Air-Water Gas Exchange of Chemicals

Air-Water Gas Exchange of Chemicals terry.bidleman@ec.gc.ca Centre for Atmospheric Research Experiments Science & Technology Branch, Environment Canada Photo: Thule Bay, Alaska Jim Milne, DRDC

Why? Air-Water Gas Exchange of Persistent Cl Organic Pollutants (POPs) Cl 71% of earth s surface is water. Gas phase important for many semivolatile chemicals terry.bidleman@ec.gc.ca Major pathway of transport for many compounds, especially in oceans and large lakes. Cl Cl Cl Cl α-hexachlorocyclohexane (α-hch) Cl Cl Cl Centre Photo: Lake for Atmospheric Superior Research Experiments Science & Technology Branch, Environment Canada Liisa Jantunen, Environ. Canada Cl Cl Cl

Henry s Law Constant H = p/c W P = partial pressure, Pa C W = water concentration, mol/m 3 K AW = C A /C W = air-water partition coefficient = H/RT Sometimes K AW is called the dimensionless Henry s Law constant, H

Getting the Henry s Law Constant 1. Calculate from saturation vapour pressure and solubility. Note: either liquid-phase or solid-phase properties can be used, because melting point corrections cancel, but don t mix! H = p o L /So L or p o S /So S 2. Calculate from octanol-water (K OW ) and octanol-air (K OA ) partition coefficients: gas phase K AW = K OW /K OA K AW K OA aqueous dissolved phase K OW organic dissolved phase

Getting the Henry s Law Constant 3. Measure directly; common method is bubble stripping. pre-saturator column air outlet Constant Temperature water bath air insulation bubbling chamber N 2 constant temperature water outlet

Gas stripping relationship V t lnc t /C o = (HG/RT)t V t lnc t /C o Slope = HG/RT C t and C o are solution concentrations at time = t and time = 0, hours. G = gas flow, m 3 /h V t = solution volume (m 3 ) at time = t time H = Pa m 3 /mol L. Jantunen & T. Bidleman, Chemosphere 62: 1689-1696 (2006); L. Sahsuvar et al., Atmos. Environ. 37: 983-992 (2003)

Problem with the gas stripping method: HLCs biased too high by adsorption to bubble-air interface More serious for highly hydrophobic (low solubility) chemicals Y.D. Lei et al., J. Chem. Eng. Data (2006)

Enhancement factors (EF) for alcohols of different carbon chain length (n c ). EF expresses the extent to which K AW values are elevated due to adsorption to the air-bubble interface. Y.D. Lei et al., J. Chem. Eng. Data (2006)

Temperature relationships for vapour pressure, solubility and Henry s law constant Ln p o L = ΔH vap /RT + b vap Ln S o L = ΔH sol /RT + b sol From: H = p o L /So L ΔH hlc = ΔH vap ΔH sol Ln H = ΔH hlc /RT + b hlc Ln p o L Ln S o L Ln H 1/T 1/T 1/T

The Henry s law constant decreases with temperature At constant partial pressure, gases are more soluble at lower temperatures. Apparent to anyone with a fish tank! Solubility of oxygen in water partial pressure = 0.21 atm (2.1 x 10 4 Pa) Temperature, o C Solubility, ml O 2 /L water 5 8.90 15 7.05 25 5.77

Salting Out (Setchenow Equation) Most organic chemicals are less soluble in salt water than in fresh water. Since H = p o /S o, adding salt raises H. Important consideration for gas exchange in oceans. Change in solubility is predictable from the Setchenow Equation Log(S o salt /So ) = K S C salt S o and S o salt are solubilities in pure water and salt solution of concentration C salt (M).

Salting Out (Setchenow Equation) Log(S o salt /So ) = K S C salt The Setchenow coefficient, K S varies with the chemical and the salt. For PCBs in seawater, K S = 0.2 0.3 What is S o salt /So? Seawater is ~0.5 M NaCl. Log(S o salt /So ) = 0.25(0.5) = -0.125 S o salt /So = 0.75, or a 25% decrease Accordingly, H should increase by 25%

Finding the Net Exchange Direction Gas exchange is a two-way street. Deposition (invasion) and volatilisation (evasion) occur simultaneously. Which pathway dominates (net exchange)? Using measured C W and C A, calculate (C W /C A ) meas Compare to C W /C A at equilibrium (1/K AW = RT/H) If (C W /C A ) meas = 1/K AW, no net exchange (equilibrium) If (C W /C A ) meas is >1/K AW, net volatilisation If (C W /C A ) meas is <1/K AW, net deposition

Finding the Net Exchange Direction The pesticide toxaphene was heavily applied in the southern U.S. during the 1960s and 70s. Use was stopped in 1986. During the usage years, toxaphene was atmospherically transported northward and deposited into Lake Superior, resulting in high concentrations in the water. Today, air concentrations of toxaphene have decreased and the pesticide is revolatilising from the lake. For details of this story, see papers by: D. Swackhamer et al., Environ. Sci. Technol. 33: 3864-3872 (1999) R. James et al., Environ. Sci. Technol. 35: 3653-3660 (2001) L. Jantunen & T. Bidleman, Environ. Toxicol. Chem. 22: 1229-1237(2003)

Finding the Net Exchange Direction Toxaphene concentrations in Lake Superior water and air in 1996 were 918 ng/m 3 (C W ) and 0.020 ng/m 3 (C A ). The Henry s law constant is 1,2 : Log H (Pa m 3 /mol) = 10.42 3209/T What is the net exchange direction of toxaphene at 22 o C? 1. (C W /C A ) meas = 918/0.02 = 4.6 x 10 4 2. Log H = 10.42 3209/295 = 0.458, H = 0.348 Pa m 3 /mol 3. 1/K AW = RT/H = (8.314 x 295)/0.348 = 7.0 x 10 3 4. Since (C W /C A ) meas is >1/K AW (step 1 > step 3) 5. Water is oversaturated, net volatilisation 1. Jantunen, L., Bidleman, T. Chemosphere, Global Change Science 2: 225-231 (2000); 2. Jantunen, L., Bidleman, T. Environ. Toxicol. Chem. 22: 1229-1237 (2003)

Gas Exchange Fugacity Approach 1-3 Fugacity (unit: Pa). For most environmental purposes consider fugacity synonymous with partial pressure. Describes the escaping tendency of the chemical. Systems in equilibrium have equal fugacities. Fugacity in air: f A = (n/v)rt = C A RT (ideal gas law) Fugacity in water: f W = C W H (recall that H = p/c W ) where C A and C W have units mol/m 3 1. Mackay, D. Environ. Sci. Technol. 13: 1218 (1979); 2. Mackay, D., Paterson, S. Ibid. 15: 1006-1014 (1981); 3. Ibid. 16: 654A-660A (1982).

Toxaphene Example Fugacity Approach f A = [(2.0 x 10-11 g/m 3 )/414 g/mol](8.314 Pa m 3 /mol K)(295 K) = 1.18 x 10-10 Pa f W = [(9.18 x 10-7 g/m 3 )/414 g/mol](0.348 Pa m 3 /mol) = = 7.72 x 10-10 Pa Since f W > f A, toxaphene is undergoing net volatilisation Another way to consider this is by calculating f W /f A f W /f A = C W H/C A RT In this example, f W /f A = 7.72 x 10-10 /1.18 x 10-10 = 6.5

Fugacity Decision Rules f W /f A > 1: net volatilisation f W /f A = 1: equilibrium (no net exchange) f W /f A < 1: net deposition In the previous example, we assumed the same air and water temperature (22 o C). What if the temperatures are different 1? f W /f A = C W H/C A RT A Henry s law constant at water temperature air temperature 1. See: Wania, F. et al., Environ. Pollut. 102: 3-23 (1998)

Summary of steps to calculate gas exchange direction 1. Measure C A and C W. C A is for the gas phase only and C W is for the dissolved phase only (no particles, colloids). Experimental challenges!

Summary of steps to calculate gas exchange direction 1. Measure C A and C W. C A is for the gas phase only and C W is for the dissolved phase only (no particles, colloids). Experimental challenges! 2. Calculate H (and K AW ) for the chemical at the water temperature. 3. Calculate (C W /C A ) meas 4. Compare with RT/H = 1/K AW. (C W /C A ) meas = 1/K AW : equilibrium (C W /C A ) meas > 1/K AW : net volatilisation (C A /C W ) meas < 1/K AW : net deposition

Two-Film Model of Gas Exchange Rates 1,2 C A bulk air: turbulence C A,S air film: diffusion Z A C W,S water film: diffusion Z W Equilibrium of surface concentrations: H/RT = C A.S /C W,S C W bulk water: turbulence 1. Whitman, W. Chem. Metal Eng. 29, 146-150 (1923) 2. Liss, P., Slater, P. Nature 247: 181-184 (1974)

Diffusion through the air and water thin films limits the gas exchange rate Fick s first law: Flux = k dc/dz Flux (mol/m 2 h) = mass transfer coefficient x conc. Gradient Concentration gradients are assumed to be linear Flux through the air (gas) film: F A = k G (C A C A,S ) Flux through the water (liquid) film: F W = k L (C W,S C W )

Diffusion through the air and water thin films limits the gas exchange rate Flux (mol/m 2 h) = mass transfer coefficient x conc. gradient Flux through the air (gas) film: F A = k G (C A C A,S ) Flux through the water (liquid) film: F W = k L (C W,S C W ) mass transfer coefficients for gas and liquid films These equations are set up such that deposition flux is positive (C A > C A,S and C W,S > C W ).

Flux through the air (gas) film: F A = k G (C A C A,S ) Flux through the water (liquid) film: F W = k L (C W,S C W ) The above equations are not useful in this form, because we don t know the interfacial concentrations. However, we assume that the interfacial concentrations are in equilibrium: H/RT = C A,S /C W,S Substituting this into the above equations, and assuming steady state (F A = F W ) leads to the next relationship:

Flux (F) = K OL (C A RT/H C W ) water concentration in equilibrium with air actual water concentration Checking units: K OL is m/h, C A and C W are mol/m 3 and RT/H is dimensionless. Thus, F has units mol/m 2 h To repeat, this equation is set up such that deposition is positive and volatilisation is negative. This is consistent with the convention used in the Great Lakes Integrated Atmospheric Deposition Network (IADN). Some papers define fluxes with the opposite sign (i.e., volatilisation positive) 1,2. 1. Bidleman, T., McConnell, L. Sci. Total Environ. 159: 101-117 (1995). 2. Wania, F. et al., Environ. Pollut. 102: 3-23 (1998).

Flux (F) = K OL (C A RT/H C W ) K OL is the overall mass transfer coefficient, which includes resistances to transfer in both the water and air films: 1/K OL = 1/k L + RT/Hk G overall resistance water film resistance air film resistance

Which resistance dominates? Depends on relative magnitude of k L, k G and H. 1/K OL = 1/k L + RT/Hk G Situation #1: Air film resistance dominates Lindane, H = 0.24 Pa m 3 /mol at 25 o C. Assume k G = 15 m/h and k L = 0.02 m/h (more later about k G and k L ). 1/K OL = 1/(0.02) + 8.31(298)/(0.24)(15) 1/K OL = 50 + 688 = 738 water film resistance (7%) air film resistance (93%)

Here, diffusion across the air film limits lindane exchange and there is essentially no concentration gradient across the water film (C w = C W,S ) C A bulk air: turbulence C A,S Equilibrium of surface concentrations: H/RT = C A.S /C W,S C W,S C W air film: diffusion bulk water: turbulence

Which resistance dominates? Depends on relative magnitude of k L, k G and H. 1/K OL = 1/k L + RT/Hk G Situation #2: Water film resistance dominates Same calculations for PCB-52. Again, assume k G = 15 m/h and k L = 0.02 m/h. H = 32 Pa m 3 /mol at 25 o C 1/K OL = 1/(0.02) + 8.31(298)/(32)(15) 1/K OL = 50 + 5.2 = 55.2 water film resistance (91%) air film resistance (9%)

Here, diffusion across the water film limits PCB-52 exchange and there is essentially no concentration gradient across the air film (C A = C A,S ) C A C A,S Equilibrium of surface concentrations: H/RT = C A.S /C W,S bulk air: turbulence C W,S water film: diffusion C W bulk water: turbulence

Which resistance dominates? Depends on relative magnitude of k L, k G and H. 1/K OL = 1/k L + RT/Hk G Intermediate situation: both films contribute Same calculations for p,p -DDT. Again, assume k G = 15 m/h and k L = 0.02 m/h. H = 1.1 Pa m 3 /mol at 25 o C 1/K OL = 1/(0.02) + 8.31(298)/(1.1)(15) 1/K OL = 50 + 150 = 200 water film resistance (25%) air film resistance (75%)

More about k L and k G k = D/Z D = molecular diffusivity; Z = film thickness Typically, k L and k G are measured for light gases, such as water vapour, oxygen and carbon dioxide. Adjustments of these k-values for higher molecular weight and lower diffusivity molecules (like PCBs, pesticides) are made using relative diffusivities and Schmidt numbers. D, cm 2 /s Sc medium H 2 0 0.256 0.60 air CO 2 0.164 0.94 air CO 2 1.77 x 10-5 559 water PCB52 0.052 2.94 air PCB52 4.27 x 10-6 2665 water

What s a Schmidt Number? Sc = ν/d D = molecular diffusivity of the chemical, cm 2 /s ν = kinematic viscosity of water = viscosity/density = (g/cm s)/(g/cm 3 ) = cm 2 /s So, ν/d is dimensionless There are models for calculating Schmidt numbers and diffusivities in air and water. See references in papers dealing with gas exchange.

More about k L and k G Mass transfer coefficients are influenced strongly by wind speed. Many experimental and theoretical relationships have been given which relate k-values to wind speed, diffusivity and Schmidt number (Sc) 1-3. Here are two: k L (cm/h) = (0.45U 10 1.64 )(Sc X /Sc carb. diox. ) -0.5 k g (cm/s) = (0.2U 10 + 0.3)(D X /D water vap. ) 0.61 Note the different units for k L and k G, which is the way they were published! The units of U 10 in these equations are m/s and D is cm 2 /s. If you want flux in mol/m 2 h, need K OL in m/h. Multiply above k L by 10-2 m/cm and k G by 10-2 m/cm x 3600 s/h. 1. Bidleman, T., McConnell, L. Sci. Total Environ. 159: 101-117 (1995) 2. Wania, F. et al., Environ. Pollut. 102: 3-23 (1998). 3. Hornbuckle, K. et al., Environ. Sci. Technol. 28: 1491-1504 (1994)

k L (cm/h) = (0.45U 10 1.64 )(Sc X /Sc carb. diox. ) -0.5 k g (cm/s) = (0.2U 10 + 0.3)(D X /D water vap. ) 0.61 From the previous table: Sc values in water: 2665 for PCB52 and 559 for CO 2. D values in air: 0.052 for PCB52 and 0.256 for water vapour. For U 10 = 4 m/s (14 km/h): k L = 2 cm/h, k G = 0.42 cm/s In common units: k L = 0.02 m/h, k G = 15 m/h

More about k L and k G Others use different relationships for k L and k G. These are based on molar volumes, or atomic diffusion volumes, of the chemicals instead of D and Sc. Results not much different from the previous equations. For details, see the following references: 1. Jantunen, L., Bidleman, T. Environ. Toxicol. Chem. 22: 1229-1237 (2003) 2. Hornbuckle, K. et al., Environ. Sci. Technol. 28: 1491-1504 (1994)

Calculating flux using fugacity terminology Flux = D OL (f A -f w ) D OL = K OL /H Checking units: K OL has units m/h D OL has units (m/h)/(pa m 3 /mol) = mol/pa m 2 h When multiplied by fugacity (Pa), the result is flux in mol/m 2 h

Completing the gas exchange calculation for PCB-52 Assume C W = 0.5 pg/l (= 500 pg/m 3 ) and C A = 2 pg/m 3 Assume lake temperature = 15 o C K AW = H/RT = 8.36 x 10-3 RT/H = 120 MW = 292 (in case you want flux in moles/m 2 h) K L and K G are 0.02 and 15 m/s K OL = 1/k L + RT/Hk G 1/K OL = 1/0.02 + 120/15 1/K OL = 50 + 8 = 58 (note mainly liquid-phase resistance) K OL = 0.017 m/s F net = K OL (C A RT/H C W ) F net = 0.017(2*120 500) = 4.4 pg/m 2 h volatilisation

Gas exchange calculation for PCB-52 fugacity approach C W = 500 pg/m 3 = 1.71 x 10-12 mol/m 3 C A = 2 pg/m 3 = 6.85 x 10-15 mol/m 3 K AW = H/RT = 8.36 x 10-3, H = 20 Pa m 3 /mol MW = 292 K L and K G are 0.02 and 15 m/s D OL = K OL /H = 0.017/20 = 8.4 x 10-4 mol/pa m 2 h f W = C W H = 1.71 x 10-12 (20) = 3.42 x 10-11 Pa f A = C A RT = 6.85 x 10-15 (8.31)(288) = 1.64 x 10-11 Pa F net = D OL (f A -f W ) F net = 8.4 x 10-4 (1.64 x 10-11 3.42 x 10-11 ) = 1.49 x 10-14 mol/m 2 h F net = 1.49 x 10-14 (292 g/mol)(10 12 pg/g) = 4.4 pg/m 2 h volatilisation

Summary of steps to calculate gas exchange flux 1. Calculate k L and k G as functions of wind speed (U 10 ), diffusivity (D), Schmidt number (Sc) or other relationships. Note that both D and Sc are functions of temperature. 2. Calculate K OL (or D OL ) using: 1/K OL = 1/k L + RT/Hk G D OL = K OL /H

You can also calculate individual air and water fluxes by simply using the two parts of the flux equation: F dep = K OL (C A RT/H) F vol = K OL ( C W ) Similarly for the equation in fugacity format. 8.5 pg/m 2 h volatilisation 4.1 pg/m 2 h deposition 4.4 pg/m 2 h net Why might you want to do this?

Net flux of 4.4 pg/m 2 h doesn t seem like much Lake Superior is huge! surface area is 82,100 km 2 = 8.21 x 10 10 m 2 4.4 pg/m 2 h x 8760 h/y x 8.21 x 10 10 m 2 x 10-15 kg/pg = 3.1 kg/y PCB-52 is only one of many PCBs in the lake. In 1992, Hornbuckle et al. 1 estimated an annual average net flux of 345 pg/m 2 h for total PCBs from Lake Superior. Total net volatilisation of 248 kg annually! 1. Hornbuckle, K. et al., Environ. Sci. Technol. 28: 1491-1501 (1994)