Chemical Kinetics
Jens Poulsen
5 x 2 hours
A B
Atkins
Quanta, matter, and Change.
Chapters 19-21.
A reaction may be investigated
on several levels…
stochiometry:
2 N2O5 ( g )  4 NO2 ( g )  O2 ( g )
rate law:
v  k  [ N 2O5 ( g )]
atomistic:
N 2O5  NO2  NO3 k a
NO2  NO3  N 2O5 k a
'
NO2  NO3  NO2  NO  O2 kb
NO  N 2O5  NO2  NO2  NO2 k c
Chapter 22: Rates of chemical
reactions. Concepts:
 Defining reaction
rates.
 Integrated rate
laws.
 Elementary
reactions.
 Consecutive
reactions.
A B v 
d [ A]
dt
 k  [ A]
[ A]  [ A]0  exp(kt)
A B C
A BC
Start with chemical kinetics
on an empirical level:
2N2O5 ( g )  4NO2 ( g )  O2 ( g )
Rate equations: How much N2O5 (g) as a function of time etc?
No information about reaction mechanism
(on an atomic scale).
Experimental techniques
concentrations can be followed by...
Measuring pressure during
chemical reaction:
2 N2O5 ( g )  4 NO2 ( g )  O2 ( g )
Measuring conductivity during
chemical reaction:
(CH 3 )3 CCl (aq)  H 2O(l )
 (CH 3 )3 COH (aq)  H  (aq)  Cl  (aq)
Example 19.1
monitoring variation in pressure.
2N2O5 ( g )  4NO2 ( g )  O2 ( g )
 : fractionof N2O5 thathas decomposed.
Amount:
N 2O5
NO2
O2
tot la
1
3
n(1   ) 2n
n n(1   )
2
2
3
p  p0 n(1   )
2
Def. of v - rate of reaction
2HOIO(aq)  IO3 (aq)  H  (aq)  HOI(aq)
v
1 d[ HOIO]
2
dt


d[ IO3 ]
dt

d[ H  ]
dt

d[ HOI ]
dt
more generally
aA  bB  cC  dD
1 d[ A]
1 d[ B] 1 d[C ]
1 d [ D]
v



(a) dt
(b) dt
c dt
d dt
Reaction order
A  B  ..  C  ...
Often has reaction rate of form
v  k [ A] [ B]  ...
m
n
Order m with respect to A,..
Overall order is m+n+..
Examples
first order reaction
C2 H6 ( g )  2CH3 ( g ) v  k[C2 H6 ( g )] T  7000 C
second order reaction
2NOBr( g )  2NO( g )  Br2 ( g ) v  k[ NOBr( g )]
2
zero order reaction
N2  3H 2 on Iron  2NH3 ( g ) v  k
Determination of rate law
A  B  ..  C  ...
v  k [ A]m[ B]n  ...
 isolation method: excess of B, C, …
m can be determined etc.
 method of initial rates: vary the conc. of A,
B, C, .. In turn and check how rate changes.
Integrated rate laws
 First order rate
v
d [ A]
dt
 k  [ A]
[ A]  [ A]0  exp(kt)
ln([A] /[ A] )  kt
straight line:
 Second order rate
0
v
d [ A]
dt
 k  [ A]
 straight line:
2
1
1
[ A]  [ A]0  kt
1
[ A]1  [ A]0  kt
v
d [ A]
dt
 k  [ A][B]
leads to
[ B] /[ B]0
ln(
)  kt  ([B]0  [ A]0 )
[ A] /[ A]0
Reactions approaching equil.
A B
BA
v  k[ A]
v  k ' [ B]
d [ A]
 k[ A]  k '[ B]
dt
Solution:
k ' k exp( (k  k ' )t )
k ' k
for
t   : [ A]eq  [ A]0
k'
k ' k
[ A]  [ A]0
[ B]eq  [ A]0  [ A]eq  [ A]0
K
[ B]eq
[ A]eq

[ B]0  0.
k
k ' k
k
k'
Relation between equilibrium constant
and rate constants.
Disturbing a system in eq:
k
A B
k'
Sudden change in k , k '
e.g. by change in T.
[ A]  [ A]eq  x [ B]  [ B]eq  x
d [ A]
 k[ A]  k '[ B ]  k ([ A]eq  x)  k ' ([B]eq  x)
dt
d [ A]
d [ A] dx
 kx  k ' x but

dt
dt
dt
dx
 (k  k ' ) x  x(t)  x(0)exp(-( k  k ' )t )
dt
The rate to new equilibrium is
given by the sum of
k, k'
1

 k  k'
Example 19.4
k'






H 2O(l ) H (aq)  OH (aq) KW  [ H ][OH ] / c
k
  37s  3710 s.
6
Calculate the rate constants
k, k'
for water autoprotolysis, given
Forward/backward reaction is first/secondorder, respectively, and T=298 K, pH=7.
BLACK-BOARD!
2
Temperature dependence of
chemical reactions.
A B C
v  k (T )[ A][B]
What can be said about k(T) ?
k (T )  A exp(Ea / RT )
Arrhenius equation
Svante Arrhenius
Nobel prize 1903.
interpretation
k (T )  A  exp(Ea / RT ) A  B  C
v  k (T )[ A][B]
*Reaction coordinate: molecular distortion
along which the reactants become products.
*Transition state = activated complex = climax
of reaction.
*Once the reactants have passed the
transition state, products
are formed.
interpretation
k (T )  A  exp(Ea / RT )
Ea: activation energy (change in potential energy).
Only molecules having kinetic energy larger than Ea
get over barrier.
exp(-Ea/RT): fraction of reactants
having enough kinetic energy to pass barrier
A: pre-exponential factor: measure of the rate of
collisions
k (T )  A exp(Ea / RT )
A B C
v  k (T )[ A][B]
 exp(-Ea/RT): fraction of reactants having
kinetic energy higher than barrier height Ea.
 A: proportional to collision frequency.
 K(T) = collision frequency times
fraction of successful collisions
= A * exp(-Ea/RT)
Accounting for the rate laws.
 Elementary reactions: The fundamental
“building blocks” of chemical reactions.
 Describes what happens on an atomic
scale.
CH 3 I  CH 3CH 2O  CH 3OCH2CH 3  I 
one ethanoate ion collides with one
methyliodide molecule and forms one iodide
ion plus one methylethylether….
Elementary reactions
 The molecularity of a reaction: # of
molecules participating.
 Unimolecular: one molecule.
 Bimolecular: two molecules.
 A unimolecular reaction is first-order.
 A bimolecular reaction is second-order.
AP
d[ A] / dt  k[ A]
A B  P
d [ A] / dt  k[ A][B]
Elementary reactions
 If we know a reaction is single-step and
bimolecular we can write down a rate
equation directly:
A B  P
d[ A] / dt  k[ A][B]
 Same applies to unimolecular reactions...
Elementary reactions
 On the other hand, consider the reaction
A( g )  B( g )  P( g )
A rate law of form
d[ A] / dt  k[ A][B]
does not imply the reaction to be simple bimolecular:
A B  P
Example.
v  k[ HBr( g )][O2 ( g )]
Consecutive elementary
reactions
ka
kb
A I  P
example: enzyme/substrate
Solving time-dependence of
[P].
ka
kb
A I  P
BLACK-BOARD!
 ka exp(kbt )  kb exp(ka t ) 
[ A]0
[ P]  1 
kb  k a


Consecutive reactions
Qualitative time-dependence of
[A], [I] and [P].
Steady State approx.
VERY IMPORTANT!!
 The higher the molecularity, the more
complex mathematics when solving rate
equations. Need approximation.
 Introduce steady state approximation:
ka
kb
A I  P
[I ]  0
d[I ]
0
dt
kb  k a
Steady state approx. continued.
More generally: all intermediates I , I ,...,I
are assumed to have negligible concentration
and rate of change of concentration:
1
[ I1 ]  0,[ I 2 ]  0,...,[ I n ]  0
d[ I n ]
d [ I1 ]
d[I 2 ]
 0,
 0,.....,
0
dt
dt
dt
2
n
Apply steady state approx.
to consecutive reaction.
ka
kb
A I  P
d[I ]
 k a [ A]  kb [ I ]  0  [ I ]  k a [ A] / kb
dt
d [ P]
 kb [ I ]  k a [ A]  k a exp( k a t )[ A]0
dt
d [ P]
 kb [ I ]  k a [ A]  k a exp(k a t )[ A]0 
dt
[ P]

[ P ]0
s
d [ P]   dtk a exp(k a t )[ A]0 
0
[ P]  [ P]0  (1  exp(k a t ))[A]0 
[ P]  (1  exp(k a t ))[A]0
Steady state example.
2 N2O5 ( g )  4 NO2 ( g )  O2 ( g )
N 2O5  NO2  NO3 k a
NO2  NO3  N 2O5 k a
'
NO2  NO3  NO2  NO  O2 kb
NO  N 2O5  NO2  NO2  NO2 k c
Two intermediates: NO and NO3.
d [ NO3 ] / dt  0  0  k a [ N 2O5 ]  (k a'  kb )[NO2 ][NO3 ]
d [ NO] / dt  0  0  kb [ NO2 ][NO3 ]  kc [ NO ][N 2O5 ]
d [ N 2O5 ] / dt  ka [ N 2O5 ]  ka' [ NO2 ][NO3 ]
 kc [ NO][N 2O5 ]
Hence..[exercise]:
d[ N2O5 ] / dt  2ka kb [ N2O5 ] /(k  kb )
'
a
Rate determining step
Example. when ka<kb then
ka
kb
A I  P
may be treated as a simple
reaction:
ka
A P
”the principle of the rate
determining step”
Proof:
When ka<kb then
 ka exp(kbt )  kb exp(ka t ) 
[ A]0
[ P]  1 
kb  k a


becomes
[ P]  1  exp(ka t ) [ A]0
 [ A]0  [ A]t
as expected
Rate determining step – in general
 If one elementary step in a reaction is
slower than others then this step controls
the rate of the overall reaction.
 The slow step is rate determining if it can’t
be sidestepped.
 Once the rate determining step is found, the
rate expression can be written down
immediately.
Example.
ka
kb
ka '
kb '
A I  P
A  I ' P
 Even if ka<< kb the upper reaction may be sidestepped if
ka<ka’ and ka<kb’. Then
ka
A I
is not rate determining.
Preequilibria.
ka


kb
A B I P
ka '
Assume A and B are in eq. then
ka
[I ]
K
K '
[ A][B]
ka
d[ P] / dt  kb [ I ]  kb K [ A][B]
Example. Analysing pre-eq. by
steady state.
 we do not assume eq.
ka


kb
A B I P
ka '
d [ P] / dt  kb [ I ]
d [ I ] / dt  kb [ I ]  k a [ A][B]  k a' [ I ]  0 
k a [ A][B]
k a kb [ A][B]
[I ] 
 d [ P] / dt 
'
kb  k a
kb  k a'
When pre-eq exists kb<ka’
ka


kb
A B I P
ka '
ka kb [ A][B] ka kb [ A][B]
d [ P] / dt 

'
kb  k a
ka'
theold pre- eq analysisresult.
Unimolecular reactions.
Lindemann Hinshelwood
mechanism.
cyclo  C3 H 6  CH 3CH  CH 2
v  k[cyclo  C3 H 6 ]
 Found to involve bimolecular step, how can it be
first order?
 The reaction is not given by an elementary first
order mechanism.
 Can be described by the Lindemann Hinshelwood
mechanism.
Lindemann Hinshelwood
d [ A ]
 ka [ A]2
dt
ka
A  A  A  A

d
[
A
]

A A  A A
 ka ' [ A][ A ]
dt
ka '

kb
A P

d[ A ]
 kb [ A ]
dt
Use steady state...
d [ A ]
 kb [ A ]  ka ' [ A][ A ]  ka [ A]2  0 
dt
ka [ A]2

[A ] 

kb  ka ' [ A]
2
k
k
[
A
]
d [ P]
d
[
P
]
 kb [ A ]  0 
 b a
dt
dt
kb  ka ' [ A]
If kb is small: kb << ka’ [A] then firstorder reaction
d [ P] kb k a [ A]

dt
ka '
On the other hand at low
pressure:
d [ P]
 k a [ A]2
dt
Since the reaction
ka
A  A  A  A
d [ A ]
 ka [ A]2
dt
becomes ”bottleneck”.
Define effective rate constant:
k k [ A]
d [ P] kb ka [ A]2

 b a
[ A]  K [ A]
dt
kb  ka ' [ A] kb  ka ' [ A]
with K 
kb ka [ A]
kb  ka ' [ A]
Lindemann Hinshelwood
mechanism gives linear plot
K
1
ka '
d [ P] 1
1
 [ A] /
 [ A] 
dt
ka
k a kb
The kinetics of complex reactions
 Chain reactions
 Explosions
 Catalysis (enzymes)
What is a chain reaction?
 Example, thermal
decomposition of ethanal: CH 3CHO( g )  CH 4 ( g )  CO( g )
 Elementary reactions
(Rice-Herzfeld):
initiation: CH 3CHO  CH 3  CHO  v  ki [CH 3CHO]
propagation : CH 3  CH 3CHO  CH 4  CH 3CO  v  k p [CH 3 ][CH 3CHO]
propagation : CH 3CO  CO  CH 3  v  k ' p [CH 3CO]
termination : 2CH 3   CH 3CH 3
 Chain carriers are CH3
and CH3CO radicals
v  kt [CH 3 ]2
Derivation of rate law:
 Steady state:
d [CH 3 ]
 ki [CH 3CHO]  k p [CH 3 ][CH 3CHO]
dt
 k p' [CH 3CO]  kt [CH 3 ]2  0
d [CH 3CO]
 k p [CH 3 ][CH 3CHO]  k p' [CH 3CO]  0
dt
 Add these together:
ki
0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
[CH 3CHO]
kt
2
Insert into
0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
2
ki
[CH 3CHO]
kt
d [CH 4 ]
 k p [CH 3 ][CH 3COH ]
dt
ki
 kp
[CH 3COH ]3 / 2
kt
As observed experimentally.
Key elements of chain reaction
 Initiation: involves formation of chain carriers
 propagation: done by chain carriers
 Termination: chain carriers are ”destroyed”.
New example:
Br2 ( g )  H 2 ( g )  2HBr( g )
Initiated by heat.
New example:
Br2 ( g )  H 2 ( g )  2HBr( g )
Complicated rate law:
d [ HBr]
k[ H 2 ][Br2 ]3 / 2

dt
[ Br2 ]  k '[ HBr]
Can be explained by postulating
a chain reaction mechanism.
Mechanism: collision induced
Initiation: Br2  M  2 Br   M
Propagation :
Br   H 2  HBr  H  v  k p [ Br][H 2 ]
Propagation :
H   Br2  HBr  Br  v  k ' p [ H ][Br2 ]
Retardation :
H   HBr  H 2  Br  v  k r [ H ][HBr]
T ermination :
Br   Br  Br2
v  kt [ Br]2
Retardation step: A product is
removed.
Chain carriers are H and Br radicals.
Steady state analysis.
Rate of formation
d [ HBr ]
 k p [ H 2 ][ Br ]  k p '[ H ][ Br2 ]  k r [ HBr ][ H ]
dt
Steady-state:
d [ H ]
 k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]  0
dt
d [ Br]
 2ki [ Br2 ][M ]  k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]
dt
 2kt [ Br]2 [ M ]  0
Adding these two equations give
2ki [ Br2 ][M ]  2kt [ Br] [ M ]  0  [ Br] 
2
ki
[ Br2 ]1/ 2
kt
Inserting the expression for Br
radical into first eq. gives
ki
kp
[ Br2 ]1/ 2 [ H 2 ]
kt
[ H ] 
k ' p [ Br2 ]  k r [ HBr]
Substituting into rate expression gives
d [ HBr]
 k p [ H 2 ][Br]  k p '[ H ][Br2 ]  k r [ HBr][H ]
dt
ki
2k p
[ H 2 ][Br2 ]3 / 2
kt

[ Br2 ]  (k r / k ' p )[HBr]
Which has the same form as the
experimental rate law provided
k  2k p
ki
kt
k '  (k r / k ' p )
Explosions
 An explosion is a rapidly
accellerating reaction arising
from a rapid increase in reaction
rate with increasing
temperature
fast reactionreleasingheat (exothermi
c)  T rises 
even faster reaction T rises even more ..
Example
O2 ( g )  2H 2 ( g )  2H 2O( g )
Has not been fully understod. # Chain carriers
grow exponentially.
initiation: H 2  H   H  v  const  vinit
propagation : H 2  OH   H   H 2O v  k p [ H 2 ][OH ]
branching: O2  H   O   HO  v  kb [O2 ][H ]
 O   H 2  H   HO  v  k 'b [O][H 2 ]
1
termination : H   wall  H 2 v  kt [ H ]
2
H  O2  M  HO2   M  v  kt '[ H ]
H , OH ,  O 
 chain carriers are
 Branching: one chain
carrier becomes two or
more.
Occurence of explosion given by explosion region
(regions of T,p).
 If T is low, rate constants are
too small.
 If p is low, the reaction rates, v,
are too low due to infrequent
collisions.
Example
Show that an explosion happens
when rate of chain branching
exceeds that of chain termination.
Answer:
Focus only on the rate of prod of H 
chain carrier as important. Rate of
change of chain carrier:
d
[ H ]  vini  k p [ H 2 ][HO]  kb [ H ][O2 ] 
dt
kb '[O][H 2 ]  kt [ H ]  k 't [ H ][M ][O2 ]
Example
Steady state approximation:
d
[OH ]   k p [OH ][H 2 ]  kb [O2 ][H ]  k 'b [O][H 2 ]  0
dt
d
[O]  kb [O2 ][H ]  k 'b [O][H 2 ]  0 
dt
k
[OH ]  2 b [O2 ][H ] /[ H 2 ]
kp
[O]  kb [O2 ][H ] / k 'b [ H 2 ]
Then chain carrier production rate is
d
[ H ]  vini  ( 2kb [O2 ]  kt  k 't [ M ][ O2 ])[ H ]
dt
Example
write:
kbranch  2kb [O2 ] measuringchain branchingrate.
kterm  kt  kt' [O2 ][M ] measuringterminati
on rate.
Then chain carrier production rate is
d
[ H ]  vini  ( kbranch  kterm )[ H ]
dt
vinit
a) [ H ] 
(1  e ( kterm  kbranch ) t ) kbranch  kterm
kterm  kbranch
vinit
b) [ H ] 
(e ( kbranch  kterm ) t  1) kbranch  kterm
kbranch  kterm
Homogeneous catalysis
 A catalyst lowers the activation energy for a
reaction.
 It is not consumed in the reaction (reforms in
the end).
 Examples are enzymes, metal catalysts etc.
 Homogeneous catalysis: the catalyst exists
in same phase as reactants.
Example:
2H 2O2 (aq)  2H 2O(l )  O2 ( g )
as catalyzed by bromide ions.

2

H 3O  H 2O H 2O  H 3O2

2
K  [ H 3O2 ] /[ H 3O  ][H 2O2 ]


2

H 3O  Br  HOBr  H 2O v  k[ H 3O ][Br ]
HOBr  H 2O2  H 3O   O2  Br 
( fast)
Notice that bromide reappears in the end.
Enzymes
 Enzymes (E) are
biological catalysts.
 Typical proteins.
 Contains an active site
that binds substrate
(S):
 Induced fit model: S
induces change in E
and
only then do they fit
together.
E  S  ES  P  E
Michaelis Menten mechanism.
Three exp. observations
 i) For a given amount of S, the rate of product
formation is proportional to [E].
 ii) For a give amount of E and low values of [S],
the rate of product formation is proportional to [S].
 iii) For a given amont of E and high values of [S],
the rate of product formation reaches a max.
value, v(max), and becomes independent of [S].
i) and ii) - but not iii) - is consistent with
E  S  P  E v  k[S ][E]
Michaelis-Menten explains i)-iii)
E  S  ES
k a , k 'a
ES  P  E kb
leads to
kb [ E ]0
v  kb [ ES] 
1  K M /[S ]0
[ E ][S ]
KM 
[ ES]
Proof.
Steady state approx.:
then
Define
[ ES] 
KM 
d [ ES ]
 k a [ E ][ S ]  k 'a [ ES ]  kb [ ES ]  0
dt
ka [ E ][S ]
k ' a  kb
k 'a  kb [ E ][S ]

ka
[ ES ]
Use [ E]0  [ E]  [ ES] [S ]  [S ]0 substratein excess
[ E]0  [ ES](1  KM /[S ])  [ ES](1  KM /[S ]0 )
then
[ ES]  [ E]0 /(1  K M /[S ]0 )
v  kb [ ES]  kb [ E]0 /(1  KM /[S ]0 )
Then i)-iii) is correct. show!
Alternative formula:
vmax
v  kb [ ES] 
1  K M /[S ]0
vmax  kb [E]0

1
1
1 KM


v vmax [ S ]0 vmax
Plot of 1/v against 1/[S] gives straight
line for MM mechanism.
Plot is called
Lineweaver-Burk plot
Determine enzyme parameters.
 Lineweaver-Burke plot
gives
vmax  kb [E]0
and
K m / vmax
v  kb [ ES] 
1
1
1 KM


v vmax [ S ]0 vmax
 From knowing total
enzyme concentration
we calc. K m , kb
 Cannot determine ka ,
vmax
1  K M /[S ]0
k 'a

Catalytic constant.
 kb gives the number
of ”turn overs” pr. unit
time. This number –
with units of pr time –
is called the catalytic
constant, k
cat
E  S  ES
ES  P  E
k a , k 'a
kb
Catalytic efficiency.

E  S
ES
ES  P  E
k a , k 'a
kb
Catalytic efficiency: measure of overall
enzym efficiency equals
effective rate constant for overall
reaction
ka
kb
v  [ES] kb  [ E ][S ]
kb 
[ E ][S ]
k 'a  kb
KM
kb

KM
 If enzyme is very
efficient then kb is
large. Then


k a , k 'a
ES  P  E
kb
E  S ES
ka

kb  k a
k ' a  kb
 ka is determined by
diffusion.

1 1
 ka  10 10 M s
8
9
Enzyme inhibition
 An inhibitor reduces rate of product
formation by binding to E, ES or both,
thereby decreasing [ES] and hence product
formation.
E  S  ES k a , k 'a
ES  P  E
kb
EI E  I , K I  [ E ][I ] /[ EI ]
[ ES ][I ]
ESI ES  I , K ' I 
[ ESI]


New rate of product formation.
vmax
v  kb [ ES] 
   ' K M /[S ]0
  1  [ I ] / K I  '  1  [ I ] / K 'I
vmax  kb [ E ]0
 Maximum velocity is no longer obtainable
when competitive inhibition occurs.
Proof:
 Introduce conservation
of enzyme molcs:
[ E ]0  [ E ]  [ EI ]  [ ES ]  [ ESI]
then
[ E ]0  [ E ]  [ ES ] '
since
K M  [ E ][S ] /[ ES ]  [ E ][S ]0 /[ ES ]
[ E ]0  K M [ ES ] /[ S ]0  [ ES ] ' 
[ ES ]  [ E ]0 /( ' K M  /[ S ]0 )
[ ES ]  [ E ]0 /( ' K M /[ S ]0  )
v  kb [ ES ] 
v  kb [ E ]0 /( ' K M  /[ S ]0 )
v  vmax /( ' K M  /[ S ]0 )
Different types of inhibition
 Case 1:   1  '  1 meaning only EI not ESI
is formed. Once EI is formed, S cannot bind
to E. Termed competitive inhibition.
Case 2:   1  '  1 meaning that only ESI
not EI is formed. I can only bind to E if S is
already present i.e. ES. ESI does not lead to
product. Termed uncompetitive inhibition.
Different types of inhibition
 Case 3:   1  '  1 meaning that both EI
and ESI are formed. Inhibitor binds to a site
different from the active site in both cases.
EI cannot bind S. Termed non-competitive
inhibition.
How to determine  ,  '
 Do a unhibited exp. with S & E and find
vmax and Km.
 Add inhibitor with known conc. and plot
Lineweaver-Burk from which  ,  ' and
thereby K I , K 'I can be determined.
Example of use of competitive
inhibition.
 To kill bacteria (B). B
has enzyme dihydropteroate synthease
producing folate which
is crucial for survival of
B.
 Active substrate paminobenzoic acid.
 Inhibitor Sulfanilamide.