Chapter 12
Chemical Kinetics
Chapter 12
Table of Contents
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Reaction Rates
Rate Laws: An Introduction
Determining the Form of the Rate Law
The Integrated Rate Law
Reaction Mechanisms
A Model for Chemical Kinetics
Catalysis
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2
Section 12.1
Reaction Rates
Objectives
• To define Chemical Kinetics
• To define Reaction Rate and illustrate how it is
determined
• To illustrate that the Instantaneous Rate is
calculated by determining the slope of the
tangent line at a given point on a curve.
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3
Section 12.1
Reaction Rates
Chemical Kinetics
• Chemical Kinetics – the area of chemistry that
studies the rate at which a chemical reaction takes
place.
• The following reaction is considered spontaneous
(favorable) and will release energy:
N2 (g) + 3H2 (g)  2NH3(g) + energy
• However, the two gas reactants may coexist
indefinitely under normal conditions.
• Not so good if you are in the fertilizer business!
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4
Section 12.1
Reaction Rates
Reaction Rate
• Change in concentration of a reactant or
product per unit time.
Rate =
=
concentration of A at time t2  concentration of A at time t1
t2  t1
 A
t
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
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5
Section 12.1
Reaction Rates
The Decomposition of Nitrogen Dioxide
2NO2(g)  2NO(g) + O2(g) (@ 300OC)
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6
Section 12.1
Reaction Rates
The Decomposition
of Nitrogen Dioxide
2NO2 2 NO + O2
Rate =  [NO2]/  t
Slope! Average if you
use the data points
or instantaneous if
you use the tangent
line at a given point
for a given time.
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7
Section 12.1
Reaction Rates
Objectives Review
• To define Chemical Kinetics
• To define Reaction Rate and illustrate how it is
determined using a slope calculation
• To illustrate that the Instantaneous Rate is
calculated by determining the slope of the
tangent line at a given point on a curve.
• Work Session: page 566 #7, 19, 21
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8
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Objectives
• To define the Rate Law Equation and its variables
• To determine the form (order) of the rate law
experimentally
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9
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
• The Rate Law shows how the rate depends on the
[concentrations] of reactants . [Reactant] units of M
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n
 k = rate constant (coefficient of parabola or m if linear)
 [NO2] = Concentration of reactant (M = mol/L)
 n = order of the reactant (form)
This form of the rate law allows you to model the
continuous curve regardless of the order (exponent).
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
Rate = k[NO2]n
• The concentrations of the products do not appear in
the rate law because the reaction rate is being
studied under conditions where the reverse reaction
does not contribute to the overall rate.
• 2NO2(g)
2NO(g) + O2(g)
• The value of the exponent n must be determined by
experiment; it cannot be written from the balanced
equation.
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11
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
Rate = k[NO2]n
Rate =  [NO2]/  t
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12
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
Rate = k[NO2]n (How to Calculate n experimentally)
• Experiment Initial [NO2]
Initial Rate(mol/L*s)
•
1
0.450 M
1.25 X 10-5
•
2
0.90 M
2.50 X 10-5
• Rate = k[NO2]n write the expression for both exp’ts
the bigger value in the numerator….
• Rate 2 = k[0.9]n = 2.50 X 10-5
• Rate1 = k[0.45]n = 1.25 X 10-5
• Simplify and solve for n
•
[2]n = 2
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
•
•
•
•
•
•
Rate 2 = k[0.9]n = 2.50 X 10-5
Rate1 = k[0.45]n = 1.25 X 10-5
Simplify and solve for n
[2]n = 2
n = 1 or in other words,
The rate law for this reaction is first order with
respect to the reactant NO2. (Important to define)
• 2NO2(g)  2NO(g) + O2(g)
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14
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Experimental convenience usually dictates
which type of rate law is determined
experimentally.
• Knowing the rate law for a reaction is important
mainly because we can usually infer the
individual steps involved in the reaction from the
specific form of the rate law. We are usually
looking for the slow step in a process.
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15
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Objectives Review
• To define the Rate Law Equation and its variables
• To determine the form (order) of the rate law
experimentally
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16
Section 12.3
The Mole
Determining
the Form of the Rate Law
Objectives
• To determine the overall reaction order of the rate
law experimentally using data from several reactants
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Overall Reaction Order
• The sum of the exponents in the reaction rate
equation.
aA + bB  cC + …?
Rate = k[A]n[B]m
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
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18
Section 12.3
The Mole
Determining
the Form of the Rate Law
• Let’s use the following EXPERIMENTAL data to
determine the order (form) for each reactant.
• NH4+ + NO2-  N2 + 2H2O
• First, write the general rate law……Rate = ??
• Exp’t Initial
Initial
Initial Rate
•
[NH4+ ]
[NO2- ]
(mol/L*s)
• 1
0.100 M
0.0050 M
1.35 X 10-7
• 2
0.100 M
0.010 M
2.70 X 10-7
• 3
0.200 M
0.010 M
5.40 X 10-7
• Remember the trick to find rate order = bigger numerator
• For order w/respect to [NH4+ ], [other] must stay constant
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19
Section 12.3
The Mole
Determining
the Form of the Rate Law
•
•
•
•
•
•
•
•
•
•
Overall Reaction Order
Remember bigger numerator!
NH4+ + NO2-  N2 + 2H2O
Exp’t Initial
Initial
[NH4+ ]
[NO2- ]
Initial Rate
(mol/L*s)
1
0.100 M
0.0050 M
1.35 X 10-7
2
0.100 M
0.010 M
2.70 X 10-7
3
0.200 M
0.010 M
5.40 X 10-7
Rate 2 = k[0.100]n [0.010]m = 2.70 X 10-7
Rate1 = k[0.100]n [0.0050]m = 1.35 X 10-7
Simplify and solve for m
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20
Section 12.3
The Mole
Determining
the Form of the Rate Law
•
•
•
•
•
•
•
•
•
•
NH4+ + NO2-  N2 + 2H2O
Exp’t Initial
Initial
[NH4+ ]
[NO2- ]
Initial Rate
(mol/L*s)
1
0.100 M
0.0050 M
1.35 X 10-7
2
0.100 M
0.010 M
2.70 X 10-7
3
0.200 M
0.010 M
5.40 X 10-7
Rate 2 = k[0.100]n [0.010]m = 2.70 X 10-7
Rate1 = k[0.100]n [0.0050]m = 1.35 X 10-7
Simplify and solve for m
[2]m = 2
The rate law for this reaction is first order in the
reactant NO2-. What about the other reactant?
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21
Section 12.3
The Mole
Determining
the Form of the Rate Law
•
•
•
•
•
•
•
•
•
•
•
With respect to NH4+ (keep other [conc] same)
NH4+ + NO2-  N2 + 2H2O
Exp’t Initial
Initial
Initial Rate
[NH4+ ]
[NO2- ]
(mol/L*s)
1
0.100 M
0.0050 M
1.35 X 10-7
2
0.100 M
0.010 M
2.70 X 10-7
3
0.200 M
0.010 M
5.40 X 10-7
Rate 3 = k[0.200]n [0.010]m = 5.40 X 10-7
Rate2 = k[0.100]n [0.010]m
= 2.70 X 10-7
Simplify and solve for n
Also first order with respect to NH4+
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22
Section 12.3
The Mole
Determining
the Form of the Rate Law
Overall Reaction Order
• The sum of the exponents in the reaction rate
equation.
aA + bB  cC + …?
Rate = k[A]n[B]m
Overall reaction order = n + m
m=1,n=1
Overall Reaction order
=2
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23
Section 12.3
The Mole
Determining
the Form of the Rate Law
• Turn to page 567 and work on #25.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Concept Check
How do exponents (orders) in rate laws
compare to coefficients in balanced equations?
Why?
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25
Section 12.3
The Mole
Determining
the Form of the Rate Law
Objectives Review
• To determine the overall reaction order of the rate
law experimentally using data from several reactants
• Work Session: # 26, 27, 28 (just find the order)
• CLICK to Slide 43 to skip Integrated Rate Law
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26
Section 12.4
The Integrated Rate Law
Half-Life of Reactions
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Section 12.4
The Integrated Rate Law
First-Order
• Rate = k[A]
• Integrated:
ln[A] = –kt + ln[A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of ln[N2O5] vs Time
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Section 12.4
The Integrated Rate Law
First-Order
• Half–Life:
t1
2
0.693
=
k
k = rate constant
• Half–life does not depend on the
concentration of reactants.
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Section 12.4
The Integrated Rate Law
Exercise
A first order reaction is 35% complete at the
end of 55 minutes. What is the value of k?
k = 7.8 x 10–3 min–1
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Section 12.4
The Integrated Rate Law
Second-Order
• Rate = k[A]2
• Integrated:
1
1
= kt +
A
 A 0
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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32
Section 12.4
The Integrated Rate Law
Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
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Section 12.4
The Integrated Rate Law
Second-Order
• Half–Life:
t1
2
1
=
k  A 0
k = rate constant
[A]o = initial concentration of A
• Half–life gets longer as the reaction progresses
and the concentration of reactants decrease.
• Each successive half–life is double the
preceding one.
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Section 12.4
The Integrated Rate Law
Exercise
For a reaction aA  Products,
[A]0 = 5.0 M, and the first two half-lives are 25
and 50 minutes, respectively.
a) Write the rate law for this reaction.
rate = k[A]2
b) Calculate k.
k = 8.0 x 10-3 M–1min–1
c) Calculate [A] at t = 525 minutes.
[A] = 0.23 M
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35
Section 12.4
The Integrated Rate Law
Zero-Order
• Rate = k[A]0 = k
• Integrated:
[A] = –kt + [A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of [A] vs Time
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Section 12.4
The Integrated Rate Law
Zero-Order
• Half–Life:
t1 =
2
 A 0
2k
k = rate constant
[A]o = initial concentration of A
• Half–life gets shorter as the reaction
progresses and the concentration of reactants
decrease.
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38
Section 12.4
The Integrated Rate Law
Concept Check
How can you tell the difference among 0th, 1st,
and 2nd order rate laws from their graphs?
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Section 12.4
The Integrated Rate Law
Rate Laws
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Section 12.4
The Integrated Rate Law
Summary of the Rate Laws
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41
Section 12.4
The Integrated Rate Law
Exercise
Consider the reaction aA  Products.
[A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units
are appropriate for each case). Calculate [A] after
30.0 seconds have passed, assuming the
reaction is:
a) Zero order
b) First order
c) Second order
4.7 M
3.7 M
2.0 M
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42
Section 12.6
Reaction Mechanisms
Objectives
• To define reaction mechanism
• To determine if a mechanism is a possible
explanation of observed behavior
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Section 12.6
Reaction Mechanisms
Reaction Mechanism
• Most chemical reactions occur by a series
of elementary steps.
• An intermediate is formed in one step and
used up in a subsequent step and thus is
never seen as a reactant or product in the
overall balanced reaction.
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44
Section 12.6
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
.
--------------------------------------------------------------------NO2(g) + NO2(g) → NO(g) + NO3(g)
.
NO3(g) + CO(g) → NO2(g) + CO2(g) intermediate?
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45
Section 12.6
Reaction Mechanisms
Elementary Steps (Molecularity) Page 550
• Molecularity – the number of species that must collide
to produce the reaction in that step.
• Unimolecular – reaction involving one molecule; first
order.
• A  products
Rate = k[A]
• Bimolecular – reaction involving the collision of two
species; second order.
• A + A  products Rate = k[A]2
• A + B  products Rate=k[A][B]
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Section 12.6
Reaction Mechanisms
Elementary Steps (Molecularity) Page 550
• Termolecular – reaction involving the collision of
three species; third order.
• A + A + B  products
Rate = [ ][ ][ ] = [ ] [ ]
• A + B + C  products
Rate =
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Section 12.6
Reaction Mechanisms
Rate-Determining Step
• A reaction is only as fast as its slowest
step.
• The rate-determining step (slowest step)
determines the rate law and the
molecularity of the overall reaction.
• Focus on the slow step…..
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Section 12.6
Reaction Mechanisms
Reaction Mechanism Criteria (must meet All 3)
• The sum of the elementary steps must give
the overall balanced equation for the
reaction. (Like Hess’s Law)
• The mechanism must agree with the
experimentally determined rate law. (Given)
• If these two conditions are met, the
mechanism is possibly correct as a
mechanism can never be proved absolutely
correct.
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49
Section 12.6
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
--------------------------------------------------------------------NO2(g) + NO2(g) → NO(g) + NO3(g) (slow)
NO3(g) + CO(g) → NO2(g) + CO2(g) (fast)
•
•
•
•
Does the sum of the elementary steps give the overall
balanced equation for the reaction?
Does the mechanism agree with the experimentally
determined rate law given as Rate = k[NO2]2 ?
Is this a possible mechanism for this reaction?
The overall rate can not be faster than the slow step.
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50
Section 12.6
Reaction Mechanisms
Decomposition of N2O5
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Section 12.6
Reaction Mechanisms
Decomposition of N2O5
Is this a possible Mechanism? 1?2?
2N2O5(g)  4NO2(g) + O2(g)
Step 1: 2(N2O5
NO2 + NO3 )
(fast)
Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Step 3: NO3 + NO → 2NO2
(fast)
•
•
•
Does the sum of the elementary steps give the overall
balanced equation for the reaction?
Does the mechanism agree with the experimentally
determined rate law??
Is this a possible mechanism for this reaction?
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Section 12.6
Reaction Mechanisms
Write the balanced reaction for the synthesis of gases
nitrogen dioxide and fluorine that forms 1 product.
____NO2 + ____F2  ____ NO2F
The experimentally determined rate law is:
Rate = k[NO2][F2]
A suggested mechanism is:
NO2 + F2  NO2F + F (slow)
F + NO2  NO2F
(fast)
Is this an acceptable mechanism?
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Section 12.6
Reaction Mechanisms
The experimentally determined rate law is:
Rate = k[NO2][F2]
A suggested mechanism is:
NO2 + F2  NO2F + F (slow)
F + NO2  NO2F
(fast)
Is this an acceptable mechanism?
• Does the sum of the elementary steps give the overall
balanced equation for the reaction?
• Does the mechanism agree with the experimentally
determined rate law??
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Section 12.6
Reaction Mechanisms
Concept Check (sometimes the Rate Law Looks
Tricky..
The reaction A + 2B  C has the following
proposed mechanism:
A+B
D
(fast equilibrium)
D+BC
(slow)
Does this rate law make sense?
rate = k[A][B]2
A has to collide with B, then with another B
Why wouldn’t rate = [B] work? Slow step.
D is really A + B
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55
Section 12.6
Reaction Mechanisms
Objectives Review
• To define reaction mechanism
• To determine if a mechanism is a possible
explanation of observed behavior
• Work Session: 49, 51
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Section 12.7
A Model for Chemical Kinetics
Objectives
• To define activation energy EA
• To understand EA on a Potential Energy
Graph
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Section 12.7
A Model for Chemical Kinetics
Arrhenius Equation
• Molecules must collide to react.
• Main Factors:
k = Ae
k =
A =
Ea =
R =
T =
 Ea / RT
rate law constant
frequency factor (Molecular Orientations)
activation energy
gas constant (8.3145 J/K·mol)
temperature (in K)
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Section 12.7
A Model for Chemical Kinetics
Activation Energy, Ea
• Energy that must be overcome to produce
a chemical reaction.
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Section 12.7
A Model for Chemical Kinetics
Transition States and Activation Energy
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Section 12.7
A Model for Chemical Kinetics
Change in Potential Energy
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Section 12.7
A Model for Chemical Kinetics
For Reactants to Form Products
• Collision must involve enough energy to
produce the reaction (must equal or
exceed the activation energy).
• Relative orientation of the reactants must
allow formation of any new bonds
necessary to produce products.
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Section 12.7
A Model for Chemical Kinetics
The Gas Phase Reaction of NO and Cl2
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63
Section 12.7
A Model for Chemical Kinetics
Objectives Review
• To define activation energy EA
• To understand EA on a Potential Energy
Graph
• Work Session: Page 571 #53
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Section 12.8
Catalysis
Objectives
• To define a catalyst
• To identify catalysis on a Potential Energy
Graph
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65
Section 12.8
Catalysis
Catalyst
• A substance that speeds up a reaction
without being consumed itself.
• Provides a new pathway for the reaction
with a lower activation energy.
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Section 12.8
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a
Given Reaction
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Section 12.8
Catalysis
Effect of a Catalyst on the Number of Reaction-Producing
Collisions
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Section 12.8
Catalysis
Heterogeneous Catalysis
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Section 12.8
Catalysis
Homogeneous Catalysis
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Section 12.8
Catalysis
Objectives Review
• To define a catalyst
• To identify catalysis on a Potential Energy
Graph
• Work Session Page 572 # 65(a, b) #68a
• Catalyst Video 1:40
• End of Chapter 12!
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Section 12.8
Catalysis
Heterogeneous Catalyst
• Most often involves gaseous reactants
being adsorbed on the surface of a solid
catalyst.
• Adsorption – collection of one substance
on the surface of another substance.
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Section 12.8
Catalysis
Heterogeneous Catalyst
1. Adsorption and activation of the reactants.
2. Migration of the adsorbed reactants on the
surface.
3. Reaction of the adsorbed substances.
4. Escape, or desorption, of the products.
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Section 12.8
Catalysis
Homogeneous Catalyst
• Exists in the same phase as the reacting
molecules.
• Enzymes are nature’s catalysts.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Types of Rate Laws
• Differential Rate Law (rate law) – shows how
the rate of a reaction depends on
concentrations.
• Integrated Rate Law – shows how the
concentrations of species in the reaction
depend on time.
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75
Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Because we typically consider reactions only
under conditions where the reverse reaction is
unimportant, our rate laws will involve only
concentrations of reactants.
• Because the differential and integrated rate
laws for a given reaction are related in a well–
defined way, the experimental determination of
either of the rate laws is sufficient.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
• Determine experimentally the power to which
each reactant concentration must be raised in
the rate law.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Method of Initial Rates
• The value of the initial rate is determined for
each experiment at the same value of t as close
to t = 0 as possible.
• Several experiments are carried out using
different initial concentrations of each of the
reactants, and the initial rate is determined for
each run.
• The results are then compared to see how the
initial rate depends on the initial concentrations
of each of the reactants.
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Section 12.7
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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Section 12.7
A Model for Chemical Kinetics
Exercise
Chemists commonly use a rule of thumb that an
increase of 10 K in temperature doubles the
rate of a reaction. What must the activation
energy be for this statement to be true for a
temperature increase from 25°C to 35°C?
Ea = 53 kJ
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80
Section 12.7
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
Ea  1 
ln(k ) =    + ln  A 
R T
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81
Section 12.7
A Model for Chemical Kinetics
Collision Model
• Molecules must collide to react.
• Main Factors:
 Activation energy, Ea
 Temperature
 Molecular orientations
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82