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Homogeneous Catalysts and Enzyme Kinetics
Lecture 1: Homogeneous CatalysisFor the reaction: 2H2O2(l) → 2H2O(l) + O2(g), species such as I-, catalase, and Fe3+ are examples of
homogeneous catalysts
...

A catalyst accelerates a reaction but undergoes no net chemical change
...

An alternative, faster reaction path that avoids the rate determining step (RDS) of the uncatalysed
reaction is provided by the catalyst
...
Most of the product will be produced by the catalysed route (as its
usually faster than the uncatalysed route) however some will be made in the uncatalysed route
...

All catalysts do not catalyse at the same rates, so the activation energy of each catalyst is measured
...
In this reaction, we can assume a simple 2-step catalysed reaction:
Step 1- R1+C→I
Step 2- R2+I→P+C
Catalyst (C) is consumed to form an intermediate which then reacts to regenerate the catalyst and
form the products
...
The rate of the catalysed
equation would be vcat=kcat[R1][R2][C]
...

Voverall= kuncat[R1][R2]+ kcat[R1][R2][C]
...
g
...

This is an example of a catalytic cycle, showing that the catalyst
reacts with reactant 1 to form the intermediate product
...


C and I are both catalytic species
...
Unlike reactants, [C]T doesn’t decrease during the reaction because the catalyst is not
consumed
...

The resting state of the catalyst is the catalytic species of the highest concentration during the
catalytic cycle
...
g
...

How much faster at 298K is the catalase catalysed reaction compared to the uncatalysed reaction?
If we assume that the pre-exponential factor, A, is the same then we can use activation energy and
the Arrhenius equation kr=Ae-Ea/RT using the values for the uncatalysed and catalysed activation
energy e
...
uncatalysed= +75kj mol-1 and catalase catalysed= +8kJ mol-1
...
8x108
...

Q1- Formulate a rate law for this reaction using the steady-state approximation
...
The rate of change of

the IO- ion is equal to its rate of formation in step 1 minus the rate of its removal in step 2 e
...

𝑑[𝐼𝑂 − ]
𝑑𝑡

= 𝑘1[𝐻2 𝑂2 ][𝐼 − ] − 𝑘2[𝐼𝑂− ][𝐻2 𝑂2 ]
...
E
...

= 0
...

However, I- is a catalyst and its total concentration is preserved throughout the reaction e
...

[I-]T=[I-]+[IO-]
...
This is how we get the concentration of the
free I- from the total I-
...
When we rearrange this, we get: [IO-]=
Now we can substitute this into the

𝑑[𝑂2]
𝑑𝑡

= 𝑘2[𝐼𝑂− ][𝐻2 𝑂2 ] equation to get:

𝑘1[𝐼− ]𝑇

...

𝑘1+𝑘2

Q2- Sketch a reaction coordinate diagram assuming IO- is a high energy intermediate and the
following: (1) Step 1 is slow and step 2 is fast and (2) Step 1 is fast and step 2 is slow
...

(1) I- is the resting state
...

Lecture 2: Reactions in Solution and Enzyme CatalysisA+B→P, in this reaction A and B encounters in solution are less frequent than they are in the gas
phase as solvent molecules are in the way
...

A and B migrate more slowly away from an encounter due to solvent molecules- cage effect
...
Even if
the encounter pair, AB, doesn’t have enough energy when it first forms, it may accumulate enough
energy to react
...
g
...
Rate of formation of AB=Kd[A][B]
AB can also react to form P with a rate constant ka e
...
AB→ P with rate law

𝑑[𝑃]
𝑑𝑡

= 𝑘𝑎[𝐴𝐵]
...
g
...


Formation of AB

Diss
...
Then we
𝑘𝑎+𝑘′𝑑
𝑑[𝑃]
𝑘𝑎𝑘𝑑
= 𝑘𝑎+𝑘′𝑑 [𝐴][𝐵]
...
g
...
If we
have a diffusion-controlled rate law, we can simplify

𝑑[𝑃]
𝑑𝑡

𝑘𝑎𝑘𝑑

= 𝑘𝑎+𝑘′𝑑 [𝐴][𝐵] to just

𝑑[𝑃]
𝑑𝑡

= kd[𝐴][𝐵]
...
If we have a activation-controlled rate law,
we can simplify

𝑑[𝑃]
𝑑𝑡

𝑘𝑎𝑘𝑑

= 𝑘𝑎+𝑘′𝑑 [𝐴][𝐵] to just

𝑑[𝑃]
𝑑𝑡

=

𝑘𝑎𝑘𝑑
[𝐴][𝐵]
...
These are multi-step reactions
...
The enzyme binds (non-covalent interactions) to the substrate (S) to form an
intermediate complex (ES)
...
An
enzyme molecule can catalyse thousands of reactions per second
...

Enzyme Kinetics- The simplest enzyme-catalysed reaction involves a single substrate going to a single
product (unireactant-uniproduct system/ uni-uni system)
...
g
...

Derivation of a rate law for an enzyme catalysed reaction- simplified mechanism assumes that the
product is formed irreversibly from ES
...

For ES to be in a steady-state then k2>>k-1 (ES is not in equilibrium with E and S)
...
Enzyme

isn’t consumed so [E]T remains constant e
...
[E]T=[E]+[ES]
...
In this reaction, we
assume P is formed irreversibly from ES
...
E
...


𝑑[𝑃]
𝑑𝑡

= 𝑘2[𝐸𝑆] and additionally [E]=[E]T-[ES]
...


𝑘1𝑘2[𝑆][𝐸]𝑇

= 𝑘1[𝑆]+𝑘−1+𝑘2
...
Km is the Michaelis
constant, which is the substrate concentration that would yield half maximum rate e
...
Vmax/2
...

Lecture 3: Enzyme KineticsM-M kinetics rate law: 𝑣 =

𝑉𝑚𝑎𝑥 [𝑆]

...
e
...


The M-M rate law gives a right rectangular hyperbola, meaning there isn’t a single value of order wrt
[S]
...


The low [S] limit: this is where the [S]< ...

The other limit is the high [S] limit where [S]>>Km
...


Above is an example of a Lineweaver-Burke reciprocal plot, which is how you linearise the MichaelisMenten rate laws
...


Problems with the L-B plot are: equally spaced increments of [S] do not give equally spaced points
on a reciprocal plot meaning points can become clustered and this messes with determining the line
of best fit, and small errors in rate are magnified on the reciprocal plot which can give significant
error to the slope
...

The turnover frequency, Kcat, is the number of catalytic cycles (turnovers) or the number of
substrate molecules transformed in one-unit time by a single enzyme molecule when the enzyme
concentration is rate-limiting
...
The higher
the value the more efficient the enzyme
...

A catalytically perfect enzyme or kinetically perfect enzyme is an enzyme that catalyses so efficiently,
that almost every time enzyme meets its substrate, the reaction occurs
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An enzyme inhibitor is any substance that reduces the rate of an enzyme catalysed reaction
...
There are 3 types of enzyme inhibition: competitive,
uncompetitive, and non-competitive
...
The inhibitor and the substrate are mutually exclusive because of true

competition
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Both I and S bind irreversibly to E to form EI and ES
...

(y-intercept is constant and slope increases
...
The effect that uncompetitive inhibition has on the vmax and km is that be Max
decreases an km decreases as well so the km/vmax ratio is unchanged
...
It occurs when the inhibitor has no
structural similarity to the substrate
...
I will bind to E and ES and S will
bind to E and EI with equal affinity
...
(y-intercept increases,
slope increases)

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Title: Homogenous Catalysts and Enzyme Kinetics (Level 2)
Description: University of Lincoln Level 2 Chemistry (Core Chemistry 2.2- Chemistry of Activated Systems and Radicals). Lecture notes on homogenous catalysts, rate equations, and enzyme kinetics inc. inhibition.

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