RUSSELL R. DICKERSON
1. HW#1 due Thrs
2. HW#2 data in excel
3. ISSUES with Chameides
Is CH4 oxidation the major source of CO?
Do steady state calculation
Way too much NO and NOx for a global average; 3 ppb this is a suburban valve
RO2 + NO -> NO2 + RO is important.
Wayne Chapt. 5,9
ID. BIOGEOCHEMICAL CYCLES AND ATMOSPHERIC BUDGETS
A. ENTHALPY FORMATION AND COMBUSTION
ID. Biogeochemical Cycles and Atmospheric Budgets
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, τ (also called residence time) is the burden divided by production (or destruction) rate.
τ = BURDEN/PROD.
After time τ (1 - e-1) of the material has been exchanged. We will derive this later.
EXAMPLE 1 - WATER
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)
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 93 to 120 cm/yr. We will use 100 cm/yr or 100 grams per cm 2 year. The strength of this sink is uncertain because of limited observations over the oceans.
Assume steady state, i.e., sources = sinks
Lifetime (τ) = 3/100 years or 11 days
Does this tell us anything useful about water in the atmosphere?
1. τ << 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 τ the higher variation in time and space (Junge, Tellus, 1974).
EXAMPLE 2 - CARBON
How much C is there in the atmosphere? We will assume all the carbon is CO2, and that the mean concentration, [CO2], is 350 ppm.
350 x 10-6 x 1.8 x 1020 x 12 = 7.6x 1017 gC (as CO2)
UNITS: [CO2] x moles air x g/mole = gC
· Major source - Respiration
· Major sink - Photosynthesis
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
· Other source - fossil fuels, volcano, oceans
· Other sink - oceans
Lifetime = burden / sources
τ = 7.6x1017
Can man make much of a change in the burden?
Total reduced (fossil and living) carbon = 142 x 10 17 g
[CO2] = 142 over 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?
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) = 1.4x10 24 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 1025 g 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 C
Take all of organic C and make CO2 out of it you produce:
142x1017 x 32 over 12 = 380 x 1017 g O as CO2
UNITS: mass C X mwt's = mass O2 consumed.
Problem for students: At what altitude is the oxygen partial pressure reduced by 3.2%?
DPO2 = 3.2%
ans. about 250 m
A. ENTHALPY OF FORMATION AND COMBUSTION
(In search of the Criterion of Feasibility)
1. First Law of Thermodynamics (Joule 1843 - 48)
dE = DQ – DW
(In METO 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 = DQ - DW + PdV + VdP
At constant pressure and if the only work is done against the atmosphere i.e. PdV work, then
DW = PdV
dHp = DQp
and DQ is now an exact differential, that is independent of path.
For example, the burning of graphitic carbon might proceed through CO:
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 DHfo in Pitts & Pitts,
Appendix I, p. 1031. The units of kcal are commonly used because DHfo is usually measured with Dewars and change in water temperature.
Note that you can do the same thing at constant volume except the result is:
DQv = dEv
2. Heat capacity:
The amount of heat required to produce a one degree change in temp in a given substance.
C = DQ/dT
Cp = (∂Q/∂T)p = (∂H/∂T)p
Cv = (∂Q/∂T)v = (∂E/∂T)v
Because DQp = dH and DQv = 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 = 1/2 R each
(every gas has 3 translational degrees of freedom)
Rotation = 1/2 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
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)?
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)
Definition: The enthalpy of formation. DHfo is the amount of heat produced or required to form a substance from its elemental constituents.
The standard conditions, represented by a super "o", 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 DH was the criterion of feasibility. Although DH tends toward a minimum, it is not the criterion. Things usually tend toward minimum in DH, 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.
DHrxn = S DHfo(products) - S DHfo(reactants)
The change of enthalpy of a reaction is fairly independent of temperature.
EXAMPLE: ENTHALPY CALCULATION
Which is hotter, an oxygen-acetylene flame or an oxygen-methane flame?
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.
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
H2O → 2H + O +221 kcal/mole
We add together the two steps:
H2O → OH + H +120
OH → O + H +101
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: C-H bond in methane
We want DHfo for the reaction:
CH4 → Cgas + 4H
any path will do (equation of state.)
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.