An Important Goal of This Chapter Is to Learn Techniques to Calculate Vapor Pressures s1

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 X ipiL* pure liquid

· 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

dmi liq = Vi liqdp - Si liqdT

Vi liqdpi - Si liqdT = Vigasdp - SigasdT

d/dT = (Sigas -Si liq)/Vigas

at equilibrium DG = DH -DS T= zero

so (Sigas -Si liq) = DHi vap/T

substituting

dpi /dT = DHi /( Vigas T)

(Clapeyron eq)

substituting Vi gas = RT/pi

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= 578 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

Both boiling points and entropies of vaporization become important parameters in estimating vapor pressures

A constant entropy of vaporization Troutons rule

Figure 4.5; at 25oC

This suggests that DvapS may tend to be constant

At the boiling point is DvapSTb constant?

const slope = DS

Tb oC DvapH DvapS

kJ mol-1 kJ mol-1K-

n-hexane 68.7 28.9

n-decane 174.1 38.8

ethanol 78.3 38.6

naphthalene 218 43.7

Phenanthrene 339 53.0

Benzene 80.1 30.7

Chlorobenzene 131.7 35.2

Hydroxybenzene 181.8 45.7

Predicting DvapSTb

Kistiakowsky derived an expression for the entropy of vaporization which takes into account van der Waal forces

· DSvap= 36.6 +8.31 ln Tb (eq 4-20)

· 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 (Mydral et al, 1996)

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

(eq 4-21)

t = S(SP3 +0.5 SP2 +0.5 ring) -1

SP3 = 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 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 (Tb obs-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 DvapH/DT = DCp Tb

DvapHT = DvapH Tb + DCpTb(T-TTb)

at the boiling point DvapH Tb= Tb DvapS Tb

for many organic compounds

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

so substituting DCp Tb= -0.8 vapDS 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

DvapS 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

t = S(SP3 +0.5 SP2 +0.5 ring) –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= 589K; CRC = 613K

At 298K, ln p* = -12.76; p = 2.87 x10-6atm =

and p*iL = 0.0022 torr

What do we get with the real boiling point of 613K ?

ln p*iL = 8.7 x10-7atm
Solid vapor pressures (is there a relationship between solid and liquid vapor pressures??)

For a solid to go directly to the gas phase (sublimation),we can say that the free energy of sublimation(DsubGi) has to be the sum of the free energy needed to go from the solid to the liquid (DfusGi) + the free energy of going from a liquid to a gas (DvapGi)

DsubGi = DfusGi + DvapGi

also

DsubHi = DfusHi + DvapHi and DsubSi = DfusSi + DvapSi

back to free energy: DfusGi = DsubGi - DvapGi

for both DvapGi and DsubGi

the change in free energy in going from one temperature to another, like say ambient to the melting point

Dphase j Gi = RT ln { pref / p*ij }

so in

DfusGi = DsubGi - DvapGi

DfusGi = RTln { pismelt / p*is } - RTln { piLmelt / p*iL }

DfusGi = RTln { p*iL / p*is }

this suggests a relationship between p*iL and p*is
Solid Vapor Pressures

DsubH = DfusH + DvapH

D fusH (s) DfusH= Tm DfusS

DfusS

DfusH/ Tm = DfusS = const?

DHsub = DHvap+ Tm D+DfusS

for any phase

; Dlnp*= -DH/R (1/T2-1/T1)

for a vapor going from the melting point to some ambient temperature:

and for a solid subliming from the melting point to some ambient temperature:

at the melting point p(l)m= p(s)m

DsubH = DvapH+ Tm DfusS

substituting

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

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, DfusS /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

Measuring solid vapor pressures

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

A chromatographic Method for measuring liquid vapor pressures

A B

Hamilton,J.Chrom.195,

75-83,1980

Hinckley et al., J. Chem.

Eng. Data, 1990;

Yamasaki, 1986

to t1 t2

retention time A = t1- to=t’1

retention volume (Vr) = column flow x the retention time and

t’1/t’2= VrA / VrB

and Vr varies inversely with the vapor pressure of a compound

VrA / VrB = p*BL/p*AL eq1

eq 2

eq 3

; eq 4

dividing eq 4 into 3 and integrating

+ const eq 5

we said that

eq 2

combining eqs. 2 and 5

eq 6

1.If we can know or can calculate the vapor pressure p*AL at different temperatures

2. And run our GC isothermally at these temperatures and obtain of t’2/t’1 and hence VrB / VrA

3.we can plot ln VrB / VrA vs p*AL , get the slope and intercept, and plug into eq 5 to obtain p*BL

Using vapor pressure and activity coefficients to estimate organic gas-particle partitioning (Pankow, Atmos. Environ., 1994)

Gas Atoxic + liquid particle à particle Atoxic +liquid particle

Kip = Aipart / (Aigas xTSP);.

TSP has units of ug/m3

Aigas and Aipart have units of ng/m3

log Kip= -log p*iL + const based on solid-gas partitioning

pi = C g p*iL (in atmospheres)

pi = ni/V RTx760 = [Aigas] RTx 760; (mmHg)

[Aigas] = [PAHigas] = ni /V in moles/liter air

multiplying Mwi and by 109 to put [PAHgas]in ng/m3

[PAHigas] = C g p*iL MWi x 109/(RT x760)

Let’s look at the mole fraction

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

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

[Aipart] = [PAHipart]

The number of moles in the particle phase is:

iMoles = [PAHipart]/ {MWi 109 } = moles/m3

We usually measure TSP as an indicator of total particle mass and the amt. of liquid in the particle phase = TSP x fom

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 liquid moles of TSP measured in( mg/m3) = fomTSP/ {MWavg 106 }

we said before

iMoles = [Partipart]/ {MWi 109 } = moles/m3

Ci = iMoles/ total moles =

[PartiPAH] MWavg / {fomTSP MWi 103}

we also said before that

[PAHigas] = Ci g p*iL MWi x 109/( 760 RT) in ng/m3

and

Kip = PAHipart / (PAHigas xTSP)

Kip = 760 RT fomx10-6/{p*iLtorr g MWavg}

p*iL here is in torr

R= 8.2x10-5 M3 atmos./{Mol K}

p* =