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

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 contribution 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 substitutions Q = qN (for distinguishable molecules, as in a solid) or Q = qN/N! (for indistinguishable 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

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 independent, 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 partition 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 proportional 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 contribution. The electronic contribution is not actually a ‘mode of motion’, but it is convenient 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 separation 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

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 present 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 approximated 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