© r. r. dickerson 201111 lecture 3 aosc 637 atmospheric chemistry finlayson-pitts chapt. 1 & 14...

© R. R. Dickerson 2011 11

LECTURE 3AOSC 637

Atmospheric Chemistry

Finlayson-Pitts Chapt. 1 & 14Seinfeld Chapt. 1 & some of 16Wayne Chapt. 5 & 9

OutlineID. Biogeochemical Cycles and Atmospheric BudgetsExamples:

1. Water2. Carbon3. Oxygen

II. Thermodynamics• Enthalpies of formation and combustion


ID. Biogeochemical Cycles and Atmospheric Budgets Definitions: Biogeochemical Cycle -The process by which an element or compound passes through the atmosphere, biosphere, and geosphere (oceans and crust). Global Budget - The total atmospheric burden of a substance and the rates of its production and destruction, or its source and sink strengths. Steady State - The condition of constant concentration of a substance in the atmosphere. Steady state implies that the sources and sinks are equal; the lifetime, (also called residence time) is the burden divided by production (or destruction) rate.

= Burden/ProductionAfter time (1 - e-1) of the material has been exchanged. We will derive this later.


Example 1 - Water The atmosphere contains 0.48% water on average. Burden = grams water in the global atmosphere: 0.48 x 0.01 x 18/29 x (5.15x1021) = 1.5x1019 g H2O

(Note mass of atmosphere and surface area on table in syllabus.) or 1.5x1019/5.13x1018 = 3 g/cm2 UNITS: mole fraction x Mwts x mass atmos. = mass water Source - evaporation - rate not easily measured Sink - precipitation - estimates range from 95 to 110 cm/yr. We will use 100 cm/yr or 100 grams per cm2 year. The strength of this sink is uncertain because of limited observations over the oceans.Assume steady state, i.e., sources = sinks


Lifetime () = 3/100 years or 11 days

Does this tell us anything useful about water in the atmosphere?

1. << transport time - more H2O in areas of strong sources. The fraction of

water in the atmosphere varies from a few percent near the surface to < 10 ppm at the tropopause.

2. Global budgets are best for compounds whose lifetimes are much longer than the transport time of ca 1 year.

3. Absolutely absurd approach w.r.t. meteorology - tells us doodley squat about the probability of rain.

4. A wide variety of units is possible

5. The longer the lifetime the more stable concentration in time and space.

NOTE: In general, the shorter the the higher variation in time and space (Junge, Tellus, 1974).


EXAMPLE 2


How much C is there in the atmosphere? We will assume all the carbon is CO2, and that the

mean concentration, [CO2], is

380 ppm. Burden 380 x 10-6 x 1.8 x 1020 x 12 = 8.2x 1017 gC (as CO2)

(350 ppm gives 7.5) UNITS: [CO2] x moles air x

g/mole = gC Major source - Respiration Major sink - Photosynthesis


But if the biosphere is in S.S. then the net is zero. The biosphere is actually slight source of CO2 to the atmosphere because of forest

destruction. Story about Brazil with rainforest and U.N. speech.

Other source - fossil fuels, volcano, oceans

Other sink - oceans

Lifetime = burden / sources

= 7.6x1017 / {(1.5 + 0.05 + 0.05 + 0.0007) x 1017} 5 yr

Can man make much of a change in the burden?

Total reduced (fossil and living) carbon = 142 x 10 17 g

[CO2] = (142 /7.6) x 350 ppm = 6500 ppm!

Yes, we can make a big increase.


EXAMPLE 3 OXYGEN

Can fossil fuel burning affect atmospheric oxygen?


Where is the oxygen?

Burden: In the atmosphere: 32/29 x 0.21 x 5.15 x 1021 = 1.2x 1021 g O as O2

UNITS: mwt's [O2] x mass atmosphere = grams O2

In the oceans as H2O: (16/18) x (1.6x1024) =

1.4x1024 g O

UNITS: mwt's x mass seas x mass O in seas

The crust doesn't count, because the exchange is very slow but we can calculate the burden anyway.

3 x 106 x 17 x 103 x 5.1 x 1014 x .47 = 1.2 x 1025g O

UNITS: density x depth x area x O by wt x O in crust.

UNITS: g/m3 x m x m2 = g


Biomass: 0.015 x 1021 grams C Organic sediments: 45 x 1021 grams CTake all of organic C and make CO2 out of it you produce:

142x1017 x 32/12 = 380 x 1017 g O as CO2

UNITS: mass C X mwt's = mass O2 consumed.

380x1017 /1.2 x 1021 = 3.2% Problem for students: At what altitude is the oxygen partial pressure reduced by 3.2%? PO2 = 3.2%

ans. about 250 m


Thermodynamics


FORMATION AND COMBUSTION

(In search of the Criterion of Feasibility)

1. First Law of Thermodynamics (Joule 1843 - 48)

dE = đQ – đW

(In AOSC 620 E = U)

The energy of a system is equal to the sum of the heat and the work.

Explain eq. of state and exact differentials. 1. Define Enthalpy (H)

dH = dE + d(PV)

dH = đQ - đW + PdV + VdP

At constant pressure and if the only work is done against the atmosphere i.e. PdV work, then: đW = PdV

dHp = DQp

and đQ is now an exact differential, that is independent of path.

Note that you can do the same thing at constant volume except the result is:

dQv = dEv


For example, the burning of graphitic carbon might proceed through CO:

H

Cgraph. + O2 → CO2 -94.0 kcal/mole

Cgraph + 1/2 O2 → CO -26.4

CO + 1/2 O2 → CO2 -67.6

___________________________________

NET Cgraph + O2 → CO2 -94.0 kcal/mole

This is Hess' law. There is a table of Hfo in Finlayson-Pitts & Pitts, Appendix I, p. 943. The units of kcal are commonly used because Hfo is usually measured with Dewars and change in water temperature.


2. Heat capacity:

The amount of heat required to produce a one degree change in temp in a given substance.

C = đQ/dT

Cp = (∂Q/∂T)p = (∂H/∂T)p

Cv = (∂Q/∂T)v = (∂E/∂T)v

Because đQp = dH and đQv = dE

For an ideal gas PV = nRT

Cp = Cv + R

Where R = 2.0 cal/moleK

The heat capacity depends on degrees of freedom.

Translation = ½ R each; every gas has 3 translational degrees of freedom.

Rotation = ½ R

Vibration = R

For a gas with N atoms you see 3N total degrees of freedom and 3N - 3 internal (rot + vib) degrees of freedom.


Equipartition principle: As a gas on warming takes up energy in all its available degrees of freedom.

Measured Heat Capacities

Cv Cp

He 3.0 5.0

Ar 3.0 5.0

O2 5.0 7.0

N2 4.95 6.9

CO 5.0 6.9

CO2 6.9 9.0

SO2 7.3 9.3

H2O 6.0 8.0

Cv = R/2 x (T.D.F.) + R/2 x (R.D.F.) + R x (V.D.F.)

Cp = Cv + R

Translational degrees of freedom - always 3.

Internal degrees of freedom = 3N - 3

Where N is the number of atoms in the molecule


Check Cv(He): 3 x R/2 = 3.0 cal/(mole K)

Cv(O2): 3 x R/2 + 2(R/2) + 1(R) = (7/2) R = 7.0 cal/(mole K)?

What's wrong?

Not all energy levels are populated at 300 K

Not all the degrees of freedom are active (vibration)

O2 vibration occurs only with high energy; vacuum uv

radiation.

at 2000K Cv (O2) approx 7.0 cal/mole K

Students: show that on the primordial Earth the dry adiabatic lapse rate was about 12.6 K/km.


IIA. Enthalpy (Heat)

1. Formation

Definition: The enthalpy of formation. Hfo is the amount of heat

produced or required to form a substance from its elemental constituents.

The standard conditions, represented by a super “0”, are a little different from those for the Ideal Gas Law: 25oC (not 0oC), 1.0 atm. and the most stable form of elements. The standard heat of formation is zero for elements. This quantity is very useful for calculating the temperature dependence of equilibrium constants and maximum allowed rate constants. It was thought for a long time that H was the criterion of feasibility. Although H tends toward a minimum, it is not the criterion. Things usually tend toward minimum in H, but not always. Examples are the expansion of a gas into a vacuum, and the mixing of two fluids.


2. Enthalpy of Reactions

The heat of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.

Hrxn = Hfo

(products) - Hfo

(reactants)

The change of enthalpy of a reaction is fairly independent of temperature.

Problem for the student: Enthalpy calculation.

Which is hotter, an oxygen-acetylene flame or an oxygen-methane flame?

REACTIONS

C2H2 + 2.5O2 → 2CO2 + H2O

CH4 + 2O2 → CO2 + 2H2O

Note: melting point iron = 1535°C.


3. Bond Energies

See Appendix III of Pitts for a table of bond energies. The quantity is actually heat not energy. Don’t confuse with free energy to follow.

Definitions:

Bond Dissociation Energy - The amount of energy required to break a specific bond in a specific molecule.

Bond Energy - The average value for the amount of energy required to break a certain type of bond in a number of species.


Example: O-H in water

We want

H2O → 2H + O +221 kcal/mole

We add together the two steps:

H2O → OH + H +120

OH → O + H +101

---------------------------------

NET +221

Bond energy (enthalpy) for the O-H bond is 110.5 kcal/mole, but this is not the b.d.e. for either O-H bond.


Example: what is the C-H bond energy in methane?

We want Hfo for the reaction: CH4 → Cgas + 4H

any path will do (equation of state).

H (kcal/mole)

CH4 + 2 O2 → CO2 + 2H2O -193

CO2 → Cgraph + O2 +94

2H2O → 2 H2 + O2 +116

2H2 → 4H +208

Cgraph → Cgas +171

-------------------------------------------------

NET CH4 → Cgas + 4H + 396 kcal/mole

The bond energy for C-H in methane is: +396/4 = +99 kcal/mole

Bond energies are very useful for "new" compounds and substances for which b.d.e. can’t be directly measured such as radical.


Why does your car get worse gas mileage in winter? Why do Europeans like diesel powered cars?


What is the energy density of common fuels?

Consider the combustion of Consider the combustion of isooctaneisooctane (C₈H₁₈), an approximation of (C₈H₁₈), an approximation of gasolinegasoline. . Density = 0.67 g/cmDensity = 0.67 g/cm33. Note that the ratio of N₂ to O₂ is 3.76.. Note that the ratio of N₂ to O₂ is 3.76.

C₈H₁₈ + 12.5 O₂ (+ 47N₂) → 8CO₂ + 9H₂O (+ 47N₂)C₈H₁₈ + 12.5 O₂ (+ 47N₂) → 8CO₂ + 9H₂O (+ 47N₂)

If we assume complete combustion, then COIf we assume complete combustion, then CO₂ and H₂O will be the only products.₂ and H₂O will be the only products.

This is a lot of This is a lot of heatheat. The heat from 114 g of gasoline is sufficient to boil 15 L of . The heat from 114 g of gasoline is sufficient to boil 15 L of water if none of the heat escapes.water if none of the heat escapes.

kJ/mole) 5200(kcal/mole124351)(589948ΔH rxn0

reactants0

fproducts0

frxn0 ΔHΔHΔH

http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29#Miscellaneous_Compounds


Ethanol: CH3CH2OH

Hf0 = -234.8 kJ mole-1 Density = 0.79 g/cm3

What is the heat of combustion?

CH3CH2OH + 7/2O2 2CO2 + 3H2O

H = – 2*94 – 3*58 –(– 234.8/4.18) = -305.8 kcal/mole

Diesel fuel: The average chemical formula for common diesel fuel is C12H23, ranging approximately from C10H20 to C15H28. This is similar to Jet-A of JP5. Density = 0.82 g/cm3

Hf0 ≈ - 70 kcal mole-1

C12H23 + 23.5O2 12CO2 + 11.5H2O

H = – 2*94 – 3*58 –(– 70) ≈ –1730 kcal/mole

Note these calculations use gaseous water; some use liquid.


FuelApprox. formula

MolecularWeight(g/mole)

Heat of Combustion(kcal mole-1)

Density(g cm-3)

Energy (heat)

DensityKcal cm-3

Gasoline(iso-octane)

C8H18 114 -1243 0.67 7.31

EthanolC2H5OH 46 -306 0.79 5.26

Diesel and Jet fuel

Appx C12H23

167 -1730 0.82 8.49

Energy density of common fuelsH * / Mwt


The USEPA mandates more ethanol in gasoline in the winter, to reduce CO emissions, although the benefits are small. You get less energy per gallon in the winter.

Europeans like diesel fuel, even though it is more expensive to build diesel engines, because it has more energy per unit volume.


Biodiesel• What is Biodiesel?

– Vegetable Oil/ Animal Fat (triglyceride) derivative• Rapeseed (canola) Oil common feed source in Europe

• Soybean Oil common feed source in United States

• How is it produced?– Transesterification:

• 1 triglyceride substituted by 3 monohydric alcohol (methanol)

Environmental Problems see Crutzen et al. (2008) Atmospheric Chemistry and Physics.


http://chemistry.boisestate.edu/people/richardbanks/organic/nomenclature/organicnomenclature1.htm

Organic Nomenclature worth remembering:

Alkanes CmHn

Examples: Methane, ethane, propane, butane...

Alkenes RC=CR’

Examples: ethene or ethylene H2C=CH2, a plant hormone and isoprene C5H8 or (CH2=CH-C(CH3)=CH2, a diene.

Alkynes

Examples: Acetylene or ethyne HC≡CH

Aromatic Hydrocarbons

Based on the benzene ring

Example: Toluene

http://chemistry.boisestate.edu/people/richardbanks/organic/nomenclature/organicnomenclature1.htm

Organic Nomenclature worth remembering:

Alcohols R-CH2OH

Example ethanol CH2CH3OH

Ethers R-O-R’

Example diethyl ether CH3CH2OCH2CH3 (good night!)

Aldehydes R-CHOExamples formaldehyde H2CO; acetaldehyde CH3CHO

Ketones R-C(O)-R’Example Acetone

Carboxylic acids R-COOHExamples Formic acid acetic acid

Formate HCOO- acetate CH3COO-

More organic nomenclature worth remembering:

Organic Halides Example

called 2-chloropropane, or 2-propyl chloride, or isopropyl chloride

*Freons

Examples: CFC-11, Trichlorofluoromethane, CFCl3

CFC-12, Dichlorodifluoromethane, CF2Cl2

Halons (used in fire extinguishers)

Example: Halon 1011 (bromochloromethane, CH2BrCl)

HCFC’s (Freon substitutes)

Example: HCFC-123, CF3CHCl2

Sulfur CompoundsExamples: CH3–SH, methanethiol.

CH3–S–CH3, dimethyl sulfide (DMS).

* See Finlayson-Pitts Chapt 13 or Seinfeld Chapt 2.

More organic nomenclature worth remembering: N compounds. More details in the lecture on the biogeochemical cycles of Nr.

Amines, RNH2

Example CH3NH2, methylamine (a primary amine).

Nitriles, Nitro Compounds and Organic NitratesExamples acetonitrile

A nearly unique marker for biomass burning.

Alkyl nitrates, such as ethylnitrate Methyl nitrite CH3ONO.Nitryl Chloride ClNO2 (ClONO).All NOx reservoirs.

Amides, RC(O)NH2

Example ethanamide

Amino acids, H2N-CH(R)-COOH building blocks of proteins and enzymes.

Urea, (NH2)2CO

© r. r. dickerson 201111 lecture 3 aosc 637 atmospheric chemistry finlayson-pitts chapt. 1 & 14...

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