Aocfch

Thus ODP, is a relative measure.

Table 5.3 presents ozone depletion potentials calculated with two-dimensional models, as presented by the World Meteorological Organization (2002). Tropospheric OH reaction is the primary sink for the hydrogen-containing species. As seen in Table 5.3, the ODPs for the hydrogen-containing species are considerably smaller than those of the CFCs, the difference reflecting the more effective chemical removal in the troposphere. Although Table 5.3 gives the ODPs of several brominated halocarbons relative to CFC-11,

TABLE 5.3 Steady-State Ozone Depletion Potentials (ODPs) Derived from Two-Dimensional Models"

Trace Gas ODP

CFC-12 (CF2C12) 0.82

CFC-113 (CFC12CF2C1) 0.90

CFC-114 (CF2C1CF2C1) 0.85

CFC-115 (CF2C1CF3) 0.40

CCU 1.20

CH3CCI3 0.12

HCFC-22 (CF2HC1) 0.034

HCFC-123 (CF3CHC12) 0.012

HCFC-124 (CF3CHFC1) 0.026

HCFC-141b (CH3CFC12) 0.086

HCFC-142b (CH3CF2C1) 0.043

HCFC-225ca (CF3CF2CHC12) 0.017

HCFC-225cb (CF2C1CF2CHFC1) 0.017

CH3Br 0.37

H-1301 (CFjBr) 12

H-1211 (CF2ClBr) 5.1

H-2402 (CF2BrCF2Br) <8.6

CH3C1 0.02

"World Meteorological Organization (2002).

the ODPs for bromine-containing compounds should really be compared relative to each other because of the strong coupling of bromine catalytic cycles to chlorine levels in the lower stratosphere. The halon that is generally selected as the reference compound for ODPs of bromine-containing halocarbons is H-1301 (CF3Br), the halon with the longest atmospheric lifetime and largest ODP

PROBLEMS

5.1a The rate coefficients for O('D) + 03—>202 and 0 + 03—>202 are given below. In the middle stratosphere, at about 30 km, T = 230 K, typical noontime concentrations of 0('D) and O are 50 molecules cm-3 and 7.5 x 107 molecules cm~3, respectively. Evaluate the rates of both 03 loss processes at this altitude. Which loss process dominates?

k0+o, = 8 x 10 12 exp(—2060/7') cm3 molecule"1 s"1

5.2a Estimate the lifetime of odd oxygen (i.e., ozone) at 20km and 45 km. It may be assumed that the photolysis rates of 03 at these two altitudes are (see Figure 4.14)

jo3 = 7 x 10~4 s-1 at 20 km = 6 x 10~3 s-1 at 45 km and that the 03 concentrations at the two altitudes are (see Figure 5.2)

[03] - 2 x 1012 molecules cm"3 at 20 km = 0.2 x 1012 molecules cm"3 at 45 km

5.3a Estimate the stratospheric lifetime of N20 at 30 km using Figure 5.6. 5.4a For the catalytic cycle

CI + 03 -U CIO + 02 ki= 2.9 x 10"11 exp(—260/T) CIO + O —> CI + 02 k2 = 3.0 x 10"11 exp(70/T)

determine the ratio of [C10]/[03] at which the rate of ozone destruction from this cycle equals that from the Chapman cycle at 20 km where T = 220 K. If the 03 mixing ratio is 4 ppm, what concentration of CIO does this correspond to?

5.5a a. Write out the catalytic ozone loss cycle that is initiated by the self-reaction of CIO, which is a key cycle for loss of polar ozone, b. The ozone abundance at 20 km altitude fell from 2.0 ppm on August 23 to 0.8 ppm on September 22. Assuming that the self-reaction of CIO is the rate-limiting step for ozone loss at 20 km, calculate the mixing ratio of CIO required to account for the observed loss of ozone. Assume T — 200 K and p = 60 mbar. Since the catalytic cycle involves a photolytic process, it occurs only during periods of daylight. Between August 23 and September 22, air in the polar vortex experiences roughly 8 h of sunlight per day.

5.6b The dominant reaction producing HC1 in the stratosphere is

Observations of CIO, HC1, and C10N02 in the midstratosphere could not, however, be reconciled with this reaction as the sole source of HC1 until it was discovered that HC1 is formed as a minor channel in the reaction of OH with CIO:

OH + CIO —► H02 + CI k = 7.4 x 10"12 exp(270/r) —► HC1 + 02 k = 3.2 x 10"13 exp(320/r)

Compare the strength of this HC1 source to that of C1 + CH4 at 20 and 35 km. Assume that [OH] = 1 x 106 and 1 x 107 molecules cm-3 at 20 and 35 km, respectively. The following conditions can be assumed:

20 km

35 km

[NO]

5 x 108

2 x 109 molecules cm-3

[o3]

3 x 1012

1 x 1012 molecules cm-3

J(h

5 x 10"4

1 x 10"3 s"1

^cu,

1.6

1.2 ppm

5.7 A The steady-state concentration of CIO in the lower stratosphere results from a balance between the CIO + N02 reaction and photolysis of C10N02. Given values of 7ciono2 at 27 and 37 km

./ciono2 = KrV at 27 km = 5 x 10~4 s_1 at 37 km evaluate the extent to which the profiles of CIO and C10N02 in Figure 5.14 agree with the steady-state relation at these two altitudes.

5.8a Calculate the number of collisions an N205 molecule makes with a stratospheric aerosol particle over a 12-h nighttime. Assume a temperature of 217 K and that the aérosol surface area per unit volume, is 0.8 x 10~8cm2cm~3.

5.9b In June 1991, Mount Pinatubo erupted, injecting 15 - 20 x 106 metric tons (t) of S02 into the stratosphere. The subsequent oxidation of S02 to sulfate aerosol dramatically increased the aerosol surface area in the stratosphere. Ground- and satellite-based instruments recorded a decrease in the stratospheric abundance of NO, and an increase in the abundance of HN03.

a. What is the cause of the decrease in NO*?

b. At 28 km, the nonvolcanic background stratospheric aerosol surface area is about 0.2 pm2cm-3. During the 6 months following the eruption, aerosol surface area was 5-10 times higher. 03 was observed to increase by 15% at this altitude. Why?

c. After the eruption, large negative changes in O3 were observed in both the lower stratosphere and in the total column. Explain.

5.10b In the stratosphere from 15 to 20 km, the cycle

is responsible for 30 to 50% of total odd oxygen loss (Cohen et al. 1994; Wennberg et al. 1994). The coupled H02/C10 and H02/Br0 catalytic cycles

where X = Cl(a) or Br(b), are responsible for about 15% of the odd-oxygen loss. The rate-determining reactions are 1 and 3a,b. Interconversion of OH and H02 also occurs by

The interconversion of OH and H02 occurs about 10 times faster (t « 10 s) than the net processes contributing to increase or decrease of HO* (x m 100 s). Show that the OH to H02 ratio owing to these reactions is given by

Using rate constants given in Appendix B, and neglecting reactions 3a and 3b, compute and plot [0H]/[H02] as a function of [03] at T — 215 K over the range of [03], (1-5) xlO12 molecules cm"3, and at [NO] =7.5 x 108 molecules cm-3.

5.11b Le Bras and Piatt (1995) have estimated the rate of tropospheric ozone depletion in the Arctic boundary layer as a result of CIO and BrO. Molecular bromine and chlorine would be photolyzed

followed by (rate constants at 250 K, mean temperature of the Arctic boundary layer in spring)

Br + 03 —> BrO + 02 k3 = 1 x 10 13 cm3 molecule"1 s"1 (reaction 3) CI + 03 —> CIO + 02 t4 = lx 10"11 cm3 molecule"1 s"1 (reaction 4)

Reconversion of CIO and BrO to halogen atoms occurs particularly through reaction with NO:

BrO + NO—>Br + N02 k5 = 2.5 x 10"11 cm3molecule"^"1 (reaction5) ClO + NO—>C1 + N02 k6 = 2 x 10-11cm3 molecule"1 s"1 (reaction6)

Ozone destruction would then be catalyzed by the following cycle:

BrO + BrO—>2Br + 02 k7 = 3.2 x 10"12 cm3 molecule"1 s"1 (reaction 7) Br + 03 —> BrO + 02 (reaction 3)

(The self-reaction of CIO leads to the C1202 dimer and not to the direct production of CI atoms.) A combined BrO-CIO cycle might also occur:

ClO+BrO—>Br+Cl + 02 £8a = 7x 10"12cm3molecule"1 s"1 (reaction8a) —>BrCl + 02 = 1.2 x 10~12cm3 molecule"1 s"1 (reaction8b) —>0C10+Br hc = 9x 10"12cm3 molecule-1 s-1 (reaction8c)

The products BrCl and OCIO are readily photolyzed

BrCl + hv —> Br + CI TBrCi^53s (00 = 70°) (reaction 9)

OCIO + hv —> O + CIO Tocio — 6s (0O = 70°) (reaction 10)

a. Show that the net reaction of the cycles involving reactions 8a and 8b is

b. Show that no net ozone loss occurs as a result of reaction 8c and thus that the total rate of loss of 03 by reaction 8 is

at c. Show that the steady-state ratio of [CIO] to [CI] is

d. Calculate the [C10]/[C1] ratio for [NO] = 0 and lOppt, [BrO] = lOppt, and [O3] = 40ppb. Using [CI] = 8 x 104 molecules cm-3, calculate the concentration of CIO corresponding to NO equal to 0 and lOppt.

e. Calculate the lifetime of an individual CIO radical against reaction with NO and BrO at 10 ppt of NO and 15 ppt of BrO.

f. Show that the rate of 03 loss from the combined BrO-CIO cycle with the CIO concentration range calculated in (d) and BrO equal to 15 ppt is (8 to 30) x 105 molecules cm~3s_1, or 0.1 to 0.34ppbh_1. (This rate would consume the initial 40ppb of 03 within 7-13 days.)

g. Show that the BrO cycle destroys 03 at a rate

5.12b The vertical S02 column in the plume from Mt. Pinatubo, integrated from the surface to the top of the stratosphere, was 3 x 1016 molecules cm~2. We wish to estimate the amount of aerosol surface area that this S02 would produce if converted to H2S04 aerosol. Assume complete conversion of this S02 to aqueous H2S04 particles of mean size 0.1pm diameter and 75% mass concentration H2S04/25% mass concentration H20. Calculate the resulting column-integrated aerosol surface area in (im2 cm 2. To convert this to an aerosol surface area concentration, assume that all the aerosol is confined to a uniform layer 5 km thick.

5.13c Annual mean global-scale mass exchange between the stratosphere and troposphere can be estimated from the budget of a long-lived trace species, such as CFC13, that is inert in the troposphere but photodissociates in the stratosphere (Holton et al. 1995). Air parcels passing upward through the tropical tropopause will have CFC13 mixing ratios characteristic of the well-mixed troposphere; air parcels passing downward from the midstratosphere to the lower stratosphere (see Figure 5.34) will have smaller mixing ratios as a result of photodissociation in the stratosphere. Let

Fc = net flux of CFC13 into the midstratosphere Fau = upward mass flux of air into the stratosphere in the tropics

and that, for the conditions above, is 1.2 x 106 molecules cm 3 s 1. h. Finally, show that the combined rate of 03 loss from both cycles is

F^ = downward mass flux of air out of the midstratosphere into the extratropical lower stratosphere = mean volume mixing ratio of CFC13 in the troposphere = mean volume mixing ratio of CFC13 transported downward across the control surface

Show that

where Mcech are the molecular weights of CFC13 and air. Conservation of mass requires that on average Fau = Faj. Observed values of cT and c,s and an estimate of Fau can be used to compute Fc. Applying the preceding equation at the lOOhPa level, using c,s ~ 0.6 cr, = 268 ppt, at this level and the estimate Fau = 2.7 x 1017 kgyr-1, compute Fc.

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