Chapter 16
Kinetics: Rates and Mechanisms of Chemical
Reactions
16-1
Kinetics: Rates and Mechanisms of Chemical Reactions
16.1 Factors that influence reaction rates
16.2 Expressing the reaction rate
16.3 The rate law and its components
16.4 Integrated rate laws: Concentration changes over time
16.5 Reaction mechanisms: Steps in the overall reaction
16.6 Catalysis: Speeding up a chemical reaction
16-2
Chemical Kinetics
The study of reaction rates, the changes in concentrations of
reactants (or products) as a function of time
Quantitative relationships through the derivation of a rate law
Kinetics can reveal much about the mechanism of a reaction.
16-3
Reaction rate: the central focus of chemical kinetics
Figure 16.1
16-4
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The wide range of reaction rates
Figure 16.2
16-5
Factors Influencing Reaction Rates
Reactant Concentration: molecular collisions are required for
reactions to occur
reaction rate a collision frequency a concentration
Physical State: molecules must mix to collide
When reactants are in different phases, the more finely divided
a solid or liquid reactant, the greater the surface area per unit
volume, the more contact it makes with other reactants,
and the faster the reaction.
Temperature: molecules must collide with sufficient energy to react
Higher T translates into more collisions per unit time and
into higher-energy collisions
reaction rate a collision energy a temperature
16-6
The effect of surface area on reaction rate
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Figure 16.3
16-7
steel nail in O2
steel wool in O2
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Collision energy
and reaction rate
Other factor: reaction
trajectory (not all collisions
with sufficient energy
are productive)
Figure 16.4
16-8
Expressing Reaction Rate
reaction rate = the changes in [reactants] or [products] per unit time;
[reactant] decreases, [product] increases
A
B
Measure [A1] at t1, then measure [A2] at t2
Average rate of reaction = - ([A2] - [A1]) / (t2 - t1)
rate is always
expressed as a
positive value
=
- D[A] / Dt
rate units = mol/L/s
Also, reaction rate = + D[B] / Dt
16-9
Table 16.1
Concentration of O3 at various times in its
reaction with C2H4 at 303 K
C2H4(g) + O3(g)
time (s)
C2H4 O(g) + O2(g)
concentration of O3 (mol/L)
0.0
3.20 x 10-5
10.0
2.42 x 10-5
20.0
1.95 x 10-5
30.0
1.63 x 10-5
40.0
1.40 x 10-5
50.0
1.23 x 10-5
60.0
1.10 x 10-5
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16-10
Different Ways to Measure Reaction Rates
Reaction rate varies with time as the reaction proceeds!
C2H4(g) + O3(g)
C2H4O(g) + O2(g)
average rate = - D[O3] /Dt = - (1.10 x 10-5 mol/L) - (3.20 x 10-5 mol/L)
60.0 s - 0.0 s
= 3.50 x 10-7 mol/L/s
Similar calculations over earlier and later time intervals (e.g., 0.0 -10.0 s,
50.0 - 60.0 s) reveals average rates of 7.80 x 10-7 mol/L/s and
1.30 x 10-7 mol/L/s, respectively.
Data show that the average rate decreases as the
reaction proceeds!
16-11
Concentrations of O3 vs
time during its reaction
with C2H4
Curved line indicates change
in reaction rate with time.
The slope of a tangent at any
point on the curve yields the
instantaneous rate at that point.
reaction rate = (instantaneous)
reaction rate
initial rates: measured to avoid
complications due to back
reactions
Figure 16.5
16-12
Other Plots to Determine
Reaction Rate
C2H4(g) + O3(g)
C2H4O(g) + O2(g)
Plots of [C2H4] and [O2]
vs reaction time
Figure 16.6
16-13
Expressing rate in terms of changes in
[reactant] and [product]
H2(g) + I2(g)
2HI(g)
rate = - D[H2]/Dt = - D[I2]/Dt = 1/2 D[HI]/Dt
or
rate = D[HI]/Dt = -2 D[H2]/Dt = -2 D[I2]/Dt
The mathematical expression for the rate, and the numerical value of
the rate, depend on which substance is chosen as the reference.
For the general reaction: aA + bB
cC + dD
rate = -1/a D[A]/Dt = -1/b D[B]/Dt = 1/c D[C]/Dt = 1/d D[D]/Dt
16-14
Sample Problem 16.1
PROBLEM:
Expressing rate in terms of changes in
concentration with time
Because it generates a nonpolluting product (water vapor),
hydrogen gas is used for fuel aboard the space shuttle and may
be used by automobile engines in the near future.
2H2(g) + O2(g)
2H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) If [O2] decreases at 0.23 mol/L/s, at what rate is [H2O] increasing?
PLAN: Choose [O2] as the reference since its coefficient is 1. For every
molecule of O2 that disappears, two molecules of H2 disappear and two
molecules of H2O appear, so [O2] decreases at one-half the rates of
change in [H2] and [H2O].
SOLUTION:
(a)
(b) 16-15
D[O2]
Dt
rate = -
1
2
= 0.23 mol/L.s
D[H2]
D[O2]
D[H2O]
1
=+
=2
Dt
Dt
Dt
D[H2O]
D[H2O]
=+ 1
;
= 0.46 mol/L.s
2
Dt
Dt
The Rate Law
The rate law expresses the reaction rate as a function of reactant
concentrations, product concentrations, and temperature. It is
experimentally determined.
The derived rate law for a reaction must be consistent
with the postulated chemical mechanism of the reaction!
For a general reaction: aA + bB + .....
cC + dD
rate law: rate = k[A]m[B]n........
k = rate constant
m and n = reaction orders (not related to a, b,...)
(for a unidirectional (one-way) reaction)!
16-16
How is the rate law determined experimentally?
The Procedure
1. Measure initial rates (from concentration measurements)
2. Use initial rates from several experiments to find the reaction orders
3. Calculate the rate constant
16-17
Reaction Order Terminology
rate = k[A]
rate = k[A]2
rate = k[A]0 = k(1) = k
first order overall (rate a [A])
second order overall (rate a [A]2)
zero order (rate independent of [A])
Examples
NO(g) + O3(g)
NO2(g) + O2(g)
rate = k[NO][O3]
second order overall
2NO(g) + 2H2(g)
rate = k[NO]2[H2]1
16-18
N2(g) + 2H2O(g)
third order overall
Properties of Reaction Orders
Reaction orders cannot be deduced from the
balanced chemical equation.
Reaction orders are usually positive integers or zero.
Reaction orders can be fractional or negative.
16-19
Sample Problem 16.2
PROBLEM:
Determining reaction order from rate laws
For each of the following reactions, determine the reaction order
with respect to each reactant and the overall reaction order from
the given rate law.
(a) 2NO(g) + O2(g)
2NO2(g); rate = k[NO]2[O2]
CH4(g) + CO(g); rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq)
I3-(aq) + 2H2O(l); rate = k[H2O2][I-]
(b) CH3CHO(g)
PLAN: Inspect the rate law and not the coefficients in the balanced chemical
reaction.
SOLUTION:
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
(b) The reaction is 3/2-order in CH3CHO and 3/2-order overall.
(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,
and 2nd order overall.
16-20
Determining Reaction Orders
O2(g) + 2NO(g)
2NO2
rate = k[O2]m[NO]n
To find m and n: A series of experiments is
conducted starting with different sets of reactant
concentrations. The initial reaction rate is measured
in each experiment.
16-21
Table 16.2
Initial rates in a series of experiments for the
reaction between O2 and NO
initial reactant
concentrations (mol/L)
experiment
16-22
O2
NO
initial rate
(mol/L.s)
1
1.10 x 10-2
1.30 x 10-2
3.21 x 10-3
2
2.20 x 10-2
1.30 x 10-2
6.40 x 10-3
3
1.10 x 10-2
2.60 x 10-2
12.8 x 10-3
4
3.30 x 10-2
1.30 x 10-2
9.60 x 10-3
5
1.10 x 10-2
3.90 x 10-2
28.8 x 10-3
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Manipulating the Data
rate 2
rate 1
k[O2]2m[NO]2n
=
k[O2]1m[NO]1n
[O2]2m
rate 2
rate 1
=
6.40 x 10-3 mol/L/s
3.21 x 10-3 mol/L/s
[O2]1m
=
([NO] is held constant)
= ([O2]2 / [O1]1)m
(2.20 x 10-2 mol/L/ 1.10 x 10-2 mol/L)m
1.99 = (2.00)m
m=1
Thus, the reaction is first order with respect to O2.
When [O2] doubles, the rate doubles.
16-23
rate 3
rate 1
=
k[O2]3m[NO]3n
k[O2]1m[NO]1n
([O2] is held constant)
via same manipulations as before:
3.99 = (2.00)n
n=2
Thus, the reaction is second order with respect to NO.
When [NO] doubles, the rate quadruples.
The rate law is: rate = k[O2][NO]2
The reaction is third order overall.
16-24
Sample Problem 16.3
PROBLEM:
Determining reaction order from initial rate data
Many gaseous reactions occur in a car engine and exhaust
system. One such reaction is as follows:
NO2(g) + CO(g)
NO(g) + CO2(g)
rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
initial rate (mol/L.s)
experiment
1
initial [NO2] (mol/L)
initial [CO] (mol/L)
0.10
0.40
0.10
2
0.0050
0.080
3
0.0050
0.10
0.20
PLAN:
0.10
Solve for each reactant using the general rate law by applying the
method described previously.
SOLUTION:
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains constant and [NO2] varies.
16-25
Sample Problem 16.3 (continued)
rate 2
=
rate 1
0.080
=
rate 3
=
0.0050
0.0050
k [NO2
]m
=
1
[CO]n
=
1
[NO2] 2
m
[NO2] 1
m
0.40
0.0050
rate 1
k [NO2]m2[CO]n2
;
0.10
16 = 4m and m = 2
k [NO2]m3[CO]n3
k [NO2
]m
0.20
n
1
[CO]n
;
=
1
[CO] 3
The reaction is
2nd order in NO2.
n
[CO] 1
1 = 2n and n = 0
0.10
The reaction is
zero order in CO.
Rate Law: rate = k [NO2]2[CO]0 = k [NO2]2
The reaction is second order overall.
16-26
Determining the Rate Constant
The rate constant, k, is specific for a particular reaction
at a particular temperature.
k = rate 1 / ([O2]1[NO]12)
3.21 x 10-3 mol/L/s
= 1.73 x 103 L2/mol2/s = k
1.86 x 10-6 mol3/L3
The units for k depend on the order of the reaction
and the time unit.
16-27
Table 16.3
Units of the rate constant, k, for several overall
reaction orders
overall reaction order
units of k (t in seconds)
0
mol/L.s (or mol L-1 s-1)
1
1/s (or s-1)
2
L/mol . s (or L mol-1 s-1)
3
L2 / mol2 .s (or L2 mol-2 s-1)
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16-28
Integrated Rate Laws: Change in Concentration
with Reaction Time
Consider a general first order reaction: A
rate = - D[A]/Dt = k[A]
Upon integration over time, we obtain:
ln ([A]0/[A]t) = kt
ln = natural logarithm
[A]0 = concentration of A at t = 0
[A]t = concentration of A at any time t
ln [A]0 - ln [A]t = kt
16-29
B
For a simple second order reaction (one reactant only):
rate = - D[A]/Dt = k[A]2
Upon integration over time, we obtain:
1/[A]t - 1/[A]0 = kt
For a zero order reaction:
rate = - D[A]/Dt = k[A]0
Upon integration over time, we obtain:
[A]t - [A]0 = -kt
16-30
Sample Problem 16.4
Determining reaction concentration at a given time
PROBLEM: At 1000 oC, cyclobutane (C4H8) decomposes in a first-order
reaction with the very high rate constant of 87 s-1 to yield two
molecules of ethylene (C2H4).
(a) If the initial [C4H8] is 2.00 M, what is the concentration after
0.010 s of reaction?
(b) What fraction of C4H8 has decomposed in this time?
PLAN:
Find [C4H8] at time t using the integrated rate law for a 1st-order
reaction. Once that value is found, divide the amount decomposed by
the initial concentration.
SOLUTION:
ln
(a)
[C4H8]0
= kt ; ln
[C4H8]t
2.00 M
[C4H8]t
2.00 M / [C4H8]t = e0.87 = 2.4
(b)
[C4H8]0 - [C4H8]t
[C4H8]0
16-31
= (87 s-1)(0.010 s) = 0.87
[C4H8]t = 0.83 M
2.00 M - 0.83 M
=
2.00 M
= 0.58
Reaction Order from the Integrated Rate Law
Graphical methods to determine reaction orders
(when you don’t have initial rate data!)
Rearranged integrated rate laws:
ln [A]t = -kt + ln [A]0
first order reaction
1/[A]t = kt + 1/[A]0
simple second order reaction
[A]t = -kt + [A]0
zero order reaction
All take the form: y = mx + b
16-32
Graphical method to determine reaction order
[A]t = -kt + [A]0
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
Figure 16.7
16-33
Graphical determination of the reaction order for the
decomposition of N2O5
nonlinear
linear!
nonlinear
Figure 16.8
16-34
Conclusion
For the decomposition of N2O5......
the reaction must be first order in N2O5.
Graphical approach can likewise be applied to reactions
involving multiple reactants.
16-35
Reaction Half-Life
Just another way to express the speed of a reaction
half-life (t1/2) = the time required for the reactant concentration
to reach one-half of its initial value
For a first order reaction at fixed conditions: t1/2 is a constant,
independent of reactant concentration
16-36
A plot of [N2O5] vs time for three half-lives
A first order
reaction
Figure 16.9
16-37
Why is t1/2 independent of reactant concentration
for a first order reaction?
ln ([A]0/[A]t) = kt
After one half-life, t = t1/2 and [A]t = 0.5[A]0. Substituting.....
ln ([A]0/0.5([A]0) = kt1/2
or
ln 2 = kt1/2
t1/2 = (ln 2)/k = 0.693/k
first order reaction: A
16-38
B
Sample Problem 16.5
PROBLEM:
Determining the half-life of a first-order reaction
Cyclopropane is the smallest cyclic hydrocarbon. Because its
60o bond angles allow poor orbital overlap, its bonds are weak.
As a result, it is thermally unstable and rearranges to propene at
1000 oC via the following first-order reaction:
CH2
H2 C
CH2
(g)
D
H3C CH
CH2 (g)
The rate constant is 9.2 s-1. (a) What is the half-life of the reaction? (b) How
long does it take for [cyclopropane] to reach one-quarter of its initial value?
PLAN:
0.693
, to find the half-life.
k
One-quarter of the initial value means two half-lives have passed.
Use the half-life equation, t1/2 =
SOLUTION:
(a)
16-39
t1/2 = 0.693 / 9.2 s-1 = 0.075 s
(b)
2 t1/2 = 2 (0.075 s) = 0.15 s
Half-Life for Zero and Second Order Reactions
t1/2 = 1/k [A]0
simple second order (rate = k[A]2)
As a second order reaction proceeds, the half-life increases.
t1/2 = [A]0/2k
zero order (rate = k)
For zero order reactions, higher initial reactant concentrations
translate into longer half-lives.
16-40
An overview of zero-order, first-order and
simple second-order reactions
Table 16.4
zero order
first order
second order
rate law
rate = k
rate = k[A]
rate = k[A]2
units for k
mol/L.s
1/s
L/mol.s
integrated rate law in
straight-line form
[A]t =
kt + [A]0
ln[A]t =
-kt + ln[A]0
1/[A]t =
kt + 1/[A]0
plot for straight line
[A]t vs t
ln[A]t vs t
1/[A]t = t
slope, y-intercept
-k, [A]0
-k, ln[A]0
k, 1/[A]0
half-life
[A]0/2k
(ln 2)/k
1/k[A]0
16-41
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Temperature and Reaction Rate
Consider the rate law for a first order reaction:
rate = k[A]
Where is the temperature dependence?
Answer: It is embodied in the rate constant, k, that is,
k depends on the temperature at which the
reaction is conducted.
What does the T-dependence of k look like?
test
reaction
R-COOR’ + H2O
ester
R-COOH
acid
organic ester hydrolysis
16-42
+ R’OH
alcohol
Dependence of k on temperature for the
hydrolysis of an organic ester
Note that both
reactant
concentrations are
held constant
k increases
exponentially!
Figure 16.10
16-43
Quantitative treatment of the effect of
temperature on k
The Arrhenius equation
k = the kinetic rate constant at T
k = Ae -Ea/RT
Ea = the activation energy
R = the universal gas constant
ln k = ln A - Ea/R (1/T)
y
= b
+
T = Kelvin temperature
mx
A = collision frequency factor
e = base of natural logarithms
ln
16-44
k2
k1
= -
Ea
R
1
T2
1
T1
Graphical determination of the activation energy, Ea
ln k = -Ea/R (1/T) + ln A
An Arrhenius plot
Figure 16.11
16-45
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More about the Arrhenius Equation
The activation energy = the minimum energy that the molecules
must possess in order for reaction to occur
The negative exponent suggests that, as T increases, the
negative exponent becomes smaller (less negative), the value of k increases,
and thus the reaction rate increases.
higher T
16-46
larger k
increased rate
Sample Problem 16.6
PROBLEM:
Determining the energy of activation
The decomposition reaction of hydrogen iodide,
2HI(g)
H2(g) + I2(g)
has rate constants of 9.51 x 10-9 L/mol.s at 500. K and 1.10
x 10-5 L/mol.s at 600. K. Find Ea.
PLAN:
Use a modification of the Arrhenius equation to find Ea.
SOLUTION:
ln
k2
k1
= -
Ea
1
R
T2
Ea = - (8.314 J/mol . K) ln
-
1
T1
Ea = - R ln
1.10 x 10
-5 L/mol.s
9.51 x 10
-9 L/mol.s
k1
T2
T1
1
-1
1
600. K
Ea = 1.76 x 105 J/mol = 176 kJ/mol
16-47
k2
1
-
-
500. K
1
-1
Information Sequence to Determine the
Kinetic Parameters of a Reaction
series of plots
of [ ] vs time
Determine slope
of tangent at t0 of
each plot
A + B
initial
rates
Compare initial
rates when [A]
changes and [B]
is held constant
and vice versa
C + D
integrated
rate law
(half-life,
t1/2)
Figure 16.12
16-48
reaction
orders
rate constant
(k) and actual
rate law
Substitute initial rates,
orders, and
concentrations into
general rate law: rate = k
[A]m[B]n
rate constant
and reaction
order
Rearrange to
linear form
and graph
Find k at
varied T
activation
energy, Ea
Find k at
varied T
Using Collision Theory to Explain the Effects
of [ ] and T on Reaction Rate
Model: A + B
products
Why are [A] and [B] multiplied in the rate law?
Consider several cases where there are a finite number
of particles each of A and B and determine the
number of possible A-B collisions.
16-49
The dependence of possible collisions on the
product of reactant concentrations
A
A
B
B
Add another
molecule of A
4 collisions
2x2=4
A
B
6 collisions
3x2=6
A
B
A
Add another
molecule of B
Figure 16.13
16-50
A
B
9 collisions
A
B
A
B
3x3=9
Temperature and Reaction Rate
increased T
increased average speed of particles
increased collision frequency
increased reaction rate
But,
most collisions fail to yield products!
Significance of activation energy: only those collisions with
energy equal to, or greater than, Ea can yield products.
Increasing T enhances the fraction of productive collisions, f.
f = e-Ea/RT
From this equation, we can see that both Ea and T
affect f, which in turn influences reaction rate.
16-51
Table 16.5
Effect of Ea and T on the fraction (f) of collisions
with sufficient energy to allow reaction
Ea (kJ/mol)
50
1.70 x 10-9
75
7.03 x 10-14
100
2.90 x 10-18
T
16-52
f (at 298 K)
f (Ea = 50 kJ/mol)
25 oC (298 K)
1.70 x 10-9
35 oC (308 K)
3.29 x 10-9
45 oC (318 K)
6.12 x 10-9
DT = 10o; f
~ doubles; reaction
rate ~ doubles!
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The effect of temperature on the distribution of collision energies
Figure 16.14
16-53
Energy-Level Diagram for an Equilibrium Reaction
A + B
C + D
Ea (forward)
Ea (reverse)
reactants
A + B
Collision Energy
Collision Energy
ACTIVATED STATE
products
C + D
Figure 16.15
16-54
The forward reaction is exothermic because the
reactants have more energy than the products.
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An energy-level diagram of the fraction of
collisions exceeding Ea
A + B
Figure 16.16
16-55
C + D
Effective Collisions
Not all collisions that occur with energy equal to, or exceeding,
the activation energy lead to products.
Molecular orientation is critical!
k = Ae -Ea/RT
the frequency factor = product of collision
frequency Z and an orientation probability
factor p (A = Zp)
16-56
Molecular orientation and effective collisions
NO + NO3
2 NO2
Figure 16.17
16-57
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Transition State Theory
Addresses the limitations of collision theory in explaining
chemical reactivity
Key principle: During the transformation of reactant into
product, one or more very short-lived chemical species
form that resemble (but are different from) reactant or product;
these transitional species contain partial bonds; they are called
transition states (TS) or activated complexes.
The activation energy is used to stretch/deform specific bonds
in the reactant(s) in order to reach the transition state.
Example reaction:
CH3Br + OH-
CH3OH + Br-
What might the TS look like for this substitution reaction?
16-58
The proposed transition state in the reaction
between CH3Br and OH-
The TS is trigonal bipyramidal; note the elongated C-Br
and C-O bonds
Figure 16.18
16-59
Reaction Energy Diagrams
Shows the potential energy of the system during a
chemical reaction
(a plot of potential energy vs reaction progress)
16-60
Reaction energy diagram for the reaction between CH3Br and OH-
Figure 16.19
16-61
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General Principles of TS Theory
Every reaction (and each step in an overall reaction) goes
through its own TS.
All reactions are reversible.
Transition states are identical for the individual forward and reverse
reactions in an equilibrium reaction.
16-62
Sample Problem 16.7
PROBLEM:
Drawing reaction energy diagrams
and transition states
A key reaction in the upper atmosphere is
O3(g) + O(g)
2O2(g)
The Ea(fwd) is 19 kJ, and DHrxn for the reaction is -392 kJ. Draw a reaction
energy diagram for this reaction, postulate a transition state, and calculate
Ea(rev).
Consider relationships between reactants, products and transition state.
The reactants are at a higher energy level than the products, and the
transition state is slightly higher in energy than the reactants.
SOLUTION:
Potential Energy
PLAN:
Ea= 19 kJ
O3+O
transition state
DHrxn = -392 kJ
411 kJ
2O2
16-63
Ea(rev) = (392 + 19) kJ =
Reaction progress
O
breaking
bond
O
O
forming
bond
O
Reaction energy diagrams and possible
transition states for three reactions
endothermic
exothermic
exothermic
Figure 16.20
16-64
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Reaction Mechanisms
The Specific Steps in an Overall Reaction
How a reaction works at the molecular level
2A + B
A + B
C
C + A
D
D
E + F
E + F
possible
steps
C and D are reaction intermediates.
All postulated steps must sum to yield the chemical equation for the overall reaction.
Mechanisms of reactions are proposed and then tested.
16-65
Elementary Reactions and Molecularity
individual steps = elementary reactions (or steps)
Elementary steps are characterized by their molecularity.
molecularity = number of reactant particles involved in the step
2O3(g)
O3(g)
3O2(g)
O2(g) + O(g) unimolecular
O3(g) + O(g)
2O2(g)
bimolecular
Termolecular elementary steps are rare!
The rate law for an elementary reaction can be deduced
from the reaction stoichiometry; equation coefficients are
used as the reaction orders.
16-66
Rate Laws for General Elementary Steps
Table 16.6
elementary step
molecularity
rate law
product
unimolecular
rate = k [A]
2A
product
bimolecular
rate = k [A]2
A+B
product
bimolecular
rate = k [A][B]
2A + B
product
termolecular
rate = k [A]2[B]
A
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16-67
Sample Problem 16.8
PROBLEM:
Determining molecularity and rate laws for
elementary steps
The following two reactions are proposed as elementary steps in
the mechanism of an overall reaction:
(1)
NO2Cl(g)
NO2(g) + Cl (g)
(2)
NO2Cl(g) + Cl(g)
NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(c) Write the rate law for each step.
PLAN: (a) The overall equation is the sum of the steps.
(b) The molecularity is the sum of the reactant particles in the step.
SOLUTION:
NO2Cl(g)
NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) (1)
2NO2Cl(g)
16-68
2NO2(g) + Cl2(g)
(b) Step (1) is unimolecular.
Step (2) is bimolecular.
(c) rate1 = k1 [NO2Cl]
rate2 = k2 [NO2Cl][Cl]
The Rate-Determining Step of a Reaction Mechanism
All of the elementary steps of a mechanism do
not occur at the same rate. Usually one step is
much slower than the others. This step is called the
rate-determining (or rate-limiting) step.
The overall rate of a reaction is related to the rate of
the rate-determining step. That is, the rate law for
the rate-determining step represents the rate law for
the overall reaction!
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An example:
NO2(g) + CO(g)
NO(g) + CO2(g)
rate = k[NO2]m[CO]n
Rate law: rate = k[NO2]2
Step 1: NO2(g) + NO2 (g)
Step 2: NO3(g) + CO(g)
NO3(g) + NO(g)
NO2(g) + CO2(g)
Rate1 = k1[NO2]2
Rate2 = k2[NO3][CO]
Step 1 is slow and rate-determining; Step 2 is fast.
NO3 is an intermediate!
Thus, if k1 = k, the rate law for Step 1 (rate-determining) is the
same as the experimental rate law!
16-70
Correlating the Mechanism with the Rate Law
Chemical mechanisms can never be proven
unequivocally! But potential chemical mechanisms
can be eliminated based on experimental data.
Three Key Criteria for Elementary Steps
1. The elementary steps must add up to the overall equation.
2. The elementary steps must be physically reasonable.
3. The mechanism must correlate with the rate law.
16-71
Reaction Energy Diagram for the two-step NO2-F2 reaction
2NO2(g) + F2(g)
2NO2F(g)
rate = k[NO2][F2]
Two transition states and
one intermediate are involved.
The first step is
rate-determining.
The reaction is exothermic
(thermodynamics).
Figure 16.21
16-72
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What happens when a fast reversible step(s) occurs prior to the
rate-determining step?
2NO(g) + O2(g)
2NO2 (g)
Rate law: rate = k[NO]2[O2]
Proposed Mechanism
Step 1: NO(g) + O2(g)
Step 2: NO3(g) + NO(g)
NO3(g)
2NO2(g)
fast and reversible
slow; rate-determining
Rate laws for the two elementary steps:
Step 1: rate1 (fwd) = k1[NO][O2]; rate1 (rev) = k-1[NO3]
Step 2: rate2 = k2[NO3][NO]
Step 2 (rate-determining) contains [NO3] which does
not appear in the experimental rate law!
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If Step 1 is at equilibrium, then....
rate1 (fwd) = rate1 (rev)
or
k1[NO][O2] = k-1[NO3]
Solving for [NO3]:
[NO3] = (k1/k-1)[NO][O2]
Substituting into the rate law for rate-limiting Step 2:
rate2 = k2[NO3][NO] = k2 (k1/k-1)[NO][O2][NO] = [(k1k2)/k-1][NO]2[O2]
Conclusion: The proposed mechanism is consistent
with the kinetic data.
16-74
Catalysis: Enhancing Reaction Rates
Catalyst: increases reaction rate but is not consumed in the reaction
A catalyst increases reaction rate (via increasing k) by lowering the
activation energy (barrier) of the reaction.
Both forward and reverse reactions are catalyzed; reaction
thermodynamics is unaffected!
The catalyzed reaction proceeds via a different mechanism
than the uncatalyzed reaction.
16-75
Energy diagram of an uncatalyzed and catalyzed reaction
Note that both reactions
exhibit the same
thermodynamics!
The catalyzed
and
uncatalyzed
reactions occur
via different
pathways.
Figure 16.22
16-76
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Uncatalyzed reaction:
A + B
product
Catalyzed reaction:
A + catalyst
C + B
C
product + catalyst
Types of Catalysts
Homogeneous: exists in solution with the reaction mixture
 Heterogeneous: catalyst and reaction mixture are in different phases

16-77
16-78
Homogeneous Catalysis
The Mechanism of Acid-catalyzed
Organic Ester Hydrolysis
R-COOR’ + H2O + H+
ester
R-COOH
acid
+ R’OH + H+
alcohol
The reaction rate is slow at neutral pH (pH 7.0), but
increases significantly at acidic pH (pH 2).
H+ ion catalyzes the reaction at low pH; what is the
molecular basis for the catalysis by H+?
16-79
Proposed reaction mechanism for the H+-catalyzed
hydrolysis of an organic ester
O
H
+
+
H O
fast
R C
R C
O
O
R'
R'
H O
resonance forms
H O
R C
R C
O
H O
R'
R C
O
R'
O
R'
resonance hybrid
H O
R C
O
H O
O H
H
R'
Figure 16.23
16-80
R C
slow, ratedetermining
step
O
O H
+
H
H+ O
all fast
R C
R'
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OH
R'
O H
Heterogeneous Catalysis: Metal-catalyzed
hydrogenation of ethylene
Ni, Pd or Pt
16-81
H2C CH2 (g) + H2 (g)
Figure 16.24
H3C CH3 (g)
The widely
separatedam
ino acid
groups that
form the
active site of
an enzyme
Figure B16.4
16-82
The enzyme, chymotrypsin
A protease (hydrolyzes proteins)
Two Models of Enzyme Action
lock and key
induced fit
Figure B16.5
16-83
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Tools of the Laboratory
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Spectrophotometric monitoring of a reaction
16-84
Figure B16.1
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Tools of the Laboratory
Figure B16.2
Conductometric monitoring
of a reaction
Figure B16.3
Manometric monitoring
of a reaction
16-85