1. Reaction Rates The term reaction rate refers to how fast a
product is formed or a reactant is consumed.
Consider the following generic reaction:
aA(g) + bB(g) cC(g) + dD(g) 1 a
Δ[A] Δt 1 b
Δ[B] Δt
Rateave = - = - 1 c
Δ[C] Δt 1 d
Δ[D] Δt = =

2. Rates are always positive quantities.
The concentration of reactants do not change
at the same rate when coefficients are
different.
Consider the following specific example:
N2(g) + 2O2(g) NO2(g)
Δ[N2] Δt 1 2
Δ[O2] Δt
Rateave = - = - 1
Δ[NO2] M = = 2 Δt s

3. Instantaneous Rate vs Average Rate
The following graph shows the decomposition
of N2 as a function of time.
There are two different rates shown in the
graph.
The instantaneous rate is the rate of
change (Δ) at a specific instant of time,
i.e. 40 s.

4. In a non-calculus environment, this is
determined by calculating the slope of a
tangent line containing that point.
The average rate is computed over a
time interval necessitating using two
different times.
Δ[N2] Δt
Δ[N2] Δt
0.30 M – 0.55 M
Rateinst
Rateinst = = - - = = -
60. s – 30. s
Rateinst =
8.3 x 10-3 M s-1

5. This is the average rate from 20. s to 60. s.
Δ[N2] Δt
0.30 M – 0.67 M
Rateavs = - = -
60. s – 20. s
Rateave =
9.2 x 10-3 M s-1

6. [N2] vs Time
average rate
instantaneous rate
[N2]
Δ[N2] Δt
Time (s)

7. Factors Affecting Reaction Rates
Chemical reactions involve the breaking of
bonds (endothermic) and the forming of
bonds (exothermic).
The factors affecting reaction rates:
The physical state of reactant rates.
When reactants are in different phases
(states), the reaction is limited to their
area of contact.

8. The concentration of reactants.
As the concentration of reactants
increase, the rate of reaction increases.
Temperature
As the temperature is increased,
molecules have more kinetic energy
resulting in higher reaction rates.
Presence of a catalyst.
Increases the rate of reaction without
being consumed.

9. Rate Laws Consider the following generic reaction:
aA(g) + bB(g) cC(g) + dD(g)
The rate law is written:
Rate = k[A]x[B]y
k is called the rate constant and is
temperature dependent.
The exponents x and y are called
reaction orders.

10. The overall reaction order is the sum of the
orders with respect to each reactant.
The reaction orders in a rate law indicate how
the rate is affected by the concentration of
the reactants.
The effect of reactant concentration can
not be predicted from the balanced
equation.
only be determined empirically!

11. The rate of a reaction will decrease as the
reaction proceeds because the reactants are
being consumed.
A very important distinction needs to be
made at this point:
The rate of a reaction depends on
concentration but k, the rate constant,
does not!
The rate constant and the rate of
reaction are affected by temperature
and the presence of a catalyst.

12. Reaction Rate and Stoichiometry
How is the rate of disappearance of N2
and O2 related to the rate of appearance of NO2 in the reaction shown below.
N2(g) + 2O2(g) 2NO2(g)
Δ[N2] Δt 1 2
Δ[O2] Δt
Rate = - = - = 1 2
Δ[NO2] = Δt

13. (b) If the rate of decomposition of N2 at an
instant is 3.7 x 10-6 M•s-1, what is the rate
of disappearance of O2 and the rate of
appearance of NO2?
Δ[N2] Δt
Rate = - =
3.7 x 10-6 M•s-1
Δ[O2] Δt =
2 × 3.7 x 10-6 M•s-1 =
7.4 x 10-6 M/s
Δ[NO2] Δt =
2 × 3.7 x 10-6 M•s-1 =
7.4 x 10-6 M/s

14. Determining Rate Laws The following data was measured for the
reaction of nitrogen and oxygen to form
nitrogen(IV) oxide.
N2(g) + 2O2(g) NO2(g)

15. Trial [N2] (M) [O2] Init Rate (M/s) 1 0.12 1.19×10-3 2 0.23 2.44×10-3
0.22
4.92×10-3

16. Determine the rate law for this reaction.
k[N2]x[O2]y =
Rate 1
2.44 × 10-3 M/s
1.19 × M/s =
k(0.12)x(0.23)y
k(0.12)x(0.12)y 2 = 2y
y = 1

17. .
Rate 3
k[N2]x[O2]y =
Rate 1
4.92 × 10-3 M/s
k(0.22)x(0.12) =
1.19 × M/s
k(0.12)x(0.12) 4 = 2x
x = 2
Rate = k[N2]2[O2]

18. (b) Calculate the rate constant.
Rate = k[N2]2[O2]
1.19 × 10-3 Ms-1 = k(0.12 M)2(0.12 M)
k = 0.69 M-2s-1
(c) Calculate the rate when [N2] = M and
[O2] = 0.16 M
Rate = 0.69 M-2s-1 × (0.047 M)2 × 0.16 M
Rate = 2.4 x 10-4 M/s

19. Integrated Rate Laws The second type of rate law is called the
integrated rate law which shows the
concentration as a function of time.
There are three possible orders for a
reaction, zeroth, first, or second.

20. Zeroth Order Kinetics The integrated rate law is given by:
[A] = -kt + [A]0
A plot of [A] vs Time
produces a straight line
with a slope equal to the
negative rate contant, -k.

21. The reaction rate is
independent of
concentration.
The reaction rate is
independent of time.

22. For 0th order kinetics:
Rate Law: Rate = k
Integrated Rate Law: [A] = -kt + [A]0
Plot needed for straight line: [A] vs T
Slope: -k
[A]0
Half-Life:
t1/2 = 2k

23. First-Order Kinetics The integrated rate law for 1st order kinetics
is given by:
ln[A] = -kt + ln[A]0
A plot of ln[A] vs Time
produces a straight line
with a slope equal to the
negative rate contant, -k.

24. The reaction rate is
directly proportional to
the concentration.
The reaction rate
decreases but not
linearly with time.

25. A first-order reaction is one in which the rate
depends on the concentration of a single
reactant to the first power.
aA(g)
Product
Δ[A] Δt
Rate = - =
k[A]
Δ[A] =
- k Δt
[A]
ln[A]t – ln[A]0 = -kt

26. ln[A]t = -kt + ln[A]0 .
y = mx + b
The slope of the straight line gives -k,
the rate constant, and ln[A]0 is the y-intercept.
For a first-order reaction, the half-life (the time
required for one-half of a reactant to
decompose) has a constant value.
ln 2 k
0.693 k
t1/2 = =

27. For a first-order reaction, the half-life is only
dependent on k, the rate constant and
remains the same throughout the reaction.
The half-life is not affected by the initial
concentration of the reactant.

28. For 1st order kinetics:
Rate Law: Rate = k[A]
Integrated Rate Law: ln[A] = -kt + ln[A]0
Plot needed for straight line: ln[A] vs T
Slope: -k
0.693
Half-Life:
t1/2 = k

29. Second-Order Kinetics
The integrated rate law for 2nd order kinetics
is given by:
1/[A] = kt + 1/[A]0
A plot of 1/[A] vs Time
produces a straight line
with a slope equal to the
rate contant, k.

30. The reaction rate is
directly proportional to
the concentration.
The reaction rate
decreases but not
linearly with time.

31. A second-order reaction is one in which the
rate depends on the concentration of a single
reactant concentration raised to the second
power or concentrations of two different
reactants, each raised to the first power.
aA(g)
Product or
aA(g) + bB(g)
Product
Δ[A] Δt
Δ[B] Δt
Rate = - = - =
k[A][B] or
Δ[A]2 Δt
Rate = - =
k[A]2

32. The integrated rate law for a second-order
reaction is given by: 1 1
[A]0 = k t +
[A]t
y = mx + b
The slope of the straight line gives k, the rate
constant, and is the y-intercept. 1
[A]0

33. .
For a second-order reaction, the half-life (the
time required for one-half of a reactant to
decompose) is double the preceding one. 1
t1/2 =
k[A]0

34. For 2nd order kinetics:
Rate Law: Rate = k[A]2
Integrated Rate Law:
Plot needed for straight line:
Slope: k 1 1 1 1 1
[A]0 = kt +
[A]
[A]
[A] 1
[A]0
vs T 1
Half-Life:
t1/2 = k
[A]0

35. Collision Theory Particles must collide for chemical reactions
to occur.
Not every collision leads to a reaction.
For a reaction to occur, an “effective
collision” must take place.
An “effective collision” consists of two
conditions.

36. The colliding particles must approach
each other at the proper angle. v1 v2 H2
Cl2 v3 v4
HCl
HCl

37. In addition, the colliding particles must
have sufficient energy.
At the point of impact, ΔKE = ΔPE.
An elastic collision is assumed in which
the law of conservation of mass-energy
applies.
The KE/molecule must be sufficient to
break the H-H and the CI-CI in both
hydrogen and chlorine.

38. Keep in mind that temperature is a
measure of the KE/molecule.
Some molecules will be traveling faster
than others while the slower molecules
will bounce off each other without
reacting.

39. A less than ideal set of conditions for a
collision. v1 v2 H2
Cl2 v3 v4 H2
Cl2

40. Factors Affecting Reaction Rates
The bonding found in the reactants.
Triple bonds are stronger than double
bonds which are stronger than single
bonds.
The temperature of the system (reactants and
products).
The initial concentration of the reactants.

41. The amount of surface area when reactants
are in more than one phase.
The use of a catalyst to provide an alternate
pathway.

42. Activation Energy In order to react, colliding molecules must
have KE equal to or greater than a minimum
value.
This energy is called the activation energy
(Ea) which varies from reaction to reaction.
The Ea depends on the nature of the
reaction and is independent of
temperature and concentration.

43. The rate of a reaction depends on Ea.
At a sufficiently high temperature, a greater
number of molecules have a KE > Ea.

44. Maxwell-Boltzmann Distribution
lower temperature
higher temperature Ea

45. For the Maxwell-Boltzmann Distribution, only
those molecules with energies in excess of Ea
react.
As shown on the graph, as the temperature
is increased:
The curve shifts to higher energies
meaning more molecules to have
energies greater than Ea.
The curve gets broader and flatter but
the area remains the same under the
curve.

46. Energy Profile for Exothermic Rxn
Minimum Energy for Reaction Ea
Energy Content of Reactants
ΔH = -
Energy Content of Products

47. Energy Profile for Endothermic Rxn
Minimum Energy for Reaction Ea
Energy Content of Products
ΔH = +
Energy Content of Reactants

48. Arrhenius Equation The Arrhenius equation illustrates the
dependence of the rate constant, k, on the
frequency factor, A, the activation energy, Ea,
the gas constant, R, and the absolute
temperature, T. k =
Ae-Ea/RT
The frequency factor, A, is related to the
frequency of collisions and the probability of
these collisions being favorably oriented.

49. Taking the ln of both sides gives:
ln k =
ln A – Ea/RT
Rearranging the equation above shows that the
ln k is directly proportional to 1/T.
ln k = -Ea/R • 1 T +
ln A
y = m x b
The equation also shows that the reaction rate
decreases as the activation energy increases.

50. An unknown gas at 300.°C has a rate constant
equal to 2.3 x s-1 and at 350.°C the rate
constant is determined to be 2.4 x 10-8 s-1.
(a) Calculate Ea for this reaction.
T1 = 300.°C = 573 K T2 = 350.°C = 623 K
k1 = 2.3 x s-1 k2 = 2.4 x 10-8 s-1
R = 8.31 J mol-1K-1 ln k2 k1 = Ea R [ 1 T1 - ] T2 1 T2

51. 2.4 × 10-8 s-1 ln =
2.3 × s-1 . Ea 1
573 K 1
623 K [ - ]
8.31 J mol-1 K-1 Ea =
2.7 × 105 J/mol

52. (b) Determine the rate constant at 425°C.k2 ln =
2.4 x s-1
2.7 x 105 J mol-1 1
573 K 1
698 K [ - ]
8.31 J mol-1 K-1 k2 =
6.2 x 10-6 s-1

53. Reaction Mechanisms A reaction mechanism consists of a series of
elementary steps indicating how reacting
particles rearrange themselves to form
products.
Elementary steps are individual steps
showing what molecules must collide
with each other and the proper
sequence of collisions.
Most reactions occur in more than one
step.

54. A reaction mechanism provides much more
information than a balanced chemical
equation.
Intermediates are species that are
formed and consumed between
elementary steps which never appear in
a balanced chemical equation.
Not all elementary steps occur at the
same rate and consequently are
designated as fast or slow steps.

55. Each elementary step has a molecularity
associated with it.
An elementary step is referred to as a
unimolecular step when one molecule
decomposes or rearranges to form a
product as shown below:
A B
A unimolecular reaction follows a
first-order rate law, i.e.
Rate = k[A]

56. An elementary step is referred to as a
bimolecular step when two molecules
collide to form a product as shown
below:
A + B C
A bimolecular reaction follows a
second-order rate law, i.e.
Rate = k[A][B]

57. An elementary step is referred to as a
termolecular step when three molecules
collide to form a product as shown
below:
A + B + C D
A termolecular reaction follows a
third-order rate law, i.e.
Rate = k[A][B][C]
Termolecular reactions are very rare.

58. Determining a Rate Expression
The slowest step in a mechanism is called the
rate determining step and determines the
rate of the overall reaction.
To determine the rate expression from a
mechanism do the following:
The rate of the overall reaction is the
same as the rate of the rate-determining
step.

59. Write the rate expression for the slowest
step.
The rate expression must only include
those species that appear in the
balanced equation.
Intermediates are species that are
produced in one step and consumed in
the next step.
Intermediates can not appear in the rate
expression because their concentrations
are always small and undectable.

60. Determine the overall reaction and the rate
expression for the suggested mechanism:
NO2(g) + NO2(g) NO3(g) + NO(g) (slow)
NO3(g) + CO(g) NO2(g) + CO2(g) (fast)
NO2(g) + CO(g) NO(g) + CO2(g)

61. Note that the NO3 is an intermediate because
it is produced in one step of the reaction and
consumed in the next.
Because the rate-determining step comes
first, the intermediate NO3 cancels out.
This is good because intermediates can
not appear in the overall reaction.
The rate expression is determined by using
the rate-determining step and is given by:
Rate = k[NO2][NO2] = k[NO2]2

62. The following reaction
2NO2(g) + O3(g) N2O5(g) + O2(g)
has as its rate law, Rate = k[NO2][O3].
Which of the following mechanisms predicts a
satisfactory rate law?
Mechanism 1:
NO2(g) + NO2(g) N2O2(g) + O2(g) (slow)
N2O2(g) + O3(g) N2O5 (g) (fast)

63. Mechanism 2:
NO2(g) + O3(g) NO3(g) + O2(g) (slow)
NO3(g) + NO2(g) N2O5(g) (fast)
For Mechanism 1:
NO2(g) + NO2(g) N2O2(g) + O2(g) (slow)
N2O2(g) + O3(g) N2O5 (g) (fast)
2NO2(g) + O3(g) N2O5(g) + O2(g)

64. The first criteria is satisfied because after
adding the elementary steps, the overall
equation for the reaction is correct.
However, the predicted rate law determined
from the rate-determining step is:
Rate = k[NO2]2
which does not agree with given rate law.

65. For Mechanism 2:
NO2(g) + O3(g) NO3(g) + O2(g) (slow)
NO3(g) + NO2(g) N2O5(g) (fast)
2NO2(g) + O3(g) N2O5(g) + O2(g)
The first criteria is satisfied because after
adding the elementary steps, the overall
equation for the reaction is correct.

66. The predicted rate law determined from the
rate-determining step is:
Rate = k[NO2][O3]
which does agrees with given rate law.
Mechanism 2 is consistent with the given rate
law.
When the intermediate is not a reactant in the
rate-determining step, mechanisms are
usually easier to work with.

67. When the intermediate is a reactant in the
rate-determining step, mechanisms are
more involved.
Determine the overall reaction and the rate
expression for the suggested mechanism:
NO(g) + Br2(g) NOBr2(g) (fast)
NOBr2(g) + NO(g) 2NOBr(g) (slow)

68. NO(g) + Br2(g) NOBr2(g) (fast)
NOBr2(g) + NO(g) 2NOBr(g) (slow)
The predicted rate law determined from the
rate-determining step is:
Rate = k[NOBr2][NO]
which is incorrect because intermediates can
not appear in the rate law.
2NO(g) + Br2(g)
2NOBr(g)

69. When this situation occurs, you must equate
the forward and reverse rates obtained from
the fast step as shown below.
NO(g) + Br2(g) NOBr2(g)
Ratef = Rater
kf[NO][Br2] = kr[NOBr2] kf
[NOBr2] =
[NO][Br2] =
k[NO][Br2] kr

70. Now you can use the rate law obtained
originally but substitute for [NOBr2].
Rate = k[NOBr2][NO]
[NOBr2] =
k[NO][Br2]
Rate =
k[NO]2[Br2]
There is no need to keep track of the individual
rate constants because they would be
incorporated into a single constant.

71. Catalysis A catalyst is a substance that increases the
speed of a reaction without being consumed.
Catalysts can be either homogeneous or
heterogeneous depending if they are in the
same phase as the reactants.
The function of a catalyst is to provide an
alternate pathway (mechanism) for a reaction
which lowers the activation energy, Ea.

72. Reviewing the Arrhenius equation,
shows that as activation energy, Ea, is made
smaller, the rate constant ,k, increases. k =
Ae-Ea/RT

73. For the given mechanism, identify the catalyst
and the intermediate.
O3 + Br O2 + BrO
BrO + O Br + O2
Remember an intermediate is produced in
one step and consumed in the other making
BrO the intermediate.
Remember a catalyst remains chemically
unaltered during the reaction making the Br
the catalyst.