Chem 3322 homework #9 solutions, out of 36 marks
Problem 1 – 5 marks
The vibrational frequency of I2 is 208 cm−1 . At what temperature will the population in
the first excited state be half of the ground state?
Solution:
We know that, since there is no degeneracy in this problem,
P (0) =
e−βE0
q
(1)
P (1) =
e−βE1
q
(2)
and
where En = (n + 1/2)~ω and where q is the partition function for the harmonic oscillator
model.
Hence, the ratio P (1)/P (0) is simply
e−β~ω
(3)
We are asked for the temperature corresponding to P (1)/P (0) = 0.5. Putting the numbers
in (convert units and multiple by hc to turn wavenumbers into energy) gives the temperature
as 432 K.
Problem 2 – 8 marks
The force constants for H2 and Br2 are 575 and 246 Nm−1 , respectively. Take the mass
of H as 1.0079 g/mol. Take the mass of Br as 79.90 g/mol. Calculate the ratio of the
vibrational state populations P(1)/P(0) and P(2)/P(0) at T = 300 K and at 1000 K for
each of H2 and Br2 . Summarize all your answers in a table.
Solution:
s
ω=
k
µ
(4)
For H2 ,
µ=
0.50395 g/mol
= 8.3685 × 10−28 kg
6.022 × 1023 mol−1
1
(5)
s
ω=
575 Nm−1
= 8.29 × 1014 s−1
8.3685 × 10−28 kg
~ω
1.05459 × 10−34 Js 8.29 × 1014 s−1
=
kB
1.38066 × 10−23 JK−1
(6)
(7)
P (1)/P (0) = eβ∆E1−0 = e−β~ω = e−~ω/kB T = e−21.1 = 6.86 × 10−10 at 300 K, and =
e−6.33 = 1.78 × 10−3 at 1000 K.
P (2)/P (0) = eβ∆E2−0 = e−2β~ω = e−2~ω/kB T = 4.71 × 10−19 at 300 K, and = 3.18 × 10−6
at 1000 K.
For Br2 ,
0.39.95 g/mol
= 6.634 × 10−26 kg
6.022 × 1023 mol−1
s
246 Nm−1
ω=
= 6.09 × 1013 s−1
−26
6.634 × 10 kg
µ=
(8)
(9)
and ~ω/kB = 465.13 K. Hence P (1)/P (0) = 0.212 at 300 K and 0.628 at 1000 K.
P (2)/P (0) = 0.045 at 300 K and 0.394 at 1000 K.
Problem 3 – 6 marks
For six identical, non-interacting harmonic oscillator molecules (energy levels given by the
harmonic oscillator model), and given the macroscopic constraint Etot = (5 + 6/2)~ω = 8~ω,
work out the probabilities of observing a molecule with energy E using the fundamental
postulate of statistical mechanics.
Solution:
particle #1 #2 #3 #4 #5 #6 # ways
For Etot = 5 + 6/2 we have
5
0
0
0
0
0
6
4
1
0
0
0
0
30
3
2
0
0
0
0
30
3
1
1
0
0
0
60
2
2
1
0
0
0
60
2
1
1
1
0
0
60
1
1
1
1
1
0
6
2
P (0) =
6 5
252 6
+
30 4
252 6
+
30 4
252 6
+
60 3
252 6
+
60 3
252 6
+
60 2
252 6
P (1) =
30 1
252 6
+
60 2
252 6
+
60 1
252 6
+
60 3
252 6
+
6 5
252 6
=
70
252
P (2) =
30 1
252 6
+
60 2
252 6
+
60 1
252 6
=
35
252
P (3) =
30 1
252 6
+
60 1
252 6
=
15
252
P (4) =
30 1
252 6
=
5
252
P (5) =
6 1
252 6
=
1
252
+
6 1
252 6
=
126
252
Problem 4 – 11 marks
a – 4 marks) For I2 , the harmonic oscillator approximation (force constant k=170 N/m)
assumes that the vibrational energy levels are equally spaced. At 300 K, calculate the ratio
of molecules in the first excited state compared to the ground state. Also calculation the
proportion of molecules in the second excited state relative to the ground state. Repeat for
the third, fourth, and fifth excited states as well.
solution:
For I2 , the reduced mass is (127 amu)(127 amu)/(127 amu + 127 amu) = 63.5 amu
roughly. We will convert to kg later. Another way to approach this problem is to write the
mass in g/mol, and then use Avogadro’s number to get rid of the 1/mol units. Either way,
p
p
ω = k/µ = (170N/m)/µ=4.016 × 1013 s−1 .
The ratio of populations in levels n and m is
P (n)
e−β~ω(n+1/2)
q
=
= e−β~ω(n−m)
−β~ω(m+1/2)
P (m)
q
e
(10)
[which answers part (b) incidently]. In general, for non-degenerate energy levels Ei , we have
P (n)/P (m) = e−β∆E where ∆E = (En − Em ).
Hence the relative populations of the excited states with respect to the ground state
(P(n)/P(0)) are:
n P(n)/P(0)
1 0.3593
2 0.129
3 0.0465
4 0.0167
5 0.00602
3
b – 1 mark) Show that you do not need q, the partition function, to calculate these
quantities. Show that, if you are only interested in relative populations (eg. how many more
molecules are in one state compared to another state), you can always calculate this without
knowing the value of q.
solution: q cancels from the ratio you set up for relative population
c – 4 marks) I2 is experimentally found to have vibrational energy levels at the following
wavenumbers above the ground state: 213.30 cm−1 , 425.39 cm−1 , 636.27 cm−1 , 845.93 cm−1
, 1054.38 cm−1 ,.. Repeat the relative population analysis you did in part (a) for the actual
molecule I2 using the experimental data.
solution:
Using e−β∆E where ∆E = (En − Em ) with m = 0 gives us the expression we want for the
ratios. We need to convert from wavenumbers to energies using E = h × c × wavenumber.
This gives
wavenumber (cm−1 ) ∆ E (J)
P(n)/P(0) n
213.30
4.24 × 10−21 0.35953
1
425.39
8.45 × 10−21 0.13
2
636.27
1.26 × 10−20 0.0473
3
845.93
1.68 × 10−20 0.0173
4
1054.38
2.09 × 10−20 0.006366 5
d – 2 marks) Compare the answers of (a) and (c) and comment.
solution:
The relative populations in (c) are all slightly higher than in (a), with the difference
getting larger as n increases. A real molecule has vibrational energy levels that get closer
together as the energy increases, eventually converging at the dissociation energy of the
molecule. The harmonic approximation has equally spaced levels based on a local analysis
(Taylor series) about the equilibrium bond length.
Problem 5 – 6 marks
Consider the energy-level diagrams shown in Fig. 1.
a – 3 marks) At what temperature will the probability of occupying the second energy
level be 0.15 for the states depicted in Fig. 1A?
Solution:
4
FIG. 1:
for this problem we have the partition function
q = 1 + e−β∆E
(11)
so that the excited state population is
e−β∆E
P (1) =
1 + e−β∆E
(12)
For P (1) = 0.15, we have
T =
−300cm−1 hc
300cm−1 6.626 × 10−34 Js 3.00 × 1010 cm s−1
=
kB ln(0.15/0.85)
2.395 × 10−23 J K−1
(13)
which works out to be T = 249 K.
b – 3 marks) Perform the corresponding calculation for the states depicted in Fig. 1B.
Before beginning the calculation, do you expect the temperature to be higher or lower than
that determined in part a)? Why?
Solution:
We discussed this in class as related to the balance between internal energy and entropy.
The double degeneracy in the excited state tips the balance (compared to part a) farther
towards the excited state for a fixed value of temperature. Hence to get the same population,
we don’t need as high a temperature as in part a.
The math is almost the same, but now we have
q = 1 + 2e−β∆E
5
(14)
so that the excited state population is
P (1) =
2e−β∆E
1 + 2e−β∆E
Putting the numbers in gives T = 178 K.
6
(15)