Lecture 2 - Forces
Reading: ‘Molecular Biophysics’, M.Daune, ‘Intermolecular and Surface
Forces’, J.Israelachvili
Study a whole series of mesoscopic forces: Van der Waals, hydrogen
bonding, electrostatics, steric and fluctuation forces, depletion forces,
hydrodynamic interactions and experimental measurements. Nature has
used all these forces to determine the interactions of biological molecules.
2.1) Van der Waals
~1kJmol-1
Act between all atoms and molecules (even neutral ones) –dispersion
forces
 an attractive dispersion force of quantum mechanical origin operates
between any two molecules, which arises form the interaction between
2
oscillating dipoles.
disp
12
V
R   
nm0
1
1
24 0 
2
1
R6
nm0
 E
n, k
n
1
2
km0
1
0
1
2
 E  E 2k  E 20 

C12
R6
is the transition dipole moment from state n to 0, R is the
distance between the two molecules, and E1n is the energy of state n.
Called a dispersion interaction because the same quantities determine the
optical properties of the molecules.
The dispersion interaction is the main cause of cohesion in condensed
matter states.
1
a) They are long range forces and can be effective from large distances
(>10nm) down to interatomic spacings (<0.2nm).
b) Forces may be repulsive or attractive. In general it does not follow a
simple power law
c) Dispersion forces tends to bring molecules together, but also tend to
mutually align or orient them.
d) Unlike gravitational and coulomb forces, van der waals forces are not
generally additive.
e) At larger separations (>10nm) the effect of the finite speed of
propagation (c) becomes important. This is the retardation effect - r-7
dependence on separation rather than r-6, and semi-infinite sheets h-3
rather than h-2.
Note that with the Van der Waals forces between two bodies the
geometry is very important. See figure.
2
WORKED EXAMPLE
QUESTION:
Many small molecules have ionization potentials I close to 2*10-18J. If
their polarizability can be modelled in terms of the Bohr atom show that
the strength of a typical van der Waals ‘bond’ is always approx. a few kT
at room temperature, irrespective of the size or polarizability of the
molecules.
ANSWER:
The strength of a van der Waals bond is given by w(r) at a separation
r=2a, where a is the molecular radius. From the Bohr atom model of
polarizability =40a3, so with r=2a into the London equ
3
  02 I
4
w(r ) 
4 0 2 r 6
W(r=2a)2*10-20J, which is a few kT independent of a or .
3
2.2) Hydrogen Bonding
Water exhibits unusually strong interaction between molecules, which
persist into the solid state.
Eg. See figure
Hydrogen bonds ~ 10-40kJmol-1
Plays a central role in molecular self-assembly, micelle formation,
biological membrane structure and determining the conformation of
proteins. They occur between a proton donor group D, which is strongly
polar FH, OH, NH, SH and a proton acceptor atom A which is slightly
electronegative F, O, N.
Clathrate structures form around hydrophobic compounds. These are
labile, but the water molecules are more ordered in the cages.
For hydrocarbons the free energy of transfer is proportional to the surface
area of the molecules. Ab initio methods to quantify the strength of
hydrogen bonds are still at the rudimentary level. One stumbling block is
the possibility for bifurcation of bonds. Another is the wide dynamic
spectra of behaviour.
O=C
NH
O=C
SFA evidence has been provided on the long range nature of the
hydrophobic effect. It is still an active area of study. The energy of
repulsion has the form
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W  W0 e  D / 
 is the decay length ~nms.
2.3) Electrostatics
Ionic bond (~500kJmol-1)
The electric ‘double layer’
Charging of a surface in a liquid can happen in two ways
i)
by the ionisation or dissociation of surface groups (eg surface
carboxylic groups -COOH-COO-+H+) which leave behind a –
vely charged surface.
ii)
By the adsorption of an ion from solution onto a previously
uncharged surface eg binding Ca2+ onto zwitterionic head groups of
lipid bilayers surfaces charging he surface +vely.
-vely
charged
surface
-
+
+
-
+
-
Water
+
+
Diffuse
Counterion
cloud
The chemical potential (total free energy per molecule) is the sum of two
terms
  ze  kT log 
 is the potential and  is the number density of counterions.
This is consistent with the Boltzmann distribution
   0 e  ze / kT
 0  e  / kT
Poisson equation for the electrostatics is.
 r  0  2    freeion
5
Solving for the density of counterions away from the surface.
Combining with the Boltzmann distribution gives the Poisson Boltzmann
equation
 ze 0   ze / kT
d 2
ze

e




 0

dx 2
 0 
When solved the PB equation gives the potential , the electric field
E=/x and counterion density  at any point in the gap between the
surfaces.
Limitations of the PB equ. at short separations include:
Ion correlation effects, finite ion effects, image forces, discreteness of
surface charges and solvation forces are involved.
Pressure between 2 charged surfaces in water: contact value theorem
P(r )  kT S r    S 
The pressure is given by the increase in the ion concentration at the
surfaces as they approach each other – contact value theorem. Valid as
long as there is no interaction between the counterions and the surfaces
(see discussion of cartilage later).
It is a very general theory valid for double layer interactions, solvation
interactions, polymer associated steric and depletion interactions,
undulations and protrusion forces.
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WORKED EXAMPLE
QUESTION:
Is the electric field near a charged surface (=0.3Cm-2) sufficiently
intense to immobilise the water molecules adjacent to it? E-field needs to
be greater than 109 Vm-1 for this to happen.
ANSWER:
 
D/2

zedx   0
D/2

0
Surface
charge
density
0
d 2
dx
dx 2
Sum up the
charges
Substitute
Poisson’s
equation
d
d
  0
  0 E
dx D / 2
dx s

0.3
ES 

 4.2 *10 8 Vm 1
12
 0 808.85 *10 
  0
Z valency
E electronic charge
D distance form the surface
 potential
x displacement
 dielectric permitivity of
free space
r relative permitivity
No
QUESTION:
The solution to the PB equation near a plane surface is:

 kT 
    log cos 2 Kx
 ze 

K 
2
ze2  0
2 0 kT
If two surfaces with =0.2Cm-2 are placed at D=2nm and K=1.34*109m-1
what is the repulsive pressure between them?
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ANSWER
From the contact value theorem we have
2
 kT 
P  kT 0  2 0   K 2  1.7 *10 6 Nm  2
 ze 
This a large value equivalent to 17 atmospheres
0 is the ion concentration
P is the pressure
K is from the solution of
Poission equation
kT thermal energy
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Electrostatic interactions
Explicitly the electrostatics can be calculated in a MD simulation.
Coulomb’s law
EC 
q1 q 2
4 0 r
 dielectric constant
Ion-dipole interaction
qm

p
O
EP  
p2q2
4 0 2 3kTr 2
Dipole-dipole
EP 
p1 p 2 K
4 0 r 3
DLVO
A surprisingly successful theory for forces between colloidal particles. It
has received confirmation from optical tweezers, light scattering
(+neutron and X-ray), coagulation and surface force apparatus.
The competition between attractive van der Waals and repulsive double
layer forces determines the stability or instability of many colloidal
systems.
Sketch the potential between colloidal particles
64kTc0* 02 h
H 121
V h 


e
2
area

12h
Van der
Waals
Component
Electrostatic
Component
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WORKED EXAMPLE
QUESTION:
For a number of colloidal systems it is found that the critical coagulation
concentration varies as the inverse sixth power of the valency z of the
electrolyte counterions ie 1/z6. Is this empirical observation, known
as the Schultz-Hardy rule, consistent with the DLVO theory?
ANSWER:
The total DLVO interaction potential between 2 spherical particles
interacting at constant potential is:
 64kTR   2  D AR
e 
W D   
2
6D



D is the interparticle distance
A Hamaker constant
 density of colloids
kT thermal energy
-1 Debye screening length
 surface potential
R radius of particle
The critical coagulation concentration occurs when both W=0 and
dW/dD=0. This leads to
 2 /    384kTD 2 e D / A
The second condition leads to D=1, this show that the potential
maximum occurs at D=-1 . Inserting this above leads to
 3 /    768kT 2 e 1 / A
Now
 2  z 2 / T
 is constant at high surface potentials (=1)
10
z 6     3T 5 4 / A 2
Therefore 1/z6
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2.4) Steric and Fluctuation Forces
> Packing constraints produce oscillatory force curves with a period
determined by solvent size and are most readily observed between two
smooth hard surfaces.
Eg between two hard surfaces
 2h   h /
e
F pack h  const * cos



Where h is the separation of the surfaces and  is the diameter of a
molecule
>Polymers at surfaces eg proteins on the surface of interacting
membranes, synovial joints, DNA on histones etc.
a) for a polymer to be effective at stabilising a colloid the solvent must
be a good solvent for the polymer.
b) Range of interaction is governed by the distance from the surface the
polymer chains extend.
c) Chains can be attached by absorption or chemical means.
D
Steric forces
The repulsive energy per unit area is roughly exponential
W D   36kTe
 D / Rg
Rg is the unperturbed radius of gyration and D is the separation between
the surfaces.
 Interacting membranes
Undulation
forces
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Entropic force per unit area between 2 surfaces
PD   kT  S ( D)   S ()
(D) volume density of molecular contacts
S(D)=1/(volumeof mode)=1/x2D
D
S()=0
.
x
2k
Eb  2b
R
Per unit area
R
Each mode occupies an area x2 and has an energy of kT
2x 2 k b 4Dk b
kT 

R
R2
‘Chord theorem’
x 2  2 RD
kT
kT
P(D )  2 
x D 2RD2
P(D ) 
kT 2
kb D 3
This has been verified experimentally.
Note that there are also peristaltic and protrusion forces between
membranes which are weaker, but also measureable.
2.5) Depletion forces
Lowering a solvents osmotic pressure creates depletion forces that drive
colloidal surfaces together. This is a trick often used to promote protein
crystallisation for structural studies.
Water soluble
Polymer chains (eg
PEG)
Depletion force
Colloidal
sphere
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For a dilute solution of polymers
N
kT
V
  PosmVdep
Posm 
Fdep
Vdep is the volume between the spheres for which the polymer is depleted.
Rg is the radius of gyration, N/V is the concentration of polymer
molecules. You want a high Mw and a high polymer concentration for a
strong depletion potential. The equation was first verified by the
measurement of the force between 2 interacting bilayer surfaces in a
concentrated dextran solution.
2.6) Hydrodynamic Interactions
These mesoscopic forces have a time scale associated with them and the
dynamics of solvents, counterion clouds and tethered polymers need to be
understood to gauge the interaction potentials.
2.7) Experimental Measurements of Intermolecular and
Surface Forces
i)
Thermodynamic data on gases, liquids and solids (PVT data,
boiling points, latent heats of vaporisation, lattice energies).
Adsorption isotherms provide information on interactions of
molecules with surfaces. Information on short range potentials.
ii)
Physical data on gases, liquids and solids (eg molecular beam
scattering experiments viscosity, diffusion, compressibility, NMR,
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x-ray, and neutron scattering) provide information on short range
interactions of molecules especially their repulsive forces.
iii)
Thermodynamic data on liquids and liquid mixtures (phase
diagrams, solubility, partitioning, miscibility, osmotic pressure)
provide information on short range solute solvent and solute-solute
interactions.
iv)
Particle detachment and peeling experiments provide information
on particle adhesion forces and the adhesion energies of solid
surfaces in contact.
v)
Measuring the force between 2 macroscopic surfaces as a function
of surface separation can provide full force law of interaction.
vi)
Surface studies such as surface tension and contact angle
measurement gives information on liquid-liquid and solid-liquid
adhesion energies.
vii)
The thicknesses of free soap films and liquid films absorbed on
surfaces can be measured as a function of salt concentration or
vapour pressure.
viii) Dynamic interparticle separation and motion in liquids can be
measured using NMR, light scattering, X-ray scattering and
neutron scattering.
ix)
Coagulation studies on colloidal dispersions, the salt concentration,
pH or temperature of suspending liquid medium is changed until
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the dispersion becomes unstable and the particles coalesce
(coagulate or flocculate).
light
Adsorbed
Liquid film
f)
Stable
colloid
e)
Unstable
colloid
piston
solution
Semipremeable
membrane
Light,
x-rays,
neutrons
salt
clear
x-rays,
neutron
s
g)
Plate-like
particles
h)
Particles
coagulated
Particles
dispersed
i)
a) adhesion measurements
b) peeling measurements
c) direct measure of force
d) contact angle measurement
e) equilibrium thickness of thin films
f) equilibrium thickness of thin adsorbed films
g) interparticle spacing in liquids
h) sheet like particle spacing in liquids
i)
coagulation studies
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