1. Chemical Kinetics
Chapter 12

2. Reaction Rates 01 2 HI(g)  H2(g) + I2(g)
Reaction Rate: The change in the concentration of a reactant or a product with time (M/s).
change in concentration divided by the change in time
Reactant  Products A  B
2 HI(g)  H2(g) + I2(g)
Chapter 12

3. Reaction Rates and Stoichiometry
What is the general rate of the following reaction ?
2 HI(g)  H2(g) + I2(g)
Rate = − 1 2
[HI] t =
[I2]
Chapter 12

4. Reaction Rates and Stoichiometry
To generalize, for the reaction
aA + bB
cC + dD
Rate* = − 1 a
[A] t
= − b
[B] = c
[C] d
[D]
*: General rate of reaction
Chapter 12

5. Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

6. Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?

7. How Do we study Rate of a reaction?
Consider the decomposition of N2O5 to give NO2 and O2: N2O5(g) NO2(g) + O2(g)
Brown
Colorless
Chapter 12

8. Reaction Rates: concentration versus time curve 03
2 N2O5(g) NO2(g) + O2(g)
Reaction Rates: concentration versus time curve 03
Average Rate = Rate between two points in time
The slope of each triangle Between two points
Chapter 12

9. Reaction Rates 2N2O5(g) 4NO2(g) + O2(g)
The rate of change for both NO2 and O2 is positive (increasing) and the rate of change for N2O5 is negative (decreasing). The calculated rate is dependent on the time points taken.

10. Instantaneous rate: Rate for specific instance in time
Slope of the tangent to a concentration versus time curve
Initial Rate
Chapter 12

11. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
393 nm
Br2 (aq)
D[Br2] a DAbsorption
Chapter 12

12. 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
The slope of a line tangent to the curve at any point is the instantaneous rate at that time
Chapter 12

13. slope of
tangent
slope of
tangent
Chapter 12

14. The Rate Law; rate = 3.50 x 10-3 s-1 [Br2]
rate a [Br2]
rate = k [Br2]
k =
rate
[Br2]
= rate constant
The Rate Law; rate = 3.50 x 10-3 s-1 [Br2]
Chapter 12

15. k = 3.50 x 10-3 s-1
Chapter 12

16. The Rate Law and Reaction Order
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
Chapter 12

17. The Rate Law and Reaction Order are Experimentally Determined
Chapter 12

18. Determine the reaction order for:
F2 (g) + 2ClO2 (g) FClO2 (g)
----
rate = k [F2]x[ClO2]y
1 vs 3
Double [F2] with [ClO2] constant
Rate doubles
x = 1
1 vs 2
rate = k [F2][ClO2]
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
The instantaneous rate at the beginning of a reaction is called initial rate
Chapter 12

19. 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]
Chapter 12

20. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
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
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
Chapter 12

21. Determine the Rate Law and Reaction Order
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Comparing Experiments 1 and 2, when [NH4+] doubles, the initial rate doubles.
Chapter 12

22. NH4+(aq) + NO2−(aq) N2(g) + 2 H2O(l)
Likewise, comparing Experiments 5 and 6, when [NO2−] doubles, the initial rate doubles.
Chapter 12

23. This equation is called the rate law, and k is the rate constant.
This means
Rate  [NH4+]
Rate  [NO2−]
Rate  [NH+] [NO2−] or
Rate = k [NH4+] [NO2−]
This equation is called the rate law, and k is the rate constant.
Chapter 12

24. Rate Laws
The exponents tell the order of the reaction with respect to each reactant.
This reaction is
First-order in [NH4+]
First-order in [NO2−]
Chapter 12

25. Rate Law & Reaction Order
The reaction of nitric oxide with hydrogen at 1280°C is: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
From the following data determine the rate law and rate constant. E x p e ri m n t [ NO ] H 2 I
iti a l Ra
(M/ s ) 1 5 .
x 1 – 3 4
Second order in NO, First order in H2
k = 1/3( ) = 250 M-2.s-1
Chapter 12

26. First-Order Reactions Concentration and Time Equation 01
First Order: Reaction rate depends on the reactant concentration raised to first power.
Rate = k[A]
A product
D[A] Dt
= k [A] -
Chapter 12

27. -(ln[A] -ln[A]0) = kt Concentration and Time Equation For
A First-Order Reactions
D[A]
= K Δt -
-(ln[A] -ln[A]0) = kt
[A]
ln[A] = ln[A]0 - kt
See next slide for proof of the formula ln
[A]0
[A]
= k t
[A] = [A]0exp(-kt)
[A] is the concentration of A at any time t
rate
[A]
M/s M =
[A]0 is the concentration of A at time t = 0
k =
= 1/s or s-1
[A] = [A]0exp(-kt)
Chapter 12

28. Integration:
Chapter 12

29. First-Order Reactionsln
[A]0
[A]
= k t
First-Order Reactions
ln[A] = ln[A]0 - kt
[A] is the concentration of A at any time t
ln[A] = ln[A]0 - kt
[A]0 is the concentration of A at time t = 0
Chapter 12

30. The reaction 2A B is first order in A with a rate constant of 2
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
[A]0 = 0.88 M ln
[A]0
[A]
= k t
[A] = 0.14 M
t = ? ln
[A]0
[A] k ln
0.88 M
0.14 M
2.8 x 10-2 s-1 =
t =
= 66 s
Chapter 12

31. What is 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
Half Life For the First Order Reaction ln
[A]0
[A]0/2 k = t½ ln
[A]0
[A]
= k t
ln2 k =
0.693
Chapter 12

32. Units of Rate Constants vs Reaction Orders
Zeroth Order Reaction:
Rate = K [A]0 = K
Units of Rate Constants vs Reaction Orders
Overall order of this reaction is third.
Chapter 12

33. What is the order of decomposition of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
What is the half life of decomposition of N2O5 ?
2N2O5(g)  4NO2(g) + O2(g)
units of k (s-1)
Therefore, decomposition is first order? t½
ln2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
Chapter 12

34. Half life of a First Order Reaction
Chapter 12

35. First-order reaction A product # of half-lives [A] 1 ½ [A]0 2 1/4 [A]03
1/8 [A]0 4
1/16 [A]0
[A] = [A]0 x (1/2)n
Chapter 12

36. First-Order Reaction 2N2O5(g)  4NO2(g) + O2(g)
Show that the decomposition of N2O5 is first order and calculate the rate constant and Half life.
k = 1.7 x s-1
t1/2 = 408 S
Chapter 12

37. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
What is Unit of k ?
rate
[A]2
M/s M2 =
k =
= 1/M•s or M-1 s-1
What is Conc. Vs time equation?
D[A] Dt
= k [A]2 - 1
[A] =
[A]0
+ kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
Chapter 12

38. Second-Order Reactions
So if a process is second-order in A,
a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.
Drive the formula for half life of a second order reaction
t = t1/2 =
t1/2
[A] 2
[A]0
Chapter 12

39. Half-life for a second-order reaction
t = t1/2 =
t1/2
[A] 2
[A]0
= kt +
[A]0 1
[A]t
[A]0 1
= kt1/2 + 2 =
t1/2
k[A]0 1
Chapter 12

40. Second-Order Reactions=
t1/2
k[A]0 1
For a second-order reaction, the half-life is dependent on the initial concentration.
Each successive half-life is twice as long as the preceding one.
Chapter 12

42. 2 NO2(g)  2NO(g) + O2(g) Example:
Is the following reaction first or second order ?
What is the value of k?
2 NO2(g)  2NO(g) + O2(g)
Figure: Table UN
Title:
Caption:
Chapter 12

43. Figure: Table UN
Title:
Caption:
Chapter 12

44. Second-Order Reactions
Figure: UN
Title:
Worked Example 12.8
Caption:
Concentration–time data are given for the decomposition of nitrogen dioxide to nitric oxide and molecular oxygen. Plot the data using the relationships from the integrated rate laws to determine if the reaction is first- or second-order.
k = 0.54 M-1 . S-1
Second-Order Reactions
Chapter 12

45. Zero Order Reaction: Rate = k Example of Zeroth Order Reaction:
Figure: Table UN
Title:
Caption:
Example of Zeroth Order Reaction:
Decomposition of N2O on hot platinum surface:
       N2O →  N2  +  1/2 O2
       Rate  [N2O]0 =  k[N2O]0 = k
       d[N2 O]/dt = k
Chapter 12

46. Reaction Mechanisms 01
A reaction mechanism is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products.
Chapter 12

47. Reaction Mechanisms 02
Single steps in a mechanism are called elementary steps (reactions).
An elementary step describes the behavior of individual molecules.
An overall reaction describes the reaction stoichiometry.
Chapter 12

48. Reaction Mechanisms NO2(g) + CO(g)  NO(g) + CO2(g)
NO2(g) + NO2(g)  NO(g) + NO3(g) Elementary
NO3(g) + CO(g)  NO2(g) + CO2(g) Elementary
NO2(g) + CO(g)  NO(g) + CO2(g) Overall
The chemical equation for an elementary reaction is a description of an individual molecular event that involves the breaking and/or making of chemical bonds.
NO3(g) is called reaction intermediate.
Chapter 12

49. Reaction Mechanisms 04
Molecularity: is the number of molecules (or atoms) on the reactant side of the chemical equation.
Unimolecular: Single reactant molecule.
Chapter 12

50. Reaction Mechanisms 05 Bimolecular: Two reactant molecules.
Termolecular: Three reactant molecules.
Chapter 12

51. Reaction Mechanisms 06
Determine individual steps , the reaction intermediates, and the molecularity of each individual step.
2N2 + O2
2N2O
Chapter 12

52. Rate Laws and Reaction Mechanisms 01
Rate law for an overall reaction must be determined experimentally.
Rate law for elementary step follows from its molecularity.
Chapter 12

53. Rate Laws and Reaction Mechanisms 02
The rate law of each elementary step follows its molecularity.
The overall reaction is a sequence of elementary steps called the reaction mechanism.
Chapter 12

54. Rate-Determining Step
The slowest elementary step in a multistep reaction is called the rate-determining step.
The overall reaction cannot occur faster than the speed of the rate-determining step.
The rate of the overall reaction is therefore determined by the rate of the rate-determining step.
Chapter 12

55. What is the equation for the overall reaction?
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1:
NO2 + NO NO + NO3
Step 2:
NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
Chapter 12

56. Determining Reaction Mechanism From The Rate Law

57. Activation Energy (Ea):
Chapter 12

58. The Arrhenius Equation
Typically, as the temperature increases, the rate of reaction increases.
2N2O5(g)
4NO2(g) + O2(g)
rate = k[N2O5]
Where is temperature dependence?
Is it hidden in k?
Chapter 12

59. The Arrhenius Equation 01
Collision Theory: A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy.
Activation Energy (Ea): The potential energy barrier that must be surmounted before reactants can be converted to products.
Chapter 12

60. Temperature dependence of Rate Constat
Adding more of the reactants speeds up a reaction by increasing the number of collisions that occur.
Collision rate = Z [A][B]
The fraction of collisions with an energy equal or more than activation energy( Ea) :
f = e-Ea/RT
Raising the temperature speeds up a reaction by providing the energy of activation to more colliding molecules.
Z is a constant related to collision frequency
Chapter 12

61. The Arrhenius Equation 02
Only the fraction of collisions having proper orientation can result to products.
This is called steric factor, p., In the above example p = 0.5
Chapter 12

62. The Arrhenius Equation
Collision rate = Z [A][B]
Where Z is a constant, related to the collision frequency .
Reaction rate = p.f.Z [A][B]
Reaction rate = k [A][B]
k = p.f.Z,
Assume p.Z = A, frequency factor
A = frequency factor
k = A. f,
f = e-Ea/RT
k = Ae-Ea/RT (p.Z )= A
Chapter 12

63. The Arrhenius Equation
K = Ae-Ea/RT
A = pZ
Figure: UN
Title:
The Arrhenius equation
Caption:
Relationship between the rate constant and the activation energy of the reaction. The term pZ is usually represented with the symbol A and is called the frequency factor.
(steric factor)
Chapter 12

64. Ea = -R . (slope) Calculating Activation Energy k = A .e-Ea/RT
Figure: UN
Title:
The Arrhenius Plot
Caption:
According to the rearranged Arrhenius equation, the activation energy for a reaction can be determined by plotting ln k versus (1/T) which yields a line whose slope is equal to (–Ea/R).
Ln k
Ea = -R . (slope)
1/T
Chapter 12

65. 2HI(g) + H2(g)  I2(g) + H2(g)
Find the activation energy for the following reaction
2HI(g) + H2(g)  I2(g) + H2(g)
Figure: Table UN
Title:
Caption:
Chapter 12

66. Calculating Activation Energy

67. Slope = -2.24 x 10 4 K Ea = -R . (slope)
Ea = - (8.314 j/K.mol) (-2.24 x 10 4 K)
Ea = 190 kj/mol
Figure: UN
Title:
Worked Example 12.11
Caption:
Plot of ln k vs. (1/T) from the data for the gas-phase decomposition of hydrogen iodide.
Chapter 12

68. Effect of Temperature on Fraction of Collisions with Activation energy
f = e-Ea/RT
Chapter 12

69. f = e-Ea/RT Effect of Temperature
Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.

70. Change of Rate Constant with temperature
If the Ea is known , we can calculate the Rate Constant when temperature is changed:
The above formula could be used to
determine the rate constant at a different temperature.
Chapter 12

71. Homework: Determination the Activation Energy
The second-order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures: Determine (graphically) the activation energy for the reaction. k (M - 1 s ) t ( C .8 7 x 3 6 .0 5 9 .2 4
Ea = 241 KJ/mole
Chapter 12

72. Catalysis 01
Chapter 12

73. Catalysis 01
A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction.
Chapter 12

74. Catalysis
Note that the presence of a catalyst does not affect the energy difference between the reactants and the products

75. Catalysis 02
The relative rates of the reaction A + B  AB in vessels a–d are 1:2:1:2. Red = A, blue = B, green = third substance C.
(a) What is the order of reaction in A, B, and C?
(b) Write the rate law.
(c) Write a mechanism that agrees with the rate law.
(d) Why doesn’t C appear in the overall reaction? 1
Chapter 12 2 1 2

76. Catalysis
Catalyst: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step.
rate-determining
step
H2O2(aq) + I1-(aq)
H2O(l) + IO1-(aq)
H2O2(aq) + IO1-(aq)
H2O(l) + O2(g) + I1-(aq)
fast step
2H2O2(aq)
2H2O(l) + O2(g)
overall reaction
Chapter 12

77. Catalysis 03
Homogeneous Catalyst: Exists in the same phase as the reactants.
Heterogeneous Catalyst: Exists in different phase to the reactants.
Chapter 12

78. Catalysis 04
Catalytic Hydrogenation:
Chapter 12

79. Mechanism of Catalytic Hydrogenation:B X Y H
Chapter 12 8

80. Mechanism of Catalytic Hydrogenation:B Y C C A X H H H H
Chapter 12 9

81. Mechanism of Catalytic Hydrogenation:B X Y H H H H
Chapter 12 9

82. Mechanism of Catalytic Hydrogenation:B Y A X H H H C C H
Chapter 12 9

83. Mechanism of Catalytic Hydrogenation:B X Y H H
Chapter 12 9

84. Mechanism of Catalytic Hydrogenation:B X Y H H
Chapter 12 9