Ch5[1] - CHAPTER The Detailed Chemistry of the Atmosphere...

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Unformatted text preview: CHAPTER The Detailed Chemistry of the Atmosphere Problem 5-1 CFZCl2 and other CFCs do not contain hydrogen or multiple bonds, so they are not attacked by OH. They also do not absorb light of wavelength regions that reach the troposphere; therefore, photochem— ical processes do not initiate CFC oxidation either. in contrast, CHZCl2 contains hydrogen, and so it will be attacked by OH and its oxidation initiated: CHZCIZ + OH' —> CHClZ' + H20 Problem 5-2 If the reaction is endothermic it will have a substantial activation energy and therefore it will be slow. Problem 5-3 Since the reaction is not fast enough to be observed, presumably its activation energy must be large. Thus, the reaction probably is quite endothermic, because most free radical reactions with large actia vation energies are quite endothermic. Problem 5—4 H2 will react with OH' because it does not absorb visible or UV—A light; because H2 contains only a single bond, the OH' abstracts one hydrogen: OH' + H2 —9 H20 + H' Since H' is a radical, it reacts with 02 by addition to it: H' + 02 —9 H00' Peroxy radicals react with NO', so HOO' + NO' —9 OH' + NO; 4] 42 Chapter 5 Adding up these three reactions, we find the overall reaction involving H, to be H2 + 03 + NO' —) HBO + NOZ' Problem 56 Since methanol contains no multiple bonds and does not absorb visible or UV’A light, it will react with OH' by hydrogen atom abstraction. First, consider that it is a hydrogen bonded to carbon that is abstracted: H,COH + OH’ —) HzeoH + H20 The carbonecentered radical reacts With 02; because by abstracting the H bonded to O, a stable molecule forms and the C—— 0 bond is converted to C =0, this process rather than 0, addition subsequently occurs: HZCOH + 02 —) HZCO + HOO’ In an alternative scenario, the OH' abstracts the hydrogen bonded to oxygen from the original methanol molecule: H,CoH + oa' —) HBCO + 1—110 In this case, the 0, molecule attacks and abstracts a hydrogen bonded to carbon because formaldehyde, with its C: 0 bond, is thereby formed. H360 + 02 —) cho + HOO' Problem 5-6 Carbon monoxide is a non’radical and does not absorb visible or UV—A light; therefore, it must react with OH'. Since it contains a C E 0 bond, the OH. adds to the CO molecule: C0 + OH' ——) H—O—C=O The radical reacts with OZ, and does so by giving up its hydrogen because thereby the O ———C bond is converted to O = C: H—o—e =o + 02 —) HOO' + O=C=O Finally, because it is a peroxy radical, HOO' is returned to OH' by reaction with NO’: HOO' + NO' —) OH' + NOZ' Adding the three steps and cancelling terms that appear in both sides, we obtain: CO + 02 + NO' —-> CO2 + NOZ' as the overall reaction. The Detailed (hemisiry of ihe Atmosphere 43 Problem 5-7 The reaction sequence begins with an H abstraction: OH' + H200 —> H20 + H—é =0 The radical HCO reacts with OZ also by H abstraction, because this produces the stable molecule CO with a triple bond: H—é=0 + 02 —> 020 + HOO' As in Additional Problem 5—6, CO is then attacked by OH°—it is a stable molecule with a multiple bond, and therefore the OH. adds to it: OH' + 00 —> H—o—é =0 Finally, 0Z abstracts the H from HOCO because the stable molecule (:0Z is thereby formed: H—0—é =0 + 02 —> HOO' + O=C=O The two HOO' return to OH' by reaction with NO' 2 [H00' + NO' —> OH' + N021 Summing the reactions, we obtain H200 + 2 NO' + 2 01 —> (:02 + H20 + 2 N0; There is no increase in the number of free radicals; an increase would occur only when photochemical decomposition ofa non~radical into two radicals is involved. Problem 5-8 Substituting H for R in the net reaction given in the text, we obtain: HZC = CH2 + OH' + 202 + NO' —> ZHZCO + HOO' + N02‘ The formaldehyde molecules undergo photochemical decomposition to give H‘ + HCO': H200 LH- + HCO' We know from our principles that 02 will immediately add to H', and the 02 will abstract the hydrogen from HCO‘ because the double bond in H — C: O is thereby converted to a triple one in CO: H' + 02 —> HOO' HCO' + 02 —> HOO' + 00 44 Chapter 5 As previously discussed, the CO adds OH and the resulting HOCO reacts with 03 by hydrogen abstraction because the number of CO bonds is thereby increased: co + oa- ——> HOCO' HOCO' + 02 ——> HOO' + co2 When twice these reactions (because 2 HZCO must be destroyed) are added to the net reaction above, and after cancellation of common terms, we obtain: chzcul + so! + NO' + 301+ 11>ch + N0; + moo- The HOO' oxidation of NO' involves 7 molecules of each: 7[HOO° + NO' ——> OH' + NOZ'] Thus the final net reaction is: H2C=CHZ + soZ + sNo- flazcol + 8N0; + 4OH° Problem 5-9 sunlight CH3(H)CO ——> CH; + HCO' HCO' + 02 ——> HOO' + co co + OH' ——> HOCO' HOCO' + oz ——> co2 + HOO' CH; + 02 ——> capo cuyoo + NO' ——> cap' + No; cago' + 02 ——> HZCO + HOO' light HZCO ——> H' + HCO' H' + 02 ——> HOO' HCO' + 02 _—> HOO' + co (:0 + OH' ——> HOCO' HOCO' + 02 ——> coZ + HOO' 6lHOO' + NO' ——> OH' + N05] Overall, the reaction is CHy(H)CO + 702 + 7NO' ——) 2C0Z + 7N0; + 4OH' The Detailed Chemistry of the Atmosphere 45 Problem 5-10 From the text, the dominant reaction at 5—8 AM is the production of NO2 from N0, and that from 8 AM—lZ noon is the production of aldehydes from hydrocarbons. Problem 5—11 The initial reaction must be sunlight HZCO —> HZ + co Since HZ contains H, we expect OH' to abstract 3 hydrogen from it: OH' + HZ —> H20 + H‘ We expect H' will add 02 H' + QZ —> HOO' The HOO' will oxidize NO‘ HOO' + NO. —> OH' + N0; From previous examples, we know OH' adds to CO and then 02 abstracts hydrogen from the radical: OH' + CO —> HOCO' HOCO' + QZ —> HOO' + COZ HOO' + NO' —> OH' + NOZ' Adding together all six steps involving radicals, we obtain HZ + C0 + 202 + ZNO' —> HZO + COZ + ZNOZ' Adding this to the first step, we obtain: HZCO + 202 + ZNO‘ H20 + coZ + 2N0; 46 Chapter 5 Problem 5~12 1th — CHZOH ioz 0 ~ 0' l H Zc CH 2OH lNO' O' H 2c — CHZOH cleaves (see text) H Zco + éHZOH log HZCO + HOO' Both HZCO C02 + H00. in all instances. HOO' + NO. —> OH. + N02. Problem 5-13 lfCH3OO' combines with HOO', we have a fouraoxygen chain that presumably will split out at least one 07 molecule caloo- + HOO' ——> [caaooooa] ——> 02 + CHtOOH Problem 5-14 Items b, c, and d do not contain “loose” oxygens. The Derailed Chemistry of the Atmosphere 47 Problem 515 From the Systematics section we find that (a) NO abstracts an oxygen atom from all species containing a loose oxygen, i.e., 03, C10, BrO, and H00. (b) O abstracts an oxygen from all species with “loose” oxygens, i.e., from 03, C10, BrO, H00, and N02. (c) Sunlight is absorbed in the UV region by all the species 03, C10, BrO, H00, and N02, and an oxygen atom is consequently detached. (d) YOOY molecules of. the type ClOOCl and (NOZ)Z and perhaps BrOOBr will form. (e) 02 is produced when two YO molecules react but result in a chain that has more than two 0 atoms bonded to each other; thus 02 is obtained when .7. H00 react and when two 0} react. Problem 5-16 (a) BrO+O —> Br+OZ (b) BrO + ClO —> Br + Cl + 02 (c) .7. BrO —> BrOOBr (d) BrO+UV —> Br+O Problem 5~17 (a) C10 + NoZ —> CIONOZ (b) ZCIO —> CIOOCI (c) ClO + UV —) C1 + O or ClOOCl + UV —> C1 + ClOO orClO + O —> Cl + 02 orClO + NO —> Cl + NOZ 48 (hupier 5 Problem 5~18 :.F.__ .qo Yes, the oxygen should be “loose” because it is joined by a single bond to an electronegative atom that possesses lone pairs. Problem 5-19 The expected reaction is abstraction of the loose oxygen from ClO' by NO': ClO‘ + NO' —-> C1' + N02' The N05 could decompose in sunlight by the loss of. atomic oxygen, or react with atomic oxygen, or with another NOZ': sunlight N02' —-«> No- + o No; + o ——9 N0- + 02 ZNOZ' —-> N204 The Cl' could abstract an oxygen from ozone or a hydrogen from methane: C1' + 03 ——> 00- + 02 C1' + CH4 —-—> HCI + CH; The cycle would destroy the ozone (by 03 + O —-> 2 01 overall) if the Cl reaction is that between Cl' and O3 and the N02' reacts with 0; if, however, the N02' decomposes by sunlight, the net reac' tion is 03 —-> 02 + O, which is followed by O + 02 ——-> 03. and so no ozone is destroyed. Box 5~1 Problem 1 (a) Only the oxygen here can be the site for the unpaired electron, which is consistent with it form- ing only one bond rather than its usual two: 2 O — H (b) Only the carbon here can be the site for the unpaired electron, which is consistent with it form; ing only three bonds rather than its usual four: H C ~— H H/ The Detailed Chemistry of the Atmosphere 49 (c) Since F and Cl must form (single) bonds to the carbon, neither can be the site of the unpaired electron, which therefore must be located on the carbon, which forms three rather than four bonds: \ C — : @/ (d) The carbon here must form four bonds, and therefore cannot be the site of the unpaired elec~ tron, and similarly for the central oxygen, which must form two bonds. The unpaired electron must therefore be located on the terminal oxygen atom, as in HOO: H l .. . H—C—O—O: 1 u u H (e) As in (d), the unpaired electron must be on the terminal oxygen: Ii H_C_@ I .. H (f) As in (d), the unpaired electron must be on the terminal oxygen. The Cl atom must form its usual one single bond and thus cannot be the site: za—é—d (g) As in (f), the unpaired electron must be on the terminal oxygen. The Cl atom must form its usual one single bond and thus cannot be the site: :Q—@ (h) The oxygen here cannot be the site of the unpaired electron, since that would require the cen— tral carbon atom to form only two bonds (as in H — C — 0). Hence the oxygen must form two bonds to carbon, and the hydrogen one bond to C. Thus the site of the unpaired electron here is the carbon atom: (i) In the best structure for NO, the oxygen atom forms two bonds to the nitrogen atom, which is the site of the unpaired electron and forms two rather than its usual three bonds: 50 Chapter 5 Additional Problems 1. Since CO is a stable molecule containing multiple bonds, it reacts with OH' by addition of. the free radical to the C atom (since addition to the 0 gives a DO bond): CO + OH' —> HOCO' (i.e., Hoé=0) The free radical HOCO' reacts with OZ, by abstraction of H since this allows the addition of a new CO bond, in C03: HOCO' + 02 —> HOO' + CO2 (i.e., O=C=O) The HOO radical reacts in the usual way with NO: HOO' + NO' —> OH’ + NOZ‘ It'NO2 photolyzes, it produces atomic O, which then adds to O2 to produce ozone: No; 3; NO' + o O + 02 —> 0} Adding up all these steps, and cancelling common terms, gives the result co + 2 02 3; co2 + 03 Thus the concentration of ozone is increased by this process involving OH' as a catalyst. 2. Since H3C—— CH3 has no double bonds, initially OH' reacts by abstracting a hydrogen from it. Then the carbon—centered radical adds 02, and the peroxyl radical oxidizes NO’: ale—CH3 1*; Hie—(Ea2 i; H3C—CHZOO' (+HZO) 1‘11; HSC —- CHZO' (+Nog) This oxygen—central radical can be transformed to one containing C: C) if one of the CHz’s H atoms is abstracted by OZ. The resulting aldehyde is then photolyzed: O photon H3C—CHZO' —1> H3C—CH=O —> Hp + aco- (+HOO') The HCO' radical has its H abstracted by 02 because a CO triple bound is formed; OH' then adds to this bond, and oxygen abstracts the oxygen to give the final product, carbon dioxide: HCO' 0—1; (:0 fl; HOCO' 0—1; co3 (+HOO’) (+i 100') The Detailed Chemistry of the Atmosphere 51 Meanwhile the methyl radical H3C' adds O2 to produce a peroxyl radical, which then oxidizes NO', and has an H abstracted to form a C = 0 bond: H3C' L) H3COO° fl) H3CO' HZC=O (+No,-) (+HOO') Formaldehyde undergoes photochemical decomposition to H', which adds 02, and to HCO', which reacts as described above: ch0 H- + HCO' fl HOO' + HOO' + co co fl) HOCO' co2 + HOO' Summing up all the reactions, we obtain: H,CCH3 + 902 + ZNO' + 3OH' —> zco2 + 2N0; + 7HOO' + H20 Finally, adding in the conversion of7 HOO' back to 7 OH' 7[HOO° + No- —> OH' + N021 we obtain the overall reaction: H,CCH3 + 902 + 9NO° —> zco2 + 9N0; + H20 + 4OH' Carbon monoxide reacts with OH' radical to produce HOCO', which then reacts with O2 to produce HOO' radicals co % Hoco- L) co2 + Hoo- If HOO' reacts with ozone, a chain of five oxygen atoms result, so two 03’s and OH result: HOO' + O3 ——) [HOOOOO]' —) OH' + 2 02 Thus the overall reaction is: c0+o, —~) coz+o2 Thus the concentrations ofO3 should be abnormally low, because it is destroyed by CO. (i) CH3CH2CH3 + OH' —> CH,(°:HCH3 + H20 (secondary carbon radical more stable than a primary carbon radical) (n) H2C=CHCH3 + OH' —> Hocazcaca3 (abstraction of an H atom from the —CH3 may also be likely) (iii) HCl + OH' —) no reaction (very strong HCl bond) 52 (hupier 5 (iv) HZO + OH'——) OH + H20 (no net reaction!) N07: -b—N=¢ imq;n_§—N=d .. /O-° HNofii_o—N\ " \u Q. The energy of the photon absorbed must be equal to or greater than the energy of the bond to be cleaved. In the case of the longest wavelength of light that will cleave the bond, the energy of a photon with this wavelength will be exactly equal to the bond energy: 9» = 119,627 /bond energy In the case of BrO', the bond energy is 235 k] mol‘lz K =119,6Z7/Z35= 509 nm This is in the visible region, specifically the green part of the spectrum. Similarly, the following can be calculated: HOO': 450 nm (visible region—blue) C10": 440 nm (visible region—blue) NOZ': 392 nm (UV—A region) ...
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This note was uploaded on 09/23/2011 for the course CHEM 380 taught by Professor Staff during the Spring '11 term at S.F. State.

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Ch5[1] - CHAPTER The Detailed Chemistry of the Atmosphere...

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