CHE 230S ENVIRONMENTAL CHEMISTRY PROBLEM SET 8 Full Solutions

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Tamsin Rogers
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1 CHE 230S ENVIRONMENTAL CHEMISTRY PROBLEM SET 8 Full Solutions Easier problems 1) Calculate the maximum wavelength of radiation required to promote dissociation of a) a dinitrogen molecule (127nm) b) a dioxygen molecule (241 nm) Use bond enthalpies of 946 and 497 kj/mol for N2 and O 2 respectively. Account qualitatively for the difference. λmax = (hc)/e Bond enthalpy for N 2 = 946 kj/mol = * J/ bond λmax = (6.626*10-34 Js x 3*10 8 m/s)/ * J =1.27 *10-7 m =127nm Bond enthalpy for O 2 = 497 kj/mol = * J / bond λmax = (6.626*10-34 Js x 3*10 8 m/s)/ * J =2.41 *10-7 m =241nm The bond enthalpy for nitrogen is higher (triple bond, bond order 3), thus it requires a wavelength of higher energy (lower wavelength) for dissociation compared to dioxygen molecule (double bond, bond order 2). 2) What is the enthalpy change for the following reaction? Is this an energetically favourable reaction? Mean bond enthalpies for C-H and O-H bonds are 414 kj/mol and 460 kj/mol respectively. (-46 kj) One hydrogen us removed from carbon and attached to oxygen. Mean bond enthalpies: -To break H-C bond: 414 kj/mol -To form O-H bond: 460 kj/mol ΔH= Σ E (bonds broken) - Σ E (bonds formed) = (1mol x 414 kj/mol) (1mol x 460 kj/mol) = -46 kj This is an energetically favourable process. 3) Which of the following species are radicals?

2 The following species are radicals according to Lewis dot diagrams: 4) The abstraction by OH of hydrogen from tricholoroethylene (TCE), a volatile organic compound used to clean circuit boards, has a rate constant of 2.3 x cm 3 molecule -1 s -1 at 300 K. Write this reaction and calculate the mean life time of TCE on a day when the concentration of OH is 2 x 10 6 molecule/cm 3. (2.5 days) C 2 H 2 Cl 3 + OH HC 2 Cl 3 +H 2 O Mean half life of TCE = 1/ (k [OH]) = 1/ (2.3*10-12 x 2*10 6 ) = 2.2*10 5 =2.5 days 5) The second-order rate constant at 25 o C for the following reaction is 1.8*10-14 cm3/molecules*sec: On a summer day in remote northern Ontario ( T= 25 o C P = 1 atm), the concentration of NO is 0.10ppbv and that of O 3 is 15 ppbv. a) Calculate these concentrations in units of molecules/cm 3. (2.46*10 9 molecules/cm 3, 3.69*10 11 molecules/cm 3 ) b) Calculate the rate of NO oxidation in units of molecules/sec*cm 3. (1.63*10 7 molecules/sec* cm 3 ) c) Show how the rate law may be expressed in pseudo first-order terms and calculate the pseudo first-order rate constant (6.64*10-3 s -1 ) a) PV = nrt P= Pa T=25C=298K Therefore, [M] = n/v= P/RT = ( Pa) / (8.315 J/(molK) x 298 K) x 10-6 m 3 /cm 3 =4.09*10-5 mol /cm 3 Therefore, [NO] =( 0.10ppbv/ 1*10 9 ) x 4.09*10-5 mol = 4.09*10-15 mol/cm 3 = 2.46*10 9 molecules/cm 3 [O 3 ] =( 15ppbv/ 1*10 9 ) x 4.09*10-5 mol = 6.13*10-13 mol/cm 3 = 3.69*10 11 molecules/cm 3 b) Rate of oxidation of NO: Rate NO = k2[no][o 3 ]

3 =1.8*10-14 cm 3 /sec*molecules x 2.46*10 9 molecules/cm 3 x 3.69*10 11 molecules/cm 3 =1.63*10 7 molecules/sec*cm 3 c)if we assume the concentration of O 3 is unchanging during the reaction (since its concentration is more than 100 times that of NO), we can treat this value as constant. We can than combine the two constants and derive the new rate constant k (pseudo first order constant with units s -1 ) Therefore, Rate NO= k *NO+ k = k [O3]=3.69*10 11 molecules/cm 3 x 1.8*10-14 cm 3 /sec*molecules = 6.64*10-3 s -1 6) a) Over the temperature range K the second order rate constant for the reaction: Is given by the expressions k H2 ( K) = 4.27 x x (T/ 298) exp(-1240/t) (Orkin V. 2006) k = 9.61x10-18 x T 2 x exp(-1457/t) (Atkinson R. ) a) Calculate the rate constant at 298K form both these expressions (6.6 * cm 3 /molec-s and 6.4 * cm 3 /molec-s) b) Fit these expressions to the form k= A exp (-Ea/RT) in order to estimate a value for the Activation energy ( 16.0 kj/mol and 16.8 kj/mol) c) Estimate the frequency factor A for both expressions (4.3*10-12 cm 3 /molec-s and 5.8 *10-12 cm 3 /molec-s) a) Orkin: k 298K = 4.27 x x (298/ 298) exp(-1240/298)= 6.6 * cm 3 /molec-s Atkinson: k 298K = 9.61x10-18 x (298) 2 x exp(-1457/298)= 6.4 * cm 3 /molec-s b) k= A exp (-Ea/RT) Ea=-ln(k/A)x RT lnk T2 =lna Ea/RT 2 lnk T1 =lna Ea/RT 1 ln(k T2 /k T1 )=-Ea/R (1/T 2-1/T 1 ) Let T 1 =330K, T 2 =259K Orkin: Ea=-8.314*330K* ln(k 259K /k 330K )/(1/(250-1) = J/mol = 16.0 kj/mol Atkinson: Ea=-8.314*330K* ln(k 259K /k 330K )/(1/(250-1) = J/mol = 16.8 kj/mol c) Orkin: A 298K = k 298K x exp (Ea/(8.314*298K)) = 4.3*10-12 cm 3 /molec-s

4 Atkinson: A 298K = k 298K x exp (Ea/(8.314*298K)) = 4.3*10-12 cm 3 /molec-s 7) Convert all the units to ppbv and find the ratio of the ozone concentration in Jakarta to the ozone concentration in Tokyo, given that ozone concentration in Jakarta is 0.015mg/m 3 and the ozone concentration in Tokyo is 20 ppbv. Assume the conditions are: pressure Pa and temperature 25C (0.38) Convert 0.015mg/m 3 to ppbv: 0.015mg/m 3 / 1000L/m 3 / 1000mg/g = 1.5*10-8 g of ozone per 1L of air 1.5*10-8 g / 48g/mol = 3.13*10-10 mol of ozone per 1 L of air Concentration in ppbv = Concentration in M x 10 9 /mixing ratio Mixing ratio is number of moles of ozone to the total number of moles. PV=nRT n= (PV)/RT =( Pa x 1*10-3 m 3 ) / (8.315 J/(molK) x 298 K) = mol of gas in 1 L Therefore, 3.13*10-10 mol /L x 10 9 / mol /L = 7.6 ppbv As a result, 7.6ppbv/20ppbv = ) a) Give the chemical formulas for CFC 113, CFC-12 and CFC-152 and the CFC numbers for CHF 2 Cl, C 2 H F 4 Cl, C 2 H 2 F 2 Cl 2 b) The C-Cl bond strength in CFC-12 is 318 kj mol -1. Estimate the wavelength range over which you would expect this reaction to be possible, and comment on your calculated result. (374 nm) c) The rate constant for the reaction of CF 3 CH 2 F with OH 8.6x10-15 cm 3 molecule -1 s -1. Assuming that this is the only pathway for elimination CF 3 CH 2 F, calculate the fraction of this substance that reaches the stratosphere, if the characteristic time for its migration to the stratosphere is ~7 years. Assume a constant value of 8 x 10 5 molec cm -3 for [OH]. (40%) a) CFC 113 C 2 F 3 Cl 13 CFC-12 CF 2 Cl 2 CFC-152 C 2 H 4 F 2 CFC-22 CHF 2 Cl CFC-124 C 2 H F 4 Cl CFC-132 C 2 H 2 F 2 Cl 2 b) λ =hc/ E = / 318 kj mol -1

5 =374 nm c) Rate of removal of CF 3 CH 2 F = rate of migration to stratosphere + rate of reaction with OH = R migration [CF 3 CH 2 F] + R rxn [OH] [CF 3 CH 2 F] Fraction that reaches stratosphere = R migration [CF 3 CH 2 F] / (R migration [CF 3 CH 2 F] + R rxn [OH] [CF 3 CH 2 F]) Since migration half life is 7yr, R migration = 1/7 = 0.14 y -1 R rxn [OH]= 8.6x10-15 cm 3 molecule -1 s -1 x 8 x 10 5 molec cm -3 =6.8*10-9 s -1 =0.21 y -1 Therefore, fraction that reaches stratosphere = 0.14 y -1 / (0.14 y y -1 ) = 0.4 = 40% More challenging problems 9) Consider the following set of reactions occurring in the troposphere: NO 2 ---> NO + O O + M + O 2 ---> O 3 O 3 + hv ----> O * 2 + O * O * + M ----> O O * + H 2 O ----> 2 OH OH + CO ----> CO 2 + H k 2 = 6x10-34 cm 3 /molec-s k 4 = 2.9x10-11 cm 3 /molec-s k 5 = 2.2x10-10 cm 3 /molec-s k 6 = 2.7x10-13 cm 3 /molec-s Assume these are the only reactions involved. The concentrations of OH and CO are 10 7 molec/cm 3 and 4.6 ppmv respectively. The concentration of M is 2.45 x10 19 molec/cm 3 and the mixing ratio of O 2 is Assume that OH, O *, O and O 3 are at steady state, and that T = 25 o C and P = 1 atm. Answer the following questions. a) Calculate the mean lifetime of OH (0.033s) b) Given that 30% of O * reacts with H 2 O, calculate the partial pressure of H 2 O. (0.056 atm) c) Calculate the concentration of O * atoms. (0.5 molec/cm 3 ) d) Calculate the concentration of O atoms. (6700 molec/cm 3 ) a) Mean lifetime of OH = ([OH] at steady state) / (sinks of OH) = [OH] / (k 6 [CO][OH]) = 1/(k 6 [CO]) [CO] = (4.6 x 10-6 molec CO/molec air) x (2.45 x10 19 molec air/cm 3 ) = 1.1 x molec/cm 3 Therefore, mean lifetime of OH = 1/ (2.7x10-13 x 1.1 x ) = s

6 b) Consider reactions R4 and R5 where O* is consumed: 30% of O* s total consumption is attributed to reaction 5: 0.3 = R 5 / (R 5 +R 4 ) = k 5 [H 2 O] / (k 5 [H 2 O] + k 4 [M]) 0.3 (k 5 [H 2 O] + k 4 [M] ) = k 5 [H 2 O] 0.3 k 4 [M] = 0.7 k 5 [H 2 O] [H 2 O]/ [M] = 0.3 k 4 / 0.7 k 5 =(0.3 x 2.9x10-11 ) / (0.7 x 2.2x10-10 ) =0.056 [H 2 O]/ [M] = P H2O / P M P H2O = 0.056(1 atm) = atm c) Since the rate constant for reaction 3 is missing, consider the net rate for [OH] production instead: -d[oh]/dt = 2 k 5 [O * ][H 2 O] - k 6 [OH][CO] At steady state: k 6 [OH][CO] = 2 k 5 [O * ][H 2 O] [O * ] = k 6 [OH][CO] / 2 k 5 [O * ][H 2 O] =[ (2.7x10-13 ) (10 7 ) ( 4.6x10-6 ) ] / 2[ (2.2x10-10 ) (0.056 x 2.45 x10 19 ) ] [O * ] = 0.5 molec / cm 3 d) Net rate of reaction for [O]: d[o]/dt = -k 2 [O][M][O 2 ] + k 1 [NO 2 ] + k 4 [O*][M] Since k1 is unknown, and there is no direct way of calculating this value, we must indirectly calculate [O] by using steady state assumptions for other species and info from part c) d[o 3 ]/dt = R2 R3; since O 3 is assumed to be at steady state, R2 = R3 d[o*]/dt = R3 R4 R5; since O* is assumed to be at steady state, R3 = R4 + R5 Also from part c) when OH is at steady state: R6 = 2R5 R5 = 0.3(R4 + R5) Therefore R6 = 2R5 = 0.6(R4 + R5)

7 But R2 = R3 = (R4 + R5) So R2 = R6/0.6 k 2 [O] [M] [O 2 ] = k 6 [OH][CO]/ 0.6 [O]= (k 6 [OH] [CO] ) / ( 0.6 k 2 [M] [O 2 ]) = [ ( 2.7x10-13 ) (10 7 ) (4.6x10-6 x 2.45x10 19 ) ] / [ (0.6) (6x10-34 )(2.45*10 19 )(0.21x 2.45x10 19 ) ] [O]= 6705 molec/cm 3 10) Using these reactions, answer the following questions: OH + TCE ----> products k TCE =??? OH + CH > products k CH4 = 8.4x10-15 cm 3 /molec-s NO 2 ---> NO + O k 1 O + M + O 2 ---> O 3 k 2 = 6x10-34 cm 3 /molec-s O 3 + hv ----> O * 2 + O * k 3 O * + M ----> O k 4 = 2.9x10-11 cm 3 /molec-s O * + H 2 O ----> 2 OH k 5 = 2.2x10-10 cm 3 /molec-s OH + CO ----> CO 2 + H k 6 = 2.7x10-13 cm 3 /molec-s a) On a summer morning with T = 25 o C and P = 1 atm, some trichloroethylene (TCE) is released to the atmosphere and found to have a mean lifetime of 6 hours. In contrast the mean lifetime of CO is 51.1 hours. Assuming that both TCE and CO are only eliminated through reaction with OH, calculate the rate constant for the reaction of OH and TCE. (2.3x10-12 cm 3 /molec-s) b) By the middle of the afternoon, the rate of O * production through reaction 3 has tripled as compared to that in the morning and the partial pressure of water has increased from 1.5 kpa to 3.0kPa (although the total pressure is still 101kPa). Calculate the mean lifetime of TCE under these afternoon conditions. Assume that the release rates of TCE, CH 4 and CO increase in the afternoon such that their concentrations remain same as in the morning. (1.1 h) a) Mean lifetime of TCE = ([TCE] at steady state)/ (sinks of TCE) Mean lifetime of TCE = 1 / (k TCE [OH]) Mean lifetime of CO = 1 / (k 6 [OH]) Mean lifetime of TCE/ Mean lifetime of CO = k CO / k TCE k TCE = k CO x (Mean lifetime of CO/ Mean lifetime of TCE) = = 2.7x10-13 x (51.1/6) = 2.3*10-12 b) Notation: A = afternoon; M = morning

8 Given: R 3 A = 3R 3 M P H2O A = 3 kpa; P H2O M = 1.5 kpa Mean lifetime of TCE in morning = 6 h (from a) From a) Mean lifetime of TCE = 1 / (k TCE [OH]) If we assume [OH] is at steady state, the following expression is derived: [OH] = 2 R 5 / (k 6 [CO] + k TCE [TCE] + k CH4[CH 4 ]) Given: [CO], [TCE], and [CH 4 ] remain constant in the morning and afternoon (Mean lifetime TCE ) A / (Mean lifetime TCE MEAN ) M = [OH] M / [OH] A (Mean lifetime TCE ) A / (Mean lifetime TCE MEAN ) M = (R 5 ) M / (R 5 ) A Assume [O*] is at steady state: R 3 =R 4 +R 5 Multiply both sides by (1/R 5 ) and rearrange R 5 = R 3 x [R 5 / (R 4 + R 5 )] R 5 = R 3 x k 5 [H 2 O]/(k 4 [M]+ k 5 [H 2 O] ) R 5 = R 3 x (k 5 P H2O ) / (k 4 P M + k 5 P H2O ) R 5 = R 3 x (k 5 * x H2O * P M )/[ (k 4 P M + (k 5 * x H2O * P M ) ] R 5 = R 3 x (k 5 * x H2O )/(k 4 + k 5 * x H2O ) x H2O (Morning) = 3 kpa /101 kpa = x H2O (Afternoon) = 1.5 kpa /101 kpa = (R 5 ) A = (R 3 ) A (2.2*10-10 x )/(0.29 * (2.2*10-10 x0.0297)) = (0.184 R 3 ) A (R 5 ) M = (R 3 ) M (2.2*10-10 x )/(0.29* (2.2*10-10 x0.0148)) =( R 3 ) M (t MEAN ) A /(t MEAN ) M = (R 5 ) M / (R 5 ) A = ( R 3 )/ x 3R 3 ) = Mean lifetime of TCE in afternoon = 6h x = 1.1 h

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

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