Physical Chemistry II
Chem 402
Spring 2011
Chapter 14 (1, 5, 7, 11, 16, 19, 25, 28, 35)
P14.1) Evaluate the translational partition function for H2 confined to a volume of 100. cm3 at
298 K. Perform the same calculation for N2 under identical conditions. (Hint: Do you need to
reevaluate the full expression for qT?)
qT  H 2  
V
3
 h2 


 2 mkT 
qT  H 2  
1
2


2


34
6.626  10 J s 




  0.00202 kg mol1 

23
1
 2 
 1.38  10 J K   298 K  
NA

 

 7.15 10
100. cm 10

m
 7.15 10
3
V
11
3
6
11
m3 cm 3 
m
3
1
2
 7.15 1011 m
 2.74 1026
V
3
3
qT  N 2    N 2   3  H 2   m  N 2   2  28.0 g mol1  2

 3

 51.8
 
1 
V
qT  H 2 
  N 2   m  H 2  
 2.02 g mol 
3  H 2 
3
qT  N 2   51.8  qT  H 2    1.42 1028
P14.5) At what temperature is there Avogadro’s number of translational states available for O2
confined to a volume of 1000. cm3?
Since the number of translational states is equal to qT:
3
6
3
3
V 1000. cm 10 m cm 
qT  6.022  10  3 
3/2

 h2 


 2 mkT 
23
 h2 


 2 mkT 
T
3/2
 1.66  1027 m3
 6.626 10

 0.0320 kg mol 
m 
2 
 1.38  10
N
34
h2
2 mk 1.66 1027
3
2
1
3

23

A
J s
2
J K 1 1.40 1018 m 2 
T  0.068 [K]
P14.7) For N2 at 77.3 K, 1 atm, in a 1-cm3 container, calculate the translational partition function
and the ratio of this partition function to the number of N2 molecules present under these
conditions.
V
qT  N 2   3

 h2 


 2 mkT 
qT  N 2  
1
2


2


34
6.626  10 J s 




  0.028 kg mol1 

23
1
 2 
 1.38  10 J K   77.3 K  
NA

 

 3.75 10
1 cm 10 m cm   1.90 10

m
 3.75 10 m 
3
V
11
3
6
2
 3.75  1011 m
3
3
11
1
25
3
Next, the number of molecules (N) present at this temperature is determined using the ideal gas
law:
1 atm  1.00 103 L 
PV
n

 1.58 104 mol
RT  0.0821 L atm mol1 K 1   77.3 K 
N = n  N A  9.49 1019 molecules
With N, the ratio is readily determined:
qT 1.90  10 25

 2.00  10 5
19
N 9.49  10
P14.11) Consider para-H2 (B = 60.853 cm–1) for which only even-J levels are available. Evaluate
the rotational partition function for this species at 50. K. Perform this same calculation for HD (B
= 45.655 cm–1).
For para-H2, only even J levels are allowed; therefore, the rotational partition function is:
qR 
  2 J  1e
  hcBJ  J 1

 6.62610
34
 1  5e

1.3810


J s 3.001010 m s 1 60.853 cm 1  6 
23

J K 1  50. K 
J  0,2,4,6,...

 6.62610
34



J s 3.001010 m s 1 60.853 cm 1  20 


1.381023 J K 1  50. K 
 9e
 ...
 1  1.35  104  ...
 1.00
Performing this same calculation for HD where both even and odd J states are allowed:
qR 
  2 J  1e
  hcBJ  J 1

 6.62610
34
 1  3e

1.3810


J s 3.001010 m s 1 45.655 cm 1  2 
23

J K 1  50. K 
J  0,1,2,3,...

 6.62610
 5e
34

1.3810


J s 3.001010 m s 1 45.655 cm 1  6 
23

J K 1  50. K 
 ...
4
 1  0.216  3.74 10  ...
 1.22
P14.16) Calculate the rotational partition function for ClNO at 500. K where BA = 2.84 cm–1, BB
= 0.187 cm–1, and BC = 0.175 cm–1.
1
1
  1  2 1  2 1 
qR 


 
 
   BA    BB    BC 
1
2
2
2
1
1
1
   0.695 cm   500. K     0.695 cm   500. K     0.695 cm   500. K  
1

1 
2.84 cm 1

 3.78  104

 
 
 
1
0.187 cm 1
 
 
 
0.175 cm 1
1
2



P14.19) Calculate the rotational partition function for oxygen (B = 1.44 cm–1) at its boiling
point, 20.8 K, using the high-temperature approximation and by discrete summation. Why should
only odd values of J be included in this summation?
First, O2 in the ground state is a triplet such that the spins are parallel. Therefore, to
ensure an anti-symmetric wavefunction with rotation, only odd J states are allowed.
In the high-temperature approximation, the rotational partition function is given by:
1
1
 1    0.695 cm K   20.8 K  
  5.02
qR  


2 1.44 cm 1 
  B  

And by discrete summation:
qR 
  2J  1e
  hcBJ  J 1

6.626 10
 3e
34



J s 3.00 1010 m s1 1.44 cm 1  2 
1.3810
23
JK
1
20.8 K 

6.626 10
 7e

1.3810
34

23

J K 1  20.8 K 
J 1,3,5,7,...

 11e
6.626 10

1.3810
34


J s 3.00 1010 m s1 1.44 cm 1  30 
23

J K 1  50 K 

J s 3.00 1010 m s1 1.44 cm 1 12 
 ...
 5.18
P14.25) Evaluate the vibrational partition function for SO2 at 298 K where the vibrational
frequencies are 519, 1151, and 1361 cm–1.
The total vibrational partition function is the product of partition functions for each vibrational
degree of freedom:
qV ,total   qV ,1  qV ,2  qV ,3 
1
1
1





 hc3 
 hc1 
 hc2 
 1 e
 1  e
 1  e





1
1


10
34
1
1
34
6.626
10
J
s
3.00
10
cm
s
519
cm
6.626
10
J
s
3.00






 

 1010 cm s1 1151 cm1 



1.381023 J K 1  298 K 
1.381023 J K 1  298 K 
 1 e
 1 e




1


34
10
1
1
 6.62610 J s 3.0010 cm s 1361 cm  



1.381023 J K 1  298 K 
 1 e

 1.09






P14.28) In using statistical mechanics to describe the thermodynamic properties of molecules,
high-frequency vibrations are generally not of importance under standard thermodynamic
conditions since they are not populated to a significant extent. For example, for many
hydrocarbons the C-H stretch vibrational degrees of freedom are neglected. Using cyclohexane
as an example, the IR-absorption spectrum reveals that the C-H stretch transitions are located at
~2850 cm–1.
a. What is the value of the vibrational partition function for a mode of this frequency at 298 K?
b. At what temperature will this partition function reach a value of 1.1?
a.
1

qV (2850 cm -1 )  
  hc
 1 e

1
 
34

6.626
10
J
s
3.00


 1010 cm s1  2850 cm1 
 


1.381023 J K 1  298 K 
 1 e


  1.00



b.


1
qV  1.1  
34
6.626

10
J
s
3.00

 1010 cm s1  2850 cm1 



1.381023 J K 1 T 
 1 e

 6.62610
34
1.1  1.1e

 6.62610
34

1.3810

1.3810

J s 3.001010 cm s 1 2850 cm 1

23
J K 1  T 


J s 3.001010 cm s 1 2850 cm 1
23








1
J K 1  T 
.1
 0.0909
1.1
 6.626 1034 J s  3.00 1010 cm s1  2850 cm1 
e

 2.398
1.38 10 J K  T 
 6.626 10 J s  3.00 10 cm s  2850 cm   1710 [K]
T
2.398 1.38  10 J K 
23
34
1
1
10
23
1
1
P14.35) NO is a well-known example of a molecular system in which excited electronic energy
levels are readily accessible at room temperature. Both the ground and excited electronic states
are doubly degenerate, and separated by ~121 cm–1.
a) Evaluate the electronic partition function for this molecule at 298 K.
b) Determine the temperature at which qE = 3.
a)
qE   gn e    2e   0  2e
n
n
 2  2e
 3.12

121 cm 1
1
K 1  298 K 
 0.695 cm


  121 cm 1

b)
  121 cm 
q E  3  2e   0  2e 
1
3  2  2e
0.5  e

0.693 
T

121 cm 1
1
K 1 T
 0.695 cm
121 cm 1
1
K 1 T
 0.695 cm
121 cm 1
 0.695 cm1 K 1  T
121 cm 1
 0.695 cm1 K 1   0.693
T  251 [K]