Order and Molecularity. If we have a very simple reaction, like
HþH?H 2 in the gas phase, this is a true bimolecular reaction and also the
order is two. Such a simple correspondence between molecularity and order
is, however, the exception rather than the rule. For example, a reaction
involving water as a reactant in a dilute aqueous solution, for instance, the
hydrolysis of an ester,
R 22 CO 22 OCH 222 R^0 þH 2 O?R 22 COOHþR^022 CH 2 OHmay effectively be first order, although it is a bimolecular reaction. This is
because the concentration of water (or the water activity, rather) does not
significantly alter during the reaction. Another point is that even a fairly
simple reaction may involve a number of elementary reactions or steps, and
one of them may be effectively rate determining, thereby also determining
the order. Consequently, the order has to be established from experimental
results, and it may be very difficult to elucidate the elementary reactions
involved. In many cases, it turns out that the order is not an integer number;
or the order may change in the course of the reaction; or it may be different
with respect to different reactants; or the order with respect to concentration
may be different from that with respect to time.
A simple (and probably oversimplified) example will be discussed, the
inactivation of an enzyme by heat treatment. At a high temperature, the
protein molecule will unfold, but if nothing else happens, it will probably
refold after cooling and thereby regain its enzyme activity. This means that
the unfolded molecule must undergo a reaction that prevents it from
refolding into its native conformation. In the simplest situation we thus have
N?U?I, where N is the native, U the unfolded, and I the inactivated state.
The second reaction will mostly involve other molecules, but we will assume
here that both the first and the second step are first-order reactions. We then
have
d½N
dt
¼k 1 ?½Nand
d½I
dt¼k 2 ?½USimple though this set of differential equations may seem, it has no simple
solution. Ifk 14 k 2 , we have effectively that all N has been converted into U
andd½I&d½U, which leads to a simple solution [i.e., Eq. (4.2)]. If, on the