Chemical Kinetics
Chemical Kinetics
Do Now:
1. TRUE / FALSE: By definition, spontaneous
reactions are fast.
2. What is meant by the term “rate of
reaction”?
3. List three factors that may affect the rate of
reaction, and indicate how/why.
Section Objectives
• Explain the rate of reaction is the
decrease in concentration of reactants or
the increase in concentration of products
with time
• Describe how reaction rates depend on
such factors as concentration,
temperature, and pressure
• Explain the role a catalyst plays in
increasing the reaction rate
Reaction Rates
• Chemical reactions don’t all happen
at the same rate…
Some reactions, like the explosion of
dynamite, are completely finished in
less than a second.
Others, like the rusting of iron, may not
completely finish for many years.
Reaction Rates
• There are two ways to measure the
rate of a reaction…
Either measure the rate at which a
REACTANT DISSAPPEARS or…
…measure the rate at which a PRODUCT
APPEARS
Reaction Rates
• Often, chemists are trying to control
the rate of a chemical reaction
Sometimes, we wish to speed a reaction
up, like when we design a quick-drying
paint formula.
Other times, we want to slow reactions
down, like when we want the paint to
resist fading for many years.
Collision Theory
• Molecules must collide in order to react.
• In a successful collision existing bonds are
broken as new bonds are formed and the
reactants are transformed into products.
• This is the collision theory of reactions.
Factors in Successful Collisions
• There are multiple factors that affect
the rate of a chemical reaction.
• Collision Frequency:
By increasing concentration and/or
temperature we can increase the
frequency at which molecules collide and
increase the rate of reaction.
WHY?
• Concentration: __________
• Temperature: ___________
• Collision Energy:
For a reaction to occur, the molecules
must collide with enough energy to form
the new bonds. Increase temperature to
increase the energy.
Greater temperature = more #
molecules with greater energy = greater
probability of molecules with E > Eact
• Collision Geometry:
For a reaction to occur, the molecules
must be oriented in the proper geometry
for the reaction to occur.
• A Catalyst can help with the geometry. It is
NOT consumed by the reaction
• Solid Particle Size:
When a solid reacts,
only particles on the
surface of the solid
are available to
react.
Solid broken up into
smaller pieces
results in greater
surface area
• Therefore more
particles are
available for
collision, and
reaction rate
increases.
Energy Barriers in Reactions
• For a chemical reaction to occur, the reactants
must collide with sufficient energy to react.
• This energy is required to achieve the transition
state required to form the products (a).
• WITHOUT sufficient energy, the reaction does not
occur (b).
Reaction Profiles
• A reaction profile shows the energy of
reactants and products during a reaction.
• The highest point on a reaction profile is the
transition state or the activated complex.
• The energy
required for
reactants to
achieve the
transition state is
the activation
energy, Eact.
• The energy
difference
between reactants
and products is
the heat of
reaction, ΔH.
Endothermic Reactions
• An endothermic reaction absorbs
heat as the reaction proceeds.
• N2(g) + O2(g) + heat  2 NO(g)
• Endothermic: reactants have
LOWER energy than products
• ΔH for
endothermic
reaction is
positive
Exothermic Reactions
• An exothermic reaction releases
heat as the reaction proceeds.
• NO(g) + O3(g)  2 NO2(g) + O2(g) +
heat
• The ΔH for an exothermic
reaction is negative.
• The products are
at a lower energy
than the reactants
•
Factors that Affect Reaction
Rate
Temperature: Higher temperature means
particles hit each other more often and with
more energy; they will react faster.
• Concentration: If the particles are more
concentrated, they will collide more and react
faster.
• Surface Area: Atoms and molecules can only
react at the surface of an object. The greater
the surface area, the faster the particles can
collide. Example: Small crystals of NaCl dissolve
in water faster than rock salt.
• Catalyst: Some substances aren’t reactants in
a reaction, but they can speed up the reaction
by being present.
Effect of a Catalyst
• A catalyst is a substance that allows a
reaction to proceed faster by LOWERING
the activation energy.
• It does this by
finding an
alternative path
for the reaction
to occur
• An inhibitor is the
opposite of a
catalyst
• A catalyst does not
change ΔH for a
reaction.
• A catalyst speeds up
both the forward and
reverse reactions.
• Examples of biological
catalysts: enzymes
• A catalyst can’t make a
reaction occur that
wouldn’t otherwise also
occur.
Kinetics
• Studies the rate at which a chemical
process occurs.
• Besides information about the speed at
which reactions occur, kinetics also
sheds light on the reaction mechanism
(exactly how the reaction occurs).
Reaction Rates
Rates of reactions can be determined
by monitoring the change in
concentration of either reactants or
products as a function of time.
Factors That Affect Reaction Rates
1. Physical State of the Reactants
 In order to react, molecules
must come in contact with each
other.
 The more homogeneous the
mixture of reactants, the faster
the molecules can react.
Factors That Affect Reaction Rates
2. Concentration of Reactants
 As the concentration of reactants
increases, so does the likelihood that
reactant molecules will collide.
Factors That Affect Reaction Rates
3. Temperature
 At higher temperatures, reactant
molecules have more kinetic
energy, move faster, and collide
more often and with greater
energy.
Factors That Affect Reaction Rates
4. Presence of a Catalyst
 Catalysts speed up reactions by
changing the mechanism of the
reaction.
 Catalysts are not consumed during
the course of the reaction.
 Enzymes are biological catalysts
1.
2.
3.
4.
The effect of increasing the partial pressures of the
reactive components of a gaseous mixture depends
on which side of the chemical equation has the most
gas molecules.
Increasing the partial pressures of the reactive
components of a gaseous mixture has no effect on
the rate of reaction if each reactant pressure is
increased by the same amount.
Increasing the partial pressures of the reactive
components of a gaseous mixture increases the rate
of reaction.
Increasing the partial pressures of the reactive
components of a gaseous mixture decreases the rate
of reaction.
1.
2.
3.
4.
The effect of increasing the partial pressures of the
reactive components of a gaseous mixture depends
on which side of the chemical equation has the most
gas molecules.
Increasing the partial pressures of the reactive
components of a gaseous mixture has no effect on
the rate of reaction if each reactant pressure is
increased by the same amount.
Increasing the partial pressures of the reactive
components of a gaseous mixture increases the rate
of reaction.
Increasing the partial pressures of the reactive
components of a gaseous mixture decreases the rate
of reaction.
12.1 – Reaction Rates
• The speed of a chemical reaction – reaction
rate – is the change in concentration of
reactants or products per unit time
• Units are usually molarity per second ([M]/s)
Average rate = Δ [A]
ΔT
**NOTE: Rate may be positive (appearance of
product) or negative (disappearance of
reactant). For convenience, we will always define
rate at positive
Calculate the
average rate
at which A
disappears
over the
time interval
from 20 s to
40 s.
SAMPLE EXERCISE 1 Calculating an Average Rate of Reaction
From the data given in the caption of figure above,
calculate the average rate at which A disappears over the
time interval from 20 s to 40 s.
PRACTICE EXERCISE
Calculate the
average rate of
appearance of B
over the time
interval from 0 to
40 s.
Answer: 1.8  10 –2 M/s
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the
concentration of
butyl chloride,
C4H9Cl, was
measured at various
times.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
The average rate of
the reaction over
each interval is the
change in
concentration
divided by the
change in time:
Average rate =
[C4H9Cl]
t
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• Note that the
average rate
decreases as the
reaction proceeds.
• This is because as
the reaction goes
forward, there are
fewer collisions
between reactant
molecules.
• WHY?
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• A plot of
concentration vs.
time for this
reaction yields a
curve like this.
• The slope of a line
tangent to the
curve at any point is
the instantaneous
rate at that time.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• All reactions slow
down over time.
• Therefore, the best
indicator of the rate
of a reaction is the
instantaneous rate
near the beginning.
1.
2.
3.
4.
Larger triangles should be used if x and y values
are three significant figures or more.
The size of the triangle is mostly a matter of
convenience; different size triangles give the
same ratio (slope).
The size of the triangle determines the number of
significant digits in your final slope calculation.
Smaller triangles should be used only near the
ends of the curve while larger triangles are used
in the center only.
1.
2.
3.
4.
Larger triangles should be used if x and y values
are three significant figures or more.
The size of the triangle is mostly a matter of
convenience; different size triangles give the
same ratio (slope).
The size of the triangle determines the number of
significant digits in your final slope calculation.
Smaller triangles should be used only near the
ends of the curve while larger triangles are used
in the center only.
SAMPLE EXERCISE 2 Calculating an Instantaneous Rate of Reaction
Using Figure given, calculate
the instantaneous rate of
disappearance of C4H9Cl at t = 0
(the initial rate).
PRACTICE EXERCISE
Using Figure given, determine the
instantaneous rate of disappearance
of C4H9Cl at t = 300 s.
Answer: 1.1  10 –4 M/s
Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• In this reaction, the
ratio of C4H9Cl to
C4H9OH is 1:1.
• Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rate =
-[C4H9Cl]
=
t
[C4H9OH]
t
Reaction Rates and Stoichiometry
• To generalize for the reaction:
aA + bB
cC + dD
1 [B]
1 [D]
1 [A]
1 [C]
=
=
Rate = −
=−
b t
d t
a t
c t
SAMPLE EXERCISE 3 Relating Rates at Which Products Appear and
Reactants Disappear
How is the rate at which ozone disappears related to the
rate at which oxygen appears in the reaction
SAMPLE EXERCISE 3 Relating Rates at Which Products Appear and
Reactants Disappear
If the rate at which O2 appears, [O2]/t, is
6.0  10–5 M/s at a particular instant, at what rate is O3
disappearing at this same time, –[O3]/t?
Solving the equation from part (a) for the rate at which
O3 disappears, –[O3]/t we have:
SAMPLE EXERCISE 3 continued
PRACTICE EXERCISE
The decomposition of N2O5 proceeds according to the
following equation:
If the rate of decomposition of N2O5 at a particular
instant in a reaction vessel is 4.2  10–7 M/s, what is the
rate of appearance of (a) NO2, (b) O2?
Answers: (a) 8.4  10 –7 M/s, (b) 2.1  10 –7 M/s
Concentration and Rate
One can gain information about the
rate of a reaction by seeing how the
rate changes with changes in
concentration.
**Generally compared by doubling
the concentration
**NOT A FIXED RULE. READ TABLE
CAREFULLY
Concentration and Rate
• Compare observed initial rate for experiments 1-4, paying
attention to concentrations of each reactant. Then compare
observed initial rate for experiments 5-8, again, paying
attention to concentrations of each reactant. What do you
notice? Then go onto questions below.
Concentration and Rate
N2(g) + 2 H2O(l)
NH4+(aq) + NO2−(aq)
Comparing Experiments 1 and 2, when [NH4+]
doubles, the initial rate doubles.
Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, comparing Experiments 5 and 6,
when [NO2−] doubles, the initial rate
doubles.
• This means
Rate  [NH4+]
Translation: rate is DIRECTLY proportionate to
concentration of NH+
Rate  [NO2−]
Translation: rate is DIRECTLY proportionate to
concentration of NO2-
Rate 
[NH+]
[NO2−]
Rate = k [NH4+] [NO2−]
Rate is directly proportionate to the
product of the concentrations of each
reactant
Rate equation shows relationship
between rate constant and each
reactant
• This equation is called the rate law, and k is
known as the rate constant.
Rate Laws
• A rate law shows the relationship between the
reaction rate and the concentrations of reactants.
• The exponents tell the order of the reaction with
respect to EACH reactant.
• The previous reaction is
First-order in [NH4+] because doubling
concentration doubles the rate. It’s a one to one
relationship.
First-order in [NO2−] because doubling
concentration doubles the rate. It’s a one to one
relationship.
Rate Laws
• The overall reaction order can be
found by adding the exponents on the
reactants in the rate law.
• This reaction is second-order overall.
RECALL: Algebraic manipulations. When
multiplying variables, exponents get
added
a.
1.
2.
3.
4.
2nd order in NO, 1st order in H2, 2nd order overall.
2nd order in NO, 1st order in H2, 3rd order overall.
1st order in NO, 1st order in H2, 2nd order overall.
1st order in NO, 2nd order in H2, 3rd order overall.
a.
1.
2.
3.
4.
2nd order in NO, 1st order in H2, 2nd order overall.
2nd order in NO, 1st order in H2, 3rd order overall.
1st order in NO, 1st order in H2, 2nd order overall.
1st order in NO, 2nd order in H2, 3rd order overall.
b.
1.
2.
Yes
No
b.
1.
2.
Yes
No
Determining Exponents
Determining Exponents for rate laws:
• If the observed rate has:
No change, the exponent is 0
Doubled, the exponent is 1
Quadrupled, the exponent is 2
SAMPLE EXERCISE 6 Determining a Rate Law from Initial Rate Data
The initial rate of a reaction
was
measured for several different starting concentrations of
A and B, and the results are as follows:
Using these data, determine (a) the rate law for the
reaction, (b) the magnitude of the rate constant, (c) the
rate of the reaction when [A] = 0.050 M and [B] = 0.100
M.
SAMPLE EXERCISE 6 continued
Hence, the rate law is
(b) Using the rate law and the data from experiment 1, we have
(c) Using the rate law from part (a) and the rate constant from part (b), we have
PRACTICE EXERCISE
The following data were measured for the reaction of
nitric oxide with hydrogen:
(a) Determine the rate
law for this reaction. (b)
Calculate the rate
constant. (c) Calculate
the rate when
[NO] = 0.050 M and
[H2] = 0.150 M.
Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5  10–4 M/s
SAMPLE EXERCISE 4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction
for which rate =
k[A][B]2. Each of the following boxes represents a
reaction mixture in which A is shown as red spheres and
B as purple ones. Rank these mixtures in order of
increasing rate of reaction.
Thus, the rates vary in the order 2 < 1 < 3.
PRACTICE EXERCISE
Assuming that rate = k[A][B], rank the mixtures
represented in this Sample Exercise in order of
increasing rate.
Answer: 2 = 3 < 1
Integrated Rate Laws
Using calculus to integrate the rate law
for a first-order process gives us
Where
[A]t
ln
[A]0
= −kt
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time,
t, during the course of the reaction.
Integrated Rate Laws
Manipulating this equation produces…
[A]t
ln
= −kt
[A]0
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y
= mx + b
First-Order Processes
Therefore, if a reaction is first-order, a
plot of ln [A]t vs. t will yield a straight
line, and the slope of the line will be -k.
Graph of:
ln [A]t = -kt + ln [A]0
ln[A]t – ln[A]0 = -kt
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to
acetonitrile.
CH3NC
CH3CN
First-Order Processes
CH3NC
This data was
collected for this
reaction at
198.9°C.
CH3CN
First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
 The process is first-order.
 k is the negative slope: 5.1  10-5 P.s−1.
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in
reactant A, we get
1
1
= kt +
[A]t
[A]0
also in the form y = mx + b
1
1
This can be rewritten as:
= kt
[A]t [A]0
Second-Order Processes
1
1
= kt +
[A]t
[A]0
So if a process is second-order in A, a
plot of 1/[A]t vs. t will yield a straight
line, and the slope of that line is k.
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to
the following:
Fill in the columns for ln [NO2]
and 1/[NO2]. Then graph
EACH line separately on the
grid provided
** verify values before graphing
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
0.01000
0.00787
0.00649
0.00481
300.0
0.00380
Time (s)
[NO2], M
ln [NO2]
1/[NO2]
0.0
0.01000
−4.610
100
50.0
0.00787
−4.845
127
100.0
0.00649
−5.038
154
200.0
0.00481
−5.337
208
300.0
0.00380
−5.573
263
Second-Order Processes
• Graphing ln [NO2] vs. t
yields:
• The plot is not a straight
line, so the process is not
first-order in [A].
Time (s)
0.0
50.0
[NO2], M
0.01000
0.00787
ln [NO2]
−4.610
−4.845
100.0
200.0
300.0
0.00649
0.00481
0.00380
−5.038
−5.337
−5.573
Second-Order Processes
• Graphing ln
1/[NO2] vs. t,
however, gives this
plot.
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
1/[NO2]
100
127
0.00649
0.00481
0.00380
154
208
263
• Because this is a
straight line, the
process is secondorder in [A].
SAMPLE EXERCISE 8 Determining Reaction Order from the Integrated Rate Law
The following data were obtained for the gas-phase
decomposition of nitrogen dioxide at 300°C,
Is the reaction first or second
order in NO2?
SAMPLE EXERCISE 8 continued
In order to graph ln[NO2] and 1/[NO2] against
time, we will first prepare the following table from
the data given:
Solve:
Now draw two separate graphs for ln[NO2] and 1/[NO2]
SAMPLE EXERCISE 8 continued
As Figure below shows, only the plot of 1/[NO2] versus time is linear. Thus, the reaction obeys a second-order
rate law: Rate = k[NO2]2. From the slope of this straight-line graph, we determine that k = 0.543 M–1 s–1 for the
disappearance of NO2.
PRACTICE EXERCISE
Figure Kinetic data
for decomposition of
NO2.
The reaction is NO2(g)
NO(g)
+ 1/2O2(g), and the data
were collected at
300°C. (a) A plot of
[NO2] versus time is not
linear, indicating that the
reaction is not first order
in NO2. (b) A plot of
1/[NO2] versus time is
linear, indicating that the
reaction is second order
in NO2.
Consider again the decomposition of NO2 discussed in the Sample Exercise. The reaction is second order in NO 2
with k = 0.543 M–1s–1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining
concentration after 0.500 h?
Answer: Using Equation given, we find [NO2] = 1.00  10–3 M
Half-Life
• Half-life is defined
as the time required
for one-half amount
of a reactant to
react.
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 0.5 [A]0.
Half-Life
For a first-order process, this becomes
0.5 [A]0
ln
= −kt1/2
[A]0
NOTE: For a first-order
process, the half-life
does not depend on [A]0.
ln 0.5 = −kt1/2
−0.693 = −kt1/2
0.693
= t1/2
k
Half-Life
For a second-order process,
1
1
= kt1/2 +
0.5 [A]0
[A]0
2
1
= kt1/2 +
[A]0
[A]0
2 − 1 = 1 = kt
1/2
[A]
[A]0
0
1
= t1/2
k[A]0
Graphs and Kinetic Data
• Plotting concentration of reactant
[REACTANT] (or other relevant value)
against time
• One method of defining rate is to use
the expression below,
Where gradient (slope or slope of tangent)
of such a graph will equal the rate
Rate =
Change in concentration of reactant
Time
Zero Order
• A straight line
shows a zero order
with respect that
reactant. The
slope of the graph
is constant.
Translation:
changing
concentration has
no effect on the
rate.
First Order
Recall: What is half life?
• A constant HALF-LIFE
graph shows a first
order with respect to
that reactant.
 Let’s consider this
graph.
Rate = [0.2] – [0.1] = 0.025
∆t
vs
Rate = [0.1] – [0.05] = 0.0125
∆t
Translation 1: the time required for EACH half-life is the same
Translation 2: when concentration is reduced by half, rate is
reduced by half. 1st order
Second order
0.25
0.2
[Reactant]
• Carefully note: curve of
graph is similar to 1st
order. BUT on closer
inspection, note
significant differences.
Let’s consider rate
Rate = [0.2] – [0.1] =0.002
∆t
vs
Rate = [0.1] – [0.05] = 0.0005
∆t
0.15
0.1
0.05
0
0
50
100
150
200
Time
250
300
350
400
Translation: Half life is NOT constant. Notice that it takes TWICE the amount of
time to reduce concentration by HALF. When concentration is reduced by half,
Rate is reduced to a quarter. That’s a factor of 4, making this a SECOND order
Plotting initial rates against
concentrations of reactants
Summary of graphical interpretations
and order of reaction. MEMORIZE
Practice (try on your own first-then
check answers on next two slides
• Data concerning the change in
concentration of a single reactant A,
in a particular chemical reaction,
are collected and tabulated below.
 By visual inspection of data,
make a prediction about the
order of reaction.
 Graph [A] versus time
 Graph ln[A] vs. time
• Based on the graphs made
 What is the order of this reaction
with respect to A? Justify your
answer
 Use your graph to calculate the
half-life for this reaction
 Calculate the rate constant for
the reaction
Time (min)
[A] in
mol/L
ln [A]
0.000
1.00
0
2.00
0.820
-0.198
4.00
0.670
-0.400
7.00
0.490
-0.713
10.0
0.370
-0.994
14.0
0.240
-1.43
20.0
0.140
-1.97
25.0
0.0800
-2.53
time vs. ln[A]
1.2
0
1
-0.5
0.8
-1
0.6
-1.5
time vs. [A]
0.4
0.2
ln [A]
[A]
time vs. [A]
0
10
20
30
time vs. ln[A]
-2
-2.5
0
0
10
20
30
-3
time
Time
1.
1st order because
a.
b.
2.
3.
half life is constant (in first graph ∆ [] from 1 to 0.490 takes 7
minutes, and from 0.490 to 0.240 takes 7 minutes.
Graph of time vs. ln[A] is a straight line with negative slope,
which corresponds to a 1st order reaction.
Half life is 7 minutes
Rate constant (see next slide)
• Solve for k using
time for half life
(7 minutes)
k = 0.693/7
k = 0.99 /sec
appropriate
units for 1st
order reaction.
Reaction Mechanisms
• A balanced chemical equation tells
us what substances are present at
the beginning of a reaction and
what is formed as the reaction
proceeds.
It does NOT tell us HOW a reaction
occurs
•NEVER assume that the coefficients are
an indication of reaction rate.
Reaction Mechanisms
• The process by which a reaction
occurs is the reaction mechanism
A reaction mechanism can describe
in detail the order in which bonds
are broken and formed and the
changes of relative positions of
atoms in a reaction
Elementary Reactions
• Reactions that occur in a single event or
step are called elementary reactions
Ultimately, ALL complex reactions are a series
of elementary reactions.
• The number of molecules that participate
as reactants in an ELEMENTARY reaction
determine the molecularity of a reaction
RECALL: All successful reactions are still
subject to the criteria of sufficient energy
and correct orientation
Elementary Reactions
• A single molecule is a
unimolecular reaction
 O3  O2 + O
• The collision of two molecules is
a bimolecular reaction
 NO + O3  NO2 + O2
• Three molecules is termolecular
 2NO + O2  2 NO2
Rare for elementary reactions.
Reaction Mechanisms
• The molecularity of elementary reactions DO
indicate the Rate Law.
• The molecularity of a process tells how many
molecules are involved in the process.
 The fewer the molecules involved, the more likely that
collisions will be in correct orientation and therefore
successful.
 Chances of correctly oriented collision (and hence
successful reactions) decrease with increasing
molecularity
Rate law?
1.
2.
3.
4.
Bimolecular
dimolecular
termolecular
Unimolecular
Rate law?
1.
2.
3.
4.
bimolecular
dimolecular
termolecular
Unimolecular
Rate Law
Rate = k[NO][Cl2]
Multistep Mechanisms
• Many balanced equations occur in
multiple steps
• For instance:
NO2 + CO  NO + CO2
• Really occurs as two steps:
NO2 + NO2  NO3 + NO
NO3 + CO  NO2 + CO2
Multistep Mechanisms
• Net equation:
NO2 + CO  NO + CO2
• TASK: Identify the intermediates in the steps
shown here
NO2 + NO2  NO3 + NO
NO3 + CO  NO2 + CO2
• The chemical equations for a multistep
mechanism must always add to give
the chemical equation of the overall
process
• Notice that NO3 is neither a reactant
nor a product in the overall reaction –
it is called an intermediate
SAMPLE EXERCISE 14.12 Determining Molecularity and Identifying Intermediates
It has been proposed that the conversion of ozone into
O2 proceeds by a two-step mechanism:
(a) Describe the molecularity of each elementary
reaction in this mechanism. (b) Write the equation
for the overall reaction. (c) Identify the
intermediate(s).
PRACTICE EXERCISE
For the reaction
the proposed mechanism is
(a) Is the proposed mechanism consistent with the
equation for the overall reaction? (b) What is the
molecularity of each step of the mechanism? (c)
Identify the intermediate(s).
Answers: (a) Yes, the two equations add to yield the equation for the reaction. (b) The first elementary reaction
is unimolecular, and the second one is bimolecular. (c) Mo(CO)5
Rate laws for Elementary Reactions
• Rate laws are ALWAYS determined
experimentally
• Elementary reactions are
significant: if we know that a
reaction is an elementary reaction,
then we know its rate law
Rate laws for Elementary Reactions
The rate law is based on its molecularity
***Coefficient indicates order of
reaction for elementary reactions ONLY.
SAMPLE EXERCISE 14.13 Predicting the Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary
reaction, predict the rate law:
Rate = k[H2][Br2]
Comment: Experimental studies of this reaction show that the reaction actually has a very different rate law:
Rate = k[H2][Br2]1/2
Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we
can conclude that the mechanism must involve two or more elementary steps.
PRACTICE EXERCISE
Consider the following reaction:
(a) Write the rate law for the reaction, assuming it
involves a single elementary reaction. (b) Is a singlestep mechanism likely for this reaction?
Answers: (a) Rate = k[NO]2[Br2] (b) No, because termolecular reactions are very rare
Rate-determining Step for a
Multistep Mechanism
• What do you do when it is not a
simple elementary reaction?
• Most reactions are composed of
two or more elementary steps
• One step is usually slower, and it
is called the rate-determining
step
• assembly line
1.
2.
3.
4.
All reactions are not elementary.
Some information must be known about the rate
constant to determine the rate law.
Concentrations of reactant must be known to
determine the rate law.
The rate law depends not on the overall reaction,
but on the slowest step in the mechanism.
1.
2.
3.
4.
All reactions are not elementary.
Some information must be known about the rate
constant to determine the rate law.
Concentrations of reactant must be known to
determine the rate law.
The rate law depends not on the overall reaction,
but on the slowest step in the mechanism.
Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does NOT depend on its
concentration.
• This suggests the reaction occurs in two steps.
Mechanisms with a Slow Initial
Step
Step 1:
Step 2:
NO2 + NO2
NO3 + CO
k1
k2
NO3 + NO (slow)
NO2 + CO2 (fast)
• Step 1 must be rate determining
and we write the rate of the
overall reaction equaling the rate
of Step 1
Rate = k1[NO2 ]2
Mechanisms with a Slow
Initial Step
Our two step mechanism:
Step 1:
NO2 + NO2
Step 2:
NO3 + CO
k1
k2
NO3 + NO (slow)
NO2 + CO2 (fast)
Overall: NO2 + CO
NO + CO2
Step 2 is much faster than Step 1,  k2 >> k1
The intermediate NO3 is slowly produced in Step 1
& immediately consumed in Step 2
As CO is not involved in the slow, rate-determining
step, it does not appear in the rate law.
SAMPLE EXERCISE 14.14 Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to
occur by a two-step mechanism:
(a) Write the equation for the overall reaction. (b) Write
the rate law for the overall reaction.
(b) The rate law for the overall reaction is just the rate
law for the slow, rate-determining elementary reaction.
Because that slow step is a unimolecular elementary
reaction, the rate law is first order:
Rate = k[N2O]
PRACTICE EXERCISE
Ozone reacts with nitrogen dioxide to produce
dinitrogen pentoxide and oxygen:
The reaction is believed to occur in two steps
The experimental rate law is rate = k[O3][NO2]. What
can you say about the relative rates of the two steps of
the mechanism?
Answer: Because the rate law conforms to the molecularity of the
first step, that must be the rate-determining step. The second step
must be much faster than the first one.
Mechanisms with a Fast Initial
Step
• It is difficult to derive the rate
law for a mechanism in which an
intermediate is a reactant in the
rate-determining step.
• The situation arises in multistep
reactions when the first step is
not rate determining.
Fast Initial Step
2 NO (g) + Br2 (g)  2 NOBr (g)
• The experimentally determined
rate law for this reaction is second
order in NO and first order in Br2:
Rate = k [NO]2 [Br2]
Mechanisms with a Fast Initial
Step
• One possibility is the reaction occurs
in a single termolecular step:
NO + NO + Br2  2NOBr
Rate = k[NO]2[Br2]
• But this is unlikely because
termolecular reactions are rare (why
are they rare?)
Fast Initial Step
• A proposed mechanism is
Step 1: NO + Br2
kf
kr
NOBr2
k2
Step 2: NOBr2 + NO 
2 NOBr
(fast)
(slow)
Step 1 includes the forward and reverse reactions.
Mechanisms with a Fast Initial
Step
• Because step 2 is the slow, ratedetermining step, the rate of the overall
reaction is determined by that step.
Rate law for the rate determining
step becomes
Rate = k[NOBr2][NO]
• We have a problem though: NOBr2 is a
an unstable intermediate, which occurs
in a low unknown concentration
Nearly impossible to isolate and measure.
14.6 – Mechanisms with a Fast
Initial Step
• How do we deal with this? We
make assumptions.
• We assume the NOBr2 is so
intrinsically unstable that it does
not accumulate to a significant
extent
Fast Initial Step
• NOBr2 can react two ways:
With NO to form NOBr
By decomposition to re-form NO and Br2
• The reactants and products of the first
step are in equilibrium with each
other.
• Therefore,
Rateforward = Ratereverse
Mechanisms with a Fast Initial Step
• Because step 2 is slow (forming
NOBr), we assume that most of
the NOBr2 falls back to NO and Br2
• So both the forward and the
reverse reactions of Step 1 are
faster than Step 2 and the forward
and reverse reactions form an
equilibrium
• What will be the rate equation for the
forward and reverse reactions of step 1?
Ratef = k1 [NO] [Br2]
Rater = k−1 [NOBr2]
• Since we know they are in equilibrium
therefore Ratef = Rater
Fast Initial Step
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
• Solving for [NOBr2] gives us
k1
[NO]
[Br
]
=
[NOBr
]
−
2
2
k1
Fast Initial Step
Substituting this expression for [NOBr2]
in the rate law for the ratedetermining step gives
Rate =
k 2k 1
[NO] [Br2] [NO]
−
k1
= k [NO]2 [Br2]
Where k is equal to k2(k1/k1-)
Mechanisms with a Fast Initial
Step
• This mechanism involves only
unimolecular and bimolecular steps,
which is far more likely than
termolecular processes
• In general, whenever a fast step
preceeds a slow one, we can solve for
the concentration of an intermediate
by assuming that an equilibrium is
established in the fast step.
•The kinetics of the reaction: 2X + Y  Z was studied and the results are below:
Exp
[X]o (M) [Y]o (M)
1
2
3
4
0.20
0.20
0.40
0.60
0.10
0.20
0.20
0.60
Initial Rate
(M/s)
7.0 X 10-4
1.4 X 10-3
1.4 X 10-3
4.2 X 10-3
1.Deduce the rate law.
2.The following 3 mechanisms have been proposed. In each case,
a)identify the intermediates
b)write the overall reaction for each mechanism
c)identify the molecularity of each step
d)write the rate law derived from each
e)which mechanism is consistent with the rate law derived in part a?
Mechanism 1:
X + Y  M
X + M  Z
Mechanism 2:
Y  M
X + M  Z
Mechanism 3:
Y  M
M + X  N
N + X  Z
(slow)
(fast)
(slow)
(fast)
(slow)
(fast)
(fast)
Catalysts
• RECALL: A catalyst is any substance that increases the
rate of a reaction while remaining chemically unchanged.
By decreasing activation energy (common definition)
• But what does this mean? How is activation energy
reduced?
Catalysts work in various ways
• Allowing reaction to occur at a lower temperature
• Helping reactants collide with correct orientation
• Providing a surface (usually through adsorption)
• Increasing (effective) concentration
• Providing an alternate reaction pathway
(mechanism)
Types of Catalysts
• Acid-Base
Reactant either loses
or gains H+, forming
a new intermediate—
result: reaction rate
increases
Acid
catalyzed
Base
Catalyzed
• Surface catalysts:
Often a metal (Ni, Pt) working in a
gaseous reactant environment.
Process
• Metal surface adsorbs gas (thin layer formed
on surface) covalent bonds in gas weakened
makes it easier to react
• Effectively “concentration” is also increased
by putting more gas molecules into a smaller
area (only on surface rather than spread
throughout container)
• Enzymes: biological catalysts.
Active sites—interact with substrate
(reactant) molecules lower activation
energy.
Induced fit mechanism—”flexible”
substrates and enzymes allow a single
enzyme to act as catalyst for many
different substrates
• Enzyme may react with substrate to form
completely new intermediate with lower energy
for the transition state.
Representing Catalysts
• Reaction Profile
• Dotted line at lower energy indicates presence of
a catalyst
 An alternate reaction pathway and/or stabilizing the
transition state
• Result: greater number of particles possess the lower Eact and
therefore will result in a successful reaction
Maxwell Boltzmann curve
• RECALL: Maxwell Boltzmann curve shows relationship
between number of particles (y-axis) and energy of
particles (x-axis).
• RECALL: Temperature is a measure of AVERAGE kinetic
energy—meaning in the same system, different particles
have variable ACTUAL energies (as shown in a Maxwell
Boltzmann curve)
• Think about it: If you add a catalyst, is the
temperature/energy of the system decreasing?
 NNNNNOOOOOOOOOOOO
• More particles are reacting because the Eact is reduced
 Translation-MORE particles in the SAME system that
are at a lower energy (temperature!!) CAN react
Maxwell-Boltzmann Curve w/ catalyst
• Notice: Presence of catalyst is denoted by a vertical line
shifting to the LEFT
 Indicates a GREATER AREA under the curve = larger number of
particles with sufficient Eact to react
 ***temperature of system has NOT changed
Maxwell-Boltzmann curve at
different temperatures
• Notice that Eact has NOT changed
 However, at the three different temperatures shown,
increasing temp flattens and shifts the curve to the
right. Therefore number of particles (i.e area under
the curve) that meets minimal Eact to react increases.
• There is NO indication of a catalyst.
The Arrhenius Equation
• The Arrhenius Equatin relates
Rate constants
Activation energy
Temperature
• Takes multiple forms (3 common)
***ALERT: NONE of these appear on the
equations and constants table
Abbreviations
•
•
•
•
•
Ea: Activation energy
R: gas constant in J/mol K
e: base of natural log
T: temperature in Kelvin
A: collision frequency factor
 Collision frequency is
considered constant for a
particular reaction over a
large temperature range,
but IS affected by the
molecularity of the reaction
• RECALL: Higher
molecularity = fewer
effective collisions.
Therefore A is smaller
Equations
k = Ae
(-Ea/RT)
• Shows k increases with
increasing temperature,
but decreases with
increasing activation
energy
Abbreviations
•
•
•
•
•
Ea: Activation energy
R: gas constant in J/mol K
e: base of natural log
T: temperature in Kelvin
A: collision frequency factor
 Collision frequency is
considered constant for a
particular reaction over a
large temperature range,
but IS affected by the
molecularity of the reaction
• RECALL: Higher
molecularity = fewer
effective collisions.
Therefore A is smaller
Equations
ln K=
_ Ea
R
1
T
+ ln A
Taking natural log of the
first equation and
rearranging leads to this
format. It is useful since
it takes the form of y =
mx + b and a plot of ln k
on the y axis again 1/T
on the x axis, will give a
slope (gradient) = -Ea/R
Abbreviations
•
•
•
•
•
Ea: Activation energy
R: gas constant in J/mol K
e: base of natural log
T: temperature in Kelvin
A: collision frequency factor
 Collision frequency is
considered constant for a
particular reaction over a
large temperature range,
but IS affected by the
molecularity of the reaction
• RECALL: Higher
molecularity = fewer
effective collisions.
Therefore A is smaller
Equations
k1
Ea
ln k = R
2
1 _ 1_
T1 T2
• this form relates rate
constants of single
reaction at two separate
temperatures and as in
the first format, notice
that as temperature
increases, so does the
rate constant
• Another form of Arrhenius equation quantitatively
describes effect of orientation factor, p, on rate
constant
k = pAe(-Ea/RT)
where p = orientation factor and
 equal to 1 for the simplest required orientations
• Very simple reactions (H + Cl  HCl) requires no particular orientation
the atoms simply need to collide with sufficient energy
 Equals numbers that are many hundreds of thousands of
times smaller for more complicated reactions that require
more intricate and precise orientations in order for
collisions to be successful.
• Complex molecules ( CO + NO2  CO2 + NO) would likely require a more
precise orientation.
– Likely to have a much lower value for p
 In short, bigger p value = easier required orientation =
larger rate constant = the faster the reaction
k1
Ea
ln k = R
2
1 _ 1_
T1 T2
14.5 - Temperature and Rate
• Generally, as
temperature increases,
so does the reaction
rate.
• This is because 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.
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.
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, E).
• The high point on the
diagram is the
transition state.
Reaction Coordinate Diagrams
• 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.
Maxwell–Boltzmann Distributions
• Temperature is
defined as a
measure of the
average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.
Maxwell–Boltzmann Distributions
• As the temperature
increases, the
curve flattens and
broadens.
• Thus at higher
temperatures, a
larger population of
molecules has
higher energy.
Maxwell–Boltzmann Distributions
• If the dotted line represents the activation
energy, as the temperature increases, so does
the fraction of molecules that can overcome
the activation energy barrier.
• As a result, the
reaction rate
increases.
Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the
expression
−E /RT
f=e
a
where R is the gas constant and T is the Kelvin
temperature.
Catalysis
• Catalyst-substance that changes the
rate of reactions, but is not
permanently changed itself
• Provides and easier way to react
• Typically lowers activation energy.
• Typically provides a completely
different mechanism for a reaction.
Homogeneous Catalyst
• A catalyst present in the same phase as
the reacting molecules
• Example: aqueous hydrogen bromide
and hydrogen peroxide.
2H2O2  H2O + O2
Heterogeneous Catalyst
• A catalyst in a different phase from the
reactant molecules.
• Catalyst is usually a solid and the
reactants gases or liquids.
Heterogeneous Catalysts Cont…
• Terminology
Adsorption: binding of the reactant
molecules to the surface of the metal.
Active Sites: places where reacting
molecules may become adsorbed.
Catalysts
• Catalysts increase the rate of a reaction by
decreasing the activation energy of the
reaction.
• Catalysts change the mechanism by which
the process occurs.
Catalysts
One way a
catalyst can
speed up a
reaction is by
holding the
reactants
together and
helping bonds to
break.
Enzymes
• Enzymes are
catalysts in
biological systems.
• The substrate fits
into the active site
of the enzyme much
like a key fits into a
lock.
Enzyme Inhibitor
• Molecule that has a strong bond with
the active site of an enzyme.
• Blocks substrate entry and prevents
reactions from occurring.