Environmental Chemistry

Chapter 0:Chemical Principals - Review

Copyright © 2009 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)

ppm = mg kg-1 = mg L-1

[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)

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

0.01 g/L x (1000 mg/g) = 10 mg/L = 10 ppm

10 ppm x 1000 ppb / ppm = 10,000 ppb

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 1: convert 0.100 M lead nitrate to ppm

Example 2: convert 0.01 g lead nitrate dissolved in 1L to 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. moleculesPV = nRT

Units for Gases

• Conversion (at 25 ºC and 1 atm) from w/v to v/v:

concentration (ppm) = concentration (mg m-3) x 24.0Molar 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

Gas phase is higher

Conversion between w/v and v/v at STP:mg/m3 = ppm * M / 24.0

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.

[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

[Convert to mg m-3 then use w/v to v/v conversion]

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 inE-Chem!

• Try pre-lab questionsbefore 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 = [X] 10-3 g

m3 mg m3

= [X] 10-3 g = [X] 10-3 mol = [X] 10-3 mol x 0.0240 m3/mol

M g/mol M m3

m3 m3

= [X] x 10-3 x 0.0240 = [X] x 24.0 x 10-6 = [X] x 24.0

M M M

M = molec. masss)

Molecules, Radicals, Ions

• Molecules are comprised of atoms bound together by chemical bonds:

e.g. CO2 and CCl2F2 unreactiveH2O2 and NO quite reactiveHO• 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.htmlVol. 1 of CCA

Demo

http://chemmovies.unl.edu/chemistry/redoxlp/A02.html

Chem Comes Alive Vol. 1

Free RadicalsMolecular 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

Water is such a stable molecule that driving force for its creation is strong – pulls H atom from methane

H2O2 + hν → 2HO• Initiation

CH4 + HO• → CH3• + H2O Propagation

CH3• + O2 → CH3O2•

HO• + HO• → H2O2 Termination

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

• 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- donorFe2+: 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 pEOxidizing environment = large +ve pE

• Reactions do not always proceed completely from reactants to products

• Chemical equilibriumrates 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

Chemical Equilibria

Henry’s Law

• 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

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

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

1 g

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/Tq/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

C(s) + 2H2(g) CH4(g)

CO2(g) + H2O(l)

ΔH

Using Hess’s law: ΔH + ΔH2 = ΔH1

ΔH = ΔH1 = ΔH2

ΔH1ΔH2

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

Standard enthalpy and free energies of formation kJ at 298K

Free energy trends parallel enthalpy trends

-ve free energy trends show the formation of these compounds are favored

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

logeK = (log10K) / (log10e)

N2 + O2 ↔ 2NOK = [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 kJΔG = -RTlnK

lnK = 2.303 logK = - ΔG / RTLogK = (-173.4 x 103 J) / (2.303 x 8.314 J/K x 298 K) = -30.4K = 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

Free Energy and Temperature

• ΔG = ΔH – TΔS and ΔG = -RTlnKlnK = - 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

• 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 (where h = Plank’s constant) λ

• 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

• 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

• Rate is comparable to chemical reaction rate• Dry deposition and wet deposition

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 TimeAtmospheric 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

• 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.