Advanced A Level Chemistry Advanced A Level Chemistry - Kinetics-Rates
revision notes Part 6
7.3 Enzyme kinetics - biological catalyzed reactions
(enzymes being biological catalysts)
Advanced A Level Chemistry Kinetics
GCSE/IGCSE rates reaction notes INDEX
on the uses of enzymes and optimum conditions
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kinetics – Biological Catalysts
Explanation and derivation of orders of reactants and how to
write the rate expression
in sequence: E
= free enzyme, S
= free substrate reactant molecule, ES
= enzyme–reactant complex, EP
= enzyme–product complex, E
= free enzyme, P
= free product
BIOLOGICAL CATALYSIS – HOW DOES AN
WHY IS THE REACTION PROFILE QUITE
'COMPLICATED' COMPARED TO MANY OTHER REACTION?
of enzyme catalysed:
E + S ==> ES
ES ==> EP
The conversion of the substrate–enzyme
complex ES into the enzyme–product EP, is
the rate determining step. The docking in of the substrate to
be held e.g.
by inter molecular forces is likely to have a low activation
energy and a relatively high probability initially, so E + S ==> ES cannot be the rds.
However, ES ==> EP
involves bond breaking and will have a much higher
activation energy giving a much lower
probability of ES ==> EP transformation.
[ES] cannot be measured directly, so a simplified
kinetics approach is to treat the formation and transformation of
ES/EP through to P as
a function of the concentrations of E and S (see the rate
expressions below, which is a big bad fudge seen here and in most textbooks!).
The situation should be dealt with via the Michaelis–Menten
equation, but this goes way above
AS–A2 chemistry demands.
EP ==> E + P
possible detailed reaction
profile is shown below.
uncatalysed profile would be a single, and much higher 'hump' (see 2nd
simplified diagram below).
is small for the formation of the enzyme–substrate complex or
intermediate ES. The energy of ES is lower than the reactants
because it involves an energy releasing interaction whether it be via
intermolecular forces or chemical bonding.
is much larger for the transformation of the ES intermediate
into the EP intermediate because it involves bond breaking in
the substrate molecule.
is still much smaller than Ea for the uncatalysed
reaction, which is not shown here, but is shown in the simplified
diagram below e.g. the uncatalysed decomposition of hydrogen
peroxide solution has an Ea of 76kJmol–1,
but with an enzyme like catalase/peroxidase, the Ea is
reduced to 30kJmol–1 AND the rate of reaction is
increased by a factor of 108, impressive
is relatively small for the breakdown of the EP intermediate into the
free product and free enzyme.
the simplified reaction profile diagram below should be good enough in an exam and shows the
Here, Ea1 is for E+S ==> ES, Ea2
is for ES ==> EP ==> E + P
and Ea3 for the uncatalysed
reaction, which with no enzyme is significantly greater.
Treating step (2) as the rds, gives the rate expression ...
rate = kES[ES],
that is the rate
of reaction = rate of transformation of ES into EP .
you cannot measure the concentration of the enzyme–substrate
rate is treated as a function of the concentrations of the enzyme and substrate, [E] and [S],
which form the ES intermediate, assuming a steady state
So we are
treating the reaction as 1st order with respect to both
enzyme and substrate, 2nd order
For a set of
experiments, using a fixed amount of enzyme, the initial rate should be
proportional to the concentration of substrate at relatively low
concentrations, so the
effective rate expression is pseudo 1st order ...
If you keep
the substrate concentration constant and vary the quantity of enzyme, the effective
rate expression is also pseudo 1st order ...
However, if the
substrate concentration is very high, the maximum number of active
centres on the enzymes are
occupied and the rate is completely controlled by the transformation
of the ES complex at its maximum possible
concentration and independently of the high substrate concentration. This means you are now dealing with the maximum possible
rate, which only depends on the amount of enzyme present,
so the rate
expression becomes zero
order with respect to the substrate,
and for a
constant [E], and varying [S] at high concentrations of S the rate
The two situations are summed up by
graph (1) below for a constant amount
of enzyme. The graph moves from 1st order kinetics to zero
is a 'simple' comparison of the effects of no inhibitor, a
constant amount of a competitive or a
non–competitive inhibitor for a constant amount of enzyme.
is the initial reaction rate for a constant amount of enzyme
and varying the substrate concentration.
is the maximum possible rate when the maximum number of active centres are
occupied with S or P.
inhibition: Although the inhibitor initially decreases the
reaction rate, the substrate can increasingly compete with the
inhibitor to occupy the 'active sites' as the substrate
concentration increases (Le Chatelier's equilibrium concentration
that the maximum reaction rate
can eventually equal
the non inhibited enzyme.
inhibition: Since the inhibitor molecule does not bind to the
'active site' of the enzyme it cannot be 'replaced' by increasing
the substrate concentration. However, because it constantly
interferes with a fraction of the enzymes, the reaction rate will be
reduced giving the lower the
which can never reach
however high the substrate concentration.
enzyme structure and
function notes for more details on inhibition
The remarkable efficiency of enzymes
- biological catalysts are superb!
For example the decomposition of
hydrogen peroxide - an unwanted harmful chemical produced in the body!
==> 2H2O(l) + O2(g)
Three activation energies (Ea) are
(i) uncatalysed: 75 kJmol-1
(ii) catalysed with colloidal
platinum: 49 kJmol-1
(iii) catalysed by the enzyme
catalase in you body: 23 kJmol-1
The numbers speak for themselves!
Both catalysts are effective in
considerably reducing the activation energy and hence greatly increasing
the speed of the decomposition reaction.
BUT, the enzyme catalase is far
superior to a class transition metal catalyst!
GCSE page describing enzyme uses and rates graphs for
concentration, pH and temperature changes.
Isomerism and Stereochemistry
Part II page there is a discussion of enzyme structure and function.
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