Download chp0-Intro

Environmental Chemistry
Chapter 0:
Chemical Principals - Review
Copyright © 2007 by DBS
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
Units Mixing Ratios
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
Question
Prove that 1 mg L-1 = 1 μg mL-1
1 mg x 1000 μg
mg
L
x 1000 mL
L
= 1 μg / mL
Question
Example 1: convert 0.100 M lead nitrate to ppm
M[Pb(NO3)2] = 331.2 g/mol
0.100 mols/L = 0.100 mols x (331.2 g/mol) / 1 L
= 33.1 g / L = 33.1 g/L x 1000 mg/g = 33100 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
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
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
Molar mass
concentration (ppt) = concentration (ng m-3) x 24.0
Molar mass
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
Conversion between w/v and v/v:
mg/m3 = ppm * M / 24.0
= 250 ppt x M / 24.5
= 2.0 x 103 ng m-3
ppm = (mg/m3)*(24.0 / M)
 Gas phase is higher
ppt = (ng/m3)*(24.0 / M)
Question
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 10-12 mols cm-3
= 3.3 x 10-12 mols cm-3 x 48 g/mol = 1.6 x 10-10 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
= 0.080 ppmv = 0.080 x 1000 ppmv / ppbv = 80 ppbv
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 0.0821 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
Lab: Mixing Ratio or ppm, ppb
• Quantities are
very important in
E-Chem!
• Try pre-lab questions
before Ozone lab
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 10-3 g
m3
mg
= [X] 10-3 g
= [X] 10-3 g
m3
= [X] 10-3 mol
M g/mol
M
m3
m3
= [X] 10-3 mol x 0.0240 m3/mol
m3
= [X] x 10-3 x 0.0240 = [X] x 24.0 x 10-6 = [X] x 24.0
M
M
M
Molecules, Radicals, Ions
•
Molecules are comprised of atoms bound together by chemical
bonds:
e.g. CO2 and CCl2F2
H2O2 and NO
HO•
•
unreactive
quite reactive
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•
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.
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 = -184.6 kJ)
Mixture is exothermic but stable at room temperature
H2 + Cl2
2HCl
Demo
Chem Comes
Alive Vol. 1
http://chemmovies.unl.edu/chemistry/redoxlp/A02.html
Free Radicals
Molecular Fragments
• 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
CH3• + O2 → CH3O2•
Propagation
HO• + HO• → H2O2
Termination
Water is such a
stable molecule
that driving
force for its
creation is
strong – pulls H
atom from
methane
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’
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•
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.
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
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
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
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)
•
The equilibrium constant is
K = c(O2)
p(O2)
•
O2(g)
(= 1.32 x 10-3 mol L-1 atm-1)
O2 at 1 atm would have molar solubility of 1.32 x 10-3 mol L-1 = 1.32 mmol/L
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 32.00 g mol-1
= 8.7 x 10-3 g L-1 x 1000 mg = 8.7 mg L-1 = 8.7 ppmv
1g
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
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•]
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
Nitrogen Oxides Kinetics
N2 + O2
2NO
• 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 = -69.8 kJ/mol
NO2 + O2 + 2H2O → 4HNO3 ΔG = -239.6 kJ/mol
Activation Energy
• Activation energy represents additional energy to drive a reaction
to the thermodynamic requirement
• Reaction proceeds with the lowest activation path
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
λ
•
(where h = Plank’s constant)
Energy increases vibrational or rotational energy, if energy exceeds bond
strength photodissociation occurs
Photochemical Reaction Rates
Rate:
d[C] = - J[C]
dt
Rate of photodissociation of H2O2 is given by
Rate = J[H2O2]
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
Residence Time
• For a 1st order reaction:
C = C0e-kt
• When t = τ = 1/k
C = C0e-1
C = C0 / e = 0.37 C0
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
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.