Solutions Exam 3
1) As you know there are two common stable conformations of ethane—staggered and eclipsed.The energy
difference between these two conformations is 12 kJ/mole, with the staggered conformation being the most
stable.In this problem you will determine what percent of the ethane molecules are in the eclipsed conformation at room temperature.(This is just like the dice problem only easier.Be careful with your units.)
a) As a warm up, draw an energy level diagram showing the energy difference between the two states of
ethane.Give the energy difference in joules per mole
All I was looking for here was two lines to show that there were two states
12000 Joules/mole
b) Construct the partition function for ethane intercoversion and write it neatly
First I must convert 12000 Joules/mole to joules/particle
Es = 80, 12 000  A< H* where A is Avagodro's number *L
12 000
Out[88]=
:0,
>
A
2
Solutions_exam_3.nb
In[89]:=
q = Sum@Exp@- Es@@iDD  H kb TLD, 8i, 1, 2<D
Out[89]=
1+ã
12 000
-
A kb T
Note that A kb is equal to R (the gas constant) = 8.314 j/K/mol now my units are all the same and
12 000
-
In[96]:=
qfinal = 1 + ã
8.314 T
1 + ã-1443.348T
c) Calculate the probability that ethane is in the eclipsed conformation at 300 K
Peclipsed = Exp@- 12 000  H8.314 TLD  qfinal
ã-1443.35T
Out[97]=
1 + ã-1443.35T
Now evaluate the previous expression at T = 300 K
In[98]:=
% . T ® 300
0.008072 = .81 %
2) The vibrational temperature, Qvib, of I2 gas molecules is 308 K. Calculate the molar vibrational heat capacity of I2 gas at 300 K.
I need from my notebook the following :
In[102]:=
U@q_D := kb T2 ¶T Log@qD; H* Internal Energy *L
Cv@q_D := ¶T U@qD; H* Constatant Volume Heat Capacity *L
-
ã
Qv
2T
qvib =
; H* Vibrational Partition Function *L
-
1-ã
In[110]:=
Out[110]=
Qv
T
CvI2 = Cv@qvibD . Qv ® 308 . T ® 300.
0.916605 kb
But recall that I am asking for the molar heat capacity which will be Avagodros number times this amount or :
0.916 R
5) One can measure the energy of a molecule’ s rotational levels using FTIR spectroscopy.For HCl the first ten
rotational levels (j = 0, 9) are : 0, 21.47, 62.495, 123.28, 205.32, 307.89, 430.89, 574.16, 737.665, 921.29 in
units of reciprocal cm.What is the approximate probability that HCl is in its ground state at 300 K?(Helpful
hints : Kb in units of reciprocal cm is 0.695 and recall that rotational levels are 2 j + 1 fold degenerate.)
Solutions_exam_3.nb
In[112]:=
Out[112]=
Es = 9 0,
21.47, 62.495, 123.28,
205.32, 307.89, 430.89, 574.16, 737.665, 921.29<
80, 21.47, 62.495, 123.28, 205.32, 307.89, 430.89, 574.16, 737.665, 921.29<
Make up a list of ten elements giving the degeneracy
In[113]:=
Out[113]=
Dg = Table@2 j + 1, 8j, 0, 9<D
81, 3, 5, 7, 9, 11, 13, 15, 17, 19<
Now construct the partition function
In[116]:=
Out[116]=
q = Sum@Dg@@iDD Exp@- Es@@iDD  H0.695 TLD, 8i, 1, 10<D
1 + 19 ã-1325.6T + 17 ã-1061.39T + 15 ã-826.129T + 13 ã-619.986T +
11 ã-443.007T + 9 ã-295.424T + 7 ã-177.381T + 5 ã-89.9209T + 3 ã-30.8921T
And now the probability
In[117]:=
Pgs = Exp@- Es@@1DD  H0.695 TLD  q . T ® 300
0.048815 = 4.88 %
3