CHM 3410 – Problem Set 10
Due date: Friday, December 3rd, in class (no late problem sets will be accepted). (NOTE: The final exam is
Monday, December 6th, from noon to 2:00pm. It will be comprehensive, and it will also include the new material we
have covered from chapters 22, 23, and 15.)
Do the following problems. Show your work.
1) The chemical reaction
Cl + O3  ClO + O2
(1.1)
plays an important role in ozone destruction. Because of this, it has been extensively studied both experimentally
and theoretically.
a) The Arrhenius constants for the reaction are A = 2.3 x 10 -11 cm3/molecule.s and Ea = 1.67 kJ/mol. Based
on this information find the value for k for reaction 1.1 at T = 300. K.
b) According to collision theory, the rate constant for a bimolecular gas phase reaction is
k = P AB (8RT/AB)1/2 exp(-Ea/RT)
(1.2)
where P is a steric factor, AB = dAB2, where dAB = (dA + dB)/2, and AB = MAMB/(MA+MB) is reduced mass.
Using collision theory find the value for k for reaction 1.1 at T = 300. K. Use Cl = 0.40 nm2, O3 = 0.55
2
nm , and P = 1 in your calculation.
c) By comparing your answers in a and b, find the value for P. Is this value reasonable? Explain.
2) The molecule ozone (O3) has three vibrations. In the harmonic oscillator model, the vibrational energy for an
ozone molecule (relative to the lowest energy state) is
E(v1,v2,v3) = v11 + v22 + v33
where
(2.1)
v1 = 0, 1, 2, …
v2 = 0, 1, 2, …
v3 = 0, 1, 2, …
are possible values for v1, v2, and v3. The values for the vibrational constants are 1 = 1110. cm-1, 2 = 705. cm-1,
and 3 = 1042 cm-1.
a) Find all of the vibrational energy levels for ozone that have E < 2000. cm-1. Label each vibrational level
by its three quantum numbers. For example, the lowest energy state is v1 = v2 = v3 = 0, and so is labeled (0,0,0) and
by our convention above has an energy E = 0.0 cm-1.
b) Find the fraction of molecules in each of the above vibrational states of ozone at T = 350. K. You may
assume that no molecules have vibrational energy greater than 2000. cm-1. Note that one way of expressing the value
for the Boltzmann constant is k = 0 69503 cm-1/K.
Also do the following from Atkins.
Exercises.
23.2b The enzyme-catalyzed conversion of a substrate has a Michaelis constant of 0.032 mol/L. The rate
of the reaction is 2.05 x 10-4 mol/L.s when the substrate concentration is 0.875 mol/L. What is the maximum
velocity (rate) of the reaction?
15.3b What is the temperature of a two level system of energy separation equivalent to 300. cm-1 when the
population of the upper state is one-half that of the lower state?
Problems.
22.6 The rate constant of the reaction
I-(aq) + H2O2(aq)  H2O() + IO-(aq)
varies slowly with ionic strength, even though the Debye-Huckel limiting law predicts no effect. Use the following
data from 25. C to find the dependence of log10kr on the ionic strength
I
0.0207
0.0525
0.0925
0.1575
kr(L/mol.min)
0.663
0.670
0.679
0.694
Evaluate the limiting value kr at zero ionic strength. What does the result suggest for the dependence of log 10 on
ionic strength for a neutral molecule in an electrolytic solution?
23.2 There are different ways to represent and analyze data for enzyme-catalyzed reactions. For example,
in the Eadie-Hofstee plot, v/[S]0 is plotted against v. Alternatively, in the Hanes plot v/[S]0 is plotted against [S]0.
a) Using the simple Michaelis-Menton mechanism, derive relationships between v/[S]0 and v and between v/[S]0 and
[S]0. b) Discuss how the values of KM and vmax are obtained from the analysis of the Eadie-Hofstee and Hanes plots.
c) Determine the Michaelis constant and the maximum velocity of the reaction from the following data (from
problem 23.1) by using the Eadie-Hofstee and Hanes plots to analyze the data. For this data, taken at T = 20. C, the
concentration of enzyme (ATPase) is 20. nmol/L.
[ATP] (mol/L)
0.60
0.80
1.40
2.00
3.00
v(mol/L.s)
0.81
0.97
1.30
1.47
1.69
Solutions.
1)
a) k = A exp(-Ea/RT) = (2.3 x 10-11 cm3/molecule.s) exp[- (1670. J/mol)/(8.314 J/mol.K)(300. K)]
= 1.18 x 10-11 cm3/molecule.s
b) W first calculate the pre-exponential term, P AB (8RT/AB)1/2 , assuming P = 1.
Since  = d2 , it follows that d = (/)1/2
So
dCl = [(0.40 nm2)/]1/2 = 0.357 nm
dO3 = [(0.55 nm2)/]1/2 = 0.418 nm
dCl-O3 = (dCl + dO3)/2 = (0.357 nm + 0.418 nm)/2 = 0.388 nm
Cl-O3 = (0.388 nm)2 = 0.473 nm2
And
Cl-O3 = MClMO3/(MCl+MO3) = (35.45)(48.00)/(35.45 + 48.00) = 20.39 g/mol = 20.39 x 10 -3 kg/mol
So
P AB (8RT/AB)1/2 = (1)(0.473 x 10-18 m2) [8(8.314 J/mol.K)(300. K)/(20.39 x 10-3 kg/mol)]1/2
= 2.64 x 10-16 m3/molecule.s = 2.64 x 10-10 cm3/molecule.s
and so
k = (2.64 x 10-10 cm3/molecule.s) exp[- (1670. J/mol)/(8.314 J/mol.K)(300. K)]
= 1.35 x 10-10 cm3/molecule.s
c) To make k the same for Arrhenius theory and collision theory, we choose P so that
P = (1.18 x 10-11)/(1.35 x 10-10) = 0.087
So about 9% of collisions have an orientation favorable for reaction.
Is this a reasonable result? Well, since an oxygen atom is transferred to the chlorine atom in the reaction,
and ozone is an elemental form of oxygen, I would expect P ~ 1. The actual result is within a factor of 10 of this.
The smaller value of P might mean that there is still a little bit of an orientation effect – perhaps the Cl atom has to
collide with one of the end oxygen atoms in ozone.
2)
The data are tabulated below. Note that to get the fraction of molecules in a particular vibrational state we
take the Boltzmann factor for that state and divide by the sum of the Boltzmann factors (and, as indicated in the
problem, assume the Boltzmann factors for levels above 2000. cm-1 are equal to zero).
Note that P(v1,v2,v3)/P(0,0,0) = exp[-(E(v1,v2,v3)/kT]
where kT = (0.69503 cm-1/K)(350. K) = 243.3 cm-1
State (v1,v2,v3)
E(cm-1)
Boltzmann factor
fraction of molecules
0
0
0
1
0
1
0.
705.
1042.
1110.
1747.
1815.
1.0000
0.0551
0.0138
0.0104
0.0008
0.0006
0.9228
0.0508
0.0127
0.0096
0.0007
0.0006
0
1
0
0
1
1
0
0
1
0
1
0
Sum =
1.0837
Notice that about 7.5% of the ozone molecules are vibrationally excited.
Exercise 23.2b
v=
So
vmax
[1 + KM/[S]0]
vmax = v [1 + KM/[S]0] = (2.05 x 10-4 mol/L.s) [1 + (0.032 mol/L)/0.875 mol/L)] = 2.12 x 10 -4 mol/L.s
Exercise 15.3 b
P2/P1 = ½ = exp(-E/kT)
Taking the logarithm of both sides gives
ln(1/2) = - ln(2) = -E/kT
and so
T=
E
=
300. cm-1
= 623. K
k ln(2)
(0.69503 cm-1/K) ln(2)
Problem 22.6
My intention was to try various ways of plotting the data (k vs I, lnk vs I, k vs I 1/2, lnk vs I1/2, and so forth)
until it looked like I found good linearity. Much to my surprise I found the first thing I plotted (k vs I) was linear, as
shown on the next page.
The best fitting line gives
kr = 0.6582 L/mol.min + (0.2269 L/mol.min) I
Therefore, at zero ionic strength kr = 0.6582 L/mol.min.
What does this imply as far as the dependence of log10k on ionic strength? Nothing, as far as I’m
concerned. The solution manual claims that the data indicate log10k is proportional to I, and in fact a plot of log10k
vs I is linear – but not noticably more linear than a plot of k vs I (see next page). So I don’t see any justification for
plotting log10k instead of k.
Plot of k vs I
k (L/mol.min)
0.71
0.7
0.69
0.68
0.67
0.66
0.65
0
0.05
0.1
0.15
0.2
0.15
0.2
I
Plot of logk vs I
-0.15
0
0.05
0.1
logk
-0.16
-0.17
-0.18
-0.19
I
Problem 23.2
a+b) From the Michaelis-Menton model we were able to show
v=
vmax
[1 + (KM/[S]0]
If we multiply both sides by the denominator on the right, we get
v [1 + (KM/[S]0] = v + KM (v/[S]0) = vmax
Subtracting v from both sides gives
KM(v/[S]0) = vmax – v
If we divide both sides by KM, we get (Eadie-Hofstee)
v/[S]0 = (vmax/KM) – (1/KM)v
So a plot of v/[S]0 vs v has
slope = - 1/KM
intercept = vmax/KM
For a second way to plot the data, again begin with
v=
vmax
[1 + (KM/[S]0]
Inverting both sides gives
1/v = (1/vmax) + (KM/vmax[S]0)
Multiplication by [S]0 gives (Hanes)
[S]0/v = KM/vmax + (1/vmax)[S]0
So a plot of [S]0/v vs [S]0 has
slope = (1/vmax)
intercept = KM/vmax
c) The data for the plots is given below
v (mol/L.s)
[ATP]0 (mol/L)
v/[ATP]0 (s-1)
[ATP]0/v (s)
0.81
0.97
1.30
1.47
1.69
0.60
0.80
1.40
2.00
3.00
1.350
1.212
0.929
0.735
0.563
0.741
0.825
1.077
1.361
1.775
From the Eadie-Hofstee plot (next page) I get
slope = - 0.9060 (L/mol) = - 1/KM
intercept = 2.0884 s-1 = vmax/KM
KM = 1.104 mol/L = 1.10 mol/L
vmax = (1.104 mol/L)(2.0884 s-1) = 2.31 mol/L.s
From the Hanes plot (next page) I get
slope = 0.4336 L.s/mol = 1/vmax
vmax = 2.31 mol/L.s
intercept = 0.4794 s = KM/vmax
KM = (2.31 mol/L.s)(0.4794 s) = 1.11 mol/L
So, which is the better method for data analysis? This is the sort of question that can be rigorously answered, but I
will use a more qualitative approach. For cases where one has to carry out an extrapolation over a large distance to
find the y-intercept it is better to determine a parameter from a value for slope than it is a value for the y-intercept.
By that standard, the Eadie-Hofstee plot is better for determining KM, and the Hanes plot is better for determining
vmax.
The calculation of confidence limits on the values for KM and vmax can be carried out using the general
methods discussed in the lab portion of this class.
Eadie-Hofstee plot
v/[ATP]o (s-1)
1.4
1.2
1
0.8
0.6
0.4
0.8
1
1.2
1.4
1.6
1.8
v (umol/L.s)
Hanes plot
[ATP]o/v (s)
2
1.6
1.2
0.8
0.4
0
0.5
1
1.5
2
[ATP]o (umol/L)
2.5
3
3.5