1. Environmental Chemistry
Chapter 0:
Chemical Principals - Review
Copyright © 2007 by DBS

2. Chemical Principals – A Review
Units
Concentration
Mole fraction/mixing ratios
Molecules, Radicals, Ions
Free Radicals
Termolecular Reactions
Other Important Radicals
Acid-Base Reactions
Oxidation and Reduction
Chemical Equilibria
Henry’s Law
Chemical Thermodynamics
Entropy and Energy
Free Energy and Equilibrium Constant
Free Energy and Temperature
Hess’s Law
Speed of Reactions
Activation Energy
Photochemical Reaction Rates
Deposition to Surfaces
Residence Time
General Rules for Gas-phase Reactions

3. Units Mixing Ratios ppm = mg kg-1 = mg L-1 Parts per million:
e.g. 10 mg F per million mg of water
= 10 tons F per million tons water
= 10 mg F per million mg water
= 10 ppmw F (or ppm/w)
[since 1000,000 mg water = 1 kg water = 1 L water]
= 10 mg F per L water (10 mg/L or 10 mg L-1 F)
= 10 ppmv F (or ppm/v)
ppm = mg kg-1 = mg L-1

4. Question Prove that 1 mg L-1 = 1 μg mL-1 1 mg x 1000 μg mg L x 1000 mL

5. Question Example 1: convert 0.100 M lead nitrate to ppm
M[Pb(NO3)2] = g/mol
0.100 mols/L = mols x (331.2 g/mol) / 1 L
= 33.1 g / L = 33.1 g/L x 1000 mg/g = ppm
Example 2: convert 0.01 g lead nitrate dissolved in 1L to ppb
0.01 g/L x (1000 mg/g) = 10 mg/L = 10 ppm
10 ppm x 1000 ppb / ppm = 10,000 ppb

6. Units for Gases Concentration units
Molecules per cubic centimeter (molec. cm-3)
Mole fraction / mixing ratios
volume analyte/total volume of sample
Molecule fraction per million or billion
e.g. 100 ppmv CO2 refers to 100 molec. of CO2 per 106 molec. of air
Partial pressure of gas expressed in units of atmospheres (atm) kilopascal (kPa) or bars (mb),
Ideal gas law relates pressure and temperature to no. molecules
PV = nRT

7. Units for Gases
Conversion (at normal temperature of 20 ºC and 1 atm) from w/v to v/v:
concentration (ppm) = concentration (mg m-3) x 24.0
Molar mass
Note: At STP of 273 K (0 C) the molar volume is 22.4
Similarly:
concentration (ppb) = concentration (μg m-3) x 24.0
concentration (ppt) = concentration (ng m-3) x 24.0

8. Question
The hydrocarbons that make up plant waxes are only moderately volatile. As a consequence, many of them exist in the atmosphere partly as gases and partly as constituents of aerosol particles. If tetradecane (C14H30, molecular weight 198) has a gas phase mixing ratio over the N. Atlantic Ocean of 250 ppt (pptv) and an aerosol concentration of 180 ng m-3, in which phase is it more abundant?
Must convert v/v to w/v
= 250 ppt x M / 24.5
= 2.0 x 103 ng m-3
 Gas phase is higher
Conversion between w/v and v/v:
mg/m3 = ppm * M / 24.0
ppm = (mg/m3)*(24.0 / M)
ppt = (ng/m3)*(24.0 / M)
From Graedel and Crutzen (1993), Chapter 1

9. Question [Convert to mg m-3 then use w/v to v/v conversion]
Express [O3] = 2.0 x 1012 molecules cm-3 as a volume mixing ratio (ppbv)
[Convert to mg m-3 then use w/v to v/v conversion]
[O3] = 2 x 1012 molecules cm-3
= 3.3 x mols cm-3
= 3.3 x mols cm-3 x 48 g/mol = 1.6 x g cm-3
= 1.6 x 10-7 mg cm-3 x (1 x 106 cm3 / m3)
= 0.16 mg m-3
= 0.16 mg m-3 x 24.0 / 48 g mol-1
= ppmv = x 1000 ppmv / ppbv = 80 ppbv
From Graedel and Crutzen (1993), Chapter 1

10. Question
Calculate the pressure of ozone in atm and in ppmv at the tropopause (15 km, 217 K), given [O3] = 1.0 x 1012 molecules cm-3, and p(total) = 0.12 atm
[O3] = 1.0 x 1012 molecules cm-3 x 1000 cm3/1 L x 1 mol/6.022 x 1023 molecules
= 1.7 x 10-9 mol L-1
pV = nRT,
p(O3) = (n/V) RT = 1.7 x 10-9 mol L-1 x L atm/mol K x 217 K = 3.0 x 10-8 atm
p(O3) ppmv = (3.0 x 10-8 atm / 0.12 atm ) x 106 ppmv = 0.25 ppmv
From Graedel and Crutzen (1993), Chapter 1

11. Lab: Mixing Ratio or ppm, ppb
Quantities are very important in E-Chem!
Try pre-lab questions before Ozone lab
Image from

12. Question
Convert mg m-3 [X] to ppm (v/v) and derive the conversion factor for mg m-3 to ppm
ppm = (mg m-3)*(24.0 / M)
[X] mg x g = [X] 10-3 g
m mg m3
= [X] 10-3 g = [X] 10-3 mol = [X] 10-3 mol x m3/mol
M g/mol M m3
m m3
= [X] x 10-3 x = [X] x 24.0 x 10-6 = [X] x 24.0
M M M

14. Molecules, Radicals, Ions
Molecules are comprised of atoms bound together by chemical bonds:
e.g. CO2 and CCl2F2 unreactive
H2O2 and NO quite reactive
HO• very reactive (Hydroxyl radical)
Free radicals – molecular fragments containing an odd number of e- (unpaired)
Bonding reqirements unsatisfied, react to form more stable state
Molecules are ‘teared apart’ via photodissociation
e.g. Cl2 + hν → 2Cl•

15. Question
Draw full and reduced Lewis structures for hydroxyl radical, chlorine monoxide radical and nitric oxide radical
Note that there are a total of 11 e- in this structure. The more electronegative atom oxygen has 8 e- in its outer shell while nitrogen has only 7 e- in its outer shell. This extremely reactive free radical seeks to obtain another e- to fulfill the octet rule and become a lower energy species.

16. e.g. a mixture of H2 and Cl2 is irradiated with UV Cl• + H2 → HCl + H•
Demo
e.g. a mixture of H2 and Cl2 is irradiated with UV
UV breaks apart a chlorine molecule Cl2 + hν → 2Cl•
Cl• + H2 → HCl + H•
H• + Cl2 → HCl + Cl•
H2 + Cl2 → 2HCl (ΔH = kJ)
Mixture is exothermic but stable at room temperature
H2 + Cl HCl

17. Demo Chem Comes Alive Vol. 1

18. Free Radicals Molecular Fragments
Water is such a stable molecule that driving force for its creation is strong – pulls H atom from methane
Once created radical attacks other molecules
Product is another radical since e- remains unpaired
e.g. CH4 + HO• → CH3• + H2O
When CH3• radical reacts with O2 to form CH3O2•
Chain reaction propagates
H2O2 + hν → 2HO• Initiation
CH4 + HO• → CH3• + H2O Propagation
CH3• + O2 → CH3O2•
HO• + HO• → H2O2 Termination

19. Termolecular Reactions
Another termination reaction is:
HO• + NO2 + M → HNO3 + M
Termolecular (3 molecule) reactions are important in atmospheric chemistry
M is an unreactive molecule e.g. N2 and O2
Energy released on chemical bond formation is removed by ‘M’

20. Other Important Radicals
Oxygen radicals
O2 + hν → 2O•
Organic oxygen radicals
RCH2• → RCH2OO• (alkylperoxy radicals)
RCH2OO• + X → XO + RCH2O•
X can be NO or SO2 or organics
Hydroxyl radical
O3 + hν → O2 + O*
O* +H2O → 2OH•
OH• + RCH3 → RCH2 + H2O
OH• + NO2 → HNO3
OH• + SO2 → HSO3 → O2 + H2O → H2SO4 + HO2•

21. Warning: The textbook is inconsistent in denoting radicals
Warning: The textbook is inconsistent in denoting radicals. In many cases it shows a “dot” to indicate the one unpaired electron. However, some examples in the textbook do not have the dot so the reader is left to assume the species is a radical. You should know that species such as OH, CH3, ClO, H3COO, and others are all radical species.

22. Acid-Base Reactions Acids react with water to form a hydrated proton
e.g. HNO3 + H2O H3O+ + NO3-
Acid is a proton donor, base is a proton acceptor
Degree of acidity,
pH = log10[H+]
Ions are stable in aqueous solution due to their hydration spheres
Free radicals in the aqueous phase can initiate many chemical reactions

23. Oxidation and Reduction
Chemical reaction involving the transfer of e- from one reactant to another
e.g. Mn3+ + Fe2+ → Mn2+ + Fe3+
Mn3+: Oxidant, e- donor
Fe2+: Reductant, e- acceptor
Two half-reactions:
Reduction: Mn3+ + e- → Mn2+
Oxidation: Fe2+ → Fe3+ + e-
Redox potential, pE is a measure of the tendency of a solution to transfer electrons:
pE = -log10[e-]
Reducing environment = large -ve pE
Oxidizing environment = large +ve pE

24. Chemical Equilibria
Reactions do not always proceed completely from reactants to products
Chemical equilibrium rates of forward and reverse reaction are equal
e.g. αA + βB ↔ γC + δD
Equilibrium constant is defined as
K = [C]γ[D]δ
[A]α[B] β

25. K = c(O2) (= 1.32 x 10-3 mol L-1 atm-1)
Henry’s Law
At a constant temperature the concentration of a solute gas in solution is directly proportional to the partial pressure of that gas above the solution
e.g. the equilibrium between oxygen gas and dissolved oxygen in water is
O2(aq) O2(g)
The equilibrium constant is
K = c(O2) (= 1.32 x 10-3 mol L-1 atm-1)
p(O2)
O2 at 1 atm would have molar solubility of 1.32 x 10-3 mol L-1 = 1.32 mmol/L

26. Question
Calculate the concentration of oxygen dissolved from air in mol L-1 and ppmv
K = c(O2)/p(O2)
c(O2) = 1.32 x 10-3 mol L-1 atm-1 x 0.21 atm
= 2.7 x 10-4 mol L-1
= 2.7 x 10-4 mol L-1 x g mol-1
= 8.7 x 10-3 g L-1 x 1000 mg = 8.7 mg L-1 = 8.7 ppmv
1 g

28. Speed of Reactions
If thermodynamically favored, speed may be crucial to importance
Reaction rate is +ve if species is created and –ve if destoyed
e.g. aA + bB → cC + dD
Rate law:
R = k[A]a[B]b
Where k is the rate constant (cm3 molecule-1 s-1), a, b etc. are reaction orders and [A] are reactant concentrations

29. Question
Consider the oxidation of carbon monoxide by the hydroxyl radical
CO + HO• → CO2 + H•
What is the rate expression for this reaction?
Rate = k[CO][HO•]

30. Activation Energy
Reactions and their rate constants are temperature dependent
Magnitude of AE determines how fast a reaction occurs
Gas-phase reactions with large AE are slow
Radical reactions are exothermic and occur faster

31. Nitrogen Oxides Kinetics
N2 + O NO
The reaction proceeds when the temperature is sufficiently high (combustion)
When temperature decreases it should drive the reaction to the left
In the atmosphere other reactions with –ve ΔG are favoured
2NO + O2 → 2NO2 ΔG = kJ/mol
NO2 + O2 + 2H2O → 4HNO3 ΔG = kJ/mol

32. Activation Energy
Activation energy represents additional energy to drive a reaction to the thermodynamic requirement
Reaction proceeds with the lowest activation path

34. Photochemical Reactions
Photochemistry – reactions initiated by absorption of photons of radiation
Electromagnetic radiation is described with its wave-like properties in a single equation
νλ = c
where ν is frequency, λ is wavelength and c is the speed of light
The energy of this radiation is quantized into small packets of energy, called photons, which have particle-like nature. Electromagnetic radiation can be pictured as a stream of photons. The energy of each photon is given by
EPHOTON = hν = hc (where h = Plank’s constant) λ
Energy increases vibrational or rotational energy, if energy exceeds bond strength photodissociation occurs

35. Photochemical Reaction Rates
d[C] = - J[C] dt
Rate of photodissociation of H2O2 is given by
Rate = J[H2O2]

36. Residence Time
Average amount of time a molecule exists before it is removed. Defined as follows:
Residence time = amount of substance in the ‘reservoir’
rate of inflow to, or outflow from, reservoir
Must be distinguished from half-life, residence time is the time taken for the substance to fall to 1/e (~37%) of the initial concentration
Important for determining whether a substance is widely distributed in the environment c.f. CFC’s and acidic gases

37. Residence Time For a 1st order reaction: C = C0e-kt When t = τ = 1/k
C = C0 / e = 0.37 C0

38. Half-Life
Time taken for the concentration in the reservoir to fall by 50%
When C = C0/2
C = C0e-kt
C0/2 = C0e-kt
e-kt = ½
τ1/2 = ln 2 / k
For photolysis
τ1/2 = ln2/J

39. Further Reading
Finlayson-Pitts, B.J. and Pitts, Jr., J.N. (1986) Atmospheric Chemistry: Fundamentals and Experimental Techniques. John Wiley, New York, 1098 pp.
Graedel, T.E. and Crutzen, P.J. (1993) Atmospheric Change: An Earth System perspective. Freeman.
Harrison, R.M., deMora, S.J., Rapsomanikis, S., and Johnston, W.R. (1991) Introductory Chemistry for the Environmental Sciences. Cambridge University Press, Cambridge, UK.