kinetics of complex reactions

Question 1

how are rate laws for elementary reactions written?

Answer 1

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

Question 2

assumptions for sequential (multi-step) reactions

Answer 2

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)

Question 3

what is the pre-equilibrium hypothesis?

Answer 3

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

Question 4

k(eq) for pre-equilibrium hypothesis

Answer 4

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

Question 5

what is the steady state assumption?

Answer 5

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

Question 6

in what cases might the steady state assumption be appropriate?

Answer 6

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)

Question 7

describe saturation kinetics

Answer 7

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

Question 8

overall kinetic scheme of the Michaelis-Menten equation

Answer 8

E + S <=> ES –> E + P

Question 9

how is the Michaelis-menten equation derived?

Answer 9

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

Question 10

expression for velocity of reaction

Answer 10

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

Question 11

Michaelis-Menten equation

Answer 11

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

Question 12

Michelis constant, K(M) expression

Answer 12

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

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

Question 13

expression for max velocity from Michaelis-Menten eqn

Answer 13

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

Question 14

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

Answer 14

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

Question 15

Michaelis-Menten eqn for LOW substrate concentrations

Answer 15

[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])

Question 16

Michaelis-Menten eqn for HIGH substrate concentrations

Answer 16

[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

Question 17

K(M) physical interpretation and derivation

Answer 17

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

Question 18

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

Answer 18

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

Question 19

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

Answer 19

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

Question 20

units of Michaelis constant

Answer 20

CONCENTRATION, mol/dm^-3

it is NOT an equilibrium constant

Question 21

how to plot Michaelis-menten equation as a straight line

Answer 21

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

Question 22

what is the apparent activation energy of a reaction?

Answer 22

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)

Question 23

what is a chain reaction?

Answer 23

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

Question 24

what are chain carriers?

Answer 24

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)

Question 25

examples of chain reactions

Answer 25

common for many reactions involving reactive species such as free radicals combustion polymerisation

Question 26

three steps in chain reactions + other kinds of steps

Answer 26

initiation propagation termination inhibition step chain branching step

Question 27

define the initiation step in a chain reaction

Answer 27

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

Question 28

define termination step in chain reactions

Answer 28

processes which destroy chain carriers usually involves a third body to carry away excess energy

Question 29

define inhibition step in chain reaction

Answer 29

consumes product, leading to a reduction in the overall rate of reaction does NOT destroy chain carriers

Question 30

define chain branching step

Answer 30

one chain carrier reacts to give rise to 2+ chain carriers can lead to ever increasing rate of reaction --> explosion

Question 31

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

Answer 31

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)

Question 32

method of approach for analysing chain reactions

Answer 32

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

Question 33

define chain length

Answer 33

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)

Question 34

what is the chain length in terms of rates?

Answer 34

l = rate of overall reaction / rate of initiation