is an index of water pollution. It is
used in water quality management and assessment, Ecology and environmental science.
BOD is not an accurate quantitative test, although it could be considered as an indication
of the quality of a water source. BOD indicates amount of decomposable organic matter
in the sewage. It enables us to determine the degree of pollution at any time in the sewage
stream. Thus it is very essential in sewage treatment.
Biological oxygen demand (BOD) of sewage is defined as amount of free oxygen
required for the biological oxidation of the organic matter under aerobic conditions at
20qC and for a period of 5 days. The unit of BOD is mg/L or ppm.
An average sewage has BOD of 100-150 mg/L
Reaction:
Micro-organisms
Organic matter + O2 o CO2 + H2O
The demand for oxygen is proportional to the amount of organic waste to be degraded
aerobically. Thus BOD approximates the amount of oxidizable organic matter present in
the solution.
The higher the BOD of a sample the higher will be pollution caused by it.
Table 6.2: BOD values of some effluents
Source of effluents
BOD (mg/liter)
Domestic sewage
320
Low
Community Kitchen Sewage
2170
Medium
Cow Shed Sewage
3010
Medium
Paper Mill Effluent
8190
High
Tannery Effluent
12360
Very High
Drinking water should have BOD preferably less than 1 ppm.
Degree of Pollution
F.E. Sem.-I Engineering Chemistry-I
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Water
Determination of BOD:
A known volume of effluent sample is diluted with a known volume of dilution
water. The diluted sample is taken in two stoppered bottles. The dissolved oxygen
content of one of the bottles is immediately determined (blank). Another bottle is
incubated at 20°C for 5days, after which unused oxygen is determined. The difference
between original oxygen content in the blank and unused oxygen of effluent water after
five days gives the BOD. Thus BOD is equal to (DOb - DOi) × Dilution Factor.
Where, BOD = Biological oxygen demand.
DOb = Dissolved oxygen present in the effluent sample before incubation.
DOi = Dissolved oxygen present in the effluent sample after incubation.
While carrying out the BOD test, water is always diluted to ensure that sufficient
oxygen is available for the complete oxidation of organic matter. BOD is the requirement
of oxygen due to biodegradable organic matter only, and is biological phenomenon
carried out more or less under natural conditions existing in water. BOD test is usually
influenced by type of micro organism, presence of toxins, pH, some reduced mineral
matters etc.
Significance of BOD:
The BOD values are useful in designing of treatment plants and calculations of waste
load. It is a measure of efficiency of operation in treatment plant. It is important in
sewage treatment as it indicates the amount of decomposable organic matter in sewage. It
is the best test in assessing the organic pollution. From the BOD value, self purifying
capacity of streams can be determined. This serves as a measure to assess the quantity of
waste which can be safely discharged into the stream. Thus its significance in pollution
control is unique since it provides the degree of pollution at any time in the sewage
stream. BOD test is the best test for assessing the organic pollution which serves as a
guideline for Regulatory Authorities to check the quality of effluents discharged into
water bodies.
Limitations of BOD:
BOD values of effluents of rayon, paper and chemical industries are much less
although they contain enough organic matter. The conclusion of presence of less organic
matter in these effluents should not be drawn from their low BOD values. Thus Bod
values should not be used as equivalent to organic load because of presence of other nondegradable organic matter.
F.E. Sem.-I Engineering Chemistry-I
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Water
6.9.2 COD - Chemical Oxygen Demand
Chemical Oxygen Demand is the amount of oxygen consumed under specified
conditions in the oxidation of organic and oxidisable inorganic matter. Most applications
of COD determine the amount of organic pollutants found in surface water (e.g. lakes and
rivers), making COD a useful measure of water quality. It is a measure of oxidisable
impurities present in the sewage. In COD test, the sample is subjected to chemical
oxidation with potassium dichromate (K2Cr2O7) which is a strong oxidizing agent. It
oxidizes both biologically oxidisable and biologically inert material in the effluent water
sample. Thus COD values are higher than BOD values as BOD measures only the
oxygen consumed by living organisms while assimilating organic matter present in the
water. COD may be defined as the amount of oxygen required for oxidation of
chemically degradable organic matter in the hot conditions with use of oxidant like
acidified potassium dichromate (K2Cr2O7) in 1½ hours. It is expressed in milligrams per
litre (mg/L) or ppm which indicates the mass of oxygen consumed per litre of solution.
Determination of COD:
A known volume (25 ml) of waste water sample is refluxed with a known excess of
standard potassium dichromate (K2Cr2O7) (1N) and dilute sulphuric acid (H2SO4) in
presence of silver sulphate (Ag2SO4) for 1½ hours. Silver sulphate is used as a catalyst to
promote oxidation of straight chain aliphatic compounds, aromatic compounds and
pyridine. The organic matter of the sample is oxidized to ammonia, carbon dioxide and
water. The unreacted potassium dichromate (K2Cr2O7) is titrated against Ferrous
Amonium Sulphate (FeSO4.(NH4)2SO4.6H2O) i.e. Mohr’s salt solution. This
experimentally measured amount of potassium dichromate (consumed) is used to
calculate the equivalent oxygen required by the waste water for degradation of the
pollutants.
Reaction:
CXHYOZ + (X + Y/4 – Z/2) O2 o X CO2 + Y/2 H2O
The unreacted dichromate solution is titrated against (FeSO4.(NH4)2.6H2O) using
ferroin as indicator. At end point, blue colour changes to wine red.
Cr2O7 -2 + 14 H+ + 6e-- o 2 Cr+3 + 7 H2O
[Fe+2 o Fe+3 + e--] x 6
−2
Cr2O7 + 14 H+ + 6 Fe2+ o 2 Cr+3 + 6Fe3+ + 7H2O
The COD of the sample can be calculated as follows:
COD =
(V1 – V2) N u 8000
mg/L
Y
F.E. Sem.-I Engineering Chemistry-I
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Water
Where, Y is the volume of effluent water sample taken for test. V1 and V2 are volume
of ferrous ammonium sulphate consumed in blank and test experiments respectively.
N is the normality of Ferrous Ammonium Sulphate.
Significance of COD:
The COD value is not affected by the presence of toxins and other unfavourable
conditions for the growth of micro organisms. It measures the effect of pollutants on
dissolved oxygen. It is taken as basis for calculation of efficiency of treatment plant. It is
important in proposing standards for discharging domestic and industrial effluents in
various kinds of water. Due to its rapid determination over BOD, it has become important
in the management and design of treatment plants.
Limitations of COD:
COD is a poor measure of strength of organic matter as oxygen gets consumed in
oxidation of some of the inorganic matter. Some organic matter like benzene does not get
oxidized by the test. COD test fails to differentiate between bio-inert and bid-degradable
materials.
Comparison:
BOD
COD
BOD of water is a measure of amount of
oxygen required for biological oxidation
of organic matter under aerobic condition
at 20 oC for a period of 5 days.
COD of water is a measure of amount of
oxygen required by organic matter in a
water sample for its oxidation by strong
oxidizing agent.
It measures the oxygen demand of bio It measures the oxygen demand for
degradable pollutants only.
biodegradable pollutants along with nonbiodegradable pollutants.
Less stable measurement method as it More stable measurement method as it
uses micro organisms which are uses potassium dichromate which oxidises
susceptible to pH, temperature and other regardless of water conditions.
variables in the water.
Slow process. It takes five days.
Fast process. It takes 2-3 hours.
BOD values are generally less than COD COD values are generally greater than
values.
BOD values.
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Water
Numericals Based on BOD
Problem 1: A 50 ml of sample contains 840 ppm of dissolved oxygen. After 5 days
the dissolved oxygen value becomes 230 ppm after the sample has been diluted to
80 ml. Calculate the BOD of the sample.
Solution:
BOD = (DOb – DOi) u Dilution Factor
ml. of sample after dilution
= (DOb – DOi) u ml. of sample before dilution
80
= (840 – 230) u 50
= 976 ppm
Problem 2: A 50 ml of water sample contains 500 ppm of dissolved oxygen. The
water sample is diluted to 100 ml. After 5 days of incubation the DO value of water
sample reduces to 400 ppm. Calculate BOD of water sample.
Solution:
BOD = (DOb – DOi) u Dilution Factor
ml. of sample after dilution
= (DOb – DOi) u ml. of sample before dilution
100
= (500 – 400) u
50
= 200 ppm
Problem 3: A 100 ml of water sample contains 600 ppm of dissolved oxygen. The
water sample is diluted to 200 ml. After 5 days of incubation the DO value of water
sample reduces to 300 ppm. Calculate BOD of water sample.
Solution:
BOD = (DOb – DOi) u Dilution Factor
ml. of sample after dilution
= (DOb – DOi) u ml. of sample before dilution
200
= (600 – 300) u
100
= 600 ppm
Numericals Based on COD
Problem 1: A 25 ml of a sewage water sample was refluxed with 10 ml of 0.25 N
K2Cr2O7 solution in presence of dil. H2SO4, Ag2SO4 and HgSO4. The unreacted
dichromate required 6.5 ml of 0.1 N ferrous ammonium sulphate. 10 ml of the same
K2Cr2O7 solution and 25 ml of distilled water, under the same conditions as the
sample, required 27 ml of 0.1 N ferrous ammonium sulphate. Calculate the COD of
the sewage water sample.
F.E. Sem.-I Engineering Chemistry-I
Solution: Given
?
Vb
Vt
N
Ve
6.43
Water
=
=
=
=
27 ml
6.5 ml
0.1 Normal
25 ml
(27 – 6.5 u 0.1 u 8)
COD =
u 1000 = 0.656 ppm
25
Problem 2: A 25 ml of a sewage water sample was refluxed with 10 ml of 0.25 N
K2Cr2O7 solution in presence of dil. H2SO4, Ag2SO4 and HgSO4. The unreacted
dichromate required 5.5 ml of 0.1 N ferrous ammonium sulphate. 10 ml of the same
K2Cr2O7 solution and 25 ml of distilled water, under the same conditions as the
sample, required 15 ml of 0.1 N ferrous ammonium sulphate. Calculate the COD of
the sewage water sample.
[Dec 2014]
Solution: Given
Vb = 15 ml
Vt = 5.5 ml
N = 0.1 Normal
Ve = 25 ml
(15 – 5.5) u 0.1 u 8 u 1000
?
COD =
= 304 ppm
25
Problem 3: A 5ml sample of waste water was refluxed with 30ml of potassium
dichromate solution and after refluxing the excess unreacted dichromate required
23ml of 0.1M FAS solution. A blank of distilled water on refluxing with 30ml of
dichromate solution required 36ml of 0.1M FAS solution. Calculate the COD value
of the waste water.
[Dec 2015]
Solution: Given
Vb = 36 ml
Vt = 23 ml
N = 0.1 Normal
Ve = 5 ml
(36 – 23) u 0.1 u 8 u 1000
?
COD =
= 2080 ppm
5
Problem 4: Calculate the COD of an effluent sample if 25cc of the effluent sample
required 8.3cc of 0.001M K2Cr2O7 for oxidation.
[May 2016]
Solution: Given
Vb - Vt = 8.3 ml
N = 0.001 Normal
Ve = 25 ml
8.3 u 0.001 u 8 u 1000
?
COD =
= 2.656 ppm
25
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6.44
Water
Problem 5: 6ml of waste water was refluxed with 25ml of K2Cr2O7 and after
refluxing the excess unreacted dichromate required 20ml of 0.1N FAS solution. A
blank of distilled water on refluxing with 25ml of K2Cr2O7 solution required 35ml of
0.1N FAS solution. Calculate the COD of waste water sample.
[Dec 2016]
Solution: Given
Vb = 35 ml
Vt = 20 ml
N = 0.1 Normal
Ve = 6 ml
(35 – 20) u 0.1 u 8 u 1000
?
COD =
= 2000 ppm
6
Problem 6: 10ml of waste water was refluxed with 20ml of K2Cr2O7 and after
refluxing the excess unreacted dichromate required 36.2ml of 0.1N FAS solution. A
blank of 10ml of distilled water on refluxing with 20ml of K2Cr2O7 solution required
46ml of 0.1N FAS solution. Calculate the COD of waste water sample. [May 2017]
Solution: Given
Vb = 46 ml
Vt = 36.2 ml
N = 0.1 Normal
Ve = 10 ml
(46 – 36.2) u 0.1 u 8 u 1000
?
COD =
= 784 ppm
10
Problem 7: 10ml of waste water was refluxed with 20ml of K2Cr2O7 and after
refluxing the excess unreacted dichromate required 26.2ml of 0.1N FAS solution. A
blank of 10ml of distilled water on refluxing with 20ml of K2Cr2O7 solution required
36ml of 0.1N FAS solution. Calculate the COD of waste water sample. [May 2018,19]
Solution: Given
Vb = 36 ml
Vt = 26.2 ml
N = 0.1 Normal
Ve = 10 ml
(36 – 26.2) u 0.1 u 8 u 1000
?
COD =
= 784 ppm
10
Problem 8: 20ml of waste water was refluxed with 30ml of K2Cr2O7 and after
refluxing the excess unreacted dichromate required 11ml of 0.1N FAS solution. A
blank of 20ml of distilled water on refluxing with 30ml of K2Cr2O7 solution required
14ml of 0.1N FAS solution. Calculate the COD of waste water sample.
[Dec 2018]
Solution: Given
Vb = 14 ml
Vt = 11 ml
F.E. Sem.-I Engineering Chemistry-I
?
6.45
Water
N = 0.1 Normal
Ve = 20 ml
(14 – 11) u 0.1 u 8 u 1000
COD =
= 120 ppm
20
6.10 SEWAGE TREATMENT BY ACTIVATED SLUDGE PROCESS
Sewage contains minerals and inorganic matter in suspension and in solution. It
contains sometimes harmful living organisms. Hence it is better to treat sewage before
releasing into the river, lake and sea. The treatment processes are as follows:
(1) Preliminary Treatment:
The principal objective of preliminary treatment is the removal of gross solids
i.e. large floating and suspended solid matter, grit, oil and greases if present in
considerable quantities. For removing inorganic matter, sewage is allowed to pass
through bar screen and mesh screens.
(2) Primary Treatment:
For removing suspended matter efficiently and economically, sedimentation process
is carried out. The sewage is treated with certain chemicals (like alum, hydrated lime etc.)
which form a floc that absorbs and entrains the suspended and colloidal particles present.
(3) Secondary or Biological Treatment:
It is essentially an aerobic chemical oxidation which includes filtartion and activated
sludge process. In this process, sewage water is filtered through specially designed
sprinkling filters so that aerobic conditions are maintained all the times. During this
process, carbon of the organic matter is converted into CO2; the nitrogen into NH3 and
finally into nitrites and nitrates.
Tricking Filters: Tricking filters are commonly used for the biological oxidation of
sewage. It is rectangular or circular in shape and about 2 meters deep. They are filled
with coarse, crushed rock or anthracite coal or broken bricks.
Sewage is delivered to the filters by means of rotating distributor. As the trickled
sewage starts percolating downwards, through the filtering medium, the aerobic bacteria
grow on the surface of aggregates by using organic matter in sewage as food. Highly
aerobic condition is maintained constantly in this design. The bacteria bring about the
biological oxidation of organic matter of sewage. The treated water comes out from
bottom. A normal trickling filter removes about 90% BOD within few hours.
F.E. Sem.-I Engineering Chemistry-I
Raw
sewage
6.46
Water
Treated
discharge
Alum
Activated
sludge
Screening
chlorination
Sedimentation
Sedimentation
Trickling
filter
Anaerobic
Primary treatment
Aerobic
Secondary treatment
Fig. 6.7 : Flow Diagram for Sewage Treatment
Activated Sludge Process: This is the much faster oxidation of organic matter in
sewage by aerobes, under highly aerobic conditions, in the presence of a part of sludge
from previous oxidation process. The sludge from the previous oxidation process is
known as activated sludge, since it contains organic matter inhabited by numerous
aerobes.
This process consists of mixing the sedimented sewage from step (2) with proper
quantity of activated sludge and the mixture is sent to the aeration tank in which agitation
is carried out for 4-6 hours by blowing air in it. Here suspended and dissolved organic
matter is oxidised efficiently by the aerobic bacteria. After aeration, the affluent is sent to
sedimentation tank, where sludge is deposited and clean water is drawn off. A part of the
sludge deposited is used for next oxidation batch and the remainder is either spread on
land as fertile matter or used for biogas or dumped in sea.
(4) Tertiary Treatment:
It is applied to decrease the load of nitrogen and phosphorous compounds present in
the effluents.
(i) Precipitation: It involves treatment of effluent with lime which reacts with
phosphorus to form calcium phosphate.
(ii) Nitrogen Stripping: It is carried out to remove ammonia gas.
(iii)Chlorination: It involves treating the water from nitrogen stripping with chlorine
so as to kill the disease causing micro-organisms present in sewage waste water.
The so treated water is finally discharged into rivers or lakes.
F.E. Sem.-I Engineering Chemistry-I
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Water
6.11 REVIEW QUESTIONS
1. What causes soft water to form lather with soap?
Ans. Refer Section 6.3.
2. Explain hardness of water?
Ans. Refer Section 6.3.
3. Differentiate between temporary hardness and permanent hardness.
Ans. Refer Section 6.3.1.
4. Why temporary hardness is called carbonate hardness?
Ans. Refer Section 6.3.1.
5. Differentiate between hard water and soft water.
Ans. Refer Section 6.3.2.
6. Explain how you would determine hardness of water using EDTA method.
Ans. Refer Section 6.5.
7. What are the problems caused by hard water in different industries?
Ans. Refer Section 6.6.
8. How the regeneration of ion exchange resin is done?
Ans. Refer Section 6.7.1.
9. What is deminiralisation? Explain its merits and demerits.
Ans. Refer Section 6.7.1.
10. What is brackish water?
Ans. Refer Section 6.8.
11. What is desalination?
Ans. Refer Section 6.8.
12. Explain reverse osmosis. Give its industrial application.
Ans. Refer Section 6.8.
13. Define BOD and COD. What is the difference between them?
Ans. Refer Section 6.9.1 and 6.9.2.
14. What is COD of waste water? How its numerical value is important?
Ans. Refer Section 6.9.2.
15. Write a note on ultrafilteration.
Ans. Refer Section 6.8.
16. Explain with the help of neat and labeled diagram the method of electro dialysis.
Ans. Refer Section 6.8.
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Water
17. With the help of diagram explain the activated sludge method to control water
pollution.
Ans. Refer Section 6.10.
6.12 UNSOLVED PROBLEMS
1. A water sample on analysis has been found to contain following impurities in
ppm: Mg(HCO3)2 = 5.84, Ca(HCO3)2 = 4.86, MgSO4 = 16.8, CaSO4 = 13.6.
Calculate the temporary, permanent and total hardness of water.
Ans. [Temporary hardness = 7ppm, Permanent hardness = 24ppm,
Total hardness = 31ppm]
2. A water sample contains the following impurities in ppm:
Mg(HCO3)2 = 73, Ca(HCO3)2 = 81, MgSO4 = 120, CaSO4 = 136.
Calculate the temporary and permanent hardness of water.
Ans. [Temporary hardness = 100 ppm, Permanent hardness = 200 ppm]
3. Find out temporary, permanent and total hardness in a sample of water with
following impurities: Ca(HCO3)2 = 81 ppm, MgSO4 = 60 ppm, MgCO3 = 84 ppm,
KCl = 30 ppm, CaCl2 = 22.2 ppm.
Ans. [Total hardness = 220 ppm, Permanent hardness = 70 ppm,
Temporary hardness = 150 ppm]
4. Calculate carbonate and non-carbonate hardness of water containing following
impurities in mg/L:
Mg(HCO3)2 = 14.6, Ca(HCO3)2 = 8.1, MgCl2 = 19, Mg(NO3)2 = 29.6.
Ans. [Carbonate hardness = 15 ppm, Non-carbonate hardness = 40 ppm]
5. Calculate total hardness of water sample containing the following impurities:
Mg(HCO3)2 = 146 mg/L, Ca(HCO3)2 = 81 mg/L, MgCl2 = 95 mg/L,
NaCl = 10 mg/L
Express the result in qCl and qFr.
Ans. [Total hardness = 250 mg/L = 25 qFr = 17.5 qCl]
6. What is the total hardness of a sample of water which has the following impurities
in mg/L: Ca(HCO3)2 = 162, CaCl2 = 22.2, MgCl2 = 95, NaCl = 20.
Express the result as qCl and qFr.
Ans. [Total hardness = 220 mg/L = 22 qFr = 15.4 qCl]
F.E. Sem.-I Engineering Chemistry-I
6.49
Water
7. Three water samples A, B and C were analyzed for their salt content. Sample A
was found to contain 168 mg magnesium carbonate per litre. Sample B was found
to contain 82 mg of calcium nitrate and 2 mg of silica per litre. Sample C was
found to contain 20 mg of potassium nitrate and 20 mg of calcium carbonate per
500 ml. Determine the hardness in all the above three water samples A, B and C.
Ans. [Sample A = 200 ppm, Sample B = 50 ppm, Sample C = 40 ppm]
8. A water sample contains 16.8 mg of MgCO3 and 6.0 mg of SiO2 per litre. Find its
temporary and permanent hardness.
Ans. [Temporary hardness = 20 ppm]
9. 50 ml of standard hard water (1.2 g of CaCO3/l) requires 30 ml of EDTA solution.
100 ml of water sample consumes 14 ml of EDTA solution. 100 ml of boiled
filtered water takes 7.0 ml of EDTA. Find temporary and permanent hardness of
water sample.
Ans. [Permanent hardness = 140 ppm, Temporary hardness = 140 ppm]
10. 0.5 g of CaCO3 was dissolved in dilute HCl and diluted to 500 ml. 50 ml of this
solution required 45 ml of EDTA solution for titration. 50 ml of hard water
sample required 15 ml of EDTA solution for titration. 50 ml of same water
sample on boiling, filtering requires 10 ml of EDTA solution. Calculate the
temporary, permanent and total hardness in ppm.
Ans. [Total hardness = 333.34 ppm, Permanent hardness = 222.23 ppm
Temporary hardness = 111.11 ppm]
11. 0.2 gm of CaCO3 is dissolved in dilute HCl and diluted to 250 ml. 25 ml of this
solution required 24.0 ml of EDTA using Eriochrome Black-T indicator. 50 ml of
a hard water sample requires 20 ml of the same EDTA. 100 ml of the water after
boiling and filtering requires 24 ml of the same EDTA. Calculate the hardness in
the water sample.
Ans. [Total hardness = 336 ppm, Permanent hardness = 201.6 ppm,
Temporary hardness = 134.4 ppm]
12. 50 ml of standard hard water containing 1 mg pure CaCO3 per ml consumed
30 ml of EDTA, 50 ml of water sample consumed 35 ml of EDTA solution using
Eriochrome Black-T indicator, 50 ml of water sample requires 21 ml of EDTA
using the same indicator. Calculate the temporary and permanent hardness.
Ans. [Temporary hardness = 466.67 ppm, Permanent hardness = 700 ppm]
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Water
13. Calculate the hardness of water sample whose 20 ml requires 30 ml of EDTA.
10 ml of calcium chloride solution, whose strength is equivalent to 300 mg of
calcium carbonate per 200 ml required 20 ml of EDTA solution.
Ans. [Hardness = 1013.25 ppm]
14. A 50 ml of sample contains 840 ppm of dissolved oxygen. After 5 days the
dissolved oxygen value becomes 230 ppm after the sample has been diluted to
80 ml. Calculate the BOD of the sample.
Ans. [BOD = 976 ppm]
15. A 50 ml of water sample contains 500 ppm of dissolved oxygen. The water
sample is diluted to 100 ml. After 5 days of incubation the DO value of water
sample reduces to 400 ppm. Calculate BOD of water sample.
Ans. [BOD = 200 ppm]
16. A 40 ml of sample contains 700 ppm of dissolved oxygen. After 5 days the
dissolved oxygen value becomes 200 ppm after the sample has been diluted to
80 ml. Calculate the BOD of the sample.
Ans. [BOD = 1000 ppm]
17. A 25 ml of a sewage water sample was refluxed with 10 ml of 0.25 N K2Cr2O7
solution in presence of dil. H2SO4, Ag2SO4 and HgSO4. The unreacted dichromate
required 6.5 ml of 0.1 N ferrous ammonium sulphate. 10 ml of the same K2Cr2O7
solution and 25 ml of distilled water, under the same conditions as the sample,
required 27 ml of 0.1 N ferrous ammonium sulphate. Calculate the COD of the
sewage water sample.
Ans. [COD = 0.656 ppm]
18. A 25 ml of a sewage water sample was refluxed with 10 ml of 0.25 N K2Cr2O7
solution in presence of dil. H2SO4, Ag2SO4 and HgSO4. The unreacted dichromate
required 5.5 ml of 0.1 N ferrous ammonium sulphate. 10 ml of the same K2Cr2O7
solution and 25 ml of distilled water, under the same conditions as the sample,
required 15 ml of 0.1 N ferrous ammonium sulphate. Calculate the COD of the
sewage water sample.
Ans. [COD = 304 ppm]
19. A 20 ml of a sewage water sample was refluxed with 10 ml of 0.1 N K2Cr2O7 solution
in presence of dil. H2SO4, Ag2SO4 and HgSO4. The unreacted dichromate required 10
ml of 0.1 N ferrous ammonium sulphate. 10 ml of the same K2Cr2O7 solution and 20
ml of distilled water, under the same conditions as the sample, required 20 ml of 0.1
N ferrous ammonium sulphate. Calculate the COD of the sewage water sample.
Ans. [COD = 400 ppm]
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6.51
Water
6.13 UNIVERSITY QUESTIONS
December 2007
1. What do you mean by hardness of water? Distinguish between alkaline and nonalkaline hardness of water.
(3 M)
Ans. Refer Section 6.3 and 6.3.1.
2. Explain the terms : (i) BOD, (ii) COD.
What is their significance?
(3 M)
Ans. Refer Section 6.9.1 and 6.9.2.
May 2008
1. A water sample contains :
Ca(HCO3)2 = 32.4 mg/L, Mg(HCO3)2 = 29.2 mg/L, CaSO4 = 13.5 mg/L.
Calculate temporary, permanent and total hardness.
(At. Wt. Ca = 40, Mg = 24, H = 1, C = 12, O = 16, S = 32).
(3 M)
Ans. Temporary Hardness = 40 ppm. Perm. Hardness = 10 ppm,
Total Hardness = 50 ppm.
2. What are carbonate and non carbonate hardness?
A standard hard water contains 15gm/litre calcium carbonate. 20 ml of this water
required 25 ml of EDTA solution. 100 ml of sample water required 18 ml of
EDTA solution. The same sample, after boiling required 12 ml of EDTA solution.
Calculate temporary hardness of water.
(6 M)
Ans. Refer Section 4.3.1 and Total Hardness = 2160 ppm,
Temporary Hardness = 1440 ppm, Perm. Hardness = 720 ppm.
3. Give brief account of reverse osmosis.
(3 M)
Ans. Refer Section 6.8.
4. Write a note on activated sludge method.
(5 M)
Ans. Refer Section 6.10.
December 2008
1. Explain the principle of EDTA method.
Ans. Refer Section 6.5.
2. Define the terms : (i) BOD, (ii) COD.
What is their significance?
Ans. Refer Section 6.9.1 and 6.9.2.
(3 M)
(5 M)
F.E. Sem.-I Engineering Chemistry-I
6.52
3. Explain Reverse Osmosis and Ultra filtration.
Ans. Refer Section 6.8.
4. Write a note on activated sludge method.
Water
(5 M)
(5 M)
Ans. Refer Section 6.10.
May 2009
1. A water sample contains:
Mg(HCO3)2 = 14.6 ppm, Mg(NO3)2 = 29.6 ppm, Ca(HCO3)2 = 8.1 ppm,
MgCl2 = 19 ppm, MgSO4 = 24 ppm.
Calculate the temporary and permanent hardness of water sample.
(At. Wt. Ca = 40, Mg = 24, H = 1, C = 12, O = 16, S = 32)
(3 M)
Ans. Temporary Hardness = 15 ppm, Perm. Hardness = 60 ppm.
2. How demineralization of water is carried out?
(6 M)
Ans. Refer Section 6.7.1.
3. 0.5 gm CaCO3 was dissolved in HCl and the solution made up to 500 ml with
distilled water. 50 ml of the solution required 48 ml of EDTA solution for
titration. 50 ml of hard water sample required 15 ml of EDTA and after boiling
and filtering required 10 ml of EDTA solution. Calculate temporary hardness of
water.
(5 M)
Ans. Temporary Hardness = 104.17 ppm.
4. Give brief account of Ultra filtration.
(3 M)
Ans. Refer Section 6.8.
5. Write a note on activated sludge method.
(5 M)
Ans. Refer Section 6.10.
December 2009
1. Calculate temporary and total hardness of a water sample containing,
Mg(HCO3)2 = 7.3 mg/L, Ca(HCO3)2 = 16.2 mg/L, MgCl2 = 9.5 mg/L,
CaSO4 = 13.6 mg/L.
(3 M)
Ans. Temporary Hardness = 15 ppm, Total Hardness = 35 ppm.
2. Explain the ion exchange process of softening of hard water. What are its
advantages and disadvantages?
(7 M)
Ans. Refer Section 6.7.1.
3. Write a note on activated sludge method.
Ans. Refer Section 6.10.
(5 M)
F.E. Sem.-I Engineering Chemistry-I
6.53
May 2010
1. Write a short note on reverse osmosis.
Water
(3 M)
Ans. Refer Section 6.8.
2. Describe the Demineralisation process. State its advantages and disadvantages.
Ans. Refer Section 6.7.1.
(5 M)
December 2010
1. Explain the principle of EDTA method.
(3 M)
Ans. Refer Section 6.5.
2. Explain the terms : (i) BOD, (ii) COD.
What is their significance?
(3 M)
Ans. Refer Section 6.9.1 and 6.9.2.
3. Write a note on reverse osmosis.
(3 M)
Ans. Refer Section 6.8.
4. 1 gm of CaCO3 was dissolved in 1 litre of distilled water. 50 ml of this solution
required 45 ml of EDTA solution for titration. 50 ml of hard water required 25 ml
of EDTA for titration. The same sample of water after boiling consumed 15 ml of
EDTA for titration. Calculate the hardness of water.
(5 M)
Ans. Total Hardness = 555.55 ppm, Perm. Hardness = 333.33 ppm,
Temporary Hardness = 222.22 ppm.
5. Write a note on activated sludge method.
(5 M)
Ans. Refer Section 6.10.
May 2011
1. A water sample on analysis has been found to contain MgCl2 = 19 ppm,
CaCO3 = 5 ppm, Ca(HCO3)2 = 29.5 ppm, CaSO4 = 13 ppm. Calculate temporary,
permanent and total hardness.
(3 M)
Ans. Total hardness = 52.77 ppm, Perm hardness = 29.56 ppm,
Temp hardness = 23.21 ppm.
2. 50 ml water sample required 7.2 ml of N/20 disodium EDTA for titration. After
boiling and filtration the same volume required 4 ml of EDTA. Calculate each
type of hardness.
(5 M)
Ans. Total hardness = 360 ppm, Perm hardness = 200 ppm,
Temp hardness = 160 ppm.
3. Explain the reverse osmosis and ultrafiltration.
(5 M)
Ans. Refer Section 6.8.
F.E. Sem.-I Engineering Chemistry-I
6.54
Water
December 2011
1. Classify the following impurities into temporary, permanent and non hardness
causing impurities. Ca(HCO3)2, MgSO4, CaCl2, Mg(HCO3)2, CaSO4, NaCl. How
many grams of CaCl2 dissolved per litre gives 150 ppm of hardness?
(3 M)
Ans. Temp hardness = Ca(HCO3)2, Mg(HCO3)2, Perm hardness = MgSO4,
CaCl2, CaSO4, Non hardness = NaCl.
2. Define BOD and COD. Give its significance.
(3 M)
Ans. Refer Section 6.9.1 and 6.9.2.
3. What is reverse osmosis? Give its applications.
(3 M)
Ans. Refer Section 6.8.
4. 20 ml of standard hard water containing 1.2 gm CaCO3 per litre required 35 ml of
EDTA. 50 ml of hard water sample required 30 ml of the same EDTA. 100 ml of
hard water sample after boiling required 25 ml of same EDTA. Calculate the
various hardnesses.
(5 M)
Ans. Total hardness = 411 ppm, Perm hardness =342.5 ppm,
Temp hardness = 68.5 ppm.
5. Write a note on activated sludge method.
(5 M)
Ans. Refer Section 6.10.
December 2012
1. Distinguish between BOD and COD.
(3 M)
Ans. Refer Section 6.9.1 and 6.9.2.
2. Calculate temporary, permanent and total hardness of water sample containing
Mg(HCO3)2 = 7.3 ppm, Ca(HCO3)2 = 16.2 ppm, MgCl2 = 9.5 ppm, CaSO4 = 13.6
ppm.
(3 M)
Ans. Temp hardness = 15 ppm, Perm hardness = 20 ppm, Total hardness = 35 ppm.
3. Write a note on activated sludge method.
(5 M)
Ans. Refer Section 6.10.
June 2013
1. Differentiate between BOD and COD.
(3 M)
Ans. Refer Section 6.9.1 and 6.9.2.
2. What is the total hardness of water sample which has following impurities in
mg/l? Ca(HCO3)2 = 162, CaCl2 = 22.2, Mg Cl2 = 95, NaCl = 20.
(3 M)
Ans. Total hardness = 220 ppm.
F.E. Sem.-I Engineering Chemistry-I
6.55
Water
3. 0.5 g of CaCO3 was dissolved in dil HCl and diluted to 500 ml. 50 ml of this
solution required 45 ml of EDTA solution for titration. 50 ml of hard water
sample required 15 ml of EDTA solution for titration. 50 ml of same water
sample on boiling & filtering reqires 10 ml EDTA solution. Calculate temporary
hardness and total hardness.
(6 M)
Ans. Temp hardness = 111.11 ppm, Total hardness = 333.33 ppm.
4. Explain demineralization of water by ion exchange method.
(5 M)
Ans. Refer Section 6.7.1.
December 2013
1. What Happens when temporary hard water is boiled? Give equation to explain.
Ans. Refer Section 6.3.1.
(3 M)
2. Two samples of water A and B were analysed for their salt content :
(3 M)
(i) Sample A was found to contain 168 mg MgCO3 per litre.
(ii) Sample B was found to contain 820 mg Ca (NO3)2 per litre and 2 mg SiO2 per
litre.
Calculate the total hardness of each sample and state which sample is more hard.
Ans. Total Hardness of A = 200 ppm, Total Hardness of B = 500 ppm. Sample B is
more hard.
3. 50 ml of standard hard water containing 1 mg of pure CaCO3 per ml consumed
20 ml of EDTA. 50 ml of the water sample consumed 30 ml of same EDTA
solution. After boiling and filtering, 50 ml of the water sample required 10 ml of
the same EDTA for titration. Calculate the total and permanent hardness of water
sample.
(6 M)
Ans. Perm hardness = 500 ppm, Total hardness = 1500 ppm.
4. Write short notes of any two :
(6 M)
(a) Reverse Osmosis, (b) Electro dialysis, (c) Ultra filtration.
Ans. Refer Section 6.8.
May 2014
1. Give the principle of estimation of hardness of water using EDTA method
(Only Equations).
(3 M)
Ans. Refer Section 6.5.
2. Classify the following salts into temporary and permanent hardness causing salts
and also calculate their hardness :
(3 M)
(a) Ca (HCO3)2 – 16.2 mg/L, (b) MgSO4 – 1.2 mg/L, (c) FeCl2 – 12.7 mg/L.
F.E. Sem.-I Engineering Chemistry-I
6.56
Water
Ans. Ca (HCO3)2 Temporary hardness = 10 ppm and
MgSO4 Permanent hardness = 1 ppm.
3. With a neat diagram explain the principle and reaction of Ion-Exchange method
of softening of water.
(5 M)
Ans. Refer Section 6.7.1.
4. (a) Define the significance of BOD and COD.
(6 M)
(b) Discuss Reverse Osmosis.
Ans. Refer Section 6.9.1 and 6.9.2, and 6.8.
December 2014
1. Explain the principle of EDTA method.
(3 M)
Ans. Refer Section 6.5.
2. 25 ml of sewage water is refluxed with 0.1N K2Cr2O7 solution in presence of
H2SO4 and Ag2SO4. The unreacted dichromate required 5.5ml of 0.1N FAS
solution. Blank titration consumed 15 ml of 0.1N FAS solution. Calculate COD of
the effluent.
(3 M)
Ans. COD = 304 ppm.
May 2015
1. Differentiate between temporary and permanent hardness.
Ans. Refer Section 6.3.2.
2. Calculate all types of hardness of water sample containing:
Ca (HCO3)2 – 81 ppm
MgSO4 – 60 ppm
Mg CO3 – 42 ppm
Ca(NO3)2 – 82 ppm.
Ans. Temporary hardness = 100 ppm and Permanent hardness = 100 ppm.
3. Write note on Ultra filtration and Reverse osmosis.
(3 M)
(3 M)
(5 M)
Ans. Refer Section 6.8.
December 2015
1. A 5ml sample of waste water was refluxed with 30ml of potassium dichromate
solution and after refluxing the excess unreacted dichromate required 23ml of
0.1M FAS solution. A blank of distilled water on refluxing with 30ml of
dichromate solution required 36ml of 0.1M FAS solution. Calculate the COD
value of the waste water.
(3 M)
Ans. COD = 2080 ppm.
F.E. Sem.-I Engineering Chemistry-I
6.57
2. Write a note on activated sludge method.
Ans. Refer Section 6.10.
Water
(5 M)
May 2016
1. Write two balanced equations to describe the changes that occur when hard water
is boiled.
(3 M)
Ans. Refer Section 6.3.1.
2. Calculate the COD of an effluent sample if 25cc of the effluent sample required
8.3cc of 0.001M K2Cr2O7 for oxidation.
(3 M)
Ans. COD = 2.656 ppm.
3. (a) Discuss the softening and regeneration reactions in the ion exchange process.
(b) Discuss the reverse osmosis method of purification of water.
(6 M)
Ans. (a) Refer Section 6.7.1.
(b) Refer Section 6.8.
December 2016
1. Distinguish between alkaline and non-alkaline hardness.
(3 M)
Ans. Refer Section 6.3.2.
2. 6ml of waste water was refluxed with 25ml of K2Cr2O7 and after refluxing the
excess unreacted dichromate required 20ml of 0.1N FAS solution. A blank of
distilled water on refluxing with 25ml of K2Cr2O7 solution required 35ml of 0.1N
FAS solution. Calculate the COD of waste water sample.
(3 M)
Ans. COD = 2000 ppm.
3. What are the advantages of ion exchange process?
(3 M)
Ans. Refer Section 6.7.1.
4. What are the industrial applications of ultrafiltration?
(2 M)
Ans. Refer Section 6.8.
June 2017
1. What are cation and anion exchangers?
(3 M)
Ans. Refer Section 6.7.1.
2. 10ml of waste water was refluxed with 20ml of K2Cr2O7 and after refluxing the
excess unreacted dichromate required 36.2ml of 0.1N FAS solution. A blank of
10ml of distilled water on refluxing with 20ml of K2Cr2O7 solution required 46ml
of 0.1N FAS solution. Calculate the COD of waste water sample.
(3 M)
Ans. COD = 784 ppm.
F.E. Sem.-I Engineering Chemistry-I
6.58
Water
3. Explain ion exchange process for softening of hard water. It’s advantages and
disadvantages.
(6 M)
Ans. Refer Section 6.7.1.
4. Write a note on ultrafiltration.
(3 M)
Ans. Refer Section 6.8.
5. Distinguish between temporary and permanent hardness of water.
Ans. Refer Section 6.3.2.
(2 M)
December 2017
1. Distinguish between BOD and COD.
Ans. Refer Section 6.9.1 and 6.9.2.
(3 M)
2. Calculate total hardness in ppm in given water sample.
(3 M)
(i) 50ml standard hard water containing 1 mg pure CaCO3 per ml, consumed 20ml
EDTA solution.
(ii) 50ml water sample consumed 30ml EDTA solution using Erio-Black T
indicator.
Ans. Hardness of water = 1500 ppm.
3. Explain Reverse Osmosis.
(2 M)
Ans. Refer Section 6.8.
4. Write a note on activated sludge method.
(6 M)
Ans. Refer Section 6.10.
May 2018
1. 10ml of waste water was refluxed with 20ml of K2Cr2O7 and after refluxing the
excess unreacted dichromate required 26.2ml of 0.1N FAS solution. A blank of
10ml of distilled water on refluxing with 20ml of K2Cr2O7 solution required 36ml
of 0.1N FAS solution. Calculate the COD of waste water sample.
(3 M)
Ans. COD = 784 ppm.
2. Calculate total hardness in ppm in given water sample.
(4 M)
(i) 50ml standard hard water containing 1 mg pure CaCO3 per ml, consumed 20ml
EDTA solution.
(ii) 50ml water sample consumed 30ml EDTA solution using Erio-Black T
indicator.
Ans. Hardness of water = 1500 ppm.
3. Distinguish between BOD and COD.
Ans. Refer Section 6.9.1 and 6.9.2.
(3 M)
F.E. Sem.-I Engineering Chemistry-I
6.59
4. Discuss reverse osmosis.
Ans. Refer Section 6.8.
Water
(2 M)
December 2018
1. Write a short note on reverse osmosis.
(2 M)
Ans. Refer Section 6.8.
2. 20ml of waste water was refluxed with 30ml of K2Cr2O7 and after refluxing the
excess unreacted dichromate required 11ml of 0.1N FAS solution. A blank of
20ml of distilled water on refluxing with 30ml of K2Cr2O7 solution required 14ml
of 0.1N FAS solution. Calculate the COD of waste water sample.
(3 M)
Ans. COD = 120 ppm.
3. Draw the diagram for demineralization process and write suitable reactions
involved in the process. What are the advantages and disadvantages of the
method?
(6 M)
Ans. Refer Section 6.7.1.
4. Explain the principle of EDTA method.
(3 M)
Ans. Refer Section 6.5.
5. Write a note on activated sludge method.
(3 M)
Ans. Refer Section 6.10.
May 2019
1. 10ml of waste water was refluxed with 20ml of K2Cr2O7 and after refluxing the
excess unreacted dichromate required 26.2ml of 0.1N FAS solution. A blank of
10ml of distilled water on refluxing with 20ml of K2Cr2O7 solution required 36ml
of 0.1N FAS solution. Calculate the COD of waste water sample.
(3 M)
Ans. COD = 784 ppm.
2. In the process of determination of hardness, standard hard water sample was
prepared by dissolving 2.5g CaCO3 and making solution up to 1 liter. 50ml of
above hard water required 45ml of EDTA. 50ml of unknown hard water sample
consumed 30ml EDTA solution using Erio-Black T indicator. The unknown hard
water sample was boiled and filtered. 50ml of this boiled solution required 20ml
of EDTA. Calculate hardness of all types of unknown hard water sample. (4 M)
Ans. Total Hardness = 1666.2 ppm
Permanent Hardness = 1110.8 ppm
Temporary Hardness = 555.4 ppm.
F.E. Sem.-I Engineering Chemistry-I
6.60
Water
3. A hard water sample contains following impurities in ppm. Mg(HCO3)2 = 150,
NaCl =77, CaCl2 = 135, MgSO4 = 85. Calculate temporary, permanent and total
hardness of the given water sample.
(3 M)
Ans. Temporary hardness = 102.74ppm
Permanent hardness= 192.45ppm
Total hardness = 295.19 ppm.
4. Write a short note on reverse osmosis.
Ans. Refer Section 6.8.
(3 M)
REFERENCES

Engineering Chemistry - Jain & Jain (Dhanpat Rai)

Engineering Chemistry - Dara & Dara (S Chand)

Engineering Chemistry - Wiley India (ISBN - 9788126519880)

A Text Book of Engineering Chemistry - Shashi Chawla (Dhanpat Rai)

Engineering Chemistry - Payal Joshi & Shashank Deep (Oxford University Press)

Concise Inorganic Chemistry - J D LEE

Essentials of Physical Chemistry - B S Bahl, Arun Bahl, G D Tuli.
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