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Lecture notes on Physical Chemistry

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PHYSICAL CHEMISTRY IN BRIEF Prof. Ing. Anatol Malijevsk´ y, CSc., et al. (September 30, 2005) Institute of Chemical Technology, Prague Faculty of Chemical Engineering4 Introduction Dear students, Physical Chemistry is generally considered to be a difficult subject. We thought long and hard about ways to make its study easier, and this text is the result of our endeavors. The book provides accurate definitions of terms, definitions of major quantities, and a number of relations including specification of the conditions under which they are valid. It also contains a number of schematic figures and examples that clarify the accompanying text. The reader will not find any derivations in this book, although frequent references are made to the initial formulas from which the respective relations are obtained. Intermsofcontents,wefollowedthesyllabiof“PhysicalChemistryI”and“PhysicalChem- istry II” as taught at the Institute of Chemical Technology (ICT), Prague up to 2005. However the extent of this work is a little broader as our objective was to cover all the major fields of Physical Chemistry. This publication is not intended to substitute for any textbooks or books of examples. Yet we believe that it will prove useful during revision lessons leading up to an exam in Physical Chemistry or prior to the final (state) examination, as well as during postgraduate studies. Even experts in Physical Chemistry and related fields may find this work to be useful as a reference. Physical Chemistry In Brief has two predecessors, “Breviary of Physical Chemistry I” and “Breviary of Physical Chemistry II”. Since the first issue in 1993, the texts have been revised and re-published many times, always selling out. Over the course of time we have thus striven to eliminate both factual and formal errors, as well as to review and rewrite the less accessible passages pointed out to us by both students and colleagues in the Department of Physical Chemistry. Finally,asthenumberofforeignstudentscomingtostudyatourinstitutecontinues togrow, wedecidedtogivethemaproventoolwrittenintheEnglishlanguage. Thistextisthe result of these efforts. A number of changes have been made to the text and the contents have been partially extended. We will be grateful to any reader able to detect and inform us of any errors in our work. Finally, the authors would like to express their thanks to Mrs. Flemrov´a for her substantial investment in translating this text.CHAP. 1: BASIC TERMS CONTENTS 24 Chapter 1 Basic terms A good definition of basic terms is an essential prerequisite for the study of any physicochemical processes. Some of these terms may be also used beyond the field of physicalchemistry,buttheirmeaningisoftenslightlydifferent. Inthischapterwewilltherefore sum up the major basic terms that will be used in the subsequent parts of this book. 1.1 Thermodynamic system The concept (thermodynamic) system as used in this book refers to that part of the world whose thermodynamic properties are the subject of our interest, while the termsurroundings is used for the remaining part of the universe. Note: Both a certain part of the real space and a certain part of the imaginary (abstract) space forming a simplified model system, e.g. an ideal gas, may be chosen as a system. Systems are classified as isolated, closed and open, based on their inter-relations with their surroundings. 1.1.1 Isolated system A chemical system exchanging neither matter nor energy with its surroundings is an isolated system.CHAP. 1: BASIC TERMS CONTENTS 25 1.1.2 Closed system Achemicalsystemexchangingenergybutnotmatterwithitssurroundingsisaclosedsystem. 1.1.3 Open system A chemical system exchanging both energy and matter with its surroundings is an open sys- tem. Example Differences between individual types of chemical systems may be demonstrated using the example of making coffee. The pot on the heater represents a (practically) closed system until the water is brought to the boil. At the boiling point, when steam is leaking from the pot, it becomes an open system. The ready-made coffee kept in a thermos bottle represents a simple model of an isolated system. 1.1.4 Phase, homogeneous and heterogeneous systems The termphase is used for that portion of the investigated system volume in which its proper- ties areconstantor continuouslychanginginspace. If asystembehaves in this way throughout all its volume, we call it a homogeneous system. If a system contains more phases, we call it a heterogeneous system. Example Let us imagine a bottle of whisky. How many phases does this system consist of?CHAP. 1: BASIC TERMS CONTENTS 26 Solution If we are, from the thermodynamic point of view, interested solely in the liquid content of the bottle, the system is homogeneous. It contains one liquid phase (a mixture of water, ethanol and some additives). If, on the other hand, we are interested in the entire content of the bottle but not the bottle itself, the system is heterogeneous. In this case it consists of two phases, liquid and gaseous, with the latter containing air and whisky vapour. If, however, we focus our attention on both the bottle content and the bottle itself, we have a heterogeneous system again, but this time it also contains other phases in addition to the gaseous and liquid ones, i.e. the glass of the bottle, its cap, label, etc.CHAP. 1: BASIC TERMS CONTENTS 27 Obr. 1.1: The volume of a system as an extensive quantity. The volume V is the sum of the volumes of the individual parts (i.e. sub-systems) I, II and III, i.e. V =V +V +V . I II III 1.2 Energy There are two basic forms of energy exchange between a system and its surroundings, heat andwork. A positive value is assigned to such energy exchange during which the system gains energy (work or heat) from its surroundings, i.e. energy is added to the system. A negative value indicates that the system passes energy (work or heat) to its surroundings, i.e. energy is subtracted from the system. 1.2.1 Heat Whentheenergyofasystemchangesasaresultofatemperaturedifferencebetweenthesystem and its surroundings (e.g. transfer of kinetic energy of disordered movement of molecules), we speak about exchanged heat. U Main unit: J. 1.2.2 Work Other forms of energy exchange, which are usually driven by some forces acting between the system and its surroundings, are called work. Based on the type of interaction between the system and its surroundings, we distinguish volume work see 3.1.2, electrical work, surface work, etc. U Main unit: J.CHAP. 1: BASIC TERMS CONTENTS 28 1.3 Thermodynamic quantities Observation of any system allows us to determine a number of its properties. The proper- ties in which we are interested from the thermodynamic point of view are called thermody- namic quantities, or, briefly, quantities. Typical thermodynamic quantities are temperature, pressure, volume, enthalpy and entropy. Neither heat nor work rank among thermodynamic quantities. Note: Terms such as thermodynamic function, thermodynamic variable, state quantity (i.e.aquantitydeterminingthestateofasystem,see1.4),statefunction,orstatevariable are used as synonyms of the term thermodynamic quantity. 1.3.1 Intensive and extensive thermodynamic quantities Letusconsiderahomogeneoussystemwithoutanyexternalforcefieldspresent. Wedistinguish betweenextensive andintensive thermodynamic quantities of a system. Intensive quantities are those whose values do not change when the system is divided into smaller sub-systems. Extensive quantities are those whose values are proportional to the amount of substance of the system at a fixed temperature and pressure (see Figure 1.1). Temperature, pressure, and composition expressed by mole fractions are typical intensive quantities. Volume, mass and the number of particles are typical extensive quantities. Note: Some quantities, e.g. the system surface, are neither extensive nor intensive. Every extensive quantity may be converted into an intensive one if we relate it to a certain constantmassofthesystem. Wethenobtainspecificormolarquantities(see3.2.5). Forevery extensive quantity X and the respective molar and specific quantities X and x we may m write X = nX , (1.1) m X = mx, (1.2) where n is the amount of substance and m is the mass of the system. S Symbols: We will use the subscript to denote molar quantities and small letters to denote m specific quantities.CHAP. 1: BASIC TERMS CONTENTS 29 1.4 The state of a system and its changes Any system may be in any moment characterized using a certain number of quantities. These quantities define the state of a given system. The degree of generality at which we observe a given system has to be taken into account at the same time. In terms of a microscopic scale, the state of a system is defined by the position and velocity of all its particles. In terms of thermodynamics, however, it is enough to know only a few quantities, e.g. temperature, pressure and composition. 1.4.1 The state of thermodynamic equilibrium The state of thermodynamic equilibrium (equilibrium state, equilibrium) is a state in which no macroscopic changes occur in the system and all quantities have constant values in time. Note: In the state of thermodynamic equilibrium, changes take place at the microscopic level. For instance, when the liquid and vapour phases are in equilibrium, some molecules continuously move from the liquid to the vapour phase and others from the vapour to the liquid phase. However, the temperature and pressure of the system do not change. The state of thermodynamic equilibrium embraces the following partial equilibria: 1 • mechanical (pressure) equilibrium—the pressure in all parts of the system is the same , • thermal (temperature) equilibrium—the temperature in all parts of the system is equal- ized, • concentration equilibrium—the concentration of the system components is the same in all parts of each phase of the system, but the composition of individual phases is usually different, • chemicalequilibrium—nochangesincompositionoccurasaresultofchemicalreactions, • phaseequilibrium—ifasystemisheterogeneous(see1.1.4),thecomponentsofitsphases are in equilibrium. 1 The osmotic equilibrium is an exception.CHAP. 1: BASIC TERMS CONTENTS 30 Note: If a system in the state of thermodynamic equilibrium occurs in an external force field, e.g. the gravitational field, the pressure is not the same in all parts of the system but it changes continuously. The concentration of the system components also changes continuously in each phase, with a discontinual change occurring at the phase boundary. 1.4.2 System’s transition to the state of equilibrium If a system is not in the state of equilibrium, its properties change in time in such a way that it tends toward equilibrium. Thermodynamics postulates that every system under invariable externalconditionsisboundtoattainthestateofthermodynamicequilibrium. Thetimeneeded forasystemtoattainequilibriumvariesconsiderably,rangingfromfractionsofasecondneeded for pressure equalization up to hundreds of years needed for glass transition to the crystalline state. Ameasureofthevelocityofasystem’stransitiontoequilibriumiscalledtherelaxation time. Example If we immerse several crystals of copper(II) sulphate pentahydrate (CuSO· H O) into a closed 4 5 2 vessel containing water, the system thus created will be in a non-equilibrium state at the be- ginning. There will be neither a phase equilibrium between the crystals and the liquid phase nor a concentration equilibrium. After some time the crystals will dissolve (phase equilibrium). If we do not mix the system, the dissolved copper(II) sulphate pentahydrate will slowly diffuse through the solution from the bottom up to the surface, and after many weeks (relaxation time), concentration in all parts of the system will become equal as well (thermodynamic equilibrium). 1.4.3 Thermodynamic process Ifthepropertiesofasystemchangeintime,i.e.ifatleastonethermodynamicquantitychanges, we say that a certain thermodynamic process takes place in the system. The term process relates to a very broad range of most varied processes, from simple physical changes such as, e.g., heating, various chemical reactions, up to complex multistage processes. Individual kinds of processes may be classified according to several criteria.CHAP. 1: BASIC TERMS CONTENTS 31 1.4.4 Reversible and irreversible processes The course of any process depends on the conditions under which the given system changes. If we arrange the conditions in such a way that the system is nearly at equilibrium in every moment, and that, consequently, the direction of the process may be reversed by even a very slight change of the initial conditions, the process is called reversible or equilibrium. A reversible process is thus a sequence of (nearly) equilibrium states of a system. However, processes in the real world are mostly such that the system is out of equilibrium at least at the beginning. These processes are called irreversible or non-equilibrium (the direction of the process cannot be reversed by any slight change of external conditions, and the process is a sequence of non-equilibrium states). An equilibrium process is thus actually a limiting case of a non-equilibrium process going on at an infinitesimal velocity. Example Infinitely slow heating or infinitely slow compression of a system may serve as an example of equilibrium processes which cannot be carried out in practice. In contrast, water boiling at a ◦ temperature of 100 C and pressure of 101 325 Pa is an example of an equilibrium process which may take place in practice. If we lower the temperature slightly, the direction of the process will be reversed and boiling will be replaced by water vapour condensation. 1.4.5 Processes at a constant quantity Inmostinvestigatedprocesses, oneormorethermodynamicquantitiesaremaintainedconstant during the whole process. These processes are mostly termed using the prefix iso- (is-), and denoted using the symbol X, with X indicating the given constant quantity. The following processes are encountered most often:CHAP. 1: BASIC TERMS CONTENTS 32 Name of the process Constant quantity Symbol Isothermal temperature T Isobaric pressure p Isochoric volume V Adiabatic heat ad Isentropic entropy S Isenthalpic enthalpy H Polytropic heat capacity C Example In the initial state, a system of a constant volume has a temperature of 300K and a pressure of 150kPa. A certain process takes place in the system, and in the final state the system’s temperature is 320K and its pressure is 150kPa. Does the process take place under a constant thermodynamic quantity? Solution The initial and the final temperatures of the system are different. Consequently, the process cannot be isothermal. Both the initial pressure and the final pressure are identical. In this case it may be, but not necessarily is, an isobaric process. The specification does not allow us to find out whether pressure changes in any way in the course of the process. However, the process is definitely an isochoric one because the system has a constant (i.e. unchanging) volume. 1.4.6 Cyclic process A cyclic process is such at which the final state of the system is identical with its initial state. In a cyclic process, changes of thermodynamic quantities are zero. Note: Heat and work are not thermodynamic quantities and therefore they are not zero during a cyclic process.CHAP. 1: BASIC TERMS CONTENTS 33 Example Let our system be a cube of ice with a mass of 1g, and the initial state be a temperature of ◦ −10 C and a pressure of 100kPa. The sequence of processes taking place in the system was as ◦ follows: thecubewasheated to0 C atwhichitmelted. Theliquid waterwas electrolyzed atthis temperature. The resulting mixture of hydrogen and oxygen was expanded to 200Pa and ignited. ◦ The water vapour resulting from the reaction had a temperature of 500 C at the end of the ◦ reaction. It was then cooled to−10 C and compressed to 100kPa. In the course of compression desublimation (snowing) occurred, and the system returned to its initial thermodynamic state. A cyclic process took place.CHAP. 1: BASIC TERMS CONTENTS 34 1.5 Some basic and derived quantities 1.5.1 Mass m U Main unit: kg. 1.5.2 Amount of substance n 23 U Mainunit: mol. 1molisN ofparticles(atoms,molecules,ions...),whereN = 6.022025×10 A A is the Avogadro constant. 1.5.3 Molar mass M −1 U Main unit: kgmol . Themolarmassisthemassofonemoleofparticles. Therelationbetween the amount of substance, mass and molar mass is n =m/M. (1.3) This relation applies to both a pure substance and a mixture. The molar mass of a mixture can be calculated using the relation k X M = xM , (1.4) i i i=1 where M is the molar mass of component i, and x is its mole fraction see 1.6.1 i i 1.5.4 Absolute temperature T U Main unit: K. ◦ Note: Temperature given in C is denoted t (t = T – 273.15) 1.5.5 Pressure p U Main unit: Pa. 5 Older units: bar (1bar = 10 Pa), atm (1atm = 101 325Pa), torr (760torr = 101 325Pa).CHAP. 1: BASIC TERMS CONTENTS 35 1.5.6 Volume V 3 U Main unit: m . 3 Older units: litre (1l = 1dm ).CHAP. 1: BASIC TERMS CONTENTS 36 1.6 Pure substance and mixture We speak about a pure substance (chemical individual) when only one kind of molecules is present in a system. When a system contains more kinds of molecules, we speak about a mixture. The substances of which a mixture is composed are its components. According to the number of components we distinguish binary mixtures consisting of only two compo- nents,ternarymixturesconsistingofthreecomponents, quarternarymixturesconsistingoffour components, etc. In addition to thermodynamic quantities used to describe pure substances (temperature, pressure, volume), the description of a mixtures also requires knowledge of the composition of all its phases, which may be expressed using one of the quantities listed below. th 1.6.1 Mole fraction of the i component x i Definition n i x = , (1.5) i n where n is the amount of substance of component i, n is the total amount of substance of all i components k X n = n , (1.6) j j=1 and k is the number of components in the mixture. U Main unit: dimensionless quantity. It follows from the definition (1.5) that the sum of mole fractions equals one k X x = 1. (1.7) i i=1 Note: Instead of mole fractions, the expression mole percent is often used, meaning 100- times the mole fractions.CHAP. 1: BASIC TERMS CONTENTS 37 Example A binary mixture contains 4moles of substance A and 6moles of substance B. Express the composition of the mixture using mole fraction and mole percent. Solution According to (1.6), the amount of substance in the mixture is n = 4+6 = 10 mol. From (1.5) we get x = 4/10 = 0.4, x = 6/10 = 0.6. The mixture contains 40mole percent of substance A B A, and 60mole percent of substance B. A mixture in which mole fractions of all components have the same value is called an equimolar mixture. Example Calculate the mole fractions of an equimolar mixture of hydrogen and oxygen, and the mole fractions of an equimolar mixture of nitrogen, oxygen and argon. Solution It follows from the definition of an equimolar mixture that 1 x =x = H O 2 2 2 and 1 x =x =x = . N O Ar 2 2 3 1.6.2 Mass fraction w i Definition m i w = , (1.8) i m P k where m is the mass of component i, and m = m is the mass of the mixture. i i i=1 U Main unit: dimensionless quantity.CHAP. 1: BASIC TERMS CONTENTS 38 The sum of mass fractions of all components equals one k X w = 1. (1.9) i i=1 We convert mole and mass fractions using the relations: w/M xM i i i i x = , w = . (1.10) i i k k X X w /M x M j j j j j=1 j=1 Example −1 Amixturecontains5gofsubstanceAwithamolarmassM =25 gmol , and15gofsubstance A −1 B with a molar mass M = 75 gmol . Calculate the mass fractions and the mole fractions. B Solution Substituting into (1.8) gives 5 15 w = = 0.25, w = = 0.75. A B 5+15 5+15 We calculate the mole fractions using the first of equations (1.10) 0.25/25 0.75/75 x = = 0.5, x = = 0.5. A B 0.25/25+0.75/75 0.25/25+0.75/75 1.6.3 Volume fraction φ i Definition • • xV V i m,i i φ = = , (1.11) i k k X X • • V x V j j m,j j=1 j=1CHAP. 1: BASIC TERMS CONTENTS 39 • • where V and V are the volume and the molar volume of a pure substance i in the same i m,i state of matter as the mixture. U Main unit: dimensionless quantity. • S Symbols: ThesymbolX willbeusedtodenotethermodynamicquantityX ofapuresubstance j j at the temperature and pressure of the mixture, with the pure substance being in the same • stateofmatterasthemixture(i.e.ifthemixtureisliquid,X willbeathermodynamicquantity j • of a pure liquid substance). If the mixture is in the solid state, the symbol will denote a pure substance in the same crystalline form as the mixture. The sum of the volume fractions of all components equals one. k X φ = 1. (1.12) i i=1 Note: For a mixture of ideal gases, the volume fraction equals the mole fraction φ =x . i i Example 3 3 Calculate the volume fractions in a solution prepared by mixing 40 cm of ethanol and 160 cm of water. Is it possible to calculate the volume of the solution based on this data? Solution From (1.11) we obtain 160 40 φ = = 0.8, φ = = 0.2. water ethanol 160+40 160+40 The volume of a solution cannot be calculated using the volumes of pure substances, but it has to be measured. In the considered mixture of ethanol and water it will be smaller than 160 + 40 3 = 200 cm .CHAP. 1: BASIC TERMS CONTENTS 40 1.6.4 Amount-of-substance concentration c i Definition c =n/V , (1.13) i i where V is the total volume of the mixture. −3 U Main unit: molm . In a pure component this quantity is identical with the amount-of- substance density. Note: Theexpressionamount-of-substanceconcentrationisusuallyabbreviatedtoamount concentration or substance concentration. The same applies to the expression amount- of-substance density. When there is no risk of ambiguity, the word concentration may be used alone. In older literature, the term molarity may be found indicating the same −3 quantity (using the unit moldm ). Example 3 A mixture of substances with a volume of 5dm contains 56g of nitrogen. Calculate its concen- tration. Solution The amount of substance of nitrogen is 56 n = = 2mol. N 2 28 From equation (1.13) we obtain 2 −3 c = = 0.4moldm . N 2 5 1.6.5 Molality m i Definition m =n/m (1.14) i solvent iCHAP. 1: BASIC TERMS CONTENTS 41 −1 U Main unit: molkg . This quantity is used mainly in connection with aqueous solutions of electrolytes. Example A total of 58.5g of NaCl has been mixed with 500g of water. Calculate the molality of sodium −1 chloride given that you know its molar mass to be M = 58.5gmol . Solution By substituting into (1.14) we obtain 58.5/58.5 −1 m = = 2molkg . NaCl 0.500 −1 The molality of the obtained solution is 2molkg .