Chapter 14 – Chemical Kinetics
The rate of a chemical reaction is the speed at
which products or formed and reactants broken
down. There factors that affect the rate of a
chemical reaction are temperature, concentration,
state of reactants, and the presence of a catalyst. A
rate law can be constructed to predict the rate at
which products form or reactants disappear. Some
chemical reactions happen in several steps called a
reaction mechanism. The slowest step of a reaction
mechanism is called the rate-limiting step. A
catalyst is a factor that can speed up a chemical
reaction. Biological catalysts are called enzymes
that are an important part of biochemical reactions
within living organisms.
Chemical Kinetics
•
Factors that affect rates of chemical reactions
o Collision Theory – In order for two
reactants to form products, they must
collide with enough kinetic energy and in
the correct orientation.
Chemical Kinetics
•
Factors that affect rates of chemical reactions
1. 2. -
3. 4. -
Chemical Kinetics
•
Factors that affect rates of chemical reactions
o Concentration – The more abundant the
reactants are, the more successful collisions
to form products.
Chemical Kinetics
•
Factors that affect rates of chemical reactions
o Temperature – The more kinetic energy the
reactants possess, the more likely a
collision will result in a successful reaction.
Chemical Kinetics
•
Factors that affect rates of chemical reactions
o Catalyst – A substance the helps the
reactants collide in the correct orientation.
o They are not consumed in a chemical
reaction.
Chemical Kinetics
•
Factors that affect rates of chemical reactions
o Inhibitors – Compete with binding sites on a
catalyst that slow, or even stop, a chemical
reaction form occurring.
Chemical Kinetics
•
Rates of chemical reactions
o
o
The speed at which a chemical reaction occurs is called
its ‘reaction rate’.
The rate of a chemical reaction is measured as the
change in concentration of reactant or products per
unit of time.
Rate = Δ Concentration / Time
= Molarity / Time
=M/s
** In chemistry [ ] means the molar concentration **
Chemical Kinetics
•
Rates of chemical reactions
o
A flask is charged with 0.100 mol of A and allowed to
react to form B according to the hypothetical gasphase reaction: A(g)  B(g).
Time (s)
0
Moles of A 0.100
40
80
120
160
0.067
0.045
0.030
0.020
a.) Calculate the number of mole of B at each time in the
table.
Chemical Kinetics
•
Rates of chemical reactions
o
A flask is charged with 0.100 mol of A and allowed to
react to form B according to the hypothetical gasphase reaction: A(g)  B(g).
Time (s)
0
Moles of A 0.100
40
80
120
160
0.067
0.045
0.030
0.020
b.) Calculate the average rate of disappearance of A for
each 40 s interval in units of mol / s.
Chemical Kinetics
•
Rates of chemical reactions
o
A flask is charged with 0.100 mol of A and allowed to
react to form B according to the hypothetical gasphase reaction: A(g)  B(g).
Time (s)
0
Moles of A 0.100
40
80
120
160
0.067
0.045
0.030
0.020
c.) What additional information would be needed to
calculate the rate in units of concentration per time?
Chemical Kinetics
•
Rates of chemical reactions
o
Instantaneous Rate versus Average Rate of Reactions

Instantaneous Rate – Refers to the rate of a chemical
reaction at a specific time.

Average Rate – Refers to the rate of a reaction over an
interval of time.

Which rate do you think is usually higher?
Chemical Kinetics
•
Rates of chemical reactions
o
The role of coefficients on reaction rates
aA + bB  cC + dD
In terms of ‘A’, the rate of reaction will be
Rate = - 1
a
Δ[A]
Δt
Chemical Kinetics
•
Rates of chemical reactions
o
The role of coefficients on reaction rates
aA + bB  cC + dD
In terms of ‘B’, the rate of reaction will be
Rate = - 1
b
Δ[B]
Δt
Chemical Kinetics
•
Rates of chemical reactions
o
The role of coefficients on reaction rates
aA + bB  cC + dD
In terms of ‘C’, the rate of reaction will be
Rate =
1
c
Δ[C]
Δt
Chemical Kinetics
•
Rates of chemical reactions
o
The role of coefficients on reaction rates
aA + bB  cC + dD
In terms of ‘D’, the rate of reaction will be
Rate =
1
d
Δ[D]
Δt
Chemical Kinetics
•
Rates of chemical reactions
o
The decomposition of N2O5 proceeds according to the
following reaction: 2N2O5(g)  4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a particular
instant in a reaction vessel is 4.2 x 10-7 M/s, what is the
rate of appearance of (a) NO2 and (b) O2?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o
o
The rate of a chemical reaction cannot be determined without
conducting an experiment.
The initial concentration of reactants are measured and the rate
of the reaction is determined.
X+YZ
Trial
Initial [X]
Initial [Y]
Rate (M/s)
1
0.0100
0.200
5.4 x 10-7
2
0.0200
0.200
10.8 x 10-7
3
0.0400
0.200
21.5 x 10-7
4
0.200
0.0202
10.8 x 10-7
5
0.200
0.0404
21.6 x 10-7
6
0.200
0.0808
43.3 x 10-7
Chemical Kinetics
• Rates Laws of Chemical Reactions
aA + bB  cC + dD
Rate = k [A]m [B]n
k = rate constant
[A] = initial molar concentration of A
[B] = initial molar concentration of B
m = reaction order exponent
n = reaction order exponent
Chemical Kinetics
•
Rates Laws of Chemical Reactions
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
•
•
Trial
Initial [NH4+]
Initial [NO2-]
Rate (M/s)
1
0.0100
0.200
5.4 x 10-7
2
0.0200
0.200
10.8 x 10-7
3
0.0400
0.200
21.5 x 10-7
4
0.200
0.0202
10.8 x 10-7
5
0.200
0.0404
21.6 x 10-7
6
0.200
0.0808
43.3 x 10-7
Notice that when the concentration of NH4+ is doubled, and NO2- is
held constant, the rate doubles.
Notice that when the concentration of NO2- is doubled, and NH4+ is
held constant, the rate doubles.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
•
•
Trial
Initial [NH4+]
Initial [NO2-]
Rate (M/s)
1
0.0100
0.200
5.4 x 10-7
2
0.0200
0.200
10.8 x 10-7
3
0.0400
0.200
21.5 x 10-7
4
0.200
0.0202
10.8 x 10-7
5
0.200
0.0404
21.6 x 10-7
6
0.200
0.0808
43.3 x 10-7
Rate = k [NH4+] [NO2-]
Since NH4+ and NO2- affect the rate proportionally, we do
not need to raise to concentration of the ions to any power.
The order of both reactants is 1.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
•
•
Trial
Initial [NH4+]
Initial [NO2-]
Rate (M/s)
1
0.0100
0.200
5.4 x 10-7
2
0.0200
0.200
10.8 x 10-7
3
0.0400
0.200
21.5 x 10-7
4
0.200
0.0202
10.8 x 10-7
5
0.200
0.0404
21.6 x 10-7
6
0.200
0.0808
43.3 x 10-7
Rate = k [NH4+] [NO2-]
The overall order of this reaction is the sum of the two
orders of the reactants. (1+1 = 2)
This is second order reaction overall.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
Example 1: The experimentally determined rate law
for the reaction 2NO(g) + 2H2(g)  N2(g) + 2H2O(l) is
rate = k[NO]2[H2].
a.) What are the reaction orders in this rate law?
b.) Does doubling the concentration of NO have the
same effect on the rate as doubling the concentration
of H2?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Units of the rate constant (k)
Units of rate constant
=
units of rate
=
(units of concentration)2
M/s
M2
= M-1 . s-1
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Example 2: The initial rate of reaction A + B  C was
measured for several different starting concentrations
of A and B, and the results are as follows;
Trial
[A]
[B]
Initial Rate
(M/s)
1
0.100
0.100
4.0 x 10-5
2
0.100
0.200
4.0 x 10-5
3
0.200
0.100
16.0 x 10-5
a.) Determine the rate law for this reaction.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Example 2: The initial rate of reaction A + B  C was
measured for several different starting concentrations
of A and B, and the results are as follows;
Trial
[A]
[B]
Initial Rate
(M/s)
1
0.100
0.100
4.0 x 10-5
2
0.100
0.200
4.0 x 10-5
3
0.200
0.100
16.0 x 10-5
b.) Determine the rate constant.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Example 2: The initial rate of reaction A + B  C was
measured for several different starting concentrations
of A and B, and the results are as follows;
Trial
[A]
[B]
Initial Rate
(M/s)
1
0.100
0.100
4.0 x 10-5
2
0.100
0.200
4.0 x 10-5
3
0.200
0.100
16.0 x 10-5
b.) Determine the rate of reaction when [A] = 0.050 M
and [B] = 0.100 M
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws
o An integrated rate law allows use to calculate the rate of a
chemical reaction during a specific time interval.
integrate rate law (first-order reactions):
ln [A]t = -kt
[A]0
or
ln [A]t - ln [A]0 = -kt
ln = natural logarithm
[A]t = concentration of A a specific time
[A]0 = initial concentration of A
k = rate law constant
t = time interval
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Law
ln [A]t - ln [A]0 = -kt
rearrange into
ln [A]t = -k t + ln [A]0
y
= m.x +
ln = natural logarithm
[A]t = concentration of A a specific time
[A]0 = initial concentration of A
k = rate law constant
t = time interval
b
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws – The graph on the left shows [A]
versus time. The graph in the middle shows the ln [A]
versus time. A linear relationship of ln [A] proves that it
is a first-order reaction.
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws
o Example: The decomposition of a certain insecticide in water
follows first-order kinetics with a rate constant of 1.45y-1 at
12°C. A quantity of this insecticide is seeps into a lake on
June 1st, leading to a concentration of 5.0 x 10-7 g/cm3.
Assuming that the lake remains at 12°C:
a.) What is the concentration of the insecticide in the lake on
June 1st of the following year?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws
o Example 1: The decomposition of a certain insecticide in
water follows first-order kinetics with a rate constant of
1.45y-1 at 12°C. A quantity of this insecticide is seeps into a
lake on June 1st, leading to a concentration of
5.0
x 10-7 g/cm3. Assuming that the lake remains at 12°C:
a.) What is the concentration of the insecticide in the lake on
June 1st of the following year?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws
o Example 1: The decomposition of a certain insecticide in
water follows first-order kinetics with a rate constant of
1.45y-1 at 12°C. A quantity of this insecticide is seeps into a
lake on June 1st, leading to a concentration of
5.0 x 10-7 g/cm3. Assuming that the lake remains at 12°C:
b.) How long will it take for the concentration of the insecticide
in the lake to drop to 3.0 x 10-7 g/cm3?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws
o Example 2: The decomposition of dimethyl ether, (CH3)2O, at
510°C is a first-order process with a rate constant of
6.8 x 10-4 s-1:
(CH3)2O(g)  CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is the
pressure after 1420 s?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws (second-order reactions)
1 = kt + 1
[A]t
[A]0
[A]0 = initial concentration
[A]t = concentration at time t
k = rate law constant
t = time
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Integrated Rate Laws (second-order reactions)
Example: The following is a second-order reaction
2NO2(g)  2NO(g) + O2(g)
k = 0.543 M-1s-1
If the initial concentration of NO2 in a closed vessel
is 0.0500M, What is the remaining concentration
after 0.500h?
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Half-Life (t1/2) – The time required for the concentration
to fall to half of its original amount.
[A]t1/2 = ½[A]0
Calculating the the half life of a first-order reaction:
t1/2 =
ln1/2 =
k
0.693
k
Chemical Kinetics
•
Rates Laws of Chemical Reactions
o Half-Life (t1/2) –
Example: Calculate the half-life for the decomposition of
the insecticide used in the previous example (slide 31).
How long does it take for the concentration of the
insecticide to reach ¼ of its original value?
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
An increase in the temperature of a chemical reaction
will also increase the rate constant, k.
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
•
An increase in the temperature is also an increase in
the kinetic energy of the reactants.
More kinetic energy = more velocity = more collisions
= more successful collisions.
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Not only do molecules need to collide with enough
kinetic energy, but they also need to be in the correct
orientation.
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Activation energy (Ea) – The minimum amount of
energy required to initiate a chemical reaction.
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Activation energy (Ea) – The minimum amount of
energy required to initiate a chemical reaction.
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Energy diagrams
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Activation Energy
•
Arrhenius Equation – Incorporates the fraction of molecules
that posses enough kinetic energy (Ea), the number of
molecular collisions, and the fraction of reactants that are in
the correct orientation.
k = Ae-Ea/RT
k = reaction constant
Ae = frequency factor constant
Ea = activation energy
R = gas law constant
T = Temperature (K)
Chemical Kinetics
•
Temperature and Rates of Chemical Reactions
•
Determining Activation Energy
ln k1 = Ea ( 1 - 1 )
k2 R ( T2 T 1 )
K1 = rate constant at T1
K2 = previously known rate constant at T2
Ea = activation energy
T1 = Temperature of reaction with k1
T2 = Temperature of reaction with k2
Chemical Kinetics
•
Reaction Mechanisms
•
•
•
A detailed step-by-step explanation as to how a
reaction occurs.
Some reactions occur only as 1 step.
Other reactions occur in several steps.
Chemical Kinetics
•
Reaction Mechanisms
•
Elementary reactions (1 step)
NO(g) + O3(g)  NO2(g) + O2(g)
• The rate of this reaction is only determined in this step.
Chemical Kinetics
•
Reaction Mechanisms
•
Multistep Reaction Mechanisms (2 or more steps)
NO2(g) + CO(g)  NO(g) + CO2(g)
This reaction is the result of two steps;
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
The sum of these reactions gives us the overall reaction.
Chemical Kinetics
•
Reaction Mechanisms
•
Multistep Reaction Mechanisms (2 or more steps)
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
2NO2(g) + NO3(g) + CO(g)  NO3(g) + NO(g) + NO2(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)
NO3(g) is called an intermediate, it is neither a reactant nor a
product.
Chemical Kinetics
•
Reaction Mechanisms
•
Example: For the reaction
Mo(CO)6 + P(CH3)3  Mo(CO)5P(CH3)3 + CO
The proposed mechanism is
Mo(CO)6  Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3  Mo(CO)5P(CH3)3
a.) Is the proposed mechanism consistent with the overall
reaction?
Chemical Kinetics
•
Reaction Mechanisms
•
Example: For the reaction
Mo(CO)6 + P(CH3)3  Mo(CO)5P(CH3)3 + CO
The proposed mechanism is
Mo(CO)6  Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3  Mo(CO)5P(CH3)3
b.) Identify any intermediate(s).
Chemical Kinetics
•
Reaction Mechanisms
•
Example: For the reaction
Mo(CO)6 + P(CH3)3  Mo(CO)5P(CH3)3 + CO
The proposed mechanism is
Mo(CO)6  Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3  Mo(CO)5P(CH3)3
b.) Identify the molecularity of each step of the
mechanism.
Chemical Kinetics
•
Reaction Mechanisms and Rate Determining Steps
o
Single step reactions – The rate of the reaction is
determined by the speed of one step.
o
Multiple Step Reactions – The rate of the reaction will
be determine by the slowest step (rate determining
step).
NO2(g) + NO2(g)  NO3(g) + NO(g)
(slow)
NO3(g) + CO(g)  NO2(g) + CO2(g)
(fast)
--------------------------------------------------------NO2(g) + CO(g)  NO(g) + CO2(g)
Chemical Kinetics
•
Catalyts
o A substance that speeds up a chemical reaction
without being consumed as a reagent.
o Catalysts speed up a chemical reaction by lowering the
activation energy.
Chemical Kinetics
•
Catalyts
o Homogenous Catalyts – Catalysts in the same state as
the reagents.
Chemical Kinetics
•
Catalyts
o Heterogeneous Catalyts – Catalysts that are not in the
same state as the reagents.