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.
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