CHE 230S ENVIRONMENTAL CHEMISTRY
PROBLEM SET 8 – Partial 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
The bond enthalpy for nitrogen is higher (triple bond, bond order 3)
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)
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])
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*cm3. (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
[M] = n/V= P/RT
b) Rate of oxidation of NO:
Rate NO = k2[NO][O3]
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)
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:
kH2(250-479 K) = 4.27 x 10-13 x (T/ 298)2.406 × exp(-1240/T)
Atkinson:
k = 9.61x10-18 x T2 x exp(-1457/T)
b)
Ea= -R * T max* ln (kTmin/kTmax)/(1/(Tmin-1))
c)
A=k x exp(Ea/RT)
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/m3 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.
Concentration in ppbv = mixing ratio x 109
Mixing ratio is number of moles of ozone to the total number of moles.
PV=nRT
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%)
b)
λ =hc/ E
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])
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 x10 19 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])
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])
[H2O]/ [M] = PH2O/ PM
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]
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
d[O*]/dt = R3 – R4 – R5
Also from part c)
when OH is at steady state:
R6 = 2R5
R5 = 0.3(R4 + R5)
10) Using these reactions, answer the following questions:
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
k1
k 2= 6x10-34 cm3/molec-s
k3
k4 = 2.9x10-11 cm3/molec-s
O* + H2O ----> 2 OH
OH + CO ----> CO2 + H
a)
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])
b)
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])
Assume [O*] is at steady state:
R3=R4+R5