Transcript Environmental Chemistry - Robert Morris University

Environmental Chemistry
Chapter 0:
Chemical Principals - Review
Copyright © 2009 by DBS
Chemical Principals – A Review
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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
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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
Units Mixing Ratios
Parts per million conversions:
Parts per billion: 1 part in 109 parts:
ppm x 1000 ppb/ppm = ppb
Parts per trillion: 1 part in 1012 parts:
ppb x 1000 ppt/ppb = ppt
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
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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
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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 25 ºC and 1 atm) from w/v to v/v:
concentration (ppm) = concentration (mg m-3) x 24.0
Molar mass
• Mixing ratio (v/v) is conserved if temperature pressure changes
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
Conversion between w/v and
v/v at STP:
mg/m3 = ppm * M / 24.0
 Gas phase is higher
ppm = (mg/m3)*(24.0 / M)
Question
Express [O3] = 2.0 x 1012 molecules cm-3 as a volume mixing ratio
(ppbv) at 25 ºC, 1 atm.
[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 mol/m3 and then to ppm (v/v) and derive
the conversion factor for mg m-3 to ppm
[X] mg x 10-3 g
m3
mg
= [X] 10-3 g
= [X] 10-3 g
M = molec.
masss)
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
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Molecules are comprised of atoms bound together by chemical
bonds:
e.g. CO2 and CCl2F2
H2O2 and NO
HO•
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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
http://dwb4.unl.edu/chemistry/redoxlp/A02.html
Vol. 1 of CCA
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
• A typical 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
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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+]
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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
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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
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Two half-reactions:
Reduction: Mn3+ + e- → Mn2+
Oxidation: Fe2+ → Fe3+ + e-
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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
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e.g. the equilibrium between oxygen gas and dissolved oxygen in water is
O2(aq) ⇌ O2(g)
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The equilibrium constant is
K = c(O2)
p(O2)
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(= 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) = K x p(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
Chemical Thermodynamics
First law: Energy is neither created nor destroyed
– Or sum of energy and mass is conserved, E = mc2
Second Law: Entropy always increases in a spontaneous
process
– Entropy is a measure of disorder
One measure of entropy is the heat energy q divided by the
Temperature
S = q/T
q/T increases in a spontaneous process. Heat energy produces
greater disorder for a cold sample (smaller T) than for a hot
sample
Enthalpy Cycles & Hess’s Law
Hess’s Law
• The total enthalpy change for a reaction is independent of the route
• The enthalpy change for the formation of CH4 cannot be determined
experimentally
• It is possible to determine the enthalpy of combustion of C, H and CH4
ΔH
C(s) + 2H2(g)
CH4(g)
ΔH2
ΔH1
CO2(g) + H2O(l)
Using Hess’s law:
ΔH + ΔH2 = ΔH1
ΔH = ΔH1 = ΔH2
Entropy and Energy
• Chemical reactions are associated with changes in entropy and
energy. Entropy limits the work that can be extracted
• The amount of energy available for work is called Free Energy
Change or ΔG
ΔG = ΔH - T ΔS
• ΔH is the enthalpy change and ΔS is the entopy change in the
reaction
Free energy trends parallel
enthalpy trends
-ve free energy trends
show the formation of
these compounds are
favored
Standard enthalpy and free energies of formation
kJ at 298K
O3, NO, NO2 have a +ve
ΔG hence they are readily
decomposed
Free Energy and Equilibrium Constant
• Rate(forward) = kf [A][B] and Rate(back) = kb [C][D]
• At equilibrium kf [A][B] = kb [C][D]
K = kf/kb
• The thermodynamic relationship between the equilibrium
constant and the free energy change is
ΔG = -RTlnK
Where R is the gas constant (8.314 JK-1 or 0.082 L atm mol-1 K-1)
ΔG = -8.314 JK-1 x T x 2.303 x logK
= -19.57 T(logK) joules = 5706 logK (at 298K)
Question
Determine the equilibrium concentration of NO in the atmosphere at
sea level and 25 °C given that for NO ΔG0 = 86.7 kJ
N2 + O2 ↔ 2NO
K = [NO]2/[N2][O2]
therefore [NO] = (K [N2][O2])0.5
At 1 atm [O2] is 0.21 atm, [N2] is 0.78 atm
ΔG for 2 NO = 2 x 86.7 kJ = 173.4 kJ
ΔG = -RTlnK
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lnK = 2.303 logK = - ΔG / RT
LogK = (-173.4 x 103 J) / (2.303 x 8.314 J/K x 298 K) = -30.4
K = 10-30.4
Substitute K, [O2] and [N2]
[NO] = (K [N2][O2])0.5 = (10-30.4 x 0.78 x 0.21)0.5 = 1.6 x 10-15.7 atm
NB: Small compared to 10-4 ppm or 10-10 atm in urban areas
logeK = (log10K) / (log10e)
Free Energy and
Temperature
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ΔG = ΔH – TΔS and ΔG = -RTlnK
lnK = - ΔH/RT + ΔS/R
Increasing temperature increases the difference between ΔH and ΔG
ΔH is negative (exothermic) lnK is positive (unless overcome by ΔS)
– Products are favoured over reactants
– With increasing temperature K becomes less positive (less
products)
ΔH is positive (endothermic) lnK is negative
– Reactants are favoured over products
– With increasing products lnK becomes less negative and the
equilibrium drives towards products
In either case the effect of raising temperature is to produce a more
even distribution of reactants and products
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
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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
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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
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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
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(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]
Deposition to Surfaces
• Rate is comparable to chemical reaction rate
• Dry deposition and wet deposition
• Vertical flux, Фi
Фi = vd[Ci]
• Where [Ci] is the
concentration at some
reference height and vd is the
deposition velocity
• Wet deposition is efficient at
cleaning the air
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 lifetime and half-life.
• Important for determining whether a substance is widely distributed in
the environment c.f. CFC’s and acidic gases
Residence Time
Atmospheric Lifetimes
• Loss mechanisms or ‘sinks’ operate on different time-scales
• It is always assumed first order:
A → products
-d[A]/dt = kRt
• Solution:
[A] = [A]0e-kt
• τ is the time it takes to reduce [A] to 1/e (37 %) of its initial value,
[A]0
• Substituting [A] = e-1[A]0 gives e-1[A]0 / [A]0 = e-kτ
kR τ = 1 or τ = 1/kR
Half-Life
• Time taken for the concentration in the reservoir to fall by 50%
• When [A] = [A]0/2
[A] = [A]0e-kt
[A]0/2 = [A]0e-kt
e-kt = ½
τ1/2 = ln 2 / k
• For photolysis
τ1/2 = ln2/J
General Rules for Gas-phase Reactions
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Reactions are limited to those that are exothermic or very slightly
endothermic
Reactions between two stable molecules, even if energetically
favorable, tend to be slow
Reactions between free radicals and molecules are common. Such
reactions generally have positive activation energies, and as shown by
the Arrhenius equation, their rates increase as tempoerature increases
Reactions between free radicals are often very fast. Many have
negative activation energies, indicating that they proceed faster at
lower temperatures
Termolecular reactions proceed faster at lower temperatures, because
as the reaction partners approach each other with decreased energy,
the transition state is more stable. The third body M can more easily
carry away ecess energy
Radical formation and the initiation of chemical reaction chains often
begins with photodissociuiation of molecules by solar ultraviolet
radiation
Gas-phase ions in the troposphere and stratosphere are rare and do
not play important chemical roles
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.