Let’s look at the mole fractionChapter 4 Vapor PressureAn important goal of this chapter is to learn techniques to calculate vapor pressuresTo do this we will need boiling points and entropies of vaporizationThe saturation vapor pressure is defined as the gas phase pressure in equilibrium with a pure solid or liquid.-fi = -i X ipiL* pure liquid -constplogTSPCClog*iLgaspart-KiH = p*iL/Ciwsat1Vapor pressure and TemperaturedGliq = dGgasfrom the 1st law H= U+PVdH = dU + Vdp+pdVdU= dq - dwfor only pdV work, dw = pdV and from the definition of entropy, dq = TdSdU = TdS –pdVfrom the general expression of free energydG = dU + Vdp+pdV- SdT-TdSsubstituting for dUdG= +VdP - SdT2The molar free energy Gi/ni = -ifor a gas in equilibrium with a liquidd-liq = d-gasd-i liq = Vi liqdp - Si liqdTVi liqdpi - Si liqdT = Vigasdp - SigasdTd/dT = (Sigas -Si liq)/Vigasat equilibrium -G = -H --S T= zeroso (Sigas -Si liq) = -Hivap/T substitutingdpi /dT = -Hi /( Vigas T) (Clapeyron eq)substituting Vi gas = RT/piRH)T(d)p(lnd;RTxTpHdTdpiiiii1constTRHplnioi13Figure 4.3 page 61 Schwartzenbach45This works over a limited temperature range w/o any phase changeOver a larger range Antoine’s equation may be usedACTBln *ipover the limits P1 to P2 and T1 to T2log( ).PPvapH T TRT T212 11 22 303If the molar heat of vaporization, -Hvap of hexane equals 6896 cal/mol and its boiling point is 69oC, what is its vapor pressure at 60oCKKKmolcalKKmolcalP 342333991303233334268967601/..)(/logP1= 578 mm Hg6Below the melting point a solid vaporizes w/o melting,that is it sublimesA subcooled liquid is one that exists below its melting point.-We often use pure liquids as the reference state-logKpLog p*i 7Molecular interaction governing vapor pressureAs intermolecular attractive forces increase in a liquid, vapor pressures tend to decreasevan der Waals forcesgenerally enthalpies of vaporizationincrease with increasing polarity of the moleculeBoth boiling points and entropies of vaporization become important parameters in estimating vapor pressures8A constant entropy of vaporization Troutons ruleFigure 4.5; at 25oCThis suggests that -vapS may tend to be constantAt the boiling point is vapSTb constant?9-Hconst slope = -STTb oC -vapH -vapS kJ mol-1 kJ mol-1K-n-hexane 68.7 28.9n-decane 174.1 38.8ethanol 78.3 38.6naphthalene 218 43.7Phenanthrene 339 53.0Benzene 80.1 30.7Chlorobenzene 131.7 35.2Hydroxybenzene 181.8 45.710Predicting vapSTbKistiakowsky derived an expression for the entropy ofvaporization which takes into account van der Waal forces-Svap= 36.6 +8.31 ln Tb (eq 4-20)-for polarity interactions Fistine proposed Svap= Kf (36.6 +8.31 ln Tb) Kf= 1.04; esters, ketones Kf= 1.1; amines Kf= 1.15; phenols Kf= 1.3; aliphatic alcoholsCalculating Svap using chain flexibility and functionality (Mydral et al, 1996)vapSi (Tb) = 86.0+ 0.04  + 1421 HBN(eq 4-21) - = -(SP3 +0.5 SP2 +0.5 ring) -111SP3 = non-terminal atoms bonded to 4 other atoms (unbonded electrons of O, NH, N, S, are considered a bond)SP2 = non-terminal atoms singly bonded to two other atoms and doubly bonded to a 3rd atomRings = # independent ringsHBN = is the hydrogen bond number as a function of the number of OH, COOH, and NH2 groupsMWNH.33COOHOHHBN21213A more complicated method:From Zhao, H.; Li, P.; Yalkowski, H.; Predicting the Entropy of boiling for Organic Compounds, J. Chem. Inf. Comput. Sci, 39,1112-1116, 1999Sb= 84.53 – 11 +.35 + 0.052 + iwhere: i = the contribution of group i to the Entropy of boiling= the molecular planarity number, or the # of non-hydrogen atoms of a molecule that are restrictedto a single plane; methane and ethane have values of 1 and 2; other alkanes, 3; butadiene, benzenes, styrene, naphthalene, and anthracene are 4,6,8,10,14 measures the conformational freedom or flexibility ability of atoms in a molecule to rotate about single bonds-= SP3 + 0.5(SP2) +0.5 (ring) –114- = symmetry number; the number of identical images that can be produced by a rigid rotation of a hydrogen suppressed molecule; always greater than one; toluene and o-xylene = 2, chloroform and methanol =3, p-xylene and naphthalene = 4, etc1517Boiling points can be estimated based on chemical structure (Joback, 1984)Tb= 198 + ---Tb-T (oK)-CH3= 23.58 K-Cl = 38.13-NH2 = 73.23C=O = 76.75CbenzH- = 26.73Joback obs(K) (K)acetonitrile 347 355acetone 322 329benzene 358 353amino benzene 435 457benzoic acid 532 522toluene 386 384pentane 314 309methyl amine 295 267trichlorethylene 361 360phenanthrene 598 613Stein, S.E., Brown, R.L. Estimating Normal boiling Points from Group Contributions, J.Chem. Inf Comput. Sci, 34, 581-587, 199418They start with Tb= 198 + Tb and go to 4426 experimental boiling points in AldrichAnd fit the residuals (Tb obs-Tb calcd)Tb= 198 + Tb19Tb(corr) = Tb- 94.84+ 0.5577Tb- 0.0007705Tb2 T b< 700 K Tb(corr) = Tb+282.7-0.5209Tb Tb>700K20Estimating Vapor Pressures2RTHdTplndvap*To estimate the vapor pressure at a temp lower then the boiling temp of the liquid we need to estimate -Hvap at lower temperatures.Assume that -Hvap is directly proportional to temp and that -Hvap can be related to a constant, the heat capacity of vaporization -Cp Tbwhere -vapH/-T = -Cp Tb-vapHT = -vapH Tb + -CpTb(T-TTb))TbT(lnRpTbΔC)TbT(1RbpTΔC)T1bT1(RbTHvapΔ*iLPln at the boiling point -vapH Tb= Tb -vapS Tb)()((TTlnRC)TTRCRSplnbpTbbpTTvap*bbiL 121)())((TTlnRCTTRCRSplnbpTbbpTTvap*iLbb 1for many organic compounds -Cp Tb/--vapS Tb ranges from -0.6 to -1so substituting -Cp Tb= -0.8-vap-S Tb)]()([TTln.TT.RS*plnbbvapTiLb80181 if we substitute -Svap Tb= 88J mol-1 K-1 and R =8.31 Jmol-1 K-1)]()( ln.ln*TTTTpbbiL58119 when using -Cp Tb/--Svap Tb = -O.8, low boiling compounds (100oC) are estimated to with in 5%, but high boilers may be a factor of two offIf the influence of van der Waal forces (Kistiakowky)and polar and hydrogen bonding effects (Fishtine’s correction factors) are appliedvapS Tb= Kf(36.6 +8.31 ln Tb)22)]()([ ln..)ln.(ln*TTTTbTfKpbbiL8018144 If we go back to:)())((TTlnRCTTRCRSplnbpTbbpTTvap*iLbb 1vapSi (Tb) = 86.0+ 0.04  + 1421 HBNand Mydral and Yalkowsky suggest that vapCpi (Tb) = -90 +2.1 in J mol-1K-1- = -(SP3 +0.5 SP2 +0.5 ring) –1