1 / 46

Chapter 6: Aqueous Phase Reactions

Chapter 6: Aqueous Phase Reactions. Introduction Gas-liquid equilibrium: Henry’s law Oxidations of SO2 Nitrate and nitrite formations Close and open systems. Liquid water in the atmosphere 大氣中水蒸氣質量濃度與溫 度 ,RH 成函數關係. Cloud type and Liquid Water Content.

odell
Download Presentation

Chapter 6: Aqueous Phase Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6: Aqueous Phase Reactions • Introduction • Gas-liquid equilibrium: Henry’s law • Oxidations of SO2 • Nitrate and nitrite formations • Close and open systems

  2. Liquid water in the atmosphere 大氣中水蒸氣質量濃度與溫 度,RH成函數關係

  3. Cloud type and Liquid Water Content wL (vol water/vol air)=10-6 L(g m-3) cloudwater wL =5*10-8~3*10-6 g m-3 fog wL =0.02~0.5 g m-3

  4. Absorption equilibria and Henry’s Law • A(g) ↔ A(aq) (6.2) • [A(aq)]=HApA (6.3) • (6.4) (6.5) • : the reaction enthalpy

  5. Gas/Aqueous-Phase Distribution Factor caq:aqueous-phase mass concentration cg :gas-phase mass concentration 假設達到Henry’s law平衡 fA=10-6HARTL=HARTwL (6.7) (6.8) (6.9)

  6. Aqueous-Phase Chemical Equilibria1. Water • H2O↔H++OH- (6.10) • (6.11) • Kw=[H+][OH-] (6.12) • pH=-log10[H+] (6.13)

  7. 2. Carbon Dioxide/Water Equilibrium • CO2(g) + H2O ↔ CO2·H2O (6.14) • CO2·H2O↔ H+ + HCO3- (6.15) • HCO3-↔ H+ + CO3- (6.16) • (6.17) • (6.18) • (6.19)

  8. (6.20) • (6.21) • (6.22)

  9. (6.23) • (6.24) • [CO2T]=H*CO2pco2 (6.25) • H*CO2 > HCO2 (6.26)

  10. [H+]=[OH-]+[HCO3-]+2[CO32-] (6.27) (6.28) • [H+]3-(Kw+HCO2Kc1pCO2)[H+] -2HCO2Kc1KC2pCO2=0 (6.29) 若已知T,即可決定Kw,HCO2,Kc1,Kc2,pCO2 就可以求出pH值

  11. 3. Sulfur Dioxide • SO2(g) + H2O ↔ SO2·H2O (6.30) • SO2·H2O↔ H+ + HSO3- (6.31) • HSO3-↔ H+ + SO3- (6.32) • (6.33) • (6.34) • (6.35)

  12. (6.36) • (6.37) • (6.38)

  13. (6.39) (6.40) • (6.41) • [S(IV)]=H*S(IV)pSO2 (6.42)

  14. 4. Ammonia/Water Equilibrium • NH3 + H2O ↔ NH3·H2O (6.46) • NH3·H2O ↔ NH4+ + OH- (6.47) • (6.48) • (6.49) (6.50)

  15. (6.51) • (6.52) • pH <8 , Ka1[H+]>>Kw

  16. 5. Nitric Acid/Water Equilibrium • HNO3(g)↔ HNO3(aq) (6.53) • HNO3(aq) ↔ NO3- +H+ (6.54) • (6.55) • [HNO3T]=[HNO3(aq)]+[NO3-] (6.56) • [HNO3(aq)]=HHNO3pHNO3 (6.57)

  17. (6.58) (6.59) • (Kn1/[H+]) >> 1 (6.60) • (6.61)

  18. 6. Equilibrium of Other Important Atmospheric Gases • Hydrogen Peroxide H2O2(aq)↔ HO2- + H+ (6.62) (6.63) • Ozone • Oxides of Nitrogen

  19. Formaldehyde HCHO(aq)+ H2O ↔ H2C(OH)2 (6.64) (6.65) (6.66) (6.67)

  20. Formic and Atmospheric Acids HCOOH(aq)↔ HCOO- + H+ (6.68) • OH and HO2 Radicals HO2(aq) ↔ O2- + H+ (6.69) [HO2T]=[HO2(aq)]+[O2-] (6.70) (6.71)

  21. Aqueous-Phase Reaction Rates • S(IV) + A(aq)↔ Products (6.72) Ra=k[S(IV)][A(aq)] (Ms-1) (6.73) Ra’=kwL[A(aq)][S(IV)]=10-6 kL[A(aq)][S(IV)] (=mol(L of air)-1s-1) (6.74) Ra”=3.6*10-6LRTRa (ppb h-1) (6.75) (%h-1) (6.76) (%h-1)(6.77) (6.78)

  22. S(VI) to S(IV) Transformation and Sulfur Chemistry SO2·H2O pH<2 • SO2+H2O=> HSO3- 2<pH<7 SO32- 7<pH

  23. Oxidation of S(IV) by Dissolved O3 • O3+SO2在氣相反應非常慢,但是在液相反應卻非常快 S(IV)+O3→S(VI)+O2 • 有文獻指出:S(IV)-O3反應速率隨著溶液離子強度的增加而成線性般的增加

  24. Oxidation of S(IV) by Hydrogen Peroxide • H2O2在大氣中對S(IV)來說是很有效的氧化物之一 • H2O2比O3更容易溶於水中 • H2O2為弱電解值 • HSO3-+H2O2↔SO2OOH-+H2O • SO2OOH-+H+↔H2SO4

  25. Oxidation of S(IV) by Organic Peroxides • HSO3-+CH3OOH+H+↔SO42-+2H++CH3OH • HSO3-+CH3C(O)OOH→SO42-+H++CH3COOH

  26. Uncatalyzed Oxidation of S(IV) by O2 • 無Fe,Mn之摧化反應很慢,且空氣中多多少少都有這些東西存在,因此不考慮無Fe,Mn摧化的反應

  27. Oxidation of S(IV) by O2 Catalyzed by Iron and Manganese 增加離子強度,此反應反而會受到抑制 • Fe(II)先氧化成Fe(III)再將S(IV)氧化 Fe(OH)3+3H+↔Fe3++3H2O • Fe3+濃度平衡於Fe(OH)3 Fe3++H2O↔FeOH2++H+ FeOH2++H2O↔Fe(OH)2++H+ Fe(OH)2+H2O↔Fe(OH)3(s)+H+ 2FeOH2+↔Fe2(OH)24+ • Fe,Mn兩者一起摧化,反應比單獨個別摧化之和之速度快了3~10倍

  28. S(IV) Oxidation by the OH Radical • HSO3-+OH→SO3-+H2O • SO32-+OH→SO3-+OH- • SO3-+O2→SO5- • SO5-+HSO3-→HSO5-+SO3- • SO5-+SO3- HSO4-+SO3-+OH- • SO5-+SO5-→2SO4-+O2 • SO5-+SO5-→S2O82-+O2 • SO4-+HSO3-→SO3-+H++SO42-

  29. Oxidation of S(IV) by Oxides of Nitrogen • 2NO2+HSO3- 3H++2NO2-+SO42- • NO2在水相中受限於微小水溶液,使得此反應對S(IV)成為次要的反應途徑 • 但在都市地區,有較高的NO2濃度則此反應 會成為重要途徑

  30. Reaction of Dissolved SO2 with HCHO • HCHO(aq)+HSO3-→HOCH2SO3- • HCHO(aq)+SO32-→-OCH2SO3- • HOCH2SO3-+OH-→HCHO(aq)+SO32-+H2O

  31. Comparison of Aqueous-Phase S(IV) Oxidation Paths

  32. Aqueous-phase nitrite and nitrate reactions NOx oxidation NO2(aq)+NO2(aq) NO2-+NO3-+2H+ 6.127 NO(aq)+NO2(aq) 2NO2-+2H+ NO(aq)+OH(aq)→NO2-+H+ NO2(aq)+OH(aq)→NO3-+H+ 但是空氣中NO,NO2濃度<1nM 6.127反應速率<0.3 幾乎可忽略

  33. Nitrogen Radicals • 在夜晚NO3radical是液相中N物種裡最有反應性的,而在白天會快速光解 • N03,N2O5極易溶於水,可能是N的主要來源 N2O5(aq)+H2O→2H++2NO3- NO3+Cl-→NO3-+Cl(aq) 但是大氣中Cl-濃度低NO3轉而與HSO3-反應 NO3(aq)+HSO3-→NO3-+H++SO3- • NO3參與許多另外的反應所產生的自由基可導致S(IV)之氧化所需之自由基連鎖反應

  34. Aqueous-phase Organic Compounds • 與OH基反應較重要 H2C(OH)2+OH(aq) HCOOH(aq)+HO2(aq)+H2O HCOO-+OH CO2+HO2+OH- HCOOH+OH CO2+HO2+H2O

  35. Oxygen and hydrogen chemical • O3(aq)+O2- OH+2O2+OH- 6.142 • OH radical 主要之消耗:與hydrated formaldehyde反應 與H2O2,HCOOH,S(IV) 主要之來源:O2-與O3反應,H2O2光解 次要之來源:NO3-光解,HO2氧化S(IV) • HO2 radical 其主要來源即OH之主要消耗,反之亦然 • HO2(aq)+O2- H2O2+O2+OH- 6.143 • SO5-+O2 HSO5-+OH-+O2 6.144 • S(IV)+HO2(aq)→S(VI)+OH 6.145 • S(IV)+O2- S(VI)+OH+OH- 6.146 • HCO-+O2-→HO2-+CO3- 6.147

  36. Dynamic behavior of solutions with aqueous-phase chemical reactions 1. Closed system • [H2O2(aq)]open=HH2O2pºH2O2 6.148 6.149 6.150 • [H2O2(aq)]closed=HH2O2pH2O2 6.151 6.152

  37. 6.153 6.154 • [HNO3]total=HHNO3pHNO3 6.155 6.156 6.157 6.158

  38. Calculation of concentration changes in a droplet with aqueous-phase reaction • [H+]+[NH4+]=[OH-]+[HSO3-]+2[SO32-]+[HSO3-]+[NO3-] [S(VI)]=[SO42-]+[HSO3-]+[H2SO4(aq)]

  39. closed 系統H2O2,O3之消耗使得S(VI)產生較少 • 源源不絕的NH3提供更多的電中性反應在open系統中 • closed系統中由於S(IV)剩下較多,所以pH較低 • 較低的pH值使S(IV)被O3氧化速率降低 • 在closed系統中S(IV)是不斷的被消耗掉 • 在open系統中由於pH降低使得[HNO3]total降低

More Related