review of statistical mechanics continued
DESCRIPTION
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 IsothermTRANSCRIPT
ChE 553 Lecture 8
Review Of Statistical Mechanics Continued
1
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
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
q=molecular partition functionqn=partition function for an individual
modegn=degeneracy of the mode
5
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
6
Oa Ue
63nV
3r
3t egqqqq (non linear
molecules) (6.77)
Oa U
e53n
V2
r3
t egqqqq (linear molecules) q = Molucuar partition function
Equations For Molecular Partition Function
7 Molecular
Equations For The Partition Function For Translational, Rotational, Vibrational Modes And Electronic Levels
8
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
q m T) aht
g B
12
x
p
(2 ?
qt 1 - 10/ ax
Translation of a
molecule of an ideal gas at a pressure PA and a
temperature T
qN
m T)h
TP
t3
g B
32
p3
B
A
(2 ? ?
qt3 10 106 7
Rotation of a linear
molecule with moment of inertia I
q I TS hr
B
n p2
228
?
where Sn is the symmetry number
qr
2 10 102 4
TB q =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
9
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)
q ETe
B
exp
? q E)e exp(
ķB
3 12 2
B a b c3r 3
n p
8πk T I I Iq =
S h
vp 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
10
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 12 2
B a b c3r 3
n p
8πk T I I Iq =
S h
vp B
1q =1-exp -h υ/k T
qT m
tg3
32
32
300K 1AMU
1.16Å3
AMU-ÅI
K300T
S4.12q 2
n
2r
11 2213
g
Å T 1amuv=2.52×10sec 300K m
33223 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
11
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 = 6Translations = 3Rotations = 2Leaves 1 vibration
12
The Translational Partition Function
13
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
14
3/ 227 2
-232
3 3 32 1034
1.66 10 kg kg m2π amu 1.381 10 300K1amu sec K 0.977
Åkg m 10 Å6.626 10sec m
3/2B3
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
15
(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 12.4300K 1amu Å S
16
q I 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 IFrom P-chem
Where
18
I rAB2
(6.C.10)
)m(mmm
BrHBrH
0.988AMU
80AMU1AMU80AMU1AMU
=
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
19
(6.C.14)
22r 2
300K 1.97amu Å 1q 12.4 24.4300K 1amu Å 1
The Vibrational Partition Function
From Table 6.6
p -3-1
B
h υ υ 300K=4.78×10k T 1cm T
20
(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
1q1-exp(-h / T)k
Derivation
Evaluation Of h For Our Case
22
Plugging (6.3.19) into (6.3.15) yields
q v
11 12 7
10exp .
.
(6.3.20)
(6.C.19)
(6.C.20)
(6.C.15)
-1p -3
-1B
h υ 2650cm 300K=4.78×10 12.7k T 1cm 300K
Substituting
(6.C.19)
Summary
qT=843/ , qr=24.4 qv=1
Rotation and translation much bigger than vibration
23
Å
3Å
Example Calculate The Molecular Velocity Of HBr
Solution
24
81300
AMUT K
V x KK
AMUAMU
x
2 52 10 300300
181
2 8 10131 2 1 2
12.sec
.sec
/ /Å Å
Derivation2
1
AB
21
13
mamu1
K300T
secÅ1052.2v
Next Derive Adsorption Isotherm
30
• 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
31
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.
32
.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:
33
)!N(S!S)1N2)...(S1)(S)(S(Sg
ao
oaoooo
Da
(6.83)
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)
34
!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.
35
NaA
aao
oNcanon )(q
!N)!N(S!S
Q
)!N-Ln(S-)!Ln(N-)!Ln(S+qLnNTA aoaoAaBs
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)
36
)TLn(QA NcanonBs
kB
kB
Use Stirling’s Approximation To Simplify Equation (6.88).
XXLnX)Ln(X!
37
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)
38
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.
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
!N1gg
a
!Nq
Qg
NgN
canon
g
Lots Of Algebra Yields
gGBg LnN-)Ln(qT
41
(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):
42
as
g
a
aog
aqq
Ln)N(SN
NLn
(6.97)
g
a
aog
aqq
)N(SNN
(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
aqq
)N(SNN
43
(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
)(CCC
44
Partition function per unit volume
Table 6.7 Simplified Expressions For Partition Functions
45
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 12 2
B a b c3r 3
n p
8πk T I I Iq =
S h
vp B
1q =1-exp -h υ/k T
qT m
tg3
32
32
300K 1AMU
1.16Å3
AMU-ÅI
K300T
S4.12q 2
n
2r
11 2213
g
Å T 1amuv=2.52×10sec 300K m
33223 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
46