Physical Chemistry II
Chem 402
Spring 2011
Chapter 13 (3, 8, 14, 18, 19)
_____________________________________________________________________
P13.3) Consider the case of 10 oscillators and eight quanta of energy. Determine the dominant
configuration of energy for this system by identifying energy configurations and calculating the
corresponding weights. What is the probability of observing the dominant configuration?
The configuration with the maximum weight is a3 = 1, a2 = 1, a1 = 3, a0 = 5, with the
corresponding weight:
10!
W
 5040
1!1!35!
Writing down all possible configurations and summing the weight results in Wtotal = 20170.
Therefore, the probability of observing the dominant configuration is:
P
W
5040

 0.25
Wtotal 20170
P13.8) Consider the following energy-level diagrams:
a) At what temperature will the probability of occupying the second energy level be 0.15 for the
states depicted in part (a) of the figure?
b) Perform the corresponding calculation for the states depicted in part (b) of the figure. Before
beginning the calculation, do you expect the temperature to be higher or lower than that
determined in part (a) of this problem? Why?
(a) According to the energy-level diagram, there is one state at  = 0 and one state at
 = 300 cm–1. Therefore:

e 
e 
0.15  p1 

  300 cm 
q
1+e 
  300 cm 1
1
1

0.15  0.85  e



  300 cm 
  300 cm 
0.15  0.15 e 
e 
1
   300 cm 1
1
  300 cm 
0.176  e 
1


1
  300 cm 1 
1.74    300 cm 1   
  0.695 cm 1 K 1  T  


T  248 [K]
(b) According to the energy-level diagram, there is one state at  = 0 and two states at  =
300 cm–1. Therefore:

2e 
2e 
0.15  p1 

  300 cm 
q
1+2e 
  300 cm 1
1
1

0.15  1.70  e



  300 cm 
  300 cm 
0.15  0.30 e 
 2e 
1
   300 cm 1
1
  300 cm 
0.088  e 
1


1
  300 cm 1 
2.43    300 cm 1   
1
1
  0.695 cm K  T  


T  178 [K]
The temperature at which p1 = 0.15 is lower for case b since the presence of two states at
the higher energy provides more opportunity for the system to populate the higher energy
state.
P13.14) Consider a collection of molecules where each molecule has two non degenerate energy
levels that are separated by 6000. cm–1. Measurement of the level populations demonstrates that
there are 8 times more molecules in the ground state than in the upper state. What is the
temperature of the collection?
The ratio of any two occupation numbers for a set of non-degenerate energy levels is
given by:
  i  j  1

k T

ai
   
e   e 
aj
i
j
The above expression suggests that the ratio of occupation numbers can be used to
determine the temperature. The ratio of a1 to a0 is 1/8. Using this information and the
separation between energy levels, the temperature is:
  2  0  1

k T

a1
e 
a0

6000. cm –1

1
 e  0.695 cm K
8
T  4150 [K]
–1
–1
1

T
P13.18) Determine the partition function for the vibrational degrees of freedom of Cl2 (  =
525 cm–1) and calculate the probability of occupying the first excited vibrational level at 300 and
1000 K. Determine the temperature at which identical probabilities will be observed for F2 (  =
917 cm–1).
For Cl2 at 300 K:
q
1

1  e   hc
p1 
e
 1

q
e
1
1 e
  hc
q
  6.6261034 J s  3.001010 cm s 1  525 cm 1  




1.381023 J K 1 300 K 



e
 1.087
  6.6261034 J s  3.001010 cm s 1  525 cm 1  




1.381023 J K 1 300 K 


1.087
 0.074
For Cl2 at 1000 K:
q
1

1  e   hc
p1 
1
1 e
e   e

q
q
  hc
1
  6.6261034 J s  3.001010 cm s 1  525 cm 1  




1.381023 J K 1 1000 K 



e
 1.885
  6.6261034 J s  3.001010 cm s 1  525 cm 1  




1.381023 J K 1 1000 K 


1.885
 0.249
To determine the temperatures at which F2 has equivalent populations in the first
vibrational excited state, we first reduce the expression for p1 as follows:
p1 
e  
e   hc

 e   hc 1  e   hc 
q
1
1  e   hc
1


Substituting in the values for F2 and solving as a quadratic equation:
p1  0.074  e
0  0.074  e
T  524 [K]
  hc  917 cm 1 
1319 K
T
e
e

2  hc  917 cm 1 
e
1319 K
T
e

21319 K 
T
21319 K 
T
Repeating this calculation to determine at which temperature p1 will be equal to the Cl2
case at 1000 K:
p1  0.249  e
0  0.249  e
T  1742 [K]
  hc  917 cm 1 
1319 K
T
e
e

2  hc  917 cm 1 
e
1319 K
T
e

21319 K 
T
21319 K 
T
P13.19) The vibrational frequency of Cl2 is 525 cm–1. Will the temperature be higher or lower
relative to I2 (see Problem P13.18) at which the population in the first excited vibrational state is
half that of the ground state? What is this temperature?
Since the energy level spacings are larger for Cl2 relative to I2, we would expect the
temperature at which a1/a0 = 0.5 to be greater for Cl2.
a1 1
  525 cm 
  e        e 
a0 2
1
1
0.5  e
0


525 cm 1


  0.695 cm 1 K T  


0.693 
525 cm 1
 0.695 cm1 K  T 
T  1090 [K]