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QUESTIONS PowerPoint Presentation

QUESTIONS

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QUESTIONS

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  1. QUESTIONS 1. What is the most abundant greenhouse gas in the Earth’s atmosphere? 2. What is the radiative equilibrium temperature of the Earth’s surface at the poles in winter? What prevents this temperature from ever being achieved? 3. How would you observe savanna and forest fires from space? (Clue: fires are hot). 4. Clouds are the wild card in predictions of future climate change. They reflect solar radiation, thereby cooling the Earth; but they also absorb in the IR, thereby warming the Earth.  Whether a cloud has a net warming or cooling effect depends mainly on its altitude. Why?

  2. CLIMATE FEEDBACK FROM HIGH vs. LOW CLOUDS Clouds reflect solar radiation (DA > 0) g cooling; …but also absorb IR radiation (Df > 0) g warming WHAT IS THE NET EFFECT? sTcloud4 < sTo4 sTcloud4≈ sTo4 Tcloud≈ To convection sTo4 sTo4 To HIGH CLOUD: WARMING LOW CLOUD: COOLING

  3. PRINCIPLES OF PHOTOCHEMISTRY

  4. SUN AND EARTH E/M SPECTRA As we saw last class…Emission = f(T) [Stefan-Boltzmann Law] NOTE: Sun Planck function actually much larger (higher T), normalized here

  5. RADIATION IN THE ATMOSPHERE Incoming radiation: solar (blackbody emission at 5800K). Radiation drives atmospheric chemistry by dissociating molecules into fragments that are often highly reactive. Photon energy per mole: Radiation of interest for tropospheric chemistry is from visible (~700 nm) to near-UV (~290 nm). The corresponding energies are sufficient to break chemical bonds such as: weak O2-O bond in ozone (~ 100 kJ/mol) moderately strong C-H bond in formaldehyde(~368 kJ/mol) nm Shorter wavelengths are attenuated as they travel through the atmosphere (by molecular O2 and N2 and ozone)

  6. MAJOR PHOTODISSOCIATING SPECIES A photon (hv) is a reactant. Reminder: λ> 290 nm only in the troposphere! NO2 + hv  NO + O < 420 nm O3 + hv  O(3P) + O2 315 < < 1200 nm O(1D) + O2 < 315 nm HNO2 + hv  NO + OH < 400 nm H2O2 + hv  2OH NO3 + hv  NO + O2 NO3 “stores” NOx at night NO2 + O HCHO + hv  HCO + H CO + H2 dominant path for > 320 nm

  7. EXAMPLE: PHOTODISSOCIATION OF OXYGEN Estimate the wavelength of light at which photodissociation of O2 into 2 ground-state oxygen atoms: O2 + hv  O + O The enthalpy for this reaction is H=498.4 kJ/mol (endothermic) So O2 cannot photodissociate at wavelengths longer than about 240 nm

  8. LIGHT: REFLECTING, SCATTERING AND ABSORBING GASES: GHGs are absorbing in the IR(as seen for example for CO2) Gases can scatter in the UV/visible  Rayleigh scattering AEROSOLS: Absorption depends on composition (eg. black carbon) Scattering explained by Mie Theory  reduction in visibility

  9. BEER-LAMBERT LAW AND OPTICAL DEPTH Beer-Lambert Law:Attenuation of radiation  This holds if number density does not vary significantly over dx I = radiation flux =cross section [cm2/molecules] n = number density l = path length (x2-x1)  = optical depth [dimensionless] = solar zenith angle (SZA) n, σ If account for angle of light transmitted to Earth’s surface (at angle): Top of Atmosphere zTOA l=zTOA/cos() 

  10. ACTINIC FLUX ACTINIC RADIATION: the integrated radiation (photon flux) from all directions to a sphere (sum of direct, scattered, reflected light) ACTINIC FLUX (J): Number of photons absorbed by species X: ()=absorption cross section [cm2/molecules] J()=actinic flux [photons/cm2/s] Fa()=number of photons abs by X [photons/cm3/s] [X] is in units of number density • Actinic flux will be modified by SZA (thus time, season, latitude), as well as by scattering and absorption by gases and particles. The absorption is mostly from stratospheric O3 (the ozone column). • Must also consider not only direct solar, but also scattered/ • reflected radiation  need surface albedo estimates • Computational codes developed to calculate actinic flux for • different constituent profiles, angles, locations

  11. COMPUTING RATES OF PHOTOLYSIS Molecule is excited into an electronically excited state by absorption of a photon: A + hv  A* The excited molecule may release the absorbed energy by any of: 1. Dissociation A*  B1 + B2 2. Direct Reaction A* + B  C1 + C2 3. Fluorescence A*  A + hv 4. Collisional deactivation A* + M  A + M 5. Ionization A*  A+ + e- The relative efficiency of each of these is described by the quantum yield (i): number of excited molecules of A* undergoing a process (i) to the total number of photons absorbed. By Stark-Einstein Law, I = 1 You may come across the “overall quantum yield of a stable product A” (A) which is defined as the number of molecules of A formed over the number of photons absorbed. A > 1 for a chain reaction

  12. COMPUTING RATES OF PHOTOLYSIS (CONT’D) Example: Two photochemical processes for formaldehyde to produce (H+HCO) or(H2+CO) thus, ()= H+HCHO + H2+CO To compute the rate of disappearance of HCHO according to 1st rxn use only H+HCHO The total rate of photolysis of X: Rate [molecules/cm3/s] j = photolysis rate constant [s-1] Note, the use of j distinguishes the photolysis rate constant from other rate constants (k).

  13. DISCOVERY OF PHOTOCHEMICAL REACTIONS IN THE ATMOSPHERE • Some observations which remained unexplained until ~1960: • NO is oxidized to NO2 in photochemical smog •  the known thermal oxidation rxn is too slow: 2NO + O2  2NO2 • Organics are rapidly oxidized during smog formation • Known before 1970, for example for propylene were rxns: • a. C3H6 + O3 products • b. C3H6 + O  products • with rates too slow compared to propylene loss rates Observed Reaction a Propylene Loss Rate Reaction b Time (min) Leighton (1961) speculated that free radicals might be formed from organics and involved in oxidation: R alkyl (formed from any hydrocarbon group, eg. CH4, C2H8) RO2 alkyl peroxy RO alkoxy OH hydoxyl HO2 hydroperoxy H hydrogen free radical

  14. DISCOVERY OF PHOTOCHEMICAL REACTIONS IN THE ATMOSPHERE (cont’d) In mid 1960’s reactions of CO or hydrocarbons with OH were found to be rapid. Chain reactions were proposed in 60’s which regenerate OH, convert NONO2 and involve HC species: CO + OH  H + CO2 (1) H + O2 + M  HO2 + M (2) HO2 + NO  OH + NO2 (3) Note the chemical effects: oxidation of CO (CO  CO2) conversion of NO  NO2 (as observed to occur in atmosphere) reactions are fast (as observed in atm) free radicals are used up, but also created in each step This sequence (1-3) is important in the clean troposphere. In the polluted air, organic species play a role similar to that of CO.

  15. GENERATION OF OH OH radical drives the daytime chemistry of both polluted and clean atmosphere What is the source of OH (and other radicals)? Major source: O3 + hv ( 320 nm)  O(1D) + O2 O(1D) + H2O  2OH Other sources: HONO + hv (< 400 nm)  OH + NO nitrous acid (photodissociation) H2O2 + hv (< 370 nm)  2OH hydrogen peroxide HO2 + NO  OH +NO2 (sources and sinks of HO2 effectively sources and sinks of OH  HOx)

  16. BASIC PHOTOCHEMICAL CYCLE OF NO2, NO AND O3 • NOx is released in combustion processes, also saw that there are natural sources, such as lightning. The following is a “fast” photochemical cycle with no net consumption or production of species: • NO2 + hv  NO + O (1) • O + O2 + M  O3 + M (2) • O3 + NO  NO2 + O2 (3) O3 hv NO2 NO O3 • What is steady state [O3]? • SS for O: R2=R1  [O] depends on [NO2] • SS for NO2: R1=R3 • SS for NO: R3=R1 • SS for O3: R3=R2  sub R1 for R2 from above: This is the photostationary state relation. The steady state concentration of ozone is controlled by the ratio of NO2 to NO here. However, 3 rxn cycle is incomplete for predicting ozone concentrations  don’t forget carbon compounds!

  17. FREE RADICAL KINETICS EXAMPLE: ETHANE PYROLYSIS C2H6 C2H4 + H2 (*) Step 1: inititation (generates free radicals) C2H6 + M  2 CH3 + M (1) (thermal initiation step) Step 2: chain propagation (free radical  free radical) CH3+ C2H6  CH4 + C2H5 (2) C2H5 + M  C2H4 + H + M (3) H + C2H6  H2+ C2H5 (4) Step 3: termination: two free radicals combine to form a stable molecule 2H  H2 (5) H + C2H5  C2H6 (6) H + C2H5 C2H4+ H2 (7) H + CH3  CH4 (8) CH3 + C2H5  C3H8 (9) 2C2H5  C4H10 (10) C2H4 C2H6 H2 • Products we expect are generated in (3), (4), (5) and (7) • “Side products are generated in (2), (8), (9), (10) • If there were no termination reaction, (1) would only need to occur once to start the “chain” • The “chain length” is the average number of times the chain sequence is repeated before a chain-propagating radical is terminated • Rate is NOT equal to k[C2H6]! Need reaction mechanism to describe the kinetics, even if net result is described by (*)

  18. FREE RADICAL KINETICS EXAMPLE: ETHANE PYROLYSIS C2H6 + M  2 CH3 + M (1) CH3+ C2H6  CH4 + C2H5 (2) C2H5+ M  C2H4 + H + M (3) H + C2H6  H2 + C2H5 (4) 2H H2 (5) H + C2H5  C2H6 (6) H + C2H5  C2H4 + H2 (7) H + CH3  CH4 (8) CH3 + C2H5  C3H8 (9) 2C2H5 C4H10 (10) • To determine overall rate: Each step is an “elementary” reaction so can be written: • R1 = k1[C2H6][M] • d[C2H6]/dt = -R1 + …. • OPTIONS: • Write mass conversation eqns for all species and solve ODEs numerically to get [C2H6](t) • Use simplifying assumptions to get analytical expression for d[C2H6]/dt Assumption: SS applies to free radicals: for H: R3 = R4 (consider only propogation reactions for now) analysis shows k3[M] << k4[C2H6] so can argue that for termination (10) is most important for CH3: R2 = 2R1 for C2H5: R3 + 2R10 = R2 + R4  sub in above, find R10=R1 Plug results into rate equation for ethane to get: rxn order = 1/2 If assume long chain lengths, can ignore initiation (R1)