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

9. Key Equations Continued
Rotation of a nonlinear
qr3
molecule with a q 3 ? 10 4 ? 10 5 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 ?
Electronic Level ? E ? q ?
exp ? ? ?
(Assuming That the e ? ? T ? q ?
exp( ? ?? E) B e
Levels Are Widely
Spaced)

10. Table 6.7 Simplified Expressions For Partition Functions
Type of Mode
Partition Function
after substituting values of kB and hp
Average velocity of a molecule
Translation of a molecule in thre dimensions (partition function per unit volue
Rotation of a linear molecule
Rotation of a nonlinear molecule
Vibration of a harmonic oscillator 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
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

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

17. Derivation Of Simplified Function

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

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

20. 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

21. Simplified Expression For The Term In The Exponent
(6.C.16)
(6.C.17)
Therefore
(6.C.18)

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

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

24. Example Calculate The Molecular Velocity Of HBr
Solution
Derivation

25. Derivation Of Expression For Molecular Velocities
Use the classical partition function (replace sums by integrals). The expectation value of the molecular velocity, v, is given by:
(6.B.1)

26. Solution Cont
Consider a single molecule whose energy is independent of position. Substituting momentum p = mass times the velocity, and canceling out all of the excess integrals yields:
(6.B.2)

27. Next Substitute For U
(6.B.3)
(6. B.5)

28. Performing The Algebra Noting  = 1/kBT Yields:
P-Chem expression for molecular velocity
(6. B.8)

29. Simplified Expression

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

31. 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

32. 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.

33. 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)

34. 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)

35. 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.

36. 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

37. 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)

38. 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

39. 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.

40. 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.

41. Lots Of Algebra Yields kB
(6.95)

42. 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)

43. 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)

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

45. Table 6.7 Simplified Expressions For Partition Functions
Type of Mode
Partition Function
after substituting values of kB and hp
Average velocity of a molecule
Translation of a molecule in thre dimensions (partition function per unit volue
Rotation of a linear molecule
Rotation of a nonlinear molecule
Vibration of a harmonic oscillator 6

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