1. Chemical Kinetics Chapter 13
PowerPoint Lecture Presentation by J. David Robertson University of Missouri
Chemical Kinetics
Chapter 13
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. LEARNING OBJECTIVES: CHEMICAL KINETICS
Writing rate expression based on a balanced equation & do calculations related to the expression.
Factors affecting rate of reaction.
Determining the rate law of a reaction based on experimental data.
CHEMICAL KINETICS
Understand the integrated rate law and how to use the formulas.
Understand the concept of catalysis.

3. CHEMICAL KINETICS Kinetics – how fast does a reaction proceed?
Rate of a chemical reaction is determined experimentally and depends on:
Nature (physical state) of the reactants
Concentration of the reactants
Temperature
Presence or absence of a catalyst
13.1

4. Factors that Affect the Rate of Reactions:
1) Physical state: molecules must mix in order to collide.
The physical state (solid, liquid, gas) will affect frequency of collisions, as well as the physical size of droplets (liquid) or particles in the case of solids.
2) Concentration: molecules must collide in order to react.
Reaction rate is proportional to the concentration of reactants. Higher concentration means more molecules will collide.

5. Factors that Affect the Rate of Reactions:
3) Temperature: molecules must collide with enough energy to react.
Raising the temperature increases the reaction rate by increasing the number of collisions, and especially, the energy of the collisions.
4) Catalyst : substances that increase rate of reaction.
Catalyst will lower the activation energy of the reaction.

6. Chemical Kinetics : Rate Expression
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = –
D[A] Dt
D[A] = change in concentration of A over time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over time period Dt
*Because [A] decreases with time, [A] is negative.
13.1

7. A B
time
rate = -
D[A] Dt
rate =
D[B] Dt
13.1

8. = – = Consider a general reaction: a A + b B → c C + d D
Rate of reaction can be measured using rate of disappearance of reactants and/or rate of appearance of products.
Rate of reaction = rate of disappearance of reactants
Δ[A] Δt 1 a
= –
Δ[B] b
= rate of appearance of products =
Δ[C] Δt 1 c
Δ[D] d

9. Rate Expression and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = –
D[A] Dt 1 2
rate =
D[B] Dt
aA + bB cC + dD
rate = -
D[A] Dt 1 a
= -
D[B] Dt 1 b =
D[C] Dt 1 c =
D[D] Dt 1 d
13.1

10. Question 1: Write the rate expression for the following reactions:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = –
D[CH4] Dt
= –
D[O2] Dt 1 2 =
D[CO2] Dt =
D[H2O] Dt 1 2
2N2O5 (g) → 4NO2 (g) + O2 (g)
13.1

11. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time
13.1

12. rate a [Br2] rate = k [Br2] k = rate [Br2] = rate constant
= 3.50 x 10-3 s-1
13.1

13. Chemical Kinetics : The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x [B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x+y)th order overall
13.2

14. Double [F2] with [ClO2] constant rate = k [F2][ClO2]
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
rate = k [F2][ClO2]
Rate Law!
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
13.2

15. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g) 1
rate = k [F2][ClO2]
13.2

16. Question 2: The table below contains data for the reaction of 2A + B → C + D. What is the rate law for this reaction?
Expt [A] [B] Initial rate (mol/L•s)
x 10-3
x 10-2
x 10-2
x 10-1
Answer:
Rate = k [A]x [B]y
To determine x, y has to be cancelled, possible options:
Expt 2
Expt 1
Expt 3 OR
To determine y, x has to be cancelled, only one option:
Expt 4
Expt 2

17. Therefore, rate law is: Rate = k [A] [B]2
Expt [A] [B] Initial rate (mol/L•s)
x 10-3
x 10-2
x 10-2
x 10-1
To determine x:
Expt 2
Expt 1 =
(0.20)x (0.05)y =
1.2 x 10-2
6.0 x 10-3
(0.10)x (0.05)y
2x = 2
 x = 1
To determine y:
Expt 4
Expt 2 =
(0.20)x (0.15)y =
1.1 x 10-1
1.2 x 10-2
(0.20)x (0.05)y
3y = 9
 y = 2
Therefore, rate law is: Rate = k [A] [B]2

18. Clue: use pairs of experimental data.
Question #3: Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) → 2SO42- (aq) + I3- (aq)
Experiment
[S2O82-], M
[I-], M
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
Clue: use pairs of experimental data.
13.2

19. Answer (to determine rate law): Rate = k [S2O82-]x [I-]y
2y = 2  y = 1
2x = 2
 x = 1
Therefore, rate Law is:
Rate = k [S2O82-] [I-]
13.2

20. Answer (to determine order of reaction):
Rate = k [S2O82-] [I-]
reaction is 1st order in S2O82-
reaction is 1st order in I-
reaction is (x+y) = 1+1 = 2nd order overall
Answer (to determine value of k):
k =
rate
[S2O82-][I-] =
2.2 x 10-4 Ms-1
(0.08 M)(0.034 M)
= 0.08 M-1s-1
13.2

21. Chemical Kinetics : The Integrated Rate Law
The Rate Law gives us an equation that relates the reaction rate to the concentration of a species in the reaction.
Since measuring reaction rates can be difficult, it it often more useful to have a relationship that directly relates the concentration of a species to the time since the reaction began.
These relationships can be found by integrating the rate law.

22. First-Order Reactions : Integrated Equations
D[A] Dt
A product
rate = k [A]
D[A] Dt
= k [A] -
rate
[A]
M/s M =
k =
= 1/s or s-1
[A]t is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
[A] = [A]0exp(-kt)
ln [A] = ln [A]0 - kt
13.3

23. Answer [A]0 = 0.88 M ln[A]t = ln[A]0 - kt [A]t = 0.14 M
Question #4: The reaction 2A  B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 80oC. How long will it take for A to decrease from 0.88 M to 0.14 M ?
Answer
[A]0 = 0.88 M
ln[A]t = ln[A]0 - kt
[A]t = 0.14 M
kt = ln[A]0 – ln[A]t
ln[A]0 – ln[A]t k
t = ln
[A]0
[A]t k = ln
0.88 M
0.14 M
2.8 x 10-2 s-1 = t
= 66 s
13.3

24. First-Order Reactions : Half-life
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2 ln
[A]0
[A]0/2 k = t½
ln 2 k =
0.693 k =
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½
ln 2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
*How do you know decomposition is first order?
From the unit of k (s-1).
13.3

25. 13.3

26. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
rate
[A]2
M/s M2 =
D[A] Dt
= k [A]2 -
k =
= 1/M•s 1
[A] =
[A]0
+ kt
[A]0 is the concentration of A at time t = 0
[A]t is the concentration of A at any time t
t½= 1
k[A]0
t½ = t when [A] = [A]0/2
13.3

27. Zero-Order Reactions [A] = [A]0 - kt rate = - D[A] Dt A product
rate = k [A]0 = k
rate
[A]0
D[A] Dt
= k -
k =
= M/s
[A] = [A]0 - kt
[A]t is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
13.3

28. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A]t = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A]t = ln[A]0 - kt 1
[A]t =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2
13.3

29. Integrated Rate Laws and Reaction Order
Slope = k
C. Zero order.

30. Question #5: The conversion of cyclopropane (C3H6), to propene occurs with a first order rate constant equal to 1.7 x 10-2 hr-1.
If the initial concentration of cyclopropane is M, what is the concentration after hrs?
(b) What fraction of cyclopropane has decomposed?
Answer:
(a) Using the integrated rate law for first order reaction:
ln [C3H6]t = ln [C3H6]0 - kt
ln [C3H6]t = ln (1.200 M) - (1.7 x 10-2hr-1)(20.00 hr)
ln [C3H6]t = =
[C3H6]t = M
(b) The fraction that has decomposed after hrs:
Fraction decomposed

31. QUESTION #6: It is found that 500 mg of chymopopsin spontaneously decomposes in a first‑order reaction to give 335 mg of chymopopsin in 2000 sec. Calculate the rate constant k (in sec-1).
Answer:
ln [CP]t = ln [CP]0 – kt
ln = ln 500 – k (2000 sec)
= – k (2000 sec)
2000 sec
6.215 – 5.814
k =
= x 10–4 sec–1

32. QUESTION #7: The decomposition of NOBr is second order, with a rate constant k = 25 L/mol•s at 20°C. If the initial concentration of NOBr is M, find the time (in sec) at which the concentration will be M.
Answer: 1
[NOBr]t =
[NOBr]0
+ kt 1
0.010 M =
0.025 M
+ kt
100 M–1 = 40 M–1 + (25 M–1s–1) t
= 2.4 s
100 M–1 – 40 M–1
 t =
25 M–1s–1

33. QUESTION #8: The decomposition of HI(g) to give H2(g ) and I2(g) is 2nd order with a rate constant k equal to 30 M-1min-1 at 443° C. How long does it take for a concentration of 2.0 M of HI (g) to drop to one‑tenth of its original value?
Answer: 1
[HI]t =
[HI]0
+ kt
[HI]0 = 2.0 M 1 10
[HI]t = x 2.0 M
= 0.2 M 1
0.2 M =
2.0 M
+ (30 M-1min-1)t
5 M–1 = 0.5 M–1 + (30 M–1min–1) t
= min
5 M–1 – 0.5 M–1
 t =
30 M–1min–1
= 9 sec.

34. COLLISION THEORY
Collision theory qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions.
It assumes that for a reaction to occur the reactant particles must collide, but only a certain fraction of the total collisions, the effective collisions, cause the transformation of reactant molecules into products.

35. This is due to the fact that only a fraction of the molecules have sufficient energy and the right orientation at the moment of impact to break the existing bonds and form new bonds.
Sufficient energy
Correct orientation

36. Collision Theory & Concentration of Reactant
High concentration (more particles) increases the number of possible collision.
As a result, the number of effective collision increases.

37. Collision Theory & Molecular Orientation
Correct orientation will result in effective collision.

38. Activation Energy & Temperature Dependence of the Rate Constant
Temperature often has major effect on reaction rate.
Generally, increasing the temperature will increase the rate of reaction.
Particles can only react when they collide. If a substance is heated, the particles move faster and so collide more frequently. That will speed up the rate of reaction.
Collisions only result in a reaction if the particles collide with enough energy to get the reaction started. This minimum energy required is called the activation energy for the reaction.
13.4

39. Activation Energy & Temperature Dependence of the Rate Constant
Exothermic Reaction
Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.
13.4

40. Temperature Dependence of the Rate Constant
Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation:
k = Ae-Ea/RT
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
13.4

41. Question #9:What is the activation energy (in kJ) of a reaction whose rate constant increases by a factor of 100 upon increasing the temperature from 300 K to 360 K?
Answer:
T1 = 300 K T2 = 360 K
Ea = J/mol
= 69 kJ/mol

42. Question #10: The activation energy of a certain reaction is 250 kJ/mol. If the rate constant at T1 = 300 K is k1 and the rate constant at T2 = 320 K is k2, then how fast is the reaction at 320 K compare to that at 300 K?
Answer:
The rate at 320 K is 526 times faster than the rate at 300 K.

43. CATALYSIS
Catalysis is the acceleration (increase in rate) of a chemical reaction by means of a substance called a catalyst, which is itself not consumed by the overall reaction.
Catalysts work by providing an (alternative) mechanism involving a different transition state and lower activation energy.
Many catalysts used in refineries and in petrochemical applications are regenerated and reused multiple times to save costs and energy and to reduce environmental impact from recycling or disposal of spent catalysts.

44. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = Ae-Ea/RT Ea k
uncatalyzed
catalyzed
ratecatalyzed > rateuncatalyzed
Ea < Ea ‘
13.6

45. Fig

46. Types of catalysis:
In heterogeneous catalysis, the reactants and the catalysts are in different phases, examples are:
Haber synthesis of ammonia, finely divided iron acts as a heterogeneous catalyst;
Ostwald process for the production of nitric acid;
Catalytic converters.
Active sites on the catalyst allow the reactants to form partial weak bonding, which are adsorbed onto the catalysts’ surface.
As a result, the bond within the molecule of a reactant is weakened sufficiently for new bonds to be created. The bonds between the products and the catalyst are weaker, so the products are released.
13.6

47. Catalysis on a Surface

48. Haber Process
Fe/Al2O3/K2O
catalyst
N2 (g) + 3H2 (g) NH3 (g)
13.6

49. The Metal-Catalyzed Hydrogenation of Ethylene

50. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
Acid catalysis
Base catalysis
In homogeneous catalysis the catalyst is a molecule which facilitates the reaction.
The reactant(s) which are coordinated to the catalyst (or vice versa), are transformed to product(s), which are then released from the catalyst.
13.6

51. Homogenous Catalysis

52. Enzyme Catalysis
Enzyme catalysis is the catalysis of chemical reaction by specialized proteins, enzymes.
An enzyme has features of both a homogenous and a heterogenous catalyst – it can provides an active surface, and it also can interact actively with the substrate.
Enzymes are extremely specific, each reaction is generally catalyzed by a particular enzyme.

53. A simplified scheme of enzyme catalysis is as follow:
E = enzyme
S = substrate
ES = enzyme-substrate complex
EP = enzyme-product complex
P = product
Two models of enzyme action are:
Lock-and-key Model
Induced-fit Model

54. Lock-and key Model
Both the enzyme and the substrate possess specific complementary geometric shapes that fit exactly into one another.

55. Induced-fit Model
The active site is continually reshaped by interaction with the substrates.