1. Chemical Kinetics Jens Poulsen 5 x 2 hours
Atkins
Quanta, matter, and Change.
Chapters

2. A reaction may be investigated on several levels…
stochiometry:
rate law:
atomistic:

3. Chapter 22: Rates of chemical reactions. Concepts:
Defining reaction rates.
Integrated rate laws.
Elementary reactions.
Consecutive reactions.

4. Start with chemical kinetics on an empirical level:
Rate equations: How much N2O5 (g) as a function of time etc?
No information about reaction mechanism
(on an atomic scale).

5. Experimental techniques concentrations can be followed by...
Measuring pressure during chemical reaction:
Measuring conductivity during chemical reaction:

6. Example 19.1 monitoring variation in pressure.

7. Def. of v - rate of reaction
more generally

8. Reaction order Often has reaction rate of form
Order m with respect to A,..
Overall order is m+n+..

9. first order reaction second order reaction zero order reaction
Examples
first order reaction
second order reaction
zero order reaction

10. Determination of rate law
isolation method: excess of B, C, …
m can be determined etc.
method of initial rates: vary the conc. of A,
B, C, .. In turn and check how rate changes.

11. Integrated rate laws
First order rate
straight line:
Second order rate

12. leads to

13. Reactions approaching equil.

14. Relation between equilibrium constant and rate constants.
Solution:
Relation between equilibrium constant and rate constants.

15. Disturbing a system in eq:
Sudden change in
e.g. by change in T.

16. The rate to new equilibrium is given by the sum of

17. Calculate the rate constants
Example 19.4
Calculate the rate constants
for water autoprotolysis, given Forward/backward reaction is first/second-order, respectively, and T=298 K, pH=7.
BLACK-BOARD!

18. Temperature dependence of chemical reactions.
What can be said about k(T) ?
Arrhenius equation

19. Svante Arrhenius Nobel prize 1903.

20. interpretation
*Reaction coordinate: molecular distortion along which the reactants become products.
*Transition state = activated complex = climax of reaction.
*Once the reactants have passed the transition state, products
are formed.

21. interpretation Ea: activation energy (change in potential energy).
Only molecules having kinetic energy larger than Ea get over barrier.
exp(-Ea/RT): fraction of reactants
having enough kinetic energy to pass barrier
A: pre-exponential factor: measure of the rate of collisions

22. exp(-Ea/RT): fraction of reactants having
kinetic energy higher than barrier height Ea.
A: proportional to collision frequency.
K(T) = collision frequency times
fraction of successful collisions
= A * exp(-Ea/RT)

23. Accounting for the rate laws.
Elementary reactions: The fundamental “building blocks” of chemical reactions.
Describes what happens on an atomic scale.
one ethanoate ion collides with one methyliodide molecule and forms one iodide ion plus one methylethylether….

24. Elementary reactions
The molecularity of a reaction: # of molecules participating.
Unimolecular: one molecule.
Bimolecular: two molecules.
A unimolecular reaction is first-order.
A bimolecular reaction is second-order.

25. Elementary reactions
If we know a reaction is single-step and bimolecular we can write down a rate equation directly:
Same applies to unimolecular reactions...

26. Elementary reactions On the other hand, consider the reaction
A rate law of form
does not imply the reaction to be simple bi-molecular:

27. Example.

28. Consecutive elementary reactions
example: enzyme/substrate

29. Solving time-dependence of [P].
BLACK-BOARD!

30. Consecutive reactions
Qualitative time-dependence of
[A], [I] and [P].

31. Steady State approx. VERY IMPORTANT!!
The higher the molecularity, the more complex mathematics when solving rate equations. Need approximation.
Introduce steady state approximation:

32. Steady state approx. continued.
More generally: all intermediates
are assumed to have negligible concentration and rate of change of concentration:

33. Apply steady state approx. to consecutive reaction.

35. Two intermediates: NO and NO3.
Steady state example.
Two intermediates: NO and NO3.

36. Hence..[exercise]:

37. Rate determining step Example. when ka<kb then
may be treated as a simple reaction:
”the principle of the rate determining step”

38. When ka<kb then becomes as expected
Proof:
When ka<kb then
becomes
as expected

39. Rate determining step – in general
If one elementary step in a reaction is slower than others then this step controls the rate of the overall reaction.
The slow step is rate determining if it can’t be sidestepped.
Once the rate determining step is found, the rate expression can be written down immediately.

40. Example.
Even if ka<< kb the upper reaction may be sidestepped if ka<ka’ and ka<kb’. Then
is not rate determining.

41. Assume A and B are in eq. then
Preequilibria.
Assume A and B are in eq. then

42. Example. Analysing pre-eq. by steady state.
we do not assume eq.

43. When pre-eq exists kb<ka’

44. Unimolecular reactions. Lindemann Hinshelwood mechanism.
Found to involve bimolecular step, how can it be first order?
The reaction is not given by an elementary first order mechanism.
Can be described by the Lindemann Hinshelwood mechanism.

45. Lindemann Hinshelwood

46. If kb is small: kb << ka’ [A] then first-order reaction
Use steady state...
If kb is small: kb << ka’ [A] then first-order reaction

47. On the other hand at low pressure:
Since the reaction
becomes ”bottleneck”.

48. Define effective rate constant:
Lindemann Hinshelwood mechanism gives linear plot

49. The kinetics of complex reactions
Chain reactions
Explosions
Catalysis (enzymes)

50. What is a chain reaction?
Example, thermal decomposition of ethanal:
Elementary reactions (Rice-Herzfeld):
Chain carriers are CH3 and CH3CO radicals

51. Derivation of rate law:
Steady state:
Add these together:

52. As observed experimentally.
Insert into
As observed experimentally.

53. Key elements of chain reaction
Initiation: involves formation of chain carriers
propagation: done by chain carriers
Termination: chain carriers are ”destroyed”.
New example:
Initiated by heat.

54. New example: Complicated rate law: Can be explained by postulating
a chain reaction mechanism.

55. Mechanism: collision induced
Retardation step: A product is removed.
Chain carriers are H and Br radicals.

56. Steady state analysis. Rate of formation

57. Adding these two equations give
Inserting the expression for Br radical into first eq. gives

58. Substituting into rate expression gives
Which has the same form as the experimental rate law provided

59. Explosions
An explosion is a rapidly accellerating reaction arising from a rapid increase in reaction rate with increasing temperature

60. Has not been fully understod. # Chain carriers grow exponentially.
Example
Has not been fully understod. # Chain carriers grow exponentially.

61. Occurence of explosion given by explosion region (regions of T,p).
chain carriers are
Branching: one chain carrier becomes two or more.
Occurence of explosion given by explosion region (regions of T,p).
If T is low, rate constants are too small.
If p is low, the reaction rates, v, are too low due to infrequent collisions.

63. Example Show that an explosion happens when rate of chain branching
exceeds that of chain termination.
Answer:
Focus only on the rate of prod of
chain carrier as important. Rate of change of chain carrier:

64. Steady state approximation: Then chain carrier production rate is
Example
Steady state approximation:
Then chain carrier production rate is

65. write: Then chain carrier production rate is
Example
write:
Then chain carrier production rate is

67. Homogeneous catalysis
A catalyst lowers the activation energy for a reaction.
It is not consumed in the reaction (reforms in the end).
Examples are enzymes, metal catalysts etc.
Homogeneous catalysis: the catalyst exists in same phase as reactants.

69. Example: as catalyzed by bromide ions.
Notice that bromide reappears in the end.

70. Enzymes Enzymes (E) are biological catalysts. Typical proteins.
Contains an active site that binds substrate (S):
Induced fit model: S induces change in E and
only then do they fit together.

71. Michaelis Menten mechanism. Three exp. observations
i) For a given amount of S, the rate of product formation is proportional to [E].
ii) For a give amount of E and low values of [S], the rate of product formation is proportional to [S].
iii) For a given amont of E and high values of [S], the rate of product formation reaches a max. value, v(max), and becomes independent of [S].

72. i) and ii) - but not iii) - is consistent with
Michaelis-Menten explains i)-iii)
leads to

73. Proof.
Steady state approx.:
then
Define
Use

74. Then i)-iii) is correct. show!

75. Plot of 1/v against 1/[S] gives straight line for MM mechanism.
Alternative formula:
Plot of 1/v against 1/[S] gives straight line for MM mechanism.
Plot is called
Lineweaver-Burk plot

77. Determine enzyme parameters.
Lineweaver-Burke plot gives
and
From knowing total enzyme concentration we calc.
Cannot determine

78. Catalytic constant.
gives the number of ”turn overs” pr. unit time. This number – with units of pr time – is called the catalytic constant,

79. Catalytic efficiency.
Catalytic efficiency: measure of overall enzym efficiency equals
effective rate constant for overall reaction

80. If enzyme is very efficient then kb is large. Then
ka is determined by diffusion.

81. Enzyme inhibition
An inhibitor reduces rate of product formation by binding to E, ES or both, thereby decreasing [ES] and hence product formation.

82. New rate of product formation.
Maximum velocity is no longer obtainable when competitive inhibition occurs.

83. Proof:
Introduce conservation of enzyme molcs:

85. Different types of inhibition
Case 1: meaning only EI not ESI
is formed. Once EI is formed, S cannot bind to E. Termed competitive inhibition.
Case 2: meaning that only ESI not EI is formed. I can only bind to E if S is already present i.e. ES. ESI does not lead to product. Termed uncompetitive inhibition.

86. Different types of inhibition
Case 3: meaning that both EI and ESI are formed. Inhibitor binds to a site different from the active site in both cases. EI cannot bind S. Termed non-competitive inhibition.

90. How to determine Do a unhibited exp. with S & E and find vmax and Km.
Add inhibitor with known conc. and plot Lineweaver-Burk from which and thereby can be determined.

91. Example of use of competitive inhibition.
To kill bacteria (B). B has enzyme dihydro- pteroate synthease producing folate which is crucial for survival of B.
Active substrate p-aminobenzoic acid.
Inhibitor Sulfanilamide.