University of Washington Department of Chemistry Chemistry ...

Lecture 17 2/25/13

University of Washington Department of Chemistry

Chemistry 453 Winter Quarter 2013

A. The Molecular Partition Function for Larger Molecules

? To construct a partition function for molecules, the partition function for

translational, rotational, vibrational, and electron motions are combined. For a

diatomic molecule this is:

q = qtrans qrot qvibqelec

( ) = V

h3

2 mkBT

3/2 ?

-vib / 2T

T e De / RT

? ? g e -vib /T

1

1 - e rot

(17.1)

? Equation 17.1 is only valid for diatomic molecules. The presence of more

bonds complicated the rotational and vibrational partition functions. The

simplest polyatomic molecule is carbon dioxide CO2, which is linear but has two bonds. With two bonds CO2 has four vibrational modes as shown in Figure 17.1:

Figure 17.1: The vibrational

modes of CO2. a) symmetric bond stretch; b) asymmetric

bond stretch; c) two bending

modes, one in the page and

one out of the page.

? Each bond vibration mode

has a different frequency so

each mode has its own

vibrational

temperature

vib, j

=

h j kB

.

For

CO2

the

vibrational

partition

function is now

-vib,1 / 2T

-vib,2 / 2T

-vib,3 / 2T

-vib,4 / 2T

4

-vib, j / 2T

q = e ? e ? e ? e = e vib

-vib ,1 /T

-vib,2 /T

-vib,3 /T

-vib,4 /T

-vib, j /T

1- e 1- e 1- e 1- e 1- e j=1

(17.2)

? The molecular partition function for CO2 is now:

( ) q

=

V h3

2 mkBT 3/2 ?

4

-vib, j / 2T

T ? 1e- e ? g e rot j=1

-vib, j /T

De / RT 1

(17.3)

? And the internal energy of CO2 is from equation 17.3:

U

=

N AkBT 2

ln q T

=

5RT 2

+

4 Rvib, j

j=1

2

+

Rvib, j evib, j /T -1

-

N

A

De

(17.4)

? The first term is the total energy from translations and rotations, both in the

classical limit.. The summation in the second term is over the four vibrational

modes. The last term is the contribution to the internal energy from electronic

motions.

? From equation 17.4 the heat capacity is:

( ) CV

= 5R + 2

4 vib, j

j=1

T

2

e-vib, j /T 1 - e-vib, j /T

2

(17.5)

? For linear polyatomic molecules we need the moment of inertia I = ?R2 ,

the

partition

function

is

qrot

=

T rot

and Urot = NkBT

or just RT for one

mole of rigid rotors. For a non-linear molecule three moments of inertia

are required to describe the rotation of the molecule. The three moments

of inertia identify the center of mass around which the molecule rotates

and are designated IA, IB, and IC. Therefore a non-linear molecule has three

rotational temperatures designated A, B, and C. The rotational partition

function for a non-linear molecule is

qrot

=

1/2

T3

A

BC

1/ 2

(17.6)

? Note there are three degrees of rotational freedom for a non-linear molecule and according to equation 17.6 the internal energy and heat capacity associated with this rotation are

U rot

=

3RT 2

; CV ,rot

=

3R 2

(17.7)

B. Statistical Description of Equilibria

? Here we learn how to use statistical principles to calculate equilibrium

constants for gas phase chemical reactions. Consider a gas phase reaction

of the following general form: AA + BB UCC + DD

(17.8)

? Now the condition for equilibrium is

A?A + B?B =C?C + D?D

(17.9)

where

the

chemical

potentials

are

?

=

A N

V ,T

.

We

have

already

shown

that using the statistical definition of the Helmholtz energy A = -kBT ln Q

and the definition of the partition function for a gas Q = qN the statistical N!

definition of the chemical potential is:

?

=

-kBT

ln

q N

(17.10)

? Now putting the statistical expression for the chemical potential from

equation 17.10 into equation 17.9 we get

A

ln

qA NA

+

B

ln

qB NB

=C

qC NC

+

D

qD ND

(17.11)

? We can remove the logarithms, rearrange equation 17.11 and divide all

terms by the volume V to get:

[[ ]] [[ ]] KC

=

ND V

NA V

D A

NC V

NB V

C B

=

D D A A

C D B B

=

qD V

qA V

D A

qC V

qB V

C B

(17.12)

? Equation 17.12 is the statistical expression for the equilibrium constant.

Using the ideal gas law we can expression the equilibrium constant in

terms of pressures

KC

=

ND V

NA V

D A

NC V

NB V

C B

=

PD kBT

PA kBT

D A

PC kBT

PB kBT

C B

=

qD V

qA V

D A

qC V

qB V

C B

(17.13)

? We now define the equilibrium constant KP as the ratio of equilibrium

pressures:

( ) ( ) KC

=

P P D C DC

P P A B AB

k T = K A + B -C - D

B

P

kBT

A + B - C - D

=

qD V

D

qC V

C

qA V

A

qB V

B

(17.14)

? We can finally write out the equilibrium constant in terms of pressure

( ) KP

=

qD V

qA V

D A

qC V

qB V

C B

k T C + D - A - B B

(17.15)

Example:

For

the

equilibrium

H2

+

1 2

O2

U

H 2O

calculate

the

equilibrium constant

KP

at

T=1000K. Calculate Go and determine if the formation of water is thermodynamically

favorable at this temperature. Assume V=1m3. The following data are available:

? For H2: vib = 6215K , rot = 85.3K , De = 457.6kJmol-1, g1 = 1.

? For O2: vib = 2256K , rot = 2.07K , De = 503kJmol-1, g1 = 3.

? For H2O there are three vibrational motions shown to the right. The three vibrational temperatures are:

vib,1 = 5360K , vib,2 = 5160K , vib,3 = 2290K .

Because water is non-linear the three rotational

temperatures are:

rot,1 = 40.1K , vib,2 = 20.9K , vib,3 = 13.4K

For water also: De=940kJmol-1, g1=1, and =2.

( ( )( ) ) ( ) Solution: KP =

1 kBT 1/2

qH2O V

1/ 2

qH2 V qO2 V

( ) qH2

= qtrans qvibqrot qelec

=

V h3

2 kBT

M e 3/2

3/ 2

-vib / 2T

N

A

1 - e-vib /T

T rot

g eDeH 2 / RT 1

( )(( ) ) ( )( ) 1m3

qH2 =

6.28?1.38?10-20 J

3/2

0.002kgmol -1

3/2

e-3.11

6.62 ?10-34 Js 3

6.02

?1023

mol

-1

1 - e-6.215

1000K 2 85.3K

eDeH 2 / RT

( ) ( )( ) =

2.55 ?10-29 J -3/2m3s-3 2.90 ?10-100

33.2 ?10-28 kg 3/2

0.045

5.86 = 4.44 ?1030 eDeH 2 /RT

( ) qO2

= qtrans qvibqrot qelec

=

V h3

2 kBT

M e 3/2

3/2

-vib / 2T

N

A

1 - e-vib /T

T rot

g eDeO 2 / RT 1

( )( ) 1m3 ( ) ( )( ) ( ) qO2 =

6.28?1.38?10-20 J

3/2

0.032kgmol -1

3/2

e-1.128

1000K

6.62 ?10-34 Js 3

6.02

?1023

mol

-1

1 - e-2.256 2

2.07K

3 eDeO 2 / RT

( ) ( )( )( ) =

2.55 ?10-29 J -3/2m3s-3 2.90 ?10-100

5.31?10-26 kg 3/2

0.362

242

3 eDeO2 / RT = 8.20 ?1035 eDeO2 / RT

( ) qH2O

=

qtrans qvibqrot qelec

=

V h3

2

M e e e 3/2 k T N 1- e 1- e 1- e B

3/ 2 A

-vib1 / 2T -vib1 /T

-vib 2 / 2T -vib 2 /T

-vib3 / 2T -vib3 /T

1/2

T3

1/ 2

DeH 2O / RT g e1

rot1 rot 2 rot 3

( )( ) 1m3 ( ) qH2O =

6.28?1.38?10-20 J 6.62 ?10-34 Js 3

3/ 2

0.018kgmol -1 6.02 ?1023 mol-1

3/2

e-2.680 1 - e-5.360

e-2.580 1 - e-5.160

e-1.145 1 - e-2.290

?

(3.14)1/2 (2)

109

( 40.1) ( 20.9 ) (13.4 )

1/ 2

e DeH

2O

/ RT

( ) =

2.55?10-29 J -3/2m3s-3 2.90 ?10-100

3.00?10-26 kg 3/2 (0.069) (0.076) (0.354) (0.886) (2221)1/2 eDeH 2O /RT

= 3.46 ?1031eDeH 2O /RT

( ) ( ( )( ) ) ( )( ) KP =

1 kBT 1/2

qH2

qH2O V V qO2

V

1/ 2

=

1.38

1 ?10-20

J

1/ 2

3.46 ?1031eDeH 2O /RT m-3 4.44 ?1030 eDeH 2 /RT m-3 82.0 ?1034 eDeO2 / RT m-3 1/2

( ) ( ) 0.851?1020 J -1

8.61?10 m e 1/2

-2+31-30-17 3/ 2 DeH 2O - DeH 2 -DeO 2 / 2 / RT

= 7.94 ?10-8 J m e -1/2 3/2 (940-458-252)/8.31

=

7.94 ?10-8 e27.7 Pa-1/2

=

8.51?104

Pa -1/ 2

1Pa 9.88?10-6 atm

1/ 2

=

27.1atm-1/2

? This problem is far more challenging than anything you will encounter in regular course work. However, given the right data, obtaining an equilibrium constant is simply a two stage process. First, you calculate the relevant partition function for each molecule. Then you plug these values into the equilibrium constant equation.

? Be careful not to let the big numbers scare you into thinking you have made a mistake. Partition functions for molecules are large numbers due to the translational degrees of freedom.

? Be careful with the dissociation energy exponent. If you use values of De in kJ per mole be sure to divide by RT not kBT. And convert the kJ to Joules because R uses Joules.

? The equilibrium constant has units because I did not divide by a standard state pressure. If I had done that the equilibrium constant would be unitless.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download