Section 3 Environmental Chemistry 1
Environmental Chemistry Definitions Chemical Reactions Stoichiometry Photolytic Reactions Enthalpy and Heat of Reaction Chemical Equilibria ph Solubility Carbonate Systems 2
Introduction Almost every pollution problem has some chemical basis From a chemical transformation or reaction or from the chemical properties of waste products Some problems involving chemical reactions: Greenhouse gases Ozone Hole Urban smog Acid deposition Water pollution 3
Definitions Atom 1 atomic mass unit (amu) = 1/12 the mass of one carbon-12 atom (1.66053886 10 27 kg Proton (charge = +1, m = 1 amu) Neutron (charge = 0, m = 1 amu) Electron (charge = 1, mass ~0) Atomic weight Isotope 4
Molecule Molecular weight Mole Number of moles = mass/molecular weight 1 g-mol = 6.022 10 23 molecules 1 lb-mol = 2.7 10 26 molecules 5
Chemical Reaction Stoichiometry Balance those chemical reaction equations! Example 1 A 1.67 10 3 M glucose solution (C6H12O6) is completely biodegraded to carbon dioxide and water. How much oxygen is required (mg/l)? C 6 H 12 O 6 + O 2 CO 2 + H 2 O Balanced equation: 6
From the amount of glucose we have in moles, it takes six times that amount in oxygen to decompose the glucose 6 1.67 10 3 mol = 0.01 mol If this is the amount of oxygen in one liter of air, then the concentration in mg/ is 0.01 mol L 32 g mol 1000 mg g mg = 320 L This oxygen demand given by stoichiometry is called the theoretical oxygen demand. 7
Theoretical Oxygen Demand (TOD): oxygen needed to fully oxidize a quantity of organic material to carbon dioxide and water Biochemical Oxygen Demand (BOD): oxygen required for oxidation of organic wastes carried out by bacteria BOD ~ TOD 8
Photodissociation Photodissociative reactions (also known as photolysis) perform key steps in atmospheric chemistry Photodissociation occurs when a molecule absorbs a photon of light and decomposes Photochemical describes a reaction or set of reactions that derive at least some of the energy needed for the reactions from sunlight 9
Energy in photons used in photochemistry are related to the wavelength of the light: E = hν = hc λ E = energy of the photon (J) h = Planck s constant (6.6 10 34 J s) = frequency (cycles/s = Hz) c = speed of light (3 10 8 m/s) = wavelength (m) 10
Example Energy per photon of light with wavelength 550 nm E = hν = hc λ E = hc λ ( )( 3 10 8 m) s 6.6 10 34 Js = 1 m 10 9 nm 550 nm = 3.6 10 19 J 11
Enthalpy We may need to determine the energy of a photon based on the enthalpy of the system Enthalpy (H) is determined by the internal energy (U) and the product of the pressure (P) and volume (V) H = U + PV For a process with constant volume: ΔU = mc V ΔT For a process with constant pressure: ΔH = mc P ΔT 12
Heat of Reaction 0 ΔH rxn = ΔH 0 f Products ΔH 0 f Reactants (The zero superscript means the enthalpy is measured at 1 atm and 298 K) Endothermic reaction: ΔH 0 rxn is positive Exothermic reaction: ΔH 0 rxn is negative where ΔH 0 f is the heat of formation 13
Example Using the enthalpy (ΔH 0 ) of a reaction calculated from the heats of formation (ΔH 0 f) of the reactants and products, calculate the maximum wavelength that can drive this photolytic reaction: O 3 + hν O 2 + O Standard enthalpies for the oxygen species are given in Table 2.1 (text). Thus, 142.9 kj/mol + energy of light = 0 kj/mol + 247.5 kj/mol The energy of the photon would be 104.6 kj/mol, which corresponds a wavelength of 1.13 µm 14
Chemical Equilibria Examples where chemical equilibria are important: Acid/base reactions affecting ph Solubility products affecting precipitation Solubility of gases in water Most liquid phase chemical reactions are reversible, more or less: aa + bb cc + dd 15
When are chemical equilibria not important? Gas phase reactions reversible, but low concentrations of the reactants means they may not interact enough to make the reversibility meaningful 16
If the forward and reverse reactions are proceeding at the same rate, the system is in equilibrium constant concentrations of species aa + bb cc + dd For, K = [ C ] c D [ ] d Equilibrium Constant [ A] a [ B] b Concentrations must be expressed in moles/liter = M Molecules may dissolve to form ions: cations (+), anions ( ) 17
Dissociation of Ions A 2 B 2A + + B 2 H 2 SO 4 2H + + SO 4 2 K is a dissociation or ionization constant: K = A+ 2 B 2 A 2 B [ ] 18
Since this K can range over several orders of magnitude, it is more convenient to use logarithmic notation px = log X X = 10 px Acid-Base ph = log H + H + = 10 ph mol 19
Pure water dissociates: H 2 O H + + OH A special equilibrium relationship applies to water: K w = H + OH = 1 10 14 (at 25 C) In neutral water, H + = OH Then, K w = H + OH = H + 2 = 1 10 14 20
K w = H + H + = 10 7 OH = H + = 1 10 14 2 ph = 7 for a neutral solution 21
Significance of ph to environmental science: Indicates whether a solution is acidic (ph < 7) or basic (ph > 7) Life is very sensitive to ph, particularly aquatic life (affects biodiversity of this group) ph affects solubility of gases and solids, which affects effects of acidity in aquatic systems Industrial wastes may have extreme ph levels, requiring neutralization 22
Example Household ammonia has ph = 11.9 at 25 C. What is [H + } and [OH ]? H + = 10 ph mol L = mol 10 11.9 L = 1.26 mol 10 12 L ph + poh = 14, so poh = 14 11.9 = 2.1 OH = 10 poh mol L = mol 10 2.1 L = 7.94 mol 10 3 L 23
Solubility Product describes dissolution of solids and precipitation of components solid aa + bb K = [ A] a [ B] b [ solid] 1 The solid portion is in a different phase than the solutes, and its concentration is irrelevant to the equilibrium. We combine [solid] with the equilibrium constant: K sp = A [ ] a B [ ] b solubility product 24
In actual situations, solid may or may not be present [ions] > Ksp, solid is present or will subsequently form [ions] < Ksp, more solid can dissolve in the solution 25
Example Fluoride (F ) in water from CaF2 dissolution: CaF 2 Ca 2+ + 2F K sp = Ca 2+ F 2 = 3 10 11 Given that there is solid present, what will the equilibrium concentration of F be? 26
For each mole of Ca 2+, there will be 2 moles of F. Let the molar concentration of Ca 2+ be represented by s Then K sp mol = 3 10 11 L = Ca2+ s = 2 10 4 mol L and s = Ca 2+ 2s = F F 2 mol = 2 10 4 L = 4 mol 10 4 L = s ( 2s) 2 = 4s 3 27
Solubility of Gases in Water Henry s Law Describes how much gas can dissolve into water (at equilibrium) X [ ] aq = K H,X P X [X]aq = aqueous phase concentration of X, in mol L KH = Henry s Law coefficient, in mol L i atm PX = partial pressure of X 28
partial pressure = concentration in vol atmospheric pressure vol 1 ppm = 10 6 atm at 1 atm Ex. Sea level pressure is 1 atm; pressure in Boulder CO is 0.8 atm (at about 6000 ft altitude) At sea level, partial pressure of oxygen (O2) is about 0.21 atm (concentration is 21%). In Boulder, it is (21%)(0.8 atm) = 0.17 atm ( ) 29
Oxygen in the human body Approximate partial pressure of oxygen in the blood is 0.13 atm. If we assume that pressure is related to altitude by: z P ( z ) = P ( H 0)e where H = 7 km What altitude is within the human body s comfort limits? Find altitude where partial pressure of oxygen is not less than the blood s partial pressure of oxygen 30
What altitude is within the human body s comfort limits? Find altitude where partial pressure of oxygen is not less than the blood s partial pressure of oxygen z Partial pressure of O 2 = ( 0.21)P ( z ) = ( 0.21)P ( H 0)e 0.13 atm = 0.21( 1 atm)e 0.62 = e z 7 km z 7 km z = ln( 0.62) 7 = 3.4 km 11000 ft 31
Henry s Law constants are temperature dependent. Ex., the solubilities of CO2 and O2 roughly double between 25 and 0 C T ( C) KH,CO2 (M atm 1 ) KH,O2 (M atm 1 ) 0 0.076 2.2 10 3 10 0.053 1.7 10 3 20 0.039 1.4 10 3 25 0.033 1.3 10 3 32
Example Leave some water on a table outside on a cold day in Denver (10 C, 1525 m) How much CO2 will dissolve in the water (in M) if its concentration in the atmosphere is 360 ppmv? KH,CO2 (10 C) = 0.0532 M/atm Patm = 0.825 atm in Denver [CO2] = 360 ppmv = 0.036% X [ ] aq = K H,X P X P CO2 = 360 10 6 0.825 atm ( ) = 2.97 10 4 atm [ CO 2 ] aq = ( 0.0532 M atm 1 )( 2.97 10 4 atm) = 1.58 10 5 M 33
Carbonate Systems Carbonates are the largest reservoir of carbon on Earth Controls ph in natural systems Four important species: Carbonic acid H2CO3 Bicarbonate ion HCO3 Carbonate ion CO3 2 Calcium carbonate CaCO3 34
Carbon dioxide dissolves in water to form CO2(aq), also stated as (CO2 H2O) This turns out to be H2CO3 This dissociates in water: CO 2 H 2 O (aq) + H (aq) + HCO 3 (aq) HCO 3 H + + CO 3 2 The carbonate ion serves as a carbon sink when it forms, bicarbonate is removed and more carbon dioxide is allowed to dissolve 35
An effective Henry s Law constant can take into account the effect of dissociation and chemical loss We also need to consider competing effects found in natural systems, such as the presence of limestone: The equilibria are: CaCO 3(s ) Ca 2+ + CO 3 2 CO 2 K 1 K 2 + H 2 O H + + HCO 3 H + 2 + CO 3 K sp CaCO 3(s ) Ca 2+ + K w H 2 O H + + OH 36
Equilibrium constants are: K 1 CO 2 + H 2 O H + + HCO 3 H + HCO 3 CO 2(aq ) = K 1 = 4.47 10 7 M = 10 6.35 M pk 1 = logk 1 = 6.35 K 2 HCO 3 H + + CO 3 2 H + 2 CO 3 = K HCO 2 = 4.68 10 11 M 3 = 10 10.3 M pk 2 = logk 2 = 10.3 K sp CaCO 3(s ) Ca 2+ + CO 3 2 Ca 2+ 2 CO 3 = K sp = 4.57 10 9 M 37
How much carbonate vs. bicarbonate? look for [CO3 2 ] / [HCO3 ] Divide H + 2 CO 3 by HCO H + 3 H + 2 CO 3 1 HCO 3 H + = 10 10.3 H + 2 CO 3 HCO 3 = 10 10.3 10 ph = 10 ph 10.3 38
Role of Organisms in Carbon Cycle Carbon enters oceans as carbon dioxide dissolution, then may be converted to carbonate or bicarbonate Then, certain organisms bind calcium to bicarbonate Calcium carbonate CaCO3 used to make shells, coral, exoskeletons, etc. Parts of dead organisms collect on sea floor and eventually become sedimentary rock largest carbon sink on Earth 39
40
Application Natural acidity of rainwater from dissolved carbon dioxide Determine the ph of rainwater that has an equilbrium amount of atmospheric carbon dioxide dissolved in it. Actual ph may be lower in polluted areas due to sulfuric, nitric, or organic acids. Get [H+] from an electroneutrality expression (charge conserved) insert each species from dissociation expressions, equilibrium relation for water, and Henry s Law 41
CO 2 H 2 O (aq) + H (aq) + HCO 3 (aq) HCO 3 H + + CO 3 2 H 2 O H + + OH 42
H + = HCO 3 + 2 CO 3 2 CO 2 H 2 O [ ] = K H P CO2 + OH HCO 3 = K CO 1 2(aq ) H + = K 1K H P CO2 H + 2 CO 3 = K HCO 2 3 H + OH = K w H + 43
Substitute for the concentrations of bicarbonate, carbonate, and hydroxyl ions in the electroneutrality expression so everything ends up in terms of [H + ]: H + = K K P 1 H CO 2 H + + 2 K K K P 1 2 H CO 2 H + 2 + K w H + H + 3 H + ( K 1 K H P CO2 + K ) w 2K 1 K 2 K H P CO2 = 0 Solving this for [H+] is not easy. Retry with a simplified version by applying: CO 3 2 HCO 3 44
H + = HCO 3 + 2 CO 3 2 + OH H + = K CO 1 2(aq ) H + + 10 14 M 2 H + H + 2 = K 1 CO 2(aq ) + 10 14 M 2 CO 2(aq ) = K H P CO2 1.3 10 5 M ph = 5.62 45