Lecture 13
Steric Interactions and Entropic
repulsion
What did we cover in the last
lecture?
The pressure exerted between charged
surfaces in an electrolyte is the sum of osmotic
pressure and dispersive terms
2 2
2
o
z eV
A
Ptot  no
exp( D) 
3
k BT
6

D


osmotic pressure
attractivedispersion
forces
In this lecture…
1) Steric (Entropic) stabilisation of particles and
surfaces
2) Surfaces with grafted molecules
3) Boltzmann form of entropy
4) Free Energy of grafted molecules
5) Entropic forces between sterically stabilised
surfaces
Further Reading
Gases, Liquids and Solids, 3rd Edition, D. Tabor,
Cambridge University Press, p326-321, 1993
Intermolecular and Surface forces, J.
Israelachvili, chapter 14
Colloidal stability
It is sometimes not possible to use
charges to keep particles and
surfaces apart in solution
This is true when using organic
solvents as the suspending
medium (e.g. in ferrofluids), as
ionic solids (electrolytes) are often
poorly soluble in these solvents
An alternative route to stabilisation is to use
Steric (or entropic) repulsion effects between
surfaces
Entropic (Steric) Forces
Surfaces can be decorated with long molecules (e.g. polymers)
Molecules are tethered to a
surface by one end and can
wave around freely in
solution. As a result they have
a large number of
configurations (directions in
which they can point)
When two surfaces come close together, the number of
configurations that the molecules can adopt is reduced. This
reduces the entropy (or measure of disorder).
The system resists this reduction in entropy by generating a
repulsive force…
Boltzmann form of Entropy
Boltzmann derived a statistical form of the entropy of a
system, S, such that
S  kB ln W
where W is the number of available
(micro) states corresponding to a
given macrostate and kB is
Boltzmann’s constant (JK-1 ). In this
case…
Macrostate → molecule tethered to
the surface
Microstate → a single orientation of
the molecule
The entropy of a tethered rod
Suppose we treat the molecule as a
rigid rod with length, l, and area, a,
at each of its ends (See OHP)
If we allow the rod to point in any
direction it can sweep out half a
sphere with a total area, A1
The number of different possible
microstates (orientations of the rod),
W, is then given by dividing ratio of
A1 to a
So the entropy (S1) of the tethered
rod becomes
4l 2
A1 
 2l 2
2
A1 2l
W

a
a
2
 2l 2 

S1  k B ln 
 a 
Confinement of tethered
molecules
When a solid surface
approaches the tethered rod,
its motion becomes restricted
and it can sweep out a much
reduced area
At a separation D between the
surfaces, the total area that can be
swept out is A2 where
The entropy of these ‘confined’
molecules is therefore
A2  2lD
 2lD 
S 2  k B ln 

 a 
How do we extract a force?
We can extract a force if we know
how the entropic contribution to the
interaction energy , U, depends upon
separation D
dU
F 
dD
The change in the Gibbs free energy of the system at
constant temperature is given by
U  H  TS
Where H and S are the change in enthalpy (energy of
interaction) and entropy caused by bringing the surfaces
together
Entropic repulsion Force per
molecule
If we consider only the entropy contribution (for now) the
force per molecule is
d (TS )
d (S )
F 
T
dD
dD
where
 2l 2 
 2lD 

S  S 2  S1  k B ln 
  k B ln 
 a 
 a 
D
S  k B ln  
l 
So the force F is
k BT
F
D
Pressure due to entropic forces
The force between the surfaces is
positive and therefore repulsive
If the separation between
tethering points is, d, then the
effective area occupied by one
molecule is d2
The force per unit area (Pressure) is P,
where
k BT
P 2
d D
k BT
F
D
Problem
Two perfectly flat surfaces are brought into close proximity
such that dispersion forces attract them towards one
another. One of the surfaces is coated with a layer of rodlike molecules of length 3nm with the highest possible
surface grafting density that allows the free end of the
molecules to rotate freely
1) Calculate the magnitude of the pressure exerted on the
surfaces due to repulsive steric (entropic) forces, at
room temperature, if the surfaces are separated by a
distance of 2.5nm
2) Calculate the equilibrium separation of the surfaces at
room temperature if the Hamaker constant is 1x10-20 J
Summary of key concepts
When the motion of molecules
near a surface is confined it
reduces the entropy of the
system
This gives rise to repulsive
force which acts to push
surfaces apart and increase
the entropy again
For rod like molecules on a
surface which occupy an area
d2, the pressure between
surfaces at a separation, D, is
given by
k BT
P 2
d D