Download Practice Problems:

Practice Problems:
For 9/16/09
From the book:
Chapter 1: 4, 6, 9, 21, 25, 28, 32, 40, 44, 46, 49, 51, 56.
Appendix E of the text has answers to even numbered questions
9. Anaerobic bacteria are able to metabolize carbon without the use of oxygen, and
usually cannot live in an oxygen environment.
21. Producers are organisms that provide their own food, usually through photosynthesis.
Consumers must consume other organisms as a food source.
25. The First Law of Thermodynamics states that the energy of the universe is
conserved. An example of an energy transformation would be burning hydrocarbons.
Chemical energy is transformed into thermal energy.
44. a.
b.
1L H 2O 1000g 10g NH4 1mol
 5.6 x104 M
6
1L 10 g H 2 O 18g
5.4x10 -6 mol CaCO3 40.08g 1 L
10 6  0.22ppm
1L
1mol 1000g
49. a. The ocean acts a sink for atmospheric carbon dioxide. The ocean’s buffer
capacity means that it is able to uptake a large amount of CO2 from the atmosphere.
Carbonaceous sediments form on the ocean floor. These sediments are a long lived
reservoir for carbon.
b. Limestone makes up a long-term reservoir for carbon. Limestone is formed in
the oceans, and is broken down when plate tectonics exposes the limestone to chemical
weathering .
c. Crude oil is made up of decayed biological materials. In the earth’s crust, crude
oil represents a long-lived reservoir for carbon.
51. Nitrogen fixation is the process that makes atmospheric nitrogen available to plants.
a. The bacteria that bring about this process can be found in soils or attached to the
roots of certain plants.
b. These bacteria initially produce ammonia.
Other problems:
Human activities can have a major influence on which of the following compounds in the
atmosphere on time scales of a few years? Explain your reasoning for each compound.
O2, CO2, N2.
In the short-term, human activities can only have a major effect on the atmospheric
concentration of CO2. The lifetime of carbon in the terrestrial
biosphere/atmosphere/surface ocean system is on the order of 70 years, by changing the
flux between any of these reservoirs, or by changing the inputs of carbon into any of
these reservoirs, we can have an influence on the atmospheric concentrations. O2 and N2
are controlled by processes that occur over millions of years. It is not probable that
human activities could affect the atmospheric concentrations of these gases over ‘short’
time scales.
Why does the ocean resist changes in pH?
The oceans have a HCO3-/CO32- buffer system that allows the uptake of acid (CO2)
without changing the pH very much.
Using the pre-industrial carbon cycle figure presented in class, calculate the lifetime of
carbon in the (atmosphere + terrestrial biosphere + soil) reservoir relative to uptake by
the ocean.
Referring to the figure: (2000+730+615)TgC / 60TgC/yr = 56yrs.
Is the following statement reasonable? Reducing atmospheric CO2 concentrations by
planting more trees is not effective because trees are a short-term reservoir for carbon.
To some degree this statement is true. It is possible to sequester carbon in the terrestrial
biosphere, however, depending on the specific ecosystem, the lifetime of the reservoir
may be anywhere from 1 year to a maximum of several thousand years. These reservoirs
would have to be well-managed. One fire could put all of the sequestered carbon back in
the atmosphere in a very short time.
A developer wants to raise atmospheric O2 concentrations by planting managed forests
that will be cut down every couple decades, sealed in weighted leak-proof containers, and
dropped into the deep ocean. Will this plan work? Explain.
No, this plan would not work. Raising atmospheric O2 levels by converting atmospheric
CO2 to O2 through photosynthesis is limited by the amount of CO2 in the atmosphere.
Using this method, one could only raise O2 levels by a theoretical maximum of 370ppmv.
This is almost nothing relative to the 20% O2 currently found in the atmosphere.
For 9/16/09
From the book:
Chapter 3: 3, 5, 8.
Chapter 4: 5, 10, 11, 12, 31, 37, 39, 40, 41.
Appendix E of the text has answers to even numbered questions
Chapter 3
3. The pressure of the atmosphere drops off exponentially with altitude (P=Poe-z/H). The
pressure at sea level is approximately 1000mbar. The pressure at the tropopause is
approximately 200mbar. Only 20% of the mass of the Earth’s atmosphere is contained in
the stratosphere.
5. The amount of water vapor in the Earth’s atmosphere generally lies between 1 and 3%
Chapter 4:
5. A free radical has one unpaired electron. A hydroxyl radical is a free radical; it has a
neutral charge. Hydroxide ion has a full octet and has a negative charge.
11. The Lewis electron structure for NO:
.. ..
: N - O.
.. ..
There is one unpaired electron because NO is a radical compound.
31. Tall smokestacks release pollution into better mixed air, and commonly above the
nighttime boundary layer. This means that the pollution is dispersed better than if it were
released near the ground. All else being equal, it is good to disperse a pollutant rather
than have it concentrated in one area. However, the best thing would be to reduce
emissions overall.
37.
NO2  h 
 NO  O
O  O2  M 
 O3  M
NO2 + H2O to make 2OH, as the book describes it, does not really happen. Additional
OH, in addition to the photolysis of O3, can be produced in the troposphere through the
photolysis of formaldehyde.
HCHO  h 
 H  HCO

 H 2  CO
HCHO  OH 
 HCO  H 2 O
HCO  O 2 
 HO 2  CO
HO2 is produced without using up OH.
39.
OH  NO2 
 HNO 3
HO 2  HO 2 
 H 2 O 2  O 2
OH  CO, or CH 4 , or VOC
41.
j1
O3  h 

O 2  O(1 D);   320nm
k2
O(1 D)  H 2O 
2OH
Other Problems:
Consider xenon, an un-reactive atmospheric gas whose mixing ratio is 900pptv. What is
the concentration of xenon at 1010 mbar (sea level) and at 200 mbar (tropopause)?
Assume the temperature of the tropopause is 190K.
Use PV=nRT
At 1010 mbar, assume T = 300K
n
PV
(1atm)(0.00 1L)
 900 

 4.06 x10 5 moles  12   3.66 x10 14 mol Xe
-1
-1
RT (0.08206L atm mol K )(300K)
 10 
 6.022x10 23 molec
3.66 x10 14 mol Xe
1 mol


  2.2 x1010 molecules cm -3

At 200 mbar, assume T = 190K
n
PV
(0.2atm)(0. 001L)
 900 

 1.28 x10 5 moles  12   1.15 x10 14 mol Xe
-1
-1
RT (0.08206L atm mol K )(190K)
 10 
 6.022x10 23 molec
1.15 x10 14 mol Xe
1 mol


  7.0 x109 molecules cm -3

Why is the troposphere well-mixed, but the stratosphere is not?
The troposphere is heated form the Earth’s surface. Turbulent convection mixes
tropospheric air vertically on time scales of a month. The stratosphere is heated from
above due the photodissociation of O2. This temperature inversion leads to very stable
stratification of the air in the stratosphere.
How can it be that the lifetime of air in the troposphere with respect to transport to the
stratosphere is 7.4 years, but the lifetime of air in the stratosphere with respect to transfer
to the troposphere is 1.3 years?
This is possible because the stratosphere only contains 20% of the mass of the
atmosphere. Even though the mass of air exchanged between the troposphere and
stratosphere is the same (conservation of mass). To calculate a lifetime, we divide the
mass in the reservoir by the mass transported. In the case of the troposphere and
stratosphere, the mass of the stratosphere is 20% the mass of the atmosphere, and the
troposphere is 80% the mass of the atmosphere.
What factors affect the production of OH in the troposphere?
Ozone concentration, water vapor concentration, pressure, and the actinic flux of light
less than 320nm, all affect the production of OH in the troposphere.
It seems that you need to have ozone to make ozone. Where does the initial ozone come
from?
The initial ozone would have been transported from the stratosphere.
Explain how NOx and HOx act as catalysts for ozone production in the troposphere.
NOx cycles between NO and NO2. Each time NO2 is formed without the oxidation of NO
by O3, there is the potential for new ozone formation. This usually occurs by the
oxidation of NO by HO2. HOx cycles between OH and HO2. OH oxidizes reduced
carbon in the atmosphere. The product of this reaction is peroxy radicals and HO2. HO2
(and peroxy radicals) facilitate the production of ozone by oxidizing NO to NO2. The
oxidation of NO also represents the reduction of HO2 back to OH; therefore, completing
the catalytic cycle.
The overall lifetime of CH3CCl3 (methyl chloroform) in the atmosphere is 5.7 years and
is controlled by oceanic uptake (40 years), transport to the stratosphere (30 years), and
reaction with OH. Calculate the lifetime of CH3CCl3 due to reaction with OH in the
troposphere. Calculate the concentration of OH if the rate constant for the reaction
between CH3CCl3 and OH is 6.7x10-15 cm3 molecules-1 s-1.
Lifetime of CH3CCl3 with respect to OH:
1
1
1
1
1
1
1
1
1
1
1
1











  OH  8.5years
 tot  ocean  strat  OH
 OH  tot  ocean  strat  OH 5.7 40 30
Concentration of OH; 8.5years = 2.7x108s
1
1
1

 [OH] 
 [OH] 
 5.6x10 5 molecules cm -3
-15
3
k[OH]
k
6.7x10 cm molecules 1s 1 (2.7 x108 s)