Chapter2. Some Thermodynamics Aspects of Intermolecular Forces Chapter2. Some Thermodynamics Aspects...

Chapter2. Some Thermodynamics Aspects

of Intermolecular Forces

Chapter2. Some Thermodynamics Aspects

of Intermolecular Forces

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2.1 Interaction Energies of Molecules in Free Space and in

Interaction potential w(r) between two molecules

a Medium.

The pair potential : interaction potential between two particles

The potential of mean force : interaction in a solvent medium.

F = - dw(r)/dr The work can be done by F w(r) : the free energy or available energy.

In considering the forces between two molecules in liquids,

Solvent effects

several effects should be involved that do not arise when the

interaction occurs in free space.

1. Two solute in solvent

1) Pair potential w(r) = solute-solute interaction energy + change in solute-solvent interaction energy + change in solvent-solvent interaction energy

2) The net force depends on the attraction(solutes – solvents)

3) Solutes behavior• In free space : attraction• In a medium : repel each other

2. The ‘structure’ of solvents molecules is perturbed.

If the free energy varies with r (solute-solute)

Solvation or Structural force

3. Solute – solvent interaction can change the properties of solute.

e.g. dipole moment, charge, etc

So, the properties of dissolved molecules may be different in different media.

4. Cavity formation by solvent

Gas

Solute Condensed medium

: consider the cavity energyexpended by the medium

Cohesive energy(self-energy) : i

: The energy of an individual molecule in a medium (gas or liquid)

= The sum of the interaction with all the surrounding molecules.Now, how are i

and w(r) related?

In the gas phase, w(r) is written like this

W(r) = C/rn for r > σ ( where n > 3)

= ∞ for r < σ σ : hard sphere diameter of the molecules

igas =

Example 1.Molecules in a liquid or solid contact with 12 other molecules(close packing) When a molecule is introduced in a own liquid

Net energy change : iliq 6w(σ)

The molar cohesive energy : U = No iliq 6 No w(σ)

And = 1/(molecular volume) = 1/[(4/3)( /2 )3]

∴

Cf. For n=6, iliq 4w() or U - 4Now()

Z ∞

W(r)4r2 dr = -4C/(n-3) n-3

iliq = Z1

2W(r)4r2 dr =

-12C

(n – 3)n

- 12

(n – 3)w()

∞

Example 2.

Solute molecule is dissolved in a solvent medium. Solute molecule is surrounded by 12 solvent molecules Solute(s) size Solvent(m) size

iliq -[6 wmm() - 12 wsm()]

2.2. The Boltzmann Distribution

The effective pair potential between two dissolved solute molecules in a medium is just the change in the sum of their free energies i as they approach each other.

When the i1 is not the same i

2 in two regions of a system.

X1 = X2 exp[-(i1 - i

2)/kT] by Boltzmann distribution

For dilute system

i1 + kTln X1 = i

2 + kTln X2 Xn : equilibrium concentrations

In many different regions of states in a system

in + kTln Xn = in equilibrium for all states n = 1,2,3, …

2.3. The Distribution of Molecules and Particles in System at

Equilibrium

Case 1.i

z + kTlnρz = io + kTlnρo

ρz = ρo exp[-(iz - i

o) / kT] since (iz - i

o) = mgz z:

altitude,ρz = ρo exp[-mgz / kT] : gravitational distribution law

Case 2. For charged molecules or ions

ρ2 = ρ1 exp[-e( ψ2 - ψ1 ) / kT] ψ1 , ψ2 : electric potentials

From case 1 and 2, the interaction energy did not arise from local intermolecular interactions, but from interactions with an externally applied gravitational or electric field.

Case 3. Two phase system If one of the phaes is a pure solid or liquid, i

1= i2 + kTlnX2

X2 = X1 exp [-(i2 - i

1) / kT] = X1 exp( -Δ i / kT)

Generally, X2 = X1 exp [-Δ i + mgΔz + eΔψ / kT]

2.4. The Van der Waals Equation of State. i

gas= - 4C/(n-3) n-3 = -A

For molecules of finite sizes, Xgas = 1/(v – B) = /(1 - B)

V = : the gaseous volume occupied per molecule

B = 43/3 : the excluded volume

∴ Chemical potential of the gas

(∂μ/∂P)T = v = 1/ or (∂P/∂)T = (∂μ/∂P)T

Now pressure P is related to ,

So,

P =

o

Z ∂∂ T

d =Zo

-A + (1 -B)

kTd

= - A2 - ln(1 - B)1

2

kT

B

= igas + kTlnXgas = -A + kT ln [ / (1 - B)]

1

For B < 1,

ln(1 - B) = -B - (B)2 + … - B(1 + B)

-B / ( 1 - B) - B / ( v - B)

1

2

1

21

2

1

2

∴ P +a

v2 ( v – b) = kT : Van der Waals equation

a = A = 2C/(n-3) n-3 and b = B = 23 / 31

2

1

2

Therefore, conceptually, the constants a and b can be thought of as accounting for the attractive and repulsive forces between the molecules.

b depends on only on the molecular size and on stabilizing repulsive contribution to the total pair potential.

Van der Waals Coefficients

Gas a(L2 atm/mole2)

b(L/mole)

Ar 1.363 0.0322

CH4 2.283 0.0428

C6H6 18.24 0.1154

CO2 3.640 0.0427

H2 0.248 0.0266

H2O 5.536 0.0305

He 0.0346 0.0237

N2 1.408 0.0391

O2 1.378 0.0318

SO2 6.803 0.0564

Xe 4.250 0.0511

a coefficients correlates with the degree of polarity of substance.The most polar of these moleculeshave the highest a coefficients.

The pressure of these gasesis most significantly affectedby intermolcular attractions

b coefficients increase with the size of the atom or molecle

2.5. The Criterion of the thermal energy kT for gauging

the strength of an interaction How strong the intermolecular attraction must be if it is to condense molecules into a liquid at a particular temperature and pressure.

igas + kT logXgas = i

liq + kT logXliq ( in equilibrium)

igas – i

liq – iliq = -kT log(Xliq/Xgas) (∵i

liq >> igas )

-kT log(22400/20)

(at STP condition, Vgas~22400cm-3,Vliq~20cm-3)

– iliq 7 kTB or -N0i

liq / TB 7N0k = 7R (TB=boiling temp.)

: the TB of liq. Is simply proportional to the energy needed to take amolecule from liq. into vap.

For one mole of molecules , Uvap =-N0iliq,

Lvap = Hvap = Uvap + PV Uvap + RTB

Lvap / TB ( Uvap / TB ) + R 7R + R = 8R 70JK-1mol-1

Lvap / TB 80JK-1mol-1 (cohesive energy iliq 9kT)

Trouton’s rule

Trouton’s rule (at the Normal Boiling Point, G°vap = 0.)

the latent heat of vaporization divided by the boiling point (in kelvin) is approximately constant for a number of liquids. This is because the standard entropy of vaporization is itself roughly constant, being dominated by the large entropy of the gas.

Boiling point of a substances provides a reasonably accurate indication of the strength of the cohesive forces or energies holding molecules together in condensed phases.

The molecules will condense once their cohesive energy i

liq with all the other molecules in the condensed phase

exceeds about 9kT.

When , iliq 6w()

∴ pair interaction energy of two molecules or particles in contact exceeds about 3/2kT, then it is strong enough to condense them into a liquid or solid.

standard reference for gauging the cohesive strength of an interaction potential

X(2) = X(1)expkT

Orientational distribution

w(r, 2) –w(r, 1)

2.6. Classification of Forces

Intermolecular forces

Purely electrostatic (from Coulomb force b.t. charges)

interaction b.t. charges, permanent dipoles,

quadruples …

Polarization forces(from dipole moments induced)

Quantum mechanical

covalent or chemical bonding , repulsive steric or

exchange interaction

the chemical potential is

=

Four forces

Hellman.Feynman theorem: "Once the spatial distribution of the electron clouds has been determined by solving the Schrödinger equation, the intermolecular forces may be calculated on the basis of straightforward classical electrostatics."