Physical Chemistry of Foods

equation(10.9), which can be written as

RTln

sr

s?



¼

2 gV

r

ð 14 : 8 Þ

wheresis solubility andVmolar volume. It gives the equilibrium radius for
a very small particle of the new phase (an embryo), wheresr/s?is the
supersaturation ratio. The lower the temperature, the greater the super-
saturation and the smaller the equilibrium radius. Putting the latter equal to
rcr, we can write the Kelvin equation as

Dm¼RTlnb¼

2 gV

rcr

ð 14 : 9 Þ

whereDmis the difference in chemical potential,bis thesupersaturation
ratio, and lnbis called thesupersaturation. Equation (14.9) can be shown to
be identical to Eq. (14.6) in the present case. The larger the supersaturation
is, the smaller the equilibrium value ofr, i.e.,rcr. It may be noted that Eq.
(14.9) has general validity, whatever the cause of the difference in chemical
potential.

Nucleation Rate. The value ofDGmaxis often considered to be an
activation free energy for nucleus formation. Moreover, it is generally
assumed that there is also a certain barrier to incorporation of a molecule
into an embryo, owing to diffusional resistance; this is usually expressed in a
molar activation free energy for transportDG*. Classical nucleation theory
then stipulates that the nucleation rate (number of nuclei formed per unit
time per unit volume) will be given by

Jhom¼N

kBT

hP

exp 

DG

RT



exp 


DGmax

kBT



¼N

kBT

hP

exp 

DG

RT



exp 

16 pg^3 Teq^2

3 ðDHVÞ^2 ðDTÞ^2 kBT

"

ð 14 : 10 Þ

whereNis the number per unit volume of molecules that can form the new
phase. As mentioned in Section 4.3.3, the factorkBT/h(of order 6? 1012 s
^1 ),
wherehPis Planck’s constant, is the maximum reaction rate possible, i.e., in
the absence of free energy barriers.

Examples. Some examples of results, experimentally observed and
interpreted by means of Eq. (14.10), are in Table 14.2. An important
conclusion that can directly be drawn from the equation is that the second