{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch2[1] - CHAPTER The Ozone Holes Problem 2~1 03 C1 —>...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER The Ozone Holes Problem 2~1 03 + C1 —> ClO + 02 03 + Br —> BrO + 02 ClO+BrO —> Cl+Br+OZ overall 2 03 —> 3 02 Problem 22 The C1202 mechanism will become more dominant. Since rateclzo2 = k [C10]2 whereas FatCClOBrO = k' [ClO] [BrO] Thus, rateC1ZQZ/rateCIQBrO = k [ClO] k' [BrO] In other words, because the rate ratio is proportional to the chlorine/bromine ratio in the strato~ sphere, as this ratio increases, the C1202 rate relative to the ClO/BrO rate will increase, and become even more important. Problem 28 In the upper stratosphere, atomic O is relatively plentiful and C10 is normally rare, so the rate for the C10 + 0 process that depends upon the first power of the CIO concentration will be much greater than one that depends on the square of the C10 concentration. Problem 2-4 If C— F bonds are stronger than O—F, then the fluorine abstraction reaction OH + CFZCl2 —> HOF + CFCl2 will be endothermic and therefore have a high activation energy and thus be slow. Thus, the reaction will not be a realistic sink for CFCs. ll 12 Chapter 2 Problem 2-5 For OH, Mechanism I of catalytic ozone destruction is: OH+O_, —> HOO+O2 HOO+O —> OH+O2 lfCF3 -— O replaces H — 0, this mechanism would be: CF30 + 03 —) CF3OO + 02 CFJOO + O —) CF30 + 02 Problem 2-6 (a) Mechanism I F+o3 —) 0F+o2 OF+O —) 1:+o2 Mechanismll F + 03 —> OF + 02 20F ——) [FOOF] —) 2F + 02 (b) Mechanism 1 here is changed to F+O3 —> FO+O2 FO+O3 —> F+ZO2 netzo5 —) 3 02 Problem 2-7 If the C—H bond energy in propane and butane is less than in methane, then the inactivation of atomic chlorine by the reactions C1 + CJH8 —> HCl + C3H7 and C1 + C4Hio —) HCl + C4H9 will have lower activation energies and thus faster rates than for C1 + CH4 —> HCl + CH}. Thus because of the faster propane and butane reactions, the steady—state concentration ofatomic Cl will decrease more than with methane and less ozone will be destroyed due to the lower concentration of active Cl. The Ozone Holes 13 Problem 2-8 Presumably CHJCl, CHZCl, and Cl—lCl3 are destroyed to a large extent (by the H abstraction accord— ing to our principles) in the troposphere since they are not candidates for regulation. Thus most of the chlorine in these molecules will find a sink in the troposphere and not have time to rise to the strato— sphere to destroy ozone. Green Chemistry Problems 1. (a) Principle 2. Alternative reaction conditions for green chemistry. (b) Principle 1. Prevention of waste. Principle 4. Preserving efficacy of function while reducing environmental damage. 2. The use of carbon dioxide as a blowing agent uses waste products from other processes. Unlike CFCS, carbon dioxide does not deplete the ozone layer. Unlike hydrocarbons, carbon dioxide does not contribute to the pollution of the troposphere. 3. Although carbon dioxide is a greenhouse gas and thus contributes to global warming, the carbon dioxide that is used as a blowing agent is captured as a waste byproduct from processes such as natural gas production and ammonia production. This waste carbon dioxide would have nor— mally been vented to the atmosphere. 4. (a) Principle 3. The design chemicals that are less toxic than current alternatives or inherently safer with regard to accident potential. (b) Principle 1. Prevention of waste. Much less harpin is used compared to conventional pesticides. Principle 4. Preserving efficacy of function while reducing toxicity. Lower toxicity (toxicity category IV). Principle 6. Energy reduction. Harpin is produced by natural fermentation, not from peth' chemical feedstocks. Principle 10. Degradation in the environment. 5. Harpin does not effect the pest directly. It is applied to the plant, which elicits the plant’s own natural defenses. 6. Harpin is readily degraded in the environment by microorganisms and UV light. 14 Chapter 2 Additional Problems 1. (a) For a gas, the number of moles in any sample is proportional to the sample’s volume, because PV = nRT. The volume here equals the height of the layer, when converted to pure ozone at 1 atmosphere and 0°C, times the surface area. Thus moles oc height. As defined in the text, 1 Dobson Unit equals 0.01 mm or 0.001 cm of ozone at 1 atmosphere pressure. Since PV is a constant for a gas sample, the product 1 atm X 0.001 cm (X surface area) can be revexpressed as 0.001 atm X 1 cm (X surface area); in other words, as a milli— atmosphere (0.001 atm) times a centimeter. Thus, 1 DU = 0.001 cm atm = 1 matm cm. (b) We are told that the volume ofa sphere is proportional to the cube of its radius. The two spheres that are involved here are the Earth, radius r, and the Earth plus the 3.5 mm of airr space above its surface that corresponds to pure ozone when the Column ozone is collapsed to standard temperature and pressure. The volume of ozone is the difference in volume be- tween these spheres: Volume of ozone = Volume of sphere of Earth + ozone — Volume of sphere of Earth alone = i11:(1-+A)3~-1117r3= irl:[(r+A)3—r3] where r is the Earth’s radius and A is 3.5 mm. Now expanding (r + A)3 = r3 + 3rZA + 3rAZ + A3 so the term in square brackets reduces to 3rZA + 3rA2 + A3 Since r > > A, the lead term, 3rZ A, will completely dominate. Thus, Volume of ozone = i1t(3rZ A) = 411tr2 A 3 Nowr = 6400 km = 6400 X 103m andA = 3.5 mm = 3.5 X 10'3m Thus,Volume = 4 X 3.14 X (6400 X 101m)Z X (3.5 X 10‘J m) = 1.8 X 1012 m3 = 1.8 X 1015L since 1 m3 = 1000 L The Ozone Holes 15 The mass of ozone can be deduced from the ideal gas law by first deducing its number of moles: n = PV/RT = 1.0 atm X 1.8 X 1015 L/0.082 Latm mOl‘l K“l X 273 K = 8.0 X 1013 moles Since each mole ofO3 has a mass of 48 grams, the total mass ofozone is 48 X 8.0 X 10’3 = 4 X 1015 grams 2. To obtain the formula, add 90 to the code number to give the three—digit integer corresponding to the number ofC, H, and F atoms, respectively. Then deduce the number ofchlorine atoms present. (a) 12 + 90 = 102, so this corresponds to CiHoni because the carbon atom forms a total of four bonds, we deduce that two chlorines must be present. Therefore, the formula is CFZCIZ. (b) 113 + 90 = 203, which gives CZHOF3. Total hydrogens + substituents in alkane deriva— tives is 2n + 2 if there are n carbons, so there must be 2 X 2 + 2 = 6 atoms total bonded to the carbons here. Thus, there are 3 Cl atoms because there are only three (H + F) atoms. The formula must be C2F3Cl3. (c) 123 + 90 = 213, which gives C2H1F3. Since, as in (b), six atoms are bonded to the car bons, there must be 6 — 1 — 3 = 2 chlorine atoms; thus the formula is CZHF3C12. (d) 134 + 90 = 224, which gives CZHZF4. As in (b), 6 atoms must be attached to two carbons, but 2 + 4 = 6, so no chlorine atoms are present. The formula is C2H2F4' 3. Write the formulas as CnHme, ignoring the chlorines, so the code number is (nmp — 90). (a) Formula rewritten is CszFo’ so code is 230 — 90 = 140. (b) Formula rewritten is ClI-IOFO, so code is 100 — 90 = 10. (c) Formula rewritten is CZHJFI, so code is 231 — 90 = 141. 4. To the code of 134, we add 90, giving 224. Thus the numbers of C, H, and F atoms should be 2, 2, and 4 respectively, yielding the formula C2H2F4’ which indeed is consistent with the con- densed formula CHZFCFj. Since the compound has an isomer, namely CHFZCHFZ, there must be an additional designation—a or b—to distinguish them. 5. The conditions under which the chlorine dimer mechanism is important are: (a) chlorine is activated on the surface of particles (b) the C1202 dimer is stable to thermal decomposition (c) most N02 is tied up as I-INO3 16 (hapter 2 At mid—latitudes, the temperature in the lower stratosphere would not be very low, so the C1202 dimer would probably ultimately thermally dissociate back to ClO before photolysis could occur: c1202 ——> 2 C10 Thus the cycle of C1 ——> ClO—-) Cl would not occur, and ozone wold not be catalytically destroyed. In addition, nitrogen dioxide would not be inactivated but available to convert ClO immedi’ ately back to the inactive ClONO2 form: ClO + NO2 —-) ClONO2 Stratospheric bromine, atom—for’atom, can destroy more ozone than chlorine since a smaller fraction of it is tied up in catalytically inactive forms. Thus more mid—latitude ozone would have been destroyed had the compounds been made using bromine rather than chlorine. The reaction step of interest is: / Cl + Br + 02 ———> further rxn ClO + BrO \ BrCl +» no rxn at night At any given time, most reactions of C10 and BrO will produce Cl and Br, which will continue the cycle and soon again appear as (:10 and BrO. However, in each cycle, a fraction of the halo; gen will enter the BrCl sink and remain as such until dawn. Thus the fraction of halogen exist— ing as BrCl will increase as time goes on and the cycle is repeated over and over, each time with less and less halogen involved. CHI:3 contains no chlorine, only H and F, so it has zero ozone—depleting potential. CHFCl2 contains chlorine, but also contains H, making it an HCFC H atom abstraction. H via hydroxyl radical attack thus provides a significant sink for this compound in the troposphere, so only a fraction of this molecule released at ground level will make it to the stratosphere, where its chlorine will eventually be released and it will destroy ozone. CF3Cl and CFCl3 are both CFCs with no tropospheric sinks, and thus most of the mass of each of them will rise to the stratosphere and catalytically destroy ozone there. Of the two, CFCl3 has 3 Cl per molecule compared to 1 in the case of CF3C1, and so CFCl3 would destroy more OZODG. Thus, the order in terms of increasing GDP is: CHF, < CH FCl: < CF,Cl < CFCIT 10. The Ozone Holes 17 Hydrocarbons contain no chlorine (or bromine), so have zero ozone’depleting potential. In terms of ozone depletion, they are ideal replacements for CFCs as aerosol propellants, allowing for the existence of aerosol products that do not harm the ozone layer. The major disadvantage is that they are highly flammable. (Recall that butane is used in cigarette lighters.) This results in a safety hazard, especially for products such as hairspray that might be used by a smoker, or around other sources of ignition such as candles. The extra type of agent that would need to be added is a flame retardant, to make the resulting aerosol spray inflammable. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern