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Fundamentals of Atmospheric Chemistry: Free Radicals and Trace Species Dynamics

This course module focuses on the fundamentals of atmospheric chemistry, emphasizing the role of trace species and free radicals. Students will learn essential concepts such as pollutants, reactive substances, and the processes that govern the release and removal of trace compounds (sources, sinks, reservoirs). Key topics include free radical chemistry, quantifying concentrations in ppm and ppb, and calculating partial pressures and number densities. Interactive field trip activities enrich the learning experience, allowing students to apply theoretical knowledge in real-world contexts.

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Fundamentals of Atmospheric Chemistry: Free Radicals and Trace Species Dynamics

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  1. CHEM 304 – 1/27/12 I.A. Fundamentals of Atmospheric Chemistry • More on “Trace” Species - Free Radical Chemistry - Concentration units (p. 28-29) FIELD TRIP MONDAY ! HW #1 e-mailed W Reading: Baird Intro, Ch. 3 (p. 91-95), and start chapter 1

  2. Some lingo regarding “Trace species” (e.g. pollutants, reactive substances, etc.) Source: reaction or process that release trace compound. Sink: Reaction or process that “permanently” removes trace compound. Reservoir: An entity (process, substance, physical feature) that temporarily holds a trace substance

  3. Which of the following are “free radicals”? (all are neutral …) CH3, ClONO2, NO3, SO3, HOSO2, Cl2O2 CH3, NO3, & HOSO2

  4. A summary for free radical reaction types: Initiation: X-Y —> X° + Y °(often light-driven) Propagation: X° + Y-Z —> XY + Z° or X° + W —> XW° Initiation: X° + Y ° —> X-Y

  5. Quantifying Concentrations: Express the Following as ppm and ppb a) XCH4 = 0.000002 b) XO3(10 km) = 5.1 x 10-8 a) 2.0 x10-6 = 2.0 ppm = 2000 ppb b) 0.051 x10-6 = 0.051 ppm = 51 ppb

  6. Quantifying Concentrations: Express the following as X’s (i.e. pure numbers). a) H2O (10km)= 36 ppm b) CH4 (40 km) = 300ppb a) 36 ppm = 36 x 10-6 = 3.6 x 10-5 b) 300 ppb = 300 x 10-9 = 3.0 x 10-7

  7. Partial pressures: Calculate the Partial Pressure of CO2(XCO2= 360 ppm) at: a) 0 km (PTOT=1 atm) b) 15 km (PTOT=122 mbar) a) PCO2 = 3.6x10-4 x 1.0 atm = 3.6 x 10-4 atm b) PCO2 = 3.6 x 10-4 x 122 mbar = 4.4 x 10-5 bar

  8. Number Densities: molec/cm3 For OH, X = 6 x 10-14, what is C in molec/cm3 at 0 km (P=1.0atm)? Calc p first, then multiply by 2.7x1019 molec/cm3atm 1.6 x 106

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