Download Understanding Chemical Reactions & Pollutants in Photochemical Smog: Air Pollution - Prof. and more Assignments Green and Environmental Chemistry in PDF only on Docsity! ATM S 212: Air Pollution Autumn 2008 Homework 3 Name:_____KEY_______________________________________ Assignment will be collected at the beginning of class on Thursday, October 23, 2008. 1. Briefly describe 3 types of chemical reactions. [6 points] 6 points possible • Binary Reaction: is a reaction where two molecules react to form (one or more) product molecules. Takes the form: DCBA +→+ • Three-Body Reaction: is a reaction where three molecules collide, two molecules react to form products and the third body, “M”, carries away excess energy. Takes the form: MABMBA +→++ • Photolysis Reaction: is a reaction that is initiated by the absorption of solar radiation. Takes the form: BAhAB +→+ ν 2. London smog is associated with cold temperatures, while L.A. smog is associated with warm temperatures (cold/warm). Pollution is highest in the morning during a London smog event, and highest in the afternoon during an L.A. smog event (morning/afternoon/evening). [4 points] 3. What are the 3 main ingredients (primary pollutants) in photochemical smog? At what time(s) of day do their concentrations peak? [4 points] 4 points possible 4 points possible • Carbon Monoxide (CO) • Nitrogen Oxide (NO) • Hydrocarbons (HC) and/or Reactive Organic Gases (ROG) All three of these species peak during the morning and afternoon rush hours with a higher concentrations during the morning rush hour as a result of the lower mixing height. 4. Name the 2 secondary pollutants in photochemical smog. Explain why they peak in the afternoon. [6 points] 6 points possible • Ozone (O3) • Nitrogen Dioxide (NO2) Both of these species peak during the afternoon because that is when the maximum in sunlight and temperature occur along with allowing some time for the primary pollutants that were emitted during the morning rush hour to be transformed. ATM S 212 Assignment 3 1 5. Box model: a) The current U.S. air quality standard for carbon monoxide (CO) is 9 ppm (averaged over 8 hours). A power plant reports a source rate (S) of CO equal to 1012 grams of CO per year. Calculate the mixing ratio [in units of grams of CO per cm3 of air] of carbon monoxide (CO) in the boundary layer (0 to 1 km altitude) within 10 km of the power plant (e.g. volume of the box (V) = 20 km x 20 km x 1 km = 400 km3 = 4 x 1017 cm3). Assume that the concentration of CO is at steady state, the mixing ratio of CO is uniform within this volume, and the lifetime (τ) of CO in the “box” is 2 hours. (Hint: first calculate the concentration of CO in the box, then convert to mixing ratio assuming the density of air is 2.5 x 1019 molecules of air per cubic centimeter. You also need to know that CO weighs 1.7 x 10-21 grams per molecule.) Would you want to live here? [15 points] 15 points possible V Sq τ*= molecule gMW cm airmolecules hours year COgS CO air 21 3 19 12 107.1 105.2 2 10 −×= ×= = = ρ τ 10 km 10 km 20 km 20k 317 35 3 3 10410400 400 12020 cm km cmkmV kmV kmxkmxkmV ×=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = = COppbvCOppmv airmolec COmolec airmolec COmolec airmolec COmolec COg COmolec airmolec cm cm COg cm COg cm hourshour COg V Sq hour COg hours day days year year COgS 4.130134.0 .101 .34.1 . .1034.1 . .1034.1 107.1 .1 .105.2 11071.5 1071.5 104 2*1014.1* 1014.1 24 1 365 110 8 8 8 2119 3 3 10 3 10 317 8 8 12 == × =× ×=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ×= × × == ×=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − − − −τ Since the air quality standard is 9 ppmv and the average in the box is 13.4 ppbv, this is a safe place to live. ATM S 212 Assignment 3 2
