Envr 210, Chapter 3,
 Intermolecular forces and
partitioning
 Free energies and equilibrium
partitioning
 chemical potential
 fugacity
 activity coef.
 phase transfer- activity coef
and fugacities
 more on free energies and
equilibrium constants
1
Much of this class deals with the
partitioning of an organic compound i
between two phases
A+B C
Keq = [C]/{[A][B]}
Keq = [ iphase1]/[ iphase 2]
When we deal with air liquid
partitioning
KiaL = Cia/CiL
Octanol-water
Kiow = Cio/Ciw
Solid-water
Kid = Cis/Ciw
We will find that often for classes of compounds
log Kid = a log Kiow + b
Why???
2
For a compound to move between one
phase and another, the intermolecular
forces that hold a molecule in one
phase need to be broken and others
reformed in the other phase
Simply this can be represented as:
1:i:1 + 2:2  1:1 + 2:i:2 (absorption)
if the phase change is from molecule i
in phase 1 to the interface or surface
between 1 and 2, then
1:i:1 + 1:2  1:1 + 1:i:2 (adsorption)
What is the nature of the bonds that
are being broken or formed??
1. Nonspecific interactions
(van der Walls interactions)
3
a. related to a compound’s
polarizability () or the extent to which
an uneven electron distribution results
in response to an imposed electronic
field on timescales of10-15 sec; the
intermolecular attraction energy is
related to the product of the s of the
interacting set of atoms…London
dispersive energies
b. dipole-induced interactions
(Debye energies) resulting from
electron distribution differences in one
molecule (carbon and oxygen bond)
inducing a charge distribution in the
adjacent molecule. The strength of the
interaction should be a function of the
dipole moment, = qr ,in the “dipole”
molecule, times the polarizability of the
“charge induced molecule.
4
c. dipole-dipole interactions:
strength of attraction proportional to
1 x 2
2. Specific interactions:
intermolecular attractions between
electron rich and electron poor sites of
corresponding molecules
hydrogen bonding between the
electron poor hydrogen of a carbon
hydrogen bond and the unpaired
oxygen electrons in an adjacent
molecule…electron donor or acceptor
interactions
In the absence of electron donor
or accepter interactions, London
dispersive energies can be used to
characterized the attractions of many
molecules to their surroundings with
respect to equilibrium partitioning
5
Consider a molecule moving from the
gas to a liquid phase, 1
i (g) + 1:1(L)  1:i:1 (L)
when i dissolves in solvent 1, the
dispersive attraction energy per
interaction , disp g is given as
(Israelachvile, 1992) as a function of
polarizability ,, and the 1st ionization
energies, I, of compounds i and solvent
1;
I= Ii+ I1 /( Ii I1)

disp g = -(3/2) I i 1 /(4 0)2
Visible light has frequencies (and its
changing electric fields) on the order of
10-15cycles /sec. A material’s ability to
respond to light is related to its index of
refraction, nDi, and nDi is related to that
material’s polarizability via the LorenzLorenz relationship
6
i /(4 0)= [n2Di -1]/ [n2Di +2]x(3Mi/4Na)
Assuming spherical molecules, and the induced
temporary dipoles distances are diameters of the
molecules (see page 64 of text)
 disp gperinteraction
n2Di  1 n2D1  1
  3I / 256 2
x 2
nDi  2 nD1  2
for a mole of interactions we need to consider the
total surface area (TSA) of the solvated molecule and
the contact area (CA) it has with solvent molecules
 disp Gfor one mole
n2Di  1 n2D1  1
  N A TSA / CA3I / 256 2
x 2
nDi  2 nD1  2
Since NA, CA, 3 and I are relatively constant,
7
If the equilibrium is dominated by
dispersive forces, this free energy,
disp G can be related to the
equilibrium of this process by
dispG = - RT ln Keq
where Keq= [ iphase1]/[ iphase 2]
For an organic gas in equilibrium with
a pure liquid the equilibrium is:
KiaL= Ciasat/CiL = Mi p*iL/[iLRT]
 disp Gfor one mole
n2Di  1 n2D1  1
  const. xTSA 2
x 2
nDi  2 nD1  2
8
disp G = - RT ln Keq
we should be able to plot
calculated ln KiaL = M p*iL/[iLRT]
vs.
n2Di  1 n2D1  1
TSA 2
x 2
nDi  2 nD1  2
for a pure solvent interacting with the
gas phase, i=1
9
Table 3.1 page 65
10
Figure 3.4page 69 new text
Partitioning in air-pure solvent vs. index of refraction term
Partitioning in Hexane?
Partitioning in water?
11
Figure 3.6 page 71 air-hexane, top, air-water, bottom
12
Chapter 3, then uses thermodynamics
to quantify molecular energies and
equilibrium partitioning
Section 3.3 starts with:
N
i   ( nGi )T , P , n
i 1
j
and
N
G( p ,T ,n1n2 ,ni ...nN )    i ni
1
How do we get to these equations and
what do they mean??
13
Chapter 3
The First Law
U2 - U1 = q - w
work
change in
internal energy
of an object
heat
reservoir
object
b
U = q1-w1











a 


U = q2-w2
14
For example one gram of H2O at 25oC is evaporated
and condensed; the condensed gram of water at 25oC
will have the same internal energy as it did previously.
If only pV work is done and the pressure of the system
is constant
w rev =
 pdV
What is the work of a reversible expansion of a mole
of an ideal gas at 0oC from 2.24 to 22.4 liters?
Wrev   VV 12 pdV
pV = nRT
Wrev

V2
V2
 nRT dV / V  nRT ln ( )
V1
V1
Wrev = 1mole x1.987 cal K-1 mole-1
x 273 K x 2.303 log (22.4/2.24)
Wrev = 1.25 Kcal mole-1
15
Internal energy, heat and work
when one mole of water is vaporized at 100oC the work is
w = p V = RT = 1.987 cal K-1 mole-1 x 373.15K
w= 741.4 cal mole-1
The energy or heat required to vaporize water at 100oC
requires energy to separate the liquid molecules;
that is 529.7 cal g-1
q = 18.02 g mole-1 x 539 cal g-1 = 9725 cal mole-1 ;
For a mole of water, the internal energy U = q - w
U = 9725 cal mole-1 - 741 cal mole-1
U = 8984 cal mole-1
16
Enthalpy
U = q - pV)
at constant pressure
q= (U2 + pV2) - (U1+ pV1)
We define U + pV as the enthalpy, H
q = H2-H1 = H
or the heat adsorbed in a process at constant
pressure
There are usually two types of calorimetric experiments used
to determine heat, one at const volume (no PV work, so U=q)
and one at constant pressure.
The heat of combustion of CO in a constant vol calorimeter
is –67.37 kcal mol-1. Calculate the enthalpy of combustion in
a const pressure.
CO(g) + ½ O2  CO2(g)
Work done is (n2-n1)RT, n2 is moles of products, n1 is moles
of reactants.
H =U + (n1-n2)RT
= -67.37 – (0.5 mol) 0.00199 kcal K-1mol-1) 298K
H = -67.6 kcal mol-1
17
Standard Heats of formation
kcal/mole
C graphite
kcal/mole
0
H2O(g)
-57.78
CO(g)
-26.45
H2O(l)
-68.32
CO2(g)
-94.05
ethene
+12.50
Benzene
+19.82
ethane
-20.24
H2C=CH2 + H2 --->
H3C-CH3
Htotal = Hf(H3C-CH3) -Hf(H2) - Hf(H2C=CH2)
H= -20.24 -(+12.5) +0
= -32.75 kcal
18
Heat Capacity
Heat Capacity, C = ratio of heat absorbed/mole to the
temperature change = q/T
At constant pressure
q = U+pV = H
Cp = dH/dT
i.e. the calories of heat
adsorbed/mole by a substance/oC
so
H= Cp(T2-T1)
At constant volume
U = q - pV
U = q
Cv = dU/dT
What is the relationship between Cp and Cv?
19
The Second Law
Lord Kelvin (1824-1907): “It is impossible by a
cyclic process to take heat from a reservoir and
convert it into work without at the same time
transferring heat from a hot to a cold reservoir”
Clausius: “It is impossible to transfer heat from a
cold to a hot reservoir without at the same time
converting a certain amount of the work to heat”
i.e. work can only be obtained from a system
when it is not at equilibrium
It can be shown (see any p-chem book) that the
max. efficiency of a sequence of isothermal and
adiabatic process is
eff = (TH-TL)/TH = (qH + qL)/qH
rearranging
qH qL

0
TH TL
20
qH qL

0
TH TL


qi
0
Ti
dqrev
0
T
define
dS = dq/T
and
S2  S1  S 

2
1
dqrev
T
at absolute zero the entropy is assumed to be zero
21
Consider 1 mole of H20 (l)---> H20 g at 100oC
SH20= dq/T = 1/Tdq = 1/T

n 1
Hdn
n 0
= dHvap/T = 9,720 cal/373K= + 26cal/degK mole
ssurroundings= a negative 26cal/degK

S total = zero
________________
When spontaneous processes occur
there is an increase in entropy
When the net change in entropy is zero the system is
at equilibrium
If the calculated entropy is negative the process will
go spontaneously in reverse.
22
Entropy cont.
S = dq/T
1. What is it about a gas that makes it have more entropy
when it is expanded, then when it is compressed or in the
liquid state?
Let’s say that in the reaction of
A ---> B
B has more entropy than A
2. What is it about B that gives it more entropy?
1st consider a box with a 4 pennies; if we place
them with heads up and then shake the box, we get:
4 heads, 0 tails
3 heads, 1 tail
2 heads, 2 tails
1 head, 3 tails
0 heads, 4 tails
# combinations
1
4
6
4
1
we might consider this to be the normal state, or
equilibrium state, because there are more combinations to
“go to”
23
At a molecular level,
A ---> B can be understood if A and B have states
with equal energies, and if B has more energy levels
to “go to” within these states. Moving to the B energy
states brings the system to higher entropy.
the average translational energy of a gas in one
direction is given by 1/2 RT
If the energy level has the form
n= n2 h2/(8ma2)
there will be a certain # of n levels for the gas as
given by the quantum #s “n” The sum energies in
each n level will be 1/2 RT
If the gas is expanded in the “a” direction, we
decrease the spacing between the energy levels,
which permits more energy states
24
Figure 6.7, Physical Chemistry, Barrow, McGraw Hill, New
York, 1963, page 147
25
The concept of free energy comes from the need
to simultaneously deal with the enthalpy energy
and entropy of a system
G = H -TS
G = U+PV - TS
dG= dU + PdV + VdP -TdS -SdT
dH = dU +pdV
at const temp and pressure
G= H -TS
What is the free energy for the process of converting
1mole of water at 100oC and one atm. to steam at
one atm.
H= H vap
Svap = 1/T  dq = Hvap/T
TS = Hvap
G= Hvap - TS
G= Hvap - Hvap= 0
26
Equilibrium Constants
dG= dU +VdP + pdV -TdS -SdT
for a reversible process
TdS = dq
dU -dq+dw = 0
so dG= +VdP -SdT
at const temp
(G/P)T = V; and if const. temp is stated all the time
dG/dP= V
dG =nRT dP/P
G2 -G1 = nRT ln(P2/P1)
At standard state
G = Go + nRT ln(P)
27
G = Go + nRT ln(P)
for a reaction
aA + bB--> cC + dD
for A
we have GA =GoA +aRT lnPA
it is the free energy of the products minus the
reactants that is of interest
G =Gprod - Greact
for reactants A and B
GAB = GoA + GoB+ RT lnPAa + RT lnPBb
For aA + bB--> cC + dD
( PC )c ( PD )d
G  G  RT ln
( PA ) a ( PB ) b
o
if the reaction goes to completion
G = zero
Go  RT ln
(PC )c (PD )d
(PA )a (PB )b
28

GO= -RT lnKeq
29
Equilibrium Constants and Temperature
30
We are now ready for the 1st two equations of
Chapter 3, Section 3.3
N
i   ( nGi )T , P , n
i 1
j
N
G( P ,T ,n1n2 ...ni ...nN )    i ni
i 1
31
Closed Systems
From the first law
dU = dq - pdV
from the definition of entropy
dS = dq/T
dU = TdS - pdV
If we differentiate by parts, i.e. separately hold dV
and dS constant
(
U
)V  T ; and ( U )S   p
S
V
dU  (
U
)V dS  ( U)S dV
S
V
Environmental systems are often
open systems, i.e. material is being
added or removed, and/or material is
reacted
If a homogeneous system contains a number of
different substances its internal energy may be
32
considered to be a function of the entropy, the
volume and the change in the # moles
dU  (
U
)V dS  ( U)S dV
S
V
k
U
U
dU  ( )V,ni dS  ( )S,ni dV   ( nU )
dni
i
S
V
S, V, n
i 1
j
k
the term
 ( nU )S, V, n
n i
.
i
is call the Chemical Potential ,
j
k
dU  TdS  pdV   ( nU )S , V , n dni
i 1
i
i
j
k
  ( nU )S, V , n
n i
i
j
k
dU  TdS  pdV   i dni
i 1
33
The complete expression for the differential of free
energy is
dG = dU +pdV+Vdp -TdS- SdT
if
k
dU  TdS  pdV   ( nU )S , V , n dni
i 1
i
j
Substituting for dU in the free energy expression
k
dG  SdT  Vdp   ( nU )S , V , n dni
i 1
i
j
or we could directly define
k
dG  SdT  Vdp   ( nG )T , Pj , n dni
j
i
i 1
k
i   ( )
i 1
U
ni S, V, n j
k
 (
i 1
G
ni T, P, n j
)
k
  ( nH )S, P, n
i 1
i
j
k
i   ( nGi )T, P, n
i1
j
34
if we rewrite the equation for the potential energy in its
integrated form
k
U  TS  pV   i ni
i 1
it can be shown that for an expansion in which
temperature, pressure and the number of elements are
proportionately increased and that the relative
proportions of the components are kept constant...
k
U  TS  pV   i ni
i 1
since H= U+ pV
k
H  TS    ini
n i
and
G= H –TS
G( P,T,n n ...n ) 
1
2
i
k
 i n i
n i
35
G( P,T,n n ...n ) 
1
2
i
k
 i n i
n i
What this says is that the total free energy of a system
is directly related to the sum of the individual chemical
potentials times the number if moles of each
contributing entity
Going back to
k
i   ( nGi )T, P, n
i1
j
what does this say about systems
that are not at equilibrium
36
37
Chemical Potentials and Pressure
If we go back to the expression for potential energy
k
U  TS  pV   i ni
i 1
dU = SdT+ TdS -Vdp-pdV+
k
k
i 1
i 1
 idni   nidi
For a closed system which only does pressure volume work
we said that
k
dU  TdS  pdV   i dni
i 1
subtracting
k
0 = SdT -Vdp+
 nidi
i 1
At constant temperature, one obtains the Gibbs-Duhem
Equation for gases
38
k
Vdpi =  ni d ig
i 1
and so for just compound i
Vdpi/ni = (dig)T
substituting for V = nRT/pi
and integrating from a partial pressure of a compound
defined as pi0 to pi
uig = RT ln pi/pi0
if our boundary conditions or limits start at
standard states

ig = oig + RT ln pi/poi
your book has elected to define one bar as the standard
state for pressure
39
What if the system is not ideal?
Van der Waal’s equation
an 2
( p  2 )( V  nb )  nRT
V
inter molecular
attraction
occupied molecular
volume
For a non-ideal system we could attempt to substitute
for V in the chemical potential relationship
Vdpi/ni = (di)T
another way is to define a parameter related to pressure
called fugacity
where by analogy f
i = oi +RT ln fi/ foi
fi =i pi
40
i is a fugacity coef.
in a mixture of gas phase compounds
pi = xi pi* the vapor pressure in bars
so what is vapor pressure???? which
your book calls pi L*
one atmosphere =1.013 bars
one atmosphere supports a 76 cm column of Hg
one atmosphere = 760 mm Hg = 760 torr
one atmosphere =1.013x106 dynes/cm2
derived from the force of mercury on 1 cm2
1 bar = 105 pascals
133.3 pascals = 1torr
fig = xig ig piL*
where xi is the mole fraction
xig 
nig
nj
j
41
Fugacitys of liquids
pi = Xi pi* (Raoult’s Law)
for two different liquids with the same
components
p1i
p2i
5%
10%
A in B A in B

2i = 1i +RT ln p2i/p1i
since p1i = x1 piL* and p2i = x2 piL*
2i = 1i +RT ln x2i/x1i
(Ideal)
similarly
2i = 1i +RT ln f2i/f1i
fi pure liquid = i pure liquid piL*
42
Where i is called an activity
coefficient
if we discuss compound i in a liquid
mixture
fiL = i Xipi*L (pure liquid)
the fugacity of compound i with respect to
the fugacity of the pure liquid can also be
written as
fi = i Xifi*L (pure liquid);
for ideal behavior of similar compounds like benzene
and toluene in a mixture, i = 1
43
If we go back to the chemical
potential with respect to a pure
liquid
i = i pure liquid +RT ln fi/f*i pure liquid
fi = i Xif*i pure liquid
so
i = i pure liquid +RT ln iXi
where iXi is called the activity, a, of the
compound in a given state with respect to
some reference state
in iXi = ai
the activity sometimes is
called the “apparent concentration”
because it is related to the to the mole
fraction, Xi or the “real” concentration via i
44
Phase Transfer Processes
i
Consider a compound, ,which is dissolved
in two liquids which are immiscible like
water and hexane.
at equilibrium
i H2O = i pure liquid +RT lniH2OXi H2O
i hx= i pure liquid +RT lni hx Xi hx
at equilibrium
 i
H2O
=i hx
RT lni hx Xi hx = RT lniH2OXi H2O
substituting
i Xi = fi /fi*L (pure liquid)
45
RT ln fi hx /fiL*(pure liquid) = RT lnfi H2O /fiL*(pure liquid)
fi hx = fi H2O
Hint For your homework: Calculate the
activity coef. i of hexane from its solubility in water.
hexane has some low solubility in water in
grams/LH2O; 1st recall we derived
RT lni hx Xi hx = RT lniH2OXi H2O
What is the activity coefficient and mol fraction of
hexane in hexane?
This gives the important result:
i
H2O=1/ X H2O
i

to calculate the iH2O we need to know Xi H2O
/
Ci = sat. conc. = Xi molar volumemix (why???)
molar vol = liquid vol of one mole (L/mol)
46
the VH2O = 0.0182 L/1 mol
Vmix =
XV ;
i
i
typically organics have a Vi of ~0.1 L/mol


Vmix 0.1 Xi + 0.0182 XH2O
=========================
Excess Free Energy, Excess Enthalpy and
Excess Entropy
Going back to
RT lni hx Xi hx = RT lniH2OXi H2O
rearranging
RT ln
xiH2O
 (RT ln iH2O  RT ln ihx )
xihx
xiH2O
'
 KH
2O / hx
xihx
'
RT ln KH
2O / hx  (RT ln  iH2O  RT ln  ihx )
we already know that
47
-RTln Keq = G
and we will call this G, 12Gi,
RT ln K i',12  ( 12Gi )
and , 12Gi= Gi1 E -Gi2 E
in our water, hexane example of i dissolving
in both
 RT ln K'H 2O / hx  ( RT ln  iH 2O  RT ln  ihx )
or
GiE1  GiE2  ( RT ln  i 1  RT ln  i 2 )
so we could therefore say
GiE1  RT ln  i 1
and we know
GEi1 = RTln i1= HEi1-TSEi1
HEi1 is the particle molar excess enthalpy
of solution
48
and
SEi1 is the partial excess entropy
In a calorimeter, we could measure HEi1
the heat required to dissolve a compound in say water
which is a measure of the total
bonding forces that have to be broken and
fromed (vdW, polar attractions)
or hexadecane,
This would then give -TSEi1 from GEi1 - HEi1
In the gas phase HEig is its heat of
vaporization and GEig we can get from its
equilibrium partitioning const.
K1aL= MiPil*/(RTiL)
49
Table 3.3, p83
50
By calculating Gi1 E -Gi2 E from Table 3.4 page 87 between two
different phases we can estimate transfer energies and Ki12
51
Almost done,
For a liquid phase a reaction
A + B--> C + D
A =  +RT ln AXA
o
A
A =  +RT ln( [A]Vmix)
o
A
’
A
B =
dGtotal= productsdn-reactantsdn
dG/dn= C+D-(A+B)
(dG/dn =G, the molar free energy)

G=C+D-A-B
Vmix [C ] c Vmix [D ] d
RT ln
Vmix [ A ] A Vmix (B) B
52
if the reaction goes to completion
G = zero
Go  RT
ln VVmix[[AC]] c VVmix [[DB]] d
mix
A
mix
B
[i]i = an activity (I)
K eq
(C)(D)

( A)(B)
in the gas phase a reaction
aA + bB--> cC + dD
c
d
(
P
)
(
P
)
Go  RT ln C a D b
(PA ) (PB )
53
Using Linear free energy
relationships (LFERs)
In environmental systems, often
the free energy describing two
phases, say water and air, for a
compound i is assumed to be
directly related to the free energy
in two different phases (which can
often be measured or determined)
12Gi = a 32Gi + const
ln K12 = ln K32 + const’
and example is the organic
semivolatile gas-particle partitioning
coefficent Kiap, or Kp,which can be
related to an air-octanol
partitioning coefficent Kiow
log Kip= a log Kiao + b
54
Kiao, the air octanol partitioning coef.
can be determined from the ratio of
Henry’s law partitioning Kiaw,or KH’
between air and water divided by the
octanol water partitioning coef., Kiow.
Kiaw and Kiaw
are usually known or can be
estimated
Kiao = Kiaw / Kiow
In addition, there are linear free energy
techniques that permit the estimation of
equilibrium constants base on molecular
structure
This is possible if one assumes that the overall
free energy of phase transfer is related to the
linear combinations of the free energies related
to the individual parts of the molecule that are
involved in the transfer.
12Gi = 12Gparts of i + special interaction
terms
LogK i12 = LogKparts of i12 + special
interaction terms
55
56