f N 2 O* + M N 2 O + M

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1 CHM 5423 Atmospheric Chemistry Problem Set 2 Due date: Thursday, February 7 th. Do the following problems. Show your work. 1) Before the development of lasers, atomic mercury lamps were a common source for UV radiation. The strongest light emission for a low pressure mercury lamp occurs at = 254. nm. a) Find the energy of one photon of light with = 254. nm. Give your answer in units of J/photon and cm -1. b) Find the energy of one mole of photons with = 254. nm. Give your answer in units of kj/mol. 2) Beer s law, as derived in class, can be written as ln(i l /I 0 ) = - Nl (2.1) where I 0 is the initial intensity of light (at a specific wavelength ), I l is the intensity after the light has traveled a distance l (in cm), is the absorption cross-section for the absorping molecule (in cm 2 /molecule), and N is the number density of absorbing molecules (in molecule/cm 3 ). Beer s law is a common method for monitoring concentrations of absorbing molecules both in the laboratory and in in the atmosphere. a) The absorption cross section for acetaldehyde (CH 3 CHO) is = 4.16 x cm 2 /molecule at = 300. nm. What is the concentration of acetaldehyde (in molecules/cm 3 ) in a system if I l /I 0 = for a pathlength l = 552. cm? Give your answer in units of molecules/cm 3. You may assume that at this wavelength acetaldehyde is the only molecule in the system that absorbs light. b) Chemists often use a different form of Beer s law A = log 10 (I 0 /I l ) = acl (2.2) where c is the concentration of absorbing molecules (in mol/l), l is the pathlength (in cm), and a is the absorption coefficient of the molecule (in L/mol. cm). Find the numerical value for the conversion factor between and a, and use it to calculate the value for a for acetaldehyde at = 300. nm. 3) Consider the UV photodissociation of nitrous oxide (N 2 O). The following primary processes could occur following the absorption of a photon by the molecule N 2 O + hc/ N 2 O* N 2 O* N 2 + O( 3 P) d N 2 O* N 2 O + hc/ f N 2 O* + M N 2 O + M In the above processes d is the primary quantum yield for each of the two photodissociation processes, f is the primary quantum yield for fluorescence, and q is the primary quantum yield for collisional quenching. a) If the above process is carried out in synthetic air (with p(air) = 1 00 atm >> p(n 2 O)) then the oxygen atom produced by the photodissociation process will quantitatively form ozone (O 3 ) by the reaction O( 3 P) + O 2 + M O 3 + M a) What will be the value for, the overall quantum yield, for N 2 O, N 2, O 2, and O 3? Give your answers in terms of the primary quantum yields for the three primary processes discussed above. b) Which of the overall quantum yields could actually be measured in an experiment? Which could not? Justify your answer. c) For which of the primary quantum yields (if any) is there sufficient information to find their value based on the overall quantum yields for a photodissociation experiment? Indicate how the overall quantum yield would be used to find the primary quantum yield when this can be done. q

2 d) Suggest additional experiments that could be done to find values for the primary quantum yields for the processes where they could not be found by the above single experiment. How would these additional eperiments allow you to find the missing primary quantum yield values? Be specific. 4) Find the value for (zenith angle) and f s (Earth-sun correction factor) for the following conditions: a) 1600 hours, latitude = 30 N, date = April 1 st b) 1100 hours, latitude = 20 S, date = December 5 th c) 1200 hours, latitude = 25 N, date = February 15 th 5) Using the method discussed in class, find the photodissociation rate constant for HO 2 NO 2 (peroxynitric acid, a reservoir species for NO 2 in the troposphere) for the following conditions: 1200 hours, latitude = 25 N, date = February 15 th (the conditions in part c of problem 4). The absorption cross-sections for this molecule as a function of wavelength are given below. You may assume the primary quantum yield for photodissociation of HO 2 NO 2 is equal to 1 at all wavelengths. (nm) (cm 2 /molecule) (nm) (cm 2 /molecule) x x x x x x x x ) Using the procedure discussed in class and the data below (given at T = 298. K), find the longest wavelength of light (in nm) capable of carrying out the following photodissociation reactions. Based on your calculations, which of the reactions (if any) be an important process in the troposphere? Which of the reactions (if any) could be an important process in the stratosphere? a) H 2 S HS + S( 3 P) b) N 2 O N 2 + O( 3 P) c) N 2 O N 2 + O( 1 D) d) CH 3 COCH 3 CH 3 COCH 2 + H( 2 S) e) CH 3 COCH 3 CH 3 CO + CH 3 species H f (kj/mol) species H f (kj/mol) H( 2 S) H 2 S O( 3 P) N O( 1 D) N 2 O 82.1 S( 3 P) CH 3 COCH HS CH 3 COCH CH CH 3 CO

3 Solutions. 1) a) E = hc = (6.626 x J s) (2.998 x 10 8 m/s) = 7.82 x J/molecule (254. x 10-9 m) Since E(cm -1 ) = E(J), then E(cm -1 ) = 7.82 x J/molecule = cm -1 Hc (6.626 x J s) (2.998 x cm/s) b) For one mole of photons. E = E N A, where N A = Avogadro s number So E = (7.82 x J/molecule) (6.022 x molecule/mol) (1 kj/1000 J) = 471. kj/mol 2) a) ln(i l /I 0 ) = - N So N = - ln(i l /I 0 ) = - ln(0.813) = 9.02 x molecule/cm 3 l (4.16 x cm 2 /molecule)(552. cm) b) The conversion factor can be found as follows ln(i l /I 0 ) = - Nl So ln(i 0 / I l ) = + Nl The conversion factor between ln and log 10 is = 2.303, and so if we divide both sides of the above equation by, we get ln(i 0 /I l ) = log 10 (I 0 /I l ) = Nl But log 10 (I 0 /I l ) = acl Therefore Nl = acl If we cancel the common factor of l, and solve for the absorption coefficient a, we get a = N c To complete the work above, we only need the conversion factor between N (number density) and c (molar concentration). And so c = 1 mol = x molecule = x molecule L L cm 3 Substituting above gives N (6.022 x molecule L/mol cm 3 ) a = 1 (6.022 x molecule L/mol cm 3 ) = (2.615 x molecule L/mol cm 3 ) The conversion factor is therefore the term in brackets above.

4 We can use this conversion factor to find the value for a for acetaldehyde at = 300. Nm a = (4.16 x cm 2 /molecule) (2.615 x molecule L/mol cm 3 ) = L/mol cm 3) a) The only process that removes N 2 O from the system is photodissociation, and so N2O = - d This process produces a molecule of N 2, and so N2 = + d Photodissociation generates an O( 3 P) atom,, whose only fate is reaction with O 2 to form O 3, and so O2 = - d O3 = + d b) We would most easily be able to measure the overall quantum yield for O 3 because we begin with no ozone in the system, and ozone concentration can be found using absorption spectroscopy. The decrease in concentration for N 2 O could be used to find the overall quantum yield for nitrous oxide, assuming that there was a significant decrease in concentration. This could be done using either absorption spectroscopy or fluorescence spectroscopy. Since there is already a large amount of nitrogen and oxygen in the system, it is not possible to measure the small changes in these concentrations due to their production or removal. c) We can find d using either the overall quantum yield for production of O 3 or for removal of N 2 O. Since d + f + q = 1 Then ( f + q ) = 1 - d However, there is not sufficient information to determine individual values for f and q. d) If we look at the three processes taking place from kinetic perspective, we can say k d = rate constant for dissociation k f = rate constant for fluorescence k q = k q [M] = rate constant for quenching (a pseudo 1 st order rate constant) d = k d k d + k f + k q [M] If we invert both sides of this expression, we get 1 = k d + k f + k q [M] = 1 + (k f /k d ) + (k q /k d ) [M] k d k d By studying the reaction at several pressures and plotting the data as suggested above, the values for k f /k d and k q /k d can be found, from which individual values for the primary quantum yields can be obtained.

5 4) Zenith angle and Earth-sun corrections can be found as follows a) 1600 hours = 0800 hours 30 N and April 1 st are in the table So = 62.1 For April 1 st the Earth-sun correction is b) 1100 hours is in the table 20 S December 5 th = 20 N June 5 th For 20 N June 1 st 1100 hours = 15.7 July 1 st 1100 hours = 15.1 So = ( ) (5 1)/(30 1) = 15.6 For the Earth-sun correction December 1 st = December 15 th = So = ( ) (5 1)/(15 1) = c) 1200 hours is in the table. We may interpolate for February 15 th and 25 N 20 N 25 N 30 N February 1 st February 15 th March 1 st The interpolation factor for February 15 th is (15 1)/(28 1) = 14/27 So = 37.2 The Earth-sun correction factor for February 15 th is ) We will do this as in the example in the notes. The data are compiled on the next page. Note that = We have interpolated for both absorption cross-section and photon flux using the data in the Chapter 2 appendix and data provided in the problem. The value obtained for the photodissociation rate constant is k d = 3.4 x 10-6 s -1. This corresponds to a halflife of t 1/2 = ln(2) = ln(2) = 57 hours (about 2 days) k d (3.4 x 10-6 s -1 )

6 range ( ) x F( ) x k d x 10 6 (nm) (cm 2 /molecule,e) (photon/cm 2. s) (s -1 ) k d (uncorrected) = n k d,n = x 10-6 s -1 k d (corrected) = f s k d (uncorrected) = (1.024) x x 10-6 s -1 = x 10-6 s x 10-6 s -1. 6) We may use the data to find the values for H rxn. The energy needed per molecule will then be E = H rxn /N A, where N A is Avogadro s number, and H rxn = [ H rxn (products) ] [ H rxn (reactants) ] If that energy comes from the absorption of a photon, then or E = hc/ = H rxn /N A = hcn A H rxn reaction H rxn (kj/mol) (nm) a b c d e

7 Reactions b, c, and e are energetically possible in the troposphere, since the cutoff for the actinic UV region of the spectrum is = 295. nm. Reaction d is close, and so might also be energetically possible. Reaction a is only energetically possible in the stratosphere (as are all of the other reactions). To actually occur, the reactant molecules would also have to absorb light, and have a primary yield for photodissociation d > 0 at at least some of the absorbing wavelengths in the actinic UV.

T(K) k(cm 3 /molecule s) 7.37 x x x x x 10-12

T(K) k(cm 3 /molecule s) 7.37 x x x x x 10-12 CHM 5423 Atmospheric Chemistry Problem Set 3 Due date: Tuesday, February 19 th. The first hour exam is on Thursday, February 21 st. It will cover material from the first four handouts for the class. Do