CHEM173Midterm1Version1Key

CHEM173Midterm1Version1Key - CHEM173/273 Chemistry 173/273...

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Unformatted text preview: CHEM173/273 Chemistry 173/273 Midterm #1 28 April 2009 K. Prather STUDENT ID: SHOW ALL WORK ON EXAM OR ZERO CREDIT WILL BE GIVEN IF NOT ENOUGH INFORMATION IS GIVEN For most of these questions, as indicated by the length of the space, only brief answers are required. If you know the answer, fill it in. If you don’t, move on to the next question. Go back to the questions for which you are less sure of the answers at the very end of the time period for the exam. 1. /1'5 2. (/15 3. /l6 4. /24 5. /30 TOTAL MIDTERM /100 CHEM173/273 1. 15 points, 5 points each part Below 30 km, where atomic oxygen is very scarce, the following catalytic cycle, which destroys O3, is increasingly important: OH + 03 9 H02 + 02 k1 = 2><10‘14 cm3molecule'ls'l H02 + 03 9 OH + 202 k; = 3><10'l6 cm3molecule'ls'1 (3) Write the rate expressions for d[02] d[03] 5 dt dt EEO—A : \gtouliosl + uJHOzJCOJ 6* 1%} ; -k‘CoHJLOJ ~ \c‘EHOJEOQ (b) Using the steady-state approximation for [OH], and [OH]SS = 2.0><105 molecules/cmS, what is the concentration of H02 in molecules/ems? 929E“ .. -LCOMtofl + MHOJCGJ =0 4* \thod HIM a. q - o to W: [Wily E L‘ (:0ng D ( 31:!0‘“) K c 171 3. wet : i-gl‘W‘ (“‘1 (c) The rate expression for the destruction of 03 can be written as d[03] dt steady state condition from (b) and [OH]SS = 2.0x105 molecules/cm3, what is the value of k? “0’1 t _ \t \[orflCofl - LLLHOJLO‘I] 7T \(LCHOIJ = \‘\[0H] : — ‘2,\( \LOHJSS [03] k: Z\L\[OH]SS .r'tk C '5' (3. : 1(‘15l0 I$\(I\o 1‘10 % "q : ‘ gyro r ‘ —k[03] . Again using the CHEMl73/273 2. 15 points, 5 points each part T -1 Recall the lapse rate I‘ = —j— and the hypsometric equation P = Poe RT (The two cannot be Z considered at the same time, the hypsometric equation here assumes constant T) (a) Plot separate, qualitative dia rams of the variation of T and P with altitude 2 (Plot 2 on the vertical axis) in th- T l’ (b) Assume z = 0 and T = 288K at sea-level, and assume a linear lapse rate F = 9.8><10'3 K/m in the troposphere, what is the ambient air temperature on the top of Mt. Everest (z = 8850 m)? - \4 ~51: = 431m; '5: A? AT: “(Qt M03 éyflm“3 : Sgt T: wr-H = [1:0‘ “ (c) Assume constant T = 270 K variation with altitude, what is the local air pressure also on the top of Mt. Everest? (g = 9.8 m/sz, R = 287 J/kgK, P0 = 1 atm) -33 P : Po ‘3 r” «(3.? “IBM Yksowy] [(184 ififiirxz'lok ““ ( l 0A1“) e .z 6.3 6441’“ CHEM173/273 3. 16 points, 5 points (a) and (c), 6 points (b) (21) Write the most important two-step mechanism for the noontime synthesis of 03 in the polluted troposphere. H01 + i/w ~——e H0 *0 W\ C ‘i' 01 *—’P 03 (b) Complete the following mechanism for the atmospheric conversion of CH4 to C02 (ONE product in each reaction is a reactant in the following reaction) CH4 + OH 9 G4; + H10 Q;_+ 02» m' 6430-6 +NO —>C_HJO_+ N02 Cfigo +029k£§1L+H02 flc’db +hv9 _Cflb_+H CH6 +oz->__CI;~+H02 CO +OH9 C02+ H (c) The rate constant for the reaction of OH with 2,2-Dimethylpentane is kOH = 3.4><10'12 cm3/molecule-s at 298 K. Assuming OH is the only oxidizing species present with [OH]SS = 2><105 molecule/cm3 , how much time is required for the concentration of 2,2-DMP to fall to 1/16 of its original concentration at 298 K? Leave your answer in seconds. ‘i'llo : H‘hh W [OHlss kOH C : L‘¥‘° S CHEMl73/273 4. 24 points total, 8 points each part (a) Write the Chapman cycle. Explain why Chapman occurs in the stratosphere. If ozone is found in the troposphere, how can OH be one possible result? 01 4 \m (1640 at.) ——v 10 W\ O 4 01 A O?» 0*03 "" 10'— (b) Write the Beer-Lambert law, and properly label the components 10, I, o, N, 1. What are the components of “optical depth?” ._ Q" H Q ‘5’ 3:2 :1: e \\ News» D“, it? / \ Msw‘ol‘vm ~Cro58 fit («A CHEM173/273 (c) Assume a diatomic molecule. From lst principles graph the form of the potential energy as a function of distance between the nuclei. Assume next that only discrete Vibrational energy states are possible. Write the equation for the vibrational energy. What are two reasons one cannot assume this ideal potential energy diagram for a real diatomic molecule? l( "( U : {lc(r —(C))’ U MF— Ew. kmw + '1.) {Q r CHEM173/273 5. Multiple Choice 30 points total, 2 points each (a) Human eyes are most sensitive to radiation in the (i) Ultaviolet (We (iii) Infrared (iv) All of the above (b) A loss of stratospheric ozone would (i) Increase CFC levels (ii) Cause global warming (iii) Increase troposphere lapse rate 1v amage I V ‘ (c) Smog is (i) Smoke + ozone (ii Small or anics dllli Smoke + f@ (iv) Smoke + organics (d) One source of acid rain is ' " 2 oxidatio 11 ' y o uoric acid (iii) Dimethylsulfide oxidation (iv) Organic nitrate formation 6 (e) A temerature inversion is characterized by a -. a; - uo‘ . li, - with increase (ii) A high lapse rate (iii) An increasing ozone concentration in the troposphere (iv) A decreasing profile of CF Cs in a stable airmass (f) Ozone mixing state is highest (i) In polluted regions with a high photolysis rate (ii) At the tropopause (iii) Int - u ' a a le of a temperature inversion (g) UVA, B, and C describe (i) Rate constants for photolysis (ii Allowed transition states l“‘ In.'|‘ (iv) Vibrational quantum nubers (h) The Ozone hole occurs (i) Over the Winter pole (ii) Due to man-made CFCs (iii) A ‘ - ' - : s - n al dynamic vortex 1v) All of the abov CHEM173/273 (j) A profile of a chemical species that is constant throughout the troposphere implies: (i) A source at the top of the atmosphere ( 1 ong lifeti - (iii) Photochemical production (iv) Both b and c (k) The gases and particles in an airmass can change their mixing state due to: (i) Coagulation (ii) Chemical reactions (iii) G phase uptake (1) The combined absorption and scattering of particles and gases is the (i) Actinic flux (iii rra lance (iv) Radiance (m) High air pollution can be caused by (i) A temperature inversion (ii) Emission of anthropogenic gases and particles (iii) D deosition (n) One possible photochemical result of an excited molecule is: (ii) Radiation (iii) Explosion (iv) Absorption (0) One ossible photophysical result of an excited molecule is: < (1) :fiadiation > (ii) Dissociation (iii) Rearrangement (iv) Absorption (p) For a chemical reaction with several possible outcomes the fraction producing a certain result is the: (i) Photolysis rate (ii) Abso tion coefficient W. 1v ‘ owed transition ...
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CHEM173Midterm1Version1Key - CHEM173/273 Chemistry 173/273...

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