CHEM/ENVS 380 F14, Midterm Exam 14 Oct 2014 Chem/Envs 380 F14 Midterm Exam Answer Key PART-­‐A. Multiple Choice Questions (5 points each): Each question may have more than one correct answer. You must select ALL correct answers, and correct answers only, to receive full credit. 1. Which of the following are redox reactions? In (b) and (e), the oxidation number of Ca is +II. a. 2Fe + O2 + 2H2O à 2Fe2+ + 4OH-­‐ b. Ca(OH)2 + SO2 à CaSO3 + H2O c. Zn + Cu2+ à Zn2+ + Cu d. NH3 + H2O à NH4+ + OH-­‐ e. CaCO3 à CaO + CO2 f. 2CO + O2 à 2CO2 g. 2Pb + O2 + 4H+ à 2Pb2+ + 2H2O h. 2NO + 2H2 à N2 + 2H2O i. HgS + O2 à Hg + SO2 2. Formation of stratospheric clouds (PSCs) in the polar vortex: a. increases the stratospheric concentration of Cl-­‐containing ODSs such as CFCs b. speeds up the conversion of reservoir molecules (gases) such as HCl into reactive gases such as Cl2 c. speeds up the conversion of ODSs such as CFCs into reservoir molecules such as HCl d. speeds up the conversion of ODSs such as CFCs into catalytically reactive form e. is a necessary step in the formation of the ozone hole over the poles 3. Select those that are true about O3 found in photochemical smog. a. It is produced in the same manner as stratospheric O3. b. It is a major primary pollutant. c. It is not a local pollutant; it is instead a regional to hemispheric pollutant. d. Its occurrence is curbed by presence of ozone-­‐destroying substances (ODSs) such as CFCs. e. None of the above. 4. The concentration xenon (Xe) in our present atmosphere is 0.087 ppm. This means that: a. the concentration of Xe in the atmosphere is 870 ppb b. there are 0.087 µmoles of Xe in 1 mol of air c. there are 0.087 L of Xe in 106 L of air at some fixed temperature and pressure d. there are 0.087 µg of Xe in 1 mol of air e. if the total pressure of the sample is 1 atm, then the pressure exerted by Xe in this sample is 0.087 µatm 5. 2.35 µg of a plant-­‐growth regulator called diphenylamine (DPA) are found in a sample of apples weighing 1.00 g. This means that: a. you will ingest 235 µg of DPA by eating 100 g of these apples b. you will ingest 235 µmol of DPA by eating 100 g of these apples c. in every liter of these apples, you will find 2.35 µL of DPA d. the concentration of DPA in these apples is 2350 ppb e. the concentration of DPA in these apples is 2.35 ppm 1 PART-­‐B. Short Answer Questions 6.
The following questions are on stratospheric O3. Note that this question continues on to the next page. a. To the right, sketch a vertical profile of O3 over the Antarctic for the time of year when the ozone hole is absent [5 pts]. Clearly label the ozone layer [2 pts]. 14 Oct 2014 Final Score __________/100 Ozone Hole Absent
50 km
stratopause
altitude
CHEM/ENVS 380 F14, Midterm Exam ozone layer 15 km
tropopause
b. Please describe the processes that produce and destroy stratospheric O3 under the conditions given in (a) [19 pts]. 0 km Ozone (arbitrary units)
Production Pathway(s) Stratospheric ozone is produced by photolysis of O2 by short-wave radiation (240 nm or shorter) and the reaction
of resulting atomic O with intact O2:
O2 + hν à 2O
O + O2 + M à O3 + M + heat
where hν represents photons with wavelengths of 240 nm or shorter
where M is a third body (typically N2 of another O2)
5 points total for this box
Destruction Pathway(s) Stratospheric ozone is naturally destroyed through (a) catalytic and (b) non-catalytic pathways.
Non-catalytic destruction occurs in the following two ways (these are the loss mechanisms in the Chapman
Cycle):
O3 + hν à O2 + O*
where hν represents photons in the UV-B or shorter range
O3 + O à 2O2
The second reaction is slow in the absence of catalysts.
Ozone is also destroyed through reaction with X catalysts: HO, NO, Cl, Br. All of these catalysts exist naturally
in unpolluted air. Under conditions where ozone hole is absent (no vortex; sunlight available), destruction cycle-1
is the primary mechanism responsible for removing ozone in the stratosphere:
O3 + X à XO + O2
XO + O à X + O2
14 points total for this box (7 points each for description of non-catalytic and catalytic cycles)
2 CHEM/ENVS 380 F14, Midterm Exam 14 Oct 2014 c. (Question 6 continued) Referring to the answers you gave in (b), which production or destruction pathway(s), if any, have been affected significantly by human activity, and how? Identify production or destruction pathway affected by human activity [4 pts]: The catalytic loss pathway has been significantly affected by human activity. Catalytic cycle-1 described in the
answer to (b) is affected, as well as cycle-2 and -3 which are responsible for the appearance of the ozone hole over
the Antarctic.
Describe specifically how human activity affects the above [5 pts]: Emissions at ground-level of ozone depleting substances (ODSs) containing chain carrier X (X being NO, Cl and
Br) have led to increased levels of these X catalysts in the atmosphere resulting in greater ozone loss. Major ODSs
include CFCs, HCFCs, halons, and N2O. ODSs are stable in the troposphere, but eventually break down (through
chemical and photochemical reactions) once in the stratosphere, releasing X catalysts. ODS emissions have resulted
in decline in overhead ozone around the globe with the exception of the tropics.
7.
d. Ozone depletion over the Arctic (N. Pole) is less severe than that over the Antarctic (S. Pole). Please provide a mechanistic explanation behind this observation. [9 pts] Three conditions must be met for the formation of an ozone hole: (1) reactive halogen gases and reservoir species
are available, (2) stratospheric temperatures are sufficiently low so as to allow formation of a vortex and polar
stratospheric clouds (PSCs), and (3) some sunlight is available. In the Antarctic, all three conditions are met each
year during towards the end of the polar night. In the Arctic, however, condition (2) is not always met. Even when
temperatures do drop low enough for vortex and PSC formation, such cold conditions tend not to last as long as they
do in the Antarctic. Formation of PSCs is critical in ozone depletion over the poles, because PSCs function as
catalysts in the transformation of reservoir molecules (such as HCl and ClONOO) into reactive gases (Cl2, HOCl).
This conversion facilitates mass destruction of ozone at the end of the polar night.
Challenge statement: “Emitting 100 kg of Halon-­‐1301 into the atmosphere does more damage to the ozone layer than emitting 100 kg of CFC-­‐11 into the atmosphere.” Please state whether or not you agree with this statement, and provide a scientific defense for your answer. circle one [2 pts]: AGREE DISAGREE provide scientific support for your selection [9 pts]: I agree with this statement, because Halons are known to have high ozone depletion potential (ODP) than CFCs. By
definition, an ozone destroying substance (ODS) with high ODP have the potential to do more damage to the ozone
layer than an equal amount of another ODS with lower ODP. Halons have high ODPs, because upon decomposition
in the stratosphere, its key ingredient Br remains predominantly in its catalytically active forms, Br and BrO, instead
of forming reservoir molecules such as HBr and BrONOO. CFCs, on the other hand, release Cl into the stratosphere,
most of which is then converted into reservoir species HCl and ClONOO.
3 CHEM/ENVS 380 F14, Midterm Exam 8.
9.
14 Oct 2014 The image to the right has been used by the City of Fort Collins, Colorado, to help reduce !
ozone pollution (http://www.fcgov.com/airquality/). Please explain the scientific basis of this message. [9 pts] What the City of Fort Collins aimed to do with the image on the right is to reduce
photochemical smog by reducing the emissions of NOx and VOCs from lawnmowers
during peak daylight hours. Ozone pollution in tropospheric air is caused by the emissions
of two primary pollutants, NOx and VOCs, in the presence of sunlight in a warm, stable
air mass. Fossil fuel combustion in internal combustion engines is a major source of both
NOx and VOCs, and these engines are found in both on-road and non-road vehicles,
lawnmowers being one of these. When you mow the lawn is critical, because the
formation of ozone from NOx and VOCs require photons in the UV-A range and shorter
which become less available toward the end of daylight hours. Challenge statement: “If we wiped out the human race, tropospheric O3 will disappear.” Please state whether or not you agree with this statement, and provide a scientific defense for your answer. circle one [2 pts]: AGREE DISAGREE provide scientific support for your selection [9 pts]: I disagree with this statement, because ozone is produced naturally in the troposphere in the absence of human
influence. The overall photochemical smog process can be written as:
NOx + VOCs (in the presence of sunlight in warm, stable air) à O3, nitric acid, more organics
The two primary ‘pollutants’ required to make this process go are NOx and VOCs, and both of these have major
natural sources (for example, NOx is produced by lightning, and VOCs are emitted by many living plants). Therefore,
as long as the physical conditions are right (warm, stable air with sunlight), ozone will be produced in the troposphere
through natural processes alone.
Resources
• Commonly used prefixes:
n = nano = 10-9
µ = micro = 10-6
m = milli = 10-3
k = kilo = 103 106 = million
109 = billion
• Some regions of the electromagnetic spectrum (in nm): 200-280, (UV-C); 280-320 (UV-B); 320-400 (UV-A); 400-750
(VIS); >750 (IR)
• Energy, E of a single photon = hc/λ, where h is the Planck’s constant (6.63 x 10-34 J s), c is the speed of light in vacuum
(3.00 x 108 m s-1), and λ is the wavelength of the photon (in meters, m).
• Avogadro’s number = 6.02 x 1023 mol-1
• Ozone destruction cycle-1 (where X denotes catalyst or chain carrier)
O3 + X à XO + O2
XO + O à X + O2
• Ozone destruction cycle-2 and -3 (cycle-2 involves Cl only; cycle-3 involves Cl and Br)
O3 + X à XO + O2
O3 + X’ à X’O + O2
XO + X’O à XOOX’
XOOX’ + hν à X + X’ + O2
4