Kinetics
The Rates and Mechanisms of Chemical Reactions
What does the word KINETIC imply to
you?
 Why should we care about KINETICS?
 What factors affect KINETICS?

FAST
SLOW
We Are Talking About Reaction
Rates

Speed of any event is measured by a change
that occurs per unit of time.

The speed of reaction (i.e. reaction rate) is
measured as a change in concentration
(Molarity; M) of a reactant or product over a
certain timescale.
◦ Time is the independent variable (x-axis) and
concentration is the dependent variable (y-axis)
◦ Reaction rate is expressed in M/s
The 3 Fundamental Questions of
Chemical Reactions
1)
What happens?
◦ Answer given by balanced chemical equation
and stoichiometry
2)
To what extent does it happen?
◦ Answer deals with chemical equilibrium which
we will study in a later unit
3)
How fast and by what mechanism?
◦ Chemical kinetics
Why? Importance Examples

Chemical Kinetics is very important for biological
(your life), environmental (our lives) and
economic (industry) processes.
◦ Biological: Large proteins (aka Enzymes) increase the
rates of numerous reactions essential to life.
◦ Environmental: The maintenance or depletion of the
ozone layer depends on the relative rates of reactions
that produce or destroy ozone.
◦ Economic: The synthesis of ammonia (NH3) from N2 and
H2 depends on rates of reactions. Fertilizer industries use
catalysts to speed up these rates for economic reasons.
Schematic: Exothermic
Process where energy is released as it proceeds. Heat is
given off to surroundings.
Reactants  Products + Energy
Schematic: Endothermic
Process where energy is absorbed as it proceeds. Heat is
consumed and surroundings become cooler.
Reactants + Energy  Products
Factors that affect KINETICS
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough
energy (activation energy) and must collide in the right
orientation.
FACTORS:
1) Concentration of reactants
2) Temperature
3) Presence of a catalyst
4) Surface area
5) Agitation
6) Nature of reactants
Factor 1: Concentration
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough
energy (activation energy) and must collide in the right
orientation.
FACTORS:
1) Concentration of reactants
If you increase concentration (Molarity), the rate of
reaction increases.
Why? There are more molecules which increases the
number of collisions altogether; however, there are
better chances that molecules will collide in the right
orientation.
Factor 2: Temperature
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough energy
(activation energy) and must collide in the right orientation.
FACTORS:
2) Temperature
Temperature is an averaged kinetic energy of molecules so if
you increase temperature, you increase kinetic energy. This
means you increase the number of collisions
Heat supplies the energy to allow the reaction to proceed
(i.e. overcoming the activation energy barrier)
Think about: Why do we refrigerate milk?
Factor 3: Presence of Catalyst
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough energy
(activation energy) and must collide in the right orientation.
FACTORS:
3) Presence of a catalyst
Catalyst assist a reaction and increase the reaction rate
without being consumed in the reaction.
Adding a catalyst decreases the activation energy which
means more molecules will have enough energy to react.
Think about: Catalytic converter, Enzymes
Factor 3: Presence of Catalyst
Factor 4: Surface Area
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or molecules
must collide with one another with enough energy (activation
energy) and must collide in the right orientation.
FACTORS:
4)
Surface Area
Increased surface areas of molecules/particles will increase the rate
of reaction. This means to break into smaller particle sizes.
More places to react give better chances for collisions in the right
orientation.
How to increase surface area? Grind or crush a mixture of
reactants.
Ex: A crushed aspirin will enter your blood stream faster than
taking it whole.
Factor 5: Agitation
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough
energy (activation energy) and must collide in the right
orientation.
FACTORS:
5) Agitation
Stirring or shaking a reaction will increase the reaction
rate.
By stirring or shaking, you are introducing energy into
the reaction and thus giving molecules/particles more
energy to react (overcome activation energy barrier).
Your mechanical energy is converted to kinetic energy.
Factor 6: Nature of Reactants
All based on COLLISION THEORY:
Collision theory: For a reaction to occur, the atoms or
molecules must collide with one another with enough
energy (activation energy) and must collide in the right
orientation.
FACTORS:
6) Nature of reactants
Reactants whose bonds are weaker have a lower
activation energy and thus a higher rate of reaction.
All chemical reactions involve bond breaking and bond
making. Bond breaking occurs on reactant side.
Collisions between reactants that require less kinetic
energy are needed to break weaker bonds (i.e. smaller
activation energy)
Rate Laws and Reaction Order
Rate Law: An equation that shows the dependence of the reaction rate on
the concentration of each reactant.
products
aA+bB
rate =
∆[A]
∆t
*[ ] means molarity
rate = k[A]m[B]n
k is the rate constant.
The values of the exponents (m and n) in the rate law must be
determined by experiment; they cannot be deduced from the
stoichiometry of the reaction.
Rates of Chemical Reactions
2 N2O5(g)
4 NO2(g) + O2(g)
∆[N2O5]
∆[NO2]
∆[O2]
1
1
rate = – 2
= 4
=
∆t
∆t
∆t
General rate of reaction:
aA+bB
dD+eE
∆[A]
∆[B] 1 ∆[D] 1 ∆[E]
1
1
rate = – a
=– b
= d
= e
∆t
∆t
∆t
∆t
The rate of reaction can be measured based on reactants or products.
The negative sign in front for reactants indicates they are consumed.
Typically rates of reactions are expressed based on reactants.
Rates of Chemical Reactions
ΔT should be Δt (time, not
temperature)
Rate Laws and Reaction Order
The values of the exponents in the rate law must be determined by
experiment; they cannot be deduced from the stoichiometry of the
reaction.
Determining a Rate Law: The
Method of Initial Rates
2 NO(g) + O2(g)
2 NO2(g)
rate = k[NO]m[O2]n
Compare the initial rates to the changes in initial concentrations.
Determining a Rate Law: The
Method of Initial Rates
2 NO(g) + O2(g)
2 NO2(g)
rate = k[NO]2 [O2]n
The concentration of NO doubles, the concentration of O2 remains constant,
and the rate quadruples.
2m = 4
m=2
Determining a Rate Law: The
Method of Initial Rates
2 NO(g) + O2(g)
2 NO2(g)
rate = k[NO]2 [O2]
Reaction Order with Respect to a Reactant
• NO: second order
• O2: first order
Overall Reaction Order
• 2 + 1 = 3 (third order)
Determining a Rate Law: The
Method of Initial Rates
2 NO(g) + O2(g)
2 NO2(g)
rate = k[NO]2 [O2]
Units for this third-order reaction:
M
rate
k=
s
=
[NO]2
[O2]
=
(M2)
(M)
1
M2 s
You can pick any experiment or trial and solve for k (should all be the
same).
Example Problem #1
[HgCl2]
0.164
0.164
0.082
a.
b.
c.
[C2O42-]
0.15
0.45
0.45
Rate (M/s)
3.2 x 10-5
2.9 x 10-4
1.4 x 10-4
Determine the rate law. What is the order of the
reaction?
Determine the rate law constant (specify the units)
What is the rate when the initial concentrations of both
reactants are 0.100 M?
Example Problem #1
[HgCl2]
0.164
0.164
0.082
a.
[C2O42-]
0.15
0.45
0.45
Rate (M/s)
3.2 x 10-5
2.9 x 10-4
1.4 x 10-4
Determine the rate law. What is the order of the reaction?
Rate = k[HgCl2]m[C2O42-]n
𝑅𝑎𝑡𝑒 2
(𝑅𝑎𝑡𝑒)1
=
𝑘(0.164)𝑚 (0.45)𝑛
𝑘(0.164)𝑚 (0.15)𝑛
𝑛
=
2.9𝑥10−4
3.2𝑥10−5
=
2.9𝑥10−4
1.4𝑥10−4
3 = 9, 𝑛 = 2
𝑅𝑎𝑡𝑒 2
𝑅𝑎𝑡𝑒 3
=
𝑘(0.164)𝑚 (0.45)𝑛
𝑘(0.082)𝑚 (0.45)𝑛
𝑚
2 = 2, 𝑚 = 1
Rate = k[HgCl2] [C2O42-]2
Overall third order
Example Problem #1
[HgCl2]
0.164
0.164
0.082
a.
b.
[C2O42-]
0.15
0.45
0.45
Rate (M/s)
3.2 x 10-5
2.9 x 10-4
1.4 x 10-4
Determine the rate law. What is the order of the
reaction?
Determine the rate law constant (specify the units)
Rate = k[HgCl2] [C2O42-]2
3.2x10-5M/s = k[0.164M][0.15M]2
𝑘=
3.2𝑥10−5 𝑀/𝑠
(0.164𝑀 𝑥 0.152 𝑀2 )
=
3.2𝑥10−5 𝑀/𝑠
(0.0037𝑀3 )
= 0.0087
1
𝑀2 𝑠
Example Problem #1
[HgCl2]
0.164
0.164
0.082
a.
b.
c.
[C2O42-]
0.15
0.45
0.45
Rate (M/s)
3.2 x 10-5
2.9 x 10-4
1.4 x 10-4
Determine the rate law. What is the order of the
reaction?
Determine the rate law constant (specify the units)
What is the rate when the initial concentrations of both
reactants are 0.100 M?
Rate = k[HgCl2] [C2O42-]2
𝑘 = 0.0087
𝑅𝑎𝑡𝑒 = (0.0087
1
𝑀2 𝑠
1
)(0.100M)(0.100M)2
2
𝑀 𝑠
= 8.7𝑥10−6
𝑀
𝑠
Example Problem #2
[NOCl]
Rate M/s
3000
5980
2000
2660
1000
665
4000
10640
a. Determine the rate law and order.
b. Determine the rate law constant.
Specify the units.
c. What is the rate when the concentration
of NOCl is 8000M?
Example Problem #2
[NOCl]
3000
2000
1000
4000
Rate M/s
5980
2660
665
10640
Determine the rate law and order.
Rate = k[NOCl]2; Second Order
Determine the rate law constant. Specify the
units.
For first trial: 6.64x10-4 (1/Ms)
For second trial: 6.65x10-4 (1/Ms)
What is the rate when the concentration of
NOCl is 8000M? 42560 M/s
Example #3
[A]
[B]
Rate (M/s)
1.50
1.50
0.32
1.50
2.50
0.32
3.00
1.50
0.64
a. Determine the rate law and order.
b. Determine the rate law constant.
Specify the units.
Example #3
[A]
[B]
Rate (M/s)
1.50
1.50
0.32
1.50
2.50
0.32
3.00
1.50
0.64
Determine the rate law and order.
Rate = k[A]; first order
Determine the rate law constant. Specify
the units.
k = 0.21 (1/s)
Example #4
[CH3COCH3]
[Br2]
[H+]
Rate (M/s)
0.30
0.05
0.05
0.000057
0.30
0.10
0.05
0.000057
0.30
0.05
0.10
0.000120
0.40
0.05
0.20
0.000310
0.40
0.05
0.05
0.000076
a. Determine the rate law and order.
b. Determine the rate law constant. Specify
the units.
Example #4
[CH3COCH3]
[Br2]
[H+]
Rate (M/s)
0.30
0.05
0.05
0.000057
0.30
0.10
0.05
0.000057
0.30
0.05
0.10
0.000120
0.40
0.05
0.20
0.000310
0.40
0.05
0.05
0.000076
Determine the rate law and order.
Rate = k[CH3COCH3][H+]; second order
Determine the rate law constant. Specify the
units.
k = 0.0038 (1/Ms)
Example #5
[S2O8-2]
[I -]
Rate (mol/L/s)
0.018
0.036
2.6 x 10-6
0.027
0.036
3.9 x 10-6
0.036
0.054
7.8 x 10-6
0.050
0.072
1.4 x 10-5
a. Determine the rate law and order.
b. Determine the rate law constant.
Specify the units.
Example #5
[S2O8-2]
[I -]
Rate (mol/L/s)
0.018
0.036
2.6 x 10-6
0.027
0.036
3.9 x 10-6
0.036
0.054
7.8 x 10-6
0.050
0.072
1.4 x 10-5
Determine the rate law and order.
Rate = k[S2O8-2]m[I-]n
𝑅𝑎𝑡𝑒 2
𝑘(0.027)𝑚 (0.036)𝑛
3.9𝑥10−6
=
=
𝑅𝑎𝑡𝑒 1
𝑘(0.018)𝑚 (0.036)𝑛
2.6𝑥10−6
1.5𝑚 = 1.5, 𝑚 = 1
𝑅𝑎𝑡𝑒 3
𝑘(0.036)1 (0.54)𝑛
7.8𝑥10−6
𝑇𝑟𝑖𝑐𝑘𝑦 (𝑐𝑎𝑛 𝑝𝑖𝑐𝑘 𝑎𝑛𝑦):
=
=
(𝑅𝑎𝑡𝑒)1
𝑘(0.018)1 (0.36)𝑛
2.6𝑥10−6
2𝑥1.5𝑛 = 3, 𝑛 = 1
Rate = k[S2O8-2][I-]
Second order
Determine the rate law constant. Specify the units.
For first trial: 0.0040 (1/Ms)
For second trial: 0.0040 (1/Ms)
Integrated Rate Laws
Graphs, reaction order and concentrations.
Graphs: concentration of
substance A vs. time

Scientists don’t like curves, they like
straight lines. SO they have to manipulate
to obtain straight line.
Determining a Rate Law: The
Method of Initial Rates
Rate Law
Overall Reaction
Order
Units for k
Rate = k
Zeroth order
M/s or M s–1
Rate = k[A]
First order
1/s or s–1
Rate =k[A][B]
Second order
1/(M • s) or M–1s–1
Rate = k[A][B]2
Third order
1/(M2 • s) or M–2s–1
*Units for reaction rates are always M/s
Zeroth-Order Reactions
For a zeroth-order reaction, the rate is independent of the concentration of
the reactant.
A
product(s)
–
rate = k[A]0 = k
∆[A]
∆t
=k
t
½
[A]
=
0
2k
Calculus can be used to derive an integrated rate law.
[A]t = –kt + [A]0
y = mx + b
[A]t
concentration of A at time t
[A]0
initial concentration of A
Zeroth-Order Reactions
A plot of [A] versus time
gives a straight-line fit and
the slope will be –k.
Half-Life: t1/2
For zeroth order:
[A]t = –kt + [A]0
[A]t1/2 = [A]0
2
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔:
[𝐴]0
= -𝑘𝑡1/2 + [𝐴]0
2
𝑡1/2 =
[𝐴]0
2𝑘
• Half-life: time for A to
reduce by 50%
First-Order Reactions: The
Integrated Rate Law
A
product(s)
–
∆[A]
= k[A]
∆t
rate = k[A]
Calculus can be used to derive an integrated rate law.
[A]t
ln
= –kt
[A]0
Using:
ln
x
y
[A]t
concentration of A at time t
[A]0
initial concentration of A
= ln(x) – ln(y)
ln[A]t = –kt + ln[A]0
y
= mx + b
First-Order Reactions: The
Integrated Rate Law
ln[A]t = –kt + ln[A]0
First-Order Reactions: Half-Life
Half-Life: The time required for the reactant concentration to drop to one-half
of its initial value.
A
product(s)
rate = k[A]
t = t1/2
[A]t
ln
= –kt
[A]0
ln
1
2
= –kt1/2
[A]
[A]0
=
t1/2
2
0.693
or
t1/2 =
k
Second-Order Reactions
A
rate =
product(s)
–
k[A]2
∆[A]
∆t
= k[A]2
Calculus can be used to derive an integrated rate law.
1
[A]t
= kt +
1
[A]0
y = mx + b
[A]t
concentration of A at time t
[A]0
initial concentration of A
Second-Order Reactions
2 NO2(g)
Time (s)
0
50
100
150
200
300
400
500
[NO2]
8.00 x 10–3
6.58 x 10–3
5.59 x 10–3
4.85 x 10–3
4.29 x 10–3
3.48 x 10–3
2.93 x 10–3
2.53 x 10–3
2 NO(g) + O2(g)
ln[NO2]
–4.828
–5.024
–5.187
–5.329
–5.451
–5.661
–5.833
–5.980
1/[NO2]
125
152
179
206
233
287
341
395
Second-Order Reactions
2 NO2(g)
2 NO(g) + O2(g)
Second-Order Reactions
Half-life for a second-order reaction:
A
product(s)
rate = k[A]2
1
[A]t
2
[A]0
= kt +
t = t1/2
1
[A]
[A]0
= kt1/2 +
=
t1/2
1
[A]0
t1/2 =
[A]0
2
1
k[A]0
Second-Order Reactions
t1/2 =
1
k[A]0
For a second-order reaction, the
half-life is dependent on the initial
concentration.
Each successive half-life is twice
as long as the preceding one.
Formulas Given
Example #1
A first order reaction has a half-life of
20 minutes.
a. Calculate the rate constant for this
reaction.
b. How much time is required for this
reaction to be 80% complete?

Example #1
A first order reaction has a half-life of 20
minutes.
Calculate the rate constant for this reaction.
ln(2)
Use: 𝑡1/2 =
and solve for k
𝑘
You get: 5.75x10-4 1/s

How much time is required for this reaction to
be 80% complete? Use: ln 𝐴 𝑡 = −𝑘𝑡 + 𝑙𝑛[𝐴]0
and solve for t. You calculated k and [A]t is 0.2
if you assume [A]0 is 1 (percentages).
You get: 2800 s
Example #2
N2O5(g)  4NO2
(g)
+ O2 (g)
The following results were collected:
[N2O5(g) ]
Time(s)
0.1000
0
A.
Determine the rate
0.0707
50
constant from the data
0.0500
100
above.
0.0250
200
B.
Calculate the
0.0125
300
concentration of N2O5(g)
after 175 seconds have
0.00625
400
passed.
Example #2
N2O5(g)  4NO2
(g)
+ O2 (g)
The following results were collected:
[N2O5(g) ]
Time(s)
0.1000
0
A.
Determine the rate
0.0707
50
constant from the data
0.0500
100
above. First order k =
0.0250
200
0.00693 1/s
0.0125
300
B.
Calculate the
concentration of N2O5(g)
0.00625
400
after 175 seconds have
passed. 0.0297M
Example #3
2C4H6(g)  C8H18(g)
The following data were collected:
[C4H6]
Time (s)
0.01000
0
A.
What is the order of the
0.00625 1000
reaction?
0.00476 1800
B.
What is the rate law
0.00370 2800
constant?
0.00313 3600
C.
How long will it take for
the [C4H6] to be 93%
gone?
Example #3
2C4H6(g)  C8H18(g)
The following data were collected:
[C4H6]
Time (s)
0.01000
0
A.
What is the order of the
0.00625 1000
reaction? Second order
0.00476 1800
B.
What is the rate law
0.00370 2800
constant? 0.061(1/Ms)
0.00313 3600
C.
How long will it take for
the [C4H6] to be 93%
gone? 21,780 s
Example #4
A second order reaction has a rate constant
of 5.99M-1s-1. The reaction initially
contains 4M [A].
A. How long will it take for the reaction to
drop to 0.17M?
B. What will the concentration be after an
hour?
Example #4
A second order reaction has a rate constant
of 5.99M-1s-1. The reaction initially
contains 4M [A].
A. How long will it take for the reaction to
drop to 0.17M? 0.94s
B. What will the concentration be after an
hour? 4.63x10-5 M
Example #5
The following data was collected.
[A]
Time(s)
A.
Determine the order
10
0
of the reaction from
8
3
the data.
B.
What is the rate law
6
6
constant?
4
9
C.
What is the half life
2
12
for the reaction?
Example #5
The following data was collected.
[A]
Time(s)
A.
Determine the order
10
0
of the reaction from
8
3
the data. Zeroth
order
6
6
B.
What is the rate law
4
9
constant? 0.67 M/s
(slope)
2
12
C.
What is the half-life
for the reaction?
7.46s
Reaction Rates and the Temperature:
Collision Theory and the Arrhenius Equation
Transition State: The configuration of atoms at the maximum in the potential
energy profile. This is also called the activated complex.
The Arrhenius Equation
Activation Energy (Ea): The minimum energy needed for reaction. As the
temperature increases, the fraction of collisions with sufficient energy to react
increases.
The Arrhenius Equation
Using the Arrhenius Equation
ln(k) = ln(A) + ln(e–Ea
ln(k) = ln(A) –
Ea
/RT)
ln(k) =
RT
–Ea
1
R
T
rearrange the equation
y
Slope intercept formula when you
plot ln(k) vs. 1/T :
.
R = 0.008314 kJ/K mol
A: collision frequency factor
=
mx
+ b
+ ln(A)
Using the Arrhenius Equation
–Ea
1
ln(k) =
R
T
t (°C)
T (K)
k (M–1 s–1)
283
556
3.52 x 10–7 0.001 80
–14.860
356
629
3.02 x 10–5 0.001 59
–10.408
393
666
2.19 x 10–4 0.001 50
–8.426
427
700
1.16 x 10–3 0.001 43
–6.759
508
781
3.95 x 10–2 0.001 28
–3.231
+ ln(A)
1/T (1/K)
lnk
Using the Arrhenius Equation
ln(k) =
–Ea
1
R
T
+ ln(A)
Example #1
Determine the activation energy and the
collision frequency factor at 700K for the
data below.
k
T (K)
0.011
700
0.035
730
0.105
760
0.343
790
0.789
810
Changes in Temperature will
change the rate constant
Increasing temperature will increase the
rate constant (thus increasing the rate).
ln k1 - ln k2 = - (Ea/R)(1/T1 - 1/T2)
Example #2
The rate constant of a first-order reaction is
0.034 at 298K. The activation energy for
this reaction is 50.2 kJ/mol.
What is the rate constant at 350K?
(if the temperature is given in Celsius, add
273 to change to Kelvin)
Example #3
At 45oC the rate constant of a first order
reaction is 0.8. At 135oC the rate
constant is 2.4. What is the activation
energy for this reaction?
Example #4
The rate of a reaction increased by a factor
of 10 when the temperature goes from
25oC to 50oC. What is the activation
energy of the reaction?
Reaction Mechanisms
Reaction Mechanism: The sequence of reaction steps that describes the
pathway from reactants to products.
Elementary Reaction (step): A single step in a reaction mechanism.
Possible: Hydrogen reacts with oxygen to
make water.
H2 + O2  H2O2
H2O2  H2O + O
H2 + O  H2O
Same reaction, other
approaches
O2  2O
H 2 + O  H2 O
H 2 + O  H2 O
H2  2H
H + O2  OH + O
H + OH  H2O
O + H 2  H2 O
How do you decide which is
right?
The reaction mechanism must result in the
overall balance reaction
The slow step must support the rate law
Reaction Mechanisms
Experimental evidence suggests that the reaction between NO2 and CO takes
place by a two-step mechanism:
NO2(g) + NO2(g)
NO(g) + NO3(g)
elementary reaction
NO3(g) + CO(g)
NO2(g) + CO2(g)
elementary reaction
NO2(g) + CO(g)
NO(g) + CO2(g)
overall reaction
An elementary reaction describes an individual molecular event.
The overall reaction describes the reaction stoichiometry and is a
summation of the elementary reactions.
Reaction Mechanisms
Experimental evidence suggests that the reaction between NO2 and CO takes
place by a two-step mechanism:
NO2(g) + NO2(g)
NO(g) + NO3(g)
elementary reaction
NO3(g) + CO(g)
NO2(g) + CO2(g)
elementary reaction
NO2(g) + CO(g)
NO(g) + CO2(g)
overall reaction
A reactive intermediate is formed in one step and consumed in a
subsequent step.
Reaction Mechanisms
Molecularity: A classification of an elementary reaction based on the number
of molecules (or atoms) on the reactant side of the chemical equation.
unimolecular reaction:
bimolecular reaction:
termolecular reaction:
O3*(g)
O2(g) + O(g)
O3(g) + O(g)
2 O2(g)
O(g) + O(g) + M(g)
O2(g) + M(g)
Rate Laws for Elementary
Reactions
The rate law for an elementary reaction follows directly from its molecularity
because an elementary reaction is an individual molecular event.
unimolecular reaction:
O3*(g)
O2(g) + O(g)
rate = k[O3]
bimolecular reaction:
O3(g) + O(g)
2 O2(g)
rate = k[O3][O]
termolecular reaction:
O(g) + O(g) + M(g)
rate = k[O]2[M]
O2(g) + M(g)
Rate Laws for Overall Reactions
Rate-Determining Step: The slowest step in a reaction mechanism. It acts as
a bottleneck and limits the rate at which reactants can be converted to
products.
1) First check to make sure the individual steps will add up to the
overall reaction.
2) In elementary steps, the coefficient of the reactants become the
power in the rate law.
3) Slow step will always be the basis for comparing experimental to
predicted rate law.
4) If slow step is the first, then everything is easy.
5) If not, then the forward rate of step above = reverse rate and some
substituting will take place.
6) Intermediates cannot be in rate law.
Rate Laws for Overall Reactions
Initial Slow Step
NO2(g) + NO2(g)
NO(g) + NO3(g)
slow step
NO3(g) + CO(g)
NO2(g) + CO2(g)
fast step
NO2(g) + CO(g)
NO(g) + CO2(g)
overall reaction
k1
k2
Based on the slow step: rate = k1[NO2]2
Rate Laws for Overall Reactions
Initial Fast Step
2 NO(g)
k1
N2O2(g)
fast step, reversible
N2O(g) + H2O(g)
slow step
N2(g) + H2O(g)
fast step
N2(g) + 2 H2O(g)
overall reaction
k-1
N2O2(g) + H2(g)
N2O(g) + H2(g)
2 NO(g) + 2 H2(g)
k2
k3
Based on the slow step: rate = k2[N2O2][H2]
Rate Laws for Overall Reactions
rate = k2[N2O2][H2]
intermediate
First step:
Rateforward = k1[NO]2
Ratereverse = k–1[N2O2]
k1[NO]2 = k–1[N2O2]
[N2O2] =
Slow step: rate = k2[N2O2][H2]
k1
k–1
[NO]2
rate = k2
k1
k–1
[NO]2[H2]
Catalyst
Since the catalyst is involved in the rate-determining step, it often appears in
the rate law.
rate = k[H2O2][I–]
H2O2(aq) +
I–(aq)
H2O2(aq) + IO–(aq)
2 H2O2(aq)
H2O(l) +
IO–(aq)
rate-determining
step
H2O(l) + O2(g) + I–(aq)
fast step
2 H2O(l) + O2(g)
overall reaction
Example #1
Overall reaction:
2NO2 + CO  NO3 + CO2
Proposed mechanism:
NO2 + NO2  NO3 + NO
(slow)
NO3 + CO  NO2 + CO2
(fast)
If the rate law is Rate = k[NO2], is this a
possible mechanism?
Example #2
For the overall reaction:
2NO + Br2  2NOBr
The proposed mechanism is:
NO + NO  N2O2 (fast equilibrium)
N2O2 + Br2  2NOBr
(slow)
Determine the rate law.
Example #3
Overall reaction:
4HBr + O2  2H2O + 2Br2
Proposed Mechanism:
HBr + O2  HOOBr
HOOBr + HBr  2HOBr
2HOBr + 2HBr  2H2O + 2Br2
Rate law: Rate = k[HBr]2[O2]
Which is the rate determining step?
Example #4
From the following mechanism, determine
the rate law.
Cl2  2Cl
(fast)
Cl + CHCl3  HCl + CCl3
(slow)
Cl + CCl3  CCl4
(fast)
Overall reaction:
Cl2 + CHCl3  HCl + CCl4