CHM 5414 Thermodynamics
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Exam #1 & Solutions
22 October 1996
- For a (1-d harmonic oscillator) energy ladder with energy spacing
=1
for simplicity, find the fractional occupancy of the first excited state if
N=5 (number of oscillators) and Q=3 (total quanta above the ground state).
Level Distributions
4 _________ _________ _________
3 ____O____ _________ _________
2 _________ ____O____ _________
1 _________ ____O____ __O_O_O__
0 _O_O_O_O_ __O_O_O__ ___O_O___
W 5!/4! = 5 5!/3!=20* 5!/(3!2!)=10 Total: 35
W/35 0.1429 0.5714 0.2857 statistical weight of each distribution
Each distribution happens to have a different occupation in each level, namely,
n1 = 0, 1, or 3 for each distribution. We need to average these occupancies
over the statistical weight of each distribution, since W* doesn't dominate.
So the average n1 is
<n1> = 
n1;W (W/35) =
0*(0.1429)+1*(0.5714)+3(0.2857) = 1.428, the average population in the first excited state.
Notice it is significantly different from the population for W* (1.000), but then 5 isn't NAv!
We were asked for the fractional population in that state, so we have to divide by N=5 to arrive at
P(1) = <n1>/5 = 1.428/5 = 0.2857 .
Compare that with the fraction of a mole of molecules that are found in the
n=1 state for the same (3/5=0.6) averge vibrational energy.
Here we can take advantage of the equal spacing of the energy ladder for the
harmonic oscillator. In the homework and in class, we explicitly summed the partition
function, z = x j = 1/(1-x)
where x=exp(h
/kT).
That sum had the convenient analytical form shown. So did the energy average!
<E> = j x j = x / [1-x] = 1 / [(1/x)-1].
(Remember that we let =1
for just this convenience. It represents no loss of generality...merely a scaling of energy units.)
Now we want to use these simple results to evaluate n1 = N x1 / z
or better still, P(1) = n1/N = x/z, knowing that <E> = 0.6, that is, we
must solve to get x from <E> and use it to find z. Nothing simpler.
Inverting the expression for <E>, we get
x = 1 / ( 1 + 1/<E> ) = 1 / ( 1 + 1/0.600 ) = 0.375
So z = 1 / ( 1 - x ) = 1 / 0.625 = 1.600
That means n1/N = P(1) = x1/z = 0.375/1.600 = 0.234
in agreement with the table in Question #2 but only 0.82 of the N=5 value (0.2857)...not
surprisingly; NAv isn't 5.
The fractional occupancy of all those levels from problem #1 (to two
significant figures...you can do better above) is:
| n | 0 | 1 | 2 | 3 | 4 | >4
|
| P(n) | 0.63 | 0.23 | 0.09 | 0.03 | 0.01 | 0.01
|
What is the molar entropy for this oscillator collection?
All entropy is entropy of mixing, and we're mixing many states with different
probabilities (or mole fractions if you want to remember it that way). So...
S = - R P(n) ln [P(n)]
S = - 8.314 J/mol K [ (.63)ln(.63)+(.23)ln(.23)+(.09)ln(.09)+(.03)ln(.03)+(.01)ln(.01)+(.01)ln(.01) ]
S = + 8.67 J/mol K
In the last 3 problems, we are going to calculate the pre-exponential
components of the water gas equilibrium constant, K. That is,
we will ignore the exp(-D0°/RT)
factor for the purposes of this exam.
Given:
| Molecule | CO | H2O | CO2 | H2
|
| Ground State | 1+ | 1A1 | 1g+ | 1g+
|
| A, B, C (cm-1) | 1.9314
| 27.877
14.582
9.285 | 0.39 | 60.8
|
| (cm-1)
| 2170
| 3657
1596
3756
| 1388
667
667*
2349 | 4395
|
| 1st Excited Eel (cm-1)
| 65075 (1 )
| 53800 (1B1)
| 46000 (1B1)
| 91690 (1u+)
|
* Not a type...the bend motion of CO2 is doubly-degenerate.
The data above are from Herzberg's compendia. They are for the "water gas" reaction,
CO + H2O 
CO2 + H2
which we'll investigate at 1000 K.
Since the temperature is so high, the rotational partition function for H2
can be approximated by its integral form. However, this comes at a price.
The value of RT at this temperature (see the conversion factor on the
back of the Periodic Table) is about 695 cm-1.
That will certainly imply that CO2's weak bends (only 667 cm-1)
are turned on! I would check the other vibrations at this temperature as
well. (hint hint)
Also, CO, CO2, and H2 are all linear; H2O is not.
KROT will have constants and temperature factors which don't
cancel out! Check your dimensions carefully, and remember that
RT=695 cm-1 at 1000 K.
Calculate KTRANS and KEL for this reaction at 1000 K.
Since n=0,
all the constants and temperature factor in KTRANS will cancel
between the z's in the numerator and denominator,
K = [ z(CO2) z(H2) ] / [ z(CO) z(H2O) ]
leaving only the mass factors, m3/2. So
KTRANS = { [ m(CO2) m(H2) ] / [ m(CO) m(H2O) ] }3/2
KTRANS = { [ 44 * 2 ] / [ 28 * 18 ] }3/2 = { 0.175 }3/2 = 0.073
Since the lowest 1st excited state (46000 cm-1) is over 66 times RT
(695 cm-1), we can ignore all the excited states. And since
all the ground states are singlets, their individual degeneracies are all 1.
Which is also the ration of their degeneracies in KEL=1.
Calculate KROT bearing in mind the cautions mentioned above.
The zROT for linear molecules is kT/
B
if all the units are mks or
zROT = 0.695 T / B
if B is in cm-1. Notice that when T=1000 K, the numerator's 695 cm-1, as expected.
For non-linear molecules, there are 3 rotation constants (A,B,C) and
the expression for all those being in cm-1 is
zROT = 1.0270 T3/2 / (ABC)1/2.
All the symmetry factors are 2 except that for CO, which is 1. So...
B(CO) 0.695 T [ABC(H2O)]1/2
KROT = --------- ------- --------------
B(CO2) B(H2) 1.0270 T3/2
1(1.9314) 695 2[(27.877)(14.582)(9.285)]1/2
KROT = ---------- ------- ------------------------------
2(0.39) 2(60.8) 1.0270(1000)3/2
KROT = (2.75) (5.72) (0.003783) = 0.060 unitless
Calculate K and the complete pre-exponential equilibrium
constant
z VIB = 1 / [ 1 - exp(-/RT) ]
if both RT and have been converted to cm-1.
So in our case,
z VIB = 1 / [ 1 - exp(-/695) ]
Plugging in from the table of molecular parameters:
| (cm-1)
| 2170 | 3657 | 1595 | 3756 | 1388 | 667 | 2349 | 4395
|
| z VIB
| 1.046 | 1.005 | 1.112 | 1.005 | 1.157 | 1.621 | 1.035 | 1.002
|
Which makes
K VIB = [ z VIB(CO2) z VIB(H2) ] / [ z VIB(CO) z VIB(H2O) ]
K VIB = [ (1.157)(1.621)2(1.035) * (1.002) ] / [ (1.046) * (1.005)(1.112)(1.005) ]
K VIB = 2.68
Thus the total pre-exponential equilibrium constant is
K = KTRANS KROT K VIB KEL
K = (0.073) (0.060) (2.68) (1.00) = 0.012
Last modified 28 October 1996