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 (f), the oxidation number of Ca is +2. all are redox reactions except b) and f) a. 2CO + O 2 à 2CO 2 [C is oxidized from +2 to +4; O is reduced from zero to -2] b. Ca(OH) 2 + SO 2 à CaSO 3 + H 2 O [oxidation states of S, O and H are unchanged at +4, -2 and +1, respectively] c. 2Pb + O 2 + 4H + à 2Pb 2+ + 2H 2 O [O is reduced from zero to -2; Pb is oxidized from zero to +2] d. 6NO + 4NH 3 à 5N 2 + 6H 2 O [N in NO is reduced from +2 to zero; N in NH 3 is oxidized from -3 to zero] e. 2N 2 + 3H 2 à 2NH 3 [N is reduced from zero to -3; H is oxidized from zero to +1] f. CaCO 3 à CaO + CO 2 [oxidation states of C and O are unchanged at +4 and -2, respectively] g. HgS + O 2 à Hg + SO 2 [O is reduced from zero to -2; Hg is oxidized from some positive value (in this case +2) to zero; S is oxidized from -2 to +4. Note that even if you don t know the exact oxidation numbers of Hg and S, because O underwent reduction, you can state with certainty that this is a redox reaction.] h. 2Fe + O 2 + 2H 2 O à 2Fe 2+ + 4OH - [O reduced from zero to -2; Fe is oxidized from zero to +2] i. Zn + Cu 2+ à Zn 2+ + Cu [Zn is oxidized from zero to +2; Cu is reduced from +2 to zero] 2. Halon- 1301 has higher ODP (ozone depletion potential) than CFC- 12. This means that: a. CFC- 12 has a shorter atmospheric lifetime than halon- 1301 b. if equal masses of halon- 1301 and CFC- 12 were to be emitted into the atmosphere, halon- 1301 will destroy more stratospheric ozone than CFC- 12 c. CFC- 12 is currently emitted at a lower rate than halon- 1301 d. at present, halon- 1301 is doing more damage to stratospheric ozone than CFC- 12 [This is a common misconception. ODP is a measure of potential to do damage. Even if an ODS has very large ODP, if it is not emitted into the air, it will do zero damage. To determine which ODS is doing most damage, one must examine the ODP along with the quantity of the ODP that is present in the atmosphere.] e. halon- 1301 reaches the stratosphere at a faster rate than CFC- 12 3. Which of the following apply to the composition of Cl- containing gases in the stratosphere in the mid- to low- latitudes, and over the Antarctic when there is no ozone hole? a. catalytically- inactive species dominate b. catalytically- reactive species dominate c. Cl- source gases are absent d. HCl is the dominant gas e. there is too much variability to make any meaningful prediction 4. A soil sample has been found to contain As (arsenic) at a level of 650 ppb. This means that: a. the concentration of As in this sample is 6.5 x 10 5 ppm b. there are 650 moles of As in 10 9 grams of soil c. there are 650 ng of As in 1 gram of soil d. there are 0.650 grams of As in 10 6 grams of soil e. the concentration of As in this sample is 0.650 ppm 1
PART- B. Short Answer Questions Final Score /100 5. Answer the following questions on O 3. This question continues on to the next page. a. To the right, sketch a vertical profile of overhead O 3 in the Antarctic for the time of year when the ozone hole is absent. Clearly label the ozone layer. [5 pts] 50 km Ozone Hole Absent stratopause altitude ozone layer b. Prior to the Industrial Revolution, stratospheric O 3 levels are expected to have been at steady state. To maintain steady state, production rates must be balanced by destruction rates. Describe the major pathways through which stratospheric O 3 was produced and destroyed prior to the Industrial Revolution. Be thorough, but concise. [14 pts] 15 km Ozone (arbitrary units) tropopause Production Pathway(s) Stratospheric ozone is produced by photolysis of O 2 by short-wave radiation (UV-C or shorter) and the reaction of resulting atomic O with intact O 2 : O 2 + hν à 2O where hν represents photons in the UV-C or shorter range O + O 2 + M à O 3 + M + heat where M is a third body (typically N 2 of another O 2 ) 4 points total for this box Destruction Pathway(s) Stratospheric ozone is naturally destroyed through catalytic and non-catalytic pathways. Non-catalytic destruction occurs in the following two ways (loss mechanisms in Chapman Cycle): O 3 + hν à O 2 + O where hν represents photons in the UV-B or shorter range O 3 + O à 2O 2 The second reaction is slow in the absence of catalysts. Ozone is also destroyed through Mechanism-I and Mechanism-II that are driven by X catalysts (HO, NO, Cl, Br) that exist naturally in unpolluted air. Mechanism-I: O 3 + X à XO + O 2 XO + O à X + O 2 O 3 + O à 2O 2 (overall reaction) Mechanism-II (X and X are both catalysts): O 3 + X à XO + O 2 O 3 + X à X O + O 2 XO + X O à X + X + O 2 2O 3 à 3O 2 (overall reaction) 10 points total for this box 2
c. Sketch a vertical profile of overhead O 3 in the Antarctic during the time of year when the ozone hole is present [5 pts]. Why doesn t the ozone hole persist throughout the year? Explain. [9 pts] The ozone hole does not persist throughout the year, because its occurrence requires conditions that are met only during the polar night and in the early Austral spring. Low temperature and pressure over the Antarctic during the polar night allows the formation of a polar vortex and polar stratospheric clouds (PSCs). Ozone destruction occurs inside the polar vortex as a combined result of chemical reactions that occur on the PCSs and subsequent photochemical reactions that take place at the end of the polar night. As the temperature warms and the vortex disintegrates and PSCs disappear, the ozone hole dissipates due to termination of the loss process, and mixing of ozone- (and NOO-) rich air into ozone- (and NOO-) poor air over the Antarctic. 50 km altitude 15 km Ozone Hole Present stratopause ozone layer is mostly gone tropopause Ozone (arbitrary units) 6. Using the information provided below and at the end of this exam, show that UV- C photons carry sufficient energy to photodissociate (a) O 2 to atomic O, and (b) O 3 to O 2 and O. [9 pts] O 2 à O + O O 3 à O 2 + O H rxn = 498 kj mol rxn - 1 H rxn = 105 kj mol rxn - 1 The above tells us that the first reaction (photolysis of O 2 ) requires more energy than the second reaction (photolysis of O 3 to O 2 + O). Therefore, if UV-C photons are sufficiently energetic to make the first reaction go, they are also energetic enough to make the second reaction proceed. Plan: We are given the energy of a mole of photons (498 kj/mol). Convert this to wavelength, and find where in the electromagnetic spectrum this lies. Solution: The relationship between the energy and wavelength of a single photon is given at the end of this exam. Scale this up to a mole by multiplying it by the Avogadro s number, N A E of a mole of photons = hcn A λ hcn λ = A E of a mole of photons Using the values given at the end of the exam for constants h, c and N A, and setting E of a mole of photons equal to 498 kj mol -1, one obtains λ = 0.240 x 10-6 m = 240 nm. This is in the middle of the UV-C range (see end of exam). Therefore, UV-C photons (specifically those in the shorter half of the UV-C range) are energetic enough to make both of these reactions proceed. 3
7. Why do halons have high ozone depletion potential (ODP) relative to other ozone depleting substances? [9 pts] Halons are unique in that they contain Br in addition to Cl. With the exception of winter-early spring at the poles when the ozone hole is present, Cl exists primarily as reservoir species (or catalytically inactive forms; HCl and ClONOO) rather than in catalytically reactive forms (Cl and ClO). In contrast, Br exists primarily in catalytically reactive forms (Br and BrO) at all times, allowing Br to destroy more O 3 than Cl when compared on an equal-mass basis. This makes halons have higher ODP relative to other ozone depleting substances. (Note that the atmospheric lifetimes of halons are not that different from those of some major CFCs.) 8. The following reactions show the oxidation of methane to CO 2 (rxn- 1), oxidation of NO to NOO by hydroperoxy radicals (rxn- 2), photodissociation of NOO (rxn- 3), and generation of O 3 (rxn- 4) during a photochemical smog event. Please answer the questions below. CH 4 + 5O 2 + NO + 2HO à CO 2 + H 2 O + NOO + 4HOO (rxn- 1) 4HOO + 4NO à 4NOO + 4HO (rxn- 2) 5NOO + hν à 5NO + 5O (rxn- 3) 5O + 5O 2 à 5O 3 (rxn- 4) a. Add up the four reactions and derive the overall reaction. [5 pts] Add up reactions and cancel common terms that show up on opposite sides of reaction. Print overall reaction in this box: CH 4 + 10 O 2 + hν à CO 2 + H 2 O + 2HO + 5O 3 b. Referring to the reactions above, describe how the production of O 3 during a photochemical smog event is fundamentally different from the production of O 3 in the stratosphere. [8 pts] As discussed in the answer to question 5 b), ozone production in the stratosphere is initiated by photolysis of O 2 by short-wavelength radiation (UV-C or shorter). Given that only limited UV-C penetrates the atmosphere to reach ground level, one would expect ozone production at ground level to occur through fundamentally different pathways, and that is indeed the case. Ozone is formed at ground level through reaction of O 2 with atomic O (rxn-4), but unlike the case in the stratosphere, atomic O is generated from the photolysis of NOO (rxn-3) in the presence of photons in the UV-A or shorter regions of the electromagnetic spectrum. NOO itself is generated by oxidation of NO by HOO (rxn-2) or ROO. HOO and ROO are byproducts of VOC oxidation in warm, stable air in the presence of sunlight (rxn-1). 4
9. Explain how driving a car with a conventional internal combustion engine is linked to acid rain. [8 pts] Internal combustion engines in conventional cars run hot enough to create NO from atmospheric N 2 and O 2 : N 2 + O 2 à 2NO NO (nitric oxide) thus formed is oxidized to NOO in the photochemical smog cycle. A major fate of NOO is conversion into nitric acid (HNO 3 ) through reaction with HO radical: NOO + HO à HNO 3 The above reaction is an example of many termination reactions that occur at the end of a smoggy day. 10. Using chemical reactions, explain why pure water becomes acidic when left to equilibrate with unpolluted air. In your description, distinguish dominant from non- dominant reactions. Assume 25 C and 1 atm pressure. Note that you are not expected to calculate the ph of this system. [8 pts] Water in equilibrium with unpolluted air is naturally acidic due to the presence of CO 2 in the air. (Note that there are also strong acids such as HNO 3 and H 2 SO 4 in unpolluted air, but you are not expected to include them here, because we have not yet discussed them in class.) The following reactions show the dissolution of CO 2 (g) into water, followed by ionization of carbonic acid that lowers the ph: CO 2 (g) + H 2 O H 2 CO 3 (1) H 2 CO 3 H + - + HCO 3 (2) HCO - 3 H + 2- + CO 3 (3) Reactions (1) and (2) dominate over reaction (3), as seen from the equilibrium constants given at the end of this exam. Adding reactions (1) and (2) give the overall reaction: CO 2 (g) + H 2 O H + + HCO 3 - The overall effect is to lower the ph from neutral (ph 7) to an acidic value. Resources Commonly used prefixes: n = nano = 10-9 µ = micro = 10-6 m = milli = 10-3 k = kilo = 10 3 10 6 = million 10 9 = 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 10 8 m s -1 ), and λ is the wavelength of the photon (in meters, m). Avogadro s number = 6.02 x 10 23 mol -1 Dissociation constants for the carbonate system (298 K): CO 2 (g) + H 2 O à H 2 CO 3 K H = 3.4 x 10-2 mol/l/atm H 2 CO 3 à H + - + HCO 3 K a1 = 4.5 x 10-7 HCO - 3 à H + 2- + CO 3 K a2 = 4.7 x 10-11 5