Let’s look at mole fractionChapter 4 Vapor Pressure An important goal of this chapter is to learn techniques to calculate vapor pressures To do this we will need boiling points and entropies of vaporization The saturation vapor pressure is defined as the gas phase pressure in equilibrium with a pure solid or liquid. • fi = γi XiPiopure liquid (old book) • fi = γi X ipi* pure liquid (new book) • log logCCTSPPconspartlo=− + tgas • KiH = P*iL/Ciwsat 1Vapor pressure and Temperature dGliq = dGgas from the 1st law H= U+PV dH = dU + VdP+PdV dU= dq - dw for only PdV work, dw = PdV and from the definition of entropy, dq = TdS dU = TdS -PdV from the general expression of free energy dG = dU + VdP+PdV- SdT-TdS substituting for dU dG= +VdP - SdT 2The molar free energy Gi/ni = μifor a gas in equilibrium with a liquid dμliq = dμgas dμliq = VliqdP - SliqdT VliqdP - SliqdT = VgasdP - SgasdT dP/dT = (Sgas -Sliq)/Vgasat equilibrium ΔG = ΔH -ΔS T= zero so (Sgas -Sliq) = ΔHvap/T substituting dP/dT = ΔH/( Vgas T) (Clapeyron eq) substituting Vgas = RT/Po dPdTHPRTxTdP dTHRo==−ΔΔ;(ln) ()1 lnPHRTconsto=− +Δ 1 3Figure 4.3 page 61 Schwartzenbach 45This works over a limited temperature range w/o any phase change Over a larger range Antoine’s equation may be used ACTBln ++=*ip over the limits P1 to P2 and T1 to T2 log().PPvapHTTRT T2121122303=−Δ If the molar heat of vaporization, ΔHvap of hexane equals 6896 cal/mol and its boiling point is 69oC, what is its vapor pressure at 60oC log/().. /760 6896 333 3242 303 199 333 3241Pcal mol K Kcal Kmol K K=−∗∗∗ P1= 395 mm Hg 6Below the melting point a solid vaporizes w/o melting, that is it sublimes A subcooled liquid is one that exists below its melting point. • We often use pure liquids as the reference state • logKp Log p*i 7Molecular interaction governing vapor pressure As intermolecular attractive forces increase in a liquid, vapor pressures tend to decrease van der Waals forces generally enthalpies of vaporization increase with increasing polarity of the molecule 8A constant entropy of vaporization Troutons rule at the boiling point ΔG = ΔH-ΔSxT = zero ΔH const slope = ΔS T ΔH/T = ΔS= const = 88J mol-1K-1 = 21 cal mol-1K-1 Kistiakowsky derived an expression for the entropy of vaporization which takes into account van der Waal forces • ΔSvap= 36.6 +8.31 ln Tb • for polarity interactions Fistine proposed ΔSvap= Kf (36.6 +8.31 ln Tb) Kf= 1.04; esters, ketones Kf= 1.1; amines Kf= 1.15; phenols Kf= 1.3; aliphatic alcohols 9Calculating ΔSvap using chain flexibility and functionality ΔvapSi (Tb) = 86.0+ 0.04 τ + 1421 HBN τ = Σ(SP3 +0.5 SP2 +0.5 ring) -1 SP3 = non-terminal atoms bonded to 4 other atoms (unbonded electrons of O, etc are considered a bond) SP2 = non-terminal atom bonded to two there atoms and doubly bonded to a 3rd atom Rings = # independent rings HBN = is the hydrogen bond number as a function of the number of OH, COOH, and NH2 groups MWNH.33COOHOHHBN2++= 101112A 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 ΔSb= 84.53 – 11σ +.35τ + 0.05ω2 + ΣΧi where: Χi = the contribution of group i to the Entropy of boiling ω = the molecular planarity number, or the # of non-hydrogen atoms of a molecule that are restricted to 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 τ measures the conformational freedom or flexibility ability of atoms in a molecule to rotate about single bonds τ= SP3 + 0.5(SP2) +0.5 (ring) –1 1314σ = 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, etc15Boiling points can be estimated based on chemical structure (Joback, 1984) Tb= 198 + Σ ΔTb ΔT (oK) -CH3 = 23.58 K -Cl = 38.13 -NH2 = 73.23 C=O = 76.75 CbenzH- = 26.73 Joback obs (K) (K) acetonitrile 347 355 acetone 322 329 benzene 358 353 amino benzene 435 457 benzoic acid 532 522 toluene 386 384 pentane 314 309 methyl amine 295 267 trichlorethylene 361 360 phenanthrene 598 613 Stein, S.E., Brown, R.L. Estimating Normal boiling Points from Group Contributions, J.Chem. Inf Comput. Sci, 34, 581-587, 1994 16They start with Tb= 198 + Σ ΔTb and go to 4426 experimental boiling points in Aldrich And fit the residuals (Tbobs-Tb calcd) Tb= 198 + Σ ΔTb 17Tb(corr) = Tb- 94.84+ 0.5577Tb- 0.0007705Tb2 T b< 700 K Tb(corr) = Tb+282.7-0.5209Tb Tb>700K 18Estimating Vapor Pressures dPdTHRTovapTln=Δ2 To estimate the vapor pressure at a temp lower then the boiling temp of the liquid we need to estimate ΔHvap at lower temperatures. Assume that ΔHvap is directly proportional to temp and that ΔHvap can be related to a constant the heat capacity of vaporization ΔCp Tb where ΔHvap/ΔT = ΔCp Tb ΔHvapT = ΔHvap Tb + ΔCp Tb(T-TTb) )TbT(lnRpTbΔC)TbT(1RbpTΔC)T1bT1(RbvapTΔHln −−−−=*iLP at the boiling point ΔHvap Tb= Tb ΔSvap Tb )()(( ln)ln*TTRCTTRCRSpbpTbbpTvapTbbiLΔ−−Δ−Δ= 1 19)())(( lnln*TTRCTTRCRSpbpTbbpTvapTiLbbΔ−−Δ−Δ= 1 for many organic compounds ΔCp Tb/ ΔSvap Tb ranges from -0.6 to -1 so substituting ΔCp Tb=0.8 ΔSvap Tb )]()([ ln..*lnTTTTRSpbbvapTiLb80181 +−Δ= if we substitute ΔSvap Tb=88J mol-1 K-1 and R =8.31 Jmol-1 K-1 )]()( ln.ln*TTTTpbbiL58119 +−= when using ΔCp Tb/ ΔSvap Tb = -O.8, low boiling compounds (100oC) are estimated to with in 5%, but high boilers may be a factor of two off If the influence of van der Waal forces (Kistiakowky)and polar and hydrogen bonding effects (Fishtine’s correction factors) are applied ΔSvap Tb= Kf(36.6 +8.31 ln Tb) )]()([ ln..)ln.(ln*TTTTbTfKpbbiL8018144 −−+=− 20If we go back to: )())(( lnln*TTRCTTRCRSpbpTbbpTvapTiLbbΔ−−Δ−Δ= 1 ΔvapSi (Tb) = 86.0+ 0.04 τ + 1421 HBN and Mydral and Yalkowsky suggest that ΔvapCpi (Tb) = -90 +2.1τ in J mol-1K-1