CHM 5423 – Atmospheric Chemistry
First Hour Exam (Take home)
October 7, 2010 (Due at the start of class on Tuesday, October 12 th)
Do the following problems. Show your work.
1. (20 points) Data for the absorption cross sections for HO 2NO2 are given on the next page and will be needed to
do this problem.
a) Based on these data and the information in the Chapter 2 handout, find the value for k d, the rate constant
for photodissociation, for the following conditions.
Latitude = 20. N
time of year = Nov. 1st
Cloudless conditions, z = 0.0 km, "best albedo"
time of day = 1200 hours
You may assume that the quantum yield for photodissociation is d = 1. at all wavelengths.
b) Photodissociation of HO2NO2 leads to two sets of photodissociation products
HO2NO2 + hc/ 

HO2 + NO2
(1.1)
OH + NO3
(1.2)
In addition, quenching and light emission are in principle competing processes
HO2NO2 + hc/ 
(HO2NO2*) + M  HO2NO2 + M
HO2NO2 + hc/ 
(HO2NO2)*
 HO2NO2 + h
(1.3)
(1.4)
In a particular experiment the overall quantum yields for disappearance of HO 2NO2 and for formation of NO2 were
measured experimentally as a function of total pressure in a mixture of air and HO 2NO2, with p(air) >> p(HO2NO2).
For pressures less than 1000 torr no fluorescence was detected. Overall quantum yields of (HO2NO2) = - 1.0 and
(NO2) = 0.8 were seen, and were, within experimental error, co nstant over the pressures used in the experiments.
Secondary reactions in the system that involved HO2NO2 and NO2 were suppressed by the addition of a small
amount of a radical scavenger to the system, and so only reactions 1.1 – 1.4 need to be considered in the production
or removal of these species.
Based on the above information find the values for 1.1, 1.2, 1.3, and 1.4, the primry quantum yields for the
above four processes. Justify your answer.
2. (30 points) In a recent paper (K. Takahashi, J-H. Xing, M. D. Hurley, T. J. Wallington, J.Phys.Chem.A 114
(2010), 4224-4231) the reaction of chlorine atoms with three three different compounds was studied. Using this
paper as a starting point answer the following questions
a) Wallington and coworkers use a PLP/VUV-LIF technique to study the absolute rate of reaction in this
paper. Briefly explain this technique.
b) In the absolute rate studies chlorine atoms are produced by the photodissociation of Cl 2 at  = 351. nm.
Find the amount of excess energy for this process (the energy above that required to break the Cl-Cl bond in Cl2).
Since the dissociation products are atoms, the excess energy should appear as translational kinetic energy in the
atoms produced. Assuming the excess energy is equally divided between the two chlorine atoms, find the average
velocity of the chlorine atoms produced by photodissociation. Assuming EK = 3/2 RT for this kinetic energy find the
approximate value of temperature corresponding to the chlorine atoms produced. Could this have any effect on the
observed kinetics of the reaction? Explain.
c) For all three compounds studies the bimolecular rate constant for the reaction increased with increasing
pressure (see Table 1 of the paper). Based on this observation it is concluded that the majority of the reaction occurs
by addition rather than by abstraction. Justify this conclusion.
d) For a recombination reaction in the simple Lindemann model an apparent bimolecular rate constant can
be defined as
kbi =
k0 [M]
1 + (k0/k)[M]
(2.1)
where k0 and k are low and high pressure limiting rate constants (as discussed in the Chapter 4 handout).
Using the data in Table 1 for allyl alcohol find experimental values for k0 and k, including correct units
(concentrations in molecules/cm3, time in s).
e) Is there any evidence from your data analysis in part c that the data deviate from the behavior expected
from the Lindemann model? Discuss possible reasons for deviations from this model, and suggest additional
experiments that could be carried out to test for such deviations.
f) In the introduction to the paper a value for the rate constant for the reaction of allyl alcohol with chlorine
atoms, measured by Rodriguez and coworkers, is given for p = 760 torr and T = 298. K. Would this rate constant
correspond to k0, k, or neither of these? Justify your answer. If you answer that their value of k should correspond
to k0 or k, compare their value to the value you obtained from the analysis of the data in Table 1, and commonent
on the agreement (or lack of agreement) observed.
Solutions.
1)
a) From Table 3.9, we get  = 34.4  for the conditions of the problem. From Table 3.8 we get f = 1.015.
The quantum yield for photodissociation is assumed equal to 1.0 throughout.
The data for the calculation of the photodissociation rate constant are given below. Interpolation is used in
giving the values for () and F()
 range (nm)
() x 1020 (cm2/molecule)
F() x 10-14 photon/cm2.s
kd x 106 s-1
296-298
298-300
300-302
2.10
1.72
1.40
0.01
0.02
0.04
0.021
0.034
0.056
302-304
304-306
306-308
1.16
0.951
0.778
0.10
0.20
0.33
0.116
0.190
0.257
308-310
310-212
312-314
0.620
0.508
0.422
0.46
0.72
0.92
0.288
0.366
0.388
314-316
316-318
318-320
0.346
0.289
0.240
1.06
1.33
1.38
0.367
0.384
0.331
320-325
325-330
330-335
0.176
0.114
0.0754
4.41
6.56
7.11
0.776
0.748
0.536
335-340
340-345
345-350
0.0477
0.0307
0.0206
6.94
7.57
7.69
0.331
0.232
0.158
k(uncorrected) = 5.579 x 10-6 s-1
k(corrected) = 1.015 (5.579 x 10 -6 s-1) = 5.7 x 10-6 s-1
b) The only processes removing HO2NO2 are 1.1 and 1.2, which remove one molecule for every photon.
Since (HO2NO2) = - 1.0, it follows that 1.3 = 1.4 = 0. Since only process 1.1 produces an NO2 molecule, and
since it produces one molecule per photon, 1.1 = (NO2) = 0.8. From this, it follows that 1.2 = 0.2.
2)
a) PLP/VUV-LIF is the following technique. A UV laser is used to provide a pulse of light to a flowing gas
mixture of N2/alcohol/Cl2, with p(N2) >> p(alcohol) >> p(Cl) (where Cl is produced by photodissociation of Cl 2).
Using flow conditions means that new reactants are continuously being added to the system and products from the
previous reactions are continuously removed. The 351 nm UV light pulse photodissociates chlorine molecules
Cl2 + h  2 Cl
that serve as the source of atomic chlorine. Because p(alcohol) >> p(Cl) pseudo-first order conditions apply.
The progress of the reaction is followed by measuring the concentration of chlorine atoms using VUV-LIF
(vacuum UV laser induced fluorescence). The VUV light pulse forms electronically excited chlorine atoms, which
subsequently emit light by fluorescence.
Cl + h  Cl*  Cl + h
Since fluorescence is isotropric the fluorescence detector can be put at 90  relative to the UVU LIF light pulse and
UV LP pulse to minimize any signal from scattered light.
b) The energy from one mole of photons at 351. nm is
E(photons) = hcNA = (6.626 x 10-34 J.s) (2.998 x 108 m/s) (6.022 x 1023 mol-1) = 340.81 kJ/mol

(351. x 10-9 m)
The energy used up in the bond breaking is
H = 2 Hf(Cl) - Hf(Cl2) = 2 (121.301) = 242.60 kJ/mol
The excess energy is therefore E = (340.81 – 242.60) = 98.21 kJ/mol
Assuming the excess energy is equally divided between the two chlorine atoms (as suggested by
conservation of momentum) then the energy per mole of chlorine atoms is 49.1 kJ/mol.
Since chlorine atoms lack rotational and vibrational energy, the energy is entirely kinetic energy, so
EK = mv2/2
v = (2EK/m)1/2 = (2E/M)1/2
where E is the excess energy per mole of chlorine atoms, and M is the atomic mass of chlorine.
So
v = [2(49100 J/mol)/(35.45 x 10-3 kg/mol)]1/2 = 1660. m/s
Setting EK = 3/2 RT, we get
T = 2EK/3R = 2(49100 J/mol)/3(8.314 J/mol.K) = 3940. K
Could this have an effect on the kinetics? Sure. You would expect the rate constant for reaction to depend
on temperature, and so you would expect a different rate of reaction for “hot” chlorine atoms. Generally speaking,
the abstraction reaction is likely faster at higher temperatures and the addition reaction is likely slower at higher
temperatures (due to the requirement of intermediate complex formation).
However, as noted in the paper, translational relaxation of the chlorine atoms for the conditions of the
experiment is rapid (t1/2 < 1 s) and so for the conditions of the experiment it is not in fact a problem.
c) When a rate constant increases with increasing pressure, that is usually a sign that the reaction proceeds
via formation of an intermediate complex. Increasing pressure makes it more likely that excess energy in the
complex will be given off by collision before the complex falls apart to regenerate the original reactants. Formation
of an intermediate complex is a characteristic property of addition reactions but not usually seen in abstraction
reactions.
d) If we invert equn 2.1, we get
1 = 1
1
kbi
k0 [M]
+ 1
k
Therefore, a plot of 1/kbi vs 1/[M] should have a slope equal to 1/k0 and an intercept equal to 1/k.
The data are given below
p (torr)
k (cm3/molecule.s)
1/p (torr-1)
1/k (molecule.s/cm3)
1.9
4.9
9.9
19.9
1.09 x 10-10
1.37 x 10-10
1.51 x 10-10
1.86 x 10-10
0.526
0.204
0.101
0.050
0.917 x 1010
0.730 x 1010
0.662 x 1010
0.538 x 1010
The data are plotted below.
Based on the plot, I get
intercept = 0.554 x 1010 molecule.s/cm3 = 1/k, so k = 1.81 x 10-10 cm3/molecule.s
slope = 0.715 x 1010 molecule.torr.s/cm3 = 1/k0 , so k0 = 1.40 x 10-10 cm3/molecule.s.torr
From the ideal gas law
N/V = pNA/RT = (1 torr) (1 atm/760 torr) (6.022 x 10 23 molecule/mol) = 3.26 x 1019 molecule/L = 3.26 x 1016
molecule/cm3
(0.08206 L.atm/mol.K) (296. K)
for the conversion factor between torr and molecule/cm2 (at T = 296. K). So
k0 = (1.40 x 10-10 cm3/molecule.s.torr)(1 torr/3.26 x 1016 molecule/cm3) = 4.30 x 10-27 cm6/molecule2.s
e)
Yes. The data seems to show systematic deviations from linear behavior (though with only four data points
one must be careful not to overinterpret the data). There are several possible reasons for the deviations
1) They may not be real, but simply an artifact of the small number of data points.
2) They may indicate a systematic experimental error. For example, diffusion out of the observation zone
would lead to an apparent loss of Cl atoms and therefore a higher apparent bimolecular rate constant. However,
Wallington and coworkers chose their conditions to aviod this problem.
3) Systemic error in the Lindemann model. We discussed in class how the Lindemann model does not work
well in the “falloff” region between bimolecular and termolecular kinetics. We did not discuss in detail the error in
the Lindemann model in this region, and so it is possible the deviations here are due to shortcomings in the kinetic
model. This could be looked into by comparing the Lindemann and Troe models in the falloff region.
4) Failure to account for abstraction. The Lindemann model assumes the reaction proceeds by addition, and
so we really ought to include a second reaction for the direct abstraction of a hydrogen from the alcohol. In fact, my
preliminary analysis of the Wallington et al. data indicates this is likely the source of the deviations in the plot.
f) Rodriguez et al. value at p = 760. torr should correspond to 1/p = 0.0013 torr -1, and so to a high pressure
limiting value k. The Rodriguez value k = 1.72 x 10-10 cm3/molecule.s is close to the high pressure limiting value
k = 1.81 x 10-10 cm3/molecule.s (though I tend to agree with Wallington that the Rodriguez value is systematically
low, based on the Wallington data in the smog chamber studies, and that the agreement here is in a
sensecoincidence).