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Title: ChE 551 Lecture 11


1
ChE 551 Lecture 11
  • Review Of Statistical Mechanics

2
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.

3
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.

4
Properties Of Partition Functions
  • Assume m independent normal modes of a molecule
  • qmolecular partition function
  • qnpartition function for an individual mode
  • gndegeneracy of the mode

5
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

6
Equations For Molecular Partition Function
7
Equations For The Partition Function For
Translational, Rotational, Vibrational Modes And
Electronic Levels
qt?1-10?/ax
q
1
qt3?106-107
qr2?102-104
Where Sn is symmetry number
8
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
oscillators zero point
energy



?
?
?
E
Electronic Level

?
?
?
?
q
exp

e
?
?
?
?
??
T
q
E)
exp(
?
(Assuming That the
B
e
Levels Are Widely
Spaced)

9
Table 6.7 Simplified Expressions For Partition
Functions
Type of Mode Partition Function 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
10
Example 6.C Calculate The Partition Function For
HBr At 300K
11
How Many Modes In HBr
  • Total Modes 6
  • Translations 3
  • Rotations 2
  • Leaves 1 vibration

12
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 Boltzmanns
constant, T is temperature and hp is Planks
constant
6.C.1
13
Simplification Of Equation 6.3.1
(6.C.2)
(6.C.3)
Combining 6.C.2 and 6.C.3
3
14
Solution Continued
  • Equation 6.C.4 gives qt recall mg81 AMU, T300K

3
3
(6.C.5)
15
The Rotational Partition Function
  • From P-chem for a linear molecule

kB
(6.C.6)
kB
Derivation
Algebra yields
16
Derivation Of Simplified Function

(6.C.7)
(6.C.8)
(6.C.9)
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 Planks constant ? is the
vibrational frequency, kB is Boltzmanns constant
and T is temperature. Note
Derivation
20
Simplified Expression For The Term In The Exponent
(6.C.16)
(6.C.17)
Therefore
(6.C.18)
21
Evaluation Of h? For Our Case
(6.C.19)
Substituting
(6.C.19)
(6.C.15)
(6.C.20)
22
Summary
  • qT843/ , qr24.4 qv1
  • Rotation and translation much bigger than
    vibration

23
Example Calculate The Molecular Velocity Of HBr
  • Solution

Derivation
24
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)
25
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)
26
Next Substitute For U

(6.B.3)
(6. B.5)
27
Performing The Algebra Noting ? 1/kBT Yields
P-Chem expression for molecular velocity
(6. B.8)
28
Simplified Expression

29
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

30
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 AkBT ln(Qcanon) to estimate chemical
    potential
  • Derive an expression for the chemical potential
    of a gas
  • Equate the two terms to derive adsorption
    isotherm

31
Solution Step 1 Calculate The Canonical
Partition Function
  • According to equation (6.72),
  • qPartition for a single adsorbed molecule on a
    given site
  • gathe number of equivalent surface
    arrangements.

32
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)
33
Next Now Account For Equivalent
  • 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)1Na!
  • (6.84)
  • Therefore, the total number of inequivalent
    arrangements will be given by
  • (6.85)

34
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.

35
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
36
Use Stirlings 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)
37
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
38
Step 4 Calculate The Chemical Potential For The
Gas
  • Next, lets calculate µs, the chemical potential
    for an ideal gas at some pressure, P. Lets
    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.

39
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.

40
Lots Of Algebra Yields
kB
(6.95)
41
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)
42
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)
43
If we want concentrations, we have to divide all
of the terms by volume
Partition function per unit volume
Memorize this equation
44
Table 6.7 Simplified Expressions For Partition
Functions
Type of Mode Partition Function 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
45
Summary
  • Can use partition functions to calculate
    molecular properties
  • Be prepared to solve an example on the exam

46
Question
  • What did you learn new today
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