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Chemistry 321 Environmental Chemistry

Atmospheric structure and composition

The Ozone Layer

Sept 21, 2010

References: Chapter 1, CB/MC Daniel Jacob "Atmospheric Chemistry

Free on-line text

http://www.as.harvard.edu:16080/ctm/publications/jacobbook/Correction:

In class, I incorrectly assigned the oxidation state of S in H3C-S-CH3 as 0.

The correct oxidation state of S in H3C-S-CH3 is -2.

My apologies for this error.

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Previous lecture

abundance of elements in Earth's crust Global biogeochemical element cycles examples: N, S, C

GBECs can change: e.g., phytoplankton fertilization

CPS-1

Main message from previous lecture: Man and other processes can

change/alter theses GBECs

CPS-1: What are the most abundant elements by mass, in the Earths

crust? Answer: Aluminum, Silicon, Oxygen

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Objectives

Following this unit, you are expected to be able to:

Define the terms troposphere, tropopause, stratosphere, stratopause convert between number densities (concentrations) and mixing ratios calculate number density of air (using ideal gas law) sketch approximate profiles of temperature, pressure, water, and ozone in

the atmosphere

apply the Beer-Lambert law to calculate absorption and scattering of light describe how absorption by oxygen and ozone attenuate sunlight within the

atmosphere

describe the potential impact of missing stratospheric ozone

Please review 1st year material: ideal gas law (partial pressures) Kinetics: rate laws; reaction mechanisms, activation energy, catalysis

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12756km

The sky is blue.

this process is called Rayleigh scattering

C. Frohlich and G. E. Shaw,Appl. Optics19, 1773 (1980).

A. Bucholtz,Appl. Optics34, 2765 (May 20, 1995)

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Example calculation of absorption and scattering (I)

Calculate the distance that light with wavelength 150 nm can travel

through an atmosphere consisting of pure O2 at STP

before it is reduced to 0.1% of its initial intensity.

Given: abs (150 nm, O2) = 1.010-17 cm2 molecule-

1

Rayleigh (150 nm, air) = 1.610-24 cm2 molecule-1

Given: I/I0 = 0.001 ( ) LI

I airRayleighabsabs

0 e001.0

+

==

nn

since e-a e-b = e-(a+b)

What's nair at STP?

use ideal gas law:PV = nRT n = number of molecules

n = n/V = P/RT n = number density of molecules

and A = Avogadro's number (6.0231023

molecules mol-1

) and R = 8.314 J mol-1

K-1

nair = (A/R) (P/T)

= (6.0231023 molecules mol-1 / 8.314 J mol-1 K-1) (1.00105 Pa/273.15 K)

= 2.651019 molecules cm-3

STP (IUPAC, 1990) 273.15 K, 1.00105 Pa

1 J = 1 N m

1 Pa = 1 N m -2

( ) ( )

cm026.0

cm102.4cm1065.2

9.6)001.0ln(L

1512

airRayleighabsabs

=

+

=

+

=

nn

Goes virtually nodistance in atmos.

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Example calculation of absorption and scattering (II)

Calculate the distance that light with wavelength 200 nm can travel

through an atmosphere consisting of pure O2 at STP

before it is reduced to 0.1% of its initial intensity.

abs (200 nm, O2) = 1.010-23 cm2 molecule-1

Rayleigh (200 nm, air) = 3.610-25 cm2 molecule-1

I/I0 = 0.001 ( ) LI

I airRayleighabsabs

0 e001.0

+

==

nn

since e-a e-b = e-(a+b)

What's nair at STP?

use ideal gas law:PV = nRT n = number of molecules

n = n/V = P/RT n = number density of molecules

and A = Avogadro's number (6.0231023

molecules mol-1

) and R = 8.314 J mol-1

K-1

nair = (A/R) (P/T)

= (6.0231023 molecules mol-1 / 8.314 J mol-1 K-1) (1.00105 Pa/273.15 K)

= 2.651019 molecules cm-3

STP (IUPAC, 1990) 273.15 K, 1.00105 Pa

1 J = 1 N m

1 Pa = 1 N m-2

( ) ( )

m250

cm105.9cm1065.2

9.6)001.0ln(L

1614

airRayleighabsabs

=

+

=

+

=

nn

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The sun emits "black body" radiation at T=5770 K.

The Earth "sees" (more or less) collimated light.

Planck equation

Spectrum of the sun in the upper atmosphereIntensity of the radiation emitted by a black bodyas a function of wavelength

D.A. McQuarrie and J.D. Simon Physical Chemistry: A molecular approach (1997), Fig. 1-2 and 1-3

Solar radiation

Q13: What colour is the maximum emission intensity of a Planck black

body at 5770K?

A = yellow

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Electromagnetic radiation: Dissociation of O2

Energy of radiation E = h = h c/

h = Planck's constant = 6.62610-34 J s = frequency (s-1)

c = speed of light (m s-1) = 2.99108 m s-1

= wavelength (nm)

Energy required to dissociate oxygen:

O2 + h 2 O, Hrxn = 498.4 kJ mol-1

Hrxn = 498.4 kJ mol-1 (1 mol / 6.0231023 molecules) = 8.310-19 J molecule-1

minimum photon energy required to dissociate 1 molecule of O2

min = hc/E = 6.62610-34 J s 2.99108 m s-1 / (8.310-19 J molecule-1)

= 240 nmQ14/CPS7:

Where in the atmosphere would you expect radiation with 240 nm?

A = high up (upper stratosphere)

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Herzbergcontinuum

Schumann-Rungebands

Schumann-

Runge

continuum

Ionizationcontinuum

O2

Absorption of radiation by O2 and O3

CB/MC

Schumann-Runge continuum/band and Herzberg continuum:

O2 + h 2 O

Source of atomic O in upper atmosphere!

Hartley band

O3Absorption spectrum of O2

absorptio

ncros

s-section

(cm

2

molecul e

-1)

O3 + h O2 + O

Sink of O3

in upper

atmosphere!

O3

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Absorption of radiation by N2, O2 and O3 in the upper

atmosphere

vacuum UV

CB/MC

Little UV B or C

at surface

Only light >300nm makes it to surface, good b/c ozone

mostly absorbed

Q15: What if there was no O3 in the stratosphere?Ans: molecules would break down (ie. DNA plants)

Finlayson-Pitts, B. J., and J. N. Pitts (2000), Chemistry of the upper and lower

atmosphere: theory, experiments, and applications, Academic Press, San Diego,

Calif.

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Impact of excess UV radiation

CB/MC

What if there was no O3 in the stratosphere?

DNA mutations, skin cancer (melanoma), death cataracts crop failure

at the

surface