1. Chemical Kinetics

2. Kinetics Reaction Rates
The speed of a chemical rxn depends on the nature of reactants and frequency of collisions between the particles
What are some factors that influence reaction rate?

3. Factors that affect reaction rates
1. The physical state of the reactants
2.The concentration of reactants
3.The temperature at which the reaction occurs
4.The presence of a catalyst

4. Reaction Rates
A rxn rate is the change in concentration of reactants or products per unit time (M/s)
Can be based on rate of disappearance of the reactant or the rate of appearance of a product

5. Reaction Rates Instantaneous rate – rate at any particular time
Average rate
– average rate of reaction
over time

6. Reaction Rates
Consider reaction A  B
Rate = -D[A]/ Dt = D[B]/ Dt

7. Reaction Rates and Stoichiometry
In general
aA + bB  cC + dD
Rate = -1/a D[A] / Dt
= -1/b D[B] / Dt
= 1/c D[C] / Dt
= 1/d D[D] / Dt

8. Reaction Rates and Stoichiometry
Consider
2HI(g)  H2(g) + I2(g)

9. Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear
(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction
2 O3(g) → 3 O2(g)?
(b) If the rate at which O2 appears, Δ[O2]/ Δt, is 6.0 × 10–5 M/s at a particular instant, at what rate is O3 disappearing at this same time, –Δ[O3]/Δ t?

10. Rate Laws
A rate law tells how the rate depends on the reactants

11. Rate Laws
aA + bB  cC + dD
Rate= k[A]m[B]n
k is the rate constant affected by T
m and n are the rxn orders

12. Rate Laws
The overall reaction order is the sum of the individual reaction orders (m+n)

13. Rate Laws
The exponents in the rate law indicate how the rate is affected by the concentration of each reactant

14. Rate Laws
A rxn order of 0 means that if the conc. doubles, the rate stays the same (no change to rate)

15. Rate Laws
A rxn order of 1 means that if the conc. doubles, the rate doubles (proportional change to rate)

16. Rate Laws
A rxn order of 2 means that if the conc. doubles, the rate quadruples (the increase factor is squared and multiplied by rate)

17. Rate Laws
The values of the exponents must be determined experimentally or based on information from the rxn mechanism
The overall balanced equation does not provide the rate law unless the rxn occurs in one step

18. Rate Laws
The units of rate constants depend on the overall reaction order.

19. Rate Laws
Zero order
Rate = k
units of k ( )

20. Rate Laws
1st order
Rate = k [A]
Units of k ( )

21. Rate Laws
2nd order
Rate=k[A]2 or Rate = k[A][B]
Units of k ( )

22. Rate Laws
3rd order
Rate = k[A]2[B]
Units of k ( )

23. Rate Laws
rate of reaction is dependent on concentration of reactants
Rate constant is dependent on T or catalysts

24. The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B, and the results are as follows:
Using these data, determine
(a) the rate law for the reaction,
(b) the rate constant,
(c) the rate of the reaction when [A] = M and [B] = M.

25. (a) Determine the rate law for this reaction.
(b) Calculate the rate constant.
(c) Calculate the rate when [NO] = M
and [H2] = M

26. The change of concentration with time
A rate law tells how the rate changes at a particular T as the reactant conc. changes
To distinguish between zero,1st, and 2nd order rxns, graphs can be plotted

27. Zero order rxns Rate = -D[A]/ Dt =k Integrated Rate Law
[A]t = -kt + [A]0
y = mx + b
A graph of concentration versus time yields a straight line with slope -k

28. First order rxns Rate = -D[A]/ Dt =k[A] Integrate Rate Law
ln [A]t = -kt + ln[A]0
y = mx + b
A graph of ln[A]t vs.
time will yield a straight
line with slope –k
and y-intercept ln[A]0

29. Second order rxns Rate = -D[A]/ Dt =k[A]2 1/[A]t = kt +1/[A]0
y = mx + b
A graph of 1/[A]t
vs. time will yield
a straight line
with slope k and
y-intercept 1/[A]0

30. Half life of rxns
t1/2 is the time required for the concentration of a reactant to drop to one-half its initial value. [A]t1/2 = ½[A]0.

31. Half life of rxns
For 1st order rxns
t1/2 = 0.693/k
For 2nd order rxns
t1/2= 1/k[A]0

32. Temperature and Rate The Collision Theory
the rate of rxn depends on the number of collisions between molecules or ions, the average energy of the collisions, and their effectiveness
Temperature, concentration, orientation, and activation energy affect collisions and rate

33. Temperature and Rate The Collision Theory
Does the general effect of concentration and T on reaction rate support the collision theory? Explain.

35. Energy Profiles (1) DHreactants
(2) Activation Energy(Ea) – the minimum E required to initiate the rxn; at the peak is an activated complex (a highly energized and unstable species)
(4) DHproducts
(3) DHrxn

36. Sample Energy Profile

37. The Arrhenius Equation
k = Ae-Ea/RT
k = rate constant Ea
R = 8.314J/mol-K T
A is frequency factor

38. Sample Exercise 14.10 Relating Energy Profiles to Activation Energies and speed of Reaction
Consider a series of reactions having the following energy profiles:
Rank the reactions from slowest to fastest assuming that they have nearly the same frequency factors.
Does Enthalpy of Rxn influence the rate of rxn?

39. Reaction Mechanisms
The process by which a rxn occurs is a rxn mechanism which will describe the order that the bonds are broken and formed and changes positions of atoms

40. Reaction Mechanisms
Elementary step – a one step process or event

41. Reaction Mechanisms
Multistep Mechanism – A net change represented by a balanced chemical equation consisting of a sequence of elementary steps

42. Reaction Mechanisms
The elementary steps in a multistep mechanism must always add to give the chemical equation of the overall process

43. Reaction Mechanisms
An intermediate is a substance that is neither a product or reactant, it is formed in one elementary step and consumed in another

44. Write the equation for the overall reaction.
Sample Exercise Determining Molecularity and Identifying Intermediates
Describe the molecularity of each elementary reaction in this mechanism.
Write the equation for the overall reaction.
Identify the intermediate(s).

45. Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates
For the reaction
the proposed mechanism is
Is the proposed mechanism consistent with the equation for the overall reaction?
(b) What is the molecularity of each step of the mechanism?
(c) Identify the intermediate(s).
Practice Exercise

46. Rate laws for elementary steps
If we know a reaction is an elementary step then the rate law is based on molecularity

47. Rate laws for elementary steps
Possible rate laws for elementary steps

48. Sample Exercise 14.13 Predicting Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary reaction, predict its rate law:

49. Sample Exercise 14.13 Predicting Rate Law for an Elementary Reaction
Consider the following reaction:
2 NO(g) + Br2(g) → 2 NOBr(g).
Write the rate law for the reaction, assuming it involves a single elementary reaction.
Is a single step mechanism likely for this reaction?

50. Rate laws for multistep mechanism
the overall rxn rate can not exceed the rate of the slowest elementary step of its mechanism

51. Rate laws for multistep mechanism
rate-determining step = slow step

52. Rate laws for multistep mechanism
If the rate determining step is the first step, then the faster step has no effect on the overall rate

53. Rate laws for multistep mechanism
If the slow step is preceded by fast steps, intermediates are formed and accumulate before being used in the slow step

54. Write the equation for the overall reaction.
Sample Exercise Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:
Write the equation for the overall reaction.
Write the rate law for the overall reaction.

55. The experimental rate law is rate = k[O3][NO2].
Sample Exercise Determining the Rate Law for a Multistep Mechanism
Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:
The reaction is believed to occur in two steps:
The experimental rate law is rate = k[O3][NO2].
What can you say about the relative rates of the two steps of the mechanism?

56. Sample Exercise 14. 15 Deriving the Rate Law for a Mechanism with a
Sample Exercise Deriving the Rate Law for a Mechanism with a Fast Initial Step
Prove the rate law is R=k[NO]2[Br 2]

57. The first step of a mechanism involving the reaction of bromine is
Sample Exercise Deriving the Rate Law for a Mechanism with a Fast Initial Step
The first step of a mechanism involving the reaction of bromine is
What is the expression relating the concentration of Br(g) to that of Br2(g)?

58. Catalysis
a catalyst is a substance that can increase or decrease the speed of a chemical rxn without a change in chemical composition by altering Ea or A and providing a different mechanism

59. Catalysis
It is not consumed in a chemical reaction and can be recovered at the end