Chapter 4.  Tropospheric chemistry

 

Seinfeld and Pandis state: “The troposphere behaves as a chemical reservoir relatively distinct from the stratosphere.”  This is an understatement.  Tropospheric chemistry involves literally hundreds of chemicals and thousands of reactions.  Most chemicals are emitted from Earth’s surface and are subject to photolysis or chemical reactions on the way to the stratosphere.  In addition, air enters the stratosphere at the tropical tropopause, which has a temperature of 190-220 K – a cold trap.  Only chemicals that do not condense or are not water soluble are able to pass through this trap in substantial amounts.

 

While stratospheric and tropospheric chemistry have many chemicals and reactions in common, the troposphere also has hydrocarbons.  To study tropospheric chemistry, we will start with simple chemical systems and move on to more complicated ones.

 

1. Tropospheric gas-phase chemistry

 

1.2. NOx photochemistry

1.2.1 NOx photostationary state

 

At l<424 nm, nitrogen dioxide is photolyzed and the resulting O atom quickly becomes ozone (for the most part).

 

NO2 + hv ® NO + O

O + O2 + M ® O3 + M

 

These two reactions produce over 90% of the ozone in the troposphere.  The remaining 10% comes from the stratosphere.

 

What are the lifetimes of NO2 and O for typical daytime conditions?  JNO2 ~ 5x10-3 s-1 so that tNO2 ~ 200 seconds.  For typical conditions at Earth’s surface (p=1000 hPa, T = 300 K), the effective bimolecular rate coefficient is kO+O2+M =1.5x10-14 cm3 molecule-1 s-1 .  To compare the speed of this reaction to that of NO2 photolysis, we must multiply this bimolecular rate coefficient by the concentration of ozone, which is roughly 4.8x1018 molecules cm-3, to give a first-order rate of 7.2x104 s-1.   Note that since the second reaction is so much faster than the first, that the production of ozone is limited by the photolysis of NO2 in this couplet of reactions.

 

But                   NO + O3 ® NO2 + O2

 

The rate expression for NO2 is:

 

d[NO­2]/dt = - JNO2 [NO2] + kNO+O3[O3][NO]

 

and

 

d[O]/dt = JNO2[NO2] – kO+O2+M [O][O2][M]

 

Look at this second equation.  Suppose we were to perturb [O].  How long would it take to come back to steady state?  The answer can be found by looking at the expression:

 

τ = { JNO2[NO2]ss + kO+O2+M [O2][M]}-1

 

JNO2 ~ 0.005 s-1  ;  [O2] ~ 0.2 x 2.5x1019 cm-3  ; 

kO+O2+M[M] ~ 6x10-34 (300/T)2.3 [2.5x1019] = 1.5 x 10-14 cm3 molecule-1 s-1

 

or τ = {0.01 + 7.5x104}-1 = 1.3 x 10-5 seconds.

 

We see that O so quickly becomes O3 that O is in steady state for any change in the other species or photolysis.

 

Therefore,       JNO2[NO2] = kO+O2+M [O][O2][M] =>

 

[O]ss = JNO2[NO2]/ kO+O2+M[O2][M]

 

We can perform the same operation to see when NO2 is in steady state.  We know at a minimum that this will occur in less than 100 seconds.

 

Substituting for [O]ss, we get that

 

[O3]ss =  JNO2 [NO2]/ kNO+O3[NO]

 

This is a famous photostationary state relation that was first derived by Leighton about 40 years ago.

 

Let’s get NO and NO2 in terms of just NO2. 

 

By knowing that           [NO] + [NO2] = [NO]o + [NO2]o

 

and                              [O3]o – [O3] = [NO]o – [NO]

 

and assuming that [NO]o = [O3]o = 0, we get the expression:

 

 

 

The reaction rate for NO+O3®NO2+O2 is about 2x10-14 cm3 molecule-1 s-1. 

Since 1 ppbv = 2.5x1010 molecules cm-3, we can rewrite this reaction rate for the surface as kNO+O3[O3] = 5x10-4 s-1.  Peak JNO2 ~ 0.006 – 0.008 s-1 at the surface, so

 

                        JNO2 / kNO+O3 ~ 10 – 15 in ppbv

 

 

Using 10 for this ratio,

 

[O3]ss = ½ {[100 + 40[NO2]o]1/2 - 10}

 

For [NO]o = 100 ppbv, [O3]ss ~ 27 ppbv

 

What does [NO] equal in steady-state?

 

We will revisit NOx – O3 chemistry after we have included some hydrocarbons and CO.

 

Important note:  Notice that the only source of oxygen atoms in this chemical system is from either ozone or NO2.  The extra oxygen atom is actually cycled between O3 and NO2.  If all the initial NOx were in the form of NO2, then in the light, the sum of NO2 and O3 remains constant.  New ozone is not actually created by NOx photochemistry alone.

           

Let’s look at some real data from a real city, Nashville, TN.

 

 

 

Does this picture look like NOx photostationary state?  If it did, then the sum of ozone and NO2 should remain constant.  What is going on here?                       

 

Oxidants other than ozone.

 

Tropospheric chemistry consists of much more than just NOx – O3 photochemistry.  Ozone is not the only tropospheric oxidant.  In fact, the most important oxidants are free radicals (odd number of electrons).  We continue the discussion of  tropospheric chemistry with the most important oxidants: OH, O3, HO2, NO3, and the halogens Cl, BrO, and I.

 

Of these OH is the most important because it is everywhere and it reacts fairly fast with many of the emissions.  I do not intend to cover the hydrocarbon reaction sequences in great detail.  Rather, I want you to get a sense of how the sequences go and which types of hydrocarbons do what.

 

1. Hydroxyl radical, OH.

 

If we go to remote enough parts of the world tropospheric chemistry simplifies.

 

Assume that we have this limited chemical set.  We know that the O3 will be photolyzed at wavelengths below about 320 nm.

 

O3 + hv ® O2 + O(1D)

 

However, O(1D) is an excited state molecule that can be quenched by collisions with N2 and O2.  This is mostly what happens, resulting in the reformation of O(3P) and thus O3.  That is:

 

O(­1D) + N2, O2 ® O(3P) + N2, O2

 

However, some collisions are with H2O, which has a mixing ratio of 0.0001 to 0.05 in the troposphere.  In this case, the reaction is exothermic, unlike the collisions with N2 and O2, and a reaction occurs:

 

O(1D) + H2O ® 2 OH

 

The greater the amount of water vapor, the faster OH production is.  This is the major source of OH in the atmosphere.

 

kN2 = 2.6x10-11 ; kO2 = 4x10-11, kH2O = 2.2x10-10.  So, if water vapor is at 1%, what is the fraction of reactions that go to OH?

 

This reaction is by far the greatest source of OH in the atmosphere.   It is, however, not the only one.  In fact, under conditions of low water vapor and ozone, such as in the upper troposphere, other sources can dominate.   In some sense, they result from species that are intermediates of previous oxidation processes.  Some examples are:

 

                                                                                    HONO + hv ® OH + NO

                                                                                    H2O2 + hv ® 2 OH

 

Nitrous acid, HONO, comes from the gas-phase reaction: OH + NO + M ® HONO + M.  Another source is the reaction of NO2 and H2O on surfaces to form HONO. 

 

Hydrogen peroxide, HOOH or H2O2, is formed in the gas-phase only by the reaction: HO2 + HO2 ® HOOH + O2.

 

For environments with high levels of ozone and alkenes, such as some cities, another important OH source is the reaction of ozone and alkenes to form a Criegee Intermediate, which can decompose into OH.  The reaction rate coefficients are not very fast and the OH yield is less than 1, but this mechanism is interesting because it can make OH during both the day and the night.

 

These are primary sources of OH.

 

At the same time, we have in the HOx family the reaction of HO2 + NO ® OH + NO2, where kHO2+NO ~ 8x10-12 cm-3 molecule-1 s-1.  HO2 is the peroxy radical, a free radical with an odd number of electrons. We might consider this reaction to be a secondary source of OH, since, HO2 is quickly in photochemical steady state with OH.  Usually, this secondary source is ~10 times larger than the primary sources.

 

OH is lost by chemical reactions with many chemicals and with surfaces, as we will see later.

 

The resulting daytime OH concentrations are ~(1-10) x 106 molecules cm-3. Nighttime concentrations should be very low,

 

2. Ozone, O3

 

90% of ozone comes from the reactions:

 

                                                                        NO2 + hv (l<420 nm) ® NO + O(3P)

 

                                                                        O(3P) + O2 + M ® O3 + M.

 

kO+O2+M[M] ~ 6x10-34 (300/T)2.3 [2.5x1019] = 1.5 x 10-14 cm3 molecule-1 s-1

 

Since O2 is about 0.2 x 2.5x1019, the lifetime of O is 1.3x10-5 s. 

 

Nothing else can compete effectively with this reaction in the lower troposphere.  The stratosphere is a different matter.  So, when we get an O atom, we have ozone.

 

Ozone is chemically destroyed and is also lost on surfaces.

 

Typical ozone mixing ratios are ~0-20 ppbv in very remote environments, such as the upper tropical troposphere; 30-50 ppbv in rural areas that are not heavily impacted to large sources; 30-150 ppbv for rural areas, like State College, that are impacted by upwind sources and for many US cities; and 30-300 ppbv for more polluted cities in the developing world. 

 

An aside:  The air quality index in the newspaper is simply the ozone mixing ratio is ppbv.

 

3.  Nitrogen trioxide or nitrate radical, NO3

 

The source of NO3 is the reaction:      NO2 + O3 ® NO3 + O2. 

An immediate sink is:                          NO3 + NO ® 2NO2

and:                                                     NO2 + NO3 + M ↔ N2O5 + M

if N2O5 is lost by reactions on surfaces.

 

This reaction is not very fast.  In addition, during the day, the lifetime of NO3 is very short (seconds).  So, it is only an issue at night and only when O3 and NOx are present.  Recent studies have shown that if there is enough O3 and NO2, such as during evening rush hour in cities, it is possible for measurable NO3 to exist during twilight hours.

 

NO3 is lost by chemical reaction, photolysis, and surface deposition.

 

NO3 mixing ratios at night are typically a few pptv to a few tens of pptv.

 

4. HO2

 

The main production of HO2 is the reactions:

 

                                                                        H + O2 + M ® HO2 + M

                                                                        HCO + O2 ® HO2 + CO

                                                                        RCH2O + O2 ® RCHO + HO2

 

A large source of H and HCO is HCHO, which is photolyzed to produce HO2:

 

                                                                        HCHO + hv ® H + HCO

                                                                        H+O2+M ® HO2 + M

                                                                        HCO + O2 ® HO2 + CO

 

This is an important HO2 (and thus HOx) source in urban environments, where there is plenty of HCHO from oxidation of hydrocarbons, and in the upper troposphere, where water vapor is only a few hundred ppm and formaldehyde is generated by both local methane oxidation and by convection.

 

The typical daytime HO2 mixing ratios are a few pptv to 100 pptv.  The nighttime mixing ratios are typically ~0 to a few pptv, with much higher values possible for certain environments.

 

5.  Cl, Br, I. 

 

The halogens, Cl, Br, and I are important mainly near their source: the ocean, which contains lots of salt.  Seaspray releases the NaCl, which can react to release the Cl.  It is not clear how important halogens are in tropospheric chemistry, except in the springtime Arctic, where Br reactions rapidly deplete the ozone.  But, we will pay some attention to these halogens, not as reactants for hydrocarbons, but as catalytic destroyers of ozone.

B. Lifetimes of typical organics in the troposphere.

 

Usually, Br, Cl, and I have mixing ratios of 0.01 to a few pptv, although they are hard to measure at the low levels and there is some controversy as to their actual mixing ratios.

 

Chemistry of the background atmosphere.

 

NOx – HOx – O3 – CO Chemistry

 

If we go to remote enough parts of the world tropospheric chemistry simplifies.  However, we can still have NOx from distant pollution or from lightning.  CO has a lifetime of a month or so, and thus reaches even the most remote parts of the troposphere. 

 

Assume that we have this limited chemical set.  We know that the O3 will be photolyzed.

 

                                                                        O3 + hv ® O2 + O(1D)

                                                                        O(1D) + H2O ® 2 OH

 

The greater the amount of water vapor, the faster OH production is.

 

                                                                        CO + OH ® CO2 + H

                                                                        H + O2 + M ® HO2 + M, which is very fast                          

                                                                        CO + OH ® CO2 + HO2

                                                                        HO2 + NO ® NO2 + OH.

 

And the net result is: 

                                                                        CO + 2O2 + hv ® CO2 + O3

 

Termination reaction:

                                                                        OH + NO2 + M ® HNO3 + M

 

We solve the rate equations for O, O(1D), OH, HO2, and O3, but not for NO, NO2, and CO.  why?  Because NO2, NO, and CO are emitted from the surface.  O3 and the others respond to these emissions in the presence of sunlight. 

 

 

O3: on production through NO2 photolysis, and on loss by cycling through NO and on loss through photolysis of O3.  This last term is often small compared to the NO loss, which gives us the photostationary state relationship.

 

OH is the “chain carrier” of this reaction mechanism.  Since OH is required to recreate the HO2, which can then react with NO to form NO2, we want to know how many times OH will cycle before it is lost to the formation of HNO3.  This number of times is called

the chain length, Lc.

 

                                                                        Lc = kOH+CO[CO] / kOH+NO2+M[M][NO2]

                       

The fast cycle between NO and NO2 determines the O3 balance; the slow cycle through OH and HO2 increases the amount of O3.  Because this process requires sunlight, it is called photooxidation.

 

We see that ozone is produced if NO2 is formed by the reaction: HO2 + NO ® OH + NO2,

since

                                                                        NO2 + hv ® NO + O

                                                                        O + O2 + M ® O3 + M

 

and these last two reactions occur within a few minutes during sunlight. 

 

So, let’s look at this system in a qualitative way.  The NOx photostationary state can be represented by the expression:

 

                                                                                hv

NO2 ↔ NO + O3

 

 

With CO and HOx present, this equation gets modified:

                                                                                                                                                           

                                                                                hv

NO2 ↔ NO + O3

 

                                                                             HO2 

 

 

 


The reaction between HO2 and NO provides another pathway for NO to get back to NO2.  At the same time, HO2 is interacting in a cycle of its own:

 

                                                                                hv

NO2 ↔ NO + O3

 

                                                     CO + OH           HO2 

 


                                                                      NO

 

 

 


So, we have the relatively fast cycle between NO and NO2 and the somewhat slower cycle between OH and HO2.  Note that the reaction with NO and HO2 effectively creates more NO2 and thus more O3.  We will discuss this in more detail when we take up methane oxidation soon.

 

 

If there is no NO around, what happens?  HO2 reacts not with NO but with O3:

 

                                                                        HO2 + O3 ® OH + 2O2

 

In this case, ozone is not produced; it is destroyed.  For, say, 40 ppbv of ozone, what level of NO is required for ozone to be produced and not destroyed?

                                                                       

 

 

 

 

We can look at the cycling of HOx in a quantitative way, as in the figure on the Reaction cycle of OH and HO2.  This region of Germany is rural, but not remote.  Note that NO is actually fairly high, similar to values found in Rock Springs, PA.  Note also that the cycling between OH and HO2 is about 5 times greater than the primary production of OH by photolysis of ozone and reaction of O(1D) with water vapor.

 

 

 

 

We can present a similar view for the reactive nitrogen chemistry in the next figure.  Note that the sources are generally through NO, with rapid conversion to NO2 and eventual removal of nitric acid (HNO3).

 

 

 

 

 

 

Finally, we can look at the diurnal variation of the chemistry in the fairly remote troposphere in the next figure.  Note the shifts in the reactive nitrogen chemistry in particular.

 

 

 

 

4.2 Variation of ozone production with [NO]. 

 

HOx, the sum of OH+HO2, has the rate equation:

 

d[HOx]/dt = P(HOx) – {2kHO2+HO2[HO2][HO2] + 2kOH+HO2[OH][HO2] + kOH+NO2+M[M][NO2][OH]}

 

where P(HOx) is the production rate (molecules cm-3 s-1) of HOx, OH or HO2 and the three expressions in brackets are three loss mechanisms for HOx.  HO2+HO2®HOOH+O2 dominates when NO is low, OH+HO2®H2O+O2 dominates when NO is about 100 pptv, and OH+NO2+M® HNO3+M dominates when NO is greater than a few hundred ppt.  In each one of these NO regimes, we can assume that only the dominant HOx loss reaction is occurring.

 

Assume that [HOx] approximately equals [HO2], that OH, HO2, and thus HOx are in steady-state, that the relationship that you derived for [HO2]/[OH] in problem 4.1, section e applies, and that P(HOx) is 107 molecules cm-3 s-1, CO is 100 ppbv, O3 is 40 ppbv initially, p = 1000 hPa, T=298 K, and RO2 is negligible.  

Using the equations in problem 4.1, section d, determine the analytical expressions for [OH], [HO2], and P(O3) in terms of  P(HOx), [NO], [CO], [O3], and rate coefficients for the three NO regimes.  For this problem, P(O3) = kNO+HO2 [NO] [HO2].

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Atmospheric chemistry of formaldehyde and NOx.

 

Formaldehyde, CH2O, is an important tropospheric molecule.  It is a common reaction product of hydrocarbon degradation and an important source of OH, especially in urban areas.  However, recent measurements have shown larger-than-expected levels of formaldehyde over the remote oceans and even in the uppor troposphere.  This wide dispersal of formaldehyde is strange, since its lifetime by photolysis is typically hours to a day. 

 

The destruction of HCHO is by photolysis and reaction with OH:

 

HCHO + hv ® H + HCO    (45%)

                    ®  H2 + CO     (55%)

 

HCHO + OH ® HCO + H2O

 

HCO + O2 ® HO2 + CO

 

The theoretical maximum amount of O3 that can be produced is equal to the sum of HCHO and NO2. 

 

Methane oxidation.

 

In the very cleanest parts of the troposphere, which essentially do not exist any more, the chemistry comes down to methane oxidation chemistry.  The reason is that methane reacts rather slowly with OH, its emissions into the atmosphere are large, and therefore it is everywhere.

 

OH and HO2 are deeply involved in almost all tropospheric chemistry.  The major OH source is the photolysis of ozone followed by the reaction with water vapor.

 

Let’s look at the oxidation of the simplest (and perhaps) most abundant atmospheric hydrocarbon (alkane): methane (CH4).  The methane oxidation process gives us a model of how other hydrocarbons are oxidized.

CH4 + OH ® CH3 + H2O                    initiation          (forms a methyl, or alkyl radical)

CH3 + O2 + M ® CH3O2 + M                                     (forms a methyl peroxy, or alkyl peroxy radical)

 

CH3O2 is analogous to HO2, only the O2 is joined to the carbon.

 

If NO is present,

CH3O2 + NO ® NO2 + CH3O                                     (forms a methoxy, or alkoxy radical)_

CH3O + O2 ® CH2O + HO2                                        (forms a carbonyl, or formaldehyde)

HO2 + NO ® OH + NO2

CH2O + OH ® HCO + H2O

CH2O + hv ® H + HCO

HCO + O2 ® HO2 + CO

 

If NO is not present, then we get the reaction sequence:

 

                                    CH3O2 + CH3O2 ® CH3OOCH3 + O2

                                    CH3O2 + HO2 ® CH3OOH + O2

 

Several termination steps are possible:

 

CH3O2 + NO2 + M ó CH3OONO2 + M

OH + HO2 ® H2O + O2

CH3O2 + HO2 ® CH3OOH + O2

OH + NO2 + M ® HNO3 + M

HO2 + HO2 ® H2O2 + O2

 

These last three steps are not termination steps unless the peroxides or acid are scavenged by cloud drops or rain.  The most important of these is the formation of the peroxides.  Otherwise, more steps can occur.

 

The net result is that:

 

CH4 + 4O2 + 2hv ® HCHO + 2O3 + H2O

 

The final products depend on the relative destruction pathways for HCHO.

 

Once NO is converted to NO2, it is most likely that O3 will be formed during daylight, even in the presence of the reaction of OH + NO2.  Thus, we can write ozone production rate as:

 

PO3 = {kHO2+NO [HO2] + kCH3O2+NO [CH3O2]} [NO]

 

Further, we can re-derive the steady-state expression for the NOx photostationary state:

 

[NO2]/[NO] = {kNO+O3 [O3] + kHO2+NO [HO2] + kCH3O2+NO [CH3O2]}/ JNO2

 

Instruments to measure NO and NO2 are part of the standard package of pollution measurements in every country.  It has often been thought that it would be possible to determine the ozone production by looking at the imbalance in the NOx photostationary state caused by kHO2+NO [HO2] + kCH3O2+NO [CH3O2].  Despite several attempts to make this calculation work, it generally doesn’t.

 

 

 

 

4.4 Atmospheric organic chemistry

 

As we move into continental environments, the number of emitted volatile organic compounds increases. 

 

The different classes of hydrocarbons react at different rates with the different oxidants.  We can go into the this in more detail in the homework, but for now, I will just state the reactions in generalities.

 

Alkanes ( all single bonds between carbons): OH, to a less extent NO3 and in the MBL, Cl  (ALK1 and ALK2)

Alkenes (at least one double bond between carbons): OH, O3, NO3, and some Cl in MBL (OLE1, OLE2, OLE3)

Alkyne (at least one triple bond between carbons): OH, Cl (MBL)

Aromatics (ring structure): OH, Cl (MBL)  (ARO1, ARO2)

Aldehydes (a double bond between a carbon and an oxygen): OH, NO3¸ HO2, Cl (MBL) (HCHO, CCHO, RCHO)

4.5 VOCs and NOx in ozone formation

 

4.5.1 General oxidation mechanism

 

Start with emissions of NO and VOCs;

In the production of peroxy radicals, an oxygen molecule is added to H or R (R= CH3, C2H5, etc…).

The reaction of RO2 with NO creates NO2.

The product RO reacts with O2, a hydrogen is extracted to form HO2, and a new R’O, usually a carbonyl, is formed.

The degration products including O3, HNO3, PAN, HCHO, and others chemicals that are more water soluble.

 

The possible pathways in the oxidation of VOCs is presented in the following figure from Seinfeld and Pandis.

 

 

Ozone formation is strongly correlated with higher temperatures for several reasons:

1.  more emissions at higher temperatures (isoprene is highly T dependent)

2.  more sunlight due to subsidence and few clouds;

3.  suppressed vertical mixing increases surface O3;

4.  some chemical reaction rate coefficients increase (small effect), but equilibria of species like PAN shift towards NO2 and the radicals.

 

4.5.2 EKMA Diagram (ozone isopleth diagram).

 

It is difficult to devise a good regulatory strategy to control ozone, which is a secondary pollutant, unless the fundamental processes and emissions are understood.

We know that NOx and VOC emissions are both involved in ozone production, so we might want to base regulatory policy on these two.  But how?

 

One way is to look at the sensitivity of ozone production for different initial mixing ratios of NOx and VOCs.  By looking at the ozone produced by a computer model that is run repeatedly for different mixing ratios of NOx and VOCs, an isopleth diagram can be generated with VOCs on the x-axis and NOx on the y-axis.  This diagram, first developed by Dodge and called the Empirical Kinetic Model Approach diagram, or EKMA diagram, was used to establish regulatory policy for the EPA.

 

Usually, the plot is not of ozone production, but of the maximum amount of ozone produced in the model after a certain period of time.  Can we derived a universal EKMA diagram?  No.  The total amount of ozone that is produced is dependent on many factors: the duration allowed for ozone production, the time of year, which affects both temperature and photolysis, the specific VOCs that are present, the terrain, which affects the ozone surface deposition, and the size and location of the box for which the model is run. 

 

Conditions: 

Is chosen for a particular region, or box, which can be moving.

The time of the run is chosen to capture the maximum ozone production.

The entire plot is generated by changing the VOC/NOx mixture while leaving the meteorological conditions constant.

 

A generic EKMA diagram is presented below.

 

 

The hydroxyl radical is the key to ozone formation.  Both VOC’s and NO2 compete for OH.  For instance if we take a standard urban VOC mix, then the rate coefficient for OH loss to VOCs is about 5.5 times slower, on a per carbon atom basis, than that to NO2.  Thus, the two rates are the same when VOC/NO2 = 5.5.

 

VOC/NO2 >5.5:  OH reacts mainly with VOCs, potentially producing more radicals, potentially increasing O3 in the presence of NO.

 

VOC/NO2<5.5:  More OH reacts with NO2 than with VOCs, thus reducing the ozone formation efficiency.

 

Regions on the plot below the ridgeline are “NOx-limited”; on this diagram, that means that regions with VOC/NOx < 5.5 is NOx-limited.

Region above the ridgeline are “VOC-limited”; on this diagram, that means that regions with VOC/NOx > 5.5 is VOC-limited.

 

When VOCs >> NOx, NOx is so low that RO2 and HO2 radicals are not propagated efficiently.  These radicals react with each other and peroxides are formed.  These peroxides are water soluble, can be taken up in cloud drops and rained out.  They can also stick on the surface by dry deposition.

 

Above the ridgeline, HNO3 is formed in greater quantities, thus inhibiting the production of O3 because OH, and thus HOx, as well as NO2, and thus NOx, is consumed in this terminal reaction.    Also, with radicals consumed, this is less ozone production and thus less OH production from ozone.  The just above the ridgeline represents the maximum in OH propagation.

 

 

Decreasing VOCs  for a fixed NOx always decreases O3.  Decreasing NOx at a fixed VOC does not always decrease O3.  If the level of VOCs increases for a fixed value of NOx, more radicals will be generated that can react with the NO and more ozone will be formed.  For a fixed amount of VOCs, however, ozone is produced more efficiently at low NO, but HO2+RO2 is competitive with HO2+ NO, so that the ozone production per HOx molecule is smaller.  As NO increases and becomes the primary reactant with HO2 and RO2, ozone production is at its peak.  At even higher NOx, NO2, which is approximately in steady-state with NO, reacts with OH, inhibiting the initiation of RO2 and HO2 formation and removing NOx.  Ozone production is thus less efficient again. 

 

We can see this behavior in the figures below.

 

 

 

The ozone production efficiency (OPE), which is the amount of ozone produced per molecule of NOx consumed, is related to the chain length, is the number of times the propagation steps occur divided by the rate of the termination reaction.  In rural regions and cities with high VOC to NOx ratios, such as Houston, TX, the OPE can be greater than 10, while during rush hour in a city like New York, it can be 1-4. 

 

OH reactivity. 

 

Because different VOCs react at different rates with OH, the importance of different VOCs in ozone production is proportional not to the concentration of the VOC, but to the product of the VOC concentration and its reaction rate coefficient with OH.  We can put all chemical species on the same basis by normalizing to a particular reaction rate coefficient.  Propene (C3H6) is chosen.

 

Prop-Equiv(j) = [Aj] (kOH+Aj / kOH+C3H6 ),  where [Aj] is in terms ppbC.

 

The following figure shows propene equivalents for different environments.

 

 

Incremental reactivity.

 

The O3 isopleth plot shows that the response to additional hydrocarbons can be highly non-linear, depending on the initial conditions.  This has brought about the concept of incremental reactivity. 

 

IR = limD[VOC]®0 D[O3] / D[VOC]

 

 

Two steps:

1.  How much RO2 is generated from the initial OH + VOC reaction (kinetic reactivity)

2.  How fast NO®NO2, OH is regenerated, products formed (mechanistic reactivity).

 

4.6 Regulatory strategies

 

The EPA website contains a tremendous amount of information about the history and strategies of the Clean Air Act.  Please peruse their site, as I am not going to cover the history in class.  However, I will go briefly through a summary technical document from their website: the EPA Draft Report on the Environment 2003.  It contains information on the National Ambient Air Quality Standards (NAAQS) required by the Clean Air Act, the trends in the these pollutants over the last 20 years, and maps of areas that are in non-compliance with the Clean Air Act.

 

The primary regulatory strategy for ozone has been to reduce VOCs.  The reason is that it is clear that reduction of NOx in the urban cores for a fixed value of VOCs will result in an increase in ozone.  I should point out that outside of urban cores, conditions quickly transition from being VOC-limited to NOx-limited.  In the past, the main efforts were to reduce the very high ozone mixing ratios in the urban cores.  This has been accomplished to a great degree.  However, at the same time, urban sprawl has spread a lower level of ozone problem over large metropolitan areas.  It is clear that a strategy that involves the reduction of both VOCs and NOx will be required to reduce ozone further to acceptable levels.

 

We can see the effects of various control strategies using a model to produce ozone isopleth diagrams for different regions within the Los Angeles basin. 

 

 

 

4.7 Atmospheric chemistry (gas-phase) of sulfur compounds.

 

Sulfur oxides.

 

The dominant gas-phase reaction is: 

 

OH + SO2 + M ® HOSO2 + M

 

HOSO2 + O2 ® HO2 + SO3

 

SO3 + H2O + M ® H2SO4 + M

                       

                        net:      OH + SO2 + H2O ® HO2 + H2SO4

 

The rate-limiting step is the reaction of OH with SO2.  The termolecular reaction rate coefficient in bimolecular form is ~9x10-13 cm3 molecule-1 s-1 for [M] = 2.5 x 1019 cm-3 and T = 300 K.  For [OH] = 4x106 cm-3, the lifetime of SO2 due to gas-phase oxidation is: τSO2 ~ 2.8 x 105 s, or about 6 days.  On the other hand, its lifetime by dry deposition in a 1 km deep layer is about 1 day, with vd ~ 1 cm s-1.

 

H­2SO4 is rapidly taken up on aerosols.  When incorporated into cloud drops and then precipitation, it becomes acid rain.  This is what we will talk about next.