Reaction Kinetics(3)

8/14/2019 Reaction Kinetics(3)

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1

Reaction

Kinetics (3) Xuan Cheng

Xiamen University

PhysicalChemistry



2

Determination of theRate Law

PhysicalChemistry

λ β α ][][][ L B Ak r =The rate law (17.48)

1. Half-life method

o A

n

Ank n

t ][log)1()1(

12loglog 10

1

102/110 −−−

−=

−(17.49)

[ ] Ano

n

k Ant 1

1

2/1 )1(

12

−

−

−

−

= (17.29)For n ≠ 1

o A

n

Ank n

t ][log)1()1(

12loglog 10

1

102/110 −+−

−=

−(17.49)

ReactionKinetics



3

半衰期法确定反应级数

用半衰期法求除一级反应以外的其它反应的级数。

以 lnt 1/2 ～ ln[A]o 作图从直线斜率求 n 值。从多个实验数据

用作图法求出的 n 值更加准确。

根据 n 级反应的半衰期通式：取两个

不同起始浓度 [A]o ， [A]o’ 作实验，分别测定半衰期为

t 1/2 和，因为同一反应，常数相同，所以：

1

1

2/1][

1

)1(

12−

−

−

−=

no A

n

Ak nt

1/ 2't

PhysicalChemistry

1

2/1

2/1

][

'][

'

−

=

n

o

o

A

A

t

t

)]/[']ln([

)'/ln (1 2/12/1

oo A A

t t n +=

o An K t ]ln [)1(lnln 2/1 −+=

ReactionKinetics



4

PhysicalChemistryDetermination of the

Rate Law2. Powell-plot method

o A A ]/[][≡α

n Ak r ][=

(17.50)

the fraction of A unreacted

t Ak no A

1][ −≡φ

[ ][ ]

[ ] t k n A A

A A

no

n

o

)1(1 11−+=

−−

For n ≠ 1 (17.28)

φ α )1(11 −=−− nn For n ≠ 1

(17.13)

[ ]

[ ]t k

A

A A

o

−=ln

For n = 1

φ α −=ln For n = 1 (17.51)

ReactionKinetics





6

PhysicalChemistry

Determination of theRate Law

3. Initial-rate method


The ratio of initial rates for runs 1 and 2

α

=

1,0

2,01,02,0

][

][/ A

Ar r

Measure r 0 for two different initial concentrations [A]0,1 and [A]0,2

while keeping [B]0, [C]0, …fixed.

α can be found

The orders β ,…λ can be found similarly

ReactionKinetics



7

PhysicalChemistry

Determination of theRate Law4. Isolation method


Make initial concentrations of reactant A much less than the

concentrations of all other species

[B]0 >> [A]0, [C]0 >> [A]0, …

The rate law becomes

α λ β α ][][][][ 00

A j L B Ak r == (17.52)λ β 00 ][][ L Bk jwhere ≡

Where j is essentially constant.

The reaction has the pseudo-order α .

The orders β ,…λ can be found

similarly.

ReactionKinetics





9

PhysicalChemistryRate Laws and EquilibriumConstants for Elementary Reactions

DC B A +⇔+

]][[]][[][

11 DC k B Ak dt

Ad −+−=

Show that for a reaction that takes place in a sequence of steps, the

overall equilibrium constant is a product of ratios of the rate

constants for each step.

It is sufficient to consider a reasonably general but simple two-stepreaction sequence, such as

F E C +⇔

F E D B A ++⇔+

(second-order in each direction, k 1, k -1)

(first-order forwarded, second-order reverse, k 2, k -2)

(overall)

]][[][][

22 F E k C k dt

C d −+−=

ReactionKinetics





11

PhysicalChemistry Reaction Mechanisms

p r o d u c t s A →

The Rate-Determining-Step Approximation

The reaction mechanism is assumed to consist of one or morereversible reactions that stay close to equilibrium during most of the

reaction, followed by a relatively slow rate-determining step, which

in turn is followed by one or more rapid reactions.

products B A →+ products A →2

productsC B A →++ products B A →+2 products A →3

unimolecular

The number of molecules that react in an elementary stepThe molecularity of the elementary reaction

bimolecular

trimolecular (termolecular)

ReactionKinetics



12

PhysicalChemistry

12 k k >>

Reaction Mechanisms

C B Ak k → → 21

t k t k ee 12 −− << 212 k k k ≈−

Suppose now that

Then whenever a B molecule is formed it decays rapidly into C.

−

+−

−= −− t k t k o e

k k

k e

k k

k AC 21

12

1

12

21][][ (17.41)

reduces to ( )t k o e AC 11][][ −−=

The formation of C depends on only the smaller of the two rate constants

B Ak → 1 is called the rate-determining step of the

reaction.

For the consecutive unimolecular reactions

ReactionKinetics



13

PhysicalChemistry

The Steady-State Approximation

Assumes that during the major part of the reaction, the rates of

change of concentrations of all reaction intermediates are

negligibly small

0][ ≈dt

te Intermediad

Reaction Mechanisms

Reactants

Products

Intermediates

Time

Conce

ntration

C B Ak k → → 21 (17.35)

0][][][

21 =−=

Bk Ak dt

Bd

][][

2

1 A

k

k B = ][][

][12 Ak Bk

dt

C d ==

ReactionKinetics



14


0001 ])[1(][][ 11 Aee Ak C t k t t k −− −=∫ =

C is formed by a first-order decay of A, with a rate constant k 1,

the rate constant of the slower, rate-determining step.

][][][

12 Ak Bk dt

C d ==

[ ] [ ]t k

o e A A 1−

= (17.38)

( )t k o e AC 11][][ −−=

The same result as before, butobtained much more quickly.

ReactionKinetics



15

PhysicalChemistry

C B →

21 k k >>−

A B →

(the rate-

determining step)

Reaction Mechanisms

DC B A k k k → → → 321

←−1k

←−2k

←−3k

Consider the following mechanism composed of unimolecular reactions

is slower than

B A ⇔ remains close to equilibrium

23 k k >> 23 −>> k k

C B ⇔ is not in equilibrium

D is rapidly formed

from C

ReactionKinetics



16


O H N H C NH H C HNO H Br

22562562 2+ → ++ ++ −

]][][[ 2−+= Br HNO H k r

Example 17.4

is observed to be

A proposed mechanism is

(17.55)

The rate law for the Br --catalyzed aqueous reaction

O H ONBr Br HNOk

222 + → + −

−+ ++ → + Br O H N H C NH H C ONBr k

22562563

++ → + 2221 NO H HNO H

k

←−1k

rapid equilib.

(17.56)slow

fast

ReactionKinetics

h i l



17

PhysicalChemistry Reaction MechanismsExample 17.4

]][][[ 2−+= Br HNO H k r (17.55)

Deduce the rate law for this mechanism and relate the observed rate

constant k in (17.55) to the rate constants in the assumed mechanism(17.56)

++ → + 2221 NO H HNO H

k

← −1k

rapid equilib.

(17.56)O H ONBr Br HNOk

222 + → + − slow

−+ ++ → + Br O H N H C NH H C ONBr k

22562563 fast

the rate-determiningstep

(1)

(2)

(3)

The formation of ONBr in (2) ]][[ 222−+= Br NO H k r (17.57)

Step (1) is near equilibrium. Equation (17.53) gives

ReactionKinetics

Ph i l



18

PhysicalChemistry Reaction MechanismsExample 17.4

b

f

c k

k

K = elementary reaction (17.53)*

]][[

][

2

22

1

11,

HNO H

NO H

k

k K c +

+

−== ]][[][ 2

1

122 HNO H

k

k NO H +

−

+ =

]][[ 222−+= Br NO H k r (17.57)

]][][)[/( 2121−+

−= Br HNO H k k k r

21,121 )/( k K k k k k c== −

]][][[ 2−+= Br HNO H k r (17.55)

Example 17.5

ReactionKinetics

Ph i l



19


More examples in using the steady-state approximation

)()(4)(2 2252 g O g NO g O N +→ ][ 52O N k r =

on the basis of the following mechanism:

Account for the rate law for the decomposition of N2O5

3252 NO NOO N ak + →

5232

'

O N NO NO ak → +

NOO NO NO NO bk ++ → +2232

252 3 NOO N NO ck → +

First identify the intermediates NO and NO3

ReactionKinetics

Ph i l



20

PhysicalChemistryReaction Mechanisms

]][[]][[][

5232 O N NOk NO NOk dt

NOd cb −=

]][[]][[][][

3232'

523 NO NOk NO NOk O N k

dt

NOd baa −−=

3252 NO NOO N

ak

+ →

5232

'

O N NO NO ak → +

NOO NO NO NO bk ++ → + 2232

252 3 NOO N NO ck → +

]][[]][[][][

5232'

5252 O N NOk NO NOk O N k

dt

O N d caa −+−=

ReactionKinetics

Ph i l



21


0]][[]][[ 5232 =− O N NOk NO NOk cb

][

][

)(][][

]][[

2

52

'32

52

NO

O N

k k

k

NO NO

O N NO

k

k

ba

a

b

c

+==

According to the steady-state approximation, set both rates equal

to zero 0][ =dt NOd 0][ 3 =

dt NOd

)(][ 'bac

ba

k k k

k k

NO +=

][

]][[][

2

523

NO

O N NO

k

k NO

b

c=

0]][[]][[][ 3232

'

52 =−− NO NOk NO NOk O N k baa

][

][

)(][

][

)(][

2

52'

2

52'3

NO

O N

k k

k

NO

O N

k k k

k k

k

k NO

ba

a

bac

ba

b

c

+=

+=

ReactionKinetics

Ph i l



22


]][[]][[][][

5232'

5252 O N NOk NO NOk O N k

dt

O N d caa −+−=

The net rate of change of concentration of N2

O5

is

][

)(][

][

)(

][][][

52'2

52

'2

'52

52 O N

k k k

k k k

NO

O N

k k

k NOk O N k

dt

O N d

bac

bac

ba

aaa

+−

++−=

)(

]['

bac

ba

k k k

k k NO

+

=

][

][

)(][

][

)(][

2

52'

2

52'3

NO

O N

k k

k

NO

O N

k k k

k k

k

k NO

ba

a

bac

ba

b

c

+=

+=

][)(

][)(

][52'52'

''52 O N

k k

k k O N

k k

k k k k k k

dt

O N d

ba

ba

ba

aabaaa

+−

+

+−−=

ReactionKinetics

Ph i l



23


][

2][

52'

52

O N k k

k k

dt

O N d

ba

ba

+−=

because 2][ 52−=O N υ

][][ 5252' O N k O N k k

k k r

baba =+=

][ 52O N k r =

ba

ba

k k

k k k

+=

'

It follows that the reaction rate is

where

ReactionKinetics

Ph sical



24

PhysicalChemistry Reaction MechanismsPre-equilibria

From a simple sequence of consecutive reactions we now turn to a

slightly more complicated mechanism:

Where C denote the intermediate.

P C B A ba k k → → + ←

'ak

This scheme involves a pre-equilibrium, in which an intermediates

is in equilibrium with the reactants.

A pre-equilibrium arises when the rates of formation of the

intermediate and its decay back into reactants are much faster than

its rate of formation of products; thus, the condition is possible

when k’ a>>k b but not when k b >>k’ a. Because we assume that A, B,

and C are in equilibrium.

ReactionKinetics

Physical i



25


We can write:

c K B A

C =]][[

][ 'a

ack

k K =

In writing these equations, we are presuming that the rate of

reaction of C to form P is too slow to affect the maintenance of the

pre-equilibrium (see the following example). The rate of formation

of P may now be written:

]][[][][

B A K k C k dt

P d cbb ==

This rate law has the form of a second-order rate law with a

composite rate constant:]][[

][ B Ak

dt

P d = '

a

bacb

k

k k K k k ==where

ReactionKinetics

Physical R i



26


Example: Analyzing a pre-equilibrium

Repeat the pre-equilibrium calculation but without ignoring the

fact that C is slowly leaking away as it forms P.

][][ C k dt P d b=

The net rates of change of P and C are

0][][]][[][ ' =−−= C k C k B Ak

dt

C d baa

ba

a

k k

B Ak C

+=

'

]][[][

where

P C B A ba k k → → +

← 'ak

]][[][

B Ak dt

P d =

ba

ba

k k

k k k

+=

'

ReactionKinetics

Physical R ti



27


Pre-equilibria

where]][[][

B Ak dt

P d =

ba

ba

k k

k k k

+=

'

]][[][

B Ak dt

P d =

'a

bacb

k

k k K k k ==where

When the rate constant for the decay of C into products is muchsmaller than that for its decay into reactants '

ab k k <<

ReactionKinetics

Physical R ti



28

Homework Physical

Chemistry

Page 592

Prob. 17.28

Prob. 17.29

Prob. 17.33

Page 593

Prob. 17.39

Prob. 17.52

ReactionKinetics

Physical



29

速率决定步骤

在连续反应中，如果有某步很慢，该步

的速率基本上等于整个反应的速率，则该慢步骤称为速率决定步骤，简称速决步或速

控步。利用速决步近似，可以使复杂反应的

动力学方程推导步骤简化。

PhysicalChemistry

Physical



30

速率决定步骤

A][B][1k r ≈慢步骤后面的快步骤可以不考虑。

只需用平衡态近似法求出第 1 ， 2 步的速

率。虽然第二步是速决步，但中间产物 C

的浓度要从第一步快平衡求。

例 1. 慢快快A B C D E+ → → →

PhysicalChemistry

例 2.

快慢快快 F E DC B A →→→→+

Physical



31

稳态近似

从反应机理导出速率方程必须作适当近似，

稳态近似是方法之一。

假定反应进行一段时间后，体系基本上处于稳态，这时，各中间产物的浓度可认为保持不

变，这种近似处理的方法称为稳态近似，一般

活泼的中间产物可以采用稳态近似。

PhysicalChemistry

Physical



32

氢与碘的反应

2 2

2 2

H I 2HI

1 d[HI][H ][I ]

2 dtr k

+ →= =

总包反应

实验测定的速率方程

分别用稳态近似和平衡假设来求中间产物 [I]

的表达式，并比较两种方法的适用范围。

2

2 2 [I]1 d[HI]

[H ]2 dt k =

PhysicalChemistry

2

2

(1) I M 2I M

(2) H 2I 2HI

+ +

+ →

反应机理：

快平衡

慢

M I M I +→+ 22

Physical



33

用稳态近似法求碘原子浓度

因为 (1) 是快平衡， k-1 很大； (2) 是慢反应， k2 很小，

分母中略去 2k2[H2] 项，得：1 2

2 22 2

1

[H ][ [H ]I [] I ]r k k k

k −= =

与实验测定的速率方程一致。

2 2

1 2 -1 2 2

1 d[I]

[I ][M]- [I] [M]- [H ][I] 02 dk k k

t = =

2 1 2

1 2 2

[I ][M][I]

[M] 2 [H ]

k

k k −

=+

2 1 2 2 22 2

1 2 2

[H ][I ][M][H ][I]

[M] 2 [H ]

k k r k

k k −

= = +

PhysicalChemistry

2

2

(1) I M 2I M

(2) H 2I 2HI

+ +

+ →

反应机理：

快平衡慢

M I M I +→+ 22

Physical



34

显然这个方法简单，但这个方法只适用于快平衡下面是慢反应的机理 , 即 k -1>>k 2 。

反应 (1) 达到平衡时：2

1 2 -1[I ][M] [I] [M]k k =

2 1

21

[I] [I ]k

k −=

2 1 22 2 2 2 2 2

1

[H ][I] [H ][I ] [H ][I ]k k

r k k k

−

= = =

用平衡假设法求碘原子浓度Physical

Chemistry

2

2

(1) I M 2I M

(2) H 2I 2HI

+ ++ →

反应机理：

快平衡

慢

M I M I +→+ 22

Physical



稳态近似法与平衡态近似法的比

较

稳态近似法 — 优点：所得最终动力学方程中包含了复合

反应中的全部动力学参数 ( 如 k 1 ， k -1 ,

k 2)优缺点

—平衡态近似法缺点：所得最终动力学方程中只有一个

动力学参数(k 2)

，而且包含在k 2 K c

的

乘积中

优点：所得动力学方程的形式简单

缺点：所得动力学方程的形式复杂

PhysicalChemistry

Reaction Kinetics(3)

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