CHM103 Introductory Physical Chemistry I PDF

NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: CHM 103 COURSE TITLE: INTRODUCTORY PHYSICAL CHEMISTRY I CHM 103: Introductory Physical Chemistry 1 COURSE DEVELOPMENT Course Developer Dr. Temi Busari UNIVERSITY OF LAGOS. AKOKA. LAGOS. Course Writer - Dr. Temi Busari UNIVERSITY OF LAGOS. AKOKA. LAGOS. Course Coordinator - A. Ikpe NATIONAL OPEN UNIVERSITY OF NIGERIA. LAGOS. Programme Leader Dr. F. Peters NATIONAL OPEN UNIVERSITY OF NIGERIA, LAGOS. National Open University of Nigeria 14/16 Ahmadu Bello WaY. Victoria Island. Lagos. AbuJa Aru1cx Office 245 Samuel Adcsujo Ademulegun Street. Central Busmess District. Opposite Arewa Suites. AbuJa. E-mail: ccntrainfo:illnou.cdu.ng URL 1\WW.nou.cdu.ng ©National Open University of Nigeria 2004 First Published 2004 ISBN 978-058-132-4 All Rights Reserved Published by Unzversity Press PLCfor National Open Univemty o(Nlgeria Contents Course Guide l MODULE 1: INTRODUCTION 6 Unit I: Units and Dimensions 6 Unit 2: S\ Prefixes 13 Unit 3: Separation Techniques 20 MODULE 2: GASEOUS STATE OF MATTER 27 Unit I: Gases 27 Unit 2: Ideal Gases 34 Unit 3: The Kinetic Theory of Gases 42 Unit 4: Real Gases 52 Unit 5: Liquefaction of Gases 63 MODULE 3: SOLID AND LIQUID STATES OF MATTER 73 Unit 1: Solid States and Solid Types 73 Unit 2: Nature of Bonds in Solids 81 Unit 3: Structures of Crystals 88 Unit 4: Liquids 94 Unit 5: Other Properties of Liquids 102 MODULE 4: SOLUTIONS AND PHASE EQUILIBRIA llO Unit 1: Types of Solution 110 Unit 2: Solutions of Solids and Gases in Liquids 117 Unit 3: Binary Liquid Solutions- 1 123 Unit 4: Binary Liquid Solutions- 2 131 Unit 5: Partially Miscible Liquids 140 Unit 6: Some Colligative Properties 149 Unit 7: Phase Rule 157 Ill Course Guide Table of Contents Page l.O Introduction 2 2.0 What you will learn in this Course 3 3.0 Course Aims '_, 4.0 Course Objectives 3 5.0 Working through the Course 4 1.0 Introduction Chemistry deals broadly with the study of matter. There are different aspects of chemistry in \\hich matter is studied. These arc Phvsical. Inorganic. Organic and Analytical. In fact. the classification has been further broken d0\\11 in such way that details of matter in terms of structure.· periodic1ty and energy can be well understood. Phvsical ChemistrY is thus one aspect. Physical Chemistry deals with a wide variety of topics from the states of matter to the reaction rates. The modem approach to chemistry is to deal with the following three aspects of matter. Structure Equilibrium Properties Ability to change In physical chenllStry. we deal with the remmnmg two aspects. namely. equilibrium properties and ability to change. Physical Chemistry is one of the timdamcmal courses for Bachelor's Degree Programme in science (B.Sc.). and science education (B Sc Ed ) in am· UniversitY in the whole world. This course will also be suitable as an elective for anyone who intends to study chemical cngincenng and other science-related professional programmes since it deals with the fundamental princ1plcs goveming chemical transfom1ations. There arc manY phYsical chemistry courses you wiII encounter whilst studying tor 'our degree. The\ arc coded as CHM ·s. This present course we :m; discussing its guide. is coded CHM 103It 1s a two Credit unit course. There arc four modules consisting of 20 units overall. The lirst module starts with a d1scussion on introduction. TillS is to give you an msight mto the plnsical and mathematical instruments that ,-ou will be expected to equip yourself with for a better performance in chemistrY in general and phvsical chemistry in particular. The second module focuses on gaseous state of matter. The studv of the properties of matter in solid. liquid and gas fom1 is essential since the substances that take part in a chemical reaction are found in one of these three states of matter. Because of the cop1ous nature of gas principle. it was found essentml that a whole module be devoted to it. The third module cxammes the sohd and liqu1d states of matter. The structural compansons. properties and nature of bonds m these two states of matter arc explamed In the fourth module. the concepts of solutions arc discussed. Different solution tv pes and especmlly the binary liquids arc concentr tcd upon. Details of the unit> in each of the module arc prov1ded 111 a table under sect1on S of this gu1de The course though adapted trom IGNOU course materials. has been grcatlv transformed and new concepts in phvsical chemistrv added to suit students in Nigeria. You \\ill be required to practice as much as possible on \'Our own. the practical and self assessment questions termed as exercise provided in each of the modules. To assist you 111 accomplishing the course objectives. there will be some laboratory facliillcs m NOU for you to use and a tutor to facilitate your self mstruct10n; you must also make tremendous effort to carry out most of the activities and questions rmsed m the modules. You are free to usc other readmg materials. suggested in each module.or an e-mail facility. telephone. and even the Internet to get more knowledge about the course. 2 The course guide provides you \\ith additional information on what the course aims. objectives and learning procedures are for ·ou to maximize the effective usc of the four modules on Introductory Physical Chemistry I. CHM 103. 2.0 What you will learn in this Course While we realize that the study of physical chemistry is quite vital and enormous. we will begin with the only introductory aspect of physical chemistry at this level. Specifically. in CHM 103. you will learn about system units. states of matter. nature of bonds. solutions and phase equilibria. The remaining aspectof physical chemistry like thermodynamics. thermochemistrY. equilibrium and electrochemistry that we are unable to cover will be addressed in future courses. And in physical chemistry CHM 224 course. other advanced concepts of thermodynamics. photochemistry. electrochemistry. ionic equilibria. etc. will be addressed during the semesters of 200 level. More importantly you will learn about the significance of units in solving problems in chemistry. appropnate usage and methods of handling apparatus and data, describing and comparing behaviours of different matter based on their stmctures and energy. 3.0 Course Aims The aim of this course is to create an in-depth awareness and develop full understanding of the concept of energy as the capacity for change in any chemical reaction and moreover that strong relationship between energy and structure of matter is the foundation of physical chemistry. Thus, the aim of this physical chemistry course is to: I. Discuss the mathematical viabilities and standard procedure required in effeclivc study of physical chemistry. 2. Explain the basic theories that govern various properties of matter. 3. Discuss some aspects of physical phase equilibria. 4. Build cases for the feasibilities of chemical reactions. 5. Demonstrate the various energy relationships in chemical reactions. 6. Give you an appreciation of system at equilibrium and fuctors affecting such systems. 4.0 Course Objectives In order to achieve the course aims. there are some overall objectives set for the course. Moreover. each module and each unit has their respective objectives which you and your course tutor must constantly refer to. so that no objective is skipped or unachieved. It is therefore important that you outline what you have been able to achieve after completing a unit and compare the list with the umt objccttves. This will help you to ascertain whether you have accomplished what ts reqUired of vou. All the module and unit objectives are specificities of the course objectives. The course objectives are stated as follows: Familiarize yourself with basic instruments for physical chemistry laboratory work: Write and apply correctly Sl units· Measure in and convert other units using Sl units: Explain and compare the main features of solid. liquid and gas states: Appreciate the reasons why real gases deviate from gas laws: Analvse the properties of completely. partially miscible and immiscible liquids: Discuss the essence of types of solutions: Analyse phase changes in matter: Describe the conditions regulating the solubility of liquid solutions. 3 5.0 Working through this Course This course contains some packages that you will be given at the beginning of the semester having satisfied the conditions of admission and registration. One of them is the course materials. The details of what constitute course materials will be discussed next. But let us remind ourselves that there are two areas expected of you to fulfil at the end of your studying this course. These are your full participation in both the continuous assessment and the final written examinations. Though there is a practical course developed for this level. it is advisable that some of the sections can be practicaliscd with the assistance of the course tutor. Your tutor will be readily accessible to issue out the exercise you are to do and submit according to spcc1fied rules and regulations from the authority. For you to be considered as successful in this course. vou will be expected to pass at average level both the continuous assessment and the final wntten examinations. Thus. certain materials have been packaged for you that will enable you prepare adequately for both the continuous assessment and the final written examinations First and foremost is the study unit materials that consists of 20 units in all. However. the units have been packaged for you in modules as shown below. Study Units Number Module Unit Title of Module I Introduction I Units aJJd Dimensions 2 Sl Prctixes 3 Separation Teclmiques 2. Gaseous State of Matter I Gases 2 Ideal Gas ' The Kinetic Theory of Gases ·' 4 Real Gases :i Liquefaction of Gases 3. Solid and Liqiud States of Matter I Solids States and Solid Types 2 Nature of Bonds in Solids 3 Structure of Crystals 4 Liquids 5 Other Properties of Liquids 4 Solutions and Phase equilibria I Types of Solutions 2 Solution of Solid and Gases in Liquid 3 Binary Liquid SolutionsI 4 Binary Liquid Solutions 2 5 Partially Miscible L1qmds 6 Some Colligative Propert1es 7 Phase Rule 4 By and large, you should be able to complete this two-Credit unit for about 15-17 weeks in a Semester. Well spread out in each of the unit are introduction to the unit. specific objectives. reading material on sub-topics. self-assessment questions (SAQs). some exercises, conclusion. summary. tutor-marked assignments (TMA) and references. Course materials Major course materials of the course are as follows· I. Course Guide: This looks like a blueprint that spells out what constitutes the course itself It also extends to the basic infommtion you require on how best to study this introductory physical chemistry. 2. Study Units: Each of these provides an overview of the content and number of units that will be covered in this course. The contents are spelt out in seven major headings: the introduction. objectives to guide what to focus on, the content. conclusion. summary of the content. references and other materials that will facilitate your understanding and the tutor-marked assignments. In the body of content are exercises that should always attract your attention after every reading. 3. Assignment files: These files contain challenging tutorial questions termed as Tutor-Marked Assignments (TMAs) that will enable you to assess yourself at the end of every assignment that will be handed out by your tutor. Since we arc operating a course unit system. your scores in these assignments will be stored. the cumulative of which shall be used to compute vour cumulative grade score n this course. 4. Presentation schedule: Certainly. the modus operandis (e.g time table. hours expected on each unit/module. assigrunent submission procedure) on how it will be self-tutored with the best monitoring techniques bv the National Open University of Nigeria will be in the information package of this schedule. Study Units Details of the study units have earlier been presented. It is spelt out in modules with corresponding units and titles. You will be expected to spend 2-3 hours in studying a unit. The exercise items are provided to enable you test your ability in whatever you have covered and to help you monitor the progress made per sub-topic in the unit. There is also the TMA packaged for you but will be given as assignments. References and other resources Apart from this study unit. some reference materials are provided as additional reading materials to support your study. It is not the responsibility ofNOU to provide them but you may be lucky to spot them in the library instituted by NOU. Instructional media As an open and distance Leaming University. several and relevant multimedia that can make leaming easy are available. Accessibility to other instructional media \\1ll be disclosed to you from tune to tune Assignment file This has been discussed earlier. It is mandatory to always tum in your assignments to any tutor assigned. Assessment You will be expected to complete at least ten assignments by the end of the course. Some of these will be in form of a project. continuous assessment (CA) written tests. Each of these ten assignments shall carry ten marks making I 00% for continuous assessments. You will also be expected to write a final examination in the course. The overall score in the course will be a sum of 40% of CA and 60% of written examination. You will be expected to have 50% in the CA and 50% in the written exanunation. anything short of this will count as a failure. MODULE 1: INTRODUCTION Unit I: Units and Dimensions Table of Contents Page 1.0 Introduction 7 2.0 Objecttves 7 3.0 Units of Physical Quantilies 7 3.[ Concept of Sl 7 3.2 Basic Units 7 3.3 Derived Units 9 4.0 Conclusion 12 5.0 Summary 12 6.0 References and Other Resources 12 7.0 Tutor-Marked Assignment 12 6 10 Introduction In your previous course of studying science in general and chemistry in particular. you would have realized that you are often engaged in problems solYing. Specifically in physical chemistry. equations containing physical quantities are eminent in problem solving. For you to be at case with these mathematical expressions and applications. you will need to understand how to represent one single physical quantity in terms of its magnitude and umts. In this Unit, we shall examine how.1ysteme mternatwnalc (SI) came about. We will study the principles of representing the units of the physical quantities. In this unit also. we shall state the SI umts for a few basic and denvcd quantt!Ics 20 Objectives By the end of this Unit, vou should be able to: Explam the need for SI units State basic and derived Sl units Define the basic and derived units Derive the dimension and the SI unit of a physical quantity using an appropnatc equation 3.0 Units of Physical Quantities 3. I Concept of Sl Till recently in the scientific world. mainly two systems of units have been in common use. One is the metric system. that is. c.g.s (centimetre. gram and second) which was more commonly used over the European continent and the other is f.p.s (foot. pound and second) prevalent in England. A common system of units helps in exchanging the scientific facts and ideas originating from different countries. It is better still if the system of units could be derived from the scientific formulae or fundamental constants. This long felt need for a common system of scientific units was realized at a meeting called General Conference on We1ght and ll1easures m 1960. At this meeting. the international scientific community agreed to adopt common units of measurement known as International System of Units. This is abbreviated as SI units from the French name. Systeme lnternatwnale. 3.2 Basic Units There are seven basic physical quantities from which all other physical quantities can be derived. Can you mention two of those basic physical units you are familiar with? You are right if length. time. mass. temperature are parts of what you stated in your answer. Each of these physical quantities has a unit termed basic unit. with a specified symbol. Table 1.1 presents in details the names of these physical quantities along with their symbols of SI units. Each of these quantities is regarded as having its own dimension. The dimensions of basic quantities are useful in defining the derived phystcal quantities. which wwould use in Sec 3.2. We will be using the symbol given in column (II) of Table 1.1 to refer to the dimensions of the basic quantities. 7 Table 1.1: Basic Physical Quantities and their Units Physical quantity Symbol of the quantity Name of SI unit Symbol of SI unit Length I metre m Mass m kilogram kg Time t second s Electric current I ampere A Temperature T Kelvin K Luminous Intensity I candela cd Amount of Substances n mole mol Note that m (italized) is the symholfi>r mass olan oh1ecrwhile m (roman) is the symbol olSlunitfor metre. Before vou proceed, have you ever thought of what these basic quantities are? Assuming you can not remember what length. mass. time. temperature. electric current. luminous of intensity and amount of substancarc. thelf definitions arc briefly stated here. Recall that length is a measure of distance or duration of time. Let us consider temperature as a measurement of the hotness or coldness of matter. Have you ewr thought of the relationship between temperature and heat'' What is heat and what is temperature'' \\I tc"'perature and pressure is called critical volume (V,). To explain the phenomena better. a series of experiments performed by Andrews "ill help vou out Andrews pcrfom1cd a series of experiments and obtained isothcm1s (p against V plots at constant temperature) for carbon (iv) ox1dc. TI1e results obtained bv· him arc shown in Fig 4.4. [J s ' ' ;--;- -- ---\ -·-------;I{ ' \ VIm Fig. 4.4: Isotherms of Carhon (IV) Ox1de Finally. let us go through the methods of liquet:J.etion of gases and study the intermolecular forces in general. At high temperature. the 1sothcrm IS a hyperbola (curve l) in accordance with Boyle's law. At low temperatures. the isothenns (the curves IL lii and IV) show considerable deviation from ideal gas bchav1our. The 1Sothem1 at 304.2K (curve II) rema111s horizontal for a certain value of pressure The 58 two falling portions, EFO and OGH. of curve II meet at point 0. The point 0 is known as the critical point; the temperature and pressure at this point arc kno\\n as criticial temperature, T, and critical pressure, p,. 3.4 Critical Constants In this section we would examine how critical temperature (T,), critical pressure (Pel and critical volume (V,) are related to Van der Waals constants. 3.4.1 Critical Constants and Van derWaals Constants The Van der Waals equation (Eq. 3.6) can be made the basis of a theoretical consideration. The curves in Fig. 4.5, known as Vander Waals isotherms, sho\\ the isotherms 'calculated on the basis of this equation. Expanding Eq. 3.6 we get, (p + a!V.,2) (V.,- b)= RT ············································ 3.11 i.e., pV.,- pb +a!V.,- abN., 2 = RT Multiplying the equation throughout by Vm2/p, we obtain, V3.,- bV.,'+ aV.,/p--ab/p = RTV.,2/p.3.12 ie., V.,3 - V.,2 (b + RT/p) +alp V.,- ab/p = 0 3.13 The cubic equation will yield three values for Vm corresponding to a given presure and temperature. All the three valus t>f V m may be real or one may be real and the other two may be complex conjugates. Isotherms III and IV do yield three values ofV., in certain ranges of p and V This is true in general for all isotherms below the critical tempemture. Curve JI corresponds to critical temperature and curves at higher temperature (such as curve I) approach the isotherm representing the Boyle's law. These theoretical curves are similar to those obtained by Andrews for C02 but a maJor difference is the wave like portion BCDEF in the theoretical curves. If experiments are performed without perturbance, then portions BC and EF are realisable, these portions represent the supersaturated vapour and superheated liquid, respectively. Fig. 4.5: fim.rier Waals isotherms The wave like portion decreases as temperature increases " 62 Unit 5 Liquefaction of Gases Table of Contents Page 1.0 IntroductiOn li4 20 Objecti,cs 64 3.1 Liquefaction of Gases li4 3 I I Lmde·s Method li4 3.1.2 Claude's Method 65 32 Intermolecular Forces lili 21 Van dcr Waals Forces lili 3.2.2 Total Interaction Energy 6X 3.2.3 Hydrogen Bonding 33 Effect of Molecular Interactions on Ph\'sical Properties 6Y 40 ConclusiOn 71 5u s ununaJ'\ 71 6.0 References and Other Resources 70 Tutor-Marked Assignment 63 1.0 Introduction In the last Unit. we examined the extent to which real gases deviate from ideal gas behaviours. Some explanations to-the behaviours of real gases using Van der Waals equation were made. In this Unit. let us examine liquefaction ()(gases. We will go through the methods of liquefaction of gases and stud' the mtermolecular forces m general. 2.0 Objectives. By the end of this Unit, you should be able to: Define liquefaction process State the princlp)es 0rliquefaction methods Explain the nature of intermolecular forces. and Discuss the effect ot mtermolccular forces on the condensation of gases into liquids and solids. 3.1 Liquefaction of Gases We have learnt about the liquefaction of air. liquefaction of petroleum gas and even coal. Let us quickly look at the concept ofliquefuction. What docs it mcan'l Do you say it is concerned with obtaining liquid from gas or solid'' You arc right. Liquefaction is a process of making a gas or solid to become liquid. How can we make a solid to become liquid''We can make a solid become liquid by melting and a gas can be changed to liquid bv condensation. The critical phenomena and our knowledge of the critical constants are pmctically useful in the liquefaction of gases. The liquefaction of air is important in the rnanufucture of nitrogen and oxygen which are both important industrial chemicals. The liquefied petroleum gas (mixture of propane and butane) is used as a domestic fuel. Liquid helium and nitrogen arc particularlv important for making the materials superconducting. Easily. liquefiable gases such as ammonia and dicWorodifluoromethanc (Freon) are used in refrigeration and air conditioning. Let us now study some methods of liquefaction of gases. It has alreadv been clarified that a gas cannot be liquefied above its critical temperature. Many substances like water, ethyl alcohol. etc. have high critical temperatures and hence exist as liquids even at room temperature. Others like ammonia. sulphur (IV) oxide, etc.. under ordinarY conditions are above their critical temperature but can be easily liquefied by cooling using freezing mixtures under moderate pressure. This implies that the freezing mtxture lowers the temperature of a substance below its critical temperature and the moderate pressure is then sufficient to liquefy the gas. On the other hand. there are many gases like oxygen. nitrogen. hydrogen and helium whose critical temperatures are much lower. Special methods are adopted to cool these gases below their criticaltempcrature. Let us study the principles of two of the common methods of liquefaction. 3. 1. I Linde's Method The method is bas -d on the principk kno\\n as Joule-Thompson effect. According to this effect when a gas under high pressure is :J.IIO\\ed to expand into a region oflow pressure, its temperature falls. The gas does not do any external work but the kinetic energy of the molecules expend somwork to overcome the attractive forces between the molecules. As a result. the temperature of the gas is lowered. A precaution is required m this process. Do vou know why'' It is because to have a eoolmg effect. a gas IS to be brought below a charactenstic temperature. kno"n as mversiOn temperature. before allowing It to expand. If the temperature of the gas IS above Ils mvcrs1on temperature. Joule· Thompson c:-;pcmsiOn results 111 heatmg 64 The schematic diagram of the equipment used is shown in Fig. 5.I Fig. 5.1: Liquefaction usmg Linde."v merhod The gas at a temperature lower than its inversion temperature is compressed usmg a compressor. This gas is then allowed to expand through a valve which results in its cooling. The cold gas is used in cooling the high-presured gas in the heat exchanger and is recirculated through the compressor. It gets cooled still further. as it expands. The c,·clc continues till the liquetied gas drops from the throttle 3.1.2 Claude's Method Claude"s method (Fig. 5.2) is more efficient than Lindc·s method. The compressed gas in the insulated vessel (i.e.. under adiabatic conditions) is partly used to do work against a piston in a cylinder and partly expanded through a valve. The cooled gas obtained by adiabatic expansion is used for cooling the incoming gas in the heat exchanger. The process is repeated till the gas is liquefied. Compr sor Fig. 5.2: Liq11efaclion usmg Claude\"s method Using the principles of Linde·s method. answer the following question· Exercise I If Hydrogen gas is allowed to undergo Joule-Thompson expansion at room temperature, it is heated but not cooled. Explam. 3.2 Intermolecular Forces In section 3.3 of Module 2. Unit 4. we studied that the real gases deviate from the ideal gas behaviour because of the presence of intemJOlccular forces. The intem10lecular forces are also responsible for the conversion of gases into liquids and solids. Van der Waals not only derived an equation to explain the behaviour of real gases but also to develop a model that would explain the behaviour of liquids. In recognition of his work. the weak intennolecular forces in liquids and solids are often called Van der Waals forces. 3.2.1 Vander Waals Fot·ces Van der Waals forces include:- (i) D1 pole..\ _,I Fig. 3.5: (a) hcp arrangement - the atoms in the thmllayer are straight above those in the lirsr layer called ABAB arrangement: (b) ccp arrangement - the atoms in rhe third layer arranged di{{erentv from those in the first.. called ABCABC. arrangement. 3.3.2 Extrinsic Semiconductors Semiconductors with impurities are called extrinsic semiconductors. The,· can be class1tied into n - type and p-t ·pe semiconductors. n-type Semiconductors When a semiconductor is doped with an impuritv having more valence electrons than those in the semiconductor. an-type semiconductor is produced. Such an impurity can donate clectron(s) to the valence band of the semiconductor. and is called a donor. Phosphoms. arsenic or antimonv (each having five valence electrons) arc examples of donor impurities added to germanium or silicon (each semiconductor having four valence electrons). The addition of donor impurity to the semiconductor provides additional energy levels and if they arc righth- related to the bands of the semiconductor. conductivity may result. That is. if the impurity contains a filiienergy level just below that of an emptY band in the semiconductor. the electrons from the impurity go to the empt,· band in the semiconductor. Hence. it becomes negatively charged (n-type). Up to certain temperature. the conductivity ofn-type senuconductor mcreascs with mcrcase in temperature. p-type Semiconductors When the impurity used for doping has less valence electrons than the semiconductors. the impurit\ can accept electron(s) from the valence band ofthe semiconductor Such an impurity IS called accepwr The addition of boron. alummlllm, galhum or indium (each having three valence electrons) to silicon or 92 gemanium (each semiconductor having four valcnct; el ctrons) is an example ofthi$ type. The e sential · · feature is that the inipurlry must containan empty energy level just above a full And in_tbc semiconductor The ek-ctrons from the full band in the semiconductor will pass to the empty level of the impurity: passage of electron from the semiconductors to impurity makes the fanner pesitively charged (P- type). The effect of temperature on the conductivity of a p-tyPe semiconductor is similar to that of n- type semiconductor. The combination of p-type and n-typc semiconductor is called p--n j!mction. The p-n junctions are used·as rectifiers. solar cells. light emitting diodes and other electronic devices. Exercise 2 Differentiate between intrinsic and extrinsic semiconductors...· ---.' _. ·-·.'......_...................... ··-· -.····':........--;j,,.- - J.'.· :-................. 4.0 Conclusion In this Unit. we have examined different crystal trpes. We focussed specifically on ionic. covalent and molecu Jar crystals. The semiconductors were also discussed. The uses of two types of semiconductors were described. For the next Unit. we will be discus-sing liquids. 50 Summary At the end of this Unit, you have learnt that Crystal solids can be classified into ionic, covalent, molecular and metallic. The ionic solids are body-centred cubic (bee) and face-centred cubic (fcc). The molecules in molecular solids are held by Van der Waals interaction. Most metals have fcc, bee and hcp lattices. Extrinsic semiconductors are insulators with impurities under normal conditions but their electrical conductivity increases with temperature. 6.0 References and Other Resources IGNOU ( 1997). States of Matter: Physical Chemistry CHE -04 New Delhi._ ·· ! -. 7.0 Tutor-Marked Assignment I. IdentifY the type of attractive forces (or bonding) mainly responsible for crystal bonding in the following cases: (i) diamond (ii) potassium bromide (iii) aluminium (i\:)helium 93....... These properties are related to the strength of intermolecular forces in liquids. We now discuss surface tension and viscosity ofliquids. Surface tension The presence of a surface in a liquid gives rise to the phenomenon of surface tension. Let's see how it arises. In the absence ofe,1emal forces. liqwds form spherical drops spontaneouslv This is facilitated by the fact that for a given volume. a sphere has a smaller surface area than any other shape. Th1s fascinating phenomenon is one of the reasons for the spherical shape of earth. sun. moon. etc. Let us explain the origin of forces operating to minimize surface area. All the molecules surrounding it attract a molecule in the interior of a liquid. It is pulled equally in all directions. It is important to note that a molecule at the surface of a liquid is attracted only by molecules belm' It (Fig. 4 2) Fig 4.2: Molecules in the hulk and on the surface of" liquid hemg attracted by neighbours Therefore. the molecules on the surface of the liquid are dra\Vn inwards trying to minimize the surface area. Because of this tendency of a surface to contract, each point on the surface of the liquid is under pressure like a stretched rubber membrane. The resistance of a liquid to increase its surface area is correlated to its surface tension. It is defined as the energy required for increasing the surface area by one unit by moving the molecules from the interior of the liquid to the surface. It is also defined as the force per unit length perpendicular to a liquid surface. Corresponding to these two dcfimtions, SJ units of surface tension arc J m _,and N m·' (which are of course, equivalent). It is represented by the Greek letter y. Increase of temperature increases the thermal motion of the molecules in a liquid. This opposes the effect ofintem1olecular forces. Thus as temperature IS raised. the surface tension decreases. The value of surface tension of some liquids are given in Table 4.1. Table 4.1: Values of surface tension (y) of some liquids at 293 K Liquid 102 X yiN m·' Water 7.28 Benzene 2.89 Carbon tetrachloride 2.64 Chloroform 2.67 Mercury 46.5 98 Some of the factors, which influence the magnitude of surface tension. are given below: Molecules having strong hydrogen-bonds have high surface tension. The surface tension of water. for example. is about three times higher than that of non-polar liquids like carbon tetrachloride. Metallic bonding also lead to high surface tension. For example. the surface tension of mercury is more than six times that of water. The dispersion forces are quite significant in molecules with large atoms and are often more important than dipole-dipole forces. In fact. surface tension of carbon tetrachloride is only slightly less than that of chloroform: the effect of London forces in the former is nearly equal to the combined effect to London and dipole-dipole forces in the latter. Intermolecular forces give rise to capillary action. It is the rise of liquids through a capillary (narrow glass) tube (Fig. 4.3a). Two types offorces- cohesive and adhesive are responsible for this property. The cohesive forces are the intermolecular forces among the molecules of a liquid that we discussed in Module 2. Unit I. Adhesive forces exist between the liquid molecules and the molecules in the capillary walls. For example. glass contains many oxygen atoms: each oxygen atom (with partial negative charge) attracts (the positive end of) a polar molecule. such as water. The adhesive forces enable water to ·wet" the glass. The adhesive forces acting upward pull up a water column inside a capillary tube when the latter is in -contact with water. The height of the water column inside the capillary tube is such that the adhesive forces acting upwards balance the cohesive forces (in the form of weight of water column) acting downwards. The height of the water column inside the capillary tube has been found to be inversely proportional to the radius of the tube. Hence only in tubes of small radius. the capillary rise is meaningful. The concave shape of the meniscus of water in a glass tube indicates that the adhesive forces of water towards the glass are stronger than its cohesive forces. A metallic liquid such as mercury (Fig. 4.3b) shows a lower level in a capillary tube and a convex meniscus. This behaviour is characteristic of a liquid in which the cohesive forces between its molecules are stronger than the adhesive forces between the molecules and glass. (a) (b) Fig. 4.3: (a): A polar liquid such as water rises in a capillary htbe -- > later has concpve meniscus in a glass (b): A metallic liquid such as mercury shows a depression of level- mercury has c/JnVe%-meniscus Viscosity Another property ofliquid that depends on intermolecular forces is viscosity; it is a measure of the resistance t.o flow. A liquid, which has higher viscosity. flows slowly. It is represented by the Greek letter TJ (eta). Its unit is Pas. It decreases with temperature. The viscosities of a few liquids are given in Table 4.2. 99 3.2 ComparJson of Liquids wlttr Gases and Su1ids We can oblnotqtiite·sharp): The diffuse diffraction pattern makes· it clear that the order m the arrangement elf partiCleS is only partial but not 96 total. Experimental data indicate that the first few neighbours of a particle in a liquip are at fairly well defined distances: the neighbours farther away are randomly distributed. Tllis 'n'teans thai the arrangement of particles in a liquid exhibits short range order and long range disorder. The number of nearest neighbours around the particles in different regions of a liquid is not the:Sain\f A model !Or the structure ofliquids is shown in Fig 4.I. :\-rav dillfa..:til)n is the s..:atl 103 4.0 Conclusion In this Unit. we have expressed a solution as a homogenous mixture of two or more substances. the three phases at which solutions are fanned namelv solid. liquid and gaseous arc also discussed. Solutions can be divided mto nme types depending upon the physical state of the solute and the solutiOn The compositiOn of the solution can be measured Pressure has onlv a small effect on the solubility of solids m liquids. The solubility of a gas varies \VIth pressure. All these two phenomena are also d1scussed m th1s Unll 5.0 Summary We can summarise what we have learnt as follows A phase is a homogeneous part of a system which is susceptible to change on certain conditions Different solutions ex1st but can be d1fferentiated mto solid solution. liquid solution and gaseous solutiOns The solubility of a gas in a liquid depends upon temperature. pressure and nature of the gas and the solvent 6.0 References and Other Resources Goldberg and Dillard. C. (I '174) C allege Chemistry 3rd Revised Edition New York Macmillan Publishmg Co IGNOU ( 1993) Solution and Phase Equilismg- Phvsical Chcmistrv 2 CHEM -04 New Delhi. 7.0 Tuto ·-Marked Assignment I The vapour pressures of pure liquids A and 8 at 300 K are 2.o x I o· Pa and o x 10' Pa. respect1vclv. Calculate the mole fract10ns of A and 8 in vapour and hqu1d phases of a solutiOn when the eqUilibrium total vapour pressure of the binary liquid solution 1s 4.5 x Ill' Pa at 300 K Assume that the liquid and the vapour are ideal. 2. The solubilitv of carbon (IV) o,;ide in water at 29X K is 3.40 x I o-: mol dm-'. The partial pressure ofCO,over the solution is 1.00 x 10' Pa. Assuming that one dm' of the solution contains UlO kg of water calculate the Henrv"s law constant for carbon (IV) o,;ide. 122 Unit 3: Qinary Liquid Solutions- 1 Table of Contents Page 1.0 Introduction 124 2.0 Objectives 124 31. Binary Liquid Solutions 124 r 3.2. Raoult's Law 124 ' 3.2.1 Ideal Solutions us 3.2.2 Non-Ideal Solutions 125 3.3. Ideal Solutions ofVolatile Solutes 127 3.4 Raoult's Law Curves 127 4.0. Conclusion 12'! 5.0. Summary 129 60 Tutor Marked Assignment 130 1.0 Introduction In Module 4, Unit 2, we considered solutions of solids and gases in liquids. This Unit is devoted to solution of liquids in liquids because you will be dealing mostly with this type of solution. And it is important to understand their behaviours. In the liquid - liquid type of solution, we will consider onlv binary liquid solution at this level of your programme. We will discuss the differences between ideal and non-ideal solutions and explain which of these obey Raoult"s law 2.0 Objective By the end of this Unit, you should be able to: Explain the concept of binary liquid solution State Raoult's Law Distinguish between ideal and non-ideal Discuss Raoult's Law curves. 3.1 Binary Liquid Solutions You must bve heard of binary compounds. What arc they'' They arc compounds that are compose'd of only two elements. These clements could be a metal and a non-metal and they can also be metals. In the cascofbinarv liquid solution. make an attempt to explain what it means. Binary liquid solutions arc liquid solutions composed of two liquids. When two liquids A and B arc mixed completely there are three possibilities: (i) Liquid A is completely miscible with liquid B in all proportions (e.g. 1\atcr and ethanol. toluene and benzene, etc.). (ii) Liquids A and Bare only partially miscible in each other (e.g. water and phenol). (iii) Liquids A and B are completely immiscible with each other (e.g., water and palm oil). We shall study completely miscible liquid systems in this Unit. The other two types will be taken up for study in Unit 5. Completely miscible liquids When liquid A is completely miscible with liquid B in all proportions. A miscible liquid solution is obtained. Let us consider examples of such liquids. Assuming water and ethanol mixed to fonn a liquid solution, a water ethanol solution is formed. Another example.is toluene and benzene. Now I want you to make a list of5 miscible liquid solutions. Many more of this type of binary liquid solution can be obtained. In the next Unit. we will examine other possibilities. Next is a discussmn on Raoult's law. 3.2 Raoult's Law Consider a binary solution containing two liquids A and B which arc completely immiscible with each other in all proportions. In such solutions. the terms. solute and solvent, can be interchanged for the two components. lnl880, the French chemist, Raoult found that when a non volatile solute is dissolved in a solvent (liquid), the vapour pressure of the solvent decreases. If the solute and the solvent are both volatile the vapour pressures of both the components arc decreased. Based on these observations. he gave the following generalization which is called Raoult's Law. 124 The partial vapour pressure of any volatile component in a solution is equal to the product of the vapour pressure of the pure component and its mole fraction in the solution. If x and x are the mole fractions of the components A and B in the liquid solution and p"A and p""' the vapour preBssures of these components respecti.vely. then accor dm ' g to Raou It,s Iaw, p A = p"A X A...............................................................,....,..... 3.1 3.2 If the vapour behaves like an ideal gas. then according to Dalton's law of partial pressure. the total vapour pressure. p. is given bY 3.2.1 Ideal solutions A solution that obeys Raoult's law at all concentrations and at all temperatures is called an ideal solution. Two liquids A and B on mixing fonn an ideal solution. If (i) The molecules of A and B have similar structure and polarity, and (ii) The intennolecular attractions between A and A. Band B, and A an B are alike. Thcnnodynamically. ideal solutions are those in which there is no volume change (6V,,,,. =0) and enthalpy change (6H"'''"' = 0) when two liquids A and B are mixed. Thus the characteristics of an ideal solution arc: (i) It must obey Raoulfs law (ii) 6H nliXIJ1!!. = 0 (iii) t.Vmtxm = 0 Some examples of nearly ideal liquid mixtures arc (i) Ethylene bromide and ethylene chloride (ii) n-hexane and n-heptane (iii) Benzene and toluene (iv) n butyl chloride and n-butyl bromide (v) Carbon tetrachloride and silicon tetrachloride. 3.2.2 Non Ideal Solutions Many of the completely miscible liquid pairs form non-ideal solutions. TI1esc solutions do not obey Raoult's law. They either show positive deviation (When the vapour pressure of the solution is higher than that of an ideal solution of the same concentration) or negative deviation (When the vapour pressure of the solution is lower than that of an ideal solutiOn of the same concentration) from Raoult's law. In such solutions. 125 When the components of a non-ideal solution are mixed, a considerable change in volume and enthalpy is noticed. Thus the characteristics of non-ideal solutions are as follows: (I) They do not obey Raoulfs law (II) till""'"" etc. 0 < nn "'v"'""' "' o Examples of non-ideal solutions showing positive and negative deviations are given in Table 3.1. Table 3.1. Liquid Pairs Showing Deviation from Raoult's Law Positive deviation Negative deviation H,O + C,H,OH H,O + HC1 CH,COCH, + C,H,OH H,O + HNO, C,H, OH + C,H, H,O + H2 SO, CH,COCH, + CS, CH,COCH, + CHCI, Let us work out an example illustrating the application of Raoult's law. Example 3 Two liquids A and B form an ideal solution at 300 K. The vapour pressure of a solution containing 1.0 mol of A and 2.0 mol ofB at 300 K is 2.0 x 10'Pa. When one more mole ofB is added to the solution, the vapour pressure of the solution is 2. I X I0' Pa. Calculate the vapour pressures of A and B in the pure state. Soll'tion Act.Jrding to Eq. 10.11. total vapour pressure of the solution is given by, 0 p -p X A.\ + p''BB X Substituting the values, we get 2 0 x I 0' Pa = p'' A or 2.0 X 10' Pa = p 0)3.0 + 2.0 p'' 8/3.0 or p",+ 2.0 p 08 = 6.0 X 10' Pa.. (l) Also 2.1 X 10' Pa = p''A or p"A + 3.0 p08 = 8.4 x 105 Pa (2) From (I) and (2). P"8 = 2.4 X 10' P\ = 1.2 X 10' Pa Exercise 1 0 I 00 mol of acetone and 0. I 00 mol of chloroform arc mixed at 308 K. At this temperature. the total vapour pressure of the solution is 3.4 7 x I 04 Pa. The vapour pressures of pure acetone and pure 126 chloroform at 308 K are 4.60 x 104 Pa and 3.92 x 10' Pa, respectively. VerifY whether this solution is ideal or not. 3.3 Ideal Solutions of Volatile Solutes Now let us consider a situation in which the two constituents of an ideal solution are both volatile. For such a composition, the vapour pressure of each constituent will be lowered in accordance with Raoult's law. For two liquids A and B. at a given temperature. PA= N.po, p 8= N8 p"8 The vapour pressure of the solution will then be the sum ofthe vapour pressures of the two constituents. Thus. solution= P + P8 0 Np A A +Np" B J3 3.4 Raoult's Law Curves According to Raoult 's law, the partial vapour pressure of each component A and B of an ideal solution is given by Eqs. 3.1 and 3.2. The partial vapour pressure of each component can be plotted against its mole fraction in the liquid phase. Such vapour pressure curves fur an ideal solution are sho"'n m Fig. 3.I. Completely miscible liquid pairs may be ideal or non-ideal. Ideal solutions obey Raoulfs law. Non-ideal solutions either show positive or negative deviation from Raoulfs law. Ideal solution can be separated into two fractions by fractional distillation - one. a pure component and, another, a constant boiling azeotropic mixture. //:N --' ---- ' '............. '': ''-.LM 'I.. '-.._ Q L T Fig. 3.1: Vapour pressure curves for ideal solution Curve 1: Total vapour pressure of solution Curve 2: Partial vapour pressure of A, PA ,.oXI' A A Curve 3: Partial vapour pressure of B, P 8 =p"aXa 127 The dotted line shows the variation of vapour pressures of the components A and B with mole fraction in the liquid phase. Thus, curve QR (III) indicates the variation of partial vapour pressure ofB with its mole fraction in the liquid solution and curve ST (II) indicates the variation of partial vapour pressure of A with its mole fraction in the liquid solution. The points, Sand R, represent the vapour pressures(P0 A and po8) of the pure components A and B respectively. The variation of total vapour pressure with respect to the mole fraction of B in the liquid solution is given by the curve SR (1). From Fig. 3. I. it can be seen that the vapour pressure of an ideal solution of composition L is given by the suln of the partial vapour pressure of A and the partial vapour pressure ofB (or OL = ML + NL) Vapour Pressure curves of solutions showing positive deviation If the molecular interactions between A and Bare weaker than the A-A or B-B molecular interactions. then the escaping tendency of the molecules of A and B from the solution becomes more than that from the pure liquids. As a result. the vapour pressure of the solution will be greater than that of an ideal solution of the same composition. Such solutions are said to show the positive deviation from Raoult's law. Mathematically. PA >pi)AXA Po >p''Ifs and P > puAxA + Pu,Xs In Fig 3.2 the dotted lines are theoretical curves showing the ideal behaviours, the solid lines are curves drav.n as per experimental values and show positive deviation from behaviour. When ethanol and cyclohexane are mixed. the curves shown in Fig. 3.2 are obtained. In ethanol. there is a strong intermolecular hydrogen bonding. When cyclohexane is added to it,the cyclohexane mole ules get in between the ethanol molecules thereby decreasing the intermolecular interactions. During the ;ormation of such a solution, heat is observed and there is a slight increase in volume. P"k: ·===----- , --- x, Fig. 3.:Z: Curves showing positive deviatwn from Raoult :,· law 128 Vapour pressure curves of solutions showing negative deviation.. If the intermolecular forces between A and B are stronger than those of A-A and B-B. the solutiOn formed by mixing A and B shows negative deviation from Raoult's law Due to stronger A-B interactions. the escaping tendency of A and B from the solution becomes less than that from the pure liquids. The vapour pressure of such a solution will be less than an ideal solution of the same composition. This behaviour is shown in Fig. 3.4. Q X, X4 =1 x,=O Fig. 3.4: Curves showing negative deviation from Raoul! s law The dotted lines are the theoretical curves representing the ideal behaviour, whereas the solid lines art the curves drawn as per experimental values and show negative deviation from ideal behaviour. When acetone and chloroform are mixed. they form hydrogen bonds with each other. As a result the intermolecular attractions between acetone and chloroform become stronger. The tendency of the molecules to escape from the solution thus decrease.The vapour pressure therefore decreases. During the formation of such solutions. heat is evolved and there is a slight decrease in volume. 4.0 Conclusion We have learnt in this Unit about the solutions and their various properties. A solution is a homogenous mixture of two or more substances. Solutions can be formed in all the three phases, namely, solid, liquid and gaseous. Solutions can be divided into nine types depending upon the physical state of the solute and the solvent. A solution which cannot dissolve more amount of a solute at a particular temperature, is said to be saturated. 5.0 Summary Specifically what you have learnt can be summarized as follows: An ideal binary solution is one which can be formed from two constituents with no evolution or absorption of heat and whose volume is the sum of the volumes of each of the constituents. Raoulfs law states that the partial vapour pressure of any volatile component in a solution is equal to the product of the vapour pressure of the pure constituent and its mole fraction in the solution. Ideal solutions obey Raoult's law while non-ideal solutions do not obey. 129 6.0 Tutoh-Marked Assignment l. State whether the following statements are true or false: (i) Raoulfs law for ideal solutions is applicable both to the liquid and the vapour phase compositions. (ii) The solubilities of all substances. i.e. solids. liquids and gases. in liquids increase with rise in temperature. 2. Which of the following liquid pairs can be completely separated into its pure components'' (I) E thana I water (II) Ethylene chloride etln-lenc bromide (III) Nitric acid water (IV) Acetone carbon disulphide (V) Carbon tetrachloride water 130 Unit 4: Binary Liquids Solutions- 2 Table of Contents Page L0 Introduction 132 2.0 Objectives 132 3. I Vapour Pressure Variation with Liquid and Vapour Composition 132 3.2 Boiling Point Diagrams 134 4.0 Concluston 139 5.0 Summary 139 6.0 Tutor-MarkedAssignment 139 13 i 1.0 Introduction In Module 4. unit 2. we considered solutions of solids and gases in liquids. In Module 4. Unit 3 we learnt about completely miscible liquids and we discussed Raoult's law.-ThisUnit will be devoted to completely miscible liquids as well but we will be concentrating on vapour pressure variation 'l'ld boiling pomts diagrams. We will emphatically examine ideal solutiOn of different compos1ti6n. 2.0 Objectives By the end of this Unit. you should be able to Describe the effect of composition on the vapour pressure of the completely miscible liquid system. Explain the effect of composition on the boiling points of solutions. Recogmse the variations in boiling points of solutions. 3.1 Vapour Pressure Variation with Liquid and Vapour Composition Consider a solution containing one mole of benzene and one mole of toluene so that the mole fraction of each component in the solution is 0.5. The vapour pressures of pure benzene and toluene at 29M K are 1.2) x 10' Pa and 3.70 x 10' Pa. respectively. According to Raoult's law· p = poe H.xc H = (125 X 104 Pa) X 05 = 6.25 X JO'Pa C6 H 6 (, 6 (, (, pc - ll8 p··· X CHs" c (3 70 x I O' Pa) x 0 5 = 1.85 X 103 Pa p (6.25x IO'Pa) +(I X5x IO'Pa) =X Ill X IO'Pa roral Mole fraction of benzene in the vapour phase 6.25 X 10' Pa = II 7? X 10 x 103 Pa Mole fraction of toluene m vapour phase 1.85 x 10' pa = 0.23 8.10xiO'Pa It is quite interesting to compare the mole fractions of benzene in the vapour and liquid phase solutions. From the calculations above. the mole fraction ofbcnzene (0.77) in the vapour phase 1s more than that in the liquid phase (0.5). Again. the mole fraction of toluene in the vapour phase (0.23) is less than that in the liquid phase (0.5). Beanng in mind that benzene is more volatile than toluene. \\e can arrive at the following generalization which is one of the forms of Konowaloff's rule: The mole fraction of a more volatile component in an ideal solutiOn IS more in the vapour phase than m the liquid phase For the purpose of comparing the compos1t1ons of the solutiOns in the liquid and the vapour phases at a g1ven total pressure. it IS worth drawmg curves of the t)·pes I and II sho\\n m Fig.41. Curve I sho\\·s the vanation of total vapour pressure with respect to the mole fraction in the liquid phase. S'mtlarly. curve II shows the variatiOn of the total vapour pressure with respect to the mole fracuo, n the 132 vapour phase. The line MN is called a tie line and it gives us the compositiOn of tho solutiOn in the liquid and vapour phases m equilibrum at a particular total vapour pressure. Note that the vapour phase curve II lies below the liquid phase curve I. This is due to the fact that the vapour phase is richer in the more volatile component than the liquid phase and this has been c'X M JU> \V ;\ 1 0,1 x M" But water is 40% by weight; hence we can write 40 8.35 x 10' Pa X 0018 kg mol·' 60 1.65 x I 0' Pa M \ orM, 8.35 x 10' Pa x0018 k"mol·'x60 " 1.65 X I 0' Pa "' 40 o.l37 kg mol·' The relative molecular mass of A137. Exercise 4 Why does a mixture of two immiscible liquids boil at a temperature which is lower than the boiling of any of the pure liquids? 3.4 Concept of Distribution Law In this section. we shall discuss the effect of the addition of a solute to pair of immiscible liquids. Water and carbon tetrachloride, for example are practically immiscible with each other but iodine dissolves both in water and carbon tetrachloride. When iodine is added to a mixture of water and carbon 146 tetrochloride at a certain temperature, iodine distributes itself between the two immiscible layers- water and carbon tetrachloride. Let the concentration of iodine in water and carbon tetrachloride be c, and c,. respectively, at a particular temperature. Then we have. 2._=K ······················································································· (3.7) c" where K is a constant temperature. If more of iodine is added to this system. it again distributes itself between the two layers. Now the concentration of iodine in both the layers will be more than that in the previous case. Let the concentrations be c',fc' 11 is again equal to the constant, K, i.e., c'lc'" = K.................................................................................... (3.8) It is evidentthat, in each case, 1 1 distributes itself between the two immiscible layers in such a way that the ratio of its concentration in the two layers is a constant at a certain temperature. It was pointed out byNernst that the ratio, c',/c'tr is constant only when the solute exists in the same molecular form. i.e. the relative molecular mass of the solute is the same in the two layers. The Nerst distribution law may be stated as follows: At a fixed temperature. a substance X distributes itself between the two immiscible solvents A and B in equilibrium with each other in such a way that the ratio of the concentrations of X in the two solvents is constant. provided that the solute X is in the same molecular state in both the solvents. Mathematically, K = Concentration of X in solvent A Concentration of X in solvent B........ (3.9) The constant K is called the distrihutio11 or partition cocflicienl of the solute between the two solvents. You can see from the data given in Table 5.2 that the value of K is fairly constant in dilute solutions. As the concentration increases, there is a deviation. Table 5.2: Distribution of I, Between H,O and CCI, IO"x [H,O] 10' X [CCI] I K = [H,O] Mol dm 3 Mol dm-1 CCI, 3.22 2.745 1.17 x I O·' 5.03 4.29 1.17X IO·' 7.63 6.54 1.17x IO·' 11.5 10.1 1.14 X 10·2 13.4 11.96 1.12 X JO·O 147 K depends upon t Nature of the solute t Nature of the solvent and t Temperature The distribution law can be derived taking into account thermodynamic considerations. This will he discussed at second level physical chemistry course. 4.0 Conclusion By and large. when two liquids are mixed. there are three possibilities. two of which are partially and completely immiscible liquid systems. We have discussed these two systems in this Unit. The miscibilit\ of partially miscible liquid pairs is said to vary with temperature. The miscibility of some paniall) miscible liquid pairs e.g. phenol-water tends to increase with increase in temperature. Thus phenol- water system possesses an upper CST Liquid pairs possess lower cs·r that is then mutual solubilit) decreases with use in temperature. 5.0 Summary At the end ofthis Unit, you have learnt that: The temperature at which a pair of partially miscible liquid becomes completely miscible is called critical solution temperature. t The temperature at which the solubility becomes complete is referred to as the critical solution temperature (CST) or the consolute temperature. t Below a certain temperature, some liquid pairs that possess lower CST are completely miscible. t The presence of an impurity in one or hotl1 ihc phases changes the CST values.. A pair of immiscible liquids boi Is at a tempe1·ature lower than the boiling points of any of the liquids Mathematically. the distribution law can be expressed as K concentration of X in solvent A/concentration of X in solvent B. 6.0 References and Other Resources IGNOU ( 1993). Solutions and Phase Equilibrium: Physical chemisl!y 3 CHE-04 pp 26-32 New Delh1. 7.0 Tutor-Marked Assignment I. Two liquids A and Bare completely immiscible with each other. Their normal boiling points are Ta and T8 respectively, and T, is less than T 8. The two liquids taken together will boil at A. T, B. T 8 C. T, + T/2 D. A temperature less than T, E. None of these temperatures 2. An immiscible mixture of an organic liquid, A and water on steam distillation boils at 372 Kat pressure of 1.00 x I 0 5 Pa. At this temperature the vapour pressure of water is 9.60 x I 0 4 Pa. The ratio of the mass of water to the mass of A in the distillate is 4: I. Calculate the relative molecular mass ofliquid A. 3. Explain the behaviour of partially miscible I iquid systems with respect to change of temperature. 148 Unit 6: Some Colligative Properties Table of Contents pa :e 1.0 Introduction 150 2.0 Objectives ISO 3.1 Relative Lowering of Vapour Pressure 150 3.1.1 Determination of Relative Molecular Mass ti·om the Lowering of Vapour Pressure 151 3.1.2 Measurement of Vapour Pressure Lowering 151 3.2 Elevation of Boiling Point 154 4.0 Conclusion 156 5.0 Summary 156 6.0 References and Other Resources 156 7.0 Tutor-Marked Assignment 156 149 1.0 Introduction In this Unit. we shall study colligative properties of the solutions containing non-volatile solutes. The colligative properties depend only upon the number of particles of the solute in the solution. They do not depend upon the nature of the solute. These properties help us in determining the relative molecular mass of a non-volatile solute. The colligative prope1ties are: · Relative lowering of vapour pressure, Elevation ofboiling point, Depression of freezing point. and Osmotic pressure. We shall study only the first two properties in this Unit. 2.0 Objectives By the end of this Unit, you should be able to: Explain what is meant by colligative properties. State Raoult's law of relative lowering of vapour pressure. Explain elevation ofboi ling point 3.1 Relative Lowering of Vapour Pressure You should recall we discussed pure and impure substances even when they are homogenous or heterogenous systems. If a pure liquid is placed in a closed container that is initially evacuated. the liquid evaporates to fill up the space above the liquid. At any given temperature when equilibrium is established. the pressure exerted by the vapours of a liquid is called the vapour pressure of the pure liquid (P", ). If a non-volatile solute is added to this liquid. the equilibrium vapour pressure (p,) over the solution is found to be less than that of the pure liquid. According to Raoult's law. the vapour pressure (p 1 ) of the solvent of an ideal solution is given by P, Po, x,................................................................................................3.1 Where x, is the mole fraction of the solvent in the solution and p", is the vapour pressure of the pure solute. Since the solute is non-volatile. its contribution towards the vapour pressure of the solution is negligible. Hence. the vapour pressure of solution containing a non-volatile solute is only due to the solvent. Since the mole fraction (x 1 ) of the solvent is less than one, it is clear ti'om Eq. 3.1 that p1 will also be less than p0 ,. The lowering of vapour pressure when a non-volatile solute is added to a solvent is given by (p0 1 - p 1). Substitution for from Eq. 3.1. we get Po,- P, Po,- Po, x,= Po, ( I - x,)......................................................,........ 3.2 Since x 1 + x,I. 3.3 Where x2 is the mole fraction of the solute. According to Eq.3.3, the lowering of vapour pressure of the solvent depends both on the vapour pressure of the pure solvent and on the mole fraction of the solute in the solution. In other words. 11 !50 depends on the nature of the solvent and on the concentration of the solute and not on the nature of the solute. Eq.3.3 can be written as Po,- P,.................................................................................................. 3.4 -.-- =x2 Po, The expression on the left hand side of Eq.3.4 is called the relative lowering of vapour pressure. Eq.3.4 also may be stated as: "The relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute''. '.fhe above statement is also called Raoult's law of relative lowering of vapour pressure. 3.f.l Determination of Relative Molecular Mass from the Lowering of Vapour Pressure Having discussed relative loweringofvapour pressure, it is possible to measure the relative molecular mass of a non-volatile solute, if we know the vapour pressure of a dilute solution containing.the non· volatile solute. Now let us consider the equation that we need to measure the relative mole£.ular mass from the lowering of vapour pressure. Assuming w2 kg of the solute having molar mass M, kg mol·' is dissolved in w kg pf the solvent having molar mass M 1 kg mol·', then from Eq.3.4 we have: Po, -·p, n,............................................................................3.5 =x=- Po, 2 n +n I 2 p0 - p w,fM 2........................................................................ 3.6 or 1 0 1 =- + w,fM 2 - 1p w, 1M, For a dilute solution, the number of moles of the solute (n,) is negligible in comparison to the number of moles of solvent (n 1). Thus Eq.3.6 reduces to · Po,-p, w,fM, w, M,..........................................................................................3.7 Po, w,IM, w, M, Thus; if we are able to know the mass of the solvent, its relative molecular mass, mass of the solute. the vapour pressure of the solution and of the solvent, the relative molecular mass of the solute can be calculated from Eq.3.7. 3.1.2 easurement of Vapour Pressure Lowering The lowering of vapour pressure 'of a solvent when a non-volatile solute is dissolved in it can be measured by the following methods: (i) Static Method In this method, the difference between the vapour pressure of the solvent and the solution is determined with the help of a differential manometer. The apparatus is as shown in Fig.6.1. It consists of two bulbs which are connected to a manometer. One arm of the manometer is connected with the bulb, A, containing the solvent and the other arm with the bulb, B, coniainingthe solution. The manometric liquid is an inert, non-volatile, low density liquid such as b-bromonaphthalene. 151 From the difference in the levels of the liquid in the two arms, the difference in vapour pressure between the solvent and the solution can be read directly. Fig. 6.1: Static Methodfin· determining the lowering of vapour pressure (ii) Dynamic Method This method is due to Ostwald and Walker. In this method (Fig.6.2), a stream of dry and purified air is first passed through a pre-weighed set of bulbs containing the solution. The air coming out of the bulbs takes up an amount of vapour which is proportional to the vapour pressure of the solution. Evidently, the loss in mass suffered by tile bulbs containing the solution is proportional to the vapour pressure of the solution. lfw, is the loss in mass suffered by the solution bulbs and p, is the vapour pressure of the solution. then we have. w-\ (l p, 3.8 -\ IV Fig. 6.2: Ostwald and Walker\apparatus for measuring the vapour pressure lowering:I11 air: 1 !11 solution hulhs: Ill/; soh·enl hulhs: r!VJ H"eighed CaC 0 IUhes. rV; cily cw· The air coming out ofthe solution bulbs is then passed through pre-weighed set of bulbs containing the solvent. As the air containing vapours of the solvent passes through the solvent. it takes up a further amount of vapour which is proportional to the difference in vapour pressures of the solvent and the solution. Evidently the loss in mass suffered by the solvent bulbs is proportional to the difference in vapour pressures oft he pure solvent and the solution. If w 8 is the loss in mass suffered by the solvent- bulbs and p", is the vapour pressure of the solvent, then we have WB a Po,- P,.................................................................................................3.9 152 From Eqs.3.8 and 3.9 we have (w, + w 8) a p, + p0 ,- P, or(w,+w 8)a p",........................................................................................ 3.10 From Eqs.3.8 and 3.10 we have. 3.11 Thus knowing w,and w 0 , the relative Jo,wering of vapour pressure can be calculated using Eq.3.11. lfthe solvent is water, the moist air coming out of the solvent bulbs may be passed through a weighed tube containing fused calcium chloride which absorbs moisture from moist air. The mass of calcium chloride tube will increase and the increase in mass will be equal to (w,+ w 8 ). Example I The vapour pressure of a solution containing 0.0 12 kg of a solute dissolved in 0. I 00 kg of water at 300 K is 3.50 X I O' Pa. Calculate the molar mass of the solute, if the vapour pressure of water at 300 K is 3.70 x 103 Pa. Solution w, O.IOOkg w, O.OI2 kg M,0.018 kg mot·' M'. p0 1 3.70 X I 0' Pa P, 3.50 x 10 3 Pa Using Eq.3.7. we have (3.70 x I 0' Pa- 3.50 x I O' Pa) = (0.0 12 kg)(O.OI8 kg mot·') 3.70x IO'Pa O.IOOkg(M,) or M (0.012 kg) (0.018 kg mol·') x 3.70 X 10' Pa ' O.IOOkg 0.20 x 10' Pa 0.040.kg mol·' In this section, we discussed the correlation between the relative lowering of vapour pressure and the mole fraction of a non-volatile solute in a solution. In the next section, we shall study how the boiling. point of a solvent is suffered when a non-volatile solute is added toot. Exercise I State whether the following statements are true or false: (i) Lowering of vapour pressure is not a colligative property. (ii) Relative lowering of vapour pressure is equal to the mole fraction of the solvent. (iii) Relative lowering of vapour pressure de(>ends upon the number of particles of the solute and the solvent. 153 3.2 Elevation of Boiling Point The temperature. at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. IS called the boiling point of the liquid. Since. according to Raoult's law, the vapour pressure of a solution contaming a non-volatile solute is less than that of its pure solvent. the boiling point of such solution will be higher than the boiling point of the pure solvem. This is because when a non-volatile solute is added to a solvent at a particular temperature. the vapour pressure of the solvent decreases. Thus, the solution has to be heated further to make its vapour pressure equal the atmospheric pressure. This can be understood from the vapour pressure curves plotted in Fig.6.3. TIK T Fig. 6.3: Eleva/ion of boiling point The curve I represents the variation of vapour pressure of the pure solvent with temperature. The curve II represents the variation of vapour pressure of a dilute solution of known concentration with temperature. The boiling point of the solvent is T, and that of the solution is T so that the elevation in boiling point is T- T, = Th. At temperature T 0, the lowering of vapour pressure of a solvent. when a non-volatile solute is added to it. is p''- p. From the vapour pressure curve of the solution. it is clear that the vapour pressure of the solution is pat T, and p" at T. Determination of Boiling Point Elevation For measuring the elevation ofboiling point. a number of methods are avaiiaule. In all these methods. care is taken to avoid superheating_ofthe liquid. The most commonly used methods are given overleaf: (i) Tbe Landsberger Method In this method, the solvent or the solution is heated lo its boiling point by passing the vapours of the solvent through them. As the vapour condense. they give up their latent heat to heat the solvent or the solution. Superheating is nearly impossible in this way because. when the hoi ling point is reached. the vapour will be in equilibrium with the liquid at atmospheric pressure and no more vapour will condense. The apparatus is shown in Fig.6.4a. It consists of flask A which contains the solvent. 154 (a) Fig. 6.4: (a) Apparatus for Landsberger method: (b) Beckmann thermometer The flask is connected to a graduated tube B which also contains the solvent. Solvent in tlask A is boiled so that the vapour pass into the solvent in B. where these vapour emerge in small bubbles through a number of holes at the end of the tube. The temperature of the solvent rise' and wlilln it becomes constant. it is noted. Any excess vapour pass from the right hand top portion (H) of the graduated tube (B) into the condenser (C). A known mass of the non-volatile solute is then added to the solvent in B. The solution is then again heated by passing the vapour of the solvent through it. When the temperature reaches a constant value. it is cooled. The passage ofvapours is stopped and the volume of the solution is noted as shown on the graduated tube. A thermometer, which is graduated in tenths of a degree and is capable of reading up to 0.0 I", is generally used. One such thermometer is the Beckmann thermometer (Fig. 6.4b ). Knowing the density of the solvent. the mass of the solvent in the solution can be calculated. The volume of the solute is supposed to be negligible. The difference between the two boiling points gives the elevation in boiling point. (2) Cottrell's Method The apparatus is as shown in Fig.6.5. It consists of a graduated tube A which contains the liquid- solvent or the solution. An in vetted funnel is placed in the tube A through which the bubbles rise. When the liquid starts boiling, a steam of liquid and vapour is poured over the bulb of a Beckmann thermometer so that the bulb of the thermometer is covered with a thin layer of the boiling liquid. After detennining the boiling point of the pure solvent, a known mass of the solute is added and the boiling point of the solution is again detem1ined. 1 [- ""' -"' 1 1 Fig. 6.5: Appurulus jiJr ('o/IJ·e/1:, me/hod !55 The difference between the t\vo readings gives the elevation in boiling point. You \Viii learn more about boiling point elevation later in your programme. Exercise 2 Explain why the boiling point of a solution is higher than that of a pure solvent. 4.0 Conclusion In this Unit. we were able to examine some colligative properties specifically the lowering of vapour pressure and elevation ofboiling point. The methods tOr determining boiling point elevation and lowering. vapour pressure were described. Later in another physical chemistry course. we shall consider other colligative prope11ies and their related measurements. 5,{) Summary At the end of this Unit, you have learnt that: Colligative properties of a solution depend upon the number of particules of a solute present in the solution. Colligative properties are independent of the nature of the solute. The collig:ative properties we discussed in this Unit are relative lowering of vapour pressure and elevation of boiling point. According to Raoult's law. relative lowering of vapour pressure is equal to the mole fraction oft he solute. Since the addition of a non-volatile solute to a solvent decrease the vapour pressure of the solvent. the boiling point of the solution containing a non-volatile solute is higherthan that of the pure solvem. 6.0 References and Other Resources IGNOU ( 1993). Solutions and Phase Equilibrium: Physical Chemistry 3 CHE-04. pp 26-32. 7.0 Tutor-Marked Assignment I. Define colligative property. Name two of them. 2. What is Raoult's law of relative lowering of vapour pressure? !56 Unit 7: Phase Rule Table of Contents Page 1.0 Introduction 158 2.0 Objectives 158 3.1 Definition ofthe Terms 158 3.1.1 Phase 158 3:1.2 Definition ofNumber of Components 159 3.1.3 Degree of Freedom 160 3.2 Phase Rule 160 3.3 Phase Diagrams 162 3.4 Stability of the Phases of a Pure Substance 163 4.0 Conclusion 164 5.0 Summary 164 6.0 References and Other Resources 164 7.0 Tutor-Marked Assignment 164 157 l.O Introduction The study of the effect of various parame ers such as pressure, temperature or composition on the physical state of chemical substances is the subject matter of phase equilibria. In the last Unit, you have studied the criteria for equilibrium. In this Unit. we shall con,ider the varioL"aspect> of chemical equilibria. This means: focusing our attention on the equilibria regarding physical state and relating physical equilibria to various parameters such as pressure, temperature and composition using phase rule. We shall define the terms such as phase. component. degrees of freedom and phase transition. We will deduce phase rule. One-component system will be studied in the light of phase rule. Gas systems will be discussed. 2.0 Objectives By the end of this Unit, you should be able to: Define the terms phase. component and degree of freedom Calculate the number of phases. components and degrees of freedom in a system State and deduce the phase rule Comment on the stability of the phases of subliming and non-subliming substances. 3.1 Definition of the Terms In 1876. Gibbs deduced a simple relationship among the number of phases in equilibrium. the number of components and the number of intensive variables known as degrees of freedom. Let us detine the terms phase, component and degrees of freedom. 3.1.1 Phase In Units 8 and 9. we examined phases of different liquid -liquid solutions. Try to recollect and define a phase as given in these Units. We can define a phase. symbol Pas a physically distinct and homogenous part of the system that is mechanically separable from other parts of the system. The definition as suggested by Gibbs is that a phase is a state of matter that is uniform throughout, not only in chemical composition but also in physical state. Example I A gas or a gaseous mixture is a single phase since there cannot be an interface between one gas and another. Air. for example. is one phase system. although it is a mixture of many gases. Example II A system of totally miscible liquids will exist in one phase only as far as the liquid phase is concerned. But since each liquid has its vapour above, the total number of phases in a system of miscible liquids is two, one for the liquid and the other for vapour. These two phases are separated by the surface of the solution in the liquid phase. · Example III A system of two immiscible liquids has a total of three phases, two for the substances in the liquid state and another for the vapour phase containing vapours of both the liquids. !58 Example IV e A crystal is a single phase. Different solids having different crystal structures constitute different phases, irrespective of the fact whether they have same chemical composition or not. A mixture of graphite and diamond constitutes two phases although both are only allotropic modifications of carbon. Next, we take up the definition of the word, component. 3.1.2 Definition of Number of Components The number of components (c) in a system is the smallest number of substances in terms of which the composition of all the phases in the system can be described separately. The number of component may be smaller than the number of chemical substances that are present in the system. The definition is easy to apply when the species do not react, for them we simply count their number. For example, phase equilibria containing ice, water and its vapour is a one-component system since the composition of each phase can be expressed in terms of the component, H,O, only. A mixture of ethanol and water is a two-component system. If the species react chemically and are at equilibrium, we have to take into account the significance of the phrase' all the phases' in the above definition. Dissociation of calcium carbonate in a closed vessel is a two-component system, although at first, it may appear to be a three-component system, namely, CaCO,, CaO and CO,. CaCO, (s) CaO(s) + CO,(g) You can see that the composition of any one species is related to the other two and the dissqciation of calcium carbonate is, therefore, a two-component system.. In the case of solid ammonium chloride being heated in a closed vessel, there is an equilibrium between the solid ammonium chloride and its vapour (which consists of ammonia and hydrogen chloride molecules). Since both the phases have the formal composition 'NH 4C I', it is a one-component system. However, if ammonia or hydrogen chloride is added to the system, the system has two components because now the relative amounts of hydrogen chloride and ammonia are not the same. lfthere are'S' susbstances (may be chemical compounds or ionic species) and 'R' relations (which include equations representing chemical equilibria or charge neutrality) among them, then the number of components (C) can be calculated using the equation,.....................................3.1 To illustrate this, we can consider two examples. If you consider a solution of glucose in water, there are two substances and hence S = 2. But there is no relation between them and so R = 0; therefore. = C 2 and this is a two component system. Let us re-examine the equilibrium of solid ammonium chloride with its vapour. HereS= 3 since there are three substances, NH,C I, NH, and HC I. But R = 2 since these are two relations as shown below: !59 (2) Concentration of NH, = Concentration of HC/ (and both of them are in gaseous phase). Therefore C = S- R = 3-2 = I. This is a one-component system. If extra HC/ (g) is added then. the second relation given above will no more hold. That is, R = I although S = 3. This is therefore a two component system, as C = S- R = 3 -I = 2. 3.1.3 Degrees of Freedom The number of degrees of freedom (F) or variance of a system is the smallest number of intensive variables. such as pressure, temperature and composition, which are to be specified to describe the state of the system completely. IfF= 0, the system is invariant and this means none of the intensive variables can be changed. e In case F = 1, the system is univariant and one of the intensive variables can be changed. For a bivariant system, F is equal to 2, and two of the intensive variables can be changed. The phase rule relates the degrees of freedom to the number of phases and the components. Before studying phase rule, let us recapitulate the criteria for equilibrium. Exercise 1 Calculate the number of phase in the following cases: (a) A closed beaker partially tilled with toluene and water. (b) A closed beaker partially filled with acetone and water. Exercise 2 Calculate the number of components in the following cases: (a) Dissociation of ammonia in a closed vessel; (b) Dissociation of ammonia in a closed vessel containing nitrogen. 3.2 Phase Rule In a heterogenous equilibrium, not influenced by gravitational or electrical or magnetic forces, the number of degrees offreedom (F) of the system is defined by the number of components (c) and the number of phases (P) according to the equation, F=C-P+2.................. 3.2 Once the values of C and P are known, F can be calculated. Let us take up some examples to understand Eq. 3.2. In the vaporization equilibrium of water (C = I), there are two phase (liquid and vapour) and hence. F = C- P + 2 = I -2 + 2 = 1. That is. specifying 160 pre,sure or temperature fixes the other. This is all the more understandable from the general statement that at a particular temperature, vapour pressure of a liquid is constant. If all the three phases. ice. water and its vapour. are to be in equilibrum, then FI -3 + 20; that is, the system is invariant. This means all the three phase of water are in equilibrium only at a fixed temperature and pressure. Any effort to change pressure or temperature results in the disappearance of one of the phases. Deduction of Phase Rule Having stated the phase rule, we shall now deduce

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