Chapter 13 The Rates of Reactions
St. John’s wort（貫葉連翹 /金絲
桃/聖約翰草）is an herb that is
thought to create a sense of
tranquility. Herbs and other
medicines have been used
through the ages to cure
disease and to relieve pain.
In many cases, the
medicine is effective
because it controls the
rates of reactions within
the body. In this chapter,
we examine the rates of
chemical reactions and the
mechanisms by which they
take place.
Assignment for Chapter 13
13.22; 13.31; 13.37; 13.61
How to Live Longer and More Healthy?
Demonstrated tips:
Reduction of food
Cold body temperature
Appropriate exercise
Good temper
All related to (slower) chemical reaction rates.
Definition
kAmC+nD
Rate=change in concentration of reactant / time interval
2HI(g) H2(g)+I2(g)
R
[ HI ]
t

[ I 2 ]
t
R
[ HI ]
t
[ HI ]
t
-1 -1
10 51
  900
mmol
L
s
100
[ HI ]
t


17 100
500
6  26
1000 300
-1
mmol L s
-1
mmol L s
-1
-1
Figure 13.1 The activity of penicillin declines over several
weeks when it is stored at room temperature in the absence of
stabilizers. The shape of this graph of concentration of
penicillin as a function of time is typical of the behavior of
chemical reactions, although the time span may vary from
fractions of a second to years.
Figure 13.2 This graph shows two examples of how the rate of
consumption of penicillin can be monitored while it is being stored.
The red line shows the average rate calculated from measurements at
0 and 10 weeks, and the blue line shows the average rate calculated
from measurements at 2.5 and 7.5 weeks. The instantaneous rate at 5
weeks is the tangent to the curve at that time (not shown).
R
[penicillin ]
t
 0.074
  0.0210
mol L-1 week -1
Figure 13.3 To calculate the instantaneous reaction rate, we draw the
tangent to the curve at the time of interest and then calculate the
slope of this tangent. To calculate the slope, we identify any two
points, A and B, on the straight line and identify the molar
concentrations and times to which they correspond. The slope is
then worked out by dividing the difference in concentrations by the
difference in times. Notice that this graph shows the concentration of
a product.
R
dC
dt
Figure 13.4 The initial rate of reaction for the decomposition
of N2O5 in five experiments. The initial rate of disappearance of
a reactant is determined by drawing a tangent to the curve at
the start of the reaction.
2N2O5(g)4NO2(g)+O2(g)
Rinitial 
dC
dt t 0
|
Figure 13.5 The plot of the initial rate of decomposition of N2O5 as a
function of initial concentration for the five samples in Fig. 13.4 is a
straight line. The linear plot shows that the rate is proportional to the
concentration. The graph also illustrates how we calculate the rate
constant, k , from the slope of the straight line.
Rinitial 
dC
dt t 0
|
Initial rate=k × initial concentration
Figure 13.6 (a) The instantaneous reaction rates for the
decomposition of N2O5 at five different times during a single
experiment are obtained from the slopes of the tangents to the line at
each of the five points. (b) When these rates (the slopes) are plotted
as a function of the concentration of N2O5 remaining, the result is a
straight line with a slope equal to the rate constant. In (b), we have
indicated the rates by redrawing the tangents.
R  k  (concentration )
a
a: order of reaction, determined by experiment.
Figure 13.7 (a) The concentration of the reactant in a zeroorder reaction falls at a constant rate until the reactant is
exhausted. (b) The rate of a zero-order reaction is independent
of the concentration of the reactant and remains constant until
all the reactant has been consumed, when it falls abruptly to 0.
R  k  (concentration)
a 0
More General Cases:
b
a

R k [ A ] [ B ] ...
There are no images
in this section
 ofathe+chapter.
Overall
order
b + ...

3

+
+
+
BrO ( aq ) 5 Br ( aq ) 6 H ( aq )

 3 Br 2 ( aq ) + 3 H 2 O ( l )
 [ BrO 3 ]
t
 k [ BrO 3 ][ Br  ][ H + ] 2
Example
Find the rate law for the disappearance of bromate ions in the reaction:
BrO3 (aq)+5Br (aq)+6H (aq)  3Br2 (aq)+3H 2O(l)
-
-
+
With respect to each reactant and determine k given the following data
Initial concentration, mol/L
Experiment
BrO3-
Br-
H+
Initial rate
(mol BrO3-)/L/sec
1
0.10
0.10
0.10 1.2×10-3
2
0.20
0.10
0.10 2.4×10-3
3
0.10
0.30
0.10 3.5×10-3
4
0.20
0.10
0.15 5.4×10-3
Example
Initial concentration, mol/L
Experiment
BrO3-
Br-
H+
Initial rate
(mol BrO3-)/L/sec
1
0.10
0.10
0.10 1.2×10-3
2
0.20
0.10
0.10 2.4×10-3
3
0.10
0.30
0.10 3.5×10-3
4
0.20
0.10
0.15 5.4×10-3
R  k[BrO3 ][Br ][H ]
-
-
+ 2
5.4 103 mol/L/s=k  0.20 mol/L  0.10 mol/L  (0.15 mol/L)2
k
5.4103 mol/L/s
0.20 mol/L 0.10 mol/L(0.15 mol/L)2
 12 L/mol/s
More Complicated Rate Laws
2O3 (g)  3O2 (g)
Rate = k
[O3 ]2
[O2 ]
1
 k[O3 ] [O2 ]
2
2SO2 (g)+O2 (g) 
 2SO3 (g)
Pt
 12
Rate = k [SO ]1/ 2  k[SO2 ][SO3 ]
[SO2 ]
3
Rk
[ A ]a [ B ]b ...
p
[ P ] ...
Order=a+b+…-p-…
Figure 13.8 (a) When sulfur dioxide and oxygen are passed over hot
platinum foil, they combine to form sulfur trioxide. (b) The sulfur
trioxide forms dense white fumes of sulfuric acid when it comes into
contact with moisture in the atmosphere. The rate law for the
formation of sulfur trioxide shows that its rate of formation decreases
as its concentration increases, so the sulfur trioxide must be
removed as it is formed if the reaction is to proceed rapidly.
2SO2(g)+O2(g)2SO3(g)
Rk
[SO 2 ]a=1 [O 2 ]b=0
[SO3
p= 1
] 2
Classroom Exercise
Find the rate law for the disappearance of persulfate ions in the reaction:
S2 O8 (aq)+3I (aq)  2SO 4 (aq)+I3 (aq)
2-
-
2-
-
With respect to each reactant and determine k given the following data
Initial concentration, mol/L
Experiment
S2O82-
I-
Initial rate
(mol S2O82-)/L/sec
1
0.15
0.21
1.14
2
0.22
0.21
1.70
3
0.22
0.12
0.98
Classroom Exercise
Initial concentration, mol/L
Experiment
S2O82-
I-
Initial rate
(mol S2O82-)/L/sec
1
0.15
0.21
1.14
2
0.22
0.21
1.70
3
0.22
0.12
0.98
R  k[S2 O8 2- ][I- ]
0.98 mol/L/s=k  0.22 mol/L  0.12 mol/L
k
0.98 mol/L/s
0.22 mol/L 0.12 mol/L
 36 L/mol/s
Integrated Rate Law
• Find the concentration of a reactant at time t
from reaction order and initial
concentration.
• First order reaction:

d [ A]
dt
 k[ A] 
[ A]t  [ A]0 e
 kt
Figure 13.9 The characteristic shape of the graph showing the
time dependence of the concentration of a reactant in a firstorder reaction is an exponential decay, as shown here. The
larger the rate constant, the faster the decay from the same
initial concentration.
[ A]t  [ A]0 e
 kt
Example
2N2O5(g)4NO2(g)+O2(g) (65 oC)
Rate =k[N2O5], k=5.2×10-3 /s
Initial concentration of N2O5 was 0.04 mol/L. What is its
concentration after 10 minutes?
[N 2 O5 ]t 600 s  [N 2O5 ]t 0 e
 (0.04 mol/L)e
 1.8 10
3
0.0052600
mol/L
 kt
How to find k (first order reaction):
[ A]t  [ A]0 e
 kt
ln[ A]t  ln[ A]0  kt
Example
Time, min
0
5.0
10.0
15.0
Concentration of cyclopropane, mol/L
1.5×10-3
1.24×10-3
1.00×10-3
0.83×10-3
Example
Time, min
0
5.0
10.0
15.0
ln[cyclopropane]
-6.5
-6.69
-6.91
-7.09
ln[ A]t  ln[ A]0  kt
Figure 13.10 We test for a first-order reaction by plotting the
natural logarithm of the reactant concentration as a function of
time. The slope of the line, which is calculated as indicated in
the illustration by selecting two points A and B, is equal to the
negative of the rate constant.
Time, min
0
5.0
10.0
15.0
slope 
ln[cyclopropane]
-6.5
-6.69
-6.91
-7.09
7.02 ( 6.56)
(13.3-1.7) min
 0.040 /min
k  0.040 /min=6.7 10-4 s-1
Classroom Exercise
2N2O5(g)4NO2(g)+O2(g)
Time, min
0
200.0
400.0
600.0
800.0
1000.0
(25oC)
Concentration of N2O5, mol/L
1.5×10-2
9.6×10-3
6.2×10-3
4.0×10-3
2.5×10-3
1.6×10-3
Find the rate constant k at this temperature.
Classroom Exercise
Time, min
0
200.0
400.0
600.0
800.0
1000.0
ln [N2O5]
ln(1.5×10-2)=-6.5022
ln(9.6×10-3)=-6.9485
ln(6.2×10-3)=-7.3858
ln(4.0×10-3)=-7.8240
ln(2.5×10-3)=-8.2940
ln(1.6×10-3)=-8.7403
-6.5
ln[N2O5]
-7.0
-7.5
-8.0
Slope = - 2.2×10-3 /min
k=2.2×10-3 /min
-8.5
-9.0
200 400 600 800 1000
Time (min)
Half-Lives of First Order
Reactions
)  kt
ln(
[ A ]0
[ A ]t
ln(
[ A ]0
1 [ A]
0
2
t1/ 2 
)  kt1/ 2 
ln 2
k

0.693
k
Figure 13.11 The half-life of a reactant is short if the first-order
rate constant is large, because the exponential decay of the
concentration of the reactant is then faster.
Figure 13.12 The half-life of a substance is the time needed for it to
fall to one-half its initial concentration. For first-order reactions, the
half-life is the same whatever the concentration at the start of the
chosen period. Therefore, it takes one half-life for the concentration
to fall to half the initial value, two half-lives for the concentration to
fall to one-fourth the initial value, three half-lives to fall to one-eighth,
and so on.
t1/ 2 
ln 2
k

0.693
k
Example
A pollutant escapes into a local picnic site. Studies
show that the pollutant decays by a first order
reaction with rate constant 3.8×10-3/h. Calculate
the time needed for the concentration to fall to (a)
one half and (b) one-fourth of its initial value.
t1/ 2 
ln 2
k

ln 2
3
3.810 / h
 180 h
t1/ 4  2t1/ 2  360 h = 15 days
Classroom Exercise
A pollutant escapes into a local picnic site. Studies
show that the pollutant decays by a first order
reaction with rate constant 3.8×10-3/h. Calculate
the time needed for the concentration to fall to
one-eigth of its initial value.
t1/ 2 
ln 2
k

ln 2
3
3.810 / h
 180 h
t1/8  3t1/ 2  540 h = 22.5 days
Second Order Integrated Rate Laws
2AA+A
R  k [ A]
2

d [ A ]t
dt
1
[ A ]t
 k[ A]t  
2
 kt + C , [ A1]0  C
[ A]t 
[ A ]0
1+ k [ A ]0 t
d [ A ]t
[ A ]t
2
 kdt
Figure 13.13 The characteristic shapes (orange and green lines) of
the time dependence of the concentration of a reactant during two
second-order reactions. The gray lines are the curves for first-order
reactions with the same initial rates. Note how the concentrations for
second-order reactions fall away much less rapidly than those for
first-order reactions do.
1
[ A]t
 kt +
1
[ A]0
Figure 13.14 A summary of the plots that give
straight lines for first- and second-order reactions.
Controlling Reaction Rates
Surface Area
Temperature
Pressure
Catalyst
Activated Complex Theory
Figure 13.15 A solid iron rod can be heated in a flame without
catching fire. However, a powder of very finely divided iron
oxidizes rapidly in air to form Fe2O3, because the powder
presents a much greater surface area for reaction with oxygen.
Figure 13.16 The Hindenburg, a hydrogen-filled airship,
caught fire as it came in to land at Lakehurst, New Jersey in
1937 after its inaugural flight across the Atlantic. The
hydrogen was ignited when the airship collided with an
electrical tower.
Temperature Dependence of
Reaction Rates
2N2O5(g)4NO2(g)+O2(g)
(25oC)
k=2.2×10-3 /s
2N2O5(g)4NO2(g)+O2(g)
(65oC)
k=5.2×10-3 /s
k  Ae
Ea
 RT
Preexponent factor
Activation energy
Figure 13.17 An Arrhenius plot is a graph of ln k against
1/T. If, as here, the line is straight, then the reaction is
said to show Arrhenius behavior in the temperature
range studied. The activation energy for the reaction is
obtained by setting the slope of the line equal to Ea/R.
SvanteA Arrhenius
(1859-1927)
k  Ae
Ea
 RT

ln k  ln A 
Ea
RT
Measuring Activation Energy
C2H5Br(aq)+OH-(aq)C2H5OH(aq)+Br-(aq)
Temperature (oC)
k (L/mol/s)
25
8.8×10-5
30
1.6×10-4
35
2.8×10-4
40
5.0×10-4
45
8.5×10-4
50
1.4×10-3
Find Ea.
ln k  ln A 
Ea
RT
Measuring Activation Energy
Temperature (oC)
T
1/T
25
298
3.35 ×10-3
8.8×10-5
-9.34
30
303
3.30 ×10-3
1.6×10-4
-8.74
35
308
3.25 ×10-3
2.8×10-4
-8.18
40
313
3.19 ×10-3
5.0×10-4
-7.60
45
318
3.14 ×10-3
8.5×10-4
-7.07
50
323
3.10 ×10-3
1.4×10-3
-6.57
k (L/mol/s)
ln k  ln A 
Ea
RT
lnk
Figure 13.18 An Arrhenius plot of the data in Example 13.7.
The slope of the line has been worked out by using points A
and B.
Temperature (oC)
T
1/T
25
298
3.35 ×10-3
8.8×10-5
-9.34
30
303
3.30 ×10-3
1.6×10-4
-8.74
35
308
3.25 ×10-3
2.8×10-4
-8.18
40
313
3.19 ×10-3
5.0×10-4
-7.60
45
318
3.14 ×10-3
8.5×10-4
-7.07
50
323
3.10 ×10-3
1.4×10-3
-6.57
k (L/mol/s)
ln k  ln A 
Slope 
Ea
R
lnk
Ea
RT
3
3.22
  0.303.22



10
/K
0.3
103 / K
3
Ea   R  Slope  8.3145 J / K/mol  ( 3.22

10
/ K)
0.3
=8.9 104 J/mol
Classroom Exercise
OH (g)+H2 (g)H2O (g)+H (g)
Temperature (oC)
k (L/mol/s)
100
1.1×10-9
200
1.8×10-8
300
1.2×10-7
400
4.4×10-7
Find Ea.
ln k  ln A 
Ea
RT
Classroom Exercise
Temperature (oC)
T
1/T
100
373
2.51 ×10-3
1.1×10-9
-19.63
200
473
2.01×10-3
1.8×10-8
-17.83
300
573
1.67 ×10-3
1.2×10-7
-15.94
400
673
1.43 ×10-3
4.4×10-7
-14.64
k (L/mol/s)
ln k  ln A 
-10
lnk
-12
Slope 
-14
Ea
R
lnk
Ea
RT
3
4.99
  1.084.99



10
/K
3
1.08
10 / K
4.99
Ea   R  Slope  8.3145 J / K/mol  ( 1.08
103 / K)
-16
=4.1104 J/mol
-18
-20
1.0 1.5 2.0
2.5 3.0
1/T (K-1) ×103
Figure 13.19 The variation of the rate constant with temperature for
different values of the activation energy. Note that the higher the
activation energy, the more strongly the rate constant varies with
temperature.
ln k  ln A 
Ea
RT
ln k '  ln A 
Ea
RT '
Ea
Ea
ln k  ln k '  ln A  RT
 (ln A  RT
')

Ea
RT
+
Ea
RT '
ln 
k
k'
Ea
R
(  )
1
T'
1
T
Example
The rate constant of sucrose hydrolysis (sucrose is
broken down into glucose and fructose) at body
temperature is 1.0×10-3 L/mol/s. The activation
energy of the reaction is 108 kJ/mol. What would
it be at 35 oC?

Ea
R
( T1'  T1 )
ln
k
k'

1.08105 J/mol
8.3145 J/K/mol

k'
k
(
1
308 K

1
310 K
)  0.27
 0.76  k '  7.6 10
4
L/mol/s
Classroom Exercise
The rate constant of the first order isomerization of
cyclopropane to propene is 6.7×10-4 /s at 500 oC. The
activation energy of the reaction is 272 kJ/mol. What
would it be at 300 oC?

Ea
R
( T1'  T1 )
ln
k
k'

2.72105 J/mol
8.3145 J/K/mol

k'
k
(
1
573 K

7
1
773 K
)  14.9
 3.84 10  k '  2.6 10
10
/s
Figure 13.20 (a) In the collision theory of chemical reactions,
reaction can occur only when two molecules collide with a
kinetic energy at least equal to the activation energy of the
reaction. (b) Otherwise they simply bounce apart.
Figure 13.21 A reaction profile for an exothermic reaction. In the
collision theory of reaction rates, the potential energy (the energy
due to position) increases as the reactant molecules approach each
other, reaches a maximum as the molecules distort, and then
decreases as the atoms rearrange into the bonding pattern
characteristic of the products and these products separate. Only
molecules with sufficient energy can cross the barrier and react to
form products.
Figure 13.22 The fraction of molecules that collide with a
kinetic energy that is at least equal to the activation energy, Ea,
is denoted by the shaded areas under each curve. The fraction
increases rapidly as the temperature is raised.
k  Ae
 E a / RT
ln k  ln A 
Ea
RT
Figure 13.23 Whether or not a reaction occurs when two species
collide in the gas phase depends on their relative orientation as well
as on their energies. In the reaction between a Cl atom and an HI
molecule, for example, only those collisions in which the Cl atom
approaches the HI molecule along a path that lies inside a cone
centered on the H atom lead to reaction, even though the energies of
collisions in other orientations may exceed the activation energy.
Figure 13.24 This sequence of images illustrates the motion of two
reactant molecules (red and green) in solution. The blue spheres
represent solvent molecules. We see the reactants drifting together,
lingering near each other for some time, and then drifting apart again.
Reaction may occur during the relatively long period of encounter. To
highlight the positions of the reactant molecules, the insets show the
solvent molecules as a solid blue background.
Activated Complex Theory
Figure 13.25 We join this sequence of images at the moment when
the reactant molecules are in the middle of their encounter. They may
acquire enough energy by impacts from the solvent molecules to
form an activated complex, which may go on to form products. Once
again, the insets highlight the reactants by showing the solvent
molecules as a solid blue background.
Activated Complex
Figure 13.26 The reaction profile for the activated complex
theory of reactions in solution, in which the reactants form an
activated complex, provided they encounter each other with at
least the activation energy.
Catalysis
•
•
•
•
•
Catalyst
Homogeneous catalysis
Heterogeneous catalysis
Poisoning of catalyst
Enzyme (living catalyst)
Figure 13.27 A catalyst provides a new reaction pathway with
a lower activation energy, thereby allowing more reactant
molecules to cross the barrier and form products. Notice that
Ea for the reverse reaction is also lowered on the catalyzed
path.
Figure 13.28 A small amount of catalyst—in this case, potassium
iodide in aqueous solution—can accelerate the decomposition of
hydrogen peroxide to water and oxygen. This effect is shown (a) by
the slow inflation of the balloon when no catalyst is present and (b)
by its rapid inflation when a catalyst is present.
2H 2 O 2  2H 2 O+O 2
KI
Homogeneous catalysis
Figure 13.29 The reaction between ethene, CH2CH2, and hydrogen
on a catalytic metal surface. In this sequence of images, we see the
ethene molecule approaching the metal surface to which hydrogen
molecules have already adsorbed: when they adsorb, they dissociate
and stick to the surface as hydrogen atoms. After sticking to the
surface, the ethene molecule meets a hydrogen atom and forms a
bond. At this stage (center), a ·CH2CH3 radical is attached to the
surface by one of its carbon atoms. Finally, the radical and another
hydrogen atom meet, ethane is formed, and the product escapes
from the surface.
CH2CH2 ·CH2CH3+H CH3CH3
Figure 13.30 The internal structure of a zeolite is like a
honeycomb of passages and cavities. As a result, a zeolite
presents a huge surface area. It can also permit the entry and
exit of molecules of a certain size into the active regions
within the holes. This zeolite is ZSM-5.
Figure 13.31 The structure of a typical catalytic converter for
an automobile exhaust. The gases flow through a
honeycomblike porous ceramic support covered with catalyst.
Figure 13.32 The lysozyme （溶菌酶）molecule shown here is a
typical enzyme molecule. Lysozyme occurs in a number of places,
including tears and the mucus in the nose. One of its functions is
to attack the cell walls of bacteria and destroy them. In this
illustration the winding ribbon represents the long chain that
makes up the molecule.
Figure 13.33 In the lock-and-key model of enzyme action, the
correct substrate is recognized by its ability to fit into the
active site like a key into a lock. In a refinement of this model,
the enzyme changes its shape slightly as the key enters.
Figure 13.34 (a) An enzyme poison (represented by the mottled green
rectangle) can act by attaching so strongly to the active site that it blocks
the site, thereby taking the enzyme out of action. (b) Alternatively, the
poison molecule may attach elsewhere, so distorting the enzyme
molecule and its active site that the substrate no longer fits.
Case Study 13 (a) A plot of Michaelis-Menten enzyme kinetics (1913). At
low substrate concentrations, the rate of reaction is directly proportional
to substrate concentration. However, at high substrate concentrations, the
rate is constant, as the enzyme molecules are “saturated” with substrate.
V
Look at the shape of the graph. At low [S],
it is the availability of substrate that is the
limiting factor. Therefore as more substrate
is added there is a rapid increase in the initial
rate of the reaction - any substrate is rapidly
mopped up and converted to product. At the
KM, 50% of active sites have substrate bound.
At higher [S] a point is reached (at least theoretically) where all of the enzyme has substrate
bound and is working flat out. Adding more
substrate will not increase the rate of the
reaction, hence the leveling out observed in
the graph.
Vmax
0.5Vmax
KM
1875-1949
Leonor Michaelis, Maud Menten 1879–1960
[S]
Case Study 13 (b) A diagram of a synapse. The triangles represent
neurotransmitters that travel from the neuron on the left to the
receptors in the neuron on the right. The concentration of
neurotransmitters in the synapse is controlled by enzymes.
Dopamine
（多巴胺）
NH2
OH
OH
Case Study 13 (c) The neurons in the human brain affect how
we think and feel and how we perceive reality, including
chemistry books.
Reaction Mechanisms
• Elementary Reaction
• Molecularity (Unimolecular, Bimolecular
Termolecular Reactions)
• Reaction Intermediate
• Rate Law
• Rate Determining Step (RDS)
• Chain Reaction
I2+H2  2HI
Figure 13.35 The chemical
equations for elementary reactions
show the individual events that
take place when atoms and
molecules encounter one another.
This illustration shows two of the
steps believed to occur during the
formation of hydrogen iodide from
hydrogen and iodine vapor. In one,
I2 + I2  I2 + I + I, a collision
between two iodine molecules
results in the dissociation of one of
them. In the second step, I + H2 
H + HI, one of the I atoms produced
in the first step attacks a hydrogen
molecule and forms a hydrogen
iodide molecule and a hydrogen
atom.
I2 + I2  I2 + I + I,
I + H2  H + HI
I2+H2  2HI
Decomposition of ozone:
Step 1: O3  O 2 + O
unimolecular reaction.
Step 2: O + O3  O 2 + O 2
bimolecular reaction.
Overall: 2O3  3O 2 or
O3 + O3  O 2 + O 2 + O 2 whose reverse is
O 2 + O 2 + O 2  O3 + O3 termolecular reaction.
• Many reactions occur by a series of elementary
reactions. The molecularity of an elementary
reaction indicates how many species are involved
in that step.
The Rate Laws of Elementary Reactions
• A unimolecular reaction depends on a single energized species
shaking itself apart and always has a first order rate law:
Xproducts Rate of disappearance of X =k[X].
• A bimolecular reaction depends on collisions between two species,
so its rate law is second order:
X+Xproducts Rate of disappearance of X =k[X]2.
X+Yproducts Rate of disappearance of X =k[X][Y].
• From elementary rate laws to construct overall rate law can be hard,
simplifications (e.g., rate determining step) are normally evoked.
Rate Determining Step(RDS)
The elementary reaction that is much
slower than the rest and governs the
overall reaction.
Figure 13.36 The rate of a reaction is controlled by the rate-determining step
(RDS). (a) If the rate-determining step is the second step, then the rate law for
that step determines the rate law for the overall reaction. The orange curve
shows the reaction profile for such a mechanism, with a high activation
energy for the slow step. The concentrations of intermediates can usually be
expressed in terms of reactants and products by taking into account the steps
preceding the RDS. (b) If the rate-determining step is the first step, then the
rate law for that step must match the rate law for the overall reaction. Later
steps do not affect the rate or the rate law. (c) If two parallel paths lead to
products, the faster one (in this case, the lower one) determines the rate of the
reaction.
Example
H 2 (g) + 2ICl(g)  2HCl(g) + I 2 (g)
Proposed mechanism:
H 2 (g) + ICl(g)  HI(g) + HCl(g) fast
HI(g) + ICl(g)  HCl(g) + I 2 (g)
Find the rate law.
slow
From Mechanism to Overall Rate Law
Step 1 : forward. H 2 (g) + ICl(g)  HI(g) + HCl(g) fast  Rate  k1[ICl][ H 2 ]
Step 1 : reverse. HI(g) + HCl(g)  ICl(g) + H 2 (g) (fast)  Rate  k '1[HI][HCl]
Step 2 : HI(g) + ICl(g)  HCl(g) + I 2 (g) (slow)  Rate  k 2 [HI][ICl]
Solution :
We may assume that in the fast stage, the forward and reverse reaction rates are the same :
k1[ICl][ H 2 ]  k '1[HI][HCl]  [HI] 
k1 [ICl][H 2 ]
k '1 [HCl]
The overall rate law then is
Rate  k 2 [HI][ICl] 
k1k 2 [ICl]2 [H 2 ]
k '1 [HCl]
2
[H 2 ]
 koverall [ICl]
[HCl] .
Classroom Exercise
NO 2 (g)+F2 (g)  F(g)+NO 2 F
Proposed mechanism:
NO 2 (g) + F2 (g)  NO 2 F + F (g) slow
F+NO 2 (g)  NO 2 F
fast
Find the rate law and the relation between rate constant k for
the overall reaction and the rate constant for the elementary reactions.
Step 1 is RDS  Rate law:
k1[NO2 ][F2 ].
Overall rate constant k =k1.
From Overall Rate Law to Reaction Mechanism
NO 2 (g)+CO(g)  NO(g)+CO 2 (g)
Rate of disappearance of NO 2  k[NO 2 ] (Exp.)
2
Proposed mechanisms:
A: NO 2 +CO  NO +CO 2
B: Step 1) NO 2 +NO 2  NO3 +NO
Step 2) NO3 +CO  NO 2 +CO 2
Overall: NO 2 +CO  NO +CO 2
Which is correct?
From Overall Rate Law to Reaction Mechanism
NO2 (g)+CO(g)  NO(g)+CO2 (g)
Rate of disappearance of NO 2  k[NO 2 ]2 (Exp.)
Proposed mechanisms:
A: NO2 +CO  NO +CO2
Rate of disappearance of NO 2  k[NO 2 ][CO]
From Overall Rate Law to Reaction Mechanism
NO 2 (g)+CO(g)  NO(g)+CO 2 (g)
Rate of disappearance of NO 2  k[ NO 2 ] (Exp.)
2
Proposed mechanism:
B: Step 1) NO 2 +NO 2  NO3 +NO slow
Step 2) NO3 +CO  NO 2 +CO 2 fast
Overall: NO 2 +CO  NO +CO 2

Rate of disappearance of NO 2  k[NO 2 ]2
From Overall Rate Law to Reaction Mechanism
NO (g)+Br2 (g)  2NOBr(g)
Rate law  k[NO]2 [Br2 ] (Exp.)
Proposed mechanism:
Step 1) forward: NO+Br2  NOBr2
Reverse: NOBr2  NO+Br2
Rate =k1[NO][Br2 ]
Rate =k1'[NOBr2 ]
Step 2) NOBr2 + NO  NOBr+NOBr Rate =k2 [NO][NOBr2 ]
Which step is RDS?
Suppose step 1 forward is RDS
Rate law: k1[Br2][NO], disagreement with experimental result.
Suppose step 2 is RDS
Rate law: k2[NOBr2][NO], again, disagreement with experimental
result.
From Overall Rate Law to Reaction Mechanism
NO (g)+Br2 (g)  2NOBr(g)
Rate law  k[NO]2 [Br2 ] (Exp.)
Proposed mechanism:
Step 1) forward: NO+Br2  NOBr2
Reverse: NOBr2  NO+Br2
Step 2) NOBr2 + NO  NOBr+NOBr
Rate =k1[NO][Br2 ]
Rate =k1'[NOBr2 ]
Rate =k2 [NO][NOBr2 ]
Which step is RDS?
Suppose step 1 forward and reverse are much faster than step 2.
k1[Br2][NO] = k1’[NOBr2]
Suppose step 2 is RDS
Rate law: k2[NOBr2][NO]= (k2k1/k1’)[NO]2[Br2]
Rate law  k[NO]2 [Br2 ], k 
k2 k1
k1 '
Case Study 13 (a) A plot of Michaelis-Menten enzyme kinetics. At low
substrate concentrations, the rate of reaction is directly proportional to
substrate concentration. However, at high substrate concentrations, the
rate is constant, as the enzyme molecules are “saturated” with substrate.
V
E+S
ES
RDS E+P
Vmax
0.5Vmax
KM
[S]
Derivation of Michaelis-Menten Equation
V
The overall rate of the reaction (v) is limited by the step ES to E + P,
and this will depend on two factors - the rate of that step (i.e. k2) and
the concentration of enzyme that has substrate bound, i.e. [ES]. This
can be written as:
The first step is fast and the forward and reverse reactions keep
equilibrium:
Vmax
(
0.5Vmax
where
The total amount of enzyme is constant:
or
KM
[S]
Figure 13.37 In a chain reaction, the product of one reaction
step is a reactant in a subsequent step, which in turn
produces species that can take part in subsequent reaction
steps.
Initiation
Propagation
Example (I)
Initiation :
Br 2    Br  + Br 
heat or radiation
Propagatio n : Br  + H 2  HBr + H 
H  + Br2  HBr + Br 
Terminatio n : Br  + Br   Br2
Figure 13.38 This flame front was caught during the rapid
combustion that occurs inside an internal combustion engine
every time a spark plug ignites gasoline vapor. This radical
chain reaction occurs in automobile engines. The products,
which are hot gases, push a piston out, initiating a chain of
events that ultimately moves the vehicle.
Example (II): Branching Chain
Reaction
Initiation :
H 2 
 H  + H 
Branching : H  +O 2  HO  +  O 
spark
Branching :  O  + H 2  HO  + H 
Assignment for Chapter 13
13.22; 13.31; 13.37; 13.61