Physical Chemistry of Foods

(singke) #1

lactose. In Figure 15.5c, the shapes indicated from left to right in fact
represent a series of crystals grown at increasing concentrations ofb-lactose.
Competition will generally be stronger when more competing
molecules are present. A prime example is crystallization of a multi-
component fat, which contains several molecules that are only slightly
different from each other, often in relatively high concentrations.


Inhibition. Some substances, however, can inhibit growth of a
crystal face when present in quite low concentrations. The simplest
explanation is that such components are surface active and adsorb onto
crystal faces. The adsorbed molecules then have to desorb before molecules
fitting the crystall lattice can be incorporated. If the inhibition is so strong as
to prevent growth of one or more crystal faces, the phenomenon is called
poisoning. Examples are given in Table 15.1 for two components strongly
inhibiting growth of some faces ofa-lactose. Many other cases have been
observed, but a specific molecular explanation is usually lacking. A
‘‘poisoning’’ material generally doesnotinhibit nucleation.
Finally, Table 15.1 confirms that an increase in supersaturation has a
very large effect on growth rate, and that the relative increase in rate
significantly varies between different faces. The value of lnbcan thus affect
crystal shape.


Question

When studying crystal growth in a dispersion of crystals in a supersaturated solution,
it can be observed that the linear growth rate is greater for small than for large
crystals. In other cases, the opposite is observed. Can you propose explanations?


Answer

(a) If the growth is diffusion limited, lowering of the concentration of solute will be
stronger near a large than near a small crystal, because of the difference in surface
area, hence in the amount of solute that can be incorporated in the crystal. The local
supersaturation then is greater for a small crystal, which will thus grow faster. A
prerequisite is that the average distance between crystals be substantially larger than
crystal size. (b) If the growth rate is reaction limited, another effect can become
dominant. The material in a small crystal has a higher solubility than that in a large
crystal, according to the Kelvin equation (10.9), because of its stronger average
curvature. Hence, local ceq will be higher and c – ceq is smaller, whereby
supersaturation is smaller near a smaller crystal and growth will be slower.
Prerequisites are that the small crystals be quite small (say, radius<0.1mm), and
that lnbbe low. Why?

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