Chapter 4 Vapor Pressure - The University of North ... · Chapter 4 Vapor Pressure ... ln + + * =...

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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. f i = γ i X i P i o pure liquid (old book) f i = γ i X i p i * pure liquid (new book) log log C C TSP P cons part l o =− + t gas K iH = P * iL /C iw sat 1

Transcript of Chapter 4 Vapor Pressure - The University of North ... · Chapter 4 Vapor Pressure ... ln + + * =...

Page 1: Chapter 4 Vapor Pressure - The University of North ... · Chapter 4 Vapor Pressure ... ln + + * = pi over the limits P 1 to P 2 and T ... benzoic acid 532 522 toluene 386 384 pentane

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 = γi XiPio

pure liquid (old book)

• fi = γi X ipi*

pure liquid (new book)

• log logC

C TSPP conspart

lo= − + t

gas

• KiH = P*iL/Ciw

sat

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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 Vgas = RT/Po

dPdT

HPRTxT

d P d TH

R

o= = −

Δ Δ; (ln ) ( )1

lnP HR T

consto = − +Δ 1

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Figure 4.3 page 61 Schwartzenbach

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

ACT

Bln ++

=*ip

over the limits P1 to P2 and T1 to T2

log

( ).

PP

vapH T TRT T

21

2 1

1 22 303=

−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 3241P

cal mol K Kcal Kmol K K

=−

∗ ∗ ∗

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 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

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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

MWNH.33COOHOHHBN 2++

=

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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 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

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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

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Tb(corr) = Tb- 94.84+ 0.5577Tb- 0.0007705Tb

2 T b< 700 K Tb(corr) = Tb+282.7-0.5209Tb

Tb>700K

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Estimating Vapor Pressures d P

dTHRT

ovapTln

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)

)TbT

(lnRpTbΔC

)TbT

(1R

bpTΔC)

T1

bT1(

RbvapTΔH

ln −−−−=*iLP

at the boiling point ΔHvap Tb= Tb ΔSvap Tb

)()(( ln)ln *TT

RC

TT

RC

RS

p bpTbbpTvapT bbiL

Δ−−

Δ−

Δ= 1

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)())(( lnln *TT

RC

TT

RC

RS

p bpTbbpTvapTiL

bb Δ−−

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..*lnTT

TT

RS

p bbvapTiL

b 80181 +−Δ

=

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

)]()( ln.ln *TT

TTp bb

iL 58119 +−=

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 *TT

TT

bTfKp bbiL 8018144 −−+=−

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If we go back to:

)())(( lnln *TT

RC

TT

RC

RS

p bpTbbpTvapTiL

bb Δ−−

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

TbT

HBNiLpTTb

ln)..(

))(177.21.2(*ln

τ

τ

250810

30 1

++

++− −=

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A vapor pressure calculation for the liquid vapor for anthracene

)]()( ln.ln *TT

TTp bb

iL 58119 +−=

Tb= 198 + Σ ΔTb ; for anthracene {C14H18} C14H18Has 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) ΔHfus= Tm ΔSfus

ΔSfus

ΔHfus/ Tm = ΔSfus=const?

T ΔHsub = ΔHvap+ Tm Δsfus

It can be shown that

amb

ambmfusiSiL T

TTRS

Pp)()(

lnln ** −Δ+=

if ΔS= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, 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

amb

ambmfusiSiL T

TTRS

pp )()(lnln ** −Δ

+=

if ΔS= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, 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

amb

ambmfusiSiL T

TTRS

Pp)()(

lnln ** −Δ+= gives

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amb

ambm

iS

iLT

TTRp

p )()log...(ln *

* −−+−=

στ 321186

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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 PP

o = 7.88 x10-4 pascals 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 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 [iApart] = [iPartPAH] We usually measure TSP as an indicator of total particle mass The number of moles in the particle phase is: iMoles = [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 Po

L MWi x 109/( 760 RT)

Kp = Apart / (Agas xTSP)

Kp = 760 RT fomx10-6/{poL γ MWavg}

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log Po(L)

logKp

slope = -1

pyrene

naphthalene

BaP

log Po(L)

logKp

slope = -1

pyrene

naphthalene

BaP

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