van der Waals’ Interactions

• Refers to all interactions between polar or nonpolar molecules, varying as r -6.

• Includes Keesom, Debye and dispersive interactions.

• Values of interaction energy are usually only a few kT.

SummaryType of Interaction Interaction Energy, w(r)

Charge-charge rQQ

o421 Coulombic

Nonpolar-nonpolar 62

2

443

r

hrw

o

o

)(_=)(

Dispersive

Charge-nonpolar 42

2

42 rQ

o )(_

Dipole-charge24 r

Qu

ocos_

42

22

46 kTruQ

o )(_

Dipole-dipole

62

22

21

43 kTruu

o )(_

Keesom

321

22

21

4 rfuu

o ),,(_

Dipole-nonpolar

62

2

4 ru

o )(_

Debye

62

22

4231

ru

o )()cos+(_

In vacuum: =1

Interaction between ions and polar molecules

• Interactions involving charged molecules (e.g. ions) tend to be stronger than polar-polar interactions.

• For freely-rotating dipoles with a moment of u interacting with molecules with a charge of Q we saw:

42

22

46 kTruQ

o )(_

• One result of this interaction energy is the condensation of water (u = 1.85 D) caused by the presence of ions in the atmosphere.

• During a thunderstorm, ions are created that nucleate rain drops in thunderclouds (ionic nucleation).

Comparison of the Dependence of Interaction Potentials on r

Not a comparison of the magnitudes of the energies!

n = 1

n = 2

n = 3n = 6

Coulombic

van der Waals

Dipole-dipole

Cohesive Energy• Def’n.: Energy required to separate all molecules in the

condensed phase or energy holding molecules in the condensed phase.

• In Lecture 1, we found that for single molecules with a potential w(r) = Cr -n, and with n>3:

1/2 to avoid double counting!

• For one mole, Esubstance = (1/2)NAE

• Esubstance = sum of heats of melting + vaporisation.

• Predictions agree well with experiment!

with = number of molecules per unit volume -3, where is the molecular diameter. So for dispersive interactions, n = 6 and C is the London constant:

3)3(

4= nn

CE

63 3

4

3

4

CC

E =

Boiling Point• At the boiling point, TB, for a liquid, the thermal energy of a molecule, 3/2 kTB, will exactly equal the energy of attraction between molecules.

• Of course, the strongest attraction will be between the “nearest neighbours”, rather than pairs of molecules that are farther away.

• The interaction energy for van der Waals’ interactions is of the form, w(r) = -Cr -6. If molecules have a diameter of , then the shortest centre-to-centre distance will likewise be .

• Thus the boiling point is approximately:

k

wTB

23

)(=

Comparison of Theory and Experiment

63

42

CNE A

mole ~

Evaluated at close contact where r = .

k

rwTB

23

)(=Note that o and C increase with .

C can be found experimentally from deviations from the ideal gas law:

RTbVV

aP =))(+( 2

Additivity of Interactions

Molecule Mol. Wt. u (D) TB(°C)

Ethane: CH3CH3 30 0 -89

Formaldehyde: HCHO 30 2.3 -21

Methanol: CH3OH 32 1.7 64

C=OH

H

C-O-HH

HH

C-CH

HH

H

HH Dispersive only

Keesom + dispersive

H-bonding + Keesom + dispersive

Problem Set 11. Noble gases (e.g. Ar and Xe) condense to form crystals at very low temperatures. As the atoms do not undergo any chemical bonding, the crystals are held together by the London dispersion energy only. All noble gases form crystals having the face-centred cubic (FCC) structure and not the body-centred cubic (BCC) or simple cubic (SC) structures. Explain why the FCC structure is the most favourable in terms of energy, realising that the internal energy will be a minimum at the equilibrium spacing for a particular structure. Assume that the pairs have an interaction energy, u(r), described as

where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", A n, are given below for each of the three cubic lattices.

SC BCC FCC A6 8.40 14.45 12.25A12 6.20 12.13 9.11

Then derive an expression for the maximum force required to move a pair of Ar atoms from their point of contact to an infinite separation.

2. (i) Starting with an expression for the Coulomb energy, derive an expression for the interaction energy between an ion of charge ze and a polar molecule at a distance of r from the ion. The dipole moment is tilted by an angle with relation to r, as shown below.

(ii) Evaluate your expression for a Mg2+ ion (radius of 0.065 nm) dissolved in water (radius of 0.14 nm) when the water dipole is oriented normal to the ion and when the water and ion are at the point of contact. Express your answer in units of kT. Is it a significant value? (The dipole moment of water is 1.85 Debye.)

,=)(6

6

12

122r

Ar

Aru

r

ze

Molecular Crystals and Response of Condensed Matter

to Mechanical Stress

3SCMP

2 February, 2006

Lecture 3

See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20

Lennard-Jones Potential• The pair potential for isolated molecules

affected by van der Waals’ interactions only can be described by a Lennard-Jones potential:

w(r) = +B/r12 - C/r6

• The -ve r -6 term is the attractive v.d.W. contribution

• The +ve r -12 term describes the hard-core repulsion stemming from the Pauli-exclusion. 12 is a mathematically-convenient exponent with no physical significance!

• The two terms are additive.

L-J Potential for Ar (boiling point = 87 K)

London Constant, C = 4.5 x 10-78 Jm6; Guessing B = 10-134 Jm12

(m)

wmin -5 x 10-22 J

Compare to: (3/2)kTB= 2 x 10-21 J

Actual ~ 0.3 nm

(Guess for B is too large!)

Intermolecular Force for Ar (boiling point = 87 K)

F= dw/dr

(m)

Intermolecular Force for Ar (boiling point = 87 K)

F= dw/dr

(m)

(m)

Weak Nano-scale Forces Can be Measured

The AFM probe is exceedingly sharp so that only a few atoms are at its tip!

Sensitive to forces on the order of nano-Newtons.

Measuring Attractive Forces at the Nano-Scale

A = approach

B = “jump” to contact

C = contact

D = adhesion

E = pull-off

Tip deflection Force

Vertical position

AB

C

D

EC

Latex Particle Packing

Tg = 20 °C

L-J Potential in Molecular Crystals

Noble gases, such as Ar and Xe, form crystalline solids (called molecular crystals) that are held together solely by the dispersive energy.

In molecular crystals, the pair potential for neighbouring atoms (or molecules) is written as

The molecular diameter in the gas state is . Note that when r = , then w = 0.

is a bond energy, such that w(r) = - when r is at the equilibrium spacing of r = ro.

w(r) = 4[( )12 -( )6]r

r

L-J Potential in Molecular Crystals

The minimum of the potential is found from the first derivative of the potential. Also corresponds to the point where F = 0.

7

6

13

12 61240

rrF

drdw

+=== [ ]-

We can solve this expression for r to find the equilibrium spacing, ro:

1212 61

.==or

)

21

41

422

426

61

12

61

61

===)(w [[( ) - ( ] ]- -

To find the minimum energy in the potential, we can evaluate it when r = ro:

Lennard-Jones Potential

rw(r)

+

-

-

ro

Potential Energy of an Atom in a Molecular Crystal

• For each atom/molecule in a molecular crystal, we need to sum up the interaction energies between all pairs (assuming additivity of the potential energies).

• The total cohesive energy per atom is W = 1/2 r

w(r) since each

atom in a pair “owns” only 1/2 of the interaction energy.

• The particular crystal structure (FCC, BCC, etc.) defines the distances between nearest neighbours, 2nd neighbours, 3rd neighbours, etc., and it defines the number of neighbours at each distance.

• This geometric information that is determined by the crystal structure can be described by constants known as the lattice sums: A12 and A6.

• For FCC crystals, A12 = 12.13 and A6= 14.45.

Cohesive Energy of Atoms in a Molecular Crystal

w(r) = 4[( )12 -( )6]r

r

So, for a pair we write the interaction potential as:

We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12 ).

From the first derivative, we can find the equilibrium spacing for an FCC crystal:

0912 6

1

6

12 .==AA

ro ( )

For each atom in a molecular crystal, however, we write that the cohesive energy is:

W = 2[A12( )12 -A6( )6]r

r

Cohesive Energy of Atoms in a Molecular Crystal

We can evaluate W when r = ro to find for an FCC crystal:

682 12

26 .==

AA- -

This expression represents the energy holding an atom/molecule within the molecular crystal. Its value is only 8.6 times the interaction energy for an isolated pair.

This result demonstrates that the dispersive energy is operative over fairly short distances, so that most of the interaction energy is contributed by the nearest neighbours. In an FCC crystal, each atom has 12 nearest neighbours!

W

Elastic Modulus of Molecular Crystals

We can model the intermolecular force using a spring with a spring constant, k. The force, F, to separate two atoms in the crystal is: F = k(r - ro).

The tensile stress t is defined as a force acting per unit area, so that:

2AF

o

ot r

rrk )(=

-

Fro

The tensile strain t is given as the change in length as a result of the stress:

o

o

ot r

rrrr

==

-

oo

FA

L

The Young’s modulus, Y, relates tensile stress and strain:

ttY

=

Connection between the atomic and the macroscopic

Y

t

t

Y can thus be expressed in terms of atomic interactions:

oo

o

o

o

tt

rk

rrr

rrrk

Y =)(

)(

==2

-

-What is k?

rW

+

-

-8.6ro

F = 0 when r = ro

orrdr

dFk ==

Elastic Modulus of Molecular Crystals

rF

+

-

ro

drdW

F =

Elastic Modulus of Molecular Crystals

Force to separate atoms is the derivative of the potential:

7

66

13

1212 612

2r

A

r

AF

drdW

+== [ ]-

So, taking the derivative again:

8

66

14

1212 671213(

2r

A

r

AdrdF

)()

= [ ]-

But we already know that: 61

6

122AA

ro =( )

So we see that: 61

12

6

2AA

ro= ( )We will therefore make a substitution for when finding k.

Elastic Modulus of Molecular Crystals

To find k, we now need to evaluate dF/dr when r = ro.

8

66

14

1212 671213(

2r

A

r

AdrdF

)()

= [ ]-

Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write:

)(== 221

8

62

14

121 22

oo

o

o

o

r

CC

r

rC

r

rCk [ ]- -

Finally, we find the Young’s modulus to be:

3212

oo r

CCrk

Y)(

== -

As ro3 can be considered an atomic volume, we see that the

modulus can be considered an energy density, directly related to the pair interaction energy.

Bulk Modulus of Molecular Crystals

We recall the thermodynamic identity: dU = TdS - PdV

The definition of the bulk modulus, B, is: TVP

VB )(=

-

This identity tells us that: SVU

P )(=

-

If we neglect the kinetic energy in a crystal, then U W.

So B can be written as:TSTS V

UV

VU

VVB ,)(+=])([= 2

2

- -

After writing V in terms of , and differentiating W, we obtain for an FCC molecular crystal:

325

12

6123

754

=)(= /

AA

AB

Theory of Molecular Crystals Compares Well with Experiments

w(ro)

Response of Condensed Matter to Shear Stress

When exposed to a shear stress, the response of condensed matter can fall between two extremes: Hookean (solid-like) or Newtonian (liquid-like)

How does soft matter respond to shear stress?

A

A

y

F

AF

s =

Elastic Response of Hookean Solids

No time-dependence in the response to stress. Strain is instantaneous and constant over time.

The shear strain s is given by the angle (in units of radians).

The shear strain s is linearly related to the shear stress by the shear modulus, G:

Gs

s

=

A

A

y

FAF

s =

yx

s

~=

x

Viscous Response of Newtonian Liquids

AF

s =

A

A

y

Fx

tx

v

=

There is a velocity gradient (v/y) normal to the area. The viscosity relates the shear stress, s, to the velocity gradient.

ytx

yv

s

==

The viscosity can thus be seen to relate the shear stress to the shear rate:

====tty

xyt

xs

The top plane moves at a constant velocity, v, in response to a shear stress:

v

has S.I. units of Pa s.

The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate:

t

= Units of s-1

Hookean Solids vs. Newtonian Liquids

Hookean Solids:

G=Newtonian Liquids:

=Many substances, i.e. “structured liquids”, display both type of behaviour, depending on the time scale.

Examples include colloidal dispersions and melted polymers.

This type of response is called “viscoelastic”.

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity

When a constant stress is applied, the molecules initially bear the stress. Over time, they can re-arrange and flow to relieve the stress:

The shear strain, and hence the shear modulus, both change over time: s(t) and

)(=)(

ttG

s

s

t

s

s ttG

)(=

)(1

Elastic response

Viscous response

(strain is constant over time)

(strain increases over time)

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity

t

s

ttG

)(=

)(1

oG1

Slope:

1

==)(s

s

s

s

dtd

We see that 1/Go (1/)

is the “relaxation time”

Hence, viscosity can be approximated as Go

Example of Viscoelasticity

Physical Meaning of the Relaxation Time

time

Constant strain applied

Stress relaxes over time as molecules re-arrange

time

teGt =)(Stress relaxation:

Typical Relaxation Times

For solids, is exceedingly large: 1012 s

For simple liquids, is very small: 10 -12 s

For soft matter, takes intermediate values. For instance, for melted polymers, 1 s.

Viscosity Sometimes Depends on the Shear Rate

Newtonian:

s

Shear thinning or thickening:

s

s s

s s

An Example of Shear Thickening

Future lectures will explain how polymers and colloids respond to shear stress.

Problem Set 2

1. Calculate the energy required to separate two atoms from their equilibrium spacing ro to a very large distance apart. Then calculate the maximum force required to separate the atoms. The pair potential is given as w(r) = - A/r6 + B/r12, where A = 10-77 Jm6 and B = 10-134 Jm12. Finally, provide a rough estimate of the modulus of a solid composed of these atoms.

2. The latent heat of vaporisation of water is given as 40.7 kJ mole -1. The temperature dependence of the

viscosity of water is given in the table below. (i) Does the viscosity follow the expectations of an Arrhenius relationship with a reasonable activation energy?(ii) The shear modulus G of ice at 0 C is 2.5 x 109 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water Go and estimate the characteristic frequency of vibration for water, .

Temp (C) 0 10 20 30 40 50(10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47

Temp (C) 60 70 80 90 100(10-4 Pa s) 4.67 4.04 3.54 3.15 2.82

3. In poly(styrene) the relaxation time for configurational rearrangements follows a Vogel-Fulcher law given as

= o exp(B/T-To),

where B = 710 C and To = 50 C. In an experiment with an effective timescale of exp = 1000 s, the glass transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment with exp = 105 s, what value of Tg would be obtained?