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 O2 respectively. Account qualitatively for
the difference.
λmax = (hc)/E
Bond enthalpy for N2 = 946 kJ/mol = 1.571 * 10-18 J/ bond
λmax = (6.626*10-34 Js x 3*108 m/s)/ 1.571 * 10-18 J
=1.27 *10-7 m
=127nm
Bond enthalpy for O2 = 497 kJ/mol = 8.253 * 10-19 J / bond
λmax = (6.626*10-34 Js x 3*108 m/s)/ 8.253 * 10-19 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?
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 10 -12 cm3 molecule-1s-1at
300 K. Write this reaction and calculate the mean life time of TCE on a day when the
concentration of OH is 2 x 106 molecule/cm3. (2.5 days)
C2H2Cl3 + OH ----- •HC2Cl3 +H2O
Mean half life of TCE = 1/ (k [OH]) = 1/ (2.3*10-12 x 2*106) = 2.2*105
=2.5 days
5) The second-order rate constant at 25 oC for the following reaction is 1.8*10-14
cm3/molecules*sec:
On a summer day in remote northern Ontario ( T= 25oC P = 1 atm), the concentration of NO is
0.10ppbv and that of O3 is 15 ppbv.
a) Calculate these concentrations in units of molecules/cm3. (2.46*109 molecules/cm3,
3.69*1011 molecules/cm3)
b) Calculate the rate of NO oxidation in units of molecules/sec*cm 3. (1.63*107 molecules/sec*
cm3)
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= 101325 Pa
T=25C=298K
Therefore, [M] = n/V= P/RT = (101325 Pa) / (8.315 J/(molK) x 298 K) x 10-6 m3/cm3
=4.09*10-5 mol /cm3
Therefore,
[NO] =( 0.10ppbv/ 1*109) x 4.09*10-5 mol = 4.09*10-15 mol/cm3
= 2.46*109 molecules/cm3
[O3] =( 15ppbv/ 1*109) x 4.09*10-5 mol = 6.13*10-13 mol/cm3
= 3.69*1011 molecules/cm3
b) Rate of oxidation of NO:
Rate NO = k2[NO][O3]
=1.8*10-14 cm3/sec*molecules x 2.46*109 molecules/cm3 x 3.69*1011 molecules/cm3
=1.63*107 molecules/sec*cm3
c)If we assume the concentration of O3 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*1011molecules/cm3 x 1.8*10-14 cm3 /sec*molecules = 6.64*10-3 s-1
6) a) Over the temperature range 250-479K the second order rate constant for the reaction:
Is given by the expressions
kH2(250-479 K) = 4.27 x 10-13 x (T/ 298)2.406 × exp(-1240/T) (Orkin V. 2006)
k = 9.61x10-18 x T2 x exp(-1457/T) (Atkinson R. )
a) Calculate the rate constant at 298K form both these expressions (6.6 * 10-15 cm3/molec-s
and 6.4 * 10-15 cm3/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 cm3/molec-s and 5.8 *1012
cm3/molec-s)
a)
Orkin:
k298K= 4.27 x 10-13 x (298/ 298)2.406 × exp(-1240/298)= 6.6 * 10-15 cm3/molec-s
Atkinson:
k298K = 9.61x10-18 x (298)2 x exp(-1457/298)= 6.4 * 10-15 cm3/molec-s
b)
k= A exp (-Ea/RT)
Ea=-ln(k/A)x RT
lnkT2=lnA – Ea/RT2
lnkT1=lnA – Ea/RT1
ln(kT2/kT1)=-Ea/R (1/T2 - 1/T1)
Let T1=330K, T2=259K
Orkin:
Ea=-8.314*330K* ln(k259K/k330K)/(1/(250-1) = 16036 J/mol = 16.0 kJ/mol
Atkinson:
Ea=-8.314*330K* ln(k259K/k330K)/(1/(250-1) = 16874 J/mol = 16.8 kJ/mol
c)
Orkin:
A298K = k298K x exp (Ea/(8.314*298K)) = 4.3*10-12 cm3/molec-s
Atkinson:
A298K = k298K x exp (Ea/(8.314*298K)) = 4.3*10-12 cm3/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 101325 Pa and
temperature 25C (0.38)
Convert 0.015mg/m3 to ppbv:
0.015mg/m3 / 1000L/m3 / 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 109/mixing ratio
Mixing ratio is number of moles of ozone to the total number of moles.
PV=nRT
n= (PV)/RT
=(101325 Pa x 1*10-3m3) / (8.315 J/(molK) x 298 K)
=0.04088mol of gas in 1 L
Therefore, 3.13*10-10 mol /L x 109/ 0.04088 mol /L = 7.6 ppbv
As a result, 7.6ppbv/20ppbv = 0.38
8)
a) Give the chemical formulas for CFC 113, CFC-12 and CFC-152 and the CFC numbers for
CHF2 Cl, C2H F4 Cl, C2H2F2Cl2
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 CF3CH2F with OH 8.6x10-15 cm3 molecule-1s-1. Assuming
that this is the only pathway for elimination CF3CH2F, 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 105 molec cm-3 for [OH]. (40%)
a)
CFC 113
CFC-12
CFC-152
CFC-22
CFC-124
CFC-132
b)
λ =hc/ E
= 119627 / 318 kJ mol-1
C2F3Cl13
CF2Cl2
C2H4F2
CHF2 Cl
C2H F4 Cl
C2H2F2Cl2
=374 nm
c) Rate of removal of CF3CH2F = rate of migration to stratosphere + rate of reaction with OH
= R migration [CF3CH2F] + R rxn [OH] [CF3CH2F]
Fraction that reaches stratosphere = R migration [CF3CH2F] / (R migration [CF3CH2F] + R rxn [OH]
[CF3CH2F])
Since migration half life is 7yr,
R migration = 1/7 = 0.14 y-1
R rxn [OH]= 8.6x10-15 cm3 molecule-1s-1 x 8 x 105 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-1 + 0.21 y-1) = 0.4 = 40%
More challenging problems
9) Consider the following set of reactions occurring in the troposphere:
NO2 ---> NO + O
O + M + O2 ---> O3
O3 + hv ----> O2* + O*
O* + M ----> O
O* + H2O ----> 2 OH
OH + CO ----> CO2 + H
k 2= 6x10-34 cm3/molec-s
k4 = 2.9x10-11 cm3/molec-s
k5 = 2.2x10-10 cm3/molec-s
k6 = 2.7x10-13 cm3/molec-s
Assume these are the only reactions involved. The concentrations of OH and CO are 10 7
molec/cm3 and 4.6 ppmv respectively. The concentration of M is 2.45 x1019 molec/cm3 and the
mixing ratio of O2 is 0.21. Assume that OH, O*, O and O3 are at steady state, and that T = 25oC
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 H2O, calculate the partial pressure of H2O. (0.056 atm)
c) Calculate the concentration of O* atoms. (0.5 molec/cm3)
d) Calculate the concentration of O atoms. (6700 molec/cm3)
a) Mean lifetime of OH = ([OH] at steady state) / (sinks of OH)
= [OH] / (k6[CO][OH]) = 1/(k6[CO])
[CO] = (4.6 x 10-6 molec CO/molec air) x (2.45 x1019 molec air/cm3) = 1.1 x 1014 molec/cm3
Therefore, mean lifetime of OH = 1/ (2.7x10-13 x 1.1 x 1014) = 0.034 s
b) Consider reactions R4 and R5 where O* is consumed: 30% of O*’s total consumption is
attributed to reaction 5:
0.3 = R5 / (R5+R4) = k5 [H2O] / (k5 [H2O] + k4[M])
0.3 (k5 [H2O] + k4 [M] ) = k5 [H2O]
0.3 k4[M] = 0.7 k5 [H2O]
[H2O]/ [M] = 0.3 k4 / 0.7 k5
=(0.3 x 2.9x10-11) / (0.7 x 2.2x10-10)
=0.056
[H2O]/ [M] = PH2O/ PM
PH2O = 0.056(1 atm) = 0.056 atm
c) Since the rate constant for reaction 3 is missing, consider the net rate for [OH] production
instead:
-d[OH]/dt = 2 k5 [O*][H2O] - k6 [OH][CO]
At steady state:
k6 [OH][CO] = 2 k5 [O*][H2O]
[O*] = k6 [OH][CO] / 2 k5 [O*][H2O]
=[ (2.7x10-13) (107) ( 4.6x10-6) ] / 2[ (2.2x10-10) (0.056 x 2.45 x1019) ]
[O*] = 0.5 molec / cm3
d) Net rate of reaction for [O]:
d[O]/dt = -k2[O][M][O2] + k1[NO2] + k4[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[O3]/dt = R2 – R3; since O3 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)
But
So
R2 = R3 = (R4 + R5)
R2 = R6/0.6
k2 [O] [M] [O2] = k6 [OH][CO]/ 0.6
[O]= (k6 [OH] [CO] ) / ( 0.6 k2 [M] [O2])
= [ ( 2.7x10-13 ) (107) (4.6x10-6 x 2.45x1019) ] / [ (0.6) (6x10-34)(2.45*1019)(0.21x 2.45x1019) ]
[O]= 6705 molec/cm3
10) Using these reactions, answer the following questions:
a)
OH + TCE ----> products
OH + CH4 ----> products
kTCE = ???
kCH4 = 8.4x10-15 cm3/molec-s
NO2 ---> NO + O
O + M + O2 ---> O3
O3 + hv ----> O2* + O*
O* + M ----> O
O* + H2O ----> 2 OH
OH + CO ----> CO2 + H
k1
k 2= 6x10-34 cm3/molec-s
k3
k4 = 2.9x10-11 cm3/molec-s
k5 = 2.2x10-10 cm3/molec-s
k6 = 2.7x10-13 cm3/molec-s
On a summer morning with T = 25oC 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 cm3/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, CH4 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 / (kTCE [OH])
Mean lifetime of CO = 1 / (k6[OH])
Mean lifetime of TCE/ Mean lifetime of CO = kCO / kTCE
kTCE = kCO 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
Given:
R3A = 3R3M
PH2OA = 3 kPa; PH2OM = 1.5 kPa
Mean lifetime of TCE in morning = 6 h (from a)
From a) Mean lifetime of TCE = 1 / (kTCE [OH])
If we assume [OH] is at steady state, the following expression is derived:
[OH] = 2 R5 / (k6[CO] + kTCE [TCE] + k CH4[CH4])
Given: [CO], [TCE], and [CH4] 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 = (R5) M / (R5) A
Assume [O*] is at steady state:
R3=R4+R5
Multiply both sides by (1/R5) and rearrange
R5= R3 x [R5 / (R4 + R5)]
R5 = R3 x k5[H2O]/(k4[M]+ k5 [H2O] )
R5 = R3 x (k5 PH2O) / (k4 PM + k5 PH2O)
R5 = R3 x (k5 * xH2O * PM)/[ (k4PM+ (k5 * xH2O * PM) ]
R5 = R3 x (k5 * xH2O)/(k4 + k5 * xH2O)
xH2O (Morning) = 3 kPa /101 kPa =0.0297
xH2O (Afternoon) = 1.5 kPa /101 kPa =0.0148
(R5)A = (R3)A (2.2*10-10 x 0.0297)/(0.29 *10-10 + (2.2*10-10 x0.0297)) = (0.184 R3)A
(R5)M = (R3)M (2.2*10-10 x 0.0148)/(0.29*10-10 + (2.2*10-10 x0.0148)) =( 0.1009 R3)M
(tMEAN)A/(tMEAN)M = (R5) M / (R5) A = ( 0.1009 R3)/ 0.184 x 3R3) = 0.183
Mean lifetime of TCE in afternoon = 6h x 0.183 = 1.1 h