Kinetics: Rates and Mechanisms
of Chemical Reactions
Kinetics: Rates and Mechanisms of Chemical Reactions
1 Factors That Influence Reaction Rates
2 Expressing the Reaction Rate
3 The Rate Law and Its Components
4 Integrated Rate Laws: Concentration Changes over Time
5 Reaction Mechanisms: Steps in the Overall Reaction
6 Catalysis: Speeding Up a Chemical Reaction
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
13.1
A
B
time
D[A]
rate = Dt
D[B]
rate =
Dt
13.1
Problem
PROBLEM:
Expressing Rate in Terms of Changes in Concentration with Time
Because it has a nonpolluting product (water vapor), hydrogen
gas is used for fuel aboard the space shuttle and may be used
by Earth-bound 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) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O]
increasing?
2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)
time
393 nm
light
Detector
393 nm
Br2 (aq)
D[Br2] a DAbsorption
13.1
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = =Dt
tfinal - tinitial
instantaneous rate = rate for specific instance in time
13.1
rate a [Br2]
rate = k [Br2]
rate
= rate constant
k=
[Br2]
= 3.50 x 10-3 s-1
13.1
Problem
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 order from the
given rate law.
(a) 2NO(g) + O2(g)
(b) CH3CHO(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-]
Measured Reaction Rates
• The rate of reaction changes with time, as the
concentrations of reactants and products
change.
• The rate constant, but not the rate, is
independent of time.
• The initial rate is the reaction rate measured
before the initial concentrations have had
time to change.
Determining Reaction Orders
• The rate law for a given reaction must be
determined experimentally. This means
determining the order with respect to each
species involved.
• The two major methods are
– Initial Rate Method
– Integrated Rate Law Method
Initial Rates for a Series of Experiments in the Reaction Between O2 and NO
Initial Reactant
Concentrations (mol/L)
Experiment
O2
NO
Initial Rate
(mol/L*s)
1
1.10x10-2
1.30x10-2
3.21x10-3
2
2.20x10-2
1.30x10-2
6.40x10-3
3
1.10x10-2
2.60x10-2
12.8x10-3
4
3.30x10-2
1.30x10-2
9.60x10-3
5
1.10x10-2
3.90x10-2
28.8x10-3
Determining Reaction Orders Using Initial Rates
Run a series of experiments, each of which starts with a different
set of reactant concentrations, and from each obtain an initial rate.
O2(g) + 2NO(g)
rate = k [O2]m[NO]n
2NO2(g)
Compare 2 experiments in which the concentration of one reactant
varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2n
rate2
=
rate1
=
k
[O2]1m[NO]1n
6.40x10-3mol/L*s
[O2]2m
2.20x10-2mol/L
=
3.21x10-3mol/L*s
=
[O2]1m
[O2]2
m
[O2]1
m
;
2 = 2m
1.10x10-2mol/L
Do a similar calculation for the other reactant(s).
m=1
Problem
PROBLEM:
Determining Reaction Order from Initial Rate Data
Many gaseous reactions occur in a car engine and exhaust
system. One of these is
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.
Experiment
1
Initial Rate(mol/L*s) 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
0.10
Phosgene, a toxic gas is prepared by reacting carbon monoxide and chlorine gas.
CO (g) + Cl2 (g)  COCl2 (g)
Experiment
Initial [CO]
Initial [Cl2]
1
1.00
0.100
1.29 x 10-29
2
0.100
0.100
1.33 x 10-30
3
0.100
1.00
1.30 x 10-29
4
0.100
0.0100
1.32 x 10-31
a)
b)
Initial Rate
Write a rate law expression for this reaction
Calculate the average value of the rate constant
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)
First-Order Rate Law
k
A B
R  k[A]
or
d [A]
 kdt or
[A]
d [A]
 k[A]
dt
[A]
t
d [A]
   kdt

[A]
[ A ]0
0
[A]
ln
 kt or [A]  [A]0 e  kt
[A]0
First-Order Rate Law
• Plot the experimental ln[A] vs. time
• If the graph is linear, the reaction is first-order.
• The slope of this line = -k
ln [A] = - kt + ln [A]0
y = mx + b
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
How do we know this is a
first order rxn?
First-Order Processes
CH3NC
This data was collected
for this reaction at
198.9°C.
Does
rate=k[CH3NC]
for all time intervals?
CH3CN
First-Order Processes
• When ln P is plotted as a function of time, a straight line results.
– The process is first-order.
– k is the negative slope: 5.1  10-5 s-1.
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln2
0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = ln2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
13.3
A plot of [N2O5] vs. time for three half-lives
Second-Order Rate Law
k
2A  B
R  keff [A]
2
or
d [A]
 keff dt or
2
[A]
d [A]
 keff [A]2
dt
[A]
t
d [A]
   keff dt
2

[A]
[ A ]0
0
 1
1 
1
1
  keff t or
 


 keff t
[A] [A]0
 [A] [A]0 
Second-Order Rate Law
• Plot the experimental 1/[A] vs. time
• If the graph is linear, the reaction is
second-order.
• The rate constant is k = slope
1
1
 kt 
[ A]
[ A]0
y = mx + b
Second-Order Half Life
1
1
 kt1 
2
[ A] 1
[ A]0
2
1
1

kt
1 
1
2
[
A
]
[ A]0
0
2
1
t1 
2
k[ A]0
Zero-Order Reactions
A
product
D[A]
rate = Dt
D[A]
=k
Dt
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
rate = k [A]0 = k
Graphical determination of the reaction order for the
decomposition of N2O5
Integrated rate law summary
1/[A]t = kt + 1/[A]0
ln[A]t = -kt + ln[A]0
[A]t = -kt + [A]0
Problem
Determining Reaction Concentration at a Given Time
PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order
reaction, with the very high rate constant of 87s-1, to two
molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is the
concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?
Problem
PROBLEM:
Determining the Half-Life of a First-Order Reaction
Cyclopropane is the smallest cyclic hydrocarbon. Because its
600 bond angles allow poor orbital overlap, its bonds are weak.
As a result, it is thermally unstable and rearranges to propene at
10000C via the following first-order reaction:
CH2
D
H3C CH CH2 (g)
(g)
H C
CH
2
2
The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How
long does it take for the concentration of cyclopropane to reach one-quarter of
the initial value?
An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions
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 =
-k t + [A]0
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 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
Temperature and Rate
• Generally, as temperature
increases, so does the
reaction rate.
• This means that k is
temperature dependent.
The Collision Model
• In a chemical reaction, bonds are
broken and new bonds are formed.
• Molecules can only react if they collide
with each other.
The Collision Model
Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bond breakage and formation.
The dependence of possible collisions on the product
of reactant concentrations
A
B
4 collisions
A
B
A
Add another
molecule of A
B
6 collisions
A
B
A
Add another
molecule of B
A
B
A
B
A
B
Activation Energy
• In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll
up the hill with enough energy, a reaction cannot
occur unless the molecules possess sufficient energy
to get over the activation energy barrier.
The effect of temperature on the distribution of collision energies
The Effect of Ea and T on the Fraction (f) of Collisions with Sufficient Energy to Allow Reaction
Ea (kJ/mol)
f (at T = 298 K)
50
1.70x10-9
75
7.03x10-14
100
2.90x10-18
T
f (at Ea = 50 kJ/mol)
250C(298K)
1.70x10-9
350C(308K)
3.29x10-9
450C(318K)
6.12x10-9
An energy-level diagram of the fraction of collisions
exceeding Ea.
Reaction Coordinate Diagrams
It is helpful to
visualize energy
changes throughout a
process on a reaction
coordinate diagram
like this one for the
rearrangement of
methyl isonitrile.
Reaction Coordinate Diagrams
• It shows the energy of the
reactants and products
(and, therefore, DE).
• The high point on the
diagram is the transition
state.
• The species present at the transition state is
called the activated complex.
• The energy gap between the reactants and the
activated complex is the activation energy
barrier.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Reaction energy diagrams and possible transition states
Problem 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 the DHrxn for the reaction is -392 kJ. Draw a
reaction energy diagram for this reaction, postulate a transition state, and
calculate Ea(rev).
Consider the relationships among the reactants, products and
transition state. The reactants are at a higher energy level than the
products and the transition state is slightly higher than the
reactants.
SOLUTION:
Potential Energy
PLAN:
Ea= 19kJ
O3+O
transition state Ea(rev)= (392 + 19)kJ =
DHrxn = -392kJ
411kJ
2O2
Reaction progress
Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression:
where R is the gas constant and T is the temperature in Kelvin .
Arrhenius Equation
Svante Arrhenius developed a mathematical
relationship between k and Ea:
where A is the frequency factor, a number that
represents the likelihood that collisions would
occur with the proper orientation for reaction.
The Effect of Temperature on Reaction Rate
The Arrhenius Equation
k  Ae 
Ea
RT
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
ln k = ln A - Ea/RT
ln
k2
k1
= -
Ea
R
T is the Kelvin temperature
A is the collision frequency factor
1
T2
1
T1
.
Graphical determination of the activation energy
ln k = -Ea/R (1/T) + ln A
Problem
PROBLEM:
Determining the Energy of Activation
The decomposition of hydrogen iodide,
2HI(g)
H2(g) + I2(g)
has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5
L/mol*s at 600. K. Find Ea.
Information sequence to determine the kinetic parameters
of a reaction
Series of plots
of concentration vs. time
Initial
rates
Determine slope
of tangent at t0 for
each plot
Plots of
concentration
vs. time
Reaction
Rate constant
orders
(k) and actual
Compare initial
rate law
rates when [A]
Substitute initial rates,
changes and [B] is
orders, and concentrations
Find k at
held constant and
into general rate law:
varied T
m
n
vice versa
rate = k [A] [B]
Integrated
rate law
(half-life,
t1/2)
Use direct, ln or
inverse plot to
find order
Rate constant
and reaction
order
Rearrange to
linear form and
graph
Activation
energy, Ea
Find k at
varied T
MECHANISMS
A Microscopic View of Reactions
Mechanism: how reactants are converted to
products at the molecular level.
RATE LAW  MECHANISM
experiment  theory
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Rate Laws for General Elementary Steps
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
Problem 8
PROBLEM:
Determining Molecularity and Rate Laws for
Elementary Steps
(1)
The following two reactions are proposed as elementary steps in
the mechanism of an overall reaction:
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.
The Rate-Determining Step of a Reaction Mechanism
The overall rate of a reaction is related to the rate of the slowest, or
rate-determining step.
Correlating the Mechanism with the Rate Law
The elementary steps must add up to the overall equation.
The elementary steps must be physically reasonable.
The mechanism must correlated with the rate law.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Reaction energy diagram for the two-step NO2-F2 reaction
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Reaction energy diagram of a catalyzed and an
uncatalyzed process
The metal-catalyzed hydrogenation of ethylene
H2C CH2 (g) + H2 (g)
H3C CH3 (g)
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
13.6
Enzyme Catalysis
uncatalyzed
enzyme
catalyzed