Study Guide
Chapter 14 Chemical Kinetics
-
University:
Seattle Pacific University -
Course:
CHM 1211 | General Chemistry I Academic year:
2022
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Views:
471
Pages:
122
Author:
Alexa Livingston
Chapter 14
Chemical Kinetics
1 Kinetics
• Chemical Kinetics is the study of the
rate at which a chemical process
occurs.
• Besides information about the speed at
which reactions occur, kinetics also
sheds light on the reaction mechanism
(exactly how the reaction occurs).
2 Factors That Affect Reaction Rates
• Physical State of the Reactants
In order to react, molecules must come in
contact with each other. If the reaction is
happening between a solid and a liquid it
will react only on the surface.
The more homogeneous the mixture of
reactants, the faster the molecules can
react.
3 Factors That Affect Reaction Rates
• Concentration of Reactants
As the concentration of reactants increases,
so does the likelihood that reactant
molecules will collide.
4 Factors That Affect Reaction Rates
• Temperature
At higher temperatures, reactant
molecules have more kinetic energy,
move faster, and collide more often and
with greater energy.
5 Factors That Affect Reaction Rates
• Presence of a Catalyst
Catalysts speed up reactions by
changing the mechanism of the
reaction.
Catalysts are not consumed during
the course of the reaction.
6 Reaction Rates
Rates of reactions can be determined by
monitoring the change in concentration of
either reactants or products as a function of
time.
7 Reaction Rates
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
In this reaction, the
concentration of
butyl chloride,
C4H9Cl, was
measured at various
times.
8 The average rate of the
reaction over each
interval is the change in
concentration divided
by the change in time:
•
Average rate =
[ X ]t2 - [ X ]t1
t 2 -t1
=
∆ [X]
∆t
∆[C4H9Cl]
Average rate =
∆t
9 Reaction Rates
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
Average rate =
[ 0.0905] - [0.1000]
50-0
=-
0.0095
= −1.9 x 10-4
50
10 Reaction Rates
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
• Note that the average
rate decreases as the
reaction proceeds.
• This is because as the
reaction goes forward,
there are fewer
collisions between
reactant molecules.
11 • calculate the average rate of appearance of B
over the time interval from 0 to 40 s.
12 Reaction Rates
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
• A plot of concentration
vs. time for this reaction
yields a curve like this.
• The slope of a line
tangent to the curve at
any point is the
instantaneous rate at
that time.
Chemical
Kinetics
13 Reaction Rates
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
• All reactions slow down
over time.
• Therefore, the best
indicator of the rate of a
reaction is the
instantaneous rate near
the beginning.
14 Average vs. instantaneous rxn
rate
• Average rxn rate
change of chemical concentration
between beginning of rxn and end ∆[A]
rate of reaction (rate) =
=
total time elapsed
∆t
• Instantaneous rxn rate (tangent to curve)
change of chemical concentration
over
a
short
period
of
time
= ∆[A]
rate of reaction (rate) =
the short time elapsed
∆t
Usually in chemistry, we are interested in the instantaneous rxn rate at the very
beginning (at t = 0)
INITIAL REACTION RATE
15 Average vs. instantaneous rxn
rate
NaHCO3 + citric acid --> sodium citrate + H2O + CO2
initial [A] = 1 tablet
2
Instantaneous
rate at t = 0
1.8
concentration (tablet/100 mL)
(initial rate)
initial [A] = 2 tablets
1.6
Average rate
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
time (minutes)
1.5
2
16 Comparing concentrations
of chemicals in a reaction
A --> 2 C
A --> B
[A]
2
2
[B]
1.8
co n cen tration (m ol/L )
concentration (m ol/L)
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
time (hours)
1.5
2
[C]
1.6
0.2
0
[A]
0
0.5
1
time (hours)
1.5
2
17 What we’re after in kinetics
•
•
•
•
•
•
We want to be able to understand and predict how much reactants
remain (or products get formed) in a certain amount of time
Change with time is rate
Rate means calculus must be used, so we need to be careful which
rate we are talking about
We will use instantaneous rate at the start of a reaction (initial rate)
Several factors affect the rate of a reaction
Initial concentration
Temperature
Amount of reactant exposed (surface area)
Presence of a catalyst
A theory (equations) that predicts kinetics must explain these
18 Calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the
initial rate).
19 Determine the instantaneous rate of disappearance of C4H9Cl at t
= 300 s.
Answer: 1.1 × 10 –4 M/s
20 Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
• In this reaction, the ratio
of C4H9Cl to C4H9OH is
1:1.
• Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rate =
-∆[C4H9Cl]
=
∆t
∆[C4H9OH]
∆t
21 Reaction Rates and Stoichiometry
• What if the ratio is not 1:1?
2 HI(g) → H2(g) + I2(g)
in such a case ,
∆ [H2 ]
∆ [ I2 ]
1 ∆ [ HI ]
Rate = =
=
2 ∆t
∆t
∆t
22 Reaction Rates and Stoichiometry
• To generalize, then, for the reaction
aA + bB
cC + dD
1 ∆[A]
1 ∆[B]
1 ∆[C]
1 ∆[D]
Rate = −
=
=−
=
a ∆t
b ∆t
c ∆t
d ∆t
23 Key points about kinetics so far
•
•
•
•
•
•
Study the vocabulary
Reaction rate has weird units
Be clear whether you are talking about average reaction rate or
instantaneous reaction rate
The equations which we will study all talk about instantaneous
reaction rate, most easily measured at the beginning of a
reaction (initial rate)
Relating rates at which products appear (+) and reactants
disappear (-) has to do with inverse stoichiometric coefficients
Reaction rate depends on several factors: initial concentration,
temperature, surface area of reactants exposed, presence of a
catalyst
24 (a) How is the rate at which ozone disappears related to
the rate at which oxygen appears in the reaction
(b) If the rate at which O2 appears, Δ[O2]/Δ t, 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?
25 SAMPLE EXERCISE 14.3 continued
The decomposition of N2O5 proceeds according to the following equation:
If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 × 10–7 M/s, what is the rate
of appearance of (a) NO2, (b) O2?
Answers: (a) 8.4 × 10 –7 M/s, (b) 2.1 × 10 –7 M/s
26 Concentration and Rate
One can gain information about the rate
of a reaction by seeing how the rate
changes with changes in concentration.
27 Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Comparing Experiments 1 and 2, when [NH4+]
doubles, the initial rate doubles.
28 Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, comparing Experiments 5 and 6,
when [NO2−] doubles, the initial rate doubles.
29 Concentration and Rate
• 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.
30 • If you want your concepts to be clearer
do the visualizing concepts part of the
problems at the back of your chapter.
31 Rate Law
• An equation that shows how the rate
depends on the concentration of the
reactants is called the rate law
• The exponents tell the order of the
reaction with respect to each reactant.
• This reaction is
First-order in [NH4+]
First-order in [NO2−]
32 Rate Laws
• The overall reaction order can be found
by adding the exponents on the
reactants in the rate law.
• This reaction is second-order overall.
33 • For a general reaction:
aA + bBcC + dD
Rate law has the form
Rate =k [A]m[B]n
where k is the rate constant and is
dependant on temperature.
34 If we know the rate law and the rate for a
set of reaction conditions we can
calculate the value of k.
35 To calculate the rate constant:
Rate law
Rate = k [A]m [B]n
Rate = k[NH4] [NO2]
For experiment 1:
5.4X10-7 M/s = k (0.0100M)(0.200M)
k=
=
5.4X10-7 M/s
k (0.0100M)(0.200M)
2.7X10 -4 /Ms
36 Exponents in the rate law
The exponent in the rate law indicates
how the rate is affected by the
concentration of each reactant.
37 SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction
for which rate = k[A][B]2. Each of the following boxes represents a
reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of
increasing rate of reaction.
38 SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction
for which rate = k[A][B]2. Each of the following boxes represents a
reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of
increasing rate of reaction.
Each box contains 10 spheres. The rate law indicates that in this case [B] has
a greater influence on rate than [A] because B has a higher reaction order.
Hence, the mixture with the highest concentration of B (most purple spheres)
should react fastest. This analysis confirms the order 2 < 1 < 3.
39 Assuming that rate = k[A][B], rank the mixtures represented in this
Sample Exercise in order of increasing rate.
Answer: 2 = 3 < 1
40 Unit for k depends on the overall
order of the reaction.
For example for a second order reaction
Unit of Rate
k =
(Units of concentration)
2
M/s
M
=
= 2 = /Ms
2
M
M s
41 (a)What are the overall reaction orders for the rate laws
described in Equations 14.9 and 14.10?
(b) What are the units of the rate constant for the rate law for Equation
14.9?
42 (a)What are the overall reaction orders for the rate laws
described in Equations 14.9 and 14.10?
2N 2 O5 (g) → 4NO 2 (g) + O 2 (g)
Rate = k [ N 2O 4 ]
CHCl3 (g) + Cl2 (g) → CCl4 (g) + HCl(g)
Rate = k[CHCl3 ] [Cl2 ]1/2
Answer:
The rate of the reaction in Equation 14.9 is first order in N2O5 and first order
overall. The reaction in Equation 14.10 is first order in CHCl3 and one-half
order in Cl2. The overall reaction order is three halves.
(b) What are the units of the rate constant for the rate law for Equation
14.9?
Answer:
Rate = k [ N 2 O 4 ]
So
Notice that the units of the rate constant change as the overall order of the reaction changes.
43 1. The reaction is first order in a reactant
If the concentration of a particular component is
doubled the rate doubles and triples when the
concentration triples.
44 2. If the reaction is second order with respect to
a component:
If the rate is second order with respect to a
particular component, doubling its
concentration increases the rate by a factor of
22 = 4
Tripling its concentration increases the rate by
a factor of 32= 9
and so forth
45 • 3. If a reaction is zero order with respect
to a component:
If a reaction is zero order in a particular
reactant, changing its concentration will
have no effect on the rate as long as
long as the reactant is present.
It is because any concentration raised
to zero is one
46 Order of reaction rate
• Rate of reaction can depend on how much
reactant is present in more than one way:
Zero order:
rate ∝ [A]0
First order:
rate ∝ [A]
Second order: rate ∝ [A]2
47 48 2N 2 O5 (g) → 4NO 2 (g) + O 2 (g)
CHCl3 (g) + Cl2 (g) → CCl 4 (g) + HCl(g)
H 2 (g) + I 2 (g) → 2HI(g)
Rate = k [ N 2 O 4 ]
Rate = k[CHCl3 ] [Cl2 ]1/2
Rate = k[H 2 ][I 2 ]
49 50 • Although the exponents in the rate law are
often the same as the coefficients in the
balanced reaction, this is not necessarily the
case.
• The values of these exponents must be
determined experimentally.
• Most reaction orders are 0, 1 or 2.However
we may encounter reaction orders in fraction.
51 • The rate of the reaction
depends on the concentration
but the rate constant , k, does
not.
• k is dependent only on temperature.
52 CH 4 + 4 Cl 2 → CCl 4 + 4HCl
Rate = -
1 ∆[CH4]
1
∆t
=
1 ∆[ HCl ]
4
∆t
Rate = - rate of disappearence of CH 4 =
1
rate of appearence of HCl
4
Rate x 4 = Rate of appearence of HCl
53 Change in Concentration with Time
From the rate law we can calculate the rate of the
reaction if we know the the rate constant and the
reactant concentration
Rate = k [A]
Rate laws can also be converted into equations that tell
us what the concentration of reactants and products
are at any time during the reaction.
54 Change of Concentration with Time for
the First Order Reactions
The first order reaction is one whose rate depends on the
concentration of a single reactant raised to the first power
reactant A product
Rate =
-Δ[A]
Δt
=
k [A]
This form of rate law, which expresses how rate
depends upon concentration is called differential rate
law.
55 56 57 Integrated Rate Laws
Using calculus to integrate the rate law
for a first-order process we get:
[A]t
ln
[A]0
= −kt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t,
during the course of the reaction.
58 Integrated Rate Laws
Manipulating this equation produces…
[A]t
ln
[A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y
= mx + b
Where m is the slope of the line and b is th ey intersept of the line.
Which is a general equation of
a straight line.
59 Integrated Rate Laws
ln [A]t = − kt + ln [A]0
…which is in the form
y
= mx + b
Where m is the slope of the line
and b is the y intersept of the
line.
Which is a general equation of
a straight line.
60 First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a
plot of ln [A]t vs. t will yield a straight
line, and the slope of the line will be -k.
61 First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
62 First-Order Processes
CH3NC
CH3CN
This data was
collected for this
reaction at 198.9°C.
63 • The experiments show this reaction to
be first order so we can write the rate
equation as
In[CH3NC]t= - kt + ln [CH3NC]o
64 • For gas we can use pressure as a unit of
concentration as from the ideal gas law equation the
pressure is directly proportional to the number of
moles per unit volume
PV = nRT
n
P = RT
V
At constant temperature
n
P α
V
65 First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
The process is first-order.
k is the negative slope: 5.1 × 10-5 s−1.
66 • How to get the natural log value on your
calculator
LN will give the natural log value.
Pressing LN will show
In ( key in the value)
and press enter
67 The initial rate of a reaction
B, and the results are as follows:
was measured for several different starting concentrations of A and
Using these data, determine
(a) the rate law for the reaction,
(b) the magnitude of the rate constant,
(c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.
68 The initial rate of a reaction
B, and the results are as follows:
was measured for several different starting concentrations of A and
Using these data, determine
(a) the rate law for the reaction,
(b) the magnitude of the rate constant,
(c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M
a.
(b) Using the rate law and the data from experiment 1, we have
(c) Using the rate law from part (a) and the rate constant from part (b), we have
69 PRACTICE EXERCISE
The following data were measured for the reaction of nitric oxide with hydrogen:
(a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when
[NO] = 0.050 M and [H2] = 0.150 M.
70 PRACTICE EXERCISE
The following data were measured for the reaction of nitric oxide with hydrogen:
(a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when
[NO] = 0.050 M and [H2] = 0.150 M.
Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5 × 10–4 M/s
71 PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8 × 10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?
72 PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8 × 10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?
Answer: 51 torr
73 Half-Life
• Half-life is defined
as the time required
for one-half of a
reactant to react.
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 1/2 [A]0.
74 Half-Life
For a first-order process, this becomes
[A]t
ln
= −kt
[A]0
0.5 [A]0
ln
= −kt1/2
[A]0
ln 0.5 = −kt1/2
−0.693 = −kt1/2
NOTE: For a first-order
process, the half-life
does not depend on [A]0.
0.693
= t1/2
k
75 • The half life of a first order reaction
remains the same throughout the
reaction.
• The concentration of the reactant keeps
becoming half of the previous value
after every half life.
76 • In the first order reaction, the
concentration of reactant decreases by
½ in each of a series of regularly
spaced time intervals, namely t1/2
• The concept of a first order reaction is
widely used in describing radioactive
decay, a first order process.
77 SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law
•
•
•
The decomposition of a certain insecticide in water follows first-order kinetics with a rate
constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1,
leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the
lake is 12°C.
(a) What is the concentration of the insecticide on June 1 of the following year?
(b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3?
78 SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law
•
•
•
The decomposition of a certain insecticide in water follows first-order kinetics with a rate
constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1,
leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the
lake is 12°C.
(a) What is the concentration of the insecticide on June 1 of the following year?
(b) How long will it take for the concentration of the insecticide to drop to 3.0 × 10–7 g/cm3?
ln [A]t = -kt + ln [A]0
Solve: (a) Substituting the known quantities into Equation 14.13, we have
Note that the concentration units for [A]t and [A]0 must be the same.
(b) Again substituting into Equation 14.13, with [insecticide]t = 3.0 × 10–7 g/cm3, gives
Solving for t gives
79 Temperature and Rate
• Generally, as temperature
increases, so does the
reaction rate.
• This is because k is
temperature dependent.
80 The Collision Model
• In a chemical reaction, bonds are
broken and new bonds are formed.
• Molecules can only react if they collide
with each other.
81 The Collision Model
Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bond breakage and formation.
82 Activation Energy
• In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll
up the hill with enough energy, a reaction cannot
occur unless the molecules possess sufficient energy
to get over the activation energy barrier.
83 C
H3C
N
C-
H3C
H3C
C
N
N
• This is a reaction where methyl isonitrile is
rearranging to acetonitrile
• This reaction passes through an intermediate
state in which the N
C is sitting sideways.
84 Reaction Coordinate Diagrams
It is helpful to
visualize energy
changes
throughout a
process on a
reaction coordinate
diagram like this
one for the
rearrangement of
methyl isonitrile.
85 Reaction Coordinate Diagrams
• It shows the energy of
the reactants and
products (and,
therefore, ∆E).
• The high point on the
diagram is the transition
state.
• The species present at the transition state is
called the activated complex.
• The energy gap between the reactants and the
activated complex is the activation energy
barrier.
86 • Temperature is
defined as a
measure of the
average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.
87 Maxwell–Boltzmann Distributions
• As the temperature
increases, the curve
flattens and
broadens.
• Thus at higher
temperatures, a
larger population of
molecules has
higher energy.
88 Maxwell–Boltzmann Distributions
• If the dotted line represents the activation
energy, as the temperature increases, so does
the fraction of molecules that can overcome
the activation energy barrier.
• As a result, the
reaction rate
increases.
89 Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the
expression
f = e−Ea/RT
where R is the gas constant (8.314J/mol-K) and T is
the Kelvin temperature. Ea is the minimum energy
needed for reaction.
90 • Arrhenious noted that the increase in
reaction rate is not linear with increase
in temperature in most reactions.
91 • Arrhenius observed that most reaction rates
depend on the following three factors
A. Fraction of molecules possessing Ea or
greater
B. number of collisions per second
C. fraction of molecules having the correct
orientation
92 Arrhenius Equation
Svante Arrhenius developed a
mathematical relationship between k
(the rate constant) and Ea:
k = A e−Ea/RT
where A is the frequency factor, a
number that represents the likelihood
that collisions would occur with the
proper orientation for reaction.
93 Arrhenius Equation
Taking the natural
logarithm of both
sides, the equation
becomes
1
ln k = -Ea (
)
+ ln A
RT
y = mx + b
Therefore, if k is determined experimentally at
several temperatures, and lnk vs 1/T is plotted
Ea can be calculated from the slope of the plot.
94 • If we know the constant k at two or more temperatures
At temperatures T1 and T2 the rate constant is k1 and k
,
ln k1
= -Ea + lnA and ln k2 = -Ea + lnA
RT1
RT2
lnk1-lnk2 = [ -Ea/RT1 +lnA] - [ -Ea/RT2+lnA]
lnk1/k2
= Ea (1/T1-1/T2)
R
95 • lnk1 = Ea (1/T2 -1/T1)
•
R
96 Reaction Mechanisms
The sequence of events that describes
the actual process by which reactants
become products is called the reaction
mechanism.
98 Reaction Mechanisms
• Reactions may occur all at once or
through several discrete steps.
• Each of these processes is known as an
elementary reaction or elementary
process.
99 Rate laws for elementary
reactions
• The rate law of an elementary reaction
is directly based on its molecularity or
concentration.
100 Reaction Mechanisms
The molecularity of a process tells how many
molecules are involved in the process.
101 • In a multistep reaction each step will
have its own rate constant and
activation energy.
• The slowest step determines the overall
rate.
Example Relay race
102 Multistep Mechanisms
• In a multistep process, one of the steps will
be slower than all others.
• The overall reaction cannot occur faster than
this slowest, rate-determining step.
103 • A balanced chemical equation often
occurs in a series of elementary reactions
• A balanced chemical reaction itself could
have a multi step mechanism
104 Slow Initial Step
NO2 (g) + CO (g) → NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
105 Slow Initial Step
NO2 (g) + CO (g) → NO (g) + CO2 (g)
•
A proposed mechanism for this reaction is
Step 1: NO2 + NO2 → NO3 + NO (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
•
The NO3 intermediate is consumed in the second step.
•
As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.
•
Since step 1 is the slow rate determining step the rate law is going to be
Rate = [NO2] 2
106 Practice exercise
Ozone reacts with NO2 to give dinitrogen pentoxide and
oxygen:
O3 (g) + 2NO2 (g) N2O5(g) + O2 (g)
The reaction occurs in the following two steps:
O3(g) + NO2 (g) NO3 (g) + O2(g)
NO3(g) + NO2(g) N2O5 (g)
The experimental rate law is
Rate = [ O3][NO2]
What can you sat about the relative rate of the two steps
of the mechanism?
107 • How many elementary
reactions are there in
the mechanism?
• How many
intermediates?
• Which step is rate
limiting?
• Which is the fastest
step?
• Exo or endothermic?
108 Problems
• How many elementary
reactions are there in
the mechanism?
• How many
intermediates?
• Which step is rate
limiting?
• Which is the fastest
step?
• Exo or endothermic?
109 Fast Initial Step
2 NO (g) + Br2 (g) → 2 NOBr (g)
• Experimentally the rate law for this
reaction is found to be
Rate = k [NO]2 [Br2]
• Because ter molecular processes are
rare, this rate law suggests a two-step
mechanism.
110 Fast Initial Step
2 NO (g) + Br2 (g) → 2 NOBr (g)
• A proposed mechanism is
Step 1: NO + Br2
NOBr2
(fast)
Step 2: NOBr2 + NO → 2 NOBr (slow)
Step 1 includes the forward and reverse
reactions.
The rate of the overall reaction depends upon the rate of
the slow step.
The rate law for that step would be
Rate = k2 [NOBr2] [NO]
But how can we find [NOBr2]?
111 Fast Initial Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO → 2 NOBr
(fast)
(slow)
• NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
• The reactants and products of the first
step are in equilibrium with each other.
• Therefore,
Ratef = Rater
112 Fast Initial Step
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
Pay attention to the fact that the reverse k
is k-1
• Solving for [NOBr2] gives us
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
113 Fast Initial Step
From before Rate = k2 [NOBr2] [NO]
From the previous equation:
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
Substituting this expression for [NOBr2] in the rate law for
the rate-determining step gives
k 2k 1
Rate =
[NO] [Br2] [NO]
k−1
= k [NO]2 [Br2]
114 • In general when ever a fast step
precedes a slow step we can solve the
problem of the concentration of the
intermediate step by assuming that an
equilibrium is established in the fast
step.
115 SAMPLE EXERCISE 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step
Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the
experimentally observed one:
•
Write down the rate law for the overall reaction.
116 PRACTICE EXERCISE
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)?
117 Catalysts
Arrhenius equation
• K = A e –E/RT
A= frequency factor and
E= Energy of activation
• Catalysts increase
the rate of a reaction
by decreasing the
activation energy of
the reaction.
• Catalysts change
the mechanism by
which the process
occurs.
118 • Homogeneous catalysts
2H2O2(aq) 2H2O (l) + O2 (g)
NaBr in an acidic solution acts as a catalyst for this reaction
2Br -(aq) + H2O2(aq) + 2 H+ (aq) Br2(aq) + 2H2O(l)
Br2(aq) + H2O2(aq)
2Br –(aq) + 2H+(aq) + O2 (g)
119 120 121 Heterogeneous Catalyst
One way a
catalyst can
speed up a
reaction is by
holding the
reactants together
and helping bonds
to break.
The reactants are
adsorbed on the
surface of the
catalyst.
122 Enzymes
• Enzymes are
catalysts in
biological systems.
• The substrate fits
into the active site of
the enzyme much
like a key fits into a
lock.
123
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Chapter 14 Chemical Kinetics
of 122
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