An Important Goal of This Chapter Is to Learn Techniques to Calculate Vapor Pressures

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 = gi XiPiopure liquid (old book)

· fi = gi X ipi* pure liquid (new book)

· KiH = P*iL/Ciwsat

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

The molar free energy Gi/ni = mi

for a gas in equilibrium with a liquid

dmliq = dmgas

dmliq = VliqdP - SliqdT

VliqdP - SliqdT = VgasdP - SgasdT

dP/dT = (Sgas -Sliq)/Vgas

at equilibrium DG = DH -DS T= zero

so (Sgas -Sliq) = DHvap/T

substituting

dP/dT = DH/( Vgas T)

(Clapeyron eq)

substituting Vgas = RT/Po

Figure 4.3 page 61 Schwartzenbach

This works over a limited temperature range w/o any phase change

Over a larger range Antoine’s equation may be used

over the limits P1 to P2 and T1 to T2

If the molar heat of vaporization, DHvap of hexane equals 6896 cal/mol and its boiling point is 69oC, what is its vapor pressure at 60oC

P1= 395 mm Hg

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

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

A constant entropy of vaporization Troutons rule

at the boiling point DG = DH-DSxT = zero

const slope = DS

DH/T = DS= 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

· DSvap= 36.6 +8.31 ln Tb

· for polarity interactions Fistine proposed
DSvap= 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

Calculating DSvap using chain flexibility and functionality

DvapSi (Tb) = 86.0+ 0.04 t + 1421 HBN

t = S(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

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

DSb= 84.53 – 11s +.35t + 0.05w2 + SCi

where:

Ci = the contribution of group i to the Entropy of boiling

w = 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

t measures the conformational freedom or flexibility ability of atoms in a molecule to rotate about single bonds

t= SP3 + 0.5(SP2) +0.5 (ring) –1

s = 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

Boiling points can be estimated based on chemical structure (Joback, 1984)

Tb= 198 + S DTb

DT (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

They start with Tb= 198 + S DTb and go to 4426 experimental boiling points in Aldrich

And fit the residuals (Tbobs-Tb calcd)

Tb= 198 + S DTb

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

0.0007705Tb2 T b 700 K

Tb(corr) = Tb+282.7-0.5209Tb

Tb>700K

Estimating Vapor Pressures

To estimate the vapor pressure at a temp lower then the boiling temp of the liquid we need to estimate DHvap at lower temperatures.

Assume that DHvap is directly proportional to temp and that DHvap can be related to a constant the heat capacity of vaporization DCp Tb

where DHvap/DT = DCp Tb

DHvapT = DHvap Tb + DCp Tb(T-TTb)

at the boiling point DHvap Tb= Tb DSvap Tb

for many organic compounds

DCp Tb/ DSvap Tb ranges from -0.6 to -1

so substituting DCp Tb=0.8 DSvap Tb

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

when using DCp Tb/ DSvap 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

DSvap Tb= Kf(36.6 +8.31 ln Tb)

If we go back to:

DvapSi (Tb) = 86.0+ 0.04 t + 1421 HBN

and Mydral and Yalkowsky suggest that

DvapCpi (Tb) = -90 +2.1t in J mol-1K-1

A vapor pressure calculation for the liquid vapor for anthracene

Tb= 198 + S DTb ; 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 ?

Solid Vapor Pressures

DHsub = DHfus+DHvap

DH fus (s) DHfus= Tm DSfus

DSfus

DHfus/ Tm = DSfus=const?

DHsub = DHvap+ Tm Dsfus

It can be shown that

if DS= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, DSfus /R= 6.78

please derive this as part of the problem set

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

if DS= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1, DSfus /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

Myrdal and Yalkowski also suggest that a reasonable estimate of Dfus Si(Tm) is

Dfus Si(Tm) + 56.5+ 9.2 t -19.2 log s)

in J mol-1K-1

substitution in to

gives

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

Using vapor pressure and activity coefficients to estimate gas-particle partitioning

Gas Atoxic + liquid particle à particle Atoxic +liquid particle

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 = C g PoL (in atmospheres)

P v = nRT/760 = [Asas] RT/760; (mmHg)

[Asas] = [GasPAH]

[GasPAH] = C g PoL MWi x 109/( 760 RT)

Let’s look at mole fraction

Ci = 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

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

Ci = iMoles/ tot moles =

[PartPAH] MWavg / {TSP MWi 103}

[GasPAH] = Ci g PoL MWi x 109/( 760 RT)

Kp = Apart / (Agas xTSP)

Kp = 760 RT fomx10-6/{poL g MWavg}