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Chapter 4 Vapor Pressure pº = Pressure of a substance in equilibrium with its pure condensed (liquid or solid) phase Why do we care? -spills -pesticide application -will lead us to Henry’s law constant KH = PoL/Csatw Kow = Csato/Csatw Air Koa = Csato/PoL A gas is a gas is a gas T, P KH Koa Po L Octanol Water NOM, biological lipids, other solvents Fresh, salt, ground, pore Kow T, chemical composition T, salinity, cosolvents Pure Phase Csato Csat w (l) or (s) Ideal behavior Ranges of pº (atm) – PCBs – 10-5 to 10-9 – n-alkanes – 100.2 to 10-16 • n-C10H22 ~ 10-2.5 • n-C20H42 ~ 10-9 – Benzene ~ 10-0.9 – toluene ~10-1.42 – Ethylbenzene ~ 10-1.90 – propyl benzene ~ 10-2.35 – carbon tetrachloride ~ 10-0.85 – methane 102.44 • Even though VP is “low”, gas phase may still be important. Phase diagram picture of three-phase diagram Ideal Gas Law pV nRT p = pressure V = volume n = moles of gas R = gas constant T = temperature (Kelvin) Thermodynamic considerations (deriving the van’t Hoff equation) consider a gas: if T or P is changed and equilibrium is re-established: d1 d 2 the change in chemical potential in the two systems is equal d1 S1dT V1dp where S = molar entropy d2 S 2 dT V2 dp and V = molar volume dp ( S1 S 2 ) S12 dT (V1 V2 ) V12 at equilibrium G12 1 2 0 G H TS substituting: dp H 12 dT TV12 for a liquid vaporizing, the volume change can be assumed to be equal to the volume of gas produced, since the volume of the solid or liquid is negligible RT Q. where did the n go? V12 Vgas 0 A. this is molar volume p dp 0 p 0 (H12 ) recall (calculus!) 2 d ln u 1 du dT RT d ln p 0 H12 The van’t dx u dx Hoff equation dT RT 2 where H12 = Hvap (gas) or Hsub (solid) = energy required to convert one mole of liquid (or solid) to gas without an increase in T. Hvap is a function of T. As T approaches the boiling point, Hvap increases rapidly At T < boiling point, Hvap increases slowly from 0-40ºC, Hvap can be assumed to be constant d ln p 0 H12 dT RT 2 integrate assuming Hvap is constant: Antoine equation A ln p B 0 T H12 ln p 0 a RT if Hvap is not constant: b ln p 0 a T c another Antoine equation Using Hvap to predict VP at other temperatures As we saw in the thermodynamics lecture: KT 2 H 1 1 ln T T KT 1 R 1 2 Specifically, pT 2 H vap 1 1 ln pT 1 R T1 T2 Note the change in slope when the substance is solid (sublimation) Hsub = Hmelt (~25%) + Hvap (~75%) still use liquid phase as reference: Hypothetical subcooled liquid = liquid cooled below melting point without crystallizing compound pºs < pºL 1,4-dichlorobenzene 3.04 2.76 -log P phenol 3.59 3.41 22’55’ PCB 7.60 6.64 22’455’ PCB 8.02 7.40 Becomes important later when we talk about solubility Molecular interactions affecting vapor pressure Molecule:molecule interactions in condensed phase (L or s) have greatest affect on VP strong interactions lead to large Hvap, low VP weak interactions lead to small Hvap, high VP Intermolecular interactions can be classified into three types: van der Waals forces (nonpolar) Polar forces Hydrogen bonding van der Waals forces • nonspecific • function of size (number of electrons) • consist of: – London dispersive energies • fleeting areas of charge – Induced dipoles “nonpolar” • areas of charge arising from interactions with a polar molecule Polar interactions: dipole-dipole interactions • permanent areas of charge on two molecules attract Hydrogen bonds Specific donors and acceptors table 4.3 Part of Table 4.3 compound (class) (H-donor) (H-acceptor) alkanes 0 0 1-alkenes 0 0.07 aliphatic ethers 0 0.45 aliphatic aldehydes 0 0.45 aliphatic alcohols 0.37 0.48 carboxylic acids 0.60 0.45 benzene 0 0.14 phenol 0.6 0.31 naphthalane 0 0.2 fluorene 0 0.2 pyrene 0 0.29 DCM 0.1 0.05 Water 0.82 0.35 Vapor Pressure Estimation Technique based on regression of lots of VP data, best fit gives: n 1 2 2 2/3 ln piL 4.49 V iL n 2 15.1( i )( i ) 14.5 * Di 2 Di size H-bonding polarizability ability pressure in Pa, where: V iL molar volume (MW/density) nDi refractiveindex refractive index (response to light) is a function of polarizability. see table 3.1, also might be available in the CRC Refractive index Difference between polarity and polarizability Trouton’s rule At their boiling points, most organic compounds have a similar entropy of vaporization: exception: strongly polar or H-bonding Svap (Tb) = 85 – 90 J/molK compounds We can be slightly more accurate with Kistiakowsky’s expression: Svap (Tb) = KF(36.6 + 8.31ln(Tb)) J/molK Tb in K (eqn 4-20) KF = 1 for most compounds At the boiling point: G 0 H vap Tb S vap So if we know Tb, we can estimate Hvap (at the boiling point) fairly accurately Table 4.2 Estimating VP at other T (need Hvap) pT 2 H vap 1 1 ln pT 1 R T1 T2 Recognize that Hvap is not constant. Especially if Tb is high (> 100ºC), the estimate of Hvap from Trouton/Kistiakowsky may not be valid at the temperature of interest. Empirically, Hvap is a function of the VP: Hvap (T1 ) a log piL (T1 ) b * FIG 4.7 From a data set of many compounds, Goss and Schwarzenbach (1999) get: Hvap (298K ) 8.80 log piL (298K ) 70.0 * Less empirically, assume Hvap is linearly proportional to T (i.e. assume that the heat capacity, Cpvap is constant: H vap (T ) H vap (Tb ) C p vap (Tb ) (Tb T ) don’t let notation confuse you. (Tb) means at the boiling point. You do not substitute this expression into the Clausius- multiply Hvap by Clapeyron equation and integrate from Tb to T: the boiling point H vap (Tb ) 1 1 ln P 0 T T R b C p vap (Tb ) T C p vap (Tb ) T 1 b ln b R T R T Recall: H vap (Tb ) Tb S vap (Tb ) substitute: S vap (Tb ) C p vap (Tb ) Tb ln p 0 1 R R T C p vap (Tb ) T ln b R T but we still need to know Cp(Tb)! generally: C p (Tb ) 0.8 S vap (Tb ) ranges from 1.0 to 0.6 and Svap(Tb)~ 88 J/molK finally! Tb Tb ln p K F (4.4 ln Tb ) 1.8 1 0.8 ln 0 T T KF is the Fishtine factor, usually 1, but in atm sometimes as high as 1.3 (see p 113) Eqn 4-33 the old edition gave (where KF =1): Tb Tb ln p 19 1 8.5 ln 0 in atm T T OK for liquids with Tb < 100ºC High MW compounds, need correction for intermolecular forces (but we don’t have their boiling points anyway!) (For refinements see equation 4-33) Can estimate boiling points, see p. 120 solids? those previous equations yielded the vapor pressure of the hypothetical subcooled liquid. How can we correct this to give the true vapor pressure of a solid? Prausnitz (1969): S fus (Tm ) Tm ps0 ln 0 1 pL R T Where Sfus(Tm) = entropy of fusion at melting point unfortunately Sfus is much more variable than Svap S fus (Tm ) (56 .5 9.2 19 .2 log ) J/molK Where = number of torsional bonds and = rotational symmetry number (see p. 125) the older edition of your book gave this simpler (but less accurate) equation: Sfus(Tm) ~ 56.5 + 10.5(n-5) J/molK Where n = number of flexing chain atoms. if n<5, then ignore this term Estimation of vapor pressures for polychlorinated biphenyls: a comparison of eleven predictive methods Lawrence P. Burkhard, Anders W. Andren, and David E. Armstrong Environmental Science and Technology 1985, 19, 500 - 507 conclusions: • non-correlative methods have poor predictive ability (error increases as VP decreases) • correlative methods requiring a set of compounds with known P are much better • best method: determine VP as function of GC retention times Determination of vapor pressures for nonpolar and semipolar organic compounds from GC retention data (Hinckley et al, 1990) • Chromatographed 2 reference compounds (eicosane and p,p’DDT) having known VP and Hvap versus a host of unknowns (PAHs, organochlorines, etc) • Isothermal runs allow determination of RRT at several T • Comparison of RRT with reference compounds allows determination of VP at given T • Comparison of changes in RRT with T and knowledge of Hvap for reference compound allows calculation of Hvap for all unknowns Problem 4.2 • In a dump site, you find an old 3-liter pressure bottle containing FREON 12 with a pressure gauge that reads 2.7 bar. (First, you realize that this gauge was not manufactured in the US.) The temperature is 10ºC. What mass of FREON 12 is in the bottle? • Also estimate the free energy, enthalpy, and entropy of condensation of FREON 12. • You find the following info for FREON 12 in the CRC: T deg C p/kPa -25 123 0 308 25 651 50 1216 75 2076 Problem 4.6 • estimate VP at 0C based on VP at 25ºC or based solely on Tb and Tm log(p) @25C Tm (degC) Tb (degC) dimethyl phthalate 0.38 5.5 283.7 2,3,7,8-TCDD -6.7 305 446.5 (hint = 4) Homework • Do problems 4.3 and 4.4 • Due 2/2/10

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posted: | 7/2/2011 |

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