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 P\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?
DH
const slope = DS
T
Tb oC DvapH DvapS
kJ mol-1 kJ mol-1K-1
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
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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
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
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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
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 need 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
so
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?
T
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
see if you can derive this
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
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 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 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 V = ni RT/760 = [Aigas] RT/760; (mmHg)
[Aigas] = [PAHigas] = ni /V
[PAHigas] = C g p*iL MWi x 109/( 760 RT)
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 ng/m3 in the particle phase of compound i
[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 ofTSP( mg/m3) = fomTSP/ {MWavg 106 }
we said before
iMoles = [Partipart]/ {MWi 109 } = moles/m3
Ci = iMoles/ total moles =
[PartiPAH] MWavg / {TSP MWi 103}
we also said before that
[PAHigas] = Ci g p*iL MWi x 109/( 760 RT)
and
Kip = PAHipart / (PAHigas xTSP)
Kip = 760 RT fomx10-6/{p* iLtorr MWavg};
p*iL here is in torr
ln p* =
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