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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.