ATM 507 Lecture 5 Text reading Chapter 4 Problem Set #2 due Sept. 20 Today s topics Photochemistry and Photostationary State Relation
Beer-Lambert Law (for the absorption of light) Used to describe the absorption of light passing through a moderately weak absorber. I(λ) = I 0 (λ) exp{-(abs. coeff.)(conc.)(path length)} Formulation from Chemistry textbooks log 10 (I 0 /I) = ε C l I 0 = incident monochromatic (or narrow) intensity I = transmitted intensity ε = molar extinction coefficient (base 10) = ε(λ, T, C) C = concentration (moles li -1 or M) l = absorption pathlength in cm
Beer-Lambert Law (cont.) Gas Phase Absorption (pressure units) I = I 0 exp (-k p l ) or ln(i 0 /I) = k p l k = absorption coefficient in atm -1 cm -1 p = absorber partial pressure in atm Gas Phase Absorption (absolute units) I = I 0 exp (-σ N l ) or ln(i 0 /I) = σ N l σ = absorption cross-section in cm 2 molecule -1 N = absorber density in molecules cm -3 For a mixture of gases I = I 0 exp [-(σ 1 N 1 + σ 2 N 2 + σ 3 N 3 + ) l ]
Absorption by a Thin Layer of Gas At one wavelength I absorbed = I 0 I = I 0 { 1 exp(- σ N l ) } Absorption = A = (I o I)/I o = 1 exp(- σ N l ) Look at a Taylor series expansion of 1 exp(- σ N l ) 1 exp(- σ N l ) = σ N l - (σ N l ) 2 /2! +(σ N l ) 3 /3! - For small values of (σ N l) keep only the first term I a I 0 (σ N l ) How much error is introduced by this approximation? Optical Depth (σ N l ) % error 0.01 1.2 0.05 5.9 0.10 12.0 0.20 24.5
Thin Layer Absorption (cont.) The approximate expression links the absorbed light to the incident light and absorption cross-section ( and N and l). If we consider a thin layer, the light absorbed per unit length is (# photons absorbed) cm -3 s -1 = I a / l = I 0 σ N l / l = I 0 (λ) σ(λ) N If we identify I 0 (λ) as the Actinic Flux F λ (this is the flux at one wavelength no dependence on wavelength interval) I abs = F λ σ(λ) N Light absorbed per unit volume per unit time = Spherically Integrated Actinic flux x Absorption Crosssection x Concentration Photons cm -3 s -1 = Photons cm -2 s -1 x cm 2 molecule -1 x molecules cm -3 These absorbed photons are the ones capable of producing the chemical change we are interested in.
Photodissociation Reactions described as Rate Expressions A + h (A*) B + C -da/dt = k phot [A] k phot = photolysis or photodissociation rate constant depends on F(λ), σ(λ), Φ(λ), etc. Recall that I abs = F λ σ(λ) [A] at wavelength λ This expression considers all absorbed photons if we restrict our consideration to only the photons effective for a particular process, the quantum yield is added to the equation I abs = F λ σ(λ) Φ(λ) [A]
Photolysis Rate Constants If the expression for I abs is integrated over wavelength, the result is a photolysis rate expression (Rate expression) = (Integral of all photons that cause reaction) k phot [A] = F(λ) σ(λ) Φ(λ) d λ [A]; or k phot = F(λ) σ(λ) Φ(λ) d λ If the quantities F, σ, and Φ are not available in functional form, but are from a table, one can approximate the integral by a sum λ k p = (F λ σ(λ) Φ(λ)) {or k p = (F(λ) σ(λ) Φ(λ) Δλ} These are important results the second form is used in Problem set #3 K p is a first order rate constant with units (time) -1. λ
Chemical Compounds found in the Troposphere which Absorb Sunlight ( = 290-600 nm) NO 2 (Fig. 4.16), NO 3, O 3 (Fig. 4.13), SO 2 Nitrites HONO, RONO Aldehydes H 2 CO, Ketones & dicarbonyls acetone, glyoxal, Nitrates HONO 2, RONO 2, (PAN) Peroxides H 2 O 2, ROOH, ROOR, (PAN) Polynuclear Aromatics* Aerosols* For most of these species, at least some absorption events cause dissociation the exceptions are marked with an *.
Chemical Compounds which are not significant absorbers in the UV-visible ( = 290-600 nm) NO N 2 H 2 O CO and CO 2 H 2 SO 4 Alkanes (parafins) Alkenes (olefins) Alcohols Organic Acids
Absorption Spectrum of NO2
Quantum Yield for NO 2 +hv NO + O 400 nm
Calculated and Measured kp for NO2
Absorption Spectrum of NO3
Photochemical Reactions in the Atmosphere Consider perhaps the central reaction cycle for tropospheric chemistry: NO 2 + hv NO + O; k 1 0.5 min -1 (depends on sun) O + O 2 + M O 3 + M; k 2 300 = 6x10-34 cm 6 molecule -2 s -1 NO + O 3 NO 2 + O 2 ; k 3 300 = 1.9x10-14 cm 3 molecule -1 s -1 (Know these reactions and understand this analysis thoroughly!) NO 2 photolysis is the major source of O 3 production in the troposphere (the other major source is transport from the stratosphere this transported O 3 provides the bulk of the tropospheric background ozone ). The species involved are NO, NO 2, O, O 2, O 3, and M where M is an air molecule, [M] is the air concentration, and [O 2 ] 0.21 [M]
Reaction Rates R 1 = k 1 *[NO 2 ] R 2 = k 2 *[O]*[O 2 ]*[M] R 3 = k 3 *[NO]*[O 3 ] NO 2 is produced in reaction 3 and lost in reaction 1 NO is produced in reaction 1 and lost on reaction 3 O 3 is produced in reaction 2 and lost in reaction 3 O is produced in reaction 1 and lost in reaction 2
For each species we can write an equation of the form: d[species]/dt = Production Loss = P - L where Production is the combined rate of all production reactions, and Loss is the combined rate of all loss reactions. 1. d[no 2 ]/dt = -k 1 [NO 2 ] + k 3 [NO] [O 3 ] 2. d[no]/dt = k 1 [NO 2 ] - k 3 [NO] [O 3 ] = - d[no 2 ]/dt 3. d[o 3 ]/dt = k 2 [O][O 2 ][M] - k 3 [NO] [O 3 ] 4. d[o]/dt = k 1 [NO 2 ] - k 2 [O][O 2 ][M]
Steady State Analysis How do we explore the behavior of this (and other) system(s) of reaction-based equations? A common tool is Steady State Analysis, also called Steady State Approximation, or Stationary State Approximation or SSA for short. Since these equations describe time rates of change, it is quite appropriate to look at the chemical lifetimes (reaction lifetimes) of the species involved and make approximations based on the relative time scales that result. For example, in the above set of reactions (system of equations), the chemical lifetime of the oxygen atom is more than 6 orders of magnitude shorter than the chemical lifetimes of NO, NO 2, and O 3. Therefore, as far as NO, NO 2, and O 3 are concerned, O reacts instantaneously that is, as soon as it is produced it reacts away and is lost. We can write this as P = L; or Production = Loss in the general equation above. (Of course, this is an approximation and only strictly true for times much longer than the O atom lifetime, but in this case that gives us a lot to work with!)
Steady State Analysis (cont.) Put another way, O atom is the most reactive species (by a lot), so we invoke SS on it: d[o] ss /dt = 0 = k 1 [NO 2 ] - k 2 [O][O 2 ][M] or k 2 [O][O 2 ][M] = k 1 [NO 2 ] at SS [O] ss = (k 1 [NO 2 ])/ (k 2 [O 2 ][M]) Plug this result into the equation for d[o 3 ]/dt: d[o 3 ] ss /dt = k 2 [O 2 ][M]{k 1 [NO 2 ]/ k 2 [O 2 ][M]} - k 3 [NO] [O 3 ] = k 1 [NO 2 ] - k 3 [NO] [O 3 ] = - d[no 2 ] ss /dt Combined with our earlier result, this yields d[o 3 ] ss /dt = d[no] ss /dt = -d[no 2 ] ss /dt The only way all three time rates of change can be equal is if they are equal to zero, which means that NO, NO 2 and O 3 are all in steady state. (If only these reactions are occurring, this result must stand!)
Photostationary State Relation or Since d[o 3 ]/dt =0, Leighton Relation [ O 3 ] SS k 1 k [ NO 3 2 ] [ NO] SS SS Very important result! Covered in Chapter 6 Tropospheric Chemistry Implications [NO 2 ] increases, or light increases [O 3 ] increases
Examine the SSA Look at the numerical lifetimes τ NO2 = 1/k 1 1/(0.5 min -1 ) = 2 min = 120 s τ O = 1/(k 2 [O 2 ] [M]) = {(6x10-34 )(.21)(2.45x10 19 ) 2 } 10-5 s τ O <<< τ NO2 and SSA is justified. Guidelines for SSA 1. Invoke SS for most reactive specie (or species) in a reaction set and solve for other species. 2. If there is any doubt, compare chemical lifetimes to justify SSA. As makes sense from τ NO2 above, with regard to ozone production in the troposphere, if τ A << 1 min, SSA is likely to be okay.
SSA Tips and Guidance When faced with a set of equations that have more unknowns than equations, use SSA on the most reactive species: Start by invoking SS for O, OH, HO 2, Cl, etc. (the radical species) If necessary then invoke SS for O 3 and other reactive species like NO 3, etc.
The Stratosphere Extends from 10-15 km to ~ 50 km At 15 km, T 210-220 K, p 100 mbar [M] 3.4x10 18 molecules cm -3 At 50 km, T 270-280 K, p 1 mbar [M] 2.64x10 16 molecules cm -3 Positive temperature gradient vertically stable molecules take years to diffuse from bottom to top of the stratosphere Key constituent and focus for our analysis is Ozone
Ozone Upper scale Temperature Lower scale
Ozone Chemistry Main source region equatorial midstratosphere (Produced by photolysis where the photons are ) Mean transport is poleward and downward greatest concentrations are in the polar lower stratosphere (exclusive of ozone holes, of course!) In the absence of losses, ozone pools in the cold & dark.
Ozone source and sink regions; main transport.
Dynamical Aspects of Strat-Trop Exchange 2 PVU contour Holton, Reviews of Geophysics, 1995
Another Look From Scientific Assessment of Ozone Depletion:1998 (WMO)
Water Vapor Water vapor is a specie that has a very large change from the troposphere to the stratosphere (i.e., larger than ozone) Troposphere wet (sometimes a percent or more 10,000 + ppmv) Stratosphere dry (a few ppmv) Part of the dryness of the stratosphere has to do with very cold temperatures at the tropopause freeze drying any ascending air.
Stratospheric Water Vapor (U2 aircraft- 9/11/1980) Water Vapor Blue Temperature - Red
Back to Ozone What do measurements show? Sometimes we want absolute ozone concentrations And sometimes we want to know how much ozone is in the column above a given point
Measurement Techniques for Stratospheric O3
Zonally Averaged [O 3 ] vs. Altitude
hole).
Total Ozone Column in Dobson Units
Absorption by ozone stops the penetration of potentially harmful UV radiation. Red line corresponds to model calculation of surface radiation if stratospheric ozone was 10% lower.