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In 1780, the famous English chemist Joseph Priestley (right) found that plants could "restore air which has been injured by the burning of candles." He.

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Presentation on theme: "In 1780, the famous English chemist Joseph Priestley (right) found that plants could "restore air which has been injured by the burning of candles." He."— Presentation transcript:

1 In 1780, the famous English chemist Joseph Priestley (right) found that plants could "restore air which has been injured by the burning of candles." He used a mint plant, and placed it into an upturned glass jar in a vessel of water for several days. He then found that "the air would neither extinguish a candle, nor was it all inconvenient to a mouse which I put into it". In other words, he discovered that plants produce oxygen. A few years later, in 1794, the French chemist Antoine Lavoisier (left), discovered the concept of oxidation, but soon after was executed during the French Revolution for being a Monarchist sympathiser. The judge who pronounced sentence said "The Republic has no need for scientists".

2 mass of the atmosphere: 5 x 10 18 kg

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5 Fast vertical mixing Mixing across tropopause is slow Fast vertical mixing Direct interface with surface T  w/ altitude (adiabatic lapse rate) Rapid vertical mixing High T because of: high E of bomb. radiation Slow vertical mixing T  w/ altitude because of h abs. by O 3 T  w/ altitude because h abs. by O 3 less important

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8 © 2012 W. H. Freeman & Co.

9 Skin cancer

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11 © 2012 W. H. Freeman & Co.

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13 14a–13

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18 (wavelength) x (frequency) = speed λν = c [3x10 8 m/s]

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20 E = h ν h = 6.6 x 10 -34 J.s Light as particles or photons With energy, E, proportional to frequency, ν

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22 ~ 450-750 nm Take 500 nm

23 Copyright © Houghton Mifflin Company. All rights reserved. 14a–23 Electromagnetic spectrum λ ν

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25 E = h ν

26 Chapman Mechanism (~1930, Sidney Chapman) O 2 + h  O + O 498 kJ/mol  <242.4 nm O + O 2 + M  O 3 + M O 3 + h  O + O 2 383 kJ/mol  <280 nm

27 Chapman Mechanism (~1930, Sidney Chapman) O 2 + h  O + O 498 kJ/mol  <242.4 nm O + O 2 + M  O 3 + M O 3 + h  O + O 2 383 kJ/mol  <280 nm

28 Chapman Mechanism (~1930, Sidney Chapman) O 2 + h  O + O 498 kJ/mol  <242.4 nm O + O 2 + M  O 3 + M O 3 + h  O + O 2 383 kJ/mol  <280 nm

29 Chapman Mechanism (~1930, Sidney Chapman) O 2 + h  O + O 498 kJ/mol  <242.4 nm O + O 2 + M  O 3 + M O 3 + h  O + O 2 383 kJ/mol  <280 nm O + O 3  2O 2 O + O + M  O 2 + M

30 Ozone production λ ≤ 240 nm λ ≤ 280 nm h h Ozone destruction

31 © 2012 W. H. Freeman & Co.

32 λ ≥ 280 nm

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39 1 Dobson unit (DU) is: 0.01 mm of pure O 3 at 1 atm. and 0 o C NB: what does 300 DU expand to at 10 -6 atm?

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41 Ozone production Photochemical kinetics λ ≤ 240 nm λ ≤ 280 nm h h WRONG EQUILIBRIUM: TOO MUCH OZONE

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43 Chlorine reservoirs © 2012 W. H. Freeman & Co.

44 Chlorine reservoirs © 2012 W. H. Freeman & Co.

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47 Recipe for ozone loss: 1.POLAR WINTER  Polar vortex  Isolates air 2.LOW TEMPERATURE  PSC (=Polar Stratospheric Clouds) (=Nitric acid – HNO 3 ) 3.SURFACE CHEMISTRY  Cl 2 (+denoxification) 4.LIGHT RETUNS  Cl*  Catalytic destruction of O 3

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50 Catalytic Decomposition of Ozone X + O 3  XO + O 2 XO + O  X + O 2 _____________________________ O 3 + O  2O 2 X = HO x (  H,  OH, HOO  ) NO x (NO ,  NO 2 ) ClO x (  Cl, ClO  )

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54 CHAPTER 02: Figure 2.1 © 2012 W. H. Freeman & Co.

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57 CHAPTER 02: Figure 2.2 © 2012 W. H. Freeman & Co.

58 NASA projections of stratospheric ozone concentrations if chlorofluorocarbons had not been banned.

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60 Mount manitoba, June 1991

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63 CHAPTER 02: Figure 2.10 © 2012 W. H. Freeman & Co.

64 PrefixMeaningAtoms in the Molecule CFCchlorofluorocarbonCl, F, C HCFChydrochlorofluorocarbonH, Cl, F, C HBFChydrobromofluorocarbonH, Br, F, C HFChydrofluorocarbonH, F, C HChydrocarbonH, C PFCperfluorocarbonF, C HalonN/A Br, Cl (in some but not all), F, H (in some but not all), C

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