《物理化学》课程教材讲义（英文）17 Statistical thermodynamics（2）applications

Statistical17thermodynamics 2applicationsInthis chapterweapplytheconceptsofstatistical thermodynamicstothecalculationofFundamental relationschemically significant quantities. First, we establish the relations between thermodynamic17.1Thethermodynamicfunctionsfunctionsandpartitionfunctions.Next,weshowthatthemolecularpartitionfunctioncanbefactorizedintocontributionsfromeachmodeofmotionandestablishtheformulasforthe17.2Themolecularpartitionfunctionpartitionfunctionsfortranslational,rotational,andvibrationalmodesofmotionandthecon-tributionofelectronicexcitation.ThesecontributionscanbecalculatedfromspectroscopicUsing statisticaldata.Finallywetumtospecificapplications,whichincludethemeanenergiesofmodesofthermodynamicsmotion,the heatcapacities of substances,and residual entropies.In the final section,wesee how to calculate the equilibrium constant of a reaction and through that calculation17.3Mean energiesunderstand some of the molecular features that determine the magnitudes of equilibrium17.4 Heat capacitiesconstantsandtheirvariationwithtemperature.17.5 Equations of state17.6 Molecular interactionsinA partition function is the bridge between thermodynamics, spectroscopy, andliquidsquantum mechanics. Once it is known, a partition function can be used to calculate17.7 Residual entropiesthermodynamicfunctions,heat capacities,entropies,and equilibrium constants.Italso sheds light on the significance ofthese properties.17.8EquilibriumconstantsChecklist of key ideasFundamental relationsFurther readingDiscussion questionsIn this section we seehowto obtain anythermodynamic function once weknowtheExercisespartitionfunction.Then we seehowtocalculatethemolecularpartitionfunction,andProblemsthrough that the thermodynamic functions, from spectroscopic data.17.1ThethermodynamicfunctionsWe have already derived (in Chapter 16) the two expressions for calculating theinternal energy and the entropy of a system from its canonical partition function,Q:alnQU-U(0)+klnQ(17.1)U-U():Tawhere β= 1/kT. If the molecules are independent, we can go on to make the substitu-tions Q=qN (for distinguishable molecules, as in a solid) or Q=qN/N! (for indistin-guishable molecules, as in a gas). All the thermodynamic functions introduced in Part 1arerelated to Uand S, so wehavea route to theircalculationfromQ

Statistical thermodynamics 2: applications In this chapter we apply the concepts of statistical thermodynamics to the calculation of chemically significant quantities. First, we establish the relations between thermodynamic functions and partition functions. Next, we show that the molecular partition function can be factorized into contributions from each mode of motion and establish the formulas for the partition functions for translational, rotational, and vibrational modes of motion and the con￾tribution of electronic excitation. These contributions can be calculated from spectroscopic data. Finally, we turn to specific applications, which include the mean energies of modes of motion, the heat capacities of substances, and residual entropies. In the final section, we see how to calculate the equilibrium constant of a reaction and through that calculation understand some of the molecular features that determine the magnitudes of equilibrium constants and their variation with temperature. A partition function is the bridge between thermodynamics, spectroscopy, and quantum mechanics. Once it is known, a partition function can be used to calculate thermodynamic functions, heat capacities, entropies, and equilibrium constants. It also sheds light on the significance of these properties. Fundamental relations In this section we see how to obtain any thermodynamic function once we know the partition function. Then we see how to calculate the molecular partition function, and through that the thermodynamic functions, from spectroscopic data. 17.1 The thermodynamic functions We have already derived (in Chapter 16) the two expressions for calculating the internal energy and the entropy of a system from its canonical partition function, Q: U − U(0) = − V S = + k ln Q (17.1) where β = 1/kT. If the molecules are independent, we can go on to make the substitu￾tions Q = qN (for distinguishable molecules, as in a solid) or Q = qN/N! (for indistin￾guishable molecules, as in a gas). All the thermodynamic functions introduced in Part 1 are related to U and S, so we have a route to their calculation from Q. U − U(0) T D F ∂ ln Q ∂β A C 17 Fundamental relations 17.1 The thermodynamic functions 17.2 The molecular partition function Using statistical thermodynamics 17.3 Mean energies 17.4 Heat capacities 17.5 Equations of state 17.6 Molecular interactions in liquids 17.7 Residual entropies 17.8 Equilibrium constants Checklist of key ideas Further reading Discussion questions Exercises Problems

59017STATISTICALTHERMODYNAMICS2:APPLICATIONS(a)The Helmholtz energyThe Helmholtz energy, A, is defined as A = U- TS. This relation implies that A(O) =U(0),so substitution for UandSbyusing eqn 17.1leads to thevery simple expressionA-A(0)=-kT ln Q(17.2)(b)ThepressureBy an argument like that leading to eqn 3.31, it follows from A = U -TS thatdA=-pdV-SdT.Therefore, on imposingconstant temperature, thepressure and theHelmholtz energy are related by p=-(0A/3V) It then follows from eqn 17.2 thatalnQ(17.3)p=kTavThis relation is entirely general, and may be used for any type ofsubstance, includingperfectgases, real gases, and liquids. Because Qis in general afunction ofthe volume,temperature,and amount of substance, eqn 17.3 is an equation of state.Example17.1Derivinganequationof stateDerive an expression for the pressure ofa gas ofindependent particles.Method Weshould suspectthatthe pressureis that given bythe perfectgas law.Toproceedsystematically,substitutetheexplicitformulaforQforagasofindependent, indistinguishable molecules (see eqn 16.45 and Table 17.3 at the end of thechapter)intoeqn17.3.AnswerForagasofindependentmolecules,Q=qN/N!withq=VIA()-(%)-()p=kT14NkTAINkTnRTVAVVTo derive this relation, we have usedA(VIA)dq3Aavaand NkT=nNkT=nRT.The calculation shows that theequation of stateofa gasof independent particles is indeed the perfect gas law.Self-test17.1Derivetheequation ofstateofa sampleforwhichQ=qNf/N,withq=ViA,wherefdepends onthevolume.[p=nRT/V+kT(lnfIaV)](c)The enthalpyAt this stage we can use the expressions for U and p in the definition H= U+ pVtoobtain an expression for theenthalpy,H,ofany substance:(alnQ(alnQH-H(0) :+kT(17.4)ava

590 17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS (a) The Helmholtz energy The Helmholtz energy, A, is defined as A = U − TS. This relation implies that A(0) = U(0), so substitution for U and S by using eqn 17.1 leads to the very simple expression A − A(0) = −kT ln Q (17.2) (b) The pressure By an argument like that leading to eqn 3.31, it follows from A = U − TS that dA = −pdV − SdT. Therefore, on imposing constant temperature, the pressure and the Helmholtz energy are related by p = −(∂A/∂V)T. It then follows from eqn 17.2 that p = kT T (17.3) This relation is entirely general, and may be used for any type of substance, including perfect gases, real gases, and liquids. Because Q is in general a function of the volume, temperature, and amount of substance, eqn 17.3 is an equation of state. Example 17.1 Deriving an equation of state Derive an expression for the pressure of a gas of independent particles. Method We should suspect that the pressure is that given by the perfect gas law. To proceed systematically, substitute the explicit formula for Q for a gas of independ￾ent, indistinguishable molecules (see eqn 16.45 and Table 17.3 at the end of the chapter) into eqn 17.3. Answer For a gas of independent molecules, Q = qN/N! with q = V/Λ3 : p = kT T = T = T = ×= = To derive this relation, we have used T = T = and NkT = nNAkT = nRT. The calculation shows that the equation of state of a gas of independent particles is indeed the perfect gas law. Self-test 17.1 Derive the equation of state of a sample for which Q = qNf/N!, with q = V/Λ3 , where f depends on the volume. [p = nRT/V + kT(∂ ln f/∂V)T] (c) The enthalpy At this stage we can use the expressions for U and p in the definition H = U + pV to obtain an expression for the enthalpy, H, of any substance: H − H(0) = − V + kTV T (17.4) D F ∂ ln Q ∂V A C D F ∂ ln Q ∂β A C 1 Λ3 D F ∂(V/Λ3 ) ∂V A C D F ∂q ∂V A C nRT V NkT V 1 Λ3 NkTΛ3 V D F ∂q ∂V A C NkT q D F ∂Q ∂V A C kT Q D F ∂ ln Q ∂V A C D F ∂ ln Q ∂V A C

59117.2THEMOLECULARPARTITIONFUNCTIONWe have already seen that U- U(O)=号nRT for a gas of independent particles (eqn16.32a),and have just shown that pV= nRT. Therefore, for such a gas,H-H(0)=号nRT(17.5)(d) The Gibbs energyOne of the most important thermodynamic functions for chemistry is the Gibbsenergy,G=H-TS=A+pV.Wecan now express this function in terms ofthe parti-tion function by combining the expressions for A and p:alnQG-G(0)=-kTIn Q+kTV(17.6)avThis expression takes a simpleformfora gas ofindependentmolecules becausepVinthe expression G= A + pV can be replaced by nRT:G-G(0)=-kTln Q+nRT(17.7)°Furthermore, because Q= qN/N!, and therefore ln Q=N ln q- In N, it follows byusing Stirling'sapproximation (ln N!Nln N-N)that wecan writeG-G(0)=-NkT In q+ kTIn N!+ nRT=-nRT Inq+kT(NIn N-N)+nRT=-nRTIn!(17.8)°Nwith N = nN.Now we see another interpretation of the Gibbs energy: it is pro-portional to the logarithm of the average number of thermally accessible states permolecule.It will turn out to be convenient to define the molar partition function, qm=q/n(withunits mol-l),forthenG-G(0)=-nRTIn m(17.9)°NA17.2 Themolecular partition functionThe energy of a molecule is the sum of contributions from its different modes ofmotion:E,=++y+e(17.10)where T denotes translation, Rrotation,Vvibration, and E the electronic contribu-tion.The electronic contribution isnot actually a‘mode of motion',but it is con-venient to include it here.The separation of terms in eqn 17.10 is only approximate(except for translation)because the modes are not completely independent, but inmost cases it is satisfactory.The separation ofthe electronic and vibrational motionsis justified provided only the ground electronic state is occupied (for otherwise thevibrational characteristics depend on the electronic state) and, for the electronicground state, that the Born-Oppenheimer approximation is valid (Chapter 11). Theseparation of the vibrational and rotational modes is justified to the extent that therotational constant is independent of the vibrational stateGiventhattheenergyis asum ofindependentcontributions,thepartitionfunctionfactorizesintoaproductofcontributions(recallSection16.2b):

17.2 THE MOLECULAR PARTITION FUNCTION 591 We have already seen that U − U(0) = 3 –2nRT for a gas of independent particles (eqn 16.32a), and have just shown that pV = nRT. Therefore, for such a gas, H − H(0) = 5 –2nRT (17.5)° (d) The Gibbs energy One of the most important thermodynamic functions for chemistry is the Gibbs energy, G = H − TS = A + pV. We can now express this function in terms of the parti￾tion function by combining the expressions for A and p: G − G(0) = −kT ln Q + kTV T (17.6) This expression takes a simple form for a gas of independent molecules because pV in the expression G = A + pV can be replaced by nRT: G − G(0) = −kT ln Q + nRT (17.7)° Furthermore, because Q = qN/N!, and therefore ln Q = N ln q − ln N!, it follows by using Stirling’s approximation (ln N! ≈ N ln N −N) that we can write G − G(0) = −NkT ln q + kT ln N! + nRT = −nRT ln q + kT(N ln N − N) + nRT = −nRT ln (17.8)° with N = nNA. Now we see another interpretation of the Gibbs energy: it is pro￾portional to the logarithm of the average number of thermally accessible states per molecule. It will turn out to be convenient to define the molar partition function, qm = q/n (with units mol−1 ), for then G − G(0) = −nRT ln (17.9)° 17.2 The molecular partition function The energy of a molecule is the sum of contributions from its different modes of motion: εi = ε i T + εi R + ε i V + ε i E (17.10) where T denotes translation, R rotation, V vibration, and E the electronic contribu￾tion. The electronic contribution is not actually a ‘mode of motion’, but it is con￾venient to include it here. The separation of terms in eqn 17.10 is only approximate (except for translation) because the modes are not completely independent, but in most cases it is satisfactory. The separation of the electronic and vibrational motions is justified provided only the ground electronic state is occupied (for otherwise the vibrational characteristics depend on the electronic state) and, for the electronic ground state, that the Born–Oppenheimer approximation is valid (Chapter 11). The separation of the vibrational and rotational modes is justified to the extent that the rotational constant is independent of the vibrational state. Given that the energy is a sum of independent contributions, the partition function factorizes into a product of contributions (recall Section 16.2b): qm NA q N D F ∂ ln Q ∂V A C

59217STATISTICALTHERMODYNAMICS2:APPLICATIONSZe-Bet-Pet-Be1-Pef 81i (all states)ZZ7Z e-Pec-pef-pet-βef(17.11)i(translational) i(rotational)i(vibrational)i(clectronic)Ee-peyZe1e-βerEe-peB(i (rotational)i (vibrational)(translational)i(clectronic)=q'qRq'gEThis factorization means that we can investigate each contribution separately.(a) Thetranslational contributionThe translational partition function ofa molecule of mass m in a container of volumeVwas derived in Section 16.2:01/2VhB(17.12)q:443(2元mkT)1/22元mNoticethatqToasT-oobecausean infinitenumberofstatesbecomesaccessibleas the temperature is raised.Even at room temperature q~2× 1028 for an O,molecule in a vessel of volume 100 cm?.The thermal wavelength, A, lets us judge whether the approximations that led to theexpression for q' are valid.The approximations are valid if many states are occupied,which requires VIA’ to be large. That will be so if A is small compared with the lineardimensions ofthe container. For H, at 25°C, A=71 pm, which is far smaller than anyconventional containeris likelytobe (but comparableto pores in zeolites or cavitiesin clathrates).For O,a heaviermolecule,A=18pm.We saw in Section16.2that anequivalent criterion ofvalidityis that Ashould be much lessthan theaverage separa-tion of the molecules in the sample.(b)Therotational contribution3Asdemonstrated inExample16.1,thepartitionfunction ofanonsymmetrical (AB)linearrotor isqR= E(2J + 1)e-βhcB(1)(17.13)The direct method of calculating qR is to substitute the experimental values of therotational energylevels into thisexpression and to sum the series numericallyExample17.2Evaluatingtherotationalpartitionfunctionexplicitly1012345678910Evaluate the rotational partition function of 'H35cl at 25°C, given that B =J10.591cm-lMethodWeuseeqn17.13and evaluateittermbyterm.Auseful relationiskT/hc=Fig.17.1The contributionstotherotationalpartition function of an HCl molecule207.22cm-at298.15K.Thesum isreadilyevaluatedbyusingmathematicalsoftware.at 25°C.The vertical axis is the valueAnswerToshowhowsuccessivetermscontribute,wedrawupthefollowingtableof(2]+ 1)e-BhcB +1), Successive termsbyusingkT/hcB=0.05111 (Fig.17.1):(which are proportional to the populationsof thelevels)pass through a maximum2340110becausethepopulation ofindividual states(2) + 1)e-0.0511(J+1)2.713.7913.683.240.08decreases exponentially,but the degeneracy..of the levels increases with J

592 17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 J Contribution Fig. 17.1 The contributions to the rotational partition function of an HCl molecule at 25°C. The vertical axis is the value of (2J + 1)e−βhcBJ(J+1). Successive terms (which are proportional to the populations of the levels) pass through a maximum because the population of individual states decreases exponentially, but the degeneracy of the levels increases with J. q =∑ i e−βεi = ∑ i (all states) e−βεT i −βεR i −βεV i−βεE i = ∑ i (translational) ∑ i (rotational) ∑ i (vibrational) ∑ i (electronic) e−βεT i −βεR i −βεV i−βεE i (17.11) = ∑ i (translational) e−βεT i ∑ i (rotational) e−βεR i ∑ i (vibrational) e−βεV i ∑ i (electronic) e−βεE i = qTqRqVqE This factorization means that we can investigate each contribution separately. (a) The translational contribution The translational partition function of a molecule of mass m in a container of volume V was derived in Section 16.2: qT = Λ = h 1/2 = (17.12) Notice that qT → ∞ as T → ∞ because an infinite number of states becomes accessible as the temperature is raised. Even at room temperature qT ≈ 2 × 1028 for an O2 molecule in a vessel of volume 100 cm3 . The thermal wavelength, Λ, lets us judge whether the approximations that led to the expression for qT are valid. The approximations are valid if many states are occupied, which requires V/Λ3 to be large. That will be so if Λ is small compared with the linear dimensions of the container. For H2 at 25°C, Λ = 71 pm, which is far smaller than any conventional container is likely to be (but comparable to pores in zeolites or cavities in clathrates). For O2, a heavier molecule, Λ = 18 pm. We saw in Section 16.2 that an equivalent criterion of validity is that Λ should be much less than the average separa￾tion of the molecules in the sample. (b) The rotational contribution As demonstrated in Example 16.1, the partition function of a nonsymmetrical (AB) linear rotor is qR =∑ J (2J + 1)e−βhcBJ(J+1) (17.13) The direct method of calculating qR is to substitute the experimental values of the rotational energy levels into this expression and to sum the series numerically. Example 17.2 Evaluating the rotational partition function explicitly Evaluate the rotational partition function of 1 H35Cl at 25°C, given that B = 10.591 cm−1 . Method We use eqn 17.13 and evaluate it term by term. A useful relation is kT/hc = 207.22 cm−1 at 298.15 K. The sum is readily evaluated by using mathematical software. Answer To show how successive terms contribute, we draw up the following table by using kT/hcB = 0.051 11 (Fig. 17.1): J 0 1 2 3 4 . . . 10 (2J + 1)e−0.0511J(J+1) 1 2.71 3.68 3.79 3.24 . . . 0.08 h (2πmkT) 1/2 D F β 2πm A C V Λ3 D F A C D F A C D F A C D F A C

59317.2THEMOLECULARPARTITIONFUNCTIONThe sum requiredbyeqn17.13(the sumof thenumbers inthe second row of thetable) is 19.9, hence q=19.9 at this temperature. Taking J up to 50 gives q=19.902.Noticethat about ten J-levels are significantly populated but the number ofpopulated states is larger on account ofthe (2J+1)-fold degeneracyofeach levelWe shall shortly encounter the approximation that qRkT/hcB, which in the presentcasegivesqR=19.6,ingood agreementwiththeexactvalueand withmuchlesswork.Self-test17.2EvaluatetherotationalpartitionfunctionforHClat0°C.[18.26]AtroomtemperaturekT/hc=200cm-.Therotational constantsofmanymoleculesare close to 1 cm- (Table 13.2) and often smaller (though the very light H, molecule,for which B= 60.9 cm-, is one exception). It follows that many rotational levels arepopulated at normal temperatures.When this is the case,thepartition function maybeapproximatedbykTqRLinear rotors:(17.14a)hcB(17.14b)Nonlinear rotors:where A,B,and Caretherotational constants ofthemolecule.However,beforeusingthese expressions, read on (to eqns 17.15 and 17.16).Justification17.1Therotationalcontribution to themolecularpartitionfunctionWhenmanyrotational statesareoccupiedandkTismuchlargerthantheseparationbetween neighbouring states,the sum in the partitionfunction can be approxim-atedbyan integral, much as we did fortranslational motion in Justification 16.2:(2J + 1)e-BhcB(+)d)Althoughthis integral looks complicated, itcan beevaluated without mucheffortbynoticingthatbecauseddaJ(J+1)e(J+I) = a(2J+ 1)e(I+I)[+djitcan alsobewrittenasdβhcBJ( J+1) 94dBhcBdjThen, because the integral of a derivative of a function is thefunction itself, weobtain11eβhcBJ(J+1)akβhcBBhcBwhich (becauseβ=1/kT)iseqn17.14a.The calculation for a nonlinear molecule is along the same lines, but slightlytrickier.First, we note that the energies ofa symmetric rotor areEjk,M,= hcBJ(J+ 1) + hc(A - B)K

17.2 THE MOLECULAR PARTITION FUNCTION 593 The sum required by eqn 17.13 (the sum of the numbers in the second row of the table) is 19.9, hence qR = 19.9 at this temperature. Taking J up to 50 gives qR = 19.902. Notice that about ten J-levels are significantly populated but the number of populated states is larger on account of the (2J + 1)-fold degeneracy of each level. We shall shortly encounter the approximation that qR ≈ kT/hcB, which in the pre￾sent case gives qR = 19.6, in good agreement with the exact value and with much less work. Self-test 17.2 Evaluate the rotational partition function for HCl at 0°C. [18.26] At room temperature kT/hc ≈ 200 cm−1 . The rotational constants of many molecules are close to 1 cm−1 (Table 13.2) and often smaller (though the very light H2 molecule, for which B = 60.9 cm−1 , is one exception). It follows that many rotational levels are populated at normal temperatures. When this is the case, the partition function may be approximated by Linear rotors: qR = (17.14a) Nonlinear rotors: qR = 3/2 1/2 (17.14b) where A, B, and C are the rotational constants of the molecule. However, before using these expressions, read on (to eqns 17.15 and 17.16). Justification 17.1 The rotational contribution to the molecular partition function When many rotational states are occupied and kT is much larger than the separation between neighbouring states, the sum in the partition function can be approxim￾ated by an integral, much as we did for translational motion in Justification 16.2: qR = ∞ 0 (2J + 1)e−βhcBJ(J+1)dJ Although this integral looks complicated, it can be evaluated without much effort by noticing that because eaJ(J+1) = aJ(J + 1) eaJ(J+1) = a(2J + 1)eaJ(J+1) it can also be written as qR = ∞ 0 e−βhcBJ(J+1) dJ Then, because the integral of a derivative of a function is the function itself, we obtain qR = − e−βhcBJ(J+1) 0 ∞ = which (because β = 1/kT) is eqn 17.14a. The calculation for a nonlinear molecule is along the same lines, but slightly trickier. First, we note that the energies of a symmetric rotor are EJ,K,MJ = hcBJ(J + 1) + hc(A − B)K2 1 βhcB 1 βhcB D E F d dJ A B C 1 βhcB 5 6 7 d dJ 1 2 3 d dJ D F π ABC A C D F kT hc A C kT hcB

59417STATISTICALTHERMODYNAMICS2:APPLICATIONSK=+JwithJ=0,1,2,..,K=J,J-1,...-,andM,=J,J-1,..-J.Insteadofcon-sidering theseranges, we can cover the same values byallowingK torangefromooto oo,with J confined to|K], [K|+1,.-,oofor each value of K (Fig.17.2).Becausethe energyis independent of M,and there are2J+1 values of M,for each value of J,5each value ofJis 2J+1-fold degenerate.It follows that the partition function00:0064AZ,e-Egr,/kT9=3J=0K=-J M,=-J2can be written equivalentlyas1Ea+erarW=0K=-00 J=[K](a)AZ(2]+ 1)e-h(B()(4-B)/KTKs00 J=[K]J=|K|J=IK((B/(2+)B+)T3223J=IKINowwe assumethatthetemperature is so high thatnumerous states areoccupied6and that the sums may beapproximated by integrals.Then4e-{hc(AB)/kT]K20-000900(2J+ 1)e-hcB(+1)/kT JdK3JIKI2As before, the integral over J can be recognized as the integral ofthe derivative ofa1function, which is the function itself, so0kTdhcB(I+1)/KT J(2J+ 1)e-heB(+1)/kKTdJ=hcB(b)JafK1Fig. 17.2 (a) The sum over J= 0, 1,2,B|K1(KI+1)/kTBJU+1)/and K=J,J-1,..,-J(depicted by theKcircles)can be covered (b) by allowingKtorangefromoo to co, with J confined to |K],hcBK/kT[K|+ 1,...,o for each value ofK.hcBIn thelast line we have supposed that |K|>1 for most contributions.Now we canwritekTe-{hc(A-B)/kT]K2-hcBk/kT dKq=hcBkTkTSynoptictable17.1*Rotational ande-ThcA/kT]k'dK1hcBvibrational temperaturesacfMoleculeModeeVK0R/KDH2633088Foranasymmetrirotor,one of theBs isreplaced by C, togive eqn 17.14bHCI43009.4123090.0531997CO,Vi0.561Auseful way ofexpressingthetemperatureabovewhich therotational approxima-V23380tion is valid is to introduce the characteristic rotational temperature, g=hcB/k.960V3ThenhightemperaturemeansT》,and under these conditions therotationalpartition function ofa linearmolecule is simplyT/eg.Some typical values of GgareFormorevalues,seeTable13.2intheDatashown inTable17.1.Thevaluefor H, isabnormallyhighand wemust becareful withsection and use hc/k=1.439Kcm.theapproximationforthis molecule

594 17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS with J = 0, 1, 2, . . . , K = J, J − 1, . . . , −J, and MJ = J, J − 1, . . . , −J. Instead of con￾sidering these ranges, we can cover the same values by allowing K to range from −∞ to ∞, with J confined to |K|, |K| + 1, . . . , ∞ for each value of K (Fig. 17.2). Because the energy is independent of MJ , and there are 2J + 1 values of MJ for each value of J, each value of J is 2J + 1-fold degenerate. It follows that the partition function q = ∞ ∑ J= 0 J ∑K= −J J ∑ MJ= −J e−EJKMJ /kT can be written equivalently as q = ∞ ∑K=−∞ ∞ ∑ J=|K| (2J + 1)e−EJKMJ /kT = ∞ ∑K=−∞ ∞ ∑ J=|K| (2J + 1)e−hc{BJ(J+1)+(A−B)K2 }/kT = ∞ ∑K=−∞ e−{hc(A−B)/kT}K2 ∞ ∑ J=|K| (2J + 1)e−hcBJ(J+1)/kT Now we assume that the temperature is so high that numerous states are occupied and that the sums may be approximated by integrals. Then q = ∞ −∞ e−{hc(A−B)/kT}K2 ∞ |K| (2J + 1)e−hcBJ(J+1)/kTdJdK As before, the integral over J can be recognized as the integral of the derivative of a function, which is the function itself, so ∞ |K| (2J + 1)e−hcBJ(J+1)/kTdJ = ∞ |K| − e−hcBJ(J+1)/kTdJ = − e−hcBJ(J+1)/kT ∞ |K| = e−hcB|K|(|K|+1)/kT ≈ e−hcBK2 /kT In the last line we have supposed that |K| >> 1 for most contributions. Now we can write q = ∞ −∞ e−{hc(A−B)/kT}K2 e−hcBK2 /kTdK π1/2 = ∞ −∞ e−{hcA/kT}K2 dK = 1/2 ∞ −∞ e−x2 dx = 3/2 1/2 For an asymmetric rotor, one of the Bs is replaced by C, to give eqn 17.14b. A useful way of expressing the temperature above which the rotational approxima￾tion is valid is to introduce the characteristic rotational temperature, θR = hcB/k. Then ‘high temperature’ means T >> θR and under these conditions the rotational partition function of a linear molecule is simply T/θR. Some typical values of θR are shown in Table 17.1. The value for H2 is abnormally high and we must be careful with the approximation for this molecule. D E F π AB2 A B C D E F kT hc A B C D E F kT hcA A B C D E F kT hcB A B C kT hcB 5 6 7 kT hcB D E F kT hcB A B C D E F kT hcB A B C D E F kT hcB A B C d dJ D E F kT hcB A B C Synoptic table 17.1* Rotational and vibrational temperatures Molecule Mode qV/K qR/K H2 6330 88 HCl 4300 9.4 I2 309 0.053 CO2 ν1 1997 0.561 ν2 3380 ν3 960 * For more values, see Table 13.2 in the Data section and use hc/k = 1.439 K cm. J 0 1 2 3 4 5 K J K J  5 4 3 2 1 0 1 2 3 4 5 J 0 1 2 3 4 5 J K | | J K | | K J 0, 0, 1, 2,. K J 1, 1, 2, 3,.  K J 2, 2, 3, 4,.  K J 1, 1, 2, 3,. K J 2, 2, 3, 4,. K (a) (b) Fig. 17.2 (a) The sum over J = 0, 1, 2, . . . and K = J, J − 1, . . . , −J (depicted by the circles) can be covered (b) by allowing K to range from −∞ to ∞, with J confined to |K|, |K| + 1, . . . , ∞ for each value of K.

59517.2THEMOLECULARPARTITIONFUNCTIONThegeneral conclusion at this stage isthatmoleculeswithlargemoments ofinertia(andhencesmallrotational constants and lowcharacteristicrotational temperatures)ortho-H,havelargerotational partitionfunctions.ThelargevalueofqRreflects thecloseness inenergy (compared with kT) of the rotational levels in large, heavy molecules, and thelargenumberofthemthat areaccessible at normal temperatures.Wemusttake care, however, not to includetoo many rotational states in the sum.Forahomonucleardiatomicmoleculeorasymmetrical linearmolecule(suchasCO)para-H,orHC=CH),a rotation through 18o°results in an indistinguishable state of themolecule.Hence,the number of thermallyaccessible states is onlyhalfthenumberthat can be occupied by a heteronuclear diatomic molecule, where rotation through180° does result in a distinguishable state. Therefore, for a symmetrical linear molecule,kTTgR(17.15a)2hcB20起01JThe equations for symmetrical and nonsymmetrical molecules can be combined intoa singleexpression byintroducing the symmetry number,,which isthe numberofFig.17.3The values of the individual termsindistinguishableorientations ofthemolecule.Then(2J+ 1)e-βhcB/U+1) contributing to thekTTmean partitionfunction ofa3:lmixture ofgR(17.15b)ortho-and para-H,.Thepartition functionhcBC0Ris the sum of all these terms. At highFora heteronuclear diatomicmolecule=l;fora homonuclear diatomicmoleculetemperatures,thesum is approximatelyequal to the sum of the terms over allora symmetrical linearmolecule,o=2values ofJ,each with a weight of .This isthesum ofthecontributionsindicatedbyJustification17.2Theoriginofthesymmetrynumberthe curve.The quantum mechanical origin of the symmetry number is the Pauli principle,whichforbids theoccupationofcertain states.Wesawin Section13.8,forexample,that H, may occupy rotational states with even J only if its nuclear spins arepaired (para-hydrogen),and odd J states only ifits nuclear spins are parallel (ortho-hydrogen).There are three states of ortho-H, to each value of J (because there arethreeparallel spin states ofthe two nuclei).To set up the rotational partition function we note that ordinary molecularhydrogen is a mixture of one part para-H, (with only its even-J rotational statesoccupied)and threeparts ortho-H, (withonlyits odd-Jrotational statesoccupied).Therefore, the average partition function per molecule isqR=(2]+1)e-βMcB(+)+(2]+1)e-BtcBX+1)oddeven)The odd-J states are more heavily weighted than the even-J states (Fig. 17.3). Fromtheillustration we see thatwewould obtain approximatelythe same answerforthepartitionfunction (thesumofallthepopulations)ifeachJtermcontributedhalf itsnormal valueto the sum.That is,thelastequation canbeapproximated asE(2J + 1)e-βBhcBJ(J+1)and this approximation is very good when many terms contribute (at hightemperatures).0 2JThe same type of argument may be used for linear symmetrical molecules inwhichidenticalbosonsare interchanged byrotation(suchasCO,).As pointed outin Section 13.8, if the nuclear spin of the bosons is 0, then only even-J states areFig.17.4 The relative populations of theadmissible.Because onlyhalf the rotational states are occupied, the rotationalrotational energy levels ofCOz.Only statespartitionfunction is onlyhalfthevalue ofthe sumobtained byallowingall values ofwith evenJvalues are occupied.The fullline shows the smoothed,averagedJto contribute (Fig.17.4).population oflevels

17.2 THE MOLECULAR PARTITION FUNCTION 595 The general conclusion at this stage is that molecules with large moments of inertia (and hence small rotational constants and low characteristic rotational temperatures) have large rotational partition functions. The large value of qR reflects the closeness in energy (compared with kT) of the rotational levels in large, heavy molecules, and the large number of them that are accessible at normal temperatures. We must take care, however, not to include too many rotational states in the sum. For a homonuclear diatomic molecule or a symmetrical linear molecule (such as CO2 or HC.CH), a rotation through 180° results in an indistinguishable state of the molecule. Hence, the number of thermally accessible states is only half the number that can be occupied by a heteronuclear diatomic molecule, where rotation through 180° does result in a distinguishable state. Therefore, for a symmetrical linear molecule, qR = = (17.15a) The equations for symmetrical and nonsymmetrical molecules can be combined into a single expression by introducing the symmetry number, σ, which is the number of indistinguishable orientations of the molecule. Then qR = = (17.15b) For a heteronuclear diatomic molecule σ = 1; for a homonuclear diatomic molecule or a symmetrical linear molecule, σ = 2. Justification 17.2 The origin of the symmetry number The quantum mechanical origin of the symmetry number is the Pauli principle, which forbids the occupation of certain states. We saw in Section 13.8, for example, that H2 may occupy rotational states with even J only if its nuclear spins are paired (para-hydrogen), and odd J states only if its nuclear spins are parallel (ortho￾hydrogen). There are three states of ortho-H2 to each value of J (because there are three parallel spin states of the two nuclei). To set up the rotational partition function we note that ‘ordinary’ molecular hydrogen is a mixture of one part para-H2 (with only its even-J rotational states occupied) and three parts ortho-H2 (with only its odd-J rotational states occupied). Therefore, the average partition function per molecule is qR = 1 –4 ∑even J (2J + 1)e−βhcBJ(J+1) + 3 –4 ∑ odd J (2J + 1)e−βhcBJ(J+1) The odd-J states are more heavily weighted than the even-J states (Fig. 17.3). From the illustration we see that we would obtain approximately the same answer for the partition function (the sum of all the populations) if each J term contributed half its normal value to the sum. That is, the last equation can be approximated as qR = 1 –2 ∑ J (2J + 1)e−βhcBJ(J+1) and this approximation is very good when many terms contribute (at high temperatures). The same type of argument may be used for linear symmetrical molecules in which identical bosons are interchanged by rotation (such as CO2). As pointed out in Section 13.8, if the nuclear spin of the bosons is 0, then only even-J states are admissible. Because only half the rotational states are occupied, the rotational partition function is only half the value of the sum obtained by allowing all values of J to contribute (Fig. 17.4). T σθ R kT σhcB T 2θ R kT 2hcB ortho-H2 para-H2 0 1 J Fig. 17.3 The values of the individual terms (2J + 1)e−βhcBJ(J+1) contributing to the mean partition function of a 3:1 mixture of ortho- and para-H2. The partition function is the sum of all these terms. At high temperatures, the sum is approximately equal to the sum of the terms over all values of J, each with a weight of 1 –2. This is the sum of the contributions indicated by the curve. 0 2 J Fig. 17.4 The relative populations of the rotational energy levels of CO2. Only states with even J values are occupied. The full line shows the smoothed, averaged population of levels

59617STATISTICALTHERMODYNAMICS2:APPLICATIONSThe same caremustbeexercisedforothertypes ofsymmetrical molecule,andforaSynoptictable17.2*Symmetrynonlinear molecule we writenumbers元aMolecule(17.16)a72H,OSome typical values of the symmetry numbers required aregiven in Table17.2.TheNH,3value (H,O)=2reflects thefactthat a 180°rotation about thebisector of theH-12CH,O-H angle interchanges two indistinguishable atoms. In NH, there are three indis-12C.Htinguishable orientations around the axis shown in (1). For CHy, any of three 120rotations aboutanyofits fourC-Hbonds leavesthe moleculeinan indistinguishable*For more values, see Table 13.2 in thestate, so the symmetry number is 3 × 4 =12. For benzene, any of six orientationsData section.around the axis perpendicular to the plane of the molecule leaves it apparentlyunchanged, as does a rotation of 18o°around any of six axes in the plane of themolecule (three of which pass along each C-H bond and the remaining three passthrough each C-C bond in the plane of the molecule).For the way that group theoryisused to identifythevalueof thesymmetrynumber, seeProblem 17.17.H(c) The vibrational contributionThe vibrational partition function of a molecule is calculated by substituting the-measured vibrational energy levels into the exponentials appearing in the definitionofq,and summingthem numerically.Ina polyatomicmoleculeeach normalmode(Section13.14)has its own partition function (provided theanharmonicitiesare sosmall that the modes are independent). The overall vibrational partition function istheproductoftheindividual partition functions,and wecan writeq=q(1)q(2)...where q'(K) is the partition function for the Kth normal mode and is calculated by10directsummationoftheobservedspectroscopiclevels.If the vibrational excitation is not too great, the harmonic approximation may bemade,and the vibrational energy levels written asqE,=(u+)hcvV=0,1,2,.(17.17)If, as usual, we measure energies from the zero-point level, then the permitted values5are e,=vhcv and the partition function isq'-Ze-βuho-E(e-Bhco)(17.18)(because eax=(e")").Wemet this sum in Example16.2 (which is no accident:theladder-like arrayoflevels inFig,16.3 is exactly the same as that ofa harmonic oscillator).05100The series can be summed in the sameway,and giveskTIhcvgv(17.19)-e-βhepFig.17.sThe vibrational partition function ofa moleculein the harmonic approximation.This function is plotted in Fig.17.5.In a polyatomic molecule,each normal modeNote that the partition function is linearlygivesrisetoapartitionfunction ofthisform.proportional to thetemperature when thetemperature is high (T)Example17.3CalculatingavibrationalpartitionfunctionExploration Plot the temperature国dependence of thevibrationalThewavenumbersofthethreenormal modesofH,Oare3656.7cm-l,1594.8cm-1contributiontothemolecularpartitionfunction for several values of theand 3755.8cm-l.Evaluate thevibrational partition function at1500K.vibrational wavennumber.EstimatefromMethod Useeqn17.19foreachmode,andthenform theproductofthethreecon-your plots the temperature above whichtributions.At1500K,kT/hc=1042.6cm-lthe harmonic oscillator is in the‘hightemperature'limit

596 17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS Synoptic table 17.2* Symmetry numbers Molecule s H2O 2 NH3 3 CH4 12 C6H6 12 * For more values, see Table 13.2 in the Data section. N H H H 1 0 5 10 10 5 0 q V kT hc / ~ 1  Fig. 17.5 The vibrational partition function of a molecule in the harmonic approximation. Note that the partition function is linearly proportional to the temperature when the temperature is high (T >> θV). Exploration Plot the temperature dependence of the vibrational contribution to the molecular partition function for several values of the vibrational wavennumber. Estimate from your plots the temperature above which the harmonic oscillator is in the ‘high temperature’ limit. The same care must be exercised for other types of symmetrical molecule, and for a nonlinear molecule we write qR = 3/2 1/2 (17.16) Some typical values of the symmetry numbers required are given in Table 17.2. The value σ(H2O) = 2 reflects the fact that a 180° rotation about the bisector of the H￾O-H angle interchanges two indistinguishable atoms. In NH3, there are three indis￾tinguishable orientations around the axis shown in (1). For CH4, any of three 120° rotations about any of its four C-H bonds leaves the molecule in an indistinguishable state, so the symmetry number is 3 × 4 = 12. For benzene, any of six orientations around the axis perpendicular to the plane of the molecule leaves it apparently unchanged, as does a rotation of 180° around any of six axes in the plane of the molecule (three of which pass along each C-H bond and the remaining three pass through each C-C bond in the plane of the molecule). For the way that group theory is used to identify the value of the symmetry number, see Problem 17.17. (c) The vibrational contribution The vibrational partition function of a molecule is calculated by substituting the measured vibrational energy levels into the exponentials appearing in the definition of qV, and summing them numerically. In a polyatomic molecule each normal mode (Section 13.14) has its own partition function (provided the anharmonicities are so small that the modes are independent). The overall vibrational partition function is the product of the individual partition functions, and we can write qV = qV(1)qV(2) . . . , where qV(K) is the partition function for the Kth normal mode and is calculated by direct summation of the observed spectroscopic levels. If the vibrational excitation is not too great, the harmonic approximation may be made, and the vibrational energy levels written as Ev = (v + 1 –2)hc# v = 0, 1, 2, . . . (17.17) If, as usual, we measure energies from the zero-point level, then the permitted values are εv = vhc# and the partition function is qV =∑ v e−βvhc# =∑ v (e−βhc# ) v (17.18) (because eax = (ex ) a ). We met this sum in Example 16.2 (which is no accident: the ladder-like array of levels in Fig. 16.3 is exactly the same as that of a harmonic oscillator). The series can be summed in the same way, and gives qV = (17.19) This function is plotted in Fig. 17.5. In a polyatomic molecule, each normal mode gives rise to a partition function of this form. Example 17.3 Calculating a vibrational partition function The wavenumbers of the three normal modes of H2O are 3656.7 cm−1 , 1594.8 cm−1 , and 3755.8 cm−1 . Evaluate the vibrational partition function at 1500 K. Method Use eqn 17.19 for each mode, and then form the product of the three con￾tributions. At 1500 K, kT/hc = 1042.6 cm−1 . 1 1 − e−β hc# D F π ABC A C D F kT hc A C 1 σ

17.2THEMOLECULARPARTITIONFUNCTION597Answer Wedraw up thefollowing table displaying thecontributions of each mode:Mode:123V/cm-13656.71594.83755.8hcv/kT3.5071.5303.602qv1.0311.2761.028Theoverall vibrational partitionfunction isthereforeq=1.031×1.276×1.028=1.353Thethreenormal modes of H,Oareat such high wavenumbers that evenat1500Kmost ofthemolecules are in theirvibrational ground state.However,there maybeso many normal modes in a large molecule that their excitation may be significanteventhougheachmodeisnotappreciablyexcited.Forexample,anonlinearmoleculecontaining10atomshas3N-6=24normalmodes(Section13.14).Ifweassumeavalueofabout 1.1 for thevibrational partition function ofone normal mode,theoverall vibrational partitionfunctionisaboutq(1.1)24=9.8,whichindicatessignificantvibrationalexcitation relativetoa smallermolecule,such as H,OSelf-test17.3Repeatthecalculation for CO2,where thevibrational wavenumbersare1388cm-,667.4cm-,and2349cm-1,thesecondbeingthedoublydegeneratebending mode.[6.79]In many molecules the vibrational wavenumbers are so great that βhcv > 1.Forexample,thelowestvibrational wavenumberofCH,is1306cm-,soBhcv=6.3atroom temperature.C-H stretches normally lie in the range 2850 to 2960 cm-l, so forthem βhc=14. In these cases,e-Bhin the denominator ofq'is verycloseto zero (forexample,e-6.3=0.002), and the vibrational partition function for a single mode is veryclose to 1 (q' = 1.002 when βhcv = 6.3), implying that only the zero-point level issignificantly occupied.Now consider the case of bonds so weak that βhcykT.When this condition issatisfied, the partition function may be approximated by expanding the exponential(e*=1+x+..-):1(17.20)a:1-(1-βhcv+..-)That is, for weak bonds at high temperatures,1kTgV-(17.21)BhcvhcvThe temperatures for which eqn 17.21 is valid can be expressed in terms of thecharacteristic vibrational temperature, v=hcv/k (Table 17.1).The valuefor H, isabnormally high because the atoms are so light and the vibrational frequency is cor-respondingly high.In terms of the vibrational temperature,high temperature meansT>y and, when this condition is satisfied,q'=T/ey(the analogue oftherotationalexpression).(d)TheelectroniccontributionElectronicenergy separationsfrom the ground state are usuallyverylarge,so formostcasesq=l.An importantexceptionarises in thecaseofatomsandmolecules having

17.2 THE MOLECULAR PARTITION FUNCTION 597 Answer We draw up the following table displaying the contributions of each mode: Mode: 1 2 3 #/cm−1 3656.7 1594.8 3755.8 hc#/kT 3.507 1.530 3.602 qV 1.031 1.276 1.028 The overall vibrational partition function is therefore qV = 1.031 × 1.276 × 1.028 = 1.353 The three normal modes of H2O are at such high wavenumbers that even at 1500 K most of the molecules are in their vibrational ground state. However, there may be so many normal modes in a large molecule that their excitation may be significant even though each mode is not appreciably excited. For example, a nonlinear molecule containing 10 atoms has 3N − 6 = 24 normal modes (Section 13.14). If we assume a value of about 1.1 for the vibrational partition function of one normal mode, the overall vibrational partition function is about qV ≈ (1.1)24 = 9.8, which indicates significant vibrational excitation relative to a smaller molecule, such as H2O. Self-test 17.3 Repeat the calculation for CO2, where the vibrational wavenumbers are 1388 cm−1 , 667.4 cm−1 , and 2349 cm−1 , the second being the doubly degenerate bending mode. [6.79] In many molecules the vibrational wavenumbers are so great that βhc# > 1. For example, the lowest vibrational wavenumber of CH4 is 1306 cm−1 , so βhc# = 6.3 at room temperature. C-H stretches normally lie in the range 2850 to 2960 cm−1 , so for them βhc# ≈ 14. In these cases, e−βhc# in the denominator of qV is very close to zero (for example, e−6.3 = 0.002), and the vibrational partition function for a single mode is very close to 1 (qV = 1.002 when βhc# = 6.3), implying that only the zero-point level is significantly occupied. Now consider the case of bonds so weak that βhc# > θ V and, when this condition is satisfied, qV = T/θ V (the analogue of the rotational expression). (d) The electronic contribution Electronic energy separations from the ground state are usually very large, so for most cases qE = 1. An important exception arises in the case of atoms and molecules having kT hc# 1 βhc# 1 1 − (1 − βhc# + · · ·)

59817STATISTICALTHERMODYNAMICS2:APPLICATIONSTq3n0510kTIe"12Fig.17.7The variation with temperature oftheelectronic partitionfunction ofan NOmolecule.Note that the curve resemblesthat for a two-level system (Fig.16.5), butrises from 2 (the degeneracy of the lowerFig.17.6The doubly degenerategroundlevel) and approaches 4 (thetotal numberelectronic level of NO (with the spin andof states)at hightemperatures.orbitalangularmomentumaroundtheaxisExploration Plot the temperaturein oppositedirections)andthedoublydependence of the electronicdegeneratefirstexcited level (withthespinpartition function for severalvalues of theand orbital momenta parallel).The upperenergy separation between twodoublylevelisthermallyaccessibleatroomdegenerate levels. From your plots,temperature.estimate the temperature at which thepopulation of the excited level begins toincrease sharply.electronicallydegenerate ground states, in which caseq=g, whereg is the degeneracyoftheelectronicgroundstate.Alkali metal atoms,forexample,havedoublydegenerateground states (corresponding to the twoorientations oftheir electron spin),sog=2.Some atoms and molecules have low-lying electronically excited states.(At highenoughtemperatures,all atomsandmolecules havethermallyaccessibleexcited states.)AnexampleisNO,whichhasaconfigurationoftheform...元’(seeImpactIl1.1).Theorbital angularmomentum maytaketwoorientationswithrespecttothemolecularaxis (corresponding to circulation clockwise or counter-clockwise around the axis),andthespin angularmomentummayalsotaketwoorientations,givingfourstates inall (Fig.17.6).The energy ofthetwo states in which theorbital and spin momentaareparallel (giving the1l/2term)is slightlygreater thanthat of the two other states inwhich they are antiparallel (giving the2l/2 term).The separation, which arises fromspin-orbit coupling (Section 10.8), is only 121 cm-1.Hence, at normal temperatures,all four states arethermally accessible.If we denote the energies of the two levels asEj/2=0and E3/2=, the partitionfunction isZg;e-βBE = 2 + 2e-βeqE=(17.22)energy levelsFigure 17.7 shows the variation of this function with temperature. At T = 0, qE = 2,because only the doubly degenerate ground state is accessible. At high temperatures,qapproaches 4 because all four states are accessible.At 25°C,q=3.1

598 17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS electronically degenerate ground states, in which case qE = gE, where gE is the degeneracy of the electronic ground state. Alkali metal atoms, for example, have doubly degenerate ground states (corresponding to the two orientations of their electron spin), so qE = 2. Some atoms and molecules have low-lying electronically excited states. (At high enough temperatures, all atoms and molecules have thermally accessible excited states.) An example is NO, which has a configuration of the form . . . π1 (see ImpactI11.1). The orbital angular momentum may take two orientations with respect to the molecular axis (corresponding to circulation clockwise or counter-clockwise around the axis), and the spin angular momentum may also take two orientations, giving four states in all (Fig. 17.6). The energy of the two states in which the orbital and spin momenta are parallel (giving the 2 Π3/2 term) is slightly greater than that of the two other states in which they are antiparallel (giving the 2 Π1/2 term). The separation, which arises from spin–orbit coupling (Section 10.8), is only 121 cm−1 . Hence, at normal temperatures, all four states are thermally accessible. If we denote the energies of the two levels as E1/2 = 0 and E3/2 = ε, the partition function is qE = ∑ energy levels gje−βεj = 2 + 2e−βε (17.22) Figure 17.7 shows the variation of this function with temperature. At T = 0, qE = 2, because only the doubly degenerate ground state is accessible. At high temperatures, qE approaches 4 because all four states are accessible. At 25°C, qE = 3.1. S S S S L L L L 2 3/2 2 1/2 121.1 cm 1 Fig. 17.6 The doubly degenerate ground electronic level of NO (with the spin and orbital angular momentum around the axis in opposite directions) and the doubly degenerate first excited level (with the spin and orbital momenta parallel). The upper level is thermally accessible at room temperature. qE 4 2 0 5 10 3 kT/ Fig. 17.7 The variation with temperature of the electronic partition function of an NO molecule. Note that the curve resembles that for a two-level system (Fig.16.5), but rises from 2 (the degeneracy of the lower level) and approaches 4 (the total number of states) at high temperatures. Exploration Plot the temperature dependence of the electronic partition function for several values of the energy separation ε between two doubly degenerate levels. From your plots, estimate the temperature at which the population of the excited level begins to increase sharply.