Chapter 2 Notes
Chapter 2 is a discussion of the problem of stratospheric ozone and its
depletion by photochemical reactions with chlorine containing molecules.
the kinetics and details of the photochemistry are quite complicated and the
chapter should be read several times to get a full appreciation of the
chemistry. Chapter 3 deals with chemistry in the troposphere. The major
regions of the atmosphere are shown below.
Aside on ozone
Ozone (O 3) is an allotrope of elemental O. Ordinary oxygen is a
diatomic molecule, O2 where
Ozone has the structure
Ozone is very reactive, acting as a strong oxidizing agent. Ozone has a
distinctive pungent odor, that is frequently encountered around electrical
machinery or thunderstorms.
The following graph shows an expanded plot of the “ozone layer”
The following graph is a quick summary of the reactions responsible for the
Antarctic ozone hole.
Health Effects of Ozone Depletion
Aside on the mathematics of ozone depletion.
Assume the transmission of UV light through the stratosphere follows
Beers law.
extinction coefficient
I0 ≡ incident intensity, I ≡ transmitted intensity
For the stratosphere
l ~ 0.34 cm
εl
1
10
λ
310 nm
290 nm
Result: a 1% decrease in l causes a 1 % decrease in T at 310 nm
10% decrease in T at 290 nm
If O3 is decreasing at 0.4% / year, the O3 loss per decade is 4%. This
translates into a 12% change in T at 300nm, and a 40% change in T at 290
nm.
Politics of ozone depletion.
The attached article from C + E News discusses on of the bad aspects
of ozone depletion, the misunderstanding of the scientific data by right wing
politicians. The article represents the mainstream of scientific thinking
about ozone.
In – Chapter Problems
Most in chapter problems in Baird are very nice. They illustrate the
points make in the chapter and expand upon them. You should work each of
them as you read the material to insure that you really understand the
material. Attached are my solutions for the problems in Chapter 2.
2-1
E = 119627/ λ (nm)
a) λ = 280, E = 427.2 kJ/mol
b) λ = 400, E = 299.1 “
c) λ = 750, E = 159.5
“
d) λ=4000, E = 29.9
“
UVC / UVB boundary
UVA / vis. boundary
vis-red / IR boundary
thermal IR
2-2
O3 O2 + O
∆H = 105 kJ/mol
E = 119627
λ
λ = 119627 = 119627 = 1139 nm
E
105
This is IR radiation
2-3
NO2
NO + O
∆H = 90.2 + 249.2 – 33.2 = 306.2
λ = 119627 = 390.7 nm
306.2
NO2
N + 2O
∆H = 2 (249.2) + 472.7 – 33.2 = 937.9
λ = 119627 = 119627 = 127.5 nm
E
937.8
2-4
Relative concentration will peak at higher altitudes because O3 is
produced in upper part of stratosphere.
2-5
O3 hv O2* + O*
∆H = 105 + 190 + 95 = 390 kJ/mol
λ = 119627 = 307 nm
390
2-6
a)
b)
c)
d)
e)
f)
g)
h)
i)
OH•
CH3•
CF2Cl•
H3COO•
H3CO•
ClOO•
ClO•
HCO•
NO•
a)
c)
b)
d)
e)
f)
g)
h)
i)
2-7
X + O3
XO hv
O + O2
mall cycle
XO + O2
X+O
O3
H2O + O
2 OH
2-8
O*
69
O
obviously a smaller activation barrier
2-9
HOH + O*
2 OH•
2-10
rate = k [O*][CH4]
3 x 10-10 [102][1011] = 3 x 103 molecules/ cm3sec
g/yr = 3 x 103 + π x 107 +
16
= 2.5 x 10-12
#s/year 6.02 x 1023
2-11
Cl2 2 Cl•
∆H = 121.7 kJ/mol x 2
λ = 119627 = 491 nm
visible light
2 x 121.7
2-12
X + O3
X′ + O3
XO + X′O
2 O3
XO + O2
X′O + O2
X + X′ + O2
3 O2
Cl• + O3
Br• + O3
BrO• + ClO•
2 O3
ClO• + O2
BrO• + O2
Br• + Cl• + O2
3 O2
2-13
The dichloroperoxide rxn will be more important due to its faster rate.
2-14
At low concentrations, the probability of forming the diheloperoxide
is too small.
2-15
Cl• + O3
OH• + O3
ClO• + HOO•
HOCl
2 O3
ClO• + O2
HOO• + O2
HOCl + O2
OH• + Cl•
3 O2
CFC – 12
12 + 90 = 102
1 C, 0 H, 2 F
CF2Cl2
CFC – 113
113 + 90 = 203
2 C, 0 H, 3 F
123 + 90 = 213
2 C, 1 H, 3 F
134 + 90 = 224
2 C, 2 H, 4 F
C2F3Cl3
2-16
CFC – 123
CFC – 134
C2F3HCl2
C2F4H2
2-17
CH3CCl3
CCl4
CH3CFCl2
230 – 90 = 140
100 – 90 = 10
231 – 90 = 141
CFC – 140
CFC – 10
CFC – 141
2-18
Reaction is energetically unfavorable, therefore slower.
2-19
a)
CF3O•
b)
CF3O• + O3
CF3OO• + O
O3 + O
CF3OO• + O2
CF3O• + O2
2 O2
(a)
F• + O3
FO• + O
O3 + O
FO• + O2
F• + O2
2 O2
(b)
F• + O3
FO• + O3
2 O3
FO• + O2
F• + 2 O2
3 O2
(c)
X + O3
XO• + O3
2 O3
XO• + O2
X + 2 O2
3 O2
2-20
2-21
It is easier to get hydrogen abstraction reactions to de-activate
chlorine.
2-22
It implies the species have short lifetimes.
Problem 2-23
yes
no
no
yes
no
2-24
a)
b)
c)
d)
e)
all except NO2•, no driving force in bond formation
all
yes
ClO•, BrO•, NO 2•
O3, HOO•, ClO• (with light), BrO• (with light)
a)
Y–O + O
BrO• + O
ClO• + BrO•
BrO• + BrO•
Y–O + sunlight
Br–O + sunlight
2-25
b)
c)
d)
2-26
Y + O2
Br• + O2
Cl• + Br• + O2
BrOOBr hν
2 Br• + O2
Y+O
Br• + O
a)
ClO• + NO 2•
b)
c)
2 ClO•
ClOOCl
ClO• + O
Cl• + O2
ClO• + sunlight
Cl• + O
ClO• + NO•
Cl• + NO 2•
2 ClO•
ClOOCl hν
2 Cl• + O2
ClO• + BrO•
Cl• + Br• + O2
sunlight
ClONO 2
2-27
should have loose oxygen in that O is bonded to F
that has one or more nonbonding electron pair.