Exam Atmospheric Chemistry and Physics June 4, 2014 at 8:00-12:00 separate sheets. 1. Aerosol and hygroscopic growth 2.

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1 Exam Atmospheric Chemistry and Physics June 4, 04 at 8:00-:00 You may use: Calculator, Table booklet (Tefyma or similar), Equation sheet of the course Answer each question (-5) on separate sheets.. Aerosol and hygroscopic growth a) Which are the two main natural sources of aerosol particles, and which is the main human activity behind particles in the atmosphere? ( p) b) Which of these sources produce mainly particles smaller or larger than µm diameter, respectively? ( p) c) Compare these three sources with respect to hygroscopic growth (water uptake). Discuss their importance in cloud formation with respect to particle size and composition. ( p) d) Why is hygroscopic growth of aerosol particles important for the climate? ( p) e) A volcanic eruption induces an aerosol layer in the stratosphere of the earth with the globally averaged optical depth δ The backscattered fraction β 0.. Calculate the radiative forcing induced by the eruption. ( p). The atmosphere a) Which is the fundamental difference between the troposphere and the stratosphere, making them regarded as different parts of the atmosphere? (p) b) Is vertical motion most common in the troposphere or the stratosphere? Why? ( p) c) The global average height of the tropopause is 4 km. Calculate the pressure at that altitude, assuming constant atmospheric temperature of 50. ( p) d) Assume that the global air flow from the troposphere to the stratosphere is equal to the opposite flow (from the stratosphere to the troposphere), being kg/year in each of the two directions. Calculate the residence times of air in the troposphere and the stratosphere. The pressure at the stratopause is hpa. ( p) e) Explain why the residence times of stratospheric and tropospheric air differ. ( p). Acidification Sulfur is emitted to the atmosphere mainly as the gas sulfur dioxide (SO). The sources are both natural and anthropogenic (resulting from human activities). SO is then oxidized in the gas and aqueous phases to eventually cause acidification. a) What are the most important natural and anthropogenic sources of sulfur? (p) b) What are the most important SO oxidation agents (molecules) in the gas and aqueous phases? Briefly describe how these oxidation agents form in the atmosphere. (p) In order to be oxidized in the aqueous phase, the gas SO must first be dissolved in water according to the following reactions: () SO(g) SO(g) O (. M atm - ) () SO(g) O SO - (. 0 - M) () SO - SO - ( M) ere, the i are equilibrium constants. c) Will the reactions - above result in a permanent acidification of the aqueous solution? Motivate your answer. (p) Oxidation of SO in the aqueous phase and thereby the acidification mainly proceeds via the bisulfite ion (SO - ), and its aqueous phase concentration [SO - ] is therefore important. d) Calculate, using reactions - above, an expression for the mole fraction of bisulfite (SO - ) out of the total amount of S(IV) dissolved in water. Then use this expression to calculate the mole fraction of bisulfite at p, 4.5 and 7. (p)

2 4. Stratospheric ozone a) Describe using Figure below how emissions of halocarbons (freoner) at the surface of Earth can affect the stratospheric ozone layer. Write the (few) reaction steps that lead to catalytic ozone loss for different compounds X, where X can be the chlorine radical as shown in the figure. Also summarize the net reaction showing the net ozone loss. ( p) Figure NOx can also destroy ozone catalytically. The most important reactions for maintaining the radical balance within the ClOx- and NOx-families in the stratosphere are: () Cl O ClO O (k cm molecules - s - ) () ClO NO Cl NO (k cm molecules - s - ) () NO O NO O (k cm molecules - s - ) (4) O NO hν NO O (k4 0 - s -, λ< 40 nm) The following concentrations are found in a parcel of air in the stratosphere: [O] 0.6 ppmv, [NOx] 0. ppbv, [ClOx].0 ppbv. The air density at this altitude is molecules cm -. Assume steady state conditions for both Cl and NO. b) Show that [Cl] << [ClO] and [NO] << [NO] for the air parcel in question. Tip: Start with the chlorine radicals, and also remember that [NOx] [NO] [NO]. ( p) c) What are the lifetimes for the radicals Cl and NO considering reactions -4 above? Is the assumption regarding steady state conditions for both Cl and NO reasonable? Motivate your answer. ( p)

3 5. Tropospheric ozone Figure describes how hydrocarbons (R) are oxidized in the atmosphere, contributing to the formation of ozone. This involves many rather complex reactions. owever, the formation of ground-level ozone can be condensed and summarized using the photostationary equilibrium for ozone, given as: hν NO O NO O () a) Use the photostationary equilibrium () and the reactions shown in the Figure to explain why hydrocarbons are needed for the formation of a photochemical smog and high concentrations of tropospheric ozone. (p) The Figure also shows the single most important tropospheric chemistry radical cycle. b) What determines whether a compound is radical or not? (p) c) Which of the gas phase compounds in the Figure are radicals? (p) d) What prevents this radical chain to continue forever, producing ever more ozone. Give one example of a reaction that will end the radical chain. (p) e) Name at least one harmful trace gas that is oxidized mainly by the radicals formed in the Figure. What is the harmful effect for the trace gas in question in terms of negative impacts on human health and/or the environment, and give the end product of the oxidation process. (p) Figure. Schematic figure of how hydrocarbons (R) are oxidized in the atmosphere, contributing to the formation of ozone.

4 Answers to exam questions of June 4, 04. Aerosol and hygroscopic growth a) Natural: Sea-spray and windblown erosion particles (dust) Anthropogenic: combustion b) Natural sources (mechanical generation): > µm Combustion: mainly < µm c) Water solubility and large size > low critical supersaturation > good cloud condensation nucleus (CCN). Sea-spray (sodium chloride): soluble and large; Windblown dust (mineral): insoluble and large; Combustion: soluble and small; > Sea-spray the best CCN d) Soluble particles grow with increased relative humidity causing increased scattering of solar radiation, thus increasing the direct climate effect of the aerosol. The hygroscopic properties of particles also affect cloud formation and thus also the indirect climate effect of aerosol. e) The albedo of an aerosol layer is given by Aa βδ, and the total albedo of two layers is obtained by AT A0 Aa(-A0), where A0 0.8 is the albedo without the aerosol layer from the volcanic eruption. The albedo change A AT A0 Aa(-A0). The radiative forcing is defined by F Fin, Fout, in the system after the change of albedo (textbook pp 5). In this case we have changed the albedo, implying that the we have the same incoming radiation in the reference system (before the eruption) > Fin, Fin, Fout,, were the latter is true because the reference system is at equilibrium. ence, we can express F Fout, Fout,. Using the climate model of the textbook (Fig. 7-): F FSA0/4 ( f)σt0 4 fσt 4 (FSAT/4 ( f)σt0 4 fσt 4 ) FSA0/4 FSAT/4 FS A/4 FSAa(-A0) /4 70*0.*0.05*( 0.8) /4.78 W/m. The atmosphere a) The temperature change with altitude. Troposphere: temperature decrease with altitude; Stratosphere: temperature increase with altitude b) The troposphere, because the temperature increase with altitude in the stratosphere (inversion) induces a strong counteracting force against vertical motion > very stable conditions c) The vertical pressure profile can be obtained from the barometric law: p(z) p0 exp(-z/), where RT/(Mag), p0 the average pressure at the surface (984 hpa), z the altitude, R the gas constant, T the temperature, Ma average atmospheric molar mass and g the gravity. p(4 km) 45 hpa. d) The pressure at any given layer is caused by the weight of the overlying air: pa mg > m pa/g. The mass of the troposphere is given by the mass of the atmosphere minus the mass of the air above the tropopause: mt (p0 ptp)a/g ( )4π( ) / kg and for the stratosphere ms (ptp psp)a/g kg Residence times: τt mt/fout / years; τs. years e) The mass of the troposphere is much larger than the stratosphere and the mass flows in both directions are equal. Therefore more time is needed to exchange the tropospheric air.

5 . Acidification a) Natural sources: The ocean (DMS), volcanoes. Antropogenic: Combustion of sulphurcontaining fossil fuels, (also smelters that extract metals from sulfide ores). b) Gas phase: O radical, formed in photochemical smog (photolysis of O). Aqueous phase: O, formed via termination of O radicals; O, formed in photochemical smog (photolysis of NO) c) No, since these reactions are all reversible. There is no oxidation taking place, and the oxidation number remains as S(IV) thoughout reactions -. When the water is removed (for instance when cloud droplets evaporate), the sulfur will repartition back to the gas phase. d) [ SO ( g) O] [ SO ( g) O] [ SO ( g) O] pso SO g p [ ( )] [ SO ] [ ] [ SO ( g) O] [ SO ] [ ] [ SO ] SO [ ] [ ] ( ) SO g O SO [ ] [ ] [ ] [ SO ] SO [ ] [ ] Let [total-s(aq)] denote the total aqueous pahse concentration of all S(IV)-containing molecules and ions. [ ] [ ] [ ] [ ] total S( aq) SO ( g O SO SO [ total S( aq) ] p SO ) p SO [ ] [ ] p SO [ ] [ ] p p SO SO p SO [ SO ] [ total S( aq) ] p SO p [ ] SO [ ] [ ] [ ] [ ] [ ] [ ] Calculate this mole fraction at p, 4.5 and 7. Use p log([ ]) to get [ ]. At p, mole fraction of SO At p 4.5, mole fraction of SO -.00 At p 7, mole fraction of SO

6 4. Stratospheric ozone a) Ozone can be destroyed in catalytic processes, which means that the compound X that is responsible for the ozone loss is recycled and not consumed. () O X O XO () O XO O X Net reaction (() ()): O O O A different (but also common) set of reactions is: () O X O XO () O XO O X Net reaction (() ()): O O X can be various molecules: X O : O (XOO) X Cl (XOClO) X NO : NO (XONO) X Br (XOBrO) The figure shows how Cl destroys ozone catalytically. The Cl radicals originate from halocarbons (freoner) that contain one or more chlorine atoms. In the troposphere, the halocarbons are very stable compounds and have a long lifetime. They can therefore slowly penetrate into the stratosphere, where they are photolyzed by UV radiation. For instance: CFCl hν CFCl Cl (initiation, UV radiation required) (ClOx family: Cl and ClO, both are radicals) b) Steady state condition for Cl: k[clo][no] k[cl][o] [ ] [ ClO] [ NO] [ O ] [ NO ] [ O ] Cl k < k x mol/mol. k k ere we have used the fact that [NOx] [NO] [NO], and thus [NOx] > [NO]. Steady state condition for NO: k4[no] k[no][o] k[clo][no] [ NO] k4 k4 [ NO ] k [ O ] k [ ClO] k [ O ] k [ ] ClO x 0. mol/mol. Since we have shown already that [Cl] << [ClO], then [ClO] [ClOx].0 ppbv. Note that you need to give the concentrations in units of molecules cm - and not in ppb or ppm. c) The lifetime for Cl molecules in the stratosphere (MCl[Cl], LCl loss rate of Cl): τ M [ Cl] [ Cl][ O ] k [ O ] Cl Cl LCl k Similarly, the lifetime for NO: [ NO] τ NO k NO ClO k 0.06 s [ ][ ] [ NO][ O ] k [ ClO] k [ O ] 0 s These lifetimes are short in comparison with the processes which control the formation and loss of Cl and NO (mainly UV radiation varying on a diurnal basis). This means that new steady state concentrations can be achieved rapidly (within approximately τ), balancing production and loss.

7 5. Tropospheric ozone a) By reacting with the O radical (detergent of the atmosphere), hydrocarbons (R) produce organic oxy (RO) and peroxy radicals (RO) that are needed to propagate the radical chain involving also the Ox radicals (O and O) in the Figure. By doing so, they help shifting the photostationary equilibrium () to the right side by converting NO to NO. (UV light is also needed to photolyze NO and O. b) A radical has one unpaired electron (an odd number of electrons) c) The radicals in the Figure are O, O, RO, RO, NO, NO. d) The radical chain is terminated, which means that two radicals react to produce one nonradical compound. (This will happen more frequently as the radical concentrations increase, since their lifetime is typically very short meaning that they will tend to react first with non-radical compounds in propagation reactions. Alternatively, sunlight (UV) is lacking (night-time) or the supply of R compounds that can be oxidized is depleted.) The important termination steps (only one of them needs not be mentioned): O NO M NO M O O O O RO O ROO O e) Only one example is needed. The main radical formed is the hydroxyl radical O, the detergent of the atmosphere. CO is lethal to humans and animals since it prevents the uptake of oxygen in our blood cells. End product: Oxidized by O to CO. C4 is a powerful greenhouse gas, even more potent than CO. End product: Oxidized by O to CO, CO and eventually CO. SO is dangerous to humans (affects respiratory system) and causes acidification of ecosystems, soils and waters via dry and wet deposition. End product: Oxidized (in gas phase) by O to sulphuric acid SO4.

8 Exam Atmospheric Chemistry and Physics June 4, 05 at 8:00-:00 You may use: Calculator, Table booklet (Tefyma or similar), Equation sheet of the course Answer each question (-5) on separate sheets.. The Atmosphere a) The atmospheric concentration of CO is since a long time increasing. Which are the main reasons for the increase? ( p) b) Fig. shows the change in CO concentration from year 960 to present at Mauna Loa, awaii. What is causing the annual variation (årstidsvariation)? ( p) c) The yearly average concentration at Mauna Loa is a good estimate of the global yearly average. Why can we use data from one place for a global estimate? ( p) d) Use the data in Fig. to calculate the increase of carbon mass in the atmosphere during the industrial era (i.e. the increase from pre-industrial concentration to the concentration year 04). The pre-industrial CO concentration was 80 ppmv. ( p) 400 Monthly average Yearly average Mauna Loa CO CO (ppmv) Year Figure. The CO concentration in the air at Mauna Loa, awaii.

9 . Climate a) The Earth emits radiation approximately like a black bodyy (Fig. a). In Fig. we see that the shape of the spectrum measuredd from a satellite deviates significantly from a spectrum of a black body. What is the reason for these deviations? ( p) b) In which wavelength region can wee (approximately) identify the temperature of the Earth s surface in Fig.? Explain. ( p) c) Explain the greenhouse effect basedd on the radiation spectrum in Fig... ( p) d) In the course we use a simple climate model where the atmosphere is s approximated by a single isothermal layer. According to this model we obtainn maximumm greenhouse effect when the absorption of the atmosphere is 00% ( ). Inn reality thee absorption can exceed 00% %. ow doess that affect the greenhouse effect?? Explain. ( p) Fig.. a) Normalized radiation intensity i from the sun and from the earth as a function of wavelength. b) The fraction absorbed by the atmosphere of radiation of a given wavelength ( ). Fig.. Radiation spectrum (full line) from the Earth measured by satellite (wavelength at the upper axis). The spectrum was taken t over North Africa. The broken lines show radiation spectra from blackbodies at several temperatures.

10 . Aerosol-water interaction a) An aerosol particle has taken up water, formed a droplet, and obtained its equilibrium size with respect to relative humidity (R). What is the R (in %) if the droplet diameter (D) is 400 nm? The particle consists of. 0-8 kg sodium chloride (00 nm dry diameter). Assume that the dissociation () of NaCl is complete and that the temperature is 98. Also assume that the molar mass, surface tension and density of the solution equals that of water. [Water: M w 8.0 kg/kmol, σ Dw 0.07 N/m, ρ 998 kg/m ; Sodium chloride: M NaCl kg/kmol] (p) b) ow will the droplet react if the R increases slightly (by %)? Will the droplet still be in equilibrium state? (p) c) Make a sketch diagram that describes the relation between R (> 95%) and droplet diameter. Mark the critical R, and critical diameter. (p) d) ow will the droplet react if the R increases above the critical level of the droplet? Describe in general terms. Motivate. (p) 4. Stratospheric ozone In the stratosphere, both O and O are photolyzed by UV-radiation (reactions () and () below). a) Why is photolysis of O not enough to protect life on Earth from hazardous UV-radiation from the sun? (p) The reactions below are the most important for the production and destruction of stratospheric ozon. ere, O denotes O( P). () O h O O () O h O O( D) Cl O ClO O O O O O NO O NO O O O O O O O O NO NO M N O 5 M O O O O O O M O M Cl C 4 Cl C NO O NO O ClO NO M ClNO M O( D) M O M NO O M NO M O O O O ClO O Cl O NO O NO O Use these reactions to write b) the Chapman mechanism for ozone production in the stratosphere (p) c) the catalytic ozone loss via the O x, NO x and ClO x families. (p) d) ow are O x, NO x and ClO x radicals formed in the stratosphere? (p) e) Which of the reactions above constitute termination steps for catalytic ozone loss propagated via the O x, NO x and ClO x radicals? (p)

11 5. Acidification The gas SO dissolves in water according too the equilibrium () SO (g) SO O SO is enry s law constant forr SO. At T 98, aqueous phase concentration of SO of.0-9 SO. Matm -, which would yield an M at a typical gas phase concentration [SO (g)] ppb at standard atmospheric pressure. Figure 4 below shows that t the dissolution of S(IV) in water is actually strongly dependent on the p of the solution. For all p>, the concentration of S(IV) in water is much greater than what iss expected from the enry s law constant SO O. For neutral p in the atmosphere, p5.6, the concentration is about 0 4 timess larger. This example with SO clearly showss that the enry s law constant is in many cases totally inadequate in describing how a gas dissolves in water. Figure 4. The effective enry s law constant * S( (IV) for dissolution of S( (IV) in water as a function of the p of the aqueous solution. a) Explain why 5.6 can be considered to be neutral p in the atmosphere. (p) b) Explain qualitatively what causes the increased solubility of S(IV). (p) c) Deduce an equation for the t effective enry s law constant * S(IV) for how gaseous S( (IV) dissolves in water (* S(IV) [S(IV)] / [SO (g)] ). This equilibrium constant should be able to qualitatively explain the p dependence in Figure 4. (p) d) In which form is most S( (IV) present in the aqueous solution at p of atmospheric relevance (-6)? (p) e) This phenomenon of strongly increased solubility of S(IV) ) is of utmost importance for the aqueous phase oxidation of S(IV) too S(VI) and thus also for the acidification of soils s and waters. Explain why. (p) 4

12 Solutions to exam in Atmospheric Chemistry and Physics June 4, 05. The Atmosphere a) Fossil fuel combustion and deforestation, in particular in the tropics b) The growth period consumes CO, whereas biological decay increases the concentration c) The residence time of CO is long compared to typical transport times in the atmosphere, implying that the concentration is similar everywhere, except close to strong sources. d) C ppmv ; m c M c n c M c n a C We need to calculate the number of moles of gas molecules in the atmosphere (n a ): The weight of the overlying air causes the pressure at a given level: m a g pa 4R p > n a m a /M a 4R p/(gm a ) > m c M c n a C M c C4R p/(gm a ), where R is the radius of the Earth ( m), g the gravitation (9.8 m/s ), M a the average molar mass of air (9.0 kg/kmol) and p the average pressure ate the surface of the Earth (984 hpa). Insert the numbers: m c kg. Climate a) The atmosphere absorbs radiation from the surface which then emits blackbody radiation at its temperature. The temperature decreases with altitude in the atmosphere, implying that less radiation is emitted to space at wavelengths where the atmosphere has strong absorption. b) In Fig b we see that the absorption of terrestrial radiation is small in the atmospheric window 8 µm in wavelength. The estimated surface temperature by comparing with blackbody radiation is approximately 0 (47 C; North Africa) c) At the wavelengths where the atmosphere absorbs/emits radiation less radiation leaves the Earth. Therefore the temperature of the entire system is elevated to reach equilibrium with the incoming radiation from the sun. d) When the absorption of the atmosphere exceeds 00%, the probability for double or multiple absorption increases. As a result, the last emission of radiation before leaving the Earth will on average be at a higher altitude. This is also a colder part of the atmosphere, thus implying that less radiation leaves the earth.. Aerosol-water interaction a) Use the öhler equation for the calculation. Result: 98.5% b) The droplet will become larger due to the increase in relative humidity (R). It will still be in equilibrium state, because the R is still lower than the critical relative humidity. c) Based on the öhler equation, see Fig. in Aerosol, Water and Clouds by Martinsson. d) The droplet will follow the öhler curve until it reaches the maximum level (critical R). Up to the critical value, the droplet will stay in the equilibrium state. If the R increases further and exceed the critical R, the droplet will leave the equilibrium state. This means that the droplet will grow without limitation, the droplet activates. This is the process for cloud droplet formation. 5

13 4. Stratospheric ozone a) Photolysis of ozone (<0 nm) can proceed also in the UVB region where there is no photolysis of O (<40 nm). The stratospheric ozone layer thus protects us from the hazardous UVB radiation (80 nm < λ < 0 nm) which is not removed by ordinary oxygen O. Answer b) The Chapman mechanism: () O h O O O O M O M () O h O O( D) O( D) M O M O O O c) Catalytic ozone loss via O x : O O O O O O O O (the latter reation does not follow the general scheme for a catalyst X which is XO O ) Catalytic ozone loss via NO x : NO O NO O NO O NO O (XO O ) Catalytic ozone loss via ClO x : Cl O ClO O ClO O Cl O (XO O ) d) Formation of O x : O O( D) O Formation of NO x : N O O( D) NO N O originates from soil decomposition processes and is stable in the troposphere. NO is also emitted directly into the stratosphere by airplanes. Formation of ClO x : Photolysis of chlorine-containing CFCs that are i stable in the troposphere. e) Termination steps for catalytic ozone loss via O x : O O O O O O O O Termination steps for catalytic ozone loss via NO x : NO O M NO M NO O NO O and NO NO M N O 5 M Termination steps for catalytic ozone loss via ClO x : Cl C 4 Cl C ClO NO M ClNO M 5. Acidification a) Dissolution of CO at present atmospheric levels (80 ppm) results in a neutral and nonacidified p of 5.6 in cloud droplets and rain water. Only a p below this value is a sign of acidification. b) Dissolution of the gas SO in water shifts the equilibrium for SO (g) to push more S(IV) into the aqueous phase than would otherwise be the case. The reactions are identical to those involving CO (exchange SO by CO ), and are given in the Equation Sheet. () SO (g) SO O ( SO. Matm -, given) () SO (g) O SO - (.0 - M) () SO - SO - ( M) 6

14 The i are equilibrium constants. SO is the traditional enry s law constant for dissolution of S(IV) in water. c) [S(IV)(aq)] [SO O] [SO - ] [SO - ]. The effective enry s law constant * S(IV) for dissolution of S(IV) in water is defined as * S(IV) [S(IV)(aq)] / [SO (g)]. The equilibrium reactions in a) give SO O SO O SO SO O SO pso SO g p SO SO SO O SO SO g O SO SO SO pso SO SO S ( IV ) SO O SO SO SO p SO SO p SO SO p SO p SO SO SO pso The effective enry s law constant * S(IV) can thus be expressed as * S( IV )( aq) S( IV )( aq) S ( IV ) SO SO g pso The factor that follows the traditional enry s law constant SO explains how the p of the aqueous solution affects the solubility of S(IV). Increased p (lower [ ]) causes this factor to increase considerably, and more S(IV) can be dissolved in water. d) As SO -. Oxidation in the atmosphere need therefore proceed via SO -. e) If the solubility of gas phase S(IV) in water would have been determined only by the traditional enry s law constant SO, then very little SO (g) would have been dissolved in water, for instance in cloud droplets. As a result, nearly all oxidation of SO would have taken place in the gas phase (via O). The lifetime for SO (g) would have been considerably higher and the problems associated with acidification would have shifted further away from the source regions. Concentrations of SO (g) would also have been higher close to the sources with negative consequences for human health. 7