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

qq

)(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