EECE 507 Take Home Exam 1

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1 Spring 27 EECE 57 Take ome Exam This homework tests your ability to integrate the concepts learned so far in quantitatively addressing an important problem in atmospheric chemistry -that of acid rain. It also tests your proficiency in obtaining and displaying the results. Overall Objective: Set up a computer program (in Matlab or similar) to calculate the variation of p and of various sulfur IV and sulfur VI species as a function of time in a: a) closed, and ii) open system, assuming constant water volume fraction in air, L, and instantaneous equilibration of the all components in air and water droplets, as discussed in Chapter 8N. Numerically obtain the solutions and compare to the results presented in Chapter 8N. Specific Tasks: a) Set up the equations for an open system and outline solution method. b) Set up the equations for a closed system and outline solution method. c) Code the equations to be solved into the computer program. d) Collect all needed parameters and input data and list them in your report. e) Compute and compare the results when possible.. Problem Statement Consider acid rain formation in aqueous droplets in a cloud at 25 C. The system contains L = -6 (m water/ m air) and initially has (at time zero, t=o) the following pollutant concentrations and p. [S(IV)] = 5 ppb [N ] = ppb [N] = 5 ppb [O] = 5 ppb [22] = ppb p = 6.7 Questions to be answered: ) Show that the above conditions correspond to the following initial gas-phase concentrations (assuming instantaneous equilibrium): =. ppb P,o PN,o Po,,. P 22 =.87 ppb = 5 ppb =.65 ppb Use the enry's constants and dissolution constants for S2, 2S, 2S 4, 22, reported in the notes. Specify the list of parameters needed to solve the problem. If some are missing, find them in the literature and notify or ask the TA. 2) We know that NO, is in the range I 8 (M/atm) to over I 2 (M/atm) for a p range of 2 to 6. This means that the equilibrium partial pressure of N, when its al concentration is ppb, is negligible. As a matter of fact its partial pressure is only

2 about 8.54 x -9 ppb. The dissolution constant for N is Kn = 5.4 M and is given as K = [+ ] [NO,- ]. Estimate the enry's constant for NO, from the above " [N ] information. ) Consider the cloud to be an open system as far as air is concerned. The only oxidation reactions taking place are those between ozone and [S(IV)] and peroxide and [S(IV)], both in the liquid phase. Calculate the rise of [S(VI)] and the drop in p during 6 minutes. Show graphically [S(VI)], [S(IV)], p and the concentrations of various dissolved species as a function of time and report the values at 6 minutes. 4) epeat the calculations considering the cloud to be a closed system and display graphically the results for [S(VI)], [S(IV)], p and the concentrations of other dissolved species and report their values at 6 minutes. 5) Discuss the results and comment what causes the differences between the closed and open systems. Which system represents better a traveling cloud? Explain. NOTA BENE: Consider only reactions in the liquid phase between ozone and S(IV) and peroxide and S(IV). Use the rate forms and rate constants reported in the notes. The electro-neutrality equation that you must satisfy is: EPO EQUIEMENTS (ONE EPO PE TEAM). Discuss the problem with your team-mates at the start and make a list of items to be done regarding equations to be solved and approach used to solve them. Tell us who was the team coordinator and which team member did specific parts of the report 2. Clearly write your equations and explain how you solved them so that we can help you locate errors if any.. Discuss the completed report with each other. Each team member should clearly indicate how many hours of work did he/she need to do his/her part of the report. Prepare no more than 5 power point slides showing the highlights of your report. For your information: Students in previous classes invested between and 2 hours each in preparing professional report for this problem. Plan your time accordingly.

3 Solution for question We know that, the concentration of S(IV) of the aqueous is calculated by K K K [ S( IV )] p [ ] p [ ] [ ] S S S 2 2 S ( IV ) Besides, the al concentration of S(IV) in both gas and aqueous phase is, p SO V 2 n S ( IV ) g.25 S, ns, aq n Lp V S, g [ S( IV )] V V V p SO V 2 g S ( IV ) LpSO V.25 2 g V S ( IV ) Lp g p.25 So the partial pressure of (atm) can be calculated by, [ S( IV )] [ S( IV )] pso 2 ( ).25 ( ).25 S IV L S IV L K K K [ ] [ ] [ ] S S S 2 S ( IV ) 2 Do the same derivation process for O, 2O2, and NO, we get [ 2O2 ] [ 2O2 ] po OL OL [ O] (ppb) 2 2 [ O 2 2] (mol/l) (the above equation can convert a ppb unit concentration to a mol/l unit concentration) [ O] [ O] po.25 O L.25 O L [ NO] [ NO] pno.25 NO L.25 NO L P ( K ) NO NO n 9 g The results of partial pressure of gas phase of each species are as follows,

4 p_ =.675e-9 atm=.675 ppb p2o2_ =.6549e- atm=.6549 ppb po_ = 5e-9 atm=5 ppb pn_ =.86e-9 atm=.86 ppb Solution for question2 From the statement, we know that Kn=5.4 M which is the dissolution constant for nitric acid. The partial pressure of NO is 8.54e-9ppb, and the al concentration of NO is ppb. The dissolution process of NO is as follows, NO ( g) NO ( aq) ( ) NO NO ( K ) n NO So we can get the following equations for each composition PNO [ NO ] ([ NO ( aq)] [ NO ]) L [ NO ( aq)] NO P NO K P [ NO ] [ ] p NO n NO NO NO NO NO n [ NO] [ NO] L.25 NO L.25 P ( K ) When the al concentration of NO is ppb, one can calculate its concentration in mol/l by using the followed equation,.25 [ O] (ppb) 2 2 [ O 2 2] (mol/l) So, we can get [ NO] [ NO] pno (atm).25 NO.25 NO L L.25[ NO ] (ppb) 9 L.25 NO 9 earrange the above equation, the modified enry s constant for NO is

5 NO.25[ NO ] (ppb).25 pno (ppb) L So according to the concentration of al NO and the partial pressure of NO, we can calculate the modified enry s constant for NO. Then we can use the followed equation to get the enry s constant for NO P NO NO ( NO n ) NO P Kn K The results is 2.24e+5 M/atm NO Besides, NO 2.24e e+2 ( )

6 Solution for question(open system) Solutions: For an open system, list all the balance equations of each species as follows: dp Q p dt V j g j rjg ( Kgalg ) j ( p j C jl / j )( L) ( C j ) in g dc jl L Q L Lr K a p C C C dt L V jl ( g lg ) j ( j jl / L j )( ) ( jin jl) L For an equilibrium system (and also QL=), C p jl j j So simplify and sum the above equation, one get d Q dt V g [( L j ) p j ] rjg LrjL ( p j p ) in j g dc dt jl r jl Because the system is open, so we can neglect the variation of pressure of each species, so the only balance equation that should be considered is the balance in the aqueous phase dc dt jl r jl Besides, the electroneutrality equation is f([ ]) From this equation, one can calculate the concentration of +, and get the P value, also evaluate the enry s constant which is dependent on P value. Calculation procedures are as follows:.first get p from the al concentration of SIV [ S( IV )] [ S( IV )] pso 2 ( ).25 ( ).25 S IV L S IV L

7 [ S( IV )] [ S( IV )] pso (atm) 2 ( ).25 ( ).25 S IV S IV L L.25[ S( IV )] (ppb) 9 L.25 S ( IV ) 2.Calculate aqueous SIV concentration, assume that the partial pressure of each species in the atmosphere is constant. KS KSKS 2 [ S( IV )] SO p [ ] 2 SO 2 2 S ( IV ) p [ ] [ ]. Solve the partial differential equation of aqueous SIV concentration So the three coupled equations that should be solved are the balance equations for SIV, SVI, and +, and the equations are nonlinear, dc S ( IV ) dt r r r with initial condition, CSIV=CSIV S ( IV ) S ( IV ), O S ( IV ), 2O2 dc S ( VI ) dt r r r with initial condition, CSVI= S ( VI ) S ( IV ), O S ( IV ), 2O2 dc k[ ][ S( IV )][ ( ) 2O2] X S VI ( kx SO 2 22O kx k2x )[ S( IV )][ O ] SO SO dt K[ ] SO [ SO O] SO pso SO pso , [ SO ] K p K p P P S S, [ SO ] K K p K K p 2 2P 2P S S 2 S S 2, As the system is open system, the concentration of O and 2O2 in the gas phase can be treated as constant value. If the system is close system, we should also list the balance equations for these two species, due to that with the promotion of reaction of sulfur, the gas phase concentration of these two species changed to get phase equilibrium between the gas and aqueous phase. [ O ] p p O O O O, [ O ] p p 2 2 2O2 2O2 2O2 2O2, 4. solve the electroneutrality equation to get the [+] and P value at every time step. f([ ]) [ ] [ N ] [ O ] [ SO ] 2[ SO ] 2[ SO ] [ SO ] [ NO ]

8 K w [ O ] [ ] KS [ S( VI)] [ SO4 ] [ ] KS 4KS KS 2 [ ] [ ] 2 KS4KS [ S( VI )] [ SO4 ] 2 [ ] KS 4KS KS 2 [ ] [ ] [ SO ] K p [ S( IV )] X P S SO [ SO ] K K p [ S( IV )] X X X 2 2P S S 2 SO 2 2 SO SO 2 SO KS [ ] K K K [ ] [ ] S S S 2 2 KSKS2 2 [ ] K K K [ ] [ ] S S S 2 2 K K K [ S( IV )] p [ ] p [ ] [ ] K P [ NO ] [ ] S S S 2 2 S ( IV ) n NO NO [ N ] 4 Ka [ N 2O] [ O ] K [ N O] K K K [ ] p 2 p 4 a 2 S S S 2 2 [ O ] [ ] [ ] [ ] KS 4KS KS 2 [ ] [ ] K P 2[ S( VI)] [ S( VI)] KS 4KS KS KS 4KS KS [ ] 2 2 [ ] [ ] [ ] [ ] n NO NO By follow the above procedures, we can get the concentration of S(IV), S(VI) and P variation with time in an open system.

9 P [S(VI)aq] (mol/l) 5 4 The value in Chapter 8N The calculated value Time(s) The value in Chapter 8N The calculated value Time(s) Table. Concentrations of dissolved species and p at different time in the open system. C_SIV(M) C_SVI(M) C_2O(M) C_SO(M) C_SO(M) C_SO4(M) C_SO4(M). 8.4E-5.E+.84E E E-6.E+.E E E-5.84E-9.728E-5.7E-6.58E E E-5.5E-4.84E E-5 9.6E E-8.5E E-5.956E-4.84E E-5 6.5E-7.6E-8.956E E E-4.84E-9.895E E-7 5.7E-8 2.5E E E-4.84E-9.664E-5.92E E E-4

10 ..56E-5.7E-4.84E-9.488E E-7 8.4E-8.69E E-5.48E-4.84E-9.5E E-7.2E-7.47E E-5.77E-4.84E-9.28E-5.877E-7.24E-7.76E E E-4.84E-9.44E-5.6E-7.45E E E E-4.84E-9.64E-5.87E-7.682E E E E-4.84E E-6.24E-7.926E E E-6 5.E-4.84E E-6.7E E E-4 C_O(M) C_2O2(M) C_N2O(M) C_N4(M) C_NOaq(M) C_NO(M) P. 4.7E E-5.26E-7.28E-4.795E E E E-5.26E-7 2.5E-4.795E-2 2.E E E-5.26E-7.454E-4.795E-2.54E E E-5.26E E-4.795E-2.252E E E-5.26E E-4.795E-2.68E E E-5.26E E-4.795E E E E-5.26E E-4.795E E E E-5.26E E-4.795E E E E-5.26E-7 7.6E-4.795E E E E-5.26E E-4.795E E E E-5.26E E-4.795E E E E-5.26E E-4.795E E E E-5.26E-7.9E-.795E E Table 2. Concentrations of dissolved species and p in the open system at 6 min No. Species concentration at 6min, M [S(IV)] 9.466E+ 2 [S(VI)] 5.E+2 [ 2O].84E- 4 [SO - ] 9.55E [SO ].7E- 6 [O ] 4.7E-5 7 [2O2] 2.595E+ 8 [ N 2O ].26E- 9 [N 4+ ].9E+ [NO (aq)].795e-6 [NO - ] 5.272E+ 2 p 5.28

11 P 6.2 P [S(IV)] vs [S(VI)] Concentration (mol/l) [S(VI)] [S(IV)] Figure. Variation of p and concentration of S(IV) and S(VI) in the closed system..4 [ 2O] [SO-] Concentration (mol/l)..2. Concentration (mol/l) [SO-2].25 [SO4-] Concentration (mol/l) Concentration (mol/l)

12 6 [SO4-2]. [O] Concentration (mol/l) Concentration (mol/l) [2O2] 2 [N4+] Concentration (mol/l) Concentration (mol/l) [NO-] Concentration (mol/l) Figure 2.Variation of concentrations of dissolved species in the closed system.

13 Solution for question4(close system) If the system is closed, the balance equations for each species are as follows: In closed system, we have P P 2 [ ] ([ SO ] [ SO ] [ 2O]) L.25 [ S( IV )] L.25 Always remember that in the unit system that we use, C jg njg (mol) Pj (atm) (mol/l).25 V (L) (J/mol K) T(K) So define as /.25 C jg g njg (mol) Pj (atm) (mol/l) V (L) (J/mol K) T(K) g So using [] as an independent variable, the balance equation for [] is as follows, d[ ] d d[ S( IV )] PSO 2 L L[ rs ( IV ), O r S ( IV ), ] 2O2 dt dt T dt Do the similar treatment for [S(VI)], [O], [2O2], we can get the ODEs(ordinary differential equations) for al concentrations of each species. d[ S( VI)] [ rs ( IV ), O r S ( IV ), ] 2O2 dt d[ S( VI)] d[ S( VI)] L L[ rs ( IV ), O r S ( IV ), ] 2O2 dt dt d[ O d ] d[ O] PO L Lr dt dt T dt S ( IV ), O d[ O 2 2] d d[ O 2 2] PO 2 2 L Lr dt dt T dt S ( IV ), 2O2 In these equations, because p is varied with the promotion of the reaction, so [S(IV)] should be treated as a independent variable [ SO O] SO pso SO pso , [ SO ] K p K p P P S S, [ SO ] K K p K K p 2 2P 2P S S 2 S S 2, Because the [O] and the [2O2] is also varied with the promotion of reaction, so the

14 balance equation for these two species should also be solved simultaneously to get the right value of concentration of O and 2O2 Besides, another equation that should be used in the calculation is the electroneutrality equation to get the concentration [+](P value). [ ] [ N ] [ O ] [ SO ] 2[ SO ] 2[ SO ] [ SO ] [ NO ] T K T K K T [ ] [ ] 2 TL [ ] TL [ ] TL [ ] N Kw S [ ] S S 2 [ ] N, 2 N KS 4KS KS 2 [ ] [ ] NO T 2[ S( VI)] [ S( VI)] [ NO ] KS 4KS KS KS 4KS KS NO TL 2 2 [ ] [ ] [ ] [ ], Some of the expressions that used in the above equations are listed below, P [ ] TL T [ S( IV )] [ ] TL P O O TL [ O ] T [ O ] [ ] O O O TL P 2O2 O 2 2 TL [ O ] 2 2 T [ O ] [ ] O O2 O 2 2 TL K P K [ NO ] [ ] n NO NO n NO NO, [ ] [ ] NO TL K [ N O] K p K T [ N ] [ ] a 2 a N N a N 4 N, [ O ] [ O ] [ O ] N TL [ O ] K / [ ] w T [ SO O] p [ SO ] 2 2 SO 2 2 TL

15 [S(VI)aq] (mol/l) P [ SO ] K T [ SO ] S 2 SO 2 TL [ ] [ SO ] K K T [ SO ] 2 S S SO 2 TL [ ] KS4KS 2 2 [ ] [ SO4 ] [ S( VI)] KS 4KS KS 2 [ ] [ ] KS [ ] [ SO4 ] [ S( VI)] KS 4KS KS 2 [ ] [ ] The two kinds of reactions that happened in the system are between SIV with ozone and SIV with peroxide. SIV with ozone, r ( k X k X k X )[ S( IV )][ O ] S ( IV ), O 2 2O 2 SO SO SIV with peroxide r S ( IV ), 2O2 k[ ][ S( IV )][ 2O2] X K [ ] P k [ S( IV )][ 2O2] X P K SO SO By solving the above equations, we get the variation of concentrations of each species with time. The results are shown below The value in Chapter 8N The calculated value The value in Chapter 8N The calculated value Time(s) Time(s) Figure.Comparison of p and concentration of S(VI) in the closed system.

16 Table. Concentrations of dissolved species and p at different time in the close system. C_SIV(M) C_SVI(M) C_2O(M) C_SO(M) C_SO(M) C_SO4(M) C_SO4(M). 8.E-5.E+.74E E E-6.E+.E E E-5.8E-9.982E E-7.22E E E E-5.72E E-6.6E-7.26E E E-6 7.9E-5.68E E-6.784E E E E-6 8.2E-5.588E-9 4.6E-6 2.9E E E E-6 8.6E-5.557E-9.56E-6.62E E E E E-5.58E-9.228E-6.74E E E E E-5.525E-9.62E-6.24E-8.52E-7 8.5E E E-5.57E E-6.57E-8.9E E E E-5.5E E-6.99E-8.24E E E E-5.55E E-6.55E-8.5E E E E-5.5E E-6.8E-8.7E E E E-5.496E E E-9.9E-7 8.6E-5 C_O(M) C_2O2(M) C_N2O(M) C_N4(M) C_NOaq(M) C_NO(M) P. 4.7E E-5.54E-7.282E-4.726E-2 4.9E E-.289E-5 4.8E-8.768E E-2 4.9E E- 6.47E E-8.96E-4.44E- 4.9E E-.299E-6.26E-8.957E E- 4.9E E-.7E-6.26E-8.977E E- 4.9E E E-7 8.9E-9.986E-4.466E- 4.9E E- 4.69E E-9.99E-4.754E- 4.9E E- 2.47E E-9.99E-4.944E- 4.9E E-.257E-7 7.6E-9.995E E- 4.9E E E E-9.996E E- 4.9E E-.47E E-9.997E E- 4.9E E-.666E E-9.998E-4 4.8E- 4.9E E- 8.67E E-9.998E E- 4.9E Table 4. Concentrations of dissolved species and p in the close system at 6 min No. Species concentration at 6min Closed system, M [S(IV)] 2.72E+ 2 [S(VI)] 8.645E+ [ 2O].496E- 4 [SO - ] 2.79E+ 5 [SO 2- ] 9.856E- 6 [SO 4- ].9E [SO 4 ] 8.6E+ 8 [O ].65E-5 9 [2O2] 8.67E-

17 P [ N 2O ] 7.49E- [N 4+ ].998E+2 2 [NO (aq)] 4.45E-5 [NO - ] 4.9E+ 4 p P Concentration (mol/l) [S(IV)] Concentration (mol/l) [S(VI)] Concentration (mol/l) [ 2O] Concentration (mol/l) [SO-] Concentration (mol/l) [SO-2]

18 .25 Concentration (mol/l) [SO4-] Concentration (mol/l) [SO4-2] x Concentration (mol/l) 4.5x -5 4.x -5.5x -5 [O] Concentration (mol/l) [2O2].x Concentration (mol/l) [N4(aq)] Concentration (mol/l) [NO-(aq)] Figure 4.Comparison of calculating results of open and closed systems. Question5 Discussion Figure 4 gives the comparison of calculating results between the open and closed systems. Obviously, the variation of concentration of different dissolved species as well as P value are ally different between the open and close system. The differences are caused by the characteristics of the two system. In an open system, mass exchange of species between the target system and the environment leads to the

19 constant partial pressure of all the species. This means that although chemical reaction happens in the aqueous phase, the environment can always give supplement to the reacted species to keep its partial pressure to be a constant value. In contrast, the close system has no mass exchange between the environment, and has no such kind of supplement. So we can see from the figure that, the concentration of each dissolved species in the open system are always higher that the value in close system. We can also see that in the aqueous phase, the concentration of [2O2](aq) is far larger that the concentration of [O](aq). This is caused by the great value of enry s constant of 2O2 (_2O2 equals to 7 M/atm, _O equals to.94 M/atm). So we can conclude that in the aqueous phase, the four form S is mainly oxidized by the peroxide. In travelling cloud, it seems that there are mass exchange between the cloud and the atmospheric environment. In this case, an open system is much better than the close system to represent the travelling cloud.