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THERMODYNAMICS - NCERT

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160CHEMISTRYUNIT 6THERMODYNAMICSIt is the only physical theory of universal content concerningwhich I am convinced that, within the framework of theapplicability of its basic concepts, it will never be overthrown.After studying this Unit, you willbe able to explain the terms : system andsurroundings;discriminate between close,open and isolated systems;explain internal energy, workand heat;statefirstlawofthermodynamics and expressit mathematically;calculate energy changes aswork and heat contributionsin chemical systems;explain state functions: U, H.correlate U and H;measure experimentally Uand H;define standard states for H;calculate enthalpy changes forvarious types of reactions;state and apply Hess’s law ofconstant heat summation;differentiate between extensiveand intensive properties;define spontaneous and nonspontaneous processes;explain entropy as athermodynamic state functionand apply it for spontaneity;explain Gibbs energy change( G); andestablish relationship between G and spontaneity, G andequilibrium constant.Albert EinsteinChemical energy stored by molecules can be released as heatduring chemical reactions when a fuel like methane, cookinggas or coal burns in air. The chemical energy may also beused to do mechanical work when a fuel burns in an engineor to provide electrical energy through a galvanic cell likedry cell. Thus, various forms of energy are interrelated andunder certain conditions, these may be transformed fromone form into another. The study of these energytransformations forms the subject matter of thermodynamics.The laws of thermodynamics deal with energy changes ofmacroscopic systems involving a large number of moleculesrather than microscopic systems containing a few molecules.Thermodynamics is not concerned about how and at whatrate these energy transformations are carried out, but isbased on initial and final states of a system undergoing thechange. Laws of thermodynamics apply only when a systemis in equilibrium or moves from one equilibrium state toanother equilibrium state. Macroscopic properties likepressure and temperature do not change with time for asystem in equilibrium state. In this unit, we would like toanswer some of the important questions throughthermodynamics, like:How do we determine the energy changes involved in achemical reaction/process? Will it occur or not?What drives a chemical reaction/process?To what extent do the chemical reactions proceed?2020-21

THERMODYNAMICS1616.1 THERMODYNAMIC TERMSWe are interested in chemical reactions and theenergy changes accompanying them. For thiswe need to know certain thermodynamicterms. These are discussed below.6.1.1 The System and the SurroundingsA system in thermodynamics refers to thatpart of universe in which observations aremade and remaining universe constitutes thesurroundings. The surroundings includeeverything other than the system. System andthe surroundings together constitute theuniverse .The universe The system The surroundingsHowever, the entire universe other thanthe system is not affected by the changestaking place in the system. Therefore, forall practical purposes, the surroundingsare that portion of the remaining universewhich can interact with the system.Usually, the region of space in theneighbourhood of the system constitutesits surroundings.For example, if we are studying thereaction between two substances A and Bkept in a beaker, the beaker containing thereaction mixture is the system and the roomwhere the beaker is kept is the surroundings(Fig. 6.1).be real or imaginary. The wall that separatesthe system from the surroundings is calledboundary. This is designed to allow us tocontrol and keep track of all movements ofmatter and energy in or out of the system.6.1.2 Types of the SystemWe, further classify the systems according tothe movements of matter and energy in or outof the system.1. Open SystemIn an open system, there is exchange of energyand matter between system and surroundings[Fig. 6.2 (a)]. The presence of reactants in anopen beaker is an example of an open system*.Here the boundary is an imaginary surfaceenclosing the beaker and reactants.2. Closed SystemIn a closed system, there is no exchange ofmatter, but exchange of energy is possiblebetween system and the surroundings[Fig. 6.2 (b)]. The presence of reactants in aclosed vessel made of conducting material e.g.,copper or steel is an example of a closedsystem.Fig. 6.1 System and the surroundingsNote that the system may be defined byphysical boundaries, like beaker or test tube,or the system may simply be defined by a setof Cartesian coordinates specifying aparticular volume in space. It is necessary tothink of the system as separated from thesurroundings by some sort of wall which may*Fig. 6.2 Open, closed and isolated systems.We could have chosen only the reactants as system then walls of the beakers will act as boundary.2020-21

162CHEMISTRY3. Isolated SystemIn an isolated system, there is no exchange ofenergy or matter between the system and thesurroundings [Fig. 6.2 (c)]. The presence ofreactants in a thermos flask or any other closedinsulated vessel is an example of an isolatedsystem.6.1.3 The State of the SystemThe system must be described in order to makeany useful calculations by specifyingquantitatively each of the properties such asits pressure (p), volume (V), and temperature(T ) as well as the composition of the system.We need to describe the system by specifyingit before and after the change. You would recallfrom your Physics course that the state of asystem in mechanics is completely specified ata given instant of time, by the position andvelocity of each mass point of the system. Inthermodynamics, a different and much simplerconcept of the state of a system is introduced.It does not need detailed knowledge of motionof each particle because, we deal with averagemeasurable properties of the system. We specifythe state of the system by state functions orstate variables.The state of a thermodynamic system isdescribed by its measurable or macroscopic(bulk) properties. We can describe the state ofa gas by quoting its pressure (p), volume (V),temperature (T ), amount (n) etc. Variables likep, V, T are called state variables or statefunctions because their values depend onlyon the state of the system and not on how it isreached. In order to completely define the stateof a system it is not necessary to define all theproperties of the system; as only a certainnumber of properties can be variedindependently. This number depends on thenature of the system. Once these minimumnumber of macroscopic properties are fixed,others automatically have definite values.The state of the surroundings can neverbe completely specified; fortunately it is notnecessary to do so.6.1.4 The Internal Energy as a StateFunctionWhen we talk about our chemical systemlosing or gaining energy, we need to introducea quantity which represents the total energyof the system. It may be chemical, electrical,mechanical or any other type of energy youmay think of, the sum of all these is the energyof the system. In thermodynamics, we call itthe internal energy, U of the system, which maychange, when heat passes into or out of the system, work is done on or by the system, matter enters or leaves the system.These systems are classified accordingly asyou have already studied in section 6.1.2.(a) WorkLet us first examine a change in internalenergy by doing work. We take a systemcontaining some quantity of water in a thermosflask or in an insulated beaker. This would notallow exchange of heat between the systemand surroundings through its boundary andwe call this type of system as adiabatic. Themanner in which the state of such a systemmay be changed will be called adiabaticprocess. Adiabatic process is a process inwhich there is no transfer of heat between thesystem and surroundings. Here, the wallseparating the system and the surroundingsis called the adiabatic wall (Fig 6.3).Let us bring the change in the internalenergy of the system by doing some work onFig. 6.3 An adiabatic system which does notpermit the transfer of heat through itsboundary.it. Let us call the initial state of the system asstate A and its temperature as TA. Let theinternal energy of the system in state A becalled UA. We can change the state of the systemin two different ways.2020-21

THERMODYNAMICS163One way: We do some mechanical work, say1 kJ, by rotating a set of small paddles andthereby churning water. Let the new state becalled B state and its temperature, as TB. It isfound that T B T A and the change intemperature, T T B–TA. Let the internalenergy of the system in state B be UB and thechange in internal energy, U UB– UA.Second way: We now do an equal amount (i.e.,1kJ) electrical work with the help of animmersion rod and note down the temperaturechange. We find that the change in temperatureis same as in the earlier case, say, TB – TA.In fact, the experiments in the abovemanner were done by J. P. Joule between1840–50 and he was able to show that a givenamount of work done on the system, no matterhow it was done (irrespective of path) producedthe same change of state, as measured by thechange in the temperature of the system.So, it seems appropriate to define aquantity, the internal energy U, whose valueis characteristic of the state of a system,whereby the adiabatic work, wad required tobring about a change of state is equal to thedifference between the value of U in one stateand that in another state, U i.e., U U 2 U 1 w adTherefore, internal energy, U, of the systemis a state function.By conventions of IUPAC in chemicalthermodynamics. The positive sign expressesthat wad is positive when work is done on thesystem and the internal energy of systemincreases. Similarly, if the work is done by thesystem,wad will be negative because internalenergy of the system decreases.Can you name some other familiar statefunctions? Some of other familiar statefunctions are V, p, and T. For example, if webring a change in temperature of the systemfrom 25 C to 35 C, the change in temperatureis 35 C–25 C 10 C, whether we go straightup to 35 C or we cool the system for a fewdegrees, then take the system to the finaltemperature. Thus, T is a state function andthe change in temperature is independent of*the route taken. Volume of water in a pond,for example, is a state function, becausechange in volume of its water is independentof the route by which water is filled in thepond, either by rain or by tubewell or by both.(b) HeatWe can also change the internal energy of asystem by transfer of heat from thesurroundings to the system or vice-versawithout expenditure of work. This exchangeof energy, which is a result of temperaturedifference is called heat, q. Let us considerbringing about the same change in temperature(the same initial and final states as before insection 6.1.4 (a) by transfer of heat throughthermally conducting walls instead ofadiabatic walls (Fig. 6.4).Fig. 6.4A system which allows heat transferthrough its boundary.We take water at temperature, TA in acontainer having thermally conducting walls,say made up of copper and enclose it in a hugeheat reservoir at temperature, TB. The heatabsorbed by the system (water), q can bemeasured in terms of temperature difference ,TB – TA. In this case change in internal energy, U q, when no work is done at constantvolume.By conventions of IUPAC in chemicalthermodynamics. The q is positive, whenheat is transferred from the surroundings tothe system and the internal energy of thesystem increases and q is negative whenheat is transferred from system to thesurroundings resulting in decrease of theinternal energy of the system.Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by thesystem. This is still followed in physics books, although IUPAC has recommended the use of new sign convention.2020-21

164CHEMISTRY(c) The general caseLet us consider the general case in which achange of state is brought about both bydoing work and by transfer of heat. We writechange in internal energy for this case as: U q w(6.1)For a given change in state, q and w canvary depending on how the change is carriedout. However, q w U will depend only oninitial and final state. It will be independent ofthe way the change is carried out. If there isno transfer of energy as heat or as work(isolated system) i.e., if w 0 and q 0, then U 0.The equation 6.1 i.e., U q w ismathematical statement of the first law ofthermodynamics, which states thatThe energy of an isolated system isconstant.It is commonly stated as the law ofconservation of energy i.e., energy can neitherbe created nor be destroyed.Note: There is considerable difference betweenthe character of the thermodynamic propertyenergy and that of a mechanical property suchas volume. We can specify an unambiguous(absolute) value for volume of a system in aparticular state, but not the absolute value ofthe internal energy. However, we can measureonly the changes in the internal energy, U ofthe system.Problem 6.1Express the change in internal energy ofa system when(i) No heat is absorbed by the systemfrom the surroundings, but work (w)is done on the system. What type ofwall does the system have ?(ii) No work is done on the system, butq amount of heat is taken out fromthe system and given to thesurroundings. What type of wall doesthe system have?(iii) w amount of work is done by thesystem and q amount of heat issupplied to the system. What type ofsystem would it be?Solution(i) U w ad, wall is adiabatic(ii) U – q, thermally conducting walls(iii) U q – w, closed system.6.2 APPLICATIONSMany chemical reactions involve the generationof gases capable of doing mechanical work orthe generation of heat. It is important for us toquantify these changes and relate them to thechanges in the internal energy. Let us see how!6.2.1 WorkFirst of all, let us concentrate on the nature ofwork a system can do. We will consider onlymechanical work i.e., pressure-volume work.For understanding pressure-volumework, let us consider a cylinder whichcontains one mole of an ideal gas fitted with africtionless piston. Total volume of the gas isVi and pressure of the gas inside is p. Ifexternal pressure is pex which is greater thanp, piston is moved inward till the pressureinside becomes equal to pex. Let this changeFig. 6.5(a) Work done on an ideal gas in acylinder when it is compressed by aconstant external pressure, p ex(in single step) is equal to the shadedarea.2020-21

THERMODYNAMICS165be achieved in a single step and the finalvolume be Vf . During this compression,suppose piston moves a distance, l and iscross-sectional area of the piston is A[Fig. 6.5(a)].then, volume change l A V (Vf – Vi )We also know, pressure forceareaVfw Therefore, force on the piston pex . AIf w is the work done on the system bymovement of the piston thenw force distance pex . A .l pex . (– V) – pex V – pex (Vf – Vi )If the pressure is not constant but changesduring the process such that it is alwaysinfinitesimally greater than the pressure of thegas, then, at each stage of compression, thevolume decreases by an infinitesimal amount,dV. In such a case we can calculate the workdone on the gas by the relation(6.2)The negative sign of this expression isrequired to obtain conventional sign for w,which will be positive. It indicates that in caseof compression work is done on the system.Here (Vf – Vi ) will be negative and negativemultiplied by negative will be positive. Hencethe sign obtained for the work will be positive.If the pressure is not constant at everystage of compression, but changes in numberof finite steps, work done on the gas will besummed over all the steps and will be equalto p V [Fig. 6.5 (b)]Fig. 6.5 (b) pV-plot when pressure is not constantand changes in finite steps duringcompression from initial volume, Vi tofinal volume, Vf . Work done on the gasis represented by the shaded area. pexdV( 6.3)ViHere, pex at each stage is equal to (pin dp) incase of compression [Fig. 6.5(c)]. In anexpansion process under similar conditions,the external pressure is always less than thepressure of the system i.e., pex (pin– dp). Ingeneral case we can write, pex (pin dp). Suchprocesses are called reversible processes.A process or change is said to bereversible, if a change is brought out insuch a way that the process could, at anymoment, be reversed by an infinitesimalchange. A reversible process proceedsinfinitely slowly by a series of equilibriumstates such that system and thesurroundings are always in nearequilibrium with each other. ProcessesFig. 6.5 (c) pV-plot when pressure is not constantand changes in infinite steps(reversible conditions) duringcompression from initial volume, Vi tofinal volume, Vf . Work done on the gasis represented by the shaded area.2020-21

166CHEMISTRYother than reversible processes are knownas irreversible processes.In chemistry, we face problems that canbe solved if we relate the work term to theinternal pressure of the system. We canrelate work to internal pressure of the systemunder reversible conditions by writingequation 6.3 as follows:Vfw rev pVfexdV Vi (pin dp ) dVViSince dp dV is very small we can writeIsothermal and free expansion of anideal gasFor isothermal (T constant) expansion of anideal gas into vacuum ; w 0 since pex 0.Also, Joule determined experimentally thatq 0; therefore, U 0Equation 6.1, U q w can beexpressed for isothermal irreversible andreversible changes as follows:1.For isothermal irreversible changeq – w pex (Vf – Vi )2.For isothermal reversible changeVfVfw rev pin dVq – w nRT ln Vi(6.4)ViNow, the pressure of the gas (pin which wecan write as p now) can be expressed in termsof its volume through gas equation. For n molof an ideal gas i.e., pV nRT 2.303 nRT log3.nRTVTherefore, at constant temperature (isothermalprocess), p Vfw rev nRTViVfdV nRT lnVVi – 2.303 nRT logVfVi(6.5)Free expansion: Expansion of a gas invacuum (pex 0) is called free expansion. Nowork is done during free expansion of an idealgas whether the process is reversible orirreversible (equation 6.2 and 6.3).Now, we can write equation 6.1 in numberof ways depending on the type of processes.Let us substitute w – pex V (eq. 6.2) inequation 6.1, and we get U q pex VIf a process is carried out at constant volume( V 0), then U qVthe subscript V in qV denotes that heat issupplied at constant volume.2020-21VfViFor adiabatic change, q 0, U wadProblem 6.2Two litres of an ideal gas at a pressure of10 atm expands isothermally at 25 C intoa vacuum until its total volume is 10 litres.How much heat is absorbed and how muchwork is done in the expansion ?SolutionWe have q – w pex (10 – 2) 0(8) 0No work is done; no heat is absorbed.Problem 6.3Consider the same expansion, but thistime against a constant external pressureof 1 atm.SolutionWe have q – w pex (8) 8 litre-atmProblem 6.4Consider the expansion given in problem6.2, for 1 mol of an ideal gas conductedreversibly.SolutionVfVs 10 2.303 1 0.8206 298 logWe have q – w 2.303 nRT log2

THERMODYNAMICS167 2.303 x 0.8206 x 298 x log 5 2.303 x 0.8206 x 298 x 0.6990 393.66 L atm6.2.2 Enthalpy, H(a) A Useful New State FunctionWe know that the heat absorbed at constantvolume is equal to change in the internalenergy i.e., U q . But most of chemicalVreactions are carried out not at constantvolume, but in flasks or test tubes underconstant atmospheric pressure. We need todefine another state function which may besuitable under these conditions.We may write equation (6.1) as U qp – p V at constant pressure, where qpis heat absorbed by the system and –p Vrepresent expansion work done by the system.Let us represent the initial state bysubscript 1 and final state by 2We

THERMODYNAMICS 163 One way: We do some mechanical work, say 1 kJ, by rotating a set of small paddles and thereby chur ning water . Let the new state be called B state and its temperature, as T B