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Chapter 10

Chapter 10 Probelms.pdf

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Department
Chemistry
Course
CHEM 302
Professor
Michael Wheeler
Semester
Fall

Description
190 PROBLEMS 10. 1 Shape of the ozone layer Consider a beam of solar radiation of wavelength λ propagating downward in the vertical direction with an actinic fl∞x I at the top of the atmosphere. Assume that O is the sole atmospheric absorber of this radiation. 2 1. Show that the O p2otolysis rate R from this radiation beam varies with altitude z as follows: z Rz() = qσC O2 na(0 I∞exp —---— σC O2Hn a)z H where σ is the absorption cross-section for O at wavelength λ , q is the 2 correspnding quantum yield for O p2otolysis, n (za is the air density, H is the scale height of the atmosphere, and C is the O mole fraction. O2 2 2. Sketch a plot of R vs. z. Comment on the shape. Explain physically why R(z) has a maximum in the atmospheric column. 10. 2 The Chapman mechanism and steady state We compare here some features of the Chapman mechanism at 20 km and 45 km altitude. Adopt temperatures of 200 K at 20 km altitude and 270 K at 45 km 18 -3 altitude, and air densities of 1.8x10 molecules cm at 20 km altitude and 4.1x1016 molecules cm -3 at 45 km altitude. The reactions in the Chapman mechanism are O + hν → OO + (1) 2 OO++ 2 M → O +3M (2) O3+ hν → O +2O (3) O 3 O → 2O 2 (4) with rate constants k2= 1x10 -33 cm 6 molecule -2s , k 3 1x10 -2s , and k 4 -12 3 -1 -1 8.0x10 exp(-2060/T) cm molecule s where T is temperature. 1. Calculate the lifetime of the O atom at 20 km and 45 km altitude. Can the O atom be assumed in chemical steady state throughout the stratosphere? 2. Assuming steady state for O atoms, calculate the O/O 3oncentration ratio at 20 km and 45 km altitude. Can we assume [O 3] ≈ [O x throughout the stratosphere? 3. Show that the mass balance equation for odd oxygen (O =x 3 + O), ignoring transport terms, can be written 191 dO[] x 2 --------=-- — [] x dt where P = 2k 1O ]2is the O xroduction rate and k = 2k k3 4k 2 O2 a ). 4. Express the lifetime of O ax a function of k and [O ]. Usxng the vertical distribution of O at the equator in Figure 10-1, calculate the lifetime of O at 20 3 x km and 45 km altitude. 5. Based on your answer to question 4, in what part of the stratosphere would you expect O tx be in chemical steady state? How would that help you test the accuracy of the Chapman mechanism in predicting ozone levels? 10. 3 The detailed Chapman mechanism We examine here some features of the detailed Chapman mechanism. Consider an air parcel at 44 km altitude, 30 N latitude, overhead Sun, T = 263 K, n = a 5.0x1016 molecules cm , and [O ] = 3.0x10 11 molecules cm .-3 The reactions involved in the mechanism are: 3 -10 -1 O2 + 2ν → O( P) k1= 6.0x10 s (1) 3 -3 -1 O 3+ hν → O +2O( P) k2= 1.0x10 s (2) 1 O 3 hν → O + 2( D) k3= 4.1x10 s3 -1 (3) O(P) ++O M → O + M k = 6.0x10 -3(T/300)-2.3cm molecule s2 -1 (4) 2 3 4 1 3 -11 3 -1 -1 O(D) + N 2 O( P) + N 2 k5= 1.8x10 exp(110/T) cm molecule s (5) 1 3 O(D) + O2→ O( P) + O 2 k6= 3.2x10 -1exp(70/T) cm molecule s-1 -1 (6) O(P) + 2 → O k = 8.0x10-12exp(-2060/T) cm molecule s1 -1 (7) 3 2 7 1. Assuming that reactions (1)-(7) are the only ones occurring in the air parcel: 1 1.1 Calculate the lifetime of O( D) in the air parcel and its steady-state concentration. 3 1.2 Calculate the lifetime of O( P) in the air parcel and its steady-state concentration. 1.3 Calculate the lifetime of Oxdue to loss by the Chapman mechanism. 2. Assuming steady state for O ,xcalculate the fraction of the total O xink in the air parcel that can actually be accounted for by the Chapman mechanism. 192 10. 4 HO -cxtalyzed ozone loss Cycling of the HO xhemical family (HO ≡ HxOH+HO 2) can catalyze O3loss in a number of ways. Consider the following reactions, each important in at least some region of the stratosphere: OH +O →O + H 2 (1) OH +HO → H OO + (2) 2 2 2 OH +O → HO + O (3) 3 2 2 HO ++ 2 M → HO + 2 (4) HO+ 3→ O +2OH (5) HO +2O → OH +O 2 (6) HO 2+ O 3 OH + 2O 2 (7) HO + HO → H O + O (8) 2 2 2 2 2 1. Find five different catalyti3 O loss cycles starting with reaction of OH. 2. Which of the reactions represent sinks for HOx? 10. 5 Chlorine chemistry at mid-latitudes o An air parcel at 30 km altitude (30 N, equinox) contains the following concentrations: 12 -3 [O 3 = 3.0x10 molecules cm 7 -3 [O] = 3.0x10 atoms cm [NO] = 7x10 molecules cm -3 9 -3 [NO 2 = 2.2x10 molecules cm 6 -3 [HO 2 = 8.5x10 molecules cm [CH ] = 2.8x10 11molecules cm -3 4 We examine the mechanism for Cl-catalyzed O los3 in this air parcel on the basis of the following reactions: Cl +O → ClO +O k = 9.5x10-12cm molecule s1 -1 (1) 3 2 1 Cl +CH → HCl +CH k = 2.6x10 -14cm molecule s1 -1 (2) 4 3 2 ClO +O → Cl +O k = 3.8x10-11cm molecule s-1 -1 (3) 2 3 NO + hν → NO +O k = 5.0x10 s -1 (4) 2 4 193 -11 3 -1 -1 ClO +NO → Cl +NO 2 k5= 4.5x10 cm molecule s (5) -11 3 -1 -1 ClO +HO → 2OCl +O 2 k6= 2.1x10 cm molecule s (6) ClO +NO M → ClNO + M k = 1.3x10-13cm molecule s1 -1 (7) 2 3 7 HOCl +h ν → OH +Cl k = 2.5x10 s4 -1 (8) 8 OH +O → 3O + O 2 2 k9= 2.8x10-14cm molecule s1 -1 (9) 1. Calculate the chemical lifetimes of Cl and ClO. Which reaction is the principal sink for each? 2. Based on your answer to question 1, explain why reaction (3) is the rate-limiting step in the catalytic cycle 3or O loss: Cl +O → ClO +O (1) 3 2 ClO +O → Cl+ O 2 (3) 3. In question 2, if ClO reacts with NO instead of with O, do you still get a catalytic cycle for3O loss? Briefly explain. 4. Write a catalytic cycle fo3 O loss involving the formation of HOCl by reaction (6). How does this mechanism compare in importance to the one in question 2? 5. Calculate the lifetime of the chemical family CxO defined as the sum of Cl and ClO. Compare to the lifetime of ClO. What do you conclude? [To know more: McElroy, M.B., et al., The changing stratosphere, Planet Space Sci., 40, 373-401, 1992.] 10. 6 Partitioning of Cl y The POLARIS aircraft mission to the arctic in the summer of 1997 provided the first in situ simultaneous measurements of HCl and ClNO 3 in the lower stratosphere (20 km altitude). These data offer a test of current understanding of chlorine chemistry. According to current models, the principal Cl y cycling reactions operating under the POLARIS conditions should be Cl +O 3 ClO +O 2 (1) ClO +NO →Cl +NO (2) 2 ClO + NO M → ClNO + M (3) 2 3 ClNO + hν → Cl +NO (4) 3 3 Cl +CH 4 → HCl +CH 3 (5) 194 HCl +OH → Cl +H O 2 (6) NO +O → N3 + O 2 2 (7) O NO + hν → NO +O (8) 2 3 with rate constants k = 8x10 -12cm molecule -1 s , k = 7x10 -12 cm molecule -1 1 2 s , k 3 1.3x10 -13 cm molecule s , k = 7x40 s , k = 3x10 5 -14cm molecule -1 -1 -13 3 -1 -1 s , and k 6 5x10
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