Chemical Kinetics
14
Chemical Kinetics
CONTENTS
Slide 2 of 56
14-1
The Rate of a Chemical
Reaction
14-2
Measuring Reaction Rates
14-3
Effect of Concentration on
Reaction Rates: The Rate Law
14-4
Zero-Order Reactions
14-5
First-Order Reactions
14-6
Second-Order Reactions
14-7
Reaction Kinetics: A Summary
General Chemistry: Chapter 14
Copyright © 2011 Pearson Canada Inc.
Chemical Kinetics
Slide 3 of 56
CONTENTS
14-8
Theoretical Models for
Chemical Kinetics
14-9
The Effect of Temperature on
Reaction Rates
14-10
Reaction Mechanisms
14-11
Catalysis
General Chemistry: Chapter 14
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How fast a reaction occurs depends on the
reactions mechanism- the step by step
molecular pathway leading from reactants
to products.
Chemical kinetics concerns how rate of
chemical reactions are measured, how they
can be predicted and how reaction rate data
are used to deduce probable reaction
mechanism.
Slide 4 of 56
General Chemistry: Chapter 14
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• Reaction rates and reaction order are
determined experimentally.
• Reaction rates and reaction order are critical
to understanding how the reaction progresses
and can be controlled.
Slide 5 of 62
General Chemistry: Chapter 12
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Although stable at room temperature, ammonium dichormate
decomposes very rapidly once ignited:
The rates of chemical reactions and the effect of temperature
on those rates are among several key concepts explored in this
chapter.
(NH4)2Cr2O7  N2(g) + 4H2O(g) + Cr2O3(s)
Slide 6 of 56
General Chemistry: Chapter 14
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Rate, or speed, refers to something that
happens in a unit of time. A car traveling
at 60 mph, for example, covers a distance
of 60 miles in one hour.
For chemical reactions, the rate of reaction
describes how fast the concentration of a
reactant or product changes with time.
Slide 7 of 56
The Rate of a Chemical Reaction
Rate of change of concentration with time.
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
[Fe2+] = 0.0010 M
t = 38.5 s
Δt = 38.5 s
Δ[Fe2+] = (0.0010 – 0) M
Rate of formation of Fe2+=
Δ[Fe2+]
Δt
0.0010 M
=
38.5 s
= 2.610-5 M s-1
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
1 Δ[Fe3+]
Δ[Sn4+]
1 Δ[Fe2+]
= =
Δt
Δt
2 Δt
2
General Rate of Reaction
aA+bB→gG+hH
Rate of reaction = negative of rate of disappearance of reactants
1 Δ[B]
1 Δ[A]
==b Δt
a Δt
= rate of appearance of products
1 Δ[H]
1 Δ[G]
=
=
h Δt
g Δt
Measuring Reaction Rates by measuring the
changes in concentration over time
H2O2(aq) → H2O(l) + ½ O2(g)
Measure the rate by monitoring
volume of O2 formed,
Or, from time to time by
chemical analysis of samples of
the reaction mixture for their
H2O2 content
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ →
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Experimental setup for determining the rate of decomposition of H2O2
or by chemical analysis of aliquots
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ →
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Initial Rate of Reaction
-(-2.32 M / 1360 s) = 1.71 
10-3
M
s-1
Rate =
-Δ[H2O2]
Δt
-(-1.7 M / 2800 s) = Rate of Reaction at time t
6.1  10-4 M s-1
The rate of the reaction is determined from the slope of a tangent line to
a concentration-time curve, this is instantaneous rate of reaction.
The Rate at which a chemical reaction proceeds is
typically influenced by the amount of each reactant
present and the temperature of the reaction vessel.
And, typically, this relationship between the
Reaction Rate and Reagent Concentration is known
as the Rate Law.
Slide 14 of 56
General Chemistry: Chapter 14
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Effect of Concentration on Reaction Rates:
The Rate Law or Rate Equation
a A + b B …. → g G + h H ….
rate of reaction = k[A]m[B]n ….
Rate constant = k
Overall order of reaction = m + n + ….
The terms [A] and [B] represent reactant molarities.
The required exponents, m, n, …are generally small, positive whole numbers,
although in some cases they may be zero, fractional, or negative.
They must be determined by experiment and are generally not related to
stoichiometric coefficients a, b, ….
That is, often m ≠ a, n ≠ b, and so on.
If a reaction has a rate equation of
rate = k[A][B][C] then it is:
A) overall second order
B) overall first order
C) overall third order
D) zero order in A
E) second order in B
Slide 16 of 56
General Chemistry: Chapter 14
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For a reaction Rate = k[A][B]2, what factor will keep
k unchanged?
A) raising temperature
B) adding inhibitor
C) increasing [A]
D) adding catalyst
The reaction has the rate law Rate = k[A][B]2.
Which will cause the rate to increase the most?
A) doubling [A]
B) lowering temperature
C) tripling [B]
D) quadrupling [A]
E) doubling [B]
Slide 18 of 56
General Chemistry: Chapter 14
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If increasing the concentration of A in a chemical
reaction causes no increase in the rate of the
reaction, then we may say:
A) A is a catalyst
B) the reaction rate is zero order in A
C) the reaction rate is zero order in [A]
D) the reaction rate is first order in [A]
E) A is not involved in the reaction
Slide 19 of 56
General Chemistry: Chapter 14
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Establishing the Order of a reaction
Method of Initial Rates
Use the data provided establish the order of the reaction
with respect to HgCl2 and C2O42- and also the
overall order of the reaction.
2 HgCl2(aq) + C2O42-(aq)
2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)
rate of reaction = k[HgCl2]m[C2O42-]n
General effect of doubling the initial concentration
of a particular reactant (with other reactant
concentrations held constant).
• Zero order in the reactant—there is no effect on the
initial rate of reaction.
• First order in the reactant—the initial rate of reaction
doubles.
• Second order in the reactant—the initial rate of reaction
quadruples.
• Third order in the reactant—the initial rate of reaction
increases eightfold.
Slide 21 of 56
General Chemistry: Chapter 14
For 2NO + O2 → 2NO2, initial rate data are:
[NO] 0.010 0.010 0.030 M
[O2] 0.010 0.020 0.020 M
rate 2.5
5.0
45.0 mM/sec
The rate law is Rate = k[NO]x[O2]y:
A) x = 1, y = 2
B) x = 2, y = 1
C) x = 1, y = 1
D) x = 2, y = 2
E) x = 0, y = 2
Slide 22 of 56
General Chemistry: Chapter 14
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The rate of a specific chemical reaction is independent of
the concentrations of the reactants. Thus the reaction is:
A) first order in A
B) second order
C) first order in the product
D) catalyzed
E) overall zero order
Another useful equation is:
Integrated Rate Law
-Δ[A]
Δt
= k
Move to the
-d[A]
infinitesimal
dt
= k
And integrate from 0 to time t
t
[A]t
- d[A]
[A]0
=  k dt
0
-[A]t + [A]0 = kt
[A]t = - kt + [A]0
Slide 24 of 56
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14-4 Zero-Order Reactions
A → products
Rrxn = k [A]0
Rrxn = k =
(0 - [A]0)
(tf – 0)
[k] = mol L-1 s-1
FIGURE 14-3
A zero-order reaction: A
products
=
[A]0
tf
14-5 First-Order Reactions
H2O2(aq) → H2O(l) + ½ O2(g)
d[H2O2 ]
= -k[H2O2]
dt
[k] = s-1
An Integrated rate Law for a First-Order Reaction
[A]t

[A]0
ln
[A]t
[A]0
Slide 26 of 56
d[H2O2 ]
[H2O2]
= -kt
t
= -  k dt
0
ln[A]t = -kt + ln[A]0
General Chemistry: Chapter 14
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Test for a first-order reaction:
Decomposition of H2O2(aq)
Slide 27 of 56
General Chemistry: Chapter 14
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14-6 Second-Order Reactions
Rate law where sum of exponents m + n +… = 2
A → products
d[A]
dt
= -k[A]2
[A]t

[A]0
d[A]
[A]2
[k] = M-1 s-1 = L mol-1 s-1
t
= -  k dt
0
1
1
= kt +
[A]t
[A]0
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FIGURE 14-6
A straight-line plot for the second order reaction A
Slide 29 of 56
General Chemistry: Chapter 14
products
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Integrating our other simple Rate Laws into their
“Linear” form provides us with:
If we do not know the order of a given reaction, we can
simply plot the data in all three “Linear” forms and see which
results in a straight-line.
Testing for a Rate Law
Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t.
2 N2O5(g) → 2 N2O4(g) + O2(g)
Time (sec)
0
200
400
600
800
1000
1200
1400
1600
1800
[N2O5] (M)
1.40
1.24
1.10
0.98
0.87
0.77
0.68
0.60
0.53
0.47
This reaction has been shown to be First Order in
N2O5; meaning the Rate Law can be written as:
Rate = k [N2O5]
or in Integral form as:
ln [N2O5] = ln [N2O5]o - k t
Thus, a plot of the Natural Log of the above
Concentration data vs. Time should give us a
straight line with a slope = k. This is in fact the
case.
Determining Orders of Reactions I) Getting the data
a) Quench the reaction, measure concentrations
b) For gas phase, measure pressure vs. time
c) Spectroscopically follow reactants/products
Etc…
II) Analyzing the data A) Reactions with one reactant:
A → products
a) Plot or analyze [A] vs. t ln[A] vs. t 1/[A] vs. t …
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General Chemistry: Chapter 14
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Zero-Order Reactions
[A]t = - kt + [A]0
[k] = mol L-1 s-1
First -Order Reactions
ln[A]t = -kt + ln[A]0
[k] = s-1
Second -Order Reactions
1
[A]t
= kt +
1
[A]0
[k] = M-1 s-1 = L mol-1 s-1
Half-Life, t½
the time taken for one-half of a reactant to be consumed.
ln
[A]t
[A]0
ln
= -kt
½[A]0
= -kt½
[A]0
- ln 2 = -kt½
ln 2
0.693
t½ =
=
k
k
14-9 Effect of Temperature on Reaction Rates
Svante Arrhenius demonstrated that many
rate constants vary with temperature
according to this equation:
k = Ae-Ea/RT
ln k = -Ea 1 + lnA
R T
k=
-Ea 1
ln k =
Ae-Ea/RT
T
-Ea 1
ln k2– ln k1 =
R
ln
R
+ ln A
k2
k1
=
1
-E
a
+ ln A - ln A
T2
R T1
Ea
1
R
T1
-
1
T2
General Chemistry: Chapter 14
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N2O5(CCl4) → N2O4(CCl4) + ½ O2(g)
-Ea
= -1.2104 K
R
-Ea = 1.0102 kJ mol-1
Temperature dependence of the rate constant k for a reaction
14-9 Effect of Temperature on Reaction Rates
Svante Arrhenius demonstrated that many
rate constants vary with temperature
according to this equation:
k = Ae-Ea/RT
ln k = -Ea 1 + lnA
R T
An ancient shipwreck has been discovered off Greenland.
An unopened 1 liter bottle of wine brought to the
surface is opened and found to smell strongly like
vinegar. The contents are analyzed later and also found
to contain 1.2 ×10-4 moles of ethanol. It is known that
ethanol decomposes to acetic acid as below:
2CH3CH2OH → CH3COOH + C2H4 + H2
Ea= 121.3 KJ/mol, k = 40.9 M-1month-1 @ 25 ◦ C
Fermentation can only produce wine with a 3M of
ethanol. The bottle was submerged in 4 ◦ C water for
centuries. How many years was the ship submerged?
Slide 42 of 56
General Chemistry: Chapter 14
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k=
-Ea 1
ln k =
Ae-Ea/RT
T
-Ea 1
ln k2– ln k1 =
R
ln
R
+ ln A
k1
k2
=
1
+E
a
+ ln A - ln A
T2
R T1
-Ea 1
R
T1
-
1
T2
General Chemistry: Chapter 14
Copyright © 2011 Pearson Canada Inc.
N2O5(CCl4) → N2O4(CCl4) + ½ O2(g)
-Ea
= -1.2104 K
R
-Ea = 1.0102 kJ mol-1
Temperature dependence of the rate constant k for a reaction
14-8 Theoretical Models for Chemical Kinetics
Collision Theory
In gases 1030 collisions per liter per second.
If each collision produced a reaction, the rate would be
about 106 M s-1, extremely rapid!
Actual rates are on the order of 10-4 M s-1.
So,
Only a fraction of collisions yield a reaction.
Activation Energy:
The minimum energy that molecules must bring
to their collisions for a chemical reaction to
occur.
For a reaction to occur there must be a
redistribution of energy sufficient to break certain
bonds in the reacting molecule(s).
FIGURE 14-11
An analogy for a reaction profile and activation energy
Collision Theory
The rate of the reaction depends on how often the
molecules with sufficient kinetic energy are likely to
collide with each other.
If activation barrier is high, only a few molecules
have sufficient kinetic energy and the reaction is
slower.
As temperature increases, reaction rate increases.
Orientation of molecules may be important.
Distribution of molecular kinetic energy
Distribution of molecular kinetic energies
These are “activated”;
the molecules whose
molecular collisions lead
to chemical reaction
FIGURE 14-9
Molecular collisions and chemical reactions
In the Arrhenius equation, ln k = -Ea/RT + ln A,
the symbol A denotes:
A) the initial concentration of A
B) the activation energy
C) the rate constant
D) a constant that represents the frequency of
collisions with the proper orientation and other
steric conditions favorable for a reaction
E) the absolute temperature
Transition State Theory
The activated complex is a
hypothetical species lying
between reactants and products
at a point on the
reaction profile called
the transition state.
A reaction profile for the reaction N2O(g) + NO(g)
N2(g) + NO2(g)
According to the collision theory in gaseous molecules,
collision frequency is ________ and reaction rate is
________ because ________.
A) low, low, molecules are so far apart
B) high, high, each collision results in a reaction
C) low, low, molecules must collide before they can react
D) high, relatively low, only a fraction of the collisions
lead to a reaction
E) low, high, molecules are moving so fast that each
reaction causes many others
14-10 Reaction Mechanisms
Step-by-step description of a reaction.
Each step is called an elementary process.
Any molecular event that significantly alters a
molecule’s energy or geometry or produces a new
molecule.
Reaction mechanism must be consistent with:
1.Stoichiometry for the overall reaction.
2.The experimentally determined rate law.
Example, SN1
Elementary Processes
Unimolecular or bimolecular.
Exponents for concentration terms are the same as the
stoichiometric factors for the elementary process.
Elementary processes are reversible.
Intermediates are produced in one elementary process
and consumed in another. Intermediates must not appear
in the overall equation or the overall rate law.
One elementary step is usually slower than all the others
and is known as the rate determining step.
The SN1 Mechanism
A mechanism with a
Slow Step Followed by a Fast Step
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
d[P]
dt
= k[H2][ICl]
Postulate a mechanism:
H2(g) + ICl(g)
slow
d[HI]
HI(g) + HCl(g)
HI(g) + ICl(g) fast I2(g) + HCl(g)
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
dt
d[I2]
dt
d[P]
dt
= k[H2][ICl]
= k[HI][ICl]
= k[H2][ICl]
FIGURE 14-14
A reaction profile for a two-step mechanism
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Molecularity:
Molecularity is the number of molecules that need to collide,
and in one step form the products.
For single step elementary reactions, Molecularity = Order
A → products; 1st order rate = k[A] Unimolecular
A → products; 2nd order rate = k[A]2 Bimolecular
A + B → prod.; 2nd order rate = k[A][B] Bimolecular
A + B + C → prod. 3rd order rate = k[A][B][C] Termolecular
Choose the INCORRECT statement.
A) The rate-determining step is always the first step.
B) A unimolecular process is one in which a single molecule
dissociates.
C) A bimolecular process is one involving a collision of two
molecules.
D) A reaction mechanism is a step-by-step detailed
description of a chemical reaction.
E) An elementary process is a step in the mechanism.
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What is the rate law for the following
mechanism?
N2O + NO → N2ONO (Slow)
N2ONO → N2 + NO2 (Fast)
A) Rate = k[N2O]
B) Rate = k[NO]
C) Rate = k[N2O][NO]
D) Rate = k[N2][NO2]
E) Rate = k[N2ONO]
What is the rate law for the following mechanism?
CH3COOC2H5 + H2O → CH3COOC2H6+ + OH- (Slow)
CH3COOC2H6+ → CH3COOH + C2H5+
C2H5+ + OH- → C2H5OH
(Fast)
A) Rate = k[CH3COOC2H5][H2O]2
B) Rate = k[C2H5OH]
C) Rate = k[CH3COOH]
D) Rate = k[CH3COOC2H5]
E) Rate = k[CH3COOC2H5][H2O]
(Fast)
What is the rate law for the following reaction and
its mechanism?
2O3 → 3O2
(overall reaction)
O3 → O2 + O.
(Slow)
O∙ + O3 → 2O2
(Fast)
A) Rate = k[O3]
B) Rate = k[O3]2
C) Rate = k[O3]2/[O2]
D) Rate = k[O3]/[O2]
E) Rate = k[O3][O2]
14-5 Catalysis
• Alternative reaction pathway of lower energy.
• Homogeneous catalysis.
All species in the reaction are in solution.
• Heterogeneous catalysis.
The catalyst is in the solid state.
Reactants from gas or solution phase are adsorbed.
Active sites on the catalytic surface are important.
14-5 Catalysis
An example of homogeneous catalysis
Catalysis on a Surface
Heterogeneous catalysis in the reaction
2 CO + 2 NO
2 CO2 + N2
Figure 14-19
Reaction profile for a surface-catalyzed
reaction
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E+S
k1
k-1
ES
k2
ES → E + P
Lock-and-key model of enzyme action
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A catalyst:
I) lowers activation energy
II) provides an alternate reaction pathway
III)
is consumed in the reaction and therefore does not appear
in the chemical equation of each mechanism
IV)
speeds a reaction
V)is heterogeneous if it is in a different phase than the reactants
A) I, III, and IV
B) I, IV, and V
C) II, III, and IV
D) II and IV
E) I, II, IV, and V
If a catalyst is added to a reaction:
I) the value of k is increased
II) the value of k is decreased
III)
the rate is increased
IV)
the rate is decreased
V)neither rate nor the constant are changed, only the order
A) I and IV
B) II and IV
C) II and III
D) I and III
E) V only
A factor that decreases the activation
energy for a reaction:
I)
decreases the rate constant
II)
increases the rate constant
III) has no effect on the rate constant
IV) makes the product yield increase
V)
might be a catalyst
A) I and IV
B) II and IV
C) I, IV, and V
D) IV and III
E) II and V
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Enzymes as Catalysts
t
E+S
k1
k-1
k2
ES → E + P
d[P]
dt
d[P]
dt
= k2[ES]
= k1[E][S] – k-1[ES] – k2[ES]= 0
k1[E][S] = (k-1+ k2 )[ES]
[E] = [E]0 – [ES]
k1[S]([E]0 –[ES]) = (k-1+ k2 )[ES]
[ES] =
k1[E]0 [S]
(k-1+ k2 ) + k1[S]
FIGURE 14-21
Effect of substrate concentration on the rate of an enzyme reaction.
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E+S
k1
k-1
k2
ES → E + P
d[P]
=
(k-1+ k2 ) + k1[S]
dt
d[P]
= k2[E]0
dt
d[P]
k1k2[E]0 [S]
k2[E]0 [S]
=
(k-1+ k2 ) + [S]
dt
k1
d[P]
dt
=
k2
[E]0 [S]
KM
d[P]
dt
=
k2[E]0 [S]
KM + [S]
FIGURE 14-21
Effect of substrate concentration on the rate of an enzyme reaction.
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End of Chapter Questions
Dimensional Analysis is your friend.
Never leave units off of a number.
You are better off leaving off the numerical part
of the number and working ONLY with the
units.
The units must correctly cancel out.
The units left after that process must be the
correct units for your answer.
Only then should you calculate.
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Kinetic equations
[A]t = - kt + [A]0
R = 0.082057 L atm mol-1 K-1
ln[A]t = -kt + ln[A]0
1
R = 8.3145 m3 Pa mol-1 K-1
1
= kt +
[A]t
[A]0
R = 8.3145 J mol-1 K-1
k = Ae-Ea/RT
ln
k1
k2
=
-Ea 1
R
T1
-
1
T2