Section NAME:
Circle Number:
10
11
80
CHEMISTRY 444, SPRING, 2011(11S)
Final Examination, May 25, 2011
81
Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY
understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of
significant figures. You may use only the "Student Handbook," a calculator, and a straight edge.
1. (10 points) (a) The critical temperature of CO2 is 31.1C. Considering CO2 to be an ideal gas, what DO NOT WRITE
is the mean free path at a pressure of 1 bar and this temperature?
IN THIS SPACE
Using an equation in the handbook, one must find the molecular diameter and the number density.
For an ideal gas, the number density is
(1 105 )(6.02211415 1023 )
=
=
= 2.3806 1025 3
(8.3144349 1 )(304.25 )
= (0.4328 109 )2 = 5.885 1019 2
p. 1________/10
p. 2________/10
p. 3________/10
p. 4________/15
The Lennard-Jones diameter of CO2 is 0.4328 nm, which gives a collision cross-section of
1
1
Substitution into the equation for the mean free path gives
=
2
=
2(5.885 1019 2 )(2.3806 1025 3 )
p. 5________/10
p. 6________/10
= 5.0472 10
8
p. 7________/15
p. 8________/10
p. 9________/15
p. 10_______/10
p. 11_______/10
p. 12_______/10
(b) The critical pressure of CO2 is 73.9 bar. Again, if one assumes that CO2 under these conditions =============
can be treated as an ideal gas (a questionable assumption), what is the predicted mean free path at
p. 13_______/10
the critical point?
(Extra credit)
=============
The only factor that changes in the calculation is the number density:
TOTAL PTS
5
23 )
(out of 135)
(73.9 10 )(6.02211415 10
=
=
(8.3144349 1 )(304.25 )
1
Then, substitution in the equation yields
=
2
=
2(5.885
1
1019 2 )(1.7593
This is only slightly larger than the diameter of the CO2 molecule.
= 1.7593 1027 3
1027 3 )
= 6.8297 1010
NAME:
CHEM 444, Final Exam, Spring, 2011, page 2
2. (10 points) The diffusion coefficient of the protein lysozyme (MW = 14.1 kg mol ) in dilute water solution is 0.10410 m
-1
s at 25C. (a) How long will it take, on average, for the protein to diffuse an rms distance of 1 m under these conditions?
-1
-9
2
For three-dimensional diffusion (which this has to be!), the rms distance travelled in some time is = 6 .
Rearranging this equation gives the following equation for the time to travel an rms distance:
(1 106 )2
2
=
=
= 1.60 103
6
6(0.104 109 2 1 )
(b) Assuming that the viscosity of this dilute solution is not affected by the presence of the protein, estimate the size of the
protein (treated as roughly a sphere of radius r)?
This assumption means that one should treat the solution as if it has the viscosity of water, which from the handbook is 8.937
millipoise at 25C. Then, one solves the Stokes-Einstein equation for the radius of the equivalent sphere:
=
6
=
1.38065051023 1 (298.15 )
6(8.937103 )(0.1 1 )(0.104109 2 1 )
= 2.350 109
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 3
3. (10 points) The interchange of forms for cyclohexane by the following process
has been measured by NMR spectroscopy. [Anet, F. A. L.; Bourn, A. J. R. J. Am. Chem. Soc. 1967, 89, 760 768.] The
rate constants, k, for the backward and forward processes are identical. The rate constants at various temperatures are
given in the table. From these data, determine the activation energy for this process and the preexponential factor.
-1
-1
k (s )
k (s )
Temperature (C)
Temperature (C)
-116.7
0.004
-81.8
4.00
-112.1
0.014
-77.9
6.40
-107.3
0.039
-61.4
64.00
-105.3
0.071
-42.4
566
-100.4
0.11
-34.4
1150
-96.9
0.27
-24.0
2490
8
One plots the natural logarithm of the rate
constant versus 1/T, as shown in the
adjacent graph. From the slope of the line,
one finds the activation energy:
6
= . .
= .
= .
4
y = -5621.9x + 30.607
ln(k/s-1)
2
The intercept is related to the
preexponential factor:
0
= (. ) = .
-2
-4
-6
-8
0.003
0.004
0.005
0.006
0.007
1/T (1/K)
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 4
4. (15 points) Given the following reaction mechanism, determine the rate expression for the production of HBr.
(1)
Br2 1 2 Br
(2)
Br +H 2 2 H + HBr
(3)
H + Br2 3 Br +HBr
k
k
k
(4) H + HBr 4 Br +H 2
k
(5) 2 Br k 5 Br2
Solution:
d [ HBr ]
(I)
= k 2 [ Br ][ H 2 ] + k 3 [ H ][ Br2 ] k 4 [ H ][ HBr ]
dt
d [ Br ]
= 0 = 2k1 [ Br2 ] k 2 [ Br ][ H 2 ] + k 3 [ H ][ Br2 ] + k 4 [ H ][ HBr ] 2k 5 [ Br ]2
dt
d [ H ]
(III)
= 0 = k 2 [ Br ][ H 2 ] k 3 [ H ][ Br2 ] k 4 [ H ][ HBr ]
dt
(II)
Adding II and III gives:
0 = 2k1 [ Br2 ] 2k 5 [ Br]2
k
[ Br ] = 1 [ Br2 ]
k
5
1/ 2
(IV)
Solving III for hydrogen radical gives:
k1 1 / 2
1/ 2
k 2 ([ Br2 ]) [ H 2 ]
k5
[H ] =
k 3 [ Br2 ] + k 4 [ HBr]
(V)
Subtracting III from I (i.e., subtracting zero from I) gives:
d [ HBr ]
= 2k 3 [ H ][ Br2 ]
dt
(VI)
Thus, the rate of production of HBr is:
1/ 2
k
1/ 2
k 2 1 ([ Br2 ]) [ H 2 ]
k
d [ HBr ]
= 2k 3 [ Br2 ] 5
dt
k 3 [ Br2 ] + k 4 [ HBr ]
1
2k 3 k 2 k1 / 2 k 51 / 2 ([ Br2 ]) [ H 2 ]
[ HBr ]
k3 + k4
[ Br2 ]
1/ 2
=
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 5
5. (10 points) The decarboxylation of a -keto acid is catalyzed by a decarboxylase. The rate of the reaction is monitored
by the production of CO2, which is produced by the decarboxylation with 1/1 stoichiometry. The initial rates are
determined for various initial amounts of the acid at a constant enzyme concentration. A graph of 1/v versus the inverse
of the keto acids concentration is shown below. Determine the Michaelis-Menten constant for this system. [HINT: For
Michaelis-Menten kinetics,
v=
[M ]
v max , where the Michaelis-Menten constant is KM.]
K M + [M ]
Rearrangement of the equation shows that
1
v
=
1
v max
KM
v max
+
1
[M ] .
The plot should be a straight line with an intercept of
1
v max
and a slope of
KM
. Hence, the Michaelis-Menten constant can be found from the
v max
following formula:
KM
= ( slope) /(intercept ) .
To calculate the constant requires estimates of the slope and intercept of the plot. The intercept appears to be 3 0.2 min
-1
micromole .
Estimation of the slope is a little harder, but it seems to be about
accuracy. This gives a slope of slope
(1.1 10 )
mole 2
3
dm min
(3.0) min
6
10 mole
=
(1.1
7.8 5.6 micromole mole
to within about 10%
3
42
min
dm
mole 2
0.1) 10 6
dm 3 min . Then, from the formula
6
KM
=
= 0.36
mol
.
dm 3
But, of course there is a range since these are estimates. Conservatively, lets call it 15%. So, the range of correct answers
is:
0.30 mol dm
-3
< KM < 0.42 mol dm
-3
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 6
6. (10 points) An experiment to determine the sedimentation coefficient of lysozyme at 20C was carried out by monitoring
the movement of the boundary layer as a function of time while the sample was centrigfuged at 55,000 rpm. The data on the
boundary distance from the pivot are given in the table. Determine the sedimentation coefficient of lysozyme. [Show all
work. In particular, show an appropriate with plot, labels. Express your answer in the proper units.]
Time (minutes)
0
30
60
90
120
150
Boundary distance from pivot (cm)
6.000
6.065
6.142
6.211
6.278
6.350
The boundary should move from the initial point by the
following equation:
0.06
() = (0)(2 )
y = 0.0004x + 1E-05
0.05
ln(x/x(0))
0.04
A plot of the logarithm of the ratio of the distance should
2
be a straight line with a slope equal to s . The plot is
shown in the adjacent graph. The slope of this plot is
-1
0.000379 min .
0.03
0.02
Next convert the rotation frequency to radians per
second:
0.01
1
(2) = 5759.6 1
60
= 5.50 104 1
0
0
50
100
Time (min)
150
200
1
60
= 1.904 1013
2
Then, one finds the sedimentation coefficient to be
=
(0.000379 1 )
(5759.6 1 )
However, the proper unit of the sedimentation coefficient is the svedberg (= 110 s). So, the final answer is
= 1.904
13
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 7
7. (15 points) In the following, carry out the mathematics requested as completely as possible.
(a)
2 () = (2 /0 )(/20 )
Find the constant, A, that normalizes the radial part of the 2s wave function of the hydrogen atom:
The normalization occurs when 2 = 1. Substitution of the function into this equation gives the following
0
equation:
2 (2 /0 )2 exp 2 = 1
0
0
One may simplify this equation by defining = /0 , which gives
3
2 0 (4 4 + 2 ) exp( ) 2 = 1
0
3
2 0 (4 2 4 3 + 4 ) exp( ) = 1
Multiplying through gives the following equation
0
Each of the three integrals in this formula can be found in Table 2.3. The result is
3
2 0 (4(2!) 4(3!) + 4!) = 1
Reducing this gives the equation
3
3
2 0 4(2!) = 2 0 (8) = 1
Rearranging this equation gives an expression for the normalization constant
1
=
3/ 2
220
(b)
Determine the simplest form of the commutator,
, 2
2
This is most easily done by carrying out the action on a general function, f:
,
() =
=
=
+ = =
=
Then, by comparison, the equivalent operator to the commutator must be
,
=
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 8
8. (10 points) Match the statements/equations in part A with the names in part B by inserting the appropriate next to the
statement.
Part A
Part B
_b.__ Original explanation of the spectrum of hydrogen
a. Borns hypothesis
_f.__ Many-electron wave functions must be antisymmetric
under the exchange of any two electrons.
b. Bohr atom
_e.__
_g.__
_d.__
=
=
c. Davisson-Germer experiment
d. DeBroglies relation
e. Heisenbergs principle
f. Paulis principle
g. Plancks relation
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 9
9. (15 points) W hat is the probability that a quantum mechanical particle in a one-dimensional box will be found in the left
quarter of the box through repeated measurements when it is in the lowest-energy state? [Show all work.]
The probability of finding a particle within a region is the integral of the square of the wave function over that region. For the
particle in one-dimensional box, this integral is
2 /4
2 /4 2
2 1
1
1
2 =
=
=
0, =
0
0
8 4
2
4
2
4
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 10
10. (10 points) (a) What is the ground-state configuration for a sodium atom?
2
2
6
1
(1s) (2s) (2p) (3s)
(b) What term(s) arise(s) from this configuration? [HINT: Include the total angular momentum.]
2
The only term that arises is S1/2
(c) What are the next two excited configurations of the sodium atom?
2
2
6
1
(1s) (2s) (2p) (3p)
2
and
2
6
1
(1s) (2s) (2p) (4s)
(d) What terms arise from these two configurations (including total angular momentum)?
2
P1/2
2
P3/2
2
and S1/2
(e) The spectrum of sodium has two rather strong transitions with wave lengths of 764.494 nm and 769.901 nm. From this
information, calculate the spin-orbit coupling constant for the term that is split.
2
2
The term that is split is the P. These transitions are from that those terms down to the S1/2 term. According to the equation
2
2
in the handbook, the spin-orbit energy of the P1/2 is - , and the spin-orbit energy of the P3/2 term is /2. So, the
difference in energy of the two terms arises from and energy splitting of 3/2. The energy difference is
1
1
769.901 764.494
=
=
= 9.186 106 1
764.494
769.901
764.494(769.901)
Energies are not usually reported in these units, so one has to convert to a more convenient unit.
107
= 91.86 1
= 9.186 106 1
But the spin-orbit coupling constant is 2/3 of this value, so
2
= 91.86 1 = 61.24 1
3
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 11
11. (10 points) For the following molecules, write the ground configuration, its term symbol, and the bond order predicted by
the aufbau principle.
Molecule
C2
F2
Ne2
O2
Ground Configuration
2
2
4
1 12 2 22 1
2
2
2
4
1 12 2 22 3 1 14
2
2
2
4
1 12 2 22 3 1 14 32
2
2
2
4
1 12 2 22 3 1 12
Term Symbol
Bond Order
g
2
g
1
g
0
g
2
1
1
1
3
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 12
12
16
12. (10 points) C O is the quintessential model of a heteronuclear diatomic molecule. From data in the handbook, predict
-2
the vibrational force constant (assuming the harmonic approximation) of this molecule (in joules m ) as accurately as you
can. Show all work.
The equation that relates the force constant to measured properties is
=
(12.00000)(15.99491)
=
= 1.138505 1023
(6.02211415 1023 )(12.00000 + 15.99491)
0 ( + )
One can calculate the reduced mass of the molecule from the exact masses of the isotopes in Table 12.2.
=
-1
From Table 12.1, the vibrational frequency is 2170.21 cm .
2
=
Rearranging the equation above, one obtains an expression for the force constant
= 2(2.99792458 1010 1 )(2170.21 1 ) = 4.08792 1014 1
provided that one uses the proper units. The value of the vibrational frequency is
= 1.138505 1026 (4.08792 1014 1 )2 = 1.90257 103 2 = 1.90257 103 2
Then, substitution gives
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NAME:
CHEM 444, Final Exam, Spring, 2011, page 13
13. (10 points, extra credit) From measurements of the differential cross section for scattering of electrons off atomic
hydrogen, it was found that the atom has an electronic charge density given by (r) = exp(-r), where and are
constants.
(a) Find the relation between and such that the total charge equals e, the charge on the electron. (Express in terms
of .) (b) Find the normalized wave function that describes this state in terms of these two parameters.
( ) 2 =
(a) If one sums over all space, then the electron must be somewhere, so one has the requirement
0
( ) 2 =
Substitution of the form into this equation gives the following result
0
2! =
3
This can be integrated to give the following relationship:
=
So, in terms of , the answer is
3
2
(b) In order for the integrand to represent the probability of finding an electron at some point, the value must be
proportional to an integral of the square of the wave function (= 1). That is, the density divided by e must be the square of
the wave function. Hence, the wave function is
() =
()
3 ()
3
=
= ( /2)
2
2
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