1

Reaction kinetics

2

A B C DA B C D

R1. The rate of reactionConsider 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)

dni/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):

dtdn

v AA

dtdn

v CC

0Sii

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

Vi

i

1v

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-3 s-1]

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

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

(R4)

i

i

5

dt

dn

Vdt

dn

Vdt

dn

Vdt

dn

VD

D

C

C

B

B

A

A

1111

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

BA

C

BA

dt

dn

dt

dn

dt

dn

dt

dn C

C

A

A

B

B

A

A 11

,11

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

dt

dn

dt

dn

dt

dn C

C

B

B

A

A 111

A B CA 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

VA

A

1v

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

(R8)

8

b) Concentration A n

VA

dt

Ad

A

1

v

AdV

dnA

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 reactionA

AA nn

0

dt

d

V

1v

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

AAA nn 0

dt

d

dt

dnA

A dt

d

Vdt

dn

V A

AA

A

1v

(R10)

(R11)

10

d) Conversion wn n

nAA 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 (0wA1). If the reactants are not in stoichiometric ratio, the conversion is different for the different reactants.

AAAA wnnn 00

dt

dwn

dt

dn AA

A 0

(R12)

The change of reactant conversion (wA) is positive!

Remember the definition of v (R4)!

11

Akdt

Ad

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

nAkdt

Ad Such kind equations express the

rate laws, they are rate equations

(R13)

(R14)

12

222222 ClSOk

dt

Cld

dt

SOd

dt

ClSOd

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:

mn BAkdt

Ad

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.

BAkdt

Ad

(R16)

(R17)

14

Second order reaction with respect to one component: 2Ak

dtAd

E.g. the thermal decomposition of nitrogen dioxide:

2NO2 → 2NO + O2

The rate equation: 2

22 NOk

dt

NOd

Or if the rate is normalized with the stoichiometric coefficient:

22

2 2 NOkdt

NOd

(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: 23

33 CHOCHkdt

CHOCHd

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] = dm3mol-1s-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 + Cl2Bimolecular 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

233

O

Ok

dt

Od

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.

3

2

O

OOK

23

OOK

O

2. O. + O3 → 2O2

2

23

2323

O

OKkOOk

dt

Od

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

2a Brkv

2b HBrkv

2b BrHkv '

HBrHkv c

2d Brkv

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

bc

21

2221

dab

Br

HBrkk1

BrHkkk2

dt

HBrd

'

/

/

/

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: Ak

dt

Bd

dt

Adv

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]).

dtkA

Ad

tA

A

dtkA

Ad

00

tkAA 0lnln

tkA

A

0

ln tkAA 0lnln

tkeAA 0 tkAA 0lnln(R21a) (R21b)

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):

tkeAAAB 100 (R22)

26

The concentrations as functions of time.

concentration

time

[A]0

[A]

[B]

27

Linear plot

ln[A]

time

ln[A]0

tg = -k

tkAA 0lnln

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.

keAA

00

2 2

1 ke 2lnk

k

2ln In case of first order reaction

the half life is independent of initial concentration.

(R23)

29

R5. Second order reactions

We study two cases:

1. The rate equation is 2Akdt

Ad

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.

2AkBAkdt

Ad

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

dtkA

Ad 2

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.)

tkA

A

A

0][

1tk

AA

0][

1

][

1

tkAA

0][

1

][

1

The integrated form of the rate equation is 2Akdt

Ad

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

tkAA

0][

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

kAA 00 ][

1

][

2

kA 0][

1

0][

1

Ak

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

(R27)

34

Note: The stoichiometric equation for second orderreactions 2A Products is written sometimes in the form

22

1Ak

dt

Ad

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.

BAkdt

Ad

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.

dtk

BAAd

(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 - xx = [B]0 - [B], [B] = [B]0 - x

dtk

xBxA

Ad

00

The differential equation:

Since d[A] = -dx,

dtkdxxBxA

00

1

37

To alter the left hand side consider the following identity:

ba

ab

ba

11

baabba

11

)(

11

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

dtkdxxBxAAB

0000

111

dtkABxB

dx

xA

dx

00

00

38

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

xA

AxA

xA

dx xx

0

000

0 0

lnln

xB

BxB

xB

dx xx

0

000

0 0

lnln

The integral of the right hand side

tkABdtkABt

00

0

00

39

So the solution of the differential equation

tkAB

xB

B

xA

A

000

0

0

0 lnln

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

tk

AB

AB

AB

0

0

00

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

tk

BA

BA

BA

0

0

00

ln][

1

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

tkBA

BA

BA

000

0 ][ln

tkBAA

B

B

A 00

0

0 ][lnln

tkBAB

A

B

A 00

0

0 ][lnln

(R29)

(R30)

41time

experimental data

BA

ln

0

0lnB

A

kBAtg 00][

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

tkAA 0lnln

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

(R21b)

44

First order

ln[A]

time

tkAA 0lnln

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

tkAA

0][

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

tkAA

0][

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]

tgdtAd

v

[A]0

t

(R31)

49

Experimentally we obtain reaction rates at different concentrations.

Consider the following general equation.

AA + BB Products

The rate equation

mBnAkdt

Bd

dt

Ad

BA

11

v

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

nAkmBnAk ,0v

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

mBkk 0, 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.

AAA

B

A

B

BB 00

where

mnAkm

AnAkmBnAkA

B

,v

So the rate equation is

mkk

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).

nAk v

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

Alnnklnln 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)

53ln[A]

experimental data

ntg

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 = vf - vb

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

tgdtAd

vt

00

[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 nAk If the form of the kinetic equation is

nAk 00v it is valid for the initial rates, too:

We change the initial contentration Ao 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).

nAkmBnAk 0,

00v

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 Bk1

k2

(R42)

59

The rate equation

BkAkdt

Ad21v

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:

2

1Kk

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 BkAk

dt

Ad21

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).

AAkAkdt

Ad 021

0221 AkAkkdt

Ad

e

e

e

e

A

AA

A

B

k

k 0

2

1 ee AkAkAk 1202

eAkkAk 2102

eAkkAkkdt

Ad 2121

62

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

eAAkkdt

Ad 21

dtkk

AA

Ad

e

21

tA

A e

dtkkAA

Ad

0

21

0

63

[A] → [A] - [ A]e

tkk

AA

AA

e

e

210

ln

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

tkA

A

0

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

Bk1

E.g. C2H5OH C2H4 + H2O (dehydratation)

C2H5OH CH3CHO + H2 (dehydrogenation)

The simplest model

CA k2

where both k1 and k2 are first order rate constants.

67

The change of concentrations in time:

AkkAkAkdtAd 2121

AkdtBd 1

AkdtCd 2

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

dtkkA

Ad 21

tA

A

dtkkA

Ad

0

21

0

tkkAA 210lnln

tkkeAA

210

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)

tkkAA 210lnln

experimental data

70

We have to find another relationship in order to determine k1 and k2 separately. Divide equation (R50) by equation (R51).

2

1

k

k

Cd

Bd

2

1

k

k

C

B

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)

tkk

eAAACB 2100 1

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

1

2

k

kBC

tkk

eAk

kB 21

01

2 11

tkk

eAk

kkB

k

kB 21

01

21

1

2 11

72

By similar consideration the time dependence of [C]

tkk

eAkk

kB 21

021

1 1

tkk

eAkk

kC 21

021

2 1

Results:

(R55)

(R56)

The sum of the three concentrations (see R52, R55 and R56) is equal to the initial concentration of A according to R54.

73

The concentrations as functions of time when k2=2k1. (Note that the sum of concentrations is always [A]0.)

concentration

time

[A]0

[A]

[B]

[C]

74

The product of one reaction becomes the reactant in the following reaction.

R9. Consecutive reactions

The simplest model

A B C k1 k2

where k1 and k2 are first order rate constants.

76

Substituting the right hand side of this expression for [A] into equation R57b we have

Bktk

eAkdt

Bd 2

101

The solution of this differential equation gives the time dependence of [B].

The general solution

Rearranging equation R59 into a standard form:

(R59)

tkeAkBk

dt

Bd 1012

(R60)

77

(R61)

In accordance with the general solution of this type differencial equations (c is a constant):

tk

etk

e.cB 21

At zero time [B]=0. With this initial condition

12

10 kk

kAc

Substituting this result for equation R61, the time dependence of concentration [B] is

78

tke

tke

kk

AkB 21

12

01

The concentration of A decreases exponentionally. [B] reaches a maximum, then decreases gradually to zero. [C] at first slowly increases, then goes through an inflection and eventually approaches [A]0. Since

(R62)

12

tk2

tk1

0 kk

ekek1AC

12

(R63)

[C] = [A]0 - [A] - [B]

We have

79

The concentrations as functions of time.

concentration

time

[A]0

[A]

[B]

[C]

tm

80

How to find the maximum of the [B] – t curve (when [B] is the highests) ?

At the maximum the first derivative of the [B] – t curve is zero.

tke

tke

kk

AkB 21

12

01

02

21

112

01

mm tkek

tkek

kk

Ak

dt

Bd

where tm is the time at the maximum of [B]. Since the first factor is not zero, the second factor must be zero.

81

Take the logarithm of both sides:

mm tkek

tkek

22

11

mm tkktkk 2211 lnln

1212 lnln kkkktm

12

1

2ln

kk

kk

tm

Consequently

(R64)

82

R10. Temperature dependence of reaction rates

We shall deal with several types of temperature dependence of reactions.

A) The rate of most of the reactions increases exponentially with temperature. The reason is that with increasing temperature more and more molecules reach the excess energy which is needed for the reaction. These types of reactions are called thermally activated reactions.

83

The reaction rate as a function of temperature for thermally activated reactions

Temperature

Reaction rate (v)

84

Reaction rate (v)

Temperature

B) Explosions. At a certain temperature the rate increases by many orders of magnitude.

The background of an explosion is very often a chain reaction with chain branching That mains, the number of the chain centers grows exponentially: autocatalysis. If there exists a reaction that catches the chain center radical, then the reaction brakes up, terminates (see Example 3, HBr reaction, chapter R3).

Such a reaction is the gas phase oxydation of hydrogen to water:

gOHOgH 222 22

Initiation:

Propagation:

Branching:

Termination:

HOHHO 2

HHH 2

OHHOHH 22

HOOHO2

22

1HwallH

MHOMOH 22

M: other molecule

222 OOHHOOH

The branching step produces more than one chain carrier radicals. At higher temperature the radicals react before reaching the wall, consequently, the branching rate is higher than 1.

The M molecule in the termination step removes the excess energy.

The thermal explosion is a very rapid reaction. At very high temperature the reaction rate increases rapidly with temperature, namely, the high pressure means at these temperatures high carrier concentration and rapid radical recombination.

88Temperature

C) Enzyme reactions. (Enzymes are biocatalysts). At higher temperatures the enzyme looses its catalytic activity.

Reaction rate (v)

89

v

Temperature

D) The rate of some reactions does not depend or just slightly depends on temperature.

E.g. Photochemical reactionsRadioactive decayReactions of radicals

90

The rate constant of a thermally activated reaction can be expressed as

TRaE

eAk Arrhenius equation (R66)

Ea [J/mol]: activation energyA: preexponential factor (its dimension is identical to that of k).

TRE

Alnkln a

Take the logarithm of both sides:

If we plot lnk against 1/T, we obtain a straight line. The activation energy can be determined from the slope:

(R67)

91

lnk-1/T function

lnk

1/T

lnA

tg = -Ea/R

experimental data

92

The activation energy can be determined if we know the rate constants at two different temperatures (k1 at T1 and k2 at T2).

11 lnln

TR

EAk a

22 lnln

TR

EAk a

1212

11lnln

TTR

Ekk a

21

12

1

2lnTT

TT

R

E

k

k a

If four of the five parameters (T1, T2, k1, k2, Ea) are known, the fifth can be calculated using this formula.

and

So

and (R68)

93

The interpretation of activation energy. The example is an exothermic reaction

Reaction path

Energy

baE

faE

U

: activation energy of the

forward reaction

faE

: activation energy of the

backward reaction

baE

U: heat of reaction (at constant volume)

94

It can be seen that molecules must attain a certain critical energy Ea before they can react.

U (≈H) is the difference between the activation energy Ea

f of the forward reaction and the activation energy Ea

b of the backward reaction.

The reaction energy is

ba

fa EEU (R69)

95

A catalyst has three important properties:1. It changes (increases) the rate of reaction.2. It does not appear in the final product.3. It does not change the position of equlibrium.

R11.Homogeneous catalysis

Consequence: Since the catalyst changes the rate of reaction but not the equilibrium point, it must accelerate the forward (f) and backward (b, reverse) reaction by the same factor.

E.g. a dehydrogenation catalyst is also a good hydrogenation catalyst.

This is a simple chemical reaction: A+BC+D

Its equlibrium constant is with concentrations

BA

DCK c

The reaction rates are:

for the forward reaction BAkv ff

for the backward reaction DCkv bb

Dividing the two reaction rate expressions:

b

fc k

kK (R70)

97

The catalyst usually provides an alternative reaction path having a lower activation energy than the uncatalyzed reaction.

Reaction path

Energy

baE

faE

U

aEThe new reaction path lowers the activation energies of both forward and backword reactions by the same amount Ea.

98

A catalyst works by first entering into a reaction with the reactant to form an intermediate complex.

The complex then either decomposes directly to the product or reacts with a second reactant to yield the product.

A + C → (AC) → Product + C

or A + C → (AC) (AC) + B → Product + C

where C is the catalyst (AC) is the intermediate complex

The figure introduces the work of a Ru containing catalyst helping oxidation (and reduction).

100

Examples in gas phase

1. Dehydratation of tertiary butanol

(CH3)3COH → (CH3)2C=CH2 + H2O

Without a catalyst this reaction is very slow below 700 K. With the addition of small amount of HBr rapid decomposition occurs at 500 K. The catalysed reaction takes place in two steps.

(CH3)3COH + HBr→ (CH3)3CBr + H2O (CH3)3CBr → (CH3)2C=CH2 + HBr

A C AC

AC CAC CProduct

101

2. Decomposition of aldehydes and ethers.

E.g. CH3CHO →CH4 +CO

In the presence of small amount of iodine the rate of this reaction increases by two order of magnitudes. The mechanism of the catalysed reaction:

CH3CHO + I2 → CH3I + HI + CO

CH3I + HI → CH4 + I2

A C AC

AC B Product

C

C2H5-O-C2H5 CH3-CH=O+C2H6→

102

Examples in liquid phase

1. Acid catalysis. The first step is the transfer of proton to the substrate BH + X → B- + HX+ . In the second step HX+ reacts further.

E. g. hydrolysis of esters, keto-enol tautomerism, inversion of sucrose

Example: hydrolysis of esters, our ester is ethyl acetate.

The reaction is detailed below.

Example: hydrolysis of esters, our ester is ethyl acetate (the reaction is detailed below).

First step, the catalyst, the hydroxonium ion builds an intermediate (transition) complex:

Mesomeric stuctures

Water addition

Proton transfer:

Ethanol is lost from the ion:

At last the catalyst, the hydroxonium ion is received:

Similar acid catalyses are for example the keto-enol tautomerism, the inversion of sucrose.

2. Base catalysis starts with the transfer of proton from substrate to catalyst:

XH + B → X- + BH+ , then X- reacts.

E. g.Isomerization and halogenation of organic compounds:

HBrArBrHBrArBrArH

22

107

R12.Kinetics of heterogeneous reactions

In a heterogeneous reaction at least one of the reactants or products is in a different phase from the other components. Typical heterogeneous reactions: Reaction of metals with acidsOxidation of solids with gaseous oxygenDecomposition of solids or liquids to gaseous productsElectrode reactions(see ppt file Electrochemical rate processes )Contact catalysis.

108

Most heterogeneous reactions take place on a solid-gas or a solid-liquid phase boundary.

A surface reaction can usually be divided in the following elementary steps:

1. Diffusion of reactants to surface 2. Adsorption of reactants at surface 3. Chemical reaction on the surface 4. Desorption of products from the surface 5. Diffusion of products away from the surface

The slowest step determines the overall reaction rate.

109

The rate of reaction is defined in a different way from the homogeneous reactions. Since the reaction takes place on the surface and not in the bulk solution, the rate is related to unit surface and not unit volume.

If A is a reactant,

dt

dn

AA

s

1

v

sm

mol2

where As is the surface area.

(R71)

110

R13. Contact catalysis

Many reactions that are slow in a homogeneous gas or liquid phase proceed rapidly if a catalytic solid surface is available.In parallel reactions different surfaces can accelerate different parallel paths (selectivity).

The kinetics of surface reactions can be treated succesfully on the basis of the following assumptions:

111

a) The rate determining step is a reaction of adsorbed molecules (step 3, elementary steps, chapter R12).b) The reaction rate per unit surface area is proportional tothe fraction of surface covered by reactant. This depends on pressure (for gases) or on concentration (for liquid solutions).

The pressure dependence of can be described by the Langmuir equation.

112

Langmuir adsorption isotherm

The adsorption isotherm is (in general) an expression that gives at constant temperature as a function of pressure (or concentration).

One of the quantitative theories is given by Langmuir. He based his theory on the following assumptions:a) is the fraction of sites occupied by adsorbed molecules and (1-) is the fraction of sites not occupied.

b) Each site can hold only one adsorbed molecule

c) The energy of adsorption is the same for all sites and does not depend on the fraction covered, .

113

Derivation of the adsorption isotherm

The rate of the surface processes is the change of the suface fraction in unit time for and unit particle density. The rate of adsorption is proportional to the gas pressure and the fraction of bare surface (1-).

1pkv aads

The rate of desorption from the surface is proportional to the fraction of covered surface .

ddes kv

(R72)

(R73)

At equilibrium

vads = vdes

da kpk 1

That means

The equilibrium constant is

1pk

kK

d

a

This equilibrium constant depends on the pressure, so it is not then true equlibrium constant according to its thermodynamical definition.

(R74)

(R76)

(R75)

115

The final forms are

pK

pK

pkk

pkk

pkk

pk

d

a

d

a

ad

a

11

pK

p

1 or 1

pK

11

is a linear function of p at very low pressures while it approaches 1 at high pressures.

1/ is a linear function of 1/p:

Using R75 and R76

(R77)

(R78a) (R78b)

116

tg=1/K

p

1

1/p

1

These figures demontrate equations R78a and R78b, respectively.

117

The simplest surface reaction is when one compound reacts on the surface (E.g. isomerization or decomposition).

It is assumed that the rate of reaction is proportional to the fraction of covered surface . We apply R78b:

pK

pkk

1v

where k is the rate constant [mol/m2s]

pkKk

111

v

1

Plotting 1/v against 1/p gives a linear function

(R79a)

(R79b)

118

tg=1/(k·K)

v

1/p

1/k

Plot for equation R79b

pkKk

111

v

1