Engineering Chemistry 9781783323555, 9781783325702, 1783323558

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Engineering

CHEMISTRY

Ritu Singh

α Alpha Science International Ltd. Oxford, U.K.

Engineering Chemistry 380 pgs.

Ritu Singh Department of Chemistry Shri Ram Murti Smarak College of Engineering and Technology Bareilly Copyright © 2020 A L P H A S C I E N C E I N T E R N AT I O N A L LT D . 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K. www.alphasci.com ISBN 978-1-78332-355-5 E-ISBN 978-1-78332-570-2 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher.

Preface Any good text book, particularly in the fast changing fields such as engineering and technology, is not only expected to cater to the current curriculum requirements of various institutions but also provide a glimpse towards the latest developments in the concerned subject and the relevant discipline. This book is written as per the syllabus of Engineering Chemistry course offered to First year B.Tech students of various Indian universities and offers easy, accurate, systematic and logical leaning of the subject. The book aims to import knowledge of the subject so that students can understand role of chemistry in the field of engineering and can develop a scientific reasoning of the subject. The book includes a large number of solved and unsolved questions and numerical so that students can understand applications of various topics at the end of each Chapter. I hope this volume will prove beneficial to its readers.

ACKNOWLEDGEMENT I am duty bound to express my thanks to the authors and publishers of all the books which have been refereed during course of preparation of this book. I deeply acknowledge my family members for their never ending encouragement. A sincere expression of thanks to publisher and entire staff of Narosa Publication House. I am grateful to God for what I am and My thanks giving is perpetual.

Contents Preface

1. Chemical Bonding

iii

1.1



1. Introduction 1.1 2. Orbital Theory 1.2 2.1 Review of Valence Bond Theory 1.2 2.2 Drawbacks of Valence Bond Theory 1.2 2.3 Molecular Orbital Theory 1.2 2.4 Let us Elaborate the Ideas of the Molecular Orbital Theory 1.3 2.5 Difference between Bonding and Anti-Bonding Molecular Orbitals 1.4 2.6 Conditions for the Combination of Atomic Orbital to form Molecular Orbitals 1.4 2.7 Molecular Orbitals Energy Level Diagram by Overlapping of px-px/py-py 1.6 2.8 Relationship between Electronic Configuration and Molecular Behavior 1.7 2.9 The Electronic Structures and Bonding Properties in Hetero 1.11 Nuclear Diatomic Molecules 3. Metallic Bonding 1.14 3.1 Nature of Bonding in Metals 1.15 3.2 Explanation of Characteristics of Metals by Electron Sea Model 1.16 3.3 Valence Band Model/Resonance Model 1.17 3.4 Band Model (Molecular Orbital Approach) 1.17 Solved Questions 1.21 Unsolved Questions 1.23

2. The Solid State

2.1



2.1 2.1 2.1 2.1

1. General Introduction 1.1 Classification of Solids 1.2 Classification of Crystalline Solids 1.3 Crystal

vi  Contents 1.4 Types of Symmetry 2.2 1.5 Space Lattice 2.3 1.6 Types of Lattice 2.3 1.7 Number of Atoms Per Unit Cell of Crystal Systems 2.4 1.8 Calculation of Atomic radius of Unit Cell 2.6 2. Coordination Number 2.7 2.1 Radius Ratio 2.8 2.2 X-Ray Studies of Crystals 2.10 2.3 Bragg’s Equation 2.10 3. Imperfections in Crystals 2.11 3.1 Classification of Imperfection 2.12 3.2 Electronic Defects 2.12 3.3 Atomic Imperfections / Point Defects 2.12 3.4 Types of Point Defects 2.12 3.5 Types of Compounds Exhibiting Schottky Defect 2.13 3.6 Frenkel Defect 2.13 3.7 Graphite 2.15 3.8 Fullerenes 2.16 3.9 Liquid Crystals 2.17 Solved Questions 2.20 Unsolved Questions 2.23

3. Nanomaterials 1. Introduction 2. The Significance of the Nanoscale 3. Tools to make Nanostructures 4. Preparation/Synthesis of Quantum Dots 5. Synthesis of Nanoparticles 6. Examples of Nanomaterials 6.1 Graphite 6.2 Creating of Buckyballs 7. Carbon Nanotubes 7.1 Types of Nanotubes 7.2 Structure 7.3 Synthesis of Nanotubes 7.4 Properties of Nanotubes 7.5 Applications of Nanotubes 8. Applications of Nano Technology Solved Questions Unsolved Questions

3.1 3.1 3.2 3.5 3.6 3.6 3.13 3.13 3.14 3.15 3.15 3.16 3.16 3.18 3.18 3.19 3.20 3.20

Contents  vii

4. Electrochemistry and Ionic Equilibrium

4.1

1. Electrochemical Cells or Galvanic Cells 4.1 1.1 Galvanic Cell 4.1 2. Salt Bridge and its Functions 4.2 3. Half Cell Reactions and Electrodes 4.3 4. Representation of a Cell or Cell Diagram 4.4 5. Emf of the Cell or cell Potential 4.5 5.1 Calculation of EMF 4.5 6. Electrode Potential (Reduction Potential) 4.6 6.1 Measurement of Electrode Potential 4.7 7. Electrochemical Series 4.9 7.1 Important Features of Electrochemical Series 4.10 7.2 Importance of Electrochemical Series 4.11 8. Types of Cells 4.12 8.1 Chemical Cells 4.13 8.2 Emf of Concentration Cell 4.14 9. Cell Potential Measurement and Thermodynamic Functions 4.15 0 9.1 Relation Between D G and the Cell Potential 4.15 9.2 Entropy of the Reaction 4.16 9.3 Enthalpy of the Reaction 4.17 10. EMF of the Cell and Equilibrium Constant of a Cell Reaction 4.17 11. The Nernst Equation 4.18 11.1 Application of the Nernst Equation 4.21 12. Battery 4.23 Solved Questions 4.25 Unsolved Questions 4.31

5. Corrosion

5.1

1. Introduction 5.1 2. Theories of Corrosion 5.1 2.1 The Electrochemical or Wet or Immersed Theory 5.1 2.2 Chemical Corrosion or Dry Corrosion or Direct Chemical Attack Theory 5.3 3. Types of Corrosion 5.6 3.1 Galvanic Corrosion 5.6 3.2 Concentration Cell Corrosion 5.7 3.3 Intergranular Corrosion 5.9 3.4 Stress Corrosion (Stress Cracking) 5.9 3.5 Microbiological Corrosion 5.11 3.6 Erosion Corrosion 5.11 3.7 Soil Corrosion 5.11

viii  Contents 3.8 Selective Leaching 3.9 Hydrogen Damage 3.10 Stray Current Corrosion 3.11 Factors Affecting Corrosion 3.12 Nature of the Corroding Environment 4. Corrosion Control 4.1 Material Selection and Design 4.2 Cathodic and Anodic Protection 4.3 Protective Coatings 5. Corrosion Inhibitors 5.1 Passivity 5.2 Oxide or Protective Layer Theory Solved Questions Unsolved Questions

6. Water Treatment

5.11 5.12 5.12 5.12 5.13 5.13 5.13 5.13 5.13 5.16 5.17 5.17 5.18 5.19

6.1

1. Introduction 6.1 1.1 Moorland Surface Drainage 6.2 1.2 Lowland Surface Drainage 6.2 1.3 Deep Well Waters 6.2 2. Effect of Water on Rocks and Minerals 6.2 3. Types of Impurities Present in Water 6.3 4. Effects of Impurities in Natural Waters 6.4 4.1 Color 6.4 4.2 Tastes and Odours 6.4 4.3 Turbidity and Sediment 6.5 4.4 Micro-Organisms 6.5 4.5 Dissolved Mineral Matter 6.6 4.6 Dissolved Gases 6.13 4.7 Silica Content 6.15 4.8 Oxidability 6.16 5. Methods of Treatment of Water for Domestic and Industrial Purposes 6.16 5.1 Sedimentation 6.16 5.2 Coagulation 6.17 5.3 Filtration 6.19 5.4 Sterilization of water 6.21 6. Removal of Dissolved Salts: Softening of Water 6.24 6.1 Lime-Soda Process 6.25 6.2 Problems on Water-Treatment by Lime-soda Process 6.30 6.3 Ion-Exchange Process of Water Softening 6.32 6.4 Natural and Synthetic Zeolites 6.32

Contents  ix 6.5 Synthetic Ion-Exchange Resins 6.36 6.6 Softening of Water by Ion-exchange 6.38 6.7 Demineralisation or Deionisation 6.39 7. Boiler Feed Waters (Water for Steam Making) 6.41 8. Boiler Troubles 6.42 8.1 Carry Over 6.43 8.2 Scale Formation 6.46 8.3 Corrosion 6.51 8.4 Caustic Embrittlement 6.53 9. Desalination of Water 6.53 Summary 6.67 Solved Questions 6.69 Unsolved Questions 6.70

7. The Phase Rule

7.1

1. Definitions 7.1 1.1 Phase 7.1 1.2 Components 7.2 1.3 Degree of Freedom (F) 7.3 1.4 Application of Phase Rule 7.7 1.5 Phase Diagram 7.7 1.6 One Component System 7.8 1.7 The Water System 7.8 1.8 The Curve OA (Vapor Pressure Curve) 7.10 1.9 The Curve OB (Sublimation Curve) 7.10 1.10 Triple Point 7.10 ′ 1.11 Metastable Curve OA 7.10 2. Polymorphism 7.11 3. The Sulphur System 7.11 3.1 Mata Stable Equilibrium 7.12 3.2 Areas 7.13 Solved Questions 7.13 Unsolved Questions 7.14

8. Fuel and Combustion

8.1

1. Introduction 8.1 2. Classification 8.1 3. Calorific Value 8.2 3.1 Units of Heat 8.2 3.2 Interconversion of the Various Units of Heat 8.3 3.3 Gross Calorific Value and Net Calorific Value 8.3

x  Contents 4. Determination of Calorific Value of Solid Andnon-Volatile Liquid Fuels 8.4 4.1 Bomb Calorimeter 8.4 4.2 Calculation of Net Calorific Value 8.6 5. Determination of Calorific Value of Gases and Volatile Liquid Fuels 8.7 5.1 Boy’s Calorimeter 8.7 5.2 Description of the Apparatus 8.7 5.3 Procedure 8.8 6. Criteria for Selecting a Fuel 8.9 7. Solid Fuels 8.9 7.1 Wood 8.9 7.2 Peat 8.10 7.3 Lignite (Brown Coal) 8.11 7.4 Bituminous Coal 8.11 7.5 Anthracite 8.12 7.6 Rank of Coal 8.12 8. Coal 8.12 8.1 Origin of Coal Formation 8.12 8.2 Composition of Coal 8.13 8.3 Analysis of Coal and its Significance 8.14 8.4 Formula Based on Proximate Analysis 8.18 9. Characteristics of Coal 8.18 10. Selection of Coal 8.19 11. Commercial Types of Coal 8.19 12. Secondary Solid Fuels 8.20 12.1 Caking Coals and Coking Coals 8.20 12.2 Requisites of a Metallurgical Coke 8.20 13. Combustion of Coal 8.21 14. Liquid Fuels 8.22 14.1 Merits and Demerits of Liquid Fuels 8.22 15. Gaseous Fuels 8.22 15.1 Advantages of Gaseous Fuels 8.22 15.2 Disadvantages 8.23 15.3 Commercial Gaseous Fuels 8.23 15.4 Natural Gas 8.23 16. Biomass 8.27 16.1 Biogas 8.27 Solved Questions 8.29 Unsolved Questions 8.37

Contents  xi

9. Spectroscopic Methods of Analysis

9.1

1. Introduction 2. Electromagnetic Spectrum 3. Lambert-Beer or Beer’s Law 4. UV Spectroscopy 5. Rotational (Microwave) Spectra 6. Infrared Spectroscopy 7. Raman Spectroscopy 8. Nuclear Magnetic Resonance (NMR) Spectroscopy Solved Questions Unsolved Questions

9.1 9.1 9.3 9.5 9.12 9.15 9.22 9.27 9.33 9.35

10. Polymers

10.1

1. Characteristics of Polymers 10.2 2. Advantages of Polymers 10.3 3. Drawbacks in Polymers as Compared with Traditional Materials 10.3 4. Where do we use these Polymeric Materials? 10.3 5. Polymer Classification 10.4 6. Thermosetting Resin 10.10 6.1 Important Thermosetting Resins 10.10 6.2 Amino Resins or Urea-Formaldehyde Resin (UF) 10.12 6.3 Melamine Formaldehyde Resins 10.13 6.4 Epoxy Resin (Araldite) 10.13 7. Thermoplastics 10.15 7.1 Poly Methyl Methacrylate PMMA (Plexi Glass) 10.15 7.2 Polyvinyl Chloride (PVC) 10.15 7.3 Polyvinyl Acetate 10.16 7.4 Polyethylene 10.16 7.5 Polycarbonates 10.17 7.6 Polyamides 10.17 7.7 Polyesters [Poly Ethylene Terephthalate (PET)] 10.19 7.8 Glyptal Resins 10.20 7.9 Poly Urethanes 10.20 7.10 Polypropylene 10.21 7.11 Poly Vinyl Alcohol 10.21 7.12 Poly Formaldehyde 10.21 7.13 Poly Tetra Flouro Ethylene (PTFE) 10.22 7.14 Poly Styrene 10.22 7.15 Poly Acrylonitrile (PAN) 10.22 7.16 Elastomers (Rubber) 10.23

xii  Contents 7.17 Synthetic Rubber 10.25 7.18 Silicone Polymers 10.27 7.19 Types of Polymerization 10.28 7.20 Mechanism of Addition Polymerization 10.29 8. Conducting Polymers 10.34 9. Biodegradable polymer 10.37 10. Organometallics 10.38 11. Classification 10.38 12. Preparation 10.41 13. Applications 10.42 14. Organometallic Compounds in Polymerization 10.42 15. Grignard Reagent 10.44 15.1 Preparation 10.44 15.2 Physical Properties 10.45 15.3 Chemical Properties 10.45 15.4 Synthetic Application of Grignard Reagents 10.46 Solved Questions 10.47 Unsolved Questions 10.49

11. Concepts of Titrimetric Analysis 1. Introduction 2. Fundamentals of Titrimetric Analysis 3. Standard Solution 4. Acid–Base Titrations 5. Redox Titrations 6. Precipitation Titrations Solved Questions Unsolved Questions

11.1 11.1 11.2 11.8 11.9 11.14 11.22 11.26 11.27

Experiments on http://www.narosa.com/books_display.asp?catgcode=978-81-8487-607-9

Index

I.1

Chapter

1

Chemical Bonding 1.  Introduction Atoms, except those of noble gases, do not have free existence. They readily combine together to form molecules. The process of combination called chemical bonding involves the union of two or more atoms through redistribution of electrons in their outer shell. With the advancement of knowledge about atomic structure it was realized that electrons in atoms were primarily involved in chemical combinations. According to the Lewis octet rule atoms of all elements have a tendency to acquire an electronic configuration similar to that of inert gases because it represents the most stable electronic configuration. All atoms having unstable or incomplete outer shell have a tendency to gain or lose electrons so as to acquire an electronic configuration of the nearest inert gas in the periodic table. It is this tendency of atoms to complete and thereafter stabilize their outermost orbital of electrons which is mainly responsible for chemical combination of the atoms. Thus, according to electronic theory of valency, a chemical bond is formed as a result of electronic interactions. It may, however be noted that a molecule is formed only when electrons of the constituent atoms interact in such a way that the potential energy is lowered, greater the lowering of potential energy, greater is the strength of the bond. According to the electronic theory of valency, the interaction of extra nuclear electrons lead to the formation of the following types of bonds: 1. Electrovalent bond / Ionic bond 2. Covalent bond 3. Co-ordinate bond 4. Metallic bond 5. Hydrogen bond Subsequently, the wave mechanical theory was developed which gave logical understanding of bonding between atoms. On the basis of wave mechanical theory, it is also possible to make fruitful predictions regarding the geometrical structure of molecule. However, the electronic theory had certain limitations as it could not explain the number of facts about the molecules such as:

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1.2  Engineering Chemistry



(i) Exact shapes of molecules (ii) Existence of molecules like PCl5, SF6, etc in which the central atom has more than 8 electrons or the existence of molecules like BF3, AlCl3, etc in which the central atom has less than 8 electrons. To explain these facts, orbital concept of covalent bond was put forward. Based upon orbital concept, two different theories have been put forward, namely, (a) Valence bond theory (b) Molecular orbital theory

2. ORBITAL THEORY 2.1 Review of Valence Bond Theory This method, applied to explain the formation of a covalent bond, was put forward by Heitler and London in 1927 and later developed by Pauling and Slater in 1931. According to this approach, only the valence electrons participate in the formation of chemical bonds, the inner orbital electrons however remain undisturbed. The main ideas of this approach are as follows: 1. A covalent bond is formed by the overlap of half filled atomic orbitals of the different atoms. 2. The two half filled atomic orbitals taking part must have electrons with opposite spin. 3. The valency of an element is the same as the number of half filled orbitals present in it. 4. The strength of the bond formed depends upon the extent of overlap. Greater the overlapping, stronger is the bond formation. The above approach is called valence bond theory, according to this approach, only the valence electrons participate in the formation of chemical bonds, the inner orbital electrons however remain undisturbed.

2.2 Drawbacks of valence bond theory

1. 2. 3. 4.

Valence band theory deals with individual atoms Only half filled orbitals take part in bonding No weightage is given to ionic structure Theory tells nothing about magnetic properties of the molecule.

2.3  Molecular Orbital Theory To explain the formation of a chemical bond and to explain the behavior of molecules like their relative bond strengths, paramagnetic and diamagnetic nature etc., another approach was developed mainly by Hund and Mulliken in 1932. according to this theory when two atomic orbitals combine or overlap with each other they lose their identity and form new orbitals called molecular orbitals. The main ideas of the theory may be summed up as follows: (i) Only those atomic orbitals can combine to form molecular orbitals which have comparable energies and proper orientation. For example, 1s can combine with 1s and not with 2s. Similarly, s can combine with pz but not with px and py, or pz can combine with pz but not with px and py (taking z-axis as the internuclear axis).

Chemical Bonding  1.3



(ii) The number of molecular orbitals formed is equal to the number of combining atomic orbitals. (iii) When two atomic orbitals combine they form two new orbitals called bonding molecular orbital and anti-bonding molecular orbital. The bonding orbital is of lower energy and greater stability than the corresponding antibonding orbital. (iv) The shapes of the molecular orbitals formed depend upon the type of the combining atomic orbitals. (v) The bonding molecular orbitals are represented by s, p, d, etc, whereas the corresponding anti-bonding molecular orbitals are represented by s*, p*, d*, etc. (vi) The filling of molecular orbital takes place according to the following rules: (a) Aufbau Principle:  States that molecular orbitals are filled in order of their increasing energies. (b) Pauli Exclusion Principle:  States that a molecular orbital can have maximum of two electrons and these electrons must have opposite spin. (c) Hund’s Rule of Maximum multiplicity:  States that pairing of electrons in the degenerate molecular orbitals does not take place until each of them has got one electron each.

2.4  Let us elaborate the ideas of the molecular orbital theory Formation of molecular orbital can be explained on the basis of Linear Combination of Atomic Orbitals [LCAO]. An atomic orbital is an electron wave; the two atomic orbitals may be in phase or out of phase. Suppose jA and jB represent the amplitude of the electron waves of the atomic orbital of the two atoms A and B respectively, then there can be two possible cases. Case I:  When the two waves are in phase (Constructive interference) so that they add up and the amplitude of the new wave is

f = jA + jB

Case II:  When the two waves are out of phase (destructive interference) the waves are subtracted from each other so that the amplitude of the new wave is



f¢ = jA – jB



Knowing that the probability is given by the square of the amplitude, we have





f2 = ((jA + jB))2 = jA2 + j2B + 2 jA jB





f¢2 = (jA – jB)2





= jA2 + j2B – 2 jA jB

i.e.

j2 > jA2 + j2B

whereas

f¢2 < jA2 + j2B

Thus, by the combination of two atomic orbitals, two new molecular orbitals are formed, one by the additive effect and other by the subtractive effect of the atomic orbital. The molecular orbital formed by the additive effect of the atomic orbitals is called bonding molecular orbital and the molecular orbital formed by the subtractive effect of the atomic orbitals is called anti-bonding molecular orbital.

1.4  Engineering Chemistry

2.5 Difference between bonding and anti-bonding molecular orbitals Bonding Molecular orbital

Anti-Bonding Molecular orbital

1. They are formed by additive effect of the atomic orbital. 2. The electron density increases in the region between the nuclei of bonded atoms. 3. Molecular orbitals have lower energy than the atomic orbitals from which they are formed. 4. They are formed when the lobes of the coming orbitals have the same sign. 5. These molecular orbitals are represented by s and p.

1. They are formed by the subtractive effect of the atomic orbitals. 2. The electron density is lower in the region between the nuclei of bonded atoms. 3. Molecular orbitals have higher energy than the atomic orbitals from which they are formed. 4. They are formed when the lobes of the combining orbitals have opposite sign. 5. These molecular orbitals are represented by s* and p*.

2.6 Conditions for the combination of atomic orbital to form molecular orbitals Only those orbitals can combine to form molecular orbitals which fulfill the following conditions: (i) The combining atomic orbitals should have comparable energies. (ii) The combining atomic orbitals must have proper orientation. i.e. taking Z-axis as inter nuclear axis, 2pz orbital of one atom can combine with 2pz orbital of another atom but 2pz orbital of one atom cannot combine with 2px or 2py orbital of another atom, or 2px cannot combine with 2py. Similarly, 2s orbital of one atom can combine with 2pz orbital but cannot combine with 2px or 2py orbital of another atom because they do not have proper orientation for overlap. (iii) The extent of overlapping should be large. Greater the overlap, greater will be the electron density between the nuclei.

2.6.1  Types of molecular orbital If two atomic orbital overlap along the inter nuclear axis, the molecular orbital formed is called s molecular orbital. If two atomic orbital overlap sideways the molecular orbital formed is called p molecular orbital. Representing the orbital in terms of signs of the wave functions, as s-orbital are spherically symmetrical their wave function has the same sign in all directions. In case of p-orbitals, one lobe is given ‘+’ sign and other ‘–’ sign. Overlapping of ‘+’ part of the electron cloud of on atom with ‘+’ part of the electron cloud of the second atom implies addition of atomic orbitals leading to the formation of the bonding molecular orbital. The overlap of ‘+’ part of the electron cloud of one atom with ‘–’ part of the electron cloud of another atom means the subtraction of the atomic orbitals leading to the formation of anti-bonding molecular orbital. Few simple cases of the combination of atomic orbitals are given below: (i) 1s with 1s. The wave functions of two 1s atomic orbitals can combine in two different ways: (a) When both have the same sign (b) When they have different sign If one of the wave functions is arbitrarily assigned a +ve sign, the other may be either + ve or – ve. The bonding molecular orbital formed is designated as s (1s), s indicating that the

Chemical Bonding  1.5 overlap is along the internuclear axis and 1s indicating that 1s atomic orbitals have combined to form the molecular orbital. The corresponding bonding molecular orbital s(1s) and anti-bonding molecular orbital is designated as s*(1s) as given below:

Fig. 1  Formation of bonding and anti-bonding molecular orbital by s-s overlapping

2s with 2s  Overlapping takes place exactly in the same way as between 1s with 1s. However, in this case the bonding and anti-bonding molecular orbitals formed are designated as s (2s) and s*(2s) respectively. (ii) 2pz with 2pz  Taking z-axis as the inter nuclear axis, the molecular orbitals formed are shown below:

Fig. 2a  Formation of bonding and anti bonding molecular orbital by pz-pz overlapping.

(iii) 2px with 2px or 2py with 2py (Fig. 2b) the bonding molecular orbital formed has been designated a p(2px) or p(2py), indicating sideways overlapping and 2px or 2py indicating that the two atomic orbitals involved are 2px or 2py. The molecular orbitals in the order of increase in energy are as follows:s(1s) s*(1s) s(2s) s*(2s) s(2pz) p(2px) = p(2py) p*(2px) = p*(2py*) s*(2pz) This order is followed from H2 molecule to N2 molecule because from H2 to N2 molecule the 2s and 2p orbitals are of similar energy, so they combine with each other to form mixed atomic orbitals which further combine giving molecular orbitals. Hence, p2px, p2py are of lower energy than s2pz but after N2 molecule the order of filling of molecular orbital is as follows.

1.6  Engineering Chemistry The order is changed after Nitrogen as the energies of 2s and 2p orbital are not same so they directly form molecular orbital.

Fig. 2b  Formation of bonding and anti bonding molecular orbital py2px-2px or 2py-2py

2.7  Molecular orbitals energy level diagram by overlapping of px-px/py-py.

Fig. 3a  Molecular energy diagram for homonuclear diatomic molecules (from H2–N2).

Fig. 3b  Molecular energy diagram for homonuclear diatomic molecules (from O2 Onward).

Chemical Bonding  1.7

2.8 Relationship between electronic configuration and Molecular behavior The important rules relating the molecular behavior with the electronic configuration are as follows: (i) Stability of molecules in terms of bonding and anti-bonding electrons:  For the stability of molecules, number of electrons in bonding molecular orbital should be greater than number of electrons in anti-bonding molecular orbital. Suppose the number of electrons present in the bonding orbitals is represented by Na and the number of electrons in anti-bonding orbitals is represented by Nb. (a) If Na > Nb, the molecule is stable. (b) If Na < Nb, the molecule is unstable. Molecule will not exist. (c) If Na = Nb, the molecule is again unstable. (ii) Stability of molecules in terms of bond order:  Bond order is defined as half the difference between the number of electrons present in the bonding and anti-bonding orbitals i.e. 1 B.O. = __ ​   ​  [Nb – Na] 2 (a) Relative stability in terms of bond order: Greater the bond order, greater is the stability of molecules. (b) Nature of bond in terms of bond order: Bond order 1, 2 and 3 means single, double and triple bonds respectively. (iii) Diamagnetic and paramagnetic nature of the molecules:  If all the electrons in the molecule are paired, it is diamagnetic in nature. On the other hand if the molecule has some unpaired electrons, it is paramagnetic. (iv) Stability of molecules in term of bond length:  Bond length is inversely related to bond order .Greater the bond length, smaller will be the bond order and lesser will be the stability.

The electronic structures and bonding properties of some of the homonuclear diatomic molecule. 1. Hydrogen molecule (H2):  Hydrogen molecule is formed by combination of two hydrogen atoms. Each hydrogen atom contains one electron, total number of electrons present in H2 molecule is 2. As each molecular orbital can accommodate two electrons and these should have opposite spin, therefore two electrons will occupy the lowest energy s (1s) molecular orbital. Hence the electronic configuration of H2 molecule will be s (1s)2.

Fig. 4  Molecular energy diagram for H2

1.8  Engineering Chemistry Bond order = 1/2 (2 – 0) = 1 i.e. two hydrogen atoms are connected by a single bond. No unpaired electron is present so the H2 molecule should be diamagnetic in nature. 2. Helium molecule (He2 ):  Electronic configuration of Helium is 1s2. Total number of electrons in He2 should be 4. Electronic configuration is s(1s)2 s(1s)*2. Bond order = 1/2 (2 – 2) = 0 i.e. no bond is formed between two Helium atoms or we can say that helium molecule does not exist but Helium exist in mono atomic state.

Fig. 5  Molecular energy diagram for He2



3. Lithium Molecule (Li2):  Electronic configuration of Li is 1s2 2s1. Hence molecular orbital configuration of Li2 molecule is s(1s)2 s(1s)*2 s(2s)2 4. Be2, B2 and C2 Molecules:  The molecular orbital configurations of Be2, B2, C2 along with their bond orders and magnetic properties are given in Table 1.

Table 1 Molecule Be2 B2 C2



Molecular Orbital Configuration 2

*2

KK s(2s) s(2s) KK s(2s)2 s(2s)*2 p (2px)1 p (2py)1 KK s(2s)2 s(2s)*2 p (2px)2 p (2py)2

Bond Order

Magnetic Character

0 1 2

Be2 does not exist Paramagnetic Diamagnetic

The symbol KK used in these configuration represents the closed K shells. 5. N2 Molecule:  Electronic configuration of N atom is 1s2 2s2 2p3. Total number of electrons present in N2 molecule is 14. Electronic configuration of N2 molecule is KK s(2s)2 s(2s)*2 s(2px)2 p(2py)2 s(2pz)2



1 Bond order = __ ​   ​  (Nb – Na) 2

1 = __ ​   ​  (8 – 2) = 3 2 (a) Nature of bond:  Bond order is 3 molecule means N2 contains triple bond. (b) Bond strength:  High value of bond order implies that it should have highest bond dissociation energy as compared to all other diatomic molecules. The observed value of 945.0k Jmol–1 is very much in accordance with the expectation.

Chemical Bonding  1.9 (c) Diamagnetic character:  Presence of no unpaired electrons indicates it to be diamagnetic which is actually found to be so. The molecular orbital configuration for N2+, N2–, N–2 2 ions along with their bond orders and magnetic properties are given in Table 2

Table 2 Molecule

Molecular Orbital Configuration

N2+ N2–

KK s(2s) s(2s)

N2–2

KK s(2s)2 s(2s*)2 p(2px)2 p(2py)2 s(2pz)2 p(2px*)1 p(2py*)1

2

*2

2

*2

KK s(2s) s(2s)

2

2

1

2

2

2

p(2px) p(2py) s(2pz)

1

p(2px) p(2py) s(2pz) p(2px*)

Bond Order

Magnetic Character

3

Paramagnetic

1 2 ​ __ ​  2

Paramagnetic

1 2 ​ __ ​  2

Paramagnetic

Fig. 6  Molecular energy diagram for N2

1.10  Engineering Chemistry Relative stabilities, Bond dissociation energy and Bond length of N2, N2+, N2–, N–2 2 species. As calculated above, the bond orders of these species are N2 = 3, N2+ = 2.5, N2– = 2.5, N2–2 = 2 As bond dissociation energies are directly proportional to the bond orders, therefore the dissociation energies of these molecular species are in the order: N2 > N2+ > N2– > N2–2 N2+ and N2– have same bond order but N2+ is more stable than N2– because N2– molecule has more electrons in anti-bonding molecular orbital in comparison to N2+. As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order. As bond length is inversely proportional to the bond order, therefore their bond lengths will be in the order: N2 < N2+ < N2– < N2–2 Oxygen molecule O2 :  One of the greatest achievements of molecular orbital approach was that it could explain the paramagnetic nature of oxygen molecule which could not be explained by valence bond approach. Electronic configuration of oxygen atom is 1s2 2s2 2p4. Total electrons in O2 molecule are 16. electrons will be filled in the various molecular orbital as has been done in the case of N2 [with the exception that in this case s(2pz) will be filled up first than p(2px) and p(2py)]. The orbital next in order of increasing energy are p*(2px) p*(2py) which have equal energy, where two electrons are added. Keeping in view Hund’s rule, one electron will enter the p*(2px) molecular orbital and the other electron will enter the p*(2py) molecular orbital. Thus, electronic configuration of O2 molecule will be KK s(2s)2 s*(2s)2 s(2pz)2 p(2px)2 p(2py)2 p*(2px)1 p*(2py)1. 1 Bond order = __ ​   ​  (8 – 4) = 2 2



(a) Nature of bond: Bond order is 2 which justify the presence of double bond. (b) Paramagnetic character: Presence of two unpaired electrons p*(2px)1 p*(2py)1 molecular orbitals accounts for its paramagnetic behavior. The molecular orbital configuration for O2+, O2– and O2– 2 ions along with their bond order and magnetic properties are given in Table 3

Table 3 Molecule

Molecular orbital Configuration

Bond Order

Magnetic Character

2

2

2

2

2

1

2.5

Paramagnetic

2

2

2

2

2

2

KK s(2s) s*(2s) s(2pz) p(2px) p(2py) p*(2px) p*(2py)1

1.5

Paramagnetic

KK s(2s)2 s*(2s)2 s(2pz)2 p(2px)2 p(2py)2 p*(2px)2* p*(2py)2

1.0

Paramagnetic

O2+ O2–

KK s(2s) s*(2s) s(2pz) p(2px) p(2py) p*(2px)

O2– 2

Chemical Bonding  1.11

Fig. 7  Molecular energy diagram for O2 molecule

Relative stabilities, Bond dissociation energy and Bond length of O2, O2 +, O2 –, O2 –2 As calculated above, the bond orders of these species are

O2 = 2 O2+ = 2.5, O2– = 1.5, O2–2 = 1

As bond dissociation energies are directly proportional to the bond orders, therefore the dissociation energies of these molecular species are in the order O2+ > O2 > O2– > O2–2 As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order. As bond length is inversely proportional to the bond order, therefore their bond lengths O2–2 > O2– > O2 > O2+

2.9 The electronic structures and bonding properties in Hetero Nuclear Diatomic Molecules In case of hetero nuclear molecules (such as NO, CO and HF) the electronic distribution in between the two atoms is not symmetrical, because the electron pair is found closer to one atom than other, due to energy consideration. This results in a polar bond, in which the electron pair is shared unequally between the two atoms. Now a polar covalent bond consisting of two electrons in an orbital of the form: Y = CA Y(A) + CB Y(B) with unequal coefficients. The relative proportions of atomic orbital Y(A) and Y(B) are CA2 and CB2 respectively. In case of homopolar covalent bond, CA2 = CB2 and for

1.12  Engineering Chemistry pure ionic bond either CA = 0 or CB = 0. During the formation of bonding molecular orbital, the atomic orbital, with lower energy makes the larger contribution; while during the formation of antibonding molecular orbital, the dominant component comes from atomic orbital with higher energy. Nitric Oxide (NO) Molecule: The electronic configuration of participating nitrogen and oxygen atoms are as follows:

N = 1s2 2s2 2p3  O = 1s2 2s2 2p4

15 Electron

The order of energy level filling is as follows; KK s(2s)2 s*(2s)2 s(2pz)2 p (2px)2 = p (2py)2 p (2px*)1



Bond order = 10 – 5/2 = 2.5

(i) Magnetic behavior; Because the NO molecule contained one unpaired electron in p(2px*) orbital so the NO molecule is paramagnetic. (ii) NO molecule is less stable than O2 or N2 because in NO molecule there is significant difference of about 250K j/mol in the energy of the atomic orbitals involved, so that combination of atomic orbitals to give molecular orbitals is less effective than in O2 or N2. As a consequence, bonds are weaker. Hence, the molecular energy diagram for NO is as given below:

Fig. 8  Molecular energy diagram for Nitric Oxide Molecule

Chemical Bonding  1.13 The molecular orbital configuration for NO +, NO – and NO +2 ions along with their bond orders and magnetic properties are given in table 4

Table 4 Molecule NO+ –

NO

+2

NO

Molecular orbital Configuration

Bond Order

KK s(2s)2 s*(2s)2 s(2pz)2 p(2px)2 = p(2py)2 2

2

2

2

2

2

2

2

1

1

KK s(2s) s*(2s) s(2pz) p(2py) = p(2px) p*(2px) p2*py 2

1

KK s(2s) s*(2s) s(2pz) p(2px) = p(2py)

Magnetic Character

3

Diamagnetic

2

Paramagnetic

2.5

Paramagnetic

CN Molecule:  The electronic configuration of participating Carbon and Nitrogen atoms are as follows:

C = 1s2 2s2 2p2  N = 1s2 2s2 2p3

13 Electron

The electronic configuration is written as KK s(2s)2 s*(2s)2 s(2pz)2 p(2px)2 = p(2py)1



Bond order = 1/2 (Nb – Na)

= 1/2 (7 – 2) = 2.5 Since CN has an unpaired electron so it is paramagnetic. Carbon Monoxide Molecule:  The electronic configuration of participating Carbon and Oxygen atoms are as follows:

C = 1s2 2s2 2p2  O = 1s2 2s2 2p4

14 Electron

The electronic configuration is written as KK s(2s)2 s*(2s)2 s(2pz)2 p(2py)2 p(2px)2. Bond order = 1/2 (10 – 4) = 3 and the CO molecule is diamagnetic in nature as it contains no unpaired electrons. Moreover, experimentally, it is found that the bond length in CO is 1.128A° and CO + it is 1.115A°. As bond order and bond length are inversely related so bond order of CO + found experimentally is 3.5 whereas that of CO is 3 i.e. electrons must be removed from anti-bonding orbitals and not from bonding orbitals. This is only possible when s2s* molecular orbital is higher in energy than s (2pz), p (2px) and p (2py) molecular orbitals. So when an electron is lost from s2s* molecular orbital bond order increases. This can be explained as follows: The stabilization and destabilization of bonding molecular orbital and anti-bonding molecular orbital is same in normal condition but in case of CO + the destabilization of anti-bonding molecular orbital is much greater than stabilization of bonding molecular orbital so much as that s*2s is of higher energy in comparison to s2pz, p2px and p2py molecular orbitals. So the correct electronic configuration for CO + is

KK s(2s)2 s(2pz)2 p(2px)2 = p(2py)2 s*(2s)1

1.14  Engineering Chemistry

Fig. 9  Molecular energy diagram for CO [a] and CO+ [b] Molecule



(i) Magnetic character: Because in the CO molecule, all the electrons are paired so it is diamagnetic molecule. (ii) CO + molecular ion is more stable than CO molecule because bond order of CO + is greater than CO. (iii) Magnetic character: Because in the CO + molecule unpaired electron is present in s*(2s) orbital so it is a paramagnetic species. Hydrogen Fluoride:  The hydrogen and fluorine have 10 electrons in total which are needed to be accommodated in the molecular orbital of HF molecule. H – 1s1 F – 1s2 2s2 2px2 2py2 2pz1 Spectroscopic evidence show that energies of 1s2 and 2s2 electrons of Fluorine do not participate in bonding. Out of three 2p orbital, only 2pz which is taken as inter nuclear axis will combine with s orbital of Hydrogen giving bonding molecular orbital. Electrons which do not participate in bonding will remain in their Atomic orbital. The F (2pz) atomic orbital contributes more than the H (1s) atomic orbital to the bonding s*(2spz) molecular orbital.

Fig. 10  Molecular energy diagram for Hydrogen Fluoride Molecule

Chemical Bonding  1.15

Electronic configuration of HF is 1s2 2s2 s(2spz)2 2py2 2px2. Bond order of HF molecule: 1/2 (Nb – Na) = 1/2 (2 – 0) = 1.

3.  Metallic Bonding The constituent particles in metals are metal atoms held together by the bonds called metallic bond. In fact, more than 80 elements in the periodic table are metals. Except mercury and gallium, which are liquids, all other metals are solid at ordinary temperature and pressure. It is electromagnetic interaction between delocalized electrons called conduction electrons and metallic nuclei within the metals, whereas most chemical bonds are localized between specific neighboring atoms, Metallic bond extend over the entire molecular structure and it is collective in nature, Metallic bond accounts for physical characteristics of metals like strength, malleability, ductility, lustre and conduction.

3.1 Nature of bonding in metals The bonding among metal atoms cannot be ionic, covalent or Vander Waal’s type, as explained below: Taking the example of Li, it is known that each Lithium atom has eight nearest neighbors electronic configuration of Li is 1s2 2s1 i.e. It has only one valence electron. Hence, the possibility of its forming electron pair covalent bonds with eight other atoms is ruled out. Thus covalent bonding is not possible in metals. (i) presence of only one kind of atoms in a metal without any electro negativity difference rules out the possibility of ionic bonding. (ii) The fact that the metallic bonds are quite strong rules out the possibility of these being held together by the weak Vander Waal’s force. Thus, the following models have been put forward to explain the nature of bonding in metals. Electron Sea Model or electron gas model:  To explain the various characteristics of metals such as conductivity, lustre, malleability, ductility etc., a model called electron sea model of electron gas model has been proposed. According to this model, as metals have low ionization energy, their valence electrons are held very loosely, i.e., they are almost like free electrons. All atoms contribute for a pool of electrons which are mobile leaving the valence electrons, the remainder portion of the metal atom is a positive ion called “kernel”. For example, in lithium, each atom contributes one electron to the pool leaving behind Li+ ions; in case of Mg, each atom contributes two valence electrons to the pool leaving behind Mg+2ions. These positive ions or kernels are held in the three-dimensional space in a definite pattern in the sea of mobile electrons. This model is also called electron gas model because the electrons are free to move in all directions like the molecules of gas. The simultaneous attractive force between the kernels and the mobile electrons is called the metallic bond.

Structure of metals (Electron Sea Model)

1.16  Engineering Chemistry

3.2 Explanation of characteristics of Metals by electron sea model The characteristic properties of metals and their explanation on the basis of electron sea model of the metallic bonding are given below: 1. Electrical conductivity:  Electrical conductivity of the metals is due to mobile electrons. In a metal crystal, the electrons are flowing equally in all directions. But when a potential difference is applied across a metal, there will be a directed flow of electrons towards the positive electrode. The directed flow of electrons carries the electric current from one point to another and therefore, the metals are known to be good conductors. 2. Effect of temperature on conductivity:  The conductivity of the metal decreases with rise in temperature. This is because with the rise in temperature, the vibrations of kernels increase which interferes with the flow of electrons. As a result conductivity decreases. In the metals, the electrons are the charge carriers while in the molten ionic compounds, the charge is carried by the ions. Further, the electrons are lighter than ions and hence are more mobile. It is because of this reason that the electrical conductivity of metals is much higher than the electrical conductivity of ionic liquids. 3. Thermal conductivity:  Thermal conductivity of the metals is because of mobile electrons. On heating a part of the metal, the kinetic energy of the electrons in that region increases. The energized electrons move rapidly to the cooler parts and give their excess kinetic energy to other electrons in the cooler part of the metal. Thus, the heat is conducted throughout the metal. 4. Metallic lustre:  The bright metallic lustre is due to the delocalized mobile electrons. When light falls on the surface of a metal, the most loosely bound electrons absorb photon of the radiant energy of visible light. Consequently, the electrons start vibrating at a frequency equal to that of the incident light. The vibrating electrons emit electromagnetic radiations in the form of light. Thus it appears as if light is reflected from metal surface and the surface gains a shining appearance which is known as metallic lustre. 5. Malleability and ductility:  Since metallic bonding is non directional metals can be a beaten into sheets or drawn into wires/tubes. And kernels can slip over each other when a deforming force is applied. 6. Hardness:  The hardness depends upon the strength of metallic bond which in turn depends upon: (i) Number of valence electrons:  Greater the number of valence electrons, greater is the attraction by the kernels and hence stronger is the bond. (ii) Size of the kernels:  Smaller the size of kernels, greater is the attraction on the valence electrons and hence stronger is the bond. 7. Opaqueness: Metals are opaque because when light falls on the crystal surface it is either reflected or absorbed but it is not transmitted.

3.3  Valence Band Model/Resonance Model Taking the example of Li metals in the lattice, each Li atom is surrounded by eight near neighbors and six next nearest neighbors. As Li atom has only one valence electron, it can form only one normal covalent bond with another Li atom.

Chemical Bonding  1.17 The only way in which Li atom can bond with all its neighboring Li atoms can be explained by postulating that the covalent bond resonates among all the neighboring Li atom and the actual structure is a resonance hybrid of all these structures. Taking the example of four Li atoms, the resonating structures may be represented as follows:

The actual bonding includes all atoms of the crystal and as a result of resonance the crystal is stabilized. In the structures III, IV, V, VI ionic forms have been shown. In these structures, Li carrying negative charge is bonded to the other Li atoms. This means that more than one valence orbital of each Li atom is involved in the metallic bonding. It is assumed that the vacant 2p orbital, which is not much higher than the 2s, is used. This is called the metallic bonding.

3.4  Band model (molecular orbital approach) According to molecular orbital approach, when two atomic orbitals of nearly same energy and same symmetry but belonging to two different atoms (e.g. two H-atom) combine or overlap, they form two molecular orbitals, one with lower energy than the combining atomic orbitals and is called the bonding molecular orbital and the other with energy higher than the combining atomic orbitals is called anti-bonding molecular orbital. Now, let us apply this theory to the bonding between metal atom, taking the example of lithium when two Li atoms come close together their 2s orbitals overlap (because they have same energy i.e., they are degenerate when they are for apart and also have the same symmetry). As a result, two molecular orbitals will be formed i.e., their degeneracy is split into two levels. The 2s electron of each Li atom will be in the lower energy bonding molecular orbitals if more than two atoms come close together, more than two atomic orbitals formed is also more. As the number of Li atoms coming close together increases, the number of atomic orbitals combing together increases and so the number of orbital formed increases which extend over all the Li atoms. However, as this happens the energy difference between the molecular orbitals becomes smaller and smaller and ultimately, we say that there is bands of molecular orbitals as shown in fig. 11. As each Li atom contributes one level to the band, therefore, number of levels in the bond is equal to the number of atoms each level of the band can accommodate two electrons of opposite spin. if N atoms contribute N levels to a band, the band is capable of holding 2N electrons. Since N Li atoms have a total of N valence electrons, this means that half of the levels would be vacant. However, as this happens, the energy difference between the molecular orbitals become smaller and smaller and ultimately we say that a band of molecular orbitals (Representing almost a continuum of energy) is formed. Each level of the band can accommodate 2 electrons of opposite spin of n atoms contribute n levels to a band; the band is capable of holding 2n electrons.

1.18  Engineering Chemistry

Fig. 11  2s orbitals of Li atoms combine to form molecular orbitals equal to the number of Li atoms combining together.



This model helps to explain the properties like conduction: (i) Conduction in half filled orbital:  For example lithium electronic configuration of Li is 1s2 2s1 i.e. it is half filled. Now when n number of 2s atomic orbitals combine we get n number of 2s molecular orbital. Molecular orbital formed is in the form of band which is half filled. If we provide energy electrons from filled portion can move to empty portion within the band and conduct electricity.

Conduction due to half filled orbitals

Chemical Bonding  1.19 Similarly we can explain conduction in Na metal (1s2 2s2 2p6 3s1). If in the crystal, N atoms of Na are present then 1s, 2s and 2p bands are completely filled and 3s is half filled so when n number of 3s orbitals overlap they form n number of 3s molecular orbital which appear as a bond. If energy provides electron from filled portion will move to empty portion. This explain the electrical conductivity of sodium. (ii) Conduction in Completely filled orbital:  For example Be, Electronic configuration of Be is 1s2 2s2. According to the electronic configuration it should not conduct electricity as it is completely filled but it is a good conductor. This discrepancy has been overcome by suggesting that the 2s orbitals are not the only atomic orbitals which overlap to form band. The 2p orbitals which are only slightly higher in energy also overlap to form empty band. The bands of 2s and 2p are so close to one another that they overlap one another but the two orbital bands can be differentiated. The band which holds the valence electrons is called valence band and on the other band the one which is slightly higher in energy and is formed by the overlap of valence atomic orbitals is called the conduction band. So if we provide energy electrons from 2s band can move to 2p band, thus conducting electricity.

Similarly, we can explain Mg crystal, the electronic configuration is 1s2 2s2 2p6 3s2. Here 3s and 3p orbitals overlap simultaneously forming bands which overlap each other. So if we provide energy, electrons from 3s band will move to 3p band theory conducting electricity. Explanation of Conductors, semi Conductors and Insulators on the basis of band theory: On application of molecular orbital theory on solids, solids can be categorized into three i.e., conductors, semi conductors and insulators. Insulators:  The energy gap (more than 3ev) between the valence band and the bottom of the conduction band is so large that enough energy is not available under normal conditions to promote the electrons from the valence band to the conduction band. The in-between region is called the forbidden zone. (Energy level diagram is represented as shown in Fig 12 (a). Conductors:  There is no energy gap between the valence band and conduction band. Electrons can easily move from valence band to conduction band Fig. 12 (c). Semi-Conductors:  The energy gap between the valence band and conduction band is small than in insulators (less then 3ev) Fig. 12 (b). The thermal energy available under ordinary conditions is sufficient to excite some electrons from valence band to conduction band and thus a small amount of current can flow through them.

1.20  Engineering Chemistry

Fig. 12  Representation of energy gaps of insulators, semiconductors and conductors

Semi conductors are of two types: 1. Intrinsic semi conductors:  Intrinsic semi conductors are in their pure form, for example, Silicon(Si), Germinium(Ge) having four electrons in their outer shell and their forbidden energy gap is less than 3eV, so at ordinary temperature electrons from valance band move to conduction band leaving behind a positive charge hole which move in crystal lattice from one site to other thus electron and hole are charge carrier of electricity in intrinsic conductors. 2. Extrinsic semi conductors:  Extrinsic semi conductors have low electrical conduction in pure state but conductance increase by the addition of trace of an element belonging to other group of the periodic table. Extrinsic semi conductors are of two types. n-Type Semiconductor:  The addition of a small amount of impurity to a semiconductor before it crystallizes is known as doping. When a small amount of Phosphorus (with 5 valence electrons) element of 5th group is added to silicon (with 4 valence electrons) the fifth valence electron remains unshared which move through the crystal under influence of electrical force. This give n-type semiconductor. These surplus electrons occupy delocalized level called donor impurity level which is just below the empty conduction band of Silicon crystal. These electrons can be easily excited to empty conduction band by application of electric or thermal energy so the crystal of Silicon starts conducting.

Chemical Bonding  1.21

p-Type Semiconductor:  These are obtained when an impurity having lesser electron than the parent insulator atoms e.g., when Boron/Aluminium (with 3 valence electrons) element of Third group is added to Silicon/Germanium (with 4 valence electrons). The 4th valency of Si/Ge is free due to electron deficiency which created a positive hole in valence band. There are as many positive holes as boron atoms. These holes occupy the level close to the filled valence band called acceptor impurity level. When we provide an energy these positive hole move to conduction band. The movement of holes through the crystal lattice of a p-type semiconductor is nothing but movement of a positive electric charge. Thus conducting electricity.

Solved Questions

1. Taking Z axis as internuclear axis, explain why 2px and 2py orbital do not combine with 2s orbital to form molecular orbital? Ans. 2px and 2py orbitals do not combine with 2s because 2px and 2py orbitals will form p bond and s orbital will form s bond. 2. Give the account of electrons which occupy the bonding orbitals in H2+, H2, O2+. Ans. 1, 2 and 15 respectively. 3. Explain why N2 has greater bond dissociation energy than N2+ whereas O2+ has greater bond dissociation energy than O2.

Ans. B.O. of N2 (3) > B.O. of N 2+ (2.5) but B.O. of O2+ (2.5) > B.O. of O2 (2). Greater the bond order, greater is the bond dissociation energy. 4. Can we have a diatomic molecule with its ground state molecular orbitals full with electrons? Give a reason for your answer. Ans. No. Because bond order becomes zero, e.g., in case of He2, Be2, Ne2, etc. Note that in H2, s1s M.O. is full but s*1s M.O. is empty.

1.22  Engineering Chemistry 5. Compare the relative stabilities of O2– and N2– and comment on their magnetic behavior. Ans. M.O. electronic configuration of O2– = KK s(2s)2 s*(2s)2 s(2pz)2 p(2px)2 p(2py)2 p*(2px)2 p*(2py)1 Bond order = 1/2 (8 – 5) = 3/2 = 1.5 M.O. electronic configuration of N2+ = KK s(2s)2 s(2s)*2 p(2px)2 p(2py)2 s(2pz)1 6. Ans.



7.

Ans. 8. Ans. 9.

Ans. 10. Ans. 11. Ans. 12. Ans. 13. Ans.

Bond order = 1/2 (7 – 2) = 5/2 = 2.5 As the bond order of N2+ > bond order of O2– , therefore, N2+ is more stable than O2–. Each of them have unpaired electron, hence both are paramagnetic. Two p-orbitals from one atom and two p-orbitals from another atom are combined to form molecular orbitals. How many MOs will result from this combination? Explain. Four molecular orbitals will be formed. One p-orbital of one atom will combine with one p-orbital of another atom to form two molecular orbitals (one bonding the other antibonding). Likewise the second p-orbital of the first atom will combine with the second p-orbital of the second atom to form two molecular orbitals (of course, it is presumed that they have proper orientation). What is the energy gap in band theory? Compare its size in conductors, semiconductors and insulators. The difference of energy between the lowest level of the conduction band and the uppermost level of the valence band is called energy gap. The size of energy gap in the order: insulators > semiconductors > conductors According to M.O. theory, what happens to the identity of the atoms after the overlap. The identity is lost after the overlap. Out of bonding and anti-bonding M.O.s which one has lower energy and which one has higher stability. Bonding M.O. has lower energy and higher stability. How is bond order related to the stability of a molecule? Higher the bond order, greater is the stability. How is bond length related to the stability of a molecule? Bond length is inversely related to stability. When is the molecule paramagnetic in nature? When it contains one or more unpaired electrons. Define bond order. Bond order is half of the difference of the electrons present in the bonding and anti-bonding molecular orbitals. i.e.,



bond order = 1/2 (Nb – Na)

14. Indicate which of the two, O2– or O2–2 has higher bond order? Ans. O2– has higher bond order. 15. What is metallic bonding? Ans. A bonding between electropositive elements

Chemical Bonding  1.23



16. Ans.

17.

Or A bond or a force which binds metal ions to the number of electrons within its sphere of influences. Or The attractive force which holds the atoms of two or more metals together in a metal crystal or in an alloy. What type of atomic orbitals can overlap to form molecular orbitals? Atomic orbitals with comparable energies and proper orientation or same symmetry and similar energy. Arrange O2, O2–, O2–2 and O2+ in increasing order of bond length.

Ans. O2– 2 > O2– > O2 > O2+

18. What is bonding molecular orbital and anti-bonding molecular orbital? Ans. The molecular orbital formed by the additive effect of the wave functions of orbitals is called bonding molecular orbital and the molecular orbital formed by subtraction of wave functions of the atomic orbitals are called anti-molecular orbital. 19. How is bond order related to dissociation energy? Ans. Dissociation energy is directly proportional to bond order. 20. What happens to the probability of finding the electron in the molecular orbitals after the combination of two atomic orbitals? Ans. Probability of finding electrons increases in the internuclear region of bonding molecular orbital, whereas decreases in the internuclear region in the anti-bonding molecular orbital.

Unsolved Questions

1. Distinguish between orbit and orbital? 2. What are the main features of valence bond theory? 3. Give reason: H2+ and H2– ions have the same bond order but H2+ ions are more stable than H2–. 4. Distinguish between bonding molecular orbital and anti-bonding molecular orbital. 5. Compare the relative stabilities of O2– and N2– and comment on their magnetic behavior. 6. Use molecular orbital theory to explain why Be2 molecule does not exist. 7. Compare the relative stability of the following species of N2 N+2, N–2, N2–2 and indicate their magnetic behavior. 8. Bond order of CO + is greater than CO. Explain. 9. Give relative stability of NO +, NO, NO –. 10. Explain analogous behavior of HF molecule with the help of molecular orbital theory. 11. What is meant by bond order? Calculate the bond order of He2+, O2– and O2–2 molecule. 12. Explain why O2 is paramagnetic in behavior. 13. Compute the bond orders for H2+, H2 and He2. 14. Explain the bond order of N2 molecule with the help of molecular orbital energy level diagram.

1.24  Engineering Chemistry

15. What is non-bonding molecular orbital? 16. What are different models put forward to explain metallic bonding? How they can explain the properties of metals? 17. Give energy sequence for molecular orbitals. 18. Describe LCAO for the formation of molecular orbital. 19. Write a short note on semiconductors. 20. Find out bond order of CO, NO, N2 and O2. 21. Can Hund’s rule be applied for filling the electrons into molecular orbitals?

Chapter

2

The Solid State 1.  General Introduction A solid is defined as that form of matter which possesses rigidity and hence possesses a definite shape and a definite volume. Unlike gases and liquids in which the molecules are free to move about and hence constitute fluid state, in a solid the constituent particles are not free to move but oscillate about their fixed positions.

1.1 Classification of Solids Solids are classified into two types, namely crystalline and amorphous. A solid is said to be crystalline. If the various constituent particles like atoms, ions or molecules are arranged in a definite geometric pattern within the solid. In other words, they are said to possess a long range order. All solid elements and compounds exist in this form. On the other hand, A solid is said to be amorphous. If the constituent particles are not arranged in any regular fashion, they may have only short range order. Amorphous solids are generally obtained when the melts are rapidly cooled e.g., glass, plastics, amorphous silica, etc.

1.2 Classification of Crystalline Solids Depending upon the nature of bonding the crystalline solids have been classified into four types as given in the table given below.

Table 1 Types of Solid 1. Ionic

Constituent Particle

Nature of Forces

Characteristics Properties

Example

Positive and negative ions

Strong electrostatic force of attraction

Hard, brittle, high melting point, soluble in polar solvents, insulator in solid state, good conductor in molten state

NaCl, KNO3, LiF, MgO

Table

2.2  Engineering Chemistry 2. Molecular

Molecules

Vander Waal’s force

3. Covalent

Atoms

Covalent bonds

4. Metallic

Atoms

Metallic bonds

Soft, low melting point, low density, volatile and insulators of heat and electricity very hard with very high melting points and insulators (except graphite) Range from soft to hard, high melting point, good conductor of heat and electricity

I2, Solid CO2, CH4, S, ice. Carbon (Diamond + graphite, Si, SiO2 All metals and some alloys

1.3 Crystal A homogeneous anisotropic substance having a definite geometrical shape with surfaces that are usually plane and having sharp edge is known as crystal. 1. Faces:  The crystals are bounded by surfaces which are usually planar and arrange in a definite pattern. These surfaces are called faces of the crystal. 2. Edge:  When two adjacent faces intersect we get an edge. 3. Angle:  When three or more edges intersect we get a solid angle. 4. Interfacial angle:  The angle between the normal to the two intersecting faces is called an inter facial angle.

1.4 Types of Symmetry In cubic systems we have three elements of symmetry. (a) Plane of symmetry:  Is an imaginary plane which passes through the centre of symmetry of a crystal and divides it into two equal halves. Such that one part is exactly the mirror image of the other.



(b) Centre of Symmetry:  Is an imaginary point within the crystal such that any line drawn through it intersects the surface of the crystal at equal distances in both directions. Centre of symmetry in a cubic system is one.



(c) Axis of Symmetry:  Is an imaginary straight line about which when the crystal is rotated, it will present the same appearance more than once during a complete revolution. If the same appearance of crystal is repeated through an angle of 360/n about an imaginary axis. The axis is called an n-fold axis.

The Solid State  2.3

1.5  Space Lattice A regular arrangement of the constituent particle (i.e atoms, ions or molecules) of a crystal in a three dimensional space. Unit Cell:  The smallest three dimensional portions of a complete space lattice which when repeated again and again in different directions produce the complete space lattice. It is a fundamental building block of the crystal lattice. The size and shape of a unit cell is delivered by the length of the edges of the unit cell. (a, b and c) and by the angle a, b and g. Based upon the dimensions of the unit cell there are seven types of unit cell. These are listed below. System 1. 2. 3. 4. 5. 6. 7.

Axial length

Cubic Teliagonal Orthorhombic Monoclinic Triclinic Trigonal Hexagonal

a a a a a a a

= = π π π = =

b b b b b b b

= π π π π = π

c c c c c c c

Axial angles a = b = g = 90° a = b = g = 90° a = b = g = 90° a = b = 90° π b a π b π g π 90° a = b = g π 90° a = b = 90° g = 120°

Examples Cu, NaCl, KCl, Diamond White Tin, SnO2, TiO2 KNO3, K2SO4, BaSO4, CaCO3 Na2SO4.10H2O, CaSO4.2H2O CuSO4.5H2O, K2Cr2O7 NaNO3, ICl, As, Sb, Bi Cinnabar, Graphite, Zn, Cd, ZnO

1.6 Types of Lattice 1. Primitive unit cell 2. Non Primitive unit cell 1.  Primitive unit cell:  Unit cell in which constituent particles are present only at the corners is called simple unit cell or primitive unit cell.

2.4  Engineering Chemistry 2.  Non Primitive unit cell:  Unit Cell in which constituent particles are not only present at the corner but may also be present at some other special positions in the unit cell. They are also known as centered unit cell. They may be divided as follows: 1. Face Centered (fcc): When the constituent particles are present not only at the corner but also at the centre of each face of the unit cell, the unit cell is called face centred unit cell.



2. Body Centered (bcc): When in addition to the constituent particles at the corners, there is one constituent particle present at the centre within the body of the unit cell. Such unit cell is called body centered unit cell.



3. End Centered:  When in addition to the particles at the corners, there are particles at the centers of the two end faces.

Every crystal system does not have all the four types of unit cells. Hence there are only 14 types of space lattices corresponding to seven crystal systems. The fourteen lattices corresponding to seven crystal system are known as Bravais lattice.

1.7 Number of atoms per unit cell of crystal systems Number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence, it helps to predict the formula of the compound.

The Solid State  2.5 1. Simple cubic: It has lattice point only at corners and each corner atom is common to eight unit cells. Hence, contribution of each atom present at each corner = 1/8

So number of atoms per unit cell is given by

8 × 1/8 = 1 atom per unit cell.

2. Body Centered cubic crystal (bcc):

Lattice point at each corner atom is common to eight unit cells. Hence, contribution of each atom present at each corner = 1/8. It has lattice point at the centre of the body (not shared). So number of atoms per unit cell is given by

8 × 1/8 + 1 = 1 + 1 = 2 atoms per unit cell.

3. Face centered cubic crystal (fcc):

Lattice point at each corner atom is common to eight unit cells. Hence, contribution of each atom present at each corner = 1/8. It has lattice point at each face (shared by 2 unit cell).

8 × 1/8 + 6 × 1/2 = 1 + 3 = 4 atoms per unit cell.

2.6  Engineering Chemistry

1.8 Calculation of Atomic radius of unit cell Atomic Radius:  Half the distance between the centers of two neighboring atoms is called atomic radius; while the distance between the centers of two corner atoms of the cube is called length of cubic edge.

(a) Atomic radius of a simple cubic structure:

If a is length of the cube edge. r is the atomic radius. Then

a = 2r, r = a/2

(b) Atomic radius of fcc structure:

If a is length of the cube edge and r is the atomic radius. Considering D ABC

AC 2 = AB2 + BC2 (r + 2r + r)2 = a2 + a2 16r 2 = 2a2

÷ 

____



__

​ 2 ​    ÷ 2a2 r = ​ ​ ___ ​ ​  = a ___ ​   ​  4 16

(c) Atomic radius of a bcc structure:



AC = r + 2r + r = 4r



AE2 = a2 + a2 = 2a2



AG 2 = AE2 + GE2



(4r)2 = 2a2 + a2

The Solid State  2.7



(4r)2 = 3a2 __



​ 3 ​     a ÷ r = ____ ​   ​    4

2. Coordination number Coordination number of any given sphere in a crystalline structure is the number of other spheres by which it is immediately surrounded. Greater the coordination number, the more closely packed up will be the structure. The coordination number may be determined as explained below. 1. Simple Cubic lattice:  In such a cell, there are six atoms which are the nearest neighbors for every corner atom. Thus, there are two nearest atoms, along ± Xaxis. Similarly, there are two nearest neighbors along ± Yaxis and two along ± Zaxis. Thus, in all, there are 2 + 2 + 2 = 6 nearest neighbors. Hence, coordination number is 6. CsCl crystallizes in simple cubiclattice. 2. Face Centred Cubic Lattice:  Atoms are present at corners as well as at each face in a plane. In this each atom is in direct contact with 12 nearest neighbors, 6 of which lie in one plane, 3 are above the plane and 3 atoms are below the plane. Hence, the coordination number is 12. Cu, Au, Al crystallize in fcc lattice.



3. Body Centred Cubic Lattice:  Atoms are present at corners as well as at the centre of the body. On drawing a sphere of this radius with any atom at the centre then it is easier to see that eight atoms fall on it. Hence, the coordination number is 8. NaCl, Cr, Fe, Mo, W crystallizes in bcc lattice.

2.8  Engineering Chemistry

2.1 Radius Ratio Ionic solids possess different coordination number, because of differences in the number and relative sizes of the cations and anions. Larger the size of cation (or anion), greater will be the number of anions(or cations) which can be packed around it. The geometrical requirement for the structure of an ionic solid in terms of relative sizes of the two oppositely charged ions is expressed by radius ratio, which may be defined as: Radius Ratio is the ratio of cation radius to that of the anion radius in an ionic solid. Thus, radius ratio is Radius of cation/Radius of anion = r+/r– Greater the radius ratio, larger is the size of the cation and hence greater is its coordination number. The relationship between the radius ratio and the coordination number and the structural arrangement are called radius ratio rule. The list of limiting radius is given in the Table 2 with structure and coordination number:

Table 2 S. No 1. 2. 3. 4. 5.

+

r __ ​  – ​  (Radius Ratio) r < 0.155 0.155 – 0.225 – 0.141 – 0.732 –

0.225 0.414 0.732 1.00

Structure Linear Planar triangle Tetrahedral Octahedral Cubic structure

Coordination number 2 3 4 6 8

Example B2O3 CuCl, CuBr, CuI Na, KBr, MgO CsI, CsBr, NH4Br

Calculation of density of a cubic crystal Let a = Edge of the unit cell, d = Density of the unit cell, M = Molar mass, Z = Number of atoms present in one unit cell, NA = Avogadro’s number and m = Mass of a single atom Then, volume of a unit cell = a3

Mass of the unit cell = (Number of atoms in the unit cell) × (mass of each atom)

= Z × m

But, mass of an atom present in the unit cell (m)

= M/NA

Therefore, density of the unit cell of a cubic crystal (d)



Mass of unit cell r = ​  ________________       ​ = i.e. Volume of unitcell



r = Z ×

/NA × a3

Example 1  A face centred cube element (atomic mass 60) has a edge length of 400pm. What is the density? Solution:  For a face centered cubic, Z = 4

The Solid State  2.9

r = Z × M/NA × a3

4  ×  60 gmol–1 = __________________________________ ​           ​ (400  ×  10 –10cm)3  × (6.023  ×  1023mol–1) = 6.23g/cm3 Example 2  Sodium has a bcc structure with nearest neighbor distance 365.9pm. Calculate its density (atomic number of sodium = 23). Solution: For a bcc structure, nearest neighbor distance (d) is related to the edge length (a) as __



​ 3 ​     a ÷ d = ____ ​   ​    2



2 d 2 a = ___ ​  __ ​ = ​ _____    ​  × 365.9 1.732 ​ 3 ​    ÷

Or

For bcc structure, Z For sodium, M Therefore, r

= = = =

422.5 pm 2 23 Z × M/NA × a3

2  ×  23gmol–1 = ___________________________________ ​           ​ (422.5  ×  10 –10cm)3  × (6.023  ×  1023mol–1) = 1.51g/cm3. Example 3  The density of KBr is 2.75g/cm3. The edge length of unit cell is 654pm. Predict the type of cubic lattice to which unit cell of KBr belongs. (N = 6.023 × 1023mol–1, atomic mass: K = 39, Br = 80) Solution:

r = Z × M/NA × a3



r  ×  NA × a3 Z = ​ ___________        ​ M

(2.75g/cm3)(654  ×  10 –10 cm)3 × (6.023  ×  1023mol–1) = ​ ___________________________________________             ​ (39  +  80) gmol–1 = 3.89 ª 4 Example 4  The density of a face centered cubic element (atomic mass = 60.2amu) is 6.2g/cm3. Calculate the length of the edge of a unit cell. (Avogadro’s constant NA = 6.023 × 1023mol–1) Solution: r = Z × M/NA × a3 a3 = ZM/rNA

4 × 60.2g/cm3 a3 = ​  _________________________          ​ 6.2g/cm3 × 6.023  ×  1023mol–1

= 64.52 × 1024 cm3

a = (64.52 × 1024 cm3)1/3 = 4.01 × 10 –8 cm = 401 pm

2.10  Engineering Chemistry

2.2  X-Ray Studies of Crystals Much of our present day knowledge about the structures of crystals at the level of molecules, atoms and ions has been revealed by their interaction with X-rays. It is known that when light falls on an object, which is of the same size range as the wavelength of light, it is diffracted. This is utilized to investigate the structure of solids. A crystal lattice is considered to be made up of regular layers or planes of atoms equal distance apart. X-rays are electromagnetic radiation of wavelength 10 –10 m. Crystal can act as grating to X-ray, if X-rays encounters a barrier of the size of the wavelength and spacing between layers of crystal is 100/200 pm. Thus, when a beam is allowed to fall on a crystal a large number of images of different intensities are formed. If the diffracted waves are in the same phase, they reinforce each other and a series of bright spots are produced. And if out of phase, dark spots are produced. On the photographic plate, from the overall diffraction patterns produced by a crystal, we can arrive at the detected information regarding the position of particles in the crystal.

2.3 Bragg’s Equation Bragg’s equation or law gives mathematical relations to determine inter atomic distance from x-ray diffraction patterns. They show that:



(a) The x-ray diffracted from atoms in crystal planes follows the law of reflection. (b) The two rays reflected by successive planes will be in phase if the extra distance travelled by the second ray is an integral number of wavelengths. A beam of X-rays when falls on the crystal surface, two successive atomic planes of the crystal are shown separated by distance d. Let the X-rays of wavelength l strike the first plane at on angle q. AM and AN are two normals. The waves reflected from different layer planes will be in phase with one another only if the difference in path length of the waves reflected from the successive planes is equal to an integral number of wavelength l. Path difference = n l ...(1) D MAB = D NAB = q In D MAB Sin q = BM/BA or BM = BA sin q = d sin q In D NAB Sin q = BN/BA or BN = BA sin q = d sin q Path difference = BM + BN = d sin q + d sin q = 2d sin q ...(2)

The Solid State  2.11 Combining (1) and (2)

nq = 2d sin q

This is Bragg’s equation (after William H Bragg and Sir William L Bragg) and n is called the order of reflection. If n = 1, 2…. etc. it is called first order reflection, second order reflection….etc. respectively. Usually, the order is kept unity (1). Example 5  X rays of wavelength equal to 0.134 nm give a first order diffraction from the surface of a crystal when the value of q is 10.5°. Calculate the distance between the planes in the crystal parallel to the surface examined. Solution:  Here, we are given l = 0.134, n = 1, q = 10.5° Applying Bragg’s equation 2d sin q = n l

nl 1 × 0.134 d = ​ _______     ​ = ​ __________      ​ 2d sin q 2 × sin 10.5

= 3.68 nm Example 6  At what glancing angle would the first order diffraction from the plane of KCl be observed, using X ray of wavelength of 154pm? The dimension of the unit cell is 315pm. Solution: For the first order diffraction,

n l = 2d sin q l = 154 pm and d = 315 pm 154 pm = 2 × 315 pm sin q Sin q = 0.244 q = 14.1°

Example 7  The distance between the two layers in a NaCl crystal is 282 pm. X-rays are diffracted from these layers at an angle of 23.0°. Assuming that n = 1, calculate the wavelength of the Xrays in nm. Solution: For the first order diffraction,

n l = 2d sin q



d = 282 pm and q = 23.0°



l = 2d sin q

= 2 × 282 pm sin 23.0 = 0.220 nm

3.  Imperfections in crystals At 0 K, there is perfect order in arrangement of atom in crystals. But with the rise in temperature some deviations occur which results in disorder. The presence of impurities also adds to the disorder. So any deviation from perfectly ordered arrangement of atoms in crystal is called imperfections. Imperfections not only modify the properties but also impart new properties to the solids.

2.12  Engineering Chemistry

3.1 Classification of imperfection

3.2 Electronic defects At 0 K, in a purely covalent crystal (e.g. Silicon) or purely ionic crystal (e.g. NaCl) the electrons are present in ground energy level. But above 0 K some electrons occupy higher energy states depending upon the temperature. So in the crystal of pure silicon, some of the electrons from the covalent bonds get thermally released and electrons deficient site called holes is created. Holes also provide electrical conductivity but the holes in an electric field move in directions opposite to that in which electrons move. Thus electrons and holes in solid give rise to electronic imperfections.

3.3 Atomic imperfections / point defects When deviations exist from the regular (or periodic) arrangement around an atom or a group of atoms in a crystalline substance the defects are called point defects.

3.4 Types of point defects Point defects in a crystal may be classified into the following three types: A–Stoichiometric defects B–Non-stoichiometric defects A–Stoichiometric defects:  If imperfections in the crystal are such that the ratio between the cations and anions remain the same as represented by the molecular formula the defects are called stoichiometric defects. These are further divided into two types as follows: i–Schottky defects:  If in an ionic crystal of the type A+ B–, equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained it is called schottky

The Solid State  2.13 defects. The schottky defect containing one pair of holes due to missing cation and anion given below.

3.5 Types of compounds exhibiting schottky defect This type of defect is shown by highly ionic compounds which have (i) High co-ordination number and (ii) Small difference in the size of cations and anions A few examples of ionic compounds exhibiting schottky defect are NaCl, KCl, KBr, & CsCl. Density of an ionic Crystal having schottky defects:  As the numbers of ions-decrease as a result of the defect, the mass increases where as the volume remains the same. Hence the density of the solid decreases.

3.6 Frenkel defect An ion is missing from its lattice site and it occupies the interstitial site, electrical neutrally as well as the stoichiometry of the compound are maintained. This type of defect is called frenkel defect. Since cations are smaller, it is more common to find the cations occupying cation interstitial sites.

Types of compounds exhibiting Frenkel defects This type of defect is present in these compounds which have 1. Low co-ordination number and 2. Large difference in the size of cations & anions. Frenkel defects are found in Silver halides due to small size of Ag+ ion. Consequences of Schottky and Frenkel defects (i) Solids having these defects conduct electricity to a small extent. (ii) Due to presence of holes the stability of crystal decreases. (iii) In frenkel defect similar charges come closer. Thus dielectric constant of the crystals increases. The above two defects are also called intrinsic defects or thermodynamic defects. B–Non-stoichiometric defects:  If as a result of the imperfections in the crystal, the ratio of the cation to the anion becomes different from that indicated by the ideal chemical formula, the defects are called non stoichiometric defects. Depending on whether the positive ions are in excess or negative

2.14  Engineering Chemistry ions are in excess in a given non-stoichiometric crystal, there are two types of defects found in non stoichiometric crystals. these are 1.  Metal excess defect:  this may occur on either of the following two ways: (a) By anion vacancies: A+ B– A+ B–

B– A+ + A

A+ B– A+ B–

B– A+ B– A+

A negative ion may be missing from its lattice site leaving a hole occupied by an electron, thereby maintaining electrical balance

The electrons thus trapped in the anion vacancy are called F-centres because they are responsible for imparting color to the crystals (F = farbe which is a german word for color). E.g. When NaCl is heated in an atomosphere of Na vapor, the excess of Na atoms deposit on the surface of NaCl crystal. Cl– ions then diffuse to the surface where they combine with the Na atoms which become ionized by losing electrons. These electrons diffuse back in to the crystal and occupy the vacant sites created by the Cl– ions. These electrons absorb some energy of the white light, giving yellow color to NaCl. Similarly excess of Li in LiCl makes it pink and excess of K in KCl makes it violet. By the presence of extra cations in the interstitial sites:  It is caused by an extra cation occupying the interstitial site electrical neutrality is maintained by an electron present in another interstitial site. A+ B– A+ B–

B–

A+

B–

+

A   A B– A+ +

–

B A+

+

A B–

–

B A+

e.g. When ZnO is heated, it loses oxygen and turns yellow due to following reactions: ZnO +    Zn+2 + 1/2O2 +2e – The excess of Zn+2 ions formed get trapped into the vacant interstitial sites and the electrons in the neighbouring interstitial site

Crystals with this type of metal excess defect contain some free electrons. Hence such materials act as semi-conductors. 2.  Metal Deficiency defects:  This defect is produced when one or more cations are missing from their lattice site and thus cation vacancies are produced. In order to maintain the electrical neutrality of the crystal, one are more nearby cations acquire extra positive charge and are thus converted into cation with higher oxidation state. Thus although the crystal is electrically neutral the number of metal ions is less then that of anions i.e. there is deficiency of metal ions. This defect is found in those ionic crystals whose positive ions show variable oxidation state. e.g. Ferrous Oxide, NiO, Ferrous Sulphide.

The Solid State  2.15 Metal deficient defect Consequences of metal deficient defects: Crystal with metal deficient defects will start behaving semi conductor. Impurity defects:  These defects arise when foreign atoms are present at the lattice site or at the vacant interstitial sites. In the former case we get substitutional solid solutions while in the latter case we get interstitial solid solutions. The formation of the former depends upon the electronic structures of the impurity while that of the latter on the size of the impurity. 1. Introducing impurity defect in covalent solids For example group 13 elements such as B, Ga, & Al and the group 15 elements such as P and As enter Ge or Si of group 14. The group 15 elements have one excess valance electron after forming the four covalent bonds normally formed by group 14 elements. The excess electrons give rise to electrical conduction. A group 13 elements which has only three valence electron forms an electron deficient bond or a hole. Such holes can move across the crystal giving rise to electrical conductivity. Thus impurity doped Si and Ge are semi conductors whose conductivity increases with increasing temperature. This is unlike metals whose conductivity decreases with increase of temperature. Doping of silicon with group 13 and group 15 elements to produce n-type and p-type semiconductor. Group 14 element doped with group 15 elements are called n type semiconductors. the symbol ‘n’ indicating that negative charge flows in them. Group 14 elements doped with group 13 elements are called p-type semiconductors. the symbol ‘p’ is used because in an electric field the holes move through the crystal like positive charge i.e. in a direction opposite to the flow of electrons. 2. Introducing impurity defect in ionic solids In case of ionic solids the impurities are introduced by adding impurity of ions. If the impurity ions are in a different valence state from that of the host ions, vacancies are created. For example if molten NaCl containing a little SiCl2 as impurity lattice sites Na+ ions are substituted by Sr+2 ion. For every Sr+2 ion thus introduced, two Na+ ions are removed to maintain electrical neutral. One of these lattice sites is occupied by Sr+2 ions and the other remains vacant. These vacancies result in the higher electrical conductivity of the solid.

Fig. 1  Impurity defect introduced by substituting Na+ ions by Sr+2 ions.

3.7  Graphite It is a polymorph of carbon, it has a crystal structure distinctly different from that of diamond and is also more stable than diamond at influent temperature and pressure.

2.16  Engineering Chemistry

Graphite offers a very good example of 2-dimensional solid. The graphite structure is composed of layers of hexagonally arranged carbon atoms (sp2 hybridized) within the layers, each carbon atom is bonded to three coplanar neighbouring atoms by covalent bonds (1.4A°). Due to covalent bonding graphite has high melting point i.e., 3700°C and it has a firm structure and high thermal stability. The fourth bonding electron participates in a weak Vander Waal’s type of bond (3.4A°) between the layers. As a consequence of these weak inter planar bonds, inter planar cleavage is facile, coefficient of friction becomes low as a result it is soapy in touch and gives rise to the excellent lubricative properties of graphite. Graphite has two forms: 1. a-graphite: it is naturally occurring having hexagonal arrangement. It is of three types: (a) Crystalline graphite (b) Amorphous graphite (c) Flake graphite 2. b-graphite: it is synthetic graphite having rhombohedral arrangement, for example graphene. Applications: 1. Used as heating elements for electric furnaces. 2. Used in batteries (crystalline graphite) 3. In metallurgical crucibles (amorphous graphite) 4. For high temperature refractors and insulations 5. In nuclear reactors as moderators 6. In Lead pencil 7. In air purifications devices.

3.8  Fullerenes Fullerenes are Archimedian solids. They have polymorphic form of carbon which exist in discrete molecular form and consist of hollow spherical cluster of carbon atoms. e.g. C20, C24, C32, C50, C60, C72 etc. Out of these C60 is most important. It has truncated icosahedrons or geodesic dome structure. Each molecule is composed of C atoms that are bonded to one another to form both hexagon and pentagon geometrically configuration. One such molecule is found to consist of 20 hexagons and 12 pentagons which are arranged such that no. two pentagons show a common side. The molecular surface thus, exhibits the symmetry of the soccer ball. The material composed of C60 molecules is known as buck minister fullerene.

The Solid State  2.17

Preparation Fullerenes are prepared by vaporizing a graphite rod in the atmosphere of helium. Mixture of fullerenes are formed which are separated by solvent extractions or by sublimation. C60 is isolated from the mixture by chromatographic technique using aluminate hexane. Properties 1. It is a mustard colored solid. 2. The color changes from brown to black with increasing thickness of fullerene films. 3. In pure form, it is an insulator but functions best as super conductor by doping with alkali metals. 4. Many higher fullerenes are chiral with increasing cage size, the number of chiral isomers increases rapidly. 5. It can be compressed to loss 30% of its volume without destroying its carbon cage structure. 6. It can form solid soft ferromagnetic substance if electron donor impurity is introduced . Applications 1. It can be use as super conductor when doped with alkali metals. 2. Carbon nanotubes and nanowires. One potential important application for these C nanotubes is the area of composite. Carbon fibers made from organic polymers are used to strength light weight high technical materials such as the carbon. Epoxy resins used in golf clubs, tennis racquets and yachts. 3. Catalysis:  C60 participates in catalysed processes is either a part of the catalyst and or a substrate. e.g. the complex h2-C60Pd (PPhh3)2 as the catalyst in the homogenous and heterogeneous hydrogenation reactions. 4. Polymerization Reactions: When C60 fullerenes pellets are exposed to a pressure at 1.2 GP at 600k for 5 hours, cycle addition occurs to give polymers.

3.9 Liquid Crystals These are substances that exhibit a phase of matter that has properties between those of a conventional liquid and those of a solid crystal. They may flow like liquids but have the molecules oriented in crystal like way. They exhibit intermediate thermodynamic phase between the crystalline solid and liquid state. These phases are called mesomorphic phases.

2.18  Engineering Chemistry Molecules of substances with a liquid crystal state are generally oblong and rigid, that is rod shaped. They are compounds of moderate size organic molecules which tend to be elongated and rod shape. Typical liquid crystalline molecules are cholesterol ester, surfactants, paraffin’s phenyl benzoates, surfactants, glycol liquids and cellulose derivatives.

Mesogen:  It is the fundamental unit of a liquid crystal that induces structural order in the crystals. Liquid crystal consists of a rigid part and a flexible part. The rigid part aligns molecules in one direction called mesogen, whereas the flexible part induces fluidity in the liquid crystals. Typical Behavior of Liquid crystals:  In materials that form liquid crystals the inter molecular forces are not the same in all directions, In some directions the forces are weaker than in other directions. As such when a material is heated the increased molecular motion overcomes the weaker forces first but its molecules remain bound by the stronger forces. Thus a molecular arrangement is produced which is random in some direction and regular in others. Classification of Liquid Crystals:  Depending upon the amount of order in the material, liquid crystal may be classified into following two categories. (i) Thermotropic liquid crystal (ii) Lyotropic liquid crystal. Thermotropic liquid crystal:  The transitions between the liquid crystals are given by the change in temperature. They are mostly used for technical applications. Thermotropic Liquid Crystals are Classified as: (i) Isotropic Phase:  Molecules are randomly aligned. They have low viscosity and often appear to be crystal clear. There is no long range positional or orientational order of the molecules. So, macroscopically appears to the like any other isotropic liquid such as water.



(ii) Nematic Phase:  It is simplest liquid crystal phase. It has high degree of long range orientation order but no translational order. They are polarizable rod. Like organic molecules have fluidity similar to liquids but they can be easily aligned by an external magnetic electric field. A twisted (aligned) nematic has the optical properties of a uniaxial crystal which makes it extremely useful in LCD (Liquid crystal displays).

The Solid State  2.19

(iii) Smectic Phase:  It is found at lower temperature than the nematic form well defined layers. That can slide over one another. They are positionally ordered along one direction. They have more degree of orientational order than nematics. There are several different categories of smectic but the two best known are: 1. Smectic A:  In which the molecules are aligned perpendicular to the layer planes. 2. Smectic C:  In which the alignment of the molecules is at some arbitrary angle to the normal. Smectics have high viscosity and this phase is generally not useful for devices.



(iv) Chiral Nematic/Cholesteric Liquid Crystals:  In this phase, the molecules twist slightly from one layer to the next resulting in a spiral formation. Cholesteric liquid crystal should have a chiral carbon.



(v) Columnar Liquid Crystals:  They are shaped like discs instead sof long rods. It is also called discotic liquid crystals.

2.20  Engineering Chemistry (vi) Lyotropic liquid crystal:  They are amphiphilic (Both hydrophilic and hydrophobic) molecules. They are often called surfactants having polar head group attached to one or more non-polar chains. When these are dissolved in an appropriate solvent they form aggregates called micelles beyond a particular concentration called critical micelle concentration. Lyotropic liquid crystals have biological applications (Biological membranes and cell membranes). The outermost layer of skin is primarily a lyotropic liquid crystal made of fatty acids. Applications 1. Used in large displays e.g. TVs and computer screens. 2. Used in small displays for the propose of digital displays on electrical appliances. 3. Also used in portable displays e.g. laptops, PDAs, mobile phones. 4. Used in telecoms devices e.g. beam stirrers, Optical correction. 5. Used in sensors. e.g. stress and temperature sensing. 6. Used in screens e.g. office privacy screens. 7. Used as solvent for the study of structure of anisotropic molecules spectroscopically. 8. Used in gas liquid chromatography because their electrical and mechanical properties between crystalline solids and isotropic liquids. 9. The cholesteric type of crystals is used for detecting termours in the body by the method called thermography. 10. Used as a commercial lubricant.

solved Questions

1. At what glancing angle would the first order diffraction from (110) plane of KCl be observed, using x-ray of wavelength of 1.54 × 10 –8 cm? The dimension of the unit cell is 3.15 × 10 –8 cm. Sol. For first order diffraction, l = 2d sin q \ 

l = 1.54 × 10 –8 d = 3.15 × 10 –8 1.54 × 10 –8 = 2 × 3.15 × 10 –8 154 sin q = _______ ​      ​ = 0.244 2 × 315

= sin 14.1°

2. Xenon crystallises in the face centred cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbour distance and what is the radius of xenon atoms? Sol. Here a = 620 pm d = ? r = ? For the face centred cubic. 620 a__ _____ d = ​ ___   ​ = ​     ​  = 438.5 pm 1.414 ​ 2 ​    ÷

The Solid State  2.21 d 438.5 r = __ ​   ​  = _____ ​   ​   = 219.25 pm 2 2 3. Sodium has a bcc structure with nearest neighbour distance 365.9 pm. Calculate its density (Atomic mass of sodium = 23). Sol. For the bcc, nearest neighbour distance (d) is related to the edge (a) as

__



​ 2 ​    ÷ d = ___ ​   ​ a 2

For bcc structure,

2 2 a = ___ ​  __  ​ d = _____ ​     ​  × 365.9 = 422.5 pm 1.732 ​ 3 ​    ÷ Z = 2

For sodium,

M = 23



Z×M r = ​ _______      ​ a3  ×  No

or

2 × 23 g mol–1 = ​  _________________________________           ​ (4.22 × 10 –10 cm)3 × (0.2 × 1023 mol–1) = 1.51 g/cm3.

4. Calculate the value of Avogadro’s number from the following data: density of NaCl = 2.165 g/cm–3. Distance between Na+ and Cl– in NaCl is 281 pm. Sol. Z = 4,  M = 58.5 g/mol

r = 2.165 g/cm–3

As distance between Na+ and Cl– = 281 pm. Edge of the unit cell = 2 × 281 = 562 pm. Z×M r = ​ _______      ​ a3 × Nc 4 × 58.5 g/mol–1 2.165 g cm–3 = ​  ____________________         ​ (562 × 10 –10 cm)3 × No No = 6.09 × 1023 mol–1 5. An element occurs in the Bcc structure with cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms of the element does 208 g of the element contain? Sol. For Bcc structure, Z = 2 Edge of the unit cell a = 288 pm r = 7.2 g/cm3

Z  ×  M r = ​ _______      ​ a3 × No 2×M 7.2 g cm–3 = ​ _________________________________           ​ (288 × 10 –10 cm)3 × (6.02 × 1023 mol–1)

2.22  Engineering Chemistry M = 51.8 g/mol. By mole concept, 51.8 g of the element contains = 6.02 × 1023 atoms. \ 208 g of the element contains 6.02 × 1023 = __________ ​   ​      × 208 atoms. 61.8 = 24.17 × 1023 atoms. 6. X–rays of wavelength equal to 0.134 nm give a first order diffraction from the surface of a crystal when the value of q is 10.5°. Calculate the distance between the planes in the crystal parallel to the surface examined. Sol. Here, we are given l = 0.134 nm, n = 1, q = 10.5°

Applying Bragg’s equation

2d sin q = n l

We get

n l 1 × 0.134 nm 2d sin q = ______ ​      ​ = 10.5° = ​ ____________       ​ 2 × sin 10.5° 2 sin q

= 3.68 nm.

7. Calculate the angle at which first order reflection will occur when x–rays of wavelength 1.54 Å are diffraction from the planes of the crystal separated by a distance of 4.04 Å. Sol. According to Bragg’s equation Here, \ or

2d sin q = n l d = 4.04 Å, l = 1.54 Å, n = 1 2 × 4.04 × sin q = 1 × 1.54 sin q = 0.191 or q = sin–1 (0.191) = 11°



8. What will be the wavelength of the x-rays which give a diffraction angle, 2 q, equal to 16.80° for a crystal if the inter planar distance in the crystal is 0.200 nm and only first order diffraction is observed. Sol. 2 q = 16.80°, i.e q = 8.40°

2d sin q = n l 2d sin q _______________ 2 × 0.20 × 0.146 l = ​ _______  = ​   ​      = 0.0584 nm n ​  1

= 5.84 × 10 –11 m

9. A sample of crystalline solid scatters a beam of x-rays of wavelength 70.93 pm at an angle of 2 q of 14.66°. If this is a second order diffraction (n = 2), calculate the distance between the parallel planes of the atoms from which the scattered beam appears to have been reflected.

The Solid State  2.23 Sol.

2(70.93 × 10 –22 m) n l d = ______ ​      ​ = ​ ________________     ​    2 sin 7.33 2 sin q

= 556.3 × 10 –12 m = 556.3 pm.

Unsolved Questions

1. 2. 3. 4. 5. 6.



7. 8. 9. 10.



11.



12. 13.



14.



15.



16.



17.



18.



19.



20.



21.



22.

Give examples of amorphous and crystalline solids. What is a body centred cube structure? Define co-ordination number. How many atoms are present in the unit cell of (a) simple cubic lattice (b) bcc (c) fcc? On the basis of nature of bonding, how can the solids be classified into different types. Explain with the help of diagrams the structural differences between three types of cubic crystals. Write Bragg’s equation. What do different symbols signify? Define space lattice and unit cell? Briefly explain Bragg’s x–ray method for the study of crystal structure. Gold crystallizes in a face centred cubic lattice. If the length of the edge of the unit cell is 407 pm. calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu. The effective radius of an iron atom is 1.42 Å. It has rock salt like structure. Calculate its density (Fe = 56 amu). The edge length of NaCl unit cell is 564 pm. What is the density of NaCl in g/cm3? A compound having bcc geometry has atomic mass 50. Calculate the density of the unit cell if its edge length is 290 pm. The compound CuCl has ZnS (cubic) structure. Its density is 3.4 g cm–3. What is the length of the edge of the unit cell? (At masses Cu = 63.5, Cl = 35.5) The density of a face centred cubic element (atomic mass = 60.2 amu) is 6.25 g cm–3. Calculate the length of the edge of the unit cell. Chromium crystallizes in a body centred cubic lattice whose density is 7.20 g/cm3. The length of the edge of the unit cell is 288.4 pm. Calculate Avogadro’s number. An element of atomic mass 98.5 g/mol occurs in fcc structure. If its unit cell edge length is 500 pm and its density is 5.22 g cm–3, what is the value of Avogadro’s constant. What is the distance between Na+ and Cl– in a NaCl crystal if its density is 2.165 g cm–3 NaCl crystallizes in the fcc lattice. KBr has fcc structure. The density of KBr is 2.75 g cm–3. Find the distance between K+ and Br–. (At mass of Br = 80.0) Use the data given below to find the type of cubic lattice to which the crystal of iron belongs: a/Pm = 280, P/g cm–3 = 7.86. A body centred cubic element of density 10.3 g cm–3 has a cell edge of 314 Pm. Calculate the atomic mass of the element. Write a note on Fullerenes.

2.24  Engineering Chemistry

23. Write a note on Graphite. 24. Explain briefly liquid crystals. 25. Define the following terms: (a) Axis of symmetry (b) Centre of symmetry (c) Inter facial angle (d) Bravais lattice (e) Plane of symmetry 26. Give applications of liquid crystals.

Chapter

3

Nanomaterials 1.  INTRODUCTION The word “nano” has a Greek origin meaning dwarf (small). Technically, the prefix nano means “one billionth” or 10–9. Therefore one nanometer is 10–9, one nano second is 10–9 sec and so on. Today’s wide spread activities in nano science and technology are actually rooted in the ideas of some leading scientists of the last century. Among them, the foremost name was Richard P. Feynman. He delivered a legendary talk entitled “There is plenty of room at the bottom” in the annual general body meeting of the American Physical society on Dec. 29, 1959 at California Institute of Technology. In that talk he discussed about the ideas of manipulating and controlling things at the atomic scale. Nano materials are defined as those materials which have structured components with size less than 100 nm at least in one dimension. Bulk materials always exhibit macroscopic physical properties. The same material at the nanoscale can have properties (e.g., optical, mechanical, electrical etc) which are very different from the properties of the material that has at the macro scale. Non-intentionally made nano materials, which refers to nano sized particles or materials that belong naturally to the environment (e.g., proteins, virus, nano particles produced during volcanic eruptions, etc.) or that are produced by human activity without intention (such as nano particles produced from diesel combustion, chimney soot etc.). Intentionally produced nano materials, which means nano materials produced deliberately through a defined fabrication process. Nano science is the study of the fundamental principles of molecules and structures with at least one dimension roughly between 1 and 100 nanometers or it is study of phenomena and manipulation of materials at atomic, molecular and macro molecular scales. Nanotechnology is the principle of manipulation by atom, through control of the structure of matter at the molecular level or is the design, characterization, production and application of structures, devices and system. Nanotechnology may also be defined as following– • Research and technology development at the 1 to 100 nm range. • Creating and using structures that have novel properties because of their small size. • Ability to control or manipulate at the atomic scale. Nanotechnology is a rapidly emerging technology with vast potential to manipulate and a substance at the nanometer level (1 nm = 10–9 m) and create new useful materials and devices with fascinating functions making the best use of the special physical properties of nano sized substances and improved products for numerous applications. It is also concerned with materials and systems

3.2  Engineering Chemistry whose structures and components exhibit novel and significantly improved physical, chemical and biological properties, phenomena and processes, because of their small nanoscale size. Generally, nano science aims to understand the novel properties and phenomena of all nano-based entities.

2. The Significance of the Nanoscale At this scale, atomic classical properties are not applicable but they enter the world of quantum mechanics. The reason for this changeover is very simple. The bulk properties of any material are merely the average of all the quantum forces affecting all the atoms. As we make things smaller and smaller, we eventually reach a point where the averaging no longer holds. The properties of materials are different at the nano scale for two reasons: 1. Nanomaterials have a relatively large surface area as compared to the same mass of material produced in a larger form. This makes material more chemically reactive and affects their strength as electrical properties. 2. Quantum effects begins to dominate the behavior of matter at the nanoscale affecting the optical, electrical and magnetic behavior of materials. So, nanomaterials have unusual electrical, optical and magnetic properties than their bulk counterpart. So, here we have some interesting examples of kind of changes which are expected in the properties. (i) Opaque substances can become transparent, e.g. Copper. (ii) Inert materials can become catalyst e.g. Platinum. (iii) Stable materials can turn combustible e.g. Aluminum. (iv) Solids can turn into liquid at room temperature e.g. Gold. (v) Insulators can become conductors e.g. silicon. To understand all the above facts, scientists have drawn information from many disciplines. Chemists are generally concerned with molecules, physicists work on the properties of matter which change at the nanoscale very rapidly. So, nanoscale physics plays a very important role in nanotechnology. Engineers are concerned with the understanding and utilization of nanoscale materials. Materials Scientists, electrical engineers, Chemical engineers and mechanical engineers, all deal with the unique properties of nano-structures. Substances that function as nanomaterials can be metals, ceramics, polymeric materials or composite materials. Their defining characteristic is very small feature size in the range of 1-100 nanometers. Nanomaterials are not simply another step in miniaturization, but a different arena entirely; the nanoworld lays midway between the scale of atomic and quantum phenomena, and the scale of bulk materials. Types of nano materials Typical Nanomaterials are summarized in Table 1 given below, which include

Table 1 1. Nanocrystals and clusters (quantum dots) 2. Nanowires and nanotubes

Diameter 1-10 nm

3. 2-dimensional arrays surface and thin films 4. 3-dimenstional arrays (superlattics)

Several nm2-pm2. Thickness 1-1000 nm Several nm in all three dimensions

Diameter 1-100 nm

Metals, Semiconductors, Magnetic materials, Ceramic oxides Metals, Semiconductors, oxides, sulphides, carbon layer metal chalcogenides Metals, Semiconductors, Magnetic materials Metals, Semiconductors, Magnetic materials

Nanomaterials  3.3

• zero dimension nanostructures such as metallic, semiconducting and ceramic nanoparticles;



• one dimension nanostructures such as nanowires, nanotubes and nanorods;



• two dimensions nanostructures such as thin films, and magnetic materials;



• three dimensions nanomaterials such as metals, semiconductors, and magnetic materials;



• Besides these individual nanostructures ensembles of nanostructures from high dimension arrays, assemblies and superlattics.

The structural features of nanomaterials are in between of those of atoms and the bulk materials. While most microstructured materials have similar properties to the corresponding bulk materials, the properties of materials with nanometer dimensions are significantly different from those of atoms and bulks materials. These properties are mainly due to the nanometer size of the materials which renders them: • Large fraction of surface atoms

• High surface energy



• Spatial confinement



• Reduced imperfections, which do not exist in the corresponding bulk materials.

The ‘surface’ dependent material properties of Nanomaterials are due to their small dimensions; and have extremely large surface area to volume ratio, which makes a large fraction of atoms of the materials to be the surface or interfacial atoms. Especially when the size of Nanomaterials is comparable to Debye length, the entire material is affected by bulk materials. For example, very active catalysts are obtained from metallic nanoparticles. The sensitivity and sensor selectivity of chemical sensors can be improved by using nanoparticles and Nanowires. The nanometer feature sizes of Nanomaterials also have spatial confinement effect on the materials, which bring the quantum effects. Nanoparticles can be viewed as a zero dimension quantum dot while various Nanowires and nanotubes can be viewed as quantum wires. The profound effects on the properties of Nanomaterials are due to their quantum confinement. The energy band structure and charge carrier density in the materials can be modified quite differently from their bulk count part and in turn will modify the electronic and optical properties of the materials. Lasers and light emitting diodes from both of the quantum dots and quantum wires are very promising in the future optoelectronics. High density information storage using quantum dot devices is also a fact developing area. Reduced imperfections are also an important factor in determining the properties of Nanomaterials. Nanostructures and Nanomaterials favors a self purification process in which the impurities and intrinsic material defects will move to near surface upon thermal annealing. This increased material perfection affects the properties of Nanomaterials. For example, the chemical stability for certain Nanomaterials may be enhanced, the mechanical properties of Nanomaterials will be better than the bulk materials. The superior mechanical properties of carbon nanotubes are well known. Due to their nanometer size, Nanomaterials are already known to have many novel properties. Many novel applications of the Nanomaterials rose from these novel properties have also be proposed. Nanomaterials are cornerstones of nanoscience and technology. Nanomaterials can exist in one dimension (e.g., surface films), two dimensions (e.g., strands or fibres), or three dimensions (e.g.,

3.4  Engineering Chemistry NsM particles). They can exist in single, fused, aggregated or agglomerated forms with spherical, tabular and irregular shapes. Nanostructured materials are classified as zero dimensional, one dimensional, two dimensional, three dimensional nanostructures.

Fig. 1  Classification of Nanomaterials; (a) 0D spheres and clusters, (b) 1D nanofibres, wires and rods (c) 2D films, plates and networks (d) 3D nanomaterials



(a) Zero-dimensional nanomaterials—Materials wherein all the dimensions are measured within the nanoscale (no dimensions, or 0D, are larger than 100nm) The best example of zero-dimensional nanomaterials is nanoparticles. Nanoparticles can: Be amorphous or crystalline Be single crystalline or polycrystalline Be composed of single or multi-chemical elements Exhibit various shapes and forms Exist individually or incorporated in a matrix Be metallic, ceramic, or polymeric (b) One-dimensional nanomaterials—Nanomaterials exist in one dimension structure. This lead to needle like shaped nanomaterials. 1-D materials includes nanotubes, nanorods and nanowires. 1-D nanomaterials can be: Amorphous or crystalline Single crystalline or polycrystalline Chemically pure or impure Stand alone materials or embedded in within another medium Metallic, ceramic or polymeric (c) Two-dimensional nanomaterials—2-D nanomaterials exhibit plate like shapes. Two dimensional nanomaterials include nanofilms, nanolayers and nanocoating. 2-D nanomaterials can be– Amorphous or crystalline Made up of various chemical compositions Used as a single layer or as multilayer structures Deposited on a substrate Integrated into surrounding matrix material Metallic, ceramic or polymeric

Nanomaterials  3.5

(d) Three-dimensional nanomaterials—Bulk Nanomaterials are materials that are not confined to the nanoscale in any dimension. These materials are thus characterized by having three arbitrarily dimensions above 100nm. Materials possess a nanocrystalline structure or involve the presence of features at the nanoscale. In terms of nanocrystalline structure, bulk nanomaterials can be composed of a multiple arrangement of nanosize crystals most typically in different orientations. In nanoscale, 3-D nanomaterials can contain dispersions of nanoparticles, bundles of nanowires and nanotubes as well as nanolayers.

3. TOOLS TO MAKE NANOSTRUCTURES A number of tools are used to make nanostructures. Here we shall consider the method of lithography. The word “Lithography” belongs to the concept on stone. A lithograph is an image (usually on paper) that is produced by making a pattern on the stone, inking the stone and then pushing inked stone onto the paper. We shall consider following lithography: 1. Nanoscale Lithography

2. Dip Lithography



3. E-beam Lithography

1. Nanoscale Lithography:  A Nanoscale lithography cannot use visible light. The reason is that the wavelength pen of visible light is at least 400 nanometers. Therefore, structure smaller than that are difficult to make directly using it. Indeed, the common methods use X-ray Lithography. For example, current computer chips normally use this Lithography, in this process, a master mask is made using Chemical method and X-rays are passed through that mask to produce the actual chip structure. One of the most straight forward techniques doing small scale Lithography is micro-imprint Lithography. This method is simple and works as rubber stamp. A pattern is made on the rubber surface (actually a rubber like silicon/oxygen polymer). The rubber surface is coated with molecular ink. The ink can then he stamped out on a metal, polymer, oxide or any other surface in small scale stamps just like a stamp rubber on simple paper. 2. Dip Pen Nanolithography:  [DPN] Dip pen nanolithography is a process of writing in the same way as we write ink lines with a fountain pen. In order to make such lines in nanoscale, it is necessary to have a nano pen. Fortunately, an atomic force microscope (AFM) tips are ideal nano pen. Atomic force microscope is a very high resolution type of scanning probe microscope. The scanning probe tip of this microscope is similar to the tip of a fountain pen. This microscope has high resolution of fraction of A0 i.e. more than 100 times better than optical diffraction limit. In DPN a reservoir of ink (atom as molecules is stored on the top of scanning probe tip). The tip is manipulated across, the surface, leaving lines and patterns behind vide fig. 2. AFM tip are relatively easy to manipulate. This fact makes DPN the techniques of choice for creating new and complex structures in small volumes. 3. E-Beam Lithography:  The light based industrial lithography is limited to creating features no smaller than the wavelength used. Although we can use smaller wavelength light but this creates side effects like blowing the features which we are trying to create. The reason is that smaller wavelength light has high energy an alternate way to solve the problem is to use electrons instead of light. This E-beam lithography may be used to make structure at nanoscale. E-beam lithography also has applications in current micro electronics manufacturing.

3.6  Engineering Chemistry

Fig. 2

4. PREPARATION/SYNTHESIS OF QUANTUM DOTS The following two main techniques are used for the preparation of quantum dots. (i) Bottom up Technique (ii) Top down Technique (i)  Bottom Up Technique:  This is a technique in which materials and devices are built up atom by atom, i.e. a technique to collect, consolidate and fashion individual atoms and molecules into the structure. This is carried out by a sequence of chemical reactions controlled by series of catalysts. This process is used widely in biology. For example, catalysts called enzymes assemble amino acids to construct living tissues that forms and supports the organs of the body. (ii) Top Down Technique:  This is a technique in which materials and devices are synthesized as constructed by removing existing material from larger entities. Therefore, in this technique a large scale object or pattern is gradually reduced in technique called Lithography. Lithography is an image that is produced by making a pattern on the inked stone onto the paper. The Lithography used may be a nanoscale Lithography or dip pen lithography or E-beam Lithography. The Lithography shines radiation through a tip to the surface coated with radiation sensitive resist. The resist is then removed and the surface is chemically treated to produce the nanostructure. The semiconductors like PbS, GaAs, CdS etc. can be synthesized in a nanometer level and they are called as semiconductor or quantum dots. Their properties like band gap, luminescence etc., always differs from their bulk counterpart.

5.  SYNTHESIS OF NANOPARTICLES Nanoparticles with size ranging from 1 nm to 100 nm can be synthesized by means of various techniques, physical, chemical, biological and self-assembly techniques are few common of them. There are six widely known methods used to produce Nanoparticles, they are: 1. Mechanical alloying or high energy ball milling 2. Sputtering 3. Plasma Synthesis 4. Inert gas condensation 5. Electro deposition 6. Sol Gel synthesis

Nanomaterials  3.7

Here some of which are discussed below:

1.  Ball Mill Method:  Ball mill method is a mechanical method. This is a special type of grinder. Ball mill is a cylindrical device used in grinding materials like ores, chemical, ceramic raw material, etc. The mill rotates round a horizontal axis and is partially filled with material to be grounded plus the grinding medium. Different materials are used as media. Industrial ball mills can operate continuously fed at one end and discharged at the other end. The grinding works on the principle of critical speed. The critical speed is that speed, after which the grinding medium (say balls which is made of chrome steel, stainless steel, ceramic or rubber are responsible for the grinding of particles) start rotating along the direction of a cylindrical device, after which no further grinding takes place. 2.  Gas Condensation:  Gas condensation was the first technique used to synthesize nanocrystalline metals and alloys. In this technique, metal or inorganic material is vaporized using thermal evaporation sources such as Joule heated refractory crucibles, electron beam evaporation devices, in an atmosphere of 1-50 m bar. In gas evaporation, a high residual gas pressure causes the formation of ultra fine particles (100 nm) by gas phase collision. The ultrafiine particles are formed by collision of evaporated atoms with residual gas molecules. Gas pressures greater than 3 mPa (10 torr) are required. Vaporization sources may be resistive heating, high energy electron beams, low energy electron beam and inducting heating. Clusters form in the vicinity of the source by homogenous nucleation in the gas phase grew by incorporation by atoms in the gas phase. It comprises of a ultra high vacuum (UHV) system fitted evaporation source, a cluster collection device of liquid nitrogen filled cold finger scrapper assembly and compaction device. During heating, atoms condense in the supersaturation zone close to Joule heating device. The nanoparticles are removed by scrapper in the form of a metallic plate. Evaporation is to be done from W, Ta or Mo refractory metal crucibles. If the metals react with crucibles, electron beam evaporation technique is to be used. The method is extremely slow. The method suffers from limitations such as a source-precursor incompatibility, temperature ranges and dissimilar evaporation rates in an alloy. Alternative sources have been developed over the years. For instance, Fe is evaporated into an inert gas atmosphere (He). Through collision with the atoms the evaporated Fe atoms loose kinetic energy and condense in the form of small crystallite crystals, which accumulate as a loose powder. Sputtering or laser evaporation may be used instead of thermal evaporation. Sputtering is a non-thermal process in which surface atoms are physically ejected from the surface by momentum transfer from an energetic bombarding species of atomic/molecular size. Typical sputtering uses a glow discharge or ion beam. Interaction events which occur at and near the target surface during the sputtering process in magnetron sputtering has advantage over diode and triode sputtering. In magnetron sputtering, most of the plasma is confined to the near target region. Other alternate energy sources which have been successfully used to produce clusters or ultra fine particles are sputtering electron beam heating and plasma methods. Sputtering has been used in low pressure environment to produce a variety of clusters including Ag, Fe and Si. Sputter means to split out or throw out and sputtering is a process in which surface atoms are physically ejected from the surface by momentum transfer from an energetic bombarding beam called potential sputtering. A high intensity laser beam is allowed to incident on target metal disc. This causes evaporation of atoms from the surface of metal. The atoms are then swept by a burst of helium in vacuum chamber through an orifice. The expansion of gas in vacuum chamber produces a cooling because it has to pass through orifice. Now the cluster of metal atoms as Nanoparticles are formed in vacuum chamber.

3.8  Engineering Chemistry

Fig. 3

3.  Radio Frequency (RF) Plasma (Ionized gas) Method:  Thermal plasma can also deliver the energy necessary to cause evaporation of small micrometer size particles. Figure 3 shows a method of Nano particles synthesis utilizing plasma generated by radio frequency heating coils. The metal is contained in a pestle in an evacuated chamber. The RF coils are wrapped around the evacuated system in the vicinity of the pestle. The evacuated chamber is provided with an opening to enter helium gas. The evacuated chamber is also provided with a cluster collection device of liquid nitrogen filled cold finger scrapper assembly. The metal is heated above the evaporation points by R.F. coils. Now helium gas is allowed to pass into evacuated chamber which forms high temperature plasma in the regions of coils. The metal vapor nucleates on the helium gas atoms. It is important to mention here those ultra-fine particles are formed by collision of evaporated atoms with residual gas molecules. They now diffuse to colder collector rod where Nano particles are formed. 4.  Chemical Method Chemical vapor Deposition (CVD) and Chemical vapor condensation (CVC) CVD is a well known process in which a solid is deposited on a heated surface by a chemical reaction from the vapor or gas phase. CVC reaction requires activation energy to proceed. This energy can be provided by several methods. In thermal CVD the reaction is activated by high temperature above 900°C. A typical apparatus comprises of gas supply system, deposition chamber and an exhaust system. In plasma CVD, the reaction is activated by plasma at temperatures between 300-700°C. In laser CVD, pyrolysis occurs when laser thermal energy heats an absorbing substrate. In photo laser CVD, the chemical reaction is induced by ultra violet radiation which has sufficient photon energy to break the chemical bond in reactant molecules. In this process, the reaction is photon activated and decomposition occurs at room temperature. Nano composite powder has been prepared by CVD. SiC/Si3N composite powder was prepared using SiH3, CH4, WF6 and H2 as a source of gas at 1400°C. Another process called chemical vapor condensation (CVC) was developed in Germany in 1994. It involves pyrolysis of vapors of metal organic precursors in a reduced pressure atmosphere. Particles of ZrO2, y2O3 and nanowhiskers have been produced by CVC method. A metallorganic precursor is introduced in the hot zone of the reactor using mass flow controller. For instance, hexamethyldisilane (CH3)3 SiNHSi(CH3)3 was used to produce SiCxNyOz powder by CVC technique. The reactor

Nanomaterials  3.9 allows synthesis of mixtures of nanoparticles of two phases or doped nanoparticles by supplying two precursors at front end of the reactor and coated nanoparticles, n–ZrO2, coated with n-Al2O3 by supplying a second precursor in a second stage of reactor. The process yield quantities in excess of 20 g/hr. The yield can be further improved by enlarging the diameter of hot wall reactor and mass of fluid through the reactor. Typical nanocrystalline materials which have been synthesized are shown in Table 2 below

Table 2  Typical nanocrystalline materials synthesized by the CVC method Precursors

Product powder

Phase as prepared

Average particle size (nm)

Surface area (m2/g)

(CH3)3SiNHSi(CH3)3

SiCxNyOz

amorphous

4

377

9

201

3.5

449

Si(CH3)4

SiC

b-phase

Al[2-OC4H9]3

Al2O3

Amorphous

Ti[I-OC3H7]4

TiO2

Anatase

8

193

Si[OC2H5]4

SiO2

Amorphous

6

432

Zr[3-OC4H9]4

ZrO2

Monoclinic

7

134

The nanoparticles of Ag can be prepared by decomposing (CH3)2 – C2H5NaAlH3 in toluene and heating the solution to 105°C for two hours. The Titanium isopropoxide is also added to act as catalyst. The size of the particles produced depends on the choice of catalyst. For example, in presence of Titanium as Catalyst 80nm particles are produced. 5.  Pulsed Laser Ablation:  The principle of laser ablation has been described in several papers. Laser ablation means the removal of material from a surface by means of laser irradiation. The term “laser ablation” is used to emphasize the nonequilibrium vapor/plasma conditions created at the surface by intense laser pulse, to distinguish from “laser evaporation,” which is heating and evaporation of material in condition of thermodynamic equilibrium. A typical schematic diagram of laser ablation is shown in figure 4. Briefly, there are two essential parts in the laser ablation device, a pulsed laser (CO2 laser, Nd-YAG laser, ArF excimer laser, or XeCl excimer laser) and an ablation chamber. The high power of the laser beam induces large light absorption on the surface of target, which makes temperature of the absorbing material increase rapidly. As a result, the material on the surface of target vaporizes into laser plume. In some cases, the vaporized materials condensate into cluster and particle without any chemical reaction. In some other cases, the vaporized material reacts with introduced reactants to form new materials. The condensed particle will be either deposited on a substrate or collected through a filter system consisting of a glass fiber mesh. Then, the collected nanoparticle can be coated on a substrate through drop-coating or screen-printing process. Williams and Coles prepared nanocrystalline SnO2 by a laser ablation technique for detection of CO, H2, and CH4. Their studies revealed that the gaseous atmosphere in which the condensation of the laser-ablated SnO2 occurs has a significant influence on the size of the nanoparticles generated. The use of Ar at the pressure of 1 mbar to replace the standard conditions employed in air at 1 bar led to a decrease in SnO2 grain size to 8 nm. Furthermore, by shortening the laser pulse from the customary 20 ms to 30 ns employing a XeCl excimer laser, a further reduction in the grain size was achieved. Their gas sensors based on nanocrystalline SnO2 powders prepared by laser ablation and gas-phase condensation route offered enhanced sensitivity to CO, H2, and CH4 compared with the materials prepared by conventional methods. Hu and his co-workers prepared nanocrystalline SnO2

3.10  Engineering Chemistry

Fig. 4

thin film using a SnO2 target and a metallic Sn target respectively for C2H5OH detection. Their results demonstrated that the oxidation of Sn into SnO2 depends strongly on the substrate temperature. Oxidation of Sn into SnO2 proceeds mainly on the substrate surface instead of in the ablation plume during the condensation of Sn species onto the substrate, even if the ambient oxygen pressure reaches 100–150 Pa. Recently, Starke and Coles reported their gas sensors prepared using laser ablated nanocrystalline metal oxides. They found that SnO2 and In2O3 are capable of detecting ozone at concentrations well below 100 ppb with response times of less than one minute. Pt doped SnO2 and, particularly, In2O3 show some cross sensitivity to NO and NO2. WO3 shows sensing properties superior to these two materials in terms of selectivity and response time but regrettably does not exhibit such high sensitivity. Their CO sensor is highly sensitive to single-figure ppm concentration with a resolution down to 1 ppm. These studies demonstrate that the laser ablated nanostructured metal oxides can greatly enhance the sensing performance of gas sensors. This method is used in the synthesis of Ag Nano particles. Ag NO3 Solution and a reducing agent are arranged in a vessel (Fig. 5). A solid disc attached with a rotating device is placed inside the solution with the help of motor. This disc is subjected to pulses from laser beam. The pulse of laser beam produces hot spots on the surface of the disc. The reaction of AgNO3 and reducing agent at these hot spots result in the formation of silver particles. These particles are separated from solution by centrifuge. The size of the particles depends on the energy of incident laser pulses as well as on the rotation speed of disc. 6. Sol Gel Method:  In addition to techniques mentioned above, the sol-gel processing techniques have also been extensively used. Colloidal particles are much larger than normal molecules or nanoparticles. However, upon mixing with a liquid colloids appear bulky whereas the nanosized molecules always look clear. It involves the evolution of networks through the formation of colloidal suspension (sol) and gelatin to form a network in continuous liquid phase (gel). The precursor for

Nanomaterials  3.11

Fig. 5

synthesizing these colloids consists of ions of metal alkoxides and aloxysilanes. The most widely used are tetramethoxysilane (TMOS), and tetraethoxysilanes (TEOS) which form silica gels. Alkoxides are immiscible in water. They are organo metallic precursors for silica, aluminum, titanium, zirconium and many others. Mutual solvent alcohol is used. The sol gel process involves initially a homogeneous solution of one or more selected alkoxides. These are organic precursors for silica, alumina, titania, zirconia, among others. Mortia et al [11-14] A catalyst is used to start reaction and control pH. Solgel formation occurs in four stages. Hydrolysis Condensation Growth of particles Agglomeration of particles Hydrolysis:  During hydrolysis, addition of water results in the replacement of [OR] group with [OH-]group. Hydrolysis occurs by attack of oxygen on silicon atoms in silica gel. Hydrolysis can be accelerated by adding a catalyst such as HCl and NH3. Hydrolysis continues until all alkoxy groups are replaced by hydroxyl groups. Subsequent condensation involving silanol group (Si-OH) produced siloxane bonds (Si-O-Si) and alcohol and water. Hydrolysis occurs by attack of oxygen contained in the water on the silicon atom. Condensation:  Polymerization to form siloxane bond occurs by either a water producing or alcohol producing condensation reaction. The end result of condensation products is the formation of monomer, dimer, cyclic tetramer, and high order rings. The rate of hydrolysis is affected by pH, reagent concentration and H2O/Si molar ratio (in case of silica gels). Also ageing and drying are important. By control of these factors, it is possible to vary the structure and properties of sol-gel derived inorganic networks. Growth and Agglomeration:  As the number of siloxane bonds increase, the molecules aggregate in the solution, where they form a network, a gel is formed upon drying. The water and alcohol are driven off and the network shrinks. At values of pH of greater then 7, and H2O/Si value ranging from 7 to 5. Spherical nanoparticles are formed. Polymerization to form siloxane bonds by either an alcohol producing or water producing.

3.12  Engineering Chemistry 2 HOSi (OR)3 Æ (OR)3 Si O Si (OR)3 + H2O or 2 HOSi (OR)3 Æ (OR)2 OH Si O Si (OR)3 + H2O Above pH of 7, Silica is more soluble and silica particles grow in size. Growth stops when the difference in solubility between the smallest and largest particles becomes indistinguishable. Larger particles are formed at higher temperatures. Zirconium and Yttrium gels can be similarly produced. Despite improvements in both chemical and physical methods of synthesis, there remain some problems and limitations. Laser vaporization technique has offered several advantages over other heating techniques. A high energy pulsed laser with an intensity flux of 106-107 W/cm2 is forced on target material. The plasma causes high vaporization and high temperature (10,000°C). Typical yields are 1014-1015 atoms from the surface area of 0.01 cm2 in a 10-8 s pulse. Thus a high density of vapor is produced in a very short time (10-8 s), which is useful for direct deposition of particles. 7.  Electrodeposition Nanostructured materials can also be produced by electrodeposition. These films are mechanically strong, uniform and strong. Substantial progress has been made in nanostructured coatings applied either by DVD or CVD. Many other non-conventional processes such as hypersonic plasma particle deposition (HPPD) have been used to synthesize and deposit nanoparticles. The significant potential of nanomaterial synthesis and their applications is virtually unexplored. They offer numerous challenges to overcome. Understanding more of synthesis would help in designing better materials. It has been shown that certain properties of nanostructured deposits such as hardness, wear resistance and electrical resistivity are strongly affected by grain size. A combination of increased hardness and wear resistance results in a superior coating performance. 8.  Arc Discharge Process Carbon nanotubes and fullerenes are formed during carbon soot formation in arc discharge process. The high temperature caused by discharging caused the carbon contained in the negative electrode to sublime and CNTs are deposited on the opposing electrode. CNTs produced by this method were initially multi walled tubes. Nanomaterials categories Nanomaterials are also classified into ten main categories in order to give a good overview of the different nanomaterials. The main material categories are defined as: 1. Nanotubes-carbon based nanomaterials 2. Nanocomposites 3. Metals and alloys 4. Biological nanomaterials 5. Nanopolymers 6. Nanoglasses 7. Nanoceramics 8. Nanocatalyst 9. Nanofilms 10. Nanomachine

Nanomaterials  3.13 In addition on the basis of phase composition, nanomaterials in different phases can be classified as, • Single phase solids include crystalline, amorphous particles and layers, etc. • Multi phase systems include colloids, aero gels, ferro fluids

6. Examples of nanomaterials



1. Creation and Use of Buckyball:  Until 1985, there were only two known forms of pure Carbon graphite and diamond. Both these substances consist entirely of carbon atoms. However, they differ greatly in their structures and physical properties. First of all we shall discuss the structure of diamond and graphite. 2. Diamond:  In diamond all carbon atoms are stacked neatly in a three dimensional array or lattice. Each carbon atom is bound to four other carbon atoms in a pattern of tetrahedrons. This structure is extremely hard (Fig. 6). shows the structure of diamond.



Fig. 6

Fig. 7

6.1 Graphite In Graphite, the carbons atoms form sheet of linked hexagons. Each carbon atom within a sheet forms strong bonds to three other carbon atoms. However, the sheets are held together by weak bonds due to vanderwaal’s forces. This means that sheets can slide over each other. Therefore, graphite is soft and greasy. Figure 7 show the structure of carbon atoms connected by covalent bonds in a sheet of graphite. In 1985, a third form of carbon was discovered. It is a hallow cluster of 60 carbon atoms shaped like a football. Just like the case of graphite, in which each carbon atom is bonded to three adjacent carbon atoms and arranged in a sphere about a nano meter in diameter. It was named, Buckminister fullerene or in short buckyballs after the famous American R. Buckminister Fuller who had already designed domes in the structures. The shape of a buckyball is shown in fig. 8. Buckyballs is the roundest and most symmetrical large molecule known in the world. The buckyballs has carbon atoms at 60 chemically equivalent vertices that are connected by 32 faces, 12 of which are pentagonal and 20 hexagonal. Today a whole family of related molecules has been discovered and comes under fullerenes.

3.14  Engineering Chemistry Larger buckyballs such as C70, C76, C70 and C84 have also been found. In addition to it, smaller Bucky-balls C20, C22, and C36 have also been identified.

Fig. 8

6.2  Creating of Buckyballs Buckyballs are created by vaporizing carbon placed between two graphite rods which are placed in low pressure Helium atmosphere in a reaction chamber. An arc is produced; tremendous energy is created which is transferred to carbon. The carbon is evaporated as dust. It is mixed with a solvent and a mixture of fullerenes is formed which can be separated by distillation, fractional distillation or chromatography.

6.2.1  Properties

1. Buckyballs are stable and withstand very high temperature and pressure. 2. The carbon atoms of buckyballs can react with other atoms and molecules leaving the stable, spherical structure still intact. 3. Buckyballs do not bond to each other they however stick together via Vanderwaal forces. 4. New molecules can be created by adding other molecules to the outside of a buckyballs and by trapping smaller molecule inside a buckyballs.

6.2.2  Uses Most of the commercial applications of buckyballs are still under developing stages. However various places where buckyballs can be used are: 1. When a buckyballs is doped by inserting the right amount of potassium or cesium into empty spaces within the crystal, it becomes a super conductor. 2. Almost every carbon atom in C60 can absorb hydrogen atom which suggest that buckyballs can be used as a better storage medium for hydrogen fuel than metal hydrides. 3. Buckyballs can deliver medicine directly to the infected regions of the body, they are involved in delivering elements for medical imaging, and they have ability to act as antioxidants on counter acting free radicals in the human body. Researchers are modifying buckyballs to fit the section of the HIV molecule that binds to proteins, possibly inhibiting the spread of virus.

Nanomaterials  3.15

4. Making bullet proof vests with inorganic (tungsten disulphide) buckyballs. 5. Buckyballs combing with nanotubes and polymers are used to produce expensive solar cells that can be formed by simply painting the surface. 6. Buckyballs based light detector is developed. 7. Buckyballs are being used for production of diamonds and carbides as cutting tools or hardening agents. 8. Buckyballs can be used to reduce the growth of bacteria in pipes and membranes in water system. 9. The antioxidant properties of buckyballs may be able to fight the deterioration of motors function due to multiple sclerosis.

7.  CARBON NANOTUBES In 1990, Richard Smalley gave the concept if buckyballs get big enough then they can become carbon cylinders. Sumio Lizima discovered these cylinders in 1991 and named them nano tubes. Carbon nano tubes are sheet of graphite rolled into a tube with bonds at the end of the sheet that close. The tube dimensions are variable and can be as small as 0.4nm in diameter & several millimeters in length. Carbon Nanotubes (CNTs) have a length to diameter ratio greater than 10,00,000. Figure 9 gives structure of carbon nano tube.

Fig. 9

7.1 Types of Nanotubes There are two main types of nano tubes: •  Single walled nano tubes (SWNTs) and Multi walled nano tubes (MWNTs).

3.16  Engineering Chemistry

7.2  Structure

Fig. 10

Figure 10 gives the structure of SWCNT AND MWCNT. The diameter of most single walled nanotubes is close to 1 nanometer with a tube length of many million times longer. A one atom thick layer of graphite is wrapped into a seamless cylinder to give the structure of SWNT. Multiwall carbon nanotubes consist of multiple concentric nanotube cylinders i.e. they consist of multiple layers of graphite rolled in on themselves to form a tube shape. A carbon nanotube is a cylinder of carbon atoms covalently bonded together some of these cylinders are closed at the ends and some are open. Each carbon atom is bonded to three other carbon atoms and forms a lattice in the shape of hexagons except near the end. For nanotubes with closed ends, where the ends start to curve to form a cap, the lattice forms pentagons. Depending on the direction of roll up of graphite hexagons, carbon nanotube can be classified as either Zigzag, armchair or Chiral as given in Fig. 11.

Fig. 11

Nanomaterials  3.17 The structure of nanotube can be specified by Chiral vector (n, m), which define how the graphite sheet is rolled up n and m are integers if the vector equation–

R = na1 + ma2

The values of n and m determine the Chirality or twist of the nanotube. (i) If m = 0, then the CNT is called Zigzag. (ii) If n = m then the nanotube is called armchair. (iii) For any other combination of n and m, nanotubes are called Chiral.

7.3  Synthesis of Nanotubes Some methods of preparation of carbon nanotubes are as given below. 1. Carbon arc method  Carbon nanotubes and fullerenes are formed during carbon soot formation in arc discharge process. The high temperature caused by the discharge causes the carbon contained in the negative electrode to sublime and the CNTs are deposited on the opposing electrode CNTs produced by this method were initially multiwall tubes (MWNTs). However, vaporized carbon in the presence of cobalt can grow SWNTs. Mixtures of components are produced in this method and requires further purification to separate the CNTs from the soot and the residue catalytic metals. Producing CNTs in high yield depends on the uniformity of the Plasma, arc and the temperature of the deposit forming on the carbon electrode. 2. Dual pulsed laser Method  Dual pulsed laser method produces pure SWCTs with high yield. With direct vaporization of a Co/Ni doped graphite rod with a high powdered laser in a tube furnace operating at 1200° SWCTs can grow easily the material produced by this method appears as mat of ropes 10-20nm in diameter and up to 100 um or more in length. Each rope consists of a bundle of SWNTs, aligned along a common axis. By varying the process parameters such as catalyst composition and the growth temperature the average nano-tube diameter and size distribution can be varied.   The arc discharge and laser vaporization techniques have drawbacks for the synthesis of small quantities of high quality SWNTs, The first is that they involve evaporation of carbon source, making scale up on an industrial level difficult and energetically expensive. The second issue relates to the fact the vaporization methods grow. SWNTs. in highly tangled forms, mixed with unwanted forms of carbon. The SWNTs, thus produced are difficult to purify manipulate and assembly for building nanotube device architectures for practical applications. 3. Chemical Catalysis Method  In this method catalytic vapor decomposition of hydrocarbon is done to produce carbon fibers. During CVD a substrate is prepared with a layer of metal catalyst particles. Pyrolysis of hydrocarbons in the presence of metal catalyst can generate MWNTs and SWNTs. MWNTs are produced at lower temperature (300-800°C) in an inert gas atmosphere and SWNTs are generated at higher temperature (600-1150°C). With a mixture of hydrogen and an inert gas like Ar being present in the chamber.

3.18  Engineering Chemistry

4. Vapor Liquid Solid Growth Method  Large scale production of nanotubes can be done by vapor, liquid solid growth. The carbon containing gas, is broken apart at the surface of the catalyst particle in VLS growth method, and the carbon is transported to the edges where it forms nanotubes at the sites of metal catalyst. The length of the tube grown in surface supported catalyst VLS system appears to be dependent on the orientation of the growing tube with the surface. By properly adjusting the surface concentration and aggregation of the catalyst particles it is possible to synthesize vertically aligned carbon nanotubes i.e. as a carpet perpendicular to the substrate.

7.4 Properties of Nanotubes





1. Carbon nano tubes are super strong. The tensile strength of carbon nanotubes is 100 times greater than steel. This is due to the fact that firstly, each carbon nanotube is one large molecule and secondly the strength provided by the interlocking carbon to carbon covalent bonds is very large. 2. Nanotubes are very elastic. The young modules of a nanotube are 5 times higher than for steel. This is due to the fact that nanotubes have a perfect structure and bond strength between carbon atoms is very strong. 3. In addition to being strong and elastic, carbon nanotubes are light weight with a density about one quarter that of steel. 4. The thermal conductivity of nanotube is very high. Conduction in nanotubes takes place due to vibration of covalent bonds holding the carbon atoms together. 5. A carbon nanotube happens to be a non-polar molecule. 6. The carbon nanotubes show negative magneto resistance (Phenomenon of change in resistance of a material due to valuation in D.C. magnetic field). 7. Carbon nanotubes are metallic as semiconducting depending upon the diameter and how they are rolled. The synthesis of carbon nanotubes generally results in a mixture of tubes, two thirds are semiconducting and one third is metallic.

7.5 Applications of Nanotubes

1. Due to its unusual current conduction mechanism, wires made from nanotubes can conduct huge amount of current with less power wastage. 2. Nanotube based transistors can operate at room temperature and are capable of digital switching using a single election. 3. Due to its great mechanical properties, nanotubes can be used to produce from every day, items like clothes, sports gear to combat bullet proof jackets & space suits. 4. Using nanoscales, nano scale electric motors are used. 5. Chemical vapors are also being detected using nanotubes. 6. Research is being done to store hydrogen in nanotubes if successful this would act us a fuel tank for hydrogen fuel cell powered cars. 7. In medical applications, the carbon nanotube can be used as a vessel for transporting drugs into the body. It is especially being used for treatment of cancer in destroying cancer cells.

Nanomaterials  3.19

8. APPLICATIONS OF NANO TECHNOLOGY 1.  In Device Technology:  Nano materials with less than 100nm size are used in microprocessors in electronic industry-smaller size allow faster processing time and also more processing power to be packed into a given area. (a) Biochips—Biochip is a miniaturized device that can be used for multiplexing, enabling analysis of different DNA/proteins simultaneously. (b) BioNEMs—Nanoelectro mechanical systems are nanoscopic devices less than 100nm in length. The NEMs fabricated with new nanomaterials acts as biofunctionalized nanoelectro mechanical systems. 2.  In Health and Medicine:  Nanotechnology has its applications in field of health and medicine called nano medicine. The approaches to nano medicine range from the medical use of nano materials to nano electronic biosensors and even possible future application of molecular nanotechnology. Nano medicine has the potential to enable early detection and prevention and to essentially improve diagnosis, treatment and follow up of diseases. A list of some applications of nanomaterials to biology and medicine is listed below: (a) Fluorescent biological labels (b) Drug and gene delivery (c) Biodetection of pathogens (d) Detection of protein (e) Tissue engineering (f) Tumour destruction (g) Separation and purification of biological molecules and cells. (h) MRI contrast enhancement 3.  In Transportation:  Nanomaterials will make car and planes to become safer & cheaper. light Nanomaterials replace heavy weight structural materials. • Reduce pollution • Cerium Oxide nanoparticles are used in diesel fuel to increase fuel efficiency. 4.  In energy and Environment:  Nanotechnology will provide sufficient energy for growing world to protect environment in which we live. • Carbon nanotubes fuel cell are being used to store hydrogen. • By nano porous filter combustion engine pollutants can be reduced. • Reduce energy consumption. 5.  In Space Exploration:  Rocket scientists are actively researching new forms of space propulsion systems because today’s rocket engine rely on chemical propulsion. • Space structure can be made much lighter and more viable. • Performance can increase using solar powered ion engines with nanotechnology. 6.  In Optics:  Electric light and fluorescent lights are in common use. Nano science has entered in the field of light emission by the use of light emitting diode (LED). 7.  Therapeutics:  Nanotechnology aids in delivery of just the right amount of medicine to the exact spot of the body that need it. Drug and gene delivery system includes organic, inorganic, polymeric

3.20  Engineering Chemistry and lipid based nanobiomaterials. These nanobiomaterials could further be engineered to be stimuliresponsive. 8.  Water:  Nanotechnology will provide efficient water purification techniques. Water from the oceans can also be converted into drinking water. 9.  Sensors:  based on nanotechnology are more sensitive and more effective. 10.  Computers:  can be made more spowerful and smaller using nanotechnology. 11.  Dendrimers are nanomaterials consisting particles between 1-100 nm, they have applications in engineering, industry, pharmaceuticals etc,. For example Maganese dendritic nanoparticles developed as MRI contrast agent increase hydrophobicity and relaxivities. 12.  Bioimaging/Molecular imaging: The use of nanoparticles has boosted the development of diagnostic agents for bioimaging. It of increased attention has been devoted to the development of nanoparticles as multimodel agents for diagnosis, imaging and theory. For example Quantum dots are used in optical imaging, for example SPIO nanoparticles are used for biomaging applications.

solved Questions 1. What are nanomaterials? Ans. Nanomaterials are defined as those materials which have structured components with size less than 100 nm at least in one dimension. 2. Give two techniques to manufacture quantum dots. Ans. The two techniques to manufacture quantum dots are (i) Top-bottom approach (ii) Bottom-top approach 3. Why properties of materials are different at nano scale? Ans. The properties of materials are different at the nano scale for two reasons: 1. Nano materials have a relatively large surface area. 2. Quantum effect begins to dominate. 4. Give significance of nano scale with the help of some examples. Ans. When a material enters nano range properties change significantly for example (i) opaque substances become transparent eg. Copper. (ii) Inert materials can become catalyst eg. Platinum. (iii) Stable material can turn combustible eg. Aluminium. (iv) Insulators can become conductors eg. Silicon. 5. What are carbon nanotubes? Ans. Carbon nanotubes are a sheet of graphite filled into a tube with bonds at the end of the sheet that close.

UNsolved Questions

1. What is nano science? 2. What is nano technology? 3. Give preparation of carbon nanotubes.

Nanomaterials  3.21

4. 5. 6. 7. 8. 9. 10.

Explain three methods for the synthesis of nanomaterials. Explain types and properties of carbon nanotubes. Explain two methods for the synthesis of nanotubes. What are Buckyballs? Give applications of Buckyballs. What is lithography? Explain applications of nanomaterials in day to day life.

Chapter

4

Electrochemistry and Ionic Equilibrium Electrochemistry is the study of the relationship between chemical and electrical phenomena and the laws of their interaction. Such interactions are observed at the boundaries between electrolytes and electrodes dipping in them. Electrochemistry is of great significance, since the laws of electrochemistry forms the basis of technical processes like electrolysis and electro synthesis. The branch holds importance due to number of reasons: (i) It is of huge economical importance because of the costly destruction caused by corrosion and the possibilities of fuel cells which generate electricity directly from fuel. (ii) It provides techniques for measuring the thermodynamic state functions and for predicting equilibrium concentration of dissolving and reaction of ion. Thus the branch of electrochemistry is of major technological importance.

1. Electrochemical Cells Or Galvanic Cells An electrochemical cell is a single arrangement of two electrodes and an electrolyte for producing an electric current due to chemical action within the cell, or for producing chemical action due to passage of electricity. Thus electrochemical cells may be used for two purposes. (i) To convert chemical energy into electrical energy. (ii) To convert electrical energy into chemical energy.

1.1  Galvanic Cell A galvanic cell, also known as voltaic cell is a device in which electrical current is generated by a spontaneous redox reaction. A simple voltaic cell is shown in fig. 1 and the reaction which takes place is:

Zn (s) + CuSO4 (aq)

ZnSO4 (aq) + Cu (s)

...(1)

A bar of zinc metal is placed in zinc sulphate solution in the left container. A bar of copper is immersed in copper sulphate solution in the right container. A salt bridge containing potassium chloride solution interconnects the solutions in the anode and cathode compartment.

4.2  Engineering Chemistry

Fig. 1  A simple galvanic cell



The oxidation occurs in the anode compartment



Zn

Zn2+ + 2e –

...(2)

The reduction takes place in the cathode compartment



Cu2+ + 2e –

Cu

...(3)

When the cell is set up, electrons flow from zinc electrode through the wire to the copper electrode. As a result, zinc dissolves in the anode solution to form Zn2+ ions. The Cu+2 ions in the cathode half cell picks up electrons and are converted to Cu atoms on the cathode. At the same time, sulphate ions from the cathode half-cell move to the anode half-cell through the salt bridge. Likewise, Zn2+ ions from the anode half cell move to the cathode half cell. This flow of ions from one halfcell to the other completes the electrical circuit which ensures continuous supply of current. As the cell operates the weight of copper rod will increase while that of zinc rod will decrease. Consequently, the cell will function till either zinc metal or copper ion is completely used up. A typical type of voltaic cell in which the salt bridge is replaced by porous pot is known as Daniel cell (as shown in fig. 2). Daniel cell resembles the above voltaic cell in all details except the Zn2+ ions and SO42– ion flow to the cathode and the anode respectively through the porous pot instead of through the salt bridge. Inspite of this difference the cell diagram remains the same.

2. Salt bridge and its functions A salt bridge consists of U shaped glass tube filled with a semisolid paste prepared by adding agaragar or gelatin to the aqueous solution of a strong electrolyte e.g. NaCl, KCl, NaNO3, NH4NO3 etc. the strong electrolyte should be chemically inert. The main functions of salt bridge are as follows:

Electrochemistry and Ionic Equilibrium  4.3

Fig. 2

It provides a low resistance electrical connection between the two half cells and allows the flow of ion from one half cell to the other half cell. Thus salt bridge completes the circuit. It keeps the solution in the two half cells electrically neutral. An equal number of SO2– 4 ions will move from the salt bridge toward zinc electrode and thus the positive charge around zinc electrode is neutralized. Similarly K+ ions will move from salt thereby neutralizing the negative charge. Therefore, the solution in two half cells are electrically neutral.

3. Half Cell Reactions and Electrodes A redox reaction is a reaction in which there is a transfer of electrons from one substance to another. Any redox reaction may be expressed as the sum of the two half reactions. One of the half reaction shows loss of electrons and other shows the gain of electrons. For example, the reaction occurring in galvanic cell.

Zn (s) + Cu+2 (aq)

Zn2+

(aq)

+ Cu (s)

...(4)

Can be expressed as the sum of the following two half cell reaction.



Cu2+ (aq) + 2e –



Zn (s)

Cu (s)

(Reduction of Cu2+)

Zn2+ (aq) + 2e

(Oxidation of Zn)

The half cell in which oxidation takes place is called oxidation half cell while the other in which reduction takes place is called reduction half cell. The reduced and oxidized substances in a half reaction form a redox couple, denoted as ox/Red. In this reaction Cu2+/Cu and Zn2+/Zn are the redox couple.

4.4  Engineering Chemistry As shown in figure1 in an electrochemical cell the two half reactions take place in two different compartments. When the reaction takes place, the electron released in one compartment travels through the external circuit and enters the cell through the other electrode. For example,

Zn (s)

Zn2+ (aq) + 2e –

(Oxidation)

Zn will lose the electrons which will travell through the circuit. The electrode where oxidation half reaction takes place is called anode. These electrons are given to the electrode which thus acquires a negative charge. When the electron flows from anode toward cathode the electrical energy is produced. Thus an electrochemical cell converts the chemical energy into electrical energy.

4. Representation of a cell or cell Diagram A cell diagram is an abbreviated symbolic depiction of an electrochemical cell. For this purpose, consider that a cell consists of two half-cells as discussed in earlier sections. In 1953, IUPAC recommended the following conventions for writing cell diagram which will be illustrated with help of zinc-copper cell. (i) A single vertical line represents a phase boundary between metal electrode and ion solution (Electrolyte). Thus the two half cells in a voltaic cells are indicated as

The metal electrode in anode half-cell is on the left of the metal ion, while in cathode halfcell it is on the right of the metal ion. Anode half-cell is written on the left and cathode half-cell on the right. (iii) In the complete cell diagram, the two half-cells are separated by a double vertical line. The zinc-copper cell can now be written as



(iv) The symbol for an inert electrode like the platinum electrode is often enclosed in a bracket. For example,



(v) The value of Emf of a cell is written on the right of the cell diagram. Thus, a zinc copper cell having Emf 1.1 V and is represented as   E = + 1.1 V

Electrochemistry and Ionic Equilibrium  4.5

5. EMF of the Cell or cell Potential An electro chemical cell is made up of two electrodes, one of these electrodes must have a higher electrode potential (higher tendency to lose electrons) than the other electrode. As a result of this potential difference, the electrons flow from the electrode at higher potential to the electrode at lower potential. A cell that has not been reached to equilibrium, can do the electrical work till a given transfer of electron is accomplished. Thus the electron transfer depends upon the potential difference between two electrodes. This potential difference is called the cell potential and measured in volts. The potential is also known as electromotive force of the cell. The magnitude of the Emf of the cell reflects the tendency of electrons to flow externally from one electrode to another. The electrons are transported through the cell solution by ions present and pass from the positive electrode to the negative electrode. This corresponds to a clockwise flow of electrons through the external circuit. Thus the Emf of the cell is given the positive sign. If the Emf acts in the opposite direction through the cell circuit, it is quoted as negative value. For example, daniel cell has an Emf 1.1 V and the copper electrode is positive (as shown in Fig. 3). This can be expressed in two ways:

Zn | ZnSO4 || CuSO4 | Cu

E = + 1.1 V.



Cu | CuSO4 || ZnSO4 | Zn

E = – 1.1 V.

Fig. 3  Illustiation of emf sign in daniel cell

The negative sign indicates that the cell is not feasible in the given direction. The reaction will take place in the reverse direction.

5.1  Calculation of Emf The Emf of a cell can be calculated from the half-cell potentials of the two cells (anode and cathode) by using the following formula:

ECell = ECathode – EAnode



ECell = ER – EL.

...(5)

4.6  Engineering Chemistry

Table 1 EMF

Cell Potential

1. Emf is the potential difference between two electrodes, when no current is flowing in the circuit.

Cell potential is the measure of the difference in potentials of the two half cells when electric current flows through the cell.

2. Emf is the maximum voltage which can be obtained from the cell.

Cell potential is always less than the maximum voltage obtainable from the cell.

3. Emf is measured by a potentiometer.

Cell potential is measured using a voltmeter.

4. Emf corresponds to the maximum useful work obtainable from the galvanic cell.

Cell potential does not correspond to the maximum useful work.

Where ER and EL are the reduction potentials of the right-hand and left-hand electrodes respectively. It may be noted that absolute values of these reduction potentials cannot be determined. These are found by connecting the half cell with a standard hydrogen electrode whose reduction potential has been arbitrarily fixed as zero. Difference between EMF and Cell Potential are listed in Table 1 given above

6. Electrode Potential (Reduction Potential) When a piece of metal is immersed in a solution of its own ions as shown is fig. 4, a potential difference is created at the interface of the metal and the solution. The magnitude of the potential difference is a measure of the tendency of electrode to undergo oxidation or reduction or tendency to lose or gain electrons. The metal and ion  (M and Mn+) represent the half cell and the reaction is half reaction. The immersed metal in an electrode and the potential due to reaction at the interface of the electrode and the solution is called the electrode potential. Thus electrode potential is the tendency of an electrode to lose or gain electrons. If the reduction takes place at the electrode, it is termed as reduction potential. M n+ + ne– M If the oxidation takes place at the electrode it is called the oxidation potential. M M2+ + 2e–

Fig. 4  Electro chemical half cell

Electrochemistry and Ionic Equilibrium  4.7 More recently, the reduction potential has been adopted by the international union of pure and applied chemistry (IUPAC) for the designation of electrode potential. When the half cell reaction is carried out at temperature of 298 K and the electrode is suspended in a solution of one molar concentration, the electrode potential is termed as the standard electrode potential and is represented by E 0. The electrode potential of an electrode is measured with respect to standard hydrogen electrode. The standard electrode potential E 0 enables one to access thermodynamic activity of various chemical substances. But there are no methods available by which we can measure its absolute value. The electrode potential of an electrode depends upon concentration of ions in solution in contact with metal. Oxidation potential of an electrode is inversely proportional to the concentration of ions whereas reduction potential is directly proportional to the concentration of ions.

6.1 Measurement of Electrode Potential A galvanic cell consists of the conjunction of two electrodes, and each making a characteristic contribution to the cell potential. Although it is not possible to determine the absolute value of the electrode potential of a single electrode, but it can be determined with respect to a reference electrode. The reference electrode used is the standard hydrogen electrode (SHE) which is assigned the value zero. Standard hydrogen electrode contains hydrogen gas in contact with H+ ion of 1M concentration at 298K and at 1 atmospheric pressure. (as shown in fig. 5).

Fig. 5  Standard hydrogen electrode

In hydrogen electrode, hydrogen gas is bubbled through a solution of hydrogen ions. The inert metal (which is often Pt) acts as a source or sink of electrons but takes no other part in the reaction other than acting as a catalyst for it. This electrode is denoted as, Pt | H2(g) | H+ (aq) The hydrogen electrode may act as either a cathode or an anode, depending on the other electrode in the cell and the spontaneous direction of the overall reaction. When the hydrogen electrode is acting as cathode, the reaction is,

2H+ (aq) + 2e –

H2(g)

(Reduction)

4.8  Engineering Chemistry

Fig. 6  Standard electrode potential of Zinc by combining with SHE



When the hydrogen electrode is acting as anode.



H2 (g)

2H+ (aq) + 2e –

(Oxidation)

To determine the electrode potential of the given electrode, the electrode is coupled with SHE. The Emf of the cell is then determined with the help of a potentiometer. The electrode potential of SHE is arbitrarily taken as zero the Emf of cell gives the value of electrode potential of the given electrode. For example when zinc electrode (Zn rod dipped in an electrode) is coupled with SHE, oxidation takes place at the zinc electrode. (Figure 6) The two half cell reactions are represented as:

Zinc electrode

Zn(s)



Hydrogen electrode 2H++ 2e –

Zn2++ 2e –

(Oxidation)

H2 (g)

(Reduction)

The Emf of cell measured by a potentiometer is 0.76 V. Therefore the oxidation potential of zinc electrode is taken as +0.76 volt and the reduction potential are taken as –0.76V. Similarly, when copper electrode (copper rod dipped in an electrolyte) is coupled with SHE, reduction takes place at copper electrode (fig. 7) H2 (g) Æ H+ + 2e –



Hydrogen electrode



Copper electrode



Emf of the cell = 0.34 V

Cu2+ + 2e – Æ Cu(s)

(oxidation) (Reduction)

Hence, the reduction potential of copper electrode is taken as + 0.34 V and oxidation potential is – 0.34 V. The negative sign of oxidation potential indicates that the oxidation at the electrode will not be spontaneous.

Electrochemistry and Ionic Equilibrium  4.9

Fig. 7  Standard electrode potential (reduction potential) of Copper by combining with SHE.

7. Electrochemical Series The standard potential of an electrode can be measured with reference to standard hydrogen electrode, which has been assigned the value zero. These electrodes can be arranged in order of their increasing reduction potential. Electrochemical series is arrangement of various electrodes in order of their standard electrode potential value. The values are reported in Table 2 for the standard electrode potential corresponding to half cell reaction in which reduction takes place. Therefore E0 values are also known as standard reduction potentials. The electrochemical series is also called activity series.

Table 2 Standard electrode potentials Elements

Electrode Reaction

E0red (Volts)

Li

Li+(aq) + e– Æ Li(s)

– 3.05

K

K+(aq) + e– Æ K(s)

– 2.93

Ba

Ba2+ (aq)

–

Ca

Ca2+ (aq)

– 2.90

–

+ 2e Æ Ca (s)

– 2.87

Na

Na+(aq) + e– Æ Na (s)

– 2.71

+ 2e Æ Ba(s)

Mg

Mg2+ (aq)

Al

Al3+ (aq)

Zn

Zn2+ (aq)

+ 2e Æ Zn(s)

– 0.76

Cr

– Cr2+ (aq) + 3e Æ Cr(s)

– 0.74

Fe

Fe2+ (aq)

– 0.44

Cd

Cd2+ (aq)

Pb Co

–

– 2.97

–

– 1.66

–

+ 2e Æ Mg(s) + 3e Æ Al(s)

–

+ 2e Æ Fe(s) –

+ 2e Æ Cd(s) –

PbSO4(s) + 2e Æ Pb(s) +

2– SO4(aq)

Co2+ + 2e– Æ Co(s)

– 0.40 – 0.31 – 0.28 Contd...

4.10  Engineering Chemistry Ni

– Ni2+ (aq) + 2e Æ Ni(s)

– 0.25

Sn

Sn2+ (aq)

– 0.14

Pb

Pb+(aq)

+ e Æ Pb(s)

– 0.13

H2

2H+ + 2e– Æ H2(g)

– 0.00

I2

–

–

+ 0.54

–

2+

+ 0.77

–

+ 2e Æ Sn(s) –

I2(s) + 2e Æ 2I 3+

Fe

Fe

Hg

Hg2+ 2(aq)

(s)

+ e Æ Fe –

+ 2e Æ 2Hg(l)

+ 0.79

Ag

Ag+(aq) + e– Æ Ag(s)

+ 0.80

Hg

Hg2+ (aq)

+ 0.85

Br2

–

+ 2e Æ Hg(l) –

Br2(aq) + 2e Æ

2Br–(aq)

+ 1.08

O2

1 __ ​   ​  O2(g) + 2H3O+ (ag) + 2e– Æ 3H2O 2

+ 1.23

Cr

Cr2O7– + 14H+ + 6e– Æ 2Cr3+ + 7H2O

+ 1.33

–

-

Cl2

Cl2(g) + 2e Æ 2Cl

+ 1.36

Au

Au3+ (aq)

+ 1.42

–

+ 3e Æ Au(s)

Mn

+ MnO–4(aq) + 8H3O(aq) + 5e– Æ Mn2+ (aq) + 12H2O

+ 1.51

F2

–

+ 2.87

F2(g) + 2e Æ

– 2F (g)

7.1 Important features of Electrochemical Series







1. Metals near the top of the series are strongly electropositive (or weakly electronegative). They lose electrons readily to give cations. 2. Lower the position of metals in the series greater is its tendency to be reduced. Thus, metals having higher position in the series are strong reducing agents and their ions are stable. On the other hand, those near the bottom of the series, are stable metals and their ions are easily reduced to the metals. Accordingly, any metal will reduce any cation below itself in the electrochemical series, if one molar solution is used. 3. Weakly electronegative metals in the upper part can displace more electronegative metals below them from their salts. For example, iron displaces copper from CuSO4 solution, Cu displaces silver from silver salt solution, silver displaces gold from gold salt solution as in photography during gold toning process. 4. Metals above hydrogen displace hydrogen from dilute acids. 5. An oxidizing agent with a higher reduction potential will oxidize any reducing agent with a lower reduction potential. For example, silver (reducing potential +0.80 V) will be deposited from its solution when treated with copper metal (reduction potential +0.34 V). 6. A reducing agent with a lower reduction potential will reduce any oxidizing agent with a higher reduction potential. For example, Li+/Li have the lower reduction potential. This means that all ions below it can be reduced by lithium.

Electrochemistry and Ionic Equilibrium  4.11

7.2 Importance of electrochemical series 1. Calculation of Standard Emf of cell (E0cell):  In order to calculate the standard Emf of the cell, the electrochemical reaction is to be split into two half reaction, viz. (i) Oxidation half reaction (ii) Reduction half reaction From the values of standard reduction potentials of the two half cells the standard Emf can be calculated as follows. Standard Emf of the cell = [Standard reduction potential of the reduction half reaction] – [Standard reduction pottential of the oxidation half reaction] or

0 (E 0cell) = (E 0right ) – (E anode )

By convention, the cathode is written on the right and anode on the left in a electrochemical cell. Thus, (E 0cell) may be written as 0 (E cell ) = (E 0right) – (E 0left). 2. Predicting whether a metal can liberate hydrogen from the acid solution or not:  The metals having lesser reduction potential compared to the SHE can liberate hydrogen from the acids (HCl, H2SO4 etc.). For example Zn, Fe, Mg and Ni release hydrogen when placed in HCl or H2SO4. In other words hydrogen will be liberated when metal has a tendency to lose electrons and H+ ions have tendency to get reduced. 3. Predicting feasibility of a redox reaction:  A redox reaction is feasible or occurs spontaneously when it fulfills the following conditions: (a) The species which has higher reduction potential undergoes reduction and the one with lower reduction potential is oxidized. (b) If value of the E0cell is positive, the reaction is spontaneous. However if E0cell comes out to be negative the forward reaction is not feasible. 4. Prediction of the anode and cathode:  The values of the electrode potential of the electrodes in a cell help in predicting the anode and cathode. An electrode with lower reduction potential can easily lose the electrons and will act as a anode. The electrode with higher reduction potential will act as cathode. 5. Predicting the oxidizing or reducing ability:  Let us consider a series of elements Cu, H2, Ni, Zn and their ions. These four elements could act as reducing agents. On the other hand, their ions Cu+2, H+, Ni+2 and Zn+2 can act as electron acceptors or oxidizing agents. If we list the respective half-reaction (or electrodes) in order of descending E0 values, we have placed the oxidizing agents in descending order of their ability to attract electrons. The arrangement in increasing strength of oxidizing agent is shown in Figure 8.

Fig. 8

4.12  Engineering Chemistry The value of E0 becomes more negative down the series. This means that Cu+2 is the best oxidizing agent (most electron attracting ion) of those in the list. That is Cu2+ shows the greatest tendency to be reduced. Conversely, Zn+2 is the worst oxidizing agent, being the least electron-attracting ion. Of the elements Cu, H2, Ni and Zn; Zn is the best reducing agent (best electron donor), since E0 for the half–reaction

Zn Æ Zn2+ + 2e E0 = + 0.76 V

has the most positive value. By the same reasoning Cu is the worst reducing agent. The half reaction potentials given in Fig. 8 above tells us that at standard condition the following reaction occur spontaneously.



Some important points concerning the table of standard reduction potential (Table 2) are: 1. On moving upward in the table more positive value of E0, better the oxidizing ability (the greater the tendency to reduce) of the ion or compound. 2. On moving downwards in the table more negative the value of E0, better the reducing ability of the ions, elements or compounds. 3. Under standard conditions, any substance in this table will spontaneously oxidize any other substance placed lower in the table.

8.  Types of Cells When various electrodes are combined, two general classes of cells are formed. (i) Chemical cells, in which the Emf is due to a chemical reaction occurring within the cell. (ii) Concentration cells, in which the Emf is due to transfer of material from one electrode to the other due to species involved. Each type of the cells may or may not involve a junction potential and it is usually said, the cell may or may not have transference. The different types of cells are shown in Figure 9.

Fig. 9

Electrochemistry and Ionic Equilibrium  4.13

8.1  Chemical Cells These cells are of following types: (i) Chemical cells without transference:  The simplest of chemical cell without transference, i.e with no liquid junction potential, consists of two electrodes and an electrolyte common to both the electrodes. A typical example of chemical cells without transference is the cell consisting of hydrogen electrode and silver electrode dipped in common electrolyte which is hydrochloric acid. Pt H2(g)  HCl (aq)  AgCl (s)  Ag In the cell, oxidation takes place at hydrogen electrode and reduction at the silver-silver electrode 1/2 H2 Æ H+ + e – (Oxidation) Ag+ + e – Æ Ag(s)



(Reduction)

(ii) Chemical cell with transference:  In Chemical cell with transference the Emf results from chemical reaction occurring within cell, but the assembly of electrode contains liquid junction between solutions of different electrolyte. The typical example is the Daniel cell in which redox couple at one electrode is Cu2+ |Cu and the other is Zn2+|Zn. Zn | Zn2+ || Cu2+ | Cu. Concentration Cells In concentration cells, Emf results from the transfer of a material from one electrode to other due to a concentration difference between the two. The difference in concentration arises due to: (i) Both the electrode dipping in the same electrolyte having different concentration. These types of cells are known as electrolyte concentration cells. For example,

(a) H2 (g, 1 atm) | H+(a1) H+ (aq) | A2 (g, 1 atm)



(b) Ag |A​g+​(a )​| A​g+​(a )​| Ag



(c) Pt |H2(P1) | HCl |H2(P2) | Pt

1

2



(ii) The electrodes may have different concentration either because they are gas electrodes operating at different pressure or because they are amalgams with different concentration. A typical concentration cell is shown in fig. 10. It consists of two silver electrodes, one immersed in 0.1 M silver nitrate solution and other in 1 M solution of the same electrolyte. The two solutions are in contact through a membrane (or salt bridge). When the electrodes are connected by a wire, it is found experimentally that electrons flow from the electrodes in more dilute (0.1 M) solution to that in the more concentrated (1 M) solution. The concentration of Ag+ ions in the left compartment is lower (0.1M) and in the right compartment is higher (1 M). There is a natural tendency to equalize the concentration of Ag+ions in the two compartments. This can be done if the electrons are transferred from the left compartment to the right compartment. This electron transfer will produce Ag+ ions in the right compartment by the half-cell reaction. Ag – e – Æ Ag+ (Left compartment)

Ag+ + e – Æ Ag

(Right compartment)

4.14  Engineering Chemistry

Fig. 10  A typical concentration cell

Thus in a bid to equalize concentration of Ag+ ions in the two compartments the cell will develop an Emf (Le Chatelier’s Priniciple) to cause the transfer of electrons equal. Eventually, the solution in two compartments will have equal Ag+ ion concentration and there will be no Emf recorded thereafter.

8.2 Emf of concentration cell Suppose the concentration in the two half-cells are C1 and C2 at 25°C, C2 being greater than C1. Then Emf, E of the concentration cell will be given by the difference between the two electrode potentials in terms of Nernst equation. 0.0591 0.0591 E = ​ E + ______ ​  n    ​ log C2   ​ – ​ E + ______ ​  n    ​ log C1   ​

[ 



] [ 

[  ]

C2 0.0591 E = ______ ​  n      ​ log ​ ​ ___ ​   ​ C1

]

...(6)

Where E is the electrode potential of the metal M and n is the valence of the ions in contact with it. For example, the Emf, E, of the concentration cell. Ag |Ag+ (1M)| |(0.1M) Ag+| Ag can be calculated by substituting the values in the above equation.

[ 

0.0591 E = ​ ______ ​  n    ​ log

]

1 M ​ _____  ​   ​ = 0.0591, here n = 1 0.1 M

8.2.1  Applications 1.  Determination of solubility of sparingly soluble salts The ionic concentration of a solution is calculated from the Emf of a concentration cell. In case of a sparingly soluble salt, the salt can be supposed to be completely ionized in saturated solution. Hence the ionic concentration is proportional to the solubility of the salt. Suppose we want to find the solubility of silver chloride. This can be done by measuring Emf, E of the cell.

Electrochemistry and Ionic Equilibrium  4.15 Ag | N/100 AgNO3 | | saturated AgCl | Ag.

The Emf of the cell at 25°C is given by the relation



0.0591 E = ______ ​   ​   log 1

C2 ​ ___ ​  C1

Here, n, the valence of Ag+ ion is 1 and the concentration of Ag+ (C2) in N/100 AgNO3 solution is 0.01 gram ion per litre.

Hence,

0.0591 E = ______ ​   ​   log 1

0.01 ____ ​   ​    C1

2.  Determination of valence The expression for the Emf, E of a concentration cell is,

C2 0.0591 E = ______ ​  n    ​ log ​ ___ ​  C1

Where n is the valence of the metallic ion in solution, while C1 and C2 are the concentration of the ions in the two half-cells. Knowing the experimental values of E, C1 and C2, n can be calculated. For example, the valence of mercury in mercurous nitrate, Hg2 (NO3)2, can be determined by the concentration cell method. The following cell is constructed. Hg |0.05 N Hg2 (NO3)2| |0.5 N Hg2 (NO3)2| Hg be

And its Emf found experimentally is 0.029 volts. Let C1 be the concentration of mercurous ion in 0.05 N Hg2 (NO3)2 in the left half cell and C2 the concentration of mercurous ion in 0.5 N Hg2 (NO3)2 solution in the right half cell. Subsitituting the values in the expression (6), we have,

0.0591 0.5 0.029 = ______ ​  n      ​ log ____ ​    ​  0.05 0.0591 0.0591 = ______ ​  n      ​ log 10 = ______ ​  n      ​



Therefore,

0.0591 n = ______ ​   ​  = 2. 0.029

9.  CELL POTENTIAL MEASUREMENT AND THERMODYNAMIC FUNCTIONS 9.1 Relation between D G0 and the Cell Potential In an electrochemical cell, chemical energy is converted into electrical energy, so it is a source of electrical energy. The net electrical work performed by a reaction is the product of electromotive force E and the quantity of electricity Q. Net Electrical work done = Q E Also, Quantity of Electricity = n F Where ‘n’ is the number of electrons taking part in reaction. F is the charge on 1 mole of electrons and is equal to 96,485C. Thus, Net Electrical Work done = n  E  F

4.16  Engineering Chemistry

Where

E = Emf of the cell.

We know that any work performed by the cell can be accomplished only at the expense of decrease in free energy occurring within the cell. Further, when the cell is operating reversibly, the electrical work done is maximum; the decrease in free energy – DG, is equal to the electrical work done. Therefore, – DG = nFE Thus the reversible ‘Emf’ of any cell is determined by the free energy change of the reaction occurring in the cell. In cell reaction, the substances involved are in their standard states (one atmospheric pressure and 298K) then, D G 0 = – nF E 0cell



...(7)



The significance of the above equation is: (i) The various thermodynamic properties of a reaction can be calculated from ‘Emf’ measurement using this equation. (ii) This equation helps to know about the spontaneity of the reaction at constant temperature and pressure. When, DG = negative The reaction is spontaneous DG = positive The reaction is non-spontaneous DG = zero The reaction is at equilibrium From equation 7, it may be concluded that for any spontaneous reaction, the cell potential will be positive, for any non-spontaneous reaction, the cell potential will be negative reaction, while for any reaction at equilibrium, E will have to be zero. The relation between spontaneity of reaction DG 0 and E 0 values are summarized in Table 3.

Table 3 G0

Reaction

E0

Spontaneous

– ve

+ ve

Non-spontaneous

+ ve

– ve

0

0

Equilibrium

9.2 Entropy of the Reaction The standard cell potential is related to standard free energy or cell function through equation– DG = – nFE The thermodynamic relation– ∂(D G) ​ ​ ______     ​  ​ = – DS ∂ T By substituting the value of – DG from equation 7

( 

( 

)

)

∂(–nEF) ​​ ​ _______     ​  ​​ ​ = – DS ∂T P

Electrochemistry and Ionic Equilibrium  4.17

(  )

∂E – nF ​​ ___ ​    ​  ​​ ​ = – DS ∂T P



(  )

∂E – DS = – nF ​​ ___ ​    ​  ​​ ​ ∂T P The temperature coefficient of cell potential gives the energy of the cell reaction.



...(8)

9.3 Enthalpy of the Reaction The enthalpy of the reaction is related with D G and TDS as–

DH = DG + TDS

Substituting the value of DG and DS.

(  ) [  (  ) ] (  )



∂E DH = – nEF + nFT ​​ ___ ​    ​  ​​ ​ ∂T P



∂E DH = – nF ​ E – T ​​ ___ ​    ​  ​​ ​  ​ ∂T P



∂E DH Ecell = – ​ ___ ​ + T ​ ___ ​    ​  ​ nF ∂T

...(9)

Whether the electrical work done is greater or less than the heat of reaction depends upon the temperature coefficient of the cell, i.e. Ecell varies with T. ∂E If ___ ​    ​ is negative than DG > DH and heat is absorbed from the surroundings during the working ∂T of the cell. ∂E If ___ ​    ​ is positive than DH > DG and heat is given out to the surrounding of the cell during ∂T working. Thus DG, DH and DS can be calculated for the cell reaction from measurements of Emf and temperature coefficient of Emf.

10. EMF OF THE CELL AND EQUILIBRIUM CONSTANT OF A CELL REACTION Consider, the general electrochemical cell reaction–

aA + bB

cC + dD

The decrease in free energy – DG for the process in given by thermodynamic equation

DG = DG 0 – RT ln Q Where DG 0 is the decrease in free energy of system when the substance are in their standard states of unit activity. Q = Reaction quotient

[C] c [D] d Q = ​ ________    ​ [A] a [B] b

The electrical energy produced during the reaction by the cell is nEF volt. – DG = nFE

4.18  Engineering Chemistry nFE = – DG 0 – RT ln Q



We know that

– D G = RT ln K

...(10)

Where K is the equilibrium constant for the cell reaction. Substituting in equation 10, we get– nFE = RT ln K – RT ln Q

Changing ln into log10, we get–



[C] c [D] d nFE = RT ln K – RT ln ​ ________  ​  [A] a [B] b 2.303 2.303 RT Ecell ​ _____      ​ RT log K ​ ________       ​ log nF nF

[C] c [D] d ​ ________    ​ [A] a [B] b

...(11)

2.303 RT At 25°C, the value of ​ ________       ​ is nF

2.303 × 8.313 JK–1 mol–1 × 298 K = ​ _____________________________           ​ n × 96500 C mol–1 0.0591 = ______ ​  n    ​ Volt

Putting the value in equation 11



[C] c [D] d 0.0591 0.0591 Ecell = ______ ​  n    ​ log K – ______ ​  n      ​ log ​ ________  ​  [A] a [B] b

The equilibrium constant of the cell can be calculated from this equation if we know the Emf of the cell and the concentration of reactants and products of the cell reaction.

11. THE NERNST EQUATION At any instant, the value of DG for a reacting system depends on the concentration of the species in solution and the reaction quotient ‘Q’. If a system is not at equilibrium, DG is not zero. As the reaction proceeds DG continously changes until the system reaches the equilibrium, at which DG is 0.

DG = DG 0 + RT ln Q



DG = – nEF



Divide this equation by nF–



0 – n FEcell = – nE cell F + RT ln Q

RT Ecell = E 0cell – ​ ____  ​ ln Q nFn

Where, E = Electrode potential E 0 = Standard electrode potential R = Gas Constant

T = Temperature F = Faraday of Electricity n = no. of electrons

Electrochemistry and Ionic Equilibrium  4.19 2.3030 RT 0 Ecell = E cell – ​ _________       ​ log Q nF



...(12)

0.0591 The value of 2.303 RT/nF at 25°C is ______ ​  n    ​ log Q 0.0591 Ecell = E 0cell – ______ ​  n      ​ log Q

For an electrode reaction–

Consider



M n+ (aq.) + ne–

M(s) Q = ​ ________     ​ n+ M (aq.) M(s) 0.0591 Ecell = E 0cell – ______ ​  n      ​ log ​ ________     ​ M n+ (aq.)

For a reaction– Zn2+ (aq) + 2e –

M (s)

Zn(s)

Zn(s) 0.0591 Ecell = E 0cell – ______ ​  n      ​ log ​ _________     ​ Zn2+ (aq.)

The activity of solids is taken as unity, for solids, (a = 1).



0.0591 1 0 Ecell = Ecell – ______ ​  n    ​ log ​ _________      ​ 2+ Zn (aq.)



0.0591 0 Ecell = Ecell – ______ ​  n    ​ log [Zn2+ (aq)]



...(13)

Therefore, for a general reaction [M(s)] is taken as unity



0.0591 Ecell = E 0cell + ______ ​  n      ​ log [M n+ (aq)]

...(14)

This equation gives the effect of concentration of M n+ ions on the potential of M n+. it can be modified as: (i)  Consider the reaction– – Fe3+ (aq) + e



Fe2+ (aq)

Hence,



[Fe2+] 0.0591 0 Ecell = Ecell – ______ ​  n      ​ log ​ ______    ​ [Fe3+]

n = 1 From this, for an electrode reaction can be written as

Oxidized State + ne –

Reduced State

Correspondingly the electrode potential is given by–

4.20  Engineering Chemistry 0.0591 Reduced State 0 Ecell = Ecell – ______ ​  n      ​ log ​ _____________        ​ Oxidized State or 0.0591 Oxidized State 0 Ecell = Ecell + ______ ​  n      ​ log ​ _____________     ​ Reduced State





(ii)  Suppose on a chlorine electrode, chlorine gas is passed at a pressure of 1 atmosphere through a solution containing Cl– ions. 1 __ ​   ​  Cl2 (g) (1 atom) + e – 2

The cell potential is–



Cl– (aq)

[  ]

0.0591 Cl – Ecell = E 0cell – ______ ​  n      ​ – ​ ​ ___   ​  ​ Cl2

Since activity of chlorine gas at 1 atmosphere pressure is taken as unity, therefore



0.0591 Ecell = E 0cell – ______ ​  n      ​ log [Cl–]

(iii)  Calculation of equilibrium constant (K) from Nernst equation. Consider the galvanic cell, Zn (s) |Zn2+ (aq) | |Cu+2 (aq) | Cu (s)



The state of equilibrium may be written by the reaction,





Zn (s) + Cu2+ (aq)

Zn+2 (aq) + Cu (s)

In equilibrium state the concentration of Zn 2+ (aq) and Cu2+ (aq) will be the equilibrium concentration and Emf of the cell will be zero. Therefore the Nernst equation is modified as,



2.3030 RT 0 = E 0cell – ​ _________       ​ log nF

[Zn2+ (aq)]equilibrium ​ ___________________          ​ [Cu2+ (aq)] equilibrium

0.0591 E 0cell = ​  _____ ​ log K n   

Where, K = equilibrium concentration at 298 K.

or

0.0591 E 0cell = ______ ​  n    ​ log K nE 0cell log K = ​ ______   ​  0.0591 nE 0cell K = Antilog ​ ______   ​  ...(15) 0.0591

[Zn2+ (aq)] equilibrium \  K = ​ ___________________          ​ [Cu2+ (aq)] equilibrium

Electrochemistry and Ionic Equilibrium  4.21

11.1 Application of the Nernst equation

(i) Nernst equation can be used to study the effect of electrolyte concentration on electrode potential

Example 1  A zinc rod is placed in a 0.1 M solution of zinc sulphate at 25°C. Calculate the potential of the electrode at this temperature, assuming 96% dissociation of ZnSO4 and E 0 (Zn+2 /Zn) = 0.76 V. Solution:  Concentration of Zn+2 (with 96% dissociation) 96 = 0.1 × ​ ____  ​ = 96 × 10 –3 M 100

The electrode reaction: Zn+2 (aq) + 2e –



According to the Nernst equation, the potential of the electrode is



Zn (s)

[  ]

RT 1 E = E 0 – ___ ​   ​ ln ​ ____ ​  +2     ​  ​ nF Zn

[ 

]

8.314 × 298 1 = (– 0.76) – ​ __________       ​ ln ​ __________ ​       ​  ​ = 0.79 V. 2  ×  96500 (96 × 10 –3)

(ii) Nernst equation can also be used for the calculation of the potential of a cell under non standard conditions.

Example 2  Calculate the potential of the following electro chemical cell at 25°C. Cu (s)  Cu+2 (aq) (0.50 M)    H+ (0.01)  H2 (0.95 atm); Pt Given: E 0cathode = 0.00 V and E 0 Solution:  The overall reaction is Cu (s) + 2H+

anode

= 0.34 V

Cu+2 (aq) + H2(g)

[Cu+2] PH2 0.059 Ecell = E 0cell – _____ ​   ​   log ​ _________  ​    2 [H+] 2

[ 

]

(0.50) (0.95) 0.0591 = (0.00 – 0.34) – ______ ​   ​   log ​ ​ __________     ​   ​ 2 (0.01)2

Ecell = – 0.449 V

(iii) Determination of unknown concentration of one of the ionic species in a cell is possible with the help of Nernst equation provided E 0 cell and concentration of ionic species is known. Example 3  Determine the concentration of Cd+2 ions in the following electro chemical cell, Fe, Fe +2 (0.1 M) Cd+2 (x m), Cd, assuming that activities are of equal concentrations. Given the emf of the cell E = – 0.02 V and E 0 = 0.04 V at 25°C. Solution:  Overall cell reaction: Fe (s) + Cd+2 (aq)

Fe +2 (aq) + Cd (s)

4.22  Engineering Chemistry

0.059 E = E 0 – _____ ​   ​   log 2

[Fe +2] ​ ______      ​ [Cd+2]

[  ]

0.059 0.1 0.02 = 0.04 – _____ ​   ​   log ​ ___ ​  x    ​  ​ 2 x = 0.00093  or  x = 0.001 M

(iv) The pH of a solution can be calculated from the measurement of emf and Nernst equation.

Example 4  The emf of a cell measured by means of a hydrogen electrode against a saturated calomel electrode at 298 K is 0.4188 V. If the pressure of the H2(g) is maintained at 1 atm, calculate the pH of the unknown solution, given potential of reference calomel electrode is 0.2415 V. The cell in this case is Pt  H2 (1 atm)  H+ (aH+)   Cl– (aCl–)  Hg2Cl2  Hg

Hydrogen electrode with potential

Calomel electrode with potential

RT E = ___ ​   ​ ln aH+  Eref = 0.2415 V F

= ER (given) RT = – 2.303 ___ ​   ​ (– log aH+ ) = 0.0591 pH = EL F

The emf of the cell is given by



E = ER – EL = Ecathode – Eanode



E = Eref – (– 0.0591 pH)

= Eref + 0.0591 pH

E – Eref ______________ 0.4188 – 0.2415 pH = ​ _______ ​  = ​   ​       = 3.0 0.0591 0.0591

(v) Nernst equation can also be used for finding the valency of an ion or the number of electrons involved in the electrode reaction.

In general, oxidized state + ne–

Reduced state

The potential of the electrode is given by

[Reduced state] RT E = E 0 – ___ ​    ​ ln ​ _____________        ​ nF [Oxidised state]

Thus by knowing E, E 0, [oxidized state], [Reduced state], and temperature, we can easily calculate n, the number of electrode involved in the cell reaction and hence the valency of an ion.

Electrochemistry and Ionic Equilibrium  4.23

12. Battery The term battery is usually applied to a group of two or more electric cells connected together electrically in series. Battery is a device which transforms chemical energy into electrical energy. Thus it acts as a portable source of electrical energy. various combinations of electrolytes and electrode are employed to make batteries for specific purposes. Thus commercial batteries can be made from different types of electrochemical cells. There are many different types of batteries ranging from the relatively large ‘flash light’ batteries to the miniature versions used for curist watches or calculators. Batteries consist of either two types of cells. 1. Primary Cells:  Primary cells are constructed so that only one continous or intermittent discharge can be obtained. Primary batteries are used as a source of dc power where the following requirements are important: • Convenience is of major importance, • The cost of a discharge is not much, • Standby power is desirable without cell deterioration during periods of nonuse for day or years A brief account of primary cell is given below Type

Anode (–)

Cathode (+)

Electrolyte

Voltage

Uses

1. Leclanche or dry cell Zn| (ZnCl2), NH4Cl |MnO2, C

Zinc case Zn(s) Æ Zn+2 (aq) + 2e–

Carbon rod in contact with carbon & MnO2 NH4+(aq) + MnO2(s) + e– Æ NH3(aq) + MnO(OH)(s)

Paste of NH4Cl, ZnCl2 (acid electrolyte)

1.5 Volts

Low drain appliances such as radios, torches, liquid crystal calculators, toys

2. Alkaline Cell Zn, ZnO|KOH(paste) |MnO2, C

Brass rod in contact with powdered Zinc Zn(s) + 2oH–(aq) Æ ZnO(s) + H2O(l) + 2oi

Steel case in contact with carbon and MnO2 MnO2(s) + H2O(l)+ e– Æ MnO(OH)(s) + – OH(aq)

Paste of 7M KOH (alkaline Electrolyte)

1 Volts

Photoelectric flash units, tape recorders, radios shavers.

3. Mercury button cell

Steel cap in contact with powdered zinc

Steel case in contact with powdered HgO. HgO(s) + H2O(l) + 2e– – Æ Hg(lg) + 2OH(aq)

Paste of KOH (alkaline electrolyte)

1.3 Volts

Watches, pacemakers, hearing aids, microphones

4. Silver button cell Zn, ZnO|KOH|(paste) Ag2O, Ag.

Steel cap in contact with powdered zinc Zn(s) + 2oH–(aq) Æ ZnO(s) + H2O(l)+ 2e–

Steel case in contact with powdered Ag2O Ag2O(s) + H2O(l) + 2e– Æ 2 Ag(s) + 2OH–(aq)

Paste of KOH (alkaline Electrolyte)

1.3 Volts

Watches, Pacemakers, hearing aids

2. Secondary cells:  A secondary cell is one in which electrodes and the electrolyte are altered by the chemical action that takes place when cell delivers the current. However, these cells can be

4.24  Engineering Chemistry restored to their original condition by facing an electric current through them. In other words, they can be recharged again and again number of times. The lead storage cells and nickel cadmium cells fall under this category. (i) The lead-acid storage cell:  A storage cell can operate both as voltaic cell and as an electrical cell. It has ability to work in both the ways to receive electrical energy and also to supply it. When it operates as a voltaic cell, it supplies electrical energy and as a result it eventually becomes ‘run down’. it then needs to be recharged. when being recharged the cell operates as an electrolytic cell.   Lead storage cell is the common example of storage cell. it is so classified because the electrolyte is an acid and the plates are largely leads. It consists of an lead antimony alloy coated with lead dioxide as cathode and spongy lead as anode. The electrolyte is a 20% solution of H2SO4 (specific gravity 1.15 at 25°C). In fact, a lead accumulator for car consist of six lead acid storage cells in series. (Fig. 11)

Fig. 11  Lead Storage Cell

Discharging:  When the lead accumulator is used for supplying electrical energy, it is said to be discharging. The lead electrode looses electrons which flow through the wire. (Fig. 11a) i.e., at anode, oxidation takes place Pb Æ Pb +2 + 2e – The so formed Pb +2 ions then combine with SO4 –2 ions Pb +2 + SO4–2 Æ PbSO4 Ø The released electron flow to the cathode, where PbO2 forms Pb +2. In other words, lead undergoes reduction at cathode. PbO2 + 2H+ + 2e – Æ Pb +2 + 2H2O The Pb +2 ions than combine with SO4 –2 ions Pb +2 + SO4–2 Æ PbSO4 – Ø

Electrochemistry and Ionic Equilibrium  4.25 The released electron flow to the cathode, where PbO2 forms Pb +2. In other words, lead undergoes reduction at cathode PbO2 + 2H+ + 2e – Æ Pb +2 + 2H2O The Pb +2 ions then combine with SO4 –2 ions Pb +2 + SO4–2 Æ PbSO4 Ø Thus, the net reaction during discharging is Pb + PbO2 + 4H+ + 2SO4–2 Æ 2PbSO4 Ø + 2H2O + Energy The voltage of each cell is about 2.0 volts at a concentration of 21.4% H2SO4 at 25°C. Recharging:  During the discharging PbSO4 is precipitated at both the electrodes. When PbSO4 covers completely both anode and cathode, the cell stops functioning as a voltaic cell. So far further use, it needs to be recharged. (Fig. 11b)

Fig. 11a  Discharging of Lead acid Cell

Fig. 11b  Recharging of Lead acid Cell

  Recharging is done by passing an external emf greater than 2 volts so that the reactions taking place during discharging are reversed. at cathode : at anode :

PbSO4 + 2H+ Æ Pb + SO4–2 PbSO4 + 2H2O Æ PbO2 + SO4–2 + 2H+ + 2e –

Net reaction during charging is:

2PbSO4 + 2H2O + energy Æ Pb + PbO2 + 4H+ + 2SO4 –2

  Obviously, during the charging cycle, the cell electrodes are restored to their original conditions (i.e., Pb and PbO2 respectively)   Generally, overcharging does not cause any damage because continued electrolysis liberates H2 and O2 at the electrodes. But excessive charging may reduce the acid level and may damage the exposed electrodes. Application of lead acid storage cells:  These cells are used for supplying current to electrical vehicles, railway, mines, laboratories, hospitals, automobiles, power station.

4.26  Engineering Chemistry

(ii) Nickel-Cadmium (Nicad) cell:  Nickel–cadmium cell consists of a cadmium anode and a paste of Ni O(OH) comprises the cathode. The cell reactions are: Cd (s) + 2OH–(aq) Æ Cd(OH)2(s) + 2e – at anode 2NiO(OH)(s) + 2H2O + 2e – Æ 2Ni(OH)2(s) + 2OH–aq at cathode

  The reaction can be easily reversed, because the reaction products, Ni(OH) 2(s) and Cd(OH)2(s) adhere to the electrode surface.   Nicad battery has fairly constant voltage. the potential of Cd anode is below the hydrogen potential. therefore, the cadmium electrode is completely inert to electrolyte.   Nicad batteries are used in electronic calculators, electronic flash units, transistors, etc.

solved Questions

1. A cell is prepared by dipping copper rod in 1M CuSO4 solution and Zn rod in 1M ZnSO4 solution, the standard reaction potentials of Cu and Zn are 0.34 and – 0.76 respectively? (i) Which electrode will be positive? (ii) How will the cell be represented? (iii) What is the cell reaction? (iv) What will be the emf of the cell? Sol. (i) Cathode is the positive electrode and has greater reduction potential than the second electrode of the cell. Therefore, the positive electrode is copper electrode. (ii) Oxidation reaction

– Cu +2 (aq) + 2e

Reduction reaction Cell represented as follows.

(aq)

+ 2e –

Cu (s)

Zn ZnSO4   CuSO4 Cu (1M) (1M)

(iii) Cell reaction is the sum of oxidation half cell reaction plus reduction half cell reaction so, it is written as follows:



Zn+2

Zn (s)

Cu (s) + ZnSO4(aq)

CuSO4(aq) + Zn (s)

E 0cell = E 0R – E L0

(iv)

= 0.34 – (– 0.76) 1.1 V.

2. Calculate the standard reduction electrode potential of Zn+2 /Zn electrode when the cell potential E 0 of the cell Zn (s) Zn+2 (1M) Cu+2 (1M) Cu (s) is 1.10 V.



E 0 Cu+2 /Cu = + 0.34 V. Zn+2 (aq) + 2e –

Sol. Oxidation reaction

Zn (s)

Reduction reaction

Cu+2 (aq) + 2e –

Cu (s)

Electrochemistry and Ionic Equilibrium  4.27 Cell reaction

Zn (s) + Cu+2 (aq)



E 0cell = E 0cathode – E 0anode



1.10 = 0.34 – E 0 Zn+2 / Zn



Cu (s) + Zn+2 (aq)

E 0 Zn+2 / Zn = – 0.76 V.

3. Two half cell reactions with their oxidation potentials are



(a) Pb (s) – 2e–



(b) Ag(s) – e–

Pb +2 (aq); E 0 = – 0.13 V. Ag+ (aq); E 0 = + 0.80 V.

Write the cell reaction and calculate its emf. Sol. Here E 0 oxidation values are given in reduction potentials E 0 values are

0

Pb +2 /Pb = – 0.13 V E 0 Ag+/Ag = + 0.80 V E 0cell = E 0cathode – E 0anode

= 0.80 – (– 0.13) = 0.93 V Oxidation reaction Pb (s)

Pb +2 (aq) + 2e –

Reduction reaction 2Ag+ (aq) + 2e – Cell reaction

Pb (s) + 2Ag+ (aq)

2Ag(s) 2Ag(s) + Pb +2 (aq)

4. Standard oxidation potential of Cd/Cd +2 and Pb/Pb +2 electrodes are – 0.40 V and – 0.13 V respectively. State whether the following cell is feasible. Sol. The standard reduction potentials of the electrodes will have sign opposite to oxidation potentials



E 0 Pb +2 /Pb = 0.13 V E 0 Cd+/Cd = + 0.40 V E 0cell = E 0red – E 0oxid

= 0.40 – 0.13 = 0.27 V Since E 0 cell is positive, the cell is feasible.

5. Can we use a copper vessel to store 1M AgNO3 solution?

Given

E 0 Cu+2 /Cu = 0.34 V; E 0 Ag+/Ag = + 0.80 V

Sol. According to the standard reduction potential values, copper will function as anode. i.e., Cu metal will be oxidized to Cu+2 and silver ions will be reduced to Ag metal and will function as cathode. Therefore, copper vessel will dissolve in AgNO3 if kept in it. Hence we cannot store AgNO3 in copper vessel. 6. What is the emf of the following cell at 25°C. Zn (s)  Zn+2 (0.2M)   Ag+ (0.002M)  Ag(g)

4.28  Engineering Chemistry Sol. The standard emf of the cell is 1.54 V.

[Ag+] 2 0.0591 E 0cell = E 0cell – ______ ​   ​   log ______ ​  +2    ​ 2 [Zn ] = 1.54 – 0.0296 log 2 × 10 –5

= 1.54 – 0.0296 [0.301 – 5] = 1.54 – 0.0296 (– 4.699) = 1.54 + 0.139 = 1.689 V. 7. A cell is formed by dipping Zn rod in 0.01 M Zn+2 solution and Ni rod in 0.5 M Ni+2 solutions. The standard electrode potentials of Zn and Ni are 0.76 V and – 0.25 V respectively. Write the cell representation, cell reaction and calculate the emf of the cell. Sol. Since E 0 (= – 0.25 V) for nickel electrode is higher than that of Zn electrode (– 0.76 V). So nickel electrode is cathode; while zinc electrode is anode. Hence, the cell is represented as: Zn (s) Zn+2 (0.01 M)   Ni+2 (0.5 M) Ni (s)

\

[ 

Ecell = ​ ​E 0​

Ni+2 /Ni

​ – ​E 0​

Zn +2 /Zn

]

​  ​

= [– 0.25 + 0.76] – 0.0296 log [0.5/0.01] = 0.51 – 0.0296 log 50 = 0.51 – 0.0296 × 1.6987 = 0.51 – 0.0503 = 0.4697 V.

8. Calculate the valency of mercurous ions with the help of the following cell:

When the emf observed at 18°C is 0.029 volt (F = 96,500) Sol. emf of the given concentration cell is given by: C2 C2 ​ ___ ​ = 0.059 log ​ ___ ​  C1 C1



2.303  RT E 0cell = ________ ​        ​ log nF



Ecell = 0.029 V, c2 = 0.01 N, c1 = 0.001 N



0.0591 2.303  RT 0.01 ______ 0.029 = ________ ​  n    ​  ln _____ ​     ​  = ​  n      ​ × 1 = 0.0591 0.001 0.059 n = _____ ​   ​  =2 0.029

Hence, valency of mercurous ion is 2 and its formula is Hg+2 2.

Electrochemistry and Ionic Equilibrium  4.29 9. The emf of the cell: Ag  0.0093N AgNO3   Saturated NH4NO3 AgNO3(x)  Ag is 0.086V at 25°C. Find the concentration (x), (2.303RT/F) = 0.059). Sol. Emf of the above concentration cell is given by 0.059 x Ecell = _____ ​   ​   log ​ ______    ​  (   n = 1) 1 0.0093 x 0.086 = 0.059 log ______ ​     ​  0.0093 0.086 x    ​   = _____ ​   ​  log ​ ______ 0.0093 0.059 x    ​   = 1.4576 = 28.68 ​ ______ 0.0093 x = 28.68 × 0.0093

= 0.267 M. 10. Find the emf of the following at 25°C. Ag Ag+ (0.01 M)   Ag+ (0.1 M) Ag Sol. Emf of concentration cell is given by

0.059 Ecell = _____ ​  n    ​ log

C2 0.059 ​ ___ ​ = _____ ​   ​   log C1 1

(n = 1 for Ag+ + e

0.1 ____ ​    ​  0.01

Ag)

Ecell = 0.059 log 10

= 0.059 V

11. The potential of a hydrogen gas electrode in a solution of an acid of unknown strength is 0.29 V at 298 K, as measured against normal hydrogen electrode. Calculate the pH of acid solution. Sol. E 0cell = E 0hydrogen – Ehydrogen = 0 V – (– 0.059 pH) = 0.059 pH

Ecell 0.29 pH = ​ _____   ​  = _____ ​     ​  = 4.9 0.059 0.059



12. Find out the pH of a solution in a quinhydrone half cell, which is coupled with standard calomel electrode. The emf of the combined cell was determined to be 0.123 V = (0.6990 V – 0.0592 V pH) – 0.2415 V. 0.6990 – 0.2415 – 0.123 Sol. pH = ​ _____________________      ​    0.0592 = 5.65

4.30  Engineering Chemistry

13. A hydrogen electrode is immersed in a solution of pH = 3.5 and other half cell is SHE. Calculate the emf of the cell, if the two half cells are connected by a salt bridge. Sol. The cell is: Pt, H2(g) H+ (pH) = 3.5   H+ (1M) H2 (g), Pt Ecell = E 0of SHE – E 0of hydrogen electrode



= 0 – (– 0.592 V pH) = 0.059 × 3.50 = 0.207 V

14. A zinc electrode is dipped in a 0.1M solution at 25°C. Assuming that salt is dissociated to 20% at this solution. Calculate the electrode potential.

​E 0​

zn +2 /zn

​ = – 0.76 V.

Zn Æ Zn+2 + 2e E 0OP = + 0.76 V.

Sol.

[ 

]

0.1 × 20 [Zn+2] = ​ ​ _______  ​    ​ ( Salt gives 20% of ions) 100 0.059 ​E​ ​ = E ​ 0​ ​ – _____ ​   ​   log10 [Zn+2] OP Zn/Zn +2 OP Zn/Zn +2 2

[ 

]

0.059 0.1 × 20 = + 0.76 – _____ ​   ​   log10 ​ ​ _______  ​     ​ 2 100 ​E​

​ = 0.81 V

​E​

​ = – 0.81 V

OP Zn/Zn +2 RP Zn/Zn +2



15. The Zinc/Siver oxide cell is used in hearing aids and electric watches. The following reactions take place. Zn Æ Zn+2 + 2e– E 0 = 0.76 V

Ag2O + H2O + 2e– Æ 2Ag + 2OH– E 0 = 0.344 V

(a) Which will be oxidized and which will be reduced? (b) Find E 0 of the Cell and DG in Joules. Sol. (a) Zn is oxidized and Ag2O is reduced. (b) E 0cell = ​E 0​Ag O/Ag​ (Red) + ​E 0​ +2​ (ox) 2

Zn/Zn

= 0.344 + 0.76 = 1.104 V

DG = – nFE° Cell.

= – 2 × 96500 × 1.104 J = – 2.13 × 105 J

16. Caluculate the electrode potential of a copper wire dipped in 0.1 CuSO4 solution at 25°C. The standard electrode potential of Copper is 0.34 volt.

Electrochemistry and Ionic Equilibrium  4.31 Sol. The electrode reaction written as reduction reaction is Cu2+ + 2e– Æ Cu so than n = 2 Applying rest equation we get.

0.0591 1 E = E 0 – ______ ​   ​   log ​ ______      ​ 2 [Cu2+] [   [Cu] = 1]

0.0591 1 = 0.34 – ______ ​   ​   log ​ ___  ​  2 0.1 = 0.34 – 0.0295 = 0.310 V For the Daniell Cell, Zn (s) + Cu2+ (ag)

Zn+2 (aq) + Cu(s)

2+ DG 0 of formation of Zn (s), Cu (s), Cu2+ (aq) and Zn (aq) are 0, 0, 64.4 kJ/mol and 154 kj/mol

respectively. Calculate the standard EMF of the Cell. Sol.

0 DG 0 = DG(products) – DG 0(Reactant)

= (– 154 + 0) – (0 + 64.4) = – 218.4 kj/mol. Since

– DG 0 = nFE 0 or DG E 0 = – ​ ___ ​  nF

– 218.4 kJ/mol. = ​ _________________        ​. 2 × 96500 Coulomb = + 1.13 V.

UNsolved Questions



1. What is the basic difference between the voltage and emf of the cell? Explain. 2. Define electrode potential. How is it different from liquid function potential? 3. Describe the construction of a simple electro chemical cell. Indicate the positive and negative electrodes? What are the reactions taking place at these electrodes? Give the overall reactwion and cell diagram. 4. Differentiate between electro chemical and electrolytic cells. 5. What is a reference electrode? Describe the construction of normal hydrogen electrode. 6. Write the (a) electrode reaction (b) cell reaction and calculate the emf of the given cell: Ag(s) Ag+ (a = 0.1)   Zn+2 (a = 0.1)  Zn

4.32  Engineering Chemistry ​E 0​Ag+/Ag​ = 0.799 V and ​E 0​

Fe +2 /Fe



​ = – 0.44 V

7. 8. 9. 10.

Why is a salt bridge used? What are different types of electrodes? Describe a concentration cell. Describe a concentration cell with transference. Derive the mathematical expression for its emf. 11. Calculate the standard emf of the following cell at 25°C, writing its half cell reaction and net cell reaction. Zn ZnSO4   CuSO4 Cu

Standard potentials of copper and zinc electrodes are + 0.34 and – 0.76 V respectively. 12. Will a reaction take place, if silver has been placed in a solution of FeSO4? Explain using the following data: ​E 0​Ag+/Ag​ = 0.799 V and ​E 0​

Fe +2 /Fe



​ = – 0.44 V

13. Calculate the emf of the cell at 25°C. Cu  Cu+2 (0.01 M)  Cu+2 (0.1 M)  Cu





14. Consider the cell: Fe  Fe +2 (0.01M)   Cu+2 (0.5M)  Cu. The standard electrode potential of iron and copper are – 0.44 V and + 0.34 V respectively. Write the cell reaction and calculate the emf of the cell. 15. What is the potential of a half cell consisting of zinc electrode in 0.01 M ZnSO4 solution at 25°C, E ° = 0.763 V. 16. Emf of the cell:

Pt, H2 (g) H+ (10 –6 M)   H+ (C = ?)  H2 (g), Pt is 0.11 V at 25°C. Calculate the value of C.

17. Predict whether the reaction: 2Ag (s) + Zn + 2 (aq)

2Ag+ (aq) + Zn (s) is feasible or not.

Chapter

5

Corrosion 1.  INTRODUCTION When metals are exposed to atmospheric conditions they react with air or water in the environment forming undesirable compounds, usually oxides. This process is called “corrosion”. Almost all metals except the less reactive metals such as gold, platinum and palladium, undergo corrosion. For example – Silver tarnishes, copper develop a green coating; lead and stainless steel lose their lustre to corrosion. Thus, corrosion may be defined as the gradual eating away, disintegration, decaying or deterioration of a metal by electrochemical reaction with its environment.

Fig. 1  Corrosion and metallurgy process

2. THEORIES OF CORROSION 2.1 The Electrochemical or Wet or Immersed theory This theory is based on Nernst theory, according to which all metals have a tendency to pass into solution. The tendency of a metal to pass into solution of its salt is measured in terms of electrode Potential. If a metal having higher electrode potential comes into contact with another metal having a lower electrode potential, a galvanic cell is set up and the metal having the lower electrode potential becomes anodic and goes into solution to a measurable extent. Greater the difference in the electrode potential of cathode and anode, greater will be the corrosion. Also the smaller the area of the anode as compared to the cathode, the more will be the attack.

Corrosion Reactions The corrosion process involves the formation of galvanic cells. The various reactions that occur during corrosion are:

5.2  Engineering Chemistry (i) Anode Reactions:  The anode is the metal which undergoes oxidation and loses electrons to the environment and passes into solution in the form of positive ions. For example: M (s) Mn+ (aq.) + ne –



Fe (s)



Fe2+ (aq.) + 2 OH – (aq.)

Fe2+ (aq.) + 2e – Fe (OH)2

This continues as long as the electrons and ions are removed from the environment. If they are not removed, the corrosion will not proceed further. (ii) Cathode Reactions:  The electrons released at the anode are carried to the cathode and are responsible for following reactions: (a) Hydrogen electrode or hydrogen type corrosion:  In acidic solution the common reaction is the reduction of H+ ions to form H2 gas H+ ions are present due to the ionization of water.

H2O



2H+ + 2e –

2H+ + O2– H2

Such type of corrosion in which hydrogen is involved is called hydrogen type corrosion (Fig. 2). At the anode, the metal ions pass into the solution as

Fig. 2  Mechanism of wet corrosion by hydrogen evolution

M M2+ + 2e – The overall reaction can be written as M + 2H+ H2 + M2+ For example, iron corrosion in an acidic solution consists of two half cell reactions. Fe Fe2+ + 2e – (Anode reaction)

2H+ + 2e –

H2 (Cathode reaction)

(b) The Water Electrode:  In oxidizing acids, the cathodic reaction is the formation of water as a by-product. (Fig. 3) O2 + 4H+ + 4e – 2 H2O

Corrosion  5.3

Fig. 3  Mechanism of wet corrosion by oxygen absorption



(c) The Oxygen Electrode: Hydrogen ion concentration is very low in neutral or basic solutions, thus evolution of hydrogen gas is not favorable. In such cases reduction of oxygen to hydroxyl ion occurs.

   e.g.

O2 + 2H2O + 4e –

4OH –

The corrosion of iron occurs by oxygen in the presence of aqueous solution. At anode, iron dissolves to form ions as Fe Fe2+ + 2e – At cathode, the electrons evolved by the above reaction will be intercepted by oxygen in the presence of H2O

2e – + 1/2 O2 + H2O

2OH– (Cathode reaction)

The overall reaction will be

2Fe + 1/2 O2 + H2O

Fe2+ + 2OH–

The OH– ion produced reactions with the positively charged Fe2+ ion forming Fe(OH)2 or rust. This process is known as rusting of iron. 1. If enough oxygen is present, ferrous hydroxide is easily oxidized to ferric hydroxide. 4Fe (OH)2 + O2 + 2H2O

4Fe (OH)3

2. This product, called yellow rust, actually corresponds to Fe2O3 ◊ H2O. If the supply of oxygen is limited, the corrosion product may be black anhydrous magnetite, Fe3O4.

2.2  Chemical Corrosion or Dry Corrosion or Direct Chemical Attack Theory Direct chemical attack includes all kinds of corrosion in which there is no appreciable flow of current through the metal for precise distance. This type of corrosion occurs mainly through the action of atmospheric gases such as oxygen, halogens, sulphur dioxide, hydrogen, sulphide, etc. However, if a

5.4  Engineering Chemistry soluble or liquid corrosion product is formed, then the metal is exposed to further attack. For example, chlorine and iodine attack silver, generating a protective film of silver halide. During detaining of tinned low-carbon steel cans using Cl2 gas at high temperature (over 120°C), tin is converted into volatile SnCl4 and hence all tin is readily removed, however dry Cl2 attacks the base metal only superficially because the FeCl3 formed on the surface protects the rest of the metal. Dry chemical corrosion is of three main types. 1.  Oxidation corrosion:  This type of corrosion is brought about by the direct chemical action of oxygen at low or high temperature on metals alkali & alkaline metals (e.g., Na, Ca, Mg etc.), generally in the presence of moisture. The reactions involved are: 2M

2 Mn+ + 2e – (Oxidation)

nO2 + 2ne –



2M + nO2

2Mn+ + 2nO2–

metal ion

2nO2– (Reduction) Oxide ion

Metal Oxide

In the case of zinc and aluminum, the outward diffusion of metal is great. As a result, the temperature of the metal is increased and the metal undergoes melting or volatilization. In high temperature oxidation of iron in dry air, three different oxide layers are formed on iron. Fe2O3 is the outermost layer, FeO is the innermost and Fe3O4 is the intermediate layer. The reactions may be written as:

4 Fe + 3O2

2 Fe2O3



3 Fe + 2O2

Fe3O4



2 Fe + O2

2FeO



The nature of the oxide film formed is very important in oxidation corrosion.



Metal + Oxygen

Metal Oxide (Corrosion product)

When oxidation starts, a thin layer of oxide is formed on the metal surface and the nature of this film decides the further action. The film may be: (i) Stable:  A stable layer is fine-grained in structure and can adhere tightly to the parent metal surface. Such a film is protective in nature, thereby shielding the metal surface. The oxide films on Al, Sn, Pb, Pt etc. are stable, tightly adhering and impervious in nature. Consequently, further oxidation corrosion is prevented. (ii) Unstable:  The oxide layer formed, decomposes back into the metal and oxygen,

Metal + Oxide

Metal + Oxygen

Consequently, oxidation corrosion is not possible in such a case. Thus, Ag, Au and Pt do not undergo oxidation corrosion.

Corrosion  5.5

Fig. 4

(iii) Volatile:  Here the oxide layer volatilizes as soon as it is formed. Thereby leaving the under lying metal surface exposed for further attack. The cause’s rapid and continuous corrosion, leading to excessive corrosion, e.g. molybolenum oxide (MoO3) is volatile.

Fig. 5



(iv) Porous:  The oxide film formed is having pores or cracks. In such a case, the atmosphere i.e., oxygen have access to the underlying surface of the metal, through the pores or cracks of the layer, thereby the corrosion continues unobstructed, till the entire metal is completely converted into its oxide.

Fig. 6

to

The formation of anodic and cathodic region on the surface of the metal can be explained due the following factors: (a) Non-uniform or rough surface of metal. (b) Presence of impurities in the bulk of metal. (c) Difference in oxygen concentration along the metal surface.

5.6  Engineering Chemistry 2.  Corrosion by other gases:  Gases like CO2, SO2, Cl2, H2S, F2 etc. cause corrosion of metals. The degree of attack depends upon the formation of protective or nonprotective films on the metal surface. For example, the AgCl layer formed from the attack decreases and protects the silver metal from further attack. In the petroleum industry steel is attacked by H2S gas at high temperature and the FeS scale is formed interfers with the normal operation of the industry. The silver metal also gets tarnished by H2S gas in the presence of moisture.

Fe + H2S



2Ag + H2S

FeS + H2 Ag2S + H2

3.  Liquid metal corrosion:  This type of corrosion is due to the chemical action of flowing liquid metal at high temperature on solid metal or alloy. It occurs in devices used for nuclear power. The corrosion involves dissolution of a solid metal or internal penetration of the liquid metal into solid metal. 4.  Cause of corrosion:  Most metals in nature exist as compounds such as sulphides, sulphates, oxides etc. because these compounds represent their thermodynamically stable state. The metals are extracted from these ores and unless the nature of the metal is substantially changed, the metal will have a natural tendency to revert back to its natural thermodynamically stable state. However, metal existing in an elemental state naturally have excellent corrosion resistance.

2.2.1  Difference Between Chemical Corrosion and Electrochemical Chemical corrosion

Electrochemical corrosion

1. It takes place in dry conditions.

It takes place in presence of wet conditions such as presence of aqueous solutions or electrolytes.

2. It takes place by direct chemical attack of environment on the metal.

It takes place through the formation of a series of galvanic cells.

3. It takes place on homogeneous and heterogeneous surface.

It takes place on heterogeneous metal surfaces only.

4. Uniform corrosion takes place.

Non-uniform corrosion takes place.

3. TYPES OF CORROSION There are a number of methods of classifying corrosion but no universally accepted terminology is in use. In broad sense, corrosion can be classified into following types:

3.1  Galvanic Corrosion Galvanic corrosion is caused when two dissimilar metals are electrically connected with each other and are exposed to an electrolyte. The two metallic parts constitute “a galvanic cell” in which the metal with the lower electrochemical potential will get corroded. For example, if Zinc and Copper are in contact with each other in presence of an electrolyte then Zinc, which is higher in electrochemical series forms the anode and gets dissolved, while Copper, which is lower in electrochemical series gets as cathode and is protected. (Fig. 7)

Corrosion  5.7 The use of Copper pipes in conjunction with iron pipes in water distribution systems has been the cause of galvanic corrosion. Larger the potential difference between the two metals, greater will be the galvanic corrosion. Galvanic corrosion is also known as differential metal corrosion.

Fig. 7  Galvanic Corrosion

3.2  Concentration Cell Corrosion Concentration cell corrosion takes place when two or more areas of a metal surface are in contact with different concentration of the same electrolyte or are exposed to an electrolyte of varying aeration.

3.2.1  Differential Oxygen Concentration Cell This is the common type of concentration cell corrosion where corrosion takes place due to “differential aeration”. Differential aeration occurs when a metallic surface is partially immersed in an electrolyte and the other part of the metal is exposed to a different concentration than the other. The anodic part suffers the corrosion. For example: If Zinc metal is partially immersed in dilute solution of NaCl and the solution is not stirred, the part above and closely adjacent to the water line is well aerated and acts as cathode. Thus, a potential is developed which causes a flow of current between two areas of the same metal. Zn dissolves at the anodic area. (Fig. 8) Zn Zn2+ + 2e – (at Anode) 1/2 O2 + H2O + 2e –

OH– (at Cathode)

Fig. 8  Differential oxygen concentration cell

5.8  Engineering Chemistry

Fig. 9

3.2.2  Drop Corrosion If takes place when a drop of an electrolyte solution is in contact with the metal surface, for example, if a drop of electrolyte is placed on iron surface, rusting occurs in the central region beneath the drop. The central region beneath the drop has lowered concentration of oxygen; Fe is oxidized to Fe2+, i.e., gets corroded. (Fig. 9) Fe

Fe2+ + 2e – (at Anode)

1/2 O2 + H2O + 2e –

OH– (at Cathode)

3.2.3  Waterline Corrosion You might have observed rusting along a line just beneath the level of the water stored in an iron tank. This is called waterline corrosion and is due to differential aeration. The area above the waterline is well aerated and acts as cathode whereas the surface below it acts as anode and hence get corroded. The reaction at the cathode and anode are same discussed in drop corrosion. (Fig. 10)

Fig. 10  Waterline corrosion occurs just underneath the meniscus and the water level

3.2.4  Crevice Corrosion Crevice or contact corrosion is the corrosion product at the region of contact of metals with metals or non-metals. For example – it occurs at the Cervice between different metallic objects such as bolts, rivets, washers, nuts, etc. The corrosion occurs at the Crevice.

Corrosion  5.9 This type of corrosion is accelerated by the deposition of dirt, debris, etc. These materials restrict the supply of oxygen underneath the covered portion which become anodic and hence suffers corrosion. Crevice corrosion can also be attributed to concentration cell effect.

3.2.5  Pitting Corrosion Pitting corrosion is a localized corrosion resulting in the formation of pits or holes. Pitting usually takes place due to cracking of the protective film on the surface of the metal. An anode is formed where the film has been broken, while the unbroken film acts as a cathode. Large cathodic areas results in severe corrosion of the anode region. Because of corrosion of smaller anodic region, small holes or pits are formed on the surface of the metal. As shown in fig. 11 at the surface of the iron bar, the more oxygenated portion is cathode whereas less oxygenated part is anode. The corrosion takes place at anode and rust is accumulated. The rust further screens the bottom of the pit from oxygen, thus increasing the rate of corrosion.

Fig. 11  Pitting corrosion at the surface of iron

3.3  Intergranular Corrosion When a metal is solidified, the solidification starts at number of randomly distributed nuclei within the molten mass. Because of random distribution of the nuclei, the planes of atom in adjacent grains are known as grain boundaries. Grain boundaries are generally more prone to corrosion attack, to the precipitation occurs at grain boundaries. This is because the precipitation of certain compounds at the grain boundaries leaves the solid metal solution adjacent to them depleted in one constituent. Thus, a potential difference is created leading to the corrosion at anodic grain boundaries. Alloys are generally more susceptible to intergranular corrosion. For example, Stainless steel is an alloy of iron, carbon and chromium. The chromium depleted region becomes more anodic and suffers corrosion. Aluminum and copper alloys are also attacked by this type of corrosion.

3.4  Stress Corrosion (Stress Cracking) The term stress corrosion takes place due to the combined effect of mechanical stress and the corrosive environment on a material. During their fabrication, these materials are subjected to different types

5.10  Engineering Chemistry

Fig. 12  General intergranular attack

of convene stress such as rolling. These areas become very reactive chemically and may be attacked even by trace of corrosion environment, resulting in the formation of cracks. Stress corrosion is usually unpredictable in nature. Even after a few years of trouble free service, a metal can suddenly crack without any warming or previous deformation. It may be noted that stress corrosion is due to the combined effect of static tensile stress and the corrosive environment. Moreover, stress cracking takes place only in highly selective environment i.e. different environments are required for different alloys for stress cracking. For example, corrosion cracking of brass takes place only in presence of traces of ammonia of amines. Some examples of stress corrosion are: (a) Season Cracking:  Season cracking due to stress corrosion is commonly observed in colddrawn brass articles. In the presence of alloying atmosphere containing traces of ammonia or amines brass show Intergranular cracking. Fissures propagate and cause cracking in the presence of high tensile stress. (b) Caustic Embrittlement of Mild Steel:  When Mild Steel is exposed to alkaline solutions at high temperature and stresses, it causes failure of the material due to caustic embrittlement. This type of corrosion mainly occurs in boilers or heat transfer equipment. Boiler water contains Na2CO3 which undergoes decomposition under high pressure.

Na2CO3 + H2O

2NaOH + CO2

The alkaline water flows into the hair cracks present at the inner surface of the boiler due to capillary action. Eventually the concentration of NaOH in the cracks crevices increase. The boiler material is thus under the two different concentrations of the electrolyte, higher inside the crack, and dilute outside. Thus a concentration cell sets in:

Anode Cathode Fe | Conc. NaOH || Dil. NaOH | Fe

The iron of the main body containing dilute NaOH behaves as cathode whereas the iron of the cracks and crevices surrounded by Conc. NaOH behaves as anode. The sodium ferroate so formed decomposes as,

Corrosion  5.11

2Na2FeO3 + 4H2O

6NaOH + Fe3O4 + H2

Fe3O4 is precipitation and NaOH is regenerated causing further corrosion.

3.5  Microbiological Corrosion Microbiological corrosion is the deterioration of materials caused directly or indirectly by microbes such as bacteria, algae, moulds of fungi, singly or in combination. These microorganisms can be aerobic or anaerobic depending upon whether they thrive in presence or absence of oxygen. (a) Aerobic Conditions  Aerobic Bacteria such as sulphur bacteria, decrease the concentrate of oxygen in the medium in contact with a metal structure. Aerobic bacteria can also produce corrosion by producing a corrosive metabolite usually an acid, either mineral or organic. The group of bacteria of the genus thiobacillus can oxidize sulphur into sulphate present in their cell to yield sulphuric acid, which cause corrosion.

3.6 Erosion Corrosion Erosion Corrosion is the result of relative movement between a corrosive fluid and metal surface. This results in the development of differential cells at these points leading to corrosion. Metals that are soft and are readily damaged due to mechanical abrasion are quite susceptible to this type of corrosion. Many types of corrosive medium moving with high velocities can cause erosion corrosion. For example, hot gases may oxidize a metal and then at high velocity blow off a protective scale around the metal.

3.7  Soil Corrosion The two factors which promote corrosion viz.: Moisture and dissolved electrolytes are present in soil, making it corrosive. Thus soil corrosion of the structure embedded in soil is due to the combined effect of the presence of

(i) Acidity in the soil



(ii) Moisture and electrolytes

(iii) Micro-organisms

(iv) Differential bio-accretion

Microbiological corrosion of the buried structures may take place in water-logged soils which give the environment in which bacteria can grow.

3.8  Selective Leaching Selective leaching is the removal of one element from a solid alloy corrosion process. Similar processes occur in other alloy system in which aluminum, iron, cobalt, chromium and other elements are removed from their alloys. Dezincification is prevalent in condenser tubes made of brass, where sea water is used as condenser water. Although both zinc and copper go into the solution but only zinc ions remain in solution and copper is re-deposited as a spongy layer. The dezincification results into a porous metal with poor mechanical properties.

5.12  Engineering Chemistry Similarly, gray cast iron also shows the effects of selective leaching in mild environments. After selective leaching of the iron, graphite network is left behind. Similarly white iron has no free carbon and is not subjected to graphitization.

3.9 Hydrogen Damage Hydrogen damage is a general term which refers to mechanical damage of a metal caused by the presence of or interaction with hydrogen. It can be classified into the following types: (a) Hydrogen Blistering:  It is caused by penetration of hydrogen into metal. Surface bulges appear on the surface of the metal due to hydrogen busting.

(b) Hydrogen Embrittlement:  Hydrogen damage embrittlement is also caused by penetration of hydrogen into a metal leading to loss of ductility and tensile strength. (c) Hydrogen Attack:  It refers to the interaction between hydrogen and a component of an alloy at high temperature. For example, the oxygen containing copper gets disintegrated in presence of hydrogen.

3.10  Stray Current Corrosion Electric current from electric system can leak into adjacent conducting structures. The metal structure such as H2O pipes, gas pipes etc. thus get corroded because these become anodic due to leakage from the electric circuits.

3.11  Factors Affecting Corrosion The rate and extent of corrosion depends upon the following factors: (i) Position in the Galvanic series:  When two metals are in direct contact in presence of an electrolyte, the metal higher up in the galvanic series suffers corrosion. The rate and extent of corrosion depends upon the difference in the position of metals in the electrochemical series. (ii) Over-Voltage:  When a metal having a high position in H2SO4, it undergoes corrosion, but the initial rate of reaction is quite slow because of high over-voltage of zinc metal, which reduce the effective potential to a small value. Thus, reduction in over-voltage of the corroding metal accelerates the corrosion rate. (iii) Purity of Metal:  Impurities in a metal cause heterogeneity and form minute galvanic cells at exposed parts, thereby the anodic part get corroded. For example, zinc metal having impurities Pb or Fe undergoes corrosion due to the formation of local galvanic cells. The rate and extent of corrosion increases with the increasing expiry and extent of impurities. (iv) Solubility of the corrosion products:  In electrochemical corrosion, if the corrosion product is soluble in the corroding medium, then corrosion proceeds at a faster rate and vice versa. (v) Volatility of the corrosion product exposes:  The metal surface to further attack and thus leads to excessive corrosion. (vi) Nature of the Surface Film:  All metals in the atmosphere get covered with a thin surface film of metal oxide. The ratio of the volume of the metal oxide to the metal is known as “Specific Volume Ratio”. The specific volume ratios of Ni, Cr and w are 1.6, 2.0 and 5.6 respectively thus tungsten get least corroded.

Corrosion  5.13

3.12  Nature of the Corroding Environment (i) Temperature:  The rate of chemical reaction and diffusion rate increases with the increase of temperature thus corrosion rate is greatly influenced. (ii) Effect of pH:  Acidic medium are generally more corrosive then alkaline and natural medium atmospheric metals like Al, Zn and Pb cause corrosion of iron in oxygen free water at slower rate until pH falls below 5.0. (iii) Conductance of the Medium:  This is an important factor for corrosion in underground or submerged structures because corrosion current depends on this factor. Hence stray currents from power leakage will damage the metal structure buried under soils of higher conductance to a greater extent than those under dry sandy soils having higher resistance. (iv) Other factors influencing corrosion include nature of anions and cations present, flow velocity or process stream, formation of oxygen concentration cells.

4.  CORROSION CONTROL Corrosion control methods may be classified into the following categories:

4.1  Material Selection and Design



1. Choice of right type of material is the main factor for corrosion control. The selection of metal should be based on its chemical properties as well as its cost and structure. 2. Heat treatment like annealing reduces internal stress and corrosion. 3. When two metals have to be in contact, they should be selected so that their oxidation potentials are as near as possible. Further, the area of more active metal i.e., the anode should be larger than that of the cathode or less active metal. 4. If an active metal is used, it should be insulated from the more cathodic area. 5. The components should be designed in such a way that a stagnant pool of liquid should not be collected. 6. Moisture should be excluded wherever practicable, no corrosion occurs without the presence of moisture. The assembly should be designed such that the moisture should be as low as possible.

4.2  Cathodic and Anodic Protection When it is impossible or impractical to alter the nature of the corrosion medium, corrosion control may be achieved by cathodic or anodic protection.

4.2.1  Cathodic Protection The principle involved in cathodic protection is to force the metal to behave like a cathode. It is applied in two ways: 1. Sacrificial anode or galvanic protection:  In this method more active metal is connected to the metal structure to be protected so that all the corrosion is concentrated at the more active metal. This sacrificial anode, typically zinc or magnesium is consumed requires replacement. Important applications of cathodic protection include (i) Protection from soil corrosion of underground cables and pipelines.

5.14  Engineering Chemistry

(ii) Protection from marine corrosion of cables, ship hulls, etc. offshore drilling platforms and water heaters.

Fig. 13  Galvanic protection using sacrificial anode

2. Impressed current cathodic protection:  In this method an impressed current is applied to convert the corroding metal from anode to cathode. This is done by applying sufficient amount of direct current from a DC source to an anode (e.g. graphite or high silicon iron) buried in the soil or immersed in the corrosion medium and connected to the corroding metal structure which is to be protected. The electrons flow to the metals, as a result the metal acts as a cathode and the auxiliary anode. This type of protection is of particular volume in cause of buried structures such as tanks and pipelines transmission line towers, marine piers, laid-up ships, etc.

Fig. 14  Impressed current cathodic protection

Corrosion  5.15

4.2.2  Anode Protection In this method the metal is passivated by applying a current in a direction that renders it more anodic. It has been applied in the case of steel and stainless steel and to some extent in the case of Fe, N, Al and Cr. For example, if iron is dipped in very concentrated HNO3, the iron rapidly and uniformly corrodes to form a thin, protective iron hydroxide coating. The coating prevents the iron from subsequent corrosion in nitric acid.

Fig. 15  Anodic protection system for tanks

4.3 Protective Coatings Protective coatings are used to isolate the anode and cathode regions. For applying any type of coating, the metal surface is prepared in the following manner: (a) grease and other surface contamination is removed; (b) oxide scales, rust and corrosion products are removed; (c) etching treatment to aid in proper adhesion or buffing or polishing is given. Protective coatings are of the following two types:

4.3.1  Metallic Coating These are the coatings of metals applied as a protective coating on center base metals. The metallic coating is usually imparted by the following methods; (i) Hot Dipping:  In this method, the metal to be coated is dipped in the molten bath of the coating metal for sufficient time and then removed along with the adhering film. For example, applying coating of low melting metals and alloys such as Zn, Pb, Sn, solder, etc. (ii) Electroplating:  In this method, the coating metal is deposited on the base metal by passing a direct current through the electrolyte solution containing a salt of the coating metal. For example, in steel, tin plating or zinc plating is applied. The metals like Au, Ag, Cr, Ni, Cu, Sr, etc. may be electroplated. (iii) Spraying:  A stream of atomized droplets is blown on the surface of the metal to be protected. Eg. in case of tanks, ships and towers. (iv) Metal cladding:  Alloy to be protected is sandwitched between two layers of protecting metal. Layers are bonded firmly by passing them through rollers under the action of heat and pressure. Cladding material are Ni, Pb, and Ag.

5.16  Engineering Chemistry

4.3.2  Non-metallic Coating Non-metallic coatings may be of organic or inorganic nature. The commonly used non-metallic coatings are, (i) Oxide Coating:  The process consists of dipping the metallic compound in an electrolyte bath of suitable composition. The process is called anodizing. In nuclear reactors, the zirconium alloy containing Sn, Fe, Cr, Ni is often used for protection of uranium fuel rods. The alloy is protected by ZrO3 formed on it by controlled oxidation. (ii) Organic Coating:  Important organic protective coatings include paints, varnishes, enamels and lacquers. When applied on cleaned metal surfaces, they act as effective inert barriers which protect the metal corrosion and also give decorative and aesthetic appeal (iii) Paints and Lacquers:  Paints and Lacquers are applied on metallic surfaces to avoid contact of air with base metals. Paint is mixture of pigment, dry oil and thinner. Lacquer is a solution of resin and plasticizer. (iv) Vitreous Enameling:  Vitreous enameling is carried out by applying a dispersion of vitreous enamel in water containing silicate and clay on the base metal surface. The enamel coating has good resistance of dilute alkalis, acids, boiling water and can withstand high temperature.

5.  Corrosion Inhibitors Inhibitors are the substances which when added to the electrolyte solution reduce the rate of corrosion. These are of two major classes: 1. Inorganic inhibitors:  Inorganic inhibitors such as silicates, chromates, borates, etc. suppress the rate of corrosion by acting on the anode. Alkaline sodium nitrites alone or in combination with other inhibitors such as phosphate have been used to control the corrosion of tankers and pipelines. Sodium benzoate has been used as an inhibitor for mild steel and in preventing corrosion in cooling system such as automobile radiators. A mixture containing 0.1% NaNO2 and 1.5% sodium benzoate has been used as inhibitors in anti-freeze solutions. Lime acts as a cathodic inhibitor by precipitating CaCO3 in water containing temporary hardness or dissolved CO2. Colloidal particles of CaCO3 having positive charge are attracted to the cathodic areas and get deposited there, thus, reducing the corrosion current. 2. Organic inhibitors:  Organic inhibitors act by different mechanisms e.g., organic colloidal inhibitors form protective layers on the metal surface by adsorption. Surface active reagents known as surfactant, containing polar groups promote spreading and oriented adhesion to the surface, thus, forming a protective film. Organic bases e.g., amines, pyridines, quinoline and their derivatives contain hydrophobic group or radicals. These positively charged cationic groups attack themselves to the cathodic areas through the nitrogen and provide inhibition. The efficiency of inhibition depends upon size and number of the alkyl groups. E.g., primary amyl amine (C5H11NH2) has been found to be more effective as inhibitor than primary ethyl amine (C2H5NH2). High molecular weight amines derived from rosin have been used as corrosion inhibitors. The product obtained by rosin amine and pentachlorophenol was used in castings for underground pipelines to protect them from corrosion by soil bacteria. Organic inhibitors such as soluble salts of rosin e.g., stearate or naphthenate have been used as inhibitors in the various metal cleaning operations like acid pickling of metals, cleaning of boilers, condensers, heat

Corrosion  5.17 exchangers, chemical equipment, pipelines, etc., to prevent undue and excess corrosion of the metal while removing rust, scales and deposits. Some vapour phase inhibitors (VPI) such as dicyclohexylammonium nitrite and cyclohexylamine carbonate are used to inhibit the corrosion in storing aircraft engines as well as during storage, packing, shipping, etc.

5.1 Passivity It is defined as a phenomenon in which metal or an alloy exhibits a much higher corrosion resistance than its position in the electrochemical series would indicate. The passivated metal can rendered active by a change in the environmental conditions. Metals which exhibit passivity include iron, chromium, nickel, titanium and alloys of these metals. However, it should be emphasized that a metal which is passive in one environment may not necessarily be so in a different environment because passivity results from a continuous reaction at a very slow rate. The passivation is not static in nature but is quite stable under certain environmental conditions. Such as oxidizing conditions, e.g., the metals or alloys of Al, Cr, Ti etc., exhibit high corrosion resistance in oxidizing environments, but are quite unstable in reducing environments. The passivity of some metals are in the order of:

The presence of oxide films may affect the initial rate of attack by acids. Iron dissolves readily in a very dilute HNO3, but as the concentration of acid is raised, its oxidizing power rises and a point is reached at which the acid directly oxidizes the metal to its oxide. The layer of oxide formed on the surface makes the iron passive. The passive iron has high, positive electrode potential and is not easily corroded. The oxide film is easily damaged by scratching, heating or because of the presence of chloride ions. Under such conditions, the metal continues to corrode. However, some passive films are not easily damaged as that of iron, and even if there are cracks in the film, these films can be readily heal themselves in an oxidizing atmosphere, for example stainless steel composed of 18%Ni. The highly protective coating of Cr2O3 is believed to be responsible for the corrosion resistance. Other examples of increased resistance to oxidation by alloying are silicon containing cast iron and silicon containing molybdenum.

5.2  Oxide or Protective Layer Theory According to this theory, the passivity of metals is due to formation of invisible oxide film on the surface. The oxide layer so formed is not soluble in acids. The thickness of this film depends on the treatment used to produce it. This theory is supported by following facts: (i) Removal of this film either mechanically or chemically makes the metal active again. (ii) When passive iron is heated in nitric acid at 75°C Fe2O3 layer begins to dissolve. (iii) Certain metals like lead derive protection from atmospheric corrosion in the presence of a film of a oxide.

5.18  Engineering Chemistry It is the oxide film that is responsible for the occurrence of passivity. But in some cases films other than those of oxides are responsible for the passivity. Therefore, any protective films other than those of oxides are responsible for the passivity. Therefore, any protective film produced during the chemical reaction is sufficient to cause passivity and thus the theory is termed as protective layer theory. The oxide film is extremely thin, transparent and invisible so long as it is in contact with the bright reflecting metallic surface. During the formation, it is assumed that oxygen reacts extremely fast with the metal and the rate of film growth is inversely proportional to the thickness of the film. Therefore, dy/dt = k/y where, y = film thickness, t = time and k = constant. On integration, we get Y 2 = 2kt + I Where I = integration constant. Experimentally it is observed that the rate of film growth does not obey the above equation. In other words the film is not necessarily uniform throughout its thickness.

solved Questions 1. Define Corrosion? Ans. Any process of deterioration and consequent loss of solid metallic materials through an unwanted chemical or electrochemical attack by its environment, starting at its surface is called corrosion. 2. What is rust? Ans. Hydrated feric Oxide (Fe2O3) with Fe(OH)3- brown rust, mixed ferrous & ferric oxides. Fe3O4 FeO.Fe2O3- black rust. 3. Formation of which types of metal oxide film cause rapid and continuous corrosion? Ans. Volatile oxide film and porous oxide film. 4. What is electro chemical corrosion? Ans. This type of corrosion occurs due to the existence of separate ‘anodic’ and ‘cathodic’ areas between which current flows through the conducting solution at anodic area oxidation occur and anodic part of metal is deployed. 5. What is galvanic corrosion? Ans. When two dissimilar metals are electrically connected and exposed to an electrolyte the metal highest in electro chemical series undergoes corrosion. 6. Iron corrodes faster than aluminium, even though iron is placed below aluminium in the electrical series. why? Ans. This is because aluminium forms a nonporous, thin, tightly adhering protective oxide film (Al2O3) on its surface and this film does not permit corrosion to occurs. 7. Wire mesh corrodes fastest at the joints. why? Ans. The joints of wire mesh is stressed so these becomes anodic w.r.t. unjoined wires. At these anodic parts, oxidation takes place and the metal is corroded fast, while the cathodic part remain unaffected.

Corrosion  5.19 8. What is meant by the term passivity? Ans. The phenomenon in which a metal or an alloy exhibits a much higher corrosion resistance than expected from the position in the galvanic serves, in called passivity. 9. What is the effect of pH on corrosion? Ans. Lower the pH (or more acidic) greater is the corrosion. 10. Name the volatile oxidation corrosion product of a metal. Ans. MoO3 (molybdenum Oxide). 11. Where the electro chemical corrosion takes place? Ans. Always at the anodic areas. 12. What is the effect of grain size of the metal on the corrosion? Ans. Smaller the grain size of the metal or alloy, greatest is its corrosion, due to increased solubility. 13. A piece of impure zinc and pure zinc are placed in a salt solution. which will corrode faster? Ans. Impure zinc, since impurities leads to the formation of local galvanic cells. 14. In a structure two dissimilar metals should not be allowed to come in contact with each other. why? Ans. Due to galvanic corrosion, the metal higher in electro chemical series undergoes corrosion. 15. Which of the following metals could provide cathodic protection to iron: Al, Zn, Cu, Ni? Ans. Al and Zn 16. Which gases in the atmosphere are likely to accelerate the rusting of iron? Ans. CO2 and the acid rain gases like SO2 & NO2 17. Why are galvanized utensils not used? Ans. Zinc of galvanized articles get dissolved in dilute food acids and forms highly toxic compounds. Hence, galvanized utensils cannot be used of or preparation and storing food stuffs. 18. Why is coating of zinc on iron is called sacrificial anode? Ans. Zinc acts as an anode w.r.t iron, so zinc is worn away before the iron. Hence, zinc is sacrified instead of iron. 19. What is the effect of CO2 on electro chemical corrosion? Ans. The corrosion is speed up because CO2 get dissolved in water producing an acidic electrolyte (H2CO3).

UNsolved Questions

1. 2. 3. 4. 5. 6.

What is corrosion of metals? Explain the basic reason of metallic corrosion. Discuss briefly the consequences of corrosion. Explain electrochemical theory of corrosion. Differentiate chemical and electrochemical corrosion with suitable examples. Discuss factors affecting rate of corrosion. Explain briefly various factors influencing corrosion.

5.20  Engineering Chemistry

7. Silver and copper metals do not undergo much corrosion like iron in moist atmosphere. Explain. 8. Explain methods for prevention of corrosions. 9. Discuss corrosion inhibitors. 10. How are metals protected against corrosion by modifying the environment? 11. How will you protect an underground pipe line from corrosion by sacrificial anodic and impressed current cathodic protection methods? 12. What are cathodic and anodic protection methods for controlling corrosion? 13. Distinguish between anodic and cathodic coating. 14. What is meant by dry corrosion? 15. Explain passivity and its significance. 16. Discuss in detail the nature of corroding environment. 17. Discuss briefly (i) Galvanic corrosion (ii) Pitting corrosion (iii) Crevice corrosion (iv) Dropline corrosion.

Chapter

6

Water Treatment “Water is one of the most abundant commodities in nature, but is also the most misused one”.

1.  INTRODUCTION One of the basic necessities of life is water. Living things exist on the earth because this is the only planet that has the presence of water. Water is necessary for the survival of all living things be it plant or animal life. Water is one of the most abundant commodities in nature but is also the most misused one. Although earth is a blue planet and 80% of its surface is covered by water, the hard fact of life is that about 97% of it is locked in the oceans and sea which is too saline to drink and for direct use for agricultural or industrial purposes. 2.4% is trapped in polar ice caps and giant glaciers, from which icebergs break off and slowly melt at sea. > 1% water is used by man for various development, industrial, agricultural, steam generation and domestic use.

Table 1 Analyses of some types of water used in industry S. No

Type of analysis

Ionic constituents and their concentration in ppm

PH

Silica ppm

Dissolved Solids ppm

Hardness in terms of CaCo3 in (ppm) alkaline

nonalkaline

Total

1.

Moorland surface drainage

Na2– O, Ca+2–7 Mg+2– 6, HCO3–15, Cl––10, SOH–2–81 NO–3 –Traces

6.7

8

77

12

30

42

2.

Lowland surface drainage

Na+ – 25 Ca+2– 63 Mg+2– 18 HCo3–160 Cl–– 48 So4+2–10 NO3– – 15

7.5

10

333

130

100

230

3.

Deep Well water

Na+ – 74, Ca+ – 48 Mg+2– 20, HCo3 –350 Cl– – 48, So4+2–10, No3 – Traces.

7.0

15

388

203

0

203

6.2  Engineering Chemistry

1.1  Moorland Surface Drainage Water from this source is fairly constant in composition. It is generally clear and colored brown. It is slightly acidic due to the presence of dissolved carbon dioxide and of weak organic acids, which renders it corrosive. Although its hardness is low, it can cause scale formation in boilers unless it is suitably treated before use. It contains some strains of iron bacteria which must be removed by chlorination to prevent their deposition in the pipe-lines. It possesses a tendency to dissolve lead and copper. This fact should be considered if the water is used for drinking purposes.

1.2  Lowland surface drainage Water from this source varies widely in composition from place to place. It is not generally colored but may contain fine mud in suspension which does not easily settle unless with the help of coagulants. Its hardness is usually high and hence can cause serious scale formation in boilers, economizers and coolers, unless the water is properly treated before use. When the water is heated in boilers, the CO2 produced from the bicarbonate ions passes off with the steam and dissolves in the condensate, forming carbonic acid which is corrosive to the mild steel. River and canal waters may get contaminated by sewage and industrial wastes, which may require preliminary treatment prior to softening.

1.3  Deep Well Waters This type of water is fairly constant in composition unless contaminated by other waters percolating through faults in the surrounding strata. When freshly drawn, this is usually colorless, clear and devoid of finely divided suspended matter and hence sparkling. This type of water may develop a brown opalescence on exposure to air due to the presence of small amounts of ferrous iron which gets converted into hydrated ferric oxide. Traces of manganese as well as H2S may also be present. In many deep well waters, the concentration of the bicarbonate is more than equivalent to the combined concentrations of Ca2+ and Mg2+ ions so that Na2CO3 may be considered to be present (wide Table-l). The hardness is then entirely alkaline hardness. The sulphate concentration is often very low. Water from shallow wells possess a composition similar to that of low-land surface drainage waters. The concentration of bicarbonate ions is less than the combined concentrations of Ca2+ and Mg2+ ions. Hence, the water contains non-alkaline hardness. Some, deep well waters contain considerable amounts of free CO2. Because of the hardness present and because of the carbonate ions produced due to thermal decomposition of bicarbonate ions, deep well waters like low-land waters, give rise to severe scale formation in boilers. The high silica content also contributes to the formation of hard scales in boilers. CO2 formed by the decomposition of bicarbonate ions, similar to lowland waters, also results in the production of acid condensate leading to corrosion.

2. EFFECT OF WATER ON ROCKS AND MINERALS

l. Dissolution:  Some mineral constituents of rocks such as NaCl and CaSO4. 2H2O readily dissolve in water. 2. Hydration:  Some minerals are easily hydrated with the consequent increase in volume leading to disintegration of the rocks in which these minerals are present. Examples are: (a) CaSO4

anhydrate

Hydration

CaSO4 .2H2O (accompanied by an expansion of 33%) gypsum

Water Treatment  6.3 (b) Mg2SiO4 Olivine.

Hydration

Mg2SiO4 XH2O Serpentine

3. Effect of dissolved oxygen:  This leads to oxidation and hydration Fe3O4

Oxidation

Magnatite

Fe2O3

Hydration

Haematite

2FeS2 + 702 + 2H2O



Limonite

2FeSO4 + 2H2SO4

Marcesite

4. Effect of dissolved CO2 (a) Water containing dissolved CO2 converts the insoluble carbonates of Ca, Mg and Fe into their relatively soluble bicarbonates CaCO3(s) + H2CO3 MgCO3(s) + H2CO3



3Fe2O3.2H2O

Ca (HCO3)2 Mg (HCO3)2

(b) Rock forming minerals like silicates and alumino-silicates of Na, K. Ca and Fe are attacked by CO2, producing soluble carbonates, bicarbonates and silica K2O. Al2O3. 6SiO2 + 2H2O + CO2

Al2O3. 2SiO2. 2H2O + K2CO3 + 4SiO2



(c) Rocks containing felspar disintegrate and charge nearby river water with dissolved salts, fine clay and silica in suspension. Thus, water collects impurities from the ground, rocks or soil with which it comes in contact. Contamination of water may also result from sewage or industrial wastes, either by actual contact when these are allowed to flow into the running water, or by percolation through the ground.

3. TYPES OF IMPURITIES PRESENT IN WATER The impurities present in natural waters may be broadly classified as follows: 1. Dissolved impurities (a) Inorganic salts e.g., (i) Cations: Ca2+, Mg2+, Na+, K+, Fe +. Al3+ and sometimes traces of Zn2+ and Cu2+ (ii) Anions: Cl–. SO42–, NO3–, HCO3– and sometimes F– and NO2– (b) Gases e.g., CO2. O2. N2, oxides of N2 and sometimes NH3, H2S (c) Organic salts 2. Suspended impurities (a) Inorganic e.g., clay and sand (b) Organic e.g., oil globules, vegetable and animal matter. Finely divided clay and silica, aluminum hydroxide, ferric hydroxide, organic waste products, humic acids, coloring matter, complex protein, amino acids, which are generally classified as albunoid ammonia. 3. Colloidal impurities:  Clay and finely divided silica colloidal particle of 104 – 106 mm size. 4. Micro-organisms:  Bacteria, fungi, algae, other micro-organisms and other forms of animal and vegetable life.

6.4  Engineering Chemistry

4. EFFECTS OF IMPURITIES IN NATURAL WATERS The various types of impurities present in natural waters impart some properties on the waters. From the point of view of industrial use, the characteristics and effects of the impurities on the water quality are discussed under the following headings: 1. Color 2. Tastes and odors 3. Turbidity and sediment 4. Micro-organisms 5. Dissolved mineral matter: (a) hardness, (b) alkalinity, (c) total solids, and (d) corrosion 6. Dissolved gases 7. Silica content and 8. Oxidability.

4.1  Color Color is found mostly in surface waters, although water from some shallow wells, springs and deep wells may also be occasionally colored. The color of natural waters range from yellowish-brown to dark brown. The colors and the materials which cause it are often objectionable in which the water and the manufactured product come into contact, e.g., dyeing, scouring and laundering. The color of natural waters is mainly due to the presence of dissolved or colloidally dispersed organic matter. The measurement of color is usually made with a tint meter and, the result is expressed in “Hazen units” or “Standard units of color”. In determining the color of water, it is the true color as expressed in the standard units, that is of interest and not the apparent color. It is known that solutions of potassium chloroplatinate tinted with small amounts of cobalt chloride give colors similar to the colors of natural waters. The color produced by 1 mg/ litre (1 ppm) of platinum (used as K2 Pt Cl6) is taken as the standard unit of color. The color determination is done only after the removal of suspended matter by centrifugation. The color standards are usually prepared by dissolving 1.2545 g of K2PtCl6 (containing 0.5 g of Pt) and 1 g of crystallized cobaltious chloride (CoCl2. 6H2O) containing about 0.248 g of cobalt in water with 100 ml of concentrated HCl and diluting to 1 liter with distilled water. This solution is deemed to possess 500 units of color, as per the American Public Health Association’s books of standard methods of water analysis. Removal or reduction of color and organic matter is generally accomplished by coagulation, settling, adsorption, filtration and sometimes super chlorination.

4.2 Tastes and Odours Most of the odours in natural waters, with the exception of H2S, and iron are organic in nature. The odours and tastes observed in chlorinated waters are due to compounds formed by the reaction of chlorine on traces of organic matter present in the water. These organic tastes and odours are usually confined to surface waters and are either very low or totally absent in deep well waters.

Water Treatment  6.5 Disagreeable odours and tastes are objectionable for various industrial processes such as beverages, food products. paper, pulp and textiles. Organic tastes and odours may be removed by means of activated carbon, aeration, or aeration followed by activated carbon treatment. The removal of inorganic odours and tastes due to H2S or iron will have to be removed by chemical methods like oxidation, chlorination or precipitation.

4.3 Turbidity and Sediment “Turbidity” is imparted to natural waters due to the presence of finely divided insoluble impurities which remain suspended in water and reduce its clarity. These suspended impurities may be inorganic in nature (e.g., clay, silt, silica, ferric hydroxide, calcium carbonate, sulphur, etc.) or organic (e.g., finely divided vegetable or animal matter, oils. fats, greases. micro-organisms, etc.). Since the different materials that causes turbidity in natural water, an arbitrary standard is used 1 mg SiO2 /L = 1 unit of turbidity. Standard suspensions of pure silica are used for measuring turbidity. However, the actual standards of turbidity are determined from the depths of liquid in a standard Jackson candle turbiditimeter, in which the flame of a candle disappears when viewed lengthwise through the tube. For instance, the depths of 72.9 cm, 2l.5 cm, 4.5 cm and 2.3 cm correspond to the turbidities of 25 ppm, 100 ppm, 500 ppm and 1000 ppm respectively. Nowadays, turbidity is conveniently measured with the help of more reliable, sensitive instrument photometers. A beam of light from a source produced by a standardized electric bulb is passed through a sample. The light emerging from the sample is then directed through a photometer which measures the light absorbed. The reading on the meter is calibrated in terms of turbidity. Formazin polymer suspension has now mostly replaced silica suspension as the standard because it provides more reproducible results. Therefore, the turbidity is now also expressed as “formazin turbidity units” (FTU). Another method used for measurement of turbidity is by light scattering. The light falling on the sample is scattered because of the turbidity present. The scattered light is then measured by putting a photometer at right angles to the path of the incident light generated by the light source. This technique of measurement of light scattered at an angle of 90° is called “nephelometry”. Accordingly the unit of turbidity is “nephelometric turbidity unit” (NTU). Tolerances of turbidity for different industries depend upon the type of industry, the grade of the product being manufactured, the nature of the turbidity present and the type of wet processing being practiced. Suspended silt and mud which cause turbidity may be objectionable in boilers and in cooling-water systems. Colloidal or dissolved organic matter causing turbidity may interfere with water-softening processes. For example, the zeolites and cation-exchangers used in water-softening processes may be coated with coagulated organic or suspended matter which leads to reduction in their efficiency. Turbidity of water may be removed by sedimentation followed by (a) coagulation and filtration. (b) coagulation and settling, or (c) coagulation, settling and filtration.

4.4  Micro-organisms Micro-organisms are more abundant in surface waters (since these come into contact with air, soil and vegetation in which the organisms originally existed), whereas in deep well waters the bacterial count is often low or even absent. The growth of these organisms in water used for industrial purposes may cause serious problems and hence effective measures have to be taken to prevent the growth

6.6  Engineering Chemistry of these organisms. Organic growths generally take place most readily in water at temperatures ranging from 10°C–35°C. Many of them form coatings in pipe lines, thus reducing their carrying capacity considerably. These coatings frequently break loose in large masses which may completely block the flow through valves, pumps, nozzles and other parts of the water distribution systems. In filters and water softeners employing granular media, the granules may become matted together by such organic growths, thus impairing their operation by lowering their flow rates and also resulting in channeling and overturning of the beds. The commonest types of living organisms that are important from the point of view of treatment are algae, fungi and bacteria, all of which often form slime with consequent fouling and corrosion. The slime surrounding the organisms causes them to adhere to metal surfaces. This leads to difficulties such as reduced heat transfers and tube blockages. The growth of marine organisms, particularly mussels, in sea-water systems may lead to serious reduction in the carrying capacity of the pipe lines. Control of algal, fungal and bacterial growths is usually achieved by chlorination. Water soluble solid sterilizing agents such as CuSO4, sodium pentachlorophenate and organic mercurials are used in special circumstances. The growth of mussels may be prevented by chlorination. Algae and other chlorophyll-containing plant need sunlight for their growth. Hence, the growth of these organisms can be prevented by storing the water in covered reservoirs. In many industrial plants, the organic growths are removed in the settling basin of the water-treatment plant. This is usually done with chlorine and coagulation, settling and filtration, to remove the remains. The use of chlorine in this way is called pre-chlorination, which is helpful in reducing the dosages of coagulant required. In many other cases, the bulk of the organic matter is first removed by coagulation, settling and filtration followed by chlorination. This method is called post-chlorination. Sometimes, both pre-chlorination and post-chlorination are used. Iron and manganese bacterial growths, known as “Crenothrix” are best prevented by removal of these metals, followed by chlorination. In the case of sulphur water, the H2S should be first removed followed by chlorination to remove the last traces of H2S and to kill any sulphate-reducing bacteria which may be present.

4.5  Dissolved mineral matter For most of the industrial uses, only the following mineral constituents are usually determined: Ca, Mg, Na, K, bicarbonate, carbonate, hydroxide, chloride, sulphate, nitrate, fluoride, silica, Fe, Mn and mineral acid. The most important quality of the dissolved mineral matter from the point of industrial application include hardness and alkalinity. Hardness Hardness was originally defined as the soap consuming capacity of a water sample. Soaps generally consists of the sodium salts of long-chain fatty acids such as oleic acid, palmetic acid and stearic acid. The soap consuming capacity of water is mainly due to the presence of calcium and magnesium ions. These ions react with the sodium salts of long-chain fatty acids present in the soap to form insoluble scums of calcium and magnesium soaps which do not possess any detergent value (cleaning tendency). 2CI7H35COONa + CaCl2

Soap (soluble)

(C17H35COO)2CaØ +2 NaCl Calcium soap (insoluble ppt.)

Water Treatment  6.7 Other metal ions like Fe +2, Mn2+ and Al3+ also react with the soap in the same fashion, thus contributing to hardness but generally, these are present in natural waters only in traces. Further, acids such as carbonic acid can also cause free fatty acid to separate from soap solution and thus contribute to hardness. However, in practice, the hardness of a water sample is usually taken as a measure of its Ca2+ and Mg2+ content. Temporary and permanent hardness When natural water is boiled, the bicarbonate ions present are decomposed to form carbonate ions and carbon dioxide is set free. The hardness so precipitated was referred to as “temporary hardness”, but this term is now referred to all the hardness associated with the bicarbonate content of the water (i.e., that determinable by titration with acid). This is due to the fact that CaCO3 and more particularly MgCO3 have appreciable, though slight, solubility in water. Ca(HCO3 )2 Mg(HCO3)2

D

CaCO3 + H20+ CO2 Mg(OH)2 +H2O+CO2 (Insoluble)

The difference between the temporary and total hardness is referred to as “permanent hardness”, since this is not removed by boiling the water. The permanent hardness is regarded as comprising of the dissolved chlorides, sulphates and nitrates of calcium and magnesium. Alkaline and non-alkaline hardness The terms, temporary and permanent hardness, are gradually being replaced by the preferred terms alkaline and non-alkaline hardness. “Alkaline hardness” is defined as the hardness due to the bicarbonates, carbonates and hydroxides of the hardness-producing metals. It is also called “carbonate hardness”. In raw water, the alkaline hardness is almost always the hardness associated with the bicarbonates. However, a treated or boiler water may also contain hardness due to small quantities of CaCO3 and Mg (OH)2 in solution. The alkalinity as measured by titration with mineral acid using methyl orange as indicator is equal to the sum of the concentrations of the bicarbonates, carbonate and hydroxide expressed in equivalents. If this alkalinity is less than the total hardness also expressed in equivalents, then the alkaline hardness is equal to the alkalinity. Conversely, when the alkalinity to methyl orange is equal to or greater than the total hardness, the alkaline hardness is equal to the total hardness”. The “non-alkaline hardness” is obtained by subtracting the “alkaline hardness” from the “total hardness”. This is also known as “non-carbonate hardness”. Estimation of hardness Hardness is usually determined by the following two methods: (1) Soap solution method Soluble soaps consist of sodium or potassium salts of higher fatty acids, such as oleic acid, stearic acid and palmitic acid. These soaps give lather with hard water only after sufficient quantity of the soap is added to“precipitate all the hardness causing metal ions present in water”. Ca or Mg (HCO3)2 + 2Cl7 H35COONa



Calcium or maginesium bicarbonate



Sodium stearate (soluble soap with detergent value)

(2Cl7 H35COO)2 Ca or Mg + 2NaHCO3 Ca or Mg stearate (insoluble soap with no detergent value)

6.8  Engineering Chemistry CaCl2 or MgCl2 + 2Cl7 H35COONa CaSO4 or MgSO4 + 2Cl7 H35COONa

(2Cl7 H35COO)2 Ca or Mg + 2NaCl (2Cl7 H35COO)2 Ca or Mg + Na2SO4

Thus after precipitation of all the hardness causing metal ions present in the hard water sample further addition of soap gives lather. The total hardness of a water sample can be determined by titrating an aliquot of the sample against a standard soap solution in alcohol. The appearance of a stable lather persisting even after shaking for about 2 minutes marks end-point. If the water sample is boiled for 30 minutes to remove the temporary hardness and then is titrated with the standard soap solution as described above, the titre value corresponds to the permanent hardness of the sample. The difference between the two measurements corresponds to the temporary hardness. (2) EDTA method This method gives more accurate results than the soap solution method. HOOC H2C

CH2COOH N–CH2–CH2–N CH2COOH

HOOC H2C

Fig. 1  Structure of EDTA

EDTA can be represented by H4Y. H4Y from is not used because of its limited solubility, again Na4Y form also not used because of its extensive hydrolysis in solution which make the solution highly alkaline Na2H2Y form is mostly used in analytical work since it can be obtained in high state of purity. NaOOCH2C

CH2COONa N–CH2–CH2–N CH3COOH

HOOCH2Cl

Ethylene diamine tetra acetic acid (EDTA) (Fig. 1) forms complexes with Ca2+ and Mg2+, as well as with many other metal cations, in aqueous solution. These complexes have 3 general formula given in Fig. 2 below: M

CHzCOO N

CH2

OOCH2 C N

CH2 –

–

OOCH2 C

CH2COO

Fig. 2  EDTA complex with a divalent metal cation, M2+. such as Ca2+, Mg2+. etc.

Thus, in a hard water sample, the total hardness can be determined by titrating the Ca2+ and Mg2+ present in an aliquot of the sample with Na2EDTA solution, using NH4Cl-NH4OH buffer solution of pH 10 and, Eriochrome Black-T as the metal indicator. The color change at the end-point is from wine red to blue. Na2H2y Æ 2Na+ + H2y–

disodium EDTA solution

Water Treatment  6.9 Mg+2 + HO2– Æ MgO – + H+ Metal indicator (Indicator blue)



(Colorless) –

Mg or (Ca)O + H2y Æ (Mg or Ca)y– + HD – + H+­ Metal From metal Free



indicator complex (wine red color)

–

disodium EDTA solution

EDTA Complex (colorless)

indicator (blue color)

Permanent hardness can be determined by precipitating the temporary hardness by prolonged boiling for about 30 minutes followed by titration with the Na2EDTA solution as above. The difference in the titre values corresponds to the temporary hardness of the water sample. Units of Hardness 1. Parts per million (ppm):  One part per million (ppm) is a unit weight of solute per million unit weights of solution. In dilute solutions of density = l, 1 ppm = l mg/liter. It is customary to express hardness in terms of equivalents of CaCO3. Hence, all the hardness causing impurities are first converted in terms of their respective weights equivalent to CaCO3 and the sum total of the same is expressed in parts per million. Equivalent of   weight of the substance causing hardness × 50 CaCO3 for a =​  ___________________________________________              ​ Equivalent weight of the substance causing hardness hardness causing substance

(Since chemical equivalent weight of CaCO3 = 50)

For instance, 136 parts by weight of CaSO4 would react with the same amount of soap as 100 parts by weight of CaCO3 (i.e.. 2 equivalents of CaCO3). Hence, in order to convert the weight of CaSO4 present as its CaCO3 equivalent, the weight of CaSO4 should be multiplied by a factor 100 50 of ____ ​   ​ or ___ ​   ​  . Similarly, if ‘a’ gms of CaCl2, ‘b’ gms of MgSO4, ‘c’ gms of Mg Cl2 ‘d’ gms of 136 68 Ca (HCO3)2 and ‘e’ gms of Mg (HCO3)2 are present in a hard water sample, each of them can be 100 100 ____ 100 100 100 converted in terms of their weight equivalent of O3 by multiplying with ____ ​   ​, ____ ​   ​ ​   ​ ____ ​   ​ and ____ ​   ​  111 120 95 162 146 respectively. The sum total of the values so obtained expressed as parts per million represents the hardness of the water sample under consideration.

100g CaCO3 = = = = = = = =

111 g CaCl2 120 g MgSO4 95 g MgCl2 162 g Ca (HCO3)2 146 g Mg (HCO3)2 148 g Mg (NO3)2 44 g CO2 136 g CaSO4

6.10  Engineering Chemistry 2. Equivalents per million (epm):  One equivalent per million is a unit chemical equivalent weight of solute per million weight units of solution. In dilute solutions of density not differing very much from unity, l epm = l milligram equivalent per litre; and in titrimetry, epm is conventionally taken as equal to 1 ml of l N solution per litre. Thus, 1 epm of Mg = 12 ppm of Mg = 50 ppm of CaCO3 = 42 ppm of MgCO3 = 73 ppm Mg (HCO3)2 = 81 ppm Ca (HCO3)2 = 68 ppm CaSO4 = 47.5 ppm MgCl2 = 55.5 ppm CaCl2 = 60 ppm MgSO4 and so on. It should be noted that for any dissolved substance, a concentration of l epm is equal to “50 ppm as CaCO3”. 3. Grains per imperial gallon (gpg):  In English system, hardness is expressed in terms of grains​ 1 l  grain = ​ _____    ​  lb  ​ per gallon (10 lbs); i.e. parts per 70,000 parts per gallon is also called as degree 7,000 Clark. Thus, 9 degrees Clark means that 9 grains in terms of CaCO3 are present per gallon of water; 106 or 9 parts are present per 70,000 parts of water. On ppm scale, it means that 9 × ______ ​    ​  = 128.57 70,000 ppm of hardness (as CaCO3) is present in the water sample.

( 

)

Inter-relationship between various units of hardness Thus, 1 epm of Mg = 12 ppm of Mg 1 ppm. = 1 mg/l = 01° French = 0.07° Clark = 0.07 grains per imperial gallon = 0.0583 grains per US. gallon = 0.02 epm as CaCO3 1° Clark = 14.3 ppm = l.43° French = 1 grain per imperial gallon = 0.833 grains per US. gallon 1 grain per US. gallon = 17.1 ppm = 17.1 mg/l = 1.2 grains per imperial gallon 1° Fr = 10 ppm = 10 mg/l = 07° Clark 1° Russian = 1 part Cal 10° parts of water 1° German = 1 part Cal 105 parts water. 10 ppm as CaO or 17.9 ppm as CaCO3. Water containing less than 150 ppm of hardness is classified generally as “good”. Those containing 150 to 300 ppm as fair and those exceeding 300 ppm as “bad” as per Bureau of Indian standard BIS. Alkalinity:  By alkalinity of water we mean the total content of those substances in water that cause an increased concentration of OH– ions upon dissociation or due to hydrolysis. The alkalinity of natural waters is generally due to the presence of HCO3–, SiO32–, HSiO3– and sometimes CO32– ions

Water Treatment  6.11 and also due to the presence of salts of some weak organic acids, known as humates, that bind H + ions as a result of hydrolysis, thereby increasing the concentration of OH– ions. In addition to the above, the alkalinity of boiler water is also conditioned by the presence of PO43– and OH– ions. Also, the presence of salts of weak acids such as silicates and borates induces buffer capacity in water and resists the lowering of pH. Surface waters containing algae and also water treated by lime-soda process may contain considerable quantities of alkalinity due to CO32– and OH–. Depending on the anion that is present in water (HCO3–, CO32– or OH–), alkalinity is classified respectively as bicarbonate alkalinity, carbonate alkalinity or hydroxide alkalinity. The maximum contaminant level for alkalinity is 200 ppm for domestic purpose as per BIS. Highly alkaline waters may lead to caustic embrittlement and also may cause deposition of precipitates and sludge’s in boiler tubes and pipes. With respect to the constituents causing alkalinity in natural waters, the following situations may arise: l. Hydroxides only 2. Carbonates only 3. Bicarbonates only 4. Hydroxides and carbonates 5. Carbonates and bicarbonates (Notes:  The possibility of hydroxides and bicarbonates existing together is ruled out because of the fact that they combine with each other as follows forming the carbonates:

OH– + HCO3– = CO3– + H2O

The types and extent of alkalinity present in a water sample may be conveniently determined by titrating an aliquot of the sample with a standard acid to phenolphthalein end-point, P, and continuing the titration to methyl orange end-point M. The reactions taking place may be represented by the following equations: OH– + H+ H2O ...(l) – + – CO3 + H HCO3 ...(2) HCO3– + H+ H2CO3 H2O + CO2 ...(3) The volume of acid run-down upto phenolphthalein end-point P corresponds to the completion of equations (1) and (2) given above, while the volume of acid run-down after P corresponds to the completion of equation (3). The total amount of acid used from beginning of the experiment, i.e., the methyl orange end-point M, corresponds to the total alkalinity present which represents the completion of reactions (1) to (3). The results may be summarized in the following Table 2, from which the amounts of hydroxides, carbonates and bicarbonates present in the water sample may be computed:

Table 2 Re su lt s of t it r a t ion of Hydroxide [OH–] phenalphthalein end point, P and methyl range end point M.

Carbonate [CO3–2] Bicarbonate [HCO3] –

P=O

Nil

Nil

M

P=M

M

Nil

Nil

6.12  Engineering Chemistry P = V2M

Nil

2P

Nil

P > V2M

[2P-M]

[2M-P]

Nil

P < V2M

Nil

2P

[M-20]

Alkalinity is generally expressed as parts per million (ppm) in terms of CaCO3. 1000 ml of N/50 acid solution a 1000 mg of CaCO3. Hence Vol.of N/ 50 acid ×  1000 Alkalinity = ​  ____________________________________             ​ Vol.of sample taken for titration mg or ppm On the basis of the analysis of water with respect of alkalinity and total hardness, the amounts of carbonate hardness (temporary hardness or alkaline hardness) and non-carbonate hardness (nonalkaline hardness or permanent hardness) present in the water can be determined as follows: 1. If the methyl orange alkalinity of the water equals or exceeds the total hardness, all the hardness is present as carbonate hardness. 2. If the methyl orange alkalinity of water is less than the total hardness, the carbonate hardness equals the alkalinity. 3. The non-carbonate hardness, under conditions in (2) above, is equal to the total hardness minus the methyl organe alkalinity. Further, on the basis of hardness and alkalinity, natural waters can be subdivided into two groups: Non-alkaline and alkaline. If the hardness is greater than alkalinity, the water is called non-alkaline. If the hardness of the water is lesser than alkalinity, the water is characterised as alkaline. Nonalkaline waters are more frequently encountered in nature, which are characterized by different kinds of hardness as follows: Ht = (Hc+Hnc) = (HCa+HMg) where Ht = total hardness, Hc = carbonate hardness Hnc = non-carbonate hardness, HCa = calcium hardness and HMg = magnesium hardness.

Chlorides Chlorides are present in water generally as NaCl, MgCl2 and CaCl2. Although chlorides are not considered as harmful as such, their concentrations over 250 mg/L impart peculiar taste to the water which is objectionable or unacceptable for drinking purposes for most people from aesthetic point of view. Hence the secondary standard for chlorides is 250 mg/L. Further, presence of unusually high concentrations of chloride in water generally indicates pollution from domestic sewage or from industrial wastewaters, presence of chlorides is also undesirable in boiler feed water. Salts like MgCl2 may undergo hydrolysis under the high pressure and temperature prevailing in the boiler, generating hydrochloric acid which causes corrosion of boiler parts.

Sulphates Sulphates are among the major anions present in natural water. When sulphates are present in excessive amounts in drinking water, they may produce a laxative or cathartic effect on the people consuming such water. The secondary maximum contaminant level (SMCL) for sulphates is 250 mg/L.

Water Treatment  6.13

Nitrates Excessive concentrations of nitrates are objectionable particularly for infants. The maximum contaminant level (MCL) for nitrates is 10 mg/L. In agricultural regions, ground water can have significant concentrations of nitrates from unused fertilizer leaching into the underlying aquifers. Surface waters can be polluted by nitrates both from discharge of municipal wastewater and from drainage from agricultural lands. Ingestion of excessive nitrates in drinking water by infants causes a disease known as “methemoglobinemia” (infant cyanosis or blue baby syndrome). In the intestines of infants (particularly those below 6 months of age), nitrates can be reduced to nitrites, which are absorbed into the blood, oxidizing the iron present in the blood, thereby resulting in cyanosis which causes a blue color to the baby. That is why, this condition is known as “Blue-baby syndrome”. Infant methemoglobinemia can be readily diagnosed by medical doctors and is treated readily by injecting methylene blue into the infant’s blood. Nitrates can be effectively removed from water with the help of strongly basic anion exchange resins. However, high operating cost and the disposal of huge quantity of waste brine from regeneration of the resin pose major limitations for this treatment process.

Fluorides Flouride is found in groundwater as a result of dissolution from geologic formations. It is particularly found in ground waters that come into contact with fluoride containing minerals such as fluorspar (CaF2), fluorapatite [Ca10F2 (PO4) 6], cryolite (Na3AIF6) and igneous rocks containing fluosilicates. Surface waters generally contain much smaller concentrations of fluoride, unless they are contaminated otherwise. Water pollution by fluorides may be caused by the contaminated domestic sewage and the run-off from agricultural lands where phosphatic fertilizers have been used. Phosphatic fertilizers may contain 0.5 to 4% of fluorine by weight as an impurity. The fluoride concentration found in some water samples is 1.5 to 6 mg/L, and in extreme cases, it may be as high as 16 to 36 mg/L. Unsightly fluorosis may be caused when the fluoride level in drinking water exceeds 4 mg/L. Optimum fluoride concentrations prescribed in public water supplies generally are in the range of 0.7 to 1.2 mg/L, depending on the annual average of maximum daily air temperature of the place, on which the consumption of the water by the people depends. Beneficial health effects have been observed where the fluoride levels are optimum. However, too low or too high concentrations of fluoride in drinking ‘water are problematic, and such situations may have to be tackled by fluoridation (addition of fluoride) or defluoridation (removal of fluoride) respectively. Absence or low concentration of fluoride in drinking water causes a high incidence of dental caries, particularly in children. On the other hand, excessive concentration of fluoride in drinking water causes “fluorosis” which is manifested in mottling of teeth, discoloration and at times, chipping of teeth.

4.6  Dissolved gases The dissolved gases occurring in various water supplies are: (i) Carbon dioxide (CO2). (ii) Oxygen (O2). (iii) Nitrogen (N2)

6.14  Engineering Chemistry (iv) Hydrogen Sulphide (H2S), and (v) Methane (CH4). (i) Carbon dioxide:  Varying amounts of free CO2 are present in natural waters. The amount of CO2 picked up by rain water from the atmosphere is very small and ranges from 0.5 to 2 ppm. Most of the surface waters, if sampled at their surfaces, contain CO2 in the range of 0 to 5 ppm. In lake waters, the CO2 concentration in the surface samples range from 0 to 2 ppm and it increases with the depth. This is due to the fact that CO2 is generated at the bottom of the lake, due to the decay of organic matter, while in the upper layers, microscopic plants consume CO2 for photosynthesis and give out oxygen. Rivers containing considerable organic matter showed as much as 50 ppm of CO2 content. Rivers receiving acid mine waters or acid wastes may also show high CO2 content. Ground waters usually contain 1 to 50 ppm CO2 while shallow wells, in areas overlain by peaty soils, may show 50 to over 300 ppm of CO2. Phenolphthalein alkalinity in natural waters Freshly sampled natural waters usually contain some free CO2. Some of them sometimes show an appreciable phenolphthalein alkalinity due to photosynthesis of the large or microscopic plants present, which under the influence of sunlight, breathe-in CO2 and breathe out oxygen. This process may continue even after the free CO2 supply gets exhausted by deriving the half-bound CO: content from the bicarbonates present, thus forming the normal carbonate and imparting phenolphthalein alkalinity to the water. Effect of CO2 on pH values When CO2 dissolves in water, it forms weakly dissociated carbonic acid, H2CO3 to some extent. If the water is free from all traces of alkali and is saturated with CO2 (about 1450 ppm at 77°F), then the pH is about 3.8. Such a low pH would not be found in natural waters. (Except for waters containing free mineral acidity) because of the presence of some bicarbonate alkalinity. (Pure distilled water in equilibrium with the CO2 content of the atmosphere will have a pH of about 5.7. However, even the slightest trace of alkalinity raises the pH, so that most distilled waters in glass vessels show a pH around 6.4). If the water contains bicarbonate alkalinity also, then its pH value depends not only on the free CO2 content, but instead, depends on the ratio of free CO2 to the methyl orange alkalinity of the water. Hence, the pH value of any water having a free CO2 may be calculated from the 5 relative proportions of the free CO2, and its bicarbonate alkalinity. Similarly, if the pH of the 5 water and its bicarbonate alkalinity are known, the free CO2 content may be calculated. Dissolved carbon dioxide, if present in boiler-feed water, produces an acid solution which leads to a general corrosive attack on the metal. Further. CO2 is also an accelerating factor in dissolvedoxygen corrosion. Hence water which contains, in addition to dissolved O2, a high CO2 content in relation to its alkalinity will be much more corrosive than water containing a low CO2 content in relation to its alkalinity. In other words, water having a dissolved O2 content is much more corrosive, if it has a low pH than if it has a higher pH. Carbon dioxide in water may be removed or reduced by means of an aerator, degasifier or vacuum deaerator. Carbon dioxide in water may also be neutralized by the addition of lime or an alkali such as caustic soda, but these procedures are limited only for raw or treated waters containing relatively small amounts of CO2. CO2 may also be partially removed from water by filtration through a neutralizing filter, employing a bed of granulated calcite. Some of this dissolves in the water forming

Water Treatment  6.15 Ca (HCO3)2 while bringing the pH up to about 7.2. Such filters are widely used in the household field of water treatment and to some extent in industry also. (ii) Oxygen:  The solubility of pure oxygen at 32°F and atmospheric pressure is 48.89 ml per litre; while under the same conditions, the solubility of nitrogen is 23.54 ml per litre. Consequently, when air is dissolved in water, the two main components, namely N2 and O2, exist in different proportions in solution than that existing in the atmosphere. According to Henry’s law, the solubility of a gas is proportional to the absolute pressure. Thus, if pressure is increased, the amount of air or O2, which can be held in solution at a given temperature, is also increased proportionally. Dissolved oxygen is very corrosive to metals like iron, steel, galvanized iron and brass, which are widely used for making vessels for conducting and holding water. Low pH values and elevated temperatures accelerate the rate of this corrosion. Inhibition or reduction of dissolved oxygen corrosion could be achieved by deaeration of the boiler feed water, vacuum deaeration, sodium sulphite treatment, treatment with sodium silicate plus caustic soda, chromate treatment and by cathodic protection. Dissolve oxygen is needed for living organisms to maintain their biological process. As per BIS domestic water must have 4-6 ppm dissolve oxygen and 0.01-0.05 ppm for boiler water. (iii) Nitrogen:  Nitrogen is a rather inert gas which has no corrosive effects on metals. In analyzing waters, nitrogen is practically never determined, since it is inert and is relatively unimportant as far as water treatment is concerned. (iv) Hydrogen sulphide:  Waters which contain sulphides are known as “sulphur waters” which are characterized by offensive odours and their marked corrosiveness. Most sulphur waters are ground waters. H2S, like CO2, when dissolved in water is feebly ionized. Oxidation of sulphides by dissolved oxygen is apparently a rather slow process. Chlorine may also be used to oxidize H2S, but the process is rather expensive on raw “sulphur waters” because it takes eight atoms of chlorine to oxidize one molecule of H2S to form a sulphate. H2S + 4Cl2+ 4H2O

H2SO4 + 8HCI

(v) Methane:  In general, well waters containing methane have been found in the glacial drift and in oil and gas well areas. While methane itself is apparently unobjectionable in drinking water, it would be advisable to aerate the water for either industrial or household use, so as to eliminate fire and explosion hazard.

4.7  Silica Content Silica content means the concentration of silicic acid, H2SiO3 (expressed as SiO2) present in water. The presence of SiO2 in boiler feed water, especially the high pressure boilers, gives rise to many difficulties in operating the power generation equipment because of the formation of silicate scales of low thermal conductivity and similar heavy deposits on the blades and nozzles of turbines. This is the reason why the water-treatment technology always includes a process ensuring partial or total desilication of raw water. In modern practice, silica is removed from raw water by using ionite exchangers. The concentration of SiO2 in natural water varies over wide limits, from 5 to 90 mg/l. It decreases with increasing mineralization (salt content) of water. Natural waters contain silica acid both in an ionic (HSiO3) and in a colloidal state. This complicates desilication chemical control of feed-water for boiler units, since only ion-dispersed silica acid responds to ion-exchange treatment processes.

6.16  Engineering Chemistry

4.8  Oxidability Oxidability to some extent is a measure of the contamination of water by organic substances. It is generally expressed in milligrams of oxygen required to oxidize under given conditions, the organic substances present in 1 kg of water. However, it should be noted that oxidability does not represent the total content of organic substances in water, since under the given experimental conditions used for the determination of oxidability, complete oxidation of the organic substances does not take place. Oxidability can also be expressed by the amount of KMnO4 (mg/kg) spent to oxidize the organic substances.

5. Methods of Treatment of Water for Domestic and Industrial Purposes Municipal water supply has been one of the most challenging problems of water technology. Water supplied by municipalities for domestic purposes must be free from pathogenic bacteria. It should be clear, colorless and pleasant to taste. It should be free from excessive dissolved salts, suspended impurities and harmful microorganisms. Water for domestic purposes should be obtained from such a source which is least contaminated by animal and vegetable matter as well as industrial effluents. Rivers, lakes and wells are the most common sources of water used by municipalities. Generally, the treatment of these waters involves removal of suspended impurities and removal of colloidal impurities if any, followed by sterilization. If the water is very hard, certain amount of softening may be needed, which is very rare. Sedimentation, coagulation, filtration and sterilization are the treatment techniques usually employed depending on the requirements of the situation.

5.1  Sedimentation Sedimentation is a process of removing relatively large particles (suspended solids) into large reservoirs of settlement tanks in which it is left for a few days or even weeks, where the suspended impurities partially sink to the bottom. The principle involved is to slow down the flow of water so that substances held up by the turbulence of fast moving water can fall gravitationally to the bottom of the tank when water flow is stilled. Periodically the accumulations of the debris are to be scraped away. In order to remove floating impurities screens of various kinds (eg., Bar screen, Band and drum screens and microstrainers) are employed. These screens also must be continuously cleaned. The rate of settling in still water at 10°C is known as the hydraulic settling value of a particle and is generally expressed in millimeters/second. During sedimentation, solid particles settle by gravity on the bottom of a settling tank in which the water being clarified is at rest or in slow horizontal or upward motion. The velocity with which a particle in water will fall under the action of gravity depends upon (i) the horizontal flow velocity of the water, (ii) the size of the particle, (iii) the specific gravity of the particles (iv) the shape of the particle and (v) the temperature of the water. Accordingly several formulae have been given to calculate the velocity of falling spherical particles in slowly moving water on the basis of which several types of sedimentation tanks have been designed. The sedimentation tanks commonly used are of horizontal flow rectangular type and circular shaped upward flow type. Sedimentation takes a long time, requires large-capacity settling tanks and cannot ensure complete removal of coarse-dispersed impurities from water. Plain sedimentation usually removes only 70 to 75% of the suspended matter.

Water Treatment  6.17

5.2  Coagulation Finely divided silica, clay and organic matter do not settle down easily and hence cannot be removed by simple sedimentation. Most of these are in colloidal form (e.g., sols, gels or emulsions) and are generally negatively charged and hence do not coalesce because of mutual repulsion. Such impurities are generally removed by chemically assisted sedimentation, in which certain chemicals are added which produce ions of right electrical charge that neutralize the oppositely charged calloidal particles and bring about their coalescence. This process is called coagulation. This permits the particles to aggregate together until a denser particle is formed which falls through still water at a reasonable rate and is called flocculation. Aluminium sulphate is the most common coagulating agent used for removing clay particles and is generally called filter alum. Other coagulants which also find application in water treatment include ferric sulphate, ferrous sulphate (copperas) chlorinated copperas, alum, ammonia alum or potash alum and sodium aluminate. Aluminium sulphate, when added to natural waters, hydrolyses to form colloidal aluminium hydroxide and an equivalent quantity of sulphuric acid as follows:

Al2 (SO4)3+6H2O

2Al(OH)3+3H2SO4

The Al(OH)3 so formed acts as a floc or coagulant, which has an enormous surface area per unit volume and removes the finely divided and colloidal impurities by neutralizing the charge on them as well as by other mechanisms like adsorption and mechanical entrainment. Thus the smaller particles join together to form denser particles which settle down to the bottom. Some bacteria and color associated with these particles also get removed simultaneously. In order to render the Al(OH)3 filterable and also in order to neutralize the H2SO4 liberated to permit the hydrolysis reaction to completion, some alkali will have to be added if the water is not sufficiently alkaline. The reaction taking place in alkaline waters may be represented as follows:

Al2 (SO4)3 + 3Ca(HCO3)2

2Al(OH)3 + 3CaSO4 + 6CO2

With water having a little or no natural alkalinity (e.g., moorland waters) an alkali such as calcium hydroxide or sodium carbonate is added; the latter is more commonly used as it does not increase the hardness of the water.

Al2 (SO4)+3Na2CO3+3H2O

2Al(OH)3 + 3Na2SO4 + 3CO2

For treatment of acidic waters, sodium aluminate can be used as a source of Al(OH)3 and it is often used in conjunction with aluminium sulphate. The reactions taking place may be represented as under: Na AIO2 + 2H2O Al2 (SO4)3, + 6Na AlO2 + 12H2O

NaOH + Al(OH)3, 8Al(OH)3 + 3Na2SO4

Thus the alkalinity due to aluminate is neutralized by the acidity of the sulphate and Al(OH)3 is thus precipitated from both the reagents. for treatment of alkaline water FeSO4 can be used

FeSO4 + H2O Fe(OH)2 + 1/2 O2

Fe(OH)2 + H2SO4 Fe(OH)3

In order to increase the efficiency of the coagulation process, coagulant aids such as lime, Fuller’s earth, bentonite clay and polyelectrolytes are also added.

6.18  Engineering Chemistry The coagulants are generally added in solution form to the water with the help of mechanical flocculators provided with slow-moving rotating baffles or stationary baffles turning the flow of water: thus ensuring a gentle contact with the water and the reagents. After any sedimentation process, especially after that utilizing a chemical flocculent, there will be a substantial reduction in the bacterial count in the water. In addition, many coagulants release oxygen to the water as the chemical transformations take place. This oxygen helps in destroying some bacteria, in breaking up some organic compounds and also in partial removal of color and taste producing organisms present in water. Thus the process of ‘x coagulation, flocculation and sedimentation has many beneficial results and is one of the most important processes of purification of water. Coagulation and settling equipment should be so designed that quick and thorough mixing of the coagulant and raw water should be achieved so that the coagulant dosage required is minimized. For this purpose, mechanical type mixers and baffled mixing troughs are widely used. Modern types of coagulation and settling equipment include the floc-forrner type and the sludge-blanket type. The floc-former type coagulation and settling equipment (Fig. 3) consists of (i) a flash mixer where the coagulant is quickly and efficiently mixed with the raw water with the help of a rotating stirrer settling (ii) rolling mix chambers where only a basin gentle mixing of the gelatinous precipitate formed to aid in the even distribution and formation of a large-particled, easily settled-floc, and (iii) settling basins where the floc is allowed to settle and the sludge thus formed is removed continuously or intermittently.

Fig. 3  Floc-former type Coagulation and settling equipment.

Water Treatment  6.19 The sludge-blanket type of water treatment equipment came into extensive use for coagulation and settling well as water. It majorly differs from the floc former type of equipment in that it filters. All of the coagulated water upwardly through a suspended sludge blanket instead of merely dropping it to the bottom by gravity. This prolonged intimate contact with the floc helps in fully utilizing its absorbing capacity which effects savings in the amount of coagulant or other adsorbents (such as activated carbon used for removing colors and odours) required. The equipment is made in both vertical and horizontal designs. For certain industrial uses where turbidities upto 10 ppm are permissible and where the raw water is turbid but not highly colored the effluent from the sludge-blanket type of coagulation and settling equipment may be used directly. However, if color removal and a greater degree of clarity are required, filtration is also necessary.

5.3  Filtration Filtration is a process of clarification of water by passing the water through a porous material, which is capable of retaining coarse impurities on its surface and in the pores. The porous material used is called the filtering medium and the equipment used for filtration is known as a filter. Filtration of water takes place due to the difference between the pressures at the of the bed of the filtering material and that underneath it. The difference in the two pressures, Dh is called the pressure drop through the filtering medium, but also the drop in pressure-through the filter proper (in the pipelines, distribution devices, etc.). This pressure drop through the filtering bed depends on the rate of filtration, height of the filtering bed, diameter (size) of the grains of the filtering material and the extent of its contamination by the trapped impurities. Greater the numerical values of these factors (excepting grain size), greater will be the pressure drop. When contamination or a filter cannot be tolerated any longer, it is taken out or service tor subsequent washing, in the course of which the impurities trapped in the filtering bed are washed out. Then the filter can be pressed into service again. The common materials used for the filtering medium are quartz sand (grain size 0.5 – 1.0 mm) containing not more than 96% SiO2, crushed anthracite (piece size 0.8 – 1.5 mm) and porous clay (piece size 0.8 – 1.5 mm). It should be remembered that when quartz sand filters are used with alkaline water, the filtrate (filtered water) gets enriched with silicic acid (due to the solubility of quartz sand in alkaline water). Slow sand filtration Water for domestic use may be filtered through large area of finely graded sand beds at a slow rate (about 2 gal/sq. ft./hour). The rate of filtration slowly diminishes due to the accumulation of sediment in the capillaries of the filter bed, and finally the rate becomes so slow that the bed must be cleaned. Cleaning is usually done by scarping the surface of the sand bed, or excavating, washing, and then relaying the entire filtering medium. Slow gravity filters are not capable of removing colloidal impurities. These can be filtered out only when the water has been chemically treated (such as alum treatment) and the slow sand filters cannot deal with the gelatinous type of precipitates which are produced in such treatments. Typical equipment for Municipal water treatment with coagulation, settling and filtration tanks is shown in Fig. 4. Rapid-gravity filtration Rapid gravity filters are capable of producing potable waters at flow-rates as high as 100 gal/sq. ft./ hour. This is achieved by using carefully graded quartz sand and collecting the filtrate as evenly as

6.20  Engineering Chemistry water

Fine Sand Coarse Sand Water inlet

Fine Gravel Course Gravel Sludge outlet Coagulation cum Setting Tank

Water outlet

Filtration Tank Chlorination

Fig. 4  Municipal water treatment with coagulation, setting and filtration tanks

possible over the entire bottom area of the sand bed which avoids undesirable channeling. The sand bed may be cleaned either by agitation with compressed air or in small units mechanically. This is followed by a flush back with clean water to wash away the accumulated impurities. Rapid Gravity filtration has the following advantages (1) The filter bed and the quality of the filtrate can be easily inspected (2) The filter is unaffected by pressure variations on either the inflow or draw-off sides, and (3) Large reinforced concrete filters can be constructed at relatively low cost. Rapid pressure filtration Pressure filters are much more widely used than gravity filters, particularly in industrial installations. Rapid pressure filters are used preferentially where filtration is to be effected in a rising main without breaking the hydraulic head, or where water from an elevated source can be passed through the filter and delivered in storage. The operation and cleaning of these filters are more or less similar to gravity sand filters. Filtration rates are of the order of 80 to > 200 gallons/sq. ft./hour. Pressure filters are manufactured in vertical and horizontal types. These filters consist of cylindrical steel shells fitted with dashed heads, containing a layer of a granular filter medium (sand or anthrafilt) supported by graded gravel (or anthrafilt), and is equipped with the required accessories e.g., piping, under drains, valves, etc., for carrying out the cycle of operations, viz., filtration, backwashing and filtering to waste. Filter installations may consist of one or more units depending upon the requirements. Several types and designs of mechanical filters are employed (Fig. 5). The most commonly used clarifying filter is the single-flow, single-bed, closed pressure filter, which is quite simple in construction (Fig. 5(a)). The filtering medium is filled upto a certain volume “a’’ of the filter, on top of which water is filled (which is known as the water cushion). Closed pressure filters operate under the pressure created by the pumps delivering the water to be clarified. In order to increase the rate of filtration, to improve the quality of clarified water and to raise the mud capacity filtering material, single-flow filters are often charged with two filtering materials of different bulk mass and grain size, as for instance, crushed anthracite (0.8 to 1.6 mm grain size) and quartz sand (0.5 to 1.0 mm grain size). The sand, being a denser material, is arranged underneath the anthracite filtering bed. Two-bed filters of this kind are often used in practice (Fig. 5(f)). In open-type mechanical filters (Fig. 5(b)) which are widely used in purification and treatment of drinking water; filtration takes place due to the pressure exerted by the column of water, h in

Water Treatment  6.21

(a)

(e)

(b)

(c)

(f)

(g)

(d)

(h)

Fig. 5  Mechanical filters

the filter. However, the relatively small head, it restricts the use of these filters at increased filtering rates. Horizontal closed pressure filters (Fig. 5(c)) are provided with larger filtering surface and hence have larger throughout per filter. However, they require more floor space when installed indoor and cumbersome to operate. Multifold mechanical filters (Fig. 5(e)) are free from such limitations. These are filled completely with filtering material, inside which are fitted the draining and distributing devices. They permit the water to be passed in several flows (three in the illustration given) and ensuring increased output of a filter as many times. In radial flow filter (Fig. 5 (g)) the filtering material is charged into the annuus formed between the filter shell (wall) and an internal tube. The water treated is delivered into the central tube which distributes it through the filtering material height wise. Water is passed in radial flows through the filtering material. The clarifying filters, with a granular filtering bed discussed above, contain a high bed of coarse-grained filtering material and a small filtering surface. In contrast to these filters, tubularelement precoat type filters (Fig. 5(h)) contain a large filtering surface, thin filtering bed (5 to 10 mm thick), and mainly use powders as filtering media (e.g., powdery or fibrous cellulose, ionite and perlite powders). Precoat-type ionite-exchanger filters are capable of operating at high temperatures (100-110°C), and hence can be used to purify hot condensate and water. These filters are considered promising for service in high-capacity steam-generator-turbine units operated at nuclear power plants. A precoat type filter with powdery perlite (porous SiO2) as filtering material is suitable for fine clarification of water. A mixture of powdery perlite and activated carbon is used for removing oil from condensate. Clarification for domestic and industrial purposes is often achieved by a combination of two to four of the different processes as listed below: (a) Sedimentation (b) Filtration (c) Coagulation + filtration (d) Coagulation + settling (e) Coagulation + settling + filtration (f) Sedimentation + coagulation + settling + filtration (g) Chlorination and (h) Special filters (e.g., activated-carbon for removing taste and odour; manganese zeolite for removing Fe and Mn; and Neutralizing filters such as graded calcite for removing CO2).

5.4  Sterilization of Water Clarification of water by sedimentation, filtration and storage removes suspended solids and also reduces the number of bacteria in the water. Total elimination of bacteria can be achieved only by sterilization.

6.22  Engineering Chemistry The organisms which should be eliminated by disinfection are divergent in character. They include (a) the enteric bacteria belonging to Salmonella, Shigella and Vibrio groups (b) the intestinal protozoa such as Entamoeba histolytica (c) some types of worms such as Schistosomes (d) Viruses such as those of infectious hepatitis and (e) Coliform organisms which indicate water pollution, though not pathogentic. Coliforms and enteric bacteria can be easily destroyed while viruses and cysts of E. histolytica are resistant to disinfection. Chlorine is the most common sterilizing agent in water treatment. It is capable of removing B. Coli and substantially reducing other bacteria. Chlorine may be added in the form of bleaching powder, or directly as a gas or in the form of concentrated solution in water. Whatever may be the method employed, the treatment should give accurate dosage, good distribution and sufficient time of contact (~ 30 minutes) so as to ensure effective sterilization. CaOCl2 + H2O Ca (OH)2 + Cl2 C2 + H2O HOCl + HCl HOCl [O] + HCl

Hypochlorous Acid

Nascent Oxygen

The nascent oxygen so liberated destroys the germs and bacteria by oxidation. The chlorine itself, the hypochlorous acid and other chlorine compounds are also believed to have powerful germicidal properties. The OCl– ions are capable of rupturing the cell membranes of the disease producing microbes. Bleaching powder has the limitations of being unstable during storage and also it increases the calcium content of water rendering it more hard. Chlorine as a sterilizing agent has the advantages of being economical, efficient, limited space requirement and convenience. Further, it does not introduce any other impurities in the water. Both chlorine as well as bleaching power when used in excess produce disagreeable odour in the water. Too much excess may cause irritation to the mucous membranes. The unpleasant taste of this excess chlorine can be removed by treatment with ammonia, which reacts with chlorine to form the tasteless compound chloramine (NH.CI). Thus, the ammoniachlorine treatment (chloramine process) is particularly useful, where traces of impurities are present (e.g., phenols) which produce unpleasant tastes when chlorine alone is used. Further, chloramine provides a more lasting effect than that of chlorine.

Cl2 + NH3





Monochloramine

2Cl2 + NH3





NH2Cl + HCl NHCl2 + 2HCl Dichloramine

NH2CI + Cl2 NH2Cl + H2O HOCl

NHCl2 + HCl HOCl + NH3 HCl + [O] nascent oxygen

The chloramine process consists of adding ammonia to water in the form of gas together with chlorine resulting in the formation of dichloramine and monochloramine as shown above. Dichloramine is a relatively stable compound and is perhaps not a sterilizing agent by itself. However, it slowly decomposes with evolution of chlorine. Thus, the addition of ammonia stabilizes the chlorine

Water Treatment  6.23 to provide a prolonged effect. This is particularly useful when the water is passed into storage after treatment. A disadvantage of chlorination is the potential formation of trihalomethanes (such as chloroform, bromodichloromethane, dibromochloromethane and bromoform) which are carcinogenic. Trihalomethanes (THMs) may be formed when chlorine combines with natural organic substances, such as decaying vegetation, etc., that may be present in water itself. One approach to tackle this problem is to remove the organics completely before subjecting it to water chlorination. In future, the actual removal of THMs from the treated water, perhaps by aeration or adsorption on activated carbon, may become necessary. Superchlorinatlon In superchlorination, a large excess of chlorine is added to the water, thereby destroying not only the micro-organisms but also the other organic impurities present. This process ensures rapid and complete sterilization and successfully used for waters derived from wells and rivers. This process is usually followed by dechlorination by NH3 or SO2.

Residual Chlorine

A Break-point chlorination is a more precisely controlled process in which just sufficient chlorine is added to oxidize all the organic matter, destroy bacteria and react with any ammonia, leaving a slight excess of free chlorine. The break-through point, i.e., the appearance of free chlorine, in the water, must be determined experimentally as follows: If chlorine is added to a sample of water and after a few minutes, the residual chlorine available in the water is estimated, it will be found that the residual chlorine in water is less than the amount added initially. This is due to the fact that some of the chlorine added initially is consumed by oxidizing bacteria and other organic matter. Now, if we take a few more aliquots of the same volume of the water sample and add increasing doses of chlorine to different samples and analyze the residual chlorine after the same interval of time (a few minutes), a curve of the type shown below is obtained (Fig. 6).

Break Point Combined Chlorine

Free Chlorine

Chlorine dose added

Fig. 6  Break-point chlorination

It can be seen from Fig. 6 that the quantity of residual chlorine increases with increasing dose of chlorine added giving a straight line until at a definite chlorine dose; a sudden decrease in the residual chlorine is noticed. This is known as the break-point after which the residual chlorine appearing more or less agrees with the chlorine dose added. The reason for such a behavior is due to the fact that some organic compounds which defy oxidation at lower chlorine concentrations, get oxidized when the break-point chlorine concentration is reached. Since it is these organic compounds

6.24  Engineering Chemistry which are generally responsible for bad tastes and odours in water, it is obvious that break-point chlorination eliminates bad tastes and odours. Determination of free chlorine in a water sample The principle involved in the estimation of free chlorine in water is that when a measured quantity of water is treated with excess of potassium iodide, the free chlorine present in the water oxidizes the corresponding amount of potassium iodide to iodine. The liberated iodine is estimated by titrating against standard sodium thiosulphate solution using starch as indicator

Cl2 + 2KI I2 + 2Na2S2O3

2KCl + I2 Na2SO4 + 2NaI

Dechlorinatlon The water treated by the process of break-point chlorination may be filtered through activated carbon, in order to remove the decomposition products formed and the excess chlorine remaining. Other methods of dechlorination include treatment with SO2 or Na2SO3.

SO2 + Cl2 + 2H2O Na2SO3 + Cl2 + H2O

H2SO4 +2HCI Na2SO4 + 2HCl

Sterilization by Ozone Ozone is a powerful disinfectant and is readily absorbed by water. Ozone being unstable decomposes as follows giving nascent oxygen which is capable of destroying the bacteria. O3 O2 + [O] However, this process is relatively expensive but has the advantage of removing bacteria, Color, odour and taste without leaving any harmful residual effects in the water being treated. Sterilization by ultraviolet radiation Ultraviolet radiation (190-380 mm) emanating from electric mercury vapor lamp is capable of sterilizing water. This process is particularly useful for sterilizing swimming pool waters. However this process cannot be economical for water works. Irradiation of water by ultraviolet light is commonly used for disinfection in food industries. Water for domestic purposes on a smaller scale may be sterilized by boiling the filtered water for about 20 minutes. Chemicals like potassium permanganate and tincture iodine are also used occasionally but chlorine tablets and bleaching powder are more commonly employed.

6. Removal of Dissolved Salts: Softening of Water The concentration of dissolved impurities mostly determines the hardness of water. The process of removing the hardness causing salts from water is called softening of water. It has been pointed out under the section on “hardness” that soft water is essential for many industries such as in textiles, laundries, paper, rayon, ice, brewing, distilleries, canning, etc. Water used for generation of steam should be perfectly soft to minimize troubles like scale formation in the boilers, which leads to loss of efficiency and even ultimate failure of the boiler tubes. The following methods are generally used for softening of water. Temporary hardness can be removed by prolonged heating of the water. The dissolved gases e.g., CO2 and O2 are also removed simultaneously.

Water Treatment  6.25

6.1  Lime-Soda Process This is the most important method of chemical water-softening. The principle involved in this process is to chemically convert all the soluble hardness– causing impurities into insoluble precipitates which may be removed by settling and filtration. In the lime-soda process, a suspension of milk of lime is added in requisite amount to the water together with a calculated amount of sodium carbonate (soda) solution. Generally about 10% excess of chemicals are added for quick completion of the reactions. The reactions taking place may be summarized as follows: (i) lime removes the temporary hardness:

Ca(HCO3)2+Ca(OH)2 Mg(HCO3)2 + 2Ca(OH)2

2CaCO3Ø + 2H20 Mg(OH)2Ø + 2CaCO3 + 2H2O

(ii) lime removes the permanent magnesium hardness:



MgCl2 + Ca(OH)2 MgSO4 + Ca(OH)2

Mg(OH)2Ø + CaCl2 Mg(OH)2Ø + CaSO4

(iii) lime removes dissolved iron and aluminium salts:

2HCl + Ca(OH)2 H2SO4 + Ca(OH)2

CaCl2 + 2H2O CaSO4 + 2H2O

(v) lime removes dissolved CO2 and H2S



Fe(OH)2Ø + CaSO4 2Fe(OH)3Ø 2Al(OH)3Ø + 3CaSO4

(iv) lime removes free mineral acids:



FeSO4 + Ca(OH)2 2Fe(OH)2 + 2H2O + O Al2 (SO4)3 + 3Ca(OH)2

Ca(OH)2 + CO2 Ca(OH)2 + H2S

CaCO3Ø + H2O CaSO4Ø + 2H2O

(vi) Soda removes all the soluble calcium permanent hardness (i.e., which is originally present as well as that which is introduced during the removal of Mg+2, Fe2+, Al3+, HCl, H2SO4 etc. by lime).



CaCl2 + Na2CO3 CaSO4 + Na2CO3

CaCO3Ø + 2NaCl CaCO3Ø + Na2SO4

Since natural waters generally contain a large proportion of temporary hardness, it is often convenient and economical to remove temporary hardness by lime treatment. Lime is rather cheap and it removes temporary hardness efficiently without introducing any soluble salts in the water. Magnesium hydroxide produced in the above reactions precipitates as an insoluble sludge. The reaction of soda with the permanent calcium hardness produces insoluble CaCO3. Addition of a coagulant such

6.26  Engineering Chemistry as sodium aluminate or alum helps in accelerating the coagulation of the carbonate sludge, which is subsequently removed by filtration. Water softened by this process contains appreciable concentrations of soluble salts, such as sodium sulphate, and cannot be used in high pressure boiler installations. Types of cold lime-soda softeners There are four basic types of cold lime soda softeners: (i) The intermittent type (batch process) (ii) The conventional type (iii) The catalyst or spiractor type    continuous processes (iv) The sludge blanket type (i) Intermittent or batch process The intermittent type of cold lime-soda softener consists of a set of two tanks which are used in turn for softening of water. Each tank is provided with inlets for raw water and chemicals, outlets for softened water and sludge, and a mechanical stirrer (Fig. 7). Raw water and calculated quantities of the chemicals are slowly sent into the tank simultaneously under agitation with the help of the stirrer. Some sludge from a previous operation is also added which forms nucleus for fresh precipitation and thus accelerates the process. Thus by the time the tank is full, the reaction is more or less complete. Stirring is stopped and the sludge formed is allowed to settle. The clear softened water is collected through a float pipe and sent to the filtering unit. The sludge formed in the tank is removed through the sludge outlet. By employing a set of tanks planned for alternate cycles of reaction and settling, continuous supply of softened water may be ensured.

Fig. 7  Intermittent cold lime-soda softener

(ii) Conventional type In this process, the raw water and the chemicals in calculated quantities are raw continuously fed from the top into an inner chamber of vertical circular tank provided with a paddle stirrer (Fig. 8). The raw water and the chemicals flowing down the chamber come into close contact because of the continuous stirring and the softening reactions take place. The sludge formed settles down to the bottom of the outer chamber from where it is periodically removed through the sludge outlet softened water rising up passes through the fibre filter where traces of sludge are removed and filtered soft water passes through the outlet provided.

Water Treatment  6.27

Fig. 8  Conventional type of lime-soda softener

Water treated by the cold lime-soda process generally produces softened water containing about 50–60 ppm of residual hardness. (iii) Catalyst or spiractor type The spiractor consists of a conical tank which is about two-thirds filled with finely divided granular catalyst (Fig. 9). The tank used may be either open (for gravity catalyst operation) or closed (for operation under pressure). In both the cases the raw water and the calculated quantities of chemicals enter the tank tangentially near the bottom of the cone and spiral upwards through the suspended catalyst bed. The catalyst employed is a finely granule (0.3 to 0.6 mm diameter) insoluble mineral substance such as graded calcite or sand or green sand. The retention time is about 8 to 12 minutes. The sludge formed during the softening reactions deposits on the catalyst grains in an adherent form and hence the granules grow in size. The softened water rises to the top from where it is drawn off.

Fig. 9  Catalyst or spiractor type cold lime-soda water softener

6.28  Engineering Chemistry The catalyst or spiractor type of continuous water softener is of interest as it gives a granular sludge which drains and dries rapidly and can be handled easily. (iv) The sludge blanket type The sludge blanket type of water treatment equipment is extensively used for coagulation and settling as well as water softening by cold lime soda process. These softeners differ from the conventional type in that the treated water is filtered upwardly through a suspended sludge blanket composed of previously formed precipitates. Thus in a single unit, all the three processes namely mixing, softening and clarification take place. In the conventional type of equipment, some of the added lime suspension is carried down in the sludge formed by the precipitates, before it has time to dissolve and react with the hardness causing impurities of the raw water and thus some of the lime is wasted. In the sludge blanket type, this does not happen because the upward filtration through the suspended sludge blanket ensures complete utilization of the added lime. With the conventional type of equipment, it is generally observed that after-precipitates or afterdeposits form on the granules or filter media employed, and in pipe lines or distribution systems carrying the filtered effluents. This usually necessitates recarbonation with CO2 to obviate the formation of such deposits. However, in the sludge blanket type of equipment, the intimate contact of the treated water with a large mass of solid phase mostly prevents super-saturation or the formation of after-deposits. This result in the production of the effluent which is clear enough (turbidity usually less than 10 mg/l) for many industrial applications, so that subsequent filtration is often unnecessary. The retention period required with sludge blanket type equipment is one hour as against four hours with the conventional units. Further, silica is removed better in sludge blanket units. The sludge blanket type of water softening equipment, owing to its higher efficiency, shorter detention period and smaller space requirements, is rapidly displacing the conventional type. Hot lime-soda process The effect of temperature on the velocity and completeness of precipitation reac involving the removal of scaleforming constituents is shown in Fig. 10. It can be seen that at 96°C the precipitation is more complete in 10 minutes than after several hours at 10°C. Thus effect is more pronounced with the precipitation of magnesium compounds. The reactions during water softening take place in very dilute solutions (about 0.001 M) and hence proceed very slowly. The rate of these precipitation reactions can be greatly accelerated by increasing the temperature, because, this not only increases the rate of the ionic reactions themselves, but also the rate at which particles of measurable size are formed. Hot lime-soda plants carry out softening at 94°–100°C which has several advantages. For efficient softening, cold lime-soda softening plants must be of considerable area and water-storage capacity, whereas hot lime-soda softeners are much more rapid in operation and therefore for a given throughput, much more compact. Elevated temperatures not only accelerate the actual chemical reactions but also reduce the viscosity of the water and increase the rate of aggregation of the particles. Thus, both the settling rates and filtration rates are increased. Thus the softening capacity of the hot lime-soda process will be several times higher than the cold process. Since the sludge formed settles down rapidly, there is no need of adding any coagulants. A smaller excess of chemicals is needed than with the cold process. Further, dissolved gases are driven out of the solution to some extent at the high temperature. The hot lime-soda process yields softened water having relatively lower residual hardness

Water Treatment  6.29

Parts per million

143.0 114.4 85.8

CaCO3

57.2

Mg(O

Ca

H)

CO

3

Mg(O

28.5 0

H)

0

0.5

2

at 96°C

at 10°C at 10°C

at 96°C

2

1.0

1.5 2.0 2.5 Time in hours

3.0

3.5

4.5

Fig. 10  Effect of temperature on rate of precipitation of CaCO3 and Mg(OH)2 from CaSO4 and MgSO4 solutions respectively

(about 17 to 34 ppm) as against the cold process (about 50–60 ppm). A typical hot lime-soda water softening unit is shown in Fig. 11, which includes a reaction cum settling tank and a filter. If the water is alkaline, filtration through sand and gravel beds might steam contaminate the water with dissolved silica, particularly if the quartz used is of inferior quality. Other filtering media used are anthracite coal, calcite and magnetite. If the precipitation is incomplete in the softening tank, “afterprecipitation” occurs in pipes, storage tanks and even in boiler itself. If slight excess of chemicals are used over that theoretically required, more rapid and more complete removal of hardness will result. But if larger excess of chemicals are used, naturally they will appear in the softened water. Lime-soda plants do not produce water of zero hardness. Raw water

Exhaust or Live steam

Chemicals

Fines Coarse

Sludge

Anthrafilt Clarified water

Sludge outlet

Fig. 11  Hot lime-soda water softener

Inter-Relationship between Hardness and Alkalinity of Water The Inter-relationship between hardness and alkalinity of water can be illustrated by the following examples:

6.30  Engineering Chemistry Notes:  Equivalent amounts of cations and anions are paired as follows: (a) The HCO3 alkalinity is first paired with Ca and then, if any alkalinity is left over, is paired with Mg2+. If any HCO3 alkalinity is still remaining it is paired with Na+. (b) Non-carbonate hardness will be present only if the total hardness is greater than carbonate hardness or total alkalinity. (c) Non-carbonate hardness is first paired with Mg+2 and then the remaining non-carbonate hardness is paired with Ca2+.

6.2 Problems on Water-treatment by Lime-soda Process On the basis of the various reactions taking place in lime-soda process given earlier the following deductions can be made: (i) One equivalent of calcium temporary hardness requires one equivalent of lime

Ca(HCO3), + Ca(OH)2

2CaCO3 + 2H2O

(ii) One equivalent of magnesium temporary hardness requires two equivalents of lime



Mg(HCO3)2 + 2Ca(OH)2

Mg(OH)2 + 2CaCO3 + 2H2O

(iii) One equivalent of calcium permanent hardness requires one equivalent of soda.

CaCO3 + Na2SO4 CaCO3 + 2NaCl

(iv) One equivalent of magnesium permanent hardness requires one equivalent of lime and one equivalent of soda.



CaSO4 +Na2CO3 CaCl2 + Na2CO3

MgSO4 + Ca(OH)2 + Na2CO3 MgCl2 + Ca(OH)2 + Na2CO3

Mg(OH)2 + CaCO3 + Na2SO4 Mg(OH)2 + CaCO3 + 2NaCl

(v) Lime reacts with HCl, H2SO4, CO2, H2S, salts of iron, aluminium, etc. Accordingly, their respective equivalents must be considered for calculating the lime requirement.



Ca(OH)2 + 2HCl Ca(OH)2 + H2SO4 Ca(OH)2 + CO2 Ca(OH)2 + H2S Ca(OH)2 + FeSO4



1 2Fe(OH)2 + H2O + __ ​    ​ O2 2



2Ca(OH)2 + Al2 (SO4)3

CaCl2 + 2H2O CaSO4 + 2H2O CaCO3 + H2O CaSO4 + 2H2O CaSO4 + Fe(OH)2 2Fe(OH)3 2Al(OH)3 + 3CaSO4



(vi) Lime, while reacting with HCl, H2SO4, MgSO4, MgCl2, Mg(NO3)2 salts of Fe, Al etc., generates the corresponding quantities of calcium permanent hardness. Accordingly, these constituents also should be considered while calculating the soda requirement. (vii) Two equivalents of HCO3 reacts with two equivalents of lime as follows: 2HCO3– + Ca(OH)2

(1 equivalents) (2 equivalents)

CaCO3 + 2H2O + CO32– (2 equivalents)

Water Treatment  6.31 It is evident that in the above reaction, 2 equivalents of CO32– are generated. Thus for every one equivalent of HCO3 present, the corresponding reduction in the dose of soda has to be t made in the calculations for soda requirement. For solving numerical problems on lime-soda requirements for softening of hard water, the following steps may be followed: l. The units in which the impurities analyzed are expressed i.e., ppm (or mg/l). grains per gallon (or degrees Clark), etc., are to be noted. 2. Substances which do not contribute towards hardness (e.g., KCl, NaCl, SiO2, Na2SO4 Fe2O3, K2SO4, etc.) should be ignored while calculating lime and soda requirements. This fact should be explicitly stated. 3. All the substances causing hardness should be converted into their respective CaCO3 equivalent, as a matter of convention and convenience. wt. of the impurity CaCO3 equivalent of a hardness causing impurity = ​  _________________        ​ × 50 Chemical equivalent (since chemical equivalent weight of CaCO3 = 50). For instance parts by weight of CaSO4 would contain the same amount of Ca as in 100 parts by weight of CaCO3. Hence, in order to convert the weight of CaSO4 as its CaCO3 equivalent, 100 50 the weight of CaSO4 should be multiplied by a factor of a or ____ ​   ​ or ___ ​   ​ . 136 68 Conversion factors for some of the impurities in water which commonly come across are given in Table 3.

Table 3

Salt (1)

Multiplication factor to convert into CaCO3 equivalent (2)

Ca(HCO3)2

100/162

Mg(HCO3)2

100/146

CaSO4

100/186

CaCl2

100/111

MgSO4

100/120

MgCl2

100/95

Mg(NO3)2

100/148

Ca

+2

100/40

+2

100/24

HCO –3

100/122

HCl

100/73

H2SO4

100/98

Mg

CO2

100/44

Al2 (SO4)2

100/114

FeSO4 7H2O

100/848

MgCO3

100/84

6.32  Engineering Chemistry Notes: (a) If the impurities are given as CaCO3 or MgCO3, these should be considered to be due to Ca(HCO3)2 or Mg(HCO3)2 respectively and they are only expressed in terms of CaCO3 and MgCO3. (b) The amount expressed as CaCO3 does not require any further conversion. However, the amount expressed as MgCO3 should be converted into its CaCO3 equivalent by multiplying with 100/84. 4. Calculate the lime and soda requirements as follows:



Temporary calcium hardness + (2 × Temporary maganesium hardness) + Perm. Mg harndess + 74 (A) Lime required for softening = ____ ​    ​   CO2 + HCl + H2SO4 + HCO3– + salts of Fe2+, 100 Al3+ etc., –NaAlO2; all expressed in terms of their CaCO3



Permanent Ca-hardness + permanent Mg 106 hardness + salts of Fe2+, Al3+ etc., + HCl + ____ (B) Soda required for softening = ​   ​    100 H2SO4 – HCO3 – : All expressed in terms of their CaCO3 equivalents



(C) If the analytical report shows the quantities of Ca2+ and Mg2+, then; 1 eq. of soda is required for Ca2+ whereas one eq. of lime and 1 eq. of soda is required for Mg2+. 5. If the lime and soda used are impure and if the % purity given, then the actual requirements of the chemicals should be calculated accordingly. Thus, if lime is 90% pure then the value 100 obtained under 4-A above should be multiplied by ____ ​   ​ to get the actual requirement of 90 lime.



Similarly, if soda is 95% pure, then the value obtained under 4-B above should be multiplied 100 by ____ ​   ​ to get the actual soda requirement. 95

6.3  Ion-exchange process of water softening As applied to water treatment, an ion-exchange process may be defined as a reversible exchange of ions between a liquid phase and a solid phase. Materials capable of exchange of cations are called cation exchangers and those which are capable of exchanging anions are called anion exchangers. Both cation and anion exchangers are widely used in industry, including preparation of boiler feed and process waters. Ion-exchangers commonly used in water treatment include: (i) Natural and synthetic zeolites (ii) Carbonaceous ion-exchangers (iii) Synthetic resins.

6.4  Natural and synthetic Zeolites The name Zeolite is derived from two Greek words (Zein + lithos) which mean “boiling stone”. The name was first used by Cronstedt, a Swedish geologist, in 1736, to a certain group of natural minerals, which released their water of hydration (or combination) in the form of steam. These natural Zeolites include:

Water Treatment  6.33

1 (a) Thomsonite (Na2O, CaO) . Al2O3. 2SiO2. 2​ __ ​  H2O 2 (b) Natrolite Na2O, Al2O3.3SiO2.2H2O (c) Laumontite CaO. Al2O3.4SiO2.4H2O (d) Harmotome (BaO, K2O) Al2O3.5SiO2.5H2O (e) Stilbite (Na2O.CaO), Al2O3.6SiO2.6H2O (f) Brewsterite (BaO, SrO, CaO).Al2O3.6SiO2.5H2O (g) Ptilolite (CaO, K2O, Na2O), Al2O3, 10SiO2.5H2O

The chemical structure of Sodium Zeolite may be represented as Na2O, Al2O3, x(2 – 10). SiO2 y(2 – 6), H2O, so that they may be regarded as hydrated sodium alumino-silicates which are capable of exchanging their sodium ions for multivalent ions of alkaline earth group and for the divalent ions of some of the metals in water. Thus, Zeolites find application in softening of water for domestic and industrial purposes. Two types of Zeolites are in common use: (a) Natural Zeolites e.g., non-porous green sands, and (b) Synthetic Zeolites which are porous and possess a gel structure. The natural zeolites are derived from green sands by washing, heating and treatment with NaOH. The synthetic zeolites are prepared from solutions of sodium silicate and aluminium hydroxide. They may also be prepared by heating together (a) China clay, felspar and soda ash and granulating the resultant mass after cooling (b) solutions of sodium silicate, Al2 (SO4)3 and NaAlO2 (c) solutions of sodium silicate and A12 (SO4)3 (d) solutions of sodium silicate and NaA1O2. Natural zeolites are more durable, whereas synthetic zeolites have higher exchange capacity per unit weight. Zeolites are now being replaced slowly by high capacity resins having greater stability. Zeolites are also known as permutits. Sodium zeolites are used in water softening and are simplistically represented as Na2Z, where Z stands for the insoluble zeolite radical framework. Since these are capable of exchanging basic radicals, these are generally known as base exchangers. When hard water passes through a bed of active granular sodium zeolite, the Ca2+ and Mg2+ ions are taken up by the zeolite and simultaneously releasing the equivalent sodium ions in exchange for them. The various reactions taking place may be indicated as follows:

Ca(HCO3)2 Mg(HCO3)2 CaSO4 MgSO4 CaCl2 MgCl2

+ + + + + +

Na2Z Na2Z Na2Z Na2Z Na2Z Na2Z

Æ Æ Æ Æ Æ Æ

CaZ + 2NaHCO3 MgZ + 2NaHCO3 CaZ + Na2SO4 MgZ + Na2SO4 CaZ + 2NaCl MgZ + 2NaCl

Relatively small quantities of iron and manganese, present as the divalent bicarbonates, may also get removed simultaneously.

Fe(HCO3)2 + Na2Z Æ FeZ + 2NaHCO3 Mn(HCO3)2 + Na2z Æ MnZ + 2NaHCO3

(Note: Soluble iron and manganese are always present in the divalent form in waters containing bicarbonate alkalinity).

6.34  Engineering Chemistry Regeneration When the Zeolite bed is exhausted (i.e., saturated with Ca2+ and Mg2+), it can be regenerated and reused. The chemicals used for regeneration are concentrated sodium chloride solution (brine), sodium nitrate, sodium sulphate, potassium chloride or patassium nitrate. Brine is most widely used on account of its cheapness, relatively low molecular weight and also because the products formed by the regeneration reactions chiefly CaCl2 and MgCl2, which are highly soluble and can be readily rinsed out from the zeolite bed.

CaZ + 2NaCl Æ Na2Z + CaCl2 MgZ + 2NaCl Æ Na2Z + MgCl2

Zeolite water softeners Zeolite water softeners are made in both pressure type and gravity type and both the types are available in automatic, semi-automatic and manually operated designs. All these units operate on the same principles, involving alternate cycles of softening run and the regeneration run. During the softening run, the water is softened by passing it through the sodium zeolite bed, where Ca2+ and Mg2+ are removed from the water and held by the Zeolite and simultaneously releasing equivalent amount of Na+ in exchange. The regeneration step comprises of (a) backwashing (b) salting (or brining) and (c) rinsing before reuse. Brine inlet Hard water

Hard water

Sodium Zeolite Fine Gravel Coarse Gravel

Softened water

Fig. 12  Zeolite softener

A Simple zeolite softener is depicted in Fig. 12 and the scheme for softening and regeneration processes are represented in Fig. 13. Advantages of Zeolite process 1. Hardness is completely removed. 2. Equipment used is compact and occupies fewer places. 3. It automatically adjusts itself to waters of different hardness. 4. It can work under pressure. Hence the plant can be installed in the water supply line itself, avoiding double pumping.

Water Treatment  6.35 Hard water

Brine +

2+

Na

Ca 2+ Mg

Ca Z and Mg Z

Ca Z and Mg Z

Softening progresses

Exhausted

Na2 Z +

Na

–

Cl

2–

SO4

–

Cl

Softened Water

Na2 Z Regenerated 2–

SO4 2+

2+

Ca Mg –

Cl

(a) Softening Cycle

(b) Regeneration Cycle

Fig. 13  Schemes for softening of water by Zeolite process



5. In this process, the hardness causing ions are simply exchanged with sodium ions. As the process does not involve any precipitation, there is no problem of sludge formation and afterprecipitation in the softened water at later stages.

Limitations 1. Water, having turbidity and suspended matter should not be directly fed to the zeolite softener because the pores of the zoelite bed will be clogged and the rate of flow will be unduly decreased. Therefore, the raw water turbidity and suspended matter should be removed (e.g., by sedimentation, coagulation and filtration) before subjecting it to the Zeolite treatment. 2. Water containing excess of acidity or alkalinity may attack the Zeolite. It is preferable to have the pH of the water passing through the Zeolite softener around 7. 3. Water containing large quantities of Fe2+ and Mn2+ when passed through the Zeolite bed are converted into their respective Zeolites which cannot be easily regenerated. 4. Hot water should not be used as the zeolite tends to dissolve in it. 5. Acid radicals are not removed by this process. Hence, the temporary hardness present in the raw water is converted to NaHCO3 which goes into soft water effluent. If such water is used in the boilers, it dissociates as follows under the boiler conditions.

NaHCO3 Æ NaOH + CO2

This NaOH may cause caustic embrittlement of boiler metal. Further, CO2 goes along with the steam and renders the condensed water acidic and corrosive. Hence, raw waters containing large amounts of temporary hardness should be subjected to prolonged boiling or lime treatment to remove temporary hardness, before subjecting to the Zeolite process. 6. Water treated by the Zeolite process contains about 25% more dissolved solids than that treated by lime-soda process. Comparison between the Zeolite process and the lime-soda process The salient features of Zeolite process and lime-soda process are summarized in the following table:

6.36  Engineering Chemistry Zeolite process

Lime-soda process

1. This process produces water of almost zero hardness. 1. This process produces water having hardness of 15 to 60 ppm depending on whether it is hot process or cold process 2. The cost of the plant and the Zeolite are higher. Hence 2. The capital cost is lower. the capital cost is higher. 3. The exhausted Zeolite bed can be regenerated with 3. The chemicals needed viz., lime, soda, and brine which is very cheap. Hence the operating cost is coagulant are consumed in the process. Hence lesser. the operating cost is higher. 4. The plant is compact and occupies less space. The size 4. The plant occupies more space. The size of the of the plant depends on the hardness of water being plant depends on the amount of water being treated. handled. 5. Cannot be used for hot water, acidic water and water 5. The process is free from such limitations. having turbidity and suspended impurities. 6. This process can operate under pressure and can be 6. This process cannot be operated under designed for fully automatic operation. pressure. 7. This process does not involve cumbersome operations 7. This process involves all the problems such as settling, coagulation, filtration. associated with settling, coagulation and filtration. 8. Water treated with the Zeolite process contains larger 8. Treated water contains lesser percentage amounts of sodium salts and greater percentage of of dissolved solids and lesser quantities of dissolved salts than the raw water since Ca2+ (eq. wt. sodium salts. 20) and Mg2+ (eq. wt. 12) are replaced by Na+ (eq. wt. 23). 9. This process adjusts itself to waters of different 9. Reagent doses must be adjusted for waters of hardness. different hardness. 10. Salts causing temporary hardness are converted to 10. Temporary hardness is completely removed in NaHCO3 which will be present in the softened water. the form. Such of insoluble CaCO3 and Mg(OH)2, a water creates problems when used as feed water in boilers. 11. No problems of after-precipitation.

11. There may be problems of after-precipitation in distribution systems and even in boilers when used as boiler feed water.

6.5  Synthetic ion-exchange resins In an ion-exchange process, a reversible exchange of ions takes place between the stationary ion-exchange phase and the external liquid mobile phase. An ion-exchange resin consists of an insoluble polymeric matrix that is permeable. Fixed charge groups and mobile counter ions of opposite charge are incorporated in the matrix. These counter ions can be exchanged for ions in the external liquid phase. The purely synthetic organic exchangers are made by: 1. Polycondensation e.g., between a substituted phenol and formaldehyde with some unsubstituted phenol to get the desired cross-linking between the linear chains. 2. Polymerisation e.g., of stryrene in presence of diviny1 benzene, to give the desired cross-linking.

Water Treatment  6.37 The functional groups are then introduced into the cross-linked resin network, either by subsequent treatment of the resin or by introducing the functional groups into the starting material itself. It is these functional groups which decide the nature of the exchanger (i.e., cationic or anionic) based on the acidity or basicity of the functional group, exchangers are further classified as follows: Types of ion-exchanger Functional groups Strongly acidic cation-exchangers – SO3H Moderately strong cation-exchangers – PO(OH)2 Weakly acidic cation-exchangers – COOH or –OH Strongly basic anion-exchanger – NR3+;∫ P+ –CH3, etc. Weakly basic anion-exchanger – NH2; – (C2H4): (NH) –y Hypothetical formulations of a typical cationic and a typical anionic resin based on styrene and divinyl benzene network are shown in Figs. 14 and 15.



Fig. 14  Hypothetical structure of a typical cation exchanger.

Fig. 15  Hypothetical structure of a typical anion exchanger.

Variation in polymer type and cross-linking affects the insolubility and life of the resin and the diffusibility of ions in an exchange process. Resins available for water treatment include cation-exchange resins of the strongly and weakly acidic types, anion-exchange resins of the strongly and weakly basic types and highly porous modifications of strong acid cation exchangers and strong base anion exchangers. These resins are available in granular or bead-like form and may be obtained in an effective size and uniformity coefficient. For effective water treatment, ion exchangers should possess the following properties: 1. They should be non-toxic. 2. They should not discolor the water being treated. 3. They should possess a high ion-exchange capacity. (It depends upon the total number of ion active groups per unit weight of the exchanger and is expressed as mill-equivalents per gram of the exchanger).

6.38  Engineering Chemistry

4. 5. 6. 7. 8.

They should be physically durable. They should be resistant to chemical attack. They should be cheap and commonly available. They must be capable of being regenerated and back-washed easily and economically. They should have a large surface area since ion-exchange is a surface phenomenon. At the same time, their resistance to flow must be compatible with hydraulic requirements. National Chemical Laboratory, Pune, developed polystyrene based cation-exchangers and melamine based anion exchangers which are available in the market with different trade names.

6.6  Softening of water by ion-exchange The ion-exchange resins used in water-softening may be used on either the sodium cycle or the hydrogen cycle (i.e., the active exchanging cationic group may be in the sodium form or hydrogen form). In the first method, hard water is passed through a bed of a cation-exchanger in sodium form: Ca

Mg

(HCO3)2 SO4 Cl2

Ca + Na2R

R + Na2

Mg

(HCO3)2 SO4 Cl2

Thus all the Ca2+ and Mg2+ are taken up by the resin in exchange of Na+. When the resin bed is exhausted, it can be regenerated with a concentrated solution of sodium chloride. Ca

Ca R + 2NaCl

+ Na2R +

Mg

Cl2 Mg

In the second method, hard water is passed through a bed of cation exchanger in hydrogen form. Ca

Ca (HCO3)2 + H2R

Mg Ca

R + 2H2O + 2 CO2 Mg Ca

SO4 + H2R

Mg

Cl2

R+H2SO4

or

2HCl

Mg

When the resin bed is exhausted, it can be regenerated with sulphuric acid or hydrochloric acid. For many industrial water supplies, chemical softening (e.g., lime soda process) is used first to remove most of the hardness followed by a cation exchange process for completing the softening process. For softening water, usually sodium cycle is preferred.

Water Treatment  6.39 Ca

Ca

SO4 + H2R

R + H2SO4 Mg

Mg

6.7  Demineralisation or deionisation for many years, distillation was the only method available for complete removal of soluble salts from water. However, with the advent of the development of commercial cation-exchange and anionexchange resins, it has been possible to produce de-ionised water of highest purity by a two-step ion exchange process. First, the hard water is passed through a cation-exchange resin in the hydrogen form



The effluent from this step is then passed through an anion exchanger bed.

Thus, all the ions present in water are removed and demineralised or deionised water is produced. Strongly basic anion-exchanger is used when silica also is to be removed. Otherwise, weakly basic anion exchangers are used generally. If CO2 removal is also desired, forced draft or vacuum degasifier unit is provided. The anion exchanger may be regenerated by treating it with dilute NaOH solution. The cation exchanger may be regenerated by passing a solution of HCl or H2SO4

A typical demineralisation unit is shown in Fig. 16. Demineralisation of water can be better achieved by employing a “mixed bed” of cationic and anionic resins. This method produces an effluent which is far superior to that produced by the two bed operation. When water is passed through a “mixed bed” consisting of anion and cation exchange resins, each pair of contrasting resin particles functions as a stage in the treatment and thus the total effect is that of a multiple-cycle-deionisation. Thus the process is highly efficient.

6.40  Engineering Chemistry Raw water To vacuum pump Cation exchanger

Acid for regeneration

Steam jacket

Anion exchanger Alkali for regeneration

Deionised water

Fig. 16  Demineralisation of water

When the resins are exhausted the bed is backwashed when the two resins are separated in different layers due to difference in their densities. Then the resins are separately regenerated, washed and mixed again by injecting air and reused for a fresh cycle. The deionisation process can also be used to purify highly acidic or alkaline waters. All ionizable impurities are removed and many commercial plants produce water having as low as 3 ppm dissolved solids and 0.1 ppm silica. However, the process is costly. The equipment and the resins as well as the regenerants are expensive as compared to those used in zeolite process. Water having turbidity reduces the efficiency of the process. Hence it is advisable to treat it first by lime-soda process and then removing the residual hardness by this process. This process is recommended for very high pressure boilers, where the specifications are more exacting. Demineralized water is also used in the manufacture of photographic materials, fine chemicals, synthetic rubber, pharmaceuticals, ceramics, explosives, plastics, paper products such as dielectrics, soluble coffee, cosmetics, liquid soaps, storage batteries, catalysts, ice, cellulose products, television tubes, for cooling purposes in broadcasting stations and diesel locomotives, power plants, oil refineries, electroplating plants and in laboratories. Distillation This method of removing hardness and scale forming constituents is used in large power stations and on steamships, where the amount of make-up water is small compared with the total amount of water evaporated in the boilers. However, until cheaper sources of energy are available, water softening by distillation is going to be an expensive proposition. Removal of iron and manganese from water Iron and manganese may be present in water as bicarbonates, sulphates, hydroxides or as chelates. The presence or iron and manganese in water even in such low concentrations as 0.1 to 0.3 ppm may be objectionable in industrial processes such as textile dyeing and finishing, paper making, rayon processing, beverage manufacture, laundering and tanning. Iron and manganese in water may be removed by the following processes: (i) By aeration followed by settling and filtration, the metals are removed in the form of Fe (OH)3 and Mn (OH)3 (ii) By ion exchange process using zeolites, carbonaceous or synthetic cation exchangers. (iii) By lime-soda process, and (iv) By manganese zeolite process

Water Treatment  6.41 Mangenese zeolite is prepared from zeolite by alternate treatments with MnSO4 and KMnO4. This results in the precipitation of the higher oxides of Mn in and around the zeolite granules. In this process, Fe and Mn are mostly removed by contact oxidation. The regeneration of manganese zeolite is performed by KMnO4 or NaMnO4 solution. The manganese zeolite is used as filter medium in pressure type filter, through which the water is filtered to remove Fe and Mn.

7.  BOILER FEED WATERS (Water for Steam Making) Water is used directly or indirectly in many industrial processes. The problem of preparing feed water for boilers is common to many industries. Its importance and complexity is obvious from the large number of technical investigations made in this field. Boilers are operated under different pressures and greater the operating pressures, more rigorous are the specifications for the feed water. The operating pressures for different types of boilers are given below: Operating pressures Low-pressure boilers … Upto 200 lb/sq. in Intermediate pressure boilers … 200 to 500 lb/sq. in High-pressure boilers … 500 to 2,000 lb/sq. in Very high-pressure boilers … 2,000 to 3,209 lb/sq. in Supercritical boilers … Above 3,209 lb/sq. in 2 (1 lb/sq. in = 0.0703 kg/cm ; 1 kg/cm2 = 14.223 lb/sq. in.) There are a large number of different types of boilers in use which may be divided into two main classes, viz., shell boilers and water tube boilers. A shell boiler essentially consists of a cylinder in which the water is contained and heated by the passage of hot gases through fines or tubes immersed in the water. The shell boilers, in turn, can be further subdivided into (a) Horizontal flue-type boilers (42.3, the Cornish boilers) and (b) Horizontal fire-tube boilers (e.g., the Locomotive and the Economic boilers). In the water-tube boilers, of which many types exist, it is the water which passes through tubes, either by forced or by natural circulation. An important difference in the two main classes of boilers mentioned above is that the simple shell boilers are not usually provided with the ancillary equipment with which the water tube boilers are fitted. Thus water tube boilers are provided with economizers and feed-water heaters (pre-boiler equipment) and with super-heaters. While designing a water-treatment process for water-tube boilers, therefore, it is essential to take into consideration the conditions existing in such equipment. Any natural source of water does not supply a perfectly suitable boiler feed water. It is generally believed that a boiler feed water should satisfy the following requirements:

Hardness Caustic alkalinity Soda alkalinity Excess soda ash

– – – –

< 0.2 ppm 0.15 - 0.45 ppm 0.15 - 1 ppm 0.3 - 0.55 ppm

External treatment of boiler feed-water make-up Steam produced in a boiler is condensed after use and returned as pure feed water to the boiler. There are inevitable losses; however, the “make-up” water must be supplied from time to time. The term “boiler feed-water make-up” means the treated water which is required to make up the water

6.42  Engineering Chemistry losses from the boiler caused by use of open steam in process work or steam otherwise lost plus the water lost in the boiler blow-down. In certain cases, there are no condensate returns or none fit for re-use in the boiler, so that the feed water is 100% make-up water. In some other cases, particularly in large power plants, surface condensers employed are capable of recovering almost all of the water evaporated and make-up may comprise of < 2% of the water fed to the boiler. All other cases lie between the above two extremes. The treatment of boiler feed waters may be broadly divided into two categories: 1. External treatment:  In this, the scale-forming and corrosive impurities are removed from the water before it enters the boiler. 2. Internal treatment:  In this, dosages of various materials are added in the boiler to react with the non-carbonate hardness, to reduce the sticking tendency of scale-forming materials and to neutralise or overcome corrosive tendencies. In many cases, both the above forms of treatment are employed. In the external treatment of boiler feed water, one or more of the following processes are usually employed: 1. Removal of “hardness” or scale forming salts (a) Lime-soda process (b) Zeolite process (c) Demineralization (d) Distillation 2. Removal of silica: Magnesia process, ferric sulphate process, anion-exchange process. Fluosilicate process. 3. Removal of suspended matter: (a) pre-treatment: sedimentation, coagulation, settling and filtration. (b) softening processes. 4. Removal of dissolved gases: CO2 and O2 (a) Decarbonation by aeration (b) Deaeration (c) Chemical treatment. 5. Removal of oil (a) Mechanical oil-separators (b) Sodium aluminate treatment. Most of these processes have been discussed earlier and others will be discussed later, along with the methods for internal treatment

8.  BOILER TROUBLES The problem of making up feed water for boilers is common to almost all industries and has been a subject of detailed study by several workers. In treatment of boiler feed water, total elimination of all the impurities is generally not attempted. Only those impurities which give rise to certain operational troubles are either eliminated or maintained within tolerable limits. The major boiler troubles caused by the use of unsuitable water are (1) carry over: priming and foaming (2) scale formation (3) corrosion and (4) caustic embrittlement. All these troubles increase with increasing operating pressure of the boilers. The limits of tolerance for boiler feed waters, suggested by the committee on water quality tolerances for industrial users of the New England Water Works Association are, given under Table 4. These tolerance limits take into account that (a) reduction in

Water Treatment  6.43 turbidity color. oxygen consumed value and total solids decrease carry over (b) reduction in hardness, silica and alumina decrease scale formation (c) elimination of oxygen, reduction of bicarbonate ions and increase in pH help in suppressing corrosion and (d) maintaining a high sulphate to carbonate ratio checks caustic embrrittlement.

Table 4  Suggested limits of Tolerances for Boiler Feed Waters Pressure, MN/m2 (Mega-newton/meter square)

Characteristic 1. Turbidity (silica scale units) 2. colour (Hazen units) 3. Oxygen consumed 4. Dissolved oxygen 5. Hydrogen sulphide 6. Total hardness (CaCO3) 7. Sulphate to carbonate ratio 8. Aluminium oxide 9. Silica 10. Bicarbonate 11. Hydroxide 12. Total solids 13. pH value (min) (a) Units for; items 3 to 12 : mg/l (b) I MN/m2 = 145 psi

0–1.0 20 80 15 2 5 80 1:1 5 40 200 50 3000-1500 8.0

1.0–1.7 10 40 10 0.2 3 40 2:1 0.5 20 100 40 2000-1500 8.4

1.7–2.3 5 5 4 0 0 10 3:1 0.05 5 40 30 1500-100 9.0

over 2.8 1 1 3 0 0 2 3:1 0.01 1 20 15 50 9.6

8.1  Carry over As steam rises from the surface of the boiling water in the boiler, it may be associated with small droplets of water. Such steam, containing liquid water, is called wet steam. These droplets of water naturally carry with them some suspended and dissolved impurities present in the boiler water. This phenomenon of carrying of water by steam along with the impurities is called “carry over”. This is mainly due to priming and foaming. The solid and liquid contaminations in the steam are expressed by steam purity and steam quality respectively. The steam purity is expressed in parts per million of the impurity present and accordingly if steam contains 2 parts of solids contamination by weight per million parts by weight of steam it is called 2 ppm steam. Similarly the liquid contamination in the steam is expressed in percentage by weight, of the steam in the mixture. Thus if the steam contains 0.2% moisture, its steam quality is reported as 99.8%. Steam used for power production is usually superheated for achieving greater efficiencies in the turbine or engine in which steam is used. Wet steam causes corrosion in the inlet ends of the superheaters. If the steam contains high percentage of moisture, the extent of superheating will decrease with the consequent reduction in efficiency of the turbine or engine. Further, the water carried over with the steam contains salts and sludges, these are carried into the super heater where they may deposit as the water evaporates. This will seriously restrict the flow of steam. Moreover

6.44  Engineering Chemistry due to the insulating effect of these deposits, the super heater tubes also may burn out. A part of the dried salts may be carried along with the steam farther and deposit on the high-pressure turbine blades or in engine valves. Even a small amount of deposit on the turbine blades decreases its efficiency considerably. In order to eliminate the bad effects of moisture, mechanical steam purifiers are often installed in the steam drums of the boiler or between the boiler and the superheater. These devices force the steam to take curved paths whereby due to centrifugal action, the moisture is thrown out of the steam. Priming is a very rapid boiling of water occurring in the boiler in such a way that some water particles are carried away along with steam in the form of spray into the steam outlet. Priming is mainly attributed to the presence of suspended impurities and to some extent to dissolved impurities in the water. Thus it may be caused by imperfect filtration or precipitation of insoluble salts from the improperly softened or unsoftened feed water. Algae and vegetable growth in tanks can also lead to priming. Feed water containing even a small quantity of scale forming salts may cause particularly if the total solids in the feed exceed 300 ppm. Priming may also be caused by: (1) steam velocities high enough to carry droplets of water into the steam pipe (2) very high water level in the boiler (3) presence of excessive foam on the surface of the water which substantially fills the foam space (4) sudden steam demands leading to sudden drop of pressure in the steam line followed by ebullition, which causes a large mass of water to be interspersed with fine bubbles and (5) faulty boiler design. Priming can be minimised by (1) good boiler design providing for proper evaporation of water maintaining uniform heat distribution and adequate heating surfaces and also providing with antipriming pipes and dash-plates, etc. (2) maintaining low water levels (3) avoiding rapid changes in the steaming rate caused by sudden steam demands (4) maintaining as low a concentration of boiler water (with respect to dissolved impurities) as is consistent with scale and corrosion prevention and (5) minimizing foaming. Foaming is the formation of small but persistent bubbles at the water surface. These bubbles are carried along with steam leading to excessive priming. According to Bancroft, foams are formed when there is a difference in concentration of solute or suspended matter between the surface film and the bulk of the liquid. Substances which increase the viscosity of the film favor production of foam. The bubbles may also be protected by finely divided solids forming a protective “shell” around each of them. Any material which lowers the surface tension of the water will collect at the interface and thus increase the foaming tendency of the liquid. In steam rising, the bubbles of steam may be stabilized because of the accumulation of soluble salts in water. Clay or organic matter in raw water, oil or grease in condensed make-up water, and finely divided particles of sludge may also cause foaming. If the steam bubble does not collapse on reaching the surface of the water, the foam may be drawn into the preheater or steam lines. The accompanying liquid film may carry along with it dissolved salts, suspended solids or other stabilizing materials which will be deposited on the cylinder walls, turbine blades or in steam lines. If the stability of the bubble is such that it breaks near the steam outlet, tiny droplets of liquid from the collapsing film may be swept along into the steam lines giving “wet steam”. Foaming (and the consequent priming) can best be prevented by removal of the foaming and stabilizing agents from the water. Clay and other suspended solids as well as droplets of oil and grease can be removed by treatment of the feed water with clarifying agents such as hydrous silicic

Water Treatment  6.45 acid and aluminium hydroxide. The concentration of salts and sludge in the boiler can be controlled by intermittent or continuous blow-down. Spiral or “cyclone baffles” or a series of baffle plates near the steam outlets help to prevent water droplets from entering the steam lines. Foaming can also be controlled by adding antifoaming agents. Some of these act by counteracting the reduction in surface tension while others reduce foaming by simple mechanical action. For instance, castor oil spreads on the surface of water and prevents foaming. This is used only for low pressure boilers. Addition of a small concentration of polyamide antifoamer alters the surface tension and leads to the formation of only large unstable bubbles at the same heat input. Antifoaming agents may also act by reducing the charge on the protective film or its components. Foam may be destroyed by the addition of another good foaming agent. The foam is finally destroyed by mutual antagonistic effect of the difference in charge (positive and negative) on the colloidal particles of the two foams. This can be illustrated by the fact that the foam obtained by a solution of an anionic detergent (e.g., Aerosol OT) and that obtained from a cationic detergent (Ethyl Cetab) will destroy each other on mixing the two solutions. For avoiding priming and foaming, the following limits on total dissolved solids (in the boiler water) are usually suggested.

Type of boiler Water tube Lancashire Vertical

Total dissolved solids, ppm 2000 to 5000 1000 to 1500 £ 3500

In order to keep down the feed water concentration within safe limits, it is customary to remove a portion of the concentrated boiler water by blow-down and replace it with fresh feed water. Blowdown may be done periodically or continuously. However, blow-down will result in loss of heat. Removal of oil Oil present in boiler water may lead to (a) the formation of heat insulating oil films, and (b) carry over (if the oil is fatty). Natural waters generally do not contain oil! Small amount of oil may come from the oil used for lubrication of the pumps. A large quantity of boiler water is obtained from the condensation of exhaust steam which usually gets contaminated with lubricating oil in the steam cylinder. In order to remove oil, the exhaust steam is passed through mechanical separators which contain a number of metal plates. As the steam passes over the metal plates, the oil droplets are retained on the metal surface. Oil is also removed by coagulation with sodium aluminate or alum and soda ash. The floc produced by the Al(OH)3 and Mg(OH)2 precipitates enmesh the oil. The floc is then removed by filtering through anthrafilt. Oil can also be removed from water by cataphoresis. Removal of Silica Silica may cause hard deposits on turbine blades due to carry over. Silica also forms thermally resistant boiler scales of magnesium or calcium silicate. Silica is removed from the boiler feed water by the following methods: 1. By the addition of magnesia (MgO) to the water after the removal of the temporary hardness.

6.46  Engineering Chemistry







2. By the addition of magnesia to the water in the boiler when magnesia silica sludge is deposited. 3. By using magnesia or dolomitic lime in hot lime soda process of water softening. Magnesia acts as Mg(OH)2. In all these processes, silica is removed by magnesia by adsorption or by the formation of magnesium silicate. 4. By the use of ferrous sulphate or sodium aluminate as coagulants. They act as Fe(OH) 2 or Mg(OH)2 and Al(OH)3 which enmesh finely suspended and colloidal impurities including oil and silica. 5. By passing the demineralised water through a strongly basic special anion-exchanger which removes silica as H2SiO3 or HSiO3 or in some other form. The silica content is usually reduced to a fraction of a ppm. 6. By converting the silica into flurosilicate by the addition of Hydrofluoric acid and then removing it by passing through an anion exchange resin. H2SiF6 + 2R3N Æ (R3N2) . H2SiF6



7. By distillation.

8.2  Scale formation In a boiler, water is continuously converted into steam. This result in the concentration of the dissolved impurities until the water becomes saturated. Then the salts start separating out from the solution in order of their solubility, the least soluble one separating out first. Some of the solids separate in the body of the liquid in the form of soft and muddy deposits or in the form of suspension which can be flushed out easily. Such deposits are known as sludges. On the contrary, some of the solids deposit on a solid surface to form a sticky and coherent scale. The formation of sludges, unless excessive, has not much deleterious effects. But, however, the deposition of solid scale on the metal surfaces of the boiler has a serious effect on the efficient operation of boilers. The commonest solids that separate from boiler water are the sparingly soluble calcium salts e.g.,CaSO4,Ca(OH)2, CaCO3, Ca3 (PO4), Ca(OH)2, and the magnesium compounds e.g., Mg(OH)2; and calcium and magnesium silicates. Some of these are deposited as sludges and others as coherent scales. The suspended matter may be coagulated owing to the increase in temperature in the boiler, or it may be entrapped by the crystallizing soluble salts. For the soluble materials to be crystallized, their respective solubility products must be exceeded. This may be caused by (1) the normal in concentration that occurs due to continuous evaporation of water in the boiler. (In a pressure boiler, the total quantity of water, equal to the capacity of the boiler, is converted into steam within 15 minutes) (2) The decrease of solubility of some dissolved substances e.g., CaSO4 or at high temperature or (3) by the reactions that produce insoluble salts, such as by the decomposition of soluble bicarbonates e.g., Ca(HCO3)2 forming insoluble carbonates (e.g., CaCO3).

2HCO –3 = CO3– + CO2 ≠ +H2O

Calcium sulphate and calcium carbonate both have negative temperature solubility curves, and may, therefore, deposit scale. CaCO3 may in certain circumstances be deposited as a scale due to decomposition of calcium bicarbonate. Calcium sulphate, calcium silicate and magnesium silicate form the most troublesome scales. Calcium and magnesium hydroxides and calcium carbonate together with iron and aluminium oxides are often entrapped in the sulphate and silicate scales during their

Water Treatment  6.47 formation. In very high pressure boilers, sodium aluminium silicate scales resembling the zeolite minerals are also noticed, particularly in regions of excessive local evaporation. Several other salts and colloidal and suspended impurities also may get deposited or entrapped in the precipitates of salts forming the scales. Disadvantages of scale formation 1. Scale is a poor conductor of heat. Its effect is like that of an insulator coating on the metal surface. This result in the reduced rate of heat transfer, and thus the evaporative capacity of the boiler will be reduced. Thus, scale formation will result in wastage of fuel and reduction in boiler efficiency. 2. Scale formation on the boiler tubes or other heated surfaces insulates the metal so well that it becomes overheated. The metal becomes soft and weak thus making the boiler unsafe particularly at high pressures. The overheating also causes burning out of the metal plates and tubes and breakdown of the expanded joints. 3. In addition to the loss of strength due to overheating, rapid reaction between water and iron occurs at high temperatures, causing additional thinning of the tube wall. 3Fe + 4H2O Æ Fe3O4 (s)+4H2 (g)



4. Since the scale acts as heat insulator, the metal of the boiler is overheated. Under the high pressure of steam existing in the boiler, the metal expands until the scale on it cracks. Sudden entry of the water through these cracks to the very hot metal causes sudden cooling of the boiler metal with the simultaneous conversion of water into steam. The sudden increase in pressure due to this large quantity of steam thus formed may lead to explosion. 5. Excessive scaling may cause clogging of tubes. Considerable quantities of sludges may also be entrapped in the scale which may reduce the water circulation and impair the efficiency of the boiler.

Prevention of scale formation Scale formation can be prevented by the following methods: (a) External treatment:  This involves removal of hardness causing impurities (such as calcium and magnesium salts) and silica from the water before entering the boiler. The various methods of external treatment of water are already discussed earlier. (b) Internal treatment:  Internal treatment consists of adding chemicals directly to the water in the boilers for removing dangerous scale forming salts which were not completely removed in the external treatment for water softening. This is mainly used as a corrective treatment to remove the slight residual hardness and also sometimes to remove the corrosive tendencies in water. This treatment is not usually applied to raw waters, except for small boilers, but it is usually practiced in larger power-stations. In modern heavy-duty high pressure boilers, water of zero hardness is required, since even an egg- shell thickness of scale may be extremely detrimental. 1. Carbonate conditioning: For a salt to be precipitated, sufficient amount of the ions forming the salt must be present so that the product of their concentrations (i.e., ionic product) exceed a limiting value known as the solubility product. Thus, for a salt like CaCO3 to be precipitated, the product of the concentration of Ca2+ and Co3–2 must exceed the solubility product of CaCO3, represented as​ K​CaCO ​. 3

6.48  Engineering Chemistry

The formation of CaCO3 may be represented by Ca+2 + CO3–2



CaCO3



Solution

when equilibrium is established, [Ca+2] [CO3–2]

(Concentration of Ca +2)

(Concentration of CO3–2)

CaCO3

...(1)

Solid

= KCaCO3

(Solubility Product of CaSO4)

at any given temperature. If under these conditions, some Na2CO3 is added, the CO32– increases. Then, in order to maintain the KCaCO3 constant, some Ca2+ will get precipitated. Similarly in a satiuated CaSO4 solution Ca+2 + SO4–2



CaSO4



Solution

[Ca+2]



(Concentration of Ca +2)

[SO4–2]

(Concentration of SO4–2)

CaSO4

...(2)

Solid

= KCaSO3

(Solubility Product of CaSO4)

Thus, if a solution is saturated with CaCO3 and CaSO4 (as occurring in boiler water conditions), both the relations (1) and (2) apply, and by dividing (1) by (2), we get [CO3–2] ​ ______  ​   > [SO4–2]

​KCaCO ​ ​ 3 ​ ______   ​ ​KCaSO ​ ​

...(3)

4

From this, it is clear that if

[CO3–2] ​ ______  ​   > [SO4–2] [CO3–2]

​KCaCO ​ ​ 3 ​ ______   ​ ​KCaSO ​ ​ 4

–2

or in other words, if > K’ [SO4 ], then only CaCO3 will be precipitated in preference to CaSO4 (because the solubility product of CaSO4 cannot be attained under these conditions). These principles are used in the carbonate conditioning. When sodium carbonate solution is added to boiler water, the [CO3–2] increases and when it becomes greater than K’ × [SO4 –2], only CaCO3 gets precipitated and CaSO4 remains in solution. Thus the deposition of scale-forming CaSO4 is prevented. Na2 CO3 + CaSO4 –> CaCO3Ø + Na2SO4 Carbonate conditioning is used only for low pressure boilers. In high pressure boilers the excess Na2CO3 might be converted into NaOH due to hydrolysis as follows:

Na2CO3 + H2O ¤ 2 NaOH + H2CO3 H2CO3 ¤ H2O + CO2≠

NaOH causes caustic embrittlement in high pressure boilers. 2. Phosphate conditioning:  Just as in the case of carbonate conditioning, if we consider a solution saturated with both Ca3 (PO4)2 and CaSO4, the solubility product equations for both salts must be satisfied.

Ca3 (PO4)2

3Ca+2 + 2PO4–3

Water Treatment  6.49

\

[Ca+2]3 [PO4–3] = ​KCa ​ (PO ) ​ 3



...(4)

4 2

(Solubility Product of Ca3 (PO4) 2

\ [Ca+2] [PO4–3] 2/3 = [​KCa ​ (PO ) ​]1/3 3 4 2 Similarly, for CaSO4 CaSO4 = Ca+2 + SO4–2 +2

[SO4–2]



\



By dividing (5) by (6) we have

[Ca ]

...(5) ...(6)

= ​KCaSO ​ ​ (Solubility product of CaSO4) 4

–3 [​KCa ​ (PO ) ​]1/3 [PO 3 4 2 4 ] _______ ___________ ​   ​   = ​         ​ or K¢ –2 [​ K ​ ​] [SO4 ] CaSO4

[PO4–3] 2/3 = K¢[SO4–2]

or

...(7) ...(8)

Therefore, as long as [PO4–] 2/3 > K¢ [SO4 –], only Ca3 (PO4)2 will be precipitated in preference to CaSO4 (because the solubility product of CaSO4 cannot be exceeded under these conditions) and hence CaSO4 scale will not form. These principles are made use in the phosphate conditioning. In this, an excess of a soluble phosphate is added to the boiler water to precipitate the residual calcium ions in the form of a nonadherent precipitate of calcium phosphate and thus the scale formation is prevented. The three sodium orthophosphates viz., Na3PO4; Na2HPO4 and NaH2PO4 have been used for phosphate conditioning as also sodium pyrophosphate (Na4P2O7) and sodium metaphosphate (NaPO3). The typical reactions of the various phosphates with the hardness represented as CaCO3 may be summarized as follows:

2 Na3 PO4 2 Na2HPO4 2 NaH2PO4 2 NaPO3

+ + + +

3 3 3 3

CaCO3 CaCO3 CaCO3 CaCO3

¤ ¤ ¤ ¤

Ca3 (PO4)2 Ca3 (PO4)2 Ca3 (PO4)2 Ca3 (PO4)2

+ + + +

3 Na2CO3 2 Na2CO3 + CO2 + H2O Na2CO3 + 2CO2 + 2H2O Na2CO3 + 2CO2

The quality of the feed water dictates the choice of a particular phosphate to be used. For instance, if the feed water tends to produce an acidic condition in the boiler, the alkaline Na3PO4 should be chosen. This treatment could be supplemented with NaOH if the required alkalinity could not be maintained with Na3PO4 alone. If the feed water produces almost the right alkalinity desired in the boiler, it is preferable to use Na2HPO4 which is practically neutral. If the boiler water becomes too alkaline, the acidic NaH2PO4 would be selected. Both sodium pyrophosphate and metaphosphate are rapidly hydrolyzed under boiler water temperatures to orthophosphate.

NaPO3 + H2O = NaH2PO4 Na4P2O7 + H2O = 2Na2HPO4

Thus, their behavior within the boiler is identical with that of orthophosphates mentioned 5 above. However, NaPO3 solutions are practically neutral, whereas NaH2PO4 solutions are acidic. Hence the former would be preferred if the use of NaH2PO4 causes feedline corrosion. The use of internal treatment combined with suitable blow-down to remove sludge has contributed largely to the operation of the modern high-pressure steam boilers without the formation of hard

6.50  Engineering Chemistry scales. However, precaution should be taken to inspect them at least once in six months and remove the scale and sludge accumulations. 3. Colloidal conditioning:  Scale formation can also be minimized by introducing into the boiler some colloidal conditioning agents such as glue, agar agar, tannin, starch and sea-weed extract. These substances act as protective colloids. They function by surrounding the minute particles of CaCO3 and CaSO4 and prevent their coalescence and coagulation. Thus, the precipitated scale-forming salts are maintained in loose suspended form which can easily be removed by blow-down operation. Thus the scale formation is prevented. 4. Calgon conditioning:  Another approach for preventing scale formation is to convert the scale forming salts into highly soluble complexes which are not easily precipitated under the boiler conditions. In order to achieve this, sodium hexametaphosphate (Na PO3) 6 or Na2 [Na4P6O18] (its trade name is calgon) is generally employed. This substance interacts with the residual calcium ions forming highly soluble calcium hexametaphosphate and thus prevents the precipitation of scaleforming salts.

Na2 [Na4P6O18] ¤ 2 Na+ + [Na4P6O18] 2– 2 Ca2+ + [Na4P6O18] ¤ 4 Na+ + [Ca2P6O18] 2–

5. Conditioning with EDTA:  Phosphate treatment fails to prevent the formation of iron oxide and cuprous depositions. Sometimes, the phosphate chemicals themselves become a source of deposit formation (e.g., iron phosphate). In distinction to phosphate treatment, the conditioning of boiler water with complexing agents can secure scale-free and sludge free operation of boiler units and sometimes, also corrosion-free operation under certain conditions. These reagents are known as complex ones or complexing agents such as ethylenediamine tetraacetic acid (EDTA) and its disodium salt (Na2EDTA) known as Trylon-B, or even its tetrasodium salt (Na4EDTA). 6. Boiler compounds: Once, during the early times of Watt’s engine, is said that workmen, after cleaning the boiler and refilling with water, hung a bag containing potatoes in the boiler to cook them. They forgot about them and put the boiler into operation after closing it. Then, when the boiler was shut down again for manual removing of the scale, it was found that the scale formation was much less, much of it having come down in the form of loose sludge. The story seems to be true because, for many years, engineers used to throw some potatoes in the boiler after every cleaning operation. Perhaps, this is the form of internal treatment of boiler saline to reduce the adherent scale formation by rendering it in the form of softer sludge which can be easily removed by blow-off. Slowly, potatoes were replaced by starch and later by tannin and other organic and inorganic materials such as soda ash and caustic soda, phosphate, coagulant such as sodium aluminate, and deoxidizer like sodium sulphite. Natural wood and plant extracts and several synthetic products have also been used. US Navy was reportedly using a boiler compound consisting of 47% anhydrous disodium phosphate, 44% soda ash and 9% starch. Castor oil compounds and various synthetic antifoam compounds were also used in locomotive boilers to reduce foaming tendency. These boiler compounds do help in reducing scale formation when properly used under proper situations. (c) Blow-off:  In spite of external and internal treatment, the concentration of impurities in the boiler water goes on increasing continuously as the generation of steam continues. In order to keep the concentration of boiler water below a set figure, a part of the boiler water is blown off to waste and fresh feed water is filled in to replace it. The blow-down (or blowoff) may be done periodically or, preferably, continuously. During the blow-down process, a

Water Treatment  6.51 pan of concentrated water containing dissolved impurities and suspended sludge is removed and the remaining saline get diluted with the fresh feed water make-up.

8.3  Corrosion Corrosion of boiler tubes, plates, economizers and pipe lines can be mainly attributed to one or more of the following factors: (a) Presence of free acids in water. (b) Acids generated as a result of hydrolysis of some salts in water. (c) Acids formed by the hydrolysis of fatty lubricating oils. (d) Presence of dissolved gases such as O2, CO2, H2S etc., in water. (e) Presence of salts like MgCl2 which directly attack the boiler metal:

MgCl2 + Fe + 2H2O Æ Mg (OH)2 + FeCl2 + H2

(f) Presence of salts like MnS2 which may generate H2SO4 due to oxidation and hydrolysis. (g) Formation of galvanic cells. Some of these factors are discussed below:

(i) Dissolved oxygen: Dissolved oxygen is the chief source of corrosion in boilers and ancillary equipment. It is highly desirable to remove the dissolved oxygen or air from softened boiler waters by physical or by chemical means. The concentration of oxygen in boiler waters should be below 0.05 ppm for low pressure boilers and less than 0.01 ppm for high pressure boilers (~ 500 psi). Oxygen enters the boiler through raw make up water and also through infiltration of air into the condensate system. When the water containing dissolved oxygen is heated in the boiler, the free gas is evolved which corrodes the metal parts under the conditions obtaining in the boiler. The solubility of a gas is directly proportional to pressure and inversely proportional to temperature (Dalton’s law and Henry’s law). These two principles are made use of in the design of mechanical deaerators. In mechanical deaerators, dissolved oxygen is removed by injecting hot feed-water, as a fine spray, into a vacuum chamber heated externally by steam. Equipments with various designs are available. For complete removal of dissolved oxygen, chemical methods of treatment are adopted. Sodium sulfite is suitable for boilers operating below 650 psi. The sulphite reacts with the dissolved oxygen to form sulphate.’

Na2SO3 + 1/2 O2 Æ Na2SO4

This method is very effective for removing oxygen in low pressure boilers but cannot be used in high pressure boilers because (i) increasing dissolved salts concentrations produce foaming and priming and (ii) sodium sulphate decomposes and liberates SO2 and/or H2S. Ferrous sulphate is also used sometimes. It reacts with dissolved oxygen giving a precipitate of Fe (OH)2 which is oxidized to Fe(OH)3

FeSO4 + 2NaOH Æ Fe(OH)2 + Na2SO4 2Fe (OH)2 + H2O + O Æ ‘2Fe(OH)3

Hydrazine, N2H4 is now extensively used to remove dissolved oxygen in high pressure boilers. The pure compound is an explosive inflammable liquid (B.Pt. 113.5°C) but the 40% aqueous solution used

6.52  Engineering Chemistry for water treatment is quite safe to handle. However, rubber gloves must be worn to guard against the danger of dermatitis in some cases. One of the important advantage of hydrazine treatment is that combination with oxygen does not produce any salts. Nitrogen and water are the only reaction products obtained.

N2H4 + O2 Æ N2 + 2H2O

The molecular weight of hydrazine and oxygen are the same (viz. 32), so that complete elimination of 1 ppm of oxygen would require only 1 ppm of hydrazine (which is 1/8 of the necessary concentration of sodium sulphite). The residual hydrazine can be measured easily by a colorimeter. The amount of hydrazine added must be closely controlled. Any excess reagent decomposes in the boiler, liberating ammonia: 3N2H4 Æ 4NH3 + N2 Dissolved ammonia can bring about corrosion of some alloys, e.g., copper alloy condenser tubes. (ii) Dissolved mineral acids:  Most of the natural waters are alkaline excepting waters from mining areas or those polluted from acidic industrial wastes or those in which wet oxidation of sulphide minerals occur. Some inorganic salts like magnesium chloride and calcium chloride are also corrosive agents. MgCl2 hydrolyzes completely at 200°C producing hydrochloric acid as follows:

MgCl2 + 2H2O Æ Mg (OH)2 + 2HCl

CaCl2 also undergoes hydrolysis but to a lesser extent. At 600°C, its hydrolysis is about 25%. Silicic acid catalyzes the reaction so that in water containing silica, appreciable quantities of HCl may be formed at lower temperatures. The HCl thus produced reacts with iron as follows:

Fe+2HCl Æ FeCl2 + H2 FeCl2 + 2H2O Æ Fe(OH)2 + 2HCl

Hence, even a small amount of MgCl2 can cause considerable corrosion of the metal. If the amount of HCl formed is small, it might get neutralized by the alkalinity present in the water, but otherwise it should be neutralized by the addition of alkali. (iii) Dissolved carbon dioxide:  Natural waters contain CO2 water containing bicarbonates release CO2 on heating. If CO2 is released inside the boiler, it will go along with the steam and as the steam condenses, the CO2 is dissolved in water forming carbonic acid. This produces intense local corrosion called pitting. CO2, along with oxygen in water, can be removed by mechanical deaeration. CO2 in water can be removed by lime treatment. Another method of removing CO2 from water is to filter it through lime stone: CaCO3 + CO2 + H2O Æ Ca (HCO3)2 but this reaction produces temporary hardness in the water. CO2 can be converted into ammonium carbonate by the addition of ammonia.

CO2 + 2NH3 + H2O Æ (NH4)2CO3

However, the excess ammonia added to the boiler feed water may go along with steam to the condenser. If some O2 is also present in the condensate, the ammonia may attack the condenser tubes made of copper. Hence, a safe limit of 10 mg of NH3/litre of the condensate is prescribed.

Water Treatment  6.53 (iv) Formation of galvanic cells:  Corrosion can also occur because of galvanic cell formation between iron and other metals, present in the alloys used in boiler fittings. This may also lead to pitting corrosion. This can be prevented by suspending zinc plates which act as sacrificial anodes.

8.4  Caustic embrittlement Caustic embrittlement is a form of corrosion caused by a high concentration of sodium hydroxide in the boiler water. It is characterized by the formation of irregular intergranular cracks on the boiler metal particularly at places of high local stress, such as riveted seams, bends and joints. It is caused by the high concentration of NaOH which is capable of reacting with steels stressed beyond their yield point. It is most likely to occur in boilers operating higher pressures where NaOH is produced in the boiler by the hydrolysis of Na2CO3 as follows: Na2CO3 + H2O 2NaOH+CO2≠ The extent of the hydrolysis increases with temperature and may reach even 90% of the carbonate present. The rate and extent of corrosion by caustic embrittlement increases with the concentration of NaOH and temperature and hence with increasing operating pressure. NaOH reacts with iron forming magnetic oxide and hydrogen. 2NaOH + Fe Na2FeO2 + H2≠ 3(Na2FeO2) + 4H2O 6NaOH + Fe3O4 + 4H2≠ – 3Fe + 4OH Fe3O4 + 4H Under normal conditions in unstressed metal a fairly continuous film of oxide is produced. When the metal is stressed beyond its yield point, the oxide coating cracks and chemical attack continues into the metal mainly along grain boundaries. This may be due partly to the energy stored there and partly to the increased emf. produced as a result of the stress. The products of the reaction viz, Fe3O4 and hydrogen also tend to favor penetration along grain boundaries. The attack is considerable under these conditions because of the large area of the metal exposed. It is observed that boiler waters containing sodium sulphate or sodium phosphate inhibit the caustic embrittlement either by crystallizing out and plugging the capillaries and crevices with the solid salts (and prevent the infiltration of NaOH) before a dangerously high concentration of NaOH has been produced or that these salts act as buffer solutions and lower the emf. to such an extent that the corrosion cannot occur. Generally, the concentrations of Na2SO4: NaOH are maintained at l:l, 2:l or 3:l for operating pressures of 10, 20 and >20 atmospheres respectively to check caustic embrittlement in boilers. However, the requirements for water conditioning and prevention of caustic embrittlement may clash with each other, because, a very high concentration of Na2SO4 may lead to the formation of CaSO4 scales. Under these conditions, Na3PO4 should be used. Na3PO4 is an effective conditioning agent and also is over 300 times as effective as Na2SO4 in suppressing embrittlement. Caustic embrittlement may also be prevented by adding lignins or tannins which help in blocking the infiltration of NaOH through the hair-cracks. Addition of NaNO3, neutralization of excess alkali is also used to prevent embrittlement.

9.  DESALINATION OF WATER Water with high levels of dissolved salt is not suitable for domestic, industrial or irrigation uses. Sea water contain 3.5% dissolved salt. Sea water can be desalinated to separate into fresh water by

6.54  Engineering Chemistry the process called desalination. The process of removing salt from sea water to make it suitable for agricultural purposes or for drinking is called desalination. The people are trying to develop the desalination process in an economical way for providing fresh water for domestic use in regions where the availability of water is limited. Desalination can be performed by two process membrane process and thermal process. Reverse osmosis (RO) and electrodialysis are two membrane processes for desalination. Reverse osmosis is a separation process forcing a solvent from a region of high solute concentration through a membrane to a region of low solute concentration by applying a pressure in excess of the osmotic pressure. The flow of water (or other solvent) through a semipermeable membrane called osmosis. Normal osmosis is a process were natural movement of solvent flow from low to high concentration through a membrane. A membrane which is permeable to solvent and not to solute is called a semipermeable membrane. When a solution is separated from pure water by a semipermeable membrane, osmosis of water occurs from water to solution. This osmosis can be stopped by applying pressure equal to osmotic pressure, on the solution. If pressure greater than osmotic pressure is applied osmosis is made to proceed in the reverse direction to normal osmosis i.e. from solution to water. The membranes used for reverse osmosis have a dense barrier layer in the polymer matrix where most separation occurs. In most cases the membrane is designed to allow only water to pass through the dense layer while preventing the passage of solutes. This process requires a high pressure to exert on the high concentration side of a membrane usually 30-250 psi for fresh and brackish water and 600-1000 psi for sea water. This process is well known for its use in desalination. it has also been used to purify fresh water for medical, industrial and domestic applications since 1970s. Electrodialysis:  Electrodialysis is used to transport salt ions from one solution through ion-exchange membranes to another solution under the influence of an applied electric potential difference. The cell consists of a feed (dilute) compartment and a concentrate (brine) compartment formed by an anion exchange membrane and a cation exchange membrane placed between two electrodes. Cation exchange membranes allow only cations to pass through whereas onion exchange membranes allow passage of only anions through them. Multiple electrodialysis cells consists of alternating anion and cation exchange membranes and electrode. Under the influence of an electrical potential difference the negatively charged ions (CI¢) migrate toward the positively charged anode. These ions pass through the positively charged anion exchange membrane, but are prevented from further migration toward the anode by the negatively charged cation exchange membrane, and stay in the C steam which becomes concentrated with the anions. The positively charged species e.g. Na¢ in the D steam migrate toward the negatively charged cathode and pass through the negatively charged cation exchange membrane, these cations also stay in the C steam, prevented from further migration towards the cathode by the positively charged anion exchange membrane. As a result of the anion and cation migration, electric current flows between the cathode and anode, equal number of anion and cation charge equivalents are transferred from the D stream into the C stream and so the charge balance is maintained in each stream. The result of the electrodialysis process is an ion concentration increase in the concentration stream with a depletion of ions in the dilute solution feed steam. E stream is the electrode stream may consist of the same composition as the feed stream (ex. NaCl or Na2SO4) depending on the stack configuration anions and cations from the electrode stream may be transported into the C stream or anions and cations from the D stream may be transported into the E stream.

Water Treatment  6.55 Cathode: Anode:

2e– + 2H2O H2O 2Cl

H2 (g) + 2OH 2H+ + 1/2 O2 +2e Cl2 +2e

Fig. 17  Removal of ions by electrodialysis

Electrodialysis systems can be operated as continuous production or batch production process. Electrodialysis is applied to deionization of aqueous solutions. The major application of electrodialysis has been the desalination of sea water as an alternative to RO for potable water production and sea water concentration for salt production. Ro is more cost-effective when total dissolved solids (TDS) are 3000 ppm or greater, while electrodialysis is more cost- effective for TDS feed concentration less than 3000 ppm.

solved Questions Type–I Problems based on Calcium Carbonate Equivalent and Hardness

1. A sample of water contains 204 mg of CaSO4 /l. Calculate the hardness in terms of CaCO3 equivalent. Sol. Hardness in terms of = (Mass CaSO4 in mg/l) × Multiplication factor CaCO3 equivalent 100 = (Mass of CaSO4 in mg/l) × ____________________________ ​           ​ 2 × chemical equivalent of CaSO4 100 = 204 × ______ ​      ​ = 150 mg/l 2 × 68 Hardness = 150 ppm. 2. How many grams of MgCl2 dissolved per litre gives 190 ppm of hardness? Sol. Hardness in terms of = (Mass of MgCl2 in mg/l) × Multiplication factor CaCO3 equivalent

100 Hardness = (Mass of MgCl2 in mg/l) × _____________________________ ​           ​ 2 × chemical equivalent of MgCl4

6.56  Engineering Chemistry Hardness × 2 × chemical equivalent of MgCl2 So mass of MgCl2 = ​ _______________________________________        ​     100 190 × 2 × 47.5 = _____________ ​   ​      100 = 180.5 mg/L = 180.5 × 10 –3 mg/L Mass of MgCl2 = 18.05 × 10 –2 gm/L 3. A sample of water on analysis was found to contain the following impurities: Impurity Ca (HCO3)2 Mg (HCO3)2 CaSO4 MgSO4

Quantity

Molecular Weight

4 6 8 10

162 146 136 120

Calculate the temporary, permanent and total hardness of water in ppm, °Fr and °Cl. Sol. Temporary hardness is due to bicarbonates of calcium and magnesium while permanent hardness is due to sulphates and chlorides of calcium and magnesium. Impurity

Quantity (mg/L) Assumed

Molecular weight

Ca(HCO3)2

4

162

Mg (HCO3)2

6

146

CaSO4

8

136

MgSO4

10

120

MF

Hardness (ppm) in CaCO3 equivalent

____ ​ 100  ​ 162 ____ ​ 100  ​ 146 ____ ​ 100  ​ 136 ____ ​ 100  ​ 120

4 × ____ ​ 100  ​ 162 6 × ___ ​ 100  ​  146 8 × ___ ​ 100  ​  136 10 × ___ ​ 100  ​  120

\

Temporary hardness = Ca (HCO3)2 + Mg (HCO3)2 100 100 = 4 × ____ ​   ​ + 6 × ____ ​   ​ = 6.578 ppm 162 146 Permanent hardness = CaSO4 + MgSO4 100 100 = 8 × ____ ​   ​ + 10 × ____ ​   ​ = 14.215 120 136 \ Total hardness = 6.578 + 14.215 = 20.793 ppm Hardness Temporary Permanent Total

Ppm

°Fr

°Cl

6.578 14.215 20.793

0.6578 1.4215 2.0793

0.460 0.995 1.4555

[\  1 ppm = 0.1 °Fr = 0.07 °Cl]

Water Treatment  6.57

Type–II Problems based on EDTA method

4. 20 ml of standard hard water (containing 15 gm CaCO3 per litre) required 25 ml EDTA solution for end point. 100 ml of water sample required 18 ml EDTA solution, while same water after boiling required 12 ml EDTA solution. Calculate carbonate and non-carbonate hardness of water. Which buffer is used in this titration and what is its pH? Sol. 1 L SHW ∫ 15 gm CaCO3 equivalent \

1 ml SHW ∫ 15 mg CaCO3 equivalent

Standardisation of EDTA 25 ml EDTA solution ∫ 20 ml 20 \ 1 ml EDT solution = ​ ___  ​ ml SHW 25 20 = ​ ___  ​ × 15 mg CaCO3 equivalent 25 1 ml EDTA solution = 12 mg CaCO3 equivalent

Calculate of total hardness

100 ml HW = 18 ml EDTA equivalent 18 \ 1 ml HW = ____ ​    ​ ml EDTA solution 100 18 = ____ ​    ​ × 12 mg CaCO3 equivalent 100 18 \ 1000 ml HW = ____ ​    ​ × 12 × 1000 mg CaCO3 equivalent 100 \ Total hardness = 2160 ppm Calculate of permanent hardness (non-carbonate hardness)

100 ml BHW = 12 ml EDTA solution 12 \ 1 ml BHW = ____ ​    ​ ml EDTA solution 100 12 = ____ ​    ​ × 12 mg CaCO3 equivalent 100 12 \ 1000 ml BHW = ____ ​    ​ × 12 × 1000 mg CaCO3 equivalent 100 \ Total hardness = 1440 ppm \

Carbonate hardness = 2160 – 1440 ppm

= 720 ppm. Buffer solution used is NH4Cl + NH4OH, whose pH is 10. 5. 50 ml of standard hard water containing 1 mg pure CaCO3 per ml consumed 20 ml of EDTA. 50 ml of a water sample consumed 25 ml of same EDTA solution, using EBT. Calculate the total hardness of water sample in ppm.

6.58  Engineering Chemistry Sol.

1 ml SHW = 1 mg CaCO3 equivalent

Step–I, Standardization of EDTA 20 ml EDTA = 50 ml SHW 50 \ 1 ml EDTA = ___ ​    ​ ml SHW 20 50 = ___ ​    ​ × 1 mg CaCO3 equivalent 20 1 ml EDTA = 2.5 mg CaCO3 equivalent Step–II, Determination of total Hardness 50 ml HW = 25 ml EDTA 25 \ 1 ml HW = ​ ___ ​ ml EDTA 50 25 ___ = ​   ​ × 2.5 mg CaCO3 equivalent 50 25 \ 1000 ml HW = ​ ___ ​ × 2.5 × 1000 mg CaCO3 equivalent 50 \ Total Hardness = 1250 ppm.

Type–III Problems based on Alkalinity of water

6. 100 ml of a raw water sample on titration with N/50 H2SO4 required 12.4 ml of acid to phenolphthalein end point and 15 ml of the same acid at methyl orange end point. Determine the type and extent of alkalinity present in the water sample. Sol. At phenolphthalein end point Volume of acid used = A = 12.4 ml Water vs Acid N1V1 = N2V2 1 \ N1 × 100 = ​ ___  ​ × 12.4  [V2 = A = 12.4 ml] 50 1 \ N1 = _____ ​       ​ × 12.4 5000 1 \ Strength of terms of CaCO3 = ​ _____     ​ × 12.4 × 50 gm/L 5000 \ Phenolphthalein alkalinity = = At methyl orange end point: Volume of acid used = = Water vs Acid N3V3 =

P = 0.124 × 103 mg/L 124 ppm. A+B 15.0 ml = V4 N4V4

Water Treatment  6.59 1 N3 × 100 = ​ ___  ​ × 15.0 50 1 \ N3 = _____ ​       ​ × 15 5000 1 \ Strength of terms of CaCO3 = _____ ​       ​ × 15 × 50 5000 \

= 0.150 gm/L \

Methyl orange alkalinity = M = 0.150 × 103 mg/L

= 150 ppm. Relation:

( 

)

1 1 P (=124 ppm) > ​ __ ​  M ​ __ ​   ​  × 150 = 75 ppm  ​ 2 2

Nature: Therefore, we have CO3–2 alkalinity = 2 (M – P) = 2 (150 – 124) = 52 ppm OH– alkalinity = (2P – M) = (2 × 124 – 150) = 98 ppm. 7. 50 ml of a sample of water required 12.5 ml of N/50 H2SO4 at phenolphthalein end point. An additional 2.5 ml of N/50 H2SO4 was used at methyl orange end point. Calculate the different types of alkalinity present in the water sample. Sol. Phenolphthalein alkalinity Volume of acid use = A = 12.5 ml Water vs Acid N1V1 = N2V2 1 \ N1 × 50 = ___ ​    ​ × 12.5   [A = V2 = 12.5 ml] 50 1 \ N1 = _____ ​     ​ × 12.5 2500 1 \ Strength of terms of CaCO3 = _____ ​     ​ × 12.5 gm/L 2500 1 \ Phenolphthalein alkalinity = _____ ​     ​ × 12.5 × 103 mg/L 2500 P = 250 ppm. Methyl orange alkalinity Further acid used = Total volume of acid used = = Water vs Acid N3V3 =

B = 2.5 A+B 12.5 + 2.5 = 15.0 ml N4V4   (A + B = V4 = 15.0 ml)

6.60  Engineering Chemistry \

1 N3 × 100 = ​ ___  ​ × 15.0 50

1 N3 = _______ ​       ​ × 15 50 × 50 1 \ Strength of terms of CaCO3 = _______ ​       ​ × 15 × 50 gm/L 50 × 50 1 \ Methyl orange alkalinity = M = _______ ​       ​ × 15 × 50 × 103 mg/L 50 × 50 \

= 300 ppm. Relation:

( 

)

1 1 P (250 ppm) > ​ __ ​  M ​ __ ​    ​ × 300 = 150 ppm  ​ 2 2

Nature: Therefore, we have CO3–2 alkalinity = 2 (M – P) = 2 (300 – 250) = 100 ppm OH– alkalinity = (2P – M) = (2 × 250 – 300) = 200 ppm. 8. 100 ml of water sample, on titration with N/50 H2SO4 using phenolphthalein as indicator, gave the end point when 10 ml of acid were run down. Another lot of 100 ml of the sample also required 10 ml of the acid at methyl orange end point. What is the alkalinity of the sample? Comment on its nature. Sol. Volume of acid used = A = 10 ml Water vs Acid N1V1 = N2V2 1 \ N1 × 100 = ​ ___  ​ × 10 [A = V2] 50 10 1 \ N1 = ___ ​    ​ × ____ ​    ​  50 100 10 1 \ Strength of terms of CaCO3 = ___ ​    ​ × ____ ​    ​ × 50 gm/L 50 100 10 1 = ​ ___  ​ × ____ ​    ​ × 50 × 103 mg/L 50 100 = 100 mg/L \ Phenolphthalein alkalinity = P = 100 ppm. Methyl orange alkalinity: Another lot of 100 ml of the sample also required 10 ml of the acid. Another lot of 100 ml of the sample also required 10 ml of the acid. Volume of acid used at = A + B = 10 ml

Water Treatment  6.61 Methyl orange end point Water vs Acid

N3V3 = N4V4   (A + B = V4 = 10 ml) 1 N3 × 100 = ​ ___  ​ × 10 50 10 1 N3 = ___ ​    ​ × ____ ​    ​  50 100

\ \

10 1 \ Strength of terms of CaCO3 = ___ ​    ​ × ____ ​    ​ × 50 gm/L 50 100 \

10 1 Methyl orange alkalinity = M = ___ ​    ​ × ____ ​    ​ × 50 × 1000 mg/L 50 100 M = 100 mg/L = 100 ppm

Relation:

P = M

Nature of alkalinity: The alkalinity of water is only due to OH– ion. OH– alkalinity = 100 ppm. 9. 100 ml of a water sample required 9.5 ml of N/50 H2SO4 for neutralization at phenolphthalein end point and further 23.5 ml of N/50 H2SO4 at methyl orange end point. Calculate the alkalinity of water in terms of CaCO3 equivalent and comment on the nature of alkalinity. Sol. Phenolphthalein alkalinity Volume of acid used at phenolphthalein end point = A = 9.5 ml Water vs Acid N1V1 = N2V2 1 \ N1 × 100 = ​ ___  ​ × 9.5   [A = V2 = 9.5 ml] 50 1 1 \ N1 = ___ ​    ​ × ​ ____    ​ × 9.5 50 100 1 \  Strength of terms of CaCO3 equivalent = ​ ___  ​ × 50 1 1 Phenolphthalein alkalinity = ___ ​    ​ × ____ ​     ​ × 9.5 50 100 P = 95 mg/L \

1 ​ ____    ​ × 9.5 × 50 gm/L 100 × 50 × 103 mg/L

Phenolphthalein alkalinity = P = 95 ppm.

Methyl orange alkalinity: Volume of acid used at methyl orange end point = A + B = 9.5 + 23.5 = 33 ml (further acid used was 23.5 ml) Water vs Acid

N3V3 = N4V4   (V4 = A + B = 33 ml)

6.62  Engineering Chemistry 1 N3 × 100 = ​ ___  ​ × 50 1 N3 = ___ ​    ​ × 50

\

33

1 ​ ____    ​ × 33 100 1 1 \  Strength of terms of CaCO3 equivalent = ​ ___  ​ × ​ ____    ​ × 33 × 50 gm/L 50 100 1 1 \ Methyl orange alkalinity = M = ___ ​    ​ × ____ ​     ​ × 33 × 50 × 1000 mg/L 50 100 M = 330 ppm Relation: 1 1 P (= 95 ppm) > ​ __ ​  M ​ __ ​   ​  × 330 = 165 ppm  ​ 2 2 Nature: \

( 

)

The alkalinity of water is due to CO3–2 and HCO3–2 ions. CO3–2 alkalinity = 2P = 2 × 95 = 190

HCO3 – alkalinity = M – 2P = 330 – 190

= 140 ppm.

Type–IV Problems based on Lime soda process

10. A sample of water contains following impurities: Mg (HCO3)2 = 73 mg/L, CaCl2 = 222 mg/L, MgSO4 = 120 mg/L, Ca (NO3)2 = 164 mg/L, calculate the quantity of lime (74% pure) and soda (90% pure) needed for softening 5000 L of water. Sol. Conversion into CaCO3 equivalent S. No.

Constituents

Amount mg/L 73

MF ____ ​ 100  ​ 146

CaCO3 equivalent 100 73 × ​ ___  ​= 50 mg/L 146

1.

Mg (HCO3)2

2.

CaCl2

222

100 ​ ___  ​

100 222 × ​ ___  ​= 200 mg/L 111

3.

MgSO4

120

100 ​ ___  ​

120 × ____ ​ 100  ​= 100 mg/L 120

4.

Ca (NO3)2

164

100 ​ ___  ​

164 × ____ ​ 100  ​= 100 mg/L 164

111 120 164

100 74 Lime required = ____ ​    ​ [2 × Mg(HCO3)2 + MgSO4 as CaCO3 eq] ________ ​      ​ × volume of 100 % purity water

100 74 = ____ ​    ​ [2 × 50 + 100] × ____ ​   ​ × 5000 100 74



= 200 mg/L × 5000 L = 106 mg = 1 kg

Water Treatment  6.63 106 100 Soda required = ____ ​   ​ [CaCl2 + MgSO4 + Ca (NO3)2 as CaCO3 eq] ________ ​      ​ × volume of 100 % purity water 106 100 = ____ ​   ​ [200 + 100 + 100] mg/L × ____ ​   ​ × 5000 L 100 90 = 2,356,000 mg = 2.356 kg 11. Calculate the quantities of lime and soda required for cold softening of 2, 00, 000 liters of water using 16.4 ppm of sodium aluminates as a coagulant. The results of the analysis of raw water and softened water are as follows: Raw water Softened water Ca+2 – 160 ppm CO3–2 – 30 ppm Mg+2 – 72 ppm OH– – 17 ppm – HCO 3 – 732 ppm Dissolved CO2 – 44 ppm 12. Write the chemical equations involved. Sol. S. No.

Constituents

Amount (in ppm)

MF

CaCO3 equivalent

Raw water ____ ​ 100  ​ 40 ____ ​ 100  ​ 24

____ ​ 100  ​× 160 = 400 40 ____ ​ 100  ​× 72 = 300 24

732

100 ​ ____   ​ 122

100 ​ ____   ​× 732 = 600 122

CO2

44

NaAlO2

16.4

____ ​ 100  ​ 44 100 ____ ​    ​ 164

____ ​ 100  ​× 44 = 100 44 100 ____ ​    ​× 16.4 = 10 164

1.

Ca+2

160

2.

Mg+2

72

3.

HCO3–

4. 5.

Treated water 1.

OH–

17

____ ​ 100  ​ 34

____ ​ 100  ​× 17 = 50 34

2.

CO3–2

30

____ ​ 100  ​ 60

____ ​ 100  ​× 30 = 50 60

74 Lime required for just softening = ____ ​    ​ [Mg+2 + HCO3– + CO2 – NaAlO2 as CaCO3 eq] 100 74 = ____ ​    ​ [300 + 600 + 100 – 10] 100 = 732.6 ppm. 106 Soda required for just softening = ____ ​   ​ [Ca+2 + Mg+2 – HCO3 as CaCO3 eq] 100

6.64  Engineering Chemistry 106 = ____ ​   ​ [400 + 300 – 600] 100 = 106 ppm. Reactions involved to get excess OH– and CO3–2 ions are

Ca

+2

Ca (OH)2 Æ Ca+2 + 2OH– + Na2CO3 Æ CaCO3 + 2Na+ Na2CO3 Æ 2Na+ + CO3–2

Lime required for excess OH– ions in treated 74 Water = ____ ​    ​ [OH– as CaCO3 eq 100 74 = ____ ​    ​ × 50 100 = 37 ppm. Soda required for excess OH– ions and CO3–2 ions in treated 106 Water = ____ ​   ​ [OH– + CO3–2 as CaCO3 eq] 100 106 = ____ ​   ​ [50 + 50] 100 = 106 ppm. Total lime required for 2,00,000 litres of

Water = [6.1 + 6.3] × 2,00,000 mg

200000 = (732.6 + 37) × ​ _______  ​    kg 104 = 769.6 × 0.2 kg = 153.92 kg. Total soda required for 200000 litres of

Water = [6.2 + 6.4] × 200000 mg

200000 = [106 + 106] × ​ _______  ​    kg 106 = 212 × 0.2 kg = 42.4 kg

12. Calculate the quantity of lime (74% pure) and soda (90% pure) required for softening 50,000 litres of water containing Mg (HCO3)2 = 50 mg/L MgCl2 = 6 mg/L Ca (HCO3)2 = 81 mg/L CO2 = 44 mg/L HCl = 73 mg/L Al2 (SO4)3 = 57 mg/L Sol. Conversion into CaCO3 equivalent

Water Treatment  6.65 S. No.

Constituents

Amount (mg/L)

1.

Mg(HCO3)2

50

2.

Ca(HCO3)2

81

3.

MgCl2

6

4.

CO2

44

5.

HCl

73

6.

Al2 (SO4)3

57

MF

CaCO3 equivalent (mg/L)

100 ____ ​    ​ 146 100 ____ ​    ​ 162 100 ____ ​   ​  95 100 ____ ​   ​  44 100 ____ ​   ​  73 100 ____ ​    ​ 114

50 × ___ ​ 100  ​ = 34.26 146 81 × ___ ​ 100  ​ = 50 162 100 6 × ​ ___  ​ = 6.31 95 44 × ___ ​ 100  ​ = 100 44 73 × ___ ​ 100  ​ = 100 73 57 × ___ ​ 100  ​ = 50 114

74 Lime requirement = ____ ​    ​ [Ca (HCO3)2 + 2 × Mg (HCO3)2 + Al2 (SO4)3 + MgCl2 + CO2 + 100 100 Volume of water HCl as CaCO3 eq] × ________ ​      ​ × ​ ______________  ​      kg % purity 106 100 50000 74 = ____ ​    ​ [50 + 2 × 34.26 + 50 + 6.31 + 100 + 100] × ____ ​   ​ × ​ ______  ​   kg 100 74 106 50000 = 374.83 × ​ ______  ​   = 18.74 kg. 106 106 Soda requirement = ____ ​   ​ [MgCl2 + Al2 (SO4)3 + HCl as CaCO3 eq] 100 100 Volume of water × ________ ​      ​ × ​ ______________  ​      kg % purity 106 106 100 50000 = ____ ​   ​ [6.31 + 50 + 100] × ____ ​   ​ × ​ ______  ​   kg 100 90 106 106 100 50000 = ____ ​   ​ × 156.31 × ____ ​   ​ × ​ ______  ​   kg 100 90 106 = 9.20 kg. 13. Calculate the amount of lime required for softening 50,000 litre of hard water containing CaCO3 = 25 ppm MgCO3 = 144 ppm CaCl2 = 111 ppm MgCl2 = 95 ppm Na2SO4 = 15 ppm Fe2O3 = 25 ppm Sol. Na2SO4 and Fe2O3 do not impart any hardness and hence they do not consume lime or soda. Conversion into CaCO3 equivalent

6.66  Engineering Chemistry S. No. 1. 2. 3. 4.

Constituents

Amount

CaCO3

25 ppm

MgCO3

144 ppm

MgCl2

95 ppm

CaCl2

111 ppm

MF

CaCO3 equivalent

100 ____ ​   ​  100 100 ____ ​   ​  84 100 ____ ​   ​  95 100 ____ ​   ​  111

100 25 × ____ ​   ​ = 25.0 ppm or mg/L 100 100 ____ 144 × ​   ​ = 171.4 ppm or mg/L 84 100 95 × ​ ____ ​ = 100.0 ppm or mg/L 95 100 ____ 111 × ​   ​ = 100.0 ppm or gm/L 111

74 Lime requirement = ____ ​    ​ [CaCO3 + 2 × MgCO3 + MgCl2 in terms of CaCO3 eq] × volume 100 of water 50000 74 = ____ ​    ​ [25.0 + 2 × 171.4 + 100] × ​ ______  ​   kg 100 106 50000 74 = ____ ​    ​ (467.8) × ​ ______  ​   = 17.31 kg. 100 106

Type–V Problems Based on Zeolite Process

14. The hardness of 40,000 litres of a sample of water was completely removed by passing it through a zeolite softener. The softener then required 150 litres of sodium chloride (Brine) solution containing 100 g/litre of NaCl for regeneration. Calculate the hardness of the water sample. Sol. Solution of sodium chloride used for regeneration contains 100 gm NaCl per litre. So NaCl contained in 150 litres of sodium chloride solution = 100 gm/L × 150 L = 100 × 150 gm = 15,000 gm of NaCl.

( 

)

100 100 15,000 gm of NaCl = 15,000 × ____ ​   ​ g CaCO3 eq  ​   MF = ____ ​   ​  ​ 117 117 = 12820.51 g CaCO3 eq = 12820.51 × 103 mg CaCO3 eq = 1.282 × 107 mg CaCO3 eq \  Hardness of 40,000 litres of water = 1.282 × 107 mg CaCO3 eq Hardness of 1 litre of 1.282 × 107 Water = ​ __________     ​  mg CaCO3 eq 40,000 = 320.5 mg CaCO3 eq Hardness of water = 320.5 mg/L = 320.5 ppm.

Water Treatment  6.67

15. An exhausted zeolite softener required 500 litres of sodium chloride solution containing 100 gm/L of NaCl for regeneration. If the hardness of water is 600 ppm, calculate the volume of water softened by softener. Sol. Sodium chloride solution used for regeneration contains 100 gm NaCl per litre. \  NaCl contained in 500 litres of sodium chloride solution = 100 gm/L × 500 L = 50,000 gm of NaCl 100 Now, 50,000 of NaCl = 50,000 × ____ ​   ​ g CaCO3 eq 117 = 4.27 × 104 g CaCO3 eq = 4.27 × 107 mg CaCO3 eq Let ‘v’ litres of 600 ppm (i.e, 600 mg/L) of water contains 4.27 × 107 mg CaCO3 eq hardness. V × 600 = 4.27 × 107

4.27 × 107 V = ​ _________  ​    Litre 600 V = 7.116 × 104 Litre.

SuMMARY 1. (a) Hardness – property of water which prevents lather formation with soap solution. (b) Types of hardness (i) Temporary hardness – due to presence of bicarbonates of calcium, magnesium and other heavy metals and carbonate of iron. (ii) Permanent hardness – due to the presence of sulphates and chlorides of calcium magnesium, iron and other heavy metals. (a) Units of hardness and inter relation. 1 ppm = 1 mg/L = 0.1° Fr = 0.07°Cl (b) Temporary hardness is removed by simple boiling. (c) Permanent hardness can be removed by (i) Zeolite process (ii) Lime soda process (iii) Ion exchange process (iv) Calcium carbonate equivalent [Mass of hardness producing substance] × 100 Calcium carbonate equivalent =​  ________________________________________________               ​ 2 × Chemical equivalent of hardness producing substance 2. EDTA method (a) Is used to determine total, temporary and permanent hardness. (b) It is complexometric titration

6.68  Engineering Chemistry

(c) EBT is used as indicator. (d) NH4Cl + NH4OH is used as buffer solution which maintains the pH at 10. (e) If standard hard water contains 1 mg CaCO3 in 1 ml distilled water and let V1, V2, and V3 are the volumes of EDTA used against standard hard water, unknown hard water and boiled water then

V2 Total hardness = ___ ​   ​ × 1000 ppm. V1 V2 Permanent hardness = ___ ​   ​ × 1000 ppm. V1

( 

)

V2 V2 Temporary hardness = ​ ___ ​   ​  × 1000 – ___ ​   ​ × 1000  ​ ppm V1 V1

3. Alkalinity of water



(i) H+ + OH– (ii) H+ + CO3–2 (iii) HCO3– + H+



4. Lime – soda process







H2O  P=A HCO3– H2O + CO2   B

  M = A + B

74 (a) Lime (purity %P) required for softening = ____ ​    ​ [Temp Ca+2 + 2 × temp. Mg+2 + perm 100 (Mg+2 + Fe +2 + Al+3) + CO2 + H+ (HCl or H2SO4) + HCO –3 - NaAlO2 in terms of V 100 CaCO3 eq] × ________ ​      ​ × ___ ​    ​ kg % purity 106 106 (b) Soda (Purity = %P) required = ____ ​   ​ [Perm (Ca+2 Mg+2 + Al+3 + Fe 100 V 100 H2SO4) – HCO –3 in terms of CaCO3 eq] × ________ ​      ​ × ___ ​    ​ kg. % purity 106 5. (a) Zeolite is – Na2O. Al2O3 X SiO2. yH2O

Where

x = 2 – 10 Y = 2 – 6

Na2Z + Ca+2 Zeolite

CaZ + 2Na+ Hardness causing ions.

(b) Exhausted zeolite is regenerated by 10% brine solution. CaZ + 2NaCl Na2Z + CaCl2 MgZ + 2NaCl Na2Z + MgCl2

Exhausted brine Zeolite

Regenerated (Zeolite)

+2

) + H+ (HCl or

Water Treatment  6.69

Unsolved Questions

1. 2. 3. 4.

Define hardness. What are the different types of hardness of water? Differentiate between hard water and soft water. Explain the causes of hardness of water. Write the constituents responsible for temporary hardness of water. Discuss the treatment method. 5. What is temporary hardness? How can it be removed? 6. Differentiate between carbonate and non carbonate hardness. 7. How is hardness expressed? Give various units of hardness. How are they related to each other? 8. Why do we express hardness of water in terms of CaCO3 equivalent? 9. Describe a method to determine temporary, permanent and total hardness of hard water. 10. Write the names of indicator and buffer solution used in the determination of the hardness of water by EDTA method. 11. Write the structure and complete name of EDTA and EBT. 12. Draw neat and labeled diagram of permutit (or zeolite) process of softening water. Discuss the chemistry involved in it. Discuss its merits over lime soda process. 13. Explain the lime soda process used for softening the hard water. 14. Define zeolite. How does zeolite function in removing the hardness of water? 15. Compare zeolite and lime soda process of softening. 16. What is the principle of EDTA titration? Briefly describe the estimation of hardness of water by EDTA method. 17. Write short notes on (i) Scale and sludge formation (ii) Caustic embrittlement (iii) Boiler corrosion (iv) Internal conditioning 18. Explain the following (i) Hard water does not give lather with soap. (ii) Alkalinity of water can’t be due to simultaneous presence of OH–, HCO3– and CO3–2. (iii) Temporary hardness is removed by boiling. (iv) Buffer solution (NH4OH + NH4Cl) is added during determination of hardness of water by EDTA method. 19. Write short notes on (i) Potable water (ii) Polished water (iii) Alkalinity of water (iv) Zeolite process and its limitation

6.70  Engineering Chemistry

20. 21. 22. 23. 24.

What are boiler troubles and what are their consequences. Prove that 1 ppm = 1 mg/L. Distinguish between temporary and permanent hardness of water. Differentiate between scale and sludge. Describe the ion exchange method of demineralisation of water.

Type–I Problems based on Calcium Carbonate Equivalent and Harness

25. A sample of water contains 81 mg/L Ca (HCO3) 2, 73 mg/L Mg (HCO3) 2, 120 mg/L MgSO4 and 136 mg/L CaSO4. Calculate the temporary and permanent hardness of the water sample. [Ans: Temporary hardness = 100 ppm, Permanent hardness = 200 ppm] 26. A sample of water on analysis has been found to contain following in mg/L Ca (HCO3)2 = 4.86, CaSO4 = 6.80, MgCl2 = 95.0, Mg (HCO3)2 = 73.0 Calculate the carbonate and non carbonate hardness also in ppm. Express your answer also in °Fr and °Cl. [Ans: Carbonate hardness = 53 ppm, 5.3 °F, 3.71 °Cl, Non carbonate hardness = 105 ppm, 10.5 °F, 7.35 °Cl]

Type–II Problems based on EDTA Method





27. 0.28 gm of CaCO3 was dissolved in HCl and the solution was made to one litre with distilled water. 100 ml of this solution required 28 ml EDTA solution on titration. 100 ml of an unknown hard water sample required 36 ml of the same EDTA solution on titration using EBT as an indicator. After boiling 100 ml of this water, cooling, filtering and then titration required 15 ml of EDTA solution. Calculate the temporary and permanent hardness of water. [Ans: Temporary hardness = 210 ppm, Permanent hardness = 150 ppm] 28. 1 L of a standard hard water sample contains 1 g of CaCO3. 100 ml of this standard hard water required 25 ml EDTA on titration. 100 ml of an unknown hard water sample required 10 ml of same EDTA on titration. After boiling 100 ml of this water, cooling, filtering and then titration, required 5 ml of EDTA solution. Calculate the carbonate and non carbonate hardness of water. [Ans: Non-Carbonate hardness = 200 ppm, Carbonate hardness = 200 ppm]

Water Treatment  6.71

Type–III Problems based on Alkalinity of water

29. 100 ml of a water sample, on titration with N/50 H 2SO 4 required 12 ml of acid to phenolphthalein end point. Further 12 ml of the same acid was used at methyl orange end point. Calculate the different types of alkalinity present in the water sample. [Ans: CO3–2 alkalinity = 240 ppm] 30. 500 ml of a water sample, on titration with N/50 H2SO4 gave a titre value of 7.5 ml at phenolphthalein end point and 20.0 ml at methyl orange end point. Calculate the different types of alkalinity present in the water sample. [Ans: OH– alkalinity = Nil, CO3–2 alkalinity = 30 ppm and HCO3– alkalinity = 10 ppm]

Type–IV Problems based on Lime–Soda process

31. Calculate the amount of lime (84% pure) and soda (92% pure) required for treatment of 10,000 litres of water containing: Ca (HCO3)2 = 40.5 ppm Mg (HCO3)2 = 36.5 ppm MgSO4 = 60.0 ppm CaSO4 = 68.0 ppm CaCl2 = 111.0 ppm And NaCl = 20.0 ppm [Ans: L = 1.10 kg S = 2.304 kg] 32. Yamuna canal water on analysis gave the following results: CO2 = 22 ppm HCO3– = 305 ppm Ca+2 = 80 ppm Mg+2 = 48 ppm Total solid = 500 ppm Calculate carbonate and non-carbonate hardness of the water sample. Also calculate the amount of lime and soda that would be required for softening the above water, if 417 ppm of FeSO4. 7H2O is used as coagulant. (At weight Ca = 40, Mg = 24, Fe = 56, S = 32) [Ans: Carbonate hardness = 250 ppm, non-carbonate hardness = 150 ppm, L = 629 mg/L, S = 38 mg/L] 33. Calculate the amount of lime (87% pure) and soda (91% pure) required to soften one million litres of water containing: Mg+2 = 25 mg/L Ca+2 = 20 mg/L – HCO3 = 150 mg/L CO2 = 30 mg/L Na+ = 10 mg/L [Ans: L = 218.9 kg, S = 33.6 kg]

Chapter

7

The Phase Rule The phase rule is an important tool used for the quantitative treatment of systems in equilibrium. It enables us to predict the conditions that must be specified for a system to exhibit equilibrium. J.W. Gibbs enunciated the phase rule in 1876 while investigating heterogeneous equilibrium. According to the phase rule, for systems in complete internal equilibrium,

F = C – P + 2

Where C is the number of components; P is the number of phase and F is the number of degree of freedom.

1.  Definitions 1.1 Phase A phase is defined as any homogeneous and physically distinct part of a system which is bounded by a surface and is mechanically separable from other parts of the system. Examples of various types of phases are: (i) A gas mixture constitutes a single phase since gases are completely miscible. (ii) Immiscible liquids constitute different phases. Thus, carbon tetrachloride and water do not mix with each other, so form two phases. (iii) A system consisting of a liquid in equilibrium with its vapor constitutes two phases. (iv) Ice-liquid water – water vapor constitutes system containing three phases. (v) Completely miscible liquids such as water and alcohol and benzene and chloroform constitute one–phase system. (vi) Consider the decomposition of calcium carbonate to CO2 and CaO. CaCO3(s)

CaO (s) + CO2(g)

Here, there are two solid phases and one gaseous phase. So it is a three-phase system. It should be noted that each phase in a heterogeneous system is homogenous in itself. When various phases are in equilibrium with one another in a heterogeneous system, there can be no transfer of energy

7.2  Engineering Chemistry or mass from one phase to another. This means that at equilibrium, the various phases must be at the same temperature and pressure.

1.2  Components The number of components of a system at equilibrium is defined as the smallest number of independently variable constituents by means of which the composition of each phase can be expressed either directly or in terms of chemical equation. We may consider a few examples. 1. Water exists in three phases, Ice

liquid

vapor

However, the composition of each phase can be expressed in terms of H2O. Hence it is one component system. Components can be given by the equation C = N – E Where E is the number of independent equations relating the concentrations of the N species and C is component. for examples (i) Consider a water system at equilibrium there are three phases

Ice

water

vapor

(Solid)

(Liquid)

(Gas)

It is a one component system because each phase can be expressed in terms of only one component. 2. Sulphur exists in four phases i.e., rhombic sulphur, monoclinic sulphur, liquid and vapor. But since the composition of each phase can be expressed in term of sulphur only, it is one component system. 3. Consider an aqueous sucrose solution. The composition of the solution phase can be expressed by specifying the amounts of sugar and water. Hence, it is a two-component system. 4. In a chemically reactive system involving reactions between various species, the situation is slightly different consider the thermal decomposition of CaCO3 in a closed vessel: CaCO3(s)

CaO (s) + CO2(g)

The system contains three phases, CaCO3(s), CaO (s) and CO2(g). However, as a result of the existence of equilibrium, the number of components is only two. If we chose them as CaO and Co2 then the composition of the solid CaCO3 can be given by CaO + CO2. If CaCO3 and CO2 are taken as the components, then the composition of solid CaO can be given by CaCO3 – CO2 and so on. In a chemically reactive system, the number of components is given by

C = N – m – n – R

Where N is the number of chemical species, m is the number of independent equilibrium conditions, n is the number of relations between concentrations due to initial concentrations due to initial conditions.

The Phase Rule  7.3 In this system we have 3 substance and they are related by one equation. So number of components are 2. Due to existence of equilibrium, composition of each of the three phases can be expressed in terms of any two constituents, CaCO3, CaO and CO2 as shown below. If CaCO3 and CaO constituents Phase CaCO3(s) CaO (s) CO2(g)

If CaCO3

(s)

Component CaCO3 + 0CaO 0 CaCO3 + CaO CaCO3 – CaO

and CO2 constituents are chosen Phase CaCO3(s) CaO (s) CO2(g)

Component CaCO3 + 0CO2 CaCO3 – CO2 0CaCO3 + CO2

If CaO and CO2 constituents are chosen

Phase CaCO3(s) CaO (s) CO2(g)

Component CaO + CO2 CaO + 0CO2 0CaO + CO2

Thus in all the cases, only two components are required to express the composition of each phase. Hence it is a two-component system and is also called binary system.

1.3  Degree of Freedom (F) The degree of freedom of a system is defined as the number of independent variables, such as temperature pressure and composition which are to be specified to describe a system completely. It is also termed as variance. Phase Rule:  The phase rule is an important generalization and a versatile tool which is related to the study of behavior of heterogeneous systems. In general, it may be stated that with the help of phase rule, it has been possible to predict quantitatively the effect of changing pressure, temperature and concentration on a heterogeneous system in equilibrium by means of a phase diagram. Gibbs Phase Rule:  Gibbs phase rule states that if a heterogeneous system is influenced by temperature, pressure and concentration and not by any action like gravity, electrical forces, magnetic forces or by surface action then the sum of number of phase (P) and degree of freedom (F) is greater than the number of components (C) by two. It is expressed mathematically as follows:

P + F = C + 2



Or

F = C – P + 2

7.4  Engineering Chemistry Derivation of the Phase Rule: We shall now derive phase rule. Consider a system of C components (C1, C2, C3....Cc) distributed between P phases (a, b, g, ....P). Assume that the passage of the component from one phase to another does not constitute a chemical reaction. The state of each phase of a system is completely specified by the two variables, temperature and pressure and also by composition of each phase. In other words, the state of each phase is specified by

T, P (x1a , x2a , xCa ), (x1b , x2b , xCb ), ...(x1P, xPb , xCP)

...(1)

Where xis are composition of the compositions of the component. The total number of variables is thus CP + 2. However, in equation (1) all the variables are not independent since in each phase, the sum of the mole fraction must equal unity, i.e.,

x1a + x2a + xCa + ... = x1b + x2b + xCb + ...,etc.

...(2)

In other words,



Sx

i, p

= 1 (i = 1, 2, 3, ...C)

For all P the phases separately. There are thus P relations of this type. Again, for complete equilibrium to exist between the phases, the chemical potential of each species must be same in each phase, i.e.,

m1, a = m1, b = m1, g = ... m1, P  m2, a = m2, b = m2, g = ... m2, P mC, a = mC, b = mC, g = ... mC, P

...(3)

We see that there are P – 1 separate equations for each component. Hence, for C components, the number for each equation is C(P – 1). Also, the equilibrium conditions for a chemical reactions require that the chemical affinity, Af for each reactions at equilibrium must be zero. i.e.,



Af, i = 0 (i = 1, 2, 3 ...,r ¢)

...(4)

This means that there is r ¢ equation of this type. Hence, total number of restricting conditions is



P + C (P – 1) + r ¢

...(5)

The degree of freedom, F, is given by difference between the number of variables required to specify the state of the system (i.e., CP + 2) and the number of restrictions imposed by their interdependence (i.e., equation 5), Thus,

F = (CP + 2) – (P + CP – C + r ¢)

= 2 + (C – r ¢) – P

...(6)

Equation 6 is known as Gibbs phase rule. If the system in non-reactive, i.e., no reaction takes place in it, then equation of type Af = 0 will be absent, i.e., r ¢ = 0. Hence, the phase rule becomes

F = C – P + 2

...(7)

The Phase Rule  7.5 We see from this equation that for a system having a given number of components, greater number of phases smaller will be the number of degree of freedom. Thus, for one component system like water, the maximum number of phases that can coexist at equilibrium is three so that for this degree of freedom is zero. In general, for a system having a given number of components where number of phases is maximum, the degree of freedom is zero. On the other hand, the system with the given number of phases, larger the number of components, greater will be the number of the degrees of freedom. In these cases, variables such as T, P and concentration must be specified in order to describe the state of the system completely. The phase rule does not tell us anything regarding the composition of matter. It merely states that a system having the same number of the degrees of freedom behave alike thermodynamically. The phase rule is an important generalization that enables us to investigate highly complex heterogeneous equilibria. Example 1  Determine phase, component and degree of freedom in the given system MgCO3(s) Solution: Number of phases Components Degree of Freedom

= = = = = = =

MgO (s)

CO2(g)

3 Substance – Relation 3–1 2 C–P+2 2–3+2 1 (System is univariant)

Example 2  Determine phase, component and degree of freedom in the given system i.e., in the dissociation of ammonium chloride. Solution:  It dissociates as follows: (i)  NH4Cl (s)

NH3(g) + HCl (g)

There are two phases, solid and gas. The system will be a one component or two components depending upon the relative quantities of HCl and NH3. (a) When ratio of NH3(g) and HCl (g) is 1:1 then we write one more equation with the above equation i.e., (ii)  Concentration of NH3 = concentration of HCl Number of Phases = 2 Number of components = Substance – Relation = 3 – 2 = 1 Degree of Freedom = C – P + 2 = 1 – 2 + 2 = 1 (Univariant) (b) When ratio of NH3 : HCl is 1:2 (i.e., not same), then we cannot write (ii) equation as in first case so, Number of components = 3 – 1 = 2

7.6  Engineering Chemistry Degree of Freedom = C – P + 2 = 2 – 2 + 2 = 2 (Bivariant) Example 3  Determine phase, component and degree of freedom in the given system N2O4(g) Solution: Number of phases Number of components Degree of Freedom

= = = = =

2NO2(g)

1 Substance – Relation 2–1=1 1–1+2 2 (Bivariant)

Example 4  Determine phase, component and degree of freedom in the given system Solution: (i) C (s) + O2(g) CO2(g)

(ii) C (s) + 1/2O2(g)

CO (g)

Number of phases Number of components Degree of Freedom

= = = = =

2 4–2 2 2–2+2 2 (Bivariant)

Example 5  Determine phase, component and degree of freedom in the system of dilute solution of H2SO4 in water. Solution: (i) H2SO4 + H2O Æ H3O + + HSO4 – (ii) H2SO4 – + H2O Æ H3O + + SO4 – (iii) Electroneutrality (Number of cations and anions are equal) Number of phases = 1 Number of components = 5 – 3 = 2 Degree of Freedom = 2 – 1 + 2 = 3 (Trivariant) Example 6  Determine phase, component and degree of freedom in an aqueous solution saturated with respect to both NaCl and KCl and is in equilibrium with the vapor phase. Solution: (i) NaCl Na+ + Cl– (ii) KCl K+ + Cl– (iii) H2O (l) H2O (v) (iv) Electroneutrality Number of phases = 4

The Phase Rule  7.7 Number of components = 7 – 4 = 3 Degree of Freedom = 3 – 4 + 2 = 1 (Univariant) Example 7  Determine the number of phases, components and degree of freedom in a mixture of N2, H2 and NH3 at equilibrium. Solution: 2N2(g) + 3H2(g) Number of phases Number of components Degree of Freedom

= = = = =

2NH3(g)

1 3–1 2 2–1+2 3 (Trivariant)

Example 8  Determine the number of phases, components and degree of freedom in the given system: Na2SO4.10H2O (s)

Na2SO4(s) + 10H2O (g)

Number of phases = 3 Number of components = 3 – 1 = 2 Degree of Freedom = 2 – 3 + 2 = 1 (Univariant) F = 0 i.e. to define the system completely, nothing has to be specified. F = 1 i.e. such a system can be completely defined by specifying either pressure or temperature. This is because on fixing one variable, the other becomes automatically fixed. F = 2 i.e. such a system can be completely defined by specifying both temperature and pressure.

1.4 Application of Phase Rule All the systems are classified on the basis of number of components present. Thus, we may have one, two, three, etc. component systems. The phase rule is perfectly applicable to these systems. The phase diagram is an important medium to clearly indicate the equilibrium conditions between different phases in a system. The phase diagram is helpful for studying and controlling various processes such as phase separation, solidification of metals, and the change of structure during heat treatment etc.

1.5 Phase Diagram Is a graphical representation of the pressure and temperature under which two or more physical states can exist together in a state of dynamic equilibrium. Phase diagram illustrates the conditions of equilibrium between various phases of a substance. If the temperature is plotted against pressure the diagram is called P-T diagram. The study of phase diagram is vital for proper understanding of the relative stability of equilibrium between phases with the help of knowledge of the number of phases and components of the system,

7.8  Engineering Chemistry phase rule enables one to predict clearly the conditions under which a system can remain in equilibrium.

1.6 One Component System The complexity of one component systems depends on the number of solid phases that can exist in the system. The simplest case is that in which only single solid phases occur. Water system and carbon dioxide system are the examples of the simplest one component system. For a one component system the phase rule equation can be written as

F = C – P + 2

= 1 – P + 2 = 3 – P

There may arise three cases.

Case 1:  When only one phase is present

F = 3 – P

= 3 – 1 = 2 (Bivariant system) It can be completely defined by specifying the two variables temperature and pressure or both temperature and pressure can be varied independently. Therefore a single phase is represented by an area on P-T graph. Case 2:  When two phases are present in equilibrium, the degree of freedom will be F = 3 – P = 3 – 2 = 1 (Monovariant or univariant system) This means that the pressure cannot be altered independently if the temperature is changed. In such a case pressure is fixed automatically for a given temperature. A two-phase system is depicted by a line on a P-T graph. Case 3:  When three phases are present in equilibrium, the degree of freedom will be

F = 3 – P

= 3 – 3 = 0 (Nonvariant or invariant system) This special condition can only be attained at a definite temperature and pressure. A three-phase system is depicted by a point on the P-T graph. This point is termed as the triple point as the three phases are in equilibrium.

1.7 The Water System Under normal conditions the water system is a three-phase and one-component system. The three phases are liquid, ice and vapor. All these are represented by one chemical entity (H2O), hence it is one component system. On the basis of experimental data obtained for the water system a plot of relationship between various phases namely solid ice, liquid water and water vapor under different conditions of temperature and pressure are shown below:

The Phase Rule  7.9

Fig. 1  The phase diagram of water system



The three forms of water (Ice, water and vapor) constitute the following equilibrium: (i) Single phase equilibrium (a) Solid (ice) – Represented by area BOC (b) Liquid (water) – Represented by area AOC (c) Gas (vapor) – Represented by area AOB (ii) Two-phase equilibrium (a) Solid (ice) Liquid (water) – Represented by curve OC



(b) Liquid (water)



(c) Solid (ice)

Gas (water vapor) – Represented by curve OA Gas (water vapor) – Represented by curve OB

(iii) Three-phase equilibrium

(a) Solid (ice)

liquid (water)

Gas (vapor) – Represented by triple point O.

The areas:  Area BOC represents the solid phase, which is thermodynamically most stable state under these conditions. Area AOC represents the liquid (water) phase in the system and similarly area AOB represents the gas phase in the system. Thus these areas represent one phase equilibria. In order to define the system completely at any point in the area, it is essential to specify both the temperature and pressure. Therefore these areas have two degrees of freedom and are called bivariant systems. It can be concluded by the phase rule equation.

F = C – P + 2

= 1 – 1 + 2 = 2 (Bivariant system) The Curves:  The curve OC (melting or fusion curve) represents the equilibrium between ice and water and it is known as melting point curve. The very little slope of the curve shows that very high pressure is required to bring about significant changes in melting point of ice. The inclination

7.10  Engineering Chemistry of the curve OC towards the pressure axis indicates that melting point of ice slightly decreases by increasing pressure. The reason for this uncommon behavior is the decrease in the volume, when ice melts into water. Thus ice transforms into denser water when the pressure is raised.

1.8 The Curve OA (Vapor pressure curve) Represents the equilibrium between the two phases – liquid water and vapor. Along this curve, water and vapor coexist in equilibrium. The curve shows the vapor pressure of liquid water at different temperatures. OA curve is also known as vapor pressure curve of water. This curve terminates at point A. This is the critical point having temperature 374°C and pressure 218 mmHg. At this critical point the liquid and vapor are indistinguishable from each other and thus only one phase in left. Critical point is a characteristic property of pure substance. When the vapor pressure is equal to 1 atmosphere, the corresponding temperature is the boiling point of water (100°C).

1.9 The Curve OB (Sublimation curve) Represents the equilibrium between ice and vapor and it is called sublimation curve of ice. It shows the vapor pressure of solid ice at different temperature. The two phases namely ice and vapor coexists in equilibrium along this curve. At the lower limit OB curve terminates at absolute temperature (–273°C), where no vapor exists. Along the curves OC, OA and OB, the numbers of phases present in equilibrium is two. F = C – P + 2 = 1 – 2 + 2 = 1 (Univariant system)

1.10 Triple Point It is a point at which the gaseous, liquid and solid phases of a substance coexist in equilibrium. For a given substance, the triple point occurs at a unique set of values of temperature, pressure and composition. In water system the point O where all the three curves OC, OA and OB meet is known as triple point. At this point all the three phases are in equilibrium which is attained at 0.0098°C temperature and 4.58 mmHg pressure. Since there are three phases and one component so the degree of freedom will be F = C – P + 2 = 1 – 3 + 2 = 0 (Non-variant system) If either the temperature or the pressure or both are changed, the three phases would no longer coexist and at least one of them would disappear.

1.11  Metastable Curve OA¢ As the water does not always freezes at 0°C, therefore if the vessel containing water and vapors it thoroughly clean and dust free, it is possible to super cool water several degrees below its freezing point. On slight disturbance the super cooled water at once changes to ice (solid). This kind of system is a Metastable system. The dotted curve OA is the continuation of vaporization curve OA. It is vapor pressure curve of super cooled water. The curve is between super cooled water and vapor as shown below Super cooled liquid

Vapor

The Phase Rule  7.11 On adding nucleus of ice (solid state), the system reverts to true stable system where the equilibrium exist between solid and vapor. Solid

Vapor

2. Polymorphism The existence of a given substance in more than one crystalline form possessing different physical properties is known as polymorphism. It occurs in number of elements and compounds, when it occurs in elements, it is referred as allotropy. Each polymorphic form constitutes a separate phase. The temperature at which one form changes in to another at a given pressure is known as the transition temperature. For example rhombic sulphur when heated under a pressure of one atmosphere changes in to monoclinic sulphur at 95.6°C. At the same time monoclinic sulphur when cooled under a pressure of one atmosphere changes in to rhombic sulphur at 95.6°C. Thus 95.6°C is the transition temperature at which one form of sulphur changes reversibly into the other. Polymorphic forms which can undergo reversible transformation into one another at the transition temperature are called enantiotropic forms. In some cases polymorphic forms do not undergo reversible transformations into one another for example diamond can be converted into graphite under suitable conditions of temperature and pressure but reverse is possible. Such forms which cannot be reversibly transformed into one another are called monotropic and the phenomenon is known as monotropy.

3. The Sulphur system Sulphur exists in two crystalline forms, rhombic and monoclinic with 95.6°C as the transition temperature at one atmosphere pressure at which they can be transformed into one another. Below 95.6°C rhombic is the stable form while above 95.6°C monoclinic is stable. Each form has its own characteristic melting point. Thus under a pressure of one atmosphere melting point of rhombic sulphur (SR) is 114°C while that of monoclinic sulphur (SM) is 120°C. The liquid form of SL undergoes change in color and viscosity when heated and ultimately exists in 4 possible phases two solids (SR & SM) one liquid (SL) and one vapor (SV) phase. However all the four phases cannot co-exist at the same time since the number of phases coexisting in a one component system cannot exceed three. In phase diagram of sulphur system : (Fig. 2) the curve AO is the sublimation curve of rhombic sulphur and gives the vapor pressure of rhombic sulphur at different temperatures. The two phases in equilibrium are rhombic sulphur and the vapor (SR SV). The equilibrium (SR SV) is monovariant. Therefore at one temperature, there can be one vapor pressure only. The point O is the transition temperature (95.6°C) at which rhombic sulphur changes in to monoclinic sulphur. O is thus a triple point at which three phases, two solids and the vapor (SR SM SV) coexist in equilibrium. This is a non-variant point. The curve OB is the sublimation curve of monoclinic sulphur at different temperatures. As the number of phases are two the system is monovariant. The point B is the melting point (120°C) of monoclinic sulphur. This is another triple point at which three phases i.e. sulphur monoclinic, liquid and vapor (SM SL SV) are in equilibrium. Thus is a non-variant point. The curve BE is the vapor pressure curve for liquid sulphur. The two phase equilibrium (SL SV) is monovariant.

7.12  Engineering Chemistry

Fig. 2  The phase diagram for the sulphur system

The curve OC is the transition curve which gives the effect of pressure on the transition temperature of rhombic sulphur in to monoclinic sulphur. The equilibrium involved is (SR SM) (both solid phases). The system is monovariant. Since the transformation of rhombic sulphur into monoclinic sulphur is accompanied by increase in volume, the increase of pressure causes a rise in the transition temperature. The curve BC is the fusion curve for monoclinic sulphur. This gives the effect of pressure on the melting point of monoclinic sulphur. The two phase equilibrium (SM SL) along the curve BC is univariant. As melting of monoclinic sulphur is accompanied by a slight increase of volume. Therefore curve BC slopes slightly away from the pressure axis. As the slope of this curve is much less than that of the curve OC, the two curves meet at the point C. Thus, C is another triple point where three phases rhombic sulphur, monoclinic sulphur and liquid sulphur are in equilibrium i.e., SR SM SL and the system is nonvariant. At C, the temperature is 151°C and pressure is 1290 atm. The curve CD is the fusion curve for rhombic sulfur. The equilibrium along this curve is SR SL . As the number of phases is 2, the system is monovariant.

3.1  Mata stable equilibrium In addition to stable equilibria discussed above there are also some metastable equilibria. Since conversion of one solid into another solid involves molecular rearrangement the process is naturally slow. Therefore unless heating is done extremely slowly there is a possibility that the first solid may not change into the second solid at the normal transition point. The first solid will exist in metastable equilibrium with its vapor. This possibility of sulphur system is represented by dotted line OA¢. If the temperature of rhombic sulphur is allowed to rise rather quickly at 95.6°C, it will persist in equilibrium with its vapor phase without changing into monoclinic sulphur several degrees above the normal transition temperature.

The Phase Rule  7.13 Thus along the line OA¢, the metastable equilibrium SR SV exist. The vapor pressure at each temperature, as can be seen, is higher than the vapor pressure of monoclinic sulphur which is a stable phase in this range. Similarly, if liquid sulphur is allowed to cool along the curve BE, the solid phase may not separate out at B unless cooling is extremely slow. The line BA’ in this case represents the metastable equilibrium between liquid sulphur and its vapor i.e., by SL SV. The point A¢, therefore represented the melting point of the metastable rhombic sulphur. This is another triple point where the three phases. SR SL SV coexists in metastable equilibrium. The temperature A¢ has been found to be about 114°C. The curve A¢C, evidently, gives the effect of pressure on the melting point of rhombic (metastable) sulphur. In other words, this is the fusion curve of metastable rhombic sulphur along which rhombic sulphur is in metastable equilibrium with the liquid.

3.2 Areas As the curves AO and OC both involve rhombic sulphur in equilibrium with the vapor, Phase in one and bounded by these lines which represents stable existence of rhombic sulphur alone. Similarly curves BE and BC both have liquid phase as common, in equilibrium with vapor in the first case and with solid monoclinic sulphur in the second case, the area bounded by these lines represented liquid sulphur as the stable phase curves. Similarly, as the curves AO, OB, and BE have vapor phase in common, the area lying below the lines represents only vapor as the stable phase. Lastly, the lines OC, OB, and BC have monoclinic sulphur as common phase the triangular area bounded by these lines represents monoclinic sulphur as stable phase. The existence of this form of sulphur, evidently is limited on all sides.

solved Questions 1. State phase rule. Ans. Provided the equilibrium between any number of phases is not influenced by gravity, or electrical, or magnetic forces, or by surface action but only by temperature, pressure and concentration, then the number of degree of freedom (F) of the system is related to the number of components (C) and of phases (P) by the phase rule equation, F = C – P + 2 for any system at equilibrium at a definite temperature and pressure. 2. What is condensed phase rule? Ans. When the pressure of the system remains constant (i.e., in which the vapor phase is not considered), then the phase rule equation becomes: F = C – P + 1. 3. A system consists of benzene and water. What is the number of phases? Ans. Two 4. How many components are present when NH4Cl is heated in a closed vessel? Ans. One 5. Give the number of components of the system: Fe (s) + H2O (l) FeO (s) + H2(g) Ans. Three 6. What is an invariant system?

7.14  Engineering Chemistry Ans. In which the degree of freedom is zero i.e., no condition is required to be specified to define the system. 7. Give an example of invariant system Ans. A system containing ice, water and water vapor in equilibrium. 8. What is the triple point of water system? Ans. Triple point is a point where all the there phases co exist, where the temperature is 0.0098°C and pressure is 4.58mmHg. 9. What is the difference between a phase and a state of matter? Ans. There are three states of matter–solid, liquid and gas. A phase is a sample of matter with definite composition and uniform properties throughout the sample. 10. What is meant by transition? Ans. The crossing of two-phase equilibrium curve in a phase diagram. 11. How many degrees of freedom are present in the following system? (i) A gas in equilibrium with its solution in a liquid. (ii) A solution of a solid in a liquid in equilibrium with solvent water. (iii) Two partially miscible liquids in the absence of vapors. (iv) I2(s) I2(g) Ans. (i) F = C – P + 2 = 2 – 2 + 2 = 2 (ii) F = C – P + 2 = 2 – 2 + 2 = 2 (iii) F = C – P + 2 = 2 – 2 + 2 = 2 (iv) F = C – P + 2 = 1 – 2 + 2 = 1 12. Can four forms of sulphur, namely, rhombic, monoclinic, liquid and vapor coexist in equilibrium at a particular temperature and pressure? Ans. No (since F = C – P + 2 = 1 – 4 + 2 = – 1, i.e., negative) 13. What type of equilibrium exists between H2O(l) and H2O(g) at a temperature below 273K. Ans. Metastable equilibrium

unsolved Questions

1. 2. 3. 4. 5.

Explain the terms-phase, component and degree of freedom. Define phase rule. Write a note on (i) Triple point (ii) Metastable state. Apply phase rule to water system. Calculate the number of degrees of freedom in the following systems. (i) 2KClO3(s) 2KCl (s) + 3O2(g)



(ii) KCl



(iii) CaCO3(s)



(iv) 2H2S (g)



(v) Na2SO4.10H2O (s)



(s)

(vi) H2O (s)

– NaCl

(s)

– H2O (l)

CaO (s) + CO2(g) 2H2(g) + S2(g) H2O (g)

Na2SO4(s) + 10H2O (l)

The Phase Rule  7.15

6. With the help of phase rule, calculate the degree of freedom of the following systems: (i) Ice and water at equilibrium (ii) Saturated solution of NaCl. 7. Water system is a representative system for explaining phase rule and phase equilibrium. Explain. 8. Draw a neat and leveled diagram of water. 9. If there is no triple point in the diagram of one component system, then what inference do you draw? 10. State the significance of triple point. 11. What is Gibbs phase rule? 12. Explain why KCl-NaCl-H2O should be regarded as a three-component system whereas KClNaBr-H2O should be regarded as a four-component system. 13. Draw a neat and labeled diagram of sulphur. 14. How many triple points are there in sulphur system? 15. What is metastable state? 16. How many phases are there in following system at equilibrium? (i) Mixture of O2 and N2. (ii) Mixture of benzene and water. (iii) A mixture of diamond powder and graphite. (iv) A finely divided sulphur graphite and sodium chloride.

Chapter

8

Fuel and Combustion 1.  INTRODUCTION Efficiency in the use and production of heat and power is one of the major engineering problems of the present day. The world’s energy demands are constantly increasing and to meet this expanding requirement we look to various kinds of fuels. Material which possesses chemical energy is known as fuel. The important fuels are carbon compounds and whole industrial society is based upon the reaction. Fuel + O2 Æ Combustion Product + Heat C (s) + O2 (g) Æ CO2 (g) + DH = – 94.1 Kcal/mole H2 + ½ O2 Æ H2O + DH = – 67.5 Kcal/mole When C and H atom from coal and oil reacts with oxygen atoms chemical energy is released, such reaction involves a rearrangement of the outer electron only at the atoms and the atomic nuclei is unaffected. A fuel may be defined as any combustible substance which is obtainable in bulk, which may be burnt in atmospheric air in such a manner that the heat evolved is capable of being economically used for domestic and industrial purposes for heating and generation of power. Fuels in the broad sense, include the stored fuels that are available in the earth’s crust, viz., ‘’fossil fuels’’ or those derived from them by an industrial process.

2.  CLASSIFICATION Fuels may be divided into two types: (i) Primary fuels which occur in nature as such, and (ii) Secondary fuels which are derived from the primary fuels. Fuels may also be classified into three groups: (a) Solid fuels, (b) Liquid fuels, and (c) Gaseous fuels.

8.2  Engineering Chemistry The examples of each of these main classes of fuels are summarized in Table 1. This classification is of practical significance because the equipment used for handling and burning of each class of fuels is usually different for these three types of fuels:

Table 1  Classification of fuels Solid Fuels

Liquid Fuels

Primary Wood, peat, lignite, brown coal, bituminous coal, anthracite, oil shales, tar sands and bitumen. Secondary Semicoke, coke, charcoal briquettes, petroleum coke, pulverized coal and colloidal fuels, solid rocket fuels such as thiokol, hydrazine, nitrocellulose, etc.

Gaseous Fuels

Primary Crude oil or petroleum.

Primary Natural gas.

Secondary Gasoline or motor spirit, diesel oil, kerosene, fuel, oils, coal tar and its fractions, alcohols and synthetic spirits.

Secondary Coal gas, coke oven gas, water gas, producer gas, carburetted water gas, oil gas, blast furnace gas, refinery oil gas, syntheses gas, acetylene and liquid petroleum gas (LPG).

3.  CALORIFIC VALUE One of the important properties of a fuel on which its efficiency is judged is its calorific value. The calorific value of a fuel is defined as the amount of heat obtainable by the complete combustion of a unit mass of the fuel.

3.1  Units of Heat The units of heat generally employed are calories, kilogram calories, British thermal units and centigrade heat units. (a) Calorie (cal) or gram calorie (g cal) For all practical purposes, the calorie or gram calorie may be defined as the amount of heat required to raise the temperature of 1g of water through 1°C (more precisely from 15°C to 16°C) 1 calorie = 4.185 joules = 4.185 × 107 ergs. (b) Kilocalorie or kilogram calorie or kilogram centigrade unit (Kcal or Kg cal or K.C.U.) This is equal to 1000 calories and is, thus the amount of heat required to raise the temperature of 1 kg of water through 1°C (more precisely from 15°C to 16°C). (c) British thermal unit is the amount of heat required to raise the temperature of 1 Pound of water through 1°F (more precisely from 60°F to 61°F).



B.Th.U. = 1,054.6 joules = 1,054.6 × 107 ergs.

(d) Centigrade heat unit (C.H.U.) The centigrade heat unit is the amount of heat required to raise the temperature of 1 Pound of water through 1°C.

Fuel and Combustion  8.3

3.2  Interconversion of the Various Units of Heat The various heat units described above can be easily interconverted on the basis that 1kg = 2.2lb and 1°C = 1.8°F, accordingly, 1 K cal = 1000 cals = 3.968 B.Th.U. = 2.2 C.H.U.

1 B. Th.U. = 252 Cals



100,000 B.Th.U. = 1 Therm.

Units of calorific value Calorific values of solid and liquid fuels are usually expressed in calories per gram (Cals/g) or Kilocalories per kilogram (K cals/Kg) or British Thermal Units per pound (B.Th.U/lb.) whereas the calorific values of gases are expressed as Kilocalories per cubic metre (K cals/m3) or British thermal units per cubic foot (B.Th.U/ft3) or C.H.U./lb or C.H.U/ft3. These units can be inter-converted as follows:

1 cal/g = 1 K cal/Kg = 1.8 B.Th.U/lb 1 K cal/m3 = 0.1077B.Th.U/ft3 1 B. Th.U/ft3 = 9.3 Kcals/m3

3.3  Gross calorific value and net calorific value The Gross Calorific Value or Higher Calorific Value is the total heat generated when a unit quantity of fuel is completely burnt and the products of combustion are cooled down to 60°F or 15°C (room temperature). When a fuel containing hydrogen is burnt, the hydrogen present undergoes combustion and will be converted into steam. As the products of combustion are cooled to room temperature, the steam gets condensed into water and the latent heat is evolved. Thus the latent heat of condensation of steam so liberated is included in the gross calorific value. The Calorific value determination by Bomb calorimeter gives the Gross or Higher Calorific Value. The Net Calorific Value or Low Calorific Value is the net heat produced when a unit quantity of fuel is completely burnt and the products of combustion are cooled to room temperature, the steam gets condensed into water and the latent heat is evolved. Thus the latent heat of condensation of steam so liberated is included in the gross calorific value. The Net Calorific Value or Low Calorific Value is the net heat produced when a unit quantity of fuel is completely burnt and the products of combustion are allowed to escape. Thus,

Net Calorific Value = Gross Calorific Value–Latent heat of condensation of the water vapor produced. = Gross Calorific Value–(Mass of Hydrogen per unit weight of the fuel burnt × 9 × latent heat of vaporization of water). 1 part by weight of hydrogen gives 9 parts by weight of water as follows: H 2 + O Æ H 2O 2g 16 18g 1g 8g 9g

8.4  Engineering Chemistry

The latent heat of steam is 587 Cal/g (or Kcal/Kg) or 1060B.Th.U./lb of water vapor produced.



Net

H C.V. = Gross C.V. – 9 × ____ ​    ​ × 587 100

= Gross C.V. – 0.09 × H × 587

Where

H = % of hydrogen in the fuel.

In actual practical use of a fuel, it is rarely feasible to cool the combustion products at the room temperature to allow the condensation of water vapor formed and utilize that latent heat; hence the water vapor formed also is allowed to escape along with the hot combustion gases.

4. DETERMINATION OF CALORIFIC VALUE OF SOLID AND NON-VOLATILE LIQUID FUELS 4.1  Bomb calorimeter The calorific value of a solid or non-volatile liquid fuel can be satisfactorily determined with a highpressure oxygen bomb calorimeter (Fig. 1). It consists of: 1. A strong cylindrical bomb made of stabilized austenitic steel which is corrosion resistant and which is capable of withstanding a pressure of at least 50 atmospheres. The bomb is provided with a gas-tight screw-cap to which a couple of stainless steel electrodes and a release valve are fitted. One of these electrodes is tubular which can therefore act also as an oxygen inlet. A small ring is attached to this electrode which acts as a support for the crucible.

Fig. 1  (1) Strong cylindrical bomb made of stabilized austenitic steel, (2) Stainless steel rod electrode, (3) Stainless steel tube electrode cum oxygen inlet, (4) Release valve, (5) Beckman thermometer, (6) Electrically driven stirrer, (7) Firing terminals, (8) Calorimeter, (9) Fuse wire, (10) Crucible, (11) Screw cap, (12) Fuel pelet, (13) Air jecket, (14) water.

Fuel and Combustion  8.5

2. 3. 4. 5.

A copper calorimeter vessel in which the bomb stands during the experiment. An outer water-jacket enclosure. A stirrer provided for stirring the water in the calorimeter at a uniform rat. A thermometer graduated in one hundredths of a degree. Fixed zero or Beckman thermometers are generally used. 6. Other necessary accessories for compressing the coal into a pellet, filling the bomb with oxygen, etc.

Procedure







1. About 0.5 to 1 g of finely ground air-dried coal (preferably compressed into a pellet) is accurately weighed into the crucible of the calorimeter. 2. A piece of fine platinum wire (0.0075 cm thick) is tightly stretched across the pole pieces of the bomb and one end of a piece of a sewing cotton thread is tied round the wire. The crucible is placed in position and the lose end of the cotton is arranged so as to be in contact with the coal. Alternatively, a longer platinum wire is used and bent into a loop so as to touch the coal pellet. 3. About 10 ml of distilled water are introduced into the bomb to absorb vapors of sulphuric and nitric acids formed during the combustion and the lid of the bomb is screwed. 4. The bomb is filled with oxygen up to 25 atmospheres pressure and the firing wires are attached to the terminals. 5. The calorimeter vessel is weighed; sufficient water is weighed to submerge the cover of the bomb. The calorimeter vessel is then kept in the outer jacket on the insulating feet provided. The bomb is then lowered in the calorimeter. The stirrer and the lid of the calorimeter vessel are placed in position and the Beckman thermometer is adjusted. 6. The stirrer is started. 7. After 5 minutes, the temperature of the water is noted to the nearest 0.002°C and 5 more readings are taken at one minute intervals. 8. At the end of the fifth minute, the electrodes are connected to a 6 to 12-volt battery to ignite the charge and readings are continued at one-minute intervals. After the maximum temperature is attained, readings are still continued until the rate of fall is uniform. 9. The stirrer is stopped and the bomb removed from the calorimeter. After allowing about half an hour for the settlement of the acid mist within the bomb. The contents of the bomb are washed into a beaker and the amounts of H2SO4 and HNO3 present in this solution are determined.

Calculations Let Weight of the fuel sample taken = m grams Higher or Gross calorific value of the fuel = L cals/gram Weight of the water taken in the calorimeter = W grams Water equivalent of the calorimeter, bomb, thermometer, stirrer, etc. = w grams Initial temperature = t1°C

8.6  Engineering Chemistry Final temperature = t 2°C Heat absorbed by the water, calorimeter, etc. = (W + w) (t 2 – t1) Heat liberated = Heat absorbed \ mL = (W + w) (t 2 – t1)

\

(W + w) (t 2 – t1) (Gross C.V.) L = ​  _____________ m     ​ 

However, for more accurate results, the following correction will have to be incorporated in the above equation: (i) Acids correction, tA, (ii) Fuse wire correction, tF, (iii) Cotton thread corrections, tT and (iv) Cooling correction, tc, Accordingly, the above equation will have to be modified as:

[(W  +  w) (t 2 –  t1 +  tc)] – [tA +  tF +  tT] L (Gross C.V.) = ​  ________________________________        ​ m 

4.2  Calculation of net calorific value Net C.V. = Gross C.V. – Latent heat of water vapor, formed during the combustion of m grams of the fuel = Gross C.V. – 0.09 × H × 587 (where H is the percentage of hydrogen present in the fuel and latent heat of steam is 587 cals/g). Corrections (i) Acids Correction (tA):  The sulfur present in the coal is converted into H2SO4 in the bomb. S + O2 Æ SO2 2SO2 + O2 + 2H2O Æ 2H2SO4; DH = – 144,00

(4 × 49) (Eq. Wt. of H2SO4 = 49)

calories

Similarly the nitrogen present in the coal and part of that in the air in the bomb are converted into HNO3 2N2 + 5O2 + 2H2O Æ 4HNO3; DH = – 57,160 (4 × 63) calories

(Eq. Wt. of HNO3 = 63)

Since the above two reactions are exothermic and since the heat thus liberated is not obtainable in practical use of coal (because SO2 and NO2 pass off into the atmosphere) correction must be made for the heat liberated in the bomb by the formation of H2SO4 and HNO3, as follows: (a) 3.6 calories should be subtracted for each ml of N/10, H2SO4 formed. (b) 1.43 calories must be deducted for each ml of N/10 HNO3 formed (as per the equations given above)

Fuel and Combustion  8.7

(ii) Fuse wire correction, (tF):  Correction has to be made for the amount of heat equivalent in calories derived from the amount of fuse wire burnt as per the instructions furnished by the supplier of the fuse wire. (iii) Cotton thread correction, (tT):  The correction for the cotton thread used for firing the charge is calculated from the weight of the dry cotton thread actually used and on the basis that the calorific value of cellulose is 4140 cals per gram. (iv) Cooling correction, (tc):  If the time taken for the water in the calorimeter to cool from the maximum temperature attained to the room temperature is x minutes and the rate of cooling is dt°/minute, then the cooling correction = x × dt. This should be added to the observed raise in temperature. For more accurate results, the cooling correction by Regnault and Pfaundler’s formula should be used. Cooling correction, (tc)

{ 

}

v¢  –  v n – 1 1 = nv + ______ ​     ​  ​  ​    ​  (t)  + ​ __ ​  (t0 – tn) – nt  ​ 2 t ¢ – t 1​ ​ = nv + KP

Where n = number of minutes between the time of firing and the first reading after the temperature begins to fall from the maximum. v = the rate of fall of temperature per minute during the period before firing. V’ = the rate of fall of temperature per minute after the maximum temperature. t and t’ = the average temperatures during the prefiring and final periods respectively. (t) = the sum of the readings during the period between firing and the start of coaling. 1 __ ​   ​  (t0 + tn) = mean of the temperatures at the moment of firing and the first temperature after 2 which the rate of change of temperature is constant, and v¢ – v K = ​ _____ ​ which is known as the “cooling constant’’ of the caloriment. t¢ – t The cooling correction so obtained should be added to the observed rise of temperature.

5. DETERMINATION OF CALORIFIC VALUE OF GASES AND VOLATILE LIQUID FUELS 5.1  Boy’s calorimeter The calorific value of gaseous and volatile liquid fuels is usually measured by Boy’s calorimeter or junker’s calorimeter. The principle involved in this method is to burn the gas at a known constant rate in a vessel under such conditions that the entire amount of heat produced is absorbed by water which is also flowing at a constant rate. From the volume of the gas burnt, the volume of water collected and the mean rise of temperature of the water, the calorific value of the gaseous fuel is calculated.

5.2  Description of the apparatus Several types of gas calorimeters are available which differ in the arrangements for burning the gas and the heat transmitting system. The Boy’s non-recording gas calorimeter (2) consists of two flat-

8.8  Engineering Chemistry flame burners ‘B’ situated in a chimney ‘C’ which forms the centre of the annual vessel ‘V’ The lower portion of “V’ is provided with a trough where water condensed from the products of combustion is collected and can be removed through the side tube ‘T’ for measurement of its volume. The products of combustion pass up to the chimney and are deflected downwards by the water cooled head ‘G’ over a spiral of similar copper tubing and finally passes out through a number of holes in the lid of the calorimeter. The inlet water passes through the outer coils downwards to return upwards through the inner coils. Finally, it flows around suitable channels on the exterior of the metal casting immediately above the chimney and passes into a mixer. The temperatures of the inlet and outlet water are measured either different thermometers. The gas flows into the burners through an accurate gas-meter.

Fig. 2  Boy’s calorimeter

5.3 Procedure The apparatus is assembled as shown in the diagram (Fig. 2). After ensuring that the assembly is leak proof, the gas is turned on and lighted. The water is turned on and the rate of flow is so adjusted that the rise in temperature of the water in passing through the calorimeter is as nearly as possible to 20°C. After the conditions are allowed to stabilize for about 45 minutes, the following readings are noted: (a) the volume of the gas burnt at a given temperature and pressure during a certain time interval (b) the amount of water passed through the cooling coils during the above time interval (c) the steady difference in temperature of the outlet and inlet water and (d) the amount of the water condensed from the products of combustion during the experiment. Also, the atmospheric pressure and the gas pressure are recorded from the barometer and manometer respectively. Further, the temperatures of the effluent gas, the ambient air and the gas meter are also noted to the nearest degree. Then the high calorific value of the gaseous fuel can be calculated from the following relation: w (t 2 – t1) Gross calorific value, J = ​ ________     ​  V

Fuel and Combustion  8.9

Where w = wt. of the cooling water passed in time t.

V = volume of gas burnt at S.T.P. in time t. t1 = temperature of the incoming water. t 2 = temperature of the outgoing water. Further, the amount of water condensed from the steam produced by burning 1 m3 of gas = m/v, and the latent heat of steam per 1 m3 of gas at 15°C = 587 Kcal)

( 



m \  Net calorific value = ​ Gross calorific value = ​ __ v ​  × 587  ​



Where  m = weight of steam condensed in time, t.

)

Since the calorific value of gases is expressed on volumetric basis, it is essential to define the conditions of temperature and pressure for the volume being referred.

6.  CRITERIA FOR SELECTING A FUEL The following characteristics are taken into consideration for the selection of a fuel for a particular purpose. 1. The fuel selected should be most suitable for the process. For instance, coke made out of bituminous coal is most suitable for blast furnace and also as a foundry fuel. 2. The fuel should posses a high calorific value. 3. The fuel should be cheap and readily available. 4. It should possess a moderate ignition temperature. Too high ignition temperatures cause difficulty in kindling while too low ignition temperatures may create safety problems during storage, transport and use of the fuel. 5. The supply position of the fuel should be reliable. 6. The velocity of combustion should be moderate. 7. The fuel should have reasonable flexibility and control. 8. The fuel should be such that a safe and clean operation is ensured. Too much smoke and obnoxious odors are not desirable. 9. It should be safe. Convenient and economical for storage and transport. 10. It should have low moisture content. 11. In case of a solid fuel, the ash content should be less and the size should be more or less uniform.

7.  SOLID FUELS The main solid fuels include wood, peat, lignite, coal, charcoal and briquetted fuels. In addition to these, certain agricultural and industrial wastes such as bagasse, spent tan, rice husk, coconut and nut shells are also employed as fuels.

7.1  Wood Wood has been the main source of fuel until recent times on account of its relatively rapid growth and production and ease of obtaining the supplies. However, large scale deforestation and the increasingly large demands of energy by the industries led to the more extensive use of other types of fuels.

8.10  Engineering Chemistry Freshly cut wood possesses greater water content (25 to 50%) than dry wood (15%) and its heating value is directly proportional to the water content. Wood mainly consists of cellulose, ligno-cellulose as well as some cell sap associated with traces of mineral ash. On the dry and ash-free basis, the average composition of wood is 50%C, 43% O, 6% H and 1% nitrogenous and resinous material. As a rule, dry wood is very combustible, easily kindled and burns with a long non-smoky flame. It gives maximum heat intensity very quickly. However, the calorific value of dry wood is only 19.7 to 21.3 MJ/kg (4710 to 5085 Kcal/kg). The ash content of wood is very low and lies in the range 0.3 and 0.6%. Wood is largely used as domestic fuel. It is rarely used in industry except for special purpose where dirt and smoke are undesirable. Due to its high flame emissivity, it is preferentially used for space heating. Wood charcoal is obtained by the destructive distillation of wood. The carbonisation is performed usually in closed retorts. The charcoal is not pure carbon because even when the carbonisation is conducted at high temperatures, it rapidly absorbs some gas and moisture. It also contains some inorganic residues derived from the wood. Charcoal was widely used for metallurgical operations formerly, but it has now been replaced by coke excepting for some special applications. The major use of wood charcoal today is for producing activated carbon which finds extensive application for decolorisation (e.g., in sugar industry), adsorption of gases and vapors and recovery of solvents from gases and air. Charcoal is also used in the production of CaC2, ferro-alloys, and special quality pig iron in small furnaces.

7.2 Peat Peat is generally considered as the first stage in the conversion of vegetable debris to coal and is produced under water-logged conditions by the action of fungi and anaerobic bacteria. Peat is generally found in high altitudes. Three main types of peat are usually distinguished: (1) upland type consisting mainly of decomposed heaths and mosses (2) lowland type derived from sedges, grasses and willows and (3) Forest peat formed from accumulations of leaves, twigs, etc. and is mostly found in tropical countries. The bog peats derived from the smaller forms of vegetation are mainly found in USSR, Ireland, U.K., Canada, Finland, Poland, Siberia, France, Germany and Italy. About 42% of the world’s peat deposits occurs in the USSR. Freshly won peat from a well-drained bog may contain even 90% of water, which can be brought down to the level of 15 to 20% by air-drying. Most of the commercial peat blocks contain about 20-25% moisture. Thus the utilization of peat depends on its economical drying. On the dry basis its calorific value is around 5450 Kcal/Kg. Peat as such is used only as a local fuel. It is not considered as an economic furl on account of the cost of drying and handling, cost of transportation as it is voluminous, relatively low calorific value and also because of its property of getting powdered during burning. It is mostly used, after briquetting with other substances, as a domestic fuel. Carbonisation of peat at low temperature (500-600°C) produces a char, oils and light spirit. Carbonisation of peat under proper conditions may yield coke and gas as well. Peat is largely used as domestic fuel in Europe but its use as such in India is limited. Dried peat is used as a fuel for the domestic appliances (e.g., cookers and space heaters), for steam rising, for thermal insulation, packing, gas purification and soil conditioning. Due to the shortage of coal, peat is used for generation of electricity in Russia. In India, peat deposits are found in Nilgiri Hills (Tamil Nadu), Sundarbans (West Bengal), Kashmir and Kolkata.

Fuel and Combustion  8.11

7.3  Lignite (Brown Coal) These belong to the intermediate stage between peat and black coals. Their moisture content lies in the range 35 to 50%. Their carbon content, on the ‘’day, ashfree basis’’, range from 60 to 75% while the oxygen is over 20%. Lignites have high volatile matter content, usually up to 48 to 50% and an ash content of 4% or more. They generally occur at shallow depths and won by open cast mining techniques. Owing to their high volatile matter, they burn with long smoky flame. When dried, they tend to disintegrate into small pieces and powder and hence much of it is lost in transport and storage in open. Their calorific value is only in the range 24.3 to 29.3 MJ/kg (5800-7000 Kcals/kg) on dry ash free basis. Lignites absorb oxygen readily on exposure to air and get ignited spontaneously. Lignites are not considered as a good fuel as such but are better than peat. Lignites are commonly made into briquettes after dehydration and marketed as such or after carbonisation. They are used as fuels in power plants and in the form of briquettes for domestic use. They are also used for production of producer gas. On further carbonisation, they give tar and ammonium sulfate. The tar on further hydrogenation gives motor spirit. Ammonium sulfate is used as fertilizer. Lignites are classified into the following types on the basis of their maturity and external characteristics: (1) earthy brown coals ‘e.g., Australian Morewel brown coal (2) wood brown coals e.g., Italian Valderno brown coal and (3) laminated lignites which are rather black. The brown varieties generally tend to darken on exposure to air. Large amounts of lignites are consumed in Germany and Russia for steam rising. Lignite deposits occur in Assam, Kashmir, Rajasthan, Tamil nadu, Travancore, Malabar coast and in M.P. Important lignite deposits of our country occur at Neyveli (South Arcot district of Tamil Nadu) which spread over 250 sq. km. and the deposits are estimated to be 200 million tonnes. The Neyveli project under the Lignite Corporation of India produces millions of tonnes of lignite briquettes for domestic and industrial use, and also urea.

7.4  Bituminous Coal These coals burn with smoky yellow flame like bitumen and their product of distillation is coal tar which is bituminous in nature. Sub-bituminous coals form a group between lignite and bituminous coal. They are harder and denser than lignite. They are black in color and possess a dull waxy luster. Its moisture content is 12 to 25% and calorific value is about 7000 Kcals/kg. Their carbon content is 75 to 83%. Their oxygen content varies from 10 to 20%. The bituminous coals are black in color with a banded appearance, laminated structure and cubical fracture. Their carbon content is 78 to 90% and volatile matter 20 to 45%. They vary widely in their properties. Their calorific value is around 8000 to 8500 Kcals/kg. They may possess the property of caking (strong, medium or weak) or they may be non-caking. They are easy to handle and good in heating qualities. They are the most widely used coals in the world. They are used for domestic and industrial purposes. They are used for steam raising, coke and gas production and by-product manufacture. The semi-bituminous coals form a group between bituminous coal and anthracite. They are characterized by low volatile matter (9 to 20%). Their carbon content is 90–93% and calorific value is around 8600 Kcals/kg. They are used for the manufacture of coke. The Gondwana coals of India are bituminous in nature and the Raniganj bituminous coals are characterized by high volatile matter. The bituminous coals of Bihar, Bengal, M.P. and Orissa are the most important deposits of our country.

8.12  Engineering Chemistry

7.5 Anthracite Anthracites are considered to be the highest rank of coal and contain the maximum percentage of carbon (92 to 96%). They are black, hard and lustrous. They have a conchoidal fracture. They have a very low percentage of volatile matter (about 6%) and hence produce a very little flame which is short, non-smoky and blue. Their calorific value is about 8600 Kcals/kg. They burn with intense local heating. Owing to their smokeless combustion, they are used for domestic heating in Canada and in the Continent. Anthracites are used in metallurgical operations, naval purposes, slow combustion stoves, central heating furnaces. Generation of producer gas, for drying malt and hops (because of its low arsenic content), and curing rubber. They are also used for special applications such as manufacture of cathode for aluminum industry and as a filter medium of water treatment. In India anthracites are found in Jammu, Darjeeling and Rajhara. The variation in average composition from wood to anthracite is summarized in the following Table 2.

Table 2  Variation in average composition from wood to anthracite during different stages of coalification Fuel

Moisture at 40°C and 60° relatives humidity

Volatile matter

C

H

N

O

Calorific value MJ/Kg (1 MJ = 238.8 Kcal/kg)

1. Wood 2. Peat 3. Lignite 4. Subbituminous coal 5. Bituminous coal 6. Semibituminous coal 7. Anthracite

25 25 18 11

75 65 50 to 56 45 to 50

50 57 67 77

6 5.7 5.0 5.0

0.5 2.0 1.5 1.8

43.5 35.3 26.5 16.2

20.9 23.0 27.2 30.2

4

20 to 45

83

5.0

2.0

10.0

36.0

1

9 to 20

90

4.5

1.5

4.0

36.4

1.5

5.6

93

3.0

0.7

3.0

35.6

7.6 Rank of Coal The different stages of coalification are called as peat-anthracite series. Each stage in the above series is considered to belong to higher rank or maturity than their respective preceding member in the series. As the coalification progresses, the percentage of carbon and hardness increases, while the percentage of hydrogen, oxygen, moisture and volatile matter generally decrease. The calorific value gradually increases from peat to semi-bituminous coal. There is a slight fall in the calorific value of anthracite because the percentage of hydrogen decreases.

8.  COAL 8.1  Origin of coal formation Coal is regarded as a fossil fuel produced from large accumulations of vegetable debris due to partial decay and alteration by the action of heat and pressure over millions of years.

Fuel and Combustion  8.13

The formation of coal is explained by the following two theories: (1) “In situ’’ theory states that coal seams are formed in the same area where vegetation grew and accumulated. The great purity of many coal seams holds testimony to this theory. (2) The “drift’’ or “transportation’’ theory contends that the coal seams are not formed where the vegetation grew and accumulated originally. These materials were drifted or transported by rivers to lakes or estuaries and got deposited there. The great thickness of coal seams supports this theory. Thus evidences are available in support of both the above theories. The various agencies responsible for the conversion of plant tissues to coal include (i) Bacteria (under water). (ii) Time (millions of years), (iii) Temperature (>300°C) and (iv) Pressure. The time required for the formation of young brown coals is of the order of 107 years while that for the most mature coals is 3 × 108 years. The vegetable matter fallen on the ground undergoes microbial degradation in presence of air and eventually gets converted to carbon dioxide and water without leaving any organic matter remaining. However, the course of decay when it is buried under water is different. The transformation of the vegetable debris to coal take place in two stages, (i) the biochemical or peat stage and (ii) the metamorphic stage during which peat is transformed into coal. The effect of temperature and pressure caused by the depth of burial on the rank of a coal is brought out by Hilt’s law which states that in any vertical section the rank of the seams increases with depth. The formation of coal from decaying plant debris to bituminous stage is explained by two alternative theories. (i) Serial evolution. This is the commonly accepted theory according to which the evolution of coals occurs through geo-chemical metamorphism of peat to anthracite as follows: (Peat Æ lignite Æ bituminous coal Æ anthracite)

(ii) Parallel evolution. This theory is based on the concept of entirely biochemical origin of coals of various ranks. According to this theory, lignites, bituminous coals and anthracite may not form a continuous series but may be the end-products resulting from the differences in the extent of the aerobic decomposition of peat, the subsequent composition of the overlying strata and the depth of the burial. This theory may be represented as follows:

8.14  Engineering Chemistry

8.2  Composition of Coal A mined sample of coal contains the coal substance intermixed with mineral constituents such as kaolin, shale, chlorides, sulfides, etc. These mineral constituents contribute to the ash content of the coal which is the residual mass obtained on combustion of coal. The coal also contains considerable amounts of free and hygroscopic moisture. The determination of moisture, volatile matter, ash and fixed carbon is known as proximate analysis which is of significance in commercial classification and industrial utilization of coal. Carbon, hydrogen and oxygen constitute the true coal substance and the properties of coal mostly depend upon these constituents. Small quantities of sulfur, arsenic and phosphorous also exist in many coal samples. If these substances are present in appreciable amounts, the coal may be unsuitable for metallurgical applications. The ultimate analysis of coal consists of determination of C, H, S, N and O. The details regarding proximate and ultimate analyses and their significance are discussed later. On the basis of difference in proportions of plant ingredients, the following types of coal are usually distinguished: Banded coals.  Bituminous coals usually possess banded structure consisting of alternate layers of very bright, laminated and dull layers. These bands are identified as Vitrain, Clarain, Durain and Fusain. Vitrain has a black brilliant, glassy luster. It breaks with a conchoidal fracture. It is very low in ash content and the ash has creamy or purplish colour. Clarain has a black and silky luster, duller than vitrain and has a tendency to break irregularly. Its ash content is more than that of vitrain and is usually reddish in color. Durain has a dull earthy luster and is hard and tough. It breaks irregularly. The ash content is higher than that of bright coal. The ash is generally grey or white and is rather infusible. Fusain is a soft black powdery material with satiny luster similar to charcoal in appearance. It occurs in thin layers. This is derived from woody tissue and is porous and soft. It is a minor constituent of most coal seams. The volatile matter is low and ask content is high which is often fusible. Splint coal  It is a type of bituminous or sub-bituminous coal having dull luster and greyish black color. It is hard, tough and has compact structure. It burns freely without swelling. It breaks with an irregular fracture.

8.2.1  Cannel coals and boghead coals Cannel is a further type of coal found in large lenticular masses in many coal seams. This has a greasy lustre and a white clayey ash. Channels and bogheads may be regarded as fossil vegetable muds deposited in water and contain algae. Cannel coal is a type of bituminous or sub-bituminous coal of uniform and compact structure and has no banded structure. It has a conchoidal or sheel like fracture. It is non-coking and yields a high percentage of volatile matter. It can be easily ignited and burns with a luminous smoky flame. Bog-head coal also is another type of bituminous or sub-bituminous coal and resembles the cannel coals in appearance and combustion properties. It is characterized by higher contents of algal remains and volatile matter.

Fuel and Combustion  8.15

8.3 Analysis of coal and its significance The composition of coal varies widely from mine to mine and hence it is necessary to analyze and interpret the results from the points of view of commercial classification, price fixation and proper industrial utilization. The results of analysis are generally reported in the following ways: 1. As used or as received basis 2. Air dried basis (i.e., in equilibrium with the laboratory atmosphere). 3. Moisture free basis (oven dried basis) 4. Moisture and ash free basis. The quality of a coal is ascertained by the following two types of analysis. 1. The proximate analysis, which includes the determination of moisture, volatile matter, ash and fixed carbon. This gives quick and valuable information regarding commercial classifications and determination of suitability for a particular industrial use. 2. The ultimate analysis, which includes the estimation of ash, carbon, hydrogen, sulfur, nitrogen and oxygen. The ultimate analysis is essential for calculating heat balances in any process for which coal is employed as a fuel.

8.3.1  Significance of proximate analysis The proximate analysis is an assay rather than true analysis, since the results have no absolute significance. However, if the assay is carried out in accordance with standard specifications, reproducible results can be obtained, thus enabling the coal to be classified and opinion formed regarding its cost, quality and probable use in a particular industry. Each constituent determined under proximate analysis has its own implication and importance in the assessment of the coal sample. Moisture.  Moisture increases the transport costs; Excessive surface moisture may cause difficulties in handling the coal. Moisture reduces the calorific value. A considerable amount of heat is wasted in evaporating the moisture available in coal during combustion. Hence high percentage of moisture is undesirable. Volatile matter.  The volatile matter is not a constituent of coal, but consists of a complex mixture of gaseous and liquid products resulting from the thermal decomposition of the coal substance. The amount of decomposition and the yield of volatile matter depends on the conditions of heating particularly the temperature. Hence, stipulated conditions should be strictly followed during its determination. The volatile matter content of a coal is related to the length of the flame, smoke forming tendency and the ignition characteristics. High volatile matter coals give long flames, high smoke and relatively low heating values. Coal with low volatile content burns with a shorter flame. Thus, the higher the volatile matter content the larger is the combustion space required. Hence, the volatile matter content of a coal influences the furnace design. Further, the % of volatile matter in a coal denotes the proportion of the coal which will be converted into gas and tar products by heat. Hence, high volatile matter content is preferable in coal gas manufacture and in carbonization plants, particularly coal with low volatile matter and high fixed carbon is preferred. The volatile matter content is more in bituminous coals than in anthracite coals. The volatile matter percentage gives some idea about coking property of the coal. Ash. The ash which is intimately interspersed within the mass of the coal is called fixed ash or inherent ash whereas the ash which occurs in different layers of the coal is known as free ash or

8.16  Engineering Chemistry extraneous ash. Only the extraneous or free ash can be removed by washing. Many Indian coals have high ash content. The nature of the ash and its amount in a coal and the softening temperature are very vital in determining the quality of a coal. Ash reduces the heating value of coal. Ash usually consists of silica, alumina, iron oxide and small quantities of line, magnesia, etc. Its composition is of considerable importance in metallurgical operations as it affects the slag and metal composition and consequently is a prime consideration in selecting the flux. When coal is used in a boiler, the fusion temperature of the ash is of particular significance. Fusion temperature of coal ash is generally between 1000°–1700°C. Ash with fusion temperatures below 1200°C is called fusible ash and that above 1430°C is called refractory ash. If the ash fuses at the working temperatures when coal is burnt on grates, it leads to the formation of clinkers (lumps of ash) which reduces the primary air supply and the efficiency of production and distribution of heat are adversely affected. Clinkers cause uneven temperature on the grates, which may contribute to further clinker formation. Some coal particles also gen embedded in the fused ash thereby causing loss of fuel. Fused particles of the ash may stick to the boiler tubes and reduce heat transfer. Removal of clinkers from grates is difficult and laborious. Ash with low melting point forms molten slag which is absorbed in the pores of the refractory lining of the boiler furnace. On account of differences in the coefficients of expansion and contraction of the refractory material and ash, the life of the refractory material might be reduced due to spalling. In view of all the above considerations, coals used in boilers should have high ash fusion temperatures. Fixed carbon.  It is reported as the difference between 100 and the sum of percentages of moisture, volatile matter and ash content of a coal. The fixed carbon content increases from low ranking coals such as lignite to high ranking coals such as anthracite. It is the fixed carbon which burns in the solid state. Hence, information regarding the percentage of fixed carbon helps in designing of the furnace and the fire box. Procedure for proximate analysis 1. Moisture. Moisture is generally determined by heating a known quantity of air-dried coal to 105°C to 110°C for one hour and calculating the loss in weight as percentage. 2. Volatile Matter. Volatile matter is determined by heating 1g of air-dried coal exactly for 7 minutes in a translucent silica crucible of specified dimensions at a steady temperature of 925°C in a muffle furnace. The loss in weight calculated as percentage minus the % moisture gives the % volatile matter. 3. Ash. Ash is determined by heating at 400°C a known quantity of the powdered sample until most of the carbonaceous matter is burnt off and then heating for 1 hour at 750°C to complete the combustion. The weight of residue remaining in the crucible corresponds to the ash content of the coal, which is reported on percentage basis. 4. Fixed Carbon. The sum total of the percentages of volatile matter, moisture and ash subtracted from 100 gives the percentage of fixed carbon.

8.3.2  Ultimate analysis Methods of determination Carbon and hydrogen are determined by burning a known weight of sample in a stream of pure oxygen in a combustion apparatus similar to that used for the analysis of organic compounds. C and H present in the sample are converted into CO2 and H2O respectively, which are absorbed separately in suitable absorption tubes. The % of C and H are calculated from the increase in weight of the respective absorption tubes.

Fuel and Combustion  8.17 Nitrogen is determined by digesting 1g of the coal sample in a Kjeldahl flask with Con. H2SO4, K2SO4 and HgSO4 when the nitrogen present is converted to ammonium salts. The sample is then made alkaline with NaOH and the liberated ammonia is distilled into a measured amount of standard acid. The residual acid is determined by back titration with NaOH. From the amount of the standard acid neutralized by the liberated ammonia, the nitrogen present in the sample is calculated. Sulfur is determined conveniently from the bomb washings obtained from the combustion of a known mass of coal in the bomb calorimeter experiment for the determination of calorific value. The washings contain sulfur in the form of sulfate from which it is precipitated as BaSO4. The precipitate is filtered, ignited and weighed. Form the weight of BaSO4 obtained the sulfur present in the coal is calculated. Sulfur can also be determined by heating the coal with Eschka mixture (2 Parts of MgO: 1 Part of Na2CO3) and estimating the sulfates produced as BaSO4. Ash is determined as described under proximate analysis. Oxygen is determined by difference as follows: % Oxygen = 100 – % of (C + H + S + N + Ash)

8.3.3  Significance Carbon and Hydrogen in coal directly contribute towards the calorific value of the coal. Higher the percentages of C and H, better is the quality of the coal and higher is its calorific value. Hydrogen is mostly associated with the volatile matter of the coal and thus influence the use of coal for the byproduct manufacture or otherwise. Nitrogen in the coal does not contribute any useful value to the coal and since it is generally present only in small quantities (~1%), its presence is not of much significance. Sulfur present in coal contributes towards the heating value of the coal but its combustion products (SO2 and SO3) have corrosive effects on the equipments, particularly in presence of moisture. Further, the oxides of sulfur are undesirable from the atmospheric pollution point of view. Sulfur containing coal is not suitable for the preparation or metallurgical coke as it adversely affects the properties of the metal. Oxygen content of coal is generally associated with moisture. The lower the oxygen content, the more is the maturity of the coal and greater is its calorific value. As the oxygen content increases, the capacity of the coal to hold moisture increases and the caking power decreases. Use of proximate and ultimate analysis in the theoretical determination of the calorific value of coal The calorific value is determined by burning 1g of coal sample in an oxygen bomb calorimeter equipment and measuring the rise of temperature thus produced in the water content of the calorimeter. However, quite often, an engineer may have to estimate the thermal efficiency of a process when the calorific value of the fuel has not been determined. In such circumstances, formulae for the calculation of calorific value from ultimate and proximate analysis are very helpful.

8.3.4  Formulae based on ultimate analysis

1. Dulong’s Formula

( 

)

0 Calorific value in B.Th.U./lb = 14,544 + 62,028 ​ H – __ ​   ​   ​ + 4,050 S 8 Where C, H, O and S represent the respective percentages of carbon, hydrogen, oxygen and sulfur. Several modifications to it have been proposed to this formula and one of them is as follows:

8.18  Engineering Chemistry

[ 

( 

]

)

0 cals 1 Gross   C.V. = ____ ​     ​ ​ 8,080 C + 34,500 ​ H – __ ​    ​  ​ + 2240 S  ​ K ​ ____ ​  100 8 kg 2. Davies Formula C O–S calorific value in B.Th.U. per lb} = (6.543 H + 406) ​ __ ​   ​  + H  – ​ _____  ​    ​ 3 8 Where C, H, O and S are their respective percentages in the coal. 3. Seyler’s Formula

( 

)

calorific value in B.Th.U. per lb} = 223.1 C + 698.6 H – 7684 + 0.45 O Where C, H and O are their respective percentages in the coal.

8.4  Formula Based on Proximate Analysis

1. Gouthal’s Formula

calorific value in B.Th.U. per lb} = 147.6 C + aV Where C is the % of carbon, V is the % of volatile matter and ‘a’ is a constant depending on V. The relation between V and a is as follows: V 1-4 10 15 20 25 30 35 40 A 270 261 210.6 196.2 185.4 176.4 171 144

2. Nakamura’s Formula

( 

)

% Ash calorific value in B.Th.U. per lb} = a ​ V – ​ ______  ​    ​ + 140.4 C, 10 Where ‘a’ depends on the % volatiles and caking prosperity as shown in table below. The deduction of Ash/10 from the volatiles is made to allow for that part of the volatiles as determined which is derived from the combined water and pyritic sulfur in the mineral matter and not from the coal substance. Caking quality

Natural moisture

V20-25

25-35

35-40

40-45

Non-caking

15

108

115

Non-caking

15

122

126

137

142

151

151

153

219-198

173

171

Caking (slight contraction) Caking (slight swelling) Caking (strong swelling)

225-219

The main use of all these formulae is that they provide a means for calculating the calorific values of coals approximately when their compositions are known but of which the samples are not available.

9.  CHARACTERISTICS OF COAL Apart from proximate and ultimate analysis, the assessment of a coal is based on the following characteristics: 1. Colour:  Lignites are brown or brownish black and the color darkens with increasing rank of the coal.

Fuel and Combustion  8.19









2. Texture:  Lignites are earthy and fibrous in structure. With increasing maturity, the coals tend to be more tough, hard and brittle. 3. Specific gravity:  The specific gravity depends upon the type of the coal and its ash content. The specific gravity increase from lignite (1.2) to anthracite (1.5). 4. Heat of combustion (calorific value):  The calorific value increases with increasing rank of the coal (except in case of anthracites whose calorific value may be lesser than semibituminous coals because of lesser percentage of hydrogen). High volatile coals having long flame have less heating value than those of low volatile short-flame coals. Coals with 20% volatiles generally have the best heating value. 5. Grind ability:  This shows the ease with which a coal can be ground and is generally expressed as grind ability index. It is a measure of the power required for grinding a coal and is of special significance for pulverized coals. The coals which can be easily pulverized have grind ability index of about 100. 6. Friability:  This is the tendency of coal to break to pieces on handling and is tested by drop shatter test. Non-friability is essential for coals used with stoker firing. Splint and cannel coal are less friable than others. 7. Caking and coking properties:  Caking is the ability of coal to form a coherent cake on carbonisation. This is an important property used to assess the value of a coal and is tested by various tests such as Swelling Number, Gray-King Assay, Roga Index and Audibert Arnu test. If the residue formed on carbonisation is strong and porous, then the coal is called coking coals which is used for preparing metallurgical coke. 8. Weathering or slaking index:  This is a measure of the tendency of a coal to break on exposure to weather or alternate cycles of dry and wet climate. 9. Bulk density:  This is an important characteristic of coal on which the design of bunkers and containers for storage of coal and for manufacturing coke depends.

10.  SELECTION OF COAL The selection of coal for different applications is mainly made on the basis of the following factors: 1. Calorific value:  Higher the calorific value, more preferred is the coal. 2. Moisture content:  A coal with low moisture content is more economical and hence is more preferred. 3. Ash content:  A coal with low ash content is preferred. Further, the ash should have high fusion temperature so that there is no danger of clinker formation. The significance of the quality, quantity, fusibility and composition of the ash has already been discussed under proximate analysis. 4. Calorific intensity and flexibility:  A coal having a high calorific intensity and flexibility is always preferred. These factors have already been discussed earlier. 5. Uniformity:  A coal having uniformity in size is better from the points of view of storage, handling and efficiency in operation. 6. Sulfur and Phosphorus content:  A coal having very low sulfur and phosphorous contents is essential for metallurgical purposes because these impurities may contaminate the metal and adversely affect its properties, P and S tend to make the metal brittle. 7. Coking quality:  Coking coals are selected for preparation of strong and porous coke suitable for metallurgical purposes.

8.20  Engineering Chemistry

11.  COMMERCIAL TYPES OF COAL





1. Steam coals:  High volatile ligneous coals and low volatile carbonaceous coals are more suitable for steam rising. They are either non-caking or slightly caking. The high volatile coals have somewhat low calorific value and require a large combustion space for efficient and smokeless combustion. These coals burn freely and are quite flexible. On the other hand, the low volatile steam coals have high calorific values. They burn freely liberating bulk of the heat in the fuel bed. 2. Gas coals:  These coals must be strongly caking. Coals with high volatile matter are preferable if the gas manufacture is the primary objective. If manufacture of hard metallurgical coke is required, coals having low volatile matter are more satisfactory. 3. House coals:  These coals should burn freely and should not be strongly caking.

12.  SECONDARY SOLID FUELS Important secondary solid fuels include charcoal, briquettes and coke. Pulverized coal may also be considered as a secondary fuel. A charcoal is the residue from destructive distillation of wood. It has limited application as a fuel in industrial practice. Briquettes are made by compressing low grade fuels with or without a binder. Briquetting provides a method of utilizing small-sized waste coal produced in mining and hence is likely to be an important method as the best coal seems get depleted. In briquetting, the fine coal is mixed with 5 to 8% of a suitable binder (e.g., asphalt pitch, coal tar, molasses, starch, gilsonite and sulfite liquor from paper industry) and compressed into briquettes under pressure. The briquettes so produced may be baked to remove some volatile matter. Coke is an important secondary fuel of industrial importance and is produced by strongly heating the coal out of contact with air. This process is known as carbonization.

12.1  Caking coals and coking coals Coals that soften on heating producing a “pasty’’ or “plastic’’ mass which fuse together yielding coherent masses impervious to air are called caking coals. Coals from which very little of such plastic material is formed are either non-caking or weakly caking. The residue so formed is called coke. If the coke so produced is hard, porous and strong, then the coal from which this coke is derived is called a coking coal. Obviously, all coking coals are caking coals but all caking coals are not coking coals.

12.2 Requisites of a metallurgical coke The quality requirements of a good metallurgical coke are given below. 1. High purity:  The best metallurgical coke should contain lowest possible percentage of moisture (< 4%), ash (< 0.5%) and phosphorous (< 0.1%). Moisture and ash reduce the calorific value. Sulfur and phosphorous in the coke may contaminate the metal and adversely affect its properties. They tend to make the metal brittle. 2. Porosity:  The metallurgical coke should be porous to provide intimate contact between the carbon and oxygen and to ensure efficient combustion of the fuel in the furnace. 3. Strength:  The coke should be strong enough to withstand the abrasion and over-burden of the ore, flux and the fuel itself in the furnace. If the coke breaks into fine particles during charging of the furnace, they may hinder the flow of gases and choke the air passages. (i) Uniformity:  The coke should be uniform and medium in size. If the lumps are too big, combustion is irregular. If they are too small, choking may result.

Fuel and Combustion  8.21 (ii) Calorific value:  The coke should possess a high calorific value. (iii) Cost:  The coke should be cheaply available near the plant site. (iv) Calorific intensity:  The calorific intensity of the fuel should be high enough to melt the metal. (v) Combustibility:  The coke should burn easily but at the same time should not be very reactive. (vi) Reactivity:  Reactivity of coke refers to its ability to react with CO2, steam, air and O2, the reactivity of the coke should not be very high. Coke of low reactivity gives a higher fuel bed temperature than what is produced by a coke of high reactivity. Coal cannot be used as a metallurgical fuel (excepting in reverberatory furnaces) because it does not have the necessary purity, porosity and strength. During the process of carbonisation from coking coals, much of the volatile matter and sulfur compounds are removed and a strong and porous coke is produced. c value is about 6300 to 9300 Kcals/m3 depending on the coal and the process used and the yield is about 150 to 330m3/tonne of the coal carbonised. Hence it is a more valuable gaseous fuel.

13.  COMBUSTION OF COAL Coal is converted into energy by combustion. 75% of the coal in the world is used in this manner for generation of steam for producing electricity. The C, H and S present in the coal undergo combustion in presence of air liberating their heats of combustion. The flue gases contain CO2 water vapor, SO2 and N2 from air. If the elementary composition of a carbonaceous fuel is known, the theoretical amount of air required for the complete combustion of the fuel can be calculated. Maximum efficiency would be obtained if the fuel is burnt completely with the theoretical quantity of air. However, in actual day to day practice, considerable amount of potential heat is lost when only the theoretical quantity of air is used and smoke formation would result in many cases. About 20% excess of air is generally used to achieve the best possible firing conditions. However, If excess air is supplied, the loss of sensible heat in the stack gases increases, due to the greater quantity of stack gases and also due to their higher temperature. The optimum amount of excess air required depends upon the design of the furnace, the fuel used and the equipment used to burn it. For a given furnace, the optimum excess air is that which keeps the sum of the potential heat loss and the sensible heat in the stack gases at a minimum. During the firing of coal on grates, air is supplied in two or more streams. The primary air sent through the fuel initiate the combustion reactions. The secondary air sent over the fuel bed helps to burn the volatile products formed by these reactions. The ‘’tertiary air’’ (heated) may be supplied at the end of the grate to ensure more complete combustion of any smoke that might have been formed. The primary air sent through the fuel bed must be evenly distributed and channels should not be allowed to form through which air escapes without reaction. Mechanical stokers are employed to achieve this objectives. Both hand firing and mechanical firing methodes are commonly used. In mechanical firing, stokers such as “overfeed’’, “underfeed’’, “crossfeed’’, “chain grate’’ and “spreader’’ are employed in practice. Where combustion of coal is carried out without grates, pulverised fuel in suspension or crushed coal in cyclone furnaces with slagging of ash are employed. Burning of coal in a fluidised bed is the latest development in this field.

8.22  Engineering Chemistry

14.  LIQUID FUELS The important liquid fuels include petroleum, petroleum products, tar, alcohols and colloidal fuels. Liquid fuels are also obtained synthetically from the hydrogenation of coal. Low boiling fractions of petroleum are used in petrol engines and higher boiling fractions in diesel engines and oil fired furnaces. Kerosene is used for heating, lighting and cooking. Liquid fuels find extensive use in domestic and industrial fields.

14.1  Merits and Demerits of Liquid Fuels Merits 1. Liquid fuels are easy to handle, store and transport, They can be transported cheaply to distant places through pipe lines. They require less volume for storage as compared to solid and gaseous fuels. Fuel oils weigh 30% less and occupy 50% less space than coal of equal heating value. 2. After burning they do not leave any apprecialbe amount of ash. Hence the problem of ash disposal is not there as in the case of solid fuels. Moreover, the problem of clinker formation is totally absent in case of liquid fuels. 3. Liquid fuels can be easily kindled. The combustion can be started or stopped at once. Maximum temperature can be attained soon unlike in the case of solid fuels. Further, the rate of combustion can be easily controlled as desired. Hence the process is under better control and uniform combustion rate can be maintained. 4. There is no need for maintaining “banked fires’’ in case of liquid fuels unlike solid fuels. Thus better fuel economy is possible in case of liquid fuels. 5. Liquid fuels are handled through pipes and one man can regulate the fuel supply to many furnaces at a time. Continuous manual feeding as in the case of solid fuels is unnecessary. Hence these are more economical. 6. Less excess air is needed in case of liquid fuels than in the case of solid fuels. 7. The furnace space required is lesser than in the case of solid fuels. 8. No danger of spontaneous combustion. If not volatile, they do not deteriorate on storage. 9. Operation is cleaner than in the case of solid fuels. Demerits 1. Liquid fuels give unpleasant odours particularly when the combustion is incomplete. 2. They are more expensive than solid fuels. 3. Possibility of losses due to evaporation during storage and leakage in containers. 4. Risk of the hazards is greater particularly in case of inflammable and volatile liquid fuels. 5. Special type of burners and sprayers are needed for efficient combustion. Careful supervision is necessary to avoid difficulties like leaking and choking which may lead to accidents.

15.  GASEOUS FUELS 15.1 Advantages of gaseous fuels

1. Gaseous fuels can be manufactured at a central place and distributed over a wide area through pipe-lines. In fact, some inferior types of coals which are unsuitable for direct use

Fuel and Combustion  8.23



2.



3.



4.



5. 6.

are converted into gaseous fuels and simultaneously the various byproducts formed are recovered. Gaseous fuels are very clean to operate. They are ashless and smokeless. Ash and smoke cause lots of difficulties while using solid and liquid fuels. They can be ignited instantaneously. As the feed of the gas to the burners can be easily controlled, the combustion process can be kept under better control. Gases are miscible with air and hence the excess air needed is much less than that required for solid or liquid fuels. This leads to fuel economy and efficiency of operation. Both oxidizing and reducing atmospheres can be easily maintained. Sensible heat from the hot flue gases can be recovered and used for preheating the incoming fuel gas using furnaces operating with regenerative principle of heat economy.

15.2  Disadvantages

1. As they are highly inflammable, special care should be taken to avoid fire hazards. 2. As they occupy large volumes, more space is required for storage. However, this problem is not there at the consumer end for “Town gas’’ which is supplied from a central place.

15.3  Commercial gaseous fuels The most important commercial gaseous fuels are as follows: Primary fuels:  Natural gas and liquefied petroleum gas (LPG) Secondary Fuels:  Wood gas, peat gas, coal gas, coke oven gas, producer gas, water gas, kipper’s gas, Winkler’s gas, Lurgi gas, synthesis gas, blast furnace gas, oil gas and refinery gas.

15.4  Natural gas Huge quantities of natural gas are associated with crude oil wells or as gas deposits which may not be associated with oil or as “fire damp’’ (from German words “feurdampf” = fire vapour) in coal measures. Natural gas is mainly composed of methane and small quantities of ethane and other hydrocarbons. If lower hydrocarbons (chiefly methane and ethane) are present, the natural gas is called “dry’’ or “lean’’. If higher hydrocarbons are present along with methane, the natural gas is called “Rich’’ or “Wet’’. It is also called “Marsh gas’’ because it majorly contains methane (“Marsh gas’’ usually seen rising from the bottom of a water covered swamp or marsh as an occasional bubble, is largely methane). Most of the natural gas used as fuel is derived from oil fields. However, sometimes, the gas, evaporated from the oil and diffused through rocks, is trapped by impervious dome shaped structure, which may be far off from the original oil deposit. This constitutes a gas field which may be under high pressure. Due to earth movement or pressure, it may suddenly escape out through a fissure. This may result in ignition on its emergence to the atmosphere because of static electricity produced by the rushing gas (the flow potential) or because of increased temperatures caused by friction. In Azerbaijan, near Baku, (USSR), the natural gas emerging out has been burning for 3000 years, which is depicted by one of their postal stamps as the “Temple of the Eternal Fires’’. The pillars of fire, cloud or smoke described in the “Old Testament’’ were probably due to the burning jets of “dry’’ and “wet’’ natural gas . Since the cause of the flames was not understood and since many ancient cultures worshipped fire, the ever-burning flames of gas were considered to be sacred.

8.24  Engineering Chemistry

The average composition of natural gas is as follows:



CH4 C2H6 C3H8 C4H10 C5+

— — — — —

88.5% 5.5% 3.7% 1.8% 0.5%

It may also contain H2, CO and CO2. “Sour’’ natural gas may contain appreciable quantities of H2S, as in Lacq (France) where it is upto 15%. The calorific value of natural gas varies from 8000 to 14000 Kcal/m3. Natural gas is subjected to various treatments before it reaches the consumer. Wet natural gas is treated to recover the vapours of liquid hydrocarbons (a) by condensation (pressure and cooling) (b) by absorption in oil and (c) by adsorption on charcoal, silica or alumina gel. The liquid so recovered is known as “casinghead gasoline.’’ The natural gas is processed further to remove undesirable components such as (1) water (2) grit and dust (3) H2S (4) CO2 (5) N2. H2S and CO2 are removed and recovered with mono-ethanolamine (NH2C2H4OH) or other ammonia derivatives: NH2C2H4OH + H2S Æ (HOC2H4NH3)2S

NH3 + H2S Æ (NH4)2S



2NH3 + H2CO3 Æ (NH4)2CO3

Natural gas is finding increasing use as a domestic and industrial fuel and as a raw material for the synthesis of methanol, formaldehyde and other chemical compounds. Natural gas is used as a source of H2. Ammonia used in one of the processes for extracting nickel from its ores, is made by reacting N2 (from liquid air) with natural gas. Sulfur is also recovered from H2S containing natural gas: 2H2S + 3O2 Æ 2 SO2 + 2 H2O 2H2S + SO2 Æ 3S + 2 H2O India’s proved reserves of natural gas are of the order of 40 km3, most of which lie in Assam and Gujarat.

15.4.1  Substitute Natural Gas (SNG) The low and medium-heat syngas produced from coal can be converted to a high-heat content gas(30 to 37 M.J/m3) similar to natural gas by the following reactions. C + H2O Æ CO + H2 gasification CO + H2O Æ CO2 + H2 water gas shift reaction (controlled to give CO:H2 = 1:3) C + CO2 Æ 2CO Boudoward reaction At sufficiently high pressures, the hydrogen from reactions (1) and (2) will hydrogenate some of the carbon to yield methane. C + 2H2 Æ CH4 CO + 3H2 Æ CH4 + H2O (methanation)

Fuel and Combustion  8.25 The gas thus produced was known originally as synthetic natural gas, but language purists argued that synthetic could not be natural, so now it is called substitute natural gas. The sulfur and carbon dioxide are removed from the gas before it is methanated. The operating pressure in the gasifier, depending upon the process, can be atmospheric to over 6.9 Mpa and the temperature can vary from 800°C to about 1650°C. The higher pressure and lower temperture result in the formation of a large amount of methane.

15.4.2  Numerical problems based on combustion and flue gas analysis The following points should be remembered in solving numerical problems based on combustion and flue gas analysis: 1. Main Objectives:  The problems are generally based on calculating: (a) The weight or volume of air theoretically required or used for the combustion of 1kg of fuel. (b) The percentage of excess air used per kg. of fuel burnt, (c) The percentage composition of flue gas obtained (by weight or by volume) by burning a known quantity of fuel, and (d) The percentage composition of the fuel burnt. 2. Concept of mole:  A mole of a substance is that quantity whose weight is numerically equal to its molecular weight. Weight of the substance, W Number of moles, n = ​ _______________________          ​ Molecular weight,  M. The “mole’’(or “mol’’) is a general unit; when expressed in grams, it is called “gram mole’’; when expressed in kilograms, it is called “kilogram mole’’ or “kilo mole’’; and when expressed in pounds, it is called “pound mole.’’ Hence, the term “mole’’ should be interpreted as “gram mole’’, “k” or pounds respectively. Also, “mol’’ is a volume unit and hence composition in terms of mols can be taken as the same as composition by volume (or composition in terms of m3) i.e., volume % = mole %. Further, 1 gram mole (g mol) of a gas at NTP occupties 22.4 litres (or dm3); 1 kilogram mole occupies 22.4 m3 and 1 pound mole occupies 359 ft3. 3. Composition and mean molecular weight of air:  The composition of air is taken as 21% of O2 and 79% of N2 (by volume); and 23% of O2 and 77% of N2 (by weight.). The mean molecular weight of air is taken as 28.95. It may be noted that air consists of 21.00 mols of O2, 78.06 mols of N2 and 0.94 mols of Ar (argon). Since argon is inert, it is considered together with nitrogen for combustion calculations. That is why, the % N2 in air by volume is taken as 79%. However, the mean molecular weight of air is taken as 28.95 on the basis of the following:

[ 

]

(21 × 32) + (78.06 × 28) + (0.94 × 39.34) ​ ​ ___________________________________       ​      ​ = 28.9522 100 Further, 1 m3 of O2 is supplied by 1 × 100/21 = 4.76m3 of air; and 1 kg. of O2 is supplied by 1 × 100/23 = 4.35 kg. of air. 4. Density of air:  The density of air at N.T.P Is 1.290 kg. m–3 or 1.290 × 10 –3 g cm–3

8.26  Engineering Chemistry

5. Minimum Oxygen Required  The minimum O2 required = (Theoretical O2 required – O2 present in the fuel) The minimum O2 required should be calculated on the basis that complete combustion is taking place according to theoretical and stoichiometric combustion reactions. In case of partial combustion, the combustion products contain CO. In case of irregular combustion, the combustion products contain both CO and O2. In such a case, the excess O2 is calculated after subtracting the amount of O2 required to burn CO to CO2. 6. Combustion of Carbon  The combustion of carbon in air may be represented as: C + (O2 + N2) Æ CO2 + N2 1 mol 1 mol 3.76 mols 1 mol 3.76 mols Or or or or or 12kg. 32kg. 107.2kg. 44kg. 107.2kg. Thus, combustion calculation can be done on mol basis or by using the stoichiometric weight relationships between the reactants and products. However, the mol method is considered to be more convenient and simpler. 7. Combustion of hydrogen  The hydrogen in coal is present as (a) Combined hydrogen in the form of H2O 2H2 = O2 = 2H2O 4 kg. 32kg. 2 × 18 = 36kg 1kg. 8kg. 9kg. (b) A vailable hydrogen: Combined hydrogen in coal present as moisture does not under go combustion. It is only the available hydrogen which is equivalent to (H–0/8) that takes part in combustion. 8. Weight of theoretical amount of air required  For complete combustion of 1 kg. of solid or liquid fuel, the theoretical amount of air required

[ 

( 

) ]

100 32 0 = ____ ​   ​   ​  ​     ​ × C + 8 ​ H – __ ​   ​   ​ + S  ​ kg. 3 ​12 ​ 8

Where C, H, O and S are the respective weights of carbon, hydrogen oxygen and sulfur present in 1 kg. of the fuel. 9. Calorific Value  If the ultimate analysis of coal is available, its calorific value may be calculated by Dulong’s formula as follows: Calorific value (Kcal/kg.) 0 = [8080 C + 34460 ​ H – __ ​   ​   ​ + 2250 S] 8

( 

)

Where C, H, O and S represent the respective weight of carbon, hydrogen, oxygen and sulfur per kg. of coal. 10. % Excess Air (Actual air used–Theoretical air) % Excess Air = ​ ____________________________           ​ × 100 Theoretical air

Fuel and Combustion  8.27

11. The mass of dry flue gases formed should be calculated by balancing the carbon in the fuel and the carbon present in the flue gases. 12. The composition of a solid or liquid fuel is usually expressed on weight basis whereas the composition of a gaseous fuel is expressed on volume basis unless otherwise stated.

16.  Biomass Biomass refers to living and recently dead biological material that can be used as a fuel or for industrial production. It is referred to material derived from recently living organisms. This includes plants, animals and their byproducts. For example, manure, cattle dung, garden waste, sugarcane waste and crop residue are all sources of biomass. Biomass include biodegradable wastes that can be burnt as fuel. Biomass is a renewable energy source based on carbon cycle. It is grown from several plants, including miscanthus, switichgrass, hemp, corn, poplar, willow, sugarcane and oil palm. Biomass is a part of carbon cycle. Carbon from the atmosphere is converted into biological matter by photosynthesis. On the death or combustion the carbon goes back to atmosphere as carbon dioxide. Thus, during a life span of animal or plant total carbon content remains same. Production of biomass is a growing industry as interest in sustainable fuel sources is growing. Biomass consists of carbon compounds which may be used as a source of energy by using either of the following methods: 1. Biomass such as cattle dung, wood, bagasse, plant wastes, agricultural wastes, dry vegetable waste, etc. is used directly in chulhas for getting energy. However, by doing so a lot of heat energy is wasted and lot of smoke is liberated, thereby causing blackening of utensils and houses. Moreover, it librates poisonous gas carbon monoxide and leaves ash as residue. 2. Biomass is converted into biogas, which is used for heating and lightening purposes. Advantages of converting biomass into biogas are: (i) Biogas production is very economical. It has been found that 1kg of dry cattle dung gives about 160 litres of gabar gas, which can supply 800kJ of heat. On the other hand, 1kg of dry dung on direct burning gives only 100kJ of effective heat. (ii) The gas has all the advantages of gaseous fuel like cleanliness, absence of smoke, flexibility, etc., It does not have poisonous gas, CO as an ingredient. (iii) It provides simultaneously yield of good manure.

16.1  Biogas Biogas refers to a gas produced by the biological breakdown of organic matter in the absence of oxygen. Biogas produced by the fermentation of biomass comprises primarily of methane and carbon dioxide. The average composition of biogas is: CH4 50-65% CO2 30-40% H2 5-10% N2 2.6% H2S Traces

8.28  Engineering Chemistry Biogas can provide a clean, easily controlled source of renewable energy from organic waste materials for a small labour input, replacing firewood or fossil fuels. Process Biogas is generated when bacteria degrade biological material in the absence of oxygen, in the process known as anaerobic decomposition. Anaerobic decomposition is a two-stage process as specific bacteria feed on certain organic materials. 1. In the first stage, acidic bacteria (such as E.coli, bacteriods etc) break down the complex organic molecules into peptides, glycerol, alcohol and the simple sugars. 2. In the second stage, a second type of bacteria (methangenoic bacteria) starts to convert these simpler compounds into methane. These methane producing bacteria are particularly influenced by the ambient conditions. The main features that influence biogas production are: pH; It is well established that a biogas plant works at an optimal pH of 7 or just above, Temperature is 32°C. Raw material for biogas: Animal dung, poultry wastes, vegetable wastes, waste paper and cotton clothes, plants wastes, human excreta, birds excreta, etc,.

Fig. 3

Manufacture of gobar gas: It is produced by anaerobic degradation of cattle dung. It is carried out in gobar gas plant which consist of well shaped underground tank (called digester) covered with a dome shaped roof, both made of bricks and cement. The dome of the digestive tank is fixed so that it acts as a gas holder for the biogas produced. At the top of the domb there is a gas outlet pipe

Fuel and Combustion  8.29 and a gas valve. On the left side of the tank there is a sloping inlet chamber and on the right side there is rectangular outlet chamber, both made up of bricks and cement. Fresh cattle dung plus water slurry is introduced from the inlet chamber while spent slurry gets collected in the outlet chamber. The inlet chamber is connected to the mixing tank; while the outlet chamber is connected to the overflow tank. Working: Slurry (made by mixing cattle dung and water in a definite proportion in mixing tank) is fed into the digestive tank via the inlet chamber, till the slurry level becomes equal to cylindrical top level. In about 50 to 60 days, the biogas plant starts functioning. During this time period, cattle dung undergoes fermentation in the presence of anaerobic bacteria with gradual evolution of biogas, which starts collecting in domed shaped space. As the time passes more and more biogas collects inside the dome, thereby exerting pressure on the slurry in the digestive tank and this in turn forces the spent slurry to the overflow tank via outlet chamber. From the over flow tank spent slurry is withdrawn periodically and used as a good manure. From time to time fresh slurry is fed to the digestive tank so as to get the regular supply of biogas. The biogas collected in the dome is taken out through outlet pipe by opening the gas valve and then used as a fuel. Fixed dome type biogas plant is cheap, since only bricks and cement are used for its construction and there is no danger of corrosion for such a plant. Biogas is considered as an ideal fuel: 1. It burns without smoke hence causes no pollution. 2. Its calorific valve is high 3. It is cheapest gaseous fuel 4. It has no storage problem Uses of biogas: (i) For cooking food (ii) As a fuel to run engines (iii) As a illuminant in villages. Limitations of biogas: It is necessary to have the gas lamp or stove or burner within 10 metres of the plant.

solved Questions Type–I 1. Classify the following fuels in to primary and secondary fuels. Lignite, Dung, Gasoline, Biogas, Kerosene. Sol. Fuels which occur in nature as such are known as primary fuels while those which are derived from the primary fuels are known as secondary fuels. Primary fuel Lignite Dung

Secondary fuel Gasoline Biogas Kerosene

2. Classify the following into conventional and non conventional energy sources: Petrol, diesel, Coal, Biogas, Solar energy, Tidal energy, Geothermal energy.

8.30  Engineering Chemistry Sol. Conventional energy sources are exhaustible and non-renewable on the other hand, the nonexhaustible and renewable energy sources are non-conventional. Conventional energy sources

Non-Conventional energy sources

Petrol Diesel Coal

Biogas Solar energy Tidal energy Geothermal energy

3. List the raw materials which can be utilized for biogas manufacture. Sol. Animal dung (mainly cow dung), poultry wastes, human excreta, plant wastes, waste papers etc are the raw materials which can be utilized for biogas manufacture. 4. Arrange the following in increasing order of their calorific value and moisture content. Peat, Lignite, Anthracite, Bituminous. Sol. Coal

Calorific value (kcal/kg)

Peat Lignite Bituminous Anthracite



5400 6500 – 7100 7000 – 8600 8650 – 8700

Increasing order of calorific value Peat < Lignite < Bituminous < Anthracite. The decreasing order of moisture content is Peat > Lignite > Bituminous > Anthracite.

Type–II Problems Based on Bomb Calorimeter 5.

The following data is obtained in a Bomb calorimeter experiment: Weight of crucible = 3.649 gm. Weight of crucible + fuel = 4.687 gm. Weight equivalent of calorimeter = 570 gm. Water taken in calorimeter = 2200 gm. Observed rise in temperature = 2.3°C. Cooling correction = 0.047°C. Acid correction = 62.6 calories. Fuse wire correction = 3.8 calories. Cotton thread correction = 1.6 calories. Calculate the gross calorific value of the fuel sample. If the fuel contains 6.5% hydrogen, determine the net calorific value.

Fuel and Combustion  8.31 Sol.

(W + w) (t 2 – t1 + Cc) – (CA + CF + CCr) HCV = ​  __________________________________          m ​

(2200 + 570) (2.3 + 0.047) – (62.6 + 3.8 + 1.6) = ​ ________________________________________        ​     (4.687 – 649) (2770 × 2.347) – 68 = ​ _________________  ​       1.038 HCV = 6197.67 cal/gm Since the fuel contains 6.5% hydrogen, LCV = (HCV – 0.09 × H × 587) cal/gm LCV = 6261 – 0.09 × 6.5 × 587 = 6261 – 343.4 LCV = 5854.28 cal/gm 6. A sample of coal containing 80% C, 15% H and 5% ash is tested in bomb calorimeter. The following results were obtained. Weight of coal burnt = 0.98 gm. Weight of water taken = 1000 gm. Water equivalent of bomb and calorimeter = 2500 gm. Rise in temperature = 2.5°C. Cooling correction = 0.02°C. Fuse wire correction = 8.0 calories. Acid correction = 50.0 calories. Assuming the latent heat of condensation of steam as 580 cal/gm, calculate the (i) higher (ii) lower calorific value of the fuel. Sol. (W + w) (t 2 – t1 + Cc) – (CA + CF) HCV = ​  _____________________________         m ​ (1000 + 2500) (2.5 + 0.02) – (50 + 8) HCV = ​  ________________________________       ​     cal/gm 0.98 = 8940.82 cal/gm



LCV = HCV – 0.09 × H × 580 LCV = 8940.82 – 0.09 × 15 × 580 cal/gm = 8940.42 – 783 LCV = 8157.82 cal/gm 7. 0.85 gm of a fuel is burnt completely in excess supply of oxygen. The increase in temperature of water in the calorimeter containing 1800gm of water was found to be 3°C. Calculate the higher calorific value of the fuel. Given that the water equivalent of calorimeter etc is 180gm. Sol. (W + w) (t 2 – t1) HCV = ​ ______________      m ​

8.32  Engineering Chemistry (1800 + 180) (3) = ​ ______________  ​      cal/gm 0.85 HCV = 6988.23 cal/gm.

8. A sample of coal containing 90% C, 8% H and 2% ash. When this coal sample was burnt in bomb calorimeter, the following results were obtained: Weight of coal burnt = 0.90 gm. Weight of water taken = 800 gm. Water equivalent of calorimeter = 2000 gm. Rise in temperature = 2.40°C. Cooling correction = 0.02°C. Fuse wire correction = 10.0 calories. Acid correction = 60.0 calories. Calculate the net and gross calorific values of the coal in cal/gm. Assume the latent heat of condensation of steam as 580 cal/gm. Sol. (W + w) (t 2 – t1 + Cc) – (CA + CF) HCV = ​  _____________________________         m ​

(800 + 2000) (2.40 + 0.02) – (60 + 10) HCV = ​  _________________________________       ​     cal/gm 0.90

(2800 × 2.42) – 70 = ​ ________________  ​       0.90 = 7451.11 cal/gm.

Type–III Problems Based on HCV and LCV (Dulong’s Formula) 9. Calculate the gross and net calorific value of coal having the following compositions: Carbon = 85% Hydrogen = 8% Sulphur = 1% Nitrogen = 2% Ash = 4% Latent heat of combustion of steam = 587 cal/g. Sol. According to Dulong’s formula

[ 

( 

)

]

0 1 HCV = ____ ​     ​ ​ 8080 C + 34500 ​ H – __ ​    ​  ​ + 2240 S  ​ cal/gm 100 8

Given that C = 85% H = 8% S = 1% Nitrogen and ash, don’t contribute to the calorific value.

[ 

(  )

]

0 1 HCV = ____ ​     ​ ​ 8080 × 85 + 34500 ​ 8 – __ ​   ​   ​ + 2240 × 1  ​ cal/gm 100 8 1 = ____ ​     ​ [686800 + 276000 + 2240] cal/gm 100 1 = ____ ​     ​ [965040] = 9650.4 cal/gm. 100 So,

...(7.1)

Fuel and Combustion  8.33

10. A coal has the following composition by weight: C = 90%, O = 3.0%, S = 0.5%, N = 0.5% and ash 2.5%. Net calorific value of coal was found to be 8490.5 kcal/kg. Calculate the percentage of hydrogen and higher calorific value of coal.

Sol. Given that

C = 90%,  O = 3%,  S = 0.5%,



LCV = 8490.5 kcal/kg.



HCV = LCV + 0.09H × 587 kcal/kg.

Or

HCV = 8490.5 + 0.09 × H × 587

HCV = (8490.5 + 52.8H) kcal/kg According to Dulong’s formula

[ 

( 

...(7.2)

]

)

3.0 1 HCV = ____ ​     ​ ​ 8080 × 90 + 34500 ​ H – ___ ​   ​    ​ + 2240 × 0.5  ​ kcal/gm 100 8

Or HCV = [7272 + 345H – 129.4 + 11.2] kcal/gm Or HCV = (7754.8 + 345H) kcal/kg From 2 and 3 we have 7754.8 + 345H = 8490.5 +52.8H (345 – 52.8) H = 8490.5 – 7754.8 1335.7 Or %H = ______ ​   ​  = 4.575% 292.2 From 2 and 4 HCV = (8490.5 + 52.8 × 4.575) kcal/kg = (8490.5 + 241 .3) = 8731.8 kcal/kg

...(7.3)

Or

...(7.4)

Type–IV Problems Based on Boy’s Gas Calorimeter 11.

The following data were obtained in a Boy’s gas calorimeter experiment: Volume of gas used = 0.1 m3 at STP. Weight of water heated = 50 kg. Temperature of inlet water = 20°C. Temperature of outlet water = 30°C. Weight of steam condensed = 0.15 kg. Calculate the higher and lower calorific value per m3 at STP. Take the heat liberated in condensing water vapour and cooling the condensate as 580 kcal.gm.

Sol. Given that V = 0.1 m3, W = 50 kg, T1 = 20°C, T2 = 30°C, Or

W(T2 – T1) 50 × (30 – 20) HCV = ​ _________       ​ = ​ _____________  ​      V 0.1 50 × 10 HCV = ​ _______  ​    = 5000 kcal/m3 0.1 M LCV = HCV – __ ​   ​  × 580 V

= 0.15 kg

8.34  Engineering Chemistry 0.15 = 5000 – ____ ​   ​ × 580 = 5000 – 3 × 290 0.1 = 5000 – 870 LCV = 4130 kcal/m3.



Type–V Problems Based on Combustion of Fuel 12. Calculate the weight and volume of air required for the combustion of 500 gm of carbon. Sol. Air contains 23% oxygen by weight while it contains 21% oxygen by volume. Combustion reaction C + O2 CO2 12  32   Weight of oxygen required for combustion of 12 gm of carbon = 32 gm 32 \  Weight of oxygen required for combustion of 500 gm of carbon = ___ ​   ​ × 500 gm 12 = 1333.3 gm = 1.33 kg 100 Weight of air required = ____ ​   ​ × 1.33 kg = 5.782 kg 23 \  32 gm of oxygen occupies 22.4 L at NTP. 22.4 \  1333.3 gm of oxygen occupies ____ ​   ​   × 1333.3 L at NTP = 933.31 L 32 100 \  Volume of required = ____ ​   ​ × 933.3 = 4444.28 L = 4.44 m3. 21

13. A gas sample has the following composition by volume:

H2 = 30%, CH4 = 5%, CO = 25%, CO2 = 11%, O2 = 10%, N2 = 19%. If 40% excess air is supplied, calculate the volume of air required for the complete combustion of 1 m3 of the fuel. Sol. Nitrogen and CO2 are incombustible matters and hence they take any oxygen during combustion. % of component 3

H2 = 30% = 0.3 m CH4 = 5% = 0.05 m3 CO = 25% = 0.25 m3

Combustion reaction

Volume of O2 required

H2 + ½ O2 Æ H2O CH4 + 2O2 Æ CO2 + 2H2O CO + ½ O2 Æ CO2

0.3 × 0.5 = 0.15 m3 0.05 × 2 = 0.1 m3 0.25 × 0.5 = 0.125 m3

Total volume of O2 required = 0.375 m3 O2 available in fuel gas 5%

Net oxygen required = 0.375 – 0.05 = 0.05 m3 = 0.325 m3

\  Net oxygen required = 0.325 m3 = 325 L

Fuel and Combustion  8.35 \  Volume of air required for the complete combustion of 1 m3 of the gas using 100 140 40% of excess air = 325 × ____ ​   ​ × ____ ​   ​ = 2166.6 L 21 100 14. A sample of coal was found to have the following % composition by weight:

 

C = 75%, H = 5.2%, O = 12.1%, N = 3.2% and ash = 4.5% Calculate (i) The minimum amount of O2 and air by weight necessary for complete combustion of 1 kg of coal (ii) Weight of air required, if 40% excess air is supplied. (iii) Gross and net calorific value of coal sample using Dulong’s formula. Sol. Given that weight of coal = 1 kg % of component

Combustion reaction

C = 750 gm

C + O2 Æ CO2

H = 52 gm

H2 + ½ O2 Æ H2O

Wt of O2 required 32 750 × ___ ​   ​ gm = 2000 gm 12 16 52 × ___ ​   ​ gm = 416 gm 2

Total weight of oxygen required = 2416 gm Oxygen present in the fuel = 121 gm

\  Net oxygen required for combustion = 2416 – 121 = 2295 gm = 2.295 kg 100 \  Minimum weight of air required for combustion = 2.295 × ____ ​   ​ = 9.978 kg 23 140 (ii) When 40% excess air is supplied weight of air required = 9.978 × ____ ​   ​ = 13.969 kg 100 O 1 (iii) Gross calorific value = ​ ____    ​ ​ 8080 C + 34500 ​ H – __ ​   ​   ​  ​ 100 8

[  [ 

( 

( 

)]

)]

12.1 1 = ____ ​     ​ ​ 8080 × 75 + 34500 ​ 5.2 – ____ ​   ​    ​  ​ = 7332 kcal/kg 100 8

Net calorific value = (HCV – 0.09 × H × 587) kcal/kg

= (7332 – 0.09 × 5.2 × 587)



= 7057 kcal/kg.

Type–VI Problem Based on Proximate and ultimate analysis

15. A sample of coal was analysed as follows: Exactly 1.40 gm was weight into a silica crucible. After heating for one hour at 105 – 110°C, the residue weight 1.10 gm. The crucible next was covered with a vented lid and strongly heated for exactly 7 minutes at 950 ± 20°C. The residue was weight 1.0 gm. The crucible was then heated without cover, until a constant weight was obtained. The last residue was found to weight 0.21 gm. Calculate the ash content, moisture content, volatile matter and percentage of fixed carbon.

8.36  Engineering Chemistry Sol. Weight of moisture in coal = 1.40 – 1.10 = 0.30 gm Weight of volatile matter = 1.10 – 1.00 = 0.10 gm Weight of ash = 0.21 gm 0.30 % of moisture = ____ ​   ​ × 100 = 21.4% 1.4 0.10 % of volatile matter = ____ ​   ​ × 100 = 7.14% 1.4 0.21 % ash = ____ ​   ​ × 100 = 15% 1.4 % of fixed carbon = 100 – (21.4 + 7.14 + 15) = 56.46%.

16. 3.00 gm of a coal sample was Kjeldahlised and NH3 gas thus evolved was absorbed in 40 ml of N/10 H2SO4. After absorption, the excess (residual) acid required 10 ml of N/10 NaOH for exact neutralization. Calculate the percentage of nitrogen in the coal sample. Sol. Method–I 1 1 Amount of H2SO4 used to neutralize = ​ ___ ​    ​ × 40 – ​ ___  ​ × 10  ​ milli equivalent the evolved 10 10 ammonia 1 = (4 – 1) × _____ ​     ​ equivalent 1000 3 = _____ ​     ​ equivalent 1000 3 Weight of nitrogen = _____ ​     ​ × 4 1000 Weight of nitrogen % of nitrogen = ​ ___________________         ​ × 100 Weight of coal sample

( 

Method–II 10 ml N/10 NaOH \  Volume of H2SO4 used to Neutralise the evolved ammonia

)

3 × 14 = ​ ________    ​  × 100 = 1.4% 1000 × 3 = 10 ml N/10 H2SO4 = 40 ml N/10 – 10 ml N/10

\  % of nitrogen

Volume of H2SO4 used × Normallity × 1.4 =  ​ _____________________________________             ​ Weight of coal sample taken



30 × 1 × 1.4 = ____________ ​         ​ = 1.4% 3.0 × 10



17. 0.8 gm of a sample of coal was burnt in excess supply of oxygen in a bomb calorimeter. The washing was then heated with barium chloride and a precipitate of barium sulphate was obtained. The precipitate was filtered, dried and weight and was found to be 0.08 gm. Calculate the percentage of sulphur in the coal sample.

Weight of BaSO4 32 Sol. % Sulphur = ​ _________________________         ​ × ____ ​    ​ × 100 Weight of coal sample taken 233

0.08 32 = ____ ​   ​ × ____ ​    ​ × 100 = 1.373% 0.8 233

Fuel and Combustion  8.37

UNsolved Questions

1. Define a chemical fuel. What are the characteristics of a good fuel? 2. Classify the following fuels into primary and secondary fuels: Coal, Coke, Natural gas, Diesel, Kerosene, Briquette, Producer gas, LPG. 3. Distinguish between natural and synthetic fuels. 4. Define calorific value of a fuel and write the units. 5. Differentiate between net calorific value and gross calorific value. 6. Describe how the calorific value of a solid fuel is determined using a bomb calorimeter. 7. Describe a suitable method for the determination of the calorific value of a solid fuel. What are the corrections needed? 8. How is the higher calorific value and the lower calorific value of a coal sample determined? 9. Give the determination of calorific value by Boy’s gas calorimeter. 10. Describe how the calorific value of coal or liquid fuel is determined by Bomb calorimeter. 11. What are chemical fuels? How are they classified? 12. What are non-conventional fuels? List a few of them. 13. Write the approximate composition and applications of biogas. 14. Give the examples of non-conventional sources of energy. 15. List the raw materials which can be utilized for biogas manufacture. How is biogas obtained from cattle dung? 16. Discuss the manufacturing of biogas. Write its merits and demerits. 17. Write short notes on: (i) Biogas (ii) Solar energy 18. Explain the process involved in the conversion of solar energy into electrical energy and heat energy in solar plaints. 19. Describe the reasons for not using solar energy at domestic and industrial levels economically in spite of the free availability of sunlight. 20. Justify that solar energy is a very good source of non conventional energy. What do you understand by renewable and non renewable fuels? 21. Discuss the process of conversion of solar energy into electrical energy. 22. Discuss p-type and n-type silicones used in solar energy plants. 23. Write a short note on biomass. 24. Explain the following: (i) Gaseous fuels are more advantageous than solid fuels. (ii) An ideal fuel should have moderate ignition temperature. (iii) Net calorific value is less than gross calorific value. 25. Distinguish between solid, liquid and gaseous fuels. 26. What are fuels? Discuss their classification. 27. Write a short note on biomass as a source of energy. How does this technique differ from biogas technique?

8.38  Engineering Chemistry

28. A sample of coal containing 89% C, 8% H, 3% ash, when tested in the laboratory for its calorific value in the bomb calorimeter, the following data were obtained: Weight of coal burnt = 0.85 gm. Weight of water taken = 650 gm. Water equivalent of bomb calorimeter = 2500 gm. Rise in temperature = 2.5°C. Cooling correction = 0.03°C. Fuse wire correction = 10 Cal. Acid correction = 50 Cal. Assuming that the latent heat of condensation of steam as 580 cal/gm, calculate (i) the gross and (ii) the net calorific values of the coal in cal/gm. [Ans:  GCV = 9305.29 cal/gm, NCV = 8887.69 cal/gm] 29. The following data were obtained in a bomb calorimeter experiment: Weight of coal burnt = 0.95 gm. Weight of water taken in calorimeter = 1000 gm. Water equivalent of bomb calorimeter = 2500 gm. Rise in temperature = 2.8°C. Cooling correction = 0.03°C. Acid correction = 60.0 calories. Cotton thread correction = 2.0 calories. Calculate the gross calorific value of the fuel sample. If the fuel contains 5% hydrogen determines the net calorific value. [Ans. GCV = 10361.05 cal/gm, NCV = 10096.9 cal/gm] 30. A gas has the following compaction by volume: H2 = 20%, CH4 = 6%, CO = 22%, CO2 = 4%, O2 = 4% and N2 = 44%, 20% excess air is used. Find the weight of air actually supplied per m3 of the gas [Ans. 2143g] 31. Calculate the weight of air needed for complete combustion of 5.0 kg of a coal containing 80% carbon, 15% hydrogen and the rest oxygen. [Ans. 7139 kg] 32. A sample of coal was found to contain following percentage composition: C = 75%, H = 5.2%, O = 12.8%, S = 1.2%, N = 3.7%, ash = 2.1%. Calculate the minimum amount of air necessary for complete combustion of 1 kg of carbon. 33. A sample of coal was analysed as fallows: 1 g of an air dried coal, sample was weighed in a silica crucible. After heating far 1 hour at 105-110°C, the dry coal residue weighed 0.985 g. The crucible was covered with a rented lid and then heated strongly far exactly 0.800 g. The crucible was then heated strongly in air, until a constant weight was obtained. The lost residue was found to weight 0.10 g calculate the proximate analysis. [Ans. Moisture = 1.5, Volatile matter = 18.5%, ash = 10.0% and fixed C = 70%] 34. 1.56 g of the coal was kjeldahlized and NH3 gas thus evolved was absorbed in 50.0 ml of 0.1 N H2SO4. After absorption, the excess (residual) acid required 6.25 ml of the 0.1 N NaOH far exact neutralization, 2.60 g of the coal sample in a quantitative analysis gave 0.1755 g of BaSO4. Calculate the percentage of N and S in the Coal sample. [Ans. N = 3.93%, S = 0.93%]

Fuel and Combustion  8.39

35. A coal sample gas the following analysis: C = 66.2%, H = 2%, O = 6.1%, N = 1.4%, S = 2.9%, moisture = 9.7% and ash = 9.5%. Determine the quantity of products of combustion, if 1 kg of coal in burnt with 25% excess air. [Ans. 12.17 kg] 36. A coal sample was found to contains the following: C = 81%, H = 4%, O = 2%, N = 10% the remaining being ash. Estimate the quantity of minimum air required for complete Combustion of 1 kg of the sample. [Ans. 10.69 kg] 37. Calculate the mass of air needed for Complete Combustion of 5 kg of coal containing 80% Carbon, 15% hydrogen and rest in oxygen. [Ans. 71.034 kg] 38. A gas used in an internal combustion engine had the following composition by volume H2 = 45%, CH4 = 36%, CO = 15% and N2 = 4%. Find the volume of air required for the combustion of 1 m3 of gas. [Ans. 4.85 m3]

Chapter

9

Spectroscopic Methods of Analysis 1.  INTRODUCTION Molecular spectroscopy is defined as the study of interaction of electromagnetic radiations with matter. It provides valuable information regarding molecular structure such as molecular symmetry, bond distances, bond angles; chemical structure such as chemical properties, electronic distribution, bond strength, molecular reactions; and thermodynamic properties.

2. ELECTROMAGNETIC SPECTRUM

Fig 1  Beam of Electromagnetic Radiation according to Wave Nature Concept

An electromagnetic radiation is a form of energy that is transmitted through space at an enormous velocity. Quantum mechanics suggests that electromagnetic radiation has a dual character, i.e., exhibiting wave as well as particle behaviour. Accordingly to wave nature of radiations, the electromagnetic wave consists of oscillating electric and magnetic fields. The vectors of electric and magnetic fields are perpendicular to the direction of propagation of radiation (Fig 1). The electromagnetic spectrum extends from the radio wave to the microwave, infrared, visible and ultraviolet regions. Beyond these are X-ray, gamma ray and cosmic ray regions shown in Figure 2.

9.2  Engineering Chemistry

Fig. 2  Various Regions of the Electromagnetic Spectrum

Table 1  Some components of the Electromagnetic Spectrum Sr. No.

Type of Spectra

Wavelength

Energy in Kcal/mol

Effect on the Molecule

1.

Ultraviolet

200-400 mm

143-82

Change in electronic energy levels within the molecule.

2.

Visible

400-800 mm

82-36

Change in electronic energy levels within the molecule.

3.

Infrared

2-16 mm

14.3-1.80

Change in vibrational and rotational movements of the molecule.

4.

Nuclear Magnetic Resonance

Induces changes in the magnetic energy levels of certain nuclei.

The wavelength of visible light range from 400mm (violet) to 800mm (red). The visible region however, is a very small part to the entire electromagnetic spectrum. Some components of the electromagnetic spectrum are listed in Table along with their wavelength, the energy associated with them, and the type of effect they are capable of producing in a molecule. Common symbols used in spectroscopy are given in Table 2

Table 2  Symbols used in Spectroscopy Symbol

Definition

g

Frequency in Hz cycles per second

l

Wavelength

Spectroscopic Methods of Analysis  9.3 mm Å Cm–1

Micrometer, same as micro (m)10 –6m Angstrom, 10 –10m or 10 –12cm Wave number, frequency in reciprocal or gl

3.  LAMBERT-BEER OR BEER’S LAW Lambert’s law states that, when a mono chromatic light is passed through a solution, the decrease in the intensity of light with the thickness of the solution is directly proportional to the intensity of incident light. It = Io e–kx



Where,



It = Io = x = k =

intensity of the transmitted light. intensity of the incident light. thickness of the medium or the length of the solution. absorption coefficient.

Beer’s law an extension of Lambert’s law states that, when a monochromatic light is passed through a solution, the decrease in the intensity of light with the thickness of the solution is directly proportional to intensity of incident light as well as the concentration of the solution.

Where,



I = Ioe E = Molar absorption coefficient C = Concentration of solution

The equation may be written.



I __ ​    ​ = e–Ecx Io



I log __ ​    ​ = – Ecx Io



Where, is called transmittance, T and, ....... and is called absorbance.



\



I – log __ ​    ​ = – log T = A Io A = Ecx E = A/cx

Where,



E = molar absorption coefficient c = molar concentration x = path length in centimeters.

The magnitude of the molar absorption coefficient is directly related to the probability of a particular transition.

9.4  Engineering Chemistry Emax = 0.87 × 1020 p.a.

Where,

P = Transition probability with values ranging from 0 to 1. a = Target area of an absorbing system.

Generally transition with Emax value more than 104 are called allowed transition and Emax value less than 104 are called “forbidden transitions”. The instrument used for absorption measurement is composed of various instruments. For example, Colorimeters or spectrometers are constructed according to the following general scheme, Fig 3. It can be studied as: 1. Light Source:  In the infrared region, the source of heat radiations is Nernst glower or glober. In the Nernst glower, the filament consists of a mixture of oxides of the cerium and thorium, which are kept at high temperature (~1500°C) electrically. However, the radiation is not monochromatic, and for used in spectrographs it must be followed by some dispersing element. In the visible – UV region, the usual sources are incandescent lamps or may types of discharge tubes such as hydrogen discharge lamps, deuterium lamps and mercury arc. In Roman spectrometer, the source is mercury arc. Detector Sample Light Source

Monochromotor Recorder

Fig. 3  A Simple Sketch of Absorption Measurement Instrument

2. The Monochromator:  Except for microwave spectroscopy, some element is necessary by means of which the radiations can be separated in space according to wavelength, after it has passed the sample. Sometimes filters transparent over limited wavelength ranges can be used in combination with gratings. 3. The Detector:  In all spectrometers, the emitted radiations from the sample must be analyzed with regard to radiation intensity as a function. The “transformer” is called the detector. In the microwave region, the detector is a crystal rectifier. At the higher frequencies in the infrared region, thermocouples are used. In the visible and the ultraviolet region the photocells are used. 4. Spectrum recording:  When there is no sample present, the detector output will be constant over the range of frequencies covered by the instrument. However, when the radiations are passed through a sample, having just two possible energy levels E1 and E2, the detector output will show a sudden fall at frequency given by V = [(E2 – E1)/h]. The spectrum has been scanned between the beginning and ending frequencies. Such a spectrum is referred to as “frequency domain”, which indicates records of detector output against frequency. It is evident from the record (11) that the ideal situation represented in (1) is seldom attained. Thus, the base line is seldom horizontal. The reasons for this are– • It is impossible to make the self life infinitely narrow. Therefore, a range of frequencies rather than a single frequency reaches the detector.

Spectroscopic Methods of Analysis  9.5

Fig. 4



• Energy transitions in atoms or molecules are not absolutely sharp but always occur over a range of frequencies. • The sensitivity of the delector is also frequency dependent.



4.  UV SPECTROSCOPY • introduciton:  UV absorption spectrum arises from the transition of electrons within a molecule or ion. When a molecule absorbs ultraviolet radiation of frequency (v), the electrons in that molecule may undergo transitions from a lower to higher energy level, the energy difference between the two states is given by. DE = hn

Where, h is plank’s constant. Also,



DE = E2 – E1 E1 = Energy of the ground state. E2 = Energy of the excited state.

We know that the total energy of the molecule is equal to the sum of the electronic, vibrational, and rotational energies. Therefore,

Etotal = Eelectronic + Evibrational + Erotational

The absorption in the UV region involves single electronic transition, from lower energy state to higher energy state. So the spectrum must contain a single discrete line. This is because the electronic transitions are also accompanied by vibrational and rotational transitions. As the energy required for these transitions is much less, so the broad band appears. In UV range, the absorbed energy produce changes in the valance electrons of the molecules. The electrons involved are– (i) p-electrons:  These are involved in unsaturated compounds such as dienes, trienes, aromatic compounds, etc. (ii) s-electrons:  These are involves the electrons of the saturated bond between carbon and hydrogen. (iii) n-electrons:  These electrons are not involved in bonding between atoms in the molecules. For example, one pair of electrons in nitrogen, oxygen, sulphur or halogens. • Electronic transitions:  When a molecule absorbs energy in the UV or visible region, its electrons are promoted from a bonding molecular orbital to an antibonding molecular orbital. The electronic transitions involved in the UV and visible region are of the following types:

9.6  Engineering Chemistry (i) s Æ s* transitions (ii) n Æ s* transitions (iii) n Æ p* transitions (iv) p Æ p* transitions s * Antibonding p* Antibonding N Nonbonding p Bonding s Bonding (i) s Æ s* transitions:  These transitions occur by the promotion of sigma electrons from the bonding to antibonding (s*) orbitals. These transitions require high energy and thus absorb radiation in far UV region (126 mm – 136 mm). These transitions are seen in saturated hydrocarbons. For example, propane shows absorption around 135 mm and methane shows l max at 121.9 mm. (ii) n Æ s* transitions:  These transitions are seen in saturated compounds, containing heteroatom in addition to s Æ s* transitions. The energy required for the transitions is less than that required for s Æ s* transitions. For example, CH3OH absorbs at ordinary UV region and l max at 183 mm, for methyl chloride l max at 173 mm and water absorbs at 167 mm. (iii) n Æ p* transitions:  Unsaturated compounds containing heteroatom such as O, N and S show this type of transition. The electron from n-orbitals are transferred to p* orbitals and require less amount of energy and occur at longer wavelength. C=O

C=O

N Æ p* transition

(iv) p Æ p* transitions:  The electron transfer occurs from the bonding p-orbital to the antibonding p* orbital. These transitions are only associated with compounds containing double or triple bond and aromatic compounds. These transitions require less energy and occur at higher wavelengths. C=O

C—O

p Æ p* transitions For example, unconjugated alkenes show an absorption bond around 170 mm to 190 mm and saturated aldehydes and lactones absorb at about 180 mm. • chromophores and auxochromes:  A group of atoms in a molecule that is responsible for the absorption of radiations called Chromophores. Chromophores are of two types– 1. Chromophores which contains p-electrons only and under p Æ p* transitions. For example, ethylenes, acetylenes, etc. 2. Chromophore which contain both p-electrons and n–electron. Such Chromophores undergo p Æ p* and n Æ p* transitions. For example, carbonyl, nitrites, azo compounds, nitro

Spectroscopic Methods of Analysis  9.7 compounds, etc. The carbonyl group in acetone shows absorption at 190 mm (p Æ p*) and 270 mm (n Æ p*). Examples of some of the molecular types in which electronic transitions occur are given in Table 3.

Table 3 Examples of Electronic Transitions lmax(nm)

Compounds s Æ s* CH4

122

CH3 – CH3 p Æ p*

135

CH2 = CH2

162

CH ∫ CH

173

(CH3)2C = O

190

CH3 – N

O

201

O

n – s* H2O:

167

n–s* (CH3)2Cl:

172

(CH3)N:

227

n Æ s* (CH3)2C:

280

Identical functional groups in different molecules will not necessarily absorb at exactly the same wavelength. The energy change depends upon the structural environments of the molecule. The position and intensity of an absorption bond of a simple Chromophore may be modified by the attachment of certain groups in the basic Chromophore system known as auxochromes. The auxochromes produce the following shifts: (a) Bathochromic Shift:  This effect causes the absorption maximum to shift towards longer wavelength (l) or lower frequencies. This is called red shift and is due to solvent effect. (b) Hypsochromic Shift:  This is the shift in which absorption maxima shifts to shorter wavelength. This is called blue shift. (c) Hyperchromic shift:  This is the effect in which intensity of absorption maxima shifts up. (d) Hypochromic shift:  This is the effect in which intensity of absorption maxima shifts down. The bathchromic shift can be brought about in two ways. (i) By Conjugation of chromophoric groups:  When two chromophoric groups present in the molecule are separated by two or more single bonds, their effect on the spectrum is usually additive and there is little interaction between the isolated chromophoric groups.

9.8  Engineering Chemistry When the two chromophoric groups are conjugated, separated by only single bond, the electron system is spread over atleast four atomic centers. The shift is of 15nm-45nm to longer wavelength, for example: l max Transition (a) 1, 3 Butadiene



(b) – C – C = C – C = O

But -2- enal

(c)

– C = C – COOH





217

p Æ p*

218

p Æ p*

320

n Æ p*

260

p Æ p*

240

n Æ p*

(ii) By attaching auxochromes to the system:  Auxochromes generally deepen the color of a Chromophore but cannot by themselves impart color to a compound. These lead to shifts in the absorption to a longer wavelength and result in increase in the intensity of the absorption peak. Some typical auxochromes are– –NH2, – NHCH3, – N(CH3) 2, – OH, – Cl, – CH3, – CN and – CONH2 etc.

These groups contain nonbonding electrons and the transitions involving non-bonding electrons are responsible for this effect. For example, benzene shows an absorption bond l max at 255nm and aniline shows l max at 280nm. NH2

lmax = 225 nm

lmax = 280 nm

• Wood ward-fieser rules for conjugated dienes and trienes Woodward and Fieser formulated certain rules for calculating the l max for dienes and trienes. These rules are: (i) The basic l max of conjugated diene is 217nm. (ii) Add 5 nm for each alkyl substitution on the doubly bonded carbon atoms. (iii) Add 5 nm for each exocyclic double bond if any. If the same bond is exoyclic to two rings 2 × 5 nm may be added. These rules can be understood by the following examples: (i) CH2 = CH – CH = CH2

217 nm

(ii) CH3–CH–CH=CH–CH3 Basic Value = 217 nm Alkyl residues = 2 × 5 = 10 nm Calculate Value = 227 nm

Spectroscopic Methods of Analysis  9.9 (iii)

CH2 = C – C = CH2 or



Basic Value = 217 nm 2 – Alkyl residues = 2 × 5 = 10 nm Calculate Value = 227 nm CH2



(iv)

C = CH2

Basic Value = 217 nm 2 – Alkyl residues = 2 × 5 = 10 nm 1 – exocyclic double bond = 5 nm Calculate Value = 232 nm Observed Value = 235 nm •  Uv spectra of aromatic compounds:  The UV spectrum of benzene shows three major maxima at 184 nm, 204 nm and 254 nm. Evidently the first maxima are in the far UV region and the other two maxima are in the weak region. Due to substitutions in the benzene ring, there is a bathochromic shift of the benzene bonds as shown below:

lmax 184 nm lmax 204 nm lmax 254 nm

OH

N(CH3)2

CH3

CH2

208 nm 260 nm

230 nm

212 nm

250 nm

280 nm

260 nm

300 nm



The value l max depends upon the nature of substituent and interaction, for example:



(i)

Cl

lmax 204 nm



(ii)

210 nm

Cl

COOH 230

COOH 242

9.10  Engineering Chemistry In the first case, there is no resonance interaction and both Cl and COOH groups show additive effect in l max. Therefore, l max for P – Chlorobenzoic acid should be 210 + 230 – 204 = 236 nm, which is very close to the observed value. On the other hand, in the second case, the – NH2 and – COCH3 groups have opposing electrical effect. Therefore, l max for P–amino acetophenone should be 230 + 246 – 272 nm. But its observed value is quite higher. The UV spectrum of diphenyl suggests considerable resonance interaction between rings as shown below:

lmax

204 nm

250 nm

If two rings carry substituents in a suitable position, the molar extinction coefficient diminished due to hindrance of the co-planarity of the rings. But hydrocarbon, like dipheny methane, has typical benzene like spectrum.

On the basis of the above discussion we may conclude that UV spectroscopy is useful only in compounds which have extended conjugation between carbon–carbon double and bonds and between in compound carbon–carbon or corbon–oxyten/nitrogen multiple bonds or between carbon–carbon double bonds and a benzene ring. The values of l max for some unsaturated compounds are given in Table 4.

Table 4  lmax of some Unsaturated Compounds Compounds Ethene Trans 3-hexene

l max (nm)

Structure CH2 = CH2

C2H5

C

C

H

H C2H5

Cyclohexene

171 184

182

1-Octene

CH3(CH2)5 – CH = CH2

177

1-Octyne

CH3(CH2)5 – C = CH

185

Propyne

CH3 – C ∫ CH

187

1, 3 Butadiene

CH2 = CH – CH = CH2

217

1, 4 Pentadiene

CH2 = CH – CH2 – CH = CH2

178

1-Butene -3yne

CH2 = CH – C ∫ CH2

228

Spectroscopic Methods of Analysis  9.11

Cis-1, 3-pentadiene

Trans-1, 3-pentadiene

CH3 H CH3 H

C

C

C

C

CH

CH2

H H CH

CH2

223

223.5

•  APPLICATION OF UV SPECTROSCOPY: 1. Qualitative Analysis:  For qualitative analysis, a curve is plotted between wavelength (l) and degree of absorption (E). Now, by comparing the absorption spectrum of known and unknown compounds, the unknown compounds may be identified. 2. Quantitative analysis:  The qualitative analysis may be carried out on the basis of Beer’s law. In this technique l max is selected for the compound then optical densities are measured for some known compounds. A straight line is obtained and from this graph, the concentration of the unknown compound is evaluated very easily. 3. Detection of Impurities:  With the help of UV absorption spectroscopy, the impurities present in organic compounds can be detected. This is because: (a) The bonds due to impurities are very intense. (b) The organic compounds may be classified into saturated & unsaturated; the former has little absorption while the latter has strong absorption bonds. 4. Determination of Molecular Weight: According to Beer’s law: Io log __ ​   ​   = C.Emax.L It



Io log ​ __ ​  It C = _______ ​       ​ Emax ×L

From the above equation, C, i.e. molar concentration of the solution, is calculated. If the amount of solute dissolved (w) is known, the molecular weight of the solute may be calculated. 5. Tautomeric Equilibrium:  UV absorption spectroscopy may be used in determining the percentage of Keto and enol forms in Tautomeric equilibrium. Consider the case of ethylacetoacetate.

The Keto form l max = 275 nm and E = 16 while enol form has l max = 244 nm and E = 1600. We can measure the proportion of tautomers present in ethyl acetoacetate from the length of 244 nm band. 6. Determination of the dissociation constant of acid and bases.

9.12  Engineering Chemistry

7. Detecting steric hindrance. 8. Determination of the structure of several vitamins.

5. Rotational (Microwave) Spectra These spectra result from transitions between the rotational energy levels of a gaseous molecule on the absorption of radiations falling in the microwave region. These spectra are shown by molecules which possess a permanent dipole moment. e.g., HCl, CO, H2O vapor, No etc. Homonuclear diatomic molecules such as H2, Cl+2, O2 etc. and linear polyatomic molecules such as CO2, which do not possess a dipole moment, do not show microwave spectra. Microwave spectra occur in the spectral range of 1–100 cm–1.

Rotational Spectra of Diatomic Molecules Assuming a diatomic molecule (AB) behaves as a rigid rotor, it means, inter nuclear bond distance does not change during rotation. Let m A and mB are the masses of two atoms A and B respectively, r is the equilibrium internuclear distance between them i.e. r = bond length. Cg is Centre of gravity; rA and rB are distance of the atoms from the Centre of gravity. The Centre of gravity is defined by the equality of the moments about it .i.e,

m ArA = mBrB

...(1)

The moment of Interia I of a molecule is defined as



I = Si  miri2

...(2)

Where ri is the distance of the ith particle of a mass mi from the Centre of gravity. Since a diatomic molecule has two atoms (or particles). I = m ArA2 + mB rB2 ...(3) = mBrBrA + m ArArB = rArB (mB + m A) ...(4) r = rA + rB ...(5) or rB = (r – rA) Hence from eq. (1) and (5) m ArA = mBrB = mB (r – rA) m Br m Ar rA = ​ _______     ​ and rB = ​ _______     ​ m A + mB m A + mB



hence



Substituting the above values of rA and rB in eq. (3) we have



m AmB2 m A2mB I = ​ _________     2 ​ r 2 + ​ _________      ​ r 2 (m A + mB) (m A + mB)2

m A mB (m A + mB) 2 = _______________ ​         ​r (m A + mB)2

...(6)

Spectroscopic Methods of Analysis  9.13 m A mB 2 = ​ _______      ​ r m A + mB = m r 2 (m AmB) Where m = ​ _______     ​ is called reduced mass of the molecule. m A + mB

Classically, the angular momentum L of a rotating molecule is given by

L = Iw,    w is angular velocity energy of a rotating molecule 1 E (g) = __ ​   ​  Iw2 2 Let J = rotational quantum number and it can have integrial values from zero onwards.

÷ 

__________

h ___ L = ​ J (J + 1) ​        ​ ​   J = 0, 1, 2, 3 2p



Since angular momentum is quantized



Now.

h2 EJ = ​ ____     ​ J (J + 1) Joule 8p2 I

J = 0, 1, 2 The energies can also be expressed in cm–1 units, which is normally reported as rotational term F(J). EJ h Thus, F(J) = ​ ___  ​ = _____ ​       ​ J (J + 1) cm–1 hc 8p2 Ic h Let ​ _____     ​ = B = Rotational Constant 8p2 Ic Hence, F(J) = BJ (J + 1) cm–1 Next, we need a selection rule to determine the radiative transitions between the rotational energy levels. The rotational transitions for a rigid diatomic molecule are governed by the selection rule. D J = ± 1 i.e. only those transitions are allowed in which the rotational quantum number changes by unity. The + sign refers to absorption and – sign to emission of radiation. Microwave spectra are usually observed as absorption spectra so that the cooperative part of the selection rule is D J = +1. For a transition taking place from J to J + 1, the rotational frequency is given by. _

​n​ (J Æ J + 1) = = = Thus _ ​v​ 0 – 1 = _ ​v​(3 Æ 4)   =

__

B (J + 1) (J + 2) – B​ ​   J (J + 1) __ 2 ​   (J + 3J + 2) – (J2 + J) B​ __ 2 ​B​  (J = 1) cm–1 __ _

_

__

2​B​,  v​ ​  = 4B, v​ ​  (2 Æ 3) = 6​B​,  __ (1 Æ 2) 8​B​  etc.

__

__

__

__ So rotational spectrum of a rigid diatomic molecule consist of a__ series of lines at 2​B​,  4​B​,  6​B​,  8​B​  etc. Evidently, these lines are equally spaced by an amount of 2​B​. 

9.14  Engineering Chemistry

Fig. [A]  Rotational Spectrum of a rigid diatomic molecule

Mechanism of Interaction between the rotating molecule with microwave radiations A molecule with permanent dipole moment generates the fluctuating electric field by rotation. When the frequency of rotation of the molecule becomes equal to the frequency of electric field of radiation, there occurs the phenomenon of resonance and hence energy can be transferred from the radiation to the molecule or from the molecule to radiation, when molecule comes back from the higher rotational energy level to lower rotational energy level. Under this condition, the molecule exhibits emission microwave spectres.

Fig. [B]  Rotation of HCl molecule having parmanent dipole moment and fluctuation in the vertical component of the dipole moment.

Spectroscopic Methods of Analysis  9.15 Thus, a molecule with permanent dipole moment generates the oscillating electric field on rotation and hence can interact with the electric field component of microwave radiation which oscillates in a plane perpendicular to the direction of propagation of the wave. It is noted that the rotational spectra are observed in the gaseous state only because the rotational energy levels are not quantized in solid and liquid states.

Applications of Rotational Spectra 1. Determination of bond length of a polar molecule. For the rotational spectrum, which consists of series of equally spaced lines, the value of B can be calculated. h Since B = _____ ​  2     ​ 8p Ic

h I = ​ ______      ​ 8p2 Bc

If we know the atomic masses, reduced mass (m) can be calculated by using the relation

m AmB m = ​ _______      ​ m A + mB

Since

I = m r 2

÷ 

__

÷ 

_______

h I r = ​ __ ​    ​ ​  = ​ _______ ​  2        ​ ​ = Internuclear distance m 8p Bcm 2. Determination of symmetry in a molecule. 3. Determination of dipole moments of gaseous molecules is also possible by recording the spectra in the presence of electric field. It is found that in the presence of external electric field (E), the splitting of rotational energy levels of the molecule occurs (starck effect). If Dv = Shift of rotational v m = dipole moment of the molecule. and E = External electric fuild. then DV µ (m × E)2. Thus, by knowing E and measuring DV, m can be determined. \

Limitations

1. By using this technique, molecule with permanent dipole moment can be analyzed. 2. The molecules to be studied must be in gaseous form so that they are free to rotate. 3. Difficulties are encountered to study large molecules because they have more than one internuclear distance which connot be easily sorted out from the determind moments of inertia.

6.  INFRARED SPECTROSCOPY

• Introduction:  IR spectroscopy is the most powerful analytical technique for the identification of chemical constitution. IR radiations range from 2mm to 16mm and are not energetic enough

9.16  Engineering Chemistry to produce electronic excitations in most organic molecules. These radiation causes stretching and bending of organic bonds. After absorption of IR radiations the molecules thus give close-packed absorption bands known as Infrared Absorption Spectrum. A number of bonds may be present in the 1R spectrum corresponding to the characteristic functional group and bonds present in the substance. • Vibrational Model of Polyatomic Molecules:  The atoms in a molecule, which may be of different types, are always in motion. This type of vibration movement is called bond stretching and the atoms remains in the same bond axis. This mode of vibration does not cause any dipole change in the symmetrical molecules e.g., O = C = O and therefore is not an infrared active vibration. In the case of polyatomic molecules, the actual vibration is complicated. However, it can be resolved into simple modes called normal modes of vibration. Each normal mode of vibration give rise to vibrational bond. The modes of vibration are (3n – 5) for linear polyatomic molecules and for a nonlinear polyatomic molecules the number of modes of vibrations are (3n – 6), where n is the number of atoms in the molecule. The various types of vibration are: 1. Stretching Vibration:  In this vibration, the distance between the two atoms increases or decreases but the bond axis remains the same. It may be symmetric or asymmetric. In the former case, movement of an atom with respect to a particular atom in a molecule is in the same direction, while in the later case, one atom approaches the central atom and other is moving away. O

O

Symmetric



O

O

Asymmetric

2. Bending Vibration:  In this case the distance between the two atoms remains constant, the position of the atom changes with respect to original bond axis. It is of two types: 1. In the plane bending, which is of two types (a) Scissoring:  In this mode of vibration two atoms approach each other. (b) Rocking:  The movement of both the atoms take place in same direction. 2. Out of plane bending which is also of two types (a) Wagging:  The two atoms move up and below the plane while the other moves down the plane with respect to central atom. (b) Twisting:  In this type, one of the atoms move up the plane while the other moves down the plane with respect to central atoms.

Spectroscopic Methods of Analysis  9.17

For example: Consider the cases of H2O and CO2. (i) Water (H2O):  Water is a tri-atomic nonlinear molecule. It has (3 × 3 – 6) = 3 normal modes of vibration.

O

O

O

H H

H

H

Symmetric stretch

H

H

Symmetric stretch

–1

Symmetric stretch

–1

V1 = 3615 cm

–1

V2 = 1595 cm

V3 = 3765 cm

All the three vibrations in water molecule involve a change in dipole moment and therefore are IR active. (ii) Carbon Di-oxide (CO2):  It is a linear molecule and has normal modes of vibration = 3 × 3 – 5 = 4. The fourth mode of vibration is identical with third except in direction and such modes are called degenerate. Thus, for linear CO2, we get two bands in the spectrum. O

C

O

O

Symmetric Stretch V = 1330 cm

O

Bending V = 667 cm



V = 2449 cm

O

–1

O

Asymmetric Stretch

–1

C

C

O

–1

C Bending V = 667 cm

O

–1

• Group Frequencies and Analysis:  A particular part of the infrared spectrum is referred to either by wavelength or more preferably by its frequency. The wave number is simply

9.18  Engineering Chemistry the number of waves per centimeter, and is equal to the reciprocal of the wavelength in centimeters. The IR spectrum may be divided into two parts: (a) Characteristic Frequencies:  The region 4000 cm–1 to 1400 cm–1 (approx.) contains the peak characteristic of individual groups. Hence, the frequencies corresponding to such groups are known as characteristic frequencies. (b) Fingerprint Region:  The region 1400 cm–1 to 650 cm–1 is referred to as the fingerprint region. This position is extremely valuable when examined with reference to other regions. For example, if alcoholic phenolic stretching frequency appears in the high frequency region and absorption also occurs at 1260 cm–1 to 1000 cm–1. This is the region usually checked for identification, since it is associated with vibrational energy changes of the molecules skeleton and so is characteristic of the compound.





Transmittance (%)

2.5 100

3

4

5

6

Wave length (nm) 7 8 9

10

12

15

80 60 40 20

Fingerprint Region Functional - Group Region

4000 3500 3000 2500 2000 1800

1600 1400 1200 1000 800

600

–1

Frequency (cm )

Table 5  Common Infrared Absorption Frequencies Functional Group



Found in

Range (cm–1)

C=O

Acyclic Ketones and Cyclonhexanones

1725 – 1705

C=O | H

Aldehydes

1740 – 1720

C=O

Cyclopentanones

1750 – 1720

C=O

Cyclobutanones

– 1775

ab-Unsaturated Ketones

1685 – 1665

ArC = O

Aromatic aldehydes and Ketones

1700 – 1680

Carboxylic acids

1710 – 1700



C=O | OH C=O | OR

Esters and S-lactones, g - electrons



C=C–C=O | OR

C=C–C=O

Aromatic and Conjugated esters

No. of Bands

1750 – 1735 – 1780 – 1760

1725 – 1710

Contd...

Spectroscopic Methods of Analysis  9.19 O || C

Anhydrides

1850 – 1800 1790 – 1740

C∫N

Nitriles

2280 – 2200

C∫C

Alkynes

2300 – 2100

OH

Alcohols (free)

3650 – 3590 – 3400 – 3200

OH

Carboxylic acids

2700 – 2500

C || O







Two bands

O

Several bands

NH

Secondary amines

3500 – 3300

One band

NO2

Nitro compounds

1570 – 1550 1370 – 1300

Two bands

NO2

Nitro compounds (Aliphatic)

1570 – 1550 1380 – 1310

Two bands

• Characteristic Infrared Bands:  The above absorption ranges of the major functions groups in organic compounds are summarized in Table 5. All the bands in the IR spectrum need not be attempted, but the presence or absence of function group such as carbonyl, hydroxyl, amino etc. is ascertained by this technique and these finding should be used in conjugation with the information obtained from UV, NMR and mass spectra. 1. Alkane and Alkane Residue:  Precise position of CH3, symmetrical and unsymmetrical vibration frequencies is well known. Since most organic molecules possess alkane residues the groups of saturated C – H absorption bands are of little diagnostic value, whereas the absence of saturated C – H absorption in a spectrum is of diagnostic value. Unsaturated and aromatic C – H stretching frequencies can be distinguished from the saturated C – H absorption since the latter occurs below 3000 cm–1 while the former gives rise to much less intense absorption above 3000 cm–1. 2. Alcohol and Phenols:  Compounds containing O – H bands exhibit a peak between 3700 and 320 cm–1. The position of this band dependent on the existence and strength of a hydrogen band. The IR spectrum of ethanol in vapor state show a sharp band near 3700 cm–1 which is described as a free hydroxyl group. However, when the spectrum is taken in solution, Sharp band is a almost replaced by a broad band at 3350 cm–1. On further increasing the concentration the band at 3700 cm–1, completely disappears, the broad band at 3350 cm–1 is characteristics of hydrogen bonded hydroxyl group and since the intensity of this band changes with concentration. It is a case of intermolecular hydrogen bonding. mm

25

3640 4000

15

CH3CH2OH –1

Cm

800

Fig. 5  The IR Spectrum of Ethanol (in vapour state)

9.20  Engineering Chemistry mm

15

3540

15

CH3CH2OH 3350

4000

Cm

–1

800

Fig. 6  The IR Spectrum of Ethanol (Concentrated Solution)



3. Absorption frequencies of Carbonyl Bond:  The absorption caused by a carbonyl group is the strongest band in the IR spectra of carbonyl compounds. The general trends of structural variation of the position of C = O, stretching frequencies may be summarized as: (i) The more electronegative the group X in the system R – CO – X, the higher the frequency. (ii) ab - unsaturation caused a lowering of frequency of 15 cm–1 to 40 cm–1, except in amides. (iii) Further conjugation has relatively little effect. (iv) Hydrogen bonding in carbonyl group causes a shift to lower frequency of 40 cm–1 to 60 cm–1. (v) When more than one of the structural influences on a particular carbonyl group is operating, the net effect is usually close to additive.   Acyclic Ketones show absorption bands at around 1725 cm–1 to 1705 cm–1 whereas in aromatic Ketones this frequency goes down by 30 cm–1 to a lower wave number. For example, the IR spectrum of acetophenone (Fig. 7) shows carbonyl absorption at 1690 cm–1.

      

Fig. 7  The IR Spectrum of acetophenone

Fig. 8  The IR Spectrum of cyclopentanone

  The IR spectrum of cyclopentanone shows absorption band at 1750 cm –1 (Fig 8) showing that decrease in ring size increase the frequency of carbonyl stretching vibration. Thus, the absorption resulting from C = O stretching of cyclohexanone (Fig. 9) and 2-cyclopentanone occur at about the same frequency.

Spectroscopic Methods of Analysis  9.21

Fig. 9  The IR Spectrum of cyclohexanone

  The position of absorption of the carbonyl stretching of esters (Fig. 10) is somewhat higher than Ketones and is dependent, as with Ketones, on conjugation, unsaturation, and ring size.   The IR spectrum (Fig. 11) in addition to showing (figure) Keto and ester carbonyl frequencies, also shows a band at 1650 cm–1 which has been ascribed to ab–hydroxyl–a, b–unsaturated ester carbonyl. mm

25

15

O

O





4000

15

O

CH3–C–CH2–C–OC2H5

CH3–C–OC2H3



mm

25

–1

800

Cm

–1

4000

Fig. 10  The IR Spectrum of Ethyl Acetate

Cm

800

Fig. 11  The IR Spectrum of Ethyl aceto Acetate

4. Carboxylic acids:  The IR spectra of carboxylic acid reveals the presence of a hydrogen bond. Comparison of the IR spectrum of acetic acid (Fig. 12) with ethanol (Fig. 5) clearly shows the absence of a free hydroxyl groups in the molecule of acetic acid. mm

25

15

O CH3–C–CH 4000

–1

Cm

800

Fig. 12  The IR Spectrum of Acetic Acid

•  Applications of IR Spectroscopy 1. Qualitative Analysis:  The qualitative analysis can easily be done by comparing the spectrum of the compound under examination and the standard one.

9.22  Engineering Chemistry

2. Quantitative Analysis:  The quantitative analysis is based on the determination of the concentration of one of the functional group of compounds being estimated, e.g., the concentration of hexagonal may be determined in the mixture of hexane and hexagonal by measuring absorption of OH bond. The following formula is used to calculate the concentration:

Where, Since, E

A = – log I/Io = Ecx

A = absorbance, I = intensity of radiation after leaving the sample under consideration, Io = intensity of radiation initially, E = absorptivity to the cell, X = path length to the sample cell, and C = concentration of solution. and x constant for a particular cell.



A µ c

With the help of a calibration curve, the concentration of an unknown sample may be determined easily. 3. Determination of Purity of the Sample:  Since impurity reduces the sharpness of the individual bands and causes an extra band to appear in the IR spectra purity of the sample may be estimated easily. 4. Presence of water in the Sample:  If the sample is crystallized by water, it may be possible that some water remains in the sample. In that case, some bands at 3600 cm–1 to 3200 cm–1, 1650 cm–1 and 600 cm–1 to 300 cm–1 region appear. 5. Study of reaction Kineties 6. Shape of molecules 7. Determination of force constant from vibrational spectrum 8. Distinguishing intra and inter molecular hydrogen bonding 9. Analysis of a mixture of aromatic hydrocarbons 10. Distinguishing positional isomers of a compound

7. Raman Spectroscopy Raman Spectroscopy deals with the scattering of light in contrast whereas other branches of spectroscopy which deal with absorption of light. Origin:  When a sample is irradiated with a beam of monochromatic light, a very small fraction of it is scattered in all directions. if the scattered light in a direction perpendicular to the incident beam is spectrographed it shows strong line (Rayleigh line) corresponding to the frequency of the incident light and weak lines on either side of it. These lines are symmetrically arrayed above and below the frequency of the Rayleigh line and are known as Anti-Stokes and stokes. The plot of these frequency shifts against their intensities is a Raman. The quantum theory of Raman effect is quite simple. Consider a photon of frequency n falling on a molecule of the collision is elastic then the scattered photon will be of same energy as the incident

Spectroscopic Methods of Analysis  9.23

photon. if on the other hand collision is inelastic the scattered photon will have either a higher or a lower energy than the incident photon. if the kinetic energy of the photon and the molecule remains unchanged before and after the collision, then from the law of conservation of energy,

hn + E = hn¢ + E¢

...(1)

where hn is the energy of the incident photon and hn¢ in the energy of the scattered photon after collision. E is the energy of the molecule (rotational, vibrational, and electronic) before collision and E¢ is the molecular energy after collision. Rearranging eq (1) we get E¢ – E ​ ______     ​  = n – n¢ h

...(2)

Consider the following cases: Case 1  n = n¢ so that E = E¢ Case 1 is referred as Rayleigh scattering, where scattered photon has the same frequency as the incident photon as given in Fig 13.

Fig. 13

9.24  Engineering Chemistry When the molecule, excited to the higher unstable vibrational state, returns to the original vibrational state, we get Rayleigh scattering. The Rayleigh scattering and the Raman scatting. Case II (a)  When the molecule excited to the higher unstable vibrational state, returns to the original vibrational state this gives rise to Raman scattering (Stokes lines). Case II (b)  When the molecule, initially in the first excited vibrational state is promoted to a higher unstable vibrational state and returns to the ground state, this again gives rise to Raman Scattering (anti-stokes lines). Thus, the Raman spectrum of a molecule consists of Stoke’s lines and anti-Stokes lines situated symmetrically about the Rayleigh line. The Rayleigh line is far more intense than the stoke’s lines which in turn have greater intensity than the anti-Stokes lines. In a conventional Raman spectroscopy, the anti Stokes lines are difficult to be observed because they correspond to the return of a molecule form the unstable excited vibrational state to the ground state and initially there are very few molecules in the excited vibrational state. Criteria for a molecule to be Raman spectra The Raman spectroscopy is observed only when a molecule shows periodic change in polarizability due to its vibrational or rotational osciallation. Pure Rotational Raman Spectra of Diatomic Molecules When the polarizability of a molecule depends on the direction in which the applied field lies, rotational transitions can also be observed by Raman effect. Since, on every revolution, the polarizability of a molecule returns to its initial value, twice the rotational selection rule in Raman spectra is therefore DJ = ± 2. The expression for rotational term F (J) is F (J) = B J(J + 1) cm–1 When a molecule makes transition with D J = ± 2 i.e., from J to J + 2 ; the scattered light emerges with a lower frequency and we get stokes lines in the spectrum. __ Stokes ​n​ (J Æ J + 2) = F(J + 2) – F(J) = B(J + 2) (J + 3) – BJ(J + 1) = B[J + 2) (J + 3) – J(J + 1] = B[J2+ 5J + 6 – J2+ J] = B[4J + 6] Where J = 0, 1, 2 ... Similarly, expression for anti-Stokes lines can be observed with the help of selection rule D J = – 2 i.e., transition from J to J – 2. When D J = 0 only Rayleigh Scattering takes place and Rayleigh line is observed at the same __ wave number as that of incident radiation (i.e., ​n​ 0). Thus, Raman lines will appear at wave numbers given by __ __ ​n​  = n​ ​  0 ± B (4 + 6) ; J = 0, 1, 2. __ Where ​n​ 0 is wave number of Rayleigh line __ __ When n​ ​   > n​ ​  0 Æ Anti-Stokes lines are observed. __ __ And when n​ ​   < n​ ​   – Stokes lines are observed __ 0 __ When J = 0 – n​ ​   = ​n​ 0 ± 6B

Spectroscopic Methods of Analysis  9.25 Hence, the first stokes and anti-stokes line will be at a separation of 6B from the Rayleigh line. However, the separation between the successive lines is 4B cm–1 as shown in Figure (14) below.

Fig. 14  Anti-Stokes and Stokes rotational Raman lines in pure rotational Raman spectrum of diatomic molecule.

From the constant separation of 4B cm–1 between the neighbourning lines in Stokes and anti-Stokes series, calculation of B is possible. Once we know the value of B, moment of inertia of the molecule. can be determined. From this bond length and bond, angles can be calculated. Vibrational Raman Spectra The selection rule for vibrational Raman effect is Dn = ± 1. When Dn = + 1, Stokes lines corresponding to low frequency of side of the incident light are observed. When Dn = –1, anti-stokes lines corresponding to high frequency side of the incident light are observed. Rotation Vibration Raman Spectrum Let us now briefly consider the rotational transition accompanying a vibrational raman transition of a diatomic molecule. The selection rules now involve the changes in both the vibrational and the rotational quantum numbers. For a diatomic molecule the selection rules are Dn = + 1 D J = 0, ± 2  Since at room temperature most of the molecules are in the ground vibrational state (v = 0), only the vibratinal transitions, v = 0 to v = 1, is of interest. The transitions with D J = 0 forms a Q branch. those with D J = +2 form an S branch and those with D J = – 2 form an O branch. The rotational Raman transitions accompanying a 0 Æ 1 vibrational transitions are shown in the figure (16) given

9.26  Engineering Chemistry below. Here D n measures the displacement from the exciting mercury line. The Q band exhibits an _ intense ‘narrow line’ at ​v​ exc that is usually unresolved while the S and Q branches form weak wings which extend to lower s higher wave numbers respectively, from the intense narrow line.

Fig. 16  The rotation-vibration transitions in the Raman spectrum of a diatomic molecule.

Experimental Raman Spectroscopy:  Intense monochromatic radiation from a source consisting of a large spiral discharge tube with mercury electrodes is allowed to fall on the cell-containing a gaseous or liquid sample. When the current discharge passes through the tube mercury emits lines in its spectrum the most intense of which at 4358Å serves as the exciting line. The scattered light is observed at right-angles to the direction of incident radiation. The detector is either a photographic plate or a photomultiplies. The horn shape of the cell helps in reducing the direct reflection of the source from the back of the cell.

Fig. 17  Experimental set up of Raman spectroscopy.

Spectroscopic Methods of Analysis  9.27 The Raman spectra of gases are generally weaker than those of liquids. It is necessary to use very long discharge lamps and cells, the latter containing mirror at both ends arranged so as to increase the effective path length of the cell. It is to be recalled here that selection rule for IR and Raman spectra is same. It does not mean that IR and Raman spectra are identical. But most of the time, they supplement each other for the determination of molecular structure. For example, if a molecular has a centre of symmetry, then IR inactive vibrational modes are Raman active and vice–versa. This is known as role of mutual exclusion. This can be exemplified by taking the example of CO2, a linear triatomic molecule with 3N – 5 = 3 × 3 – 5 = 4 vibrational modes. Another example is that of the planes dichloroethylene C2H2Cl, which consist of cis and transconfiguration in equilibrium proportions. Only the trans-configuration has a centre of symmetry. Thus, the coincident frequencies observed in IR and Raman spectra of the sample can be assigned to this cis–configuration. In this manner the analyses of the mixture of this cis and trans–isomers can be carried out. Again, for CS2, all the vibrations that are Raman active are infared inactive and vice-versa, whereas for N2O the vibrations are simultaneously Raman and IR active. From the spectral data we conclude that CS2 has a centre of symmetry where as N2O has no centre of symmetry. Thus the CS2 structure is of the type S – C – S while the N2O structure must be N – N – O rather than N – O – N. Raman shift D V generally lies within the far and near infrared regions of the spectrum. But it is not necessary that all Raman lines will have corresponding infrared bands and vice-versa Raman Spectrum is different from Infrared spectrum in the following respects: Raman spectra

Infrared Spectra

1. It results due to the scattering of light by vibrating and or rotating molecules

It results due to the absorption of light by vibrating molecules.

2. Change in polarizability is essential condition.

Change in dipole moment is essential conditon.

3. Water can be used as a solvent.

Water cannot be used as a solvent.

4. Concentrated solutions are used for increasing the intensity of weak Raman lines.

Dilute solutions are preferred.

5. Homonuclear diatomic molecules exhibit spectra.

Homonuclear diatomic molecules do not exhibit IR spectra.

6. Uses visible or ultraviolet radiations, hence the walls of the sample cell can be made of glass or quartz.

Uses IR radiations, hence the walls of the sample cells must be made of special materials that are transparent to IR radiations.

8.  NUCLEAR MAGNETIC RESONANCE (NMR) SPECTROSCOPY This spectroscopy method deals with the interaction between an oscillating magnetic field of electromagnetic radiation and the magnetic energy of nucleus of a molecule. The magnetic energy levels are produced by keeping the nucleus in a magnetic field the energy required for transitions is available in radio frequency region (4 to 900MHz). •  Spin of nuclei and angular momentum:  Magnetic field is also associated with moving electrical charges. The nuclear spin may vary from one nucleus to another. The spin of nucleus is represented

9.28  Engineering Chemistry by I called the spin quantum number. Spin quantum number is associated with the angular momentum by the following relation: where,

_______ h I = ​÷I (I   + 1) ​  ​ ___  ​  2p

h, is the Plank’s constant.

•  Rules for Nuclei showing NMR Spectroscopy:  Nuclei with an even number of protons and an even number of neutrons (i.e., even atomic and mass number) have hero spin and they do not give rise to NMR signals. For example Cl12, O16, S32 etc. Nuclei with an odd number of protons and an odd number of neutrons (i.e., odd atomic and even mass number) have integral spins. For example the nuclei H2, N14 have I = 1 and for B10, I = 3. Nuclei with odd mass numbers have half-integral spin, for example H1 and N15 have I = 1/2 and O17 have I = 5/2 etc. •  Principle of H1 NMR Spectroscopy:  Consider the nucleus of a hydrogen atom (proton). It behaves as a tiny spinning bar magnet because of posses both electric and magnetic spin. Like any other spinning charged body, it will generate a magnetic field. Applied magnetic field

Precessional orbit Spinning nucleus

Fig. 18

If the sample contain many identical nuclei with spin quantum number I = 1/2, it can have two possible orientations (2I + 1), i.e., 1/2 or –1/2 or a and b nuclear spin. When the external magnetic field is applied the magnetic nuclei (protons), like the spinning top, will process around the axis of an applied external field. It precessionals frequency is given by:

V µ Bo

The precessional frequency V is directly proportional to the strength of the external field Bo. In the presence of an external magnetic field, two changes can occur. (i) The precessional proton can adopt two principal orientations with respect to the applied external field. (a) Aligned with the field, corresponding to low energy state, a – state. E1 = 1/2 gBnH

Spectroscopic Methods of Analysis  9.29 (b) Opposed to the field, corresponding to high energy state, b – state. E2 = – 1/2 gBnH (ii) There will be more a – spins than b – spins. This small imbalance will result in net magnetization. When we irradiate the precessing nuclei with a beam of radio frequency energy of the correct frequency the proton in the low energy state (a) may absorb this energy and move to the higher energy state (b). This transition from one energy state (a) to the other energy state (b) is called the flipping of the proton. The energy of transition is given by: DE = hv

The precessing proton will absorb energy only if the precessional frequency is the same as the frequency of the radio frequency beam. The energy absorbed produces a signal at the detector and this signal is amplified and recorded as a band in the spectrum. An NMR spectrum is plotted between the absorption between signal at the detector and the strength of the magnetic field. It may be pointed out here that the character of NMR spectra is not solely determined by nuclear properties but is also affected by the environment in which it is present and in the environment we have electrons revolving around the nucleus.

Instrumentation For the observation of the absorption of radiation by protons, difficulty in setting up the arrangement is encountered because (1) of obtaining the magnetic field of sufficient high field strength, and (2) uniformity of the field. In contrast to most other absorption methods, in NMR spectroscopy the frequency of radiation is kept constant and the magnitude of applied magnetic field is varied, until absorption of radiation by protons occurs. This slight variation in magnetic field is achieved by passing current through wire coils, wrapped round the two pole pieces of the magnet. Absorption of radiation is detected on the radiofrequency bridge (a type of wheatstone bridge arrangement), which goes out of balance, when absorption of radiation occurs and the resulting signal is amplified and recorded instantaneously on the chart Fig. 19 showing schematically the basic arrangement for NMR spectrometer.

Fig. 19  A Schematic Arrangement of NMR Spectrometer

9.30  Engineering Chemistry

Chemical Shift Chemical shift may be defined as a shift in position of a spectrum peak due to a small change in environment. When a molecule is placed in a magnetic field, its electrons are caused to circulate, thereby producing a induced magnetic field, which may be either: (i) Oppose the applied field:  Evidently, the effective field experienced by the protons is diminished. Under this state, proton is said to be shielded. The shielded protons in turn shift the absorption position upfield in the NMR spectrum. This effect is also called diamagnetic shift. Consequently, the value of the applied field necessary to bring the nucleus in the resonance will be more. For example, acetylic protons exhibit this type of chemical shift. (ii) Reinforce the applied field:  Evidently, the effective field experienced by the protons turn shift the absorption position downfield in the NMR spectrum. This effect is also called paramagnetic shift. Consequently, the value of the applied field necessary to bring the nucleus in the resonance will be less. For example, ethylenic and aldehydic protons exhibit this type of chemical shift. Expression for chemical shift: It is not possible to measure in absolute term the precessional frequency for a group of nuclei. Usually, chemical shift is measured with respect to some standard. TMS [tetramethy silane (CH3) 4Si] is taken as standard. Usually, chemical shift positions are expressed in delta (d) units, which is equal to differences in ppm from TMS signals. Thus: go d = gs – ​ __ ​  g gs = frequency of the sample go = frequency of TMS g = operating frequency. Most chemical shift values have to be found to lie between 0 and 10. So TMS has been assigned d value of 0. TMS is used as a standard compound to carry out NMR spectra for the following reasons; 1. It has 12 protons but only one signal is obtained as their environment is same. 2. It is soluble in all organic solvents without showing molecular interaction with the sample. 3. It is volatile so need not be removed.

Study of NMR spectra NMR spectra may be studied under the following headings: (a) Number of signals:  A set of protons with the same chemical environment are called equivalent protons. All equivalent protons form a type or a kind of protons and the NMR spectrum consists of as many signals as there are kinds of protons. (a, b, c and d protons in following listed examples). For example: Sr. No.

Compound

Number of Signals

H 1.

H

C

H

One Signal (Four Equivalent Protons)

H Contd...

Spectroscopic Methods of Analysis  9.31 (a)

2.

(b)

Two Signals

C H3– OH

3.

Three Signals (a)

4.

H

H (a)

(a)

C=C

CH3

5. (b) (a)

H

C=C

CH3

6.

C1

H

One Signal

H H(c)

Four Signals

H(d) (b)

C=C

H

Three Signals

H(c)

(c)CH3 H(d) 7. (a)

H

H(a)

H

H(b)

(b)

Four Signals



(b) Position of signal (Chemical Shift):  The number of signals in an NMR spectrum tells the number of the sets of equivalent protons in the molecule. The protons have a precessional frequency by ~60MHz. However, the precessional frequency of all the protons in the same as the exact value depends upon the chemical environment of the protons. Since the shift in frequency depends upon the chemical environment it is called chemical shift. For example, case of a molecule of methanol CH3OH, there are two types of precessional frequencies of the proton in the spectrum. The differences in precessional frequency are very small but can be detected by high resolution instruments



(c) Splitting of Signals/ Spin-Spin coupling:  Each signal in NMR spectra represents one set of protons in a molecule. However, in certain molecules, instead of a signal peak, a group of peaks (multiple) is absorbed. The number of lines observed in the NMR signal for a group of protons is not related to the number of protons in the group but it is related to the number of protons in the neighboring group. The simplest rule to find the multiplicity of the signal from a group of protons is (n + 1), where ‘n’ is the number of neighboring protons. This can illustrated by the following examples:



(i) Ethyl Alcohol (a)

(b)

(a)

CH3– CH2– OH

It contains three types of protons, namely three methyl protons, two methylene protons and one hydroxyl proton. Hence, three NMR signals are obtained under low resolution. This is due to the spin interaction or coupling between the adjacent sets of non-equivalent protons. The adjacent methylene protons as shown in the following table. The intensities of the lines

9.32  Engineering Chemistry in the triplet are in the ratio of 1 : 2 : 1, corresponding to the number of like orientation in each case. Orientation of Protons

Number of like Orientations degeneracy

Total Spin

(i) aa i.e., ≠≠

1

1/2 + 1/2 = 1

(ii) ab, ba i.e., ≠Ø, ≠Ø

2

1/2 + 1/2 = 1

(iii) bb i.e., ØØ

1

– 1/2 – 1/2 = – 1

a-represents clockwise spin and b-anticlockwise spin. Similarly, we can draw the four possible orientations of the three methyl protons, which are responsible for splitting of the signal into a quarter with the intensity ratio 1 : 3 : 3 1; see table. Orientation of Protons

Number of like Orientations degeneracy

(iv) ≠ ≠ ≠

1

(v) ≠ Ø Ø ≠ Ø ≠ ≠ ≠ Ø

3

(vi) Ø Ø ≠ Ø ≠ Ø ≠ Ø Ø

3

(vii) Ø Ø Ø

1

Thus, the relative intensities of the individual lines of the multiple correspond to the numerical coefficient of the lines in the binomial expression where, n = number of protons. (ii) CH3-CH2-O-CH3 a b c This has three types of protons. Splitting of signals is as follows; a = 2 + 1 = 3 b = 3 + 1 = 4 c = 1 (iii) CH3-CH2-CH2-CHO a b c d This has four types of protons. Splitting is as follows; a = 2 + 1 = 3 b = (3 + 1) (2 + 1) c = (2 + 1) (1 + 1) d = 2 + 1 = 3 (iv) CH3-CH2-OH(aq) a b c This molecule has three types of protons but because the medium is aqueous c proton is involved in hydrogen bonding so b and c protons are not present in environment of each other. Splitting is as follows;

Spectroscopic Methods of Analysis  9.33 a = 2 + 1 =3 b = 3 + 1 =4 c = 1 1 How to interpret H NMR spectra? Following points may be considered for the interpretation of NMR spectra: (1) Number of signals tells the number of different environments in the molecule. (2) The relative areas of different peaks corresponds to the number of protons present in different environment is the molecule. In other words, area under each NMR signal is proportional to the number of H-atoms present in the environment. (3) Multiplicity of lines in an NMR signal for a group of protons is dependent on hydrogen atoms present in the neighbouring groups. If there are n protons, then multiplicity of signals = (n + 1). (4) The intensities of signals are asymmetric about the midpoint of the group. Pascal’s triangle gives the relative intensities of various signals in the NMR spectrum

Applications of NMR spectrum Informations obtained from NMR spectrum are given below: 1. The number of signals in the spectrum indicates types of protons in the compound. 2. Integration of traces of spectrum gives the relative number of protons of different types. 3. Chemical shift gives the types of structure associated with particular protons. 4. Positions of the peaks indicate the types of electronic environment of each kind of protons. 5. The splitting patterns give the number of protons on atoms adjacent to the group, which contains protons, whose resonance is being measured. 6. The intensities of signal give the number of protons of each kind.

solved Questions 1. What is molecular spectroscopy? Ans. The study of interaction of electromagnetic radiation with matter. 2. What is electromagnetic radiation? Ans. The emission and propagation of electromagnetic waves. 3. What are electromagnetic waves? Ans. A wave comprising of two interdependent mutually perpendicular transverse waves of electric and magnetic field. 4. What is the wavelength range of electromagnetic waves? Ans. Wavelengths of 10 –15 to 103. 5. Names the forms of electromagnetic radiations. Ans. Gamma rays, Xrays, ultraviolet radiations, visible light, infrared radiation, microwaves and radiowaves.

9.34  Engineering Chemistry 6. What is meant by absorption spectroscopy? Ans. Absorption spectra is obtained when a sample selectively absorbs certain frequencies of radiation from a range of frequencies available. The appearance of such a spectrum in the visible range is a series of dark lines on a colored background. 7. “An IR spectrum is often characterized as molecular finger prints”. Justify this statememt. Ans. This is because different functional groups produce recognizable peaks at nearly definite positions in IR spectra. Most of the peaks can be assigned due to the presence of specific group in the molecule. So IR spectra can also be used in establishing the identity of the compounds. 8. Which among CH2 =CH2, cis CH2 =CHCH=CH2, and trans CH2 =CHCH=CH2 can give max absorption peak at about 274 mm? Give reason. Ans. trans CH2 =CHCH=CH2 because it has extended conjugation. 9. What is infrared radiation? Ans. Electromagnetic radiation in the range of 0.75-1000µm. 10. What is ultra violet radiation? Ans. Electromagnetic radiation in the range of 200nm–340nm. 11. What is ultra violet spectroscopy? Ans. Method of detecting ultraviolet absorption of aromatic groups or conjugated bonds, including polymers in a suitable solvent. 12. What is infrared spectroscopy? Ans. Routile analytical tools for detection of functional groups by infrared absorption in molecules. 13. What are chromophores? Ans. Groups which are responsible for the colour of dyestuffs, e.g., C=C, C=O, C=N, N=N etc. 14. What type of excitation can take place in CH3NO2 in UV spectroscopy? Ans. p Æ p* and n Æ p*. 15. Name the spin active nuclei which can be studied by NMR spectroscopy. Ans. 1H, 13C, 13N, 19F and 31P 16. What property of nuclei is involved in NMR spectroscopy? Ans. Atomic nuclei having odd mass number. 17. 13C is NMR active whereas 12C is not. Why? Ans. 13C has nuclear spin whereas 12C does not have nuclear spin. 18. What is nuclear magnetic resonance? Ans. When atomic nuclei having spin is placed in a strong magnetic field the nuclear magnetic moment can take up certain discrete orientations, each corresponding to a different energy state. Transitions between these energy states can be induced by the application of radiofrequency radiation. This is known as nuclear magnetic resonance. 19. What is nuclear magnetic spectroscopy? Ans. A routine analytical tool for detecting atomic nuclei with spin in molecules by absorption at resonance.

Spectroscopic Methods of Analysis  9.35 20. What is the advantage of reporting chemical shift on the d scale. Ans. Chemical shift reported on the d scale is independent of applied field. 21. What is chemical shift? Ans. It is the difference between its resonance frequency and that of reference standard (TMS).

Unsolved Questions IR

spectroscopy 1. What is the difference between absorption and emission spectra? 2. Distinguish between atomic and molecular spectroscopy. 3. Explain Beer’s Lamberts law. 4. What are electromagnetic radiations? Give the regions of electromagnetic radiations in the increasing order of frequency. 5. What type of molecules gives pure rotational spectra? 6. Explain how a vibrational rotational spectrum is obtained. 7. What do you understand by fundamental bands and overtone bands? 8. Distinguish between the following from their IR spectra. (i) Ethanol and dimethyl ether (ii) Primary, secondary and tertiary amides 9. What modes of vibrations are active in IR absorption spectra and why taking CO molecule show the various types vibrations which will be observed in IR spectra. 10. Three test tubes contain three ketones–one of them absorbs near 1785, one near 1750 and one near 1710 cm–1. What types of ketones are they? 11. How many vibrational modes are there in a linear and a non-linear molecule having n atoms? Explain with the help of degrees of freedom. 12. What are different kinds of electronic transitions? 13. Distinguish between the following pairs from their IR spectra. (i) 2 propanol and propanone. (ii) Ethanol and dimethyl ether. (iii) Ethylamine and acetamide.

UV–Visible Spectroscopy 1. What are absorption spectra? 2. Write short note on the electronic transitions caused by energy absorbed in the U.V. region. 3. Why molecules absorb in UV visible region? What are the types of electronic transitions that can occur in the molecules? 4. Give one example each of s Æ s*, n Æ s*, n Æ p and p Æ p* transitions. 5. How will you distinguish between benzene and anthracene by UV spectroscopy? 6. Write a note one (i) Chromophores (ii) Auxochromes 7. Define the following terms: (i) Batho chromic shift (ii) Hypo chromic shift

9.36  Engineering Chemistry

(iii) Hyper chromic shift (iv) Hypso chromic shift 8. Explain various applications of ultra voilet spectroscopy? 9. Write a note on scope on UV spectra. 10. Which of the following will have higher l max in UV spectroscopy; 1,3 pentadiene and 1,4 pentadiene.

NMR Spectroscopy 1. Which of the following nuclei do not show nuclear magnetic resonance? 1 H, 2H, 12C, 13C, 14N, 15N, 16O, 19F, 31P 2. Describe briefly the theory of NMR spectroscopy. 3. What do you understand by the positions of the signals in an NMR spectrum? How many signals are expected in each of the following compounds? (a) Propane   (b) Ethanol  (c) Ethyl methyl ether  (d) Ethyl acetate  (e) Butanol 4. What is meant by the term chemical shift? 5. Discuss the factors which affect the magnitude of the chemical shift? 6. What do you understand by the terms splitting of signals? 7. Explain the term spin – spin coupling. 8. Explain why NMR spectrum of benzene is observed at lower field where as that of acetylene is observed at a higher field strength. 9. How many kinds of H1 are there in: (i) CH3 – CH2 –= CH3  (ii) CH2 = CH2   (iii) CH3 – CH = CH2  (iv) C6H5 – CH3 10. Explain de shielding due to hydrogen bonding. 11. Determine which of the following molecules will show spin – spin coupling in their NMR spectra. If the coupling splitting is observed, give the multiplicity of each kind of proton. (a) ClCH2CH2Cl

(b) H Br



(c) H Br

C=C

C=C

H Br Cl H



(d) CH3 – C = O



(e) CH3 CH3

12. 13. 14. 15. 16.

C=O

An organic compound C3H6O contains a carbonyl group > C = O. How will its NMR spectrum decide whether it is an aldehyde or a ketone? State and explain the principle of NMR spectroscopy. Explain the term coupling constant with examples. Why TMS is used as internal standard in NMR spectroscopy? Predict the NMR spectrum of the compound CH3COCH2C ∫ CCH3.

Spectroscopic Methods of Analysis  9.37 17. What is the multiplicity of each proton in (i) Iso butyl bromide (ii) Butyl bromide 18. Give the structure consistent with the following NMR data: C3H5Cl3 (a) Singlet, d 2.20, 3H (b) Singlet, d 4.20, 2H Rotational Spectra 1. Explain applications of Rotational spectra. 2. Give the principle of Rotational spectra. 3. Explain Rotation in diatomic molecules. 4. She pure rotational spectrum of CN molecule in gaseous phase shows series of equally spacel lines with interspacing 3.8 cm–1. Calculate the internclear distance of CN molecule. Raman Spectra 1. How is Raman spectra different from other spectra? 2. Explain Rayleigh Scattering and Raman Scattering. 3. Explain Rotational-Viluational spectra in Raman. 4. How is Raman spectra different from IR Spectra?

Chapter

10

Polymers Polymers form the backbone of the modern civilization and are the chief products of the modern chemical industries. In almost all walks of life, starting from household utensils, clothes, furniture etc., to automobiles to space aircrafts and biochemical and surgical operations, polymers are extensively used. A polymer may be defined as high molecular weight compound formed by the combination of large number of one or more types of small units or molecular of low molecular weight. The simple molecules from which the repeating structural units are derived are called monomers and the process by which these simple molecules i.e. monomers are converted into polymers is called polymerization.

nCH2 = CH2 Ethylene

—(CH2 — CH2)— n Polyethylene

In this example ‘n’ monomer units combine to form polymer. Hence the degree of polymerization is n’. Degree of polymerization is related to molecular weight of monomer (m) and polymer (M) in general, according to this relation.

n = M/m

Therefore the molecular weight of polymer depends upon the degree of polymerization The number of bonding sites present in a monomer is known as its functionality.

10.2  Engineering Chemistry e.g. (i) CH2 = CH2 When double bond breaks, there are two active sites or bonding sites at which bonding may take place and hence it is a bi functional monomer. H

H

C=C H



* * C–C

* – Two Bonding sites Functionality = 2

H

(ii) HOOC – (CH2) 4 – COOH Æ Adipic acid Æ Bi functional (iii) H2N – (CH2) 6 – NH2 Æ Hexamethylene diamine Æ Bi functional

(iv) – X – HOOC(CHOH)2COOH — 4 bonding sites tartaric acid — Ethyl alcohol C2H5OH — 1 bonding site

Functionality and Degree of Polymerisation The structure of polymers formed, depends upon the functionality. We get linear, branched or three dimensional cross-linked polymers. In case of bi functional monomer linear or branch chain polymer is obtained. When a tri functional monomer is mixed in small amount with bi functional monomer branched chain polymer is obtained. Tri functional or poly functional monomers give rise to cross linked polymers. For a substance to act as a monomer it must be bi functional in nature i.e., it must have at least two bonding sites or reactive sites.

1.  Characteristics of polymers









• Polymeric molecules are very big molecules. Their average molecular weights may approach 105 or more. That’s why, they are also known as macromolecules. They are in fact submicroscopic particles. • Polymers are semi-crystalline materials. It means they have both amorphous and crystalline regions. In fact, polymers have regions of crystallinity, called crystallites, embedded in amorphous regions. Crystallites provide strength and hardness and the amorphous regions provide flexibility to the polymeric material. • The intermolecular forces in polymers can be vander waal’s forces, dipole-dipole attractions or hydrogen bonding. These intermolecular forces are in addition to covalent bonds which connect the repeating units into a macromolecule. • The chemical, electrical, optical, mechanical and thermal properties of polymers depend on (i) size and shape of the polymers, and (ii) the presence or absence of characteristic intermolecular forces. These parameters not only determine the properties of the polymers, but also the performance of these materials in a given applications. • Polymers show time-dependent properties. They show “creep”. • Polymers are combustible materials. • Generally, polymers are thermal and electrical insulators. • Polymeric materials are tailor-made and are easily mould able even into complex shapes with reproducible dimensions with a minimum of fabrication and finishing cost.

Polymers  10.3 Polymers contribute to national economy in terms of performance, reliability, cost-effectiveness and high added value.

2. Advantages of polymers Advantages offered by polymers in comparison with metals, wood, ceramics and other conventional materials: (a) Polymers have low densities, this leads to production of light weight products, hence costs of transportation is low and general handling are easy. (b) Polymers have low absolute strength and stiffness but their specific strength and specific stiffness values are favorable. (c) Polymers have excellent resistant to corrosion. They do not require any protective covering and the maintenance of exposed surfaces is easy and economical. (d) Polymers are usually thermal and electrical insulators. (e) Some polymers, especially rubbers are inherently flexible. They have excellent elasticity and damping qualities. (f) Polymeric materials are easily moulded even into complex shapes and reproducible dimensions with a minimum of fabrication and finishing cost. This is because they require low processing temperatures, hence energy requirements are less (hence small amount of energy required). (g) Polymeric materials have the ability to take variety of colors, shades, etc. which does not fade easily since coloring is not usually restricted to the surface but is throughout the mass. (h) Polymeric materials are tailor-made. Depending on the need they can be synthesized as: (i) Transparent or opaque (ii) Rigid or flexible (iii) Brittle or tough (iv) Malleable or elastic and (v) Fibers, Elastomers and Plastic

3.  Drawbacks in polymers as compared with Traditional materials In comparison to traditional materials, polymeric materials have (a) Lower strength and stiffness, they easily get deformed under load, (b) Temperature limitations in service, they have low heat resistance and at low temperature, embrittlement occurs. (c) Time dependent properties. They show “creep” (d) They are combustible but metals and ceramics are not.

4. Where do We use these polymeric materials?

(a) We live in homes and move about in vehicles that are increasingly made of them. (like automobile wind screen, engine and body parts, floor panels etc.); (b) We wear them by using synthetic fabrics like tereylene;

10.4  Engineering Chemistry

(c) (d) (e) (f) (g)



(h) (i) (j) (k) (l) (m) (n) (o)

We sit and stand on them (by making furniture from them); They are used in leisure industry; We sleep on them (by using foamed matrices and bed sheets); We eat and drink from them (by using plastic crockery); We see sights far away from us in time and space and we hear sound with their help (by making electrical appliances like TV and radio cabinets and telephone body parts); We turn knobs, pull switches and grasp handles made from them; Damaged organs can be repaired or replaced with their help; We use them for purification of water; By their use blind people can be made to see and cripples to walk; In the form of coating they prevent rusting and decay of traditional materials; We employ them for electrical insulation for protecting ourselves against electric shock. We also employ them from heat and sound insulation in refrigeration and air conditioning. We also make overhead tanks and pipes to convey water, gases etc., from them.

5. Polymer Classification In this modern world, polymers are an integral part of everyone’s life style. They have very diverse structures and applications ranging from domestic articles to sophisticated scientific and medical instruments. These materials have penetrated into almost all the applications like transport, agriculture, construction, education etc. These materials are used as fibers, rubber, plastics, adhesives, paints etc. In fact, the very existence of life is virtually the formation, transformation and decomposition of biopolymers viz. carbohydrates, proteins and nucleic acid therefore, in view of their importance, a proper understanding of polymeric materials is very essential. Polymers can be classified into several ways as described below:

5.1 [A] On the Basis of structure of polymer there are three types of polymers: (i) Linear chain   (ii) Branched chain   (iii) Cross linked polymer (i) Linear Chain:  Linear chain polymers are those in which monomers are joined like links in a chain. They have high density, high tensile strength and high melting point because of well packed structure. For example, Poly ethylene, Nylons, Polyesters etc. Monomer structure Linear chain

(ii) Branched Chain Polymer:  They are the polymers having side chains or branches in addition to main chain, branching hinders the close packing of the polymeric chains and hence are less tightly packed in comparison to the linear chain polymers. Branched polymers have low melting point, density and tensile strength as compared to linear polymers. e.g. LDPE, Glycogen, Starch etc.

Polymers  10.5

(iii) Network or Cross Linked Polymers:  Polymers in which monomers are cross linked together in all the three dimensions to form three-dimensional network type of structure are known as network or cross linked polymers. These are generally obtained by the polymerization of poly functional monomers. Cross linked polymers are hard, rigid and brittle because of network structures. e.g. Bakelite

Cross linked linear polymer

5.2 [B] On the Basis of Stereochemistry or Tacticity: Alkenes of the type H2C = CHZ may undergo polymerization to form polymer – (H2C – CHZ) n– in which function group Z are on the (i) Same side of the main carbon chain or (ii) Alternate side of the main carbon chain or (iii) Randomly arranged along the main carbon chain. The difference in the configuration of polymers is known as tacticity. Depending upon the tacticity polymers are of three types.

(i) Isotactic  (ii) Syndiotactic  (iii) Atactic

(i) Isotactic:  Polymers, in which all the functional group Z, is on the same side of the main carbon chain, are known as Isotactic polymers. e.g. Cis – Polyisoprene.

ZH

H

ZH

ZH

Z

or

H

HH

HH

HH

H

Isotactic Polymer

H H H H

H Z H Z

H

H

Same side

10.6  Engineering Chemistry

(ii) Syndiotactic:  Polymers where functional group ‘Z’ is present in alternating fashion on the main carbon. e.g. Gutta Purcha. H

H

ZZ

HH

or

HH

H

HH

HH

H H H Z

H Z H Z

H

H

H

Z

(iii) Atactic:  Polymer where the functional group Z is arranged randomly along the main carbon chain. e.g. Polypropylene. HH

HZ

H

ZZ

H

or

H

HH

HH

H

H H H Z H Z H

H H H Z H H H

5.3 [C] On the basis of type of monomers there are two types of polymers; (i) Homopolymer (ii) Copolymer (i) Homopolymer: A polymer formed by repeated combination of one type of monomer is known as homopolymer. For example PE, PMMA, PVC etc. nA

[– A ]n

(ii) Copolymer:  A polymer formed by repeated combination of two or more types of monomer is known as copolymer. For example; Nylon 6, 6, Terylene, Vinyon. Depending upon the arrangement of A (monomer) and B (monomer) in the polymeric chain copolymers may be subdivided into following types: (a) Alternating copolymers:  Monomers are arranged alternatively in the polymeric chain.

nA + nB

[ A–B–A–B–A–B–A [

For example; Polyester, Polyamide. (b) Random copolymers: Monomers A and B are arranged randomly. nA + nB

[ A–A–B–A–B–B–A [

For example; Poly[ethylene-ran-(vinyl acetate)] (c) Block copolymers:  One block of one type of monomer is followed by the block of other type of monomer which is then followed by the first block.

Polymers  10.7 nd

II nA + nB

nd

Block

II

Block

[ A–A–A–B–B–A–A–A–B–B [ st

st

I Block

I Block

For example, Polystyrene-block-polybutadiene. (d) Graft copolymers:  One type of monomer segment is grafted on the backbone chain of another type of monomer segment. –

B

–

B

–

nA + nB

A–A–A–A–A–A B

– B

– B

graft Chain

–

Back Bone Chain

For example, Polybutadiene-graft-polystyrene.

5.4 [D] On the basis of mode of synthesis polymers are classified into two: (i) Addition Polymers  (ii) Condensation Polymers (i) Addition Polymers:  Polymers which are formed by the polymerization of monomers without elimination of atoms or groups are known as addition polymers. They are also known as chain growth polymers. Polymerization takes place in the presence of initiators. For example; Ethylene (ii) Condensation Polymers:  Polymers which are formed by the polymerization of monomers with elimination of small molecules like NH3, H 2O, HCl, CH3OH etc are known as condensation polymers. They are generally formed by different types of monomers and are also called step growth polymers. Initiators are not required in condensation polymerization. For example, Terylene.

Table 1  Difference between Addition and Condensation Polymerization Addition Polymerization

Condensation Polymerization

1. The process of polymerization in which monomers undergo polymerization to form polymer without elimination of any atom or group.

1. The process of polymerization in which monomers undergo polymerization to form polymer with elimination of small molecules like NH3, H2O, CH3OH etc.

2. Only one type of monomers is generally involved.

2. More than one type of monomers is involved.

3. The molecular formula of repeating unit is exactly same as that of starting monomer.

3. The molecular formula of repeating unit is different from that of starting monomer.

4. Molecular weight of polymer rises rapidly.

4. Molecular weight of polymer rises gradually.

5. Alkenes and their derivatives are generally used as monomers.

5. Bi functional monomers are generally used as monomers.

10.8  Engineering Chemistry 6. This is also known as chain growth polymerization.

6. This is also known as step growth polymerization.

7. Initiator is needed.

7. No initiator is used.

8. Addition polymerization gives rise to homopolymers.

8. Generally copolymers are formed.

9. For example, PE, PVC, PMMA, etc.

9. For example, Nylon 6,6, Bakelite.

5.5 [E] On the Basis of action of heat, polymers there are of two types; (i) Thermoplastic (ii) Thermosetting (i) Thermoplastic:  These are polymers which gets soften on heating and hardened on cooling without or with very little change in their properties. (ii) Thermosetting:  These are polymers which once hardened cannot be softened again.

Table 2  Difference between Thermoplastic and Thermosetting Thermoplastic

Thermosetting

1. Polymers which are softened on heating and hardened on cooling.

1. Polymers which once hardened can’t be softened again.

2. They can be moulded again and again.

2. They can’t be moulded.

3. They are generally polymerization.

formed

by

addition

3. They are generally formed by condensation polymerization.

4. They have linear structure and different chains are held by weak Vander waal’s force of attraction.

4. They have cross linked three dimensional network type of structure.

5. On heating secondary bonds break (vander waal’s force of attraction) hence they get softened.

5. On heating, they undergo hardening because of taking place of any leftover polymerization.

6. They are usually soft, weak and less brittle.

6. They are usually hard, strong and brittle.

7. They are soluble in some organic solvents.

7. They are insoluble in organic solvents.

5.6 [F] On the basis of origin: Polymers can be broadly classified into two types: (i) Natural Polymers  (ii) Synthetic Polymers (i) Natural Polymers:  Polymers which are obtained from the nature (animal or vegetable origin) are called natural polymers. e.g. Starch, Cellulose, Proteins, Natural rubber etc. (ii) Synthetic Polymers:  Polymers which are prepared in laboratory. They are of two types. (a) Organic polymers:  Main chain of polymer is made of carbon. For example. PVC, PE, Nylon, PTFE etc. (b) Inorganic polymers:  Main chain of polymer is made of elements other than carbon. For example, silicon polymers.

Polymers  10.9

e.g. Pe, PVC, PMMA, Nylons, Polysterene

e.g. Phenolic resin, Epoxyresin, Bakelite

5.7 [G] On the basis of molecular force: Polymers are three types of polymers: (i) Elastomers  (ii) Fibres  (iii) Plastics (i) Elastomers:  The group of polymers exhibiting high degree of elasticity (reversible elongation 300% to 1000%) stretching at least twice of their original length under the action of tensile force and recover their dimensions after the removal of the applied force, just like rubber are known as elastomers. Elastomers have long coiled chains, amorphous structure and a few widely separated cross links. The polymeric chains are held together by weakest inter molecular forces. When such polymers are subjected to external force the coils open up partially and get aligned in the direction of elongation. The aligned chains have a natural tendency to revert back in their original state as soon as the force is removed. Example: Natural rubber [cis-poly isoprene] is elastomer. The average initial modulus of elasticity ranges from 10 to 102 psi. There are two types of elastomers. (a) Natural elastomers. For example, natural rubber (b) Synthetic elastomers. For example, Synthetic rubber like Buna S, Buna N, neoprene, Butyl rubber. (ii) Fibres:  These are polymers having thread like structure with high aspect ratio (length to diameter ratio). l d

Aspect ratio = length/diameter = l/d In case of fibres the minimum l:d ratio is 100:1. They have linear, unbranched symmetrical chains aligned in the direction of length giving rise to higher tensile strength and crystallinity. The initial modulus of elasticity is of the order of 103 to 104 psi. Fabrics are knitted or woven from yarns which in turn are formed by spinning of a number of fibres. Fibres are classified as: (a) Natural fibres:  Depending upon the source of origin they may be classified as plant fibres and animal fibres. Plant fibres – Cotton, Linen Animal fibres – Wool, Silk

10.10  Engineering Chemistry (b) Artificial fibres:  Some important class of artificial fibres are (i) Polyamides for example, Nylons (ii) Polyesters for example, Terylene (iii) Vinyls for example, Acrylnitrile, Saran (iv) Rayons for example, Cellulose acetate Rayon The properties of fibres are also affected by the presence of hydrogen bonding. (iii) Plastics:  The polymers which can be moulded into desired shapes by the application of heat and pressure in the presence of catalyst are known as plastics. Plastics word is derived from Greek word “Plastics’ which means ‘fit for moulding’. The term plastic must be differentiated from resin. 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.

6. Thermosetting Resin These polymers undergo permanent change on heating i.e. the changed occurred in structure of polymers is chemical. These polymers are cross linked or heavily branched molecules which on heating undergo extensive cross linking and again become infusible. Cross linked polymers are always thermosets due to the presence of three-dimensional networks.

6.1 Important thermosetting resins Phenolic resins or phenoplasts are condensation polymerization products of phenolic derivatives like phenol, resorcinol with aldehydes like formaldehyde, furfural. Most important member of this class is Bakelite or phenol formaldehyde resin. It is prepared by condensing phenol with formaldehyde in the presence of acidic/alkaline catalyst. HCHO + H+ Æ C+H2OH acid catalyzed. Methylolation:  When phenol and formaldehyde react together, the first step is the entry of methyl (CH2OH) groups in ortho and Para positions to the hydroxyl group. Reaction takes place in the presence of acid or alkali. Depending upon phenol formaldehyde ratio, various phenol alcohols may be formed.

Polymers  10.11 Novalac Formation:  In the presence of acid catalyst, when the P/F (phenol formaldehyde) ratio is less than unity, the methylol derivatives condense with phenol to form a linear polymer with little methylol group. The product is thermoplastic in nature and known as novalac.

Resole Formation:  In the presence of alkaline catalysts with P/F (phenol formaldehyde) ratio greater than 1, the methylol phenols condense to form linear structure called resoles. The resoles have excess of methylol groups capable of further reaction during continued heating. Resoles, therefore have limited shelf life. However the reaction may be stopped at the deserved step by cooling. OH HOH2C

OH CH2OH +

CH2OH

OH HOH2C

OH CH2

Resoles

Bakelite:  Further heating of novalac, in the presence of hexa methylene tetramine as curing agent produces 3 dimensional, cross linked network polymer (Bakelite) which is hard, rigid and insoluble solid. Cross linked of resoles can be done by heating in either neutral of just acidic conditions. Cross linked product from resole is called resites. OH

OH CH2

CH2

CH2

CH2

CH2 OH

OH CH2

CH2

CH2 OH Bakelite

CH2 OH

10.12  Engineering Chemistry Uses:  They have moulding applications. They are widely used in making telephone parts, cabinets for radio, television and automobile parts, for making electric insulator parts like switches, plugs boards, heater handle, for producing brake linings, abrasive wheels and sand paper, used in varnishes, paints and protective coating, in production of ion exchange resin for water softening.

6.2 Amino resins or Urea-Formaldehyde Resin (UF) This is prepared by condensation reaction between 2 parts of urea and 1 part of formaldehyde in neutral or alkaline conditions, in stainless steel vessel at about 50°C. Primary products are mono and dimethylol ureas.



A fully cross linked urea formaldehyde resin can be represented as: — N – CH2 – N – CH2 – N – CH2 – N – CH2 C=O

C=O

— N – CH2 – N – CH2 – N – CH2 – N – CH2 O=C

O=C

— N – CH2 – N – CH2 – N – CH2 – N – CH2

Uses:  It shows remarkable chemical resistance and good adhesion, works as very tough materials, used in industrial flooring foams, potting material for electrical insulation, as a reinforcing material

Polymers  10.13 in many of the fibres reinforced plastics, used for surface coating and as laminating and casting material, used as decorative articles like plates, drinking glasses, dishes, etc.

6.3 Melamine Formaldehyde Resins Melamine and formaldehyde react at about 80°C to give hexa methylol melamine, which on heating in the presence of acids gives cross linked polymer called melamine resin. NH2 N

N

NHCH2OH +

N

3HCHO

NH2

NH2 N Melamine

N

HON2CHN

NHCH2OH N Methyalol derivative

The methylol derivative is further condensed with melamine to give a linear polymer. This linear polymer reacts with excess of formaldehyde producing a cross linked melamine formaldehyde resin. NHCH2NH N

NHCH2NH

N N

N HNCH2NH

N N

HNCH2NH N

HNCH2HN N HNCH2HN

N N

NHCH2NH

N N

NHCH2NH

Uses:  Uncross linked (known as amino resins) used in the paper industry to improve the wet strength of paper, finds use as sizing agents and textile finishing resins, in the plywood industry they are used as adhesives, used in the manufacture of decorative laminates and lacquers.

6.4 Epoxy Resin (Araldite) These resins derived their name from the fact that monomer and the polymer (before cross linking) contains epoxide groups. In facts, chemically, epoxy resins are polyethers. They are prepared by condensation of bis-phenol A and excess of epichlorohydrin so that each end of the low-molecular weight polymer has epoxy groups. This helps in the cross linking of the polymer during thermosetting of the resin.

10.14  Engineering Chemistry

Final curing is done by adding di- and poly-anhydrides and acids (which react with the hydroxyl groups and epoxide rings) or by adding diamines (which react with the epoxide groups). The formation of three dimensional, cross linked structure in epoxy resins (by using diamine) can be written in following ways:

Uses 1. In surface coating particularly for making skid-resistant surfaces for highways, road junctions and round about. 2. Is used as adhesives for glass, metals etc. and are popularly known as araldite. 3. Due to low water absorption tendency, dimensional stability, good heat and electrical resistance make them very good material for electronic applications particularly in mouldings containing inserts and encapsulation.

Polymers  10.15

7.  Thermoplastics These are polymers which can be easily softened repeatedly when heated and hardened when cooled with little change in their properties. The inter molecular force of attraction is more than elastomer and less than that of fibres. There is no cross linking between the chains of polymers. Some common examples are polyethene, polystyrene, polyvinyl chloride, polyvinyl acetate, polyesters and nylons etc. Some thermoplastic polymers are discussed below:

7.1 Poly methyl Methacrylate PMMA (Plexi glass) It is also known as Lucite or Perspex commercially. It is an addition polymer formed from the monomer methyl methacrylate. Polymerization of PMMA is carried out by bulk or suspension radical polymerization using peroxide as initiator.

It is amorphous due to the bulky pendent group (– COOCH3) present. PMMA can be thermally deploymerised to yield the entire amount of monomer. It is an acrylic polymer. Uses:  Because of flexible glassy nature it is used in making high fixtures, aircraft, windows, plastic signboards, artificial eyes, teeth, acrylic emulsion paints, transparent bottles moulds, tubes and optical fibres etc.

7.2 Polyvinyl Chloride (PVC) It is one of the cheapest and most widely used plastic. PVC is prepared by the polymerization of vinyl chloride by heating its water emulsion in presence of a small amount of benzoyl peroxide or H2O2 as catalyst under pressure in an autoclave.

Presence of chlorine atoms on the alternate carbon atoms of PVC causes an increase in the inter molecular attraction which increases the hardness and stiffness of polymer. Uses:  Used in an insulating material, making hand bags, bathroom curtain, kitchen upholstery, used in membrane separation technology. Chlorinated PVC increases the heat distortion temperature of PVC which can be used for the manufacture of hot water pipe. It is widely used synthetic polymer.

10.16  Engineering Chemistry

7.3 Polyvinyl Acetate It is prepared by emulsion or suspension polymerization of vinyl acetate in the presence of benzoyl peroxide or acetyl acetate as catalyst.

Polymerization

CH2 = CH CO – O – CH3

CH2 – C H COOCH3 n Poly Vinyl Acetate

It is insoluble in water and is saponified with great difficulty and hence is not absorbed by the digestive system. Polyvinyl acetate is used as a basic material for chewing gums and surgical dressings. Uses:  The production of water based emulsion paints, making chewing gums and surgical dressings making records, finishing textiles and other fabrics, PVA adhesives are used for bonding of paper leather textiles and woods etc.

7.4 Polyethylene Polyethylene is the simplest addition homo polymer. Monomer is ethylene



There are two types of polyethylene (i) Low density polyethylene (ii) High density polyethylene

(i) Low density polyethylene:  (High Pressure method) Low density polyethylene is prepared by the polymerization of high purity ethylene at high pressure (1000–1500 atm) at a temperature of 250°C in the presence of initiator O2. This reaction takes place by free radical mechanism.

(ii) High density polyethylene (HDPE):  High density polyethylene is obtained by polymerization of ethylene in presence of Zeigler Natta catalyst at lower temperature and pressure. Zeigler Natta catalyst used is TiCl4 + (C2H5)3Al. nCH2 = CH2

6 – 7 atm 60 – 70°C

—(CH2 — CH2— )n HDPE

Polymers  10.17

Difference between LDPE and HDPE LDPE

HDPE

1. LDPE stands for low density polyethylene. 2. Density is 0.91–0.925 g/cc. 3. Branching hinders the parallel arrangement of polymeric chains and hence its crystallinity is low (about 50%). 4. Obtained by free radical polymerization. 5. It has low softening point 115°C. 6. Structure

1. HDPE stands for high density polyethylene. 2. Density is 0.95–0.97 g/cc. 3. Crystallinity is greater than 80% because of linear chain structure.

7. It is used for making squeeze bottles, bags, toys, etc.

7. For making pipes, food tube, containers, gas piping, drug packaging, etc.

4. Obtained by Zeiglar Natta polymerization. 5. It has high softening point 135°C. 6. Structure

7.5 Polycarbonates They are derivatives of phenol and carbonic acid. It is prepared by interaction of diphenyl carbonate with bisphenol-A[2,2-bis(4-hydroxyphenyl) propane] also called lexan, Merlon.

Uses:  It is transparent plastic so it can be used to make safety goggles, making telephone parts and machinery parts, as electrical insulators in electronics and electrical industries.

7.6 Polyamides They are a group of polymers which contains the amide (-CONH-) linkage in the main polymer chain. These are synthetic linear aliphatic polyamides which are capable of fiber formation. Nylon is generic name for synthetic polyamides, capable of forming fibers. Polyamides are prepared by the polycondensation between dicarboxylic acids and diamines. Typical polyamides are describes below: Nylon 6, 6:  Prepared by poly condensation of hexa methylene diamine with adipic acid. By heating equimolar proportion of the two reactants, a polymeric salt is formed.

10.18  Engineering Chemistry nNH2 – (CH2)6 – NH2 + n HOOC – (CH2)4 – COOH Hexamethylene diamine

Adipic acid

N – (CH2)6 – N – C – (CH2)4 – C H

H O

O n

+ nH2O

nylon 6,6

Nylon 6:  This polymer can be made either from e amino caproic acid or e caprolactum by self condensation. Nylon 6 is also known as perlon



The monomer caprolactum is prepared by the following synthetic route:

Nylon 6, 10:  This is prepared by the condensation between hexa methylene diamine and sebacic acid. It is generally not used as a fiber, but it is suitable for monofilaments.

Nylon 11:  It is prepared by self condensation of w amino decanoic acid.

Polymers  10.19 O

H Self n – [H2N – (CH2)10 – COOH] Condensation w amino decanoic acid

— N – (CH2)10 – C — n Nylon 11

Uses:  Nylon 6, 6 is used as a plastic as well as fiber, manufacture of automobile tyre cards, ropes, threads, textile fiber for use in tresses, Nylon 6, 6 being a tough plastic, is used as a substitute for metals in gears and bearings etc. Nylon 11 is used as textile fibers, used as moulding purposes for gears. Nylon 6, 10 is not used much as synthetic fiber, but is used in making brushes and bristles. Kelvar:  It is aromatic polyamides similar to nylon. It is prepared by the condensation polymerization of terephthalic acid dichloride and 1,4-diaminebenzene.

Uses:  (a) Tyres, brakes, clutch lining and other car parts; (b)Bullet proof vests; (c) Motorcycle helmets; (d) Aerospace and air-craft industries.

7.7 Polyesters [Poly ethylene Terephthalate (PET)] Polyesters have the general formula with ester linkages as follows: — C – R – O – C – (R¢) – C — n O O



R = CH2 methylene group R¢ = C6H5 group

They are formed by condensation reaction between an acid and an alcohol each containing at-least bi-functionality. The most common example of the polyester is that obtained from ethylene glycol and terephthalic acid which is known as terylene, dacron and mylar etc,

10.20  Engineering Chemistry

Uses:  It has good fibre forming property with high tensile strength and excellent crease resistant garments. Terycot is a blend of Terylene with cotton, used as films in the manufacture of magnetic recordings tapes and aluminum sheets.

7.8 Glyptal resins Its is made by condensation polymerization of polybasic acids (or anhydride) with polyhydric alcohols (like glycerol) in the presence of heat and catalyst. O O

C

O + n OH – CH2 – CH – CH2 – OH

O Phthalic anlydride

O

C – O – CH2 – C – CH2 – O – C

C n

H

OH

–O–C O

C–O O

Uses:  In paints, enamels, varishes, car panels, car body, binding material for cements, circuit break insulators, switches and gears.

7.9 Poly urethanes Poly urethanes (poly isocyanate) resins are produced by the reaction of a di isocyanate with a compound containing at least two active hydrogen atoms such as diol or diamine. The principal linkage in poly urethane is — NH – C – O — n O

The poly addition between di-isocyanate for example toluene di isocyanate with diol or triol is used to prepared poly urethanes. Uses:  The poly urethane foams or U-foam have the uridest, used in coatings, adhesives, elastomers, in insulation of wire.

Polymers  10.21

7.10 Polypropylene It is lightest known industrial polymer. It is prepared by the polymerization of propylene using Zeigler Natta catalyst. n CH2 = CH CH3

Zeigler – natta Catalyst

— CH2 – CH — n CH3

Uses:  Have excellent mechanical and dielectric properties. 1. Used in appliances such as refrigerators, radios and TVs. 2. It is used for producing package films, pipes storage tanks, seat covers monofilaments and ropes. 3. It is moisture resistance. 4. It is insoluble in many solvents on heating above the melting point. it can be dissolved in aromatic solvents. 5. Being highly crystalline, polypropylene exhibits high stiffness, hardness and tensile strength.

7.11 Poly Vinyl alcohol It is obtained by the hydrolysis of polyvinyl acetate in the presence of Benzoye peroxide as catayst.

Uses:  Fibres made from PVA are found to possess a higher water absorption property. PV alcohol fibres have excellent dimensional stability.

7.12 Poly Formaldehyde It can be obtained by cationic polymerization of formaldehyde or trioxane

10.22  Engineering Chemistry Uses:  Due to high degree of hardness and rigidity it is useful as engineering plastics.

7.13 Poly Tetra Flouro Ethylene (PTFE) It is commercially known as Teflon. Polymerization is carried out by the emulsion method using peroxide initiators under high pressure. It is also polymerized with redox initiators such as hydrogen peroxide and ferrous sulphate. Polymerization

n CF2 = CF2 Teflon

– CF2 – CF2 – n PTFE

Teflon is a linear, inert polymer. It is inert because of the presence of highly electronegative fluorine atoms. There are very strong attractive forces between different chains due to which it is highly dense polymer. Uses: 1. Used for making gaskets and seals, used as insulating material for transformer cables, wires and fittings, as non lubricating bearings, as a non stick coating for pots and pans.

7.14 Poly Styrene It is also known as polyvinyl benzene, has the following structure.

CH = CH2

Styrene can be polymerized by radical, free radical method. n CH2 = CH Styrene C6H5

Polymerization Benzoyl Peroxide

— CH2 – CH — n C6H5 Polystyrene

Because of the presence of bulky phenyl groups it is amorphous, atactic, linear and chemical inert. Uses:  It is used in manufacturing the articles such as moulded containers, bottles, lids, jars, toys, foam, radio and television cabinets and many household items.

7.15 Poly acrylonitrile (PAN) PAN, also called poly vinyl cyanide. CH2 = CH

Free radical Polymerization

CN acrylonitrile

— CH2 – CH — CN

Polymers  10.23 It is produced from acrylonitrile by the free radical polymerization technique using peroxide initiators. Uses:  It is used to produce PAN fibres. The copolymer of acrylonitrile with butadiene called nitrile rubber has great importance. ABS polymers are made by copolymerizing acrylonitile, butadiene and styrene.

CN — CH2 — CH



x

CH2 – CH = CH – CH2

y

CH2 – CH — z

They are structural plastics or rubber possessing good strength and toughness.

Uses:  Telephones, pipes, moulded articles (like furniture, suit-cases, etc.), packing containers.

7.16 Elastomers (Rubber) Elastomers are the polymers which we capable of being stretched rapidly at least 150% of their original length without breaking and return to their original shape on release of stress. (a) It should be a high polymer. (b) It must be cross linked and amorphous. (c) It should have flexibility of chains. (d) Its glass transition temperature should be low and flow temperature should be high as possible. Properties of Typical Elastomers are 1. They must stretch rapidly and considerably when stretched reaching high elongation (500 – 1000%). 2. They must exhibit high tensile strength and high stiffness when fully stretched. 3. They must shape back to their original length, once the stretching force is withdrawn or the stress is released. Natural rubber is mainly extracted from the tree, ‘Hevea Basilienesis’. The rubber is obtained from latex collected from the cuts made in the bark of the tree. Latex is an emulsion, of poly hydrocarbon droplets in an aqueous solution and looks like milk. The negatively charged particles of poly hydrocarbons stabiles the emulsion. Chemically, natural rubber is cis poly isoprene 2 methyl 1, 3 butadiene) CH2 = C – CH = CH2 CH3

Processing [Composition of Latex water = 60%, Rubber = 35%, Protein, Enzymes and Nucleic acid = 3%, Fatty Acid & Ester = 11% and Inorganic Salts = 0.5%]

10.24  Engineering Chemistry

Gutta Percha:  It is obtained from the mature leaves of dicopsis gutta and palagum gutta tree. Gutta percha has the same empirical formula as the rubber. But it is trans poly isoprene while rubber is cis poly isoprene. H2C H 3C

C=C

(CH2)n

H2C

H

H 3C

Cis - 1, 4 polyisoprene (natural rubber)

C=C

H (CH2)n

Trans 1, 4 polyisoprene (Gutta percha)

Draw Backs of Raw Rubbers 1. It is too weak due to low tensile strength of 200 kg/cm2. 2. It becomes soft at high temperature and brittle at low temperature i.e. it is plastic in nature (Temperature range is 10 – 60°C). 3. It has large water absorption capacity. 4. It is not resistant to mineral oils and organic solvents. 5. Due to air oxidation, it deteriorates and hence its durability is decreased. 6. It is attacked by the oxidizing agents like nitric acid, conc. H2SO4. Chromic acid. 7. When two fresh surfaces of raw rubber are pressed together they coalesce to form a single piece. This process is known as tackiness. Processes for improving the properties of rubber:  The desired properties in rubber can be obtained by compounding it with certain chemicals. Mastication:  When rubber is selected to mechanical stresses arising from high speed stirring or milling it undergoes considerable molecular degradation or fragmentation. The process is called mastication.

Polymers  10.25 Vulcanization:  It is a process by which rubber is converted from plastic condition to elastic condition by heating it with certain chemicals like sulphur, sulphur mono chloride (S2Cl2), hydrogen sulphide. Raw rubber is heated with sulphur at 100–140°C. Chemically sulphur adds to the double bonds of different polymeric chains providing cross linking between them and hence it stiffens the raw rubber. The degree of vulcanization depends on the amount of sulphur added. In tyre industry, about 3 to 4% sulphur and filler like carbon is added. If the percentage of sulphur is increased beyond 30%, full saturation of the bonds takes place resulting in a rigid non flexible rubber called ebonite. CH3

CH3

CH2 – C = CH – CH2 – CH2 – C = CH – CH2 – + CH2 – C = CH – CH2 – CH2 – C = CH – CH –

+

S

CH3

CH3 unvulcanized rubber

CH3

CH3

CH2 – C – CH – CH2 – CH2 – C – CH – CH2 – Sulphur cross link.

S

S

S

S

CH2 – C – CH – CH2 – CH2 – C – CH – CH2 – CH3

CH3 vulcanized rubber

Properties of Vulcanized Rubber 1. It has good tensile strength and extensibility. 2. Resistance of vulcanized rubber gets much improved. 3. Water absorption capacity of vulcanized rubber is considerably lowered. 4. It is better electrical insulator. 5. Vulcanized rubber has much lower elasticity. 6. Vulcanized rubber is resistant to organic solvents such as benzene, CCl4 etc. 7. The working temperature for vulcanized rubber is –40 to 110°C as compared to raw rubber which has working temperature range of –10 to 60°C.

7.17 Synthetic Rubber SBR (Buna –S– rubber) Styrene butadiene rubber is first commercially made synthetic rubber. It is a copolymer obtained by the copolymerization of butadiene (75%) and styrene (25%) in an emulsion system at 50°C in the presence of cumene hydro peroxide as catalyst.

10.26  Engineering Chemistry CH2 = CH – CH = CH2 + CH2 = CH (75%) (25%) C6H5 cumene hydro peroxide – CH2 – CH = CH – CH2 — CH2 – CH C6H5

n

Its resilience is poor than that of natural rubber. It has high abrasion resistance and high load bearing capacity. Uses:  It is used in manufacturing of automobile tyres-rubber soles, Motor tyre, Shoe soles, Footwear components, Insulation of wire and cables, Gaskets, Adhesive and Floor tiles. Buna N or Nitrile Rubber (NBR) or GR-A The nitrile rubbers are polymers of butadiene and acrylonitrile in the ratio of 75% and 25% respectively. It is prepared by emulsion polymerization using cumene hydro peroxide as initiator. Hydrogen peroxide and ferrous sulphate can also be used as initiators vulcanization is done by conventional means by adding sulphur.

Due to the presence of cyano group it is less resistant to alkalis than natural rubber. It has good tensile strength, abrasion resistance good heat resistance. It possesses excellent resistance to heat, sunlight, acids and salt. Uses:  Nitrile rubber are used for oil resistance, Conveyer belts, Lining of tanks, Gaskets, Printing rollers, Oil resistance form, Automobile parts and high altitude aircraft components and Adhesive. Neoprene or GR-M Rubber This is also known as polychloroprene or poly chloro butadiene or duprene. Neoprene is prepared by polymerization of chloroprene which is 2 chloro 1,3 butadiene. Cl CH2 = C – CH = CH2 Chloroprene



Polymerization

Cl — CH2 – C = CH – CH2 — n Neoprene

The polymerization of chloroprene takes place through free radical emulsion polymerization.

Polymers  10.27 The vulcanization of neoprene is different because it can be vulcanized by heat alone. Neoprene can also be vulcanized using ZnO or MgO. Neoprene is closely related to the natural rubber. It is superior to natural rubber as far as ageing and effects of high temperature are concerned. It is soluble in polar solvents due to the presence of chloro group. It has excellent oil resistance through inferior to nitrile rubber but better than natural rubber or SBR. Uses 1. It is used in oil resistant wire and cable coating, industrial houses, shoes soles, latex is used for the manufacture of gloves and coated fabrics, for making tubes for carrying corrosive gases and oils, used in manufacture of sponges, converges belts, lining of reaction vessels etc.

7.18 Silicone Polymers Like carbon silicone has a valency of four. There are many compounds of silicon which apparently resemble carbon compound e.g. CH4, SiH4. Unlike carbon, however, silicon cannot form Si = Si double bonds. Hence the polymers of silicon are produced through condensation. The Si – Si linkage is thermally unstable in the case of large molecules and hence silicon forms polymer mainly through silicon bonds represented as – Si – O – Si –. Silicon is organo silicon polymers. They are prepared from alkyl chloro silanes obtained from Grignard reagent and silicon tetra chloride (SiCl4). CH3MgCl + SiCl4 Æ CH3SiCl3 + MgCl2 Methyl tri chlorosilane

2CH3MgCl + SiCl4 Æ (CH3)2SiCl2 + 2MgCl2 Di methyl di chloro silane

3CH3MgCl + SiCl4 Æ (CH3)3SiCl + 3MgCl2 Tri methyl chloro silane

The di methyl di chloro silane can also be prepared by passing methyl chloride over powdered silicone in the presence of Cu catalyst at 275° – 375°C. 2CH3Cl + Si

Cu

(CH3)2 SiCl2

Hydrolysis of the chlorides yields the corresponding silanols which polymerise by condensation. For example dichloro dimethyl silane on hydrolysis gives the diol which reacts as below.

10.28  Engineering Chemistry Since O an active (– OH) group is left at each end of the chain, polymerization continues at both the ends and ultimately form a linear thermoplastic polymer. CH3

CH3

CH3

CH3

— O – Si – O – Si — O – Si – O – Si – CH3

CH3

CH3

CH3

Properties and uses of different types of Silicons 1. Silicon Rubber:  Silicon resins on compounding with filler e.g. finely divided silica, alumina and ferric oxide, followed by vulcanization by means of organic peroxides give silicon rubber.

2. Silicon Resin:  Silicon used for coating electrical wires to give water proof high temperature insulation. They are mixed in paints and enamel to make them resistant to the effects of high temperature.



3. Silicon Grease:  It is obtained by mixing silicon fluids with fillers such as powered silica or metallic soaps. They are used as lubricants in aeroplane since they do not freeze at low temperature as –40°C and do not melt at 200°C.



5. Silicon Fluids:  They are highly stable and non volatile even on heating. They retain their viscosity over a wide range of temperature which makes them valuable as hydraulic fluids.

7.19 Types of polymerization 1. Addition or chain growth polymerization is a reaction that yields a product, which is an exact multiple of the original monomeric molecule. Such a monomeric molecule, usually contains one or more double bonds, which by intermolecular rearrangement, may make the molecule bifunctional. The addition polymerization reaction must be instigated by the application of heat, light, pressure or a catalyst for breaking down covalent bonds of monomers. Thus: H

H

n C=C H

H

rearrangement (heat, light n pressure or catalyst)

Ethene (monomer)

H

H

Polymerizalion

H H

C=C

C–C

H

H H

H

Bi functional molecule

Polyethene [PE]

2. Condensation or step polymerization may be defined as a reaction occurring between simple polar group containing monomers with the formation of polymers and elimination of small molecules like water, HCl, etc. For example, hexamethylene diamine and adipic acid condense to form a polymer, nylon 6, 6.

Polymers  10.29 H

H

H

OH H

OH

n N – C –N + C – C – C H

H6H

Hexamethylene diamine

O

–2nH2O Condenlation Polymerization

H 4 O

H

H

—N – C – N – C – C – C — H

H 6H

O

H 4 O

n

Poly hexamethylene adipate (Nylon 6, 6)

Adipic acid

Thus, condensation polymerization is an intermolecular combination, and it takes place through the different functional groups (in the monomers) having affinity for each other. When monomers contain three functional groups, it may give rise to a cross linked structure polymer. The types of functional groups, which are most important in the condensation reactions, are: 1st Group

2nd Group

Product

Example

Hydroxyl – OH

Carboxyl –COOH

Polyester

Terylene

Amino –NH2

Carboxyl – COOH

Polyamide –NH–CO–

nylon 6, 6.

Hydroxyl – OH

Isocyanate OCN–

Polyurethane –CO–NH–

Spandex fibre

3. Copolymerization is the joint polymerization of two or more monomers species. High molecularweight compounds obtained by copolymerization are called copolymers. For example, butadiene and styrene copolymerize to yield GR-S rubber.

7.20 Mechanism of Addition polymerization Addition polymerization reactions proceed by a chain reaction mechanism consisting of three important stages: 1. Initiation or the formation of an active center. 2. Propagation or the formation of a polymer having active centres. 3. Termination or the removal of the active centre. Depending upon the type of the active centre formed during addition polymerization different mechanism can be suggested: 1. Free Radical Polymerization Free radical polymerization consist mainly of three steps:

10.30  Engineering Chemistry 1. Initiation A reagent that is either itself a free radical O2 molecule, or that which easily decomposes to form free radicals, such as organic peroxides or aliphatic azo compounds are used as radical initiator. Commonly used radical initiator are: 1. Benzoyl peroxide 2. Azo-bis isobutyronitrile (CH3)2 – C – N = N – C – (CH3)2 CN

2(CH3)2 – C + N2 CN

CN

These reactions can be represented as, R2 Æ 2R where R2 is a free radical initiator and R. is a free radical (FR) The FR(R.) adds to a molecules of monomer producing another FR R + CH2 = CH X



R–CH2 – CH X

3. Chain propagation. The FR, RCH2CHX, further adds to a monomer molecule forming another FR, and successive addition of monomer to these radicals results in the formation of larger chain radicals.

The chain propagation results in the change of p–bonds into s bonds. This step always occurs with the liberation of heat. In vinyl polymerization free radicals attach monomers at the methylene group, giving head-to-tail addition. This addition is also favored because this carbon atom is less sterically hindered than the other carbon atoms. This addition also leads to reasonance stabilization. For example, CH = CH2

RCH2 – CH

RCH2CH

RCH2CH

R +

etc

Polymers  10.31 Chain termination: The growing polymer chains can be terminated in the following two ways: (i) Coupling or Recombination:  It involves the addition of two growing radical chains to form a saturated macromolecules. R — CH2 – CH — CH2 – CH + n1 X X

R — CH2 – CH — CH2 – CH n1 X X

R — CH2 – CH — CH2 – CH – CH — CH2 – CH — CH2 — R n1 n2 X X X X



(ii) Disproportionation:  It involves the transfer of a hydrogen atom from one growing chain to another with the formation of an unsaturated end group on the chain by losing the hydrogen atom. R — CH2 – CH — CH2 – CH + n1 X X

CH – CH2 – CH – CH2 — R n2 X X

+

CH2 – CH2 – CH – CH2 — R n2 X X

R — CH2 – CH = CH = CH n1 X X

Ionic polymerization This type of polymerization may be anionic when electron withdrawing groups such as chlorine are present or cationic when electron donating groups such as methyl are present. These reactions occur at very low temperature as the initiation reaction requires only small activation energy. 1. Anionic polymerization:  Monomers containing electronegative substituent’s such as acrylonitrile, methylmethacrylate and styrene are polymerized by initiators. The initiators used in liquid ammonia, alkali metals such as sodium, potassium and lithium in liquid ammonia, alkali metal alkyls, triphenylmethyl sodium and Grignard reagents. Mechanism:  It consists of following three steps: (a) Chain initiation. In this step the initiator produces an anion which adds to the monomer molecule to generate a carbanion as an active centre for chain growth. K–NH2 Æ K+ + NH2–

10.32  Engineering Chemistry

(b) Chain propagation



(c) Chain termination

Other examples include, 1. Synthesis of polymethylmethacryate



2. Synthesis of polystyrene

Polymers  10.33





3. Cationic polymerization:  Ethylene derivatives containing electron donating groups and other unsaturated monomers readily polymerize in the presence of very small amount of acids and friedal craft catalysts (AlCl3, BF3, SnCl4 etc,). The mechanism proceeds via the formation of carbonium ion. For example, polymerization of styrene in the presence of SnCl4. mechanism occurs in following steps: (a) Chain initiation. It involves the formation of active carbonium ion center.

(b) Chain propagation. This step involves addition of monomer molecules with regeneration of carbonium ion. ≈

–

[SnCl4] CH2 – CH – CH – CH + nCH2 = CH C6H5

C6H5

C6H5

≈

–

[SnCl4] – CH2 – CH – CH2 – CH – CH2 – CH n C6H5 C6H5 C6H5



(c) Chain termination:  The Chain gets terminated by elimation of initiator. –

≈

[SnCl4] – CH2 – CH – CH2 – CH – CH2 – CH n C6H5 C6H5 C6H5

– CH2 – CH – CH2 – CH – CH = CH + SnCl2 n C6H5 C6H5 C6H5 Polystyrene

Other examples include, polymerization of ethylene derived compounds using BF3 (lewis acid) and H2O as catalyst. (a) Chain initiation

10.34  Engineering Chemistry



(b) Chain propagation



(c) Chain termination

8.  Conducting Polymers Conventional polymers are poor conductor of electricity, because of non availability of large number of free electrons in the conduction process. Within past several years, polymeric material has been synthesized which possess electrical conductivity as per metallic conductors. Such polymers are called conducting polymers. Conductivities as high as 1.5 × 107ohm–1m–1 have been attained in these polymeric materials. On a volume basis, this value is equal to one-fourth of the conductivity of copper, or is twice its conductivity on the basis of weight. Different types of conducting polymers are discussed below: 1. Intrinsically conducting polymer or (a) conjugated p electron conducting polymers. In this polymer backbones or associated group consist of delocalized electron pair or residual charge. Such polymers essentially contain conjugated p electrons backbone, which is responsible for conducting electrical charge. In an electric field, conjugated p electrons of the polymer gets excited, thereby can be transported through the solid polymeric material. Overlapping of orbitals of conjugated p electrons over the entire backbone results in the formation of valence bands as well as conduction bands, this extends over the entire polymer molecule. Presence of conjugated p electrons in a polymer increases its conductivity to a larger extend. Important commercially produced conducting polymers: (i) Polyacetylene polymers e.g., poly-p-phenylene, polyquinoline, polyphenylene-covinylene, poly-m-phyenylene sulphide, etc. (ii) With condensed aromatic rings, e.g., polyaniline, polyanthrylene etc. (iii) With aromatic heteroaromatic and conjugated aliphatic units, e.g., polypyrrole, polythiophene, polyazomethine, polybutadienylene, etc.

Polymers  10.35 Structures of some of the above given polymers are given below: (a) Poly (p-phenylene)



(b) Poly (p-phenylene Vinylene)



(c) Trans-Polyacetylene



(d) Polythiophene



(e) poly Pyrole N



H N

N H

H N

N H

(f) Polyaniline H N

N N

N H



(b) Doped conducting polymers are obtained by exposing a polymer to a charge transfer agent in either gaseous phase or in solution. Intrinsically conducting polymers (ICP) possess low conductivity but these possess low ionization energy and high electron affinities, so these can be easily oxidized or reduced. Consequently, the conductivity of (ICP) can be increased by creating either positive or negative charges on the polymer backbone by oxidation or negative charges on the polymer backbone by oxidation or reduction. This technique called doping is of two types:

10.36  Engineering Chemistry

(i) P-doping involves treating an intrinsically conducting polymer with lewis acid, thereby oxidation process takes place and positive charges on the polymer backbone are created. Some of the common p-dopents used are I2, Br2, AsF5, PF6, etc. For example: (C2H2)n + 2FeCl3 Æ (C2H2) +FeCl4 – + FeCl2 2(C2H2)n + 3I2 Æ 2[(C2H2)n+ I3]



(ii) N-doping involes treating an intrinsically conducting polymer with a lewis base thereby reduction process takes place and negative charges on the polymer backbone are created. Some of the common N-dopant used are Li, Na, Ca, tetrabutyl ammonium, FeCl3, etc. For example:



2. Extrinsically conducting polymers are those polymers whose conductivity is due to the presence of externally added ingredient in them. These are of following types: (i) Conductive element filled polymer is a resin or polymer filled with conducting elements such as carbon black, metallic fibres, metal oxides, etc. In this, the polymer acts as a binder to hold the conductive elements together as a solid entity. These polymers possess reasonably good bulk conductivity and are generally low in cost, light in weight, mechanically durable, and strong and easily process able in different forms, shapes and sizes. Generally special grade conducting carbon black is used as filler due to its (a) very high surface area (b) high porosity and (c) filamentous properties. Minimum concentration of conductive filler in the polymer to start the conducting process is called percolation threshold. (ii) Blended conducting polymer is product obtained by blending a conventional polymer with a conducting polymer either by physical or chemical change. Such polymers can be easily processed and possess better physical or chemical and mechanical properties. 3. Coordination conducting polymer (inorganic polymer) is a charge transfer complex containing polymers obtained by combining a metal atom with a polydentate ligand. Although the degree of polymerization in such a polymer is small, yet they exhibit corrosion characteristics, and are usually undoubtable in most solvents.







Applications: Conducting polymers are finding increased use as they are light weight, easy to process and have good mechanical properties. Some of the important applications of the conducting polymers are:

Polymers  10.37





1. In rechargeable light weight batteries based on perchlorate doped polyacetylene-lithium system. These are about 10 times lighter than conventional lead storage batteries. Such batteries are sufficiently flexible to fit a variety of designed configuration. 2. In optically display devices based on polythiophene. When the structure electrically biased (1 to 3V), the optical density of the film changes, i.e. its colour changes. Such electrochromic systems produce coloured display devices (LED). 3. In wiring in aircrafts and aerospace components. 4. In telecommunication systems. 5. In antistatic coating for clothing. 6. In electronic devices such as transistors and diodes. 7. In solar cells, drug delivery system for human body, etc. 8. In photo voltaic devices, 9. In electromagnetic screening material. 10. In molecular wires and molecular switches. 11. In button type battery. These batteries are long lasting and rechargeable. 12. In biosensors and chemicals sensors, which convert chemical information into a measurable electrical response.

9. Biodegradable polymer Biodegradation is caused by attack of ester groups by non-specific esterases produced by ground microflora combined with hydrolytic attack or it is the breakdown of polymer by microbial organism in to smaller components. The polymer of molecular weight below 9000 is accessible to bacteria for degradation. The essentials for biodegradation process are (a) Microorganism (b) Environment e.g. Temperature, pressure, moisture, O2, light (c) Substrate (biodegradable polymer): It should have following essential features 1. It should have suitable functional group which is susceptible to hydrolysis or oxidation by microorganism 2. Greater hydrophilicity 3. Low molecular weight 4. Less crystalline structure Biodegradable polymers do not have solid waste and they do not cause marine pollution. They can enter geochemical cycles over They are of two types: (a) Naturally occurring polymers: e.g. Polysaccharides – starch, cellulose Proteins – gelatin, casein Polyesters – poly hydroxyl alkanoates Others – lignin, natural rubbers

10.38  Engineering Chemistry (b) Synthesized biodegradable polymer: e.g., are 1. Polyalkylene esters 2. Poly lactic acid 3. Poly amide esters 4. Polyvinyl eaters Applications 1. Biodegradable polymers have medical applications such as in absorbable surgical implants and sutures, controlled release of drugs, absorbable skin grafts and bone places to support the body recovery systems. 2. In agriculture, Time controlled biodegradable polyolefins are used in agriculture for mulching, netting, twine, etc. They have yielded considerable economic dividends in increasing crop yields and reducing crop management costs. Controlled release of fertilizers, pesticides, etc. is other important applications. 3. Waste management applications. 4. Used in manufacture of shampoo bottles e.g. Poly B-hydroxyl butyrate.

10. Organometallics Organometallic chemistry is an important subdiscipline that bridges the fields of organic and inorganic chemistry together. The term organometallics is usually restricted to those compounds in which a metal atom (including non metallic elements like boron, silicon, phosphorus, arsenic etc. which are less electronegative than carbon) is bonded directly to a carbon atom of at least one organic group. Organometallic compounds contain atleast one direct metal-carbon bond. The bond may be simple covalent or p-dative or even predominantly ionic. Organometallic compounds would include various hydrocarbon derivatives such as carbonyls and carbides. Carbon metal bond

Where M may be Mg, Ca, Na, Li, Al, Sn, Pb, etc. The reactivity of organometallic compounds depends upon percentage ionic character of carbon metal bond. The greater the percentage ionic character of carbon metal bond the higher is the reactivity of the compound examples of organometallics compounds are Grignard reagent, tetraethyl lead, Ferrocene etc.

11.  Classification

Polymers  10.39

On the basis of carbon metal bond:  Following three types 1. Ionic organometallics: The organometallic compounds of highly electropositive metals are usually ionic. Thus, the alkali and alkaline earth derivatives, with the exception of those of lithium, beryllium and magnesium as well as lanthanides and actinides are predominantly ionic. Such compounds consist of hydrocarbon part (negative charged) and positively charged metal ions held together by electrostatic forces of attraction e.g. C10H8Na, KC5H5. Metal is of group IA and IIA. 2. Bonded covalent organometallics: These are simplest and most studied organometallic compounds that contain classical covalent bond shared by the metal and the organic group each contributing a single electron. These compounds are readily formed by representative elements og 13, 14 and 15 in addition to group 12 i.e. zinc, cadium and mercury. e.g., Al (CH3)3, Pb (C2H5)2, Zn (C6H5)2. 3. Electron Deficient organometallic compounds:  Some organometallic derivatives of beryllium (BeR2), magnesium (MgR2), aluminium (AlR3) are polymeric. These compounds are associated by electron deficient or polycentric bond. Trimethyl aluminium, [Al(CH3)3] 2 is a dimer. Another example of electron deficient compound is diethyl beryllium [Be(CH3)2] n which has a polymeric structure. This compound utilizes one sp3 hybrid orbital of each methyl group and all of the sp3 orbital of beryllium to form bridge Be—CH3—Be bond. H3 C

CH3 CH3

Al

CH3 CH3

Al C H3

Structure of [Al(CH3)]2 CH3 CH3

s

CH3

Al

Al

s

CH3

CH3

CH3

CH3

CH3 Al

Al CH3

CH3 CH3

Banana shaped as three centre Bond.

CH3 Three centred Bonds in [Al(CH3)3]2



4. Transition metal organometallic compounds: Transition metals are well known to form a variety of organometallic compounds with unsaturated organic compounds. In these compounds, the transtion metal forms bond to more than one carbon atoms of the same organic compound. This type of bonding is of considerable importance in transition metal chemistry. The interaction primarily occurs between the p-orbitals of organic ligand and the appropriate p or d-orbitals of the transition metal atom.

10.40  Engineering Chemistry Unsaturated organic compounds capable of forming this important class of organometallic compounds include (i) 2-electron donor (alkenes) (ii) 3-electron donors (p-ally groups (iii) 4-electron donors (butadiene) (iv) 5-electron donar (cyclopentadiene) (v) 6-electron donors (benzene).



Some of the typical organometallic compounds in this category are the following: –

CH2

Cl K

+

Pt

CH2

Cl Zeise salt + K [PtCl3 (C2H4)]

Fe

Cr

Cl

Ferrocene (C5H5)2Fe

Dilenzene chromium (C6H6)2 Cr

Sandwich structures

It is important to note that in this type of bonding each carbon atom must be sp2 hybridized so that it becomes a part of the planar carbon atom network. There are four Organometallic compounds of transition metals: 1. Aryl/alkyl complexes:  Simple organometallic compounds of transition metals prepared by reaction like e.g., TiCl4 + 4 LiCH3

dry ether

Ti(CH3) 4 + 4 LiCl

2. Alkenes complexes:  The complexes contain alkenes as Ti bonding legends (two-electron donor or three-electron donor) e.g. Ziese’s salt K2 [PtCl4] + C2H4

K [PtCl3 (C2H4)] + KCl

3. Sandwitch bonded complexes: Complexes in which transition metal is sandwiched with symmetrical delocalized organic system e.g. Ferrocene (p C5H5)2 Fe. Fe is sandwiched by cyclopentadienyl system. 4. Mixed complexes: Complexes which contain both olefinic as well as symmetrically delocalized organic system. Transition metal attached to ligand (which can be an element O2, N2, compound NO, CO or ion CN–. Metals are in lowest oxidation state they are also called p acceptor ligand because metal carbonyls have vacant p orbital in addition to lone pair. Due to excessive negative charge on central metal atom, the back donation of electrons from metal to vacant p orbital of CO takes place e.g. V(CO) 6, Ni(CO) 4, Mn2 (CO)10, Fe3 (CO)12.

Polymers  10.41

12. Preparation

1. Direct reaction of metals:  Metals (M) react directly with alkyl halides (RX) in ether leading to the formation of alkyl/aryl metal halides.



M + RX M + C6H5 X

Grignard reagents are synthesized by this method: Mg + CH3I



RMX C6 H5 MX dry ether

2. Using alkylating agent:  Here alkylating agents such as Grignard and lithium reagents, aluminum and mercury alkyls etc are used to make organometallics compounds PCl3 + 3C6H5MgCl



CH3MgI

P (C6H5)3 + 3MgCl2

3. By metal exchange reaction:  Reactions between organometallic compounds of one metal with organometallic compound of other metal. R2Hg + 2Na 4PhLi + (CH2 =CH) 4Sn



2RNa + Hg (CH2 =CH2) 4 Li + Ph4Sn

4. Metathesis:  Reactions of organometallic compounds with metal halides RM + R¢M¢

R¢M + RM¢

e.g., 3CH3Li + SbCl3

(CH3)3Sb + LiCl

5. Hydrometallation reactions:  Addition of metalhydride to alkene and akynes H H M–H +

M–C–C–H

C=C

H H

The most important examples concern B, Al and Si. Tri isobutyl aluminium is normally made in a single stage process by reacting Br2AlH with isobutene under mild pressure. Br2 AlH + Me2 C = CH2

Br3 Al

Similarly,

(C2H5)2 AlH + C2H4

(C2H5)3 Al (Hydro alumination)

6. Transmetallation reactions:  The reaction between a metal and an organometallic compound of another metal. M + RM¢

RM + M¢

10.42  Engineering Chemistry

This general method may be applied to M = Li Be Al, Ga, Sn, Pb, Bi, Zn, Cd. RM are found among the heavy elements (Hg, Tl, Pb and Br). Few examples are given below: Zn + Me2Hg P + Ph3Br



Me2Zn + Hg Ph3P + Br

7. Preparation of p complex (a) Zeise’s salt:



[PtCl3 (C2H5)] K– + KCl

CH2 = CH2 + K2 (PtCl4) (b) Ferrocene:





FeCl2 + 2C5H5MgBr



Fe + 2 C6H5



(c) Ni + 4CO

325 K

Ni(CO) 4



(d) Fe + 5CO

HT/HP

Fe(CO)5

Fe(C6H5)N2 + 2 MgBrCl 300°C

Fe(C5H5)2

13. Applications

1. 2. 3. 4. 5.

Tetra ethyl lead is used as anti-knock compound Used to enhance agriculture production e.g. C2H5HgCl acts as fungicide Aryl arsenic compounds are used as chemotherapentic agents Used to prepare organic compounds e.g. Grignard reagent Wilkinson’s catalyst [(PPh3)3] RhCl is used for selective hydrogenation of certain double bonded compounds

C=C

+ H2

Cl (PPh3)3

–C–C– H H



6. In heterogeneous catalysis: e.g. Ziegler Natta catalyst it permits stereo regular polymerization of alkenes. i.e. unchanged and stereo regular i.e. isotactic polymer is obtained. These polymers have high density and high melting point due to efficient packing.

14. Organometallic compounds in polymerization Zeiglar natta is used as polymerization catalyst of ethylene. It is combination of organometallic compound (co catalyst) which consists of a complex of triethyl aluminum and transition metal chloride, TiCl4 (TiCl3 is a better catalyst). When these are mixed in a hydrocarbon solvent generally heptane, a heterogeneous precipitate develops which catalyzes polymerization of alkenes at low temperature and low pressure. Mechanism involved in Ziegler Natta polymerization is not certain but polymerization occurs at localized active sites on the catalyst surface and that organometallic component activates the site by alkylation of transition metal atom on the surface. Two mechanisms are proposed:

Polymers  10.43

1. Bimetallic mechanism:

Complexation of Ti atom through alkyl group is due to empty orbital on Ti atom. This is followed by ionization of transition metal alkyl bond and formation of 6 membered transition state and insertion. The alkyl group goes to olefin and a new metal alkyl bond is bond

2. Monometallic mechanism:

The active site is a penta coordinate transition metal atom with an empty orbital. The monomer is complexed to this atom. An insertion reaction then occurs with the alkyl group (R) migrating to the monomer olefin and a shifting of vacant octahedral position. Polymer chain migrates back so that the vacant site resumes its original position which is necessary to preserve stereoregularity.

10.44  Engineering Chemistry

15. Grignard reagent It is an organometallic compound discovered by Victor Grignard and is used as synthetic reagent. Almost all classes of organic compound can be prepared from Grignard reagent. The general formula for Grignard reagent is R – Mg – X or R – MgX or RMgX. Where R – alkyl, alkenyl, alkynyl or aryl group and X – Cl, Br. I.

15.1 Preparation By the action of alkyl halides on magnesium metal in the presence of dry ether (anhydrous diethyl ether).

R – X + Mg

Diethyl ether



Alkyl halide





CH3 – I + Mg

Di ethyl ether





R – Mg – X Grignard reagent



CH3MgI

Methyl Magnesium Iodide

Among the halides, the ease of formation of Grignard reagent is





Iodide > Bromide > Chlorine

The order of alkyl halide is:   CH3X > C2H5X > C3H7X> …….

For a given alkyl group (say R). The case of formation of the Grignard reagent (or an organometallic lithium compound) except alkynyl halides. These can be made, but special methods are required.

Grignard reagent is produced by dropping a solution of the alkyl halide in dry ether into the reaction flask containing magnesium ribbon suspended in dry ether. Solvents which can be used are diethyl ether, Et2O or tetrahydrofuran, dimethoxyethane, dioxane. Ether solvent is best suitable for the preparation of Grignard reagents as it is not only unreactive with magnesium but also stabilize the Grignard reagent by forming lewis acid base comlpexs In the preparation of Grignard reagent, apparatus and reactants should be absolutely dry. Even traces of moisture/impurities prevent the formation of grignard reagent as grignard reagent reacts with water forming alkanes. CH3CH2MgBr + H2O

CH3CH3 + Mg(OH)Br

Polymers  10.45 Inorganic product Mg(OH)Br is referred as basic bromide. Moreover, Grignard reagent reacts with oxygen giving peroxide species which are converted into corresponding alcohols on hydrolysis. Mechanism:  The formation of Grignard reagent by the treatment of alkyl halides with magnesium metal occurs via free radical mechanism.

2R – X + Mg



R – R + MgX2



Mg + MgX2



2 Mg•X



R – X + Mg•X



R• + MgX2



R• + Mg•X



RMgX

Role of Ether:  Grignard reagent is usually prepared in diethyl ether (CH3CH2OCH2CH3) as solvent. The two Ether Oxygen atoms complex with the Magnesium atom, stabilizing the reagent.

15.2 Physical properties

1. Non volatile, colorless solids 2. They are explosive in nature, therefore, for synthetic purpose the Grignard reagent are always prepared and used in ether solution.

15.3  Chemical properties The C–Mg bond in Grignard reagent is covalent but highly polar. The alkyl groups in Grignard reagent being electron rich can act as carbanion or nucleophile. They would attack polarized molecules at point of low electron density. Thus characteristics reactions of Grignard reagent are nucleophilic substitution and nucleophilic addition reactions. Nucleophilic substitution:  The Grignard reagent undergo substitution reaction by Following mechanism: R–MgX + A – B

R – A + MgX (B)

Nucleophilic addition reaction:  Carbonyl compounds are attacked by Grignard reagents to form addition products which on acid–hydrolysis yield alcohols, acids, and esters etc. In addition reaction positive part of the Grignard reagent combines with the oxygen atom of the carbonyl group. The negative part of Grignard reagent goes to the carbon atom of the carbonyl group.

10.46  Engineering Chemistry

15.4 Synthetic application of Grignard reagents 1. Formation of Alkanes:  The compounds like water, alcohol and amines which contain active hydrogen react with Grignard reagent to form alkanes.





RMgX + H–O–H

R–H + Mg(OH)X.



CH3MgCl + HOH

CH4 Mg(OH)Cl.

2. Reaction with alkyl halides:  Grignard reagent reacts with alkyl halides to form higher alkanes/alkenes.





CH3 I – MgI + CH3CH2 – Br

CH3CH2 CH3 + MgI (Br)

3. Formation of higher alkynes:  Terminal alkynes react with Grignard reagent to form alkylmagnesium halides which on subsequent treatment with alkyl halides form higher alkynes.



4. Reaction with epoxides:  Grignard reagent react with carbons of epoxides which results in ring opening and formation of an alcohol. Reaction occurs at the least substituted ring carbon of the epoxides.



5. Reaction with aldehyde and ketones to give primary/secondary and tertiary alcohols. (a) Formaldehyde reacts with Grignard reagent to give addition products which on protonation yield – primary alcohols.



(b) Other aldehyde reacts with Grignard reagent to give addition products which on protanation yield secondary alcohols.



Polymers  10.47

(c) Reaction with Ketones:  Grignard reagent reacts with ketones to give addition products which on protonation form tertiary alcohols.



6. Formation of Ketones:  Grignard reagent reacts with acid chlorides and cyanides to form Ketones. (a) Reaction with acid chlorides:



(b) Reaction with cyanides:  Grignard reagent reacts with cyanides to give addition products which on hydrolysis with dilute HCl yield ketones.

7. Formation of carboxylic acid:  Grignard reagent reacts with carbon dioxide to give addition products which on protonation yield carboxylic acids.

8. Formation of organometallic compounds:  Grignard reagent reacts with inorganic halides to form other organometallic compounds.

4C2H5MgBr + 2PbCl2

(C2H5) 4Pb + 4MgBr (Cl)



4CH3MgI + SiCl4

(CH3) 4Si + 4MgI (Cl)

solved Questions 1. Give common name and IUPAC name of monomer of natural rubber. Ans. Isoprene, 2-1,3-butadiene 2. What is a copolymer?

10.48  Engineering Chemistry Ans. Polymer whose repeating structural units are derived from two or more types of monomer units are called copolymers. For example Polyester, Bakelite, SBR etc. 3. What does PMMA stands for? Ans. Poly(methyl methacrylate). 4. Give the chemical name for teflon. Ans. Poly(tetrafluoroethylene). 5. Write the monomers of Bakelite. Ans. Phenol and formaldehyde. 6. Define addition polymers. Ans. The molecules of same or different monomers simply add on one another leading to formation of a macromolecule in which the molecular formula of the repeating unit is the same as that of the starting monomer. The polymer thus formed is called addition polymers. 7. Could a copolymer be formed in both addition and condensation or not? Explain with example. Ans. Copolymers can be formed both by addition and condensation polymerization. For example, styrene butadiene rubber is a copolymer which is obtained by addition polymerization while nylon 66 is a copolymer obtained by condensation polymerization. 8. What is the function of sulphur in vulcanization of rubber? Ans. During vulcanization, natural rubber is heated with suphur. The function of sulphur is to introduce sulphur cross links between polymer chains thereby imparting more tensile strength, elasticity and resistance to abrasion. 9. Write names of monomers of terylene or Dacron. Ans. Ethylene glycol and terephthalic acid or it is methyl ester. 10. What is thermoplastics polymer? Ans. Polymers which softens on heating and hardens on cooling. They have a linear polymer and different chains are held by weak vander waal’s force of attraction. 11. Give examples of thermosetting polymers. Ans. The examples of thermosetting polymers are Bakelite, Urea formaldehyde resin, Melamine etc. 12. On basis of molecular forces, how will you classify polymers? Ans. On the basis of force of attraction polymers are classified as elastomers, fibres and plastics. 13. What are monomers of SBR? Ans. The monomers of SBR are butadiene and styrene. 14. What are conducting polymers? Ans. Conventional polymers are bad conductors of electricity but polymeric material can be synthesized which possess electrical properties as per metallic conductors. Such polymers are called conducting polymers. Their electrical conductivity is as high as 1.5 × 107 ohm–1m–1. 15. What are organometallic compound? Ans. The term organometallics is usually restricted to those compounds in which metal atom is bonded directly to a carbon atom of atleast one organic group.

Polymers  10.49 16. What are Grignard reagent? Ans. Grignard reagents are synthetic reagents used to prepare almost all classes organometallic compounds. The general formula for Grignard reagent is R-Mg-X. 17. Give the formula of ferrocene. Ans. Fe (C5H5)2. 18. What is wilkinson’s catalyst? Ans. Formula for wilkinson’s catalyst is [(PPh3)]RhCl and is used as catalyst for selective hydrogenation of certain double bonded compounds. 19. What is PAN? Ans. Poly acrylonitrile. 20. What is kelvar? Ans. It is an aromatic polyamide similar to nylon. It is prepared by condensation polymerization of terephthalic acid dichloride and 1,4 diaminebenzene.

unsolved Questions

• • • • • • • • • • • • • • • • • • • • • •

What does PVC stand for? Name a synthetic polymer which is an ester? Name the polymer which is used for making non stick utensils. How is nylon 66 synthesized? Give the name and structure of neoprene rubber. Give the monomers of SBR. Why is Bakelite a thermosetting polymer? Give chemical name of Teflon. What does PMMA stands for? Write equation for the preparation of poly acrylonitrite. How is nylon 6, 10 prepared? What is the difference in the structure of natural rubber and gutter percha? Define Fibres. Differentiate between addition and condensation polymerization. Give monomers of glyptal. What are organic polymers? Give the structure of few organic polymers. What is the difference between a monomer and a polymer? Classify polymers on the basis of occurrence and chemical nature. What are graft and block copolymers? What is an addition polymer? Give five examples. What are vinyl monomers? Define the terms: (a) Degree of polymerization and functionality.

10.50  Engineering Chemistry

• Distinguish between (i) Thermoplastics and thermosetting (ii) Homopolymers and copolymers (iii) LDPE and HDPE • Give the structure of following polymers: (a) Nylon 6, 6 (b) Terylene (c) Orlon (d) Chloroprene • What are biopolymers? Give three examples. • Write two uses of the following polymers: low density polyethylene, polystyrene, Nylon 6, epoxy resin. • Write structures of phenol formaldehyde resin, urea formaldehyde resin, polyamide. • Explain the mechanism of the polymerization of ethylene monomers on Zieglarnatta catalysts. • Write a note on conducting polymers. • Write the mechanism of cationic and anionic polymerization. • How is Bakelite formed? Explain the reactions with equation? • Describe vulcanization of rubber. • Explain the structure of ion exchange resin.

Chapter

11

Concepts of Titrimetric Analysis 1.  Introduction The term titrimetric analysis refers to quantitative chemical analysis carried out by determining the volume of a solution of accurately known concentration which is required to react quantitatively with a measured volume of a solution of the substance to be determined. The solution of accurately known strength is called the standard solution. The weight of the substance to be determined is calculated from the volume of the standard solution used and the chemical equation and relative molecular masses of the reacting compounds. The term volumetric analysis was formerly used for this form of quantitative determination but it has now been replaced by titrimetric analysis. It is considered that titrimetric analysis expresses the process of titration better and ‘volumetric analysis’ is likely to be confused with measurements of volumes, such as those involving gases. In titrimetric analysis the reagent of known concentration is called the titrant and the substance being titrated is termed the titrand. The alternative name has not been extended to apparatus used in the various operations; so the terms ‘volumetric glassware’ and ‘volumetric flasks’ are still common, but it is better to employ the expressions ‘graduated glassware’ and ‘graduated flasks’. The standard solution is usually added from a long graduated tube called a burette. The process of adding the standard solution until the reaction is just complete is called a titration, and the substance to be determined is titrated. The point at which this occurs is called the equivalence point or the theoretical or stoichiometric end point. The completion of the titration is detected by some physical change, produced by the standard solution itself (e.g., the faint pink color formed by potassium permanganate) or, more usually, by the addition of an auxiliary reagent, known as an indicator, alternatively some other physical measurement may be used. After the reaction between the substance and the standard solution is practically complete, the indicator should give a clear visual change (either a color change or the formation of turbidity) in the liquid being titrated. The point at which this occurs is called the end point of the titration. In the ideal titration the visible end point will coincide with the stoichiometric or theoretical end point. In practice, however, a very small difference usually occurs, this represents the titration error. The indicator and the experimental condition should be selected so the difference between the visible end point and the theoretical end point is as small as possible.

11.2  Engineering Chemistry

For using in titrimetric analysis a reaction must fulfill the following conditions: (i) There the reaction should be relatively fast. (Most ionic reactions satisfy this condition). In some cases the addition of a catalyst may be necessary to increase the speed of reaction. (ii) There must be an alteration in some physical or chemical property of the solution at the equivalence point. (iii) An indicator should be available which, by a change in physical properties (color of formation of a precipitate), should sharply define the end point of the reaction, if no visible indicator is available, the detection of the equivalence point can often be achieved in other ways. (a) Measuring the potential between and indicator electrode and a reference electrode (potentiometric titration). (b) Developing the titrant by electrolysis (coulometric titration). (c) Measuring the current which passes through the titration cell between an indicator electrode and a depolarized reference at a suitable applied e.m.f. (amperometric titration). Titrimetric methods are normally capable of high precision and wherever applicable they posses obvious advantages over gravimetric methods. They need simpler apparatus and are generally performed quickly; any tedious and difficult separation can easily perform. There are many areas in which titrimetric procedures are invaluable. The main advantages of these methods include– (i) The precision (0.1%) is better than most instrumental methods. (ii) Methods are usually superior to instrumental techniques for major component analysis. (iii) When the sample throughout is small, simple titration are often preferable. (iv) Unlike instrumental methods, the equipment does not require constant recalibration. (v) Methods are relatively inexpensive with low unit costs per determination. (vi) They are often used to calibrate and/or validate routine analysis using instruments. (vii) The method can be automated. There are, however, several disadvantages to classical titrimetric procedures, these include– (i) They are normally less sensitive and frequently less selective than instrumental methods. (ii) When a large number of similar determinations are required, then instrument methods are usually much quicker and often cheaper than the more labor intensive titrimetric methods. Nevertheless, despite the widespread popularity of instrumental techniques, it can be seen from the above discussion we find that there is considerable scope for the use of titrimetric procedures, especially for practicing laboratory skills.

2.  Fundamentals of Titrimetric Analysis In titrimetric analysis reactants are taken in the form of solution and the volume of standard solution required to react completely with a known (definite) volume of the unknown solution is determined. The strength of the unknown solution can be determined using normality equation. Common Terms used in Titrimetric Analysis Following are some commonly used terms in volumetric analysis: 1. Titration is the process of adding the standard solution from burette to the known volume of unknown solution until the chemical reaction is just complete.

Concepts of Titrimetric Analysis  11.3

2. Standard solution is a solution of definite concentration or a solution of known strength. 3. Standardization is a process where concentration of a solution is determined by the known concentration of a solution. 4. Titrant is the unknown solution. It is generally taken in a conical flask. 5. Titrand is a solution of accurately known concentration. This standard solution is generally taken in a burette and run into an unknown solution till the reaction is just complete. 6. Indicator is a substance which helps in the visual detection of the completion of the reaction in the titration by undergoing a change in its color forming turbidity or precipitate. 7. End Point is the exact stage at which the indicator gives its visual change. 8. Equivalence or stoichiometric point is the point at which the reaction is just complete. 9. Titration error is the smallest difference between equivalence point and end point. This difference is because an indicator always produces the visual change either a little before or after the equivalence point.

Concentration or Strength of Solution:  Concentration or strength of a solution if the mass of a solute dissolved in the unit volume of a solution. It is expressed in several ways. The most common of them are as under: 1. Grams per unit volume, e.g. g/l, mg/ml. etc. If 5g of solute is dissolved in 250ml of solution, the strength is 20g/l. 2. Percentage consists of expressing strength in terms of grams or ml of solute per 100g or 100ml of solution. For example, if WA is the mass of the compound A and WB is the mass of component B in a solution, then mass percentage of

WA A = ________ ​      ​ × 100 WA + WB

Volume percentage in case of a liquid dissolved in another liquid. It is convenient to express the concentration in volume percentage. The volume percentage is defined as the volume of the component per 100 parts by volume of the solution. Sometime, we express the concentration as weight/volume. For example, a 10% solution of sodium chloride (weight/volume) means that 10g of sodium chloride are dissolved in 100ml of solution. 3. Parts per million, when a solute is present in very minute amounts (trace quantities), the concentration is expressed in parts per million abbreviated as ppm. It is the parts of a component per million parts of the solution. It is expressed as ppm

Mass of component A A = ___________________ ​           ​ × 106 Total mass of solution

For example, suppose a litre of public supply water contains about 3 × 10 –3g of chlorine. The mass percentage of chlorine is 3.0 × 10 –3 ​ _________  ​    × 100 = 3 × 10 –4 1000 The parts per million (ppm) parts of chlorine is 3 × 10 –3 × 106 = ​ _____________  ​      =3 1000

11.4  Engineering Chemistry Thus, instead of expressing concentration of chlorine as 3 × 10 –1% it is better to express as 3 ppm. Atmospheric pollution in cities due to harmful gases is generally expressed in ppm though in this case the values refer to volumes rather than masses. For example, the concentration of SO2 in Delhi has been found to be as high as 10 ppm. This means that 10 cm3 of SO2 are present in 106 cm3 (or 103L) of air. 4. Molarity:  it is the number of moles of solution dissolved per litre of the solution. It is represented as ‘M’. Thus, a solution which contains one gram mole of the solute dissolved per litre of the solution is regarded as one molar solution. For example, 1M NaCO3 (molar mass = 106) solution has 106 g of the solute present per litre of the solution.

Moles of solute Molarity = ​ _______________________         ​ Volume of solution in litres

It is convenient to express volume in cm3 or ml so that

Moles of solute Molarity = ​ ____________________________           ​ × 1000 Volume of solution (in ml or cm3)

( 1 Litre = 1000ml)

If nB moles of solute are present in Vml of solution, then nB Molarity = ​ ___ ​  × 1000 V Moles of solute can be calculated as

Mass of solute Moles of solute = ​ ______________________         ​ Molecular weight of solute

Molarity is one of the common measure of expressing concentration which is frequently used in the laboratory. However, it has one disadvantage it changes with temperature because of expansion or contraction of the liquid with temperature. 5. Molality:  It is the number of moles of the solute dissolved per 1000g (or 1kg) of the solvent. It is denoted by m. Mathematically:

Mass of solute Molality (m) = ​ ____________________         ​ Weight of solvent in Kg Moles of solute Or = ​ ___________________         ​ × 1000 Weight of solvent gam

If nB moles of solute are dissolved in W gram of solvent, then nB Molality = ​ ___  ​ × 1000 W From the discussion of molarity and molality, if is evident that in molarity we consider the volume of the solution white in molality we take the weight of the solvent. Therefore, the two are never equal. Molality is considered better for expressing the concentration as compared to molarity because the molarity changes with temperature because of expansion or contraction of the liquid with

Concepts of Titrimetric Analysis  11.5



temperature. However, molality does not change with temperature because mass of the solvent does not change with change in temperature. 6. Mole Fraction: It is the ratio of number of moles of one component to the total number of moles (solute and solvent) present in the solution. It is denoted by x. Let us suppose that a solution contains n A moles of solute and nB moles of the solvent. Then,

Mole fraction of solute

nA (x A) = ​ _______     ​ n A + nB

Moe fraction of solvent

nB (xB) = ​ _______     ​ n A + nB

The sum of mole fraction of all the components in solution is always equal to one as shown below: nA nB x A + xB = ​ _______     ​ + ​ _______     ​ = 1 n A + nB n A + nB Thus, if the mole fraction of one component of a binary solution is known, than that of the other can be calculate. For example, the mole fraction x A is related to xB as

x A = 1 – xB or xB = 1 – x A

It may be noted that the mole fraction is independent of temperature. 7. Normality:  It is number of gram equivalents of the solute dissolved per litre of the solution. It is denoted by N.

Number of gram equivalents of solute Normality (N) = ​ ________________________________            ​ Volume of solution in litres

Or

Number of  gram equivalents of solute Normality (N) = ​ ________________________________           ​ × 1000 Volume of solution in ml

Gram equivalent of solute can be calculated as

Mass of solute Gram equivalents of solute = ​ ______________        ​ Equivalent mass

Like molarity, normality of a solution also changes with temperature. Relationship between normality and molarity of solution. The normality and molarity of solution are related as For acids,

Molar mass Normality = Molarity × ​ ______________        ​ Equivalent mass Normality = Molarity × Basicity

Where basicity is the number of H+ ion that a molecule of an acid can give in solution. For bases, Normality = Molarity × Acidity Where acidity is the number of OH– ions that a molecule of base can give in solution.

11.6  Engineering Chemistry

8. Formality:  It is number of formula masses of the solute dissolved per litre of the solution. It is represented by F.



Number of formula masses of solute Formality = ​ _______________________________           ​ Volume of the solution litre

The term formality is used to express the concentration of ionic substance. The ionic compounds such as NaCl, KNO3.CuSO4 etc. do not exist as discrete molecules. The sum of the atomic masses of various atoms constituting the formula of the ionic compound is called gram formula mass instead of molar mass. Types of Titrimetric Analysis 1.  Acid Alkali, Acid–Base or Neutralization Titrations:  These are those analysis in which the two solutions used are acid and alkali respectively. These reactions involve neutralization i.e. the combination of hydrogen and hydroxyl ions to from water. H+ + OH– Æ H2O The process of determining the strength of unknown by titrating with a standard alkali solution is called acidimetry. Acid base of pH indicators are either weak organic acids or weak organic bases. They possess different colors in dissociated and undissociated forms and also in different pH conditions. Phenolphthalein and methyl orange the most commonly used pH indicates. 2.  Oxidation-Reduction or Redox Titrations:  These are those titrations in which the two solutions used are the solutions of oxidizing and reducing agents respectively. The important oxidants are KMnO4, K 2CrO7, Ce(SO4) 2, I2, K 2BrO3 and K 2IO3 while the common reductants are FeSO4, (NH4)2SO4.6H2O (Mohr’s salt), SnCl2, Na2S2O3, As2O3 and TiCl3. The reduction of KMnO4 and K2Cr2O7 by FeSO4 may be represented as;

MnO4 – + 8H+ + 5e – Æ Mn2+ + 4H2O

(Reduction)

(Oxidant)



[Fe2+ Æ  Fe3+ Æ e –] × 5



MnO4 – + 8H+ + 5e2+ Æ Mn2+ + 4H2O + 5Fe3+

(Oxidation)

(Reductant)

(Redox Reaction)

The indicators used in redox titrations are called redox indicator. They are mainly of two types: (i) Specific redox indicators:  They react specifically with oxidizing or reducing agents, e.g., Starch, thiocyanate, etc. Starch gives a dark blue color with iodine, an oxidant. Thiocyanate gives a blood red color with ferric ion which is a reductant. (ii) True redox indicators:  They exhibit difference colors in the oxidized and reduced forms. Therefore, they mark sudden change in the oxidation potential near the equivalence point in a redox titration, just as acid base indicators mark the sudden change in pH during acid base titrations. Ferroin, Methylene blue, Ferricyanide etc. are examples of some redox indicators. (iii) Iodine Titrations (Iodimetry and Iodometry):  They are also redox titrations. They are studied separately for the sake of simplicity. Titrimetric analysis involving the use of iodine is of two types.

Concepts of Titrimetric Analysis  11.7

(a) Direct method or Iodimetric method involve the use of a standard iodine solution to titrate easily oxidizable substance, e.g., the titration of AS2O3 and I2. AS2O3 + 2I2 + 2H2O

AS2O5 + 4HI

They are of limited application because I2 is a weaker oxidant than MnO4, Cr2O7–2 and Ce +4 salts. (b) Indirect method or Iodometric technique are those titrations in which I2 liberated form iodide (say KI) is titrated against standard hype or sodium thiosulphate (Na2 S2 O3) solution. e.g. the titration of Cu SO4 and hypo of K2 Cr2 O7 and hypo of KMnO4 etc. 2Cu2+ + 2 I– Æ 2 Cu+ + I2 ≠ Cr2O7–2 + 14 H+ + 6 I– Æ 2 Cr3+ + 7 H2O + 3I2 ≠ 2S2O32– + I2 Æ I2 S4O62– + 2 I– Iodine solution can be used as a self indicator. However to get a better result 1% such starch solution is used as an indicator with which even traces of I2 can give a deep blue color due to the formation of an adsorption complex called starch iodide, (C24H40O20)I3. (iv) Precipitation Titrations or Precipitimetry:  Precipitimetry involves the combination of cations and anions (other than H+ and OH–) so as to produce a precipitate or turbidity due to the formation of an insoluble salt and the strength of unknown solution is determined by complete precipitation with the help of standard solution. e.g. The titration of AgNO3 and KI, Na2CrO4, KCNS or NaCl, where AgI (yellow), AgCl (white) precipitate out. Precipitation Titration is also called argentometric titrations as the one of the solutions of either known or unknown concentration has Ag+ ions. Three types of such titrations are mentioned below: (a) Mohr’s Method:  Mohr’s procedure was given in the year 1856 for the determination of CI– or Br– in neutral solution. It is a little inferior to Fajan’s and Volhard’s method for the determination of CI– ions. (b) Volhard’s Method:  It was given in the year 1878 for the determination of Ag in diluted HNO3 by direct titration with standard KSCN or NH4SCN solution in the present of Fe3+ ions or ferric alum as indicator. The end point is judged by the formation of a soluble blood red colored [Fe (SCN)3] with the Ist slight excess of thiocyanate ions. The Cl– or Br– ions are detected indirectly by this method. (c) Fajans’s Method  involves the chloride content determination of a sample by direct titration with standard AgNO3 solution using dichlorofluorescein as an adsorption indicator. The formation of a clear supernatant solution of a colored complex ion on a colored secondary precipitate or of a colored adsorption compound indicates the end point in precipitimetry. (v) Complexometric Titration or Compleximetry:  They are the titrations involving the following formation of a complex ion. Only a few reactions of complexation can be used as a basis for volumetry partly due to the lack of s suitable indicator and partly because in some cases varying mixture of products are obtained. A large number of estimations have been carried out by using complexing agents mostly ethylene diamine tetra acetic acid (EDTA).

11.8  Engineering Chemistry

3. Standard Solution In titrimetric analysis, the required solutions must be standard. A standard solution is one whose strength (concentration) is known. Standard solution of accurately known concentration need to be stored correctly. They may be classified into four types. (i) Reagents solution which are of approximate concentration. (ii) Standard solution which have a known concentration of some chemical. (iii) Standard reference solution which have a known concentration of a primary standard substance. (iv) Standard titrimetric solution which have a known concentration (determined either by weighing or standardization) of a substance other a primary standard. Primary and Secondary Standards:  In titrimetry certain chemicals are used frequently in defined concentration as reference solution. They are known as primary standards or secondary standards. Primary standard is a compound of sufficient purity from which a standard solution can be prepared by direct weighing of a quantity of it, followed by dilution to give a defined volume of solution. A primary standard should satisfy the following requirements:

(i) It must be easy to obtain to purity, to dry and to preserve in a pure state. (ii) The substance should be unaltered in air during weighing. This condition implies that if should not be hygroscopic, oxidized by air, or affected by carbon dioxide. The standard should maintain an unchanged composition during storage. (iii) The substance should be capable of being tested for impurities by quantitative and other tests of known sensitivity. (iv) It should have a high relative molecular mass so that the weighing errors may be negligible. (The precision in weighing is ordinarily 0.1 – 0.2 mg; for an accuracy of 1 part in 1000, it is necessary to employ samples weighing at least about 0.2 g).   In practice, an ideal primary standard is difficult to obtain, and a compromise between the ideal requirements is usually necessary. The substance commonly employed as primary standards are indicated below: Acid–Base reactions:  Sodium carbonate Na2 CO3. Sodium tetra borate Na2 B4 O7, potassium hydrogen phthalate KH (C8H4O4), Potassium hydrogen iodate [KH (IO3)2] Complex formation reactions:  Pure metals (e.g. Zinc, magnesium, copper and manganese) and salt depending upon the reaction used. Precipitation reactions:  Silver nitrate, Silver, Sodium Chloride, Potassium chloride, and Potassium Bromide. Oxidation-Reduction reactions: Potassium dichromate K 2Cr2O7, potassium bromate KBrO3, Potassium iodate KIO3, Potassium hydrogeniodate [KH (IO3)2], arsenic (III) oxide As2O3, and pure iron. A Secondary Standard is a substance which may be used for standardizations, and whose content of the active substance has been found by comparison against a primary standard. A secondary standard solution is a solution in which the concentration of dissolved solute has not been determined from

Concepts of Titrimetric Analysis  11.9 the weight of the compound dissolved but by reaction of a volume of the solution against a measured volume of a primary standard solution. Standardization of Solutions:  The process of determining the exact strength of a standard solution is known as standardization and may be accomplished in several ways. an accurately weighed amount of the chemical is transferred to a volumetric flask and diluted to the mark with solvent. This procedure is not generally used in preparing standard solution of acids and bases. The usual technique is to weigh an amount of solute appearing the exact quantity which, upon dissolving in water will give the desired normality. The exact normality and date of analysis are placed on the label of the container. The problem can be minimized in this instance by protecting the solution from carbon dioxide by using soda–lime absorption tubes. If a solution of definite normality is required, such as exactly 0.1N sulphuric acids, a solution is prepared which is slightly stronger than 0.1N. The solution is standardized and the diluted with the exact volume of water to produce the desired normality. The expression

ml × N (acid1) = ml × N (acid2)

is used to calculate the final volume of the solution which has the desired normality. The final solution is then standardized. In standardizing a solution, an accurately weighted sample of the primary standard is dissolved in water and titrated to the end point with the solution being standardized. From the volume of titrant consumed the sample weight of primary standard and its equivalent that is all quantitative procedures an accuracy of at least 1 part in 1000 burette readings should be maintained. Therefore, weighing should be carried out to the fourth decimal place, and burette readings should be made to the nearest 0.02ml.

4. Acid–Base Titrations Before discussing acid-base titrations we must be familiar with certain general concepts of acid and bases. These are described in following sections. Theory of Acid-Base Indicators:  In the acid alkali titrations, the equivalence point is detected by the use of substance called acid base indicators which change their color at or near the equivalence point. The actual color that produces in solution depends on the pH of the solution. The change in color of the indicator is not sudden and abrupt but takes place in a small pH range. Let us examine briefly how the acid-base indicators works. (i)  Ostwald Theory:  The first attempt in the theory of indicators was made by Oswald. The unionized form of the acid indicator HIn or the basic indicator inOH had one color and the corresponding ionized form has another colour. The ionization equilibrium in aqueous solution may be written as H3O + + In–



HIn + H2O



And

In OH



Let us apply the law of mass action to the ionization equilibrium of an acid indicator HIn. The ionization constant of the indicator is given by



OH– + In+

[H3O +] [In–] Kind = ​ __________     ​  [HIn]

11.10  Engineering Chemistry Or

[HIn] [H3O +] = K ind ​ _____  ​  [In–] [Unionized form] [H3O +] = K ind ​ _______________       ​ [Ionized form]

In acid solution presence of excess of H3O + ions will suppress the ionization of the acid indicator and [In–] will be small. On the other hand, in alkaline solution the ionization equilibrium is shifted to the right increasing the concentration. Indicator will exist mainly in the ionized form and the colour of the ionized form become apparent. (ii) Quinonoid Theory:  The colour in most of the naturally occurring organic compounds has been attributed to be due to the presence of quinonoid structure.

Concepts of Titrimetric Analysis  11.11 This may be illustrated changes by reference to phenolphthalein where (I) represents the structure of phenolphthalein which in the presence of dilute alkalis changes to structure (II). This monoanion then yield a red resonating dianion (III). If phenolphthalein is treated with excess of concentration of alkalis the red colour first produced disappears owing to the formation of carbinol form (IV). Similarly, the two forms of methyl orange are: Methyl orange in Presence of alkalis (Non quinonoid structure - yellow) O 3S

N

OH

O 3S

–

N

H 3O

N

N +

CH3 CH3

+ CH3

N

N –

H

CH3

Methyl orange in Presence of acids (Quinonoid structure - Red)



Table contains a selected list of some indicators along with their colour change intervals.

Table 1  Some indicators and their characteristic pH range. Indicator

pH range

Colour in Acidic solution

Coresol red m-cresol purple Methyl orange Methyl red Phenol red Cresol red Thymol blue Phenolphthalein Alizarin yellow

0.2 – 1.8 1.2 – 2.8 3.1 – 4.4 4.2 – 6.3 6.8 – 8.4 7.2 – 8.8 8.0 – 9.6 8.3 – 10.0 10.1 – 12.1

Alkaline solution

Red Red Red Red Yellow Yellow Yellow Colorless Yellow

Yellow Yellow Yellow Yellow Red Red Blue Red Orange red

By suitably mixing certain indicators the colour change may be made to extend over a considered pH range. Such a mixture is known as universal indicator. These are mostly employed for determining the appear pH of solution. It converse a pH range of 3 – 11 and gives colour changes at different pH values as shown below. pH

3

4

5

6

7

8

Colour

Red

Orange

Orange

Yellow

Yellowish

Greenish Blue Violet

9

10

11 Reddish

11.12  Engineering Chemistry Acid–Base Titration in Aqueous Media:  The changes in hydrogen ion concentration accompanying the addition of a base to an acid are important for analytical purpose. A plot of pH against the volume of base (or acid) added from the burette is called a pH titration curve. Changes in pH values in the neighborhood of the equivalence point are of importance and enable us to select an indicator which will give the least titration error. (a) Titration of a Strong Acid by a Strong Base:  Shows the variation of pH with the volume of alkali added during the titration of an acid like HCl with a base like NaOH. It will be observed that pH of the solution starts increasing slowly at first, then rise more rapidly until at the equivalence point there is a sharp increase in pH for small volumes of alkali added. After the equivalence point increase in pH is small on addition of excess alkali.

pH 7.0

ml of alkali added Fig. 1  Titration of 1.0M hydrochloric acid with 1.0M sodium hydroxide.



(b) Titration of a weak acid by a strong base:  The titration curve of weak acid with a strong base (CH3COOH Vs NaOH) is depicted in fig 2. As sodium acetate is a salt of a weak acid and a strong base if gives excess of OH– ions in aqueous solution thereby increasing the pH at equivalence point beyond 7. Hence phenolphthalein or thymol blue would be satisfactory indicators. For titration of acid like boric acid against sodium hydroxide the titration curve does not show any sharp rise in pH. The pH, therefore goes on increasing gradually without showing a sharp increase. Thus very weak acids can not be successfully titrated. 14 12 10 8 pH 6 7.0 4 2 0

40 80 ml of alkali added Titration of 100ml of 0.1 N acetic acid with 0.1 N sodium hydroxide 0

Fig. 2

Concepts of Titrimetric Analysis  11.13

(c) Titration of a Weak Base by a Strong Acid:  The titration curve of a weak base against a strong acid (ammonium hydroxide Vs hydrochloric acid) is depicted in fig. 3. The pH at equivalence point is this case lies in the acid range i.e., below 7. Thus the pH change at equivalent lies in range 6 to 4. Thus the indicator having pH range on acidic side i.e., methyl red or methyl orange can be used successfully.

Fig. 3



(d) Titration of a Weak Acid by a Weak Base:  The titration curve for a weak acid say, acetic acid, by a weak base like NH3 is shown in fig 4. Hence no sharp end point can be obtained with the commonly employed indicators. However, a mixed indicator which shows a sharp color change over a limited pH range may sometimes be used. Such titration is usually avoided. 14 12 10 pH

8 6 4 2 0

0

40 80 ml of alkali added Titration curve for the titration of acetic acid with ammonium hydroxide Fig. 4

11.14  Engineering Chemistry

5 Redox Titrations The chemical reaction involving oxidation reactions are more widely used in titrimetric analysis than acid base, complex formation, or precipitation reactions. Oxidation is defined as the loss of one or more electrons by an atom (deelectronation), molecule or ion, while reduction is the gain in electrons (electronation). Thus oxidation and reduction occur simultaneously. The redox reaction may also be referred to as electron transfer reactions. Consider the reaction, 2 HgCl2 + Sn Cl2 Æ Hg2Cl2 Ø + Sn Cl4 Here the element Sn is said to be oxidized from Sn2+ (in SnCl2) to Sn4+ (in SnCl4) i.e., an increase in oxidation state is oxidation. The element that causes oxidation must itself decrease in oxidation state, here the oxidation state of Hg has changed state +2(Hg2+ in HgCl2) to 1 (Hg+ in Hg2Cl2), thus decrease in oxidation state is reduction. Here the electrons pass from Sn2+ to Hg2+, hence SnCl2 is a reducing agent while HgCl2 is an oxidizing agent. If a reagent acts both as oxidizing as well as reducing agent it is said to be undergoing auto oxidation or disproportionation. Many redox reactions can satisfy the requirement for use in titrimetric analysis provided that equilibrium is reached immediately following each addition of titrant and also that the indicator is capable of locating the end point with a reasonable accuracy. In such titrations a table of standard electrode potential serves as a guide to equilibrium condition as no other information such as the rate or mechanism of reaction is known generally the course of a titration reaction is followed by plotting a graph of potential versus volume (cm3) of the titrant. Theory of Redox Titration Curves:  In constructing titration curves in redox reaction titration it is costomary to plot of graph of emf of E cell (against SCE) versus volume of titrant. This is necessary because most redox indicators are sensitive but they are themselves oxidizing or reducing agents and hence the change in potential of the system is considered during titration let us consider a typical titration curve by considering the potential of the oxidation reduction system at the equivalence point (Eeq). The nernst equation states. [reductant] RT E = E 0 – ​ ___  ​ log ​ _________    ​  nF [oxidant] For the simple reaction of Fe2+ + Ce4+

Fe3+ + Ce3+

At equilibrium the electrode potentials for the two half reaction is the same.

4+ ​E​Ce  ​  + ​EFe ​3+​ ​ = Esystem



This is the potential of the system. For the redox indicator the potential is the same as for the

4+ system Endicator = ​E​Ce  ​  + ​EFe ​3+​ ​ = Esystem



We have the half cell



SCE‌‌‌‌||Ce4+ Ce3+:Fe3+ Fe2+ || Pt Where SCE = Saturated calomel electrode



4+ We use ​E​Ce  ​  or ​EFe ​3+​  while calculating Esystem



At equivalence point we calculate potential thus.

Concepts of Titrimetric Analysis  11.15 And also by

Ce3+ Eeq = E0Ce4+ – 0.0591 ​ ____    ​ Ce4+

[Fe2+] Eeq = E0Fe3+ – 0.0591 log ​ _____    ​ 3+ [Fe ] Adding both equations [Ce3+] _____ [Fe2+] 2Eeq = E0Ce4+ + E0Fe3+ - 0.0591 log ​ ______    ​ ​     ​ 4+ 3+ [Ce ] [Fe ] Since at the equivalence point [Fe3+] = [Ce3+] and [Ce4+] = [Fe3+]

We reduce the equation to Eq = ​E​0  4+​ + ​E​0  +3​ ​  Ce

F​e​ ​

We can calculate the equilibrium concentration of the reacting species from the equivalence point potential.

Fig. 5

Some Oxidising Agents as Titrants: 1. Ceric Sulphate:  It is very good oxidizing agent used with 1 – 10 phenanthroline as an indicator Ce4+ Æ Ce3+ + e –. Electrons from the 4f orbital are removed. The rate of reaction is affected by the nature of the solvent and by the process of complex formation Ce (IV) during reaction exists as an anionic complex in media of H2SO4. HNO3 and HClO4. The stability is limited in HCl medium with formal potential of 1.88V. Sodium oxalate (Na2C2O8) is used as the primary standard for cerium (IV) in sulphuric acid.

11.16  Engineering Chemistry

2. Potassium Permanganate:  It is a powerful oxidizing agent no special indicator is required to note the end point. Its tendency to oxidize the chloride ion is the only limitation. It is employed in acid (0.1N) media as MnO4 – + 8H+. MnO4 – + 8H+ + 5e –

Mn2+ + 4H2O E0 = 1.51V.

The oxidation reaction is rapid only for H2C2O4 but it is slow at room temperature for others hence heating is required. The permanganate end point is not permanent and slowly fades out due to the reaction. 2MnO4 – + 3Mn2+ + 2H2O 5MnO2 + 4H+ Coloured Colourless The aqueous solution is not stable and oxidizes water in the following way.

4MnO4 – + 2H2O + 6H+

2Mn2+ + 10CO2 + 8H2O

3. Potassium dichromate:  This reaction proceeds as follows; Cr2O72– + 14H+ + 6e –   Cr3+ + 7H2O E0 = 1.33V. It has limited application in comparison with KMnO4 or Ce (IV) due to lesser oxidising characteristics and slowness of some of its reaction K 2Cr2O7 is indefinitely stable and inert towards HCl. It is mainly used for the analysis of iron (III) with reaction. 6Fe2+ + Cr2O72– + 14H+   6Fe3+ + 2Cr3+ + 7H2O There are several other applications for this titrant. 4. Potassium Bromate:  It is a strong oxidising agent. It reacts as BrO3– + 6H+ + 6e –   Br– + 3H2O E0 = 1.44V. In acidic solutions BrO3– reacts with Br– (aq) + 3H2O. KBrO3 is a primary standard and is stable indefinitely. Postassium bromate finds extensive application in organic chemistry e.g., titration with oxine. Most titrations involve back titrations with arsenous acid. 5. Potassium Iodate:  It has extensive use in analytical chemistry. IO3– + 5I– + 6H+ Æ 3I2 + 3H2O and the reaction in Andrew’s titration is IO3– + CI– + 6H+ + 4e – Æ ICI + 3H2O E0 = + 1.20V. Redox indicators:  Several types of indicators used in redox titrations are: (i) Self indicators:  The KMnO4 solution are quite deeply coloured and a slight of this reagent in a titration is easily detected. Thus in the titration of oxalic acid hydrogen peroxide etc., with KMnO4, as soon as the reaction is complete, and a drop of the latter is in excess, indicating that the reaction is complete and the end point has reached. (ii) Specific indicator:  This is a substance which reacts in a specific manner with one of the reagents in a titration to exhibit a colour. Thus starch produces a deep blue colour with iodine, SCN– (usually KSCN or NH4SCN are used) produces a red colour with Fe (III) ion. (iii) External (or Spot test) indicator:  These were usually employed when no internal indicators were known. For example, Fe(CN) 63–, Ferric cyanide ion is still used to detect Fe(II) ion by the formation of a deep blue green complex (Turnbull’s blue) on a spot plate outside the titration mixture. Thus K3Fe(CN) 6 is used in the titration of Fe(II) with K2Cr2O7 solution in acid medium 2[Fe(CN) 6]3– + 3Fe2+ Æ Fe3 [Fe(CN) 6] 2 Ferro–Ferri cyanide (deep blue to green precipitate)

Concepts of Titrimetric Analysis  11.17 Structure of Some Redox Indicators 1. Diphenylamine:  Diphenylamine was one of the earliest known internal redox indicators for the titration of Fe(II) with K2Cr2O7 solution. This indicator is soluble in water with difficulty and tungstate and Hg(II) chloride interfere with its actions. The reduced form is colourless and the oxidised form a deep violet. The mechanism of colour change has been shown as followes, using disphenylamine as an example.



2. Ferroin:  The well known redox is ferroin, the Fe(II) complex of 1, 10 – phenanthrolin:

Each N-atom in 1,10 phenanthroline possesses a lone pair of electrons which is shared with Fe(II) ion. Since the coordination number of Fe(II) is 6, three such organic molecules bind themselves to the Fe(II) ion to Fe(II) ion to form a blood red chelate complex ion. The colour of the Fe(III) complex is pale blue, and hence a sharp colour takes places when Fe(II) is oxidised to Fe(III) in the presence of 1,10 phenanthroline.

(Phen)3 Fe3+ + e

(Phen)3 Fe2+ E0 = 1.06V.

Light Blue

Dark Red

The indicator is prepared by mixing equivalent quantities of Fe(II) sulphate and 1,10 phenanthroline. This complex chelate is ferroin.

11.18  Engineering Chemistry Complexometric Titrations Introduction:  The formation of a complex during a chemical reaction may serve as the basis of a titrimetric assay. The complex formed should be soluble, stable (but slightly disscociated) and posses a stoichiometric composition at the equivalence point (complexometric titrations). The groups (anion/ neutral) attached to the central cation (or a metal) are called ligands. The number of bonds formed by the central cation, is called the coordination number of the metal. A complex formation reaction may also be viewed as a Lewis acid-base reaction. For example, the formation of a stable complex ion Ag(CN)2–:

The bond formed between the central cation and the ligand is usually covalent but the bonding interaction may be coulombic attraction also. Ligands such as H2O, NH3, CN– and Cl– etc. Which possess only one electronegative donor atom (O, N, S, C or Cl– etc), containing only one unshared pair of electrons are called as bi, tri, penta, hexa, dentate ligands respectively. The process of ring formation is called chelation. Thus the chelation between Cu2+ and two molecules of ethylenediamine is:

Chelating Agents:  It may be noted that in the formation of [Cu (NH3) 4] 2+ not all of the added ammonia is used up in one step to form this complex, rather all the lower complex species such as, CuNH32+, Cu (NH3)22+, Cu (NH3)32+ are still present in appreciable concentration [i.e., these have not changed into Cu (NH3) 42+]. Schwarzenbach (1945) realized that the difficulty arising from the formation of lower complexes can be overcome by the use of chelating agents as titrant. Thus, for example, the compound triethelene (trien), can satisfy co-ordination number 4 of copper in one step. CH2 H2N

CH2 NH CH2

Cu NH CH2

H2N CH2

CH2

2+

Concepts of Titrimetric Analysis  11.19 However, only a few metals such as cobalt, nickel, zinc, cadmium and Hg (II) form stable complexes with ligands as trien. Schwarzenbach noted that certain chelating agents containing both oxygen and nitrogen are especially effective in forming stable chelates with a wide variety of metals. The most important of these is ethylene diamimetetra acetic acid, EDTA with the structure.

Since it has been shown that in EDTA, two N–atoms are held in the form of Zwitter ions, EDTA is potentially a hexadentate ligand which may co-ordinate with a metal ion through its two N-atoms and 4-carboxyl groups. IR spectra and other measurements indicate that EDTA forms octahedral complexes with metal ions such as CO2+, Ca2+, Cu (II) and Fe (II) etc. The structure of CO2+ -EDTA complex is:

To void its long name, usually abbreviation likes H4Y (Free acid form), H4Y– , H2Y22– are used and then the complexes because CoY2–, CuY2– and FeY2– etc. In some cases EDTA may function as a quinquedentate or quadridentate ligand with one or two of its carboxyl groups remain free not attaching to the metal. In general the reaction with M2+ ion may be:

M2+ + H2Y2–

MY2– + 2H+



M3 + H2Y2–

MY – + 2H+



Mn + H2Y2–

(MY)(n–4)+ + 2H+

11.20  Engineering Chemistry The disodium salt, Na2H2Y, affords the complex forming ion H2Y22– in aqueous solution, it react with all the metals in a 1:1 ratio. In each case, one mole of the complex forming ion H2Y2– reacts with one mole of the metal ion and releases two moles of hydrogen ion. Titrations involving EDTA as titrate are referred to as chelometric titrations. The more stable the complex, the lower the pH at which an EDTA titration of the metal ion under study may be performed. Below are indicated minimum pH values for the existence of selected metal EDTA complexes.

Table 2 Minimum pH at which complexes exist 1–3 4–6 8 – 10

Metals Zr4+, HF4+, Th4+, Bi3+ and Fe3+ Pb, Cu, Zn, Co, Ni, Mn, Fe, Cd, Sn (all bivalent) and Al3+ Ca2+, Sr2+, Ba2+, and Mg2+

It is thus evident in general that M2+ EDTA complexes are stable in alkaline or slightly acidic solution, while M3+ or M4+ EDTA complexes exist in solutions of much higher acidity. Titration Curves:  Complexometric titration curves are constructed and consists of a plot of the negative logarithm of the metal ion concentration (pM) versus milliliters of titrant; a point of inflexion occurs at the equivalence point; in some cases this sudden increase may exceed 10PM units. These are analogous to acid base titration curves and likewise are helpful in knowing the feasibility of a complexometric titration and in choosing a proper metal ion sensitive indicator.

Fig. 6

Such indicators form complexes with specific metal ions, which differ in colour from the free indicator and hence a sudden change of colour occurs at the equivalence point.

Concepts of Titrimetric Analysis  11.21 Types of EDTA Titrations:  EDTA titrations have been performed for nearly all common cations. The important types of EDTA titrations of metal ions are: 1. Direct Titrations:  The direct titration with EDTA may be performed on at least 25 cations using metallochromic indicators. The metal ion in solution is buffered to the desired pH (say to pH = 10 with NH4 + aqueous NH3) and titrated directly with the standard EDTA solution. The total hardness of water (Ca + Mg) is determined by direct titration with EDTA using Eriochrome Black T or calmagite indicator. 2. Substitution (or Replacement) Titrations:  Such titrations are used for metal ions. Which do not react with a metal indicator, or for metal ions which form EDTA complexes that are more stable than those of other metals such as Mg2+ and Ca2+, The metal cation Mn+ to be determined is treated with Mg– EDTA complex, when the reaction occurs: Mn+ + MgY2–

MY(n – 4)+ + Mg2+

Mg2+ + EDTA2– Æ Mg – EDTA The amount of Mg2+ liberated is equivalent to the cation present and can be titrated with standard EDTA solution using a suitable metal indicator, such as calmagite. 3. Alkali Metric Titrations:  When a solution of disodium salt (EDTA), Na2H2Y, is added to a solution of metal ions, complexes are formed and two equivalents of hydrogen ions are set free. Mn+ + H2Y2–

[MY](n – 4)+ + 2H+

These liberated hydrogen ions can be titrated with a standard solution of NaOH using a usual acid base indicator to locate the end point. 4. Miscellaneous Methods: Silver and gold cannot be titrated complexometrically but by the exchange reactions between the Ni(CN) 42– ion and Ag+ or Au+, the Ni2+ ions are set free and hence can be determined: Ni (CN) 42– + 2Ag+

2[Ag (CN)2] – + Ni2+

Such reactions take place with sparingly soluble silver salts AgSCN, AgCl, AgBr and AgI etc., and hence provide a method for the determination of these anions. The anion is first precipitated as the silver salt and is then dissolved in a solution of Ni (CN) 42–, the equivalent amount of Ni2+ is set free and is determined by immediate titration with EDTA using murexide or bromopyrogallolred as an indicator. Indicators for Complexometric (EDTA) Titration:  At present a wide variety of good indicators are available and usually the visual titrations are easier to perform such indicators are also referred to as metallochromic indicators. The metal indicator complex should be sufficiently stable, otherwise due to dissociation, a sharp colour change is not obtained. However, the metal indicator complex must be less stable than the metal EDTA complex, thereby ensuring that at the end point, EDTA can remove metal ions from the metal indicator complex. Some important indicators of this type are described here.

11.22  Engineering Chemistry



1. Calmagite:  This is, 1 – (1 – hydroxyl – 4 – methyl – 2 phenylazo) – 2 – naphthol – 4 – sulphonic acid (II). Its structure (II) is given below which is similar to (I). This is stable in aqueous solution and may be used in place of Eriochrome Black T.



2. Murexide:  It was probably the first metal ion indicator to be used in EDTA titrations. It is the ammonium salt of purpuric acid and the anion has the structure (III). The 0.5g of dye staff is suspended in 100cm3 water, shaken well and allowed to settle. The clear supernatant liquid is used as the indicator.

Fig. 6

6. Precipitation Titrations Introduction:  The precipitation method is based on titration with the use of reactions accompanied by formation of sparingly soluble compounds. Although many such reaction are known, only a few of them can be used in titrimetric analysis. They must satisfy a number of conditions, namely: (a) The precipitate must be practically insoluble. (b) The titrations results should not be distorted appreciably by adsorption effects. (c) It must be possible to detect the equivalence point during the titration. These conditions restrict severely the range of reactions which are suitable for volumetric analysis. The most important methods are those based on precipitation of insoluble silver salts in accordance with the equation.

Concepts of Titrimetric Analysis  11.23 Ag+ + X– Æ AgX Ø Where X– represents Cl–, Br–, I–, CNS–, etc. Such methods constitute a special section of titrimetric analysis, known as argentometry. Certain other precipitation reaction are also sometime used; for example precipitation of Zn2+ as the complex salt K3Zn3 [Fe (CN) 6] 2, or of PO43– as the double ammonium uranyl phosphate (UO2) NH4PO4, etc. Precipitation titrations are closely allied to volumetric determinations based on react of complex formation, such as 2CN– + Ag+ Æ [Ag (CN2)] – Hg++ + 4I– Æ [HgI4] 2– Or reactions giving rise to weakly dissociated salts such as HgCl2, HgBr2, Hg (CNS)2, etc. In addition to precipitation of chlorides and bromides as AgCl and AGBr, they can be determined by the mercurimetric method by means of the reactions. 2NaCl + Hg (NO3)2 Æ HgCl2 + 2NaNO3 2KBr + Hg (NO3)2 Æ HgBr2 + 2KNO3 A serious obstacle to the use of many reactions of complex formation in volumetric analysis is the fact that the same cation and ligand can form complexes differing in composition, i.e, with different proportions of metal and ligand. This makes the reaction complex as in such cases they do not conform to the same stoichiometric equation. Principle of Precipitation:  solubility, which is dependent on the solvent and temperature, is the concentration of the dissolved solute in moles per litre, when the solution is in equilibrium with a solid solute. This repeating pattern depends upon the molecular structure of the compound. The solute molecules are held together in that pattern by intermolecular forces of attraction. Now in order to dissolve a solid, these forces of attraction must be overcome so that solute–solute attraction is replaced by solute–solvent attraction. During precipitation, however, the opposite condition is aspired for, where the intermolecular forces between the molecules of product are high and solute– solute forces replace the solute solvent forces. Solubility Product and Precipitation:  Consider an aqueous solution of a slightly soluble salt BA in equilibrium with excess of the solid at constant temperature. The equilibrium can be represented by: BA(s)

B+ + A–

Where BA(s) represents the solid phase, in dilute aqueous solution essentially no undissociated BA will be present in the solution. Since the activity of solid is constant, the equilibrium constant for (i) may be written as:

Ksp = [B]+ [A–]

Ksp is the solubility product which is a constant for solute solvent and temperature. In a more complex case

11.24  Engineering Chemistry Bm An(S) Here,

mBn + nAm–

Ksp = [Bn+] m [Am–] n

Solubility product is of importance as it permits the calculations of one of the ion concentration if the other is known. Factors Influencing Solubility of the Precipitate:  Precipitation is the most versatile method in gravimetric analysis. We mainly consider factors affecting solubility of precipitate. The important parameters are temperature nature of solvent interfering ions pH, effect of hydrolysis, effect of complexes etc. 1. Temperature:  Solubility increases with temperature. Better precipitation occurs in hot solution, but while filtering one must take cares not to filter the solution hot if precipitation is temperature dependent. 2. Nature of Solvent:  Inorganic salts are more soluble in water, decrease in solubility in the organic solvent can be used to separate two substances. 3. Effect of pH:  The solubility of the salt of a weak acid depends upon the pH of the solution e.g., oxalates where H+ ion combines with C2O4 –2 ion to from H2C2O4 thereby control of pH has been used in qualitative analysis e.g., metals forming less soluble sulphides (group ll) are precipitated by H2S in 0.10M HCl. Then pH is increased for precipitation of group III metals. 4. Effect of Hydrolysis:  If a salt of a weak acid is dissolved in water, it will result in a change in [H+], with the result that the cationic species of a salt undergo hydrolysis and this will also increase the solubility. 5. Effect of Complex:  The solubility of sparingly soluble salt is a function of the concentration of the substance which forms complexes with the cation of the salt e.g., NH3 influence AgCl precipitated. A reaction which yields a precipitate can be made the basis for titration, if the reaction is fast and quantitative, the end point is clear. Unlike in gravimetry we can’t afford to wait till precipitation is complete. Further the solubility product must be sufficiently low so that precipitation is quantitative within experimental error when only a drop of excess of reagent is added. All such reactions are generally classified on the basis of the type of indicators used to defect the end point. Factors Affecting the Solubility of Precipitates Titration Curves:  Titration with precipitating agent are used for estimation of analystes, provided the equilibria are rapid and method for detection of end point is available. Titration curves base on pH-values are useful for deducing the properties required of an indicator and titration error that is likely to cause. Consider the titration of chloride ions with standard silver nitrate solution. A titration curve is plotted by PCl against the volume of silver nitrates in figure 7. At the beginning of the titration the PCl is 1. As the addition of titrant continues, part of Cl– is removed from solution of AgCl precipitate and we can determine the concentration of Cl– remaining. It will be seen by examining the chloride ion concentration that there is a marked change in this value towards the equivalence point.

Concepts of Titrimetric Analysis  11.25

Fig. 7

Such precipitation titration in which silver nitrate is used as a reagent are of high importance and are called argentiometric titration.

Detection of End Points Mohr’s Method:  It involves the titration of sodium chloride with silver nitrate using dilute potassium chromate as indicator. Let us consider the titration of 0.1M NaCl with 0.1M AgNO3 in the presence of few ml of dilute K2CrO4 solution.

Ksp (AgCl–) = 1.2 × 10 –10 = [Ag+] [Cl–] Ksp (AgCrO4) = 1.7 × 10 –12 = [Ag+2] [CrO4 –2]

As the AgNO3 solution is added, AgCl will precipitate first. We usually expect the salt with smaller solubility product to precipitate first, but this is true only if both salts dissociate to yield the same number of ions. Mohr’s procedure also applies to the determination of bromide in the similar manner. It is necessary that Mohr’s method should be applied only to neutral or slightly alkaline solution i.e., within pH 6.5 to 9. In the acidic solution the following reaction occurs: 2CrO4 –2 + 2H+

2HCrO4 –

Cr2O72 + H2O5

Here, HCrO4 – acts as an acid. In markedly alkaline solutions, silver hydroxide (Ksp = 2.3 × 10 –8) may be precipitated. Fajan’s Method:  K.Fajan, through his studies on nature of adsorption, introduced a useful type of indication for precipitation titrations, these indicators are either acid dyes e.g. fluorescein, eosin, etc. or basic dyes e.g., rhodamine series. The property a colloidal precipitate to adsorb its own ions which are in excess is made use of in this case. When a sodium chloride solution is titrated with silver nitrate, the silver chloride precipitates with chloride ions which are initially in excess.

11.26  Engineering Chemistry Thus, the chloride ions form the primary adsorbed layer, which in turn will hold the secondary adsorbed layer of oppositely change Na+ ions. –

NO3 +

Na

Na

Cl +

–

Cl

–

Cl AgCl Cl

–

Na

+

+

–

NO3

–

Na

+

– Fluoresceinate +

– Fluoresceinate

Ag

+

Ag

Ag – AgCl NO3 + + Ag Ag –

NO3

+

Ag AgCl + + Ag Ag

– Fluoresceinate

– Fluoresceinate

solved Questions 1. What is alkalinity? Ans. Alkalinity is amount of OH– present in a given water sample which can be increased by hydrolysis and dissociation of carbonates and bicarbonates, respectively 2. Give types of neutralization titration. Ans. Neutralization titrations are of four types– Strong acid and Strong base Strong acid and weak base Weak acid and Strong base Weak acid and weak Base 3. What is buffer solution? Ans. Buffer solution resists the change in pH. 4. Give the complexes formed in EDTA method for softening water. Ans. Two complexes are formed in EDTA method first is of wine red color formed between EBT and cations causing hardness and second is formed between EDTA and cations causing hardness. 5. Why precipitation titration is also called argentometric titration? Ans. One of the two solutions in precipitation titration is having silver ions in it so precipitation is also known as argentometric. 6. Give the conditions for precipitation to take place. Ans. For precipitation to take place ionic product should be greater than solubility product. 7. What is the pH range of phenolphthalein and methyl orange? Ans. pH range of phenolphthalein is 8.1 to 9.8 and methyl orange is 3.1 to 4.4.

Concepts of Titrimetric Analysis  11.27 8. Why is internal redox titration better than external redox titration? Ans. Internal redox titration is better than external redox titration because in external redox titration the solution containing Fe +2 is wasted as we put the glass rod into the solution to check the progress of titration. 9. Give the role of KI in iodine titration. Ans. Iodine is sparingly soluble in water so KI is added which reacts with iodine forming a complex KI3 which liberates iodine during course of reaction. 10. What is EDTA? Ans. EDTA is Etheylene diamine tetraacetic acid. It is a polydentate ligand which functions as a complexing agent and it has 6 donar atoms. 11. Why acid is added in iodine titration? Ans. Acid is added in iodine titration to liberate iodine. 12. Why phenolphthalein is colorless in acidic medium? Ans. Phenolpthalein is colorless in acidic medium due to common ion effect of H+ ion of acid. 13. What is the difference between end point and equivalence point? Ans. End point is the point where physical changes are observed (color of indicator changes) and equivalence point is the point where a reaction gets completed. 14. Which precipitate is first formed and why in precipitation titration. Ans. White precipitate of AgCl is first formed because its solubility product is smaller than silver chromate which is the second precipitate formed in the titrations.

UNsolved Questions

1. Distinguish between qualitative and quantitative analysis. 2. Explain types of titration. 3. Distinguish between (a) End point and equivalence point (b) Primary standard solution and secondary standard solution (c) Acidimetry and alkalimetry (d) Self, internal and external indicator. 4. Give pH range of phenolphthalein and methyl orange. 5. Distinguish between iodometry and iodimetry. 6. Give the structure of EDTA and EBT. 7. Give the structure of phenolphthalein and methyl orange. 8. Explain Ostwald’s theory of acid base indicators. 9. Gibe types and examples of acid base titrations. 10. What is precipitation titrations? 11. Why do we use buffer in complexometric titration? 12. Explain Redox titration. 13. What are colors of AgCl and Ag2CrO4 in precipitation titration?

11.28  Engineering Chemistry

14. 15. 16. 17.

Distinguish KMnO4 titration and K2Cr2O7 titration. What is double salt? How is it different from complexes? What do you mean by titrimetric analysis? Define the different terms used in the analysis. What is indicator? Explain different types of indicator.

Experiments http://www.narosa.com/books_display.asp?catgcode=978-81-8487-607-9

Index Araldite 10.13, 1014 Battery 4.23, 4.26 Beer’s Law 9.3, 9.11 Bond Theory 1.2 Buckyballs 3.13, 3.14, 3.15 Calorific Value 8.2, 8.3, 8.6 Calorimeter 8.4, 8.8 Caustic Embrittlement 5.10, 6.53 Cell 1.1, 1.3, 4.15, 4.17 Chemical Bonding 1.1 Chemical Corrosion 5.3, 5.4, 5.6 Classification of imperfection 2.12 Coal 8.12, 8.13, 8.18 Combustion 3.1, 3.19, 8.1, 8.4 Component System 7.2, 7.5, 7.8, 7.11 Corrosion 5.1, 5.2, Crystal 1.16, 2.1, 2.10 Crystalline Solids 2.1 Electrochemical Cells 4.1, 4.23 Electrochemical Series 4.9, 4.10, 5.6 Electromagnetic Spectrum 9.1, 9.2 Frenkel defect 2.13 Fuel 8.1, 8.2 Fullerenes 2.16, 2.17 Galvanic Cell 4.1, 4.7, 5.6 Gases 1.1, 6.13 Graphite 2.15, 3.13 Impurities 6.3, 6.4 Ion-exchange 6.32, 6.38, 6.39 Lattice 2.3, 3.16 Leaching 5.11, 6.22 Melamine 6.38, 10.13 Metallic Bonding 1.14, 1.16

Mineral matter 6.6, 8.18 Molecular orbital theory 1.2, 1.3, 1.19 Nanomaterials 3.1, 3.2, 3.3 Nanoscale 3.1, 3.2, 3.4, 3.5 Nanotubes 3.15, 3.16, 3.18 Odours 6.4, 6.5 Orbital Theory 1.2, 1.3 Organometallics 10.38, 10.39 Poly acrylonitrile (PAN) 10.22 Polyamides 10.10, 10.17, 10.17 Polymeric Materials 3.2, 10.3, 10.4 Polymers 10.1, 10.2, 10.3 Polymorphism 7.11 Polypropylene 10.6, 10.21 Polyvinyl Chloride (PVC) 10.15 Protective Coatings 5.13, 5.15 Radius Ratio 2.8 Redox Titrations 11.6, 11.14 Resin 6.36, 10.10 Sediment 6.5, 6.16, 6.17 Silica Content 6.15, 6.46 Softening of Water 6.24, 6.38 Solid Fuels 8.9, 8.20 Spectroscopy 9.15, 9.22, 9.27 Steam Making 6.41 Sterilization of water 6.21 Sulphur system 7.11, 7.12 Surface Drainage 6.1, 6.2 Symmetry 2.2, 2.16 Titrations 11.9, 11.14, 11.22 Titrimetric Analysis 11.1, 11.2, 11.3, 11.6 Water Treatment 6.1, 6.30 Zeolites 6.32, 6.33