Intermolecular Forces and Partitioning

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

Much of this class deals with the partitioning of an organic compoundibetween 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???

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)

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.

c. dipole-dipole interactions: strength of attraction proportional to 1x 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

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 , dispg 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)

dispg = -(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 Lorenz-Lorenz relationship

1

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)

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

Since NA, CA, 3 and I are relatively constant,

1

If the equilibrium is dominated by dispersive forces, this free energy, dispG 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]

dispG= - RT ln Keq

we should be able to plot

calculated ln KiaL = M p*iL/[iLRT]

vs.

for a pure solvent interacting with the gas phase, i=1

Figure 3.6 page 71air-hexane, top, air-water, bottom

Chapter 3, then uses thermodynamics to quantify molecular energies and equilibrium partitioning

Section 3.3 starts with:

and

How do we get to these equations and what do they mean??

Chapter 3

The First Law

U2 - U1 = q - w

work

change in

internal energy heat

of an object

reservoir

object

b

U = q1-w1

U = q2-w2

a

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

wrev =  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?

pV=nRT

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

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

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
Standard Heats of formation

kcal/molekcal/mole

C graphite 0H2O(g)-57.78

CO(g)-26.45 H2O(l)-68.32

CO2(g)-94.05ethene+12.50

Benzene+19.82ethane -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

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?
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

define

dS = dq/T

and

at absolute zero the entropy is assumed to be zero

Consider 1 mole of H20 (l)---> H20 g at 100oC

SH20= dq/T = 1/Tdq = 1/T

= 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.

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 ---> BB 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:

# combinations

4 heads, 0 tails1

3 heads, 1 tail4

2 heads, 2 tails6

1 head, 3 tails4

0 heads, 4 tails1

we might consider this to be the normal state, or equilibrium state, because there are more combinations to “go to”

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

Figure 6.7, Physical Chemistry, Barrow, McGraw Hill, New York, 1963, page 147

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
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)

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

if the reaction goes to completion

G = zero

GO= -RT lnKeq

Equilibrium Constants and Temperature

We are now ready for the 1sttwo equations of

Chapter 3, Section 3.3

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

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 considered to be a function of the entropy, the volume and the change in the # moles

the term is call the Chemical Potential , .

The complete expression for the differential of free energy is

dG = dU +pdV+Vdp -TdS- SdT

if

Substituting for dU in the free energy expression

or we could directly define

if we rewrite the equation for the potential energy in its integrated form

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...

since H= U+ pV

and

G= H –TS

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

what does this say about systems that are not at equilibrium

Chemical Potentials and Pressure

If we go back to the expression for potential energy

dU = SdT+ TdS -Vdp-pdV+

For a closed system which only does pressure volume work we said that

subtracting

0 = SdT -Vdp+

At constant temperature, one obtains the Gibbs-Duhem Equation for gases

Vdpi =

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

What if the system is not ideal?

Van der Waal’s equation

inter molecularoccupied molecular

attractionvolume

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

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igpiL*

where xi is the mole fraction

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*

Where i is called an activity coefficient

if we discuss compound iin a liquid mixture

fiL = i Xipi*L (pure liquid)

the fugacity of compound iwith 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

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, Xior the “real” concentration via i

Phase Transfer Processes

Consider a compound,i ,which is dissolved in two liquids which are immiscible like water and hexane.

at equilibrium

iH2O = 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)

RT lnfi hx /fiL*(pure liquid) = RT lnfi H2O/fiL*(pure liquid)

fihx = 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 hxXi hx= RT lniH2OXi H2O

What is the activity coefficient and mol fraction of hexane in hexane?

This gives the important result: iH2O=1/Xi H2O

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)

the VH2O = 0.0182 L/1 mol

Vmix =  Xi Vi ;

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 hxXi hx= RT lniH2OXi H2O

rearranging

we already know that

-RTln Keq = G

and we will call this G, 12Gi,

and , 12Gi= Gi1 E -Gi2 E

in our water, hexane example of i dissolving in both

or

so we could therefore say

and we know

GEi1 = RTln i1= HEi1-TSEi1

HEi1is the particle molar excess enthalpy of solution

and

SEi1 is the partial excess entropy

In a calorimeter, we could measure HEi1the heat required to dissolve a compound in say water or hexadecane, which is a measure of the total bonding forces that have to be broken and fromed (vdW, polar attractions)

This would then give -TSEi1 from GEi1 - HEi1

In the gas phaseHEig is its heat of vaporization and GEig we can get from its equilibrium partitioning const.

K1aL= MiPil*/(RTiL)

Almost done,

For a liquid phase a reaction

A + B--> C + D

A = oA +RT ln AXA

A= oA +RT ln(’A [A]Vmix)

B =

dGtotal= productsdn-reactantsdn

dG/dn= C+D-(A+B)

(dG/dn =G, the molar free energy)

G=C+D-A-B

if the reaction goes to completion

G = zero

[i]i = an activity (I)

in the gas phase a reaction

aA + bB--> cC + dD

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

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

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