Chapter 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 =

o γi XiPi pure liquid (old book)

• fi =

γi X

• log

* ipi pure liquid (new book)

C part = − log Plo + cons t C gas TSP

* sat • KiH = P iL/Ciw

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Vapor 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

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The molar free energy Gi/ni = μi for a gas in equilibrium with a liquid dμliq = dμgas dμliq = VliqdP - SliqdT VliqdP - SliqdT = VgasdP - SgasdT dP/dT = (Sgas -Sliq)/Vgas at equilibrium ΔG = ΔH -ΔS T= zero so (Sgas -Sliq) = ΔHvap/T substituting dP/dT = ΔH/( Vgas T) (Clapeyron eq) substituting

dP ΔHPo = ; dT RTxT

Vgas = RT/Po

d(ln P) = − d( 1T)

lnPo = −

ΔH R

ΔH 1 + const R T

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Figure 4.3 page 61 Schwartzenbach

4

5

This works over a limited temperature range w/o any phase change Over a larger range Antoine’s equation may be used

lnp*i =

B +A T+C

over the limits P1 to P2 and T1 to T2 P log P2 1

=

ΔHvap ( T2 − T1) 2.303RT1T2

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 cal / mol (333K − 324K ) = 2.303∗199 . cal / Kmol∗333K∗324K P1

P1= 395 mm Hg

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Below 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

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Molecular 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

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A constant entropy of vaporization Troutons rule at the boiling point ΔG = ΔH-ΔSxT = zero

ΔH const slope = ΔS

T -1 -1 ΔH/T = ΔS= const = 88J mol-1K = 21 cal mol-1K

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 9

Calculating 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

HBN =

OH + COOH + .33 NH2 MW

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11

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A 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

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σ = 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, etc

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Boiling 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

acetonitrile acetone benzene amino benzene benzoic acid toluene pentane methyl amine trichlorethylene phenanthrene

Joback (K) 347 322 358 435 532 386

obs (K) 355 329 353 457 522 384

314 295 361

309 267 360

598

613

Stein, S.E., Brown, R.L. Estimating Normal boiling Points from Group Contributions, J.Chem. Inf Comput. Sci, 34, 581-587, 1994

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They 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 17

Tb(corr) = Tb- 94.84+ 0.5577Tb0.0007705Tb2 T b< 700 K Tb(corr) = Tb+282.7-0.5209Tb Tb>700K

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Estimating Vapor Pressures d ln P o ΔHvapT = dT RT 2 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) * = lnPiL

ΔHvapT ΔCpT ΔCpTb Tb Tb 1 1 b( b − )− (1 − )− (ln ) R Tb T R T R T

ΔHvap Tb= Tb ΔSvap Tb

at the boiling point

ln p iL =( *

ΔS vapTb R

−

ΔC pTb R

ΔC pTb Tb Tb )(1 − ) − (ln ) T T R

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ΔS vapTb

ΔC pTb

ΔC pTb Tb =( − )(1 − ) − (ln Tb ) R R R T T for many organic compounds * ln piL

ΔCp Tb/ ΔSvap Tb ranges from -0.6 to -1 so substituting ΔCp Tb=0.8 ΔSvap Tb

ln p *iL =

ΔSvapTb R

[1.8(1 − Tb ) +0.8(ln Tb )] T

T

if we substitute ΔSvap Tb=88J mol-1 K-1 and R =8.31 Jmol-1 K-1

ln

* piL

Tb Tb =19(1 − ) +8.5(ln )] T T

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 piL =− K f ( 4.4 + ln Tb )[1.8(

Tb T − 1) −0.8(ln b )] T T

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If we go back to: * ln piL

=(

ΔS vapTb R

−

ΔC pTb R

ΔC pTb Tb Tb )(1 − ) − (ln ) T T R

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

* =− (21.2 + 0.3τ +177 HBN )(1 − Tb ) ln piL T

Tb + (10.8 + 0.25τ )ln T

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A vapor pressure calculation for the liquid vapor for anthracene

ln

* piL

Tb Tb =19(1 − ) +8.5(ln )] T T

Tb= 198 + Σ ΔTb ; for anthracene {C14H18}

C14H18

Has 10 =CH- carbons at 26.73oK/carbon And 4, =C< , carbons 31.01OK/carbon

Tb= 589; CRC = 613K At 298K, lnP = -12.76; p = 2.87 x10-6atm = and p = 0.0022 torr What do we get with the real boiling point of 613K ?

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Solid Vapor Pressures ΔHsub = ΔHfus+ΔHvap

ΔH fus (s)

ΔSfus

ΔHfus= Tm ΔSfus ΔHfus/ Tm = ΔSfus=const?

T ΔHsub = ΔHvap+ Tm Δsfus

It can be shown that

ln

* piL

= ln PiS*

( ΔS fus ) (Tm − Tamb ) + R Tamb

if ΔS= const = 56.4 J mol-1K-1 and R=8.31 J mol1 -1 K , ΔSfus /R= 6.78 please derive this as part of the problem set

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What is the solid vapor pressure for anthracene Using the correct boiling point we determined the liquid vapor pressure to be 8.71x10-7 atmospheres

* * = ln piS + ln piL

( ΔS fus ) (Tm − Tamb ) R Tamb

if ΔS= const = 56.4 J mol-1K-1 and R=8.31 J mol1 -1 K , ΔSfus /R= 6.78 ln 8.71x10-7 = ln p*iS+ 6.78 (490.65-298)/298 -13.95 – 4.38 = ln p*iS 7.8x10-9= p*iS

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Myrdal and Yalkowski also suggest that a reasonable estimate of Δfus Si(Tm) is Δfus Si(Tm) + 56.5+ 9.2 τ -19.2 log σ) in J mol-1K-1

substitution in to * = ln PiS* + ln piL

ln

* piL * piS

( ΔS fus ) (Tm − Tamb ) R Tamb

gives

− (6.8 + 1.1τ − 2.3 log σ ) (Tm − Tamb ) = R Tamb

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Using Sonnefeld et al, what is the sold vapor pressure for anthracene at 289K log10 p*iS = -A / T + B;

p*iS is in pascals

101,325 pascals = 1atm A= 4791.87 B= 12.977 log10 p*iS = -4791.87 / T + 12.977 log10 p*iS = -16.0801 + 12.977 = -3.1031 Po = 7.88 x10-4 pascals P

p*iS = 7.88 x10-4 /101,325 = 7.8x10-9 atm

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Using vapor pressure and activity coefficients to estimate gas-particle partitioning Gas Atoxic + liquid particle Æ particle Atoxic +liquid particle

Apart

Agas

Kp = Apart / (Agas x Liq) = Apart / (Agas xTSP) Liq and TSP has unit of ug/m3 Agas and Apart have units of ng/m3 P = Χ γ PoL (in atmospheres) P v = nRT/760 = [Asas] RT/760; (mmHg) [Asas] = [GasPAH]

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[GasPAH] = Χ γ PoL MWi x 109/( 760 RT)

Let’s look at mole fraction

Χi = moles in the particle phase of i divided by total moles particle phase

Usually we measure ng/m3 in the particle phase of compound i i

i

[ Apart] = [ PartPAH] We usually measure TSP as an indicator of total particle mass The number of moles in the particle phase is: i

Moles = [PartPAH]/ {MWi 109 } = moles/m3

The average number of moles in the particle phase requires that we assume an average molecular weight for the organic material in the particle phase, MWavg

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total molesTSP( μg/m3) = TSP/ {MWavg 106 }

Χi = iMoles/ tot moles = [PartPAH] MWavg / {TSP MWi 103} [GasPAH] =

Χi γ PoL MWi x 109/( 760 RT)

Kp = Apart / (Agas xTSP) o Kp = 760 RT fomx10 /{p L -6

γ MW

avg

}

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BaP

log Kp

slope = -1

pyrene naphthalene

log Po(L)

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