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# kinetic by niusheng11

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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 N 2O5 ( 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 kc
Chapter 22: Rates of chemical
reactions. Concepts:
 Defining reaction   A B v 
d [ A]
 k  [ A]
dt
rates.
 Integrated rate           [ A]  [ A]0  exp( kt)

laws.                 A B C
 Elementary
reactions.          A B C
 Consecutive
reactions.
on an empirical level:
2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )

Rate equations: How much N2O5 (g) as a function of time etc?

(on an atomic scale).
Experimental techniques
concentrations can be followed by...

Measuring pressure during
chemical reaction:
2 N 2O5 ( 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.
2 N 2O5 ( g )  4 NO2 ( g )  O2 ( g )
 : fraction of N 2O5 that has decomposed.

Amount :      N 2O5        NO2     O2          l
tota
1         3
n(1   ) 2n         n n(1   )
2         2
3
p  p0 n(1   )
2
Def. of v - rate of reaction
2 HOIO (aq)  IO  3 (aq)  H  (aq)  HOI (aq)

1 d [ HOIO ]   d [ IO3 ]   d[H  ]   d [ HOI ]
v                                  
2      dt           dt       dt           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 H 6 ( g )  2CH 3 ( g ) v  k[C2 H 6 ( g )] T  700 0 C

second order reaction
2 NOBr ( g )  2 NO ( g )  Br2 ( g ) v  k[ NOBr ( g )]                2

zero order reaction
N 2  3H 2 on Iron  2 NH 3 ( 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)

straight line: ln([ A] /[ A] )  kt
0

 Second order rate
1               1
[ A]  [ A]0  kt
d [ A]
v              k  [ A]   2

dt
 straight line:                               1
[ A]1  [ A]0  kt
d [ A]
v               k  [ A][ B]
dt

[ B] /[ B]0
ln(             )  kt  ([ B]0  [ A]0 )
[ A] /[ A]0
Reactions approaching equil.
A B         v  k[ A]
BA          v  k '[ B]

d[ A]
 k[ A]  k '[ B]
dt
Solution:
k ' k exp((k  k ' )t )
[ A]  [ A]0                                 for       [ B]0  0.
k ' k

k'
t   : [ A]eq  [ A]0
k ' k
k
[ B]eq  [ A]0  [ A]eq  [ A]0
k ' k
[ B]eq       k
K               
[ A]eq       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 ' )t )
k
dt

The rate to new equilibrium is
given by the sum of
k, k'

1
 k  k'

Example 19.4
k'
                                             2
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/second-
order, respectively, and T=298 K, pH=7.

BLACK-BOARD!
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 3OCH 2 CH 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 bi-
molecular:
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(kat ) 
[ P]  1 
                               [ A]0

          kb  k a             
Consecutive reactions

Qualitative time-dependence of
[A], [I] and [P].
VERY IMPORTANT!!
 The higher the molecularity, the more
complex mathematics when solving rate
equations. Need approximation.
ka              kb
A I  P

d[ I ]
[I ]  0          0   kb  ka
dt
More generally: all intermediates I , I ,..., I        1   2   n

are assumed to have negligible concentration
and rate of change of concentration:
[ I1 ]  0, [ I 2 ]  0,... , [ I n ]  0

d [ I1 ]      d[ I 2 ]           d[ I ]
 0,           0,....., n  0
dt           dt                 dt
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]             s


[ P ]0
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
2 N 2O5 ( 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 kc

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 [ N 2O5 ] / dt  2k a kb [ N 2O5 ] /(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(kat ) 
[ P]  1 
                               [ A]0

          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
A I  P

ka '   kb '
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

[I ]         ka
K               K '
[ A][B]         ka
d [ P] / dt  kb [ I ]  kb K [ A][B]
Example. Analysing pre-eq. by
 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'

the old pre - eq analysis result.
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

ka
d [ A ]
A  A  A  A                     k a [ A]2
dt
ka '
d [ A ]
A  A  A  A                k a ' [ A][ A ]
dt

kb             
             d[ A ]
A P                   kb [ A ]
dt
d [ A ]
 kb [ A ]  k a ' [ A][ A ]  k a [ A]2  0 
dt
       k a [ A]2
[A ]                    
kb  k a ' [ A]

d [ P]                   d[ P] kb ka [ A]2
 kb [ A ]  0       
dt                       dt    kb  ka ' [ A]

If kb is small: kb << ka’ [A] then first-
order 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
d [ A ]
A  A  A  A                k a [ A]2
dt

becomes ”bottleneck”.
Define effective rate constant:
d [ P] kb k a [ A]2      k k [ A]
                 b a            [ A]  K [ A]
dt     kb  k a ' [ A] kb  k a ' [ A]
kb k a [ A]
with K 
kb  k a ' [ A]

Lindemann Hinshelwood
mechanism gives linear plot

1            d [ P] 1    1  ka '
K         [ 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   v  kt [CH 3 ]2

 Chain carriers are CH3
Derivation of rate law:
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

ki
0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
2
[CH 3CHO]
kt
Insert into
ki
0  ki [CH 3CHO]  kt [CH 3 ]  [CH 3 ] 
2
[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 )  2 HBr ( g )
Initiated by heat.
New example:
Br2 ( g )  H 2 ( g )  2 HBr ( 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]
Termination :   Br   Br  Br2    v  kt [ Br]2

Retardation step: A product is
removed.
Chain carriers are H and Br radicals.
Rate of formation
d [ HBr]
 k p [ H 2 ][Br]  k p '[ H ][Br2 ]  kr [ HBr][H ]
dt

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

ki
2ki [ Br2 ][M ]  2kt [ Br] [ M ]  0  [ Br] 
2
[ Br2 ]1/ 2
kt

Inserting the expression for Br
ki        1/ 2
kp         [ Br2 ] [ 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
ki
k  2k p               k '  (k r / k ' p )
kt
Explosions
 An explosion is a rapidly
accellerating reaction arising
from a rapid increase in reaction
rate with increasing
temperature

fast reaction releasing heat (exothermic)  T rises 
even faster reaction  T rises even more  ..
Example

O2 ( g )  2 H 2 ( g )  2 H 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 ]
 chain carriers are     H , OH ,  O 
 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.
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
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 ] measuring chain branching rate.

kterm  kt  kt' [O2 ][ M ] measuring terminati on rate.

Then chain carrier production rate is
d
[ H ]  vini  (kbranch  kterm )[H ]
dt
vinit
a ) [ H ]                  (1  e ( kt e rm  kbranc h ) t ) kbranch  kterm
kterm  kbranch
vinit
b) [ H ]                  (e ( kbranc h  kt e rm ) 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:
2 H 2O2 (aq)  2 H 2O(l )  O2 ( g )
as catalyzed by bromide ions.
                 
H 3O  H 2O H 2O  H 3O2
2

K  [ H 3O2 ] /[ H 3O  ][H 2O2 ]
                                 
H 3O2  Br   HOBr  H 2O v  k[ H 3O2 ][Br  ]
HOBr  H 2O2  H 3O   O2  Br     ( fast)

Notice that bromide reappears in the end.
Enzymes   E  S  ES  P  E

 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.
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 ka , k 'a


ES  P  E kb

kb [ E ]0          [ E ][ S ]
v  kb [ ES ]                   KM 
1  K M /[S ]0          [ ES ]
Proof.
d [ ES]
Steady state approx.:                      ka [ E ][S ]  k 'a [ ES]  kb [ ES]  0
dt
k a [ E ][ S ]
then             [ ES ] 
k 'a  k b

k 'a  kb [ E ][ S ]
KM                
Define                         ka       [ ES ]

Use [ E ]0  [ E ]  [ ES ] [ S ]  [ S ]0 substrate in excess

[ E ]0  [ ES ](1  K M /[S ])  [ ES ](1  K M /[S ]0 )
then

[ ES ]  [ E ]0 /(1  K M /[S ]0 )

v  kb [ ES ]  kb [ E ]0 /(1  K M /[S ]0 )

Then i)-iii) is correct. show!
Alternative formula:              vm ax  kb [E ]0

vm ax
v  kb [ ES ]                   
1  K M /[S ]0
1   1     1 KM
    
v vm ax [ S ]0 vm ax

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   vm ax  kb [E ]0          v  kb [ ES ] 
vm ax

1  K M /[S ]0
and
K m / vm ax               1

1

1 KM
v vm ax [ S ]0 vm ax
 From knowing total
enzyme concentration
we calc. K m , kb

 Cannot determine k a ,     k 'a
Catalytic constant.
 k b gives the number
of ”turn overs” pr. unit        
time. This number –
E  S ES k a , k 'a
with units of pr time –    ES  P  E kb
is called the catalytic
constant, k
cat
Catalytic efficiency.

E  S  ES         k a , k 'a
ES  P  E             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      E  S ES k a , k 'a

large. Then               ES  P  E kb
ka
           kb  k a
k 'a  kb

 ka is determined by
diffusion.                                1 1
 k a  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  vm ax /( ' 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
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 dihydro-
pteroate synthease
producing folate which
is crucial for survival of
B.
 Active substrate p-
aminobenzoic acid.
 Inhibitor Sulfanilamide.

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