Engineering Chemistry: : A Textbook [1 ed.] 9781783324309, 9781783323647

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ENGINEERING CHEMISTRY A TEXTBOOK

ENGINEERING CHEMISTRY A TEXTBOOK

Praveen P. Singh Vishal Srivastava Sudhanshu Kanaujia

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

ENGINEERING CHEMISTRY A Textbook 216 pgs.

Praveen P. Singh Department of Chemistry United College of Engineering and Research Allahabad Vishal Srivastava Department of Chemistry United College of Engineering and Management Allahabad Sudhanshu Kanaujia Department of Chemistry United College of Engineering and Research Allahabad Copyright © 2018 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K. www.alphasci.com ISBN 978-1-78332-364-7

E-ISBN 978-1-78332-430-9 Printed from camera-ready copy provided by authors.

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 This book has been exclusively designed for B. Tech. (First/ Second Semester) students of Dr. A. P. J. Abdul Kalam Technical University (AKTU) and other universities. The matter given in this book consists of both theory and practical. All topics are explained in simple language supported by neat diagrams and solved numerical problems. The subject matter has been presented in a comprehensive, lucid and systematic manner which is easy to understand even for self study. The author believes that learning by solving problems gives more competence and confidence in the subject. Keeping this in view, sufficiently large number of varied problems with solutions and exercises are given in each chapter. It is the fond hope of the authors that the book will be highly useful to both the student community and the teachers. We are highly thankful to our colleagues for their discussions and suggestions. The authors highly appreciate the work of the publishing staff of Narosa Publishing House Pvt. Ltd., New Delhi who handled the project promptly and intelligently. Every precautions has been taken for accuracy and clarity of the subject matter. The authors will gratefully acknowledge suggestions and comments from the readers for further improvements of the book. Praveen P. Singh Vishal Srivastava Sudhanshu Kanaujia

Contents Preface

v

1. Chemical Bonding

1.1

2. Solid State

2.1

1.1 Introduction 1.1 1.2 Molecular orbital theory 1.2 1.3 LCAO 1.2 1.4 Conditions for the combination of atomic orbitals 1.4 1.5 Types of molecular orbitals 1.4 1.6 Energy level diagram of molecular orbitals 1.5 1.7 Difference between bonding molecular orbitals and antibonding molecular orbitals 1.7 1.8 Bond order 1.7 1.9 Significance of bond order 1.7 1.10 Molecular orbital diagram of some homonuclear, diatomic molecules/ions 1.8 1.11 Molecular orbital diagram of some heteronuclear, diatomic molecules/ions 1.14 1.12 Comparison of valence bond and molecular orbital theories 1.21 1.13 Metallic bond 1.21 1.14 Classification of materials based on band theory 1.23 Exercise 1.24 2.1 Introduction 2.2 Characteristics of solid state 2.3 Types of solids 2.4 Space lattice or crystal lattice 2.5 Unit cell 2.6 Types of unit cell in cubic system 2.7 Number of atoms in a unit cell 2.8 Calculation of density of unit cell 2.9 Bragg’s law 2.10 Imperfections in solid 2.11 Types of imperfections in solid 2.12 Point imperfections (zero dimensional effects) 2.13 Line imperfections (one dimensional effects) 2.14 Plane imperfections (two dimensional effects) 2.15 Graphite 2.16 Fullerene Exercise

2.1 2.1 2.1 2.2 2.3 2.3 2.4 2.5 2.6 2.10 2.11 2.11 2.14 2.15 2.17 2.18 2.19

viii Contents 3. Liquid Crystals and Nanomaterials

3.1

4. Polymers

4.1

3.1 Introduction 3.2 Classification of liquid crystals 3.3 Applications of liquid crystals 3.4 Nanomaterials 3.5 Different types of nanomaterials 3.6 Physical and chemical properties of nanomaterials 3.7 Major classes of nanomaterials and their benefits Exercise 4.1 Introduction 4.2 Classification of polymers 4.3 Types of polymerisation reactions 4.4 Ziegler-Natta polymerisation 4.5 Synthesis, properties and uses of commercially addition polymers 4.6 Step growth polymers 4.7 Commercial condensation polymers 4.8 Natural Rubber 4.9 Vulcanisation of rubber 4.10 Synthetic rubbers 4.11 Preparation of synthetic rubbers 4.12 Silocon rubbers 4.13 Molecular weight of polymers 4.14 Biodegradable polymers 4.15 Conducting polymers 4.16 Composites 4.17 Materials used in polymer composites 4.18 Classification of polymer matrix composites 4.19 Organometallic compounds 4.20 Grignard Reagents 4.21 Preparation of Grignard reagents 4.22 Reactions of Grignard reagents 4.23 Mechanism of Grignard reagents 4.24 Applications of Grignard reagents 4.25 Applications of organometallic compounds as catalyst in polymerisation 4.26 Applications of organometallic compounds as catalyst in organic synthesis Exercise

5. Electrochemistry

5.1 Introduction 5.2 Galvanic cell 5.3 Measurement of electrode potential 5.4 Nernst equation 5.5 Batteries 5.6 Corrosion 5.7 Effects of corrosion 5.8 Factors affecting corrosion 5.9 Types of corrosion 5.10 Corrosion control methods Exercise

3.1 3.2 3.4 3.4 3.4 3.5 3.5 3.6

4.1 4.2 4.5 4.10 4.11 4.13 4.14 4.19 4.20 4.20 4.21 4.22 4.23 4.23 4.24 4.26 4.27 4.27 4.28 4.30 4.31 4.31 4.31 4.32 4.33 4.33 4.34

5.1

5.1 5.1 5.3 5.5 5.6 5.10 5.11 5.11 5.12 5.14 5.17

Contents  ix 6. Engineering Materials

6.1 Cement 6.2 Composition of Portland cement 6.3 Raw materials 6.4 Functions of ingredients of cement 6.5 Manufacture 6.6 Calcination in rotary kiln 6.7 Formation and packing of cement 6.8 Setting of cement 6.9 Concrete 6.10 Reinforced concrete construction (RCC) 6.11 Advantage of RCC over plain concrete 6.12 Plaster of paris 6.13 Preparation 6.14 Properties 6.15 Uses 6.16 Lubricant 6.17 Classification of lubricants 6.18 Mechanism of lubrication 6.19 Properties of lubricants 6.20 Functions of lubricant Exercise

7. Water

7.1 Introduction 7.2 Hardness of water 7.3 Units of hardness 7.4 Disadvantage of hard water 7.5 Techniques for Water Softening 7.6 Reverse osmosis 7.7 Advantages of Reverse Osmosis Exercise

6.1

6.1 6.1 6.1 6.2 6.2 6.2 6.4 6.4 6.4 6.5 6.5 6.5 6.5 6.5 6.6 6.6 6.6 6.7 6.8 6.9 6.10

7.1

7.1 7.1 7.2 7.2 7.2 7.10 7.11 7.17

8. Phase Rule

8.1

9. Fuel

9.1

8.1 Introduction 8.2 Phases 8.3 Component 8.4 Degrees of freedom 8.5 Significance of triple point 8.6 Phase diagram of water system 8.7 Phase diagram of sulphur system 8.8 Advantages of phase Rule 8.9 Limitations of phase Rule Exercise 9.1 Introduction 9.2 Classification of fuels 9.3 Characteristics of good fuel 9.4 Calorific value 9.5 Units of heat

8.1 8.1 8.2 8.3 8.4 8.4 8.5 8.7 8.7 8.9 9.1 9.1 9.2 9.2 9.3

x Contents 9.6 Classification of calorific value 9.7 Determination of Calorific Value Using Bomb Calorimeter 9.8 Coals 9.9 Analysis of coal 9.10 Combustion of fuel 9.11 Calculation of air required for combustion 9.12 Biogas 9.13 Constituents of biogas 9.14 Raw materials for biogas 9.15 Manufacture of dung (or gobar gas) gas 9.16 Uses of biogas 9.17 Advantages of biogas 9.18 Limitation of biogas 9.19 Biomass Exercise

9.3 9.4 9.8 9.9 9.11 9.12 9.13 9.13 9.13 9.14 9.15 9.15 9.15 9.15 9.16

10. Spectroscopy

10.1

10.1 Spectroscopy 10.2 Electromagnetic spectrum

10.1 10.1



10.3 Origin of electronic spectra 10.4 Ultraviolet and visible spectroscopy

10.2 10.3



10.5 Lambert-Beer’s law 10.6 Electronic transitions

10.3 10.4

10.7 Chromophore 10.8 Auxochrome

10.5 10.5



10.6 10.7

10.9 Absorption and intensity shifts 10.10 Applications of ultraviolet and visible spectroscopy

10.11 Infrared (IR) spectroscopy 10.12 Types of vibrations 10.13 Number of fundamental vibrations 10.14 Finger print region 10.15 Applications of infrared spectroscopy 10.16 Proton nuclear magnetic resonance (PMR or 1H NMR) spectroscopy 10.17 Instrumentation 10.18 Shielding and Deshielding 10.19 Number of signals in 1H NMR spectrum 10.20 Splitting of signals 10.21 Chemical shift 10.22 TMS (Tetra methyl silane) 10.23 Coupling constant 10.24 Applications of NMR spectroscopy Exercise

10.7 10.7 10.9 10.10 10.10 10.11 10.11 10.12 10.13 10.14 10.15 10.15 10.16 10.17 10.17

Contents  xi 11. Engineering Chemistry Practical

11.1



11.1 To determine the alkalinity of the given water sample.

11.1



11.2 Determination of temporary and permanent hardness in water sample using EDTA as standard solution

11.4



11.3 Determination of available chlorine in bleaching powder. 11.6 11.4 Determination of chloride content in water sample.

11.8




11.5 Determination of iron content in the given water sample by Mohr’s method 11.6 pH- metric titration 11.7 Viscosity of an addition polymer like polyester by viscometer

11.9 11.11
 11.12




11.8 Determination of iron concentration in sample of water by calorimetric method

11.15




11.9 Element detection and functional group identification in organic compounds

11.16



11.10 Preparation of Bakelite and Urea formaldehyde resin

11.20





Index

I.1

Chapter – 1

Chemical Bonding Syllabus: Molecular orbital theory and its applications in diatomic molecules, band theory of solids.

1.1

INTRODUCTION

The attractive force which holds various constituents (atoms, ions etc.) together in different chemical species is called a chemical bond. Since the formation of chemical compounds takes place as a result of combination of atoms of various elements in different ways. Three different types of bond may be formed, depending on the electropositive or electronegative character of the atoms involved. Electropositive elements + Electronegative elements

Ionic bond e.g. NaCl

Electronegative element + Electronegative elements

Covalent bond e.g. HCl

Electropositive elements + Electropositive elements

Metallic bond e.g. Li metal

There are theories of chemical bonding to explain geometry, energy and other molecular parameters. Kossel, Lewis, Sidwick and other have given electronic theory of valency to explain ionic, covalent and coordinate bonds. Two other theories are valence bond theory (VBT) and molecular orbital theory (MOT). Valence bond theory was given by Heitler, London, Slater and Pauling. According to this theory, atoms retain their identities after formation of chemical bond. Also the theory was unable to explain paramagnetic nature of oxygen molecule, bonding in metals and polar nature in covalent molecules. Molecular orbital theory (MOT) is good for understanding bonding in general. It is more difficult to learn, but predicts the actual properties of molecules better than valence bond theory (VBT). Molecular orbital theory actually predicts electron transitions because of the differences in the energy levels of orbitals in the molecule. Molecular orbital theory has been more correct in numerous instances and for this reason it is preferred.

1.2 ENGINEERING CHEMISTRY: A Textbook

1.2

MOLECULAR ORBITAL THEORY

Molecular orbital (MO) theory was developed by F. Hund and R. S. Mulliken in 1932. The salient features of this theory are: 1.

The electrons in the molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.

2.

The atomic orbitals are of comparable energies and proper symmetry combines to form molecular orbitals.

3.

While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric.

4.

The number of molecular orbitals formed is equal to number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbitals (BMO) and other is called antibonding molecular orbital (ABMO).

5.

The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.

6.

Just as electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecular is given by a molecular orbital.

7.

The molecular orbitals like atomic orbitals are filled in accordance with the Aufbau principle obeying the Pauli's exclusion principle and the Hund's rule.

1.3

LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO)

According to wave mechanics, the atomic orbital can be expressed by wave functions (ψ’s) which represent the amplitude of electron waves. These are obtained from the solution of Schrodinger wave equation. However, since it can be solved for any system containing more than one electron, molecular orbitals which are one electron wave functions for molecular are difficult to obtain directly from the solution of Schrodinger wave equation. To overcome this problem, an approximate method known as 'linear combination of atomic orbitals' (LCAO) has been adopted. Let us apply this method to the homonuclear diatomic hydrogen molecule. Consider the hydrogen molecule consisting of two atoms A and B. Each hydrogen atom in the ground state has one electron in 1s orbital. The atomic orbitals of these atoms may be represented by A and B. Mathematically, the formation of molecular orbitals may be described by the linear combination of atomic orbitals that can take place by

Chemical Bonding 1.3 addition and by subtraction of wave functions of individual atomic orbitals as shown below: MO = A ± B Therefore, the two molecular orbitals  and * are formed as:  = A + B * = A - B The molecular orbital  formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital * formed by the subtraction of atomic orbital is called antibonding molecular orbital as depicted in Fig. 1.1.

Increasing Energy

Antibonding orbital higher energy than that of atomic orbitals

* = A - B

B Atomic orbital

A Atomic orbital   A   Molecular orbitals

Bonding orbital lower energy than that of atomic orbitals

Fig. 1.1: Formation of bonding () and antibonding (*) molecular orbitals by the linear combination of atomic orbitals A and B centered on two atoms A and B respectively. Qualitatively, the formation of molecular orbitals can be understood in terms of the constructive and destructive interference of the electron waves of the combining atoms. In the formation of bonding molecular orbital, the two electron waves of the bonding atoms reinforce each other due to constructive interference while in the formation of antibonding molecular orbitals, the electron waves cancel each other due to destructive interference. As a result, the electron density in a bonding molecular orbital is located between the nuclei of the bonded atoms because of which the repulsion between the nuclei is very less while in case of an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei. In fact there is a nodal plane (on which the electron density is zero) between the nuclei and hence the repulsion between the nuclei is high. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilize a molecule. Therefore, a bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have

1.4 ENGINEERING CHEMISTRY: A Textbook combined to form it. In contrast, the electrons placed in the antibonding molecular orbitals destabilize the molecule. This is because, the mutual repulsion of the electrons in these orbitals is more than the attraction between the electrons and the nuclei, which causes a net increase in energy. It may be noted that the energy of the antibonding orbital is raised above the energy of the parent atomic orbitals that have combined and the energy of the bonding orbital has been lowered than the parent orbitals.

1.4

CONDITIONS FOR THE COMBINATION OF ATOMIC ORBITALS

The linear combination of atomic orbitals to form molecular orbitals takes place only if the following conditions are satisfied. 1.

The combining atomic orbitals must have the same or nearly the same energy. This means 1s orbital can combine with another is orbital but not with 2s orbital because the energy of 2s orbital is appreciably higher than that of 1s orbital. This is not true if the atoms are very different.

2.

The combining atomic orbitals must have the same symmetry about the molecular axis. By convention z-axis is taken as the molecular axis. It is important to note that atomic orbitals having same or nearly the same energy will not combine if they do not have the same symmetry. For example. 2pz orbital of one atom can combine with 2pz orbital of the other atom but not with the 2px or 2py orbitals because of their different symmetries.

3.

The combining atomic orbitals must overlap to the maximum extent. Greater the extent of overlap, the greater will be the electron- density between the nuclei of a molecular orbital.

1.5

TYPES OF MOLECULAR ORBITALS

Molecular orbitals of diatomic molecular are designated as  (sigma), π (pi) and  (delta) etc. In this nomenclature, the sigma () molecular orbitals are symmetrical around the bond axis while pi (π) molecular orbitals are not symmetrical. For example: The linear combination of 1s orbitals centered on two nuclei produces two molecular orbitals which are symmetrical around the bond axis. Such molecular orbitals are of the sigma () type and are designated as  1s and *1s. If internuclear axis is taken to be in the z-direction it can be seen that a linear combination of 2pz orbitals of two atoms also produces two sigma molecular orbitals designated as 2pz and *2pz. Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the presence of positive lobes above and negative lobes below the molecular plane. Such molecular orbitals are labeled as π and π*. A π bonding MO has larger electron density above and below the inter-nuclear axis. The π* anti- bonding MO has a node between the nuclei. Following molecular orbitals are formed when atomic orbitals combine.

Chemical Bonding 1.5

Table 1.1 Atomic Orbitals

Molecular orbitals Bonding

Antibonding

σ1s σ2s σ2pz π2px π2py

σ*1s σ*2s σ*2pz π*2px π*2py

1s + 1s 2s + 2s 2pz + 2pz 2px + 2px 2py + 2py

1.6

ENERGY LEVEL DIAGRAM OF MOLECULAR ORBITALS

We have seen that 1s atomic orbitals on two atoms form two molecular orbitals designated as 1s and *1s. In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals on two atoms) give rise to the following eight molecular orbitals: MOs

*2s

*2pz

π*2px

π*2py

Bonding

MOs

2s

2pz

π2px

π2py

Increasing Energy

Antibonding

1s 1s 1s Atomic orbital

1s Atomic orbital

1s

Increasing Energy

(a) _

2pz

2pz Atomic orbital

2pz Atomic orbital 2pz Molecular orbital (b)

1s Antibonding sigma molecular orbital

+ 1s

1s Molecular orbital

1s

+ 2pz

_

_ + 2pz

1s Bonding sigma molecular orbital _

 + 2pz

Antibonding sigma molecular orbital _

_ + + + 2pz 2pz

-

+ 2pz Bonding sigma molecular orbital

Increasing Energy

1.6 ENGINEERING CHEMISTRY: A Textbook + + _ _ _ 2px 2px

2px

2px

2px Atomic orbital

_ + _ + 2px Antibonding  molecular orbital +

+ + + _ _

Atomic orbital 2px

2px 2px

Molecular orbital (c)

_

2px Bonding  molecular orbital

Fig. 1.2: Contours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals (b) 2pz atomic orbital (c) 2px atomic orbital The energy levels of these molecular orbitals have been determined experimentally from spectroscopic data for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order of energies of various molecular orbitals of O2 and F2 is given below: 1s

1s < 



< 2s <   s   pz
C1 Now for Zn2+ / Zn electrode, n = 2 [Zn2+ + 2e- → Zn]

E(cell) =

0.1 0.059 log = 0.0296 log 10 0.01 2

= 0.0296 × 1 = 0.0296 V Example 5.2. Calculate the e.m.f. of a galvanic cell at 250C, when the concentration of ZnSO4 and CuSO4 are 0.001 M and 0.1 M respectively. The standard potential of the cell is 1.2 volts. Solution: Cell is: Zn (s) | Zn2+   || Cu2+   | Cu (s) E(cell) = E(cell) 

[Cu2+] 0 059 log 2 [Zn2+]

= 1.1 + 0.0296 log (0.1/ 0.001)

5.10 ENGINEERING CHEMISTRY: A Textbook = 1.1 + 0.0296 × 2 = 1.1 + 0.592 = 1.1592 V Example 5.3. Why do electrochemical cell stop working after some time? Solution: An electrochemical cell produces electrical energy at the cost of redox reaction. When the redox reaction is completed, the cell stop working, since it is now incapable of undoing redox reaction any more. Example 5.4. Why does a dry cell becomes dead after a long time, even if has not been used. Ans. Acidic NH4Cl slowly corrodes the zinc container of the dry cell, even when the cell is not in the use. Hence, dry cell becomes dead after a long time, even it is not used. Example 5.5. Why is salt bridge used in the construction of cell? Ans: A wire cannot be used to connect two electrodes internally in an electrochemical cell, because it produces a voltage drop. Consequently, accurate electric cell measurement cannot be done. In order to make accurate electric measurements in solution, a salt bridge is employed. Salt bridge completes the electrical circuit, thereby permitting the ions to move from one solution to another without direct contact (or mixing) of the two solutions. Moreover, it maintains electrical neutrality of the solutions in the two half-cells. Example 5.6. The emf of a concentration cell gradually decreases why? Ans: The concentration cell produces electric energy, due to transfer of metallic ions from the solution of a higher concentration to the solution of lower concentration. As this process continues, the difference in concentration between two half- cell concentration cells decreases. Consequently, the emf of the cell also gradually decreases, since the emf is given by:

E(cell) =

2.3.3 RT nF

log

C2 C1

and the ratio C2/ C1 goes on decreasing gradually and finally the cell stops working when C2 becomes equal to C1.

5.6. CORROSION “Corrosion is the process of gradual deterioration of a metal from its surface due to the unwanted chemical or electrochemical interaction of metal with its environment”. It causes enormous damage to buildings, bridges, ships and to all objects made of metals especially that of iron. We lose crores of rupees every year on account of corrosion. The most familiar example of corrosion is rusting of iron exposed to the atmospheric conditions. The chemistry of corrosion is quite complex but it may be considered essentially as an electrochemical phenomenon.

Electrochemistry 5.11

2 Fe(s) +

3 O2(g) + x H2O(l) 2

Fe2O3 . x H2O(s) (Chemical composition of rust)

Fig. 5.7: Corrosion of iron in atmosphere

5.7. EFFECTS OF CORROSION Effects of corrosion are briefly given below: 1. Production related consequences: i. Loss of useful properties of metal and thus loss of efficiency. ii. Decrease in production rate, because efficiency is less and replacement of corroded equipment is time consuming. iii. Increase in maintenance and production cast. iv. Contamination of product. 2. Health and safety related consequences: i. Unpredictable failure of machinery, sometimes lead to loss of life. ii. Contamination of drinking water. iii. Leakage of toxic liquid or gas. iv. Leakage of inflammable gas from corrodes pipe or instrument which can cause fire hazards.

5.8. FACTORS AFFECTING CORROSION The rate of corrosion is affected by following factors. 1. The more reactive metals are more prone to corrosion. 2. Air and moisture are quite helpful in corrosion. The presence of gases like CO2 and SO2 in air makes it still faster. e.g., if iron is kept in vacuum, then no rusting is formed. 3. Presence of impurities helps in setting up a corrosion cell and increases the rate of corrosion. e.g., pure iron does not rust.

5.12 ENGINEERING CHEMISTRY: A Textbook 4. 5.

Strains in metal also help in corrosion. e.g., in iron articles, rusting is more pronounced on the areas having bends, dust, scratches, nicks and cuts. The presence of electrolytes also makes the corrosion process faster, e.g., iron rust more rapidly in saline water in comparison to pure water.

5.9. TYPES OF CORROSION Corrosion of the metals occurs by the attack of surrounding materials on the surface of metal. Corrosion is classified as: 1. Dry corrosion or direct chemical corrosion 2. Wet corrosion or immersed corrosion or electro-chemical corrosion.

5.9.1. Dry corrosion or direct chemical corrosion This type of corrosion occurs mainly through the direct chemical action of environment/ atmospheric gases such as oxygen, halogen, hydrogen sulphide, sulphur dioxide, nitrogen or anhydrous inorganic liquid with metal surfaces in immediate proximity. There are three main types of chemical corrosion. i. Oxidation corrosion ii. Corrosion by other gases iii. Liquid metal corrosion i. Oxidation Corrosion: This type of corrosion is brought about by the direct action of oxygen at low or high temperatures on metals, usually in the absence of moisture. At ordinary temperatures, metal, in general, are very slightly attacked. However, alkali metals (Li, Na, K, Rb, etc.) and alkaline earth metals (Be, Ca, Sr, etc.) are even rapidly oxidised at low temperatures. At high temperatures, almost all metals (except Ag, Au and Pt) are oxidised. The reactions in the oxidation corrosion are: 2M

2 Mn+ + 2n eMetal ions

n O2 + 2n e2

(Loss of electrons)

n O2- (Gain of electrons) Oxide ions

or

2M+

n O2 2

2 Mn+

+

n O2-

Metal ions Oxide ions Metal oxide ii. Corrosion by Other Gases: Dry gases such as SO2, CO2, Cl2, H2S, F2 etc., also cause corrosive effect on the metals. The extent of corrosion effect depends mainly on the chemical affinity to the metal and gas involved. Like in case of oxidation corrosion by atmospheric air, hence also the degree of attack depends on the formation of protective or non protective layers, on the metal surface. i.

If the layer formed is non-porous or protective, the extent of attack decreases, because the layer formed protects the metal surface from further attack. For example, AgCl layer resulting from the attack of chlorine on silver metal.

Electrochemistry 5.13 ii.

If the layer formed is porous or non-protective, the total metal surface is gradually decayed. e.g. , dry Cl2 gas attacks on tin forming volatile tin(iv) chloride, which volatilizes immediately from the metal surface, thereby leaving fresh metal surface for further attack. Similarly, at high temperature, hydrogen sulphide attacks steel forming scale of FeS, which hinders the normal operations in petroleum industry.

iii. Liquid Metal Corrosion: This type of corrosion has been found to occur in devices used for nuclear power. Liquid metal corrosion occurs due to the chemical action of flowing liquid metal at high temperatures on solid metal or alloy. This type of corrosion results in weakening of the solid metal either (i) dissociation of a solid metal by a liquid metal, or (ii) internal penetration of the liquid metal into the solid metal.

5.9.2. Wet Corrosion or Electrochemical Corrosion This type of corrosion occurs: (i) where a conducting liquid is in contact with metal or (ii) when two dissimilar metal or alloys or either immersed or dipped partially in a solution. This corrosion involves flow of electron current between the anodic and cathodic areas. At anodic area, oxidation reaction (i.e., liberation of free electrons) takes place, so anodic metal is destroyed by either dissolving or assuming combined state (such as oxide, etc.). Hence, corrosion always occurs at anodic areas. At anodic areas: M Mn+ + ne(Oxidation) On the other hand, the cathodic reaction consumes electrons with either by: (a) evolution of hydrogen or (b) absorption of oxygen, depending on the nature of the corrosive environment. (a) Evolution of hydrogen: This type corrosion occurs, usually in acidic environments. Considering metal like Fe, the anodic reaction is dissolution of iron as ferrous ions with the liberation of electrons. Fe Fe2+ + 2e(Oxidation) These electrons flow through the metal, from anode to cathode, where H+ ions (of acidic solution) are eliminated by hydrogen gas. 2H+ + 2eThe overall reaction is:

H2 Fe + 2H+

(Reduction) Fe2+ + H2

Thus, this type of corrosion causes “displacement of hydrogen ions from, the acidic solution by metal ions.” Consequently, all metals above hydrogen in the electrochemical series have a tendency to get dissolved in acidic solution with simultaneous evolution of hydrogen. Note: It may be pointed here that in hydrogen evolution type of corrosion, the anodes are, usually very large areas; whereas cathode are small areas. (b) Absorption of Oxygen: Rusting of iron in neutral aqueous solution of electrolytes (like NaCl solution) in the presence of atmospheric oxygen is a common example of this type of corrosion. The surface of iron is, usually coated with at thin film of iron oxide. However, if this iron oxide film develops some cracks, acidic

5.14 ENGINEERING CHEMISTRY: A Textbook areas are created on the surface; while the well metal parts act as cathodes. It follows that the anodic areas are small surface parts; while nearly rest of the surface of the metal forms large cathodes. At the anodic areas of the metal (iron) dissolve as ferrous ions with liberation of electrons. Fe Fe2+ + 2e(Oxidation) The liberated electrons flow from anodic to cathodic areas, through iron metal, where electrons are intercepted by the dissolved oxygen as:

1 O2 + H2O + 2e2

2OH

(Reduction)

The Fe2+ ions (at anode) and OH ions (at cathode) diffuse and when they meet, ferrous hydroxide is precipitated. Fe2+ + 2OH

Fe(OH)2

(i) If enough oxygen is present, ferrous hydroxide is easily oxidised ferric hydroxide. Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 This product, called yellow rust, actually corresponds to Fe2O3 . H2O. (ii) If the supply of oxygen is limited, the corrosion product may be even black anhydrous magnetite, Fe3O4. In other words, any corrosion process can be termed as an electrochemical process in which cathode and anode are formed on the metal surface and an electrolyte (water or salt or acid) must be present to permit ionic flow to form corrosion products. Note: 1. It may be pointed out that through the two reactants Fe2+ and OH- originate from the anode and cathode respectively, but their combination occurs, more commonly near the cathode, because the smaller Fe2+ diffuse more rapidly than the larger OH- ions. So corrosion occurs at the anode, but rust is deposited at or near the cathode. 2. An increase in oxygen content has two effects: (i) it forces the cathodic reaction to the right, producing more OH- ions, and (ii) it removes more electrons and, therefore, accelerates the corrosion at the anode. Each of these effects, in-turn, supply more reactants for the rust-forming reaction. Evidently, presence of oxygen greatly accelerates both corrosion and rust-formation, with the corrosion occurring at the anode, but the rust forming at or near the cathode.

5.10. CORROSION CONTROL METHODS There are varieties of corrosion control techniques as the corrosion depends on the nature of environment. The important corrosion control methods are given in the following sections: 1.

Design and material selection

2.

Protective coatings

Electrochemistry 5.15 3.

Cathodic protection

4.

Anodic protection

5.

Corrosion inhibitors

5.10.1. Design and material selection The designing of the material should be such that corrosion if it occurs is uniform and does not result in intense and localized corrosion. Some of the important design principles are: 1.

If two different metals have to be in contact, they should be selected and their oxidation potential is as close as possible.

2.

Heat treatment (annealing) helps to reduce internals stresses and reduce corrosion.

3.

The anodic metal should not be painted or coated when in contact with a different cathodic metal, because any crack in coating would lead to it rapid localized corrosion.

4.

If moisture of electrolyte solution is present, suitable inhibitors should be employed.

5.

The contact between dissimilar metals in the presence of corrosive solution should be avoided. If this principle is not obeyed, then corrosion is localized on the more active metal, while the less reactive metal remains protected.

6.

If an active metal is used, it should be insulated from more cathodic metals.

7.

Whenever, the direct joining of different metals, is unavoidable, an insulating fitting may be used in between them to avoid direct metal- metal electrical contact.

8.

Impingement attack can be reduced by careful filtration of suspended solids from the liquid stream and by preventing turbulent flow.

9.

By adjusting the acidity or alkalinity of the environment, corrosion can be controlled. If control of pH is not possible, inert coating and inactive metals are used to control the pH.

5.10.2. Protective coatings These are used to prevent corrosion at the surfaces of materials. Protective coatings include metallic coatings, chemical conversion coatings, ceramic coatings and organic coatings.

5.10.3. Cathodic protection The corrosion of metal takes place at the anodic region where as at the cathodic region, metal is unaffected. The principle of cathodic protection involves the elimination of anodic sites and conversion of the entire metal into cathodic site. This can be achieved by providing electrons from an external source so that the specimen always remains cathode. This technique of offering protection to a specimen against

5.16 ENGINEERING CHEMISTRY: A Textbook corrosion by providing electrons from an external source is called cathodic protection. It can be achieved by the following two methods: i. Sacrificial anode method: In this method, the metal structure is converted into a cathode by connective into a more active metal, which acts as an auxiliary anode. The most commonly used auxiliary anodes are Zn, Mg and Al. These metals being more active, acts as anode and undergo preferential corrosion, thus protecting the metal structure. Since the anode metals are sacrificed to protect the metal structure, this method is known as sacrificial anode method. New auxiliary anodes replace exhausted sacrificial anodes as and when required. For examples, (a) Magnesium block connected to underground pipelines. (b) Magnesium bars are fixed to the sides of the ships. ii. Impressed current method: Another method of providing cathodic protection is by applying a direct current larger than the corrosion current. The protected metal is made cathodic by connecting it to the negative terminal of a DC source. Positive terminal is connected to an inert anode like graphite. The metal structure being cathode does not undergo corrosion.

5.10.4. Anodic protection The prevention of corrosion by impressed anodic current method is called anodic protection. Few metals like Ti and alloys like steel, when made as anode exhibits passivity by forming their oxide layers. In case of metals, such as Ti, Ni, Cr and their alloys, application of suitable anodic current makes them passive and decreases their rate of dissolution. The potential required to protect the metal can be obtained from potential current curve. The advantage of the method is that it requires a small current. However, the corrosion rate cannot be reduced to zero as in cathodic protection. A drawback of the method is that it cannot be applied to metals that do not passivate. This method of anodic protection is utilized in the transportation of concentrated acids.

5.10.5. Corrosion inhibitors These inhibitors reduce corrosion by retarding either anodic or cathodic reactions. i. Anodic inhibitors: Anodic reaction is the oxidation of metals, that is, for example in rusting of iron, Fe2+ ions are formed at anode. Anodic inhibitors prevent the formation of Fe2+ ions (anodic reaction), and thereby retard the corrosion process. This is achieved by the addition of anions such as chromate, tungstate, molybdate, etc., which combine with metal ions formed at anodic region, forming sparingly soluble salts. These salts are deposited on the anodic sites forming protective films, which acts as barriers between metal surface and corrosion medium, prevent further anodic reaction and hence corrosion. ii. Cathodic inhibitors: These act by inhibiting the cathodic reactions which involve the liberation of hydrogen in acidic solutions and OH- ions in alkaline solution. These can be of two types:

Electrochemistry 5.17 a)

Organic cathodic inhibitors such as amines, mercaptans, thioureas, sulphoxides form a protective layers on cathodic regions, prevent the evolution of hydrogen; thus decreasing the rate of corrosion.

b) Inorganic cathodic inhibitors such as sulphates of Mg, Mn, Ni, and Zn are used in neutral or alkaline medium. These inhibitors react with OH- ions liberated at cathode, forming insoluble hydroxide, which form protective film over cathodes areas and prevent corrosion.

Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

Write short note on electrode potential. What is standard electrode potential? Describe the construction and working of galvanic cell. Explain the function of salt bridge. What do you understand by electrochemical series? Discuss the working principle of secondary batteries. Derive Nernst equation for the calculation of cell emf. Describe the construction of lead storage battery. What is electrochemical corrosion? Write down the mechanism involved in the said reaction. Describe the process of galvanization of iron? How does it prevent the corrosion of iron? Explain the mechanism of hydrogen evolution and oxygen absorption in electrochemical corrosion. Give figures. What is corrosion? Describe the mechanism of electrochemical corrosion by (i) Hydrogen evolution (ii) Oxygen absorption Explain sacrificial anodic and impressed cathodic protection method for prevention of corrosion. Explain why pure metal rod half immersed vertically in water starts corroding at the bottom? Discuss any four method of corrosion control. Define the term corrosion. Discuss in brief the electrochemical theory of corrosion. What are the factors which affect corrosion? What is corrosion of metals? Explain the basic reason of metallic corrosion. How is corrosion prevented by cathodic protection? Define corrosion. Write a short note on corrosion inhibitors. Discuss the mechanism of electrochemical corrosion.

Chapter – 6

Engineering Materials Syllabus: Setting and hardening of cement, applications of cement. Plaster of paris. Lubricants- Classification, mechanism and applications.

6.1. CEMENT Cement is the one of the most important building material at the present time. It is used in the construction of buildings, roads, bridges, dams etc. It was discovered, in 1824, by an English Mason, Joseph Aspdin who observed that when strongly heated mixture of limestone and clay was mixed with water and allowed to stand, it hardened to a stone like mass which resembled Portland rock a famous building stone of England. Since then the name Portland cement has been given to a mixture, containing high percentage of lime with silica, iron oxide, alumina, etc. Cement is a dirty greyish heavy powder containing calcium aluminates and silicates. The silicates and aluminates which form more than 90% of the cement are: 1. Tricalcium Silicate 2. Dicalcium Silicate 3. Tricalcium Aluminate 4. Tetracalcium Aluminoferrite Of these, tricalcium silicate is most important.

3CaO.SiO2 2CaO.SiO2 3CaO.Al2O3 4CaO.Al2O3.Fe2O3

6.2. COMPOSITION OF PORTLAND CEMENT The approximate composition of Portland cement is: 1. Calcium oxide CaO, 50-60%; 2. Silica SiO2, 20-25%; 3. Alumina Al2O3, 5-10%; 4. Magnesium oxide MgO, 2-3%; 5. Ferric oxide Fe2O3, 1-2% and 6. Sulpher trioxide SO3, 1-2%. For a good quality cement, the ratio of silica (SiO2) to alumina (Al2O3) should be between 2.5 and 4 and the ratio of lime (CaO) to the total of the oxides of silicon (SiO2) aluminium (Al2O3) and iron (Fe2O3) should be as close as possible to 2.

6.3. RAW MATERIALS The important raw materials needed for the manufacture of cement are: 1.

Limestone – This provides lime.

6.2 ENGINEERING CHEMISTRY: A Textbook 2. 3.

Clay – This provides alumina and silica. Gypsum.

6.4. FUNCTIONS OF THE INGREDIENTS OF CEMENT 1.

Lime is the principal constituent of cement. Its proportions must be properly regulated. However, excess of lime reduces the strength of cement, because it makes the cement to expand and disintegrate. On the other hand, presence of lesser amount of lime than required also reduces the strength of cement and makes it quick setting.

2.

Silica imparts strength to cement.

3.

Alumina makes the cement quick setting. Excess of alumina, however, weakens the cement.

4.

Calcium sulphate (gypsum) helps to retard the setting action of cement. It actually enhances the initial setting time of cement.

5.

Iron oxide provides: (i) colour, (ii) strength and (iii) hardness to the cement.

6.

Sulphur trioxide, in small proportion is desirable. When present in small amount, it imparts soundness to cement. However, its excess reduces the soundness of cement.

7.

Alkalis, if present in excess, cause the cement efflorescent.

6.5. MANUFACTURE These are two processes in use for the manufacture of cement 1. Dry process 2. Wet process 1. Dry Process: The process is used when limestone is hard in nature. The lime stone is first broken into small pieces. It is then mixed with clay in proper proportions. The mixture is finally pulverised to such an extent that is passes through 100 mesh sieve. This homogeneous mixture is known as raw meal. 2. Wet Process: The process is used when limestone and clay both are soft in nature. The clay is washed with water in wash mill to remove foreign materials like flint. The powdered limestone is mixed with clay paste in the ratio of 75% (limestone) and 25% (clay). The mixture is finally ground and made homogeneous. The homogeneous paste is known as slurry. It contains 40% (approximately) water.

6.6. CALCINATION IN ROTARY KILN It involves calcination of the mixture of limestone and clay. Dry mixture from the dry process or the slurry from the wet process is fed into the rotary kiln (Fig. 6.1) for calcination. Rotary kiln consists of a steel cylinder lined with fire bricks. It is 45-120 m long and has about 3 m internal diameter. The kiln is inclined at an angle of about 15 degrees to the horizontal and it rotates on its axis at the rate of half to one revolution per minute. The charge is introduced at upper end and it travels down as the kiln rotates.

Engineering Materials 6.3

Fig. 6.1: Rotary kiln for manufacture of portland cement. The charge is heated by burning fuels like pulverized coal or, fossil fuel, or oil or natural gas. The hot fuel gases are made to enter at the lower end of the kiln with the help of a blower. As the charge moves forward, it meets higher temperatures till it reaches the lower end where the temperature is highest (around 1400°C). The charge takes 2 to 3 hours to cover the journey in the kiln. Reactions taking place in the rotary kiln in three major zones are as follows: 1.

Maximum Temperature Zone (1000°C-1500°C): The main reactions between lime, alumina and silica takes place near the lower end of kiln resulting in formation of calcium silicates and aluminates as shown below. 2CaO + SiO2

2.

3.

→ 2CaO.SiO2 Dicalcium silicate 3CaO + Al2O3 → 3CaO.Al2O3 Tricalcium aluminate 3CaO + SiO2 → 3CaO.SiO2 Tricalcium silicate 4CaO + Al2O3 + Fe2O3 → 4CaO.Al2O3.Fe2O3 Tetracalcium alumino ferrite Moderate Temperature Zone (up to 800°C): At 100°C, the removal of free moisture takes place. At 500°C, kaolin (Al2O3, 2SiO2.2H2O) present in clay breaks up into amorphous silica and alumina. Average Temperature Zone (up to 800°C-1000°C)

Limestone decomposes to form lime and carbon dioxide. CaCO3 → CaO + CO2

6.4 ENGINEERING CHEMISTRY: A Textbook Moreover, due to high temperature in this zone, about 20-30% of mass melts and combines with solids mass to form pebbles (which vary in size) and are called cement clinkers. The burnt gases (containing a good amount of dust) are made to pass through a dust chamber, where most of dust is retained and gases are allowed to escape through chimney. The hot clinkers from the rotary kiln go to coolers which consist of series of tubes parallel to kiln in which air is forced through. As a result of this, the clinkers cool down while the air gets heated. The hot air is then used for the combustion of the fuel as an economy measure.

6.7. FORMATION AND PACKING OF CEMENT The cooled clinkers are mixed with 3-5% of gypsum CaSO4.2H2O, along with certain dispersing and water proofing agents and then ground to a fine powder. Since pure finely powered, clinkers set very rapidly, therefore, gypsum has to be added to retard the setting of cement. The powdered material is passed through fine sieves and finally it is sent to the automatic packing machines where it is packed in jute or polythene bags.

6.8. SETTING OF CEMENT When cement is mixed with water and left as such for some time, it becomes a hard mass. This is known as setting of cement. The setting of cement is complicated process, the exact mechanism of this setting process is not known. It is believed that various aluminates and silicates present in the cement form hydrates with water which separate in the form of gel 3CaO.Al2O3 + 6H2O

→

3CaO.Al2O3.6H2O Colloidal gel 2CaO.SiO2 + xH2O → 2CaO.SiO2.xH2O Colloidal gel 4CaO.Al2O3.Fe2O3 + 6 H2O → 3CaO.Al2O3.6H2O + Fe2O3.CaO Colloidal gel At the time some Ca(OH)2 and Al(OH)3 are formed as precipitates due to hydrolysis. 3CaO.SiO2 + H2O → Ca(OH)2 + 2CaO.SiO2 3CaO.Al2O3 + 6H2O → 3Ca(OH)2 + 2Al(OH)3 Gypsum combines with tricalcium aluminate to form calcium sulpho-aluminate, 3CaO.Al2O3 + 3CaSO4 + 2H2O → 3CaO.Al2O3.3CaSO4.2H2O This reaction slows down the setting. The gels formed start losing water partly by evaporation and partly by forming hydrates with anhydrates constituents. This results in the formation of a hard mass of Ca(OH)2 that binds the particles of calcium silicate together, while Al(OH)3 fills the interstices resulting in hardening of the mass. Setting of cement is an exothermic process. Hence, cement structures have to be cooled during setting by sprinkling water.

6.9. CONCRETE A mixture of cement, sand, gravel or small pieces of stone and water is known as concrete. It sets to an exceedingly hard structure. It is mainly used for construction

Engineering Materials 6.5 of floors. If the cement concrete is filled in an around a wire netting or skeleton of iron rods and allowed to set, the resulting structure is known as reinforced concrete. These structures have great strength and are used for construction of roofs, bridges etc.

6.10. REINFORCED CONCRETE CONSTRUCTION (R.C.C.) The combination of steel and concrete produces structure, called Reinforced concrete construction (R.C.C.), which can bear all types of loads. Reinforced concrete work is mostly used in floor-beams, piers, lintels, girders, arches, slabs, bridges, etc.

6.11. ADVANTAGE OF R.C.C. OVER PLAIN CONCRETE 1.

R.C.C. is easier to make and cast into any desired shapes, which can bear all types of loads.

2.

It possesses greater rigidity, moisture and free resistance.

3.

Steel reinforcement also tends to distribute the shrinkage cracks, thus preventing the formation of large cracks.

4.

Its maintenance cast is practically negligible.

1 H2O OR 2CaSO4. H2O 2 (CALCIUM SULPHATE HEMIHYDRATE)

6.12. PLASTER OF PARIS, CaSO4·

6.13. PREPARATION It is hemihydrates of calcium sulphate. It is obtained when gypsum, CaSO4·2H2O, is heated to 393 K. 2 (CaSO4.2H2O)

2 (CaSO4).H2O + 3 H2O

Above 393 K, no water of crystallization is left and anhydrous calcium sulphate, CaSO4 is formed. This is known as ‘dead burnt plaster’.

6.14. PROPERTIES Plaster of Paris is a white powder. It has a remarkable property of setting with water. On mixing with an adequate quantity of water it forms a plastic mass that gets into a hard solid in 5 to 15 minutes. It is due to this reason that it is called plaster. The setting process is exothermic. The addition of common salt accelerates the rate of setting, while a little borex or alum reduces it. The setting of Plaster of Paris is believed to be due to rehydration and its reconversion into gupsum. 2 CaSO4. H2O + 3 H2O 2 CaSO4. 2H2O The effect of heat on gypsum or the dihydrate presents a review of interesting changes.

6.6 ENGINEERING CHEMISTRY: A Textbook

CaSO4. 2H2O (Gypsum)

CaSO4. 2H2O

1200C

(Orthrhombic dihydrate)

CaSO4. 1/2H2O

2000C

CaSO4

Plaster of Paris

At very high temperature (11000C) anhydrous calcium sulphate decomposes into calcium oxide and sulphur trioxide.

CaSO4

11000C

CaO + SO3

6.15. USES The largest use of Plaster of Paris is in the building industry as well as plasters. It is used for immobilising the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

6.16. LUBRICANT Any substance introduced between two moving or sliding surfaces with a view to reduce the frictional resistance between them is known as a lubricant. The process of reducing fractional resistance between moving/ sliding surfaces by the introduction of lubricants in between them is called lubrication.

6.17. CLASSIFICATION OF LUBRICANTS Lubricants can be broadly classified, on the basis of physical state, as follows: 1. Liquid lubricants or lubricating oils 2. Semi-solid lubricants or grease 3. Solid lubricants.

6.17.1. Lubricating Oils Lubricating oils reduce friction and wear between two moving/ sliding metallic surfaces by providing a continuous fluid film in between them. Lubricating oils are further classified as: i. ii. iii.

Animal and vegetable oils Mineral or petroleum oils Blended oils (synthetic emulsion)

6.17.2. Greases or Semi- Solid Lubricants Lubricating greases is a semi-solid, consisting of a soap dispersed throughout liquid lubricating oil. Greases are prepared by saponification of fat with alkali, followed by adding hot lubricating oil while under agitation. Greases have higher shear or frictional resistance than oils and, therefore, can support much heavier loads at lower speeds. They also do not require as much attention unlike the lubricating liquids. But greases have a tendency to separate into oils and soaps.

6.17.3. Dry lubricants or solid lubricants The materials which despite being in the solid phase are able to reduce friction between two surfaces sliding against each other without the need for a liquid

Engineering Materials 6.7 medium. They offer lubrication at temperatures higher than liquid and oil-based lubricants operate. The four most commonly used solid lubricants are Graphite, Molybdenum disulfide, Hexagonal boron nitride (white graphite), Tungsten disulfide.

6.18. MECHANISMS OF LUBRICATION There are mainly three types of mechanism by which

6.18.1. Fluid-film or Thick-film or Hydrodynamic lubrication In this, the moving/ sliding surfaces are separated from each other by a thick-film of fluid (at least 1000 Å thick), so that direct surface to surface contact and welding of junctions rarely occurs. The lubricant film covers/ fills the irregularities of the sliding/ moving surfaces and forms a thick layer in between them, so that there no direct contact between the materials surfaces (Fig. 6.2). This consequently reduces wear. The resistance to movement of sliding/ moving parts is only due to internal resistance between the particles of the lubricant moving over each other.

Fig. 6.2: Fluid film lubrication or Hydrodynamic lubrication Delicate instruments, light machines like watches, clocks, guns, sewing machines, scientific instruments, etc. are provided with the types of lubrication. Hydrocarbon oils are considered to be satisfactory lubricants for fluid-film lubrication.

6.18.2. Boundary lubrication or Thin-film lubrication When a continuous film of lubricant cannot persist and direct metal to metal is possible due to certain reasons. This happens when: i. ii. iii. iv.

A shaft starts moving from rest. The speed is very low. The load is very high Viscosity of the oils is too low. Under such conditions, the clearance space between the moving/ sliding surfaces is lubricated with an oil lubricant, a thin layer of which is adsorbed, (i.e., surface attached) by physical or chemical forces or both are both the metallic surfaces. These adsorbed layers avoid direct metal to metal contact. The load is carried by the layers of the adsorbed lubricant on the both metal surfaces (Fig. 6.3).

6.8 ENGINEERING CHEMISTRY: A Textbook

Fig. 6.3: Boundary lubrication Vegetable and animal oils (glycerides of higher fatty acids), graphite or molybdenum disulphide either alone or as stable suspension in oil is also used for boundary lubrication. These minerals form films on the metal surfaces, which possess low internal friction and can bear compression as well as high temperatures.

1.18.3. Extreme pressure lubrications When the moving / sliding surfaces are under very high pressure and speed, a high local temperature is attained and under such conditions, liquid lubricants fail to stick and many decompose and even vaporize. To meet these extreme pressure conditions, special additives are added to mineral oils. These are called “extreme pressure additives”. These additives from one metal surfaces more durable films, capable of withstanding very high loads and high temperatures. Important additives are organic compounds having active radical or groups such as chlorine (as in chlorinated esters), sulphur (as in sulphurized oils) or phosphorus (as in tricresyl phosphate).

6.19. PROPERTIES OF LUBRICANTS 1.

Viscosity is the property of a liquid or fluid by virtue of which it offers resistance to its own flow. Viscosity should not be too low or too high. (Viscosity is inversely proportional to temperature).

2.

Flash - Points and Fire - Points: Flash Point is the lowest temperature at which the oil lubricant gives off enough vapour that ignites for a moment, when a tiny flame is brought near it, while fire point is the lowest temperature at which the vapour of the oil burn continuously for at least five seconds, when a tiny flame is brought near it.

3.

Oiliness of a lubricant is a measure of its capacity to stick on to the surfaces of machine parts, under conditions of heavy pressure or load. For high pressure, high oiliness oil should be used. It is important for extreme pressure lubrication.

4.

Cloud and Pour points: When oil is cooled slowly, the temperature at which it becomes cloudy or hazy in appearance, is called its cloud point; while the temperature at which the oil ceases to flow or pour, is called its pour point.

5.

Volatility: Good lubricant should have low volatility. It is measured by vaporimeter.

Engineering Materials 6.9 .

6.

Emulsification is the property of oils to get intimately mixed with water, forming a mixture, called emulsion.

7.

Carbon residue: Normally lubricants consist of high percentage of carbon containing compounds. Lubricants decompose due to raise in temperature and deposit carbon creating problems to internal combustion engines and Air compressors. A good lubricant should deposit least amount of the carbon.

8.

Decomposition stability: Lubricating oils must be stable to decomposition at the operating temperatures by oxidation, hydrolysis and pyrolysis.

9.

Aniline point: The minimum equilibrium solution temperature for equal volumes of aniline and oil sample. Good lubricating oil should have higher aniline points. Higher A.P means higher percentage of paraffinic hydrocarbons and hence lower percentage of aromatic hydrocarbons.

10. Precipitation Number: The percentage of asphalt present in oil. Precipitation Number is used to differentiate the different classes of lubricants. 11. Saponification number: Number of milligrams of KOH required to saponify 1g of oil. 12. Mechanical stability: At very high pressures of operation, the stability of a lubricant is judged by four balls extreme pressure lubricant test. 13. Neutralization Number: Is a scale to determine the amount of acidic or basic constituents of oil. Acid Number is the amount of KOH required in milligrams to neutralize the fatty acids in 1g of oil. Good lubricating oil should possess acid value less than 0.1.

6.20. FUNCTIONS OF LUBRICANT 1.

It reduces surface deformation, wear and tear, because the direct contact between the rubbing surfaces is avoided.

2.

It reduces loss of energy in the form of heat. In other words it acts as coolant.

3.

It reduce waste of energy i.e., to increase efficiency of machines.

4.

It reduces expansion of metal by local fractional heat.

5.

It avoids seizure of moving surfaces, since the use of lubricant minimizes the liberation of frictional heat.

6.

It avoids are reduces unsmooth relative motion of the moving/ sliding parts.

7.

It reduces the maintenance and running cast of the machine.

8.

It also, sometimes, acts as a seal. For example, lubricant used between piston and the cylinder wall of internal combustion engine acts as a seal, thereby preventing also the leakage of gases under high pressure from the cylinder.

6.10 ENGINEERING CHEMISTRY: A Textbook

Exercise Long Answer Type Questions 1. Write the chemistry of setting and hardening of cement. 2. Name the raw materials necessary for the preparation of Portland cement. 3. Discuss the chemical changes that occur during the setting of cement. 4. Explain the manufacture of cement in detail. 5. How will you manufacture Portland cement by wet process? 6. Write note on concrete and RCC. 7. Define the term lubricants. Mention their important functions. Explain and discuss the significance of any two properties of lubricants. Short Answer Type Questions 1. Explain the heat of hydration of cement. 2. What is Plaster of Paris? Write its use. 3. Discuss the classification of lubricants with suitable examples. 4. What are chief functions of lubricants? Very Short Answer Type Questions 1. Write a note on lubricants. 2. Write a short account on solid lubricants. 3. What are the functions of lubricants? 4. Discuss in brief lubrication.

Chapter – 7

Water Syllabus: Hardness of water, Disadvantage of hard water, Techniques for water softening; Calgon, Zeolite, Lime-Soda, Ion exchange resin, Reverse osmosis, Water treatment method for boiler feed by internal process.

7.1. INTRODUCTION A major part of all living organisms is made up of water. Human body has about 65% and some plants have as much as 95% water. It is a crucial compound for the survival of all life forms. It is a solvent of great importance. The distribution of water over the earth’s surface is not uniform. The estimated world water supply is given in Table. Table 7.1: Estimated World Water Supply Source % of Total Oceans Saline lakes and inland seas Polar ice and glaciers Ground water Lakes Soil moisture Atmospheric water vapour Rivers

97.33 0.008 2.04 0.61 0.009 0.005 0.001 0.0001

It is a colourless and tasteless liquid. The high heat of vaporisation and heat capacity are responsible for moderation of the climate and body temperature of living beings. It is an excellent solvent for transportation of ions and molecules required for plant and animal metabolism. Due to hydrogen bonding with polar molecules, even covalent compounds like alcohol and carbohydrates dissolve in water.

7.2. HARDNESS OF WATER The soap consuming capacity of water is called hardness of water. The hardness of water is due to the presence of soluble salts of calcium, magnesium and other heavy metal ions, which reacts with soap giving insoluble salts as follows: Ca2+ + 2RCOONa

(RCOO)2Ca  + 2Na+

Mg2+ + 2RCOONa

(RCOO)2Mg  + 2Na+

7.2 ENGINEERING CHEMISTRY: A Textbook Due to these reactions water cannot form lather with soap. Hardness of water is of two types: (i) temporary hardness, and (ii) permanent hardness. 1.

Temporary Hardness: This type of hardness is due to presence of soluble bicarbonate salt of calcium and magnesium (e.g. Ca(HCO3)2, Mg(HCO3)2). Temporary hardness can be removed by boiling as follows: Ca(HCO3)2 Mg(HCO3)2

2.

Heat Heat

CaCO3  + CO2  + H2O Mg(OH)2  + 2CO2 

Permanent Hardness: This type of hardness is due to presence of soluble salts of calcium and magnesium other than bicarbonates (e.g. CaSO4, MgSO4, CaCl2, MgCl2, FeSO4). It does not remove by boiling of water.

7.3. UNITS OF HARDNESS The hardness of water is expressed in following units. 1. Parts per million (ppm): ppm is the parts of calcium carbonate equivalent hardness per 106 parts of water. 2. Milligrams per litre (mg/L): It is the number of milligram of CaCO3 equivalent hardness present per litre of water. 1mg/L = 1mg of CaCO3 equivalent hardness per litre of water But 1L of water weight = 1000gm = 1000 × 1000 mg 1mg/L = 1mg/106 mg = 1ppm 3. Degree French (oFr): It is the parts of CaCO3 equivalent hardness per 105 parts of water. 4. Degree Clark (oCl): It is the parts of CaCO3 equivalent hardness per 70,000 parts of water. These are related to each other as follows: 1ppm = 1mg/L = 0.1o Fr = 0.07o Cl 1o Fr = 10ppm = 10mg/L = 0.7oCl 1o Cl = 14.3ppm = 14.3mg/L = 1.43oFr

7.4. DISADVANTAGE OF HARD WATER Hard water does not give lather with soap. It is, therefore, unsuitable for laundry. It is harmful for boilers as well, because of deposition of salts in the form of scale. This reduces the efficiency of the boiler.

7.5. TECHNIQUES FOR WATER SOFTENING 7.5.1. External Treatment of Boiler Feed Water To remove temporally and permanent hardness simultaneously certain methods like Permutit or Zeolite process, lime-soda process and ion exchange process.

Water 7.3 1. Permutit or Zeolite process Permutit or Zeolite is an inorganic resin having chemical formula Na2O.Al2O3.xSiO2.yH2O (Where x = 2 – 10 and y = 2 – 6). It is also represented as Na2Ze (where Ze = O.Al2O3.xSiO2.yH2O). Zeolite is hydrated sodium alumina silicate, capable of exchanging reversibly its sodium ions for hardness producing ions in water. Zeolites are also known as permutits. Zeolites are of two types. (i). Natural zeolites are non-porous. For example natrolite, Na2O.Al2O3.4SiO2.2H2O. (ii). Synthetic zeolite are porous and possess gel structure. They are prepared by heating together china clay, feldspar and soda ash. Such zeolite possesses higher exchange capacity per unit weight than natural zeolite.

Fig. 7.1 Zeolite process for water softening Permutit or Zeolite exchanges sodium ions with hardness causing ions present in hard water. When hard water is passed through zeolite resin, the sodium ions of zeolite are replaced by hardness causing ions present in water as follows: Na2Ze + Ca(HCO3)2

CaZe + 2NaHCO3

Na2Ze + Mg(HCO3)2

MgZe + 2NaHCO3

Na2Ze + CaCl2

CaZe + 2NaCl

Na2Ze + MgCl2

MgZe + 2NaCl

Na2Ze + CaSO4

CaZe + Na2SO4

Na2Ze + MgSO4

MgZe + Na2SO4

Regeneration When all the sodium ions are replaced by hardness causing ions, then it is said to be exhausted and it stops water softening process. It is regenerated by passing 10% solution of NaCl (called brine) as follows:

7.4 ENGINEERING CHEMISTRY: A Textbook CaZe + 2NaCl

Na2Ze + CaCl2

MgZe + 2NaCl

Na2Ze + MgCl2

Advantages of Zeolite Process i. It removes hardness in water almost completely. ii. There is no formation of precipitates as impurities. iii. It is more efficient and less skilled process. iv. The process is self-adjusting with variation of hardness in water. Limitations of Zeolite Process i. Turbid water cannot be softened, because it blocks the pores of zeolite. ii. If hard water is acidic then it destroys the zeolite. iii. Hot water may dissolve the zeolite. iv. The soft water obtained from zeolite process cannot be used as boiler feed water. 2. Lime-Soda Process In this process the soluble calcium and magnesium salts in water are chemically converted into insoluble compounds, by adding calculated amounts of lime [Ca(OH)2] and soda [Na2CO3]. Calcium carbonate [CaCO3] and magnesium hydroxide [Mg(OH)2] so-precipitated, are filtered off. The lime-soda process may be broadly classified into two categories: i.

Cold Lime-Soda Process: In this process, calculate quantity of chemicals is mixed with water at room temperature. Since precipitation does not take place at room temperature completely A small quantity of coagulant like alumina, ferrous sulphate, aluminium sulphate are added for complete precipitation.

Fig. 7.2 A continuous type hot lime soda process

Water 7.5 ii.

Hot Lime-Soda Process: This method is used specially for boiler feed water. This process is more efficient than cold lime soda process the softening reactions are made to occur almost at the boiling point of water the high temperature of water increases the rate of precipitation, decreases the viscosity of water and eliminates the need of adding coagulants. A continuous type hot lime soda process is described by the Fig. 7.2.

Chemical reactions involved in lime soda process Reaction

Constituent Lime removes the temporary (Ca and Mg) hardness Lime removes the permanent hardness

Ca(HCO3)2 + Ca(OH)2

2CaCO3  + 2H2O

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

2CaCO3  + Mg(OH)2 + 2H2O

MgCl2 + Ca(OH)2 CaCl2 + Na2CO3 MgSO4 + Ca(OH)2 CaSO4 + Na2CO3 FeSO4 + Ca(OH)2 2Fe(OH)2 + H2O + O2 CaSO4 + Na2CO3 Al2(SO4)3 + 3Ca(OH)2 3CaSO4 + 3Na2CO3

Lime removes f ree mineral acids

Soda removes all the soluble Ca permanent hardness

Required

Mg(OH)2  + CaCl2

2L

L+S

CaCO3  + 2NaCl Mg(OH)2  + CaSO4

L+S

CaCO3  + Na2SO4 Fe(OH)2  + CaSO4 2Fe(OH)3 

L+S

CaCO3  + Na2SO4 2Al(OH)2  + 3CaSO4

L+S

3CaCO3  + 3Na2SO4

2HCl + Ca(OH)2

CaCl2 + 2H2O

CaCl2 + Na2CO3

CaCO3  + 2NaCl

H2SO4 + Ca(OH)2

CaSO4 + 2H2O

CaSO4 + Na2CO3

CaCO3  + Na2SO4

CO2 + Ca(OH)2

CaCO3  + H2O

H2S + Ca(OH)2

CaS  + 2H2O

CaCl2 + Na2CO3

L

CaCO3  + 2NaCl

CaSO4 + Na2CO3

CaCO3  + Na2SO4

NaAlO2 + 2H2O

Al(OH)3  + NaOH

L+S

L+S

L

S -L

2NaOH is eq to Ca(OH)2

Calculation of the Amount of Lime and Soda Required From the above chemical reactions it becomes clear that only required amount of lime and soda are to be added. The calculations and amount of lime and soda are based on the amounts of the salts present in water. Now 100 parts by mass of CaCO3 are equivalent to: (i) 74 parts of Ca(OH)2, and (ii) 106 parts of Na2CO3. ∴ Lime requirement for softening Temp. Ca2+ + 2  Temp. Mg2+ + Perm. (Mg2+ + Fe2+ + Al3+) 74 + CO2 + H+ (HCl or H2SO4) + HCO3- - NaAlO2 = 100 all in terms of CaCO3 eq

7.6 ENGINEERING CHEMISTRY: A Textbook ∴ Soda requirement for softening =

106 100

Perm. (Ca2+ + Mg2+ + Al3+ + Fe2+) + H+ (HCl or H2SO4) - HCO3all in terms of CaCO3 eq

Advantages of Lime-Soda Process i. It is very cheap and economical. ii. Lesser amount of coagulants will be required, if this process is combined with sedimentation and coagulation. iii. Iron and manganese are also removed from water. iv. The number of pathogenic bacteria in water is considerably reduced, due to alkaline nature of treated water. Disadvantages of Lime-Soda Process i. Careful operation and skilled supervision is required in lime-soda process for softening of water. ii. Formation of large amount of sludge creates its disposal problem. iii. Lime-soda softeners do not produce water of zero hardness. iv. Softened water from lime-soda softeners is not good for boilers. 3. Ion Exchange Process This process is based on exchange of cations and anions present in hardness causing molecules of water with ion exchange resin. In this method, firstly we describe about ion exchange resins and after that about the process. The ion exchange resins are insoluble, cross linked, long chain organic polymers with a micro porous structure, and the ‘functional groups’ attached to the chains are responsible for the ion exchanging properties. It is categorised into two types viz., Cation exchange resin and anion exchange resin. Resins containing acidic functional groups (-COOH, SO3H, etc.) are capable of exchanging their H+ ions with other cations, which comes in their contact; whereas those containing basic functional groups (-NH2 = NH2 as hydrochloride) are capable of exchanging their anions with other anions, which comes in their contact. The ion exchange resins may be classified as: i. Cation exchange resin (RH+) are mainly styrene-divinyl benzene copolymers, which on sulphonation or carboxylation, become capable to exchange their hydrogen ions with the cations in the water. CH2

CH

CH2

CH

CH2

SO3-H+ CH2

CH

SO3-H+

CH2

CH

CH2

SO3-H+ CH

CH2

CH

CH2

SO3-H+

Acidic or cationic exchange resins (sulphonate form)

Water 7.7 ii. Anion exchange resins (R'OH) are styrene divinyl benzene or amine formaldehyde copolymers, which contain amino or quaternary phosphonium or tertiary sulphonium groups as an integral part of the resin matrix. These, after treatment with dil. NaOH solution, become capable to exchange their OH- anions with anions in water. CH2

CH

CH2

CH

CH2

CH2NMe2+OHCH2

CH

CH2

CH

CH2NMe2+OH-

CH

CH2

CH2NMe2+OHCH2

CH

CH2

CH2NMe2+OH-

Basic or anionic exchange resins (hydroxide form) In exchange process hard water is passed through cation exchanger where hardness causing ion is exchange by H+ ion as follows: 2RH+ + Ca2+

R2Ca2+ + 2H+

2RH+ + Mg2+

R2Mg2+ + 2H+

The water coming out from cation exchanger passes through anion exchanger, where the anions are exchanged by OH- as follows:

ROH- + Cl-

RCl- + OH-

2ROH- + SO42-

R2SO42- + 2OH-

2ROH- + CO32-

R2CO32- + 2OH-

H+ and OH- ions (released from cation exchange and anion exchange columns respectively) get combined to produce water molecule.

H+ + OH-

H2O

Thus, the water coming out from the exchanger is free from cations as well as anions. Ion-free water is known as deionized or demineralized water. Regeneration When cation and anion exchange resins get exhausted, then they are regenerated as follows: 1.

Cation exchange resin is regenerated by passing dil. HCl or dil. H2SO4. The regeneration can be represented as:

7.8 ENGINEERING CHEMISTRY: A Textbook R2Mg2+ + 2H+

2RH+ + Mg2+

R2Ca2+ + 2H+

2RH+ + Ca2+

The column is washed with deionised water and washing (which contains Ca2+, Mg2+, etc. and Cl- and SO42- ions) is passed to sink or drain. 2.

Anion exchange resin is regenerated by passing dil. NaOH solution The regeneration can be represented as:

RCl- + OHR2SO42- + 2OH-

ROH- + Cl2ROH- + SO42-

R2CO32- + 2OH-

2ROH- + CO32-

The column is washed with deionised water and washing (which contains Na+, SO42-, etc. or Cl- and SO42- ions) is passed to sink or drain. The regenerated ion exchange resins are then used again. The typical diagram of ion exchange equipment is given as follows:

Fig.7.3 Ion exchange equipment Advantages of Ion Exchange Resin i. The process can be used to soften highly acidic or alkaline water. ii.

It produces water of very low hardness (2 ppm). So it is very good for treating water for use in high pressure boilers.

Disadvantages of Ion Exchange Resin i.

The equipment is costly and more expensive chemicals are needed.

ii.

If water contains turbidity, then the output of the process is reduced. The turbidity must be below 10 ppm. If it is more, it has to be removed first by coagulation and filteration.

Water 7.9

7.5.2.

Water Treatment Method for Boiler Feed by Internal Process

In many industries water is used for generation of steam in boilers. The water obtained from natural sources in sometimes supplied as such without treatment. The boiler feed water requires following conditions: 1. The hardness of water should be below 0.2 ppm. 2. The caustic alkalinity should be in the range of 0.15 to 0.45 ppm. 3. The soda alkalinity should from 0.45 to 1.0 ppm. 4. Excess soda ash should be 0.3 to 0.5 ppm. If boiler feed water contains excess of impurities then it causes the following effects. 1. Scale or sludge formation 2. Boiler corrosion 3. Priming and foaming 4. Caustic embrittlement 5. Scale due to presence of silica Scale and Sludge Formation in Boilers Sludge is a soft, loose and slimy precipitate formed within the boiler. Sludge can easily be scrapped off with a wire brush. It is formed at comparatively colder portion of the boiler and collects in areas of the system, where the flow rate is slow or at bends. Sludge are formed by substances which have greater solubility in hot water than in cold water, e.g., MgCO3, MgCl2, CaCl2, MgSO4, etc. Scales are hard deposits, which stick very firmly to the inner surfaces of the boiler. Scales are difficult to remove, even with the help of hammer and chisel. Scales are the main source of boiler troubles. Formation of scales may be due to: 1.

Decomposition of calcium bicarbonate: Ca(HCO3)2

CaCO3  + H2O + CO2  Scale

2.

However, scale composed chiefly of calcium carbonate is soft and is the main cause if scale formation in low pressure boilers. But in high pressure boilers, CaCO3 is soluble.

CaCO3 + H2O 3.

Ca(OH)2 (soluble) + CO2 

Deposition of calcium sulphate: CaSO4 is soluble in cold water, but almost completely insoluble in super-heated water. Consequently, CaSO4 gets prsecipitated as hard scale on the heated portion of boiler. This is the main cause of scale in the high- pressure boilers.

Note: Calcium sulphate scale is quite adherent and difficult to remove, even with the help of hammer and chisel. 1. Hydrolysis of magnesium salts: Dissolved magnesium salts undergo hydrolysis (at prevailing high temperature inside the boiler) forming magnesium hydroxide precipitate, which forms a soft type scale, e.g.,

7.10 ENGINEERING CHEMISTRY: A Textbook MgCl2 + 2H2O

Mg(OH)2  + 2HCl  Scale

2.

Presence of silica: (SiO2), even present in small quantities, deposits as calcium silicate (CaSiO3) and/ or magnesium silicate (MgSiO3). These deposits stick very firmly on the inner side of the boiler surface and are very difficult to remove. One important source of silica in water is the sand filter.

Fig. 7.4 Scale and sludge in boilers Caustic embrittlement Development of certain types of cracks resulting from excessive stress and chemical attack leading to boiler failures is called embrittlement. Chemicals, like NaOH are believed to responsible for embrittlement in a steam boiler operation. These cracks appear like a brittle fracture and hence called caustic embrittlement. Some of the important methods used in the internal treatment of water softening include calgon conditioning, colloidal conditioning, carbonate conditioning, phosphate conditioning, etc. Calgon’s Method Sodium hexametaphosphate (Na6P6O18), commercially called ‘calgon’, when added to hard water, the following reactions take place. Na6P6O18 M2+ + Na4P6O182-

2Na+ + Na4P6O182[Na2MP6O18]2- + 2Na+

(M = Mg, Ca)

The complex anion keeps the Mg2+ and Ca2+ ions in solution. This method is not applicable for the prevention of iron oxide and copper deposition.

7.6. REVERSE OSMOSIS Osmosis describes flow of the solvent from dilute to concentrated solution through a semipermeable membrane. Whereas reverse osmosis describes the flow of solvent in opposite direction i.e. from concentrated solution to dilute solution across a semipermeable membrane by applying hydrostatic pressure in excess of osmotic

Water 7.11 pressure. The cellulose acetate and the more recently used polymethacrylate and polyamide membranes do not allow the solute to pass, while the solvent is forced through and collected as a pure solvent in a direction as shown in Fig.7.5.

Fig.7.5 Reverse Osmosis

7.7. ADVANTAGES OF REVERSE OSMOSIS i. ii. iii. iv. v. vi.

Easy maintenance and economical, as the membrane lifespan is high. Reverse osmosis possesses a distinct advantage of removing ionic as well as non-ionic, colloidal and high molecular weight organic matter. It removes colloidal silica, which is not removed by demineralization. The main tenancy cost is almost entirely on the replacement of the semipermeable membrane. The life time of membrane is quite high, about two years. The membrane can be replaced within a few minutes, thereby providing nearly uninterrupted water supply.

Example 7.1. A sample of hard water has hardness 500 ppm. Express the hardness in 0Fr and 0Cl. Solution Since, 1ppm = 0.10Fr Hence, 500 ppm = 0.1 x 500 = 50 0Fr Again, 1ppm = 0.07 0Cl 500ppm = 0.07 x 500 = 350Cl Example 7.2. Calculate temporary and total hardness of water sample of water containing Mg(HCO3)2 = 9.3mg/L, Ca(HCO3)2 = 17.4mg/L, MgCl2 = 8.7mg/L and CaSO4 = 12.6mg/L. Solution

7.12 ENGINEERING CHEMISTRY: A Textbook To calculate the temporary and total hardness in water sample we have to express the hardness causing salt into CaCO3 equivalent. Strength in term of CaCO3 equivalent =

strength of hardness causing salt  equivalent wt. of CaCO3 Equivalent wt. of hardness causing salt

Temporary hardness = CaCO3 equivalent of [Mg(HCO3)2 + Ca(HCO3)2] =

9.3  50 73

+

17.4  50 81

= 6.37 + 10.74 = 17.11 mg/ L Permanent hardness = CaCO3 equivalent of [MgCl2 + CaSO4] =

8.7  50 47.5

+

12.6  50 68

= 9.16 + 9.26 = 18.42 mg/ L Total hardness = 17.11 + 18.42 = 35.53mg/ L

Example 7.3. 100 mL of water sample has a hardness equivalent of 12.5 mL of 0.08N MgSO4. What is its hardness in ppm. Solution. N1V1 (water sample) = N2V2 (MgSO4 solution) N1 = ?, N2 = 0.08, V1 = 100mL V2 = 12.5 mL N1 × 100 = 0.08 × 12.5 =

0.08  12.5

= 0.01N

100

Hardness of water sample = Normality × Equivalent wt. of CaCO3 = 0.01 × 50 = 0.5g/L Hardness in ppm = 0.05 × 1000 = 500 ppm Example 7.4. Calculate temporary and permanent of a water sample which analysed as: Ca(HCO3)2 = 21.0mg/L, Mg(HCO3)2 = 25.0 mg/L, CaCl2 = 16.4 mg/L, MgCl2 = 5.2mg/L(Odd sem. 2011 - 2012) Solution. Strength in term of CaCO3 equivalent =

strength of hardness causing salt  equivalent wt. of CaCO3 Equivalent wt. of hardness causing salt

Water 7.13 Temporary hardness = CaCO3 equivalent of [Mg(HCO3)2 + Ca(HCO3)2] =

25  50 73

+

21  50 81

= 17.12 + 12.96 = 30.08 mg/ L Permanent hardness = CaCO3 equivalent of [MgCl2 + CaSO4] =

5.2  50 47.5

+

16.4  50 55.5

= 5.45 + 14.77 = 20.22 mg/ L

Example 7.5. Calculate the quantity of lime and soda required for softening 50,000 litres of water containing the following salts per litre. Ca(HCO3)2 = 9.2 mg; Mg(HCO3)2 = 7.9 mg; CaSO4 = 15.3 mg; MgSO4 = 15.0 mg; MgCl2 = 3.0 mg and NaCl = 4.3 mg. Solution: Constituent

Amount mg/L

Multiplication Factor

CaCO3 Equivalent mg/L

Ca(HCO3)2

9.2

100/162

5.68

Mg(HCO3)2

7.9

100/146

5.41

CaSO4

15.3

100/136

11.25

MgSO4

15.0

100/120

12.5

MgCl2

3.0

100/95

3.16

NaCl

4.3

Neglected

-

∴ Lime requirement = 74 [Ca(HCO3)2 + 2Mg(HCO3)2 + MgSO4 + MgCl2 in terms 100 of CaCO3 eq.] 74 = [5.68 + 2 (5.41) + 12.5 + 3.16] mg/L = 23.39 mg/L 100 Lime required for softening 50,000 litres = 23.40 × 50,000 mg = 11.70 g = 1.17 kg. Soda requirement = 106 [CaSO4 + MgSO4 + MgCl2 as CaCO3 eq.] 100 106 = [11.25 + 12.5 + 3.16] = 28.5 mg/L 100

Soda required for softening 50,000 litres = 28.52 × 50,000 mg = 14.26 g = 1.426 kg

7.14 ENGINEERING CHEMISTRY: A Textbook Example 7.6. Calculate the amount of lime and soda required per litre for chemical treatment of water containing: Ca2+ = 80 ppm; Mg2+ = 32 ppm; HCO3- = 195 ppm; FeSO4.7H2O added as coagulant = 73.5 ppm. Solution Conversion to CaCO3 eq. Constituent Amount (ppm)

Multiplication Factor

CaCO3 Equivalent (ppm)

Ca2+

80

100/40

200

Mg2+

32

100/24

133.3

HCO3-

195

100/122

159.8

FeSO4.7H2O

73.5

100/278

26.44

Lime requirement = 74 [Mg2+ + HCO3- + FeSO4.7H2O as CaCO3 eq.] 100

= 74 [133.3 + 159.8 + 26.44] ppm = 236.6 ppm

100 106 Soda requirement = [Ca2+ + Mg2+ + FeSO4.7H2O - HCO3- as CaCO3 eq.] 100 = 106 [200 + 133.3 + 26.44 – 159.8] = 211.94 ppm 100

Note: If analytical report shows the quantities of Ca2+ and Mg2+, then 1 eq. of lime and 1 eq. of soda is required for Mg2+, whereas 1eq. of soda is required for Ca2+. In other words, the ions Ca2+ and Mg2+ are treated as permanent hardness due to calcium and magnesium. Example 7.7. Calculate the amount of lime (92% pure) and soda (98% pure) required for treatment of 30,000 litres of water whose analysis is as follows: Ca(HCO3)2 = 40.5 ppm; Mg(HCO3)2 = 36.5 ppm; CaSO4 = 34.0 ppm; MgSO4 = 30.0 ppm; CaCl2 = 27.75 ppm and NaCl = 10.0 ppm. Solution Conversion to CaCO3 eq. Constituent Amount (ppm)

Multiplication Factor

CaCO3 Equivalent (ppm)

Ca(HCO3)2

40.5

100/162

25

Mg(HCO3)2

36.5

100/146

25

CaSO4

34.0

100/136

25

MgSO4

30.0

100/120

25

CaCl2

27.75

100/111

25

NaCl

-

-

-

Water 7.15 Lime requirement = 74 [Ca(HCO3)2 + 2Mg(HCO3)2 + MgSO4 in terms of CaCO3 100

eq. × volume of water] = 74 [25 + (2 × 25) + 25] mg/L × 30,000 L = 74 × 30,000 L = 2220 g 100

Since lime is 92% pure, therefore lime required will be = 100 × 2220 g = 2.413 kg 92

Soda requirement = 106 [CaSO4 + MgSO4 + CaCl2 as CaCO3 eq.] 100 106 = [25 + 25 + 25] mg/L × 30,000 L = 2385 g 100

98% pure soda, required for softening 30,000 L = 100 98

× 2385 = 2433 g = 2.433 kg

Note: If the lime and soda are impure, then the actual requirement is calculated accordingly. In the above example as lime and soda are 92% and 98% pure respectively, therefore, the values obtained in each case are multiplied by purity factor ( 100 for lime; = 100 for soda). 98

92

Example 7.8. Calculate the amount of lime (74% pure) and soda (92% pure) required for treatment of 20,000 litres of hard water containing, MgCO3 = 84 mg/L; CaCO3 = 40 mg/L; MgCl2 = 95 mg/L; CaCl2 = 111 mg/L; Mg(NO3)2 = 37 mg/L and KCl = 30 mg/L. Solution: Conversion to CaCO3 eq. Constituent

Amount (ppm)

Multiplication Factor

CaCO3 Equivalent (ppm)

MgCO3

84

100/84

100

CaCO3

40

100/100

40

MgCl2

95

100/95

100

CaCl2

111

100/111

100

Mg(NO3)2

37

100/148

25

KCl

-

-

-

Lime requirement = 74 [2 × MgCO3 + CaCO3 + MgCl2 + Mg(NO3)2 in terms of 100

CaCO3 eq.] × volume of water × purity factor = 74 [2 × 100 + 40 + 100 + 25] mg/L × 20,000 L × 100 = 7300g = 7.3 kg 100

74

7.16 ENGINEERING CHEMISTRY: A Textbook Soda requirement = 106 [MgCl2 + CaCl2 + Mg(NO3)2 in terms of CaCO3 eq.] × 100

volume of water × purity factor = 106 [100 + 100 + 25] mg/L × 20,000 L × 100 = 5184 g = 5.2 kg 100

92

Note: CaCO3 and MgCO3 are regarded as being present in the form of their bicarbonates, only their weights have been expressed in terms of CaCO3 and MgCO3. So the lime requirement for MgCO3 is 2 eq. as in the case of Mg(HCO3)2 whereas for CaCO3 it is one eq. No soda is required for either Ca or Mg if they are given as CaCO3 and MgCO3. Example 7.9. A zeolite softener was 95% exhausted, when 10,000 L of hard water was passed through it. The softener required 150 L of NaCl solution of strength 50 g NaCl/L of solution to regenerate. What is the hardness of water? Solution: 10,000 L of hard water = 150 L of NaCl solution = 150 L × 50 g/L of NaCl = 7500 g of NaCl = 7500 × 100 g of CaCO3 eq. 58.5  2 = 6410.26 g of CaCO3 eq 1 L of hard water = 6410.26 g of CaCO3 eq/10,000 = 0.6410 g of CaCO3 eq. = 641 mg of CaCO3 eq Hence hardness of water = 641 mg/L Example 7. 10. A zeolite softener was completely exhausted and then regenerated by passing 200 litres of NaCl solution containing 100 g per litre of NaCl. How many litres of a sample of water of hardness 500 ppm can be softened by the softener. Solution: 200 litre of NaCl = 100 L × 200 g/L = 20000 g of NaCl = 20000 × 100 g of CaCO3 eq. 58.5  2 = 17094 g of CaCO3 eq. Now 500 mg of hardness is present in = 1 litre of water 17094 × 103 mg of CaCO3 eq. hardness is present in = 1 × 17094 × 103 L 500 = 34188 litre of hard water. Hence, the softener can soften 34188 litres of hard water.

Water 7.17

Exercise 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

13.

14.

15.

What is hardness of water? Describe ion exchange process for making soft water from hard water. What do you mean by the term permutit? Describe the softening of water by permutit process. Describe Zeolite Process for making soft water from hard water. Write a short note on softening of water by Zeolite Process. Explain why Alkalinity of water cannot be due to simultaneous presence of OH-, CO32- and HCO3-? Water contains 408 mg CaSO4 per litre. Calculate the hardness is litres of CaCO3 equivalent. Give a brief account of the treatment of boiler feed water of calgon process. What are scales? What are their disadvantages? What is calgon conditioning? How is it better than phosphate conditioning? Explain scale and sludge formation in boilers. How are they removed? 100 mL of water sample has a hardness equivalent of 12.5 mL of 0.08 N MgSO4. What is its hardness in ppm? An exhausted Zeolite softener was regenerated by passing 150 litres of NaCl, having a strength of 150 g/L of NaCl. How many litres of hard water sample, having hardness of 600 ppm, can be softened, using the softener? A water sample contains the following impurities: Ca2+ = 20 ppm, Mg2+ = 18 ppm, HCO3- =183 ppm and SO42- = 24 ppm. Calculate the lime and soda needed for softening (Ca = 40, Mg = 24). Calculate the amount of lime and soda required for the treatment of 15,000 litres of water, which analysed as follows: temporary hardness = 20 ppm; permanent hardness = 15 ppm; permanent Mg hardness = 10 ppm. Write a note on reverse osmosis.

Chapter – 8

Phase Rule Syllabus: Phase Rule and its application to one component system (water and sulphur).

8.1. INTRODUCTION The Phase rule was proposed by J.W. Gibbs in 1874. According to the phase rule, “for any heterogeneous system in equilibrium the sum of number of phases and degrees of freedom is greater than the number of components by two.” Mathematically, P+F=C+2 F=C–P+2 Where, P – Number of phases F – Degrees of freedom C – Number of component

8.2. PHASES The physically distinct, homogenous and mechanically separable part of heterogeneous system in equilibrium is called a ‘Phase’. For example, water at freezing temperature exists in three forms in equilibrium as follows H2O (s)

H2O (l)

H2O (g)

It is three phase system. System can be classified into two categories based on the number of phases.

8.2.1. Homogeneous System A system containing one phase only (P = 1) is called homogeneous system. Examples 1. Any gaseous mixture- air 2. Two miscible liquids- water + alcohol 3. An aqueous solution- sugar solution 4. Homogeneous solid solution- Mohr’s salt 5. Any pure substance- O2, benzene, ice

8.2.2. Heterogeneous System A system containing more than one phases (P > 1) is called heterogeneous system. Examples 1. Two immiscible liquids- H2O + oil (P = 2)

8.2 ENGINEERING CHEMISTRY: A Textbook 2. 3. 4.

Saturated sugar solution (P = 2) Mixture of solids- MgCO3 + MgO (P = 2). Here MgCO3 (s) and MgO (s) but different physical and chemical properties and makes separate phase. Freezing water mixture (P = 3) CaCO3 (s)

6.

H2O (g)

H2O (l)

H2O (s)

5.

CaO (s) + CO2 (g)

(P = 3)

8.3. COMPONENT The number of components of a system in equilibrium is defined as the minimum number of independently variable constituents, which are required to express the composition of each phase. For example, i.

In the water system H2O (s)

H2O (l)

H2O (g)

It is one component system because the constituent of each phase can be expressed in term of water only. ii. The sulphur system consists of four phases, rhombic, monoclinic, liquid and vapour, the chemical components of all phases is S. Hence it is one component system. iii. In the dissociation of NH4Cl in a closed vessel, NH4Cl(s)

NH3(g) + HCl(g)

NH4Cl(g)

The proportions of NH3 and HCl are equivalent and hence, the composition of both phases (solid and gaseous) can be expressed in terms of NH4Cl alone. Hence, the number of component is one. However, if NH3 or HCl is in excess, the system becomes a two component system. i. A system of saturated solution of NaCl consists of solid salts, salt solution and water vapour. The chemical composition of all the three phases can be expressed in terms of NaCl and H2O. Hence, it is a two component system. ii.

In the thermal decomposition of CaCO3 CaCO3 (s)

CaO (s) + CO2 (g)

The composition of any phase can be expressed in terms of at least any two of the independent variable constituents, CaCO3, CaO and CO2. Suppose CaCO3 and CaO are chosen as the two components, then the composition of different phases is represented as follows: CaCO3 solid phase CaCO3 = CaCO3 + 0Ca CaO solid phase CaO = 0CaCO3 + CaO CO2 gaseous phase CO2 = CaCO3 - CaO Thus it is a two component system. i. In the dissociation reaction, CuSO4.5H2O(s)

CuSO4.3H2O(s) + 2H2O(g)

Phase Rule 8.3 The composition of each phase can be represented by the simplest components, CuSO4 and H2O. Hence, it is a two component system. Number of components of a system may alternatively be defined as the number of chemical constituents of the system minus the number of equations relating to those constituents in equilibrium state. For example: Dissociation of KClO3 in a closed vessel: Following equilibrium exists: 2 KClO3(s)

∴

2 KCl(s) + 3 O2(g)

N0. of constituents = 3 Keq =

Now

[KCl]2[O2]3 [KClO3]2

= [O2]3

[Active mass of solid is taken constant] No. of equations relating the concentration of constituents = 1. Hence, number of components = 3 – 1 = 2 i.e., it is a two component system.

8.4. DEGREES OF FREEDOM The minimum number of variable factors such as Temperature (T), Pressure (P) and Concentration (C/V) that can be independently changed without altering the state of system in equilibrium is called degree of freedom. In other words, the minimum number of variable factors, such as Pressure, Temperature and Concentration, which should be fixed in order to define the system completely, is called Degrees of Freedom (F). Degree of system may be 0, 1, 2…… For Examples: i. In case of water system, Ice (s)

Water (l)

Vapour (g)

If all the three phases are present in equilibrium, then no conditions need to be specified, as the three phases can be in equilibrium only at particular temperature and pressure. The system is therefore, zero variant or non-variant or invariant or has no degree of freedom. If condition (e.g., temperature or pressure) is altered, three phases will not remain in equilibrium and one of the phases disappears. ii.

For the system consisting of water in contact with its vapour,

Water (l)

Water vapour (g)

We must state either the temperature or pressure to define it completely. Hence, degree of freedom is one or system is univariant. iii.

For a system consisting of water vapour phase only, we must state the values of both the temperature and pressure in order to describe the system completely. Hence, the system is bivariant or has two degree of freedom.

iv.

For the system consisting of, NaCl (s)

NaCl-water (aq)

Water vapour (g)

8.4 ENGINEERING CHEMISTRY: A Textbook we must state either the temperature or pressure, because the saturation solubility is fixed at a particular temperature or pressure. Hence, the system is univariant.

8.5. SIGNIFICANCE OF TRIPLE POINT Triple point is a characteristic property of pure substances. It is the condition of temperature and pressure at which the three phases exist in equilibrium. It marks the lowest temperature at which liquid can exist. At triple point, P=3, C=1 and therefore, F=1-3+2=0 (Zero).

8.6. PHASE DIAGRAM OF WATER SYSTEM Under normal conditions the system water is a three phase and one component system. The three phases are liquid, ice and vapour. All these are represented by one chemical entity (H2O), hence it is one component system. On the basis of experimental data obtained for water system a plot of relationships between various phases namely solid ice, liquid water and water vapour under different conditions of temperature and pressure are shown in Fig. 8.1. C Fusion Curve

Critical Pressure A

218 atm Solid Ice

Liquid Water Vapour Pressure Curve

1 atm O 0.58 mm

Pressure (Not to scale)

Water Vapour

Triple Point A’ Sublimation Curve

Critical Temperature

B 0 0.0075

100

Temperature, 0C (Not to scale)

Fig. 8.1. Phase diagram of water system The phase diagram consist of 1.

Areas AOB, BOC and AOC

2.

Curves OA, OB, OC & OA’

3.

Triple point O

374

Phase Rule 8.5 The salient features of phase diagram are given as below S. No. 1

Areas /Points/ Curves Curve OA

Name Vapour pressure curve Sublimation curve

Phases in equilibrium

vapour

Liquid

Degree of freedom (Variance) 1 (univariant)

2

Curve OB

3

Curve OC Area AOB

Fusion curve -------

Vapour

2 (bivariant)

Area BOC Area AOC

-------

Solid

2 (bivariant)

---------

Liquid

2 (bivariant)

4 5 6 7

Point O

Triple Point

8

Curve OA’

Metastable curve (Supercooled and unstable state)

Solid

Liquid

Liquid

Liquid

Solid

vapour

1 (univariant)

Solid

1 (univariant)

vapour

0 (non variant)

vapour

1 (univariant)

8.7. PHASE DIAGRAM OF SULPHUR SYSTEM It is one component, four phase system. The four different phases are Rhombic Sulphur (SR), Monoclinic Sulphur (SM), Liquid Sulphur (SL), Vapour Sulphur (Sv). As all the four phases can be represented by only one chemical entity ‘Sulphur’ (S), it is one component system. The phase diagram consists of 1.

Six stable curves AB, BC, CD, BE, EG & CE

2.

Three metastable curves FB, FC & FE

3.

Four triple points B, C, E & F

4.

Four areas ABCD, DCEG, BCE & ABEG

8.6 ENGINEERING CHEMISTRY: A Textbook G

SR/L 1290 atm

E Solid Rhombic (SR) SR/SM

Liquid (L)

SM/L

1 atm

D

Solid Monoclinic (SM) Pressure (atm)

C

F SR/V

B

L/V

SM/V

A

Vapour (V) B.P. 95.6

120

114 Temperature,

165

444.6

0C

Fig. 8.1. Phase diagram of sulphur system Summary for salient features of sulphur system is given as follows S. No.

4

Areas/ Points/ Curves Curve AB Curve BC Curve CD Curve BE

Sublimation curve of SR Sublimation Curve of SM Vapour pressure curve of SL Transition curve

5

Curve CE

6

Curve EG

1 2 3

Name

Phases in equilibrium SR

SV

SM

SV

SL

SV

SR

SM

Fusion curve of SM

SM

SL

Fusion curve of SR

SR

SL

Degree of freedom (Variance) 1 (univariant) 1 (univariant) 1 (univariant) 1 (univariant) 1 (univariant) 1 (univariant)

Phase Rule 8.7 7

Point B

First triple point

SR

SM

SV

8

Point C

Second triple point

SL

SM

SV

9

Point E

Third triple point

SR

SM

SL

10

Point F

SR

SM

11

Region ABCD

Metastable triple point ---------

12

Region DCEG

--------

SL

2 (bivariant)

13

Region BCE

---------

SM

2 (bivariant)

14

Region ABEG

---------

SR

2 (bivariant)

15

Curve FB

Meta vaporization curve of SR

SR

16

Curve FC

Meta vaporization curve of SL

SL

SV

1 (univariant)

17

Curve FE

Meta fusion curve of SR

SR

SV

1 (univariant)

SV

SV

SV

0 (non variant) 0 (non variant) 0 (non variant) 0 (non variant) 2 (bivariant)

1 (univariant)

Note: The degree of freedom cannot have a minus value, therefore out of four possible phases, only three are present at a time. It is not possible for a single component system to have four phases together at equilibrium.

8.8. ADVANTAGES OF PHASE RULE 1. 2. 3. 4. 5. 6.

Phase rule is applicable to both physical and chemical equilibrium. It is applicable to macroscopic systems and hence no information is required to regarding molecular/ micro structure. It is a convenient method to classifying equilibrium states in terms of phases, components and degree of freedom. The behaviour of system can be predicted under different conditions. According to phase rule, different systems behave similarly if they have same degree of freedom. It predicts under given conditions whether a number of substances taken together would remain in equilibrium as such or would involve inter-conversion or elimination of some of them.

8.8 ENGINEERING CHEMISTRY: A Textbook

8.9. LIMITATIONS OF PHASE RULE 1. 2. 3. 4.

Phase rule is applicable only for those systems which are in equilibrium. It is not of much of use for those systems which attain the equilibrium state very slowly. Only three degree of freedom viz. temperature, pressure and composition are allowed to influence the equilibrium systems. Under the same conditions of temperature and pressure, all the phases of the system must be present. It considers only the number of phases, rather than their amounts.

Examples 8.1. Explain why KCl – NaCl – H2O should be regarded as a three component system whereas KCl – NaBr – H2O should be regarded as four component system? Solution: Number of components of a system (C) = Number of chemical constituents of the system (N) - Number of equations relating to those constituents in equilibrium state (E) For KCl – NaCl – H2O system N = 3 (KCl, NaCl, H2O) E = 0 (there is no number of equations relating to those constituents in equilibrium state) C=3–0=3 For KCl – NaBr – H2O system N = 5 (KCl, NaBr, NaCl, KBr, H2O) E = 1 (KCl + NaBr

KBr + NaCl)

C=N–E=5–1=4 Examples 8.2. Is it possible to have a quadruple point on a phase diagram for a one component system? Solution: From the phase rule, F = C – P + 2 For a quadruple point, P = 4 F = 1 – 4 + 2 = -1 That is, F has negative value, which is ridiculous. Thus the answer is no. Examples 8.3. Write the number of component, phases and degrees of freedom of the following reaction. N2O4(g)

2NO2(g)

Solution: Number of components, C = N – E = 1 – 0 = 1 P=1 F=C–P+2=1–1+2=2

Phase Rule 8.9

Exercise 1. 2. 3. 4. i. ii. 5.

What is Gibb’s phase rule? Define the term; phase, component and degree of freedom. Draw the phase diagram of sulphur system and also give the significance of triple point. With the help of phase rule diagram, show how it is possible to have super cooled water? Define and explain the terms involved in phase rule. Draw a neat labelled diagram of water system and explain the areas, curves and points in this system. Is it possible to have a quadruple point in a phase diagram for one component system? Determine the number of degree of freedom in each of the following systems:Liquid water and water vapour in equilibrium. Liquid water and water vapour in equilibrium at 1 atmosphere pressure. Define the terms: phases, component and degree of freedom, write the number of component, phases and degrees of freedom of the following reaction N2O4(g)

2NO2(g)

Chapter – 9

Fuels Syllabus: Fuels-classification of fuels, Determination of calorific values, Analysis of coal, Biogas and Biomass.

9.1. INTRODUCTION Fuel is a combustible substance, containing carbon as main constituent, which on proper burning gives large amount of heat, which can be used economically for domestic and industrial purposes. Wood, charcoal, coal, kerosene, petrol, diesel, producer gas, oil gas, etc. are some of the fuels. During the process of combustion of a fuel (like coal), the atoms of carbon, hydrogen etc. combines with oxygen with the simultaneous liberation of heat at a rapid rate. This energy is liberated due to the “rearrangement of valency electrons” in these atoms, resulting in the formation of new compounds (like CO2, H2O, etc.) are called as products of combustion. These new compounds have less energy (or heat content) in them and, therefore, the energy (or heat) released during the combustion process is the difference in the energy of the reactants (C, H, and O, etc. of fuel) and that of the products formed.

Fuel + O2 High energy contents

Products + Heat Low energy contents

The primary or main sources of fuels are coals and petroleum oils, the amounts of which are dwindling day-by-day. These are stored fuels available in earth’s crust and are generally called as ‘fossil fuels’.

9.2 CLASSIFICATION OF FUELS Fuels are classified on the basis of either occurrence or physical state.

9.2.1. On the Basis of Occurrence On the basis of occurrence, fuels are classified as follows – primary or natural fuels and secondary or artificial or manufactured fuels. 1.

Primary or Natural Fuels: Fuels which occur in nature as such and can be used directly without processing or after little processing which does not change the chemical composition of parent fuel are known as primary fuels. For example, fossil fuels like wood, coal, petroleum, natural gas, crude oil, etc.

2.

Secondary or Artificial or Manufactured Fuels: The fuels which are derived from the primary fuels by chemical processing are known as secondary fuels. For example, coke, charcoal, kerosene, coal gas, diesel, etc.

9.2 ENGINEERING CHEMISTRY: A Textbook

9.2.2. On the Basis of Physical State On the basis of physical states, fuels are classified as solid fuels, liquid fuels and gaseous fuels. 1. Solid Fuels: Those fuels which satisfy the characteristic features of a solid substance are called solid fuels. 2. Liquid Fuels: Those fuels which satisfy the characteristic features of a liquid substance are called liquid fuels. 3. Gaseous Fuels: Those fuels which satisfy the characteristic features of a gaseous substance are called gaseous fuels. Examples of Fuels Based on Classifications Occurrence Physical state Solid Fuels Liquid Fuels Gaseous Fuels

Natural Fuels

Artificial Fuels

Wood, coal, oil shale Petroleum Natural gas

Charcoal, coke, straw, briquettes Oils, coal tars, alcohols Coal gas, producer gas, water gas, hydrogen, acetylene, oil gas

9.3. CHARACTERISTICS OF A GOOD FUEL 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

A good fuel should possess high calorific value. An ideal fuel should have moderate ignition temperature. The fuel should have low moisture content. A fuel should have low content of non-combustible matter. A fuel must be renewable in nature. A fuel must be easily accessible. A fuel must be easy to handle. A fuel must be convenient to store and safe to transport. A good fuel must be economical. Combustion should be easily controllable. The fuel should burn in air with efficiency, without much smoke.

9.4. CALORIFIC VALUE Calorific value of a fuel is the total quantity of heat liberated, when a unit mass (or volume) of the fuel is burnt completely. Units of Calorific Value: Units of calorific value for solid, liquid and gaseous fuels are given below.

Fuels 9.3 Units of calorific values Solid/ Liquid Fuels Calories/ gm kcal/ kg B.T.U./ lb

System CGS MKS B.T.U.

Gaseous Fuels Calories/ cm3 kcal/ m3 B.T.U./ ft3

9.5. UNITS OF HEAT 1.

Calorie: It is the amount of heat required to raise the temperature of one gram of water through one degree centigrade (15-16oC). 2. Kilocalorie (or Kilogram Centigrade Units): It may be defined as the quantity of heat required to raise the temperature of one kilogram of water through one degree centigrade. Thus: 1 kal = 1,000 cal 3. British Thermal Unit (B.Th.U. or B.T.U.) It is defined as the quantity of heat required to raise the temperature of one pound of water through one degree Fahrenheit (60-61oF). This is the English system unit. 1 B.Th.U = 252 cal = 0.252 kcal or 1kacl = 3.968 B.Th.U. 4. Centigrade Heat Unit (C.H.U): It is the quantity of heat required to raise the temperature of one pound of water through one degree centigrade. 1 kcal = 3.968 B.Th.U. = 2.2 C.H.U.

9.6. CLASSIFICATION OF CALORIFIC VALUE The calorific value has been classified on the basis of combustion product. There are two types of calorific values depending on the fact whether the steam produced as a result of combustion is allowed to cool down at room temperature or is allowed to escape. Higher or Gross Calorific Value (HCV or GCV): The total amount of heat produced, when unit mass/ volume of the fuel has been burnt completely and the product of the combustion have been cooled to room temperature (i.e., 15oC or 60oF). Lower or Net Calorific Value (LCV or NCV): The net heat produced, when unit mass/ volume of the fuel is burnt completely and the product are permitted to escape. LCV = HCV – Latent heat of water vapour formed LCV = HCV – Mass of hydrogen × 9 × latent heat of steam or

NCV = GCV - Mass of hydrogen × 9 × latent heat of steam

because 1 part by mass of hydrogen produces 9 parts by mass of water. The latent heat of steam is 587 kcal/ kg or 1,060 B.Th.U./ lb of water vapour formed at room temperature (i.e., 15oC).

9.4 ENGINEERING CHEMISTRY: A Textbook

9.7.

DETERMINATION OF CALORIFIC VALUE USING BOMB CALORIMETER

Bomb calorimeter is used to find the calorific value of solid and liquid fuels. Bomb Calorimeter Bomb calorimeter consists of a cylindrical stainless steel vessel known as bomb and a lid is there screwed firmly on the bomb (Figure 9.1). The lid contains two electrodes and an oxygen inlet valve. One electrode is provided with a small ring which contains a stainless steel or nickel or silica crucible. The bomb is placed in a copper calorimeter containing a known weight of water. The copper calorimeter is provided with an electrically operated stirrer and a Beckmann’s thermometer, which can read accurately temperature difference up to 1/ 100th of a degree. This calorimeter is surrounded by an air jacket and then a water jacket to prevent heat losses due to radiation.

Fig. 9.1 Bomb calorimeter Working: A known mass (about 0.5 to 1.0 g) of the given fuel is taken in clean crucible. The crucible is then supported over the ring. A fine magnesium wire, touching the fuel sample, is then stretched across the electrodes. The bomb lid is tightly screwed and bomb filled with oxygen to 25-atmospheric pressure. The bomb is then lowered into copper calorimeter, containing a known mass of water. The stirrer is worked and initial temperature of water is noted. The electrodes are then connected to 6-volt battery and circuit completed. The sample burns and heat is liberated. Uniform stirring of water is continued and the maximum temperature attained is recorded. Calculation Let x = mass in g of fuel sample taken in crucible

Fuels 9.5 W = mass of water in the calorimeter w = water equivalent in g of calorimeter, stirrer, thermometer, bomb, etc. t1 = initial temperature of water in calorimeter t2 = final temperature of water in calorimeter L = higher calorific value in fuel in cal/ g Heat liberated by burning of fuel = xL  Heat absorbed by water and apparatus, etc. = (W + w) (t2 – t1) According to the law of conservation of energy Heat liberated by the fuel = heat absorbed by water and apparatus, etc. ∴ xL = (W + w) (t2 – t1) or

HCV of f uel (L) 

(W + w) (t2 - t1) x

cal/ g (or kcal/ kg)

To calculate net calorific value following procedure is required If H = percentage of hydrogen in fuel

H2 + 1/2O2 2 gm 1 gm

H2O 18 gm 9 gm

i.e., weight of water produced from 1 gm H2 = 9 gm Then, 9H/ 100g = Mass of H2O from 1 g of fuel = 0.09 H g ∴ Heat taken by water in forming steam ∴ LCV = HCV – Latent heat of water formed LCV = (L – 0.09 H × 587) cal/ g (Latent heat of steam = 587 cal/g) Corrections: To get more accurate results, the following corrections are applied: i. Fuse wire correction (CF): The heat liberated, as measured above, includes the heat given out by ignition of the fuse wire. ii.

Cotton thread correction (CT): If fuel is ignited with the help of a cotton or cotton thread, it will give extra heat and that must be subtracted from the obtained value. This is calculated from the weight of the dry cotton thread used.

iii.

Acid correction (CA): Fuels containing S and N are oxidised, under high pressure and temperature of ignition, to H2SO4 and HNO3 respectively.

S + 2 H + 2 O2 2 N + 2 H + 3 O2

H2SO4 + Heat 2 HNO3 + Heat

Formations of these acids are exothermic reactions. So the measured heat also includes the heat given out during the acid formation. The amount of these acids is analysed from washing of bomb by titration; while H2SO4 alone is determined by precipitation as BaSO4. The correction for 1 mg of S is 2.25 cal; while for 1 mL of N/ 10 HNO3 formed is 1.43 cal.

9.6 ENGINEERING CHEMISTRY: A Textbook iv.



Cooling correction CC): Time taken to cool the water in calorimeter from maximum temperature to room temperature is noted. From the rate of cooling (dto/ minute) and the actual time taken for cooling (t minutes), the cooling correction of dt × t is added to the rise in temperature. L=

(W + w) (t2 - t1 + cooling correction) - [Acid + f use corrections + thread correction] Mass of f uel (x)

cal/ g (or kcal/ kg)

Note: Dulong’s formula for calculating the calorific values is given as: Gross calorif ic values (GCV) =

(

[

]

)

O 1 + 2,240S kcal/ kg 8080C + 34,500 H 100 8

Where, C, H, O and S are the percentage of carbon, hydrogen , oxygen and sulphur respectively.

[

]

9 H  587 kcal/ kg Net calorif ic values (LCV) = HCV 100 = [HCV - 0.09H  587] kcal/ kg

Example 9.1. The following data is obtained in a bomb calorimeter experiment: Weight of fuel = 1.029g Weight of water in calorimeter = 2200g Water equivalent of calorimeter = 570g Observed rise in temperature = 2.3oC o Cooling correction = 0.047 C Acid correction = 62.2 calories Fuse wire correction = 3.8 calories Cotton thread correction = 1.6 calories Calculate the GCV of fuel sample. If the fuel contains 6.0% of hydrogen, determine NCV. (Latent heat of steam = 580 cal/g) Solution: Given that x = 1.029 g W = 2200g w = 570g t2 – t1 = 2.30C 0 CC = 0.047 C CA = 62.6 cal CT = 1.6 cal CF = 3.8 cal GCV =

(W + w) (t2 - t1 + cooling correction) - [Acid + f use corrections + thread correction] Mass of f uel (x)

GCV =

(2200 + 570) (2.3 + 0.047) - (62.6 + 3.8 + 1.6) 1.029

cal/ g (or kcal/ kg)

= 6251.88 cal/g

NCV = GCV – 0.09 × H × 580 = 6251.88 – 0.09 × 6 × 580 = 5938.68 cal/g Example 9.2. A sample of coal contained 92%C, 5% H and 3% ash. When this coal was tested in a laboratory for its calorific value in a bomb calorimeter, the following data were obtained: Weight of coal burnt = 0.95g Weight of water taken in calorimeter = 2000g Water equivalent of calorimeter = 700g Rise in temperature = 2.480C 0 Cooling correction = 0.02 C Fuse wire correction = 10 cal Acid correction = 60 cal

Fuels 9.7 Calculate the net and gross calorific values of the coal in cal/g (Latent heat of steam is 580 cal/g) Solution. Weight of fuel (x) = 0.95g Water in calorimeter (W) = 2000g Water equivalent of calorimeter (w) = 700g Rise in temperature (t2 – t1) = 2.48oC Cooling correction (CC) = 0.02oC Acid Correction (CA) = 60 cal Fuse wire correction (CF) =10 cal Since GCV =

(W + w) (t2 - t1 + cooling correction) - [Acid + f use corrections + thread correction] Mass of f uel (x)

GCV =

(2000 + 700) (2.48 + 0.02) - (60 + 10) 0.95

cal/ g (or kcal/ kg)

= 7031.58 cal/g

NCV = GCV – 0.09 × H × 580 = 7031.58 – 0.09 × 5 × 580 = 6770.58 cal/g Example 9.3. 0.72g of fuel containing 80% carbon when burnt in air in a bomb calorimeter, increases the temperature of water from 27.30C to 29.1 0C. If the calorimeter contains 250 g of water and its water equivalent is 150 g. Calculate HCV of fuel. Give answer in kJ/kg. Solution: Given Weight of fuel (x) = 0.72g Water in calorimeter (W) = 250g Water equivalent of calorimeter (w) = 150g Initial temperature t1 = 27.3oC Final Temperature t2 = 29.1oC HCV of f uel  HCV of f uel 

(W + w) (t2 - t1) x

(250 + 150) (29.1 - 17.3)

HCV of f uel 

0.72

cal/ g

= 1,000 cal/ g

4.18  1000  10-3 = 4180 kJ/kg 10-3

Example 9.4. A coal has following composition by weight: C = 92%, O = 2%, S = 0.5% N = 0.5% and ash = 1.5 %. The NCV of coal was found to be 9,430 kcal/ kg. Calculate the percentage hydrogen and GCV of coal. The latent heat of steam is 587cal/g. Solution: According to Dulong’s formula for GCV

9.8 ENGINEERING CHEMISTRY: A Textbook

[ [

] ( ) ( )

1 O + 2,240S kcal/ kg 8080C + 34,500 H 100 8 2 1 Gross calorif ic values (HCV) = 8080  92 + 34,500 H - 8 + 2,240  0.5 kcal/ kg 100

Gross calorif ic values (HCV) =

]

= [7433.60 + 345 H – 86.25 + 11.20] kcal/ kg = [7358.55 + 345 H] --- (1) Since, GCV = (Net calorific value + 0.09H × 587) kcal/ kg = 9430 + 0.09H × 587 GCV = 8490.5 + 52.2H ---(2) From (1) and (2) we get [7358.55 + 345H] = [9430 + 52.83 H] 292.17 H = 2071.45 H = 7.09% From equation (1) we have GCV = (7358.55 + 345 H) kcal/ kg = 7358.55 + (345 × 7.09) kcal/ kg = 9804.6 kcal/ kg Example 9.5. Calculate the GCV and NCV of coal having the following compositions: C = 85%, H = 8%, S = 1%, N = 2% and ash 4% and latent heat of steam is 587 cal/g. Solution. O = 100 – (85 + 8 + 1 + 2 + 4) = 0 According to Dulong’s formula for GCV

[ [

] ( ) ( )

1 O + 2,240S kcal/ kg 8080C + 34,500 H 100 8 0 1 Gross calorif ic values (HCV) = 8080  85 + 34,500 8 - 8 + 2,240  1 kcal/ kg 100

Gross calorif ic values (HCV) =

]

= 6868 + 345 + 22.4 = 7335.4 cal/ g NCV = GCV - 0.09  H  587 = 7235.4 - 0.09  8  587 = 6812.76 cal/ g

9.8. COALS Coal is a highly carbonaceous matter that has been formed as a result of alternation of vegetable matters (e.g., plants) under certain favourable conditions. It is chiefly composed of C, H, N and O, besides non-combustible inorganic matter. Therefore the various types of coals are: 1. 2.

Peat: The average composition is C = 57%, H = 5.7%, N = 2%, O = 35.3%. Lignites (brown coals): The average composition is C = 67%, H = 5%, N = 1.5%, O = 26.5%.

Fuels 9.9 3.

Bituminous coals (common coals): The average composition is C = 77-90%, H = 5%, N = 2%, O = 4-16.2%. Anthracite: The average composition is C = 93.3%, H = 3%, N = 0.7%, O = 3%

4.

9.9. ANALYSIS OF COAL In order to assess the quality of coal, the following two type of analysis are made. A. Proximate analysis B. Ultimate analysis

9.9.1. Proximate Analysis It is the simplest type of analysis of coal which includes the determination of moisture, volatile matter, ash and fix carbon. This analysis gives valuable information for accessing the application of a fuel for a particular domestic and industrial use. i.

Moisture: A known weight of air-dried coal sample is heated at 105-110oC for one hour in a silica crucible. The crucible is then cooled into desiccators and weighted.

Percentage of moisture = ii.

Loss in weight

 

Wt. of coal sample taken

Volatile matter: The coal sample is heated at 925oC ± 20oC in an electric furnace (muffle furnace) for seven minutes. The crucible is cooled in desiccators and weighted. The percentage of volatile matter is calculated by using the formula.

Percentage of volatile matter =

Loss in weight due to removal of volatile matter Wt. of coal sample taken

 

iii.

Ash: The non-combustible material left after the burning of coal is known as ash. The coal sample is heated gradually at 700 ± 50oC for half an hour. The crucible is then taken out, cooled and weighted. The ash content is calculated by using the formula. Weight of ash lef t   Percentage of ash = Wt. of coal sample taken

iv.

Fixed carbon: Fixed carbon in the coal can be determined as follows:

Percentage of f ixed carbon = 100 -  of (moisture + volatile matter ash)

9.9.2. Ultimate analysis This analysis involves in the following determinations: 1.

Carbon and Hydrogen: about 1-2 g of accurately weighed coal sample is burnt in a current of oxygen in a combustion apparatus. C and H of the coal are converted into CO2 and H2O respectively. The gaseous products of combustion are absorbed respectively in KOH and CaCl2 tubes to known weights. The increase in weights of these are then determined.

9.10 ENGINEERING CHEMISTRY: A Textbook CO2 44

C + O2 12

H2O 18

H2 + 1/2O2 2

2.

2KOH + CO2

K2CO3 + H2O

CaCl2 + 7H2O

CaCl2



Percentage of C =

Increase in weight of KOH tube  12  100 Weight of coal sample taken  44

and

Percentage of H =

Increase in weight of CaCl2 tube  2  100 Weight of coal sample taken  18

Nitrogen: About 1 g of accurately weighed powdered coal is heated with concentrated H2SO4 along with K2SO4 (catalyst) in a long necked flask (called Kjeldahl’s flask). After the solution become clear, it is treated with excess of KOH and the liberated ammonia is distilled over and absorbed in a known volume of standard acid solution. The unused acid is then determined by back titration with standard NaOH solution. From the volume of acid used by ammonia liberated, the percentage of N in coal is calculated as follows: Percentage of N =

3.

Volume of acid used  normality  1.4 Weight of coal sample taken

Sulphur: Sulphur is determined from the washings obtained from the known mass of coal, used in a bomb calorimeter for determination of a calorific value. During the determination, S is converted into sulphate. The washings are treated with barium chloride solution, when barium sulphate is precipitated. This precipitate is filtered, washed and heated to constant weight. Percentage of S =

Weight of BaSO4 obtained  32  100 Weight of coal sample taken in bomb  233

4.

Ash determination is carried out as in proximate analysis.

5.

Oxygen: It is obtained by difference Percentage of O = 100 – Percentage of (C + H + S + N + ash)

Example 9.6. A sample of coal was analysed as follows: Exactly 1.51 g of coal sample was weighed into a silica crucible. After heating for 1 hour at 110oC, the residue weighed 1.415g. The crucible was the heated strongly for 7 minutes at 950oC. The residue was weighed 0.528g. The crucible was then heated until a constant weight of residue was obtained. The last residue was found to be 0.254g. Calculate the percentage of above analysis. (Even sem. 2010 – 2011) Solution. Percentage of moisture =

Loss in weight Wt. of coal sample taken

 

Fuels 9.11

= Percentage of volatile matter = =

1.51 - 1.415 100 = 6.29  1.51 Loss in weight due to removal of volatile matter Wt. of coal sample taken 1.415 - 0.528  100 = 58.4 1.51 Weight of ash lef t

Percentage of ash = =

 

Wt. of coal sample taken

 

0.254  100 = 16.82 1.51

Percentage of f ixed carbon = 100 -  of (moisture + volatile matter ash) = 100 - (6.29 + 58.74 + 16.82) = 18.15%

Example 9.7. 3.25g of coal was Kjehldalized and NH3 gas thus evolved was absorbed in 45 ml of 0.1N H2SO4. To neutralize excess acid, 11.5 ml of 0.1 N NaOH was required. Determine the percentage of nitrogen in fuel. Solution. Percentage of N = =

Volume of acid used  normality  1.4 Weight of coal sample taken (45 - 11.5)  0.1  1.4 = 1.44% 3.25

Example 9.8. 0.2g of coal sample was analysed in bomb calorimeter for sulphur. The weight of BaSO4 precipitate was found to be 0.05g. Calculate the percentage of sulphur in the coal sample. Solution. Percentage of S = Percentage of S =

Weight of BaSO4 obtained  32  100 Weight of coal sample taken in bomb  233 0.05  32  100 0.2  233

= 3.45%

9.10. COMBUSTION OF FUEL Combustion is a chemical process accompanied by the liberation of heat and light. It is necessarily an exothermic reaction. For example

C(s) + O(g)

CO2(g)

 = -97 kcal

To ensure complete combustion, the substance must be brought to its ignition temperature. Ignition temperature is defined as the minimum temperature at which the substance ignites and burns without further addition of heat from outside.

9.12 ENGINEERING CHEMISTRY: A Textbook

9.11. CALCULATION OF AIR REQUIRED FOR COMBUSTION The amount of oxygen and air required for complete combustion of a given quantity of the fuel can be calculated on the basis of following principles. 1.

Substances always undergo complete combustion by well defined overall chemical reactions i.e., substances always combine in definite proportions. For example the combustion of carbon is represented by the following balanced equations;

C(s) + O(g) By weight 12 g 32 g By volume 1 Vol 1 Vol 2.

CO2(g) 44 g

 = -97 kcal

1 Vol

In actual practice, the combustion is carried out in presence of air. Therefore, from the amount of oxygen required (as calculated on the basis of above reaction), amount of air is calculated. The basis of this calculation is the fact that air contains 21% of oxygen by volume and 23% of oxygen by weight.

100 = 4.76 m3 of air. 21 100 Similarly 1 kg of oxygen is supplied by 1 × = 4.76 kg of air. 23 Thus 1 m3 of oxygen is supplied by 1 ×

3.

Mean molecular weight of air is taken as 28.94 g mol-1

Example 9.9. Calculate the weight and volume of air for complete combustion of 3 kg of carbon. Solution: C(s) + O(g)

CO2(g)

12 g 32 g

Since, weight of oxygen required to burn 12 kg of carbon = 32 kg Therefore, weight of oxygen required to burn 1 kg of carbon = 32 = 2.67 kg 12 32 Weight of oxygen required to burn 1 kg of carbon = × 3 = 8 kg 12 Weight of air required to burn 3 kg of carbon = 8 × 100 = 34.78 kg of air 23 Again, 32g of air occupy 22.4 litres of volume at NTP Therefore, 8 x 1000g of air will occupy = 22.4  8  1000 = 5600 litres 32 Thus, the volume of air required for complete combustion of 3 kg of carbon = 100 × 5600 = 2666.7 litres or 26.67 m3 21 Example 9.10. Calculate the volume of air required for complete combustion of 1 m3 of gaseous fuel having the composition: CO = 48%; CH4 = 8%; H2 = 40%; C2H2 = 2%; N2 = 1% and remaining being ash.

Fuels 9.13 Solution: Constituent

Amount

Combustion Reaction

CO

0.48

CO + 0.5 O2 1 mole 0.5 mole

CH4

0.08

CH4 + 2O2 1 mole 2 mole

H2

0.40

C2H2

0.02

CO2 CO2 + 2H2O

H2O

H2 + 0.5 O2 1 mole 0.5 mole C2H2 + 5/2O2 1 mole 5/2 mole

2CO2 + H2O

Volume of O2 Required 0.48 × 0.5 = 0.24 m3 0.08 × 2 = 0.16 m3 0.40 × 0.5 = 0.20 m3 0.02 × 5/2 = 0.05 m3

Total volume of oxygen required = 0.24 + 0.16 + 0.20 + 0.05 = 0.65 m3 Volume of air required = 0.65 × 100 21

9.12. BIOGAS Biogas is produced by the degradation of biological matter by the bacterial action (of anaerobic bacteria) in the absence of free oxygen. Examples i. Natural gas is a biogas, which results after a long period decay of animal and vegetable matters, buried inside the earth. ii. Gobar gas (or dung gas), which is produced by the anaerobic fermentation of cattle dung. iii. Biogas can also be produced from the sewage waste and other organic wastes.

9.13. CONSTITUENTS OF BIOGAS The average composition of biogas is: i. CH4 (methane) – 50-60% (a combustible gas). ii. CO2 – 30-40% (a non-combustible gas). iii. H2 – 5-10% (a combustible gas). iv. N2 – 2-6% (a non-combustible gas). v. H2S – trace (a combustible gas). Out of these, the constituent methane (an extremely good fuel) makes biogas as an excellent fuel.

9.14. RAW MATERIALS FOR BIOGAS Animal dung, poultry wastes, vegetable wastes, waste paper and cotton clothes, plant waste (grass, husk, leaves, skins, weeds), human excreta, bird’s excreta, etc.

9.14 ENGINEERING CHEMISTRY: A Textbook

9.15. MANUFACTURE OF DUNG (OR GOBAR GAS) GAS It is produced by the anaerobic degradation of cattle dung. It is carried out in gobar gas plant which consists of well-shaped underground tank (called digester). Covered with dome-shaped roof both made of brick and cement. The dome of the digester is fixed so that it acts as gas holder (or gas strong tank) for the biogas produced. At the top the dome, there is a gas outlet pipe and gas valve. On the left side of the digester, there is a sloping inlet chamber and on the right side, there is a rectangular outlet chamber, both made of brick and cement. Fresh cattle dung + water slurry is introduced from the inlet chamber; whole spent dung 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.

Fig. 9.1. Gobar gas Working: Slurry (made by mixing cattle dung and water in equal proportion in mixing tank) is fed into the digester tank via the inlet chamber, till the slurry level becomes nearly equal to the cylindrical top level. In about 50-60 days, the biogas plan starts functioning. During this time period, the cattle dung undergoes fermentation in the presence of anaerobic bacteria with gradual evolution of biogas, which starts collecting in dome-shaped space. As the time passes, more and more biogas collects inside the dome, thereby exerting pressure on the slurry in the digester tank and this in-turn forces the spent slurry to the overflow tank via outlet chamber. From the overflow tank, the spent slurry is withdrawn periodically and used as good manure. From time to time, fresh slurry is fed to the digester so as to get regular supply of biogas. The biogas collected in dome is taken out through the outlet pipe by opening the gas valve and then used as fuel gas.

Fuels 9.15 Note: Fixed-dome type biogas plant is also known as Janta gobar gas plant. Such a plant is quite cheap, since only bricks and cement are used for its construction. There is no danger of corrosion of such a plant.

9.16. USES OF BIOGAS i. ii. iii.

For cooking food. As a fuel to run engines. As an illuminate in villages.

9.17. ADVANTAGES OF BIOGAS i. ii. iii. iv. v. vi.

Its calorific value is high. Biogas production is very economical. The gas has all advantages of gaseous fuel like cleanliness, absence of smoke, flexibility, etc. It does not contain poisonous gas, CO, as an ingredient. It provides simultaneously excellent yield of good manure. It involves no storage problem, since biogas can be supplied through pipes directly from the biogas plant.

9.18. LIMITATION OF BIOGAS It is necessary to have the gas lamp or stove or burner within 10 meters of the plant.

9.19. BIOMASS Biomass is the waste organic matter (mostly from the dead plants and animals) which used either as a source of energy (by burning or biogas production) and/ or as a chemical feedstock. Biomass refers to all materials that are produced by photosynthesis and potentially useful as energy sources and for the production of organic chemicals. Example: Wood, cattle dung, bagasse (remaining part of sugarcane) poultry wastes, vegetable wastes, waste paper, waste cotton clothes, plant wastes (grass, husk, leaves, skins, weeds), human excreta, bird’s excreta, dead animals, sewage, etc. Biomass consists of carbon compounds which may be used as source of energy by causing either of the following methods: 1.

Biomass such as wood, cattle dung, bagasse, plant wastes, agricultural wastes, dry vegetable wastes, etc. is used directly in Chulhas for getting energy. However, by doing so a lot of heat energy is wastes and lot of smoke etc. is liberated, thereby causing blackening of utensils and houses. Moreover, it liberates poisonous gas carbon monoxide and leaves ash as residue.

2.

Biomass is converted into biogas, which is used for heating and lighting purposes. By burning biogas much large amount of heat is liberated.

9.16 ENGINEERING CHEMISTRY: A Textbook

Exercise Long answer questions: 1. 2. 3. 4. 5. 6. 7. 8.

9.

10.

11. 12. 13.

14.

15.

What is meant by calorific value of fuel? What is difference between gross calorific value and net calorific value? How the calorific value of a solid fuel is determined using bomb calorimeter experiment? List the raw materials which can be utilised for bio-gas manufacture. How the bio-gas is is obtained from cattle dung? What are chemical fuels? How they are classified? Give suitable examples of each class. Define gross calorific value of chemical fuel. Describe with a neat diagram, how it is determined by bomb calorimeter. Explain construction and working of bomb calorimeter. Why net calorific value (NCV) is less than gross calorific value (GCV)? The following data were obtained in a bomb calorimeter experiment. Weight of coal = 0.85 g, weight of water taken = 750 g, water equivalent of calorimeter = 2000 g, rise in temperature = 0.30oC, acid correction = 0.03oC. If the sample contains 10% H, calculate the net and gross calorific value. A sample of coal was found to have the following percentage composition: C = 75%, H = 5.2%, O = 12%, N = 3.2% and ash = 4.5%. (Given GCV of carbon = 8,080 kcal/ kg, Hydrogen = 34,500 kcal/ kg and Sulphur = 2,240 kcal/ kg) A gas has the following composition by volume H2 = 32%, CH4 = 14%, N2 = 40%, O2 = 14%. If 25% excess air is used, find the weight of air actually needed for combustion of this gas. Describe the proximate and ultimate analysis of fuels. What is biogas? How biogas is produced? Describe it with the help of a diagram. Explain the process of bio-gasification. 3.12 g of coal was kjeldahlized and NH3 gas thus enclosed was absorbed in 50 mL of 0.1 NH2SO4. After absorption, the excess of acid required 12.5 mL of 0.1 N NaOH for neutralization. Calculate the % of nitrogen in given coal sample. In an experiment in a bomb calorimeter, a solid fuel of 0.90 g is burnt. It is observed that increase of temperature is 3.8oC of 4000 g of water. The fuel contains 1% of H. Calculate the HCV and LCV value (Water equivalent of calorimeter = 385 g, latent heat of steam = 587 cal/ g. A sample of coal was found to have the following percentage composition: C = 75%, H = 5.2%, O = 12.1% and ash = 4.5%. Calculate the minimum amount of air is required for complete combustion of 1 kg of coal sample.

Fuels 9.17 Short answer questions: 1. Why gaseous fuels are more advantageous than solid fuel? 2. Give examples of non-conventional source of energy. 3. Write short notes on biogas and solar energy. 4. What are characteristic of good fuel? 5. Give the composition of biogas with the help of a diagram. Explain a biogas plant. 6. Write a short note on biogas as a source of energy. 7. What parameters are determined in the proximate analysis of coal? Explain each. Write the properties of a good fuel. 8. Discuss the classification of fuels. 9. What is the grass and net calorific value? 10. What are the different types of coal? 11. Name different forms of coal and arrange them in ascending order of % of carbon. 12. Give the composition of biogas.

Chapter – 10

Spectroscopy Syllabus: Elementary ideas and simple applications of UV, Visible, IR and 1HNMR spectral Techniques.

10.1. SPECTROSCOPY It is a necessary tool, used for structure determination of organic molecule. Spectroscopy may be defined as the study of the quantized interaction of electromagnetic radiations with matter.

10.2. ELECTROMAGNETIC SPECTRUM Electromagnetic spectrum covers a wide range of electromagnetic radiation from cosmic rays to radio waves. Electromagnetic radiation is radiant energy having the properties of both particles and waves. A particle of electromagnetic radiation is called a photon. We may think of electromagnetic radiation as photons traveling at the speed of light. Because electromagnetic radiation has both particle-like and wave-like properties, it can be characterized by either its frequency (υ) or its wavelength (λ) Frequency is defined as the number of wave crests that pass by a given point in one second. Frequency has units of hertz (Hz). Wavelength is the distance from any point on one wave to the corresponding point on the next wave. Wavelength is generally measured in micrometers or nanometers. One micrometer (μm) is 10-6 of a meter; one nanometer (nm) is 10-9of a meter. The frequency of electromagnetic radiation, therefore, is equal to the speed of light (c) divided by the radiation’s wavelength:

c = 

c = 3  1010 cm/ s

Short wavelengths have high frequencies, and long wavelengths have low frequencies. Frequency () in Hz

high f requency 1019 Cosmic rays 10-6 short wavelength

1015

1017

- rays

Ultraviolet light

x- rays 10-4

low f requency

10-1

1010

1013 Visible light

0.4 Wavelength () in m

Inf rared radiation 0.8

Microwaves 102

105 Radio waves NMR 106

1010 long wavelength

10.2 ENGINEERING CHEMISTRY: A Textbook The relationship between the energy (E) of a photon and the frequency (or the wavelength) of the electromagnetic radiation is described by the equation  = h  hc 

Where, h = proportionality constant called Planck’s constant, named after the German physicist who discovered the relationship. c = velocity of light υ = wavelength Wave Number: It can be defined as the number of waves which can pass through per unit length in cm. It is reciprocal of wave length. 1    cm 

10.3. ORIGIN OF ELECTRONIC SPECTRA When electromagnetic radiations are passed through an organic compound, then some of wavelengths are absorbed while other remains unaffected. A molecule can only absorb radiation of certain frequency, if there exists within a molecule an energy difference of magnitude ΔE which is equal to hυ. Suppose E1 and E2 are the two possible energy states of a system. The energy difference between the two states is given by ΔE. ΔE = E2 - E1 The energy transition from E1 to E2 or from E2 to E1 corresponds to the absorption or emission of energy exactly equivalent to the energy of the wavelength absorbed or emitted. The absorbed or emitted energy is the form of electromagnetic radiation, the frequency of which is given by υ = ΔE/h Hz The energy change or the frequency of the electromagnetic radiation, emitted or absorbed can be recorded with the help of an instrument called spectrometer. Thus spectra so obtained can be broadly classified into two categories. i.

Emission Spectrum: When there is a transition from a state of high energy to a stage of lower energy, the excess energy is emitted as a photon of energy hυ provided, E = hυ. The spectrum recorded will be emission spectrum. Emission spectra are produced by heating a substance directly in a flame or heating it electrically and then passing the emitted radiation through a prism.

ii.

Absorption Spectrum: When there is a transition from a state of lower energy to a stage of higher energy, the energy absorbed will be exactly equal to the energy difference ΔE. The spectrum recorded will be absorption spectrum. UV, visible, IR and NMR spectra are examples of absorption spectra.

Spectroscopy 10.3

E2

E2 E = h

E = h

E1

E1 (b)

(a)

Fig. 10.1 Energy transition for the (a) absorption and (b) emission of electromagnetic radiation Summary of Spectroscopic Techniques in Organic Chemistry Spectroscopy Radiation absorbed Effects on the molecule Ultraviolet visible

(uv)

Infrared (ir)

Nuclear magnetic resonance (nmr)

Uv-visible (λ = 200-750 nm)

Change in electronic energy levels within the molecule

Infrared (λ = 2.5-16 μm)

Changes in the vibrational and rotational movements of the molecule

Radio υ, 60-500 MHz

Induces changes in the magnetic properties of certain atomic nuclei

10.4. ULTRAVIOLET & VISIBLE SPECTROSCOPY U.V. and visible spectroscopy deals with the recording of the absorption of radiations in U.V. and visible region of electromagnetic spectrum. U.V. region extends from 10-400 nm. Since, U.V. and visible spectroscopy involves electronic transitions, so it is also called as “Electronic spectroscopy.” It is used for elucidating nature of conjugated multiple bonds or aromatic rings.

10.5. LAMBERT – BEER’S LAW Lambert’s law states that the fraction of incident monochromatic radiation absorbed by a homogeneous medium is independent of the intensity of the incident radiation. Beer’s law states that absorption of monochromatic radiation by a homogeneous medium is proportional to the number of absorbing molecules. log10

I0  A   cl I

I0 = Intensity of incident radiation I = Intensity of radiation transmitted through sample solution. A = Absorbance or optical density  = Molar extinction coefficient c = conc. of solute in mole/litre l = Path length of sample in cm.

10.4 ENGINEERING CHEMISTRY: A Textbook The U.V. visible spectrophotometer records a U.V. or visible spectrum as a plot of wave length of absorbed radiation versus the intensity of absorption in terms of absorbance A or molar extinction coefficient. Example 10.1. For a solution of camphor in hexane in a 10 cm cell, the absorbance A was found to be 2.52 at 295 nm with εmax 14. What is the concentration of camphor? Solution: A = εcl A = 2.52, ε = 14, l = 10 cm ∴ 2.52 = 14 × c × 10 14  10 = 1.8  10-2 mole/ litre 2.52

c=

10.6. ELECTRONIC TRANSITIONS According to molecular orbital theory the excitation of molecule by absorption of radiation in U.V. and visible region involves promotion of electrons from a bonding or non-bonding orbital to anitbonding orbitals. The following four types of electronic transition involves in U.V. visible regions:

(i)    (ii) n   (iii)    (iv) n  

E



Antibonding  orbitals



Antibonding  orbitals

n

n  

 

n     

Non-bondibg n orbitals Bonding  orbitals Bonding  orbitals

  

Fig. 10.2. Relative energies of electronic transitions The usual order of energy required for various electronic transitions is

    n        n   (i)    * Transition : The promotion of an electron from sigma bonding orbital to antibonding sigma orbital is called as    * transition. It occurs at high energy level since 𝜎 bonds are very strong in nature. Example: alkanes only.

C

C

  

C

C

Spectroscopy 10.5 (ii) n   * Transition: The promotion of an electron from nonbonding orbital to antibonding sigma orbital is called as n   * transition. Example: Alkyl Halide, Alcohols etc.

n   X

C

X

C

(iii)    * Transition: The promotion of an electron from pi bonding orbital to pi antibonding orbital is called as    * Transition. It requires lower energy. Example: Alkene, Alkynes

C

  

C

C

C

(iv). n   * Transition: The promotion of an electron from a non-bonding orbital to pi antibonding orbital is called as n   * transition. It occurs at lowest energy. Example: Carbonyl compounds, nitriles etc.

n   C

O

C

O

10.7. CHROMOPHORE A covalently unsaturated group responsible for absorption in the U.V. or visible region is known as a chromophore. For example: C=C, C  C , C  O , C  N ,

N  N , NO2 etc. If a compound absorbs light in the visible region (400-800 nm), only then it appears coloured. Thus, a chromophore may or may not impart colour to a compound depending on wheather the chromophores absorbs radiation in the visible or UV region. Chromophores like C=C or C  C having π electrons undergo π  π* transitions and those having both π and non-bonding electrons, e.g. C  O, C  N , or N  N , undergo π  π*, n  π* and n  σ* transitions. Since the wavelength and intensity of absorption depend on the number of factors, there are no set rule for the identification of a chromophore.

10.8. AUXOCHROME A covalently saturated group which, when attached to chromophore, changes both the wave length and intensity of the absorption maximum is known as auxochrome, e.g. NH 2 , OH , SH , halogans etc. Auxochrome generally increase the value of λmax as well as εmax by extending the conjugation through resonance. These are also called colour enhancing group. An auxochrome itself does not show absorption above 200

10.6 ENGINEERING CHEMISTRY: A Textbook nm. Actually, the combination of chrromophore and auxochrome behaves as a new chromophore having different value of λmax and εmax. For example, benzene shows λmax 256 nm, εmax 200, whereas aniline shows λmax 280 nm, εmax 1430 (both increased). Hence, NH2 group is an auxochrome which extends the conjugation involving the lone pair of electrons on the nitrogen atom resulting in the increased values of λmax and εmax.

10.9. ABSORPTION AND INTENSITY SHIFTS 10.9.1. Bathochromic Shift The shift of an absorption maximum to a longer wave length due to the presence of auxochrome is called as Bathochrimic shift or “Red shift.” NH2

256 n.m.

280 n.m.

10.9.2. Hypsochromic Shift The shift of an absorpation maximum to shorter wave length is called as Hyposochrmic shift or Blue shift. It is due the loss of conjugation.

10.9.3. Hyperchromic Effect An effect, which leads to an increases absorption intensity is called as “Hyperchromic Effect.”

10.9.4. Hypochromic Effect An effect which leads to a decrease in absorption intensity is called as “Hypochromic Effect.

Spectroscopy 10.7

10.10. Applications of Ultraviolet and Visible Spectroscopy Some important applications of UV and visible spectroscopy to organic chemistry are summarised as follows. 1. 2. 3. 4. 5. 6. 7. 8. 9.

Detection of functional group (Chromophore). Detection of conjugation and elucidation of its nature. Study of extent of conjugation. Distinction between conjugated and unconjugated compounds. Study of steric strain. Determination of configurations of geometrical isomers. Study of tautomerism Confirmation of suspected phenols and aromatic amines. Study of structural features in different solvents.

10.11. INFRARED (IR) SPECTROSCOPY Infrared spectroscopy allows us to determine the kinds of functional groups a compound has. It deals with the recording of the absorption of radiations in the infra red region of the electromagnetic spectrum. The position of a given infrared absorption is expressed in terms of wave number v  . The infrared region is 2.1 -15 or 4000-667 cm-1 is of greatest practical use to organic chemist. The absorption of infrared radiation by a molecule occurs due to quantized vibrational and rotational energy changes, so it is also called as vibrational– rotational spectra.

10.12. TYPES OF VIBRATIONS Types of Vibrations Stretching Vibrations

Symmetrical Stretching

Asymmetrical Stretching

Bending Vibrations

Scissoring

Rocking

Wagging Twisting

10.12.1. Stretching vibrations In this type of vibrations, the distance between two atoms increases or decreases but the atoms remains in the same bond axis. It is of two types: (a). Symmetrical Stretching: In this mode of vibration the movement of atoms with respect to common central atom is simultaneously in the same direction along the same bond axis.

10.8 ENGINEERING CHEMISTRY: A Textbook H

H C

(b). Asymmetrical Stretching: In this vibration, one atom approaches the central atom while other departs from it. H

H C

10.12.2. Bending Vibrations In this type of vibrations the positions of the atoms change with respect to their original bond axis. It is of four types: (a). Scissoring: In this mode of vibrations the movement of atoms is in the opposite directions with change in their bond axis as well as bond angle they form with central atom. H

H

C

(b). Rocking: In this vibration the movement to atom takes place in the same directions with change in their bond axis. H

H

C

(c).Wagging: In this vibration two atoms simultaneously move above and below the plane with respect to central atom. H

H

C

Spectroscopy 10.9 (d).Twisting: In this mode of vibration one of the atom moves up and the other moves down the plane with respect to central atom. H

H

C

Table 10.1 Important Infrared Stretching Frequencies Type of Bond γ‫( ־‬cm-1) C C C C

N C C N

2260-2220 2260-2100 1680-1600 1650-1550 1600, 1500-1430

C

O

C

O

C

N

1780-1650 1250-1050 1230-1020

O-H (alcohol) O-H (carboxylic acid) N-H C-H

3650-3200 3300-2500 3500-3300 3300-2700

10.13. NUMBER OF FUNDAMENTAL VIBRATIONS In case of a non-linear molecule, three of the degrees of freedom describe rotation and three describe translation. Thus, the remaining (3n – 3 - 3) = 3n – 6 degrees of freedom are its vibrational degrees of freedom of fundamental vibrations, because Total degrees of freedom (3n) = Translation + Rotational + Vibrational degrees of freedom Example: A non-linear molecule ethane (C2H6), the vibrational degree of freedom can be calculated as: Number of atoms (n) = 8 Total degrees of freedom (3n) = 3 × 8 = 24 Rotational degrees of freedom = 3 Translational degrees of freedom = 3 Therefore, vibrational degrees of freedom = 24 – 3 – 3 = 18 Thus, theoretically there should be 18 absorption bands in the IR spectrum of ethane. Example: In case of benzene (C6H6), the number of vibrational degrees of freedom can be calculated as follows:

10.10 ENGINEERING CHEMISTRY: A Textbook Number of atoms (n) = 12 Total degrees of freedom (3n) = 3 × 12 = 36 Rotational degrees of freedom = 3 Translational degrees of freedom = 3 Therefore, vibrational degrees of freedom = 36 – 3 – 3 = 30 Thus, theoretically there should be 30 absorption bands in the IR spectrum of benzene. In case of linear molecule, only two degrees of freedom describe rotation and three describe translation. Thus, the remaining (3n – 2 – 3) = 3n – 5 degrees of freedom are vibrational degrees of freedom or fundamental vibrations. Example: The number of vibrational degrees of freedom for the linear carbon dioxide molecule can be calculated as follows: Number of atoms (n) = 3 Total degrees of freedom (3n) = 3 × 3 = 9 Rotational degrees of freedom = 2 Translational degrees of freedom = 3 Therefore, vibrational degrees of freedom = 9 – 2 – 3 = 4 Since, each vibrational degree of freedom correspond to a fundamental vibration and each fundamental vibration correspond to an absorption band, for carbon dioxide molecule there should be four theoretical fundamental bands.

10.14. FINGER PRINT REGION I. R. spectrum covers a wide range from 4000-667 cm-1. It can be divided into two regions i.e. 4000-1400 cm-1, it is the region where most of the functional groups shows absorption bands, it is called as functional group region. The region from 1400-667 cm-1 is called as finger print region. The overall pattern in the finger print region is characteristic of compound as a whole because each compound shows a unique pattern in this region. So, compounds can be effectively identified by comparing their finger print region.

10.15. APPLICATIONS OF INFRARED SPECTROSCOPY Some important applications of UV and visible spectroscopy to organic chemistry are summarised as follows. 1. 2. 3. 4. 5. 6. 7.

Detection of functional groups. Confirmation of the identity of compounds. Estimation of the purity of samples. Study of hydrogen bonding. Calculation of force constants. Determination of orientations in aromatic compounds. Study of the progress of reactions.

Spectroscopy 10.11

10.16. PROTON NUCLEAR MAGNETIC RESONANCE (PMR OR 1H NMR) SPECTROSCOPY NMR spectroscopy involves nuclear magnetic resonances which depend on the magnetic property of atomic nuclei. Thus NMR spectroscopy deals with nuclear magnetic transitions between magnetic energy levels of the nuclei in molecules. NMR signals were first observed in1945 independently by Prucell at Harvard and Bloch at Stanford. The first application of NMR to the study of structure was made in 1951 and ethanol was the first compound thus studied. In 1952, Prucell and Bloch won the Nobel Prize in Physics for their discovery. NMR spectroscopy helps to identify carbon-hydrogen framework of an organic compound. The hydrogen nuclei were the first nuclei, studied by nuclear magnetic resonance hence acronym NMR is generally mean to 1H NMR. It operates at radio frequency (R.F) region. Certain nuclei have allowed spin states of +1/2 and -1/2 and this property allows them to be studied by NMR. Examples of such nuclei are 1H, 13C, 15N, 19F, 31P.

10.17. INSTRUMENTATION NMR signals are detected by spectrometers. The NMR spectrometer detects the signals and displays them as a plot of signal frequency versus intensity. The term nuclear magnetic resonance comes from the fact that the nuclei are in resonance with radiofrequency radiation. The schematic diagram of a NMR spectrometer containing the following components is given in Fig. 10.3.

Fig. 10.3 Schematic diagram of a NMR spectrometer

10.12 ENGINEERING CHEMISTRY: A Textbook

10.18. SHIELDING AND DESHIELDING Under the influence of applied magnetic field, electrons surrounding a nucleus start to circulate perpendicular to the applied magnetic field (H0), and so they generate a secondary magnetic field called induced magnetic field (σ H0) which oppose the applied magnetic field (H0) in the region of the nucleus, e.g. proton. Thus, the nucleus experiences a weaker magnetic field (Heff) than the applied magnetic field (H0), and it is said to be shielded. This type of shielding is termed diamagnetic shielding and its effect is termed as shielding effects, i.e. Heff = H0 - σ H0 Where σ is screening or shielding constant. Nucleus (e.g., proton)

Circulation of electrons

Induced magnetic field due to circulation of electrons around nucleus Applied magnetic field (H0)

Fig. 10.4 Diamagnetic shielding of nucleus by circulating electrons If the induced field reinforces the applied field, then the field experienced by the proton is greater than the applied field. Such a proton is said to be deshielded and this effect is termed as deshielding effect. Compared to a naked proton, a shielded proton requires higher field strength, whereas deshielded proton requires lower field strength for transition. Thus, shielding shifts the absorption position upfield, whereas deshielding shifts the absorption position downfield and these effects are termed as shielding and deshielding effects, respectively. At 300 MHz, an unshielded naked proton absorbs at 70,459 gauss. But a shielded proton requires a stronger field. For example, if a proton is shielded by 1 gauss when the external field is 70,459, the effective magnetic field experienced by the proton is 70,458 gauss. At this field proton will not absorb energy by flipping. If the external field is increased to 70,460 gauss, the effective magnetic field experienced by the proton will be 70,459 (70,460 - 1 = 70,459). At this field proton will absorb energy which brings this proton into resonance.

Spectroscopy 10.13 Thus, magnetic field must be increased slightly (in comparision to naked proton) above 70,459 gauss (at 300 MHz) for resonance of a shielded proton. If all protons were shielded by the same amount, they would all be in resonance at the same combination of ferquency and magnetic field. Fortunately protons in different chemical environments are shielded by different amounts. Magnitude of shielding depends on the electron density around proton. The higher the electron density around the proton, the higher is the shielding and thus higher is the field needed for flipping. For example CH3 protons of CH3-Cl is more shielded than CH3F because electronegativety oof fluorine is more than the chlorine. Thus CH3 protons of CH3F absorbs at lower field than the CH3 protons of CH3Cl. electrons shielding

effective field 70,458 300 MHz

effective field 70,459

300 MHz

300 MHz

does not absorbs

absorbs H0 70,459 gauss nacked proton absorb at 70,459 gauss

absorbs

H0 70,459 gauss Shielded proton feels less than 70,459 gauss

H0 70,460 gauss Stronger applied field compensates for shielding

Fig. 10.5 Shielded protons

10.19. NUMBER OF SIGNALS IN 1H NMR SPECTRUM Protons present in the same environment are chemically equivalent proton and will give same signal. Br

a H3C

H3C CH

b

c

CH2

CH2

a

Br

(3 signals) CH3 a

CH2

CH3 a

OH

c

b

CH2

a

O

(3 signals)

O

CH3

a

(1 signal) CH2

CH2

a

b

CH2

CH3

(3 signals)

CH3 a

a

(2 signals)

(3 signals) CH3 a

CH3

b

b

c

OH

c

CH3 a

COOH

b

(2 signals) CH3 OCO a (3 signals)

CH2

b

CH3

c

10.14 ENGINEERING CHEMISTRY: A Textbook Cl

HCOOCH2 a b

CH3

H3C

c

(3 signals) a

H

a

a

H

a

a H

Ha

aH

H a H

H

(1 signal)

a

H

H H

H

a Ha

a H

Ha

aH H

a

a

H

Ha

H

a

a

(1 signal) a

a H

H C

a

b H

C

CH3 C

H

H

a

a

(1 signal)

C H

H

d

c

(4 signals) a

b H

CH3 C

CH3

b

(2 signals)

a

a

CH

a

H

H C

C Cl

H

a

b C

Cl

H

c

c

(3 signals)

(3 signals) Cl

a

H3C

a

CHO

H

H

H

b

b

b

(2 signals)

a

H

NH3

(2 signals)

10.20. SPLITTING OF THE SIGNALS Splitting is caused by protons bonded to adjacent (i.e., directly attached) carbons. The splitting of a signal is described by (n+1) rule, where N is the number of equivalent protons bonded to adjacent carbons. By “equivalent protons,” we mean that the protons bonded to an adjacent carbon are equivalent to each other, but not equivalent to the proton giving rise to the signal.

Spectroscopy 10.15 Keep in mind that it is not the number of protons giving rise to a signal that determines the multiplicity of the signal; rather, it is the number of protons bonded to the immediately adjacent carbons that determines the multiplicity. For example, the signal for the a protons in the following compound will be split into three peaks (a triplet) because the adjacent carbon is bonded to two hydrogens. The signal for the b protons will appear as a quartet because the adjacent carbon is bonded to three hydrogens, and the signal for the c protons will be a singlet. O

CH3 a

CH2

b

C

OCH3

c

Multiplicity of the signals Number of equivalent proton causing splitting

Multiplicity of the signal

0 1

Singlet Doublet

2 3

Triplet Quartet

4

Quartet

5 6

Sextet Septet

10.21. CHEMICAL SHIFT Shielding and deshielding shifts in the NMR absorption position are called chemical shifts because they arise from the circulation of electrons in chemical bonds. The chemical shift indicates how far the signal is from the TMS peak. The most common scale for chemical shift is δ scale. TMS is assigned at zero position on δ scale.

  chemical shif t (ppm) =

distance downf ield f rom TMS (Hz) operating f requency of the spectrometer (MHz)

Most proton chemical shifts fall in the range from 0 to 10 ppm.

10.22. TMS (TETRA METHYL SILANE) It is used as reference compound in NMR spectroscopy. It is highly volatile in nature. The methyl protons of TMS are in a more electron-dense environment than are most protons in organic molecules, because silicon is less electronegative than carbon. Consequently, the signal for the methyl protons of TMS is at a lower frequency than most other signals (i.e., it appears at 0 ppm value on δ scale.

10.16 ENGINEERING CHEMISTRY: A Textbook CH3 H3C

Si

CH3

CH3

Tetramethylsilane TMS

10.23. COUPLING CONSTANT The distance, in hertz, between two adjacent peaks of a split NMR signal is called the coupling constant (denoted by J). The coupling constant for Ha being split by Hb is denoted by Jab The signals of coupled protons (protons that split each other’s signal) have the same coupling constant; in other words, Jab = Jba. Coupling constants are useful in analyzing complex NMR spectra because protons on adjacent carbons can be identified by identical coupling constants. Approximate Values of Chemical Shifts for 1H NMR Type of proton Approximate Type of proton Approximate chemical shift chemical shift (δ (δ ppm) ppm) (CH3)4Si

0

9-10

O H

C

-CH3

0.9

RNH2

1.5-4

-CH2-

1.3 1.4

ROH ArOH

2-5 4-7

CH

O

C C

CH3

1.7

C

O

OH

10-12

O

C

CH3

2.1

C

NH2

5-8

CH3

2.3

R O CH3

3.3

H

6.5-8

I

C

H

2.4-5

Spectroscopy 10.17

10.24. APPLICATIONS OF NMR SPECTROSCOPY 1. 2. 3. 4. 5. 6. 7.

Identification of structural isomers. Detection of aromaticity. Detection of hydrogen bonding. Distinction between cis-trans isomers and conformers. Detection of partial double bond character. Quantitative analysis. Magnetic Resonance Imaging

NMR has become an important tool in medical diagnosis because it allows physicians to probe internal organs and structures without invasive surgical methods or the harmful ionizing radiation of X-rays. When NMR was first introduced into clinical practice in 1981, the selection of an appropriate name was a matter of some debate. Because some members of the public associate nuclear processes with harmful radiation, the “N” was dropped from the medical application of NMR, now known as magnetic resonance imaging (MRI). The spectrometer is called an MRI scanner. MRI scans, therefore, can sometimes provide much more information than images obtained by other means. The 1991 Nobel Prize in chemistry was awarded to Richard R. Ernst for two important contributions: FT–NMR spectroscopy and an NMR tomography method that forms the basis of magnetic resonance imaging (MRI).

Exercise 1. 2. 3. 4. 5.

Why – carotone absorbs light in visible region. What are auxochromes. Why and how auxovhtome increases colouring power of chromphore? Discuss the types of electronic transition in UV – visible spectroscopy. Define the term Chromophore and Auxochrome in UV spectroscopy. Why λ max for the diene(I) is observed at a lower n.m. than (II)? C

(I)

6. 7. 8.

C

H

H

(II)

For a XY2 bent molecule show various types of stretching and bending vibrations in I.R.? What is the importance of finger print region I.R. spectroscopy? Two isomers A and B of molecular formula C3 H 6O , gives I.R. absorption bond at 1650 cm-1 and 1710 cm-1 respectively. Assign structural formula to isomers A and B?

10.18 ENGINEERING CHEMISTRY: A Textbook 9.

10.

11. 12. 13. 14.

15. 16. 17. 18.

An organic compound of molecular formula C7 H 6O shows absorption peaks at 3010, 2700, 1600, 1580, 1520, 1480 and 1270 cm-1 in it’s I.R. spectrum. Suggest it’s structure. How will you distinguish between the following pairs of compounds on the basis of infrared spectroscopy? (i). CH3COOH and CH3COOC2H5 (ii). C2H5OH and C2H5O C2H5 Why TMS is used as standard reference in NMR spectroscopy? How many NMR signals are obtained in CH3CHOHCH2CH3? What is shielding and deshielding? How many NMR signals would you expect in the NMR spectra of the following compound: Chlorobutane, 2-Chlorobutane, Cyclohexane, nButanol and Acetic acid. How many NMR signals are obtained in 2-Bromopropane? How many NMR signals are obtained in Cyclobutane and 2-Chloropropane Show the number of expected 1H NMR signals and their splitting pattern in CH3OCOCH2CH3 and HCOOCH2CH3? What are equivalent and non equivalent protons in NMR spectroscopy? A compound having molecular formula C4H9Br gave the following set of 1H NMR signals: δ 1.04 (6H, d); δ 1.95 (1H, m) and δ 3.33 (2H,d). Giving proper reasons suggest a structure consistent with the above data.

Chapter – 11

Engineering Chemistry Practical LIST OF EXPERIMENTS 1. Determination of alkalinity in the given water sample. 2. Determination of temporary and permanent hardness in water sample using EDTA as standard solution. 3. Determination of available chlorine in bleaching powder. 4. Determination of chloride content in water sample. 5. Determination of iron content in the given water sample by Mohr’s method. 6. pH- metric titration. 7. Viscosity of an addition polymer like polyester by viscometer. 8. Determination of iron concentration in sample of water by calorimetric method. The method involves the use of KCN as a colour developing agent and the measurements are carried out at λ max 480 nm. 9. Element detection and functional group identification in organic compounds. 10. Preparation of Bakelite and Urea formaldehyde resin. EXPERIMENT NO. 11.1 Object: To determine the alkalinity of the given water sample. Apparatus required: Burette, Pipette, Conical flask, volumetric flask and funnel. Chemicals required: N/50 HCl solution, phenolphthalein and methyl orange solutions. HO

Methyl orange pH range 3.1 - 4.4

Phenolphthalein pH range 8.3 - 10.0

OH

CH3 +Na-O S 3

N

N

N CH3 O

O

Theory: the alkalinity of given water sample may be estimated by titrating the water sample by N/50 HCl solution using phenolphthalein and methyl orange solution as internal indicators. They reactions occur during titration are as follows:

11.2 ENGINEERING CHEMISTRY: A Textbook

OH- + H+ CO3- - + H+ HCO3 + H+

H2O HCO3-

M

P

H2CO3

H2O + CO2

Alkalinity of water may be due to the presence of (i) hydroxides only, (ii) carbonates only, (iii) bicarbonates only, (iv) Hydroxides and carbonates, (v) Carbonate and bicarbonates. Hydroxides and bicarbonates cannot coexist due to following reaction:

OH   HCO3  CO3   H 2O Procedure 1. Add distilled water into the volumetric flask (250 mL) containing supplied water sample upto the mark. 2. Pipette out 25 ml of solution from volumetric flask into the conical flask (250 mL). 3. Add 5 drops of phenolphthalein indicator into conical flask, the solution become pink. 4. Titrate this solution against N/50 HCl solution until the solution become colourless at the end point. 5. Add 5 drops of methyl orange indicator into conical flask, the solution become yellow. 6. Titrate this solution against N/50 HCl solution until the colour of solution become pink. 7. Repeat the procedures until the concordant reading are obtained. Observation S. Volume of No. water sample taken (mL) 1. 2. 3.

25 25 25

Volume of HCl used (mL) Phenolphthalein end point (P) -

Titre value for phenolphthalein …………………….. mL. Titre value for methyl orange ……………………… mL. Calculation for phenolphthalein end point: Volume of water sample = 25 mL

N HCl used = Burette reading (x) 50 Therefore, N1V1  N 2V2 Volume of

Sample = N/50 HCl

Methyl orange end point (M) -

Engineering Chemistry Practical 11.3

1 x 50 x NP  25  50 Strength in term of CaCO3 equivalent N P  25 

x  50 g/lit. = P g/litre 25  50

NP 

For methyl orange end point :

N1V1  N 2V2 Sample N/50 HCl

N M  25  NM 

1  Burette reading ( y ) 50

y 25  50

Strength in term of CaCO3 equivalent

NM 

y  50 =M g/litre 25  50

Conditions for constituent of alkalinity S. No. Conditions OH  (ppm) 1. 2. 3. 4. 5.

P=0 P=M P=½M P>½M P