chemical kinetics-reaction mechanisms and complex reactions(notes)

7/30/2019 Chemical Kinetics-Reaction Mechanisms and Complex Reactions(Notes)

1/41

Reaction Mechanisms and Complex

Reactions

Reversible first-order reaction

Ak1

k1B

Reversible has a different meaning in kinetics than in

thermodynamics!

All chemical reactions are reversible to some extent,

but ifK>

100, this fact can usually be ignored for pur-

poses of chemical kinetics.

d[A]

d t= fAvf+bAvb

f

A=1, vf= k1[A]bA=+1, vb= k1[B]d[A]

d t=k1[A]+k1[B]

PChem BA 8.1


2/41

d[B]

d t= fBvf+bBvb

d[B]

d t= k1[A]k1[B]

Add the two rate equations:

d[A]

d t +d[B]

d t =k1[A]+k1[B]+k1[A]k1[B]d

d t

[A]+ [B]

= 0

[A](t)+ [B](t)= const= [A](0)+ [B](0)=C[B](t)=C [A](t)d[A]

d t=k1[A]+k1{C [A](t)}

d[A]

d t= k1C (k1+k1)[A]

equilibrium:

d[A]eq

d t= 0

PChem BA 8.2


3/41

0=k1[A]eq+k1[B]eq

[B]eq[A]eq

= k1k1

=Kc

C [A]eq[A]eq

= k1k1

=Kc

[A]eq

=

k1

k1+k1C

d[A]

d t= k1C (k1+k1)[A]

=(k1+k1)

[A] k1k1+k1

C

d[A]

d t =(k1+k1)[A] [A]eq[A](t)[A]0

d[A]

[A] [A]eq=(k1+k1)t

ln

[A](t) [A]eq[A]0

[A]eq

=(k1+k1)t

[A](t) [A]eq[A]0 [A]eq

= exp(k1+k1)t

PChem BA 8.3


4/41

[A](t)= [A]eq+

[A]0 [A]eq

exp

(k1+k1)t

relaxation methods: relaxation = return of a system

to equilibrium

Relaxation methods are used to measure rate con-

stants. Apply a sudden perturbation to the equilib-

rium, e.g., a temperature jump or a pressure jump,

and monitor the relaxation.

Let x be the deviation from equilibrium. Then for a re-

versible first-order reaction A B, our results showthat

x(t)= [A](t) [A]eq= x(0)exp(t/)with

= 1k1+k1

i.e., after a temperature jump at t= 0, the system re-laxes exponentially with a rate 1/ = k1+ k1. Com-bine the measurement of the equilibrium constant

PChem BA 8.4


5/41

Kc = k1/k1 with a measurement of the relaxationtime to obtain the forward and backward rate con-

stants k1 and k1.

Parallel first-order reactions

Assume the reactions are irreversible for the moment.

Ak1

BA

k2Cd[A]

d t=k1[A]k2[A]

d[B]

d t = k1[A]d[C]

d t= k2[A]

add the three equations:

dd t

[A]+ [B]+ [C]= 0

[A](t)+ [B](t)+ [C](t)= const

PChem BA 8.5


6/41

= [A](0)+ [B](0)+ [C](0)= [A]0

if initially only the reactant A is present

conversion of reactant into products:

[A](t)

=[A]0 exp{

(k1+

k2)t}

d[B]

d t= k1[A]= k1[A]0 exp{(k1+k2)t}

[B](t)= k1[A]0k1+k2

1exp[(k1+k2)t]

[C](t)

=k2[A]0

k1+k2 1exp[(k1+k2)t]yield (t)

YB

irr

=[B]()

[A]0 =k1

k1+k2Y

Cirr =

[C]()[A]0

= k2k1+k2

PChem BA 8.6


7/41

selectivity

Sirr= [B](t)[C](t)

= k1k2

for all times

What happens if the reactions are reversible?

A

k1

k1 B

Ak2

k2C

d[A]

d t=k1[A]+k1[B]k2[A]+k2[C]

d[B]d t

= k1[A]k1[B]d[C]

d t= k2[A]k2[C]

[A](t)+ [B](t)+ [C](t)= const= [A]0

can eliminate on variable, say [C]:

[C](t)= [A]0 [A](t) [B](t)

PChem BA 8.7


8/41

Consider the case t. The reaction reaches equi-librium, i.e., the concentrations no longer change with

time:

d[A]

d t=k1[A]+k1[B]k2[A]+k2[C]= 0

d[B]

d t =k1[A]

k

1[B]

=0

d[C]

d t= k2[A]k2[C]= 0

k1[A]eq= k1[B]eqk2[A]eq= k2[C]eq[B]

eq

[A]eq= k1

k1=Kc,1

[C]eq

[A]eq= k2

k2=Kc,2

[A]eq+ [B]eq+ [C]eq= [A]0

[A]eq+Kc,1[A]eq+Kc,2[A]eq= [A]0[A]eq=

[A]0

1+Kc,1+Kc,2

PChem BA 8.8


9/41

[B]eq=Kc,1[A]0

1+

Kc,1+

Kc,2

[C]eq=Kc,2[A]0

1+Kc,1+Kc,2yield (t)

Y

Brev=

[B]eq

[A]0 =Kc,1

1+Kc,1+Kc,2

YC

rev=[C]eq

[A]0= Kc,2

1+Kc,1+Kc,2selectivity

Srev=[B]eq

[C]eq= Kc,1

Kc,2= k1

k1 k2

k2= k1

k2 k2

k1

Srev=Sirr k2k1

The selectivity

S = [B][C]

= k1k2=Sirr

PChem BA 8.9


10/41

will be observed for short times, as along as the back-

ward reactions are negligible: The reaction is un-

der kinetic control.

For long times, the selectivity approaches the equilib-

rium value

S=

[B]

[C] tKc,1

Kc,2 =S

rev

The reaction is under thermodynamic control.

Example: k1 = 1 s1, k1 = 0.01s1, k2 = 0.1s1, k2 =0.0005s1

Initially B is formed ten times faster than C, however

Kc,1=1 s1

0.01s1= 100

Kc,2= 0.1s10.0005s1

= 200

and C is the more stable product.

PChem BA 8.10


11/41

1000 2000 3000 4000 5000 6000 7000

t

0.2

0.4

0.6

0.8

1.

conc

Figure 1: Parallel first-order reversible reactions. [A]:

solid line, [B]: dashed line, [C]: dot-dash line.

PChem BA 8.11


12/41

1 2 3 4 5 6 7 8 9 10

t

0.2

0.4

0.6

0.8

1.

conc

Figure 2: Parallel first-order reversible reactions: Ini-

tial stage. [A]: solid line, [B]: dashed line, [C]: dot

dash line.

PChem BA 8.12


13/41

0.01 0.1 1 10 100 1000 10000

t

0.2

0.4

0.6

0.8

1.

conc

Figure 3: Parallel first-order reversible reactions: con-

centrations versus logt. [A]: solid line, [B]: dashed

line, [C]: dot-dash line.

In the above parallel-reaction mechanism C can be

converted into B only via A. What happens if there isthe additional reaction

Ck3

k3B

(An example are the isomerization reactions be-

tween ortho-xylene, meta-xylene, and para-xylene:

o-xylene m-xylene, m-xylene p-xylene, p-xyleneo-xylene.) Let also assume that the three steps are

PChem BA 8.13


14/41

the mechanism of the reaction, i.e., all reactions are

elementary.

Is it then possible to establish equilibrium by the

cyclic reaction mechanism below?

A

B

C!

'

eee

d[A]

d t=k1[A]+k2[C]

d[B]

d t= k1[A]k3[B]

d[C]

d t =k3

[B]

k2

[C]

[A](t)+ [B](t)+ [C](t)= const= [A]0stationary state:

k1

[A]=

k2

[C]=

[C]

[A] =k1

k2

k1[A]= k3[B] =[B]

[A]= k1

k3

PChem BA 8.14


15/41

k3[B]= k2[C]

which follows from the previous two equations

[B]= k1k3

[A]

[C]= k1k2

[A]

[A]+ k1k3

[A]+ k1k2

[A]= [A]0

[A]= [A]01+ k1

k3+ k1

k2

We have determined the stationary concentrations

[A], [B], and [C] in terms of [A]0 and the three rateconstants.

This stationary state is not an equilibrium state!! It

violates the principle of detailed balance.

The principle of microscopic reversibility:

In a system at equilibrium, any molecular pro-

cess and the reverse of that process occur, on

the average, at the same rate.

PChem BA 8.15


16/41

implies the principle of detailed balance:

In a system at equilibrium, each collision has itsexact counterpart in the reverse direction, so

that the rate of every chemical process is ex-

actly balanced by that of the reverse process.

=In a system at equilibrium, every elementary forward

reaction is balanced by its own backward reaction.

In other words, in a system at equilibrium each ele-

mentary reaction must be individually at equilibrium.

Principle of detailed balance implies that all elemen-

tary reactions must be reversible.

Application to the scheme

Ak1

k1B

Ak2

k2C

Ck3

k3B

PChem BA 8.16


17/41

equilibrium:

k1[A]eq= k1[B]eq = Kc,1= [B]eq[A]eq

= k1k1

k2[A]eq= k2[C]eq = Kc,2 =[C]eq

[A]eq= k2

k2

k3

[C]eq=

k3

[B]eq=

Kc,3=

[B]eq

[C]eq =k3

k3

The product of the second and third equilibrium con-

stants is

Kc,2Kc,3=[C]eq

[A]eq

[B]eq

[C]eq =[B]eq

[A]eq =Kc,1

=

k1

k1= k2

k2

k3

k3

or

k1k2k3= k1k2k3PChem BA 8.17


18/41

This condition on the six rate constants has been de-

rived by considering chemical equilibrium, but rate

constants are concentration independent. Therefore,

this relation is general for this mechanism; it also

holds for all nonequilibrium conditions. We see that

the rate constants can never be chosen independent-

ly, and the cyclic mechanism is impossible, since it

requires k1=

k2

=k3

=0, which is forbidden by the

relation.

Consecutive first-order reactions (series reactions)

Ak1B k2C

Assume the reactions are first order and irreversible.

d[A]

d t=k1[A]

d[B]

d t = k1[A]k2[B]d[C]

d t= k2[B]

PChem BA 8.18


19/41

Assume again [A](0) = [A]0, [B](0) = 0, [C](0) = 0, andadd the three equations:

[A](t)+ [B](t)+ [C](t)= const= [A]0[A](t)= [A]0 exp(k1t)d[B]

d t= k1[A]0 exp(k1t)k2[B]

[B](t)= k1[A]0k2k1

{exp(k1t)exp(k2t)}

[C]= [A]0 [A](t) [B](t)

= [A]0

1+ 1k2k1

[k1 exp(k2t)k2 exp(k1t)]

Consider the case: k2 k1. Then

exp(k2t) exp(k1t)

and

k2k1 k2

PChem BA 8.19


20/41

[C] [A]01+1

k2 (k2)exp(k1t)

= [A]0{1exp(k1t)}

The formation ofC depends onlyon the rate constant

of the first step A

B.

AB is the rate-determining step.

Motivation for the

Steady-State Approximation: After an initial induction

period, the concentration and the rate of change of

all reaction intermediates are negligibly small.

Illustrate for the consecutive scheme:

[B](t)

[A](t) =

k1[A]0

k2k1{exp(k1t)exp(k2t)}

[A]0 exp(k1t)= k1

k2k1{1exp[(k2k1)t]}

PChem BA 8.20


21/41

For k2 k1:

[B](t)

[A](t) k1

k2{1exp[k2t]

0 for t1/k2

}= k1k2

[B](t)= k1k2

[A](t) [A](t)

d[B]

d t =k1

k2

d[A]

d t d[A]

d t

If we apply directly the steady-state approximation

(SSA), also known as the quasisteady-state approxi-

mation (QSSA), to the intermediate B, we obtain

0 d[B]d t

= k1[A]k2[B]

=

[B]=

k1

k2[A]

d[C]

d t= k2[B]= k2

k1

k2[A]= k1[A]

PChem BA 8.21


22/41

= k1[A]0 exp(k1t)

[C](t)= [A]0{1exp(k1t)} (same as above)

Application: decomposition of ozone

2 O3 3 O2

empirical rate law (M is an inert gas):

v=12

d[O3]

d t= 1

3

d[O2]

d t= k[O3]

2[M]

k[O2][M]+ [O3]

proposed mechanism:stoichiometric number

O3+Mk1O2+O+M 2

O2+O+Mk1O3+M 1

O+

O3

k2

2 O

21

2 O3+2 M+O2+O+M+O+O3

PChem BA 8.22


23/41

2 O2+2 O+2 M+O3+M+2 O2

2 O3 3 O2

intermediate O: steady-state approximation

d[O]

d t= 0

Intermediates do not appear in the rate law!!

d[O]

d t= k1[O3][M]k1[O2][O][M]k2[O][O3]= 0

[O]

=

k1[O3][M]

k1[O2][M]+k2[O3]d[O3]

d t=k1[O3][M]+k1[O2][O][M]k2[O][O3]

=k1[O3][M]+

k1[O2][M]k2[O3]

[O]

=k1[O3][M]++

k1[O2][M]k2[O3] k1[O3][M]

k1[O2][M]+k2[O3]

PChem BA 8.23


24/41

= 1k

1[O2][M]

+k2[O3]

k1[O3][M]k1[O2][M]k1[O3][M]k2[O3]+

+k1[O2][M]k1[O3][M]k2[O3]k1[O3][M]

= 2k1k2[O3]2[M]

k1[O2][M]+k2[O3]=2 k1[O3]

2[M]k1k2

[O2][M]+ [O3]=2v with k= k1, k= k1/k2

empirical rate law

If initially we have pure ozone, then k1[O2][M] k2[O3] and

d[O3]

d t=2k1[O3][M]

= the first step is rate-determiningThe reaction is self-inhibiting in O2, v as [O2]

PChem BA 8.24


25/41

Another approximation method

Pre-equilibrium or fast-equilibrium approximation

A+B k1k1

Xk2C

k1 k2, k1 k2

A, B, and Xare in equilibrium:

k1[A][B]= k1[X]

[X]= k1k

1

[A][B]=K[A][B]

(Note: K=Kc!!)d[C]

d t= k2[X]= k2K[A][B]= k[A][B]

k= k2K=k2k1

k1

PChem BA 8.25


26/41

enzyme reactions

Michaelis-Menten mechanism

E+S k1k1

(ES)k2 P+E

d[P]

d t= v= k2[(ES)]

Apply the steady-state approximation:

d[(ES)]

d t= k1[E][S]k1[(ES)]k2[(ES)]= 0

[(ES)]

=

k1[E][S]

k1+k2 =

[E][S]

KM

KM=k1+k2

k1Michaelis constant

d[P]

d t= k2

KM[E][S]

conservation of total enzyme concentration:

[E]0= [E]+ [(ES)]

PChem BA 8.26


27/41

= [E]+ [E][S]KM

= [E]

1+ [S]KM

[E]= [E]01+ [S]

KM

d[P]

d t =k2

KM [E][S]

= k2KM [E]0

1+ [S]KM

[S]

d[P]

d t= v= k2[E]0[S]

KM+

[S]

Michaelis-Menten rate law

[Figure: rate versus substrate concentration, Fig. 8.13]

At high substrate concentrations, [S]KM,

d[P]d t

k2[E]0[S][S]

= k2[E]0,

the reaction is zero order, and the rate reaches its

PChem BA 8.27


28/41

maximum:

vmax= k2[E]0

The Michaelis-Menten rate law can be written in terms

of vmax:

d[P]

d t= v= vmax

1+KM/[S]

We rearrange this expression into a form that is

amenable to data analysis by linear regression:

1

v= 1

vmax+

KM

vmax

1

[S]

A Lineweaver-Burk plot is a plot of the reciprocal of

the initial velocity v0 versus the reciprocal of the ini-

tial substrate concentration [S]0 at fixed enzyme con-

centration. If the enzyme obeys Michaelis-Menten ki-

netics, then the plot should yield a straight line, with

PChem BA 8.28


29/41

slope KM/vmax, that intercepts the ordinate axis at

1/vmax and the abscissa at

1/KM.

[Figure: Lineweaver-Burk plot, Fig. 8.14]

At low substrate concentrations, [S]KM,

d[P]

d t

k2[E]0[S]

KM = k[S],

the reaction is first order.

The turnover number or catalytic constant of an

enzyme, kcat, is the number of catalytic cycles

(turnovers) performed by the active site in a given

time interval divided by the duration of that inter-

val. It has units of a first-order rate constant, and for

the Michaelis-Menten mechanism is given by k2, the

rate constant for release of product from the enzyme-

substrate complex.

kcat= k2=vmax

[E]0

PChem BA 8.29


30/41

The catalytic efficiency, , of an enzyme is given by

= kcatKM

The higher the value of , the more efficient is the

enzyme. For the Michaelis-Menten mechanism

= k2k1+k2

k1

= k1k2k1+k2

max= k1 if k2 k1

= rate of formation of the enzyme-substrate complex;

diffusion-limited: < 108 109 M1 s1 for enzyme-sizedmolecules at room temperature

catalase (decomposition of hydrogen peroxide) =4.0

108 M1 s1: catalytic perfection

Diffusion control

reactions in solutions

PChem BA 8.30


31/41

reaction between two solute molecules A and B: the

two molecules approach by diffusion to form an en-

counter pair: {AB}

[Figure: processes in involved in solution phase reactions]

kinetic scheme:

A+Bkd

kd {AB}ka

products

kd: rate constant for diffusive approach; kd: rateconstant for separation of the encounter pair; ka: rate

constant for conversion of encounter pair into prod-

ucts

apply steady-state approximation to the encounter

pair

d[{AB}]

d t= kd[A][B] (kd+ka)[{AB}]= 0

[{AB}]= kd[A][B]kd+ka

PChem BA 8.31


32/41

reaction velocity

v= ka[{AB}]= kakd[A][B]kd+ka

thus the rate constant of the bimolecular reaction

A

+B

k2

products

is given by

k2=v

[A][B]= kakd

kd+ka

two-limiting cases:

(i) diffusion-controlled limit: ka kd, separation ofA and B is relatively difficult, e.g., the solvent has a

large viscosity , or the reaction has a small activa-

tion energy = rate-determining step is the diffusiveapproach of the reactants; once they are in the sol-

vent reaction cage, reaction is assured

k2= kdPChem BA 8.32


33/41

analysis of diffusion of molecules in liquids and use of

the Einstein relation and Stokes law:

k2= kd=8RT

3

for small molecules in aqueous solutions at room tem-

perature, kd

1091010 M1 s1

(ii) activation-controlled limit: ka kd, reactionswith large activation energies; Ea 20kJmol1 for re-actions in water,

k2= kakd

kd

the second factor is the equilibrium constant KAB for

the formation of the encounter pair; thus

k2= kaKAB

rate is determined by the equilibrium concentration

of encounter pairs and the rate of passage over the

activation energy barrier

PChem BA 8.33


34/41

Gas phase reactions

kinetic theory of Gases

average speed

c=

8RT

M

1/2

R=NAk, k Boltzmann constant

R

M= NAk

NAm= k

m

c

= 8kT

m1/2

m mass of a particle

elastic collisions

determine how often a specific particle collides with

other particles in the gas

since this is a representative particle, we can assume

that it moves with the average speed c

PChem BA 8.34


35/41

we can replace the moving collision partners by sta-

tionary particles, if we replace the average speed c

by the relative average speed crel:

c=

8kT

m

1/2, crel=

8kT

1/21

1

mA+

1

mB

, = reduced mass

for identical particles

1

= 2

m= = m

2

crel=2c

[Figure: collision tube]

number of stationary particles inside the collision

tube: N0 crelt, N= N/V number density, 0 =d

2

= elastic collision cross-section, for collisions be-tween identical particles, d= 2rcollision frequency z = average number of collisions

PChem BA 8.35


36/41

of a particle per unit time

z=N0 creltt

=N0crelz=

2N0c

z=

2N

V0c

z=2 PkT

0c; (PV=nRT= NNA

RT=N kT)

mean free path l = the average distance a molecule

travels between two successive collisions

l= c 1z= c

2N0c= 1

2(N/V)0= kT

2P0

total number of collisions per unit volume per unit

time for identical particles:

ZAA=1

2zNA=

0c2NA

2=0

4kT

mA

1/2NA

2

PChem BA 8.36


37/41

for mixture of gases A and B, total number of colli-

sions per unit volume per unit time for dissimilar par-

ticles:

ZAB=0

8kT

1/2NANB

for dissimilar particles: 0

=d2 with d

=rA

+rB

if all collisions were reactive, then the kinetic equa-

tion for the number density of Afor the reaction

A+B P

would be

dNA

d t=ZAB

and for the molarity

d[A]d t

=ZAB/NA

where NA is Avogadros number

PChem BA 8.37


38/41

so we obtain

v= kAB[A][B]=08kT

1/2NA[A][B]

kAB=0

8kT

1/2NA

most collisions will not be reactive, since collisionsmust occur with sufficient energy in the gas phase to

give rise to a reactive event

[Figure: collisions in the gas phase, Fig. 7.11]

before molecules can get close enough to react, they

must overcome an energy barrier a

only molecules with sufficient kinetic energy along

the line of centers AB to surmount this energy barrier

will react

[Figure: line-of-centers energy, Fig. 7.12]

fraction of reactive collisions

f= exp(Ea/RT), Ea=NAa

PChem BA 8.38


39/41

v=08kT

1/2

NAexp(Ea/RT)[A][B]

kAB=0

8kT

1/2NAexp(Ea/RT)

the predicted rate constant follows the Arrhenius law

kAB= Aexp(Ea/RT)

the pre-exponential factor A is given by

Ath=08kT

1/2

NA

collision theory predicts that the pre-exponential fac-

tor is weakly temperature dependent

Ath

T

for many reactions, this temperature dependence is

swamped by the strong temperature dependence of

the exponential term

PChem BA 8.39


40/41

the following Table (from M. J. Pilling and P. W. Seakins,

Reaction Kinetics, Oxford University Press, Oxford,

1995) compares the predicted and experimental val-

ues of A for some reactions, the quantity P is defined

as

P= AexpAth

Rxn T Ea 1011Aexp 10

11Ath P

1 600 0 10 2.1 4.8

2 300 0 0.24 1.1 0.22

3 470 102 0.094 0.59 0.16

4 800 180 1.24105 7.3 1.7106

T is in K, Ea in kJmol1

, A in Lmol1 s1

1: K+Br2KBr+Br2: CH3

+CH3

C2H6

3: 2NOCl 2 NO+Cl24: H2+C2H4C2H6PChem BA 8.40


41/41

except for Rxn 1, the theoretical values are too large,

for Rxn 4 by more than a factor of 105

P< 1 indicates that the relative orientation of the mol-ecules is important in reactive collisions; P is known

as the steric factor and is generally several orders of

magnitude smaller than 1

kAB= P0 8kT

1/2 NAexp(Ea/RT)orientation does not explain values of P > 1, whichwould seem to imply that the molecules react faster

than they collide

P> 1: long range attractive interactions

chemical kinetics-reaction mechanisms and complex reactions(notes)

Documents