1. Chapter 13
Chemical Kinetics

2. The study of reaction rate is called chemical kinetics.
Reaction rate is measured by the change of
concentration (molarity) of reactants or products
per unit time.

3. − Molarity of A: [A], e.g. [NO2]: molarity of NO2
reactants  products,
[reactant]↓ and [product]↑
Unit:
mol·L−1·s−1 ≡ M·s−1

4. 2NO2 (g)  2NO (g) + O2 (g) − = r(NO2) rate is a function of time
0 s → 50 s:
50 s → 100 s:

5. 2NO2 (g)  2NO (g) + O2 (g) − = r(NO2) − rate is a function of time
50 s → 100 s:
50 s → 100 s:

6. 2NO2(g)  2NO(g) + O2(g)

7. a A + b B  c C + d D
r =
r does not depend upon the choice of species

8. 2N2O5(g)  4NO2(g) + O2(g) r(N2O5) = 4.2 x 10−7 M·s−1
What are the rates of appearance of NO2 and O2 ?

9. H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
Example page 567
Consider the following balanced chemical equation:
H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the concentration of I– dropped from M to M.
Calculate the average rate of this reaction in this time interval.
(b) Predict the rate of change in the concentration of
H+ (that is, [H+]/t) during this time interval.

10. Consider the general reaction aA + bB cC and the
following average rate data over some time period Δt:
Determine a set of possible coefficients to balance this
general reaction.

11. All the rates in this table are average rates.−
= r(NO2) −
All the rates in this table are average rates.

12. Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min
2:00 pm
2:16 pm
3:16 pm
0 mile
Barnesville
16 miles
Griffin
56 miles
Atlanta
Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min
Average Speed from G to A = 40 miles ÷ 60 min = 0.7 mile/min
Instantaneous speed at green spot
Instantaneous speed contains more information

13. l
Atlanta
56 miles
Griffin
20 miles
16 min
76 min Δl Δt
Barnesville
0 mile t
0 min
Instantaneous speed at the red point = slope of the red solid line =

14. Reaction rate is a function of time

15. Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst

16. a A + b B  c C + d D r = k [A]m [B]n
Differential rate law: how r depends on concentrations
m, n: reaction order, mth order for A, nth order for B
(m+n): overall reaction order
k: rate constant: depends on temperature, but not
concentrations
m and n must be measured from experiments. They can
be different from the stoichiometry.

17. (1) 2N2O5(g)  4NO2(g) + O2(g) r = k[N2O5]
(2) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2
(3) H2(g) + I2(g)  2HI(g) r = k[H2][I2]

18. aA + bB  cC +dD r = k [A]m [B]n Units
overall reaction order (m+n) ⇌ unit of k

19. (1) 2N2O5(g)  4NO2(g) + O2(g) r = k[N2O5]
(2) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2
(3) H2(g) + I2(g)  2HI(g) r = k[H2][I2]

20. r = k [NH4+]m [NO2−]n NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
How to find the rate law by experiment: method of initial rates
NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
Experiment Number
Initial [NH4+] (M)
Initial [NO2−] (M)
Initial Rate (M·s−1) 1
0.100
0.0050
1.35 x 10−7 2
0.010
2.70 x 10−7 3
0.200
5.40 x 10−7
r = k [NH4+]m [NO2−]n

21. z = f (x,y) How does the change of x affect z?
A very common method to investigate how each factor
affects the whole system:
Change one thing at a time while keep the others constant.
z = f (x,y)
How does the change of x affect z?
How does the change of y affect z?

22. r = k [NH4+]m [NO2−]n NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
How to find the rate law by experiment: method of initial rates
NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
Experiment Number
Initial [NH4+] (M)
Initial [NO2−] (M)
Initial Rate (M·s−1) 1
0.100
0.0050
1.35 x 10−7 2
0.010
2.70 x 10−7 3
0.200
5.40 x 10−7
r = k [NH4+]m [NO2−]n

23. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Experiment Number
Initial [NO] (M)
Initial [H2] (M)
Initial Rate (M·s−1) 1
0.10
1.23 x 10-3 2
0.20
2.46 x 10-3 3
4.92 x 10-3
Determine the differential rate law
Calculate the rate constant
Calculate the rate when [NO] = M and [H2] = M

24. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
again
Experiment Number
Initial [NO] (M)
Initial [H2] (M)
Initial Rate (M·s−1) 1
0.10
1.23 x 10-3 2
0.30
3.69 x 10-3 3
1.11 x 10-2
Determine the differential rate law
Calculate the rate constant
Calculate the rate when [NO] = M and [H2] = M

25. A + B  C
Experiment Number
Initial [A] (M)
Initial [B] (M)
Initial Rate (M·s−1) 1
0.100
4.0 x 10−5 2
0.200 3
1.6 x 10−4
Determine the differential rate law
Calculate the rate constant
Calculate the rate when [A] = M and [B] = M

26. NO2(g) + CO(g)  NO(g) + CO2(g)
EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
From the data, determine:
(a) the rate law for the reaction
(b) the rate constant (k) for the reaction

27. Use the data in table to determine
1) The orders for all three reactants
2) The overall reaction order
3) The value of the rate constant

28. r = k [BrO3−]m [Br−]n [H+]p
r1 = k (0.10 M)m (0.10 M)n (0.10 M)p = 8.0 x 10−4 M · s−1
r2 = k (0.20 M)m (0.10 M)n (0.10 M)p = 1.6 x 10−3 M · s−1
r3 = k (0.20 M)m (0.20 M)n (0.10 M)p = 3.2 x 10−3 M · s−1
r4 = k (0.10 M)m (0.10 M)n (0.20 M)p = 3.2 x 10−3 M · s−1

29. one quiz after lab Relationship among reaction rates as expressed
by different species.
r =
overall reaction order (m+n) ⇌ unit of k
Method of initial rates:
table of experimental data  rate order, k, rate at other
concentrations.

30. a A + b B  c C + d D r = k [A]m [B]n
Differential rate law: how r depends on concentrations
Differential rate law: differential equation
How concentration changes as a function of time 
integrated rate law

31. A  Products First order reaction differential rate law:
First order reaction integrated rate law: or
integrated rate law: how concentration changes as a
function of time.

32. [A] is the molarity of A at t
First order reaction integrated rate law:
[A] is the molarity of A at t
y = mx + b
Plot ln[A] vs. t gives a straight line
Slope = −k, intercept = ln[A]0

33. N2O5(g)  2NO2(g) + ½ O2(g) [N2O5] (M) Time (s) 0.1000 0.0707 50
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400
Use these data, verify that the rate law is first order in N2O5,
and calculate the rate constant.
k = 6.93 x 10−3 s−1

34. Read a similar Example 13.3 on page 575

35. Using the data given in the previous example, calculate [N2O5]
at 150 s after the start of the reaction. M
[N2O5] (M)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

36. Practice on Example 13.4 on page 576 and
check your answer

37. The half-life of a reaction, t1/2, is the time required for a
reactant to reach one-half of its initial concentration.
[N2O5] (M)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

38. The half-life for first order reaction:
The half-life for first order reaction does NOT depend on
concentration.

39. [N2O5] (M)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

40. A Plot of [N2O5] versus Time for the Decomposition Reaction of N2O5

41. A certain first order reaction has a half-life of 20.0 s.
Calculate the rate constant for this reaction.
How much time is required for this reaction to be 75 %
complete?
a) k = s−1
b) k = 40.0 s

42. Try Example 13.6 and For Practice 13.6
on page 579 and check your answers

43. A  Products Second order reaction differential rate law:
Second order reaction integrated rate law:
integrated rate law: how concentration changes as a
function of time.

44. Plot 1/[A] vs. t gives a straight line
Second order reaction integrated rate law:
y = mx + b
Plot 1/[A] vs. t gives a straight line
Slope = k, intercept = 1/[A]0

45. The half-life for second order reaction:
The half-life for second order reaction depends on initial
concentration.

46. A certain reaction has the following general form: A  B
At a particular temperature [A]0 = 2.80 x 10−3 M,
concentration versus time data were collected for this reaction,
and a plot of 1/[A] versus time resulted in a straight line with a
slope value of 3.60 x 10−2 M−1·s−1
Determine the (differential) rate law, the integrated rate law,
and the value of the rate constant.
b) Calculate the half-life for this reaction.
c) How much time is required for the concentration of A to
decrease to 7.00 x 10−4 M ?

47. For first order reaction, show that
(show your work, do not copy the question)
For first order reaction, show that
from the integrated rate law

48. A  Products Zero order reaction differential rate law:
Zero order reaction integrated rate law:
integrated rate law: how concentration changes as a
function of time.

49. Plot [A] vs. t gives a straight line
Zero order reaction integrated rate law:
y = mx + b
Plot [A] vs. t gives a straight line
Slope = −k, intercept = [A]0

50. The half-life for zero order reaction:
The half-life for zero order reaction depends on initial
concentration.

51. The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g) takes place on a Platinum Surface

52. The decomposition of ethanol on alumina surface
C2H5OH(g)  C2H4(g) + H2O(g) was studied at 600 K.
Concentration versus time data were collected for this reaction, and a plot of [C2H5OH] versus time resulted in a straight line with a slope value of −4.00 x 10−5 M·s−1
Determine the (differential) rate law, the integrated rate law,
and the value of the rate constant.
b) If the initial concentration of C2H5OH was 1.25 x 10−2 M,
calculate the half-life of this reaction.
c) How much time is required for all the 1.25 x 10−2 M C2H5OH
to decompose ?

54. Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst

55. aA + bB  cC +dD r = k [A]m [B]n
k: rate constant: depends on temperature, but not
concentrations

56. Collision model: molecules must collide to react.
A Plot Showing the Exponential Dependence of the Rate Constant on Absolute Temperature
Collision model: molecules must collide to react.
Not all collisions lead to products
T ↑  v ↑  kinetic energy = ½ mv2 ↑

57. T ↑  f ↑ Ea ↑  f ↓ Fraction of molecules whose Ek > Ea is
activation energy
Prentice Hall © 2003
Chapter 14
Fraction of molecules whose Ek > Ea is
T ↑  f ↑
Ea ↑  f ↓

58. molecular orientation
Another factor needs to be taken into account
molecular orientation

59. Several Possible Orientations for a Collision Between Two BrNO Molecules BrNO + BrNO  2NO +Br2

60. T ↑  k ↑ Ea ↑  k ↓ Arrhenius equation A: frequency factor
steric factor p ≤ 1
Arrhenius equation
A: frequency factor
How k depends on T
T ↑  k ↑
Ea ↑  k ↓

61. Ea = 1.4 x 105 J/mol At 550 °C the rate constant for
CH4(g) + 2S2(g)  CS2(g) + 2H2S(g)
is 1.1 M·s−1, and at 625 °C the rate constant is
6.4 M·s−1. Using these value, calculate Ea for this reaction.
Ea = 1.4 x 105 J/mol

62. Try Example 13.8 on page 585 and
check your answers

63. h
The ball can climb over the hill only if its kinetic energy is greater
than Ep = mgh, where m is the mass of the ball, h is the height
of the hill, and g is gravitational acceleration.

64. BrNO + BrNO  2NO +Br2 exothermic reaction
(a) The Change in Potential as a Function of Reaction Progress (b) A Molecular Representation of the Reaction
BrNO + BrNO  2NO +Br2
O N Br
exothermic reaction

65. Reaction Mechanism
Chemical Equation
Reactants  Products

66. + NO2 + CO  NO + CO2 Step 1: NO2 + NO2  NO3 + NO
Step 2: NO3 + CO  NO2 + CO2 +
NO2 + CO  NO + CO2
NO3: intermediate, does not
appear in overall reaction

67. Whole process is called the reaction mechanism.
NO NO NO3 NO
NO CO NO CO2
Whole process is called the reaction mechanism.
Each single step is called an elementary reaction/step.
An elementary reaction is a single collision.

68. For an elementary reaction aA + bB  cC + dD r = k[A]a[B]b
Not for overall reaction!
Overall reaction: r = k[A]m[B]n , find m and n by experiments
The number of molecules that react in an elementary reaction is called the molecularity of that that elementary reaction (not applicable to overall reaction).

69. 1 ― unimolecular 2 ― bimolecular 3 ― termolecular
Chapter 13, Unnumbered Table 2, Page 600
1 ― unimolecular
2 ― bimolecular
3 ― termolecular

70. What determines the rate of an overall reaction?

71. Team A
Team B
Slowest step: rate determining step

72. Ep Which step is the rate determining step? 2nd exothermic or
endothermic? Ep
Ea, 2
intermediate
Ea, 1
Reactants ∆E
Products
Reaction progress

73. Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst

74. Catalyst is a substance that speeds up a reaction without
being consumed itself.
Catalyst changes the reaction mechanism through a lower
activation energy pathway.

75. Energy Plots for a Given Reaction

76. homogeneous
catalyst
heterogeneous

77. + 3O2(g)  2O3(g) 2NO(g) + O2(g)  2NO2(g) 2NO2(g)  2NO(g) + 2O(g)
Homogeneous catalyst: same phase as reactants.
3O2(g)  2O3(g)
2NO(g) + O2(g)  2NO2(g)
2NO2(g)  2NO(g) + 2O(g)
2O2(g) + 2O(g)  2O3(g)
light +
3O2(g)  2O3(g)
NO(g): catalyst;
NO2(g), O(g): intermediates

78. Heterogeneous catalyst: different phase from reactants.
The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g) takes place on a Platinum Surface

79. These Cookies Contain Partially Hydrogenated Vegetable Oil
C = C
+ H2 
− C − C − −

80. milk sugar = lactose