kinetics of complex reactions Flashcards

1
Q

how are rate laws for elementary reactions written?

A

in differential form
in terms of rate of loss of reagents (-ve) or gain of products (+ve)

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2
Q

assumptions for sequential (multi-step) reactions

A

if one rate constant is much greater than another, the slowest step is the rate determining step
the reaction can be simplified to depending only on the step w the smallest k
(only for first order - 2nd order is also concentration dependent)

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3
Q

what is the pre-equilibrium hypothesis?

A

for mechanisms where the species involved in the RDS are in equilibrium w reagents
assumes species remain in equilibrium, ie. species –> products is slow enough to not disturb equilibrium and is therefore rate determining

often used for intermediates involving simple protonation/de-protonation, fast compared to making/breaking bonds

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4
Q

k(eq) for pre-equilibrium hypothesis

A

k(eq) = k(1) / k(-1) = [eq prod]/[eq reac]

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5
Q

what is the steady state assumption?

A

for sequential reactions A–>B–>C where process 2 much greater than process 1, conc intermediate B changes very little, assume constant
d[B]/dt(SS) = 0
greatly simplifies

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6
Q

in what cases might the steady state assumption be appropriate?

A

generally appropriate for reactive intermediates (high in energy) which are slow to form but consumed at once

NOT appropriate for species that accumulate / decay during the reaction

NOTE: rate laws derived from SSA only valid when reaction is at SS (not initial/completion)

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7
Q

describe saturation kinetics

A

for a fixed quantity of enzyme, the rate of the catalysed reaction first increases linearly then levels of to a maximum value as the amount of substrate increases

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8
Q

overall kinetic scheme of the Michaelis-Menten equation

A

E + S <=> ES –> E + P

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9
Q

how is the Michaelis-menten equation derived?

A

d[ES]/dt = 0 (SSA) (construct eqn from elementary steps)
sub [E]0 = [ES] + [E] to find expression for [ES]

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10
Q

expression for velocity of reaction

A

V = k(cat) [ES]
essentially rate of catalysed reaction

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11
Q

Michaelis-Menten equation

A

V = k(cat) [E]0 [S] / [S] + K(M)

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12
Q

Michelis constant, K(M) expression

A

K(M) = (k(cat) + k(-1)) / k(1)

in terms of rate constants from the elementary steps of the Michaelis-Menten equation

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13
Q

expression for max velocity from Michaelis-Menten eqn

A

V(max) = k(cat) [E]0

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14
Q

what does the graph of velocity vs substrate concentration look like? (for Michaelis-Menten saturation kinetics)

A

from 0, very steep initial increase (approximately linear), then slowing down to flatten out (“saturation”)

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15
Q

Michaelis-Menten eqn for LOW substrate concentrations

A

[S] &laquo_space;K(M)
hence denominator of MM eqn approximates to K(M)
resulting in expression for V(low[S]) linear in terms of [S] and [E}0

corresponds to LINEAR region of saturation kinetics graph (rate vs [substrate])

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16
Q

Michaelis-Menten eqn for HIGH substrate concentrations

A

[S]&raquo_space; K(M)
expression approximates to
V(max) = k(cat)[E]0
which is independent of [S] and hence flattening of the graph

virtually all enzyme is present as enzyme-substrate complex

17
Q

K(M) physical interpretation and derivation

A

writing MM eqn w V(max),
V = (V(max) [S]) / ([S] + K(M))
consider concentration S for which V = V(max)/2

K(M) = S
the concentration of substrate which gives a velocity of half the max

18
Q

what does a large value of K(M) mean?

A

NOT a strong affinity between enzyme and substrate, as a large conc of substrate is needed to achieve half the maximum rate

19
Q

what does a small value of K(M) imply?

A

strong affinity between enzyme and substrate, so very little substrate is needed to reach velocity V(max)/2

20
Q

units of Michaelis constant

A

CONCENTRATION, mol/dm^-3

it is NOT an equilibrium constant

21
Q

how to plot Michaelis-menten equation as a straight line

A

invert both sides + rearrange
plot 1/V against 1/[S] as a straight line

aka. Lineweaver-Burke plot, USED TO FIND K(M) and k(cat) for known [E]0

22
Q

what is the apparent activation energy of a reaction?

A

form rate law from elementary reactions

assume each elementary rate constant obeys Arrhenius

sub into rate law –> overall “Arrhenius” law for the whole reaction

Apparent acivcaiton energy is a composite of the activation energies of the three steps

rate constants can be expressed by Arrhenius eqn
k(T) = A exp(-Ea/RT)

23
Q

what is a chain reaction?

A

cyclic reactions, the output of one elementary reaction is the input of the next

24
Q

what are chain carriers?

A

the species which carry forward the chain reaction
they are the INTERMEDIATES written in the elementary steps

(only a small number of chain carriers needed to produce many product molecules)

25
Q

examples of chain reactions

A

common for many reactions involving reactive species such as free radicals

combustion

polymerisation

26
Q

three steps in chain reactions
+ other kinds of steps

A

initiation
propagation
termination

inhibition step
chain branching step

27
Q

define the initiation step in a chain reaction

A

the reaction which generates the chain carrier
usually by collision with some unspecified molecule, or by photodissociation (absorption of sufficiently energetic photons)

28
Q

define termination step in chain reactions

A

processes which destroy chain carriers

usually involves a third body to carry away excess energy

29
Q

define inhibition step in chain reaction

A

consumes product, leading to a reduction in the overall rate of reaction

does NOT destroy chain carriers

30
Q

define chain branching step

A

one chain carrier reacts to give rise to 2+ chain carriers

can lead to ever increasing rate of reaction –> explosion

31
Q

what can the form of a rate law tell us about the mechanism?

A

fractional power often indicative that radicals are involved

[product] in denomination means that the product inhibits the reaction (ie. goes slower w buildup of product)

32
Q

method of approach for analysing chain reactions

A

construct elementary reactions

assume that reactive intermediates are AT STEADY STATE
construct eqns using rate of change of the species = 0

construct overall equation for rate of change of product
substitute previous eqns into this one to find expression in terms of reaction species only (not intermediates)

find differential expression for product in terms of reacting species

33
Q

define chain length

A

the number of times a reaction occurs before termination

(the number of times the closed cycle of steps which produces products is repeated per carrier)

34
Q

what is the chain length in terms of rates?

A

l = rate of overall reaction / rate of initiation