Transcript che 377 lectures - Classnotes For Professor Masel's Classes

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
m
q   gnqn
n 1
q=molecular partition function
qn=partition function for an individual
mode
gn=degeneracy 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
q  q t  q r  q V 
3
3n a  6
3
g e e   U O
(non linear
molecules)
(6.77)
q  q t  q r  q V 
3
2
3n a  5
g e e   U O
(linear
molecules)
q = Molecular
Molucuar partition function
7
Equations For The Partition Function For Translational, Rotational,
Vibrational Modes And Electronic Levels
Type of Mode
Translation of a
molecule of an ideal gas
in a one dimensional
box of length ax
Approximate Value of
Partition Function the Partition Function
for Simple Molecules
1
2


(2 ?mg B?BTT) a1x2
qt ? 2πmg k BT
hp
q=
t
hp

3
2

3
2?
(2 ?mg ?
qt
B T) ? ?
B T
?
?
?
B
3 g
3
?
N
PA ?
hp
3
?1 - 10/ ax
qqqtt1-10/a
=1
x
2πm k T
Translation of a
3
molecule of an ideal gasq t =
hp
at a pressurePA and a
temperature T
2?2
8
IBT?
T
8

I
2
B T
8πIk
2
2 qq  ?
B
r
2
r
Rotation of a linear q r = SnShnph2p 2
Sn Shisp the
molecule with moment
where
n
Wheresymmetry
Sn is symmetry
number
number
of inertia I
 


qt 10 107
67
? 106 ? 10
q3t 3
q 10 -104
24
q2r r?2102 ? 10
8
Key Equations Continued
Rotation of a nonlinear
3
1
8πk BT  2  Ia I b Ic  2

molecule with a
3
qr =
3
moment of inertia of aI
Sn  h p 
Ib, I,c about three ,
orthogonal axes
10 ? 4
10
qqr3?10
105
3
r
Electronic Level
(Assuming That the
Levels Are Widely
Spaced)
9
?
e
v

E
?
?
q ? exp
qe =exp

ķ
 ? ?/kT ??BT
?
B
B
?
5
?1 ?3
qq v1-3
1
Vibration of a harmonic
q
=
oscillator when energy v 1-exp -h υ/k T
p
B
levels are measured
where  is the
relative to the harmonic
oscillator’s zero point vibrational frequency
energy

4
qe =exp
?-β
?? E) 
q ? exp(
e
Table 6.7 Simplified Expressions For
Partition Functions
Type of Mode
Partition Function
Average velocity of a
molecule
Rotation of a linear molecule
Rotation of a nonlinear
molecule
Vibration of a harmonic
oscillator
10
Å  T   1amu 
v=2.52×1013


 
sec  300K   mg 
1
12
 8k T 
v=  B 
 πm 
g 

Translation of a molecule in
thre dimensions (partition
function per unit volue
qt3 =
 2πmg k BT 
h 
3
2
3
p
qr2 =
8πIk B T
Sn  h p 
3
Sn  h p 
3
1
2
T  2  mg 
3 1.16 
qt  3 
 

Å  300K   1AMU 
3
2
 12.4  T 
I

2

q r  
 2

 Sn  300K  Å - AMU 
2
1
8πk BT  2  Ia Ib Ic  2
q 3=
r
Partition Function
after substituting values of kB and
hp
3
1
qv =
1-exp  -h p υ/k BT 
3
3
 43.7   T  2  Ia Ib Ic  2
qr3 = 

  36
3 
 Sn   300K   1Å -AMU 
qv =
1
 
υ
 300K 
1-exp  - 

-1 
  209.2cm  T 
2
Example 6.C Calculate The Partition
Function For HBr At 300°K
Data for Example 6.C

bond length
mH
mBr
2650 cm-1
1.414Å
1 AMU
80AMU
Calculate the a) translational, b) rotational,
c) vibrational partition function for HBr.
Data is given above.
11
How Many Modes In HBr
Total Modes = 3N
Translations = 3
Rotations = 2 (linear molecule)
Rotations = 3 (non linear molecule)
Whatever left is vibrations
12
Total Modes = 6
Translations = 3
Rotations = 2
Leaves 1 vibration
The Translational Partition
Function
From Pchem
qt3 =
 2πmg k BT 
h 
3
3
2
6.C.1
p
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
13
Simplification Of Equation
6.3.1
32
 mg   T 
q =
 

1amu
300K




3
t
32
 2π x 1amu x k B x 300K 


3
hp


3/ 2
(6.C.
2)
 2π×1amu×k B ×300K 
hp
3
3/2
=
2


 1.66 1027 kg  
-23 kg  m 
300K


 2πamu  

 1.38110

1amu
sec2 K 




3
3
2
10

34 kg  m   10 Å 
6.626

10

 

sec   m 

(6.C.
3)
Combining 6.C.2 and 6.C.3
32
14
32
m


T
0.977


g
q3t = 
3
 

1amu
300K
Å


 
3/ 2

0.977
Å3
Solution Continued
Equation 6.C.4 gives qt recall mg=81
AMU, T=300°K
32
32
 81amu   300K 
q =
 

 1amu   300K 
3
t
(6.C.5)
15
0.977 712

Å
Å
3
3
The Rotational Partition
Function
From P-chem for a linear
molecule 2 8 2 I 
kBB T
qr 
Sn h 2p
(6.3.6)
(6.C.6)
where qr is the rotational partition
function, I is the moment of inertia, kBB is
the Boltzmann’s constant hp is Plank’s
constant, T is temperature and Sn is a
“symmetry number” (1.0 for HBr).
Algebra yields
I
 T 
 1 
q  12.4 


2 
300K
1amu

Å
S


 n 
2
r
Derivation
16
Calculation of Rotation
Function Step : Calculate I
From P-chem
I   rAB 
Where
2
(6.C.10)
=
mH mBr 1AMU 80AMU 
 0.988AMU
(mH  mBr ) 1AMU  80AMU
I   0.988 1.414Å   1.97amu  Å 2
2
18
(6.C.13)
Step 2 Calculate qr2
Substituting in I from equation (6.C.13)
and Sn = 1 into equation 6.C.9 yields
2

1
300K
1.97amu

Å


2
q r  12.4 
    24.4

2
 300K   1amu  Å   1 
(6.C.14)
19
The Vibrational Partition
Function
From Table 6.6
qV
1

1-exp(-h p /kBT)
(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: h p υ
υ  300K 
-3 
Derivation
=4.78×10 

-1 
k BT
1cm
T



20
Evaluation Of h For Our
Case
-1

-3 2650cm   300K 
=4.78×10 

  12.7
-1
k BT
 1cm
  300K 
hpυ
(6.C.19)
Substituting
Plugging(6.C.19)
(6.3.19) into (6.C.15)
(6.3.15) yields
1
qv 
 10
.
1  exp  12.7
(6.3.20)
(6.C.20)
22
Summary
3
Å
qT=843/ , qr=24.4 qv=1
Rotation and translation much bigger
than vibration
Å
23
Example Calculate The Molecular
Velocity Of HBr
Solution
T 
13 Å 


v  2.5210
sec  300 K 
1
2
 1 amu 


 m AB 
1
2
Derivation
  81AMU
T  300K
Å  300K
V  2.52 x10


sec  300K
13
24
1/ 2
 1AMU 


 81AMU
1/ 2
Å
 2.8x10
sec
12
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 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),
N
Qcanon
 gaq .
N
q=Partition for a single adsorbed
molecule on a given site
ga=the 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:
So !
g a (So )(So 1)(So 2)...(So N a 1)
(So N a )!
D
(6.83)
33
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:
So!
ga 
(So  N a )! N a !
(6.85)
34
Step 1b: Combine To
Calculate
Combining equations (6.72) and (6.85)
N
Q canon 
So !
Na
(q A )
(S o  N a )!N a !
(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:
N
k
As  BB TLn(Q canon )
(6.87)
Combining equations (6.86) and (6.87) yields:
As kBBTNa Lnq A +Ln(So!)-Ln(Na !)-Ln(So -Na )!
(6.88)
36
Use Stirling’s Approximation To Simplify Equation
(6.88).
Ln(X!)XLnXX
For any X. If one uses equation (6.89) to
evaluate the log terms in equation (6.88),
one obtains:
As kBBTN a Lnq A +So LnSo -N a LnNa -(So -N a )Ln(So -N a )
(6.90)
37
Step 3: Calculate The Chemical Potential Of The
Adsorbed Layer
The chemical potential of the layer, µs is
defined by:
 A s 

 s 
 N s So,T
(6.91)
substituting equation (6.90) into equation
(6.91) yields:
s k
B TLn(N a )-Ln(So -N a )-Lnq A 
(6.92)
38
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,:
1
ga 
Ng!
(6.93)
substituting equation (6.93) into equation (6.91) yields:

q 

N !
Ng
Q
N
canon
g
g
(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


 g kB
T
Ln(q
)
LnN
B
G
g
(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:
s a
(6.96)
substituting equation (6.92) and (6.95) into equation (6.96) and rearranging
yields:


 qa 
N
a
Ln 
Ln
 N g (So  N a ) 
 qg 


 
(6.97)
Taking the exponential of both sides of Equation (6.97):
Na
q
 a
N g (So N a ) q g
42
(6.98)
Note That Na Is The Number Of Molecules In
The Gas Phase
Na
qa

N g (So N a ) q g
Na is the number of adsorbed molecules and (SoNa) 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:
A g SA (ad)
Consequently:
(6.99)
qa
KA 
qg
(6.100)
43
If we want concentrations, we have to divide all
of the terms by volume
Ca
q

Cg (Cs ) q
'
a
'
g
Partition function per unit volume
44
Table 6.7 Simplified Expressions For
Partition Functions
Type of Mode
Partition Function
Average velocity of a
molecule
Rotation of a linear molecule
Rotation of a nonlinear
molecule
Vibration of a harmonic
oscillator
45
Å  T   1amu 
v=2.52×1013 

 
sec  300K   mg 
1
12
 8k T 
v=  B 
 πm 
g 

Translation of a molecule in
thre dimensions (partition
function per unit volue
qt3 =
 2πmg k BT 
h 
3
2
3
p
qr2 =
8πIk B T
Sn  h p 
3
Sn  h p 
3
2
T  2  mg 
3 1.16 
qt  3 
 

Å  300K   1AMU 
1
2
3
2
 12.4  T 
I

2

q r  
 2

 Sn  300K  Å - AMU 
2
1
8πk BT  2  Ia Ib Ic  2
q 3=
r
Partition Function
after substituting values of kB and
hp
3
1
qv =
1-exp  -h p υ/k BT 
3
3
 43.7   T  2  Ia Ib Ic  2
qr3 = 

  36
3 
 Sn   300K   1Å -AMU 
qv =
1
 
υ
 300K 
1-exp  - 

-1 
  209.2cm  T 
Summary
• Can use partition functions to calculate
molecular properties
• Be prepared to solve an example on the
exam
46