LECTURE 1
AOSC/CHEM 637
ATMOSPHERIC CHEMISTRY
Russell R. Dickerson

 	  OUTLINE
 	  I. INTRODUCTION (W. Ch. 1; F-P. & P. Ch. 1)
 	  A. Basic Chemistry and Physics

PERSPECTIVE
Imagine you are a visitor from a planet around the star Vega ("Vagarians" are an unusual species, filled with an insatiable scientific curiosity, but pernicious vacillators).  You have come to visit the sun because it is one of the few stars in the galaxy that supports planets.  True to form, you can not make up your mind which planet to explore.  Faced with an important decision, the logical next step is to gather scientific data.  Let's look at the most prominent features of first four planets around the sun.  We would like to know something about the planets before we land.  The temperature of a planet can be measured remotely by looking at the emission spectrum.  The wavelength of maximum emission (max) gives the "ideal black body" temperature: 

(max)= a/T

where T is the temperature in K and the constant a has the value 2.900 x 10-9 mK.  This is called Wein's Displacement law.

	The second through fourth planets are all about the same size, and all about the same temperature, but the contents of their atmospheres are dramatically different.  Venus and Mars are covered by CO2, but the earth is covered by a mixture of nitrogen and oxygen.  Why?
	Oxygen and nitrogen can not co-exist in thermodynamic equilibrium.  There must be some force on the third planet from the sun that uses energy, presumably solar energy, to maintain the content of the atmosphere in spite of the forces of inorganic chemistry and Gibb's free energy.  It would otherwise look exactly like Venus or Mars: full of CO2.  That life must exits on such a planet is immediately obvious to a Vagarian, who would single out this planet for further study.  We, however, may not be so perceptive and will be doing a homework problem to prove what would happen to our atmosphere were it allowed to reach thermodynamic equilibrium.
	The key to the chemistry of the atmosphere of the earth is life.  The biosphere controls to a large extent the make-up of the atmosphere.  The atmosphere may be in fact an extension of the biosphere - like the fur on a dog (or better said at this University, the shell of a terrapin), not really a living part of the animal, but a protective cover produced by the organism to protect it from the elements.  One of the most important elements that the atmosphere protects us from is ultraviolet radiation.  In fact, until there was oxygen and thus ozone in the atmosphere there was no life on the surface of the earth.  For a look at this idea (metaphor really) of the atmosphere as an extension of the biosphere read: Gaia: A New Look at Life on Earth, or even better Revenge of Gaia, by J. E. Lovelock.

WHY STUDY THE CHEMISTRY OF THE ATMOSPHERE?

 	AIR POLLUTION
 		health
 		ecology
 		materials
 		visibility

 	REMOTE ATMOSPHERE        
 		climate - "greenhouse" effect, cloud formation
 		ozone - UV
 		nutrient cycling, N, S, trace metals

                          WHAT`S IN THE ATMOSPHERE?
    
                    (viewgraph)

 	GAS PHASE
 		atoms - molecules - ions - radicals
 		most of atmos. unreactive
 		Nobel gases esp. dull
 		composition const. to 80 km
 		here concerned with trop. and strat.
 		main concern 0.0002 % trace gases         

 	RADICALS:
 		OH  not OH-  (hydroxyl not hydroxide)
 		HO2
 		NO3 not NO3-                
 		etc.
 		low conc. high reactivity
 		major sources and sinks

 		[H2O] = 0 to 3%

UNITS OF CONCENTRATION - GAS PHASE

  MOLE FRACTION
		    X, %, [ ], ppm, ppb, ppt
    examples
      same as volume fraction for ideal gases
      independent of temperature and pressure!

   MASS PER UNIT VOLUME
      best for particles - dust
      filter paper and flow meter - g/m3

   PARTIAL PRESSURE  
      atm = 1013 bar = 1.013 x106 dynes/cm2 = 760 torr = 14.5 psi
      at one atm total, partial press (in atm) same as mixing ratio

   NUMBER DENSITY
     at STP there are 2.7 x1019 molecules/cm3; often written just cm-3

                           UNITS OF CONCENTRATION - AQUEOUS PHASE

    MOLES PER LITER
        M = Molar
        Other units forbidden!

      HENRY'S LAW
   Combines two systems, i.e. gaseous and liquid.
                                    (viewgraph)

                 SUMMARY
     MIXING RATIOS: ppm, ppb, ppt
     WIEN'S DISPLACEMENT LAW
     IDEAL GAS LAW
     HENRY'S LAW
     ACID BASE CHEMISTRY
     EQUILIBRIUM CONSTANTS

 				Outer		Radius		Mass 		Atmos.
 	# PLANET		Temp.		(km)		(g x10-26)
 	 -------------------------------------------------------------------------------------------------
 		
 	1 MERCURY		440 K		2500	  	3.9 		NONE
 		
 	2 VENUS		244 K		6200	 	55.0		CO2
 		
 	3 EARTH		253 K		6370	 	60.0		N2/O2
 		
 	4  MARS		216 K		3400	  	7.3		CO2

UNITS OF CONCENTRATION

	 MIXING RATIOS EXAMPLE
	____________________________________________________
 
	MOLE FRACTIONS*		1/5 = [O2]
 
 	PERCENT 				1% = [Ar]

 	PARTS PER MILLION		1.7 ppm = [CH4]
 	PARTS PER BILLION		120 ppb = [O3]
 	(ambient air quality standard - 
 	maximum allowable 1 hr conc.) 

 	PARTS PER TRILLION 		100 ppt = [CCl2F2]
    _______________________________________________________

 *For an ideal gas volume fraction is the same as mole fraction.

	MASS PER UNIT VOLUME EXAMPLES	_____________________________________________________________

 	MICROGRAMS PER CUBIC METER 
	
Total suspended particulate mater
 	National ambient air quality standard -
 		(maximum allowable annual mean)  		75 g/m3

	Fine (respirable) particulate matter
	NAAQS annual mean					15 g/m3
		  Daily maximum				65 g/m3

 	NANOGRAMS PER CUBIC METER		30 ng/m3 = C 
  		(graphitic C content of clean Arctic air)

UNIT CONVERSION

   EXAMPLE 1

     Express 1.0 g/ m3  SO2 in units of ppb.  Assume STP.

      [SO2] = 1 x10-6 g SO2 in 1.0 m3;  Mole wt. SO2 = 64 g/mole.

{1x10-6/64 } / {1000/22.4}  =  3.5 x 10-10  =  0.35  ppb

 {g / g/mole } / { L/L/mole }  =  { moles } / { moles }  =  mixing  ratio    

EXAMPLE 2 (Seinfeld's technique)

 	Ideal gas law:       PV = nRT

where	P = press. in Mb
   	T = temp. in K
  	V = vol. in m3
   	R = 8.314 x10-2  Mb m3 /K 
 	M = mol. wt. of gas of interest in g/mole

      for the concentration of species C:
   [C] (ppm) =  {RT} / {PM}  x  [C]  (g/m3)
                                                         
     for 1.0  g/m3 SO2 at  STP:

[SO2] =  {8.314 x10-2 *273} / {1013*64} * 1.0  =  3.5 x10-4  ppm

    or:  3.5 x 10-4  * 103  ppb/ppm  =  0.35  ppb.

AQUEOUS PHASE CONCENTRATION:

UNITS: moles per liter or molar (M)

Take one mole of salt (58.5g NaCl) and pour it into an empty 1.00-l flask.  Fill with distilled water and presto a 1.00-M soln of NaCl.

This is also a soln 1.00 M in Na+ and 1.00 M in Cl-.

For sulfuric acid take 98 g H2SO4 and repeat above procedure.

You now have a 1.0-M solution of sulfate (SO42- ) and a 2.0-M solution of H+

H2SO4  =  2H+ +  SO42-

NORMAL (N):  1.0 M H2SO4 is 2.0 N in H+ because sulfuric acid is diprotic.

HENRYS LAW

     The mass of a gas that dissolves in a given amount of liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. This law does not apply to gases that react with the liquid or ionize in the liquid.

 	Gas				Henry's Law Constants
 					    (M /atm at 298 K)     ____________________________________________________________

 	OXYGEN				1.3 x10-2
 	(O2)

 	OZONE 				1.3 x10-3
 	(O3) 

 	NITROGEN DIOXIDE		1.0 x10-2
 	(NO2) 

 	CARBON DIOXIDE			3.4 x10-2
 	(CO2)

 	SULFUR DIOXIDE			1.3
 	(SO2)

 	NITRIC ACID (effective) 		2.1 x10+5
 	(HNO3) 

 	HYDROGEN PEROXIDE 		7.1 x10+4
 	(H2O2)
    
 	ALKYL NITRATES 			2
 	(R-ONO2)

See http://dionysos.mpch-mainz.mpg.de/~sander/res/henry.html for a complete table of Henry’s Law Coefficients.  The temperature dependence of Henry’s Law coefficients is usually represented with van’t Hoff’s Equation where H is the enthalpy of dissolution in kcal mole-1 or kJ mole-1.  See Seinfeld page 342.

(∂ lnKH/∂T)p = ∆H/(RT2)

KH = KHo exp [-∆H/R(To-1 - T-1)]

Post Script: Henry’s Law Example

     What would be the pH of pure rainwater in Washington, D.C. today?  Assume that the atmosphere contains only N2, O2, and CO2 and that rain in equilibrium with CO2.
                                          
     Remember
H2O =  H+ + OH-

[H+][OH-]   =  1x10-14

pH  =  -log[H+]

In pure water pH = 7.0.

[CO2] =  350 ppm

Today’s barometric pressure is 993 Mb = 993/1013 atm = .98 atm

Partial pressure, PCO2, = 350 x10-6  * .98 = 3.47 x10-4 atm 

[CO2]aq  = PCO2  =  3.4 x10-2 x  3.47  x10-4 =   1.18 x10-5 M

In water CO2 reacts slightly:

CO2 + H2O = H2CO3

H2CO3 = H+ + HCO3-

{[H+][ HCO3-]} / [ H2CO3]  = Keq = 4.3 x10-7

We know that [H2CO3] = 1.18 x10-5 M and [H+]  [HCO3-]

Thus
[H+]  = {(Keq)[H2CO3]} = 2.25 x10-6

 pH  = 5.6

Second Example

	Can alkyl nitrates, R-ONO2, be removed from the atmosphere by rain (wet deposition)?  Consider the relative amount of an alkyl nitrate in the gas phase vs. the aqueous phase in a cloud.  If most of the alkyl nitrate is in the aqueous phase, than precipitation must be important.  We need the following information:

•	Henry's Law Coef. (KH) for R-ONO2  2 M/atm at 298 K (Luke et al., 1989).

•	A thick cloud has 1.0 g liquid water per cubic meter.

•	The typical temperature of a cloud is near 0 C.

•	The typical altitude of a cloud is about 5 km thus the pressure is about 0.5 atm.

•	The most alkyl nitrate one might find in the atmosphere over a continent is about 1.0 ppb.

First we apply Henry's law to find out what the aqueous concentration of R-ONO2 would be is the cloud is in equilibrium with the vapor phase.

               [R-ONO2 ]aq = KH x [R-ONO2 ]gas x Ptotal

Where [R-ONO2 ]gas must be expressed in partial pressure, atm.

  = 2.0 x 10-9 x Ptotal

  = 2.0 x10-9 x 0.5

[R-ONO2 ]aq  = 10-9 M

How do we compare this to the gas phase concentration?  Change both values into moles/m3.

[R-ONO2 ]aq = 10-9 x 10-3 = 10-12

UNITS:  moles/L(water) x L(water)/m3 (air) = moles /m3

 [R-ONO2 ]gas = 10-9 x 103 x 0.5/22.4 = 2.2 x 10-8

UNITS:  L (R-ONO2 )/ L (air) x L/m3 x atm/(L atm/mole) = moles/m3

We see that the vapor phase concentration is 22,000 higher than the aqueous phase concentration. Rainout will still matter, however, if  R-ONO2 reacts in solution and thus is removed.  This is the case for SO2 in water containing H2O2 where H2SO4 is produced, but aqueous reactions of R-ONO2 with species commonly found in rainwater are as yet unknown.  This implies that alkyl nitrates may have a residence time long enough to be important in regional or global atmospheric chemistry.
	For species X, a general solution to the "rain out" question is given by an expression for the ratio of moles of gas-phase X to moles of aqueous-phase X in a given volume of air.

Xaq /Xgas = KH x LWC x 2.24 x 10-5

Where KH is the Henry's Law coefficient in M/atm, and LWC is the liquid water content in g/m3.  This equation is valid at 273 K; to correct for temperature multiply the right side by (T/273).  The equation above shows that the ratio is independent of pressure and concentration.  For alkyl nitrates this ratio is about 4.4 x10-5.  For a species with a Henry's law coefficient of 4 x102, about 1% will go into a cloud with a LWC of 1g/m3.