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ChE 553 Lecture 8

ChE 553 Lecture 8. Review Of Statistical Mechanics Continued. Objective. Review how to calculate the partition function for a molecule Calculate the partition function for adsorption on a surface Use result to derive Langmuir Adsorption Isotherm.

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ChE 553 Lecture 8

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  1. ChE 553 Lecture 8 Review Of Statistical Mechanics Continued

  2. Objective • Review how to calculate the partition function for a molecule • Calculate the partition function for adsorption on a surface • Use result to derive Langmuir Adsorption Isotherm

  3. Last Time We Started Stat Mech To Estimate Thermodynamic Properties • All thermodynamic properties are averages. • There are alternative ways to compute the averages: state averages, time averages, ensemble averages. • Special state variables called partition functions.

  4. Properties Of Partition Functions • The partition functions are like any other state variable. • The partition functions are completely defined if you know the state of the system. • You can also work backwards, so if you know the partition functions, you can calculate any other state variable of the system.

  5. Properties Of Partition Functions Assume m independent normal modes of a molecule q=molecular partition function qn=partition function for an individual mode gn=degeneracy of the mode

  6. How Many Modes Does A Molecule Have? • Consider molecules with N atoms • Each atom can move in x, y, z direction  3N total modes • The whole molecule can translate in x, y, z  3 Translational modes • Non linear molecules can rotate in 3 directions 3 rotational modes 3N-6 Vibrational modes • Linear molecules only have 2 rotational modes 3N-5 vibrational modes

  7. Equations For Molecular Partition Function

  8. Equations For The Partition Function For Translational, Rotational, Vibrational Modes And Electronic Levels Approximate Value of Type of Mode Partition Function the Partition Function for Simple Molecules 1 ? ( 2 m T) a 2 ? Translation of a ? q 1 - 10/ a g B x ? q t x t h molecule of an ideal gas p in a one dimensional box of length a x 3 ? ? ? ( 2 m T) 3 2 ? q T ? g B Translation of a ? ? 3 6 7 t B ? ? ? q 10 10 t 3 ? ? N P h molecule of an ideal gas A p at a pressure and a P A temperature T ? 2 8 I T ? B ? 2 q r 2 Rotation of a linear ? ? S h 2 2 4 q 10 10 n p r molecule with moment where S is the n of inertia I symmetry number qt1-10/ax q =1 qt3106-107 qr2102-104 Where Sn is symmetry number

  9. Key Equations Continued Rotation of a nonlinear qr3104-105 molecule with a ? ? 3 4 5 q 10 10 r moment of inertia of I a , I , I , about three b c orthogona l axes qv1-3 Vibration of a harmonic ? ? q 1 3 v oscillator when energy levels are measured where  is the relative to the harmonic vibrational frequency oscillator’s zero point energy ? ? ? E Electronic Level ? ? ? ? q exp ? ? ?? e ? ? T q exp( E) ? (Assuming That the B e Levels Are Widely Spaced)

  10. Table 6.7 Simplified Expressions For Partition Functions 6

  11. Example 6.C Calculate The Partition Function For HBr At 300°K

  12. How Many Modes In HBr Total Modes = 6 Translations = 3 Rotations = 2 Leaves 1 vibration

  13. The Translational Partition Function From Pchem Where qt is the translational partition function per unit volume, mg is the mass of the gas atom in amu, kB is Boltzmann’s constant, T is temperature and hp is Plank’s constant 6.C.1

  14. Simplification Of Equation 6.3.1 (6.C.2) (6.C.3) Combining 6.C.2 and 6.C.3 3

  15. Solution Continued Equation 6.C.4 gives qt recall mg=81 AMU, T=300°K 3 3 (6.C.5)

  16. The Rotational Partition Function From P-chem for a linear molecule kB (6.C.6) kB Derivation Algebra yields

  17. Calculation of Rotation Function Step : Calculate I From P-chem Where (6.C.10)  = (6.C.13)

  18. Step 2 Calculate qr2 Substituting in I from equation (6.C.13) and Sn = 1 into equation 6.C.9 yields (6.C.14)

  19. The Vibrational Partition Function From Table 6.6 (6.C.15) where qv is the vibrational partition function, hp is Plank’s constant  is the vibrational frequency, kB is Boltzmann’s constant and T is temperature. Note: Derivation

  20. Evaluation Of h For Our Case (6.C.19) Substituting (6.C.19) (6.C.15) (6.C.20)

  21. Summary qT=843/ , qr=24.4 qv=1 Rotation and translation much bigger than vibration

  22. Example Calculate The Molecular Velocity Of HBr Solution Derivation

  23. Consider adsorption on a surface with a number of sites Ignore interactions Calculate adsorption concentration as a function of gas partial pressure Next Derive Adsorption Isotherm

  24. Solution Method • Derive an expression for the chemical potential of the adsorbed gas as a function of the gas concentration • Calculate canonical partition function • Use A=kBT ln(Qcanon) to estimate chemical potential • Derive an expression for the chemical potential of a gas • Equate the two terms to derive adsorption isotherm

  25. Solution Step 1: Calculate The Canonical Partition Function According to equation (6.72), q=Partition for a single adsorbed molecule on a given site ga=the number of equivalent surface arrangements.

  26. Step 1A: Calculate ga Consider Na different (e.g., distinguishable) molecules adsorbing on So sites. The first molecule can adsorb on So sites, the second molecule can adsorb on (So-1) sites, etc. Therefore, the total number of arrangements is given by: (6.83)

  27. Next: Now Account For Equivalent arrangements • If the Na molecules are indistinguishable, several of these arrangements are equivalent. • Considering the Na sites which hold molecules. If the first molecule is on any Na of these sites, and the second molecule is on any Na-1 of those sites, etc., the arrangement will be equivalent. The number of equivalent arrangements is giving by: Na(Na-1)(Na-2)…1=Na! (6.84) Therefore, the total number of inequivalent arrangements will be given by: (6.85)

  28. Step 1b: Combine To Calculate Combining equations (6.72) and (6.85) (6.86) where qa is the molecular partition function for an adsorbed molecule.

  29. Step 2: Calculate The Helmholtz Free Energy The Helmholtz free energy at the layer, As is given by: (6.87) Combining equations (6.86) and (6.87) yields: (6.88) kB kB

  30. Use Stirling’s Approximation To Simplify Equation (6.88). For any X. If one uses equation (6.89) to evaluate the log terms in equation (6.88), one obtains: kB (6.90)

  31. Step 3: Calculate The Chemical Potential Of The Adsorbed Layer The chemical potential of the layer, µs is defined by: (6.91) substituting equation (6.90) into equation (6.91) yields: (6.92) kB

  32. Step 4: Calculate The Chemical Potential For The Gas Next, let’s calculate µs, the chemical potential for an ideal gas at some pressure, P. Let’s consider putting Ng molecules of A in a cubic box that has longer L on a side. If the molecules are indistinguishable, we freeze all of the molecules in space. Then we can switch any two molecules, and nothing changes.

  33. Step 4: Continued There are Ng! ways of arranging the Ng molecules. Therefore,: (6.93) substituting equation (6.93) into equation (6.91) yields: (6.94) where Ag is the Helmholtz free energy in the gas phase, and qg is the partition function for the gas phase molecules.

  34. Lots Of Algebra Yields kB (6.95)

  35. Step 5: Set g = a To Calculate How Much Adsorbs Now consider an equilibrium between the gas phase and the adsorbed phase. At equilibrium: (6.96) substituting equation (6.92) and (6.95) into equation (6.96) and rearranging yields: Taking the exponential of both sides of Equation (6.97): (6.97) (6.98)

  36. Note That Na Is The Number Of Molecules In The Gas Phase Na is the number of adsorbed molecules and (So-Na) is the number of bare sites. Consequently, the left hand side of equation (6.98) is equal to KA, the equilibrium constant for the reaction: Consequently: (6.99) (6.100)

  37. If we want concentrations, we have to divide all of the terms by volume Partition function per unit volume

  38. Table 6.7 Simplified Expressions For Partition Functions 6

  39. Summary • Can use partition functions to calculate molecular properties • Be prepared to solve an example on the exam

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