Phenomenon that changes reaction rate - 2

To consider the kinetics of the reaction, we symbolize: - acid strength is [H 3 O + ]

- base force is [OH - ]

For example:S + HA SH + + A -

Because an acid always has some basic properties and vice versa a base always has acidic properties.

When the catalyst shows acidity, we have: v H+ = k H+ .[S].[H 3 O + ]

When the catalyst exhibits basic properties, we have: v OH- = k OH- .[S].[ OH-] Besides, there are reactions that do not require catalysts, we have: v o = k o .[S]

Then the overall velocity of the process: v c = v H+ + v OH- + v o

= k H+ .[S].[H 3 O + ] + k OH- .[S].[ OH-] + k o .[S]

= [S]. (k H+ .[H 3 O + ] + k OH- .[ OH-] + k o ) = k . [S]

with k = k H+ .[H 3 O + ] + k OH- .[ OH-] + k o : also called the general velocity constant

(1) (2) (3)

Let the above components be (1), (2) and (3) Approximate analysis:

 In case the catalyst is a strong acid:component (1) >> (2), (3)  can ignore (2), (3) Then: k = k H+ . [H 3 O + ]

 log k = lg k H+ + lg[H 3 O + ]

If the acid is given, we know: lg k H+ = const = A

log k = A - pH mt



 In case the catalyst is a strong base:component (2) >> (1), (3)  can ignore (1), (3) Then: k = k OH- . [OH - ] (*)

Knowing that the dissociation reaction: 2 H 2 O [H 3 O + ] + [OH - ] occurs at a fast rate so

has equilibrium constant K D .

K D =

[H 3 O + ]. [OH - ]

[H 2 O] 2

K

Because [H 2 O] = 1 

K D = [H 3 O + ]. [OH - ] [OH - ]=

K

D

[H 3 O + ]

Replace (*):

k = k OH- .

D

[H 3 O + ]

lg k = lg k OH- + lg K D - lg[H 3 O + ]

lg k = B + pH mt

Let B = lg k OH- + lg K D = const 

In summary: the dependence of the rate constant on the pH of the medium is shown in the diagram

Figure 1.

For acid-base catalyzed reactions without proton and hydroxyl ions, the overall rate constant equation is replaced by a more general equation:

k = k o + k i [A i H] + k j [B j ]

that is, the concentration of all reactants present.

Figure 1: Dependence of lg k by pH

a_a' : catalyst is acid b_b' : catalyst is base

c_c' : stage 1: catalyst is acid stage 2: catalyst is base

d_d' : stage 1: no catalyst stage 2: catalyst is base

e_e' : stage 1: acid catalyst stage 2: no catalyst

h_h' : stage 1: catalyst is acid stage 2: no catalyst stage 3: catalyst is base

k_k' : no catalyst

lg k

k k'

a d'

c b'

d

e c'

h h'

e'

a'

b

pH

IV. Kinetics of acid-base catalyzed reactions

The task of kinematics:

 Hypothesize a reaction mechanism

 From there find the kinematic equation

If the kinetic equation matches reality, then the hypothesized mechanism is the correct one and from this mechanism a new catalyst can be found.

If the kinetic equation is not suitable, a new mechanism must be assumed and another kinetic equation must be found.

1/ The catalyst is acid

Reaction: S 1 + S 2

AH P 1 + P 2

Assuming the reaction mechanism is divided into stages:

S 1 + AH

S 1 H + + S 2 P 2 H + + A -

k 1

k 2

k 3

S 1 H + + A - (1)

P 1 + P 2 H + (2)

P 2 + AH (3)

Or:

P 2 H + + H 2 Ok 4

P 2 + H 3 O + (4)

H 3 O + + A -

k 5

HA + H 2 O (5)

each other

The reaction mechanism is: (1), (2), (3)

or: (1), (2), (4), (5)

Stages (3) and (4), (5) are catalyst regeneration stages according to two different mechanisms.

The overall rate of the reaction is the rate of the slowest step. Assume the following cases:

Case I: stage (1) slow, stage (2), (3) fast

Then: v c = v 1 = k 1 . [S 1 ].[AH]

v c = f([AH])

Comment: if the catalyst concentration increases, the reaction rate increases.

Case II: stage (2) is slow, stage (1), (3) is fast

Then: v c = v 2 = k 2 . [S 2 ].[S 1 H + ]

Because stage (1) is fast, we have the equilibrium constant of stage (1) as follows:



I

K  S 1 H  .  A 

 S 1  .  AH 

 SH

 K

 S 1  .  AH 

I

 A 

1

Find [A - ]: from the equilibrium conditions of the catalytic protonation stage with the solvent (H 2 O) of

acid. We have:

AH + H 2 O A - + H 3 O +

Equilibrium constant K a :

 HO  .  A 

3

 AH 

K a 

 AH 

 A 

a

3

 K  ​​HO



 SH   K I.  S 1 .  AH  K I.  S  .  HO 

1 K  AH K 1 3





aa

3

COUGH

v k . K I  S  .  S  .  HO 

K

c 2 1 2 3

a

Note: v c is proportional to the acid strength. If you want to increase the reaction rate, you must use a catalyst with strong acid strength.

Case III: stage (3) is slow, stage (1), (2) is fast

Then: v c = v 3 = k 3 . [P 2 H + ].[A - ]

Because stage (2) is fast, we have the equilibrium constant of stage (2) as follows:

 P 2H  .  P 



   

1

 SH  .  S 

1 2

1

2

K II

 SH  .  S 

P 2 H

K II

 P 1 

2

Replace [S 1 H + ] in hypothesis II into:

1

I

 SH  K

 S 1  .  AH 

 A 

 PH  K

 S 2 . K

II

I

 P 1 

 S  .  AH 

A 

.1 

v c k 3 . K I

. K II

 S  .  S  .  AH 



. 1 2

P 1

v c 

f ⎛ ⎜  AH ⎞ ⎟

 P 

⎝1 ⎠

Comment: v c depends on the concentration of the acid catalyst and the concentration of the product formed, however the presence of the product reduces the reaction rate.

Case IV: stage (4) is slow, stages (1), (2) and (5) are fast.

Stage (5) is not slow because it is a neutralization reaction. Then: v c = v 4 = k 4 . [P 2 H + ]

Replace [PH + ] in hypothesis III: 

 

 SH  .  S 

1 2

2 P 2 H

KII .

 P 1 

Replace [S 1 H + ] in hypothesis II into:  SH  K I .  S  .  HO 

K

1 1 3

a

K

 PH  K I. K II.  S 1  .  S 2 .  HO 

2

3

a P 1 

v k

.K I. K II. S 1  .  S 2 .  HO 



3

v f ⎛  H 3 O ⎞

K

c 4

a P 1 

c ⎜ P  ⎟

1 ⎠

⎝

Comment: v c depends on the acid strength of the acid catalyst and the concentration of the product formed, however the presence of the product reduces the reaction rate.

2/ The catalyst is a base:

Reaction: S 1 H + S 2

BP 1 + P 2 H

Assuming the reaction mechanism is divided into stages:

Or:

S 1 H + B

S 1 - + S 2 P 2 - + BH + P 2 - + H 2 O OH - + BH +

k 1

k 2

k 3

k 4

k 5

S 1 - + BH + (1)

P 1 + P 2 - (2)

P 2 H + B (3)

P2H + OH - (4)

B + H 2 O (5)

The reaction mechanism is: (1), (2), (3)

or: (1), (2), (4), (5)

Assume the following cases:

Case I: stage (1) slow, stage (2), (3) fast

Then: v c = v 1 = k 1 . [S 1 H].[B] v c = f([B])

Comment: if the concentration of base catalyst increases, the reaction rate increases.

Case II: stage (2) is slow, stage (1), (3) is fast

Then: v c = v 2 = k 2 . [S 2 ].[S 1 - ]

Because stage (1) is fast, we have the equilibrium constant of stage (1) as follows:



 S 1 .  BH  S 1 H  .  B 

K I 

 S 1 H  .  B 

S 1 K I .

 BH 

Find [BH + ]: from the equilibrium conditions of the catalytic protonation stage with the solvent (H 2 O) of

base. We have:

Equilibrium constant:

B + H 2 O BH + + OH -

K b 

 BH  .  OH 

 B 

 BH

  K . B 

b 

OH

 S  K I.  S 1 H  .  B K I SH  .  OH 

1 K  B K 1

[



bb

OH

v k

. K I .  SH  .  S

 .  OH 

v f   OH  

K

c 2 1 2 c

b