ChE 452 Lecture 18 Review Of Statistical Mechanics 1.
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Transcript of ChE 452 Lecture 18 Review Of Statistical Mechanics 1.
ChE 452 Lecture 18 Review Of Statistical Mechanics
1
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.
2
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.
3
Properties Of Partition Functions
Assume m independent normal modes of a molecule
q=molecular partition functionqn=partition function for an
individual modegn=degeneracy of the mode
4
m
1nnnqgq
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 modes3N-6 Vibrational modes
Linear molecules only have 2 rotational modes3N-5 vibrational modes
5
Equations For Molecular Partition Function
6
Oa U
e
63n
V
3
r
3
t egqqqq (non linear molecules)
(6.77)
Oa U
e
53n
V
2
r
3
t egqqqq (linear molecules) q = Molucuar partition function Molecular
Equations For The Partition Function For Translational, Rotational, Vibrational Modes And
Electronic Levels
7
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 ax
qm T) a
ht
g B
1
2x
p
(2 ?
qt 1 - 10/ ax
Translation of a
molecule of an ideal gas at a pressure PA and a
temperature T
q
N
m T)
h
T
Pt3
g B
3
2
p3
B
A
(2 ? ?
qt3 10 106 7
Rotation of a linear
molecule with moment of inertia I
qI T
S hrB
n p2
228
?
where Sn is the symmetry number
qr
2 10 102 4
TBq =1
32
g B3t 3
p
2πm k Tq =
h
1
2g B
tp
2πm k Tq =
h
Where Sn is symmetry number
qt3106-107
qr2102-104
qt1-10/ax
Key Equations Continued
8
Rotation of a nonlinear molecule with a
moment of inertia of Ia
, Ib, Ic, about three orthogonal axes
qr
3 10 104 5
Vibration of a harmonic oscillator when energy
levels are measured relative to the harmonic oscillator’s zero point
energy
where is the vibrational frequency
qv
1 3
Electronic Level (Assuming That the Levels Are Widely
Spaced)
qE
TeB
exp
?
q E)e exp(
ķB
3 1
2 2B a b c3
r 3
n p
8πk T I I Iq =
S h
v
p B
1q =
1-exp -h υ/k T
qr3104-105
qv1-3
e Bq =exp - /k T eq =exp -β
Table 6.7 Simplified Expressions For Partition Functions
9
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
1 2
B
g
8k Tv=
πm
32
g B3t 3
p
2πm k Tq =
h
3 1
2 2B a b c3
r 3
n p
8πk T I I Iq =
S h
v
p B
1q =
1-exp -h υ/k T
qT m
t
g3
32
32
300K 1AMU
1.16
Å3
AMU-Å
I
K300
T
S
4.12q
2n
2r
11 2213
g
Å T 1amuv=2.52×10
sec 300K m
33
223 a b c
r 3 3n
I I I43.7 Tq =
S 300K 1Å -AMU
v
-1
1q =
υ 300K1-exp -
209.2cm T
6
Example 6.C Calculate The Partition Function For HBr At 300°K
10
Data for Example 6.C 2650 cm-1
bond length 1.414Å mH 1 AMU mBr 80AMU
Calculate the a) translational, b) rotational, c) vibrational partition function for HBr. Data is given above.
How Many Modes In HBr
Total Modes = 3N Translations = 3 Rotations = 2 (linear molecule) Rotations = 3 (non linear molecule) Whatever left is vibrations
Total Modes = 6
Translations = 3
Rotations = 2
Leaves 1 vibration
11
The Translational Partition Function
12
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
32
g B3t 3
p
2πm k Tq =
h
Simplification Of Equation 6.3.1
13
3/ 2
27 2-23
2
3 3 32 1034
1.66 10 kg kg m2π amu 1.381 10 300K
1amu sec K 0.977
Åkg m 10 Å6.626 10
sec m
3/2
B3
p
2π×1amu×k ×300K=
h
3/ 23 2 3 2g3 B
t 3p
m 2π x 1amu x k x 300KTq =
1amu 300K h
(6.C.2)
(6.C.3)
3 2 3 2g3
t
m T 0.977q =
1amu 300K Å
Combining 6.C.2 and 6.C.3
3
Solution Continued
Equation 6.C.4 gives qt recall mg=81 AMU, T=300°K
14
(6.C.5)
3 2 3 23t
81amu 300K 0.977 712q =
1amu 300K Å Å
3 3
The Rotational Partition Function
From P-chem for a linear molecule
2r 2
n
T I 1q 12.4
300K 1amu Å S
15
qI T
S hr2 B
n p2
8 2
(6.3.6)
where qr is the rotational partition function, I is the moment of inertia, B is the Boltzmann’s constant hp is Plank’s constant, T is temperature and Sn is a “symmetry number” (1.0 for HBr).
(6.C.6)
kB
kB
Algebra yields Derivation
Calculation of Rotation Function Step : Calculate I
From P-chem
Where
17
I rAB2
(6.C.10)
)m(m
mm
BrH
BrH
0.988AMU
80AMU1AMU
80AMU1AMU
=
2 2I 0.988 1.414Å 1.97amu Å
(6.C.13)
Step 2 Calculate qr2
Substituting in I from equation (6.C.13) and Sn = 1 into equation 6.C.9 yields
18
(6.C.14)
22r 2
300K 1.97amu Å 1q 12.4 24.4
300K 1amu Å 1
The Vibrational Partition Function
From Table 6.6
p -3-1
B
h υ υ 300K=4.78×10
k T 1cm T
19
(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:
Vp B
1q
1-exp(-h / T)k
Derivation
Evaluation Of h For Our Case
21
Plugging (6.3.19) into (6.3.15) yields
q v
1
1 12 710
exp ..
(6.3.20)
(6.C.19)
(6.C.20)
(6.C.19) (6.C.15)
-1p -3
-1B
h υ 2650cm 300K=4.78×10 12.7
k T 1cm 300K
Substituting
Summary
qT=843/ , qr=24.4 qv=1
Rotation and translation much bigger than vibration
22
Å
3Å
Example Calculate The Molecular Velocity Of HBr
Solution
23
81
300
AMU
T K
V xK
K
AMU
AMUx
2 52 10300
300
1
812 8 1013
1 2 1 212.
sec.
sec
/ /Å Å
Derivation2
1
AB
21
13
m
amu1
K300
T
sec
Å1052.2v
Next Derive Adsorption Isotherm
29
Consider adsorption on a surface with a number of sites
Ignore interactions Calculate
adsorption concentration as a function of gas partial pressure
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
30
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.
31
.qgQ Na
Ncanon
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:
32
)!N(S
!S)1N2)...(S1)(S)(S(Sg
ao
oaoooo
Da
(6.83)
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)…1=Na!
(6.84)
Therefore, the total number of inequivalent arrangements will be given by:
(6.85)
33
!N)!NS(
!Sg
aao
oa
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.
34
NaA
aao
oNcanon )(q
!N)!N(S
!SQ
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)
35
)TLn(QA NcanonBs
)!N-Ln(S-)!Ln(N-)!Ln(S+qLnNTA aoaoAaBs kB
kB
Use Stirling’s Approximation To Simplify Equation (6.88).
XXLnX)Ln(X!
36
For any X. If one uses equation (6.89) to evaluate the log terms in equation (6.88), one obtains:
)N-)Ln(SN-(S-LnNN-LnSS+qLnNTA aoaoaaooAaBs
(6.90)
kB
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)
37
To,Ss
ss N
A
Aaoas Lnq-)N-Ln(S-)Ln(NT kB
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.
38
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.
39
!N
1g
ga
!N
g
NgN
canon
g
Lots Of Algebra Yields
gGBg LnN-)Ln(qT
40
(6.95)
kB
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):
41
as
g
a
aog
a
q
qLn
)N(SN
NLn
(6.97)
g
a
aog
a
q
q
)N(SN
N
(6.98)
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:
g
a
aog
a
q
q
)N(SN
N
42
(ad)g ASA
g
aA q
qK
(6.99)
(6.100)
If we want concentrations, we have to divide all of the terms by
volume
'g
'a
sg
a
q
q
)(CC
C
43
Partition function per unit volume
Memorize this equation
Table 6.7 Simplified Expressions For Partition Functions
44
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
1 2
B
g
8k Tv=
πm
32
g B3t 3
p
2πm k Tq =
h
3 1
2 2B a b c3
r 3
n p
8πk T I I Iq =
S h
v
p B
1q =
1-exp -h υ/k T
qT m
t
g3
32
32
300K 1AMU
1.16
Å3
AMU-Å
I
K300
T
S
4.12q
2n
2r
11 2213
g
Å T 1amuv=2.52×10
sec 300K m
33
223 a b c
r 3 3n
I I I43.7 Tq =
S 300K 1Å -AMU
v
-1
1q =
υ 300K1-exp -
209.2cm T
6
Summary
Can use partition functions to calculate molecular properties
Be prepared to solve an example on the exam
45
Question
What did you learn new today
46