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KINETICS
study 7.5 Ester hydrolysis
RCOOR'(aq)
+ H2O(l) ===> RCOOH(aq) + R'OH(aq)
RCOOR'(aq)
+ OH–(aq) ===> RCOO–(aq) + R'OH(aq)[Author
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rates of reaction factors:
KINETICS of alkaline hydrolysis of an ester
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Advanced Level Chemical Kinetics Notes
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Case
study 4.5 Ester hydrolysis
Explanation and derivation of orders of
reactants and how to write the rate expression
-
Esters can be
hydrolysed by (i) water alone (but very slow), but is (II)
catalysed by acids (H+(aq) specifically) and
(iii) alkalis (OH–(aq) specifically). The
latter hydrolysis (iii), is sometimes called a saponification reaction.
-
Hydrolysis with water
alone is usually too slow to obtain meaningful rate data.
-
However, with a fixed
amount of acid catalyst e.g. HCl(aq), it is possible to follow
the reaction by alkali titration and show that ...
-
rate = k1[RCOOR'(aq)],
that is 1st order with respect to ester, however,
-
if the concentration
of acid catalyst is varied, things are more complicated,
-
and the 2nd
order expression, rate = k2[RCOOR'(aq)][H+(aq)]
then applies.
With alkali, 2nd
order kinetics are observed overall, and the reaction can be followed by
titrating the remaining alkali with acid.
Phenolphthalein (Pk(ind)
= 9.3), would be a suitable indicator because one of the reaction products
is the salt of a weak carboxylic acid (approx. pH 9). The product of the
titration is neutral sodium chloride.
i.e.
rate = k2[RCOOR'(aq)][OH–(aq)]
Note
(i)
In general for ester
hydrolysis, you find experimentally the kinetics gives 1st order with respect to each reactant, 2nd order overall.
(ii) Except if the
hydrolysis is in just water as the solvent.
In this case, because
the water is effectively constant (the molarity of pure water is
55), overall the kinetics is 1st order i.e. theoretically the rate
expression is 2nd order,
rate = k[RCOOR'(aq)][H2O(l)]
so the experiment
kinetics rate expression is actually 1st order.
rate = k1[RCOOR'(aq)]
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