1. Copyright © 2015 R.R. Dickerson1 Introduction to Kinetics AOSC 620 Fall 2015 R. Dickerson See Finlayson-Pitts Chapter 5 Seinfeld and Pandis Chapters 4 & 9

2. Copyright © 2013 R.R. Dickerson2 Kinetics A. Rates, rate constants, and reaction order. Thermodynamics tells us if a reaction can proceed and gives equilibrium concentration. Kinetics tells us how fast reactions proceed. If thermodynamics alone controlled the atmosphere it would be dissolved in the oceans as nitrates - we would be warm puddles of carbonated water. 1. First Order Reactions A  PRODUCTS EXAMPLES 222 Rn -> 218 Po +  N 2 O 5  NO 2 + NO 3 (forward reaction only)

3. Copyright © 2013 R.R. Dickerson3 The red line describes first order loss with a rate constant of 1 min -1 The blue line is the rate of formation of the product. minutes

4. Copyright © 2013 R.R. Dickerson4 Radon is important source of indoor air pollution, and N 2 O 5 is nitric acid anhydride, important in air pollution nighttime chemistry. The rate equations take the form: d[Prod.]/dt = k[A] = -d[React.]/dt For example: d[Po]/dt = k Rn [Rn] = -d[Rn]/dt Where k is the first order rate constant and k has units of time -1 such as s -1, min -1, yr -1. We usually express concentration, [Rn], in molecules cm -3 and k in s -1.

5. Copyright © 2013 R.R. Dickerson5 Another example, nitric acid anhydride. d[NO 2 ] /dt = k [N 2 O 5 ] and d[N 2 O 5 ]/dt = -k [N 2 O 5 ] Integrating ln ([ N 2 O 5 ] t / [ N 2 O 5 ] o ) = -k  t If we define the starting time as zero: [N 2 O 5 ] t / [N 2 O 5 ] o = exp(-kt)

6. Copyright © 2013 R.R. Dickerson6 The rate constants for these first order reactions are: k Rn = 0.182 days -1 k N2O5 = 0.26 s -1 (at room temperature)

7. Copyright © 2013 R.R. Dickerson7 2. Second Order Reactions A + B  PRODUCTS EXAMPLES NO + O 3  NO 2 + O 2 HCl + OH  H 2 O + Cl Examples of the rate equations are as follows: d[NO]/dt = -k[NO][O 3 ] d[Cl]/dt = k[OH][HCl] Units of k are {conc -1 time -1 }. 1/(molecules/cm 3 ) (s -1 ) = cm 3 s -1

8. Copyright © 2013 R.R. Dickerson8 For our second order kinetics examples Units of k are {conc -1 time -1 }. 1/(molecules/cm 3 ) (s -1 ) = cm 3 s -1 Rate constants have the following values: k NO-O3 = 1.8x10 -14 cm 3 s -1 k HCl-OH = 8.0x10 -13 cm 3 s -1

9. Copyright © 2013 R.R. Dickerson9 3. Third Order Reactions A + B + C  PRODUCTS d[A]/dt = -k[A][B][C] Examples 2NO + O 2  2NO 2 O + O 2 + M  O 3 + M

10. Copyright © 2013 R.R. Dickerson10 2NO + O 2  2NO 2 O + O 2 + M  O 3 + M M is any third body (usually N 2 ) needed to dissipate excess energy. From the ideal gas law and Avg's number: Where M o is the molecular number density at STP. Third order rate constants have units of conc -2 time -1. These are usually (cm -3 ) -2 s -1. k NO-O2 = 2.0 x 10 -38 cm 6 s -1 k O-O2 = 4.8 x 10 -33 cm 6 s -1

11. Copyright © 2013 R.R. Dickerson11 USEFUL IDEA: For the following reaction: A + B  C + D d[C]/dt = k f [A][B] - k r [C][D] At steady state d[C]/dt = 0, by definition. Thus: { k f }/{ k r } = {[C][D]}/{[A][B]} = K eq

12. Copyright © 2013 R.R. Dickerson12 Halflife and Lifetime Definition: Halflife, t 1/2 the time such that: [A] t1/2 / [A] 0 = 1/2 Definition of lifetime or residence time, , comes from kinetics, where k is the first order rate constant with units of time -1. We know that: [A] t /[A] 0 = exp(-k t) The lifetime, , is when t = 1/k so   1/k We can link half-life and lifetime: t 1/2 = ln(2)/k  0.69/k

13. Copyright © 2013 R.R. Dickerson13 For radon 222 ( 222 Rn) the lifetime is 5.5 days, but the half-life is only 3.8 days. For second order reactions we need pseudo first order conditions For example: NO + O 3  NO 2 + O 2 k = 1.8 x 10 -14 cm 3 s -1 ASSUME: [O 3 ] >> [NO] and d[O 3 ]/dt ~ 0.0 Let: = 50 ppb (a reasonable value for air near the surface).  NO = = 1/{1.8 x 10 -14 x 50x10 -9 x 2.5x10 19 } = 44 s CONCLUSION: any NO injected into such an atmosphere (by a car for example) will quickly turn into NO 2, if there are no other reactions that play a role. We will call k[O 3 ] the pseudo first order rate constant.

14. Copyright © 2013 R.R. Dickerson14 For example: O + O 2 + M  O 3 + M k = 4.8x10 -33 cm 6 s -1 ASSUME: d[O 2 ]/dt = d[M]/dt = 0.0 We know that [O 2 ] = 0.21 and that [M] ~ [O 2 ] + [N 2 ] ~ 1.00. At RTP P 02 = 0.21 atm and P M ~ 1.0 atm. Therefore the lifetime of O atoms is  = [ 4.8x 10 -33 x 0.21 x 1.0 x (2.5x10 19 ) 2 ] -1 = 1.6x10 -6 s VERY SHORT! For third order reactions we must assume that two components are constant.

15. Copyright © 2013 R.R. Dickerson15 Example 2 Same reaction at stratospheric temperature and pressure. P 30km ~ P 0 exp(-30/7) = 0.014 atm  = [4.8x10 -33 x 0.21 x 1.0 x (0.014 x 2.5x10 19 ) 2 ] -1 = 2.1x10 -3 s The lifetime of an O atom is is still short, but it is a thousand times longer than in the troposphere! The pressure dependence has a major impact on the formation and destruction of tropospheric and stratospheric ozone.

16. Copyright © 2013 R.R. Dickerson16 If two of the reactants in a third order reaction are the same, we can derive a useful expression for the rate of loss of the reactant. A + A + B  PROD For a great excess of B: d[A]/dt = -(2k[B])[A] 2 [A] -2 d[A] = -(2k[B])dt Integrating from 0 to time t [A] -2 d[A] = -(2k[B])dt -[A] t -1 + [A] 0 -1 = -(2k[B])t [A] t -1 = 2k[B]t + [A] 0 -1 Now we can calculate the concentration at any time t in terms of the initial concentration and the rate constant k.

17. Copyright © 2013 R.R. Dickerson17 Transport and Evolution of a Pollution Plume from Northern China: A Satellite-based Case Study Can Li 1, Nick Krotkov 2, 3, Russ Dickerson 1, Zhanqing Li 1, 4, and Mian Chin 2 1 AOSC, UMD 2 GSFC, NASA 3 GEST, UMBC 4 ESSIC, UMD Why China? A lot of emissions… Local, regional, and global effects Why transport and evolution? Key factors determining the large- scale impact of air pollution (e.g., conversion from SO 2 to sulfate aerosols, adding CCN to the system) Why satellites? Transport episodes associated with synoptic weather system, and are of regional scale – new satellite sensors provide great spatial coverage, daily observation, and sensors of different strengths can be combined

18. Copyright © 2013 R.R. Dickerson18 On April 5,2005 on the aircraft: Lots of SO 2 Lots of dust ~1 hr later and from space: Lots of SO 2 Lots of dust

19. Copyright © 2013 R.R. Dickerson19 Trajectory model projects movement of the plume, satellite sensors take snapshots every day Trajectory Projection Satellite Snapshots AMF Correction to operational product, combination of satellites and models, and MODIS AOD data SO 2 lifetime: 1-4 day SO 2 to sulfate: ~0.1-0.2 increase in AOD near plume core in one day Uncertainties? Details? Suggestion? Please come to see our poster.

20. The comparison demonstrates that operational OMI algorithm can distinguish between heavy pollution ( April 5 ahead of cold front ): SO2 OMI SO 2 AQUA- MODIS RGB

21. …and background SO 2 conditions (on April 7 behind cold front) SO2 OMI SO2 MODIS RGB

22. SO 2 lifetime estimate By following the path of the air mass, and correcting for the altitude effect, the amount of SO 2 lost to dry deposition and chemical reactions was approximated. Ln (C/C 0 ) = -kt The mass decreases from the 5 th to the 7 th looks to be first order with a lifetime of ~4 d. The mass on April 8 shows no change from April 7 because the air mass was very disperse (large box) and mixed with other plumes.

23. SO 2 + OH  prod (H 2 SO 4 ) If attack by OH is the only sink for SO 2, and if the lifetime is ~4 d =~ 3x10 5 s. Lifetime =  = ([OH]*M*k) -1 k = 2x10 -12 cm -3 s -1 Molecular number density of OH = 1.4x10 6 cm -3