1. Chemical Box Models Markus Rex Alfred Wegener Institute Potsdam Germany (1) Basic concepts, simplified systems (Saturday) (2) The O x, NO y /NO x, HO x, Cl y /ClO x systems (Monday) (3) Application for polar ozone loss studies (Thursday)

2. Chemical Models Goal: Calculate the time evolution of the concentration of chemical species. E.g. for the a species A this can be done... fixed grid points wind... for an individual air mass: Lagrangian formulation -> box model chemical changesdynamical changes production lossadvection divergence   air mass with trajectory wind => In our box model we focus on chemistry alone ! disappears if vmr is used instead of conc. subscale mixing and molecular diffusion are neglected... at a fixed location: Eulerian formulation -> grid point model (e.g. 1d, 2d, 3d)

3. What governs chemical reactions in a system of species ? To build a chemical box model we need to understand what governs chemical reactions in the atmosphere and how to interpret published thermodynamical and kinetic data Thermodynamics of chemical processes Chemical thermodynamics describe the energetic balance of chemical processes. This tells us which processes can principally occur and whether they consume or or release energy. Kinetics of chemical processes Reaction kinetics describe at which rate chemical reactions occur.

4. Reaction thermodynamics The energetic balance of reactions Goal: calculate the energy consumed or released by a chemical reaction In general reactions can only occur if they have a positive energy balance e.g. the photolysis of ozone O 2 + h -> O + O Reaction thermodynamics tell how much energy the photon need to supply for this reaction.

5. Process at constant volume: Things are simple ! energy supplied to the system (  Q) by the photons is just the change of 'internal energy' (  U).'internal energy'  U =  Q(constant volume) In the atmosphere reactions occur at constant presure => Things are less simple ! If number of molecules is changed by the reaction, the volume changes and produces mechanical work (in case of positive volume change a negative contribution to energy balance of the system):  W = - p.  V  U =  Q +  W Enthalpy: H = U + pV With H, things again become simple:  H =  Q(constant pressure) Process at constant pressure:

6. Process at constant volume: Things are simple ! energy supplied to the system (  Q) by the photons is just the change of 'internal energy' (  U).'internal energy'  U =  Q(constant volume) In the atmosphere reactions occur at constant presure => Things are less simple ! If number of molecules is changed by the reaction, the volume changes and produces mechanical work (in case of positive volume change a negative contribution to energy balance of the system):  W = - p.  V  U =  Q +  W Enthalpy: H = U + pV With H, things again become simple:  H =  Q(constant pressure) Process at constant pressure: Relevant forms of internal energy for atmospheric chemical processes kinetic energy energy of rotation energy of vibration potential energy in the molecular bonds potential energy of the electron shell chemical reactions chemical reactions chemical reactions Collisions permamnent exchange by collision

7. Process at constant volume: Things are simple ! energy supplied to the system (  Q) by the photons is just the change of 'internal energy' (  U).'internal energy'  U =  Q(constant volume) In the atmosphere reactions occur at constant presure => Things are less simple ! If number of molecules is changed by the reaction, the volume changes and produces mechanical work (in case of positive volume change a negative contribution to energy balance of the system):  W = - p.  V  U =  Q +  W Enthalpy: H = U + pV With H, things again become simple:  H =  Q(constant pressure) Process at constant pressure:

8. Standard enthalpy of formation (  H f 0 ) By definition  H f of the elements in their most stable form (N 2, O 2, H 2, etc) is zero Obviously the the enthalpy of formation varies with pressure and temperature. But the variation is small and in practice  H f for standard conditions (  H f 0 ) can be used even at stratospheric conditions. Enthalpy is a state function I.e. it does not matter how we get from a state A to a state B, the net amount of enthalpy needed (or released) by this transition is the same => The enthalpy needed to produce a chemical species from the elements is an individual fixed property of the species. We call it: "enthalpy of formation" (  H f )  H f for all relevant chemical species is listed in the literature. E.g. JPL2003 A  BCBC ACAC reactants A products C   -C =  H A-B +  H B-C   -C  H A-B HB-CHB-C

9. Standard enthalpy of formation (  H f 0 ) Lies in a deep enthalpie valley. Difficult to get out there => very passive molecule carries a huge amount of enthalpie => enough to break most bonds examples Can the following reactions occur ? (1) O + H 2 O -> HO + HO (2) O( 1 D) + H 2 O -> HO + HO (59.6 - 57.8) (9.3 + 9.3) 1.8 < 18.3 NO ! (104.9 - 57.8) (9.3 + 9.3) 47.1 > 18.3 YES ! JPL, 2002

10. Reaction enthalpy (H r ) The enthalpy consumed (or released) by a chemical reaction (  H r ) is the difference between the enthalpy of formation for all products and all reactants:  H r =  H f (products) -  H f (reactants) In general the reaction can only occur if H r is negative (enthalpy is released) or the missing enthalpy is supplied by photons. Example: photolysis of O 3 (1)O 3 + hv -> O + O 2  H r =  H f (O) +  H f (O 2 ) -  H f (O 3 ) = 25.5 kcal/mol = 107 kJ/mol = 1.78 10 -19 J/molecule => photon max = 1120 nm O 1 D production (2)O 3 + hv -> O( 1 D) + O 2 => photon max = 340 nm

11. Reaction kinetics Goal: Describe in a quantitative way how concentrations of species change with time due to chemical reactions

12. 1st order reactions Reactions with only one reactant. E.g. photolysis:...... is a differential equation for [A](t). The number of molecules of A lost in a volume per unit of time is proportional to the concentration of A. Hence, changes of [A] with time are described by: Changes of concentrations in chemical systems are described by a system of differential equations. A "chemical box model" is nothing else than the numerical solver for this system.

13. 1st order reactions...more This "one species" / "one reaction" system has a simple analytical solution: [A](t) = A 0. e -kt exponential decay Reaction rate: R = k. [A] (or "photolysis rate") Reaction constant: k (for photolysis often called J-value or "photolysis frequency") Lifetime:  = 1/k is the e-folding time of the exp. decay  or, as more general definition:  = [A]/R => after , [A] would have been consumed if R would remain constant Line with constant initial slope (constant loss rate)

14. 1st order reactions...still more The key is to determine J: Dissociation Quenching Fluorescence The rate of photolysis processes in our volume is determined by: the rate of absorption processes  the actinic flux I( )  the absorption cross section  ( ) the fraction of dissociation vs. other processes (quenching,...)  quantum yield  ( )..........................................................from radiation transfer calc.....................................from lab. studies...........................................................from lab. studies For the Box model we get:

15. Actinic flux In near UV and vis: weakly dependent on altitude In the UV: strongly dependent on altitude

16. Variations of photolysis frequencies with altitude Which of these species photolyses at longer wavelengths than the others ? < 300 nm < 410 nm

17. Sunrise as seen by different molecules In the visible: At 20 km altitude sun climbs above the horizont at ~95 deg sza => abrupt sunrise at ~95 deg sza In the UV: at 90-95 deg sza the sun is still hidden behind the ozone layer ! => slow sunrise between ~90-85 deg sza visible light: little attenuation UV: near complete absorption in the ozone layer

18. What does that mean for the chemistry ? NO 2 + h -> NO + O ClONO 2 + h -> ClO + NO 2 Which photolysis occurs at longer wavelengths ? < 410 nm < 300 nm Wennberg et al., 1994

19. 2nd order reactions Reactions with two reactants:

20. 2nd order reactions Reactions with two reactants: The rate of the reaction is determined by: collision frequency: proportional to [A]. [B] (...and proportional to sqrt(T), usually neglected) fraction of collisions that result in a reaction: -steric factor (slightly negative temperature dependence) -activation energy needed to form AB* (if high => strong positive T dependence) E a : Activation energy (R: gas constant) E a and A are determined in the lab by plotting ln(k) vs. 1/T "Arrhenius plot" (=> slope is E a /R) Sometimes this leads to negative E a. For these reactions: => E a is very small => T dependency dominated by steric factor

22. Reaction systems Example: Three species, four reactions The system O / O 2 / O 3 and the Chapman reactions The evolution of the concentrations in the system is described by the continuity equation (one for each species): This is a set of coupled differential equations ! Here:

23. Simplified systems (1) All production and loss terms are constant In general numerical models are needed to solve the set of differential equations that describe a system of interest. Here we look at two simplified cases first, that can be solved analytically:, P i and L i all constant =[A] e steady state solution transient solution dissappears with e-folding time 1/  L i with 'steady state' e transient solution Lifetime: controls how long the system needs to reach steady state, is dominated by the shortest individual lifetime in the system

24. Simplified systems, continued (2)All production and loss terms are periodic (same frequency) or constant e.g. diurnal cycle or seasonal cycle solution has the general form:  (t) is a periodic function with the same frequency as the forcing stationary solution ("diurnal steady state") transient solution dissappears with 1/ 

25. One constant loss process L Production: Harmonic diurnal cycle, i.e. period = 1 day Production term P t [days] P [mol day -1 m -3 ] Example Forcing t [days] diurnal steady state rapidly reached strong diurnal variation of concentration no lag to forcing slow decay of transient solution virtual no diurnal variation of concentration Solution (a): Lifetime (1/L) << period of forcing lifetime = 0.05 days Solution (b): Lifetime (1/L) >> period of forcing lifetime = 20 days midnight midday

26. One constant loss process L Production: Harmonic diurnal cycle, i.e. period = 1 day Production term P t [days] P [mol day -1 m -3 ] t [days] weaker diurnal variation of concentration dirunal variation lags the forcing (we will come back to that) Shape of curve unchanged Absolute concentrations ten times larger Example, continued Forcing Solution (c): Lifetime (1/L) ~ period of forcing lifetime = 2 days Solution (d): Lifetime (1/L) ~ period of forcing lifetime = 2 days, 10 times faster production midnight midday

27. Still same example, the stationary solutions midnight midday short lifetime (0.05 days) strong diurnal variation maximum concentration ~midday long lifetime (20 days) weak diurnal variation maximum concentration ~sunset ! intermediate lifetime (2 days) some diurnal variation maximum concentration ~afternoon !

28. From these measurements alone, what do we learn about the rate of (1) ? What can we say about the rate of (2) ? => not much ! => The lifetime of ClO with respect to (2) is much shorter than one day => The reaction consumes much more than 25 pptv ClO per day. Atmospheric measurements of ClO Dominating reactions: (1) production:ClONO 2 + hv -> ClO + NO 2 (2) loss:ClO + NO 2 -> ClONO 2

29. You should now be able to set up the system of differential equations that describes the chemistry for a given set of reactions and use lab data to calculate the relevant kinetic parameters. You should also know the fundamental behaviour of such systems under simplified conditions. Tomorrow:... how to make a box model out of this... real systems that actually exist in the atmosphere