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Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 13
Chemical
Kinetics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
Ectotherms
• Lizards, and other cold-blooded creatures, are
•
ectotherms – animals whose body temperature
matches their environment’s temperature
When a lizard’s body temperature drops, the
chemical reactions that occur in its body slow
down
 as do all chemical reactions when cooled
• This causes the lizard to become lethargic and to
•
slow down
Chemical kinetics is the study of the factors that
affect the rates of chemical reactions
 Such as temperature
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Chemical Kinetics
• The speed of a chemical reaction is called its
reaction rate
• The rate of a reaction is a measure of how
fast the reaction makes products
or uses reactants
• The ability to control the speed of a chemical
reaction is important
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Defining Rate
• Rate is how much a quantity changes in a
given period of time
• The speed you drive your car is a rate –
the distance your car travels (miles) in a
given period of time (1 hour)
 so the rate of your car has units of mi/hr
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Defining Reaction Rate
• The rate of a chemical reaction is generally
measured in terms of how much the concentration
of a reactant decreases in a given period of time
 or product concentration increases
• For reactants, a negative sign is placed in front of
the definition
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at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
CA  C2A2C1A 1
Rate  
 
t 2 t 2t1 t1
tt

44 0 8 
.25
Rate  
 0.025
1616 0 0
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 1X 1
ZX  Z2X2Z
Rate
Rate  
 
t 2 t 2t1t1
t t

708 

1
Rate
0625
Rate  
00..0625
116600
Copyright  2011 Pearson Education, Inc.
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2


CA  C2AC
1A 1
2
 
Rate  
t 2 t12 t1
tt

62 44 
Rate  
 0.125
0.125
3232
 1616 
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


ZX Z2XZ


X1
1
2
Rate
Rate     
t 2 t12 t1
tt
26 1 7 0.0625
Rate

Rate  
 0.0625
3232
 1616 
Copyright  2011 Pearson Education, Inc.
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2
at t = 48
[A] = 0
[B] = 0
[C] = 8
at t = 48
[X] = 5
[Y] = 5
[Z] = 3
CA  C2A2C1A 1
Rate  
 
t 2 tt21 t1
tt

80 6 2
.125
Rate  
 0.0125
48 3232
48
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

X2Z
X1


ZX Z2
1
Rate  
Rate

t 2 tt21 t1
tt
52
6 

3
Rate  
.0625
Rate

0.00625
48 32
 32
48
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Hypothetical Reaction
Red  Blue
In this reaction,
one molecule of Red turns
into one molecule of Blue
The number of molecules
will always total 100
The rate of the reaction can
be measured as the speed
of loss of Red molecules
over time, or the speed of
gain of Blue molecules
over time
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Hypothetical Reaction
Red  Blue
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Hypothetical Reaction
Red  Blue
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Reaction Rate Changes Over Time
• As time goes on, the rate of a reaction
generally slows down
because the concentration of the reactants
decreases
• At some time the reaction stops, either
because the reactants run out or because the
system has reached equilibrium
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Reaction Rate and Stoichiometry
• In most reactions, the coefficients of the balanced
•
equation are not all the same
H2 (g) + I2 (g)  2 HI(g)
For these reactions, the change in the number of
molecules of one substance is a multiple of the
change in the number of molecules of another
 for the above reaction, for every 1 mole of H2 used, 1
mole of I2 will also be used and 2 moles of HI made
 therefore the rate of change will be different
• To be consistent, the change in the concentration of
each substance is multiplied by 1/coefficient
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Average Rate
• The average rate is the change in measured
concentrations in any particular time period
linear approximation of a curve
• The larger the time interval, the more the
average rate deviates from the instantaneous
rate
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Hypothetical Reaction Red  Blue
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H2
I2
HI
StoichiometryThe
tellsaverage
us that rate
for is
every 1 mole/L
H2 used,
theofchange
in the
2 moles/L of concentration
HI are made. in a
time period
Assuming a 1given
L container,
at
10 s, we used 0.181 moles of
Inthe
theamount
first 10 s,
H2. Therefore
of the
[H2] ismoles)
−0.181=M,
HI made is 2(0.181
0.362 moles.so the rate is
At 60 s, we used 0.699 moles
of H2. Therefore the amount
of HI made is 2(0.699 moles)
= 1.398 moles.
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average rate in a given
time period =  slope of
the line connecting the
[H2] points; and ½ +slope
of the line for [HI]
the average rate for the
first 80 s is 0.0108 M/s
the average rate for the
first 40 s is 0.0150 M/s
the average rate for the
first 10 s is 0.0181 M/s
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Instantaneous Rate
• The instantaneous rate is the change in
concentration at any one particular time
slope at one point of a curve
• Determined by taking the slope of a line
tangent to the curve at that particular point
first derivative of the function
for you calculus fans
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H2 (g) + I2 (g)  2 HI (g)
Using [H2], the
instantaneous
rate at 50 s is
Using [HI], the
instantaneous
rate at 50 s is
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Example 13.1: For the reaction given, the [I] changes
from 1.000 M to 0.868 M in the first 10 s. Calculate the
average rate in the first 10 s and the [H+].
H2O2 (aq) + 3 I(aq) + 2 H+(aq)  I3(aq) + 2 H2O(l)
solve the rate
equation for the rate
(in terms of the
change in
concentration of the
given quantity)
solve the rate
equation (in terms of
the change in the
concentration for the
quantity to find) for
the unknown value
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Practice – If 2.4 x 102 g of NOBr (MM 109.91 g)
decomposes in a 2.0 x 102 mL flask in 5.0 minutes,
find the average rate of Br2 production in M/s
2 NOBr(g)  2 NO(g) + Br2(l)
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Practice – If 2.4 x 102 g of NOBr decomposes in a 2.0 x 102 mL
flask in 5.0 minutes, find the average rate of Br2 production
2 NOBr(g)  2 NO(g) + Br2(l)
2 mL,
Given: 240.0
5.45 MgBr
NOBr,
,
3.0
200.0
0.2000
x
10
sL, 5.0
5.0min.
min.
2
Find: [Br2]/t, M/s
Conceptual
 mole Br2
M
Plan:
Rate
 min
s
}
Relationships: 2 mol NOBr:1 mol Br2, 1 mol = 109.91g, 1 mL=0.001 L
Solve:
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Measuring Reaction Rate
• To measure the reaction rate you need to be
able to measure the concentration of at least
one component in the mixture at many points in
time
• There are two ways of approaching this problem
1. for reactions that are complete in less than 1
hour, it is best to use continuous monitoring of
the concentration
2. for reactions that happen over a very long time,
sampling of the mixture at various times can be
used
 when sampling is used, often the reaction in the
sample is stopped by a quenching technique
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Continuous Monitoring
• Polarimetry – measuring the change in the
•
degree of rotation of plane-polarized light caused
by one of the components over time
Spectrophotometry – measuring the amount of
light of a particular wavelength absorbed by one
component over time
 the component absorbs its complementary color
• Total pressure – the total pressure of a gas
mixture is stoichiometrically related to partial
pressures of the gases in the reaction
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Sampling the Reaction Mixture at
Specific Times
• At specific times during the reaction, drawing off
aliquots, (samples from the reaction mixture), and
doing quantitative analysis
 titration for one of the components
 gravimetric analysis
• Gas chromatography can measure the
concentrations of various components in a mixture
 for samples that have volatile components
 separates mixture by adherence to a surface
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Methods for Determining Concentrations
in a Mixture
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Methods for Determining Concentrations
in a Mixture
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Methods for Determining Concentrations
in a Mixture
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Factors Affecting Reaction Rate:
Nature of the Reactants
• Nature of the reactants means what kind of reactant
molecules and what physical condition they are in
 small molecules tend to react faster than large molecules
 gases tend to react faster than liquids, which react faster
than solids
 powdered solids are more reactive than “blocks”
more surface area for contact with other reactants
 certain types of chemicals are more reactive than others
e.g. potassium metal is more reactive than sodium
ions react faster than molecules
no bonds need to be broken
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Factors Affecting Reaction Rate:
Temperature
• Increasing temperature increases reaction rate
chemist’s rule of thumb—for each 10 °C rise in
temperature, the speed of the reaction doubles
for many reactions
• There is a mathematical relationship between the
absolute temperature and the speed of a reaction
discovered by Svante Arrhenius, which will be
examined later
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Factors Affecting Reaction Rate:
Catalysts
• Catalysts are substances that affect the speed
•
of a reaction without being consumed
Most catalysts are used to speed up a reaction;
these are called positive catalysts
catalysts used to slow a reaction are called negative
catalysts.
• Homogeneous = present in same phase
• Heterogeneous = present in different phase
• How catalysts work will be examined later
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Factors Affecting Reaction Rate:
Reactant Concentration
• Generally, the larger the concentration of
reactant molecules, the faster the reaction
increases the frequency of reactant molecule
contact
concentration of gases depends on the partial
pressure of the gas
higher pressure = higher concentration
• Concentrations of solutions depend on the
solute-to-solution ratio (molarity)
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The Rate Law
• The rate law of a reaction is the mathematical
relationship between the rate of the reaction and the
concentrations of the reactants
 and homogeneous catalysts as well
• The rate law must be determined experimentally!!
• The rate of a reaction is directly proportional to the
concentration of each reactant raised to a power
• For the reaction aA + bB  products the rate
law would have the form given below
 n and m are called the orders for each reactant
 k is called the rate constant
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Reaction Order
• The exponent on each reactant in the rate law
•
•
is called the order with respect to that reactant
The sum of the exponents on the reactants is
called the order of the reaction
The rate law for the reaction
2 NO(g) + O2(g)  2 NO2(g)
is Rate = k[NO]2[O2]
The reaction is
second order with respect to [NO],
first order with respect to [O2],
and third order overall
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Sample Rate Laws
The bottom reaction is autocatalytic because a product affects the rate.
Hg2+ is a negative catalyst; increasing its concentration slows the reaction.
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Example: The rate equation for the reaction of NO with ozone is
Rate = k[NO][O3]. If the rate is 6.60 x 10−5 M/sec when [NO] =
1.00 x 10−6 M and [O3] = 3.00 x 10−6 M, calculate the rate constant
Given:
Find:
Conceptual
Plan:
Relationships:
[NO] = 1.00 x 10−6 M, [O3] = 3.00 x 10−6 M,
Rate = 6.60 x 10−6 M/s
k, M−1s−1
Rate, [NO], [O3]
k
Rate = k[NO][O3]
Solve:
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Practice – The rate law for the decomposition of
acetaldehyde is Rate = k[acetaldehyde]2.
What is the rate of the reaction when the
[acetaldehyde] = 1.75 x 10−3 M and the rate constant
is 6.73 x 10−6 M−1•s−1?
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Practice – What is the rate of the reaction when the
[acetaldehyde] = 1.75 x 10−3 M and the rate constant is
6.73 x 10−6 M−1•s−1?
Given: [acetaldehyde] = 1.75 x 10−3 M, k= 6.73 x 10−6 M−1•s−1
Find: Rate, M/s
Conceptual k, [acetaldehyde]
Rate
Plan:
Relationships: Rate = k[acetaldehyde]2
Solve:
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Finding the Rate Law:
the Initial Rate Method
• The rate law must be determined experimentally
• The rate law shows how the rate of a reaction
•
depends on the concentration of the reactants
Changing the initial concentration of a reactant
will therefore affect the initial rate of the reaction
then doubling the
initial concentration
of A quadruples
doubles
does
notthe the
change
initial
reaction
the initial
rate
reaction rate
if for the reaction
A → Products
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Rate = k[A]n
• If a reaction is Zero Order, the rate of the reaction
is always the same
 doubling [A] will have no effect on the reaction rate
• If a reaction is First Order, the rate is directly
proportional to the reactant concentration
 doubling [A] will double the rate of the reaction
• If a reaction is Second Order, the rate is directly
proportional to the square of the reactant
concentration
 doubling [A] will quadruple the rate of the reaction
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Determining the Rate Law When there
Are Multiple Reactants
• Changing each reactant will effect the
overall rate of the reaction
• By changing the initial concentration of one
reactant at a time, the effect of each
reactant’s concentration on the rate can be
determined
• In examining results, we compare
differences in rate for reactions that only
differ in the concentration of one reactant
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below
Comparing Expt #1 and Expt #2,
the [NO2] changes but the [CO]
does not
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate
constant for the reaction NO2(g) + CO(g)  NO(g) +
CO2(g)
given the data below
n = 2, m = 0
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below
Substitute the
concentrations
and rate for
any experiment
into the rate
law and solve
for k
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
2.
3.
0.10
0.20
0.20
0.10
0.10
0.20
0.0021
0.0082
0.0083
4.
0.40
0.10
0.033
Rate  k[NO2 ]2
for expt 1
0.0021
M
s


 k 0.10 M
2
0.0021 Ms
1
1
k

0.21
M

s
0.01 M2
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Practice – Determine the rate law and rate constant for
the reaction NH4+ + NO2− N2 + 2 H2O
given the data below
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Practice – Determine the rate law and rate constant for
the reaction NH4+ + NO2− N2 + 2 H2O
given the data below
Rate = k[NH4+]n[NO2]m
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Finding the Rate Law:
Graphical Methods
• The rate law must be determined experimentally
• A graph of concentration of reactant vs. time can
•
•
be used to determine the effect of concentration
on the rate of a reaction
This involves using calculus to determine the
area under a curve – which is beyond the scope
of this course
Later we will examine the results of this analysis
so we can use graphs to determine rate laws for
several simple cases
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Reactant Concentration vs. Time
A  Products
insert figure 13.6
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Integrated Rate Laws
• For the reaction A  Products, the rate law
•
depends on the concentration of A
Applying calculus to integrate the rate law
gives another equation showing the
relationship between the concentration of A
and the time of the reaction – this is called the
Integrated Rate Law
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Half-Life
• The half-life, t1/2, of a
reaction is the length
of time it takes for the
concentration of the
reactant to fall to ½ its
initial value
• The half-life of the
reaction depends on
the order of the
reaction
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Zero Order Reactions
• Rate = k[A]0 = k
constant rate reactions
• [A] = −kt + [A]initial
• Graph of [A] vs. time is straight line with
•
•
slope = −k and y–intercept = [A]initial
t ½ = [Ainitial]/2k
When Rate = M/sec, k = M/sec
[A]initial
[A]
time
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First Order Reactions
• Rate = k[A]1 = k[A]
• ln[A] = −kt + ln[A]initial
• Graph ln[A] vs. time gives straight line with
slope = −k and y–intercept = ln[A]initial
used to determine the rate constant
• t½ = 0.693/k
• The half-life of a first order reaction is
constant
• When Rate = M/sec, k = s–1
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ln[A]initial
ln[A]
time
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The Half-Life of a First-Order Reaction
Is Constant
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Rate Data for
C4H9Cl + H2O  C4H9OH + HCl
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C4H9Cl + H2O  C4H9OH + 2 HCl
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C4H9Cl + H2O  C4H9OH + 2 HCl
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C4H9Cl + H2O  C4H9OH + 2 HCl
slope =
−2.01 x 10−3
k=
2.01 x 10−3 s-1
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Second Order Reactions
• Rate = k[A]2
• 1/[A] = kt + 1/[A]initial
• Graph 1/[A] vs. time gives straight line with
slope = k and y–intercept = 1/[A]initial
used to determine the rate constant
• t½ = 1/(k[A0])
• When Rate = M/sec, k = M−1s−1
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1/[A]
1/[A]initial
time
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Rate Data For
2 NO2  2 NO + O2
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Rate Data For
2 NO2  2 NO + O2
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Rate Data Graphs for
2 NO2  2 NO + O2
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Rate Data Graphs for
2 NO2  2 NO + O2
ln(PNO2) vs. Time
4.8
4.6
4.4
ln(pressure)
4.2
4
3.8
3.6
3.4
3.2
3
2.8
2.6
2.4
0
50
100
150
200
250
Time (hr)
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Rate Data Graphs for
2 NO2  2 NO + O2
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Example 13.4: The reaction SO2Cl2(g)  SO2(g) + Cl2(g) is first order
with a rate constant of 2.90 x 10−4 s−1 at a given set of conditions.
Find the [SO2Cl2] at 865 s when [SO2Cl2]initial = 0.0225 M
Given: [SO2Cl2]init = 0.0225 M, t = 865, k = 2.90 x 10-4 s−1
Find: [SO2Cl2]
Conceptual
Plan:
[SO2Cl2]init, t, k
[SO2Cl2]
Relationships:
Solution:
Check: the new concentration is less than the original,
as expected
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Practice – The reaction Q  2 R is second order in Q.
If the initial [Q] = 0.010 M and after 5.0 x 102 seconds
the [Q] = 0.0010 M, find the rate constant
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Practice – The reaction Q  2 R is second order in Q.
If the initial [Q] = 0.010 M and after 5.0 x 102 seconds
the [Q] = 0.0010 M, find the rate constant
Given: [Q]init = 0.010 M, t = 5.0 x 102 s, [Q]t = 0.0010 M
Find: k
Conceptual
Plan:
[Q]init, t, [Q]t
k
Relationships:
Solution:
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Graphical Determination of
the Rate Law for A  Product
• Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs.
•
time allow determination of whether a reaction
is zero, first, or second order
Whichever plot gives a straight line determines
the order with respect to [A]
if linear is [A] vs. time, Rate = k[A]0
if linear is ln[A] vs. time, Rate = k[A]1
if linear is 1/[A] vs. time, Rate = k[A]2
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Practice – Complete the Table and
Determine the Rate Equation for the
Reaction A  2 Product
What will the rate be when the [A] = 0.010 M?
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Practice – Complete the Table and
Determine the Rate Equation for the
Reaction A  2 Product
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Practice – Complete the table and
determine the rate equation for the
reaction A  2 Product
Because the graph 1/[A] vs. time is
linear, the reaction is second order,
Rate  k[A]2
k  slope of the line  0.10 M–1 s –1
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Relationship Between
Order and Half-Life
• For a zero order reaction, the lower the initial
concentration of the reactants, the shorter the
half-life
 t1/2 = [A]init/2k
• For a first order reaction, the half-life is
independent of the concentration
 t1/2 = ln(2)/k
• For a second order reaction, the half-life is
inversely proportional to the initial concentration –
increasing the initial concentration shortens the
half-life
 t1/2 = 1/(k[A]init)
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Example 13.6: Molecular iodine dissociates at 625 K with a firstorder rate constant of 0.271 s−1. What is the half-life of this
reaction?
Given: k = 0.271 s−1
Find: t1/2
Conceptual
Plan:
k
t1/2
Relationships:
Solution:
Check: the new concentration is less than the original,
as expected
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Practice – The reaction Q  2 R is second order in
Q. If the initial [Q] = 0.010 M and the rate constant is
1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init
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Practice – The reaction Q  2 R is second order in
Q. If the initial [Q] = 0.010 M and the rate constant is
1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init
Given: [Q]init = 0.010 M, k = 1.8 M−1s−1
Find: t1/2, s
Conceptual
Plan:
[Q]init, k
t1/2
Relationships:
Solution:
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The Effect of Temperature on Rate
• Changing the temperature changes the rate
•
constant of the rate law
Svante Arrhenius investigated this relationship
and showed that:
where T is the temperature in kelvins
R is the gas constant in energy units, 8.314 J/(mol•K)
A is called the frequency factor, the rate the reactant
energy approaches the activation energy
Ea is the activation energy, the extra energy needed
to start the molecules reacting
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Activation Energy and the
Activated Complex
• There is an energy barrier to almost all
•
reactions
The activation energy is the amount of energy
needed to convert reactants into the activated
complex
aka transition state
• The activated complex is a chemical species
with partially broken and partially formed bonds
always very high in energy because of its partial
bonds
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Isomerization of Methyl Isonitrile
methyl isonitrile rearranges to acetonitrile
for the reaction to occur, the
H3C─N bond must break, and a
new H3C─C bond form
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Energy Profile for the
Isomerization of Methyl Isonitrile
As the reaction
begins, the
the activated
frequency
activation
energy
is C─N
the
weakens
number
is
complex
the bond
difference
ofismolecules
a in
enough
for the
the
that begin
energy
chemical
between
species
to form
the
CNcomplex
group
activated
reactants
with
partial
and
bonds
thetoin
start
to rotate
a given
activated
period
complex
of time
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The Arrhenius Equation:
The Exponential Factor
• The exponential factor in the Arrhenius equation is a
number between 0 and 1
• Tt represents the fraction of reactant molecules with
sufficient energy so they can make it over the energy
barrier
 the higher the energy barrier (larger activation energy), the fewer
molecules that have sufficient energy to overcome it
• That extra energy comes from converting the kinetic
energy of motion to potential energy in the molecule when
the molecules collide
 increasing the temperature increases the average kinetic energy of
the molecules
 therefore, increasing the temperature will increase the number of
molecules with sufficient energy to overcome the energy barrier
 therefore increasing the temperature will increase the reaction rate
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Arrhenius Plots
• The Arrhenius Equation can be algebraically
solved to give the following form:
this equation is in the form y = mx + b
where y = ln(k) and x = (1/T)
a graph of ln(k) vs. (1/T) is a straight line
(−8.314 J/mol∙K)(slope of the line) = Ea, (in Joules)
ey–intercept = A (unit is the same as k)
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Example 13.7: Determine the activation energy and
frequency factor for the reaction O3(g)  O2(g) + O(g)
given the following data:
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Example 13.7: Determine the activation energy and
frequency factor for the reaction O3(g)  O2(g) + O(g)
given the following data:
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Example 13.7: Determine the activation energy and
frequency factor for the reaction O3(g)  O2(g) + O(g)
given the following data:
slope, m = 1.12 x 104 K
y–intercept, b = 26.8
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Arrhenius Equation:
Two-Point Form
• If you only have two (T,k) data points, the
following forms of the Arrhenius Equation can
be used:
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Example 13.8: The reaction NO2(g) + CO(g)  CO2(g) + NO(g)
has a rate constant of 2.57 M−1∙s−1 at 701 K and 567 M−1∙s−1
at 895 K. Find the activation energy in kJ/mol
Given: T1 = 701 K, k1 = 2.57 M−1∙s−1, T2 = 895 K, k2 = 567 M−1∙s−1
Find: Ea, kJ/mol
Conceptual
Plan:
T1, k1, T2, k2
Ea
Relationships:
Solution:
Check: most activation energies are tens to hundreds
of kJ/mol – so the answer is reasonable
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Practice – It is often said that the rate of a
reaction doubles for every 10 °C rise in
temperature. Calculate the activation energy
for such a reaction.
Hint:
make
and
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T1 = 300 K, T2 = 310 K
k2 = 2k1
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Practice – Find the activation energy in kJ/mol for
increasing the temperature 10 ºC doubling the rate
Given: T1 = 300 K, T1 = 310 K, k2 = 2 k1
Find: Ea, kJ/mol
Conceptual
Plan:
T1, k1, T2, k2
Ea
Relationships:
Solution:
Check: most activation energies are tens to hundreds
of kJ/mol – so the answer is reasonable
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Collision Theory of Kinetics
• For most reactions, for a reaction to take place,
the reacting molecules must collide with each
other
 on average about 109 collisions per second
• Once molecules collide they may react together
or they may not, depending on two factors –
1. whether the collision has enough energy to
"break the bonds holding reactant molecules
together"; and
2. whether the reacting molecules collide in the
proper orientation for new bonds to form
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Effective Collisions
Kinetic Energy Factor
For a collision to lead
to overcoming the
energy barrier, the
reacting molecules
must have sufficient
kinetic energy so that
when they collide the
activated complex
can form
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Effective Collisions
Orientation Effect
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Effective Collisions
• Collisions in which these two conditions are
•
•
met (and therefore lead to reaction) are
called effective collisions
The higher the frequency of effective
collisions, the faster the reaction rate
When two molecules have an effective
collision, a temporary, high energy
(unstable) chemical species is formed – the
activated complex
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Collision Theory and the Frequency
Factor of the Arrhenius Equation
• The Arrhenius Equation includes a term, A,
•
called the Frequency Factor
The Frequency Factor can be broken into two
terms that relate to the two factors that
determine whether a collision will be effective
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Collision Frequency
• The collision frequency is the number of
•
collisions that happen per second
The more collisions per second there are, the
more collisions can be effective and lead to
product formation
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Orientation Factor
• The orientation factor, p, is a statistical term
•
•
•
•
relating the frequency factor to the collision
frequency
For most reactions, p < 1
Generally, the more complex the reactant
molecules, the smaller the value of p
For reactions involving atoms colliding, p ≈ 1
because of the spherical nature of the atoms
Some reactions actually can have a p > 1
generally involve electron transfer
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Orientation Factor
• The proper orientation results when the atoms
•
are aligned in such a way that the old bonds can
break and the new bonds can form
The more complex the reactant molecules, the
less frequently they will collide with the proper
orientation
 reactions where symmetry results in multiple
orientations leading to reaction have p slightly less
than 1
• For most reactions, the orientation factor is less
than 1
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Molecular Interpretation of
Factors Affecting Rate – Reactant Nature
• Reactions generally occur faster in solution than in pure
substances
 mixing gives more particle contact
 particles separated, allowing more effective collisions per second
 forming some solutions breaks bonds that need to be broken
• Some materials undergo similar reactions at different rates
either because they have a (1) higher initial potential energy
and are therefore closer in energy to the activated complex, or
(2) because their reaction has a lower activation energy
 CH4 + Cl2  CH3Cl + HCl is about 12x faster than CD4 + Cl2 
CD3Cl + DCl because the C─H bond is weaker and less stable than
the C─D bond
 CH4 + X2  CH3X + HX occurs about 100x faster with F2 than with Cl2
because the activation energy for F2 is 5 kJ/mol but for Cl2 is 17 kJ/mol
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Molecular Interpretation of
Factors Affecting Rate – Temperature
• Increasing the temperature raises the
•
•
average kinetic energy of the reactant
molecules
There is a minimum amount of kinetic energy
needed for the collision to be converted into
enough potential energy to form the activated
complex
Increasing the temperature increases the
number of molecules with sufficient kinetic
energy to overcome the activation energy
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Molecular Interpretation of
Factors Affecting Rate – Concentration
• Reaction rate generally increases as the
concentration or partial pressure of reactant
molecules increases
except for zero order reactions
• More molecules leads to more molecules with
sufficient kinetic energy for effective collision
distribution the same, just bigger curve
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Reaction Mechanisms
• We generally describe chemical reactions with an
•
•
•
•
equation listing all the reactant molecules and
product molecules
But the probability of more than 3 molecules
colliding at the same instant with the proper
orientation and sufficient energy to overcome the
energy barrier is negligible
Most reactions occur in a series of small reactions
involving 1, 2, or at most 3 molecules
Describing the series of reactions that occurs to
produce the overall observed reaction is called a
reaction mechanism
Knowing the rate law of the reaction helps us
understand the sequence of reactions in the
mechanism
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An Example of a Reaction Mechanism
• Overall reaction:
•
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
Mechanism:
1. H2(g) + ICl(g)  HCl(g) + HI(g)
2. HI(g) + ICl(g)  HCl(g) + I2(g)
• The reactions in this mechanism are
elementary steps, meaning that they cannot
be broken down into simpler steps and that the
molecules actually interact directly in this
manner without any other steps
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Elements of a Mechanism
Intermediates
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
•
•
•
Notice that the HI is a product in Step 1, but
then a reactant in Step 2
Because HI is made but then consumed, HI
does not show up in the overall reaction
Materials that are products in an early
mechanism step, but then a reactant in a later
step, are called intermediates
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Molecularity
• The number of reactant particles in an
•
•
elementary step is called its molecularity
A unimolecular step involves one particle
A bimolecular step involves two particles
though they may be the same kind of particle
• A termolecular step involves three particles
though these are exceedingly rare in elementary
steps
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Rate Laws for Elementary Steps
• Each step in the mechanism is like its own
•
•
little reaction – with its own activation energy
and own rate law
The rate law for an overall reaction must be
determined experimentally
But the rate law of an elementary step can be
deduced from the equation of the step
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g) Rate = k1[H2][ICl]
2) HI(g) + ICl(g)  HCl(g) + I2(g)
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Rate = k2[HI][ICl]
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Rate Laws of Elementary Steps
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Rate Determining Step
• In most mechanisms, one step occurs slower than the
other steps
• The result is that product production cannot occur any
faster than the slowest step – the step determines the
rate of the overall reaction
• We call the slowest step in the mechanism the rate
determining step
 the slowest step has the largest activation energy
• The rate law of the rate determining
step determines the rate law
of the overall reaction
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Another Reaction Mechanism
NO2(g) + CO(g)  NO(g) + CO2(g)
1. NO2(g) + NO2(g)  NO3(g) + NO(g)
2. NO3(g) + CO(g)  NO2(g) + CO2(g)
Rateobs = k[NO2]2
Rate = k1[NO2]2 Slow
Rate = k2[NO3][CO] Fast
The first step is slower than the
second step because its
activation energy is larger.
The first step in this
mechanism is the rate
determining step.
The rate law of the first step is
the same as the rate law of the
overall reaction.
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Validating a Mechanism
• To validate (not prove) a mechanism, two
conditions must be met:
1. The elementary steps must sum to the overall
reaction
2. The rate law predicted by the mechanism
must be consistent with the experimentally
observed rate law
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Mechanisms with a Fast
Initial Step
• When a mechanism contains a fast initial step,
•
the rate limiting step may contain intermediates
When a previous step is rapid and reaches
equilibrium, the forward and reverse reaction
rates are equal – so the concentrations of
reactants and products of the step are related
and the product is an intermediate
• Substituting into the rate law of the RDS will
produce a rate law in terms of just reactants
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An Example
k1
1. 2 NO(g)  N2O2(g)
Fast
2. H2(g) + N2O2(g)  H2O(g) + N2O(g)
Slow Rate = k2[H2][N2O2]
3. H2(g) + N2O(g)  H2O(g) + N2(g)
Fast
k−1
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
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Example 13.9: Show that the proposed mechanism for the
reaction 2 O3(g)  3 O2(g) matches the observed rate law
Rate = k[O3]2[O2]−1
k1
1. O3(g)  O2(g) + O(g)
Fast
k−1
2. O3(g) + O(g)  2 O2(g) Slow
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Rate = k2[O3][O]
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Practice – Mechanism
Determine the overall reaction, the rate
determining step, the rate law, and identify all
intermediates of the following mechanism
1.
2.
A + B2  AB + B
A + B  AB
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Slow
Fast
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Practice – Mechanism
Determine the overall reaction, the rate
determining step, the rate law, and identify all
intermediates of the following mechanism
1.
2.
A + B2  AB + B
A + B  AB
Slow
Fast
reaction 2 A + B2  2 AB
B is an intermediate
The first step is the rate determining step
Rate = k[A][B2], same as RDS
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Catalysts
• Catalysts are substances that affect the rate of a
•
reaction without being consumed
Catalysts work by providing an alternative
mechanism for the reaction
 with a lower activation energy
• Catalysts are consumed in an early mechanism
step, then made in a later step
mechanism without catalyst
mechanism with catalyst
O3(g) + O(g)  2 O2(g) V. Slow
Cl(g) + O3(g)  O2(g) + ClO(g)
Fast
ClO(g) + O(g)  O2(g) + Cl(g)
Slow
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Ozone Depletion over the Antarctic
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Molecular Interpretation of
Factors Affecting Rate – Catalysts
• Catalysts generally speed a reaction
• They give the reactant molecules a different
path to follow with a lower activation energy
 heterogeneous catalysts hold one reactant
molecule in proper orientation for reaction to occur
when the collision takes place
 and sometimes also help to start breaking bonds
 homogeneous catalysts react with one of the
reactant molecules to form a more stable activated
complex with a lower activation energy
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Energy Profile of a Catalyzed Reaction
polar stratospheric
clouds contain ice
crystals that catalyze
reactions that release
Cl from atmospheric
chemicals
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Practice – Mechanism
Determine the overall reaction, the rate
determining step, the rate law, and identify all
catalysts and intermediates of the following
mechanism
1. HQ2R2 + R−  Q2R2− + HR Fast
2. Q2R2−  Q2R + R−
Slow
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Practice – Mechanism
Determine the overall reaction, the rate
determining step, the rate law, and identify all
catalysts and intermediates of the following
mechanism
1. HQ2R2 + R−  Q2R2− + HR Fast
2. Q2R2−  Q2R + R−
Slow
reaction HQ2R2  Q2R + HR
R− is a catalyst
Q2R2− is an intermediate
The second step is the rate determining step
Rate = k[Q2R2−] = k[HQ2R2][R−]
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Catalysts
• Homogeneous catalysts are in the same
phase as the reactant particles
Cl(g) in the destruction of O3(g)
• Heterogeneous catalysts are in a different
phase than the reactant particles
solid catalytic converter in a car’s exhaust system
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Types of Catalysts
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Catalytic Hydrogenation
H2C=CH2 + H2 → CH3CH3
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Enzymes
• Because many of the molecules are large and
complex, most biological reactions require a
catalyst to proceed at a reasonable rate
• Protein molecules that catalyze biological
reactions are called enzymes
• Enzymes work by adsorbing the substrate
reactant onto an active site that orients the
substrate for reaction
1) Enzyme + Substrate  Enzyme─Substrate Fast
2) Enzyme─Substrate → Enzyme + Product
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Slow
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Enzyme–Substrate Binding:
the Lock and Key Mechanism
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Enzymatic Hydrolysis of Sucrose
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