The driving force for crystallizati

The driving force for crystallization in a supersaturated solution is the difference in chemical potential,
, between a supersaturated solution and a saturated
solution, and is given by (for ideal solutions):
sol ¼RgT:ln; ð2Þ
whereRg is the universal gas constant and Tis the
temperature. It should be noted that unless the phase
behavior of the many TAG families usually present in a
typical fat is known and supports it, making an
assumption of ideal solution behavior is not necessarily
valid.
Relative supercooling refers to the degree to which the
sample has been cooled,T, with respect to the melting
temperature,TM, of the crystallized sample and is given
by:
T¼ðTTMÞ: ð3Þ
Supercooling is usually required for the crystallization
of one or more components from a melt; as opposed to
a solution. The supercooling of the sample also provides
a thermodynamic driving force for the formation of
nuclei, and a chemical potential that drives nucleation
may be defined as:
melt¼H
TMT
TM
; ð4Þ
whereHis the enthalpy of fusion.
Nucleation (via the chemical potentials provided by
either supercooling, supersaturation, or a combination)
occurs via bimolecular reactions which lead to the formation of ordered domains (Kloek, 1998). Beyond a
certain size, further addition of molecules to such
ordered domains result in a decrease in the Gibbs free
energy of the system, and therefore when such ordered
domains grow beyond a critical size,r , a nucleus is
formed. As described by Kloek (1998)and Lyklema
(1991), the classical nucleation theory described by
Volmer (1939)has a number of shortcomings. Firstly, a
macroscopic model is applied to the microscopic
ordered domains. Furthermore, Gibbs free energy
changes generated from equilibrium thermodynamics is
used in the classical theory as a measure of the activation Gibbs energy; which is strictly a kinetic parameter.
Regardless, the classical nucleation theory is used in the
lipid area extensively, and is therefore reproduced here.
The Gibbs free energy change due to the formation of
an ordered domain is given by:
G¼GSSþGVV; ð5Þ
whereGSis the change in the surface free energy (due
to surface tension),GVis the change in free energy of
the system per unit volume (due to the enthalpy of
fusion), andVis the volume of the ordered domain.
G¼; ð6Þ
whereis the surface energy andSis the surface area of
the ordered domain. Therefore, for an ordered domain
considered spherical, with radiusr:
GSS¼4r
2
; ð7Þ
S¼4r
2
and
V¼
4
3
r
3
: ð8Þ
Therefore,
G¼4r
2
þ
4
3
r
3
GV: ð9Þ
For there to be a net decrease in the free energy of the
system, the ordered domain must attain a critical size of
r , where Gis maximum and is referred to as the
activation energy of nucleation, G (as mentioned
before, this is a kinetic parameter). One can calculate
this value ofr by differentiatingEq. (9)with respect to
r, and equating to zero:
8r 3þ4r 4GV¼0; ð10Þ
therefore
4r 32þr GV ½¼0; ð11Þ
and
r ¼
2
GV
: ð1