Environmental Engineering Chapter 17 Homework How do nitrogen and chlorine catalyze the destruction of ozone?

Ozone Depletion

Learning Objectives

After reading this chapter, students should be able to:

• Know what the effects of UVA, UVB, and UVC radiation are on life.

• Know that UVC radiation is mostly absorbed by ozone.

• Know the general shape of the ozone profile.

• Realize how the solar zenith angle affects the effectiveness of the ozone shield.

• Understand the chemistry behind ozone production.

• Understand the importance of the production and destruction of odd oxygen to

ozone levels.

• Know how the Chapman mechanism works.

• Understand how catalytic cycles involving nitrogen and chlorine decrease ozone

concentrations.

• Realize that anthropogenic fluxes of chlorofluorocarbons into the atmosphere are

increasing Cl concentrations and that this increase is leading to decreased ozone

concentrations.

• Understand the importance of polar stratospheric clouds in the formation of the

ozone hole.

• Explain the seasonal cycles in Antarctic ozone concentrations.

• Realize what implications ozone depletion has for the mid-latitude areas of the

planet.

• Discuss the success of the Montreal Protocol and subsequent treaties.

• Realize some of the problems with the freon substitutes currently being used.

Review Questions

1.) What are the three categories of UV radiation? Which of these are considered to

be biologically harmful?

2.) What is ozone column depth? In what units is it measured?

CHAPTER

17

2

3.) What are the four chemical reactions that comprise the Chapman mechanism?

Which ones affect odd oxygen?

These are the four reactions in the Chapman mechanism, with those producing

odd oxygen circled and those eliminating odd oxygen boxed:

4.) What is a catalyst?

5.) How do nitrogen and chlorine catalyze the destruction of ozone?

Nitrogen eliminates ozone through the following cycle:

6.) What role do polar stratospheric clouds play in the formation of the Antarctic

ozone hole?

7.) Why is a springtime hole observed over the Antarctic but not usually over the

Arctic?

The springtime hole results from a release of Cl from cloud surfaces as the clouds

heat up. It does not occur in the Arctic because the polar vortex there is not as

3

8.) What is the long-term trend in mid-latitude ozone column depths? Can it be

explained by the observed increase in stratospheric chlorine and bromine?

,

9.) What strategies have been adopted for reducing or eliminating the use of freons?

The Montreal Protocol and its amendments have limited the production of freons.

Critical-Thinking Problems

1.)

a. According to Figure 17-5, the maximum ozone column depth occurs at high

northern latitudes during late winter. The column depth there is 460 DU. How

many molecules of ozone per square centimeter does this correspond to?

1 DU is equivalent to 0.001 atm-cm. The book tells us that 1 atm-cm is equal to

2.687 x 1019 molecules/cm2. Thus,

b. The minimum ozone column depth, which occurs in the tropics, is 240 DU. How

many ozone molecules are in a 1-cm2 vertical column there?

This calculation is done in the same way:

4

c. What was the approximate ozone column depth (in Dobson units) over Wallops

Island, Virginia, at the time of the measurements shown in Figure 17-3? (Hint:

Ozone column depth is ozone concentration [in molecules/cubic centimeter]

multiplied by the height of the column. You may want to use a different average

ozone concentration for the troposphere and the stratosphere.)

We will follow this suggestion, and calculate the ozone column depth separately

for the troposphere and stratosphere, and then add these two column depths

together to get a column depth for the atmosphere. To calculate the column

depth, we will multiply the concentration of ozone in one part of the atmosphere

by the height of that layer.

The concentration of ozone in the stratosphere is ~ 3.0 x 1012 molecules/cm3, and

the height of the stratosphere where O3 is high is ~ 30 km = 0.3 x 107 cm. Thus,

Thus, the total column depth is:

2.) The absorption of solar ultraviolet radiation of a given wavelength l by

atmospheric ozone follows Beer’s Law:

a. The wavelength region where changes in the solar UV flux have the most

potential to do harm is around 290 nm, in the UVB range. The absorption

coefficient of ozone at this wavelength is about 10 atm-1-cm-1. The average

column depth of ozone from the ground up to the top of the atmosphere is about

0.3 atm-cm. By what factor is the incident solar UV flux at 290 nm attenuated

today – that is, what is the current value of F/F0 at ground level? Evaluate your

answer for a solar zenith angle of 45°.

b. If the ozone column depth were to be reduced by 1% as a consequence of

increasing concentrations of chlorofluorocarbons, what would be the resulting

percent increase in the ground-level UV flux at 290 nm? What about for an ozone

decrease of 10%? 50%? Assume a solar zenith angle of 45° in each case.

6

We can find this by dividing the equation for F/F0 for the decreased ozone levels

by the equation for F/F0 for the current ozone levels:

For a 1% decrease in ozone levels, Nfuture – Npresent = -0.003 atm-cm, so the ratio

of the F values will be:

For a 10% decrease in ozone levels, Nfuture – Npresent = -0.03 atm-cm, so the ratio

of the F values will be:

7

c. Suppose that we decided that we could tolerate a 10% increase in the UV flux at

290 nm for θ = 45°, but no more. What would be the maximum percentage

decrease in ozone that we could allow? (Hint: The increase of the exponential

function is called the natural logarithm, abbreviated as ln. If y = exp(x), then x =

ln(y).)

This time, we will be solving for the change in N. A 10% increase in the UV flux

implies that:

We can now solve for the change in N:

The % change in N is:

3.) New York City is at 43° N. Miami, FL, is at 25° N. In March, when most

colleges have their spring break and students go to Florida, the ozone column

depth is about 280 DU over Miami and 359 DU over New York. Using the data

8

First, examine the figure below (which is not to scale):

This figure shows that the solar zenith angle for each city (at local noon at equinox)

should have the same value as its latitude does. Therefore, we can use the latitude values

for θ in the F/F0 equation. For New York City, which has 359 DU of ozone, this means

F/F0 is given by:

9

and for Miami, FL, which has 280 DU of ozone, this means that F/F0 is given by:

Resource Guide

Video/Film:

Ozone 9-12

The Oxygen Partnership: Air and Evolution

Films for the Humanities and Sciences

More than simply a blanket of breathable gas, the atmosphere acts as both a heat engine

and a solar shield. This program investigates the relationship between the evolution of

life on Earth and the appearance of oxygen in the atmosphere, the development of the

ozone layer, and the implications of shifts in the balance of atmospheric components. (20

minutes, color)

Ozone: Chlorofluorocarbons and Ozone Depletion

Films for the Humanities and Sciences