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) [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 • • 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 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 • 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 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 • 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) ⇌ O2(g) • The equilibrium constant is K = c(O2) p(O2) • (= 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 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 • • • • • 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 • • • • 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] 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 • • • • • • • 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.