Chapter 6 Heat capacity, enthalpy, & entropy

1 Chapter 6 Heat capacity, enthalpy, & entropy1 By eq. & IntroductionIntegration of Eq. ( ) between the states ( 2, )and ( 1, )gives thedifference between the molar enthalpies of the two states asIn this lecture, we examine the heat capacity as a function of temperature, compute the enthalpy, entropy , and Gibbs free energy, as functions of then begin to assess phase equilibria constructing a phase diagram for a single component (unary) system.( )( )( )( )( )2- Empirical rule by Dulongand Petit (1819) : Cv 3R(classicaltheory: avg. E for 1-D oscillator, = kT, E = 3N0kT = 3RT)- Calculation of Cvof a solid element as a function ofT by the quantum theory: First calculation by Einstein (1907)- Einstein crystal a crystal containing n atoms, each of whichbehaves as a harmonic oscillator vibrating independentlydiscrete energy = +12 a system of 3n linear harmonic oscillators(due to vibration in the x, y, and z directions) THEORETICAL CALCULATION OF THE HEAT CAPACITYThe Energy of Einstein crystal( )( )3 Using, = +12 & eq.

2 Into THEORETICAL CALCULATION OF THE HEAT THEORETICAL CALCULATION OF THE HEAT CAPACITY where = , givesandin which case( ) Differentiation of eq. with respect to temperature at constant volume5 Defining = : Einstein characteristic THEORETICAL CALCULATION OF THE HEAT CAPACITY 0 0the Einstein equation goodat higher T, the theoretical values approach zero more rapidly than do the actual values.( )6 Problem: althoughtheEinsteinequationadequatelyrep resentsactualheatcapacitiesat highertemperatures,thetheoreticalvaluesa pproachzeromorerapidlythandotheactualval ues. Thisdiscrepancyis causedbythefactthattheoscillatorsdonotvi bratewitha singlefrequency. In a crystallatticeas a harmonicoscillator,energyis expressedas = 2+ (n = 0,1,2,.. )Einsteinassumedthat is const.

3 Forall thesameatomsin theoscillator. Debye sassumption(1912) : therangeof frequenciesof vibrationavailableto theoscillatorsisthesameas thatavailableto theelasticvibrationsin a THEORETICAL CALCULATION OF THE HEAT CAPACITY: the maximum frequency of vibration of an oscillator7 Integration Einstein s equation in the range, 0 THEORETICAL CALCULATION OF THE HEAT CAPACITY obtained the heat capacity of the solidwhich, with x=h /kT, gives( ) Defining = = : Debye characteristic T (Debye frequency)= = Debye sequationgivesanexcellentfit to The value of the integral in Eq. ( ) from 0 to infinity is , and thus, for very low temperatures, Eq. ( ) THEORETICAL CALCULATION OF THE HEAT CAPACITY( ): Debye 3law for low-temperature heat stheory: Noconsiderationonthecontributionmadeto theheatcapacitybytheuptakeofenergybyelec trons( absolutetemperature) At high T, where the lattice contribution approaches the Dulongand Petit value, the molar Cvshould vary with T asin which bTis the electronic By experimental measurements, THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES: Normally fitted10 For a closed system of fixed composition, with a change in T from T1to T2at the const.

4 Pi) H = HT2, P HT1, P = T1T2 CpdT: His the area under a plot of ii) A + B = AB chem. rxnor phase change at const. T, P HT, P = HABT, P HAT, P HBT, P: Hess law H < 0 exothermic H > 0 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION( )( )11 Enthalpy changeConsider the change of statewhere ( )istheheatrequiredtoincreasethetemperatu reof onemoleof solidAfrom 1to 2at constantpressure.( ) ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION12or where( )conventionassignsthevalueof zeroto H of elementsin theirstablestatesat ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITIONex) M(s) + 1/2O2g = MOs at 298K 298= ,298 ,298 12 2 ,298= ,298as ,298& 2 ,298=0 by : FortheoxidationPb +12O2= PbOwithH of12moleofO2gas ,1moleofPb(s)at 298K(= 0 byconvention)ab:298 T 600 K, whereHPb(s)= 298 TCp,Pb(s)dT ac:298 T 3000K, whereH12O2(g)=12 298 TCp,O2(g)dT.

5 HPbOs,298 K=-219,000 Jde:298 T 1159 KwhereHPbOs,T= 219 ,000 + 298 TCp,PbO(s)dT ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION14 WithH of12moleofO2(g)and1moleofPb(s)at298K(=0 byconvention)f : H of12moleofO2(g)and1moleofPb(s)at : H of 1moleofPbO(s)at ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION15 From the data in Table ,and, thus, from 298 to 600 K ( , )WithT=500K, H500 K= 217 ,800 JIn Fig . ,h:H of 1 moleof ( )at of 600K and600to 1200K, ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION16 In Fig. , ajkl: H of 1 mole of Pband 1 mole of O2(g), and hence HT is calculated from the ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION17 ThusThis gives 1000= 216,700 at = ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION18If theT of interestis higherthantheTmof boththemetalanditsoxide, ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION19 If thesystemcontainsa low-temperaturephasein equilibriumwitha high-temperaturephaseattheequilibriumpha setransitiontemperaturethenintroductiono fheattothesystem(theexternalinfluence)wo uldbeexpectedto increasethetemperatureof thesystem(theeffect)byLe Chatelier sprinciple.

6 However,thesystemundergoesanendothermicc hange,whichabsorbstheheatintroducedatcon stanttemperature,andhencenullifiestheeff ectof theexternalinfluence. Theendothermicprocessis themeltingof someof thesolid. A phasechangefroma low- to a high-temperaturephaseis alwaysendothermic,andhence Hforthechangeis alwaysa positivequantity. Thus Hmis alwayspositive. ThegeneralEq. ( ) canbeobtainedas ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION20 Subtraction givesorand integrating from state 1 to state 2 gives( )( )Equations ( ) and ( ) are expressions of Kirchhoff s ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION21 The 3rd law of thermodynamics: entropy of homogeneous substance at complete internalequilibrium state is 0 at 0 a closed system undergoing a reversible process ,6. 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS( )At const.

7 P,As T increased, ( )the molar S of the system at any Tis given by( )226. 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICSWhy? by differentiating Eq. ( ) G = H TS with respect to T at constant P:From Eq. ( )thusdG= -SdT+ VdP 0 as T (1906) T. W. Richards (1902) found experimentally that S 0 and Cp 0 as T 0. (Clue for the 3rdlaw) 0236. 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS(i) Cp= iCpi 0means that each Cpi 0 (solutions) by Einstein & Debye (T 0, Cv 0)(ii) S = iSi 0 means that each Si , every particles should be at ground state at 0 K, ( th= 1)every particles should be uniform in concentration ( conf= 1). Thus, it should be at internal equilibrium. Plank statementthus, th= conf= 1 24 If( G T)Pand( H T)P 0asT 0, S& CP 0 as T 0 Nernst sheattheoremstatesthat forall reactionsinvolvingsubstancesin thecondensedstate, Sis zeroat theabsolutezeroof temperature Thus,forthegeneralreactionA + B = AB, = = 0 = 0andif and areassignedthevalueof zeroat 0 K,thenthecompoundABalsohaszeroentropyat 0 K.

8 Theincompletenessof Nernst stheoremwaspointedoutbyPlanck,whostatedt hat theentropyof anyhomogeneoussubstance, whichis in completeinternalequilibrium, maybetakento bezeroat 0 K. 6. 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS25 Glasses- noncrystalline, supercooledliquidsliquid-like disordered atom arrangements frozen into solid glassy state metastable- 0 0, depending on degree of atomic order Solutions- mixture of atoms, ions or molecules- entropy of mixing- atomic randomness of a mixture determines its degree of order: complete ordering : every A is coordinated only by B atoms and vice versa: complete randomness : 50% of the neighbors of every atom are A atoms and 50% are B 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS thesubstancebein completeinternalequilibrium:26 Even chemically pure elements-mixtures of isotopes entropy of mixingex)Cl35 Cl37 Point defects- entropy of mixing with vacancyEx) Solid CO Structure6.

9 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS27 Maximumvalueif equalnumbersofmoleculeswereorientedin oppositedirectionsandrandommixingof thetwoorientationsoccurred. FromEq.( ) themolarconfigurationalentropyof mixingwouldbeusing Stirling sapproximation,measured value: J/mole K : requires complete internal equilibrium6. 5 THEDEPENDENCEOFENTROPYONTEMPERATUREANDTH ETHIRDLAWOFTHERMODYNAMICS28 The Third Law can be verified by considering a phase transition in an element such as where & are allotropes of the element and this for the case of sulfur: 6. 6 EXPERIMENTALVERIFICATIONOFTHETHIRDLAWFor the cycle shown in Fig. the Third Law to be obeyed, IV=0, which requires thatwhere29 In , a monoclinicformwhichis K K Themeasuredheatcapacitiesgive6. 6 EXPERIMENTALVERIFICATIONOFTHETHIRDLAW30 Assigninga valueof zerotoS0allowstheabsolutevalueof theentropyof anymaterialto bedeterminedas6.

10 6 EXPERIMENTALVERIFICATIONOFTHETHIRDLAW andmolarentropiesarenormallytabulatedat 298K, whereWith the constant-pressure molar heat capacity of the solid expressed in the formthe molar entropy of the solid at the temperature T is obtained asWhen T> 31 Richard s rule (generally metal) (FCC) , (BCC)Trouton s rule (generally metal)-more useful!! 88J/K(for both FCC and BCC)From FCCFrom BCC6. 6 EXPERIMENTALVERIFICATIONOFTHETHIRDLAW326 . 6 EXPERIMENTALVERIFICATIONOFTHETHIRDLAWB ecauseofthesimilarmolarS ofthecondensedphasesPbandPbO, it is seenthat S forthereaction,is very nearly equal to 12 , 2at 298K S isofsimilarmagnitudetothatcausedbythedis appearanceof thegas, , of12moleof O2(g)33(i) Fora closedsystemof fixedcomposition,witha changeof P at const. T,6. 7 THEINFLUENCEOFPRESSUREONENTHALPYANDENTRO PY(dH=TdS+VdP)Maxwell s equation ( ) gives ( ) = ( ) andThus34 The change in molar enthalpy caused by the change in state from (P1, T) to (P2, T) is thus6.