1. CHAPTER 4 ENVIRONMENTAL FATE

2. This chapter serves as a basis to identify the hazards associated with different substances used and produced in the chemical process, including raw materials, products and or byproducts. It would then be possible to do follow up with an exposure assessment and a dose-response assessment which are needed to perform risk characterization Introduction

3. Substance Classification Tree What Substances? Physical + Chemical Properties Old Analyses Estimating Exposure And Environmental Effects Classifying the Substances based on risk Performing P2 on the substances...

4. Chemical Properties Used to Perform Environmental Risk Screenings Environmental Process Relevant Properties Dispersion and Fate Volatility, density, melting point, water solubility, effectiveness of waste, water treatment. Persistence in the Environment Atmospheric oxidation rate, aqueous hydrolysis rate, photolysis rate, rate of microbial degradation, and adsorption. Continued on the following slide

5. Environmental ProcessRelevant Properties Uptake by Organisms Volatility, Lipophilicity, Molecular Size, Degradation Rate in Organism. Human Uptake Transport Across Dermal Layers, Transport Rates Across Lung Membrane, Degradation Rates within the Human Body. Toxicity and other Health Effects Dose-Response Relationships. Chemical Properties Used to Perform Environmental Risk Screenings

6. Distinguishes gas and liquid partitioning Using the substance’s structure, it can be estimated by : Where: T b : normal boiling point (at 1 atm) (K) n i : number of groups of type i in the molecule, g i : contribution of each functional group to the boiling point Corrected using : T b (corrected) = T b – 94.84 + 0.5577*T b + 0.0007705*(T b ) 2 (T b  700K) (4.2) T b (corrected) = T b + 282.7 – 0.5209*T b (T b > 700K) (4.3) T b = 198.2 + Σ n i g i (4.1) Boiling Point

7. Example : Boiling Point Estimation Estimate the Normal Boiling Point for diethyl ether. Diethyl ether has the molecular structure CH 3 -CH 2 -O-CH 2 -CH 3 Group -O- 2(-CH 3 ) 2(-CH 2 ) g i contribution 25.16 2(21.98) 2(24.22) Solving : The actual boiling point for diethyl ether is 307.65 K

8. a)Using equation 4.1 : T b (K)= 198.2 + Σ n i g i T b (K)= 198.2 + 2(21.98) + 2(24.22) + 25.16 T b = 315.76 b ) Using equation 4.2 : T b (corrected) = T b – 94.84 + 0.5577*T b - 0.0007705*(T b ) 2 T b (corr) = 315.76 – 94.84 + 0.5577(315.76) - 0.0007705(315.76) 2 T b (corrected) = 320.2 K Example : Boiling Point Estimation (Continued)

9. Melting Point Distinguishes solid and liquid partitioning. Can be estimated using the substance’s boiling point : Where : T m : Melting Point in Kelvins. T b : Boiling Point in Kelvins. (4.4) T m (K) = 0.5839 * T b (K)

10. Example : Melting Point Estimation Estimate the Melting Point for diethyl ether. Solving : Using equation 4.4 to calculate the T m : T m (K) = 0.5839 * T b (K) T m (K) = 0.5839 * 307.65 K T m = 179.634 K

11. Vapor Pressure Higher Vapor Pressure = Higher Air Concentrations Can be estimated using the following equations : ln P vp = A + B/(T - C)(4.5) Where : T = T b at 1 atm ln(1 atm) = 0 = A + B/(T b – C)(4.6) ln P vp (atm) ={[A(T b – C) 2 ] / [0.97*R*T b ]}*{1/(T b – C)-1/(T – C)} (4.7) the parameters A and C can be estimated using : C = -18 + 0.19 T b (4.7a) A = K F *(8.75+ R ln T b )(4.7b)

12. Where : P vp : vaporization pressure (atm). T : absolute temperature and T b is the boiling point at 1 atm. A and C are empirical constants. B : a parameter related to the heat of vaporization. K F : a correction factor. R : gas constant ; 1.987 L-atm K -1 mol -1 T m : melting point (K). For solids : ln P = -(4.4 + lnT b ) * {1.803*[(T b /T)- 1)] - [0.803*ln (T b /T)]} - 6.8(T m /T-1) (4.8) Vapor Pressure (continued)

13. Example : Vapor Pressure Estimation Estimate the Vapor Pressure for diethyl ether Using the predicted value of 315.76 K: C = -18 + 0.19T b = -18 + 0.19(320.2) = 41.9944 A = K f (8.75 + R ln T b ) = 1.06 [8.75 + 1.987 ln(320.2)] = 21.3962 ln P vp = {[A(T b – C) 2 ] / [0.97*R*T b ]}*{1/(T b – C) - 1/(T – C)} = {[21.39(315.76-41.99) 2 ] / [0.97(1.987)(315.76)]}*{1/(273.76) – 1/(256)} Ln P vp = -0.6677; P vp = 0.5128 atm = 389.79 mm Hg. (4.7.a) (4.7.b) (4.7) Repeating the calculation for the experimental boiling point leads to a vapor pressure estimated of P vp = 0.6974 atm = 530.06 mm Hg.

14. Octanol-Water Partition Coefficient Describes partition between an aqueous phase and it’s suspended organic phases. Can be estimated using the substance’s structure : log K ow = 0.229 + Σ n i f i (4.9) log K ow (corrected) = 0.229 + Σ n i f i + Σ n j c j (4.10) Where: K ow : Octanol-Water Partition Coefficient. n i : number of groups i in the compound. f i : factor associated with the group i n j : number of groups j in the compound that have correction factors. c j : correction factor for each group j

15. Example : Octanol-Water Partition Coefficient Estimation Estimate the Octanol-Water Partition Coefficient for diethyl ether. Solving : Using equation 4.9 : log K ow = 0.229 + Σ n i f i log K ow = 0.229 + 2(0.5473) + 2(0.4911) + (1.2566) log K ow = 1.0492 ≈ 1.05 therefore K ow = 11.2 Group -O- 2(-CH 3 ) 2(-CH 2 ) f i contribution -1.2566 2(0.5473) 2(0.4911)

16. Bioconcentration Factor Describes partitioning between aqueous and lipid phases in living organisms. Higher bioconcentration factors = higher quantity of bioaccumulation in living organisms Can be calculated using : log BCF = 0.79*(log K ow ) – 0.40 (4.11) log BCF = 0.77*(log K ow ) – 0.70 + Σ j j (4.12) Where : BCF : Bioconcentration Factor. K ow : octanol-water partition coefficient. j j : correction factor for each group.

17. Example : Bioconcentration Factor (BCF) Estimation Estimate the Bioconcentration Factor for diethyl ether. Solving : Using equation 4.9 we obtain log K ow : log K ow = 0.229 + Σ n i f i log K ow = 1.0492 ≈ 1.05 Using equation 4.11 we can calculate BCF : log BCF = 0.79*(log K ow ) – 0.40 log BCF = 0.79* (1.05) – 0.40 log BCF = 0.4295 therforeBCF = 2.6884

18. Water Solubility Used to assess concentrations in water Can be calculated using : Log S = 0.342 – 1.0374 logK ow – 0.0108 (T m –25) + Σh j (4.13) Log S = 0.796 –0.854 logK ow – 0.00728 (MW) + Σh j (4.14) Log S = 0.693 – 0.96 los K ow – 0.0092 (T m –25) – 0.00314 (MW) + Σh j (4.15) Where : S : water solubility (mol/L). K ow : octanol-water partition coefficient. T m : melting point (ªC). MW :s the molecular weight of the substance. h j is the correction factor for each functional group j.

19. Example : Water Solubility Estimation Estimate the Water Solubility for diethyl ether. Solving : Equation 4.9 gives the log K ow ≈ 1.05 Using equation 4.14 we can calculate the S : Log S = 0.796 –0.854 logKow – 0.00728 (MW) + Σhj Log S = 0.796 – 0.854(1.05) – 0.00728(74.12) + 0.0 Log S = -0.6403 Therfore : S = 0.2289 mol/L. = 16.966 g/L = 16,966.068 mg/ L

20. Henry’s Law Constant Describes the affinity for air over water. Can be determined using : -log H = log (air-water partition coeff) = Σ n i h i + Σ n j c j ( 4.19) Where : H : dimensionless Henry’s Law Constant. n i : number of bonds of type i in the compound. h i : bond contribution to the air-water partition coefficient. n j : number of groups of type j in the molecule. c j : correction factor for each group.

21. Example : Henry’s Law Constant Estimation Estimate the Henry’s Law Constant for diethyl ether. H H H H H-C-C-O-C-C-H H H H H Expressed as a collection of bonds, diethyl ether consists of 10 C-H, 2 C-C bonds, and 2 C-O bonds. The uncorrected value of log (air to water partition constant) is given by : -log H = log (air-water partition coefficient) = 10(-0.1197) + 2(0.1163) + 2(1.0855) = 1.2066 log H -1 = 1.2066

22. Soil Sorption Coefficient Used to describe the Soil-Water Partitioning. Can be estimated by : log K oc = 0.544 (log K ow ) +1.377 (4.16) log K oc = -0.55 (log S) + 3.64 (4.17) log K oc = 0.53 1 χ + 0.62 + Σ n j P j (4.18) Where : K oc : Soil Sorption Coefficient (μg/g of organic carbon (to μg/mL of liquid)). K ow : Octanol-Water Partition Coefficient. S : Water Solubility. 1 χ : first order Molecular Connectivity Index (from literature-appendix ). n j : number of groups of type j in the compound. P j : correction factor for each group j.

23. The first step in calculating 1 χ is to draw the bond structure of the molecule. For example, isopentane would be drawn as: CH 3 H 3 C-CH-CH 2 -CH 3 The second step is to count the number of carbon atoms to which each carbon is attached. Each C-C bond is given a value of 1 and δ i, is the parameter that defines the quantity of carbon atoms connected to a carbon atom i. The diagram below gives the δ i, values for the different carbon atoms. CH 3 H 3 C-CH-CH 2 -CH 3 (1)(1) (1)(1) (1)(1)(3)(3)(2)(2) Molecular Connectivity Index Calculations

24. The third step is to identify the “connectedness” of the carbons connected by the bond (δ i, δ j ). For isopentane, these pairs are: The value of 1 χ can then be calculated using the equation : 1 χ = Σ(δ i * δ j ) -0.5 (4.19) For isopentane, 1 χ = (1/√3) + (1/√3) + (1/√6) + (1/√2) = 2.68 CH 3 H 3 C-CH-CH 2 -CH 3 (2,1) (1,3) (3,2) Molecular Connectivity Index Calculations (continued) (1,3)

25. Example : Soil Sorption Coefficient Estimation Estimate the Soil Sorption Coefficient for diethyl ether. Solution : The molecular structure for diethyl ether is : CH 3 -CH 2 -O-CH 2 -CH 3 Using previously calculated values for log K ow (estimated at 1.0492) and log S (estimated at -0.6384) we can estimate the soil sorption coefficients using equations 4.16 and 4.17 : log K oc = 0.544 (log K ow ) + 1.377 = 1.9482 log K oc = -0.55 (log S) + 3.64 = 3.99

26. Using the molecular connectivity we can also estimate the soil sorption coefficient : First the molecular connectivity index is calculated using eq. 4.19 : CH 3 -CH 2 -O-CH 2 -CH 3 (molecular structure) 2(C-C), 2(C-O), 2(1, 2), 2(2, 2)(connection pairs) therefore : 1 χ = 2(1/√2) + 2(1/√4) = 2.414 Using equation 4.18 to calculate the soil sorption coefficient : log K oc = 0.53 1 χ + 0.62 + Σ n j P j log K oc = 0.53 1 χ + 0.62 + Σ n j P j = 0.53(2.414) + 0.62 + (-1.264) log K oc = 0.63542 therefore : K oc = 4.32 Example : Soil Sorption Coefficient Estimation

27. Where to look up this information... http://www.chem.duke.edu/~chemlib/properties.html http://www.library.vanderbilt.edu/science/property.htm http://www.library.yale.edu/science/help/chemphys.html

28. What do the different Properties mean? Adapted from the Green Engineering Textbook

29. Estimating Environmental Persistence and Ecosystem Risks To be discussed : – Atmospheric Lifetimes – Aquatic Lifetimes – Overall Biodegradability – Ecosystems

30. Estimating Atmospheric Lifetimes One way to estimate the atmospheric lifetime of a compound is to analyze the rate of oxidation of the substance, specifically the hydroxyl radical reaction rate. Group contributions is again one of the approaches that can be taken to estimate this property. Using examples, we will show how to estimate reaction rates and half lives while using the appropriate correction factors.

31. Example : Atmospheric Lifetime Estimation Dimethylsulfide (DMS, CH 3 SCH 3 ) produced by phytoplankton degredation is thought to be the major source of the sulfate and methanesulfonate aerosol found in the marine boundary layer. The primary objective of this research effort is to determine the detailed mechanism of, and final product yields from, the OH initiated gas phase oxidation of DMS. At the low NOx levels that are characteristic of the remote marine boundary layer, reaction with OH is the initial step in DMS oxidation. OH + CH 3 SCH 3 ⇒ Products (1)

32. The OH initiated oxidation of DMS proceeds via a complex, two channel, mechanism involving abstraction (1a) and reversible addition (1b, -1b). This can be described by the reaction sequence: CH 3 SCH 3 + OH ⇒ CH 3 SCH 2 + H 2 O (1a) CH 3 SCH 3 + OH + M ⇔ CH 3 S(OH)CH 3 + M (1b, -1b) CH 3 S(OH)CH 3 + O 2 ⇒ Products (3) Because of this complex mechanism the effective rate coefficients for reaction (1) and its deuterated analog, reaction (2) depend on the partial pressure of O 2 at any total pressure. OH + CD 3 SCD 3 ⇒ Products (2) The two channel reaction mechanism implies that in the absence of O 2 we measure k 1a, the abstraction rate. As we add O 2 the effective rate increases until we measure a limiting rate (k 1a + k 1b ).

34. Estimating Aquatic Lifetimes One way to estimate the aquatic lifetime of a compound is to analyze the rate of hydrolysis of the substance. The rate of hydrolysis can be estimated by : log (hydrolysis rate) = log (hydrolysis rate of a reference compound) + Constant * σ Therefore log (hydrolysis rate) = A + Bσ (4.20) Where : A is rxn and compound class specific(depends on the reference rxn chosen) B is rxn and compound class specific (depends on type of rxn considered) σ is a structural parameter commonly used in linear free energy relationship.

35. Estimating Overall Biodegradability It is difficult to do an overall biodegradability analysis. It can be estimated using : Where : a n is the contribution of the functional group (see table ). f n is the number of different functional group. MW is the molecular weight. I is an indicator of aerobic biodegradation rate. Different Values (of I) represent different life times : I = 3.199 + a 1 f 1 + a 2 f 2 + a 3 f 3 +... + a n f n + a m MW (4.21) I value54321 Expected degradation rateHoursDaysWeeksMonthsYears

36. Example : Overall Biodegradability Estimation Estimate the Biodegradation Index for diethyl ether. Solution : Molecular weight of diethyl ether : MW = 74.12 g/mol Using equation 4.21, the index can be calculated : I = 3.199 + a 1 f 1 + a 2 f 2 + a 3 f 3 +... + a n f n + a m MW I = 3.199 + (- 0.0087) - 0.00221(74.12) = 3.0267 Therefor a lifetime of WEEKS

37. Estimating Ecosystem Risks Compare the Fish, Guppy and Daphnids mortalities for an acrylate with log K ow = 1.22 (e.g. ethyl acrylate). Guppies log (1/LC 50 ) = 0.871 log K ow – 4.87 (4.22) log (1/LC 50 ) = 0.871(1.22) – 4.87 = -3.80738 LC 50 = 6417.74 µmol/L. Daphnids log LC 50 = 0.00886 – 0.51136 log K ow (4.23) log LC 50 = 0.00886 – 0.51136(1.22) = -0.6149992 LC 50 = 0.242 millimoles/L = 242 µmol/L.

38. Fish log LC 50 = -1.46 – 0.18 log K ow (4.24) log LC 50 = -1.46 – 0.18(1.22) = -1.6796 LC 50 = 0.021 millimoles/L = 21 µmol/L. The concentrations yielding 50% mortality are: Guppies (14 day): 6417.74 µmol/L. Daphnids (48 hour): 0.242 millimoles/L = 242 µmol/L. Fish (96 hour): 0.021 millimoles/L = 21 µmol/L. Estimating Ecosystem Risks Continued

39. Environmental Fate and Exposures Example : If chemicals are released into a river upstream of a water treament plant, what factors need to be taken into account to estimate the potential danger to the community. What fraction of the chemicals are: - Absorbed by river sediments.- Volatilized into the air. - Taken up by living organisms.- Biodegraded. - Reacted with other compounds.- Removed in the treatment process.

40. Classification of Substances Based on Risk By examining the table XX, we can use the calculated properties to qualitatively quantify the risk associated with the different substances Three main criteria are normally considered in the classification of the substances : persistence, bioaccumultion and toxicity. There do not exist a given set of regulations or guidelines on quantifying risk, but the above parameters are used in the process.

41. Available Ressources EPA (persistent, bioaccumulating and toxic substances) : http://www.epa.gov/pbt/aboutpbt.htm http://www.epa.gov/opptintr/pbt/ Pollution Prevention, Waste Minimization and PBT Chemical Reduction : http://yosemite.epa.gov/R10/OWCM.NSF/0d511e619f047e0d882565 00005bec99/6ad9c10eb8a06bc288256506007def78?opendocument Environment canada (existing substances evaluation) : http://www.ec.gc.ca/substances/ese/eng/psap/psap_2.cfm