Meteo 465 – The Middle Atmosphere

Mid-term Open-book Exam

 

 

Assigned:        10:00 AM, 07 March 2003

Due:                10:00 AM, 24 March 2003

 

 

You should be able to answer all the questions using the material in the chapter “Stratospheric Chemistry  - Perspectives in Environmental Chemistry”, your notes, and a reference text for things like atomic masses, air densities, etc..  You may consult texts, websites, or other documents.  However, please do not consult each other. 

 

1.     (15 points)  I posed the following question in class; now I will pose it again on this exam.  We know that heterogeneous chemistry at low temperatures causes rapid ozone loss at wintertime high latitudes, yet air gets as cold in the stratosphere just above the tropical tropopause as it does in the wintertime polar lower stratosphere.  In a few hundred words, or less, present your case for why you do or don’t think that ozone loss in the tropical lower stratosphere is important. 

 

The tropical tropopause does get as cold as 190 K, especially over Micronesia in January.  In addition, there is evidence that very tenuous Tropical Stratospheric Clouds do form.  However, several factors work against massive stratospheric ozone loss in this region.

  1. Lower stratospheric air has just entered from the tropical troposphere.  Since the stratospheric lifetime of the CFCs is months to years, very little inorganic chlorine (HCl, ClONO2, ClO) is available for conversion to reactive forms.  The same is true for bromine.
  2. The prime region for O3 production from O2 photolysis is at higher altitudes.  Very little ozone is produced in the cold tropical tropopause region.
  3. The stratospheric transit time is typically 1-4 years for any air parcel.  Even if O3 were destroyed in the tropical lower stratosphere, the O3 chemical lifetime is much shorter than the characteristic transit time along a path through the stratosphere.  Thus, because chemistry is faster than transport somewhere along the path, any decrease in the small amount of O3 produced in the tropical lower stratosphere will be compensated for production later on.

 

2.     (20 points)  It is now clear from the figure on atmospheric heating vs. altitude that ozone absorption is mainly responsible for the solar heating near 50 km, much to even my surprise in class. 

a.  Using the equation on solar heating by ozone and molecular oxygen in the web-based notes on Stratospheric Heating and Cooling, demonstrate that this heating is indeed due to ozone and that it makes sense that that its maximum heating occurs near 50 km. 

b.  Show that O3 is more important at 50 km than O2 by estimating the relative heating rates of O3 and O2.

To demonstrate the ozone heating, we must put numbers in the equation:

 

dT/dt = (r cp)-1 òl I {sO2(l) [O2]  + sO3(l) [O3] }dl

 

Both the absorption cross sections and I have wavelength dependences.  In addition, I, [O3], [O2], and r have altitude dependences.  We can use Figure 1.8 from the Chemistry chapter to get I at different wavelengths and altitudes, and use JPL to get the cross sections for O3 and O2.  We must estimate some values, but these estimates turn out to be unimportant.  Thus, the integral becomes sums over wavelengths, where I used 20 nm bins:

 

dT/dt = (r(z) cp)-1 S[ I(z,l) {sO2(l) [O2](z)  + sO3(l) [O3](z) }]Dl

 

 

This crude model worked amazingly well.  Ozone is clearly responsible for the heating at 50 km, while O2 plays a rather minor role there.

3.     (25 points)  The fast photochemistry of OH and HO2.  Equations & figures are in Stratospheric Chemistry – Perspectives in Environmental Chemistry.

(a.) Write down the rate equations for OH and HO2 using equations 1-22, 1-26 through 1-31, 1-34 (and assuming that the production of HOx (OH+HO2) is due to: O3 + hn ® O(1D) + O2, followed by O(1D) + H2O ® 2OH.  Ignore reactions 1-23 and 1-25.  Assume that HOCl rapidly photolyzes.  Remember that reaction 1-31 is much faster than the others.

(b.) Assume that HOx production equals HOx loss by reaction 1-34.  Show that the reactions that exchange HOx between OH and HO2 are fast compared to HOx production and loss using Figure 11.

(c.) Derive the steady-state expression, equation 1-32.

(d.) Simplify the expression to contain only the dominant terms at 45 km and at 15 km.

(e.) There are three ozone-destroying catalytic cycles involving HOx that are given on pages 23 and 24.  Write down the ozone loss rate expression (d[O3]/dt = ….) for these HOx catalytic cycles.

(a.)

 

 

(b.) Since HOx production and loss are equal, we can use the OH+NO2 curve in Figure 11 to evaluate the rates of these processes.  We see that they vary from ~2x103 cm-3 s-1 at 15 km to ~104 cm-3 s-1 at 30 km to less than 103 cm-3 s-1 at 45 km.  With the exception of HO2+O below 25 km, all the other reactions are at least ~5 times greater that OH+NO2 for the entire altitude range.  We can therefore safely ignore primary production and loss in examining the balance between OH and HO2.

 

(c.) We assume that the OH+CO is not important except at very low and very high altitudes.  To derive the steady-state expression 1-32, assume that d[HO2]/dt = 0. We could to the same thing for OH.  By simple algebra,

 

 

(d.) 45 km:  From Figure 11, HO2+O3, HO2+NO. and HO2+ClO << HO2+O.  Thus,

 

15 km:  From Figure 11, HO2+O, and HO2+ClO << HO2+O3 ~ HO2+NO.  Also, OH+O must be much less than OH+O3.  Thus,

 

 

(e.) Recall that HO2+NO®NO2 + OH actually ends up creating O3 because NO2 photolyzes to make O, which makes O3.  Thus, some of the OH reactions balance this reaction in the cycle, which is a null cycle for O3 loss.  So, we must consider only the HO2 reactions that cause O3 loss, realizing that some of the OH reactions balance this:

 

 

We have assumed that HO2+ClO is slower than HOCl+hn.

 

 

4.     (20 points)  Ozone loss in the polar stratosphere.

(a.) On page 27, it states “In the lower stratosphere, where [M] = 2x1018 cm-3, cClO = 1 ppbv, cBrO = 7 pptv, the loss rate of ozone can approach 1 to 3 percent per day in the sunlit parts of the vortex”.  Verify this statement with calculations.

(b.) As ClO gets tied up in ClONO2 (due to photolysis of HNO3, followed by ClO+NO2+M®ClONO2+M), the ozone destruction by ClO+BrO becomes more important.  Assuming the cBrO = 7 pptv and that f (photolysis) = 0.8, at what value of cClO are the two ozone destruction terms in equation 1-62 equal?

(c.) Polar ozone loss requires sunlight to break apart Cl2O2 and species containing bromine.  At the same time, if any HNO3 remains in the gas-phase, it too is photolyzed, although at a much lower rate than Cl2O2.  Explain why the photolysis of HNO3 is still effective at slowing polar ozone loss, even though the photolysis frequency is so small compared to the photolysis of Cl2O2.  Use numbers to support your thoughts.

(d.) Why is Figure 21 not sufficient to prove that the Antarctic Ozone Hole is caused by chlorine?  Think about what constitutes proof that a process is occurring?

 

(a.)  The ozone loss is given by the expression:

 

 

where fphotlysis is the fraction of ClOOCl that photolyzes and does not thermally decompose, while fBrO is the fraction of the reaction that does not form OClO, which photolyzes to give O back.

 

Assume T = 200 K.  Thus, for [M]=2x1018 cm3 molecule-1 s-1, kClO+ClO = 1.3x10-13 cm3 molecule-1 s-1, and kClO+BrO  = 2.9x10-12exp(220/200) + 5.8x10-13exp(170/200) = 1.0x10-11 cm3 molecule-1 s-1.  We find the concentrations by multiplying the mixing ratios by [M].

so,

 

where we assume 12 hours of sunlight so that ClOOCl can be photolyzed.

 

At these altitudes, O3 is a few ppm (See Figure 21 in the chapter).  Thus, we see that ozone is lost at a few percent per day.

 

(b.) To find out, set the two terms equal to each other:

 

 

(c.) The photolysis of ClOOCl is quite fast, about 10-3 s-1, as I said in my e-mail.  In Figure 12, we see that the photolysis of HNO3 is 3x103/1010 = 3x10-7 s-1. (We divide the rate by the concentration of HNO3 to get the photolysis frequency.)  This number is 3300 time bigger.  However, the amount of HNO3 is usually as much as 10 times more than that of ClO.  Every molecule of HNO3 results in the formation of NO2, which immediately reacts with ClO to form ClONO2.  Thus for each ClO molecule, it goes through its ozone-destroying cycle 330 times before being tied up with NO2 to form ClONO2.  Thus, a few times 103 ClO molecules are tied up into ClONO2 each second, or about 108 cm3 (0.05 ppbv) per day.  In ~20 days, all the ClO will be effectively tied up.

 

(d.) Figure 21 provides a snap-shot of the ozone and ClO inside and outside the polar vortex, but some other phenomenon could have been responsible for elevating ClO and depressing O3.  By taking several snap-shots over the ozone hole’s evolution, we can calculate the observed O3 loss rate and compare it with the calculated O3 loss rate.  They agree to about a factor of two, which gives us confidence that we understand the major loss processes that are responsible for the Antarctic Ozone Hole.

 

5.  (20 points) The End-of-Mission statement from the 1994 Airborne Southern Hemisphere Ozone Experiment / Measurements for Assessing the Stratospheric Effects of Aircraft  (ASHOE/MAESA) campaign using the NASA ER-2 high altitude aircraft contains the following statement:

“Polar air nearly devoid of ozone was sampled in-situ for the first time inside the vortex near the 400 K potential temperature surface (about 16 km altitude) on 10 and 13 October.  Mid-October is late in the period of severe ozone depletion over Antarctica.  This air contained little O3 (<0.4 ppmv), little or no ClO (<100 pptv), high NO (1 ppbv), low NOy (2 ppbv), and HCl equivalent to estimates of total inorganic chlorine (not in the form of CFCs or CH3Cl) (2.6 ppbv).”

 

Given that O3 is low, show with words and chemical equations why each other chemical species are observed to have the values that they do. 

Recall that most inorganic chlorine is in the forms of Cl, ClO, and ClOOCl as ozone is being destroyed.  What is the time constant for Cl, ClO, and ClOOCl to be converted into HCl when O3 drops to 0.4 ppmv? (hint: Write down the formation rate for HCl and assume that all inorganic chlorine is initially in the forms of Cl and ClO and that Cl and ClO are in steady-state.)

 

When most of the O3 is destroyed, there is a shift in which reactions are most important.  First, NO increases relative to NO2 because the reaction NO+O3®NO2+O2 slows down while the reaction NO2+hn®NO+O remains about the same.  Because ClO+NO®Cl+NO2 is now faster, ClO preferentially reacts with NO and not ClO.  Because ClO+NO®Cl+NO2 is now faster and Cl+O3®ClO+O2 is now slower, the balance of reactive chlorine shifts toward Cl.  Hence, the rate of HCl formation by the reaction Cl+ CH4®HCl+CH3 increases, and reactive chlorine is rapidly shifted into HCl.  This increased formation of HCl thus rapidly decreases the amount of reactive chlorine and thus the chlorine ozone-destroying catalytic cycles.  It also explains the observations.

 

The reaction that determined that lifetime of reactive chlorine in this situation is:

Cl + CH4 ® HCl + CH3.

 

Thus, Cl first-order loss rate is (kCl+CH4[CH4]).  From Figure 6.04b in the NASA GSFC Electronic Textbook, we see that CH4 is about 1 ppm in March at Northern high latitudes.

This is equivalent to about 2x1012 molecules cm-3.  From JPL, at lets say 220 K, the reaction rate coefficient for Cl+CH4 is kCl+CH4 = 1.1x10-11exp(-1400/220) = 2x10-14 cm3 molecule s-1. 

 

We are not interested in the Cl lifetime alone, but rather, the ClOx lifetime.  We can find this by considering the relationship between Cl and ClOx.  Consider the reactions that are dominant in this environment:

ClO + NO® Cl+NO2

Cl+ O3 ® ClO+O2

Cl + CH4 ® HCl + CH3

 

In steady-state,

 

We see that the lifetime of ClOx is quite short once ozone has been lost.  Only heterogeneous reactions on PSCs can reactivate the chlorine, and that is even limited by the lack of ClONO2.  Thus, the system simple stops.