Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone

PROBLEMS/SOLUTIONS
1.

Consider Figure 3.1 and explain why, at low altitudes, the number density of ozone is quite large
while the mixing ratio is very small.

Solution
The number density of ozone is relatively quite large near the Earth’s surface due to a number of additional
ozone generating processes, both anthropogenic and natural. One notable anthropogenic source (see
Chapter 4) is from photochemical smog events. The corresponding ozone mixing ratio however is very
small, since at low altitudes the concentrations of nitrogen and oxygen gases are at their highest values,
and ozone on a relative scale to all gases present is very small.

2.

Draw Lewis structures for ozone and for dioxygen. Using the data given below, qualitatively compare
the bond enthalpies, bond orders, and bond lengths of these two compounds.
O2 (g) → 2O (g)

ΔHº = + 498.4 kJ

O2(g) + O (g) + M → O3 (g) + M ΔHº = – 106.5 kJ
Solution
dioxygen

ozone

..O......O
.. ..
formal charges

0

0

(resonance)

..O......O . ..O..
.. ..
0 +1

–1

....O... ..O......O
–1

+1

0

Bond enthalpy: (see Appendix B.3)
O = O (O2 (g)) D = 497 kJ mol-1
O – O (average of many poly-atomic molecules) D = 146 kJ mol-1
Standard Enthalpy of formation values: (see Appendix B.2)
ΔHfº O3 (g) = 142.7 kJ mol-1
ΔHfº O (g) = 249.2 kJ mol-1

Enthalpy of reactions provided above:
ΔHºrxn for O3 (g) = – 105 kJ mol-1
ΔHºrxn for 2 O (g) = 498 kJ mol-1

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone

Based on the enthalpies provided, the double bond of dioxygen gas is stronger than the resonance bonds in
ozone. Since ozone is represented by a resonance structure, the actual bond enthalpies for ozone are likely
closer to the average value of the single and double bonds that are shown in the resonance structures above.
Comparing the bond enthalpies of dioxygen and ozone, yields the conclusion that it would take 498 kJ mol 1
to break the one dioxygen bond. On the other hand, to successively break the oxygen bonds in ozone
would require approximately 302 kJ mol-1 . This would suggest a bond enthalpy for ozone of approximately
D = 300 kJ mol-1. The reported range of wavelengths capable of breaking the ozone is from λ = 230 - 320
nm (Chapter 3). This suggests that the amount of energy required is in the range of 370 to 520 kJ mol -1.
Bond order:
In ozone there are six electrons that are involved in two bonds giving an average of 3 electrons per bond or
a bond order of 1.5. In dioxygen there are four electrons in one bond giving a bond order of 2.

Bond length:
Oxygen would have a shorter bond length than ozone.
The single bond length of O – O is 132 pm (Chemistry and chemical reactivity, Kotz & Treichel 4th ed., p.
401), and the oxygen double bond (O = O) for the element O 2 is 120.8 pm (CRC Handbook 60th ed. p. F217). Since ozone consists of a resonance between the single and double bond, the actual length likely is
close to the average between these two values, approximately 126 pm. A reported value of bond lengths in
molecule of ozone can be found in Chemistry and chemical reactivity, Kotz & Treichel 4th ed., p. 392 and is

127.8 pm.
3. It has been suggested that the loss of ozone in the stratosphere could lead to a negative feedback that
might allow more ozone to be produced. Explain why such a feedback is possible. (This ‘self-healing’
does in fact, occur, but only to a very small extent.)
Solution
The loss of O3 high in the stratosphere can be compensated for by increased penetration of highly energetic
solar radiation deeper into the stratosphere and as a result, cause an enhanced rate of ozone generation in
the lower stratosphere. Thus creating more ozone, which then absorbs the UV radiation and prevents it
from penetrating any further through the atmosphere. This type of compensation is called ‘self-healing’
because in this case the increased production of ozone is a result of the initial loss of ozone.
Average ozone concentrations in Jakarta, Indonesia have been reported to be 0.015 mg m-3 and in
Tokyo, Japan are 20 ppbv. What is the approximate ratio of these two values when expressed in the
same units?
Solution
4.

Use PV=nRT and assume conditions of Pº, 25ºC, and 1.00 L for the comparison of values and the
conversion of units (P = 1.01325 x 105 Pa, T = 298 K, R = 8.315 J K-1 mol-1 and V = 1.00 L = 1.00 x 10-3
m3)
n = 0.04089 mol

(total number of mol of gas per L of air)

Unit conversions for ozone; convert the Jakarta concentration of 0.015 mg m-3 to the ppbv units reported for
Tokyo.

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Solutions: Environmental Chemistry - a global perspective 4th Edition

Chapter 3:

Stratospheric chemistry – ozone

given

calculated below

0.015 mg m-3 of O3

from:

to:

7.6 ppbv of O3

Convert to moles per litre:
0.015 mg m-3 ÷ 1000 L m-3 ÷ 1000 mg g-1 = 1.5 x 10-8 g L-1
Therefore there are 1.5 x 10-8 g of O3 per L of air.
Divide by the molar mass (M.M.) of ozone, 47.9982 g mol-1, giving:
1.5 x 10-8 g ÷ 47.9982 g mol-1 = 3.13 x 10-10 mol of O3 per L of air.
The ratio of the number of moles of ozone to the total number of moles is the mixing ratio as a fraction, but
when multiplied through by 109 it can be correctly expressed as a mixing ratio with the units of ppbv.
Mixing ratio = 3.13 x 10-10 mol ÷ 0.04089 mol x 1 x 109 = 7.6 ppbv
Ratio of Jarkarta O3 concentration : Tokyo O3 concentration

(both in ppbv)

7.6 ppbv / 20 ppbv = 0.38
(Note: unit conversion to mg m-3 yields:

5.

0.015 mg m-3 / 0.039 mg m-3 = 0.38)

Using data from Table 3.4, identify the rate-determining step in the catalytic cycle involving hydrogen
and hydroxyl radicals, and determine the overall rate of ozone destruction as a consequence of this
cycle. (Note that the calculation applies to reactions occurring at 30 km only.)

Solution (refer to Table 3.4)
Catalytic cycle involving hydroxyl and hydrogen radicals for ozone destruction:
Reaction 1 •H + O3 → •OH + O2

k = 1.9 x 10-11 molecule-1 cm3 s-1

•OH + O → •H + O2
------------------------------------Net reaction
O 3 + O → 2 O2

k = 2.3 x 10-11 molecule-1 cm3 s-1

Reaction 2

Concentration of species from the data in Table 3.4:
O3
•H
•OH
O

=
=

=
=

2.0 x 1012 molecule cm-3
2.0 x 105 molecule cm-3
1.0 x 106 molecule cm-3
1.0 x 109 molecule cm-3

Rate1 = k[•H][O3]
= 1.9 x 10-11 molecule-1 cm3 s-1 x 2.0 x 105 molecule cm-3 x 2.0 x 1012 molecule cm-3
= 7.6 x 106 molecule cm-3 s-1

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone

Rate2 = k[•OH][O]
= 2.3 x 10-11 molecule-1 cm3 s-1 x 1.0 x 106 molecule cm-3 x 1.0 x 109 molecule cm-3
= 2.3 x 104 molecule cm-3 s-1
The second step (reaction 2) is the rate-determining step because it has the slower rate. The overall rate of
ozone destruction would be equal to Rate2 = 2.3 x 104 molecule cm-3 s-1.
6.

A catalyst may be defined as a substance that enhances the rate of a chemical reaction without being
consumed in the process. By this definition, a catalyst would have an infinite lifetime. The ozone
decomposition catalysts, however, have finite lifetimes. What are possible sinks for removal of the

stratospheric catalysts, NO and •Cl?

Solution
NO can be removed by reaction with the hydroxyl radical which is present in the stratosphere at a
concentration of about 1.0 x 106 molecule cm-3 to produce nitrous acid.
•NO + •OH + M → HNO2 + M
One of the largest sinks for the chlorine radical is HCl, which can be formed in the stratosphere in at least
two ways.
•Cl + •H → HCl
where •H has a concentration of about 1x105 molecule cm-3.
or

•Cl + CH4 → HCl + •CH3

Methane or other similar organic species may leak into the stratosphere where the chlorine radical can
abstract a proton and form HCl.
For both radicals (•NO and •Cl) the products HCl and HNO2 can slowly leak back into the troposphere
where they could be dissolved in water and subsequently ‘rained out’ for removal from the atmosphere.
Therefore, there are processes by which both •NO and •Cl are removed in the stratosphere. Nevertheless,
they can still be considered as catalysts for the destruction of ozone, because they are not consumed in the
specific processes of removing odd oxygen species.
7. Calculate the enthalpy change (Appendix B.2) in the following reaction:

N 2O(g)  O(g)  2NO(g)
(a) when the oxygen is derived from photolysis of nitrogen dioxide;
(b) when it is derived from photolysed ozone.
Comment on the results in terms of the tropospheric and stratospheric lifetimes of nitrous oxide.
Solution
a)


ΔHºf (reaction) = 2 x ΔHºf (NO(g)) – [ ΔHºf (N2O(g)) + ΔHºf (O(g))]

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone

ΔHºf (reaction) = 2 x 90.3 – [82.1 + 249.2]
ΔHºf (reaction) = – 150.7 kJ mol-1
b)

ΔHºf (reaction) = 2 x ΔHº f (NO(g)) – [ ΔHºf (N2O(g)) + ΔHºf (O(g)) + Ee(O)]

Note: the extra energy for the excited oxygen Ee(O) is about 190 kJ mol-1
ΔHºf (reaction) = 2 x 90.3 – [82.1 + 249.2 + 190]
ΔHºf (reaction) = – 340.7 kJ mol-1
8. Using bond energies (Appendix B.3), explain the reaction sequence of ozone-destroying capability in
the stratosphere of hydrocarbons containing the halogens: Br > Cl > F.
Solution
Bond enthalpies can be obtained in Appendix B.3:
kJ mol-1
C – Br
276
C – Cl
338
C–F
484

Of the three bonds, the weakest bond is C – Br and it therefore requires the smallest amount of energy to
break. Longer wavelength radiation penetrates deeper into the atmosphere and will cause the formation of
•
Br, which can react first with ozone where it is more concentrated (lower in the stratosphere). The
strongest bond C – F requires shorter wavelengths (higher energy), which are available at higher altitude
where ozone is not as concentrated. So when the •F radical is produced it will not have as much ozone
available to react with. Any bromine-containing compounds that are present in the stratosphere are potent
ozone destroyers. Chlorine falls between these two and its radical •Cl is also formed where ozone
concentrations can be significantly affected.
9.

HCFC-123 has been proposed as a substitute for CFC-11. How would you expect the following
environmental properties to compare?
(a) tropospheric lifetime;
(b) combustibility;
(c) ozone depletion potential (ODP);
(d) greenhouse gas properties (this can be answered after reading Chapter 8).

Solution
(a) Tropospheric lifetime: HCFC-123 is a more reactive gas than CFC-11 because it contains a H – C
bond. Increasing hydrogen content reduces inertness. It will therefore not be as long lived in the

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone


troposphere as compared to CFC-11 which is a relatively less reactive gas and it eventually will leak into
the stratosphere to a greater extent.
(b) Combustibility:
non-combustible.

Because of the C – H bond, HCFC-123 will be somewhat combustible. CFC-11 is

(c) Ozone Depletion Potential (ODP): HCFC-123 is expected to have a lower ODP since there are only
two chlorine atoms within its structure, compared with three for CFC-11. As well, since HCFC-123 is
more reactive in the troposphere, the likelihood of the same amount of HCFC-123 (as CFC-11) reaching the
stratosphere is not as great and therefore would also act to reduce the ODP.
(d) Greenhouse gas properties: It is expected that HCFC-123 would contribute less to the warming of the
atmosphere because it has a shorter tropospheric lifetime.
10. CFC-114b has a tropospheric lifetime of 189 years. Would you expect CFC-115 to have a longer or
shorter lifetime? Why?
Solution
CFC-114 molecular formula CF2ClCF2Cl
contains: 2 carbons, 0 hydrogens, 4 fluorines, and 2 chlorines.
CFC-115 molecular formula CF2ClCF3
contains: 2 carbons, 0 hydrogens, 5 fluorines, and 1 chlorine.
It would be expected that CFC-115 would have a longer tropospheric lifetime then 189 years since the C –
F bond is stronger than the C – Cl bond and it would be more resistant to breaking down. (See Table 3.1).
Increasing the proportion of fluorine at the expense of chlorine tends to produce highly stable compounds.
11. HCFC-22 is one of the compounds recommended as a replacement for CFCs. Use the following
information to determine its reaction rate and tropospheric residence time. Compare the residence time
with that of CFC–12 which is about 102 y.
[•OH] = 6.6 x 105 molec cm-3 (remains constant)
[HCFC-22] = 0.10 ppbv
k = 4.0 x 10-15 cm3 molec-1 s-1
Solution

Convert the units of ppbv to molec cm-3 for the [HCFC-22]
x / 4.09 x10-5 mol x 1x109 = 0.1 ppbv (where 4.09 x 10-5 mol is the number of moles of gas in 1 cm3)
x = [HCFC-22] = 4.09 x 10-15 mol cm-3

(x 6.022 x 1023)

= 2.46 x 109 molec cm-3
reaction rate

= k [•OH] [HCFC-22]

reaction rate

= 4.0 x 10-15 cm3 molec-1 s-1 x 6.6 x 105 molec cm-3 x 2.46 x 109 molec cm-3
= 6.49 molec cm-3 s-1

Tropospheric residence time

= steady state concentration / reaction rate
= 2.46 x 109 molec cm-3 / 6.49 molec cm-3 s-1

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone
= 3.79 x 108 s


(convert to years, 3.1536 x107 s/y)

= 12 years
The residence time of HCFC-22 is about 1/8th the time compared to CFC-12.
12. The following proposal to repair the ozone layer has been made. 1 The suggestion is to inject ‘negative
charges’ into the lower stratosphere, and these would react with CFCs to produce harmless products.
From your knowledge of basic chemistry, indicate if this process would be theoretically possible, and
discuss the practical requirements of it.
Solution
A possible chemical reaction sequence (using CFC-113 as an example) is:
Cl F
Cl – C – C – Cl
F F

Cl F
+ hν → Cl – C – C•
F F

Cl• +

+ Cl•

e- → Cl-

It seems possible that the reaction sequence shown above could work, theoretically, and this is what the
authors of the idea intended. The practical requirements however seem quite problematic. Given the very
low concentration (mixing ratios) of CFCs (Table 3.1) in the stratosphere, it would be difficult to produce
and distribute the quantity of charge necessary to effect change on a global scale. The rate of mixing in the
stable stratosphere is very small.
The authors of the original article ‘Observation of charge-induced recovery of ozone concentrations after

catalytic destruction by chlorofluorocarbons’ used CFC-l3 as the chlorofluorocarbon example. They
suggested that the removal of the chlorine radicals would occur through the transfer of charge from an ion
or electron to form the chloride ion. Once formed, it would not have the ability to continue in the catalytic
destruction of the ozone.
13. Consider the following data for total atmospheric column ozone measurements (as DU) at three
locations around the Earth, obtained using the TOMS system in 2001.
January 15

April 15

July 15

October 15

Tierra del Fuego (Chile, Argentina)

323

261

339

206

Nairobi (Kenya)

234

273


266 (Aug. 15)

266

Kiev (Ukraine)

321

420

314

273

Assume that these are typical of values that might be obtained in any other year, and discuss the trends
as you move down the columns and along the rows, in terms of your knowledge of stratospheric ozone
behaviour.
1

Chem. Eng. News, May 23 (1994), 36 and Wong, A.Y., D.K. Sensharma, A.W. Tang, R.G. Suchannek,
and D. Ho, Observation of charge-induced recovery of ozone concentration after catalytic destruction by
chlorofluorocarbons, Phys. Rev. Lett., 72 (1994), 3124.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone


Solution
The January 15 column shows the expected trend of typical values being higher in the Southern and Northern
Hemispheres and lower near the Equator. October 15th’s trend shows the possible influence of the spring-time
‘ozone hole’ over Antarctica, causing the DU value for Tierra del Fuego to be much lower, while the other two
locations have DU values that are still reasonable the same as earlier.
The trends going across the rows for each location, indicate that it is only in the Southern Hemisphere where
ozone loss is significant and only at specific times of the year. Both the Equatorial and Northern Hemisphere
show reasonably consistent DU values throughout the year.

14. According to the original Montreal Protocol, Australia’s ozone cap on CFCs and related compounds
was set at 548 tonnes (t) after 1996 (this value was later modified as described in the text). In 1998, use
of CFCs had been completely phased out and amounts of HCFCs were reported to be:
HCFC-22

2820 t

HCFC-134a

1700 t

HCFC-141b

442 t

HCFC-123

35 t

Do these quantities meet the original terms of the agreement?

Solution
Yes, the new compounds, when calculated with CFC-11 equivalents meet the original terms of the
agreement.
ODP
HCFC-22
2820 t x 0.05 = 141 t
HCFC-134a
1700 t x 0.0
=
0t
HCFC-141b
442 t x 0.12 = 53 t
HCFC-123
35 t x 0.01 =
0.4 t
-------total
= 194.4 t
15. The synthesis and decomposition of ozone can be described using ‘oxygen-only chemistry’ as shown
in the sequence of reactions 3.2–3.5. Explain the nature and significance of the change from O 2 and O
species in Reaction 3.3 to O2* and O* in Reaction 3.4.
Solution
The significant difference between O2 and O in Reaction 3.3 and O2* and O* in Reaction 3.4 is the energy
state that these species are in. Ozone is only capable of producing either both ground state species or both
excited state species. The energy required to produce ground state species is low, only 106.5 kJ mol -1. The
corresponding wavelength of 1123 nm is not able to be absorbed by ozone (see ozone absorption crosssection in Figure 3.3) and this process of producing the ground state oxygen species does not occur.
However, the reverse Reaction (3.3) occurs readily. In Reaction 3.4 higher energy photons are able to

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 3:

Stratospheric chemistry – ozone

photo-dissociate ozone and produce the excited state oxygen species. This allows for the absorption of the
wavelengths of radiation in the range of 230-320 nm to be absorbed before reaching the surface of the
Earth. As well, the excited species, one produced, are able to react and become involved in other chemical
processes. One significant process is the formation of the hydroxyl radical as a result of O* reacting with
water vapour.
16. What are the main natural and anthropogenic sources for the production of HOx, NOx, and ClOx radical
species? Describe any connections between the natural and anthropogenic sources of the various
species.
Solution
Natural

Anthropogenic

HOx

water

smog, methane

NOx

N2O (lightning)

N2O (farming)


ClOx

CH3Cl (biological),

CFCs

17. Although the name of a null or ‘do nothing’ cycle, related to ozone depletion, sounds in effect harmless,
describe the potentially harmful effects that are associated with these types of cycles.
Solution
The null cycles allow for the accumulation of radical materials in a combined form that can be unleashed at
some time in the future. They do not account for actual removal of the harmful species. A consequence of
the release of these stored radicals is the development of the Antarctic ‘ozone hole’.
18. Use the webpage given in Additional Resources 7 and review the monthly and annual changes in the
ozone over the South Pole. How has the ‘ozone hole’ changed over the past 10, 20 and 30 years?
For classroom discussion, no solution provided. Is the Montreal protocol working? Is there clear evidence of
any trends? What are some of the limitations from this data? Use the web-based resource and show in class how
the ozone over the South Pole has evolved over time.
Extra Problem:
19. Using the webpage given in Additional Resources 7 investigate the state of ozone above where you live.
How does the ozone concentration vary during the past year. How does this compare to the variation
seen each year at the South Pole?
No Solution provided.
For classroom discussion. What can you do to reduce the impact we have on the ozone in the stratosphere?

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