Reaction kinetics

1

Reaction kinetics

2









R1. The rate of reaction

Consider the following reaction

A and B are the reactants, C and D are the products, the (greek)-s are the stoichiometric coefficients.

For the definition of the rate of consumption for reactant and rate for formation for product we take the derivative of the number of moles with respect to time,

(R1)

dn_i/dt (amount of substance converted in unit time) decreases for reactants.

For reactants like For products like

A simple reaction without parallel reactions and further (consecutive) reaction of the products has the form (S maybe product or reactant):

dt v_A   dn^A

dt v_C  dn^C

0 S_i

i i 

 

According to this equation the stochiometic coefficients of reactants are defined negative, those of products are defined positive.

(R2a) (R2b)

(R3)

4

dt dn V

i i



  v 1

The quantities defined in (R2a) and (R2b)depend on volume, as well. Therefore in the definition of rate of reaction (v) we divide these type expressions both by volume and by the stoichiometric coefficient:

[mol dm

s

]

The values of the stoichiometric coefficients influence the measure of the conversion.

Since the signs of dn_i/dt and are always the same, v is always positive.

(R4)

i

i

5

dt dn V

dt dn V

dt dn V

dt dn V

D D

C C

B B

A

A  

 

 

   



1 1

1 1

The rate of reaction can be expressed with the help of any component:

This expression is unambigous, as mentioned, if - there is strict connection among the stoichiometric coefficients,

- there are no side or parallel reactions, - the products do not react further.

(R5)

6

In case of parallel and consecutive reactions (see later) we can only define the concentration changes of the different components separately.

For parallel reactions ⁽supposing volume V=constant⁾

C A B

C B

A 

 



  ^dn_dt ^dn_dt ^dn_dt ^dn_dt^C

C A

A B

B A

A   



1 , 1

1

1  

For consecutive (further) reactions, supposing the volume as constant

dt dn dt

dn dt

dn _C

C B

B A

A  



1 1

1  



A  

B  

C

In these cases a different rate of reaction is defined for each component.

(R6)

(R7)

7

The rate of reaction can be expressed with a) amount of substance

b) concentration

c) extent of reaction d) conversion

a) Amount of substance

dt dn V

A A



  v 1

The following formulas are all for reactant A (_A is negative),

(R8)

8

b) Concentration

 

 

dt A d

A



  v 1

  ^A

V d

dn



As it was already mentioned, v is positive because both _A and d[A]/dt are negative (concentration of reactant decreases in time).

(R9)

9

c) Extent of reaction

A A

A n

n

  ^{ 0}

dt d V

 

 1 v

The extent of reaction () is also defined as a positive quantity. Its unit is mol (both the numinator and the denominator are negative).

A A

A n

n  ⁰  

dt d dt

dn

A  A  

dt d V

dt dn

V

A A

A







 ^{  }

 1 v

(R10)

(R11)

10

d) Conversion w n n

A An A

A

 ⁰ 

0

dt dw V

n _A

A A 



 

0

v

The conversion of a reactant shows what fraction of that reactant has been converted (0w_A1). If the reactants are not in stoichiometric ratio, the conversion is different for the different reactants.

A A

A

A n n w

n  ⁰  ⁰ 

dt n dw

dt

dn _A

A   A⁰ 

(R12) The change of reactant conversion (w_A) is positive!

Remember the

definition of v (R4)!

11

 

 

dt A

d  



R2. The concept of order of reaction

It has been found that the rate of most of the reactions depend on the concentrations of the reactants. Usually the rate can be expressed in a power equation:

k is called the rate constant, n is the order of reaction.

First order reaction: n=1;

E.g. decomposition of sulfuryl chloride:

SO₂Cl₂ →SO₂ +Cl₂

 

 

dt A

d  

 Such kind equations express the rate laws, they are rate equations

(R13)

(R14)

12



 

  





dt Cl d dt

SO d

dt Cl SO

d    



The order of reaction is not necessarily equal to the stoichiometric coefficient. E.g. the decomposition of nitrogen pentoxide:

2N₂O₅ →4NO₂ +O₂ is found to be a first order reaction.

Explanation: the reaction takes place in two steps:

N₂O₅ → N₂O₃ +O₂ - slow N₂O₃ + N₂O₅ → 4NO₂ - fast

The first slow step determines the overall reaction rate.

(R15a) (R15b)

13

Consider the following reaction with two reactants:

_AA + _BB → Products

The following rate equation can be set up:

 

   

dt A

d  



Note that the rate equation is not always normalized with the stoichiometric coefficient.

Here n is the order with respect to A, m is the order with respect to B and n+m is the overall reaction order.

The following rate equation shows a reaction which is first order with respect to A, first order with respect to B and second order overall.

 

   

dt A

d   



(R16)

(R17)

14

Second order reaction with respect to one component:

 

 

dt A

d  



E.g. the thermal decomposition of nitrogen dioxide:

2NO₂ → 2NO + O₂

The rate equation:









dt NO

d  



Or if the rate is normalized with

the stoichiometric coefficient:









dt NO

d  



(The value of the rate constant differs by a factor of 2.) (R18)

15

The order of reaction is not always an integer. E.g.

the decomposition of acetaldehyde:

CH₃CHO → CH₄ + CO

The rate equation:









dt

CHO CH

d  



In this case the order of reaction is 3/2.

The fraction value of the reaction order refers that the reaction proceeds in several steps or parallel reactions run.

16

The unit of rate constant depends on the order of reaction.

First order [k] = s^-1

Second order [k] = dm³mol^-1s^-1

n order [k] = (time)^-1(concentration)^1-n

17

R3.Reaction molecularity and reaction mechanism

Many reactions proceed through a number of steps from initial reactants to final products. Each of the

individual steps is called an elementary reaction.

The molecularity indicates how many molecules of reactants are involved in the elementary reaction.

Unimolecular e.g. SO₂Cl₂ → SO₂ + Cl₂ Bimolecular e.g. NO + O₃ → NO₂ + O₂ Trimolecular reactions are very rare.

The molecularity is integer, 1, 2, (higher are very rare).

18

Reaction mechanism

On the one hand reaction mechanism means the sequence of elementary reactions that gives the overall chemical change. On the other hand reaction mechanism means the detailed analysis of how chemical bonds in the reactants rearrange to form the activated complex. An activated complex is an intermediate state that is formed during the conversion of reactants into products.

1. example: decomposition of ozone 2O₃ → 3O₂ The following rate equation was found experimentally

   

 

2 3 3

O k O

dt O

d  



19

Mechanism:

1. Rapid decomposition of ozone to O₂ and atomic oxygen → equilibrium.

2. The slow, rate determining step is the reaction of atomic oxygen with ozone.

1. O₃ → O₂ + O

.    

 

K O 



   

 

O  K 

2. O

.

+ O₃ → 2O₂

       

 

2 3 2

3 2

3

O K O

k O

O dt k

O

d      



k

20

3. example: formation of hydrogen bromide from hydrogen and bromine

2. example: decomposition of N₂O₅ (see example below equation R18)

H₂ + Br₂ → 2HBr

This is not a simple bimolecular reaction but a chain reaction of radicals.

Chain initiation Br₂ → 2Br

.

Chain propagation Br

.

+ H₂ → HBr + H

.

H

.

+ Br₂ → HBr + Br

.

Chain inhibition H

.

+ HBr → Br

.

+H₂ Chain termination 2Br

.

→ Br₂

 

a Br k

v 

  

b Br H k

v 

  

b H Br

k v  ^'

 



k v  _c

 

d Br k

v 

21

The reaction is initiated by bromine radicals from the thermal dissociation of Br₂.

The chain propagation steps regenerate the bromine radicals, ready for another cycle. In this step HBr is generated and also hydrogen radicals for the inhibition step.

The chain inhibition step removes the H radicals, slowing down the chain propagation.

22

Considering the mechanism desribed above, the following rate equation can be derived (no need to memorize):

Notice that [HBr] occurs in the denominator, so that the rate is inhibited by the [HBr] while [H₂] (in the numinator) initiate the reaction.

(R19)

 

   

  ^ _{ } ^

2 b

c

12 2 2

2 1 d a

b

Br k HBr k

1

Br H

k k

k 2 dt

HBr d

' /

/ /





 

23

R4. First order reactions

In a first order reaction the rate is proportional to the concentration of the reactant.

A → B (+ C +…) Examples:

SO₂Cl₂ → SO₂ + Cl₂ 2N₂O₅ → 4NO₂ +O₂

The rate equation:

    ^k   ^A

dt B d dt

A

d   



 v

where k is the rate constant, (its dimension is time^-1.) (R20)

24

Separate the variables and integrate from t = 0 (conc. = [A₀]) to time t (conc. = [A]).

    ^{A A k dt}

d     

  



^{  }

A

dt A k

A d

0 0

  ^{A  ln}   ^A

^{  k  t}

ln

    ^{A A}

^{  k  t}

ln ^ln   ^{A  ln}   ^A

^{  k  t}

    ^{A  A}

^{ e  k  t}

^ln     ^{A  ln A}

^{ k  t}

25

The concentration of the reactant A decreases exponentially with time. The concentration of the product (B) if it is formed in stoichiometric ratio (one molecule of B is formed from one molecule of A):

        

 

 

   







 A A A e k t

B

1

26

The concentrations as functions of time. concentration

time [A]₀

[A]

[B]

27

Linear plot ln[A]

time ln[A]₀

tg = -k



    ^{A  ln A}

^{ k  t}

ln

experimental data

(R21b)

28

If we plot ln[A] against time and the data points lie on a staight line (within experimental errors), the reaction is of first order. The slope of the line gives the rate

constant.

Half life is the time required to reduce the concentration of reactant to half its initial value.

    ^A

^{ A}

^{ e  k  }

2 2

 1



 k 

e k    ln 2

k 2

 ln



the half life is independent of initial concentration.

(R23)

29

R5. Second order reactions

We study two cases:

1. The rate equation is

  ^k   ^{A 2}

dt A

d  



This rate equation applies when

a) The reaction is second order with respect to one component.

E.g. 2NO₂ → 2NO + O₂ See example below

equation R18!

(R24)

30

b) The reactants of a bimolecular reaction are in stoichiometric ratio.

A + B → Products but [A]₀ = [B]₀ so [A] = [B] at any time.

  ^k     ^{A B k}   ^{A 2}

dt A

d     



We have the same rate equation in cases a) and b).

  ^A   ^{A k dt}

d  



We integrate from t = 0 (conc. = [A]₀) to time t (conc. = [A]).

(R25)

31

(The integral of 1/[A]² is -1/[A] +const.)

 

 

t A k

A

A





 



 



 

]

[

1 k t

A

A    



]

[ 1 ]

[ 1

t A k

A    ]

[ 1 ]

[ 1

The integrated form of the rate equation is

 

 

dt A

d  



If we plot the reciprocal of the concentration of the reactant against time, we obtain a straight line in case of a second order reaction. The slope is the rate

constant.

(R26)

32

Linear plot

1/[A]

time 1/[A]₀

tg ^{= k} experimental data

t A k

A    ]

[ 1 ]

[ 1



(R26)

33

The dimension of k is (concentration)^-1 ·(time)^-1, Its unit is usually dm³/mol.s^-1

Half life : [A] = [A]₀/2







 k

A A

[ ]

1 ]

[ 2





 k A ]

[ 1

]

[ 1

A

 k



The half life of second order reactions depends on the initial concentration.

(R27)

34

Note: The stoichiometric equation for second order reactions 2A  Products is written sometimes in the

form

    ²

2

1 k A

dt A

d  



Comparing this with the equation R25, the relation- ship between the two rate constants (it is the same reaction!):

k = 2k´.

35

2. The second order is the sum of two first

orders (A + B  Products) and the reactants are not in stoichiometric ratio.

  ^k     ^{A B}

dt A

d   



It is difficult to solve this differential equation because both [A] and [B] are variables (but their changes are not independent).

The next 4 slides contain the derivation (Deriv1) of a complicate equation. It is not compusory to memorize it. However, the result is important.

    ^{A d}   ^{A  B   k  dt}

(R28a) (R28b)

36

Deriv1. Solution of the differential equation R28.

We introduce a variable x, which is the difference of [A]₀ and [A]. Hence it is also the difference of [B]₀ and [B].

x = [A]₀ - [A], [A] = [A]₀ - x x = [B]₀ - [B], [B] = [B]₀ - x

   

 ^A

^{ x d}  ^{ A}    ^B

^{ x}  ^{  k  dt}

The differential equation:

Since d[A] = -dx,

 

 ^A

^{ x}  ^{1 }    ^B

^{ x}  ^{ dx  k  dt}

37

To alter the left hand side consider the following identity:

b a

a b

b

a 

 

 1

1 



 

  

 

 b  b a a b a

1 1

) (

1 1

If we substitute [A]₀-x for a and [B]₀-x for b, the differential equation takes the form

      ^B

^{ 1 A}

^ _ ^  ^A

^{1  x}  ^    ^B

^{1  x}  ^ _ ^{  dx  k  dt}

 

           ^{B A}  ^{k dt}

x B

dx x

A

dx      

 





 



0

38

We integrate from 0 to t and from 0 to x. The two terms of the left hand side can be integrated separately.

 

         

  ^{A A x}

x x A

A

dx

x

 





 



0 0 0

0 0

ln ln

 ^{B dx x}      ^{B x}  

  ^{B B x}

x

 





 



0 0 0

0 0

ln ln

The integral of the right hand side

   

 ^{B A}  ^{k dt}      ^{B A}  ^{k t}

t













 

0

0 0

39

So the solution of the differential equation

           ^{B A}  ^{k t}

x B

B x

A

A    

 



0 0

0

ln

ln

Rearrange this equation considering that [A]₀-x =[A]

and [B]₀-x =[B]

     

    ^{B B A A k t}

A

B  



 



0 0 0

] ln [

1

If we multiply both the first and the second factor of the left hand side by -1, we get to the final form.

40

     

    ^{A A B B k t}

B

A  



 



0 0 0

] ln [

1

We can rearrange this equation to make is suitable for a linear plot

   

     ^A   ^B  ^{k t}

B A

B

A    





0 0 0

0

[ ]

ln

     

   ^A   ^B  ^{k t}

A B B

A  



 

0

0

[ ]

ln ln

     

   ^A   ^B  ^{k t}

B A B

A  



 

0

0

[ ]

ln ln

(R29)

(R30)

time 41

experimental data



   

ln

   

ln B A

 

 ^{A B}  ^k

tg   [ ]





Plot ln([A]/[B]) as a function of time

42

R6. Determination of reaction order and rate constant

If a reaction of unknown kinetic parameters is studied, the primary experimental data are concentrations as function of time.

1. Integral methods. We assume an order of reaction (1 or 2) and check if the experimental data fulfil the

corresponding rate equation.

43

First order

ln[A]

time

    ^{A  ln A}

^{ k  t}

ln

If the experimental data sit on this like straight line, the reaction is of first order.

(R21b)

44

First order

ln[A]

time

    ^{A  ln A}

^{ k  t}

ln

If the experimental data do not sit on the R21b like straight line, the reaction is not of first

order.

(R21b)

45

Second order

1/[A]

time

t A k

A    ]

[ 1 ]

[ 1

If the experimental data sit on a this kind straight line, the reaction is of second order.

(R26)

46

Second order

1/[A]

time

t A k

A    ]

[ 1 ]

[ 1

In this case the experimental data do not sit on a straight line, the reaction is not of second order.

(R26)

47

The integral method can be used e.g. for deciding if the reaction is of first order or second order. The linear plot also produces the rate constant.

In more complicated cases (when the rate depends on more than one concentration or the order of the

reaction is not an integer) the integral method cannot be used.

2. Differential methods. From the concentration - time data we determine the reaction rates by graphical or numerical derivation.

48

The concentration of a reactant as a function of time.

concentration

time

[A] 

  _{  }



 tg

dt A v d

[A]₀

t

(R31)

49

Experimentally we obtain reaction rates at different concentrations.

Consider the following general equation.

_AA + _BB  Products The rate equation

    ^k     ^{A n B m}

dt B d dt

A d

B A













  

1 v 1

How to simplify this expression?

(R32)

50

a)We want to determine n (the order with respect to A).

We apply B in great excess so that its concentration can be regarded as constant :

[B] ~ [B]₀

    ^{A n B m k}   ^{A n}

k    



v

In this case the rate equation has only two

constants (k´ and n) to be determined. If we want to determine m, we apply A in great excess

  ^{B m}

k

k

 

where

(R33) (R34)

51

b) If we want to determine n+m (the overall order of

reaction), we start the reaction with stoichiometric ratio of A and B. The stoichiometric ratio of reactants holds during reaction.

    ^A     ^A

A B A

B

  

 





 B

B

where

    ^{A n B m k}   ^{A n}   ^{A m k}   ^{A n m}

k

A

B

    

 



 













v 



So the rate equation is

m k

k

A B





 



 

 

,



(R35)

(R36a) (R36b)

52

In both cases (namely a and b) we could simplify the rate equation so that it has two constants (namely a

rate constant and an exponent, n or n+m).

  ^{A n}

k 

 v

Experimentally concentration – reaction rate data pairs are obtained. Take the logarithm of this expression.

  ^A

ln n

k ln

ln v   

If we plot ln v against ln[A], the points are on a straight line, if our model is close to real case. The intercept is ln k, and the slope is n.

(R37)

(R38)

ln[A] 53

experimental data



n tg  

lnv as a function of ln[A]

ln v

ln k

54

3.The method of initial rates. In some cases the

products or one of the products influence the flow of the reaction.

a) Equilibrium reactions. The apparent rate is the

difference of the rate of forward and backward reaction.

v = v

- v

The kinetic parameters of the forward reaction can be determined when the rate of the backward reaction is negligible, i.e. at the very beginning of the reaction.

b) Autocatalytic reactions. One of the products acts as a catalyst (see later)

(R39)

55

The concentration of a reactant as a function of time.

time



  _{  }

 

 



 





dt tg A v d

t 0

0

[A]

[A]₀

56

The initial rates are determined at various [A]₀

concentrations. The kinetic parameters are calculated from [A]₀ – v₀ data pairs.

 

v  k  A n If the form of the kinetic equation is

  ^{A n}

k

v

 

it is valid for the initial rates, too:

We change the initial contentration A_o and plot ln v₀ against ln[A]₀, the points are on a straight line. The intercept is lnk, and the slope is n. The initial reaction velocity is determined measuring the concentration during the initial part of the reaction.

(R40a) (R40b)

57

E.g. with two reactants

The method of initial rates can be used for determining the order with respect to one component.

In each measurement the concentrations of all the other reactants are kept constant (so that they can be merged into the rate constant) and only the concentration of the selected reactant (A) is varied (see equation R41).

    ^{A n B m k}   ^{A n}

k

v     

Because [B]₀ is the same in all cases. So if lnv is

plotted against ln[A]₀ the result is a straight line, the intercept is lnk’, the slope is n.

(R41)

58

In many reactions the equilibrium is practically on the product side, the reaction „goes to completion”.

R7. Opposing reactions

In other cases a considerable concentration of reactants remain when equilibrium is reached (e.g.

hydrolysis of esters). The reaction rate slows down as we approach equilibrium. In such cases two opposing reactions have to be taken into consideration.

The simplest model is the following (e.g.

isomerization), k₁ and k₂ are first order rate constants

A

B

k₂ (R42)

59

The rate equation

  ^k   ^{A k}   ^B

dt A d

2

v   

 

At equilibrium

v = 0, k₁·[A]_e = k₂·[B]_e ,

where [A]_e and [B]_e are the equilibrium concentrations of A and B, respectively.

The equilibrium constant in terms of concentrations:

   

K k

k A

B

e e





So the equilibrium constant is equal to the ratio of rate constants of the forward (k₁) and backward (k₂) reactions, respectively.

(R43)

(R44)

(R45)

60

If we measure the equilibrium concentrations, the ratio of k₁ and k₂ can be determined.

We need another relationship if we want to determine k₁ and k₂. This comes from the solution of the kinetic

equation

  ^k   ^{A k}   ^B

dt A d

2

1

 





Deriv 2

To reduce the number of variables, we make use of the fact that the sum of concentrations is always [A]₀. (The number of molecules does not change during reaction.)

[B] = [A]₀ - [A], [B]_e = [A]₀ - [A]_e

(R46)

(R47)

61

Before integration we change the last term (k₂·[A]₀).

  ^k   ^{A k}      ^{A A} 

dt A

d    



  _ ^k

^k

_   ^{A k}

  ^A

dt A

d     



         

 

e e

A

A A

A B k

k 





2

1

^k

^   ^A

^{ k}

^   ^A

^{ k}

^   ^A

  ^A  ^{k k}    ^A

k









  _ ^{k k} _   ^A  ^{k k}    ^A

dt A

d      



62

We integrate this differential equation from [A]₀ to [A]

and from 0 to t

  _ ^{k k} _      ^{A A}



dt A

d    



     

   ^{k k}  ^dt

A A

A d

e







 

     

 

 

 



_ ^{ }  ^  ^ 

A e

dt k

A k A

A d

0 2

1

0

63

[A] → [A] - [ A]_e

   

    ^{A A}

^{  A A}

^{   k}

^{ k}

^{  t}

ln

This formula is similar to the equation for a simple first order reaction:

    ^{A A}

^{  k  t}

ln

The solution is the following.

[A]₀ → [A]₀ - [ A]_e [A]₀ → [A]₀ - [ A]_e

k → k₁ +k₂

(R48)

See (R21) ,

and

are the substitutions.

64

ln([A]-[A]_e)

time tg = -(k₁+k₂)

 experimental data

We can determine k₁+k₂ if we plot ln([A]-[A]₀) against time.

ln([A]₀-[A]_e)

65

The concentrations as functions of time when K =2 (R45). (Note that the sum of concentrations is always [A]₀):

concentration

time [A]₀

[A]

[B]

66

Sometimes a substance can react more than one way.

R8. Parallel reactions

k₁

B

E.g. C₂H₅OH  C₂H₄ + H₂O (dehydratation)

C₂H₅OH  CH₃CHO + H₂ (dehydrogenation) The simplest model

A

C

2

where both k₁ and k₂ are first order rate constants.

67

The change of concentrations in time:

  ^k   ^{A k}   ^A  ^{k k}    ^A

dt A

d      



  ^k   ^A

dt B

d 



  ^k   ^A

dt C

d 



The solution of equation (R49) is identical to the rate equation of a simple first order reaction (R28), the only difference being that the rate constant is the sum of k₁ and k₂.

(R49) (R50) (R51)

68

    ^{A A  k k  dt}

d  





    ^{ } 



^{   }

A

dt k

A k A d

0 2

1

0

  ^{A  ln}   ^A

^{ }  ^k

^{ k}

 ^{ t}

ln

    ^{A  A}

^{ e   k}

^{ k}

^{  t}

ln[A] is a linear function of time. (k₁+k₂) can be determined from the slope.

Time dependence of the concentration of A

(R52)

69

Linear plot

ln[A]

time ln[A]₀

tg = -(k₁+k₂)



    ^{A  ln A}

^  ^k

^{ k}

 ^{ t}

ln

experimental data

70

We have to find another relationship in order to

determine k₁ and k₂ separately. Divide equation (R50) by equation (R51).

    ^{C k k}

d B

d   

  ^{C B  k k}

So the ratio of k₁ and k₂ is equal to the ratio of the two concentrations.

Time dependence of concentrations of B and C The sum of concentrations is always [A]₀. (The number of molecules does not change during this reaction.)

[A]₀ = [A] + [B] + [C]

(R53)

(R54)

71

Using equation (R52)

          ^{ } _



 



    









 C A A A e k k t

B

1

If we want the time dependence of [B], eliminate [C] using equation (R53).

   

1 2

k B k

C  

    ^{ } _



 



    



 



 



 

 A e k k t

k

B k

1

2

1

1

      ^{ } _



 



    



 



 



 



 

 A e k k t

k k B k

k

B k

1 2 1

1

2

1

1