The rate of production of OH from 03 photolysis is Pqh = 2^4 [O (' D ) 1 [H20], which gives
Once an 0(JD) is formed, it is either quenched back to ground-state O or reacts with H20 to yield 2 OH radicals. The number of OH radicals produced per O('D) formed is the ratio of twice the rate of reaction 4 to the rate of reaction lb:
(k3[M)+k4[H20})[0(lD)} _ 2A:4[H20] ~ k3[M] + MHzO] 2£4[H2Q] ^ 2fe4^H2o h[M] k3
At 298 K, k4/k3 = 7.6. We can evaluate 6oh as a function of RH at the surface at 298 K:
At 80% RH, close to 40% of the O('D) formed leads to OH radicals.
The hydroxy] radical does not react with any of the major constituents oi the atmosphere, such as N2, 02, C02, or H20. yet it is the most important reactive spccics in the troposphere. Indeed, OH reacts with most trace species in the atmosphere, and its importance derives from both its high reactivity toward other molecules and its relatively high concentration. Were OH simply to react with other species and not in some manner be regenerated, its concentration would be far loo low, in spite of its high reactivity, to be an important player in tropospheric chemistry. The key is that, when reacting with atmospheric trace gases, OH is generated in catalytic cycles, leading to sustained concentrations on the order of 106 molecules cm-3 during daylight hours.
By using a chemical mechanism to simulate tropospheric chemistry, it is possible to estimate the atmospheric concentration of OH. Highest OH levels are predicted in the tropics, where high humidities and strong actinic fluxes lead to a high rate of OH production from 03 photolysis to O('D). In addition, OH levels are predicted to be about 20% higher in the Southern Hemisphere as a result of the large amounts of CO produced by human activities in the Northern Hemisphere that act to reduce OH through reaction with it. Hydroxy 1 radical levels are a factor of — 5 higher over the continents than over the oceans. Confirmation of the levels predicted by the chemical mechanisms is usually based on balancing budgets of species that are known to be consumed only by OH. For example, methyl chloroform (CH3CCI3) is removed from the atmosphere almost solely by reaction with OH, so the global average concentration of OH determines the mean residence time of CH3CCI3. From the history of methyl chloroform emissions, which arc entirely anthropogenic and are known to reasonable accuracy, and its present atmospheric level, it is possible to infer its residence time and then to compare the OH level corresponding to that residence time to the level predicted theoretically from tropospheric chemical mechanisms (Prinn et al. 1992). A global mean tropospheric concentration of OH of 1.0 x 106 molecules cm"3 is a good value to use for estimations.
Can Tropospheric Ozone Levels Be Explained by Downward Transport from the Stratosphere? Before we begin the study of tropospheric chemistry, we pose the question—can tropospheric 03 levels be explained solely on the basis of the O3 flux from the stratosphere to the troposphere? We can address this question indirectly through the OH radical (Jacob 1999). The principal source of the OH radical in the troposphere is 03 photolysis (Section 6.1). The principal sink of tropospheric OH is reaction with CO and CH4. Given estimated global budgets for CO and CH4, we can first determine the quantity of OH in the atmosphere needed to oxidize the emitted CO and ChU.
An estimate of the global sink of ChLj by OH reaction was given in Table 2.10 as 506 Tg CH4 yr '. The corresponding estimate for the global sink of CO, given in Table 2.14, is 1920Tg CO yr These translate to
CH4: 506Tg CH4 yr"1 |g3.2 x 10I3mol CH4 yr"1 CO: 1920Tg CO yr 1 g 6.9 x IOnmoI CO yr"1 Total OH needed - 1014mol OH yr"1
Can this amount of OH can be supplied by photolysis of the 03 that is transported down from the stratosphere? The global stratosphere to troposphere llux of 03 is estimated as 1-2 x 1013 mol 03 yr"1. Each 03 molecule can produce, at most, twoOH molecules, so the maximum OH production rate is 2 — 4 x 1013 mol OH yr~1. This OH production rate, solely from stratospheric 03, is anywhere from a factor of 2.5 to 5 too low to balance the global CH4 and CO budgets. We conclude on the basis of this simple calculation that there must be a major in situ source of 03 in the troposphere.
6.2 BASIC PHOTOCHEMICAL CYCLE OF N02, NO, AND 03
When NO and N02 are present in sunlight, ozone formation occurs as a result of the photolysis of N02 at wavelengths <424nm:
There are no significant sources of ozone in the atmosphere other than reaction 2. Once formed, 03 reacts with NO to regenerate N02:
Let us consider for a moment the dynamics of a system in which only these three reactions are taking place. Let us assume that known initial concentrations of NO and N02, [NOI,) and [N02]o. in air are placed in a reactor of constant volume at constant temperature and irradiated. The rate of change of the concentration of N02 after the irradiation begins is given by
Treating  as constant, there are four species in the system: N02, NO, O, and 03. We could write the dynamic equations for NO, O, and O3 just as we have done for N02. For example, the equation for [O] is
Since the oxygen atom is so reactive that it disappears by reaction 2 virtually as fast as it is formed by reaction 1, one can invoke the pseudo-steady-state approximation and thereby assume that the rate of formation is equal to the rate of disappearance:
jNCh (NO2] = ¿2 [O] [O2] [M] The steady-state oxygen atom concentration in this system is then given by rnl _jnoJNQ2]
Note that [01 is not constant; rather it varies with [N02] in such a way that at any instant a balance is achieved between its rate of production and loss. Thus, the oxygen atom concentration adjusts to changes in the N02 concentration many orders of magnitude faster than the N02 concentration changes, so that on a timescale of the N02 dynamics, [O] always appears to satisfy (6.5).
These three reactions will reach a point where N02 is destroyed and reformed so fast that a steady-state cycle is maintained. Let us compute the steady-state concentrations of NO, N02, and 03 achieved in this cycle. (The steady-state concentration of oxygen atoms is already given by (6.5).) The steady-state ozone concentration is given by
This expression, resulting from the steady-state analysis of reactions 1-3, has been named the photo stationary state relation. We note that the steady-state ozone concentration is proportional to the [N02]/[N0] ratio. We now need to compute [N02] and [NO]. These are obtained from conservation of nitrogen
and the stoichiometric reaction of 03 with NO:
Solving for , we obtain the relation for the ozone concentration formed at steady state by irradiating any mixture of NO, N02,03, and excess 02 (in which only reactions 1-3 are important):
We will see later that a typical value of j'no, A3 expressed in mixing ratio units is lOppb, so we can compute the ozone mixing ratio attained as a function of the initial mixing ratio of N02 with [O3]0 = [NO]0 = 0:
If, on ihe other hand, [NCb]^ = [OjJq = 0, (hen [O3] = 0. This is clear since in the absence of N02 there is no means to produce atomic oxygen and therefore ozone. Thus the maximum steady-state ozone concentration would be achieved with an initial charge of pure N02. The mixing ratios of ozone attained in urban and regional atmospheres are often greater than those in the sample calculation. Since most of the NO* emitted is in the form of NO and not N02, the concentration of ozone reached, if governed soleiy by reactions 1-3, cannot account for the actual observed concentrations. It must be concluded that reactions other than 1-3 are important in tropospheric air in which relatively high ozone concentrations occur.
Characteristic Time for the Photochemical NO* Cycle to Achieve Steady State The photochemical NO, cycle can be written concisely as the reversible reactions
where reaction 1 is the rate-determining step of the forward reaction. The rale equation for [O3I is ffl =jnoJN02]-*3[NO][Qj] The characteristic relaxation time to steady state is based on the reverse reaction;
At 298 K, ij = 1.9 x 1Q-Ucm3 molecule-1 s_l. Let us evaluate t at an NO mixing ratio of 1 ppb at 1 aim. Thus, [NO] = i(T9[M] = (I0"9) (2,5 x !019) = 2.5 x ¡O10 molecules cm-3, and
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