A Tale of Two Molecules: How the Heat Capacities ... - ACS Publications

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Cite This: J. Chem. Educ. XXXX, XXX, XXX−XXX

A Tale of Two Molecules: How the Heat Capacities of N2(g) and F2(g) ̈ Expectations Fail to Differ At High Temperature and Why Naive Explain These Differences: A Spreadsheet Exercise for Physical Chemistry Students Arthur M. Halpern* and Robert J. Noll Department of Chemistry and Physics, Indiana State University, Terre Haute, Indiana 47809, United States

J. Chem. Educ. Downloaded from pubs.acs.org by IDAHO STATE UNIV on 04/03/19. For personal use only.

S Supporting Information *

ABSTRACT: A spreadsheet-based exercise for students is described in which they are challenged to explain and reproduce the disparate temperature dependencies of the heat capacities of gaseous F2 and N2. For F2, Cp,m increases from 300 K, reaches a maximum at 2200 K, and then decreases to 74% of the maximum value at 6000 K, while Cp,m for N2 rises monotonically in this temperature range. Students compare this behavior with the prediction of the equipartition principle and then test four models with increasing sophistication, from closed-form expressions of the harmonic/ anharmonic oscillator−rigid/nonrigid rotor to calculations using discrete sums over rovibrational states, all using molecular spectroscopic constants. Fundamental principles of statistical thermodynamics are used to calculate the partition functions, which yield the internal energy, heat capacity, and third-law entropy of the molecules. They encounter the need to include the role of quasibound rotational states that arise from the rotational barrier to molecular dissociation and to employ methods for truncating the sums used in the partition functions. In this exercise, students confront the fact that, for N2, Cp,m continues to rise, reaching a maximum at 16,000 K that is 1.8 times the equipartition value. The inclusion of electronically excited states in the partition function is discussed. KEYWORDS: Upper-Division Undergraduate, Physical Chemistry, Computer-Based Learning, Thermodynamics, Statistical Mechanics

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INTRODUCTION This article presents an exercise for students to explore in detail the methods used to calculate the heat capacities of gaseous F2 and N2 under standard conditions (1 bar, ideal gas) over a wide range of temperatures, up to several thousand kelvins. Students start with spectroscopic molecular constants and, using fundamental precepts of statistical thermodynamics, calculate heat capacities and other thermodynamic quantities for F2 and N2, which present markedly different behavior at T > 2000 K. In carrying out this exercise, readily performed using a scientific spreadsheet application such as Microsoft Excel, students will gain deeper insights into “real world” statistical thermodynamical calculations, most notably the importance of properly accounting for rovibrational molecular states in calculating the partition function. This project will additionally acquaint students with several advanced concepts, such as quasibound rotational states that arise from the rotational barrier to dissociation of a diatomic molecule, the importance in taking onto account the molecular dissociation limit, and the role of higher electronic states in affecting the value of the heat capacity of a molecule. It will also enable students to understand how the authoritative heat capacities reported by © XXXX American Chemical Society and Division of Chemical Education, Inc.

NIST are obtained. This exercise is best presented to students after they have a basic grasp of diatomic molecular spectroscopy, thermodynamics, and statistical mechanics. Chemistry students often use heat capacity data available from the NIST-JANAF Thermochemical Tables, the authoritative source of such information. They typically use the data without questioning it. The visual inspection of the temperature dependence of the heat capacities, Cp(T), of F2 and N2 will present a challenge to students’ understanding of heat capacities. The theoretical background and practical spreadsheet-based exercise presented in this article afford them the opportunity to delve into the basics and the details of thermochemical calculations and will guide them to a deeper understanding of thermodynamics and applied statistical mechanics. Importantly, this learning experience will allow them to closely reproduce the NIST-JANAF Cp(T) data, thereby enabling them to make sense of what they observe. Received: January 11, 2019 Revised: March 14, 2019

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DOI: 10.1021/acs.jchemed.9b00029 J. Chem. Educ. XXXX, XXX, XXX−XXX

Journal of Chemical Education

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BACKGROUND Many chemistry students first learn about heat capacity by performing simple calorimetric experiments, for example, by measuring the temperature change due to immersing a block of hot metal into water.1 In some experiments, the identity of the metal2 or its heat capacity3 is determined by resorting to the Dulong and Petit law. A deeper learning experience from such experiments may be provided by discussing the equipartition of energy theorem, first proposed in 1845.4 According to the equipartition theorem, each degree of freedom (the squared terms in position and momentum appearing in the classical expression for the total energy of an atom or molecule) at 1 thermal equilibrium contributes 2 kT to the total molecular energy, where k is the Boltzmann constant and T the absolute temperature.5 By application of the theorem to solid metals and a view of the metal sample as a “single giant molecule” in a perfect crystal,6 each atom is imagined as a harmonic oscillator moving between neighboring atoms along each spatial dimension. In this picture, each dimension of oscillation contributes two quadratic terms to the total energy (e.g., for the x-coordinate, px2/2m and kx2/2). Thus, the total molar energy, Um, would be 3RT. Since (dUm/dT)V = Cv,m, it follows that Cv,m = 3R = 24.943 J/K mol, close to the values reported by Giauque and Meads for Al and Cu at 300 K, i.e., 23.43 and 23.71 J/K mol, respectively.7 Inasmuch as the theorem naively predicts Cv,m to be a constant value, one would expect the heat capacity to be independent of temperature. Students must subsequently refine this understanding, learning that the Dulong−Petit law only holds at sufficiently high temperatures; heat capacities actually decrease with decreasing temperature, approaching zero at 0 K. For example, Cv,m for Al and Cu decreases to half the 300 K values cited above at 92 and 74 K.7 This concept is explored in an experiment published in this Journal in which students determine the molar heat capacities of Al and Cu between 89 and 94 and 300 K.8 To apply the equipartition principle to ideal gases, including diatomic molecules, and to consider the temperature dependence of the heat capacity, one tallies for the internal energy two degrees of freedom for vibrational motion, two degrees of freedom for rotational motion, and three degrees of freedom for translational energy.9,10 The expected value of Cv,m is thus 3.5R, or 29.101 J/K mol. Scientists in the latter half of the 19th century were aware that, not only did the heat capacity of metals decrease from their “equipartition” values at lower temperatures, but also the heat capacities of gaseous diatomic molecules were markedly less than 29 J/mol K. For example, Cv,m values for H2, N2, and Cl2 at 300 K are about 20.5, 20.8, and 24.6 J/K mol, respectively. Presumably being aware of such discrepancies, which were beyond experimental uncertainties, Lord Kelvin, in a lecture before the Royal Institution, in London, in 1900, remarked that “the beauty and clearness of the dynamical theory which asserts that heat and light to be modes of motion are obscured by two clouds”.11 Partington explains, “one of the clouds concerned radiation, the other was concerned with the equipartition of energy, which Lord Kelvin did not accept”.12 In the same year, Planck was able to resolve the so-called ultraviolet catastrophe relating to blackbody radiation through the ad hoc assignment of multiples (quanta) of radiation energy. The discrepancy in heat capacities is now usually explained to students by the notion that the vibrational degrees of freedom of the molecules

are not involved because of the incommensurability of the energy spacings between the quantized vibrational states and kT, the thermal energy, at room temperature and below. (Henceforth, the symbol k denotes the Boltzmann constant.) This concept may be used to rationalize the increase in heat capacity with increasing temperature and the leveling off at sufficiently high temperatures. However, the temperature dependence of heat capacities of diatomic molecules even deviates from this more sophisticated understanding. For example, graphs of Cp,m vs T of F2 and N2 between 300 and 6000 K, shown in Figure 1, show much more complicated

Figure 1. Plot of Cp,m/R vs T: solid blue line, N2(g); solid red line, F2(g); dashed black line, equipartition value (9/2)R. Data from the NIST-JANAF Tables (refs 14 and 15).

̈ behavior. Contrary to the nai ve expectations of the equipartition theorem, even when augmented by vibrational energy quantization, Cp,m for F2 rises between 300 K and ca. 1100 K, where it exceeds the equipartition value (4.5R), then reaches a maximum at 2200 K, and then falls to ca. 0.77 of it at 6000 K. On the other hand, for N2, Cp,m rises monotonically, rising to slightly above the equipartition value at ca. 3700 K, and continues to increase. As will be shown later in this article, Cp,m increases further, reaching a value that is 1.8 times the equipartition value, and then decreases. Such unexpected, and different, behavior of these two molecules desires explanation.

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THE CHALLENGE At the outset of the work, instructors should stress three important points: (1) Although the energy of a system may be calculated using a theoretical model, e.g., the equipartition principle, from which Um = (7/2)RT for a diatomic molecule, the heat capacity (the subject of this exercise) can be experimentally obtained by measuring the derivative of Um, i.e., Cv,m = (∂Um/∂T)v. (2) Thermodynamic quantities for gas phase species, such as entropies, heat capacities of molecules, and equilibrium constants of many gas phase reactions are often obtained through computational means rather than direct experimental measurements, which could be difficult or impossible to carry out at very high temperatures. As Herzberg noted,13 such values “calculated from the spectroscopic data are often more accurate than those determined by direct thermal measurements”. (3) Although calculations may be performed for systems at very high temperatures (several B



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thousands of kelvins), conditions under which thermal dissociation may be significant for some species,14 the calculation deals only with the molecular species of interest, for example, F2 and N2, the molecules studied in this article. Instructors might begin this project by challenging students to explain the Cp,m trends in Figure 1, which displays authoritative values of Cp,m for F2 and N2 between 300 and 6000 K.15,16 The dashed line denotes Cp,m/R = 4.5, the value predicted by the equipartition theorem. They should be asked the following questions: “Why does Cp,m increase for both molecules at lower temperatures? Why does Cp,m seem to level off above the equipartition value for N2, but decrease above ca. 2000 K for F2?” Among the learning outcomes of this exercise is the students’ ability to answer these questions in depth. Additional learning goals include understanding the limitations of the simple models they have studied, such as the harmonic oscillator and rigid rotor, in accounting for molecular vibrational and rotational energy, and mastering the statistical mechanical principles as well as practical spreadsheet calculational skills needed to calculate Cp,m vs T as shown in Figure 1. This exercise might be carried out during a lab session or it may be given to students as a special enriching project. They can divide the labor and help each other learn new material, thus deepening their knowledge of the subject matter.

reproduce the Cp,m(T) values for F2 and N2. The models advance through a logical progression of sophistication, or complexity, on the basis of the spectroscopic constants and the equations used to express the internal (rotational−vibrational) energy of the molecule, Eint, and in the method used to calculate the partition function, zint. Values of Sint, Uint, and Cv,int are obtained from the temperature dependence of zint using fundamental principles of statistical thermodynamics. The translational contributions to these thermodynamic quantities are then included to provide Stot, Utot, and Cv,tot. Finally, Cp,m is obtained from Cv,tot + R (ideal gas). It is evident from Figure 1 that the equipartition theorem utterly fails to account for the data. Students will recognize the need to consider that rotation and vibrational energies are not continuous but are represented as arrays of discrete energy values. The simplest model that encompasses this energy quantization is the harmonic oscillator rigid rotor (HORR), which they encountered in their physical chemistry courses.21,22

THE APPROACH With the knowledge that the modes of energy of an atom or molecule are not continuously variable, but are quantized, it can be made apparent to students why the equipartition theorem fails to account for observables such as heat capacities. This is especially true at lower temperatures where the separation of discrete energy states is incommensurably larger than thermal energy (kT). Students are well aware of energy quantization, having likely been introduced to this idea even in precollege courses and are familiar with the energy level expressions associated with translational and internal (rotational and vibrational) degrees of freedom in college courses. The statistical mechanical principles and methods employed in this article are covered in physical chemistry texts and statistical mechanics books.17−20 A schematic outline of the computational approach is shown in Scheme 1. Four models are considered and tested to

where υ̃e is the harmonic vibrational frequency, with the tilde designating units of wavenumbers (cm−1), and v is the vibrational quantum number with values 0, 1, 2, 3... The energy levels for the rigid rotor are given by

Model 1. The Harmonic Oscillator Rigid Rotor (HORR)

The quantum mechanical vibrational energy expression for the harmonic oscillator, in units of wavenumbers, cm−1, is Evib = (v + 1/2)υẽ

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(1)

Erot = J(J + 1)Bẽ

(2)

In this expression B̃ e is the rotational constant (cm ), and J is the rotational quantum number (J = 0, 1, 2, 3...). From Scheme 1, the next step is to express the partition function for these degrees of freedom. The total molecular partition function, z, is the sum of the relative populations in all thermally accessible molecular states −1

z=

∑ gi exp(−Ei/kT ) i

(3)

where gi is the degeneracy (number of quantum states for a given energy level) of the ith level.17−20 From eq 1, the vibrational partition function is ÄÅ É ∞ ÅÅ −(v + 1/2)hcυẽ ÑÑÑ ÑÑ z vib = ∑ expÅÅÅ ÑÑ ÅÅ kT ÑÑÖ ÅÇ (4) v=0

Scheme 1. Block Diagram of the Steps Used To Calculate Thermodynamic Properties of F2 and N2

since, for diatomic molecules, g(v) = 1. In these calculations, it is convenient to express the molecular energies in kelvins, e.g., (hc/k)υ̃e = c2υ̃e ≡ Θvib.10 Here, c2 = 1.4388 K s/cm−1, and Θvib is the characteristic vibrational temperature (K).23,24 For F2 and N2, Θvib = 1,318.9 and 3,393.5 K, respectively.23,24 The rotational partition function is, from eqs 2 and 3 ÄÅ É ∞ ÅÅ J(J + 1)Bẽ ÑÑÑ Å ÑÑ zrot = ∑ (2J + 1)expÅÅÅ− ÑÑÑ Å kT ÅÇ ÑÖ (5) J=0 in which 2J + 1 is the degeneracy of the Jth level and k is 0.695036 cm−1/K (the conversion factor 5.03412 × 1022 cm−1/J has been used). The characteristic rotational temperature is Θrot = 1.4388Be. The infinite sum in eq 4 can be expressed as17−19 C


Journal of Chemical Education z vib =

1 1 − exp( −Θvib /T )

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To see how well the HORR model works in reproducing the results for F2 and N2 in Figure 1, students can apply their calculus skills to eqs 6, 7, and 10 to express Cp,m(T) in closed form. They will find that ÉÑ ÄÅ 2 ÅÅ exp(Θvib /T ) ij Θvib yz 3 ÑÑÑÑ ÅÅ zzz C p,m(T ) = ÅÅR + R jjj + R ÑÑ + R 2 ÅÅ 2 ÑÑÑÖ k T { [exp(Θvib /T ) − 1] ÅÇ | li Θ y2 o o exp(Θvib /T ) o ojj vib zz =R m + 3.5} j z 2 o o j z o ok T { [exp(Θvib /T ) − 1] (13) ~ n Alternatively, students may calculate through to ln zint and then perform two numerical differentiations (eq 10), rather than derive eq 13 in closed form. Cp,m, obtained from eq 13, is plotted between 300 and 6000 K in Figure 2 for F2 and N2. These calculations, along with all

(6)

while the rotational sum in eq 5 may be replaced by an integral over J with limits of 0 and ∞ which yields17−19 zrot =

T T + 2Θrot σ Θrot

(7)

The quantity σ is the rotational symmetry number and has a value of 2 for a homonuclear diatomic molecule. The resulting factor of 1/2 in eq 7 represents the average of the nuclear spin statistical weights for even and odd J values, and its inclusion in the rotational partition function is a requirement of the Pauli exclusion principle.25,26 Details of nuclear spin statistics are discussed by Herzberg.27 The values of Θrot for F2 and N2 are 1.2808 and 2.8751 K, respectively.23,24 With these expressions for zvib and zrot, students can calculate the thermodynamic properties of interest for the internal degrees of freedom (see Scheme 1). The total internal molecular partition energy is the product z int = zrotz vib (8) as there are no terms involving v and J that cannot be factored. The partition function is related to the desired molar thermodynamic quantities, Um,int, Cv,m,int, and Sm,int, through the following relationships:17−20 Um,int = RT 2

d ln z int i d ln zrot zy i d ln z vib zy zzz + RT 2jjjj zzz = RT 2jjjj dT d T k { k dT {

(9)

Cv,m,int =

dUm,int

dT ij d2 ln zrot d2 ln z vib yzz zz = RT 2jjj + j dT 2 dT 2 z{ k d ln z vib yz i d ln zrot zz + 2RT jjjj + dT z{ k dT

Figure 2. Plot of Cp,m/R vs T for F2 and N2: solid red line, F2 NISTJANAF; solid yellow line, F2 HORR (model 1); solid blue line, N2 NIST-JANAF; solid green line, N2 HORR (model 1); dashed black line, equipartition value. (10)

others described in this article, are described in detail in the document and workbooks in the Supporting Information (SI). Students will readily observe that, for F2, the inclusion of quantized rotational and vibrational energy levels in the partition functions for the HORR model does not comport with the NIST-JANAF data. However, for N2, the HORR model is in qualitative agreement with the NIST-JANAF values. Additionally, this model produces similar Cp,m plots for the two molecules in that both rise rapidly and then level off, approaching the equipartition value, but it does not account for the rise above the equipartition value (Cp,m/R = 4.5) for either molecule.

and Sm,int =

Um,int T

+ R ln z int

(11)

From eq 10 it is evident that Cv,m,int depends on the first and second derivatives of zint with respect to temperature and is thus sensitive to small changes in zint. Also, only Sm,int depends directly on the numerical value of ln zint. To obtain the results shown in the bottom middle box of Scheme 1, students must also add the translational energy contributions Um,trans, Cv,m,trans, and Sm,trans to Um,int, Cv,m,int, and Sm,int. Because the translational energy level spacings are so small, Um,trans and Cv,m,trans may be represented as (3/2)RT and (3/2)R.17−19 The translational entropy Sm,trans of an ideal gas is readily obtained from the Sackur−Tetrode equation17−19

Model 2. The Anharmonic Nonrigid Rotor (AHNRR)

The next level of sophistication stems from students’ likely realization that the HORR model does not take into account anharmonicity, vibration−rotation interaction, or centrifugal distortion. To accommodate these needed adjustments, the students will now advance to model 2, the anharmonic nonrigid rotor (AHNRR). This model is represented by adding several terms to eqs 1 and 2, which are then combined to give the total internal energy of the molecule (in cm−1 units), viz.28

Sm,tr = R[1.5 ln M + 2.5 ln(T /K ) − ln(P /bar) − 1.1517] (12)

where M is the molar mass of the molecule in g/mol and T is in kelvins. Because P = 1 bar, the third term in eq 12 is zero. Finally, students obtain Cp,m from the ideal gas relationship Cp,m = Cv,m + R so they can compare their results with the NIST-JANAF values.15,16 D



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Table 1. Definitions of the Quantities in Equation 14a u

Quantity Definition

υ ̃ − 2υ ̃ x c2 e T e e

(

y B ̃ − α̃ / 2 c2 e T e

)

(

)

−1

a

c2 is the second radiation constant, 1.4388 K/cm .

+ (v + 1/2)3 υẽ ye + J(J + 1)Bẽ (14)

in which anharmonicity is accounted for by the terms containing υ̃exe and υ̃eye. The rotational−vibrational coupling constant, α̃e, has the effect of decreasing the value of Bẽ because of the increased average internuclear separation at higher v, a characteristic of an anharmonic potential (note that Bẽ = h/8π2cμr2e , where μ is the reduced mass of the molecule). The centrifugal distortion constant, D̃ , accounts for bond stretching due to higher rotational energy; these last two constants account for rotor nonrigidity. The six molecular constants in eq 14 are available from the NIST Web site.23,24 The next step is to express the partition function, zint. The result, in compact form, is given by Pitzer29 z int

δ

x

αẽ Bẽ − αẽ / 2

υẽ xe υẽ − 2υẽ xe

| li Θ y 2 o o exp(Θvib /T ) o ojj vib zz Cp,m = R m + 3.5} j z 2 o j z o ok T { [exp(Θvib /T ) − 1] o n ~ ÄÅ ÅÅÅ 4γ δu 2eu(eu + 1) + RÅÅÅ + (eu − 1)3 ÅÅÇ y ÉÑ u u Ñ 2 u 4ue − 4e + 2u + 4 Ñ ÑÑ + 2xu e ÑÑ 4 u ÑÑÖ (e − 1)

Eint(v , J ) = (v + 1/2)υẽ − (v + 1/2)2 υẽ xe − (v + 1/2)J(J + 1)αẽ − J 2 (J + 1)2 D̃

γ D̃ Bẽ − αẽ / 2

(17)

Alternatively, ln zint can be calculated and then numerically differentiated twice, with substitution into eq 10. Figure 3 shows plots of Cp,m/R for the AHNRR model for F2 and N2 along with the respective NIST-JANAF values.

ÄÅ ÉÑ ÅÅ y y 2 ÑÑÑ 2γ δ 1 2xu Å Å ÑÑ = + u + u + + Å1 + σy(1 − e−u) ÅÅÅÅÇ y e −1 3 15 ÑÑÑÑÖ (e − 1)2 (15)

in which σ = 2 and y, γ, δ, x, and u, which are dimensionless, are defined in terms of molecular constants (Table 1). It should be noted that 1/y and 1/(1 − e−u) represent the respective partition functions of the rigid rotor and the harmonic oscillator (cf., eqs 7 and 8) and the middle three terms in the bracket account for anharmonicity, rotor nonrigidity, and rotation−vibration coupling (eq 14). The last two terms account for replacing the sum over rotational levels by an integral. Students should confirm that, for the harmonic oscillator rigid rotor model, υ̃exe = υ̃eye = α̃e = D̃ = 0 and eq 14 reverts to the sum of eqs 1 and 2. Accordingly, zint, as expressed in eq 15, becomes equivalent to the product of eqs 6 and 7, i.e., eq 8. The objective now is to obtain Cp,m(T) for the AHNRR model, and this task requires obtaining the first and second derivatives of ln zint, as indicated in eq 10. Fortunately, Davidson provides the needed expression as a correction term to Cp,m for the HORR model (eq 13).30 The result is ÄÅ ÅÅ 4γ δu 2eu(eu + 1) + Cp,m,corr = RÅÅÅÅ ÅÅÇ y (eu − 1)3 ÑÉ 4ue u−4eu + 2u + 4 ÑÑÑ ÑÑ + 2xu 2eu ÑÑ (eu − 1)4 (16) ÑÖ

Figure 3. Plot of Cp,m/R vs T for F2 and N2: red solid line, F2 NISTJANAF; yellow solid line, F2 AHNRR (model 2); blue solid line, N2 NIST-JANAF; green solid line, N2 AHNRR (model 2); dashed black line, equipartition value.

In comparing Figures 2 and 3, students will readily observe that although the AHNRR results are in closer agreement with the NIST-JANAF values for N2, the calculations for F2 not only fail to account for the decrease in Cp,m at higher temperatures but also trend to values that are considerably higher than the equipartition prediction. Students should be asked to speculate why, for F2, the outcome of the more realistic AHNRR model does not improve the comparison with the NIST-JANAF values relative to the HORR model. Model 3. The Truncated Harmonic Oscillator Rigid Rotor

Students should think about what physical properties of F2 could be different from those of N2 that could explain the decrease in Cp,m. At this point, the instructor might state that, with a dissociation energy (relative to the zero-point vibrational level), D0, of 154.4 kJ/mol, the single F−F bond is considerably weaker than the triple N−N bond with D0 = 941.63 kJ/mol.15,16,31 Students will probably grasp that this difference is significant in explaining the different Cp,m(T) characteristics of the two molecules. The next question for

This term is added to the right-hand-most expression in eq 13 to obtain the desired result for model 2 E



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them to consider is how to account for these observations in modeling the Cp,m(T) curves. For a hint, the instructor might point out that the closedform expressions that give Cp,m for the HORR and AHNRR models (eqs 13 and 17) are based on a sum over vibrational states and an integral over rotational states in which v and J both go from zero to infinity. Having students examine a potential energy diagram for a diatomic molecule will help them realize that, at high temperatures, the relative population of rovibrational states near the top of the potential energy well becomes larger as temperature increases, yet as constituted, models 1 and 2 fail to account for the f inite number of accessible states in a real molecule. Also, importantly, there would seem to be no rovibrational energy levels for Eint(v, J) > D0. These considerations will lead students to understand that accurate sums of states must instead be truncated. They can test this key idea using a simple approach based on the HORR model. The truncated HORR model can be readily tested using Excel. Students create an array of energies based on eqs 1 and 2 and appropriate molecular constants for F2. The key idea is to identify v and J values consistent with the constraint that Eint(v, J) < D0. First, they can set J = 0 and Eint(v, J = 0) = D0 and solve eq 1 to find that v ≤ 14. Then they create 15 arrays, one for each value of v (including 0), and calculate for each v the rovibrational energies (eqs 1 and 2), where J spans from 0 to a chosen upper value, e.g., J = 130. By inspecting these energies, they will determine for each v the maximum value of J that satisfies the constraint. For example, the following v, Jmax values will be found, (0, 120), (1, 115), (2, 106), ... (14, 10). For a given value of v, the rotational partition function is calculated by employing eqs 3 and 14 J = Jmax, v

zv =

∑ J=0

ÄÅ É ÅÅ E (v , J ) ÑÑÑ ÑÑÑ (2J + 1)expÅÅÅ− int ÅÅ kT ÑÑÑÖ ÅÇ

Figure 4. Results of the Truncated HORR (model 3) for F2: solid red line, F2 NIST-JANAF; solid yellow line, F2 HORR (model 1); solid blue line, F2 truncated HORR (model 3); dashed black line equipartition value.

molecule. The high temperature properties of Cp,m(T) for N2 will be discussed later in this article. Model 4. The Truncated Anharmonic Nonrigid Rotor

The question now arises about how to truncate the sum over states for F2 (or N2) using a more realistic but truncated AHNRR model, i.e., using an energy expression such as the one in eq 14. One possibility is to follow the same approach that was used to truncate the sum over states for the HORR model, i.e., with the constraint that Eint(v, J) < D0. However, although that method will provide a further improvement relative to the HORR model, it will not satisfactorily reproduce the NIST-JANAF data. As pointed out by Gurvich et al.32 (as cited by Chase15,16), to obtain a correct count of all levels that contribute to the heat capacity, one must also consider energy levels lying above the asymptote (dissociation energy) of the electronic potential energy curve.33 Such energy levels arise as a result of a rotational barrier to dissociation, a consequence of the conservation of angular momentum, as discussed below.

(18)

The entire internal partition function is obtained as the sum of the zv terms in eq 18 over the vibrational states, from v = 0 to v = vmax: v = vmax J = Jmax, v

v = vmax

z int =

∑ v=0

zv =

∑ ∑ v=0

J=0

ÄÅ É ÅÅ Eint(v , J ) ÑÑÑ Å ÑÑ (2J + 1)expÅÅ− Ñ ÅÅ kT ÑÑÑÖ ÅÇ

Rotational Barrier to Dissociation and Quasibound Rotational States

To understand rotational barriers to dissociation, the resulting quasibound rotational states, and why these states must be included in the partition function, let us first consider the electronic potential energy curve for a diatomic molecule. Such a potential may conveniently be expressed as a Morse function34

(19)

Then, Cp,m is found using eq 10, to which the translational contribution (1.5R) is included, as in the previous models, and then converted from Cv,m to Cp,m by adding an additional addend of R. The mechanics involved in obtaining Cp,m(T) for model 3 are demonstrated in the Supporting Information. The results are shown for F2 in Figure 4. Students can confirm that a total of 1,186 energy levels are included in this calculation of zint. Students will be impressed and encouraged to see that, for F2, truncating the HORR energy levels dramatically improves the Cp,m(T) calculation relative to the untruncated model, finally creating the correct shape of first increasing and then decreasing heat capacity and bringing the results into closer alignment with the NIST-JANAF data. However, the Cp,m(T) curve decreases much too sharply above ca. 2000 K. The outcome of this calculation implies that higher temperatures are needed to observe a similar decrease in Cp,m(T) for N2 because of the much higher dissociation energy of that

V0(r ) = Dẽ [1 − e−β(r − re)]2

(20)

in which the subscript 0 signifies J = 0, r the internuclear separation, and D̃ e the dissociation energy (in cm−1 units) of the molecule relative to the bottom of the potential well. The minimum position of the well is at r = re, and β is related to molecular and physical constants previously defined, viz.33

ij 2π 2cμ yz υẽ zz υ ̃ = β = jjj j hDẽ zz e 2re(Bẽ Dẽ )1/2 (21) k { −1 For F2, β = 2.9272 Å , which is worked out in the Supporting Information workbook. For a rotating molecule, J > 0, and a centrifugal energy term must be added. This term, Vcorr(r), is due to the kinetic energy of molecular rotation and arises because angular momentum is 1/2

F



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conserved. Classically, this term would be L2/2μr2, where L is the classical angular momentum. The quantum mechanical analogue is given by

The rotational barrier to dissociation decreases with increasing J and becomes zero when the minimum and maximum coalesce to a single inflection point in the VJ(r) curve. For the v = 0 state of F2, Jmax is about 170, as is shown in the Supporting Information workbook. The corresponding VJ=170(r) curve is also shown in Figure 5. The value of Jmax decreases with increasing v and establishes the truncation point for the rotational sum in the partition function for each v.

ij h yz i1y i1y Vcorr(r ) = jjj 2 zzzJ(J + 1)jjj 2 zzz = Bẽ re2J(J + 1)jjj 2 zzz j 8π cμ z r k { kr { k {

(22)

or Vcorr(r ) =

16.858J(J + 1) ij 1 yz jj 2 zz μ kr {

Bringing the Quasibound Rotational States into the Full Partition Function (23)

The NIST-JANAF thermochemical calculations for F2, N2, and other molecules are based on the discrete sums over rovibrational states that are truncated at the respective Jmax(v) levels. At lower temperatures (e.g., 300 K), a careful accounting of states near the dissociation limit is not needed. Our discussion of quasibound rotational states is to qualitatively explain their existence; determining these Jmax truncation points for each vibrational level used to calculate the partition functions is beyond the scope of this article.28,35 With model 4, we present the results of similar calculations and provide students with the opportunity to perform such calculations themselves using the Supporting Information workbooks, which are designed to be transparent and readily understood. After seeing the improved outcome of the truncated HORR calculation, students will grasp the need to refine this calculation to incorporate the potential energy expression that accounts for vibrational anharmonicity and rotational nonrigidity, e.g., by using eq 14 to express the energies used in the sum over states calculation. The model 4 calculations use the sum over rovibrational levels based on the Jmax(v) values published by Gurvich et al.32 and cited by Chase.15,16 For example, the Jmax(v) values for F2 and N2 are given by the expressions Jmax(v) = 178(1 − v/23) for F2 and Jmax(v) = 260 − 250v/58 for N2. For these two molecules, the vibrational states used in the calculation of the partition functions are for vmax = 22 for F2 and vmax = 27 for N2. Figure 6 portrays a schematic diagram of the rovibrational energy levels used in the sum over states calculations of the partition functions that helps students visualize the summation process. Considering the 2J + 1 degeneracies, the number of rovibronic states involved is significant; for these calculations, a total of 2,136 and 5,650 rovibrational energy levels are used for F2 and N2, respectively. The calculations reported by Chase use experimental spectroscopic vibrational energies for F2 and a sixth order power series for N2.15,16 The calculations in this article, however, are more accessible to students, being based on eq 14 and the NIST molecular spectroscopic constants for the molecules.23,24 The Supporting Information workbooks for these calculations are interactive and contain a macro that enables the efficient “harvesting” and assembly of the results. The thermodynamic properties calculated are Um, Sm, and Cp,m and are readily available for students to inspect and graph. Figure 7 shows a plot of the numerical Cp,m(T) results obtained from model 4 between 300 and 6000 K. For N2 the agreement between the calculated and NIST-JANAF values is excellent; in fact, the two curves are nearly superimposable. In the case of F2, the calculated values also closely follow the NIST-JANAF data. Students should reflect on the improvement of the truncated AHNRR model in accounting for the shape of the ̈ ones previously Cp,m vs T curve relative to the more naive

−1

In eq 23, Vcorr(r) is expressed in cm and μ and r in g/mol and Å, respectively. Although eqs 22 and 23 represent rotational kinetic energy, eq 23 is added to the potential energy because it represents energy that must be “tied up” in rotational motion and therefore not available for internuclear (vibrational) motion. Therefore, the total potential function for a rotating molecule is the sum of eqs 20 and 23 VJ(r ) = Dẽ [1 − e−β(r − re)]2 +

16.858J(J + 1) ij 1 yz jj 2 zz μ kr {

(24)

Figure 5 will help students understand how eq 24 accounts for the rotational barrier to dissociation (for v = 0) and gives

Figure 5. Potential energy curves for F2: solid black line, the Morse function (eq 20); solid red line, the centrifugal potential (Vcorr, eq 23) for a value of J = 129; solid blue line, the total potential function (eq 24 with J = 129) showing the barrier to dissociation; solid yellow line, the total potential (eq 24) for a value of J = 170; dashed black line, De (12,920 cm−1).

rise to the quasibound rotational states that lie above D̃ e. The Supporting Information contains interactive Excel workbooks for F2 and N2 that enable students to select J and observe the corresponding changes in the graph. They may also determine the locations of the potential minima and maxima (and their respective potential energy values) for a given J by finding values of r for which dVJ/dr = 0. For example, a plot of eq 24 for F2 and J = 129 is shown in Figure 5. The extrema are at 1.529 Å (13,893 cm−1) and 2.389 Å (17,492 cm−1), respectively. Thus, the rotational barrier associated with J = 129 lies 3,677 cm−1 above the dissociation limit, D̃ e, of 13,814 cm−1 and can therefore support a number of quasibound rotational states. G



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Figure 6. Schematic diagram of the rovibrational energy levels in the sum over states calculation for the partition functions used in model 4.

molecules, as indicated in eqs 9, 11, and 12. These calculations and graphs of Sm(T) are available in the Supporting Information workbooks. It is instructive to illustrate here the success of their calculations in reproducing the NIST-JANAF values. Examples are shown in Table 2 for 300 and 6000 K. Table 2. Values of Sm (J/K mol) for F2 and N2 at 300 and 6000 K 300 K Calculateda F2 N2 a

Figure 7. Plots of Cp,m vs T for F2 and N2 from calculations of the truncated AHNRR (model 4). For F2: solid blue line, NIST-JANAF; solid yellow line, AHNRR (model 4). For N2: solid black line, NISTJANAF; solid red line, AHNRR (model 4).

200.899 191.722

6000 K

NIST-JANAF b

200.983 191.789c

Calculateda

NIST-JANAF

309.867 292.975

309.789b 292.984c

This work, model 4. bReference 15. cReference 16.

Cp,m Values at Higher Temperatures

Over the interval 300−6000 K, the difference between the Cp,m(T) curves for F2 and N2 could be largely explained by the smaller dissociation energy of F2 and the observation that the population of bound and quasibound rovibrational states near the dissociation limit therefore becomes significant at lower temperatures for F2 relative to N2. The key idea is that, near the dissociation limit, the number of rovibrational states available for thermal population decreases, causing a concomitant decrease in Cp,m with increasing temperature. Students might logically reason, therefore, that, in the limit of very high temperature, Cp,m should approach zero. The data

tested. One of the key lessons is that the proper truncation of rovibrational states for the internal partition function is needed to obtain Cp,m values at higher temperatures. Third-Law Entropies

Although the main objective of this article is to show how students can both understand and calculate with high accuracy and precision the heat capacity data for F2 and N2, students can also readily calculate the third-law entropies of the H



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published by Gurvich et al.36 shows that, for F2, Cp,m indeed continues to decrease above 6000 K; for example, between this temperature and 10,000 K, Cp,m further decreases from 12.815 to 23.936 J/K mol. Another conclusion students may reach is that similar behavior should be observed for N2, but at higher temperatures. Although the NIST values of Cp,m do not go beyond 6000 K, Gurvich et al.37 provide this information for N2 up to 20,000 K. A plot of Cp,m(T) for N2 using the data from the NIST-JANAF Table (300−6000 K) and from Gurvich et al.36 (5000−20,000 K) is shown in Figure 8.

in Figure 8. These Cp,m curves capture well the tale of these two molecules and how more sophisticated ideas are needed to understand them.

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CONCLUSION After exploring four models of increasing sophistication and complexity, students will successfully reproduce, with impressive accuracy, the seemingly disparate Cp,m(T) graphs for F2 and N2 in Figure 1. They will also understand how to calculate thermodynamic quantities using the principles of statistical mechanics along with molecular spectroscopic constants. They will learn that, fundamentally, there really is not a tale of two molecules, after all, since their calculations of Cp,m for F2 and N2 employ the same ideas and tools. Model 4 uses the same equation to express the rovibrational energy levels, differing only in the molecular constants used, and the same approach for truncating the respective sums over rovibrational states to obtain the partition functions. They will realize that the underlying cause that explains the difference in the Cp,m temperature dependencies of F2 and N2 can be attributed to the lower dissociation energy of F2.

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ASSOCIATED CONTENT

S Supporting Information *

The Supporting Information is available on the ACS Publications website at DOI: 10.1021/acs.jchemed.9b00029.

Figure 8. Plot of plots of Cp,m vs T. For N2: solid black line, NISTJANAF (300−6000 K);16 solid green line, calculated value, this work using the truncated AHNRR, model 4 (5000−11,000 K); solid blue line, Gurvich et al. (5000−20,000 K).36 For F2: solid red line, NISTJANAF (300−6000 K);15 solid purple line, Gurvich et al. (5000− 10,000 K).35

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Students will be surprised to observe that Cp,m for N2 does not decrease but markedly increases above ca. 6000 K, reaching a maximum of ca. 67 J/K mol at 16,000 K. This is quite a ̈ expectation of the equipartition departure from the naive principle, i.e., 37.41 J/K mol. The green curve shows the extension of the calculations described for model 4 up to 11,000 K and lends credibility to the expectation that Cp,m for N2 should decrease at higher temperature. The next, and final, question that students confront in this study is why does Cp,m continue to increase, and, in fact, by so much. One must go back to basics. The total partition function must contain contributions from all energy modalities available to the system, not only those from translation, rotation, and vibration, but also higher electronic states. The calculations described up to this point consider only the ground electronic states of F2 and N2 (i.e., zelec = 1). For N2, there are three higher lying electronic states that become thermally accessible at higher temperatures and have an appreciable effect on the Cp,m sum over states calculation above ca. 4800 K. These states, designated as A3Σu+, B3Πg, and W3Δg, have triplet spin multiplicities and lie at 49,755, 59,307, and 59,380 cm−1 above the ground electronic state, X1Σ+g .24 They must be included in the partition function for N2 using the same techniques as described in model 4. Accordingly, the molecular constants for these states are used in such calculations. It is worthwhile to point out to students that, for F2, the first excited electronic state23 lies at 93,099 cm−1 and thus may be disregarded in the calculation of the total partition function at these temperatures. This information helps them understand why Cp,m decreases to temperatures up to 10,000 K as shown

Excel workbooks for F2 and N2 and macro file (ZIP) Further discussion and documentation of the workbooks (PDF, DOCX)

AUTHOR INFORMATION

Corresponding Author

*E-mail: [email protected]. ORCID

Arthur M. Halpern: 0000-0002-2211-2826 Robert J. Noll: 0000-0002-6854-9951 Notes

The authors declare no competing financial interest.

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ACKNOWLEDGMENTS The authors acknowledge many helpful discussions with Charles J. Marzzacco. R.J.N. thanks ISU for sabbatical support and Prof. E. Marinero and the School of Materials Engineering at Purdue University for hosting his sabbatical leave.

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REFERENCES

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