needed for the reaction is present in low concentration and has to diffuse
from another compartment, viz., the air above the food, to the oxidizable
components. This will naturally slow down the reaction.
Monoglycerides dissolved in oil droplets can hardly react with water to
be hydrolyzed into glycerol and fatty acids, and certainly not when the
reaction has to be catalyzed by an enzyme (an esterase), which is in the water
phase. Possibly, the reaction proceeds at the interface between oil and water,
which usually means that it will be slow; but if a suitable enzyme adsorbs onto
the said interface, the reaction may in fact be quite fast. If one of the reactants
is immobilized at a particle surface, this will in general slow down the rate, if
only because the diffusion coefficient for that reactant is effectively zero.
In plant and animal cells, many enzymes are compartmentalized, and
several are also immobilized, greatly slowing down reactions. After the cells
have been mechanically damaged, some reactions may proceed fast. A well-
known example is the rapid enzymatic browning of apple tissue after the
apple has been cut; here, the cutting allows an enzyme, polyphenoloxidase,
to reach its substrate, mainly chlorogenic acid.
Slowness of Reactions. In foods, we are often concerned with
reactions that are very much slower than the reactions studied in the
chemical or biochemical laboratory. For instance, the maturation of
products like hard cheese, wine, and chutney may take years, and the
maturation is ultimately due to chemical reactions, often enzyme catalyzed.
Slow quality loss or deterioration of foods is ubiquitous. An example is loss
of vitamins; when we accept a loss of 10%per year and assume it to be due
to a first order reaction, this would imply that its rate constant would be
given by ln 0: 9 ¼k 63656246 3600, ork& 3 : 3? 10 ^9 s^1 , a very small
rate constant. We may also consider loss of available lysine due to Maillard
reactions. Taking a very simplified view, this is ultimately due to a second-
order reaction between a reducing sugar and the exposed lysine residues of
protein. Assume that we have 10%glucose (say in orange juice) and that 1%
of it would be in the reducing, i.e., open-chain, form, this gives a molarity of
about 0.005. Assume 1%protein, 7%of the residues of which are lysine, of
which 80%would be reactive, this gives a molarity of about 0.0047, to which
must be added some free lysine, say 0.0003 molar. If we accept a lysine loss
of 1%in 3 months, i.e., 6? 10 ^5 molar per 3 months, we have
d½L
dt¼
6? 10 ^5
9162463600
¼ 8? 10 ^12 s^1 ¼k 60 : 00560 : 005yieldingk& 3? 10 ^7 L?mol^1 ?s^1. Such a low rate constant would imply a
fairly large activation free energy, i.e., 136 kJ?mol^1 according to Eq. (4.12).