Transcript Vapor Pressure - Department of Environmental Sciences

Chapter 4
Vapor Pressure
pº = Pressure of a substance in equilibrium
with its pure condensed (liquid or solid) phase
Why do we care?
-spills
-pesticide application
-will lead us to Henry’s law constant
KH = PoL/Csatw
Kow = Csato/Csatw
Koa = Csato/PoL
Air
A gas is a gas is a gas
T, P
Koa
KH
Octanol
Po L
Water
Fresh, salt, ground, pore
T, salinity, cosolvents
Csat
w
Kow
Pure Phase
(l) or (s)
Ideal behavior
NOM, biological lipids,
other solvents
T, chemical composition
Csato
Ranges of pº (atm)
– PCBs – 10-5 to 10-9
– n-alkanes – 100.2 to 10-16
• n-C10H22 ~ 10-2.5
• n-C20H42 ~ 10-9
–
–
–
–
–
–
Benzene ~ 10-0.9
toluene ~10-1.42
Ethylbenzene ~ 10-1.90
propyl benzene ~ 10-2.35
carbon tetrachloride ~ 10-0.85
methane 102.44
• Even though VP is
“low”, gas phase may
still be important.
Phase diagram
picture of three-phase
diagram
Ideal Gas Law
pV  nRT
p = pressure
V = volume
n = moles of gas
R = gas constant
T = temperature (Kelvin)
Thermodynamic considerations
(deriving the van’t Hoff equation)
consider a gas: if T or P is changed and
equilibrium is re-established:
d1  d 2
the change in chemical potential in the two
systems is equal
d1  S1dT  V1dp
where S = molar entropy
d2  S2 dT  V2 dp and V = molar volume
dp ( S1  S 2 ) S12


dT (V1  V2 ) V12
at equilibrium
G12  1   2  0
G  H  TS
substituting:
dp H12

dT TV12
for a liquid vaporizing, the volume change can be
assumed to be equal to the volume of gas produced,
since the volume of the solid or liquid is negligible
V12  Vgas
RT
 0
p
Q. where did the n go?
A. this is molar volume
dp
p (H12 )

2
dT
RT
d ln p 0 H12

dT
RT 2
0
0
recall (calculus!)
The van’t
Hoff equation
d ln u 1 du

dx
u dx
where H12 = Hvap (gas) or Hsub (solid)
= energy required to convert one mole of liquid
(or solid) to gas without an increase in T.
Hvap is a function of T.
As T approaches the boiling point, Hvap increases rapidly
At T < boiling point, Hvap increases slowly
from 0-40ºC, Hvap can be assumed to be constant
d ln p 0 H12

dT
RT 2
integrate assuming
Hvap is constant:
Antoine equation
A
ln p    B
T
H12
0
ln p 
a
RT
0
if Hvap is not constant:
b
ln p  
a
another Antoine equation
T c
0
Using Hvap to predict VP at other
temperatures
As we saw in the thermodynamics lecture:
K T 2 H  1 1 
  
ln

KT 1
R  T1 T2 
Specifically,
pT 2 H vap  1 1 
  
ln

pT 1
R  T1 T2 
Note the change in slope when the substance is solid
(sublimation)
Hsub = Hmelt (~25%) + Hvap (~75%)
still use liquid phase as reference:
Hypothetical subcooled liquid
= liquid cooled below melting point without crystallizing
-log P
compound
pºs
<
pºL
1,4-dichlorobenzene
3.04
2.76
phenol
3.59
3.41
22’55’ PCB
7.60
6.64
22’455’ PCB
8.02
7.40
Becomes important later when we talk about solubility
Molecular interactions
affecting vapor pressure
Molecule:molecule interactions in condensed phase
(L or s) have greatest affect on VP
strong interactions lead to large Hvap, low VP
weak interactions lead to small Hvap, high VP
Intermolecular interactions can be classified into
three types:
van der Waals forces (nonpolar)
Polar forces
Hydrogen bonding
van der Waals forces
• nonspecific
• function of size (number of
electrons)
• consist of:
– London dispersive energies
• fleeting areas of charge
– Induced dipoles
• areas of charge arising from
interactions with a polar molecule
“nonpolar”
Polar interactions:
dipole-dipole interactions
• permanent areas of charge on two molecules
attract
Hydrogen bonds
Specific
donors and acceptors
table 4.3
Part of Table 4.3
compound (class)
alkanes
1-alkenes
aliphatic ethers
aliphatic aldehydes
aliphatic alcohols
carboxylic acids
benzene
phenol
naphthalane
fluorene
pyrene
DCM
Water
 (H-donor)  (H-acceptor)
0
0
0
0.07
0
0.45
0
0.45
0.37
0.48
0.60
0.45
0
0.14
0.6
0.31
0
0.2
0
0.2
0
0.29
0.1
0.05
0.82
0.35
Vapor Pressure Estimation Technique
based on regression of lots of VP data, best fit gives:

ln piL*  4.49 V iL

 
2/3
size
 n 1  

   15.1( i )( i )  14.5
 n  2  
2
Di
2
Di
2
polarizability
H-bonding
ability
pressure in Pa, where:
V iL  molar volume (MW/density)
nDi  refractiveindex
refractive index (response to light) is a function of polarizability.
see table 3.1, also might be available in the CRC
Refractive
index
Difference
between
polarity and
polarizability
Trouton’s rule
At their boiling points, most organic compounds have a
similar entropy of vaporization:
exception: strongly
Svap (Tb) = 85 – 90 J/molK
polar or H-bonding
compounds
We can be slightly more accurate with Kistiakowsky’s
expression:
Svap (Tb) = KF(36.6 + 8.31ln(Tb))
J/molK Tb in K (eqn 4-20)
KF = 1 for most compounds
At the boiling point:
G  0  Hvap  Tb Svap
So if we know Tb, we can estimate Hvap (at the boiling
point) fairly accurately
Table 4.2
Estimating VP at other T (need Hvap)
pT 2 H vap  1 1 
  
ln

pT 1
R  T1 T2 
Recognize that Hvap is not constant.
Especially if Tb is high (> 100ºC), the
estimate of Hvap from Trouton/Kistiakowsky
may not be valid at the temperature of
interest.
Empirically, Hvap is a function of the VP:
Hvap (T1 )  a log piL* (T1 )  b
FIG 4.7
From a data set of many compounds, Goss and
Schwarzenbach (1999) get:
Hvap (298K )  8.80log piL* (298K )  70.0
Less empirically,
assume Hvap is linearly proportional to T (i.e.
assume that the heat capacity, Cpvap is constant:
H vap (T )  H vap (Tb )  C p vap (Tb )  (Tb  T )
substitute this expression into the ClausiusClapeyron equation and integrate from Tb to T:
H vap (Tb )  1 1 
  
ln P 
R
 Tb T 
C p vap (Tb )  T  C p vap (Tb )  T 

 1  b  
ln b 
R
R
 T 
T 
0
don’t let notation
confuse you.
(Tb) means at
the boiling point.
You do not
multiply Hvap by
the boiling point
Recall:
Hvap (Tb )  Tb Svap (Tb )
substitute:
 Svap (Tb ) C p vap (Tb )   Tb 
  1  
ln p  


  T
R
R


C p vap (Tb )  T 

ln b 
R
T 
0
but we still need to know Cp(Tb)!
generally:
C p (Tb )  0.8  Svap (Tb ) ranges from 1.0 to 0.6
and Svap(Tb)~ 88 J/molK
finally!
  Tb 
 Tb 
ln p   K F (4.4  ln Tb )1.8  1  0.8 ln 

 T 
 T
0
KF is the Fishtine factor, usually 1, but
sometimes as high as 1.3 (see p 113)
in atm
Eqn 4-33
the old edition gave (where KF =1):
 Tb 
 Tb 
ln p  19  1    8.5 ln 
 T
T 
0
in atm
OK for liquids with Tb < 100ºC
High MW compounds, need correction for
intermolecular forces (but we don’t have their boiling
points anyway!) (For refinements see equation 4-33)
Can estimate boiling points, see p. 120
solids?
those previous equations yielded the vapor pressure of
the hypothetical subcooled liquid.
How can we correct this to give the true vapor
pressure of a solid?
Prausnitz (1969):
S fus (Tm )  Tm 
ps0
ln 0  
  1
pL
R
T

Where Sfus(Tm) = entropy of fusion at melting point
unfortunately Sfus is much more variable than Svap
S fus (Tm )  (56.5  9.2 19.2 log )
J/molK
Where  = number of torsional bonds and
 = rotational symmetry number (see p. 125)
the older edition of your book gave this simpler
(but less accurate) equation:
Sfus(Tm) ~ 56.5 + 10.5(n-5)
J/molK
Where n = number of flexing chain atoms.
if n<5, then ignore this term
Estimation of vapor pressures for polychlorinated
biphenyls: a comparison of eleven predictive methods
Lawrence P. Burkhard, Anders W. Andren, and David E.
Armstrong
Environmental Science and Technology 1985, 19, 500 507
conclusions:
• non-correlative methods have poor predictive ability (error
increases as VP decreases)
• correlative methods requiring a set of compounds with
known P are much better
• best method: determine VP as function of GC retention
times
Determination of vapor pressures for
nonpolar and semipolar organic compounds
from GC retention data (Hinckley et al, 1990)
• Chromatographed 2 reference compounds (eicosane
and p,p’DDT) having known VP and Hvap versus a host
of unknowns (PAHs, organochlorines, etc)
• Isothermal runs allow determination of RRT at several T
• Comparison of RRT with reference compounds allows
determination of VP at given T
• Comparison of changes in RRT with T and knowledge
of Hvap for reference compound allows calculation of
Hvap for all unknowns
Problem 4.2
• In a dump site, you find an old 3-liter pressure bottle
containing FREON 12 with a pressure gauge that
reads 2.7 bar. (First, you realize that this gauge
was not manufactured in the US.) The temperature
is 10ºC. What mass of FREON 12 is in the bottle?
• Also estimate the free energy, enthalpy, and entropy
of condensation of FREON 12.
• You find the following info for FREON 12 in the
CRC:
T deg C p/kPa
-25
123
0
308
25
651
50
1216
75
2076
Problem 4.6
• estimate VP at 0C based on VP at 25ºC
or based solely on Tb and Tm
log(p) @25C Tm (degC) Tb (degC)
dimethyl phthalate
0.38
5.5
283.7
2,3,7,8-TCDD
-6.7
305
446.5
(hint  = 4)
Homework
• Do problems 4.3 and 4.4
• Due 2/2/10