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Chemistry 3303 HCl Stat. Mech. Winter 2012 Spectroscopy and Statistical Thermodynamics: Revisiting the HCl/DCl IR Lab Introduction and Theory This experiment has been developed by Bret Findley and Steven Mylon of the Department of Chemistry and Physics at Saint Michael’s College, Colchester, Vermont, USA. Statistical thermodynamics bridges the micro and macro worlds by using the concepts of quantum mechanics and spectroscopy to understand and predict thermodynamic properties. The data from the rotational-vibrational spectrum of HCl that you acquired in a Chemistry 2302 laboratory exercise provides you with the important molecular level information for HCl that will enable you to predict the thermodynamic properties of the bulk HCl gas. One of the fundamental concepts in statistical thermodynamics is the molecular partition function, 𝑞𝑞, which represents the number of states, or microstates, thermally accessible to the molecule. Calculating 𝑞𝑞 requires information about the states available to the molecule. From your study of quantum mechanics, you know that ideal gas molecules have several possible modes of energy: electronic, rotational, vibrational, translational and nuclear spin. The total number of thermally accessible states, 𝑞𝑞 , is equivalent to the product of the thermally accessible states in each mode and is given by equation 1. Each mode has a different number of thermally accessible states associated with it, however, note that most substances at room temperature have only one accessible electronic and nuclear spin state (the ground state); consequently, the molecular partition functions for these states equal 1. This leads to a simplified expression for 𝑞𝑞 . 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 (at room temperature) Equation 1 This exercise is divided into three parts. The theory for each part is present below. Before attempting any of the work in the results section, please read the corresponding theory section carefully. In section A you will begin by considering the vibrational and rotational energy levels together. Then you will consider the translational energy levels in section B. In section C, you will use the results from A and B to calculate some desired bulk thermodynamic properties, which is the objective the exercise. It is assumed that you have a good working knowledge of Microsoft Excel, which should be valid since it has been used extensively in Chemistry 2301 and 2302. Don’t hesitate to consult your lab instructor if you have any questions about using Excel. Chemistry 3303 HCl Stat. Mech. Winter 2012 PART A: Calculating the rotational-vibrational partition function and the rotational-vibrational contribution to the internal energy This section requires information about the rotational and vibrational energy levels available to H35Cl. This information can be determined from the spectroscopic constants (𝜈𝜈 , 𝜈𝜈 𝑥𝑥 , 𝐵𝐵 , 𝛼𝛼 and 𝐷𝐷 ) that were calculated in “The Infrared Vibrational-Rotational Spectrum of HCl and DCl” laboratory experiment performed in 2302. From that experiment, recall that, using the rigid-rotor-linear-harmonic-oscillator model, the vibrational-rotational energy levels for H35Cl can be calculated using: 𝐸𝐸 𝜈𝜈, 𝐽𝐽 = 𝜈𝜈 𝜈𝜈 + 1 + 𝐵𝐵 𝐽𝐽 𝐽𝐽 + 1 2 Equation 2 Where 𝜈𝜈 is the fundamental vibrational wavenumber, 𝜈𝜈 is the vibrational quantum number, 𝐵𝐵 is the equilibrium rotational constant and 𝐽𝐽 is the rotational quantum number. In reality, H35Cl is neither a harmonic oscillator nor a rigid rotor. Spectroscopists add terms to equation 2 to correct for anharmonicity, 𝜈𝜈 𝑥𝑥 𝜈𝜈 + , rotational-vibrational coupling, 𝛼𝛼 𝜈𝜈 + centrifugal distortion, 𝐷𝐷 𝐽𝐽 𝐽𝐽 + 1 . Thus, equation 2 becomes: 𝐸𝐸 𝜈𝜈, 𝐽𝐽 = 𝜈𝜈 𝜈𝜈 + 1 1 − 𝜈𝜈 𝑥𝑥 𝜈𝜈 + 2 2 + 𝐵𝐵 𝐽𝐽 𝐽𝐽 + 1 − 𝛼𝛼 𝜈𝜈 + Equation 3 𝐽𝐽 𝐽𝐽 + 1 , and 1 𝐽𝐽 𝐽𝐽 + 1 − 𝐷𝐷 𝐽𝐽 𝐽𝐽 + 1 2 Use of spectroscopic constants determined previously and this equation will allow for the calculation of rotational-vibrational energy terms (levels). Now the rotational-vibrational partition function, 𝑞𝑞 , can be calculated using: 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 = Equation 4 𝑔𝑔 𝑒𝑒 This quantity represents the number of rotational-vibrational states that are thermally accessible at the temperature considered. Here 𝑔𝑔 represents the degeneracy of each energy level, 𝛽𝛽 is the inverse of the product of the temperature and Boltzman’s constant, 1 𝑇𝑇𝑘𝑘 , and 𝜀𝜀 is the energy (in Joules) of each level relative to the ground vibrational state: 𝜀𝜀 = ℎ𝑐𝑐 𝐸𝐸 𝜈𝜈, 𝐽𝐽 − 𝐸𝐸 0,0 Equation 5 Thus, 𝜀𝜀 is zero for the ground state. Provided that the ground state is non-degenerate (it is in this case), 𝑔𝑔 𝑒𝑒 is equal to one for the ground state. This makes sense as the ground state is Chemistry 3303 HCl Stat. Mech. Winter 2012 always accessible, even at 0 Kelvin. At higher temperatures, more microstates are accessible and, therefore, 𝑞𝑞 > 1. Once you have calculated 𝑞𝑞 at the desired temperature, the probability that each level is occupied can be calculated using equation 6. Finally the rotational vibrational − 𝑈𝑈, (0), can be determined using contribution to the molar internal energy, 𝑈𝑈, equation 7 (where 𝑁𝑁 is Avogadro’s Number): 𝑝𝑝, = 𝑔𝑔 𝑒𝑒 𝑞𝑞 Equation 6 𝑈𝑈, − 𝑈𝑈, 0 = Equation 7 𝑁𝑁 𝑝𝑝 𝜀𝜀 PART B: Calculating the translational partition function and the translational contribution to the internal energy Because of the assumption that HCl behaves as an ideal gas, one can think of each HCl molecule as moving independently (not affected by other molecules). Another assumption, that an HCl molecule is a free particle-in-a (one dimensional)-box, allows for its translational energy, 𝐸𝐸 , to be calculated using: 𝐸𝐸 = 𝑛𝑛 ℎ 8𝑚𝑚𝐿𝐿 Equation 8 Where 𝐿𝐿 is the length of the box, and 𝑛𝑛 = 1, 2, … Recall the wavefunction for a particle in a one dimensional box, sin , must have some non-zero values within the box if the particle is not there; therefore, the ground state energy level is 𝑛𝑛 = 1, not 𝑛𝑛 = 0. The energy relative to the ground translational state is then: 𝜀𝜀 = . In principle, the translational partition function can be calculated in a similar fashion to the method used for 𝑞𝑞 using: 𝑞𝑞 = . At 𝑒𝑒 room temperature the spacing of the translational energy levels for an HCl molecule in one can be closely approximated by a dimension is so small that successive terms from 𝑒𝑒 continuous function. Thus, the summation can be replaced by the integral below, which is evaluated from 𝑛𝑛 = 0 to 𝑛𝑛 = ∞ due to the convenience of its solution: Chemistry 3303 HCl Stat. Mech. 𝑞𝑞 = 𝑒𝑒 Winter 2012 𝑑𝑑𝑑𝑑 = Equation 9 2𝜋𝜋𝜋𝜋 𝐿𝐿 ℎ 𝛽𝛽 This approximation does not cause any appreciable error, even though there is no 𝑛𝑛 = 0 for the particle-in-a-box system. In three dimensions, equation 9 becomes: 𝑞𝑞 = 2𝜋𝜋𝜋𝜋 ℎ 𝛽𝛽 Equation 10 𝑉𝑉 = 𝑉𝑉 Λ Where 𝑉𝑉 is the volume of the box, Λ is the thermal wavelength of HCl, and 𝑚𝑚 is the molecular mass of 1H35Cl (in kg). The thermal wavelength can be calculated using: Λ= ℎ 𝛽𝛽 2𝜋𝜋𝜋𝜋 Equation 11 The separation distance between molecules in one mole of 1H35Cl at STP is approximately 𝑉𝑉 or 2 × 10 m, which is much greater than Λ, ~2×10 m, under these conditions; therefore, the use of this equation is justified. Note that the thermal wavelength has units of length. This is equivalent to the thermal de Broglie wavelength 𝜆𝜆 = ℎ 𝑝𝑝 for the moving particle using the average of the absolute value of its momentum. Differentiating 𝑞𝑞 with respect to 𝛽𝛽 at constant 𝑉𝑉 allows one to calculate the translational energy contribution to the molar internal energy of the system by using: 𝑈𝑈, = − 𝑁𝑁 𝑞𝑞 Equation 12 𝜕𝜕𝑞𝑞 𝜕𝜕𝜕𝜕 This form of equation 11 is equivalent to the form presented for 𝑈𝑈, − 𝑈𝑈, 0 in equation 7, but proves more conventional for translational energy. PART C: Calculating Bulk Thermodynamic Properties The calculation of the molar internal energy 𝑈𝑈, − 𝑈𝑈, 0 of HCl is the easiest thermodynamic property to calculate as it is merely the sum of the two contributions calculated in parts A and B: Chemistry 3303 HCl Stat. Mech. Winter 2012 𝑈𝑈 − 𝑈𝑈 0 = 𝑈𝑈, + 𝑈𝑈, − 𝑈𝑈, 0 Equation 13 Calculating the absolute molar entropy of HCl, 𝑆𝑆 , requires the molar internal energy and 𝑄𝑄, the canonical partition function as in: 𝑆𝑆 = 𝑘𝑘 ln 𝑄𝑄 + . The system of HCl molecules is a macroscopic system containing one mole of molecules. In the case of entropy, we are concerned with the number of ways the system can be arranged at a particular temperature. Therefore, instead of considering the molecules individually, we consider a collection of replications of the system, which all have the same volume and number of molecules and are in thermal equilibrium with each other. Such a collection of macroscopic systems is known as a canonical ensemble and the number of thermally accessible states for the system is known as the canonical partition function, 𝑄𝑄. Because each molecule has a probability distribution of microstates, one might expect the number of thermally accessible states for the system to be equivalent to the product of molecular partitions for each molecule, 𝑄𝑄 = 𝑞𝑞 ; however, because HCl molecules are all the same, they cannot be distinguished from one another and this equation over counts the number of thermally accessible states for the system. To account for the indistinguishable nature of the molecules in our system, the canonical partition function takes the form: 𝑄𝑄 = ! = ! 𝑆𝑆 = 𝑘𝑘 ln , and the molar entropy equation becomes: − 𝑈𝑈 0 𝑈𝑈 𝑞𝑞 + 𝑁𝑁 ! 𝑇𝑇 Equation 14 Stirling`s approximation for large factorials ln 𝑛𝑛! = 𝑛𝑛 ln 𝑛𝑛 − 𝑛𝑛 is typically used here to reduce equation 14 to a more convenient form: 𝑆𝑆 = 𝑅𝑅 ln 𝑞𝑞 − ln 𝑁𝑁 + 1 + Equation 15 − 𝑈𝑈 0 𝑈𝑈 𝑇𝑇 Once the molar internal energy 𝑈𝑈 − 𝑈𝑈 0 and 𝑆𝑆 are calculated, the molar enthalpy, 0 , are easily found using definitions from 𝐻𝐻 − 𝐻𝐻 0 , and Gibbs free energy, 𝐺𝐺 − 𝐺𝐺 introductory thermodynamics: − 𝐻𝐻 0 = 𝑈𝑈 − 𝑈𝑈 0 + 𝑃𝑃𝑉𝑉 𝐻𝐻 Equation 16 𝐺𝐺 − 𝐺𝐺 0 = 𝐻𝐻 − 𝐻𝐻 0 − 𝑇𝑇𝑆𝑆 Equation 17 However, these quantities are not typically used in thermodynamics. Instead, the much more and common values of the standard molar enthalpy and Gibbs free energy of formation, ∆𝐻𝐻 ∆𝐺𝐺 , are used. Chemistry 3303 HCl Stat. Mech. Winter 2012 The calculations involving ∆𝐻𝐻 require 𝐻𝐻 − 𝐻𝐻 0 for 1H2 and 35Cl2 gases. For 1H2, 𝐻𝐻 − 𝐻𝐻 0 can be determined quite simply by using an analogous Excel template created for HCl and inputting hydrogen`s spectroscopic constants from the literature (ref.1). This is fairly straightforward, however, due to hydrogen`s symmetry, nuclear spin statistics need to be considered. Recall that the Pauli Exclusion Principle states that the total wave function for a system of fermions (particles with half integral spin) must be antisemmetric. Because the 1H nucleus is a fermion, 1H2 can occupy rotational states with even 𝐽𝐽 (𝐽𝐽 = 0, 2, 4, … ) only when its nuclear spins are paired (para-hydrogen). It can occupy rotational states with odd 𝐽𝐽 (𝐽𝐽 = 1, 3, 5, … ) only when its nuclear spins are parallel (ortho-hydrogen). If this is not immediately clear, ask yourself whether the rotational wave functions for even 𝐽𝐽 (𝐽𝐽 = 0, 2, 4, … ) are symmetric or antisemmetric functions. These rotational wave functions turn out to be symmetric and therefore must be multiplied by the antisymmetric nuclear spin functions for the total wave function to be antisymmetric as required by the Pauli Exclusion Principle. The rotational wave functions for odd 𝐽𝐽 (𝐽𝐽 = 1, 3, 5, … ) are antisymmetric requiring multiplication by any two of the three symmetric spin functions for the total wave function to be antisymmetric. The weighting of ortho- and parahydrogen (3:1) does make a difference in the calculations due to the large spacing between the rotational-vibrational energy levels of 1H2. Although similar symmetry exists in 35Cl2 the spacings between rotational-vibrational levels are so small at room temperature that one may ignore the difference in ortho and para contributions. Additionally, because there a so many thermally accessible energy levels at room temperature, a much larger Excel worksheet would be required. While, in principle, this is not a problem, 35 Cl2 is not the focus of this exercise. It is therefore more time-efficient to use 5 2 𝑅𝑅𝑅𝑅 for − 𝑈𝑈 0 in the calculation of 𝐻𝐻 − 𝐻𝐻 0 and the literature value of 𝑆𝑆 for 35Cl2 at 298.15 K. 𝑈𝑈 You will finally calculate ∆𝐻𝐻 by using the dissociation energy from the ground vibrational state, 𝐷𝐷 (ref.1) and 𝐻𝐻 − 𝐻𝐻 0 for each substance. In introductory chemistry courses you were taught that one can calculate ∆𝐻𝐻 using bond or dissociation energies of the molecules = 𝑁𝑁 1 2 𝐷𝐷, + 𝐷𝐷, − 𝐷𝐷, does give a good result; involved and indeed ∆𝐻𝐻 − 𝐻𝐻 0 values accounts for the thermal energy in each substance and however, including 𝐻𝐻 improves the result. Therefore the equation becomes: = 𝐻𝐻 − 𝐻𝐻 0 ∆𝐻𝐻 −1 2 𝐻𝐻 − 𝐻𝐻 0 + 𝑁𝑁 1 2 𝐷𝐷, + 𝐷𝐷, − 𝐷𝐷, + 𝐻𝐻 − 𝐻𝐻 0 Equation 18 Values of 𝐷𝐷 for all three substances could, in principle, be determined using the spectroscopic constants of each species, but it is more time-efficient to use the literature values in this case. Once the enthalpy of formation is calculated for 1H35Cl, the molar Gibbs free energy of formation can be calculated using: Chemistry 3303 HCl Stat. Mech. Winter 2012 ∆𝐺𝐺 = ∆𝐻𝐻 − 𝑇𝑇∆𝑆𝑆 Equation 19 where ∆𝑆𝑆 is the change in entropy for the formation reaction. Experimental There is no conventional laboratory work for this experiment. You will use your results from a previous HCl/DCl IR experiment (performed in Chemistry 2302) to calculate some bulk thermodynamic properties for HCl (or DCl). In theory, this exercise could be completed for all of the isotopic species studied in the original experiment (H35Cl, D35Cl, H37Cl and D37Cl), however, for the sake of time you will just consider one. During this 3-hour lab period use Excel to perform the calculations outlined in the results section. You will need your laboratory report entitled “The Infrared Vibrational-Rotational Spectrum of HCl and DCl” from Chemistry 2302. Please bring this report with you on the day that you are scheduled to complete this lab. Even though you are not performing a conventional laboratory experiment, please don’t hesitate to ask your instructor for help with any of the calculations. Results The procedure below should only be performed for the 1H35Cl isotopic species. Even if your calculations in the previous lab were for DCl (deuterium chloride) you should have obtained spectroscopic constants for 1H35Cl from your partner and included them as part of your report. Use these constants, where appropriate, in the following exercise. If you do not have these constants they may be obtained from Herzberg or Sime (refs.1 or 3). DO NOT use spectroscopic constants for DCl in this experiment!! To complete the exercise, please note that in addition to the spectroscopic constants calculated in your 2302 lab report, you will need to use data for H 2 and 35Cl2 that can be obtained from the references listed (ref.1). PART A: Calculating the rotational-vibrational partition function and the rotational-vibrational contribution to the internal energy 1. Begin setting up your Excel worksheet for hydrogen chloride by creating cells for values that you will use throughout this exercise as shown in Table 1. You will need the various constants (the speed of light, Boltzmann’s constant, etc.) listed and the spectroscopic constants (𝜈𝜈 , 𝜈𝜈 𝑥𝑥 , 𝐵𝐵 , 𝛼𝛼 and 𝐷𝐷 ) from the lab you completed previously. If you did not calculate one or more of these constants in your previous lab you may use the literature values (refs. 1 or 3). Chemistry 3303 HCl Stat. Mech. HYDROGEN CHLORIDE Constants 𝐶𝐶 (m/s) 𝑒𝑒 (C) 𝑘𝑘 (J/K) 𝑅𝑅 (J/K mol) ℎ (J/s) 𝑁𝑁 (1/mol) 299792460 1.60218E-19 1.38065E-23 8.31447 6.62608E-34 6.02214E23 Spec. Constants (cm1 ) 𝜈𝜈 𝜈𝜈 𝑥𝑥 𝐵𝐵 𝛼𝛼 𝐷𝐷 Winter 2012 Other Relevant Data 𝑇𝑇 (K) 𝛽𝛽 (1/J) 𝑚𝑚 (g/mol) 𝑚𝑚 (g/mol) Table 1 – Initial Excel spreadsheet setup for hydrogen chloride 2. Set up an additional table for calculations relevant to the rotational-vibrational internal energy. See the suggested column headings in Table 2. Then use equation 3 and your spectroscopic constants to calculate the vibrational-rotational energy terms of H35Cl, 𝐸𝐸 𝜈𝜈, 𝐽𝐽 . Do this for the first 16 rotational levels in the ground vibrational state and the first 10 rotational levels in the first excited vibrational state. While the choice to stop at these rotational levels is somewhat arbitrary, it probably makes little difference to the results if these numbers were, for instance, 18 and 12 or even higher. Why is this true at the temperature considered? Calculation of Erot-vib (J/mol) 𝜈𝜈 𝐽𝐽 𝐸𝐸 𝜈𝜈, 𝐽𝐽 𝜀𝜀 2𝐽𝐽 + 1 0 0 0 1 etc. etc. 𝑔𝑔 𝑒𝑒 Prob. 𝐸𝐸 (J) 𝐸𝐸 (J/mol) Table 2 – Suggested headings for Excel calculations associated with the rotational-vibrational contribution to the internal energy 3. Use equation 5 to calculate the energy 𝜀𝜀 , in Joules for each rotational-vibrational energy level relative to the ground rotational vibrational state. 4. Include the degeneracies of each energy level as a separate column in Excel. Recall that vibrational states are non-degenerate, but rotational levels for a diatomic molecule are 2𝐽𝐽 + 1 -fold degenerate. 5. Once you have calculated both the degeneracy and the energy of each term relative to the ground state, you are now prepared to calculate the rotational-vibrational partition function, 𝑞𝑞 , using equation 4. The partition function is calculated by summing over the 𝑔𝑔 𝑒𝑒 terms in a separate column (see Table 2) and then sum these values in a cell somewhere else on the same worksheet. 6. Use equation 6 and your answers from step 5 to calculate the probability, 𝑝𝑝, , that each vibrational-rotational energy level 𝑖𝑖 is occupied at 298.15 K. Use a separate column labeled “Prob.” as shown in Table 2. Consider your calculated results for the probability that each state is occupied. Why is it not necessary for you to calculate the probabilities of more states? Chemistry 3303 HCl Stat. Mech. Winter 2012 7. Multiply the individual energies, 𝜀𝜀 , by the probability that this energy level is occupied. This yields the contribution of each energy level for an average single molecule to the total molecular energy. As suggested in Table 2, the column labeled “𝐸𝐸 (J)” is created for the case of a single HCl molecule. In the next column, multiply your values by Avogadro’s number to obtain the contribution of each energy level to the molar internal energy in J/mol. Finally, sum the entire ensemble of energy levels to determine the entire vibrational and rotational contribution to the molar internal energy. Include this last result in a separate table, perhaps with the rotational-vibrational partition function as in Table 3 below. Summary for 𝑼𝑼𝒐𝒐𝒎𝒎,𝒓𝒓𝒓𝒓𝒓𝒓𝒗𝒗𝒗𝒗𝒗𝒗 − 𝑼𝑼𝒐𝒐𝒎𝒎,𝒓𝒓𝒓𝒓𝒓𝒓𝒗𝒗𝒗𝒗𝒗𝒗 (𝟎𝟎) 𝑞𝑞 𝑈𝑈, − 𝑈𝑈, (0) (J/mol) 𝑅𝑅𝑅𝑅 (J/mol) % difference Table 3 – Suggested table for the results of the rotational – vibrational contribution to the internal energy Part B: Calculating the translational partition function and the translational contribution to the internal energy 1. Before performing the calculations for this section, set up a table similar to Table 4. Calculation of 𝑼𝑼𝒐𝒐𝒎𝒎,𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 Λ (m) 𝑉𝑉 (m3) 𝑞𝑞 𝑞𝑞 𝜕𝜕𝑞𝑞 𝜕𝜕𝜕𝜕 (J) 𝑈𝑈, (J/mol) 3 𝑅𝑅𝑅𝑅 (J/mol) 2 % difference Table 4 – Suggested Excel table for the results involving the translational contribution to the molar internal energy 2. First calculate the thermal wavelength, Λ, at room temperature (298.15 K) as shown in equation 11 and the molar volume at 1.00 bar (1.00 bar = 10 5 Pa) assuming ideal gas behavior. Then use equation 10 to calculate 𝑞𝑞 . 3. You can now calculate the total molecular partition function using 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 . Recall that 𝑞𝑞 𝑞𝑞 = 1 at the temperatures we are considering. Chemistry 3303 HCl Stat. Mech. Winter 2012 4. Differentiate 𝑞𝑞 with respect to 𝛽𝛽 at constant 𝑉𝑉. Then use your result to calculate the translational energy contribution to the molar internal energy by using equation 12. Compare your result with 3 2 𝑅𝑅𝑅𝑅, the result you would expect from what you learned in the thermodynamics unit in Chemistry 2301. Part C: Calculating Bulk Thermodynamic Properties 1. Set up a table in Excel similar to the one suggested below. Note that the bottom three quantities, Δ𝑆𝑆 , Δ𝐻𝐻 and Δ𝐺𝐺 , all require information about H2 and Cl2 before they can be calculated. Bulk Thermodynamic Properties Calculated 𝑈𝑈 − 𝑈𝑈 (0) (J/mol) 𝑆𝑆 (J/mol) 𝐻𝐻 − 𝐻𝐻 (0) (J/mol) 𝐺𝐺 − 𝐺𝐺 (0) (J/mol) Δ 𝑆𝑆 (J/mol K) Δ𝐻𝐻 (kJ/mol) Δ𝐺𝐺 (kJ/mol) Literature % difference N/A N/A N/A N/A Table 5 – Suggested Excel table for summary of results related to calculation of bulk thermodynamic properties 2. Calculate 𝑈𝑈 − 𝑈𝑈 (0) using equation 13. Compare your result with 5 2 𝑅𝑅𝑅𝑅, the values expected from classical thermodynamics and calculate the percent difference. 3. Calculate the molar entropy, 𝑆𝑆 , using equation 15 and compare it to the value reported in Table A.8 located in appendix A of Mortimer (your textbook). Calculate the percent difference between the calculated and textbook values. − 𝐻𝐻 (0) and 𝐺𝐺 − 𝐺𝐺 (0) as indicated in equations 16 and 17. 4. Calculate 𝐻𝐻 5. Calculate 𝐻𝐻 − 𝐻𝐻 (0) and 𝑆𝑆 for H2 using the following steps: a. Copy your entire worksheet for H35Cl and paste it into a new worksheet in the same workbook (i.e. keep it part of the same Excel file) b. In your first table, change the heading to “Hydrogen Gas” and replace the spectroscopic constants of HCl with those of H2 as found in Herzberg (. Also change the mass labels and some of the quantities under “Other Relevant Data” as needed. c. In a table similar to Table 2 of your Excel worksheet, create an additional column in order to properly account for ortho and para hydrogen as shown below. Chemistry 3303 HCl Stat. Mech. Calculation of Erot-vib (J/mol) 2𝐽𝐽 𝜈𝜈 𝐽𝐽 𝐸𝐸 𝜈𝜈, 𝐽𝐽 𝜀𝜀 +1 0 0 0 1 etc. etc. Winter 2012 Para H2 Ortho H2 Prob. 𝐸𝐸 (J) 0.25×𝑔𝑔 𝑒𝑒 0.75×𝑔𝑔 𝑒𝑒 𝐸𝐸 (J/mol) Table 6 – Table for calculations involving the rotational-vibrational contribution to the internal energy for H2 gas d. As mentioned previously, para-hydrogen can only occupy states with even 𝐽𝐽’s and ortho-hydrogen can only occupy states with odd 𝐽𝐽’s. The weighting of para to ortho states is 1:3; therefore, an additional weighting factor should be included (see Table 6). One can incorporate this weighting into the calculation of 𝑞𝑞 : 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 = 0.25× 𝑔𝑔 𝑒𝑒 + Equation 20 0.75×𝑔𝑔 𝑒𝑒 e. Also include the weighting factor, 𝑤𝑤 = 0.25 𝑜𝑜𝑜𝑜 0.75, to calculate the probablility, 𝑝𝑝, , that each vibrational-rotational energy level 𝑖𝑖 is occupied at 298.15 K as: 𝑝𝑝, = 𝑤𝑤 𝑔𝑔 𝑒𝑒 𝑞𝑞 Equation 21 f. The rest of the calculations for 𝐻𝐻 − 𝐻𝐻 (0) and 𝑆𝑆 of H2 follow the same pattern as for HCl and should require very little work. − 𝐻𝐻 (0) and 𝑆𝑆 for 35Cl2 in order to calculate 6. Additionally, you will need to calculate 𝐻𝐻 and ∆𝐺𝐺, . However, as described above, it is preferable to use some ∆𝐻𝐻, approximations in order to save time: − 𝑈𝑈 0 a. Use 5 2 𝑅𝑅𝑅𝑅 for 𝑈𝑈 equation 16. and then calculate 𝐻𝐻 − 𝐻𝐻 0 for 35Cl2 using b. Obtain the literature value for 𝑆𝑆 of 35Cl2 at 298.15 K. 7. Obtain the literature values for the dissociation energy of HCl, H2 and Cl2 from the ground vibrational state, 𝐷𝐷 , by looking them up in the literature. Then use equation 18 . to calculate ∆𝐻𝐻, Chemistry 3303 HCl Stat. Mech. Winter 2012 8. Calculate the change in entropy for the formation reaction and then use equation 19 to . calculate ∆𝐺𝐺, Discussion Using the data obtained in a previous lab exercise, you have determined vibrational and rotational partition functions for HCl. You have used the particle-in-a-box model to arrive at the translational partition function for HCl. Armed with these partition functions, you were then able to calculate thermodynamic quantities such as the internal energy and absolute entropy. You did the same for H2 using method described above except with literature values for the spectroscopic constants. Finally, you were able to calculate the bulk thermodynamic properties ∆𝐻𝐻 and ∆𝐺𝐺, for H35Cl by employing these methods. In your discussion, comment on your results and answer the following questions: 1) Compare the values of 𝑞𝑞 and 𝑞𝑞 . Which mode of motion has the greater number of thermally accessible states? Which mode of motion (translational, rotational or vibrational) has the fewest thermally accessible states? Briefly explain your answers. 2) Why can the electronic partition function for H35Cl be ignored in our treatment. 3) Why can the summation for the translational partition function be replaced by integration? 4) How will changes in temperature affect your results? Comment on which partition functions are most affected by incremental changes (±100 K) changes in temperature. 5) For the sake of time, we only considered H35Cl instead of combinations of other isotopes. How would changes in atomic isotopes affect your results? Which partition functions would be affected most? 6) By making the jump from the microscopic to the macroscopic world, we compared our calculated results for the H35Cl molecules to bulk thermodynamic data found in typical physical chemistry texts. To which of these isotopes do these literature values correspond? How might changes in isotope affect the dissociation energies involved in the calculation of ∆𝐻𝐻 ? You might want to consider the Morse potentials and their zero point energy levels for the species involved. 7) As you know from the thermodynamics unit of physical chemistry, the molar heat 𝜕𝜕𝑈𝑈 capacities at constant volume, 𝐶𝐶, , and constant pressure, 𝐶𝐶, , are 𝜕𝜕𝜕𝜕 and 𝜕𝜕𝐻𝐻 𝜕𝜕𝜕𝜕 respectively. Use your excel worksheets to estimate these values of 𝐶𝐶, and 𝐶𝐶, at 298.15 K for H35Cl. Compare your result of 𝐶𝐶, with that found in your textbook. Chemistry 3303 HCl Stat. Mech. Winter 2012 References 1. Herzberg, Gerhard. Molecular Spectra and Molecular Structure, Volume 2. D. Van Nostrand Company, New York, 1953. (on reserve for Chem 2302 at QEII Library) 2. Mortimer, Robert G. Physical Chemistry, 3rd ed., Elsevier Academic Press, San Diego, 2008. (One copy is available in C-3041) 3. Sime, Rodney J. Physical Chemistry: Methods, Techniques, Experiments, Saunders College Publishing, Philadelphia, 1990.