HW #2 Solutions

5.1

To get answer, just plug+chug:

R1 = 610-9s-1[O3]

R2 = 7.710-8 s-1[O3]

Interpretation: Look at k1 and k2. We’re dealing with the same collision cross-section here, for the most part. However, k2 is 105 times less efficient than k1, on a collision-by-collision basis, implying only the most energetic of collisions will cause (2) to happen. This is balanced by the fact that there are about 106 more O around than O(1D). So R2 proceeds faster than R1.

5.2

The solution to this problem is basically Eq. (5.8) in S+P. Read the discussion from pp. 144-146 to understand exactly what the problem is asking. You can also read the following discussion below. If you use S+P Eq. 5.8, be aware that, in my opinion, they made a mistake by a factor of 2. Because we didn’t derive (5.8) in class, you can read my interpretation of its derivation below.

There is a big distinction to make between the lifetime of O3 and that of Ox. O3 is continually created and destroyed by photolysis and recombination with O2. This yields a relatively short lifetime = 1/jO3. If this was what the problem wanted, you’d be done. But the problem asks for the lifetime of Ox.Since neither reaction in the null cycle is a sink of Ox, jO3 doesn’t play directly into the sink rate. Theparenthetical statement “… Ox (i.e. O3)” is confusing here, and I’m sorry if it led you astray. The reason this statement is in there is becauseOx= O + O3is dominated by the more abundant O3, so a loss of an Ox is basically the loss of an O3. (see discussion at beginning of p. 146 in S+P). But in terms of lifetime, the distinction between whether we’re talking about O3 the molecule or Ox the family member, is incredibly important (See Fig 5.2). Furthermore, the problem doesn’t tell you which sinksof Ox to consider, which will depend on the concentrations of NOx, Clx, HOx, etc. The reference to Figure 5.2, which basically restricts itself to the Chapman Cycle, suggests we can ignore the other Ox loss reactionswhen determining the lifetime for this problem.

I will now restate the problem as: Consider the lifetime of Ox in the Chapman cycle at 20 and 45 km given the values of jO3 and [O3] listed above. Hint: use the known facts that we can apply PSSA to [O], and that [O] < [O3].

The only sink for Ox in the Chapman mechanism is from reaction (4), which kills 2 Ox per reaction.

O + O3 2O2

So we have

where we have used the approximation that [Ox] = [O] + [O3] [O3]. Likewise, we use Eq. (5.7) due the standard PSSA for [O]. Note that I include a factor of 2 here to account for the loss of the O, whereas S+P eq. 5.8 does not. I will argue that both losses are equally relevant for the long-term lifetime of O3.

This yields,

So now we just plug things in, using Table 5.1 for [M](z) and T(z) to get Ox = 10 years at 20 km, and 7.5 hours at 45 km. O3 in the lower stratosphere has such a long lifetime, the main factor controlling its abundance there is transport; O3 in the upper stratosphere is locally created + destroyed, and thus the equilibrium abundance can be calculated from the rate equations without considering transport..

Problem: Using collision theory, estimate the efficiency of reaction (4) – in what fraction of cases does a collision between O and O3 lead to destruction of the species? Assume T = 300K, and reasonable values for atomic radii from the CHF3 problem last week.

O3 + O2O2 k4 = 8.0x10-12exp(-2060/T) (re. 4)

We’ll assume everything is the same except the reduced mass that determines the mean relative molecular speed. The reduced mass of the above reaction is 12, whereas the reduced mass of problem 3.4 was 13.7. We end up with a 7% increase in the collision frequency.

RC = (3.8 x 10-10 cm3 molecule-1 s-1)[O][O3]

which compares to the reaction rate of 8.3x10-15[O][O3]. So you can see that this is also a very improbable reaction per collision (1 reaction per 50,000 collisions), due in large part to the high energy barrier evident in the Arrhenius term.

5.4

See S+P pp. 162-164 to understand the context for this problem. Since we didn’t discuss this in class, the answer was supposed to be something you could derive yourself. The solutions don’t really rely on the text.

We consider the case where the destruction rates of Ox through the reactions

ClO + O  Cl + O2; k1 = 3.0 x 10-11 exp(70K/T) cm3 s-1

O3 + O  2O2; k2 = 8.0x10-12exp(-2060/T) cm3 s-1

are equal. Setting the respective rate equations to be equal yields,

[Note: In this case, I interpret both equations as destroying two odd-oxygens, since the O atom in ClO was originally obtained from an O3 through the reaction

Cl + O3 ClO + O2; k1 = 2.9 x 10-11 exp(-260/T) cm3 s-1

So I don’t have the same problems with double-counting reaction (2) as I had in problem 5.2. For this purpose, I think of ClO being the “cousin who didn’t marry well” relative of the odd-oxygen family, even though it’s not strictly part of Ox.]

At this point, we just plug and chug. For T = 220 K, we have [ClO]/[O3] = 1.7 x 10-5. If O3 mixing ratio were 4ppm, then ClO’s would be 6.7x 10-5 ppm = 0.067 ppb = 67 ppt. Figure 5.14 shows that ClO is negligible at 20 km. Instead, most of the Cl is wrapped up reservoir species.

5.6

Compare the reaction rates of the following two HCl production pathways at 20 and 35 km.

Cl + CH4 HCl + CH3 ; k1 = 9.6 x 10-12 exp(-1360/T)

OH + ClO  HCl + O2; k2 = 3.2 x 10-13 exp(320/T)

R1 = k1[Cl][CH4]

R2 = k2[OH][ClO]

We will compare by taking the ratio of these two rates.

Likewise, we can use Eq. (5.30) for the equilibrium between Cl and ClO

Finally, we use PSSA on [O] to yield a messy equation for R1/R2

At this point, we have everything we need to plug + chug. At 20 km, R1 = 13R2. At 35 km, R1 = 1.8 R2. Thus Reaction 2 becomes a significant source of HCl in the upper stratosphere.

5.8

Z = vN2O5Rp2NA

where MN2O5 = 0.108 kg/mole

Total surface area = 4Rp2NA . The answer is 1.8 collisions per 12-hours.

5.10

This problem was more intimidating to look at than to do. It’s basically just a PSSA problem. Themain task is to find the PSSA equilibrium for [OH]/[HO2] considering only the reactions that involve their cycling. These are reactions (1), (2), and (6). [(3a,b) and (4a,b), when combined, also yield a cycle in HOx, but we are later told to ignore these, so I will from the start. This could have been a nasty problem if you also had to compute [XO]/[X] to determine the reaction rate for 3a,b for a given reservoir of X.] If we limit ourselves to these reactions only, we have

d[OH]/dt = - d[HO2]/dt = -R1 + R2 + R6

where these reactions do not consider those that destroy or create HOx. In PSSA, we assume the changes in OH are slow relative to the cycling, and this yields

Below, find the plot requested in the problem. Note that the “recharging” of NO into NO2 is important both for the cycling of HOx but also for the eventual creation of more Ox. It is a competing pathway that recycles both Ox and HOx.