Chapter 12 - Kineticswunderchem.weebly.com/uploads/5/2/4/2/52423897/chapter_12_-_ki… · Section...
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Chapter 12
Chemical Kinetics
AP*
Section 12.1
Reaction Rates
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Reaction Rate
� Change in concentration of a reactant or product
per unit time.
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
[ ]
2 1
2 1
concentration of A at time concentration of A at time Rate =
A=
−−
∆
∆
t t
t t
t
Section 12.1
Reaction Rates
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The Decomposition of Nitrogen Dioxide
Section 12.1
Reaction Rates
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Instantaneous Rate
� Value of the rate at a particular time.
� Can be obtained by computing the slope of a line
tangent to the curve at that point.
Section 12.1
Reaction Rates
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� Reactant rates are negative.
� Product rates are positive.
� Rate is not constant
� Typically it decreases with time
Tidbits
Section 12.1
Reaction Rates
Determining the Rate of Reaction in Terms of Products
� Consider the reaction
� NO is produced at the same rate as NO2 is consumed
� Rate of NO production is twice the rate of O2
production
� Is the rate of NO positive or negative?
2 22NO ( ) 2NO( ) + O ( )→g g g
Section 12.2
Rate Laws: An Introduction
Rate Law
� Shows how the rate depends on the concentrations of
reactants.
� For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
� k = rate constant
� n = order of the reactant
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Section 12.2
Rate Laws: 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.
� The value of the exponent n must be determined by
experiment; it cannot be written from the balanced
equation.
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Section 12.2
Rate Laws: An Introduction
Rate Constant
� Definition of reaction rate in terms of the
consumption of NO2
� Definition of the reaction rate in terms of
production of O2
22
[NO ]Rate = – = [NO ]
∆∆
nkt
22
[O ]Rate = – = [NO ]
∆∆
n' k't
Section 12.2
Rate Laws: An Introduction
Rate Constant (Continued)
� As two moles of NO2 molecules are consumed for
every O2 molecule produced,
Or
and
� Value of the rate constant depends on how the
rate is defined
2 2
Rate = 2 rate
[NO ] = 2 [NO ]
2
×
= ×
n n
'
k k'
k k'
Section 12.2
Rate Laws: 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.
� The type of rate law used depends on what types of
data are easiest to collect
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Section 12.2
Rate Laws: 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 12.2
Rate Laws: 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.
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Section 12.3
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
Determining the Form of the Rate Law
Method of Initial Rates
� The initial rate is determined for each experiment 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 12.3
Determining the Form of the Rate Law
Overall Reaction Order
� The sum of the exponents in the reaction rate equation.
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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Section 12.3
Determining the Form of the Rate Law
How do exponents (orders) in rate laws compare to
coefficients in balanced equations?
Why?
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CONCEPT CHECK!CONCEPT CHECK!
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Determining a Rate Law
� The reaction between bromate ions and bromide
ions in acidic aqueous solution is given by the
following equation:– – +
3 2 2BrO ( ) + 5Br ( ) + 6H ( ) 3Br ( ) + 3H O( )aq aq aq l l→
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Determining a Rate Law (Continued)
� Determine the orders for all three reactants, the
overall reaction order, and the value of the rate
constant
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Solution
� General form of the rate law
� Step 1 - Determine the value of n
� Use the results from Experiments 1 and 2
� Thus, n = 1
– – +
3Rate = [BrO ] [Br ] [H ]n m pk
–3
–4
Rate 2 1.6 × 10 mol / L s (0.20 mol / L) (0.10 mol / L) (0.10 mol / L) = =
Rate 1 8.0 × 10 mol / L s (0.10 mol / L) (0.10 mol / L) (0.10 mol / L)
n m p
n m p
k
k
⋅
⋅
0.20 mol / L2.0 = = (2.0)
0.10 mol / L
n
n
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Solution (Continued 1)
� Step 2 - Determine the value of m
� Use the results from Experiments 2 and 3
� Thus, m = 1
–3
–3
Rate 3 3.2 × 10 mol / L s (0.20 mol / L) (0.20 mol / L) (0.10 mol / L) = =
Rate 2 1.6 × 10 mol / L s (0.20 mol / L) (0.10 mol / L) (0.10 mol / L)
n m p
n m p
k
k
⋅⋅
0.20 mol / L2.0 = = (2.0)
0.10 mol / L
m
m
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Solution (Continued 2)
� Step 3 - Determine the value of p
� Use the results from Experiments 1 and 4
� Thus, p = 2
� The overall reaction order is n + m + p = 4
–3
–4
Rate 4 3.2 × 10 mol / L s (0.10 mol / L) (0.10 mol / L) (0.20 mol / L) = =
Rate 1 8.0 × 10 mol / L s (0.10 mol / L) (0.10 mol / L) (0.10 mol / L)
⋅⋅
n m p
n m p
k
k
2
0.20 mol / L4.0 =
0.10 mol / L
4.0 = (2.0) = (2.0)
p
p
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Solution (Continued 3)
� The rate law for the reaction
Rate = k[BrO3–][Br–][H+]2
� The value of the rate constant k can be calculated
from the results of any of the four experiments
� For Experiment 1, the initial rate is 8.0×10–4
mol/L·s– – +
3[BrO ] = 0.100 , [Br ] = 0.10 , and [H ] = 0.10 M M M
Section 12.3
Determining the Form of the Rate Law
Interactive Example 12.1 - Solution (Continued 4)
� Using the values from the initial rate in the rate
law
� Reality Check
� Verify that the same value of k can be obtained from
the results of the other experiments
–4 2
–4 –4 4 4
–43 3
–4 4 4
8.0 × 10 mol / L s = (0.10 mol / L) (0.10 mol / L) (0.10 mol / L)
8.0 × 10 mol / L s = (1.0 × 10 mol / L )
8.0 × 10 mol / L s = = 8.0 L / mol s
1.0 × 10 mol / L
⋅
⋅
⋅⋅
k
k
k
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
Integrated First-Order Rate Law - Key Points
� The equation expresses the influence of time on
the concentration of A
� The initial concentration of A and the rate constant k
can be used to calculate the concentration of A at any
time
� The equation is of the form y = mx + b
� y = ln[A], x = t, m = –k, and b = ln[A]0
� For a first-order reaction, a plot of ln[A] versus t is
always a straight line
Section 12.4
The Integrated Rate Law
Integrated First-Order Rate Law - Key Points (Continued)
� The equation can be expressed in terms of a ratio
of [A] and [A]0
0[A]ln =
[A]kt
Section 12.4
The Integrated Rate Law
Plot of ln[N2O5] vs Time
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Section 12.4
The Integrated Rate Law
� Calculate [N2O5] at 150 s after the start of the
following reaction:
� Use the following information:
Interactive Example 12.3 - First-Order Rate Laws II
2 5 2 22N O ( ) 4NO ( ) + O ( ) →g g g
Section 12.4
The Integrated Rate Law
Interactive Example 12.3 - Solution
� Information available
� [N2O5] = 0.0500 mol/L at 100 s
� [N2O5] = 0.0250 mol/L at 200 s
� Formula required to calculate [N2O5] after 150 s
� Where t = 150 s, k = 6.93×10–3 s–1, and [N2O5]0 =
0.1000 mol/L
2 5 2 5 0ln [N O ] = – + ln [N O ]kt
Section 12.4
The Integrated Rate Law
Interactive Example 12.3 - Solution (Continued)
� Note that this value of [N2O5] is not halfway between
0.0500 and 0.0250 mol/L
–3 –1
2 5 =150
2 5 =150
ln ([N O ] ) = – (6.93×10 s ) (150 s) + ln(0.100)
= –1.040 – 2.303 = – 3.343
([N O ] ) = antilog (–3.343) = 0.0353 mol / L
t
t
Section 12.4
The Integrated Rate Law
First-Order
� Time required for a reactant to reach half its original
concentration
� Half–Life:
k = rate constant
� Half–life does not depend on the concentration of
reactants.
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12
0.693 = t
k
Section 12.4
The Integrated Rate Law
Figure 12.4 - A Plot of [N2O5] versus Time for the
Decomposition of N2O5
Section 12.4
The Integrated Rate Law
Interactive Example 12.4 - Half-Life for a First-Order
Reaction
� A certain first-order reaction has a half-life of 20.0
minutes
a. Calculate the rate constant for this reaction
b. How much time is required for this reaction to be
75% complete?
Section 12.4
The Integrated Rate Law
Interactive Example 12.4 - Solution (a)
� Solving equation (12.3) for k gives
–2 –1
1/2
0.693 0.693 = = = 3.47 × 10 min
20.0 mink
t
Section 12.4
The Integrated Rate Law
Interactive Example 12.4 - Solution (b)
� We use the integrated rate law in the form
� If the reaction is 75% complete, 75% of the reactant
has been consumed, leaving 25% in the original form
0[A]ln =
[A]kt
0
[A] × 100% = 25%
[A]
Section 12.4
The Integrated Rate Law
Interactive Example 12.4 - Solution (b) (Continued 1)
�It takes 40 minutes for this particular reaction to
reach 75% completion
0
0
–2
0
–2
[A][A] 1 = 0.25 or = = 4.0
[A] [A] 0.25
[A] 3.47 × 10ln = ln(4.0) = =
[A] min
ln(4.0)
3.47 × 10 = = 40 minmin
kt t
t
Section 12.4
The Integrated Rate Law
A first order reaction is 35% complete at the end of
55 minutes. What is the value of k?
k = 7.8 × 10–3 min–1
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Section 12.4
The Integrated Rate Law
Second-Order
� Rate = k[A]2
� Integrated:
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of ACopyright © Cengage Learning. All rights reserved 39
[ ] [ ]0
1 1= +
A Akt
Section 12.4
The Integrated Rate Law
Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
Section 12.4
The Integrated Rate Law
Characteristics of Equation (12.5)
� A plot of 1/[A] versus t will produce a straight line
with a slope equal to k
� The equation shows how [A] depends on time and
can be used to calculate [A] at any time t,
provided k and [A]0 are known
Section 12.4
The Integrated Rate Law
Second-Order
� Half–Life:
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.
Copyright © Cengage Learning. All rights reserved 42
[ ]12
0
1 =
At
k
Section 12.4
The Integrated Rate Law
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 × 10-3 M–1min–1
c) Calculate [A] at t = 525 minutes.
[A] = 0.23 MCopyright © Cengage Learning. All rights reserved 43
EXERCISE!EXERCISE!
Section 12.4
The Integrated Rate Law
Example 12.5 - Determining Rate Laws
� Butadiene reacts to form its dimer according to
the equation
4 6 8 122C H ( ) C H ( )g g→
Section 12.4
The Integrated Rate Law
Example 12.5 - Determining Rate Laws (Continued)
a. Is this reaction a first order or second order?
b. What is the value of the rate constant for the
reaction?
c. What is the half-life for the reaction under the initial
conditions of this experiment?
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (a)
� To decide whether the rate law for this reaction is
first order or second order, we must see whether
the plot of ln[C4H6] versus time is a straight line
(first order) or the plot of 1/[C4H6] versus time is a
straight line (second order)
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (a) (Continued 1)
� The data necessary to make these plots are as
follows:
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (a) (Continued 2)
� The reaction is second order
� The rate law for this
reaction is:
A plot of ln[C4H6] versus t A plot of 1/[C4H6] versus t
4 6
2
4 6
[C H ]Rate = –
[C H ]
∆∆
=t
k
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (b)
� For a second-order reaction, a plot of 1/[C4H6]
versus t produces a straight line of slope k
� In terms of the standard equation for a straight line, y
= mx + b, we have y = 1/[C4H6] and x = t
� Slope of the line can be expressed as follows:
4 6
1Δ
[C H ]ΔSlope = =
Δ Δ
y
x t
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (b) (Continued)
� Using the points at t = 0 and t = 6200, we can find the
rate constant for the reaction
–2
(481 – 100) L / mol 381 = slope = = L / mol s
(6200 – 0) s 6200
= 6.14 × 10 L / mol s
⋅
⋅
k
Section 12.4
The Integrated Rate Law
Example 12.5 - Solution (c)
� The expression for the half-life of a second-order
reaction is
� In this case k = 6.14×10–2 L/mol · s (from part b) and
[A]0 = [C4H6]0 = 0.01000 M (the concentration at t = 0)
� The initial concentration of C4H6 is halved in 1630 s
3
1/2 –2 –2
1 = = 1.63 × 10 s
(6.14 × 10 L / mol s) (1.000 ×10 mol / L)⋅t
1/2
0
1 =
[A]t
k
Section 12.4
The Integrated Rate Law
Half-Life for a First-Order Reaction versus Half-Life of a
Second-Order Reaction
� t1/2
� First-order reaction - t1/2 depends on k
� Second-order reaction - t1/2 depends on k and [A]0
� Half-life
� First-order reaction - A constant time is required to
reduce the concentration of the reactant by half
� Second-order reaction - Each successive half-life is
double the preceding one
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:
k = rate constant
[A]o = initial concentration of A
� Half–life gets shorter as the reaction progresses and the
concentration of reactants decrease.
Copyright © Cengage Learning. All rights reserved 55
[ ]0
12
A =
2t
k
Section 12.4
The Integrated Rate Law
How can you tell the difference among 0th, 1st, and 2nd
order rate laws from their graphs?
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CONCEPT CHECK!CONCEPT CHECK!
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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Section 12.4
The Integrated Rate Law
Consider the reaction aA � Products.
[A]0 = 5.0 M and k = 1.0 × 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
Copyright © Cengage Learning. All rights reserved 59
4.7 M
3.7 M
2.0 M
EXERCISE!EXERCISE!
Section 12.5
Reaction Mechanisms
Reaction Mechanism
� Most chemical reactions occur by a series of steps
� Example - The reaction between nitrogen dioxide
and carbon monoxide involves the following
steps:
� Where k1 and k2 are the rate constants of the
individual reactions
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2 2 3
3 2 2
NO (g) + NO ( ) NO ( ) + NO( )
NO ( ) + CO( ) NO ( ) + CO ( )
k
k
g g g
g g g g
→
→
1
2
Section 12.5
Reaction Mechanisms
Reaction Mechanism (Continued)
� In the reaction, NO3 is an intermediate
� Intermediate: Species that is formed and consumed during
the reaction but is neither a reactant nor a product
� Each of the reactions is called an elementary step
� Elementary step: A reaction whose rate law can be written
from its molecularity
Section 12.5
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the Reaction
of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
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Section 12.5
Reaction Mechanisms
Elementary Steps (Molecularity)
� Unimolecular – reaction involving one molecule; first
order.
� Bimolecular – reaction involving the collision of two
species; second order.
� Termolecular – reaction involving the collision of three
species; third order. Very rare.
Copyright © Cengage Learning. All rights reserved 63
Section 12.5
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.
Copyright © Cengage Learning. All rights reserved 64
Section 12.5
Reaction Mechanisms
Reaction Mechanism Requirements
� The sum of the elementary steps must give the overall
balanced equation for the reaction.
� The mechanism must agree with the experimentally
determined rate law.
Copyright © Cengage Learning. All rights reserved 65
Section 12.5
Reaction Mechanisms
Table 12.7 - Examples of Elementary Steps
Section 12.5
Reaction Mechanisms
Example 12.6 - Reaction Mechanisms
� The balanced equation for the reaction of the
gases nitrogen dioxide and fluorine is
� Experimentally determined rate law is
2 2 22NO ( ) + F (g) 2NO F( )g g→
2 2Rate = [NO ] [F ]k
Section 12.5
Reaction Mechanisms
Example 12.6 - Reaction Mechanisms (Continued)
� Suggested mechanism for this reaction is
� Is this an acceptable mechanism?
� Does it satisfy the two requirements?
1
2
2 2 2
2 2
NO + F NO F + F (slow)
F + NO NO F (fast)
k
k
→
→
Section 12.5
Reaction Mechanisms
Example 12.6 - Solution
� First requirement for an acceptable mechanism
� The sum of the steps should give the balanced
equation
� The first requirement is met
2 2 2
2 2
2 2 2
2 2 2
NO + F NO F + F
F + NO NO F
2NO + F + F 2NO F + F
Overall reaction : 2NO + F 2NO F
→
→
→
→
Section 12.5
Reaction Mechanisms
Example 12.6 - Solution (Continued)
� Second requirement
� The mechanism must agree with the experimentally
determined rate law
� The overall reaction rate must be that of the first step
� The first step is bimolecular, so the rate law is
� The second requirement is met
� The mechanism is acceptable as both
requirements are satisfied
1 2 2Rate = k [NO ][F ]
Section 12.5
Reaction Mechanisms
Decomposition of N2O5
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Section 12.5
Reaction Mechanisms
Decomposition of N2O5
2N2O5(g) � 4NO2(g) + O2(g)
Step 1: N2O5 NO2 + NO3 (fast)
Step 2: NO2 + NO3→ NO + O2 + NO2 (slow)
Step 3: NO3 + NO → 2NO2 (fast)
Copyright © Cengage Learning. All rights reserved 72
2( )
Section 12.5
Reaction Mechanisms
The reaction A + 2B � C has the following proposed
mechanism:
A + B D (fast equilibrium)
D + B � C (slow)
Write the rate law for this mechanism.
rate = k[A][B]2
Copyright © Cengage Learning. All rights reserved 73
CONCEPT CHECK!CONCEPT CHECK!
Section 12.6
A Model for Chemical Kinetics
Collision Model
� Molecules must collide to react.
� Main Factors:
� Activation energy, Ea - Energy that must be overcome
to produce a chemical reaction.
� Temperature
� Molecular orientations
� Kinetic energy is changed into potential energy
as the molecules are distorted during a collision
to break bonds and rearrange the atoms into
product moleculesCopyright © Cengage Learning. All rights reserved 74
Section 12.6
A Model for Chemical Kinetics
Transition States and Activation Energy
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Section 12.6
A Model for Chemical Kinetics
Change in Potential Energy
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Section 12.6
A Model for Chemical Kinetics
Relation between Effective Collisions and Temperature
� The fraction of effective collisions increases
exponentially with temperature
Section 12.6
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.
Copyright © Cengage Learning. All rights reserved 78
Section 12.6
A Model for Chemical Kinetics
The Gas Phase Reaction of NO and Cl2
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Section 12.6
A Model for Chemical Kinetics
Arrhenius Equation
A = frequency factor
Ea = activation energy
R = gas constant (8.3145 J/K·mol)
T = temperature (in K)
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/ =
−a
E RT
k Ae
Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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( )aE 1
ln( ) = lnR T
−
k + A
Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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Section 12.6
A Model for Chemical Kinetics
Interactive Example 12.7 - Solution (Continued 2)
� Based on the plot
� Determine the value of Ea by solving the following
equation
4ln( )Slope = –1.2 10 K
1
∆= ×
∆
k
T
Slope a–E
R=
Section 12.6
A Model for Chemical Kinetics
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
Copyright © Cengage Learning. All rights reserved 84
EXERCISE!EXERCISE!
Section 12.7
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.
Copyright © Cengage Learning. All rights reserved 85
Section 12.7
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway
for a Given Reaction
Copyright © Cengage Learning. All rights reserved 86
Section 12.7
Catalysis
Effect of a Catalyst on the Number of Reaction-Producing
Collisions
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Section 12.7
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 12.7
Catalysis
Heterogeneous Catalysis
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Section 12.7
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 12.7
Catalysis
Homogeneous Catalyst
� Exists in the same phase as the reacting molecules.
� Enzymes are nature’s catalysts.
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Section 12.7
Catalysis
Homogeneous Catalysis
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