Metallurgical thermodynamics and kinetics numerical 9788121939645

Citation preview

METALLURGICAL THERMODYNAMICS KINETICS AND LLL Dr. S. K. DUTTA Prof. A.B. LELE

METALLURGICAL THERMODYNAMICS KINETICS & NUMERICALS INCLUDING PROBLEMS IN IRON MAKING, STEEL MAKING AND EXTRACTIVE METALLURGY For Students of B.E./B.Tech. Metallurgical and Materials Engineering

Dr. S. K. Dutta &

Prof. A. B. Lele Metallurgical and Materials Engineering Department, Faculty of Technology & Engineering,

M § University of Baroda, Vadodara (Gujarat)

Lx im

Cnn

S$.CHAHNLC FUBLISHING

S Chand And Company Limited (ISO 9001 Certified Company)

S Chand And Company Limited (1SO 9001 Certified Company) Head Office: Block B-1, House No. D-1, Ground Floor, Mohan Co-operative Industrial Estate,

New Delhi — 110 044 | Phone: 011-66672000 Registered Office: A-27, 2" Floar, Mohan Co-operative Industrial Estate, New Delhi — 110 044

Phone: 011-49731800

www.schandpublishing.com; e-mail: [email protected] Branches Ahmedabad

©

Ph: 27542389, 27541965; ahmedabad @schandpublishing.com

Bengaluru

©

Ph: 22354008, 22268048; [email protected]

Bhopal

: Ph: 4274723, 4209587; [email protected]

Bhubaneshwar

:

Ph: 2951580; [email protected]

Chennai

©

Ph: 23632120; [email protected]

Guwahati

:

Ph: 2738811, 2735640; [email protected]

Hyderabad

©

Ph: 40186018; [email protected]

Jaipur

©

Ph: 2281317, 2291318; [email protected]

Jalandhar

:

Ph: 4645830; [email protected]

Kochi

©

Ph: 2576207, 2576208; [email protected]

Kolkata

©

Ph: 23357458, 23353914; [email protected]

Lucknow

©

Ph: 4003633; [email protected]

Mumbai

©

Ph: 25000297; [email protected]

Nagpur

:

Ph: 2250230; [email protected]

Patna

©

Ph: 2260011; [email protected]

Ranchi

©

Ph: 2361178; [email protected]

Sahibabad

©

Ph: 2771238; [email protected]

© S Chand And Company Limited, 2012 All rights reserved. No photocopying or storing transient or incidental to breach of this will entail

part of this publication may be reproduced or copied in any material form (including it in any medium in form of graphics, electronic or mechanical means and whether or not some other use of this publication) without written permission of the copyright owner. Any legal action and prosecution without further notice.

Jurisdiction: All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, Tribunals and Forums of New Delhi, India only. First Edition 2012

Reprints 2017, 2018, 2019 Reprint 2020

ISBN : 978-81-219-3964-5

Product Code : HIMTK62THRM10ENAA120

PREFACE Thermodynamics and Kinetics 1s a basic science and has a wide applicability in all divisions of Metallurgical and Materials Engineering. This book covers the undergraduate curriculum in Metallurgical Thermodynamics as prescribed for degree courses and allied professional courses in Metallurgical and Matenals Engineering. Part I of this text book Metallurgical Thermodynamics and Kinetics, covers almost all the important basic concepts, derivations and numericals (about 40 solved problems in SI units) for graduate and diploma engineering students. Even post-graduate students, engineers and researchers

can brush up their understanding, at a glace about Metallurgical Thermodynamics and Kinetics. Part II of the book covers Metallurgical Numericals, which include problems in Ironmaking, Steelmaking and Extractive Metallurgy (about 60 solved problems) as a part of Tutorials in Metallurgical Engineering. Needless to say that although few books are available, none discusses the subject matter of Metallurgical Thermodynamics in easy to understand format. Based on our experience of teaching

this subject at undergraduate level for three decades, we have made a sincere attempt to include the application of thermodynamics to the numerous reactions of metallurgical interest in the form of suitable numerical problems. In spite of taking all the possible care, there may be some errors or mistakes left out unnoticed. If so, please feel free to interact with us. In moulding and casting this text book, we poured our long

experience in it and also the information from several sources. We are indebted to one and all, from whose valuable knowledge we have been benefited.

We are confident that this text book will make the subject matter of Metallurgical Thermodynamics and Kinetics more simple and easy to understand. We wish to thanks the publisher S. Chand & Co

Ltd, New Delhi for their continuous support in preparation and publication of this text book within a short period.

Dr. S.K. Dutta Prof. A.B. Lele (September 2011)

Disclaimer : While the authors of this book have made every effort to avoid any mistake or omission and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and 5. Chand does not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. 5.Chand and the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything farming part of the contents of this publication. 5. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

(iii)

CONTENTS Page Number PART —- | : METALLURGICAL THERMODYNAMICS & KINETICS | Chapter 1: Introduction

3-8

1.1 Laws of Thermodynamics, 4 1.2 Basic Terms Used in Thermodynamics, 5 1.3 Properties of a System, 6 1.4 Reversible and Irreversible Changes, 7 1.5 Equilibrium, 7 1.6 Isothermal and Adiabatic Change or Process, 8 Chapter 2: Energy and First Law of Thermodynamics 2.1 Energy, 9

9-13

2.2 First Law of Thermodynamics, 10 2.3 Measurement of Energy Change, 11 2.4 Energy Change in Terms of Partial Derivations, 11 Examples, 12 Chapter 3: Heat Capacity, Enthalpy and Heat of Reaction 3.1 Heat Capacity, 14 3.1.1 Heat Capacity at Constant Pressure and Volume, 14

14-27

3.1.2 Heat Capacity in Terms of Energy, 14 3.1.3 Dependence of Heat Capacity on Temperature, 16 3.14 Importance of C, and C,,, 17 3.2 Enthalpy, 17 3.2.1 First Law in Terms of Enthalpy, 18 3.2.2 Enthalpy Change for a Process at Constant Pressure, 18 3.2.3 Enthalpy Change for a Substance with Temperature, 19 3.3 Enthalpy Change due to Chemical Reactions, 20 3.3.1 Heat of Reaction, 20 3.3.2 Heat of Formation, 20 3.33 Heat of Combustion, 21 3.3.4 Heat of Transformation or Latent Heat, 21 3.3.5 Heat of Solution, 21 3.3.6 Heat Balance, 21 3 4 Hess’ Law and Kirchoff’s Law, 22 3.4.1 Hess’ Law, 22

3.4.2 Kirchhoff’s Law, 23 Examples, 24 Exercise, 27

Chapter 4: Second Law of Thermodynamics 4.1 Second Law of Thermodynamics, 28 42 Entropy, 28 4.2.1 Entropy is a State Property, 29 4.2 2 Entropy Change for a Reversible and Irreversible Process, 29

4.2 3 Entropy of a Substance, 29 4.3 Combined Expression of 1% and 2 Law of Thermodynamics, 31 4.4 Thermodynamics Equation of State, 32 Examples, 34 Exercise, 36

v)

28-36

Chapter 5: Free Energy and Third Law of Thermodynamics 5.1 Helmholtz Free Energy, 37 5.2 Gibbs Free Energy, 38 5.3 Free Energy of a Substance, 38 5.4 Free Energy of a Reaction, 38 5.5 Free Energy as Criteria of Equilibrium, 39

37-46

5.6 Partial Derivative of Free Energy, 39 5.7 Maxwell's Relation, 40

5.8 Third Law of Thermodynamics, 41 5.8.1 Entropy at Absolute Zero and Lack of Internal Equilibrium, 42 5.8.2 Experimental Verification of the 3" Law, 42 Examples, 43 Exercise, 45

Chapter 6: Fugacity, Activity and Equilibrium Constant

47-54

6.1 Fugacity, 47 6.2 Activity, 47 6.3 Standard State, 48 6.4 Equilibrium Constant, 48 6.4.1 Importance of Equilibrium Constant, 50 6.5 Lechatelier Principle, 50 6.6 Gas — Solid Reaction, 50 Examples, 52 Exercise, 54

Chapter 7: Inter—relations Between Thermodynamics Variables 7.1 Gibbs — Helmholtz Equation, 55 7.2 Van't Hoff Equation, 56 7.3 Integration of Van’t Hoff Equation and Sigma Function, 57 7.4 Clausius Clapeyron Equation, 58 7.4.1 Liquid — Vapour Equilibrium, 58 7.4.2 Solid — Vapour Equilibrium, 60

55-63

7.4.3 Solid — Liquid Equilibrium, 60 7.4.4 Solid — Solid Equilibrium, 60 7.5 Trouton’s Rule, 61

Examples, 61 Exercise, 63

Chapter 8: Solutions and Partial Molar Quantities 8.1 Solutions, 64

8.1.1 Atom Fraction, 64 8.1.2 Mol Fraction, 65

8.2 Ideal and Non-ideal Solutions, 66 8.2.1 Raoult’s law, 66

8.22 Henry's law, 67 8.3 Sievert’s Law, 67 8.4 Partial Molar Quantities and Gibbs-Duhem Equation, 68

8.4.1 Methods for Obtaining Partial Molar Quantities from Molar Quantity, 69 8.4.2 Methods for obtaining G, G, , G, etc., 70 8.4.3 General Inter-relation ofPartial Molar Quantities, 71

8.4.4 Integration of the Gibbs-Duhem Equation, 72 8.5 Thermodynamics of Mixing of Solutions, 73 (vi)

64-82

8.6 Regular Solution,74 8.7 Excess Partial Molar Quantities, 74

8.7.1 Excess Partial Molar Free Energy, 74 8.8 1wt % Standard State, 75 8.9 Solute — Solute Interaction in Dilute Multi-Component Solution, 76 8.10 Activities in Concentrated Liquid Metallic Solutions, 78 8.11 Activities in Industrial Liquid Metallic Solutions, 78 8.12 Chemical Potential, 79 Examples, 80 Exercise, 82

Chapter 9: Phase Rule and Phase Diagrams

83-87

9.1 Phase Rule, 83

9.2 Heterogeneous Equilibria, 83 9.3 Free Energy-Composition and Temperature-Composition Diagrams, 85 9.3.1 Corelation of Free Energy vs Composition and Temperature vs Composition Diagrams, 86 9.3.2 Metastable Phase, 86

Chapter 10: Free Energy -Temperature Diagram 10.1 Ellingham Diagram for Oxides, 88 10.2 Determine of Point O, H, C and Equilibrium Gas Pressure, 91 10.3 Metal — Sulphide System, 92 10.4 Predominance Area Diagram, 94

88-95

Exercise, 95

Chapter 11: Reaction Kinetics,

96-115

11.1 Types of Reactions, 96 11.2 Rate of Reaction, 97 11.2.1 Effect of Condition on Rate of Reaction, 97 11.3 Order of Reaction, 98 11.3.1 Zero Order, 99

11.3.2 First Order, 100 11.3.3 Second Order, 100

11.3.4 General n th Order, 102 11.4 Determination of Order and Rate Constant of a Reaction, 102

11.4.1 Method of Integration, 102 11.4.2 Half— Life Method, 103 11.4.3 Differential Method, 104 11.4.4 Initial Rate Method, 104 11.5 Reaction Rates for Homogeneous Reactions, 105 11.5.1 Law of Mass Action, 105 11.5.2 Effect of Temperature on Rate of Reaction (Arrhenius Equation), 106 11.6 Reaction Rates for Heterogeneous Reactions, 108 11.6.1 Chemical Reaction Control, 108

11.6.2 Interfacial Reaction Control, 108 11.7 Diffusion, 110 11.7.1 Fick’s Law of Diffusion, 110 (vii)

11.8 Analysis of Rate Data or Kinetic Data Analysis, 111 Examples, 111 Exercise, 115 Chapter 12: Characterization of Slag 12.1 Composition and Function of Slag, 116

116-120

12.2 Structure of Molten Slag, 117 12.3 Breakdown of Silicate Network, 119 12.4 Thermodynamic of Slag — Metal Reaction, 119 Chapter 13: Electrometallurgy

121-125

13.1 Introduction, 121

13.2 Laws of Electrolysis, 122 Examples, 123 Exercise, 124

PART - Il : NUMERICALS

IN PROCESS METALLURGY

A. Ironmaking Examples, 132 Charge Calculations, 132 Reduction, 138

131-145

Calcination, 143 Rate Measurement,

145

Exercise, 145 B. Steelmaking Examples, 150 Charge Calculations, 150 Thermodynamic Calculations, 157 Rate Measurement,

149-162

161

Exercise, 163 C. Extractive Metallurgy Examples, 167 Thermodynamic Calculations, 167 Charge Calculations, 174 Exercise, 178 Appendix

165-177

179-183

Table Al: Physical Properties of Some Elements, 179 Table A2: Most Common Units, 179 Table A3: Some SI Derived Units, 180 Table A4: Prefixes used in SI Units, 180 Table AS: Conversion Factors, 180 Table A6: Values of Selected Constants, 181

Table A7: Standard Heat of Formations and Entropies and Free Energies of the Substances, 182 Table A8: Standard Gibbs Free Energies of Reactions, 183 Bibliography 185 Index

187-188

(viii)

Notations Helmholtz’s free energy Atomic number

Activity Heat capacity Concentration of component 1 Heat capacity at constant pressure Heat capacity at constant volume Change in heat capacity at constant pressure Change in heat capacity at constant volume

Mass diffusion coefficient or mass diffusivity Energy / equivalent weight

Infinitesimal change in energy Finite change in energy Electronic charge Faraday’s constant Fugacity

Gibb’s free energy Standard Gibb’s free energy Partial free energy of a substance Change of free energy

LH

Real

[=]

dH

Finite change in free energy (for a reaction) Standard Gibb’s free energy change Enthalpy Infinitesimal change in enthalpy Finite change in enthalpy (heat of reaction) Standard enthalpy change Current

species Equilibrium constant / rate of reaction Temperature in Kelvin Mass Molecular weight of component i Atom fraction

YR

Total number of gm-atoms Pressure

vg

Bibs

ad

mRIQ

Activity quotient Heat absorbed

Infinitesimal heat absorbed Gas constant Radius Entropy Infinitesimal change in entropy Finite change in entropy Standard entropy change Temperature

(ix)

t V Ww xX Z

Time Volume Work / weight Mol fraction Electrochemical equivalent

z

Valency

Abbreviations: atm std (s) 0)

Atmosphere Standard Solid state Liquid state

(g) (Hm)

Gaseous state Hot metal

(x)

PART | Metallurgical Thermodynamics & Kinetics (Problems in Ironmaking, Steelmaking and Extractive Metallurgy) 100

80—

—20

B

-

£

=

8S

8 8

E

5

60 —

—40

=

=

EN

oN

Wustite

|

40

9

20 Hematite

200

400

| 600

1 800

3Fe,0, + CO — 2Fe,0, + CO, I | 1000 1200

Temperature, °C

1400

[=:] =]

be]

Introduction

CHAPTER

First let’s begin with, what is Metallurgy?

Metallurgy is the science and art of extracting metals from their ores, refining them and preparing them for use. Prof. F. D. Richardson classified Metallurgy in three basic divisions (Figure 1.1): (I)

(I) (ITI)

Process Metallurgy,

Chemical Metallurgy, and Physical Metallurgy.

Fig 1.1: Classification of Metallurgy. Process Metallurgy deals with momentum,

heat and mass transfer, mechanism of solids, math-

ematical modeling and computer applications. Chemical Metallurgy is application of chemistry in Metallurgy i.e. chemical methods used on

ore to extract of metal. Physical Metallurgy is application of physics in Metallurgy. It deals with solid material only, not with liquid metal or slag. It deals with the physical aspects of metal i.e. physical properties of metals, crystal structure; effect of impurities, alloying elements; heat treatment etc. On the basis of applications (applied sectors) of Metallurgy, it can be further classified (dotted

lines in Figure 1.2): (A) Extractive Metallurgy

(B) Metal Processing

(C) Material Science

Fig 1.2: Classification of Metallurgy (on the basis of application).

3

4

Metallurgical Thermodynamics,

Kinetics & Numericals

Extractive Metallurgy is the combination of process metallurgy (I) and chemical metallurgy (II). Metal Processing is the combination of process metallurgy (I) and physical metallurgy (III).

Material Science is the combination of chemical metallurgy (II) and physical metallurgy (III). Now coming to the definition of thermodynamics, dictionary meaning of thermodynamics is “the science of the relations between heat (thermo) and mechanical work (dynamics i.e. motion)”. But it might more usefully be defined as the study of the changes in energy accompanying chemical and physical changes, and helps to predict changes which have not been observed. More specifically, thermodynamics is the study of the energies involved in a reaction and,

therefore, provides information regarding the driving force behind a reaction. By studying the energy requirements, it is possible to determine the optimum conditions necessary to provide the desired reaction.

Thermodynamics has wide applications in all branches of Metallurgy: Extractive Metallurgy deals with extraction of metals from ores and refining of metals. (Study of mineral dressing, ironmaking, steelmaking, extraction of non-ferrous metals etc.)

Physical and Mechanical Metallurgy deals with properties, their testing and fabrication of metals and alloys. (Study of microstructure, phases, testing of properties, mechanical working, cold and hot working, rolling, forging, and extrusion.) Main applications of Thermodynamics:

» + * » »

Prediction of process feasibility, Calculation of equilibrium composition of coexisting phases, Properties of metallurgical solutions, Phase equilibria, phase diagrams, Vaporization,

* Electrometallurgy, = Interfacial phenomena, » Calculation of heat requirements of processes, i.e. heat balance. Usefulness:

+ Thermodynamics has non-atomic approach, i.e. it does not bother about atomic and microscopic details of a substance. + It is concemed only with initial and final macroscopic states and arrives at results. + Hence it is a simple and powerful tool for quantitative calculations and feasibility of a reaction. » It is the basic principle and foundation of the Metallurgical Science and Technology. Limitations: + It cannot predict about structures of materials. + It cannot predict mechanism of the reaction (whether diffusion control or chemical control) and rates of reaction (i.e. kinetics). Chemical and Metallurgical Thermodynamics: * Chemical Thermodynamics means thermodynamics as applied to chemical reactions and equilibria, using the concepts of entropy and free energy. * Metallurgical Thermodynamics means application of chemical thermodynamics to metallurgical processes in extractive metallurgy, phase equilibria, phase transformation etc. 1.1

LAWS OF THERMODYNAMICS 1st Law: Energy cannot be created or destroyed, but it can be converted from one form to another (this law 1s similar to the law of conservation of energy).

Introduction

5

2nd Law: (7) The older statement as applied to thermodynamics of heat engine is “heat cannot be transferred from low temperature to high temperature without the aid of external agency.” (This is apparently a common experience). Thus the law states the irreversible nature of spontaneous heat flow. (if) New statement is in terms of the concept of entropy with respect to reversible and irreversible processes, 1.€., for a spontaneous (non-equilibrium), irreversible change, the entropy (8) ofan isolated system always increases (1.e., AS = + ve, where AS=8 a — 5...) 3rd Law: The entropy of any homogeneous substance, which is in complete internal equilibrium (a condensed state) may be taken as zero at the absolute zero temperature (i.e, S,=0at T= 0K). Entropy means measure of randomness, measure of degree of irreversibility; i.e., it increases with temperature. Zeroth Law: It is called zeroth law which indicates that it is more elementary than even the first

law. This law is for bodies which are in equilibrium. If two bodies sayA and B are in equilibrium with C, then A and B are in equilibrium with each other. The equilibrium between two bodies involves the equality of temperature, pressure and chemical activity.

1.2

BASIC TERMS USED IN THERMODYNAMICS Reactor: A reactor is the physical apparatus in which chemical reactions take place. Reaction mixture: The reaction mixture is the entire material within the reactor. It consists of reactant and product species, and in some instances it may include catalytic substances or inert species. System and Surrounding: Any portion of the universe selected for consideration is called system. Rest of the universe outside the system is known as surrounding. e.g. Reduction of Hematite (Fe,O,) by hydrogen gas, 800°C

Fe,0, (s) + 3H, (g) ——— 2Fe (s) +3H,0 (g) This system consists of solid hematite, solid iron, hydrogen gas and water vapour. There are three kinds of systems: (1) Isolated, (i1) Closed and (111) Open. 1. An Isolated System: A system enclosed by impermeable walls that permit neither exchanges of energy nor transfer of matter. i.e,An isolated system does not exchange matter and energy with its surrounding. 2. A Closed System: A system enclosed by impermeable walls that does not permit the transfer of matter but allows transfer of energy. i.e., A closed system exchanges only energy but not matter with its surrounding. 3. An Open System: A system bounded by permeable walls that allow the transfer of both matter and energy across the walls. i.e., An open system exchanges both matter and energy with its

surrounding. Open system is encountered in fluid flow. Thermodynamic relations are derived by assuming closed or isolated systems, particularly the closed system. Hence, the term system, use in the text, means the closed system. e.g. Materials inside a closed metal container constitutes a closed system. Heat can be exchanged through the metallic wall but not any matter.

Apart from above kinds of systems, there are another two forms of systems: (7) Homogeneous and (if) Heterogeneous. (i) Homogeneous: A system which is uniform throughout (i.¢., chemically) and generally made

up of a single phase, e.g. liquid metal, slag etc. (if) Heterogeneous: A system which is not uniform throughout and generally consists of two or more phases, ¢.g. mixture of liquid metal and slag, liquid metal and gas etc.

6

Metallurgical Thermodynamics,

Kinetics & Numericals

The State of a System:

It is defined at any instant by specifying state variables or properties namely, temperature (T), pressure (P), volume (V), composition, viscosity, surface tension. This does not include atomic

properties or atomic arrangements. e.g. The state of a system consisting of 10 gms of water may be defined further by saying that, itis at 298 K (25°C) and one atmospheric pressure and not under the influence of electric, magnetic, gravitational or other kind of external field. Thus all other properties are fixed and may be found by experiment or referring the literature. Equation of State:

There are various interrelations between the state properties, mainly temperature (T), pressure (P), and volume (V). These are called equation of states.

One of the basic equation of states is PV = RT. In thermodynamics it is assumed that all the pure substances and ideal or perfect gases obey this equation of state. (It is applicable for one mole of an ideal gas). 1.3

PROPERTIES OF A SYSTEM State properties can be divided into two types: (i) Extensive and (ii) Intensive. (i) The extensive properties (i.e, variables) of the system vary with its amount or size. (if) The intensive properties of the system are independent of its amount or size.

In common use the word property refers to an intensive property. Comparison of extensive and intensive properties:

—

Extensive Properties:

Intensive Properties:

. Those properties of the system that vary with its size Or mass.

1. Those properties of the system that are

e.g. mass (M), volume (V), energy (E), enthalpy

(H), entropy (8), free energy (G) e tc of a sub-

e.g temperature (T), pressure (P), density (p), viscosity, refractive index, molar

stance.

energy etc. of a substance.

. These properties for a system are additive. 1.e., Total volume of a system is sum of all the volumes of the component parts.

independent of its size or mass.

. These are not additive and hence a value

may be assigned at each point in the system.

i.c., Total temperature of a system is not the sum of the temperature of the component parts. T # T +T+T,... also,

LFS]

P#£P+P+P,.

. Depends on mass of a system

e.g. 10 gms of solid copper at 273 K (0°C) and 1 atm. occupy a volume of 1.12 cm?) where as

1 Kg of the same substance has a volume 100 times larger.

. Not depends on mass of a system. e.g. If two substances (10 gms and 1 Kg mass) are heated, the temperature raise will be the same (T) for both at a

particular time.

The ratio of any two extensive properties is independent of the total mass and hence this ratio 1s intensive property. e.g. All the specific properties like specific volume (i.e., volume per unit mass) and the molar properties like molar volume or molar energy — fall in this category (i.e., for one mol)

Introduction

1.4

7

REVERSIBLE AND IRREVERSIBLE CHANGES (i) (ii)

The change or transformation in a system may be reversible (under equilibrium) or irreversible (i.e., non-equilibrium or spontaneous). If the force in the system and the external force opposing the change differ by an infinitesimal (i.e., very small) amount, the process is reversible, because its direction can be changed by a slight change in the external force; e.g. thus for a gas, having pressure P, under expansion; if the external pressure is (P - 8P) with 8P — 0 at each stage of the process than the process of a gas expansion in reversible. But if the external pressure is 1 atmosphere and gas expands from 5 atm to 1 atm, the process is irreversible.

(iii) A reversible process passes through a number of stages of equilibrium, artificially arranded, 1.e., Reversible process: equilibrium — non-equilibrium — equilibrium —» non- equilibrium — equilibrium.

(iv)

While the irreversible process is a spontaneous process, 1.¢., non-equilibrium —» equilibrium. Irreversible process is a natural process which involves spontaneous movement of a

system from a non-equilibrium to an equilibrium state. But reversible process is slow (v)

1.5

and impractical. But the concept allows us to handle many practical problems. The thermodynamics, we study, is for reversible processes only because they correspond to maximum efficiency and a simple generalized theoretical treatment can be easily given to them.

EQUILIBRIUM The equilibrium state means it is a state of rest 1.c., no further reaction occurs. The system has come into equilibrium with respect to variables, i.e., the magnitude of variables

remains same throughout the system. Thus, when a system is in a state of equilibrium, no changes occur in its thermodynamic state without the intervention of an external agency.

A chemical equilibrium corresponds to equal rates of forward and reverse reaction in a closed system consisting of the reactants and products. e.g. The closed system consisting of water at 373 K

(100°C) in contact with steam at | atmospheric pressure

forms a equilibrium system. Hence, in a

given time interval equal numbers of water molecules evaporate and steam molecules condense in this system. If slight increase in pressure will cause condensation, while a slight rise in temperature will cause evaporation. The thermodynamics, we study, is for equilibrium or reversible reaction. Types of Equilibrium: Mechanical Equilibrium: When the pressure within a system is the same is in a state of mechanical or pressure equilibrium, Thermal Equilibrium: When the temperature is uniform throughout a equilibrium. Chemical Equilibrium: When the components in the system no longer act i.e. the absence of net reaction is chemical equilibrium. Thus when the equilibrium, the rate of the forward and reverse reaction is same, resulting

at all points, the system system it is in thermal have a tendency to rereaction is in chemical in no net consumption

or production of components. Also for a chemical equilibrium, the chemical potentials (pn) of all components are same (Le, 1, = 1, = |, ).

Thermodynamic equilibrium is a total equilibrium consisting of all three equilibriums. When a system 1s in a state of complete equilibrium (1.e., mechanical, thermal and chemical), it follows that all the parts or subsystems that compose it must also be in equilibrium. Pressure must be uniform throughout the system, the temperature must be uniform, and the chemical potential of each of the components must be constant throughout, no matter whether the system is composed of one or many phases.

Metallurgical Thermodynamics,

1.6 1.

ISOTHERMAL AND ADIABATIC CHANGE OR PROCESS Isothermal

Adiabatic

Temperature remains constant throughout the process and there is a heat exchange (8q) between the system and the surrounding so as to achieve constant temperature 1.€. some heat does enter or leave the system, hence

Temperature keeps on changing throughout the process and there is no heat exchange (8g) between the system and the surrounding, 1.e., no heat enters or leaves the system, hence 6q = 0.

Since temperature energy (E) is also 0, putting this into dq — &w, we get 6q

1s constant, internal constant, hence dE = 1% law equation: dE = = dw.

Since 8q = 0, putting this into 1% law equa-

1.e., All heat absorbed = work done on the

In this case, since E is state variable (i.c., definite quantity, independent of the path followed).

3q#0.

2.

tion: dE = 8q — 8w, we get dE =—8w. i.e., Energy change = work done.

system. In this case, both q and w are not state variables (i.e., depends on path followed.) 3.

Kinetics & Numericals

Characteristic equation for an ideal gas under going isothermal change, PV, =P, V

Characteristic equation for an ideal gas under going adiabatic change,

PV/'=PV, wherey=C,/C_ 22 *

TO

«Same

Initial Point

0

Isothermal

Adiabatic

V—

Fig. 1.3: P-V relationship for 1sothermal and adiabatic expansion of an ideal gas starting from same initial stage.

4.

Since vy is greater than unity, the slope of the adiabatic change is greater (as shown in Figure 1.3) than that of isothermal change.

CHAPTER 2.1

Energy and First Law of Thermodynamics

ENERGY

The energy of a body can be defined as the capacity for doing work. Energy can be translated into work that can be measured. Different forms of energy: e Kinetic energy of the body in motion (2 mv?)

e

Potential energy of the body due to it’s position (mgh)

® ® eo eo so

Dynamic energy = kinetic energy + potential energy Mechanical energy = work done = force x distance Heat energy [Electrical energy Chemical energy

eo

Surface energy

» » ss o *

Nuclear energy (Geothermal energy Wind energy Tidal energy Solar energy.

(a) Relative energy and internal energy

There is nothing like absolute energy because energy of a system is a function of an arbitrarily chosen standard state. Thus it is only relative energy of a system and hence only energy difference can be measured. For Metallurgist and Chemist, kinetic energy (associated with motion) is of least interest and it

rarely appears in the system in which they are interested. Assuming that kinetic energy of the system is negligibly small or sensibly constant, hence the total energy other than kinetic energy is called internal energy or simply energy (E). (b) Energy, a state property Let us consider a system (Figure 2.1) which passes from state A to state B by path 1 (i.e, A1B) and it return to state A via the same path (B1A). Since this forms a cycle, the energy change will be zero, when a system passes through a cycle, AE = 0. This can be experimentally proved by measuring all heat and work effects (i.¢., g and #) when system passes through cycle (i.c., AE =g-W=0). Therefore,

AE, \g TAEy

= AE,

or,

=0

+ k22)

AE, pn =—AEq,, (2.2) In general, the energy change involved in passing from any state A to any state B is the negative of the energy change from state B to state A. 9

10

Metallurgical Thermodynamics,

A

Kinetics & Numericals

Path- 1

Path - 2

Fig 2.1: States A and B of a system

Now consider the system passing from A to B via path 1 (1.e., A1B) and then state B to A via path 2 (i.e, B2A), as shown in Figure 2.1. Since this also forms a cycle, the energy change will be zero. Therefore, AB gph AB = AE =0 (2:3) Substituting from eq. (2.2) ineq. (2.3):

AE TAFE = 0 or AEy,=AEg,, - (2.4) Thus, from experimentally proved fact that AE of a system in passing through a cycle is zero, it is found that AE in passing from one state to another is independent of the path followed, means AE is a state property.

2.2

FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is nothing but a statement of the law of conservation of energy

1.¢., neither any energy is generated not any energy is destroyed; only transfer of energy from one form to other form take place. Consider a closed system, that is free to do work (W) on the surrounding and is also exchanging heat (q) with them. Careful experiments revealed that q (quantity of heat absorbed) = W (work done) for many processes, apparently violating the law of conservation of energy. In order to make these findings confirm to the law of conservation of energy, the concept of internal energy (E) was proposed. The internal energy is the stored energy in the system and it may be written as: Internal energy = kinetic energy + potential energy +energy of atoms/molecules + energy of interaction amongst atoms/ molecules. The internal energy is a state function or state property. Therefore the change in the internal energy is characteristic of the mnitial and final states of the system, and does not depend on the path taken to bring about change. According to the first law, the change in the internal energy (AE) of the system is equal to the quantity (q—W). Therefore, AE=q-W (2.9) where AE = finite change in the internal energy = E, — FE, g = amount of heat absorbed by the system from the surroundings W = work done by the system upon the surroundings.

For an infinitely small (i.e, extremely small) change, the first law can be expressed as follows: dE =8q — 6W (2.6) where dE = an infinitely small change in the internal energy of the system dg = an infinitely small quantity of heat absorbed by the system SW = an infinitely small quantity of work done by the system.

Energy and First Law of Thermodynamics

11

Eqs (2.5 & 2.6) are the statements of the first law of thermodynamics.

Eq (2.6) can be further written as: dE = 8g — PdV —8W' walZ 7) where 8 = PdV + 5" i.e., the work done against an external pressure (PdV) and all other forms of work (8 W") which includes work against external electrical, magnetic or gravitational fields etc. But usually system

considered is such that it does work, other than that against pressure; hence, unless any thing is stated, we shall always consider §#' = 0.

Thus,

dE =38g— PdV

(28)

Eq (2.8) is also the statement of the first law of thermodynamics. Significance of the first law of thermodynamics:

s

ltis based on the law of conservation of energy.

e

It brought in the concept of internal energy.

e

It separates heat interaction and work interaction between the system and the surrounding as two different terms.

e

It treats internal energy as a state property. Hence change of internal energy would depend

only on the initial and final states of the system, and not on the path by which the system has moved from one state to another. 2.3

MEASUREMENT OF ENERGY CHANGE (AE) The only way to measure AE of the system is by measuring g and W. Since, AE=qg—-W=gq—-PdV (2.5) i.e., there is no such thing as energy meter and only way to determine AE of a system is by measuring all the heat and work exchanges of the system with the surroundings. These are measured by posting measuring instruments all along the periphery of the system (i.e., to measure P, V, T etc).

Most of the experiments are designed to minimize the number of measuring instruments required by making the conditions at various points on the periphery of the system, as much alike as possible.

2.4

ENERGY CHANGE (dE) IN TERMS OF PARTIAL DERIVATIONS

It is found from experience that, for a closed homogeneous system, any state variable may be represented as a function of three state variables only. Hence, to represent energy, three state variables are chosen (e.g. P, V, and T). So that E of such a system is a function of P, ¥, T.

ie.

E=f(P,V,T)

(29)

Now if the system obeys an equation of state (1.e., PV = RT), then P, V, or T in the functional expression for energy can be eliminated by one of the following equations of state.

Thus E=f(P, V),E=f(P,T)

or

E=f(V.T)

(2.10)

Hence dE may be expressed in terms of partial derivatives by using fundamental theorem of partial derivatives, as follows:

If E=f(P,V), then

a=(35) 5P

®+ (5) dv 3 Jp

(2.11)

If E=f(P,T),

then

dE = (5%)

(3)

dT

(2.12)

12

Metallurgical Thermodynamics,

Kinetics & Numericals

If E=f(V,T), then dE =

(5) av

(5) oT

12.13)

Examples Ex 2.1: If ¥=f(P, T) and independent of the total differential dV, then prove that:

E11)

[14

Solution: Since

V=f(P,T

«(B)

av ov Therefore, erefore, dV=|— [= )| dP+| 5— )| dT

22

Again, V is constant, hence dV=0

av 0 = I

So eq (2) becomes:

av dP il

dar

(3)

Again, differentiate with respect to T of eq (3) at constant V:

CATA EA8T “sp flsT

.

Jp

~@

[57 Therefore,

5P

| — | =— aT

8T ),

Proved.

14 | | 8P

Ex 2.2: The equation of state of a gas is given by the expression: [ pP 5

Prove that:

Jv -»)]

= RT, where a, b are constants.

oT) (82) (3) __, SP

IN%

ST

Solution: Since

I 3 6%

Therefore, T= [G

21

bh

Jo | =RT Vb

=

Js

Vb JJ"

Differentiate of eq (2) w.r.t. P at constant V:

: eq (1): : (7+(5%ZN) We can write

(1)

BT (75)

R

V-b J)

|= | = (22)

R

’)

--(2) ~-(3)

Energy and First Law of Thermodynamics

Therefore,

P=

13

(= NG )

Lo

(4)

Differentiate of eq (4) wrt. V at constant T*

ot ov

{

14

—2a

vb) HG 2a(V —b)

1

(Vb) I

v

) He)

(5)

Again from eq (1): [7+(=)] (V-b)=RT

Or

pr+

(2) (2|p? |-@rer Vv)

(6)

Differentiate of eq (6) w.r.t. T at constant P:

7)

(3 oT

Ve

SCARF Js

| FER i i | a 87

R

(MD

2) V3

From egs (3), (5) and (7), We get: aT

oP

ar

&pP

ov

8T

Proved.

Heat Capacity, Enthalpy and Heat of Reaction

wee 3.1

HEAT CAPACITY The heat capacity (C) of a substance is defined as the amount of heat required to raise its temperature by one degree. In other words, heat capacity is the ratio of the heat increment (8q) to the temperature rise (dT), as both approaches zero. i.e.,

c={5a Bl

LGD

Specific heat is the heat capacity of a unit mass of the substance. Atomic heat capacity is the heat capacity of one atom of an element. Molar heat capacity 1s the heat capacity of one mol of a substance or compound. Generally heat capacity is expressed in calories per degree centigrade of one gram atom, or one gram mole, 1.e., Cal / °C.mole (CGS Unit), or J / K.mol (SI Unit). 3.1.1

Heat Capacity at Constants Pressure and Volume Since 8q is incomplete differential because it is not a state variable (i.e., it depends on path fol-

lowed). Hence for a given change or a process, dq can be determined, only if the path of change is specified, i.e, for

Cor =)

to be definite, should always carry a subscript indicating the particular

manner in which the temperature 1s raised. Experimentally it is most convenient to determine the heat capacity under the condition of constant pressure (usually atmospheric pressure) and heat capacity at constant pressure is denoted by C,. Theoretically it is casier to determine the heat capacity under constant volume and this heat capacity at constant volume 1s denoted by C,, 3.1.2

Heat Capacity in Terms of Energy

3.1.2.1 Heat Capacity at Constants Volume (C,) Considering equations (2.8 and 2.13):

dE = dq

PdV

Since,

E=f(V,T)

Therefore,

dE

ATE)

[1

£5

_— E ) dav

8E

+

=— |

dT

L213)

Combining these two equations (2.8 and 2.13),

3q—PdV= Therefore,

dE 8g os= PdV + (5

|

dav +

dE E— |

i

av dl

dav +

w

ar

dT sal (3.2)

Heat Capacity, Enthalpy and Heat of Reaction

15

Now considering a process at constant volume i.e, dV'=0 So eq 3.2 become:

aodg

[3

(&)

8) _(3E

or

..(33)

(#5)

By definition, C), = (a)

(3.4)

Hence, Cp, = (5)

L(3.5)

3.1.2.2 Heat Capacity at Constants Pressure (C;) Considering equations (2.8 and 2.12):

dE =8q — PdV Since, Therefore

(2.8)

E=f(P,T)

:

dE = (55)

5P

2+

(5)

irl.

(2.12)

dT

Combining these two equations (2.8 and 2.12),

ro

(i) + (a7),

3E SE Sq-PdV=|— | dP+|—| SE &P

or

ar

SF dT 8T )p

8q=PdV+|—|

..(3.6)

dP+|—=|

Now considering a process at constant pressure i.€., dP =0 So eq (3.6) become:

8g =PdV+

oF

|—

oT

~(7)

| dT

Now dividing both sides by dT at constant pressure:

"

By definition, a

(sr) ="er)dq (7),

ence,

Csi

es

op fi

14 st

AO (39)

8T ),

J,

i

SE *\57 J,

.. (3.10)

: oF ; OPV The expression P | —— | can also be written as | —— OT Jn oT Jp Therefore, 3.1.2.3 Relation Between

Wek e know thattha

ory

Ce; ==] oT

SE

S(E+PV)

) +=] [= ) == 57

)

(3.11)

Cp, and C},

OF C,=|— El

pg =f lar al) ATof er L eS

(35) .. (3.10)

16

Metallurgical Thermodynamics,

Therefore,

[14

3)

C,— C, =P 2

=

J §

FE

Kinetics & Numericals

3.12)

Since E=fV, T), Therefore, dE =

SE

14

dV

+

os

dT

T

(E-FUELE)

Now dividing both sides by dT at constant pressure,

oer LEE EH) (E113)

=

Putting this value (eq 3.13) to above eq (3.12), we get

on

Equation (3.14) may be applied to an ideal gas, which obeys equation of state: PV =RT Therefi erefore,

.

xT

=

23

LL

3.15 (3.15)

14 R | =| — ar Je AP

Differentiating eq (3.15) at constant pressure, | —

(3.16)

Again in case of ideal gas, energy is a function only of temperature and it does not change with

volume at constant temperature, i.e.,

3E (7) =0

(317)

Now putting the values of eqs (3.16 and 3.17) in the eq (3.14), we get

R Cpo=Cp=[P+0] x (7) =R Therefore for ideal gas,

Cp—Cp=R

AER)

3.1.3 Dependence of Heat Capacity on Temperature Experimentally it is found that different substances require different amounts of heat energy to raise their temperature by a given temperature interval. e.g. 1 gm of mild steel requires X, KJ to raise its temperature by 1°C e.g. | gm copper requires X,, KJ to raise its temperature by 1°C

e.g. 1 gm silica brick requires X; KJ to raise its temperature by 1°C Now if the temperature of the material is raised to 1273 K (1000°C), these values of heat required may raise by as much as 50%. Hence the temperature at which the measurement of the heat capacity was made must be stated because heat capacity or even specific heat generally varies with the temperature. Experimental determinations give empirical relationships between heat capacities and tempera-

ture as:

Cp=a+bT+cT™

(319)

Heat Capacity, Enthalpy and Heat of Reaction

17

Where a, b, ¢ are constants for a substance over a particular temperature range. The values of these constants are listed in standard book. * The salient features of the above equation (3.19) are:

1. Many times the final term (i.e., cT 2 ) is omitted, 2. This equation is frequently used for solid metals and compounds at elevated temperature. 3. 4.

Insome of the cases where data is not sufficient, as for liquid metals, C,, is assumed constant. Since this equation is purely empirical, its use is strictly limited to the range of temperature

for which the data of a, b and c for a particular substance was derived.

t Cp

(JK)

T(K) —=800 Fig 3.1: G vs T plot

Figure 3.1 shows the variation of heat capacity with temperature for Nickel in temperature range 0 to 800 K (- 273 to 527°C) Thus it shows gradual increase in C,, with T, except for a sharp discontinuity at 600 K (327°C). This discontinuity is due to the magnetic transformation of Nickel at 600 K (327°C). Thus such discontinuities occur in such plots wherever there is a phase change or order-disorder transformations.

3.1.4 Importance of Coand C,, C, is rather more important and mostly used in metallurgical thermodynamics than C,, because most of the chemical reactions in metallurgical processes take place at constant pressure (i.e., one atmospheric pressure) and not at constant volume. The value of Cp, is always greater than C,, (Cp, > C)).

Since

C),=

Considering

5 aT ns

aT

and =

oe

Cp, =P Ld + a 8T Jp \8T Jp

oT Jn

, then for Cp, still have extra term of P

ud

oT

Je

.

Hence Cp >C),.

3.2 ENTHALPY Enthalpy 1s a Greek word meaning worm. As mention earlier, heat capacity at constant pressure: Cp, =

(E+

PV)

aT)

(3.11)

Thus when the system undergoes a change at constant pressure, the quantity (E + PV) occurs

very frequently and hence it is convenient to denote it by a single symbol H and to give it a special name enthalpy or heat content. Hence, enthalpy,

H=E+PV

3:20)

* 0. Kubaschewski and C. B. Alcock: Metallurgical Thermo-Chemistry, 5% Ed (Revised), Pergamon Press, Oxfort (1989).

18

Metallurgical Thermodynamics,

Kinetics & Numericals

Since E, P and V are state variables, H is also a state variable. Hence, like energy change (dE), enthalpy change (dH) for any cyclic process involving a thermodynamic substance is zero. Secondly, enthalpy is an extensive property of the system, as it depends on the amount of the system and hence, like a volume, enthalpy values are additive. Thus, we have, Cp,= a

-(57) 3

(32D) /d

3.2.1 First Law in Terms of Enthalpy

The first law of thermodynamics can be expressed in terms of the enthalpy instead of energy. Since, H=E+PFPV «1(320) By differentiating, dH =dE + PdV + VdP Now according to first law: dE = dg — PdV—6W" Since there 1s no other external work, so 8" = 0 ; therefore dE = 8g — PdV

wn (3.22) (2.7) (3.23)

Putting eq (3.23) in eq (3.22), we get: dH

=8q — PdV + PdV + VdP

dH=3q + VdP

.. (3249)

Equation (3.24) is the first law of thermodynamics in term of enthalpy.

3.2.2 Enthalpy Change for a Process at Constant Pressure If the process is carried out at constant pressure, than dP = 0; putting this value in the above equation (3.24) we have: dH = 8g (325 Hence, AH=g (3.26) Thus for a constant pressure, change in enthalpy of the system (dH) is equal to the heat absorbed by the system from the surrounding or heat exchange between the system and the surrounding (64) at constant pressure. Hence, the alternative name to enthalpy is hear content. The enthalpy (H) of a substance can be defined as its heat energy content. Chemical reactions are carried out either to obtain a useful product or to get chemical energy. e.g., combustion of coal: C (5) + 0, (g) = CO, (g) + heat evolved (— 393 KJ/mol). This process involves conversion of chemical energy into heat energy. Chemical reactions occur with an evolution of heat, are known as exothermic reactions. Energy content is lower down at the end of the reaction. Heat is lost by the system, i.e. heat is given to the surrounding.

AH = Hg) = Hpi == VE,

(SINCE Hing < Higigiar )

(3.27)

i.e. Energy content is lower at the end of the reaction. Hence, a negative (—) sign is used for change of enthalpy (AH) in case of exothermic reactions. On the other hand, the extraction of zinc at 1373 K (1100°C) absorbed the heat from the surrounding: ZnO + C= Zn + CO - heat absorbed (+350 KJ/mol).

Chemical reactions occur with absorption of heat, are known as endothermic reactions. System

gains heat from the surrounding, AH = Hy = Hypygn =F ve,

(since Hy > Hy)

(3.28)

i.e., Energy content is higher at the end of the reaction. Hence, a positive (+) sign is used for

change of enthalpy (AH) in case of endothermic reactions. The change in enthalpy (AH) is given by: The change in enthalpy for chemical reaction,

AH =H, — Hy AH=Y H, , . -YH,

= A(3:29) (3.30)

Heat Capacity, Enthalpy and Heat of Reaction

19

The change in enthalpy for standard state [i.e., at 298 K (25°C) and 1 atm pressure]:

AH

= FB rE

Ws

- (3:31)

Case I The enthalpy of a pure element is generally taken to be zero at the reference temperature (298 K)

and | atm pressure, providing the element is in its normal physical state under the condition considered.

cg.

F208 c apive ~ 0 H298.10, 1 = 0 Ha re 01 = 0 Hs pig ay =O

But

HC 630 (1g gy) = 60-84 KI/mol,

since under the specified conditions [at 630 K (357°C) mercury 1s not a liquid rather than a gas,

ie, Hg(l) > Hg(g)at630K,

AH'g. -=60.84 Ki/mol.

Case II The enthalpy of a compound is taken to be its enthalpy change of formation. eg. C(s)+0,(g)=CO,(g) AH, =-393 KJ/mol,

So

198 1005] = AH'r, 298 CO, = -393 KJ/mol.

The standard data books furnish values of H” or AHO, for compounds usually at 298 K and | atm pressure. The enthalpy change, AH, for a reaction is constant under specific conditions. The change of enthalpy (AH) 1s dependent upon: 1) temperature, 11) pressure, 11) physical states of reactant and products, and 1v) amount of substances reacting. 3.2.3 Enthalpy Change for a Substance with Temperature . As shown earlier that

oH Esl 8T

AEE)

J.

Hence, dH=Cp dT (3:32) (I) Enthalpy change when substance is heated from T to T, at constant pressure without changing of state or phase transformation. Integrating eq (3.32): Hr

To

Hy,

7

| dH =[Cpar T

Therefore, Hy ~Hy = [Cpdrl

53%

T

(IT) Enthalpy change when substance is heated from T, to T, at constant pressure with change of state or phase transformation (i.e., melting, vaporization or other phase transformation). In this case the above equation becomes: H

T;

Ty

T

I

Tf

Ty

r-Hr = [CogdT+Ly+ [CoqdT +L, + [Cpl

where I, >T,>T>T,

(3.34)

ie, T, is greater than temperature of fusion and temperature of boiling.

T,= temperature of fusion (i.e., solid to liquid) T, = temperature of boiling (i.c., liquid to gas)

L,= latent heat of fusion L, = latent heat of boiling

20

Metallurgical Thermodynamics,

Kinetics & Numericals

In above eqs (3.33 and 3.34), if H; C,, and other standard values of the substances are known, then FH, for the substance can be easily ‘calculated. (IIT) For metallurgical thermodynamic calculations, it is convenient to take room temperature (25°C 1.e., 298 K) as reference temperature. Hence, when a substance is heated from 298 K to TK,

without any change of state, then eq. (3.33) becomes: T

H' ~ Hy, = [ Cpdl

(333)

29% Where

T,=298K

and T, =T, at standard state.

He

for pure element is assumed to be zero at standard state.

€E Ho

Hoc =0 H0,0=% Hog bey =0 for compounds and alloys are equal to the heat of formation of compound at 298 K. C+0,5>C0, AH’0 00 —=—393KI/mol

eg

Hence, Haan Coxe) =

208 £00, =~ 393 KJ / mol

3.3 ENTHALPY CHANGE DUE TO CHEMICAL REACTIONS There are various ways in which enthalpy changes for reactions may be stated more specifically. Depending on the types of reaction, the enthalpy change is termed as:

1. 2. 3. 4. 5.

Heat Heat Heat Heat Heat

of of of of of

Reaction (AH) Formation of Compound (AH) Combustion of a Substance (AH) Transformation Solution

3.3.1 Heat of Reaction (AH,) A chemical reaction may generate heat or absorb heat from the surroundings. Accordingly, it is

called exothermic reaction or endothermic reaction. This heat effect is measured by heat of reaction (AH) or enthalpy changes for a reaction, when the amounts of reactants react completely. e.g. 3Fe,0, (s) + CO (g) =2Fe;0, (s) + CO, (g) AH, =-2.67 KJ/mol The above reaction takes place at 1000 K (727°C) and at constant one atmospheric pressure. Reaction evolves heat (2.67 KJ/mol). Heat of reaction (AH,) can be calculated by: AH_= [(Sum of heat content of the products) — (Sum of heat content of the reactants)] - (3.36)

= p> Hp roduets Wi p Hi eactant

e.g., for above reaction: AH, = [( 2AHg,_, + AHpg ) — (BAH, + Alp )] 3.3.2 Heat of Formation (AH) It is defined as enthalpy change when 1 mol of the compound is formed from its constituent’s elements in their stable forms, at 298 K and 1 atm pressure. e.g.

2Al1(s) + 3/20, (g) = ALLO, (s) Since

AH =%H, .

—¥YH,

AH

£ALO;

—

— 1700 KJ/mol

= {AH,, , —(H,,+3/2HO, )} = {AH Ao, — (0+3/2x0)} — AH, o, = AH;=~ 1700 KJ/mol

The absolute value of the heat content (H) of an element is not known and hence the general practice is, values of heat content of the elements are taken as zero at their standard states (i.e., at 298 K

and | atm pressure). That is why, in above calculation the values of Hyg, ,) and Hay , are taken as zero. Thus, heat of formation of a compound at 298 K and | atm pressure may be defined as heat

Heat Capacity, Enthalpy and Heat of Reaction

21

content of that compound i.e. AH £aL0; = Mags apo, 3.3.3

Heat of Combustion (AH) It is defined as enthalpy change when 1 mol of the burnt in oxygen. e.g. C(s)*+0,(g)= CO, (g) AH, 5, Mg (s) + 2 0, (g) = MgO (s) AH_ These values of heat of combustion of C and Mg oxides (i.e., CO, and MgO ).

substance (element / compound) is completely =—393.5 Kl/mol 4, =— 605 KJ/mol are also heat of formation of their respective

3.3.4

Heat of Transformation or Latent Heat When one of the components of a reaction undergoes allotropic transformation, fusion or evaporation within the temperature range considered, the latent heat is added of the transformation must be included in the heat capacity relation at the transformation temperature. This latent heat is added to the enthalpy change for the transformation of a reactant and subtracted for the transformation of a product without change of temperature. It 1s defined as enthalpy change when 1 mol of the substance undergoes a specific physical change i.e., melting, evaporation, allotropic modification etc. It is also known as latent heat of transformation. Heat supply for solid to liquid (i.e., melting) without change of temperature, it is known as latent heat of fusion (L,) and liquid to vapour (i.e. evaporation)

transformation without change of temperature, it is known as latent heat of evaporation (L,). e.g.

Zn(s) — Zn(l) at693K, Zn(l) — Zn(g) at 1180K,

L,, =7.28 KI/mol L_, e, Zn =114.22 KJ/mol

3.3.5 Heat of Solution It is defined as enthalpy change when one substance dissolves in another i.c., when 1 mol of solute is added to form a solution of particular concentration. It depends on concentration of the solution 1.e., heat of solution increases with decrease in concentration of the solution. 3.3.6

Heat Balance

For a steady state, heat input = heat output «13.37 For a system undergoing a process or reaction at constant volume (i.e., dV = 0) eq.(2.8) (1.e., dE = 8q — PdV) becomes: dE = dq +1338)

Hence, AE=q (3.39) In view of the first law, the energy change for any process is zero if we include the energy changes of the system and the surroundings. Hence, for a process of constant volume (1.e., isometric process), the energy gain by the system must equal the heat lost by the surroundings or the energy lost by the system must equal the heat gained by the surroundings. Thus for a process, heat balance may be prepared in which energy gained by the system are entered on one side and the heat lost by the surroundings on the other side. Then totals on two sides can be compared and a principal unknown term can be found out from such heat balance. Processes taking place at constant volume are few compared to those at constant pressure, usually atmospheric pressure. For the process taking place at constant pressure, eq (3.25) show that: dH = 3q , and hence, AH=gq --(3.40) In this case, the enthalpy increase of the system must equal the heat lost by the surroundings (i.e. heat gain by the system, according to eq 3.37). A heat balance may be prepared in which the increases

in enthalpy of the system are tabulated in one column and the losses of heat by the surroundings are tabulated in other column. Any lack of balance of the two columns is due to experimental error.

22

3.4

Metallurgical Thermodynamics,

HESS’ LAW AND KIRCHHOFF'S By definition, enthalpy,

Kinetics & Numericals

LAW H=E+PFV

+320)

Since E, P and V are state variables, H is also a state variable and hence independent of the path followed during a change. i.e., the value of H at a particular state (at given P and constant T) is same, irrespective of the path followed by the substance to arrive at that state. Based on the above and the fact that enthalpy a state property, two specific correlations are established: [. Hess’ Law, and 11. Kirchhoft’s Law. These are originally proposed for calculation of heat of reaction (AH,). 3.4.1 Hess’ Law Hess’ Law states that for an isothermal process the enthalpy change for a reaction is the same whether it takes place in one or several stages, i.¢., the heat of reaction (AH,) depends only on the

initial and final states of the process. Suppose AX and BX are two compounds, where 4, B and X are elements. Then to find out AH, of the reaction which is occurred at temperature, 7. AX+B=A+BX (3.41) This reaction (3.41) may be considered as a combination of the following two reactions at a temperature, T. A+ X=4X AH; ,» (3.42)

B+X=BX

AH,

Eq (3.42) can be rewrite as:

(3.43)

’

AX=A+X -AH Now by adding eqs (3.44 and 3.43), we get eq (3.41).

(344)

Therefore,

(3.45)

AH,= AH; py — AH;

Hess’ law allows us to caleulape the heat of reaction, AH, of any reaction by suitable combining those of some heat of formation reactions at the same temperature (1.e., 1sothermal process). Initial and final temperature should be same, in between whatever path it follows no matter. The great utility of this law is in calculating enthalpy changes for reactions which cannot be carried out experimentally. e.g. The reaction between methane and oxygen to produce CO, and H,0 can proceed by two distinctly different paths. (a) Direct oxidation or single stage process:

(i) CH, +20, = CO, + 2H,0

(b) Indirect oxidation or three (i) CH, =C + 2H, (iii) C+ 0, = CO, (iv) 2H, +0, =2H,0 Therefore,

AH, =~ 890 kJ/mol

stages process: AH, ,= 77 kJ/mol AH, —393 kJ/mol 2H, = 2(-287) kJ/mol

CH, +0,=C0, + eh AH = AH, + AH ;+2AH , =77 +(-393) + (- 574) = — 890 kJ/mol .

The salient features of Hess's law are as follows: 1. Asshown above, the principle of algebric addition, subtraction etc of chemical equations is applicable to the energy changes or enthalpy changes also. 2. Also in many cases, the chemical equation can be multiplied by suitable coefficients and then added or subtracted to get a desired equation. The value of state variables (AE, AH etc) should also be manipulated accordingly, 3. Itis applicable only to isothermal process i.¢., initial and final temperature should be the same, in between whatever path 1t takes, no matter.

Heat Capacity, Enthalpy and Heat of Reaction

4.

23

This law is of great help in thermodynamic calculations of the reactions that are difficult for expenmental study. T

State 1

0

wis (4.3)

irreversible process in an isolated system will occur with an increase in entropy of thermodynamics). we find that entropy becomes an index of a system at equilibrium versus a spontaneous (i.c., irreversible) change.

4.2.3 Entropy of a Substance Let us consider gradual heating of a substance from temperature T, to T, at constant pressure in a reversible manner (i.e., when the temperature of the substance is T that of the heating source is slightly higher, T+ dT; where dT — 0). For a constant pressure by definition:

+)

~ (3.1)

Therefore, 8q,,, = CpdT

.. (4.4)

Since

RY)

ai] fh 8 E: = lp

dg= (%%] =|

5 [dT =CpdnT

Cp

.. (4.5)

30

Metallurgical Thermodynamics,

Kinetics & Numericals

1. When a substance is heated from a temperature 7, to T, in a reversible manner at constant pressure, without any change of state (e.g. solid = solid), the entropy change: T, c,

ds=S; —S; = | (5 Jer

(4.6)

From this equation (4.6) entropy of a substance at temperature T, can be calculated knowing entropy of the substance at temperature T, and C,, value over a temperature range T, to T,. Similarly, if the substance is heated from room temperature (298 K) to T K in a reversible manner at constant pressure, without any change of state:

dS=Sp~Sy5 T

T

= |

D208

SeT

298

ar

47 (4.7)

For a substance in its standard state:

dst=gt.. i

80

(SeT

dT

n=l

48

(4.8)

T

Equation (4.8) is applicable for pure solid or liquid. 2. Entropy change with change of state: * If change of state occurs during heating of a substance i.e., solid to liquid at a temperature T, from T, , then additional terms must be introduced for the entropy change accompanying it.” Thus,

I, §.

T:

where

9%

iol T)

-

T7

Se) Le (Serie)

IE

Lis

I.

lf

[Cro |dT —

.. (49)

LT

T,,, = Melting point of the substance in K

T, = Temperature at which substance is heated from 7), suchas 7,>T, Cp (5) = Heat capacity of a substance in solid state Cp yy = Heat capacity of a substance in liquid state L, = Latent heat of fusion (Le., heat require in actual melting of the Substance. + In the above case, if the heating further involves vaporization (i.e., liquid — gas), the entropy change will be:

|

j

(55m

T,

T

J | [5 Jor+( 5 J I (Fre Jr

(4.10)

Where substance is heated from initial temperature T, to final temperature T, (,>1,>7,>T,) T,, = Melting point of the substance in K T,, = Boiling point of the substance in K

Cp) = Heat capacity of a substance in solid state Cp) = Heat capacity of a substance in liquid state Cp (y= Heat capacity of a substance in gaseous state L, = Latent heat of fusion

L, = Latent heat of evaporation. Melting of a solid is accompanied by a moderate entropy increase while vaporization involves a relatively large entropy increase. On the other hand, solidification and condensation are accompanied

Second Law of Thermodynamics

3

by corresponding entropy decrease. e.g. AS = 10.5 J/K/mol

Zn (s)

> Zn (])

AS =98.3 J/K/mol

s Zn (g)

a1)

Thus, gaseous products form from solid or liquid reactants, the entropy change is more. Zn (s) + H,580, (aq) = Zn80, (aq) + H, (g), AS =+ve

(4.12)

[ 0 — 1, change of gas volume ] C(s) + O,(g) = CO, (g), AS=0 [1 — I, change of gas volume] Zn (s)+% 0, (g) =Zn0 (s), AS=~ve [¥2 — 0, change of gas volume ] 3. Entropy change for a chemical reaction: When a reversible chemical reaction taking place at (1) constant temperature:

rev.

AS =

(4.13) (4.14)

(4.15)

: (11) constant temperature and pressure: AS = =

AH 5 (5 )

(4.16)

For a chemical reaction , change of entropy :

AS = [sum of the entropies of products] — [sum of the entropies of reactants] =X Sbroduct “oh SReactant

- (4.17)

If the substances are in standard state, AS is known as standard change of entropy for reaction and is denoted by ASE or AS’, 4 Thus,

AS,

= Iz s’ T, Products

x

s! T, Reactant)

ha (4.1 8)

[Note: The values of AHO, and AS, are given in Tables in standard reference books assuming that the substances taking part in the reaction are in standard state.] The entropy change of a reaction is generally evaluated at constant temperature and pressure by eq. (4.18). The entropy change of a reaction at temperature T, can be found out similar to eq (4.8): T-

AS’

Tx

= As

T

+ i (55) ar T

Knowing the value of AS’; = AS’,

(4.19)

wd

from the standard reference books:

T:

AS’

Ta

= ASO

208

+

f

(Fer

a

(4.20)

298

Where AC, =(T Cp product — 2G, Resictant) 4.3 COMBINED EXPRESSION OF THE 1°! AND 2" LAWS OF THERMODYNAMICS For a reversible process and a closed system, if there is no other work done except that against the pressure, then according to :

1 law:

2" Jaw:

dE = 6g — PdV

s5=(%

or 8q = TdS

Now putting the value of 8q [from eq. (4.1)] in eq. (2.7), we get: dE =TdS — PdV

(27)

GET) (EZ TY

This [eq (4.21)] is the general form of the combined expression of the 15 and 2™ laws of thermodynamics.

Again by definition:

H=E+PV

.. (3.20)

32

Metallurgical Thermodynamics,

By differentiating:

Kinetics & Numericals

dH = dE + PdV +VdP

(422)

Now combining eqs (4.21 and 4.22): dH =(TdS — PdV) + PdV +VdP = TdS + VdP

.. (4.23)

This expression [eq (4.23)] is an alternative combined expression of the 1% and 2" laws of thermodynamics. 4.4 THERMODYNAMIC

(I) Since

EQUATION OF STATE

E=f(I.V)

Theref erefore,

dE =

a 37

dT +

Ld 7

4.24 (424)

dv

Again from combined expression of the 1% and 2" laws of thermodynamics: dE=TdS

—PdV

(4.21)

Now equating eqs (4.21 and 4.24):

ras—pav- SE) oT ar +E) 5

av

oF SE TdS=|— | dT +|— | dV + PdV aT 14

or

Therefore, dS = E | 7)

7

JE 7) + par

(425)

This eq (4.25) is of the form : d= = Mdx +Ndy where dz is exact and M and N are function of x and y.

.. (4.26)

If z is a function only of x and y, then d= is exact inx and y, Then ==f(x) Therefore,

dz= &

J dx + E

J dy

(427)

By comparison eq (4.27) with eq.( 4.26) we get:

M=

(52)[Z| and= N= v-(5) ly

By differentiating: 5] (2)

i

d

Theref erefore,

ov) MfBYE)| (=) se \& )], eit 3

on

(428)

dy

)|

2

= =|

(8a "ax

,

430 E20 (4.31) 431

HE

-

N= (+]2), vr]

(433)

By comparison eq (4.25) with eq. (4.26) we get:

and

(429)

Second Law of Thermodynamics

33

Rat)

.

NE)

.

By differentiating with —

to V at constant T to eq. (4.32) :

By differentiating with rising to T at constant V to eq. (4.33) :

By comparison eq (4.34 and 4.35) with eq. (4.31) we get: =

aN (5)

(4.36)

=

=

2

2

2

recor (5): fl Ltd or) EL (5) Hw) [ro Hw ar) [e)57)7] “r (ELE) EEL) (G7) thio 2-7 (22) -(E] |

4638

This eq (4.38) is known as thermodynamic equation of state, it correlate E, P, T and V.

(II) Since

H=f(P,T)

Therefore,

dH=|—

(439)

6H | dP+| &P

8H | dT 87 Jp

—

(4.40)

Again from combined expression of the 1% and 2" laws of thermodynamics: dH =TdS+ VdP Now combining eqs (4.23 and 4.40):

oH

TdS+VdP=|—

dsS=

Againif Then

8H

| dP+|

LE

+=

dz=Mdx + a OM oN - F» J -(& )

oH

— 1

| dT 8H Kl — | —-V

.. (4.23)

(441

dP

.. (4.42

.. (4.26) (431)

34

Metallurgical Thermodynamics,

Kinetics & Numericals

Now comparing eq (4.42 and 4.26), then similar type eq (4.31) we get:

«el «oe

ce

V=

.. (4.44)

o 1]

or

FEEUASEE) HEE) GE) HEE EHEE TEHE

Therefore,

|T ad oT

Jp

+

oe ar

This eq (4.44) is also known as thermodynamic equation of state. It correlate H, P, Tand V. Examples

Ex 4.1 If the value of Cp=a+bT+ eT, find the expression for 8; — 8g in terms of T. Solution. Entropy change for a substance heated from 7 to T, at constant P in a reversible manner is given by: T, 2

Sr, -8p, = (G/N dT T

Let 7,=298K and 7, = TK T

T

Therefore,

[[(a+bT+cT>)/T)dT

S;— Sy = [(Cp/ T)dT=

298 T

298

=[[(a/Ty+ b+cT3]dT

298 =[ainT+bT-(cT2)]

T

298 Therefore, S;— Sys = [{a In(T/298)} + b (T'— 298) — (c/2){ T™2~29872)] Ex 4.2 Copper melts at 1083°C, and its heat of fusion is 12.97 kJ/mol. Calculate the change of

entropy at melting point of copper. Solution.

Since AS’ =(AH® /T_)=[(12.97 x 10>}/(1083 + 273)] = 9.565 J/K/mol Ex 4.3 Calculate the standard entropy change for the following reaction at 25%: Cr,0, (5) + 3C (5) = 2Cr (5) + 3CO (g)

Given: Sa, Cri03 (= 81.17 VK/mol Tam. ce = 5-69 JK/mol 8508 cr y= 23-76 J/K/mol

oy 298,C0(g) = 197.90 J/K/mol

Second Law of Thermodynamics

35

Solution.

Since

AS” 208 = PRY 298, Product -35° 208, Reactant [(2s° 298, cris) T25 208, coy)~ (5 208, cryo5 9) + 38°

=[(2%23.76 +3 x 197.9) — (81.17 +3 x 5.69)]

208,C( 8)

=(641.22 — 98.24) = 542 98 J/K/mol

Ex 4.4 Calculate the standard entropy change of solid copper (in J/K/mol) at 1073°C from the following data:

5500, cue = 8.0 calldeg/mole, Cp, ) = 541+ 1.5 x 107 T cal/deg/mole Solution. I,

80, =8" + [(Cp/ Dy dT

sil)

T1

Here 7, =300 K, T,=1073 +273 = 1346 K Since 1 cal =4.184 J, so a Cu = 8-0 cal/deg/mole = (8 x 4.184) = 33.47 J/K/mol

Cp

cur) = 541+ 15x10 T cal/deg/mole= (5.41 + 1.5 x 10° T) x 4.184 = 2264+628x 107 T J/K/mol

From eq (1) we get:

1346 §0p,=3347+[(22.64+ 6.28 x 10° T) (dT/T)

300 1346

1346

=3347 + 22.64 [(dT/T) + (6.28 x 107) [ dT

300

300 1346

=3347+ 2264 [InT] 300

1346 + (6.28 x 107) [T] 300

=74.0 J/K/mol

Ex 4.5 Calculate the entropy of liquid iron at its melting point, 1808 K, given that for iron: L,= 15.4 KJ/mol, Sn: Fe(s)™ 27.2 J/K/mol and C, = 25.2 J/K/mol. Solution. I2

8, =n + C/T) + [(C/

Here,

dT

—(D)

I,

T,=298 K and T, = 1808 KX

1808

Therefore, 5% gp = 27.2 +[( 154 x 10%)/ 1808] + [(25.2/T)dT

298 =272+852+ 252[InT)

1808 298

=81.15 J/K/mol

Ex 4.6. Zinc melts at 420°C and its standard entropy at 25°C is 41.63 J/K/mol. Calculate the standard entropy of Zinc at 750°C.

36

Metallurgical Thermodynamics,

Given:

L., =7.28kl/mol,

Kinetics & Numericals

C, Zn" 22.38 + 10.04 x 10° T J/K/mol

Cp zn = 31.38 J/K/mol Solution.

Fret

T,

Guu

I,

TIE AEA)

T

Here T,=298 K, T,=750+273=1023K,

NC

DAT

Ti

T,=420+273=693K

693

1023

5% 02 =41.63+ [(22.38+10.04 x 10° T/T) dT +(7280/693)+ [(31.38/T)dT 298

693

=41.63 +2238 [InT]

693

+10.04x 103 [T]

298

693

+(10.51)+ 31.38 [InT]

298

1023 693

= 87.22 J/K/mol Prob 4.1 Copper boils at 2575°C, and the heat of vaporization is 304.4 kJ/mol. Calculate the change of entropy of vaporization.

[Ans: 106.88 J/K/mol]

Prob 4.2 Iron melts at 1536°C at 1 atm pressure, and its heat of fusion is 13.81 kJ/mol. This is a reversible process at constant temperature and pressure. Calculate the change of entropy

at melting point of iron. [Ans: 7.634 J/K/mol] Prob 4.3 Calculate the entropy of Cu at 1800 K, given that for Cu: T = 1356 K,

Cp=24.5 J/K/mol (298 to 1800 K), Sg cys) = 33.3 J/K/mol.

[Ans: 86.95 J/K/mol]

Prob 4.4 The temperature of liquid gold is 1127°C , it is cooled to room temperature (27°C). Find out the change of entropy of gold from 1127°C to room temperature. Given : Melting point of Au= 1063°C, L;=12.76

Cp aus = 23-68 +5.19 x 107 T Jideg/mol G P, Au (l) =29.29

J/deg/mol

Prob 4.5 Prove that C,.— C= C, — Cy -{

kJ/mol [Ans: 51.59 J/deg/mol]

=

CHAPTER

Free Energy and Third Law of Thermodynamics

Most of the reactions of chemical and metallurgical interest are performed at constant temperature and pressure. Some reactions operating at constant temperature and volume are also studied. A system under these conditions is called a closed system, in which there is no mass transfer but only

the transfer of energy. Now according to 1% law of thermodynamic: and according to 2*

dE = 8q— 8W

law of thermodynamic:

dS = =

or

..(2.6) (4D

8g = TdS

Combining above equations: dE =TdS—8W Rr Now since efficiency of the reversible process is maximum, i.¢., in other words, reversible process yields maximum work, 8W in eq (2.6) correspond to maximum work. Thus, maximum work (§W) = — (dE — TdS)

(5.2)

This maximum work = mechanical work + non-mechanical work. Mechanical work = PdV and non-mechanical work = chemical work in chemical reaction or electrical work 1n electrochemical cell.

If there is no other work than mechanical work for reversible process, then dW = PdV and eq (5.1) become:

dE = TdS— PdV

(5.3)

In heat engines: mechanical work done is of main interest; and in chemical thermodynamics: non-mechanical work done is of main interest. For knowing the extent of work the system can per-

form, we have to find out the free energy. There are two kinds of free energies: 1. Helmholtz Free Energy

2. Gibbs free energy 5.1 HELMHOLTZ FREE ENERGY Anew function 4 known as Helmholt= free energy or isothermal work content is defined by the relation:

A=E_TS

(5.4)

TS basically represent bound energy, which can not be utilized for work. It gets dissipated as heat. By differentiating eq (5.4) at constant temperature (i.e., for isothermal process):

dA=dE — TdS Now comparing eqs (5.2 and 5.5), we get:

8W=—dA 37

(5.5) ..(5.6)

38

Metallurgical Thermodynamics,

Kinetics & Numericals

Here SW represents reversible maximum work done by the system (i.e., mechanical work and non-mechanical work together); and d4 represents decrease in isothermal work content. Thus the function 4 acts like a store of work or energy available for doing work for the system. Hence, when work dW 1s done, 4 decreases by dA. 5.2 GIBBS FREE ENERGY The above term dWrefers to total work i.e. mechanical (PdV) and non-mechanical works com-

bined. But in chemical thermodynamics we are mainly interested only in non-mechanical work; 1.e., Reversible non-mechanical work = 6 WW — PdV

= (dE — TdS) — Pav =— (dE + PdV — TdS) =— (dH - TdS) (Since

[from eq 5.2] (37

H=E+ PV and dH = dE + Pd} at constant pressure).

New function G is known as Gibbs free energy or simply free energy, which is defined by the relation: G=H-T8§ By differentiating eq (5.8) at constant temperature and pressure: dG =dH-TdS

..(5.8) SEE

Now comparing eqs (5.7 and 5.9): Reversible non-mechanical work = 8W — PdV = - dG (5.10) = decrease in {Tee energy Hence, the function G acts as a store of non-mechanical work or energy available to the system for doing non-mechanical work; i.c., when the system does non-mechanical work (§W — PdV), the

free energy of the system decreases by dG. AG is a measure of the work obtainable from a reversible, isothermal process occurring at constant

pressure and gives a direct indication of the possibility of chemical reaction. Since H, T and § are the state variables, and G = H— TS; hence G is also a state variable.

5.3 FREE ENERGY OF A SUBSTANCE Like H and §, free energy (G) values of all the substance can be calculated from heat capacity data (ie, C, values). If we consider heating of a substance from T| to T, in a reversible manner without change of state, then: T

dG = dH-TdS=

T

Cc

[Cp arr (Sear T

(5.11)

I

5.4 FREE ENERGY OF A REACTION The free energy change for a reaction is the difference between the sum of the free energies of the products and that of the reactants at constant temperature. Thus for a reaction at temperature 7:

AG =ZG ,pquets = ZC egctants =[ZH,-TZXS ]-[ ZH, - TLS] = [ZH,-ZH,]-T[Z5,- 25]

(5.12) [since G=H-T15]

Therefore, AG=AH-TAS (5.13) If the substances are in standard states (i.e, pure solid, pure liquid and gas at atmosphere pressure), then the standard free energy change is as follow:

At temperature T: AG" = AH? — TAS®,

At temperature 298 K: AG®,g, = AH, — TASC,

5H)

(315)

Free Energy and Third Law of Thermodynamics

39

If the reaction involves heating from temperature T, to T, in a reversible manner without change of state, then: T

T

8 + | ac, dl ]-T [ASy, + J(5

AG, = AH — TAS,= [AH

Tn

=|AHg 1 —TASy

1

Th

(5.16)

CeT Var

|+ (ac, ar—T| h

Jer ]

5

5.5 FREE ENERGY AS CRITERIA OF EQUILIBRIUM The free energy is important because: + It is a measure of energy available for doing non-mechanical work (i.e., other than PdV),

+ It serves as a criterion for equilibrium. For a closed system undergoing, a irreversible change 1.¢., spontaneous, the free energy decreases

because entropy Increases, AG =AH -TAS

(5.13)

Therefore, AG =— ve, since AS = + ve and AH = 0 (unchanged). For a closed system undergoing, a reversible change i.c., equilibrium, free energy remains unchanged and entropy change is also zero.

AG=AH-TAS Therefore, AG = 0, since AS = 0 and AH = 0 (unchanged). In a closed system, the irreversible changes i.e, spontaneous tend to carry the system to a state of minimum free energy (and maximum entropy) which is the state of equilibrium. That means, a system

of substances reacting chemically attains equilibrium state when free energy attains the minimum value. Thus, free energy becomes a criterion of chemical or phase equilibrium. The state of equilibrium can be found out by: » setting up an equation of free energy, » differentiating that equation with respect to a variable, + putting the differential dG equal to zero. Thus finally we can conclude that: if AG = 0, the system is in equilibrium, if AG = — ve, the reaction tends to proceed spontaneously in forward direction, if AG = + ve, the reaction tends to proceed spontaneously in the opposite direction. 5.6 PARTIAL DERIVATION OF THE FREE ENERGY Helmholtz free energy: A=E-TS Differentiating this equation: dA =dE—TdS—-8dT

Again

sera) ASAT)

dE = TdS — PdV

Putting value of dF to eq (5.17): dA = TdS— PdV— TdS — SdT =— PdV — SdT Again Gibbs free energy: G=H-TS =E+PV-T§

(53)

(5.18) ..(5.8) [since H= E+ PV] ..(5.19)

Now differentiating this equation: dG = dE + PdV + VdP — TdS— S5dT Since dE = TdS— PdV Therefore, dG = (TdS — PdV) + PdV + VdP — TdS — 5dT = VdP - dT

(5.20) (3.21)

40

Metallurgical Thermodynamics,

Kinetics & Numericals

Egs (5.18 and 5.21) are also forms of the combined statements of 1% and 2™ laws of thermodynamics. (1) At constant T (and varying P), eq 5.21 becomes: dG = VdP =/A(322)

and

(E51 =V oP

(5.23) Te

(2) At constant P (and varying 7), eq 5.21 becomes: dG =— SdT

28)

and

£3.25)

=) =-5 &T Jo

It 1s generally convenient to eliminate S from this eq (5.25) as follows: By definition, G= H-TS

Therefore,

—S= (5

(5.26)

Putting this value of § in above eq (5.25):

8G) 3 Jp

(G-H \ T

G-H Threrefore dG = (Fr)

or

TdG-GdT=-HdT

«+ (32T)

Now dividing both sides by T? in above eq (5.27):

LATA T

ii

d Li) T

or

(7

Hd

x) :

i

J T

=H

(528)

T )lp 5.7 MAXWELL’'S RELATIONS We know, if z=f(x,y)

d== Mdx +Ndy

Then

5]

oy

),

-(&)

Lor),

(4.26) (4301)

Apply eq (4.26) to the following four forms of the combined equations of 1 and 2™ laws of thermodynamics:

dE=TdS- PdV dH =TdS+ VdP dA =—-PdV - SdT dG = VdP— SdT

(5.3) (4.23) NCAT) (3.20)

P, V and Tare state variables by their very nature. E is a function of state according to 1™ law and S is a function of state according to 2 law. H, A and G are function of state, since they are defined in terms of these variables which themselves are function of state.

Free Energy and Third Law of Thermodynamics

a1

Apply relation of eq (4.31) to the above equations we get:

or) ~{s) wl (x) dT)

_

(oP

8Ty

(&FV

(5.29) (5.30)

i) (5) &P\

(8S

(5.31)

or) =) Vy

_

(aS

(5.32)

These above equations (5.29 to 5.32) are commonly known as Maxwells relations. These equations are used to calculate thermodynamics parameters.

5.8 THIRD LAW OF THERMODYNAMICS The 3" Law of Thermodynamics may be stated as the entropy of any homogeneous substance, which is in complete internal equilibrium, may be taken as zero at the absolute zero. The 3™ Law of Thermodynamics is the finding of the Nernst who has given Nernst heat theorem as follows: (1) Original statement of Nernst heat theorem: By definition, for a reaction, we have:

(5.9) (333) (525

dG =dH -TdS

=dH-T@EAG/3T),

[Since

8G I=) =-5 oT Jp

_

84G) __ oT Jn 8 AG

Now if 5%

|

.(5253)]

is a finite quantity, then at absolute zero (i.e., T= 0K): Pp

(356) _, 87 ), (534)

Hence, from eq. (5.33) we get: AG =AH

EE

Now differentiating both sides with respect to T at constant P:

.. (5.35)

oT

This means that AG and AH are not zero at absolute zero, but their curves of AG vs T and AH vs T meet and both have the same slope at absolute zero. 8 AG hd8T ki approach h zero Hence Original statement of Nernst heat theorem was an

)

at absolute zero, i.e, (i) Since or

5

lim r=o|

AG

nl) &T

"

3 AH

r—o|

&T

=lim|——| lp

=0

(5252)

J,

and at absolute zero

(536)

3

lim

ool

346

BT

Jp

=0,

therefore -AS=10

(337)

42

Metallurgical Thermodynamics,

; & AH (if) Since a

Kinetics & Numericals

=AC,

(3.50)

/id

and at absolute zero

lim| 70

SAH —— | =0 8T )p

therefore AC, =0

(538)

(II) Latest statement of Nernst heat theorem: For any reaction it was observed that AS approaches zero, near the absolute zero Nemst had

generalized these observations in the form of the latest Nernst heat theorem which states that for all reactions involving substances in condensed state, AS is zero at the absolute zero.

This is also known as another form of the 3 law. Thus, if the chemical reaction is: A+B=A4B where A and B are elements, AB 1s compound.

(539)

Then, AS=5,,=S§,— 5, = 0 at absolute zero, according to Nernst heat theorem.

That is, if entropies of the elements §, and §,, are assigned zero values at absolute zero, then the entropy of the compound §, , is also zero.

5.8.1 Entropy at Absolute Zero and Lack of Internal Equilibrium According to 3" Law of Thermodynamics, for homogeneous substance which is in complete internal equilibrium (that means most ordered and stable structure), §; = 0. But in actual practice it is found that, 5, # 0 because there is a lack of internal equilibrium in the substance at absolute zero. Because as solid is cooled from high temperature to absolute zero, the more random atomic arrangement characteristics of high temperature are frozen and the unique most order atomic arrangement

of the lower temperature is not actually obtained. But for the purpose of chemical reactions, this phenomena is neglected and we always take Sj, = 0. Generally there is no lack of internal equilibrium in case of solutions, 5.8.2 Experimental Verification of the 3™ Law ee

aaa

| |

1] —_—

|] I

i |

IL]

I

!

i Temp

| I I

v 0

, |

Reacted ——

Fig 5.1: The entropy change of a reaction at the absolute zero.

Figure 5.1 shows a method of through which the system moves composed of only reactants at the 1¥ step of the cycles: Consist

evaluating entropy change of a reaction at absolute zero. The cycle is shown in Figure 5.1. The stating point of the cycle is a system absolute zero. in heating only the reactants to some temperature T. From previ-

ous experience it is known that T is the temperature at which the reactants are in equilibrium with the products. 2nd step of the cycles: The reaction is allowed to proceed completion at temperature 7.

Free Energy and Third Law of Thermodynamics

43

39 step of the cycles: The products are cooled to the absolute zero. 4" step of the cycles: It is the imaginary step, in which the reverse reaction is allowed to proceed. The total entropy change over the cycle is zero, Le, AS; + AS + AS; + AS =0 (5.40) or AS, =— (AS, + AS, + AS) (541) Since the entropy change during the first three steps can be measured experimentally, the entropy change at the absolute zero can be determined by this eq (5.41).

Examples Ex 5.1 Calculate change of free energy for the following reduction reaction at 500 K: CuO (s) + H, (g) = Cu (s) + H,0 (g)

Given: AH’, =— 87 kl/mol, AS? = 47 J/K/mol Solution.

Since AG’, = AH, ~ TAS’, =— 87000 — (500 x 47) =— 87000 — 23500 =— 110500 J/mol =—110.5 kJ/mol Ex 5.2 Given the following data, determine which metal has the greater probability of oxidation in presence of steam at 827°C and 1 atm pressure:

(i) NiO (s) +H, (g) = Ni(s) + H,0 (g), AG =(-2301-4259T)] (i) 1/3Cr,04 (s) + H, (g) = 2/3 Cr (s) + H,0 (g), AG’ = (126,566 —30.67T) J Solution. Temperature = 827 +273 = 1100 K Taking reverse of eq (i) : Ni (s) + H,0 (g) = NiO (s) + H, (g),

AG" = (2301 +4259 T)=2301 +4259 x 1100 = 49.15 kJ Taking reverse of eq ii): 2Cr (5) + 3H,0 (g) = Cr,04 (5) + 3H, (g)

3AGY = 3 (—126,566 + 30.67 T) = 3 (~126,566 + 30.67 x 1100) =— 278.48 k] Since free energy change of Ni oxidation is (+) ve value, so Ni will not be oxidized at all. Whereas

free energy change of Cr oxidation is (-) ve value, so Cr will be oxidized in presence of steam at

827°C and 1 atm pressure. Hence, Cr has the greater probability of oxidation in presence of steam at 827°C and 1 atm pressure than Ni.

Ex 5.3. Calculate the standard free energy change of the reaction:

Ni (s) + %0, (2) = NiO (s), At 327°C from the following data;

AH

vio

=— 240.58 Ki/mol,

8% \; (,=29.79 J/K/mol,

S°08, nio5) = 38-07 J/K/mol,

S108, 0 gy = 205.09 J/K/mol

Cp nig) = 25:23 + 43.68 x 107° T° ~ 10.46 x 10 T J/K/mol Ch ope = 29-96 + 4.18 x 107° T — 1.67 x 10° T™ J/K/mol Cp nio 9 = 54.01 J/K/mol Solution.

Since .

Since

AG’, = AH, — TAS, AH,

0

_

550 =ZAH ar

=[AH

0

0

T=327+273=600K 0

500 produce == AH 20g pencrant 10

298, NiO (5) — (H

70

298%, Ni (s) +%H

298, 02(g) i

= [240.58 — (0 + 0)] =— 240.58 kJ/mol

a4

Metallurgical Thermodynamics,

—

Again since

_

AS, ST

0

Kinetics & Numericals

298, Product ~ zs 0 298, Reactant

0

0

=[S 208, Ni0) (5) — is 208, Ni(s) T “(8

0

298, 0 (2) JH

= [38.07 — { 29.79 + % (205.09)}] = — 94.265 J/K/mol

AC,

=

( Cy, Product -2Cy Reactant)

= [Coniow = Comin * (Cy oy] =[54.01—{(2523+43.68 x 10° T>— 10.46 x 107 T) +14(29.96 + 4.18 x 10° T— 1.67 x 10° T2)}] =[138-4368 x 10° T2 +837 x 10° T+ 0.835 x 10° T*] AH, 0 go me= AH, li} 0 +

| A Cp, @T

= — 240580+ |

[13.8-43.68 x 10° T2 +837 x 107 T+ 0.835 x 10° T2] dT

208

=-2378929] FT Similarly AS®, 0 =

600

—

AS", 0 jog + [

AS, /T

298

600

=-94265+

|

[13.8 43.68 x 10~° T2 + 8.37 x 107 T+ 0.835 x 10° T2] (dT/T)

298 =— 87.645 J/K/mol

Since AG’; = AH". — TAS, =-237892.9 — 600 x (— 87.645) =—-185305.9 J =— 185.31 kJ/mol

Ex 5.4 Find out the feasibility for the following reductions at 500 K by using the given data:

(i) CuO (s) + H, (g) = Cu (s) + H,0 (g), AH, = — 87 ki/mol, AS’, = 47 J/K/mol (i) ZnO (s) + H, (g) = Zn (s) + H,0 (g), AH, = 104 kI/mol, AS’, = 60 J/K/mol Solution.

Since

0) @)

AG", = AH’, — TAS",

AG, =- 87000 - 500 x 47 =~ 110500 J/mol AG, = 104000 — 500 x 60 = 74000 Jmol

Since AG @ 1S —ve value for the 1% case, hence the reduction of CuO by H, gas is feasible at 500 K; but in case of 2™ AG", ayy 1s tve value, hence the reduction of ZnO by H, gas is not feasible at 500 K. Ex 5.5 Using the following values of free energy changes, calculate AG, for the following reactions:

() (i)

Given:

Fe,0,(s)+2Al(s) = ALO; (s) + 2Fe (5) Fe,0,(s)+3Ca(s)=3Ca0 (s) + 2Fe(s)

A

Solution.

.

Since

(kJ/mol) for different compounds:

Fe,0, (s) =— 741, ALO, (s) =— 1576, CaO = — 604 ns

AG, 208 = TAG

0

gq. Product ~ x AG 506. Reactant

()

AG’

(if)

AG’, 105 =3AG5 000 — AG 05 pe 0, = 3(~604) — (— 741) =— 1071 kJ/mol

10a =AG"0 410, ~ AG ogg £0, = —1576 — (— 741) = 835 kl/mol

Free Energy and Third Law of Thermodynamics

45

Ex x 5.6. 5.6. Using Using M Maxwell's II's relation relati || yar =5 | =| 5¢

prove

|

iE] {EY tha

oF |. =| C,

Nor J,

Solution.

Multiplying and deviding by &T to the right hand side of the above given relation:

VY (SET) (BK) [AT 8s J, \8ras ), \ar J. as ),

| =i)

dq

Now at constant pressure, P, we know that

C, = (55) /d

Therefore

8g = Cp dT

AgaindS=

Therefore

5

rl

c r

2

= pe

[22 |=| =£

Cf

|ar

| a8

2 3 : 14 Putting this value in above eq (1) we get: | == | =| as Jp There erefore,

TY | =

\ =

(TT c,

oF Nat |,

p

rove

T Yor =— | == Pp

d

Prob 5.1 Calculate change of free energy for the following reaction at 727°C and 1 atm pressure:

MoO; (s) + 3H, (g) = Mo (s) + 3H,0 (g) Given:

(i) Mo (s) + 3/2 0, (g) = MoO, (5),

(ii) Hy (2) +%0,()=H,0 (8),

AG

0, = —502.08 kJ/mol

AG’) »=—190.372 kl/mol

Also comment on the reduction of MoO, (s) by H, (g) at 727°C and 1 atm pressure. [Ans: —69.036 kI/mol of Mo, Reduction is feasible.] Prob 5.2 During reduction of Cr,0, by C, the carbide is formed. a) Find out whether the for-

mation of carbide during the reduction is easier than reduction of oxide. b) Find out the minimum temperature above which reactions are possible in both cases under standard conditions.

Given: (1) 2/3 Cr,0, + 18/7 C = 4/21 Cr.C, +2 CO, AG®, = 490574—35459T J (if) 2/3 Cr,0, +2 C=4/3 Cr +2 CO, AG’, = 523418434853 T J

[Ans: Carbide formation is much easier than reduction of oxide to form Cr. Min. temp of carbide and Cr formations are 1111 and 1229°C.] Prob 5.3 Is the reaction Fe,0,(s)+3 H, (g) = 2 Fe (s) + 3 H,0 (1), feasible at 25%C and 1 atm pressure 7

Given: AH yo 1 y =—241.789 kJ/mol, AH’, ©

=-822.156 kJ/mol

46

Metallurgical Thermodynamics,

0 295

HOM) ==

188.7 J/K/mol, 8°5

108, 1 (9) = 130-58 J/K/mol, S

298, Fe (3) 298, FeaO3

Kinetics & Numericals

=27.15 J/K/mol = 89.95 J/K/mol (5) [Ans: Reaction is not feasible ]

Prob 5.4 Calculate AG? for the following reactions at the stated temperature:

ZnO (s)+C(s)=Zn(g)+CO(g)

atl1373K

AH’ ;,, = 349.9 kJ/mol and AS”,,, = 285 J/K/mol (ii)

Fe(s)+%0,(g)=

FeO(s)

at298K

AH", =— 265.4 kJ/mol and AS®,,,, =— 71 J/K/mol [Ans: — 41.41 kJ/mol and —244.24 kJ/mol] Prob 5.5 For the following reactions at 500 K:

(i)

(i)

PbS(s)+0,(g)= Pb(5)+SO, (g)

PbS (s)+2PbO(s)=3Pb(s)+ SO, (g)

Calculate: (a) AH, given data:

AH, (K/mol) AS, (/K/mol)

Lb) AS’, , and (¢) AG,

PbS (s) —94 91 [ Ans: (a) —202,

0, (g) 0 205 236 kJ/mol;

of the above reactions from the following

Pb (s) 0 65

SO, (g) 296 249

(b) 217, 18 J/K/mol;

(¢)-211,

PbO (s) 219 68 127.5 kJ/mol]

Fugacity, Activity and weer

Equilibrium Constant

6.1 FUGACITY Fugacity indicates the escaping tendency of the component or a substance. High gas pressure

indicates the tendency of the gas molecules to escape outside the container. Similarly, high fugacity would indicate a greater tendency of a component or a substance to escape, dissolve, intermix or react.

The combined statement of 15 and 2" law of thermodynamics, in term of Gibbs free energy, is given by: dG = VdP — SdT

(5.19)

Now at constant temperature, d7=0,

Hence above equation becomes: dG = VdP

5:1)

Since one mol of an ideal gas obeys equation of state as: PV'= RT, Therefore,

V=

(5

Hence, eq (6.1) becomes: dG =

P

RT F)

dP=RTd (InP)

6.2)

This eq (6.2) is applicable for an ideal gas. If the gas is not ideal, then pressure, P is replaced by f. Hence, where (i) (if)

eq (6.2) can be written as: dG =RTd (Inf) .. (6.3) the function fis known as fugacity of the gas. For an ideal gas: f= P, at all pressures, hence eq 6.2 is applicable. For non-ideal or real gases: non-ideal gases approach ideal behavior as the pressure is decreased ie.,P => 0, f— P. Thus, for non-ideal gas, f= P, only at low pressures.

6.2 ACTIVITY It has been observed that the thermodynamic variables (i.e, H, S, G etc) of a substance in pure state are different from those in a state of solution, i.e., when pure element goes into solution state;

the state of aggregation or dispersion changes and A, §, G also undergoes a change. This means H, Sand G changes with change in composition apart from their change with pressure and temperature. The concept of activity offers an index to measure these changes in /, S and G, when pure substance goes into a state of solution.

-

Activity is defined as the ratio of fugacity of the substance in its actual state (i.e., arbitrary state) to its fugacity in its standard state. Thus,

a7

48

Metallurgical Thermodynamics,

Kinetics & Numericals

where a = activity, f= fugacity of the substance in the actual state, f° = fugacity of the substance in the standard state. Therefore, f=fCa Now eq (6.3) becomes; dG = RT d (Inf)=dG=RTd In (fa) =RT(dln f°+dlna) Since f° is constant at constant temperature, hence d In f° =0 Therefore, dG=RTdhna Integrating this eq (6.7) from standard state to actual state (i.e., any arbitrary state):

(6.5) ...(6.6) (6.7)

[dG =RT[d(na) a°

a®

Therefore, G-G°=RTIna—RTIna® Since activity at the standard state is unity (i.e, a" = 1),

..(6.8)

So AG=G-G"=RTInha (6.9) Thus the concept of activity denotes the change in free energy ( AG ) when a substance goes from standard state (i.e., pure state) to a state of a solution (i.e. actual state). Generalizing it for species i, we have:

G, — G =RT

Therefore,

Ing

where

G; =G;"+RT

Ina,

(6.10) (6.11)

G; = partial free energy of 1 th species (in actual state of 7) G° = free energy of i th species (in standard state of 7)

6.3 STANDARD STATE The standard state is the stablest state of the pure substance at the same temperature as that of actual or arbitrary state and at one atmospheric pressure. 1. For solid and liquid: The standard state chosen is pure solid or pure liquid substance in its most stable form at the specified temperature and one atmospheric pressure. e.g. (a) Foran Fe— C solid solution at 1700 K, the standard state of iron is pure § — Fe where as for same solid solution at 1400 K, the standard state of iron is y-Fe. (b) For Cu— Ni solution at 1600 K the standard state of Cu and Ni are pure liquid Cu and pure liquid Ni. The activity of a solid or liquid in its standard state is equal to unity i.e.,

for pure solid or liquid, a= 1. 2. For ideal and non-ideal (i.e. real) gas: The standard state chosen for the ideal gas is at the

specified temperature and one atmospheric pressure; and that for the non-ideal gas at specified temperature and unit fugacity (f= 1). i.e., (a)

The activity of an ideal gas which is equal to its partial pressure is unity,

a=FP=1 (b)

The activity of non-ideal gas which is equal to its fugacity is unity,

a=f=1 6.4 EQUILIBRIUM CONSTANT Let us consider the general chemical reaction at constant temperature and pressure:

Ai Be Ae =dB+eRB+0,

.. (6.12)

where, the capital letters are general symbols for chemical elements or species, and small letters are the numbers of gram atoms or gram moles of each elements.

Fugacity, Activity and Equilibrium Constant

49

The overall free energy change for this reaction can be written as:

AG =EG nis

~ EG nis

= [(dGp +eGg +...) — (bGy +¢Gp +.....)]

We know, G; =G®+RTIna,

(6.13)

(6.14)

Now putting these values of G, [eq (6.14)] in the eq (6.13): AG=[{d(GS+RTIna,) +e(GZ + RTInag)+ ... } = {b(G,° + RT In ay) + (G+ RTIna.)+ =[{dG," +eG +...) —(bG," + ¢G "+ ...)} + RT {(In a; +lnag +...)

—(Ina+mas+..)}]

....}]

(6.15)

Now if the reactants and products are in their standard state, then standard free energy changes at temperature TK: AG" =[(dG," + eG" +...) = (bG," + cG" + ...)] (6.16)

Subtracting eq (6.16) in eq (6.15), we get: d

e

AG=AG®+RTn == ag

Ge

ives

=AG°+RTInQ where

LL

.

(6.17) (6.18)

apap

Q = activity quotient = | ————— ag .ag’...

The equation (6.18) was derived by assuming an isothermal condition, i.e., the same temperature for reactants and products. It is further assumed that the reaction is isobaric i.e., the initial and

final pressures are the same, and also that thermodynamic equilibrium prevails; i.e., the reactants and products are in equilibrium with each other, the actual free energy change is zero at equilibrium. Since (AG) rp =0 Hence, eq (6.18) becomes: AG° =—RT In 2l, =—RTInk (6.19)

where k = equilibrium constant = value of activity quotient at equilibrium BE

HPL

|

ap d .ag"

e

(6.20)

=n / dg

dpe...

q

k can be calculated from activities of reactants and products at equilibrium.

The equation (6.18) is the basis of predictions of feasibility of reaction. AG is the driving force for chemical reaction. Reaction tends to go in the direction which results in a decrease in free energy which means that the feasibility of reaction may now be defined as follows: « If AG is negative, a reaction is spontaneous or feasible. » When AG is positive, the reaction is not feasible in forward direction; but the reaction is feasible in backward direction.

If AG is zero, an equilibrium mixture is obtained; 1.e., equal feasibility for both the forward and reverse reactions. The equation (6.19) is used to calculate the equilibrium condition or end-point of a reaction.

Thermodynamic predictions and calculations can be made if the following conditions are satisfied: » The process should take place isothermally (i.e., initial and final temperature should be the same) and the temperature should be known,

» The standard free energy change, AG®, should be available, = Activity versus composition relations for all species involved should be known.

50

Metallurgical Thermodynamics,

Kinetics & Numericals

6.4.1 Importance of Equilibrium Constant Equilibrium Constant (k) is the ratio of product of activities of chemical products to that of the reactants. The importance of equilibrium constant: 1. Ifk is large, i.e., greater than 1, equilibrium lies well to the right since products predominate in the equilibrium mixture, 2 If kis small, 1e, less than 1, equilibrium lies to the left since reactants predominate. Hence, a reaction is feasible:

if AG £0; ork = 1.

(6.21)

6.5 LECHATELIER PRINCIPLE It is applicable to equilibrium system and it states that if a system in equilibrium is subjected to external constraint (i.¢., increase of T, P or V), the equilibrium of the system (i.e. direction of the

reaction) is shifted in such a way so as to counteract the external constraint. e.g. a reaction; C+CO,=2C0, AH" =+ve In this case, volume of the product is more than that of the reactants and it is also endothermic reaction. » Increase in temperature will favour the forward reaction due to its endothermic nature. » Increase in pressure will reduce the rate of forward reaction and will increase that of reverse reaction. Since PV =RT, if P is increased V should be decreased at constant T; hence reaction 1s favoured in the direction where there is a decrease in volume. 6.6 GAS — SOLID REACTION Consider the reduction of solid metal oxide (MO) by CO gas by reversible reaction:

()H)MO (s) +CO(g)

«

dy. a

Equilibrium constant, k, = | ——— 2

M(s)+CO,(g)

|=|

Avo- Yco

Since

...AG,

ay-P,

—~5%

(6.22)

(623)

Aymo-Pco

Arp, = Peo, ad arp =Pep ;

Again M and MO are pure solid in standard state, hence a,, and a, are equal to one.

Therefore,

Pp k= | ==

(629)

Pco Reaction (6.22) can be obtained from the combination of the two following standard reversible reactions:

(i) 2 MO (s) « 2 M (5) + O, (2) (ii) 2 CO, (g) + 2CO (2) +O, (8)

. ...

AG, AG,

.. (6.25) (6.26)

Equilibrium constants for the above reactions are k, and k, respectively. Peo

2

-P Hence, k, = p,, and k; = Ea

(627)

Peo,

Since AG, for the reaction (6.22) is not available in standard reference book, but AG, and AG,

for the reactions (6.25 and 6.26) are available in standard reference book. Hence AG, can be calculated from AG, and AG. By rearranging reactions (6.25 and 6.26), and adding to get reaction (6.22): 2MO (5) = 2M(s)

+02 (g)

2CO (g)+0,(g) ~2C0,(g)

cu

AGES

... —AG,

or

2MO(s)+2CO(g) «2M (s)+2CO,(g) MO (s) + CO (g) «> M (s) + CO, (g)

.. (6.22)

Therefore,

AG, ='% (AG, — AG,)

.. (6.28)

Fugacity, Activity and Equilibrium Constant

51

If AG, is — ve, the forward reaction is possible i.e, reduction of metal oxide will oceur. We know that : AG=AG°+RT hk (6.18) Since the value of AG of a reaction depends on the initial and final (i.e., equilibrium) state, so we can write that:

AG=AG°+ RT (Ink'-1nk) Hence the above reactions (6.25 and 6.26), we can write that: AG, =AG®, + RT (Ink',- Ink,) = AG", + RT (In py, — Inp,,) and AG, = AG" +RT (Ink’y—Ink,) Peo

= AG®, + RT | In|

7 Hi

SR

-P

=222

|-nf

Peo

2

22

Pco,

-P

"22

(6.29) -. (6.30) [From eq (6.27)]

(631)

Peco,

where p'y, , P'cq. Poo, are the partial pressures of the gaseous components in initial state; and Poy

Peo» Peo, are the partial pressures of the gaseous components in the final state i.e., equilibrium state. From eq (6.28), by putting the values of ; AG, = (AG, - AG,)

=% [{AG,, + RT (Inp'y, — In py )} 2

2

~AG®, +RT| In Foot

In SEs

Peo,

Peco,

= Lag —Ac)+ Lar of 2 2 2 Pco, 1

nf 2 Peo, ’

cop Rr] of -00 4 | FOO. 2

Peco,

(6.32)

Peco,

Since AG = (AGS — AGS) This is the general equation (6.32) for the reduction of metal oxide or oxidation of metal by a

gaseous reactant. Hence conditions for gas — solid reaction are as follows: » Reduction of metal oxide: AG; AG,

”

Peo

>|

Peco, =

Peo,

foo

Peo,

Oxidation of metal: AG, >0, Peo

or

Peo


|

Peo,

ie,

_

Peo Peo,

(633) ke

AG, is+ve;s0AG, > 1.87 x 107" mm i.c. , partial pressure of oxygen in the surrounding is very much higher than the equilibrium partial pressure of oxygen in contact with Zr and ZrO, at 1727°C. Hence there1at no possibility of the decomposition of zirconia (ZrO, ) crucible at 1727°C under a vacuum of 107° mm of Hg. Ex 6.3. Calculate the equilibrium constant for the reaction at 750°C:

NiO (5) + H, (g) = Ni (5) + HL,O (g)

Given:

Ni(s) + 20, (g)=NiO (s), AG? =—244555+9853T J H, (g) + 20, (g) = H,0 (g), AG, =-246,438+5481T J] Can pure Ni rod be annealed at 750°C in an atm containing 95% H,0 and 5% H, by volume without oxidation ? Solution.

NiO (s)=Ni(s)+ % 0, (g),

—AG,

H,(g)+ %0,(2) =H,0(g). AG, NiO (s) + H, (g) = Ni (s) + H,0 (g), AG’=AG", -AG", T=750+273=1023K

Fugacity, Activity and Equilibrium Constant

53

Therefore, AG®=[ (~ 246,438 + 54.81 T) — (— 244,555 + 98.53 T)] = [(— 246,438 + 244,555) + (54.81 — 98.53) T) =[-1883 43.72 T] = [— 1883 — 43.72 x 1023] =—46608.56] Since, AG’=—RTInk=-8314

x 1023 x Ink

Ink=[AG®/ (- 8.314 x 1023)] = 5.48 Therefore, k= 239.845

Since k=(Pyo/ Py), = 239.845 Again, (Py! Py, y= 95/5)=19 Since

239845>> 19

Therefore, (Py,/ Py)eq >> (Puyo! Py) on Hence there is no oxidation on pure Ni rod during annealing at 750°C in present of 95% H,0 and 5% H, . If there is any oxide present on the surface of the rod that will also be reduced by the 95% H,O and 5% H, gas. Ex 6.4 Find out the percent of CO gas require to reduce FeO at 65 0°C and 1.5 atm pressure. Given: FeO + CO = Fe + CO, AGY=-225521+240T J/mol Solution. T=650+273=923K AG’ =-225521+240T =-22552.1 + 24.0 x 923 =— 400.1 J/mol Since, AG'=—RTInk=-8314x 1023 x Ink

Ink=[AG®/ (- 8.314 x 923)] = 0.052 Therefore, k= 1.054 Peg, + Pop =15atm,

let Pp, =x

Therefore, Peo, =(1.5-x) Since k=(Pep,/ Pog) =[(1.5-x)/x] Therefore, x = P, = 0.73 atm and Peo, =(1.5-0.73)=0.77 atm %CO require = [(0.73 x 100) / 1.5] = 48.67% %CO, require =[(0.77 x 100) / 1.5] =51.337% Ex 6.5. Calculate equilibrium constant for the following reaction at 298 K and 1 atm pressure: SnO (5) + CO, (g) = Sn0O, (5) + CO (g) Given: (i) 238n(s)+0,(g) =25n0 (5), 7s A =-514k] (if) Sn (5) + 0, (g) = SnO, (5), AG’, 29g =— 320k]

(iif)

@)

Solution.

C (5) + 0, (2) = CO, (g), AG’; 1s =—395KkJ

2C(5)+0,(g)=2C0 (2). AG", ps =-274KJ 2800 (s)=2Sn(5)+ 0, (2),

2C0,(g)=2C(s)+20,(g),

2C(s)+0,(g)=2CO0 (g), 2Sn(s)+20,(g) = 2500, (s),

—AG’, 15 - 246%, 208

AG’, 5g 2AG" re

28n0 (5) +2 CO, (g) =2 Sn0, (s) + 2CO(g) Therefore, ZAG, pop = [~ AG?)05 —2AGY yp + AG? sop +2AG°, 5, |

= [~(=514)= 2 x (= 395) + (~ 274) + 2 x (- 520)] =[1304- 1314] =—10K]J Hence AG’, 5, =—5 kJ/mol

Since, AG’=-RT Ink =-8314x298 x Ink

54

Metallurgical Thermodynamics,

Kinetics & Numericals

Ink=[AG"/(-8.314 x 298)] = 2.018 Therefore, k= 7.52 Ex 6.6. For the reaction: C (5) + CO, (g) = 2C0 (g) at 1120 K and I atm, the equilibrium mixture contains 92.63% CO and 7.37% CO, by volume. Calculate equilibrium constant for the reaction. Also find out partial pressure of CO, if P.,, is change to 107~ atm. Solution.

Since equilibrium constant, k= [ (Po,)” / Peo, 1 = [ (0.9263) /(0.0737)] = 11.64 Again Po, = [ (Pep) /k]=[(107)?/11.64]=8.59 x 107 atm

Prob 6.1. What is the standard free energy change of the following reaction at 1000°C ? PbO + Fe =Pb + FeO Also comment upon the feasibility of this reaction.

Given: AG’, = — 184 ki/mol and AG’, = — 364 kJ/mol at 1000°C. [Ans: —180 kJ/mol, Reaction 1s feasible | Prob 6.2. An atmosphere containing 15% CO, , 5% CO and 80% N, . Can this atm oxidize Ni at 1000 K ?

Given:

Ni(s)+

%20,(g)=NiO(s),

k, =576x 107 at 1000 K

CO (g) + % 0, (8) = CO, (g), k,=1.68 x 10'°at 1000 K [Ans: Ni will not be oxidized at all ]

Prob 6.3. For the reaction: ZnO (s) + CO (g) =Zn (s / 1) + CO, (g)

AH 40 709 =65 KJ/mol, AH’, 0,

= 180.9 kJ/mol

AS’, 300, 7n (5 = 13-7 J/k/mal, as’, 1200, zn 0 = 288.6 J/k/mol Calculate the (i) free energy change and (if) equilibrium constant at 300 and 1200 K. Also find out the direction in which reaction will be feasible at 300 and 1200 K from the above calculation, when all reactants and products are in their standard states. [Ans: (i) 60.89 kJ/mol, — 165.42 kJ/mol; (ii) 2.4987 x 107", 1.59 x 107] Prob 6.4. For the reaction: 3 Fe (s) +4 H,0 (g) =Fe,0, (s) + 4H, (g) at 1173 K, AG", =—14.76kl. Calculate: (a) equilibrium constant at 1173 K, (b) equilibrium pressure of hydrogen, if water vapour is 6.5 x 10™ atm.

[Ans: (a) k=4.54, (b) P, = 9.489 x 107 atm ]

Prob 6.5. Calculate equilibrium (CO / CO, ) ratio for reduction of FeO at 900°C according to the following reaction: FeO (s) + CO (g) = Fe (5) + CO, (g).

Given: (i) C(s)+0, (2) =C0, (2),

AG’ =-39413-084x10>T kJ

(i) 2C(5)+0,(g)=2CO (2), AG, =-22343-0.175T

kJ

(if) 2Fe (s) +0, (g) =2 FeO (s), AG’, =— 52593 + 0.128 T kJ

[Ans: 2.087 ]

Inter-relations Between

Thermodynamics CHAPTER

Variables

7.1 GIBBS - HELMHOLTZ EQUATION It gives relation between AG, AH and T

The combine statement of 1st and 2nd laws of thermodynamic in terms of free energy is dG = VdP — 8dT Now at varying T and constant Pie. dP =0 Therefore,

dG=-S4T

8G —| 3T

or

E19)

=—8

Jo

+ (5.23)

Applying this eq (5.23) to a chemical reaction, ney 87 Jp

AS

(7.1) Th

Now for a chemical reaction, we have standard relation as:

Substituting — AS

AG=AH-TAS from eq (5.12) by eq (7.1): AG=AH+T

(512)

rd

aT

AH — AG dAG = — |@i=29)]

or

(7.2

J.

IE) dT

EL)

This is known as the Gibbs — Helmholtz equation, and it can be written in the following forms for standard state: AH® —AG® dAG® = — Baal

dT

(71.4)

Therefore, Td AG® — AG° dT= — AH dT dAG®

Divided by T? on the both sides: o

Therefore,

or

d

IS—

iG

T

|

i

iT

AG® fe T

=—

AH® 5— ‘4

[dT

a

=—

y

—

T

T

AH?

5:75)

55

56

Metallurgical Thermodynamics,

Kinetics & Numericals

This is another form of the Gibbs — Helmholtz equation. 1

Since

dT

d =

=

5[L

Therefore,

= AH"

T)

(7.6)

le

This is the third form of the Gibbs — Helmholtz equation. Uses of the Gibbs — Helmholtz equations:

»

To calculate AH" from given value of AG” for a reaction,

*

To calculate AG” at any temperature, if AG® at some other temperature is known for the same reaction.

7.2 VAN'T HOFF EQUATION This equation give relationship between k, AH® and 7. In other words, it is the variation of equilibrium constant with temperature,

The Gibbs — Helmholtz equation [eq (7.6)], as above, is

= AHP

ox)

(1.6)

3

Also, as studied earlier,

AG"=—-RTInk

(6.18)

Now combined above two equations: d(—R

(7)

In k)

— ARP

a7

al 21

Therefore,

din &)

aL

-{%

(7:8)

INF

This is known as Van’t Hoff equation. This shows that the effect of temperature on equilibrium depends on the value of the change of enthalpy for the reaction. Eq (7.8) also can be written as:

d (In k) 7]

Cll

eas)

r—

Inter—Relation Between Thermodynamics Variables

Therefore,

° al)

57

= (5

dr

(7.10)

RT?

This is another form of Van’t Hoff equation.

7.3 INTEGRATION OF VAN'T HOFF EQUATION AND SIGMA FUNCTION (Z) It gives relation between k, Cp and T.

oslo)

The Van’t Hoff equation (eq 7.8) is

d (In k) =

AH®

(18)

T

or,

—Rd (Ink)= AH” d (7)

(ELT)

Since standard heat of reaction (AH?) is not independent of temperature, hence to integrate this above eq (7.11), it must be converted to a form in which there is no term dependent on temperature. From the basic relation: (8 (i)

AH oT

a (i) Cp=a+bT—cT" 2

A = AC), and

8 AH Therefore, AC, = (

=Aa+ AbT— AcT™? lp or d(AH) = (Aa + ABT — AcT 2)dT For standard state: ~~ d (AH®) = (Aa + ABT — AcT 2)dT By integrating: ste

=aar+

ABT? | 5

(Ac

F(5)am

(E12) ATA3) (7.14)

TLE)

Where AH? is an integration constant. Putting the value of AH? ineq (7.11):

— Rd (Ink) = -

PK)(7) + AH J 7)

(7.16)

wor PER 2) Since 7

1

J 1)

, putting this value in eq (7.16):

[mmf]

Again by integrating:

~Rink= ~AalnT)- iE = 2 ar

ot, T

7.17 -- (7.17)

Where 7 is an integration constant.

|= +1 (5H == he

or, -Rlnk+AalnT+|—2

(7.18)

58

Metallurgical Thermodynamics,

Kinetics & Numericals

The left hand side of this eq (7.18) is commonly designated as I and 1s called Sigma function.

Thus

E=—Rnk+AalnT+

Ab T Ac =H) 2 27

+ (T19)

It is found from eq (7.18) that the equilibrium constant of a reaction at any temperature can be

calculated from the heat capacity (Cp) values of the reactants and products, provided the equilibrium constant values at two different temperatures are known. i.e., if k, at temperature T' and £, at temperature T, are given for a reaction, k, at temperature 7, can be found out for the same reaction; if C,, values in terms of a, b and c are known for reactants and products. Here to get Aa, Ab and Ac, Since AC, = LC pads Ch reactants:

By putting C,, values of reactants and products we get AC; which is compared with

ACp=Aa + AbT— AcT™, to get Aa, Ab and Ac. 7.4 CLAUSIUS — CLAPEYRON EQUATION Vapours of the substances in equilibrium with liquid or solid do not behave as perfect or ideal gases (i.e, one which obeys: PV'= RT). Hence, the variation of vapour pressure with temperature is studied by Clausius — Clapeyron., From the Maxwell’s relation, we get:

£1451] ELEC)

Theref erefore,

EEa ar = oid

ores]

[since 1 8q=AH] =AH

..(7.20) 7.20

Where AH = heat of transformation, AV = molar volume change associated with transformation. This eq (7.20) is known as Clausius — Clapeyron equation. Clausius — Clapeyron equation is useful for calculating the effect of change of pressure on the equilibrium transformation temperature of a pure substance. Clausius — Clapeyron equation is applicable to any phase change i.e., fusion, vaporization, sublimation, allotropic transformation, etc.

Application of Clausius — Clapeyron equation to phase changes is as follows: 1. Vaporization, i.e., liquid — vapour equilibrium 2. Sublimation, 1.e., solid — vapour equilibrium

3. Fusion, i.e, solid — liquid equilibrium 4, Allotropic transformation, i.e., solid — solid equilibrium. 7.4.1 Liquid — Vapour Equilibrium When some liquid is present in a container some vapour will form at a given temperature. After some time liquid and vapour will come to equilibrium (as shown in Figure 7.1).

Vapour

Fig 7.1: liquid — vapour equilibrium.

Inter—Relation Between Thermodynamics Variables

59

Applying eq (7.20), to a liquid — vapour equilibrium:

dP J irl

AH el

(121)

af: | || TW)

Where AH,, is the heat of vapourization or latent heat of evaporation, Viap is the molar volume of vapour, and V,, q 1s the molar volume of liquid. Since V,,, >> Vi. , so value of Vis negligible with respect to value of V.,.

H

ence, eq

7.21 & 721b

| iE ar)”

becomes:

[=

ei Aly

Assuming that the vapour behaves as an ideal gas, the volume Ay

ET(7.22)

|TW,,) may be related to:

RT [=]

123)

Substituting this value of Vigp (£0 7.23) to £q (7.22), we get: dP

_(

PAH,

dT’

724

RT

~A1)

PY 1)_(Hy di

\P)

[dn P)]

or

B=

\ RT? (AH,

= (Fe

(1.25)

This eq (7.25) is another form of Clausius — Clapeyron equation. Case I: Integration without taking limits of P and T (Graphical method): If we assume as an approximation, that AH}, remains constant (this is not really true because AH varies with temperature). Now taking integration of eq (7.25), we get:

Jd P) “5 : J(= Jor AH, Therefore, In P= —| =

+E

.. (7.26)

where C is an integration constant. 1 Now if In P vs =)

is plotted (as shown in Figure 7.2), the plot will be straight line

(e.g y=mx+c).

InP

1T

Therefore, slope of the line (m)=—

Fig 7.2: InP vs (1/T) AH, R

} hence AH, can be calculated from slop of the line

and intercept of the line with y axis will be the value of constant C.

60

Metallurgical Thermodynamics,

Kinetics & Numericals

Case II: Integration by taking limits of P and T (Numerical method): Eq (7.25) may also be titrated by taking limits of P, and P, corresponding to temperature T, and T, respectively. Assuming that the temperature range is small enough, so that AH, is independent of temperature, integration of eq (7.25) yields:

Jaa p) = & Or

(InP,~

(3)

InP) =—

ii 3)

Hh

fH

BE

JT

%

(727)

or

Eq. (7.27) maybe used to calculate the vapour pressure at any temperature if the vapour pressure at another temperature and the mean heat of vaporization over that temperature range are known.

7.4.2 Solid — Vapour Equilibrium On the basis of assumptions similar to those made in liquid — vapour equilibrium, an expression similar to eq (7.25) may be also obtained for solid - vapour equilibrium:

252+

AH = RE

dr .

.

By integration:

(7.28)

AH

InP=-

Fra

+C

= (729)

Where AHg is the heat of sublimation and C is integration constant. 7.4.3 Solid - Liquid Equilibrium Applying the Clausius — Clapeyron equation [eq (7.20)] to solid — liquid equilibrium:

E

E F

— |=] =% dT TAV AH

Or

ill 9 Pe dr

Theref erefore, Assuming that AH; (7.32), we get:

(1.30)

TWig

SA

(131)

— Voor)

dP == [aH Vig — Vou)

(arT

..- 732 (1.32)

and AV are not changed with temperature, now taking integration of eq A

AH

aP=|—L—

A

Vig

ii Therefore, (P,—P))= | ———

Vet)

n

i Jor

h

(nL -InT;)

Tr

+: (733)

(Vig ii Vet)

Eq (7.33) may be applied to calculate the change in melting point of a metal with change of pressure.

7.4.4 Solid — Solid Equilibrium The rate of change of transition temperature (at which two crystalline forms of a solid are in equilibrium) with pressure 1s given by an equation similar to earlier one.

Inter—Relation Between Thermodynamics Variables

61

If «1s the stable form above the transition temperature, T and P 1s the stable form below the transition temperature, T. Then applying the Clausius — Clapeyron equation [eq (7.20)] to solid — liquid equilibrium:

dP)

[ AH,

[& J

x

dP)

Ne

|

htt)

AH,

(& |r|

a r

where AH, is the heat of transition; ¥ and Va are the molar volumes of the indicated forms, all measured at temperature T. By integration: P=

AH,

(In) +C

("a=")

(736)

where C is an integration constant. 7.5 TROUTON'S RULE Trouton’s rule states that the ratio of latent heat of evaporation (AH) to the normal boiling temperature (7) is constant for all liquids, and is approximately equal to 21 cal/deg/mole. AH,

5

Le.,

21 cal/deg/mole.

b

ATA

= 87.864 J/k/mol

(7.38)

[since 1 cal =4.184 J] This rule is only an approximate one, and is not followed by all liquid metals. However, it can be used to get AH}, from know boiling point temperature of a metal.

Examples Ex 7.1. Calculate the standard enthalpy and entropy changes at 298 K for the reaction:

2Cu (s) +2 0,(g) = CuO (5) AG

=-169452-164

Tlog T+12343 T J/mol

Solution Since AG®=-169452-164T

log T+

12343 T J/mol

(I)

Therefore, (AGYT) =(-169452/T)— (164 x 0.4342 In 7) + 123.43

(2)

[since log

X= 0.4342 In X]

Now differentiate eq (2) wrt. T:

[d(AG®IT) /dT] = (169 452 /T?) —(7.12/T)

(3)

From Gibbs — Helmholtz equation:

[{d(AG® IT)}/dT] =— (AH® IT?)

(4)

Now combining eqs (3) and (4):

—(AHPIT ) = (169 452/T%)— (7.12/T) Therefore,

AH°=-169452+7.12T

(5)

Now at 298 K:

AFC, =—169 452+ 7.12 x 298 = — 167.33 kJ/mol

Since Therefore,

AG =-169452-164T log T+ 12343 T J/mol AG, =-169452—-16.4 x 298 log 298 + 123.43 x 298

=-144.76 kJ/mol

J/mol

62

Metallurgical Thermodynamics,

Kinetics & Numericals

Again,

AG? = AH, — TAS"

Therefore,

AS, = [(AH’,g, — AG, )/T] = [{ ~167330.24 — (-144761 85)}/ 298] = 75.73 J/k/mol

Ex 7.2. The vapour pressure of liquid Titanium at 2227°C is 200 N/m’. The heat of vaporization at the normal boiling point of Ti is 435.14 kJ/mol. Calculate its normal boiling point. Solution. Let us assume that the normal boiling point of Ti is T,. At this temperature vapour pressure is

equal to 1 atm (i.e., 101325 N/m?). Now from Clausius — Clapeyron equation:

(InP,

or

InP) =— (AH, /R) [ (/Ty) - (/T})]

In (P, /P;) =~ (AH R) [(1/Ty) ~ (1/ Ty)]

Here,

T, =2227 +273 = 2500 K, P, = 200 N/m’

(1)

T,=T,? and P, = 101325 N/m’ Now putting these values in eq (1):

Therefore, In (101325 / 200) = [(~435140 / 8.314) x {(1/ T,) — (1/2500)} ] Hence, T, = 3558 K = 3285°C Ex 7.3. The equilibrium constants for the decomposition of NiO (NiO = Ni + 12 O,) are 1.514

x 107" and 2.355 x 107% at 527°C and 727°C respectively. Calculate the value of the equilibrium constant at 627°C from the following data:

Cp pio = 46.78 + 8.45 x 107 T J/K/mol Cpr =29.71 +4.184 x 107 T—9.33 x 10° T™ J/k/mol Cp o, =29.96+4.184 x 107 T— 1.67 x 10° T~ J/k/mol Solution.

Heat capacity change for the reaction: ACy, =

(XC produc ~ Cp, Reactant)

- (1)

=HChnigs "2 (Ch oo, )} ~ Cr mio 5) =[{(29.71 +4.184 x 10° T-933 x 10° T?) + % (29.96 + 4.184 x 10° T

We know:

— 167% 10°T2)} — (46.78 + 8.45 x 10° T)] =-209-2174x 102 T — 10.165 x 10° T= J/k/mol AC, =Aa+AbT—AcT™>

fo 1) = (3)

Now eq. (2) can be compared to the eq (3):

Hence, we get: Aa=—2.09, Ab=-2.174 x 107

and Ac = 10.165 x 10°

We know from Sigma function:

— Rink + Ag InT + (AB2)T — (Ac2T2 ) = (AH/T) +1

(@)

Here at T,= 527 + 273 = 800 K and k; = 1.514 x 107" From eq (4):

Therefore, 8.314 In (1.514 x 107") —2.09 In 800 — (2.174 x 1072) x 800 ~[(10.165 x 10%)/2 n (800) ] = (AH,/800) + I Hence, (AH /800) + I= 191.502 Similarly, T,=727+273 = 1000 K, and k,=2.355 x 107 From eq. (4) we get: (AH /1000) + I =-129.99

(5) ..(6) NG)

Now from eqs (6) and (7), we get: AH = 246040]

(8)

Substitute this value in eq 7: I =—-116.05

cl

Now

I,=627+273=000K

3)

Inter—Relation Between Thermodynamics Variables

63

From eq. (4) we get: 83141Ink,-2.091n 900 — (2.174 x 107%/2) x 900 —[ (10.165 = 10%) /2 x (900)? 1= (246040 / 900) 116.05 Therefore, In ky =—20.83 ie, ky; =9.012 x 107°

Prob 7.1 The normal boiling point of Zinc is 907°C. assuming that Zinc follows Trouton’s rule.

Calculate its vapour pressure at 800°C,

[Ans: 0.35 atm or 35318 N/m?] Prob 7.2 Apply the Clausius — Clapeyron equation to calculate the vapour pressure of liquid

copper at 2000°C. Given: Boiling point of Cu is 2595°C and heat of vaporization is 305.43 kJ/mol.

[Ans: 0.035 atm or 3543.8 N/m’]

CHAPTER

Solutions and Partial Molar Quantities

8.1 SOLUTIONS Solution may be defined as a homogeneous phase composed of different chemical substances. Nearly all chemical substances of metallurgical interest (i.e. solid, liquid, or gas; dilute or concentrated) are regarded as solutions. e.g. pure copper, brass, molten metal, molten slag, flue gases etc. A solution is assumed as a homogeneous one.

Solutions are mainly classified into two types: » Aqueous solutions: These are used in inorganic and physical chemistry, mostly at room temperature or low temperatures. e.g. NaCl in water to make NaCl solution. It is a solution prepared with water as a base. * Nonagueous solution: Metallurgical processes are mostly carried out at high temperature. So all metallurgical solutions are nonaqueous and inorganic solution. Hence metallurgical solution means inorganic, nonaqueous solutions for high temperature system. Metallurgical solution can be further classified: 1. Metallic solution: Single phase alloys (e.g. Ag — Cu, Fe — Ni etc) are metallic solutions. 2. Non-metallic solution: A molten slag is an oxide solution , may contain compounds like 810,, CaO, FeO, and Al,O, etc or sulphide solution which contain FeS, Cu,S ete.

No material (either element or compound) is absolutely pure. It will have some impurities, may be in very small quantities. Even at such low concentrations they may affect some properties significantly. e.g. (7) Liquid steel dissolves hydrogen gas to the extent of few parts per million ( ppm; 1 ppm = 0.0001 %). Even this low concentration of hydrogen causes fine cracks on the surface of

solid steel. (if) Liquid copper dissolves some oxygen during processing; even a concentration less than 0.1 % decreases ductility and electrical conductivity of Cu. Hence for high conductivity Cu, oxygen should be removed to a very low level to get OFHC copper. (iif) In semi-conductors even traces of impurities of the order of parts per billion (1/ 10°) may

affect their performance. For all practical purposes the composition of a solution is expressed in terms of percentage by weight, and for theoretical purposes, it is convenient to express composition in mol or gram atomic

basis; i.e., mol fraction or atom fraction. Afom fraction is used for solid or liquid metals and mol [fraction is used for oxides, slags and other compounds. 8.1.1 Atom Fraction Atom fraction of a component in a solution is defined as the ratio of total number of gram atoms of that component, and total number of gram atoms of all components present in a given amount of solution. 64

Solutions and Partial Molar Quantities

65

Hence, atom fraction (N,) of component 1 in a solution will be:

N=1

3 |< n,

AS(8.1)

i Where n, = total number of gram atoms of component 1. Yon i

= total sumation of gram atoms of total / components present in the solution.

Gram atom (n) =

Weight of a component in

Ta

=

a

Atomic weight of that component in gms

(82)

Conversion (weight % to atom % or vice-versa) To convert given composition in weight % to atom fraction, it is convenient to consider a total mass = 100 gms. So that the number of gm atoms of each component (n,, n, etc) contained in that is: % 1 y=

% 2

5)

ny, =

%2)

etc

(8.3)

where % 1. % 2 etc are the weight % of component 1, 2 etc and 4, , 4, etc are the atomic weights of component 1, 2 etc.

. Hence, atom fraction of component 1: N, =

%1

4vi

2%) 01

.-(8.4)

e.g. in case of binary solution of A and B, the weight % and atom % are given as follows:

eid

FEAT

(atom%A x At wt of A)x 100

| (atom% A x Atwtof A + atom% B x At wt of B)

45

ASD

wt % A Atom % 4 =

At wt of A wt % A

Atwtof

2

wt % B

A

= 100

..(8.6)

At wt of B

8.1.2 Mol Fraction Similar to atom fraction, mol fraction of a component in a solution is defined as the ratio of number of gm mol of that component, and total number of mol of all components present in a solution. i.e., m

Foe 1

(8.7)

pk 7% i

Where n, = total number of gm mol of component 1,

Sn i

= total number of gm mol of total i components present in a solution i.e., solutes plus

solvent. Other composition units used in analytical work for aqueous solution are; (/) Molality: number of mol of solute per 1000 gms of solvent,

66

Metallurgical Thermodynamics,

Kinetics & Numericals

(if) Molarity: number of mol of solute per one litre of solution,

(iif) Normality: gm equivalents of solute per one litre of solution. i.e., equivalent wt in gms = (atomic wt / valency) 8.2 IDEAL AND NON-IDEAL SOLUTIONS

8.2.1 Raoult's Law There are two statements of Raoult 5 law as follows: I. The original empirical statement of Raoult’s law, which is obeyed by limited number of binary

solutions is that the partial pressure of the vapour of any component of an ideal solution is equal to the product of its mol fraction and the vapour pressure of the pure component at the same temperature as that of ideal solution.

Hence, Bi=X Py ..(8.8) where, P, is the partial pressure of the vapour of any component, A of an ideal solution, X, is its mol fraction, and P°, is the vapour pressure of the pure component. However, as studied earlier, if the vapour behaves as an ideal gas,

Then P,=f,and P°,=7°,

where fis the fugacity of vapour.

Therefore,f, = Xf", Now by definition, |

..(8.9)

Ju

=a,

0

where ais the activity of component A.

A

Therefore,

a, =X,

(8.10)

IT) Another statement of Raoult’s law is the activity of a component equal to the mol fraction of that component, i.e., a, =X, (as shown in Figure 8.1).

Raoult's Law —

Xy—> Fig 8.1. Raoult’s law curve 100

=

Xg

I]

+ ve f

~

EN

Ideal Solution — ve

|_Non-ideal Solution

0

X,

—*

100

Fig 8.2. Ideal and non-ideal solutions

An ideal solution is one which obeys Raoult’s law over all temperatures and pressures. Figure

8.2 shows the activity vs mol fraction plot for binary system of A and B components. 1) Since ideal solution obeys Raoult’s law and hence lines for component A and B are diagonal. 11) For non-ideal

Solutions and Partial Molar Quantities

67

solution the activity (a) vs mol fraction (X) curve is above or below the diagonal i.e., departure from Raoult’s law may be + ve or — ve. 8.2.2 Henry's Law Henry's law states that the activity of a component (i.e., solute) is proportional to its mol fraction. cg a, aX, (as shown in Figure 8.3). Therefore, a, =vX, ..(8.12) where 7 is called activity coefficient or activity quotient.

—

Ideal Solution

——MNon-ideal Solution [Henry's

Law

Xp—>

Fig 8.3: Henry's law curve Hence, 7, is defined as the ratio of activity of i to its mol fraction i.e., y, =

(5 ) i

i

(i) For ideal solution: y,= 1. (if) For non-ideal solution: y; > | — + ve deviation ¥;

[ET

9

gt Jol

i00Y

\ en I

t

gr ou. wh

Free Energy — Temperature Diagrams

91

e.g., 2Pb0(s) + C(s) — 2Pb(s) + CO,(g) at 100°C, ZnS(s) + C(s) — Zn(g) + CO(g) at 1200°C. 6. Reduction of MO by reducing agent, other than C: It is not necessary that two curves of M — MO should always intersect each other for reduction, as in the case of reduction by C. e.g., reduction of Cr,0, by Al or Pb at 1200°C, from the Figure 10.1, we get 4/3 Cr+ 0,=2/3 Cr,0, AG®, =—120 kcal/mole of O,

4/3 Al+ 0,=2/3 41,0, 2 Pb+ 0,=2Pb0O

AG, =—190 kcal/mole of O,. AGS, =-40 kcal/mole of O,.

Case I: 2/3Cr,0,=4/3 Cr+ 0, AG®, = 120 kcal/mole of 0,

4/341 + 0,= 2/3 AL,0, AG, =-190 kcal/mole of O, By adding, (I) 2/3 Cr,0, + 4/3 Al = 4/3 Cr + 2/3 Al,0, AG, =120-190=-T70 kcal/mole of O, Again

case Il: 2/3 Cr,0,=

43 Cr+ 0, AG®=120 kcal/mole of 0,

2Pb+0,=2Pb0O

AG

= — 40 kcal/mole of O,.

By adding, (II) 2/3 Cr,0; + 2 Pb=4/3 Cr+ 2 PbO AG = 120-40 = 80 kcal/mole of 0, From above calculation we can say that Cr,0, can be reduced by Al (since AG®, on =e value), but not by Pb (since AG® = +ve value). From diagram we can say that those metal oxide (Cr,0,) line above the line of metal oxide (41,0), that metal (4/) can reduced the above

metal oxide (Cr,0,); not vice- versa. 7. To compare stability of oxides: If the line of M — MO, which is in lower position than other, that MO is more stable. Stability of oxide increases with lowering the position of oxide. 41,0, is more stable than Cr,0, and PbO, again Cr,0, is stable than PhO etc.

10.2 DETERMINE OF POINT O, H, C AND EQUILIBRIUM GAS PRESSURE One may note a logarithmic scale for p, at the right hand side of the diagram (Fig 10.1). This

allows us to obtain the equilibrium Po, of eq (10.3) [ Le, AG®, =RT'In p, | by a geometrical construction directly from the diagram. The procedure is shown in Figure 10.2.

Pal Pio

0 keal/mole okN 2C0 + 0, = 2C0, HL-

[wt % Si] + eg™" [wt % Mn] +e’ [wt % P]

Therefore,

=—0.028x0.04+0.11 =0.08134 fo=1206

From eq 1: hg =

[Wt %

x 0.8 + 0.063 = 0.1 +(-0.026) x 0.5+ 0.029 = 0.04

S]= 1.206 x 0.04 = 0.048

Ex B8. Find out activity of O content in liquid steel during reaction of Si at | 600°C. Si content in liquid is 0.8%.

Given Solution.

log kg; = [(29700/ T)— 11.24] Si+20=(Si0,)

0

Ksio, = { (a5, / [ag] [ao] }

(2)

T=1600+273=1873 K

Therefore,

log kg;, = [(29700 / 1873) 11.24] =4.617

kgio, = 41391.77

Considering formation of SiO, is pure form, so (ag, ) = 1 and [ag] = fg; [wt% Si] =[wt% Si] since for dilute solution, fg; = 1 From eq 2:

[aol = (ago) / { kgio, * [W1% Sil} = 1/ {4139177 x 0.8} = 3.0199 x 10° Therefore,

[ao] = 5.495 x 107

Ex B9. Calculate the concentration of oxygen in molten iron at I 600° Cin equilibrium with pure FeQ. Given:

Solution.

So

Since

log kp, = — (6372 E+

2.73

FeO (1) = Fe (I) + O (wt%)

Ne)

kee = {[.). [8M (ape0)}

Be)

ag, = | (for pure FeO) and a, = 1 (assume pure Fe)

Again [h J. =f]. [W,]=[W,] Now from eq. (2) we get: T=1600+273=1873

(If we consider for dilute solution, f, = 1) kp =[W]

«: (3)

K

log kg, = «(6372 / T) + 2.73 =[ (6372 / 1873) + 2.73 | =— 0.672 Therefore kp, =0213 Hence, the concentration of oxygen in molten iron at 1600° C, [W,] =0.213% Ex B10. Calculate the residual oxygen content of liquid iron containing 00.1 wt% Si in equilibrium with solid silica at 1600°C., Given: Si(l) +0, (g) = S10, (5), AG] =—947.68 + 0.199 T kJ/mol

0, (g) = 2[0] (1 wt% std state), AG, =—233.47— 0.006 T kJ/mol Si (I) = [Si] (1 wt% std state), AG]

=-11924-0.053T

kJ/mol

eg = 032,60 =-0.24,¢,°=-0.20, ande,” =-0.14 Solution.

[Si] (1 wt% std state) = Si (1)

—- AGS

2[O] (1 wt% std state) = O, (g)

—AGS

Si (1) + 0, (g) = SiO, (s),

AGS

[Si] (1 wi% std state) + 2[0] (1 wi% sid state) = SiO, (5) AG’, Therefore

~~ AGS =AG

- AGS -AGS

158

Metallurgical Thermodynamics,

Kinetics & Numericals

=(—947.68 +0.199T)— {(-233.47-0.006 T ) + (119.24 — 0.053 T )}] =-59497+0.258T kl/mol [T=1600+273=1873K] =(—594.97 + 0.258 x 1873) = 1000 J/mol =— 111736 J/mol Since AGY=-RT Ink Or -111736 =-8314 x 1873 x Ink Or Ink=7.175

logf,=e.° [wt % 0] + ¢,5 [wt % Si] = (-0.2) x (Wt% 0) + (= 0.14) x (0.1) =-02wt% 0-0.014

log fg, = eg, [Wt % Si] + eg” [wt % O] = 0.32 x 0.1 + (= 0.24) x ( wt% O) =0.032-024 wt% O

Equilibrium constant, k = [ag /( hg, . B)] =[1/ (hg; . b°)]

(since ag, =1)

Again h=f. wt%i Therefore k= [U{(fg;. wt% Si) . (f, . wt% 071] Taking both side log: log k =—[{log fg; + log (wt% 8i)} + 2 { log f, + log (wt% 0)}] =—[{(0.032 - 0.24 wt% O) + log (0.1)} + 2{(— 0.2 wt% O — 0.014) + log (wt% 0)}] =—[{(0.032 - 0.24 wt% O) + (-1)} + {(- 0.4 wt% O — 0.028) + 2 log (wt% 0)}] = [0.996 + 0.64 wt% O — 2 log (wt% O)] Since value of wt% O is very small, so value of log (wt% O) can be taken as negligible. [since In

x=2.303 log x; In k=

7.175 =2.303 log &,

Therefore, log k=(7.175/2.303)=3.1155] So log k= 0.996 + 0.64 wt% O = 3.1155 Therefore, wt% 0 = 3.31 Ex B11. Find out the oxygen concentration in liquid steel after degassing at 1600°C and 1.0 torr fwt% CJ] =0.2. Given: (i) log ko, = (6120/T) + 0.15 (ii) log kp = (1168/T) + 2.07 Solution. T=1600+ 273 =1873 K, 760 torr= 1 atm, so 1.0 torr = (1 / 760) atm 2[0]=0, «(5 Equilibrium constant, ko=1po,/ (hy)] (2) Since hy =f, x wt% i =wt% i, due to f; = 1

From eq. (2): wi% O = \/(Po, kb) = ko * «Po, Since

ky=

Again

0)

Jeg)

log ko, =(6120/T) +0.15=(6120/ 1873) + 0.15=3 417

Therefore k,, = 2615.082, now putting this value to eq (3): wit%e O =k, x

\[py,

=2615.08 x (1/760) = 94.86 %

This high value of oxygen content in steel is impossible, hence oxygen can not remove by this way. Therefore oxygen in steel can be remove by react with carbon.

[C]+[0]= {CO}

Equilibrium constant, kw, = [Po/(he * ho)] = [ Pag/(Wt% C x wt% 0)] Againlog key =(1168/T)+2.07= (1168/1873) +2.07 = 2.693 Therefore

kop =493.853; since [wt % C]=0.2 and p., = (1/760)

(4)

(5

159

Numericals in Process Metallurgy

Putting these values in eq (5):

Therefore wt% O = [po (Wt% C x ko)] = [(1/760)/(0.2 x 493.853)] =1.332x10°% % Ex

B12. Find out wt % QO in liquid steel to remove C from HM (content 3% C, 1.2% Si, 0.8%

Mn, 0.04% S, 0.4% P) at 1600°C. Given log kop, =[(1056 / T) +2.131] ©=0.14,eM=-0012,¢"=0051,¢ 5=0.046, ¢ 5 =0. 08 =-0.13,e M"=-0.021,¢,=0.07,¢5=-0.133,¢ 5 =-0.131 Solution. ~~ I'=1600 +273 =1873K So, 10g ke = [(1056/1873) + 2.131] = 2.6948 Therefore,

Again,

k.,=495.22

logf, =e [wt % S]+e.° [wt% C] +e’ [wt % Si] +e

M [wt % Mn] +¢_" [wt % P]

=0.046 x 0.04 + 0.14 x 3.0 + 0.08 x 1 2 + (0.012) x 0.8 + 0.051 x 0.4 =0.5286 Therefore,

f,=3.378

Similarly, log f, =¢_% [wt % S] + ¢, [wt % C] + ¢_% [wt % Si] + ¢M" [wt % Mn] + ¢_" [wt % P] = (0.133) x 0.04 + (—0.13) x 3.0 + (-0.131) x 1.2 + (0.021) x 0.8 + 0.07 x 0.4 =-0.54132 Therefore, f, = 0.2875 We know that k., = {p_/h xh}

Therefore, h. x h, = {p., | kee }= {Vkeo}

So, Therefore,

{sincep_, = | and h; =f; x [wt% 1]}

fox [Wt% C] x f, x [wt% 0] = {1/kpo) [wt% 0] = {1/kq,} * {1/(f, xf, x [wt% C])} = {1/(495.22 x 0.2875 x 3.378 x 3.0)} =1/1442.8359=6.93 x 107

Ex B13. Consider deoxidation of molten steel by Al at 1600°C. The bath contains 1% Mn, 0.1% C. The final oxygen content is to be brought down to 0.001 wt%. Calculate residual Al content of molten steel assuming that [Al] — [O] — (41,0,) equilibrium is attained.

Given:

logk, = {(~ 58473 / T) + 17.74)

cate = 0,e,°= 0.091, ¢5° ==-68, eg '=0 eo =-0.021,e,5=-045,¢,°=-02,¢,*'=0 (AL,O;) =2 [Al] +3 [0]

Solution. Therefore, k

Now

{hl

(sincea,,

=a: Wal lo: Wl

0.091 x0.1+(-6.6) x 0.001 +0xW,

= 0.0091 — 0.0066 = 0.0025

fyy= 1.006

Similarly, log f, = ep ™ x wt% Mn + eo" x wt% C + “He x wt% O + eo x wi% Al

= (0.021) x 1 +(~ 0.45) x 0.1 + (~ 0.2) x 0.001 + 0 x W,,

So

= —0.021 - 0.045 — 0.0002 = -0.0662

f= 0.859

T=1600+273=1873 K

(

5 = 1, for pure Al ,05)

logfy, =e," x¢Wi% Mn + en x Wi% C+e, x wt% 0+ I x Wt% Al =0x1+

So

hl’ (a Ao). }={[h Me [ho].

log k,, = {(=58473/T) + 17.74} = {(~58473 /1873) + 17.74} =— 13.48

- (2)

160

Metallurgical Thermodynamics,

Hence

ky, =332x107™"

From eq (2) we get:

k, =[fy;- Wy 7 Ifo - AY

Kinetics & Numericals

Therefore, 3.32 x 107% =[1.006 x W,, ]*. [0.859 x 0.001]® So [1.006 x W,, 1 =5237 x 10

Therefore

[Wy I’ =5.175xx 107 [W,]1=7.193 x 1073

Hence residualAl content of molten steel is 7.193 x 107% Ex B14. Calculate the chemical potential of nitrogen in liquid steel at 1600°C. Steel content 0.5% C, 0.5% Mn, 0.2% F, 0.01%N.

Given: [wt% N] fy, = ky (py,)"” Where

log k=

{(~188.1/T) — 1.246}

ef=025¢/=0¢"=0051,¢/M=-002

Solution.

T=1600+273=1873K log ky = {(~188.1/1873) — 1.246} = 1.3464 Therefore,

Again, Therefore,

ky, =0.045

log fy; = eC wt% C + ey wi% N +e Wit% P + eM" Wt% Mn =025%05+0x0.01+0.051 x 02+ (0.02) x 0.5=0.1252 fy, = 1.334

Since ky (py)"” = [Wi% N] £,=0.01 x 1.334 =0.01334 Therefore, (py,)"” =0.01334/0.045 = 0.296 So, Py, = 0.0876 Chemical potential of nitrogen in liquid steel

=p, =RTInpy,

=8.314 x 1873 x In (0.0876) = 37.865 kd/mol of N, Ex B15 Calculate the oxygen potential of liquid steel in contact with a molten slag at 1 600°C. Assume equilibrium partitioning of oxygen between slag and metal.

Given: Solution.

1. [0], + Fe (1) = FeO (1), k, = 4.35 at 1600°C. 2.0, (g) =2[0],, , k, = 6.89 x 10°at 1600°C. T=1600+273=1873K

Therefore,

hy =(1/k)=(1/4.35)=0.23

ky = [ago/(ag - ag) =(1/hy)

Again Or

[since ap,= 1 and ag, = 1, pure at std state]

ky= [(ho)'/Py]

Po, = (ho)1ky] = [ (0.23)°/(6.89 x 106)] = 7.67 x 107

Therefore, oxygen potential of quid steel = fg, =RTInp

=8.314 x 1873 x In (7.67 x 107) =-290979.63 J/mol =—290.98 kJ/mol RATE MEASUREMENT Ex B16. Calculate the rising velocity of a 1.5 um slag particle, rising through a stagnant liquid steel at 1600° C, given slag density is 3000 kg / mi’ , the density of liquid steel is 7600 kg / m’ and viscosity is 7 cP. (Given g = 9.81 m/s”) Solution. T=1600+273=1873K

161

Numericals in Process Metallurgy

According to Stock’s Law:

u=[{2 1 (pp -p

el / (9 pl

where u = velocity of the slag particle, m/s 1 = radius of the slag particle, m p, = density of the slag particle, kg/m’ p, = density of the liquid steel, kg/m’ g = acceleration due to gravity,m/s’ 1 = viscosity, kg/m.s Here

Py ===3000 kg/m”

3

and p, := 7600 kg/m

3

r= (1.5/2)um=075x 10°m; g=9.81 m/s’ p=7cP=7x%10"2P=7 x 107 kg/ms Putting these values in the above equation:

u=[{2 x (0.75 x 107%) x (3000 — 7600) x 9.81} / (9 x 7 x 107)]

= [—(50766.75 = 1072) / (63 x 107%)] =— 805.82 x 107° = —8.06 x 107 m/s Negative sign of velocity shows that slag particle goes in upward direction. Ex B17. Molten steel is teemed from a ladle through a 5 cm diameter nozzle at its bottom. If the volume of the molten steel in the initial stage is 2 mt’ and the linear velocity of discharge is 90 cm / s. Calculate the time required to empty the ladle completely. Solution. According to the law of conservation of mass: [(Amount of fluid entering the reactor per unit time) — (Amount of fluid leaving the reactor per unit time)] = (Amount of fluid retained in reactor per unit time) (1) Amount of fluid entering the reactor per unit time =p, .u,_ A, (2)

Amount where p, Amount where V

of fluid leaving the reactor per unit time = p,.u,. A, -(3) u, and A are the density, velocity of the fluid and area of the reactor respectively. of fluid retained in reactor per unit time = V (dp / dt) (4) is the volume of the reactor and (dp / dt) is the rate of change of density of fluid with time.

Putting the values of eqs (2 to 4) in eq (1):

piu. A = pau, A, =V (dp / dt)

i 19)

Since in this case there is no fluid entering to the reactor, i.e, p,.u,. A; =0 The density of the liquid steel may be taken as constant, so that eq (4) may be written as: p (dV / dt) Hence eq (5) become: — p,.u,. A, =p (dV/dt) ordV=—u, A, dt

..(6) 2:07) (8)

Now integrating between t = 0 and t = t, where V=V and V,=0 respectively. |7

1

Vv

0

[dV =u, Ag|dt

Therefore V=u,. Ay t ort=[V/(u,. A,)]

Here F=2m’,u,= 90 cm/s =0.9 m/s and 4, = [(T/4) (0.05)] = 1.96 x 10° m” . Hence t= [2/(0.9 x 1.96 x 107)] = 1131.77 s = 18.86 minutes.

«=: (9)

162

Metallurgical Thermodynamics,

Kinetics & Numericals

Prob Bl. LD converter is charged with 10 tonne of hot metal containing 4.0 % C, 1.5 % Si, 1.0% Mn and rest Fe. The impurities are oxidized and removed from hot metal by lancing pure oxygen into the converter. One quarter of the carbon is oxidised to CO, and three quarters to form CO. Find out (i) total amount of oxygen gas require (in Nm? ), (if) the volume of product gases. Mention clearly, assumptions made. { Ans: (i) 607 Nm’, (if) 747 Nm® (25% CO, + 75% CO)] Prob B2. Estimate the activity of S in hot metal containing 0.04% S, 4.0% C, 2.0% 81, 1.0% Mn, at 1600°C in 1 wt % standard state.

Given: e° = 0.24, ¢,> = 0.066, e,° =— 0.028, and eM" = 0.025

[Ans: 0.46]

Prob B3. Find out activity of oxygen content in liquid steel during reaction of Mn at 1600°C.

Mn content in liquid is 0.8%. Given

log ky, = [(12440/T)- 5.33]

[Ans: 0.061]

Prob B4. Justify which of the following reactions is responsible for the de-oxidation of liquid steel (wt% C = 0.2) at 1850 K and 1 atm pressure.

Given: (i) 20 = 0, (g); (5)

C+0=CO(g);

log k= (6120/T) + 0.15 logkyy=(1168/T)+2.07

[Ans : 1% reaction yield wt% O = 2871.5 % , which is impossible; hence O remove by 2" reaction, wt% O = 9.945 x 107 %]

C: Extractive Metallurgy Run of Mines Ore

Mineral Dressing

Extraction Process

Pure Metal

Refining Process

Impure Metal

165

Numericals in Process Metallurgy

| WORKED EXAMPLES | Ex.C1. undergoing equilibrium Given:

THERMODYNAMIC CALCULATIONS Hydrogen is passed through a furnace at 800°C, containing FeO (solid), which is reduction to form Fe. Find out the percentage of utilization of H, gas, assuming that prevails. AG® = 8368 J/mol at 800°C.

Solution. FeO (s) + H, (g) = Fe (s) + H,0 (g), Therefore, equilibrium Constant (k) = [(ag, * ay,0)/(@gep * ap,)]

AG® =8368 J/mol at 800°C

Consider, Fe and FeO are pure solids, so ap, = 1 and a, =1 at std state;

Ayo = Puyo ad k= (pu,0/ Py)

Hence,

ay, =py,

AG® = 8368 J/mol =—RT Ink =-8314 x 1073 x Ink Ink=-0.938, Therefore, k=0.3914 = (py o/py,) Since total pressure is one atmosphere, Pi, T Pio = 1 Let Pu, =X. Puyo = 1-x Therefore, [(1 —x)/x]=0.3914; so x=0.7187

Hence, utilization of H, gas = 71.87 % Ex.C2. Find the permissible pct. of water vapour in hydrogen gas to reduce Cr,0, at 1 200°C, without oxidation of chromium.

1/3Cr,0,+H,=2/3Cr+H,0, AG® =126566-30.67 T J/mol Solution. Equilibrium Constant (k) = [ {(a¢,)”* * (ay,0 )}{(ace,0)" * (ay)}] Consider, Cr and Cr,0, are pure solids, soa, = | and a, _,, =| 73

ay.0=Puyypanda, =p, k= (pu, / Py,) AG® = 126566 — 30.67 T J/mol =-RT Ink 126566 — 30.67 x 1473 =—-8314 x 1473 x Ink

Hence, Since Or, Therefore,

Ink=-6646,

Since, Let

Therefore,

at std state;

so

k=1299x

107

Py, T Puyo = 1 Ph, =X, 80 Pho = l-x

[(1-x)/x]=1.268 x 107

Hence,

x=09987,

so

1-x=1298x

107?

The permissible pct. of water vapour in hydrogen gas = 0.1298 %

Hence if the pet. of water vapour in hydrogen gas is more than 0.13% then product Cr will be

oxidized at 1200°C.

Ex.C3. If 41,0, is take place at 15 00°C ? Given: (1) 2Al (1) (2) Hy(g) Solution.

heated at 15 00°C, in an atmosphere of hydrogen gas. Is there any reduction If not what is the minimum temperature required ? + 3/2 0, (g) = ALO, (s), AG", =—1679876+321.79 T J/mol + 1/20,(g)= H,0(g) AG®,=-251876.8+ 58.32 T J/mol

Write reversed the Eq (1): ALO, (s) = 2 Al (1) + 3/2 0, (g), — AG", 3 x Eq (2): 3H, (g) + 3/20,(g)= 3H,0(g), 3AG°, Addition of Eqs (3 & 4): ALO, (s) + 3H, (g) =2 Al ())+ 3 H,0 (g),

(3) (4d) (5)

Therefore, AG"; =— AG", + 3AG", = —«(— 1679876 + 321.79 T) +3 (-2518768 +5832 T) =0242456-146815T (6)

166

Metallurgical Thermodynamics,

Kinetics & Numericals

Since T=15004+273=1773K From Eq (6): AG, = 9242456 146.815 T =924245.6 — 146.815 x 1773 = 663942.61 J/mol Considering systems in pure state. At 1500°C free energy value is (+), hence ALO; is not reduced. The minimum temperature required for reduction, AG" =0 So, 9242456 -146815T=0 Therefore, T=6295 31 K ie, 6022.31°C Hence minimum 6022°C is required to reduce Al,O, by hydrogen gas. Ex CA. Find the minimum temperature for the reduction of PbO by carbon, under standard

condition to form CO. Given:

1) 2PbO (s) =Pb(s) + 0, (g), AG", =442248.8 220.08 TJ 2)C(s)+ 0,(g)=CO, (g). AG", =-3941328-084T J]

Solution. 2 PbO (s)=2Pb (5) + 0, (g), AG", C(s)+

0,(g)=CO0,(g)

AG",

2PbO + C(s)=2Pb (5) + CO, (g), AG"; =AG", +AG", Therefore, AG®, = (442248 8 220.08 T) + (- 394132.8 -0.84T) = 48116-22092 T The minimum temperature require , AG”; = 0 ie, 48116 -22092T=0

Therefore,

T=217.78 K i.e. - 55.2°C

Hence the minimum temperature for the reduction of PbO by carbon is — 55.2°C. Ex C5. Calculate the purity of hydrogen gas that must be maintained in a reaction chamber at 800°C to carry out the reduction: CrCl, (s) + H, (g) = Cr (5) + 2 HCI (g).

Given: (i) H, (g) + Cl, (g) = 2 HCl (g), AG®, =—202.924 kJ (if) Cr (s) + Cl, (g) = CrCl, (s), AG®, =— 264 847 kJ Solution.

CrCl, (s) = Cr (s) + Cl, (g), — AG", H, (g) + Cl, (g) = 2 HCl (g), AG?

CrCl, (s) + H, (g) = Cr (s) + 2 HCl (g), AG® = (AG®, — AG",) Therefore, AG® = [202.924 — (-264.847)] = 61.923 kJ Equilibrium Constant (k) = [{( a.) * (aye Yu! {Agr

* (ag )}]

Consider, Cr and Cr,0, are pure solids, so ac, = | and ag,¢,, =1 at std state; aye, = Pyc, and ay, =py, Assume Py, + Pyc, =1

Hence,

k= Pcl) =[(1-x)x] T=800C=1073K

Since Therefore Hence

AG*=-RTInk 61923 =-8314x

1073 xInk

Ink=-6.941

k=9.67x10"*=[(1 —x)’/x] x —2.0009%x+1=0 Since roots = [{— b= (” — 4ac)}/2a] Therefore, x=[ {2.00096 = V (2.00096) —4} / 2] = 1.031 or 0.9695 or

[Letpy, =x]

167

Numericals in Process Metallurgy

% of Hydrogen = 103.1 or 96.95

Hence 96.95 % purity of hydrogen gas must be maintained. Ex C6. Reduction of WO, with hydrogen gas is done to form metallic W at 1400°C. What is the maximum concentration of water vapour which could be allowed in reducing atmosphere? Given that: (i) W(s) + 0, (g) = WO, (5), AG" =—-564003.2 + 162.84 T J/mol (ii) Hyg) + 1/20,(g) = H,0(g) AG", =-251876.8 + 58.32 T Jimol Solution. Write reversed the Eq (i): WO, (s)=W (s) + O, (g) —AG®,

2 x Eq (ii): 2H, (g) +0, (g) = 2H,0 (g) 2AG°, Therefore Since So Since

WO, (s) + 2H, (g) = W (s) + 2H,0 (g) AG" = (2AG®, — AG") AG°=[2(-251876.8 + 58.32 T) — (—564003.2 + 162.84 T)] =60249.6 — 46.2 J/mol T=1400°C = 1400 + 273 = 1673 K AG®=- 17043 J/mol AG®=-RT Ink

Therefore — 17043 =-8314 Hence

x 1673 x Ink

Ink=1225and

Again

k=(Pyo/ Py)

k =3.405

Therefore (Py, / Py) = k= v (3.405)= 1.845 Assuming, Py, + Puo =1 atm. Let

Pho =x,

Py, =1l-x

So [x/ (1 -x)]=1.84 Therefore x = 0.6484 = 64.84% Hence, maximum 64.84% water vapour could be allowed in reducing atmosphere. Ex C7. During reduction of Cr,0, by carbon the carbide is formed. (i) Find out whether the formation of carbide during the reduction is easier than reduction of oxide. (ii) Find out the minimum temperature above which reaction is possible in borh cases, under standard conditions. Given that: 2/3Cr,0,+

18/7

C= 4/21 Cr,C, +2 CO

23Cr,0,+2C=43Cr+2C0

4G, 4G,

=490.57-035T

=52342-035T

kJ

kJ

Solution. At equilibrium, Therefore Again, Therefore

AG", =49057-035T

=0

T=1401.63 K = 1128.63'C AG", =52342-035T

=0

T'=1495.49 K = 1222.49°C

Hence, (i) the formation of carbide is much easier than reduction of oxide, since it is taking place at lower temperature. (if) The minimum temperature for carbide formation and reduction of oxide are above 1129°C and 1222°C respectively. Ex C8. Chlorination of TiO, is done by chlorine gas with or without presence of carbon. Reactions are as follows: (i) TiO, + 2Cl, = TiCl, + 0, (ii) Ti0, + 2C + 2Cl, = TiC, + 2CO Compare the free energies values of the above two reactions at 1200 K and state which reaction is more favorable from the two. Given that : (i) TiO, = Ti + O,, AG, =914.204-0175T J (i) icl,= Ti + 2Cl,, AG", = 756.049 + 7.53 x 107% TiogT-0.145T J (iii) 2C + 0, = 2C0, AG®, = -223.426-0.175T J

168

Metallurgical Thermodynamics,

Kinetics & Numericals

Solution. Ti0,=Ti+0,,

AG®,

Ti+ 2C1, = TiCl Fr AG", (7)

TiO, +2CL, = TiCl,

Therefore,

+0,

AG®| = AG®, - AG®,

AG", =[(914204 -0.175T )— (756.049 + 7.53 x 10° TlogT-0.145T)] =158.155-7.53

x 10° T logT-003T

=

x 107 = 1200 x

158.155-7.53

[since T=

1200 K]

log 1200 -0.03 x 1200

=94.331J Hence, this reaction (I) is not possible due to its free energy value is positive.

Again

TiO, =Ti +0, 2C+0,=2C0, Ti +2Cl, =TiCl,,

AG, AG, ~AG®,

(i) TiO, +2C+2ClL,=TiCl,+2CO AG°;=AG" +AG’ -AG’, Therefore, AG”) = [(914.204 0.175 T) + (- 223.426 — 0.175 T ) — (756.049 + 7.53

x 107 Tlog T— 0.145 T )] =[(914.204 — 0.175 x 1200 ) + (— 223.426 — 0.175 x 1200 ) — (756.049 +7.53 x 107 x 1200 x log 1200 — 0.145 x 1200 )] =-65271-273.823=-339.09J Hence reaction (II) is more favorable, as its free energy value is negative. Ex CY. ZnO (5) + C (5) = Zn (g) + CO (g) Find out the minimum temperature for above reduction, if the total pressure i) 1 atm, (ii) 0.1 atm. What will be the gas composition in each case ?

Given that:

ZnO (s) =Zn (g) + 1: 0, (g), C(s)+¥% 0,(g) = CO(g)

AG® = 481.16-021T kJ AG®,=—111.71-0.086 T kJ

Solution.

ZnO (s)=Zn(g) +

O,(g), AG",

C(s)+"% 0,(2)=CO(g),

AG’

ZnO (s) + C (s) = Zn (g) + CO (g) AG® AG® =AG", + AG", =(481.16-021 T)+(- 111.71 -0086T) = 36945-029 T Equilibrium constant, k = [(a, . 8,0) / (87,5 - 3: )] = (Py, - Pg) [az = 8c = 1] (7) Since P_; = 1 atm = (P,, +P) Due to one mol of each gas is formed, so P,, =P, =0.5 atm Therefore, k=025

Since

(369.45

AG°=—RTInk

0.296 T) x 10° =— 8.314 x Tx In (0.25) =— 8.314 x T x (— 1.386) = 11.526 T

Therefore, T= 1201.36 K = 928°C (i7) Since Pj; = 0.1 atm = (P,, + P;) Due to one mol of each gas is formed, so Py, = Pp, = 0.05 atm Therefore,

Since

k=25x10"

AG°=-RT Ink

(36945-0296 T) x 10°=-8314 x Tx In (2.5 x 10%) =—49 813 T Therefore, T=

1068.35 K = 795°C

Both cases gas composition 50:50.

169

Numericals in Process Metallurgy

Ex C10. For the reaction: CO, + C=2C0, AG® = 170707.2— 17447 T J (a) Find the temperature at which graphite will be in equilibrium with CO and CO, at 1 atm pressure containing i) 30% CO, ii) 60% CO.

(b) Find similar values for a total pressure of 0.1 atm. (c) Find composition of CO and CO, in equilibrium with graphite at

Solution.

CO, +C=2C0

1000 K.

Equilibrium constant, k= (Pro) / Peo,] Total pressure = 1 atm = Pp + Pp, (a) (1) Since 30% CO, i.e, Po, = 0.3, therefore Peo, =1-03=07

Hence, k= (Po) / Peo I=1 (0.3)%/0.7] = 0.1286 and In k=— 2.051 Since AG°=—-RT Ink=—8314xT x (-2.051)=17.054 T Again AG" =1707072-17447T=17054T Therefore,

T=891.31 K = 618.31°C

(if) Since 60% CO, i.e., P, = 0.6, therefore Peo, =1-06=04

Hence, k = [(Peo)’ / Peo, 1 = [(0.6)°/ 0.4] = 0.9 and In k= - 0.105 Since AG°=—RTInk=—8314x Tx (-0.105)=0.876 T Again AG®=1707072-17447T=0.876 T Therefore,

T=973.55K = 700.55°C

(6) Total pressure = 0.1 atm = Pp, + Peo,

(7) Since 30% CO, i.e, Po, = 0.03, therefore P, = 0.1 - 0.03 = 0.07 Hence, k= (Peo) /Pco,) =[(0.03)%/0.07] = 0.01286 and In k =— 4.354 Since AG°=-RTInk=-8314 x Tx (-4354)=36.198T Again AG°=170707.2-17447T=36.198T Therefore,

T=810.31

K =537.31°C

(if) Since 60% CO, i.e., Pep, = 0.06, therefore Pro, =0.1-0.06=0.04 Hence, k= [18 =[(0.06)*/0.04] = 0.09 and In k= — 1.094 Since AG°=-RTInk=-8314x Tx (-1.094)=9.097 T Again AG®=170707.2- 17447 T=9.097 T Therefore,

T = 929.95 K = 656.95°C

(¢) (i) at T=1000 K Since AG®=170707.2— 17447 T =170707.2 — 174.47 x 1000 =— 3762.8 Again AG®=—RTInk=—8314x 1000 x Ink=—8314 x Ink Therefore,

Hence

Since

—37628=-8314xInk

In k=0.4526 and k= 1.572

k= [(Peg) Peo, 1= 1.572

Let Pi, =x and total pressure = | atm = Pp + Py

Or

;50 Pp =1-x

(1 -x)]=1572 FL +1.572x-1572=0

Since roots = [b+ J(b—4ac)}/ 2a] Therefore, x = [-1.572 = V{(1.572) = 4 (-1.572)}]/2 = 0.6938 Percent of CO = 69.38% and percent of CO, = 30.62%

Ex C11. 1f a furnace atmosphere contains 16% CO,, 4% CO and 80% N,,. Does that atmosphere oxidize the nickel at 1000K ?

Given: Ni (s) + % 0, (g) = NiO (s), k, = 5.76 x 107 CO (g) + % 0, (g) = CO, (g), k, = 1.68 x 10'°

170

Metallurgical Thermodynamics,

Kinetics & Numericals

Solution.

Equilibrium Constant, k, = [(ay,0)/{(ay,) * (a)

= [1/po)""]

Consider, Ni and NiO are pure solids, so a,; = | and a, =1 at std state; and a, = py,

Therefore (p,,)"> = [1/k,] =[1/(5.76 x 10" )] = 1.736 x 107° Hence Po,”=3.014x 107

Again

Therefore,

= (Peo) /{Pco) * (ap,)"*}1=1.68 x 10"

en)

Re) ]1=1.68 x 10" x (a, Je

1.68 x 10'% x 1.736 x 10° = 291.67

Actual in the furnace atm [( peg) (Peg) 1= (0. IT; [0.04)=4 Since [(Peg,) / {oly > [Peo {Peo)],. ; therefore the furnace atm does not oxidized the nickel metal.

Ex C12. For the reaction: C (s) + CO, (g) = 2CO (g) at 1100 K and I atm, the equilibrium mixture contains 91.6% CO and 8.4% CO, i volume. Calculate equilibrium constant for the reaction and the partial pressure of CO, gas ifrpartial pressure of CO is changed to 10 ~ * atm, Solution. Equilibrium Constant, k = (Peo) /[{Pco, ) x(apil= (Peo) {Peo

= [(0.916)*/ (0.084)] = 9.989 k=9.989=( 107 (peo) Peo, = [10747 /(9.989)1= 1.0 x 107

Again Therefore

Ex C13. What is the standard free energy change of the following reaction at 1000°C? PbO + Fe = Pb + FeO

Also comment upon the feasibility of this reaction.. Given:

AG"

=— 94.14 kJ/mol at 1000°C

AGS, =— 177.82 kJ/mol at 1000°C

Solution.

Standard free energy change of the reaction, AG® = ZAG", Therefore AG” = [(AG®p, TAG, ) —(AG®po + AG )]

— ZAG”

Reactant

=[(0— 177.82) — (— 94.14 + 0)] = — 83.68 kJ/mol

[Since AG", =0and AG", =0] The standard free energy change of the reaction is negative, hence the reaction is feasible if all the species are at their standard states.

Ex C14. Find out free energy change for the reduction of TiO, by Al at 500 K. Given: AGH, =— 91402827 + 174.77 T J/mol AGY) 0, =-1697699.8 15.69 TLog T +385.85T Solution.

J/mol

Reaction: 3 TiO, + 4Al = 3Ti + 2AL0,

Free energy change of the reaction: AG®, =Y AG®, ,. —¥ AG,

@,=[(30G; + 20GS, 1) - (BAGH, +4AG3,)] {Bx 0)+2 x (- 169769981569 T Log T + 385.85 T)} — {3 x (— 91402827 +17477T) +4 x 0)}] [since AG’; =0and AG, =0]

—653314.79-3138Tlog T+24739T At T=500 K Therefore AG 500 -=-653314.79 — 31.38 x 500 log (500) + 247.39 =x 500 =—695661.63 + 123695 =—571966.63 1=-571.97 kJ

Hence reaction is feasible at temperature 500 K.

171

Numericals in Process Metallurgy

Ex C15. What is the standard free energy of the following reaction at 1000 K.

270, + C+ 24l = 2 + ALO, + CO AG pp, =— 91402827 + 174.77 T Jmol AG yo, =— 169769981569 TLog T +385.85T J/mol AG =-111712.8 - 87.65 T J/mol Solution. Free energy change of the reaction: AG®, = ZAG, — ZAG" Reactant AG, =[(2AG°; + AG’, TAG) = (2AG py + AG + 2AG°, )] =[{(2*0)+(—1697699.8 15.69 TLog T +385.85T)+(—111712.8 —-87.65T)} —{2 x (— 91402827 + 174.77 T) + (1 x 0) + (2 x 0)}] [since AG"; =0 ,AG°. =0 and AG® | = 0] AG® =1864394 1569 TLog T-5134T At T'=1000K AG’, 1499 =—79766.06 J =—T79.77 kJ Hence reaction is feasible at temperature 1000 K.

Given:

Ex C16. Find the change in the percentage of hydrogen in a mixture of 75% H,, 25% CH in equilibrium with carbon, if the total pressure changes from I atm to 0.1 atm. What is the temperature of this equilibrium at 1 aim pressure? Given: CH, =C+2H,, AG®=901652-10945T J

Solution. Equilibrium constant, k = [(Py,) Pen]

[Taking a, = 1]

Total pressure = 1 atm = Py, + Py,

Since 75% H, and 25% CH, i.c., P, = 0.75, and Py, = 0.25 Hence, Since Therefore, Hence

k=[(Py)"/Pcy,] =[ (0.75) / 0.25] =2.25, and In k= 0.8109 AG°=-RTInk=-8314xTxInk 901652-10945T =—8314xTx08109=-674T T = 877.88 K = 604.88°C

Now total pressure changes from 1 atm to 0.1 atm So

Let Since Therefore, Or Or

Since Or Therefore, Hence

Py + Pey, =0.1 atm

Po 5% Poy — 0-H) AG*=-RT Ink 901652-10945T =—8314xTx Ink 90165.2 109.45 x 878 =— 8.314 x 878 x Ink Ink=08130rk=225

(Since T = 878 K)

k=[(Py)1Pey) = [¥/(0.1 -x)] =2.25 +2250-0225=0 Roots = [{—b = (b — 4ac)}/2a] x=[{-2.25+V (2.25% - 4. 1.(- 0.225)}/2.1] x=0.096, so H, =9.6%

Therefore the change in the percentage of hydrogen in a mixture of 75% H, , 25% CH, in equilibrium with carbon, if the total pressure changes from 1 atm to 0.1 atm is 9.6%. Ex C17. Solid TiO, is being converted into gaseous TiCl, by treatment with Cl, gas in presence

of carbon. Calculate the thermodynamic efficiency of utilization of Cl, gas at 1 000°C. Total pressure is 1 atm.

Given.

TiO, (s) + 2 C (s) + 2 Cl, (g) = TiCl, (g) + 2 CO (g), AG" =-318 kJ /mol at 1000°C.

172

Metallurgical Thermodynamics,

Solution.

T=1000C.=1273K

Since

AG'=—RTInk

Therefore,

318000 =—-8314

So

Kinetics & Numericals

x 1273 x Ink

In k= 30.046, hence k= 1.119 x 10" 2

Since

2