chapter14_lecture_finalJRB.ppt

Copyright McGraw-Hill 2009
14.1 Reaction Rates
• Kinetics: the study of how fast
reactions take place
• Some reactions are fast
(photosynthesis)
• Some reactions are slow (conversion of
diamond to graphite)

Importance of Studying
Reaction Rates
• Speed up desirable reactions
• Minimize damage by undesirable
reactions
• Useful in drug design, pollution control,
and food processing

Rate of Reaction
• Expressed as either:
– Rate of disappearance of reactants
(decrease or negative)
OR
– Rate of appearance of products (increase
or positive)

Average Reaction Rate

• Equation A  B
• rate =
• Why the negative on [A]?
[A] [B]
or
t t
 

 

Br2(aq) + HCOOH(aq)2Br(aq) + 2H+(aq) + CO2(g)
Note: Br2 disappears over time

Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)

Calculate Average Rate
• Avg. rate =
• Avg. rate =
2 2
[Br ] [Br ]
final initial
t t
final initial


= 3.80
0.0101 0.0120
50 0
M M
s s
M / s





Average Rate
• Average rate depends on time interval
• Plot of [Br2] vs time = curve
• Plot of Rate vs [Br2] = straight line

Instantaneous Rate
• Instantaneous: rate at a specific
instance in time (slope of a tangent to
the curve)

Rate Constant
• Using data from Table 14.1 - what can
you conclude?
250
50
Time (s)
1.75 x 105
0.00596
3.52 x 105
0.0101
Rate (M/s)
[Br2]

Rate Constant
Answer:
• When the [Br2] is halved; the rate is
halved
• Rate is directly proportional to [Br2]
• rate = k [Br2]
• k = proportionality constant and is
constant as long as temp remains
constant

Rate Constant
• Calculate the value of the rate constant for
any set of data and get basically the same
answer!
• k = rate / [Br2]
5
3.52 x 10 /s 3
3.5x 10 /s
0.0101
M
k
M


 

Stoichiometry and Reaction
Rate
• When stoichiometric ratios are not 1:1
rate of reaction is expressed as follows
General equation:
aA + bB  cC + dD
rate =
1 [A] 1 [B] 1 [C] 1 [D]
a t b t c t d t
   
    
   

Rate
• Write the rate expression for the
following reaction
2NO(g) + O2(g)  2NO2(g)

Rate
4PH3(g)  P4(g) + 6H2(g)
If molecular hydrogen is formed at a rate of
0.168 M/s, at what rate is P4 being
produced?
3 4 2
1 [PH ] 1 [P ] 1 [H ]
4 1 6
t t t
  
  
  

Rate
4PH3(g)  P4(g) + 6H2(g)
2
1 [H ] 1
(0.168 /s) 0.028 /s
6 6
M M
t

 

4
1 [P ]
0.028 /s
1
M
t




14.2 Dependence of Reaction
Rate on Reactant
Concentration
• Rate law expression
For the general equation:
aA + bB  cC + dD
rate law = k[A]x[B]y
k = proportionality constant
x and y = the order of the reaction with
respect to each reactant

Order
• Exponents represent order
• Only determined via experimental data
– 1st order - rate directly proportional to
concentration
– 2nd order - exponential relationship
– 0 order - no relationship
• Sum of exponents (order) indicates
overall reaction order

Experimental Determination of
Rate Law
• Method of initial rates - examine
instantaneous rate data at beginning of
reaction

rate = k[F2]x [ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 3: [ClO2] is held constant
1st order with respect to [F2] (rate and M directly related)
2
2
[F ]
0.20
3 2
[F ] 0.10
1
M
M
 
[rate] 3
2.4 10 /s
3 2
3
[rate] 1.2 10 /s
1
M
M


 



rate = k[F2]1[ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 2: [F2] is held constant
2
2
[ClO ]
0.040
2 4
[ClO ] 0.010
1
M
M
 
1st order with respect to [ClO2] (rate and M directly related)
[rate] 3
4.8 10 /s
2 4
3
[rate] 1.2 10 /s
1
M
M


 



rate = k[F2]1[ClO2]1 overall order = 2
Find k (use any set of data)
4
[rate] 2.1 10 /s 2 1
9.3 s
2 2
[A] [B] (0.10 ) (0.015 )
M
k M
M M

  
   

Determining Rate Law
8.4 x 102
0.030
0.10
3
4.2 x 104
0.015
0.20
2
2.1 x 104
0.015
0.10
1
Initial
Rate
(M/s)
[B] (M)
[A] (M)
Exp.

What is Different?
• In experiment 1 and 2; [B] is constant;
[A] doubles and rate doubles - the
reaction is 1st order with respect to [A]
• In experiment 1 and 3; [A] is constant;
[B] doubles but the rate quadruples!
This means that the reaction is 2nd
order with respect to [B]

Calculate the Rate Constant
• Rate = k[A] [B]2
• The rxn is 1st order w/ respect to [A]
• The rxn is 2nd order w/ respect to [B]
• The rxn is 3rd order overall (1 + 2)
4
[rate] 2.1 10 /s 2 1
9.3 s
2 2
[A] [B] (0.10 M) (0.015 )
M
k M
M

  
   

14.3 Dependence of Reactant
Concentration on Time
• First-Order reactions may be
expressed in several ways
• Example: A  products
rate = k[A]
[ ]
rate
t

 

A

Integrated Rate Law
(First order)
y = mx + b
When the two expressions are set equal to
each other,we get an expression that can be
rearranged in the form of a straight line.
   0
ln ln
A kt A
  

Graphical Methods
• Given concentration and time data,
graphing can determine order

Integrated Rate Law
(First order)
• For a 1st order reaction, a plot of ln [A]
vs time yields a straight line
• The slope = k (the rate constant)

Try Graphing
132
300
150
250
170
200
193
150
220
100
284
0
P (mmHg)
Time (s)

Graphing
• Plot ln [Pressure] on y-axis and time on
x-axis
• If the plot is a straight line, then the
integrated rate law equation can be
used to find the rate constant, k, or the
slope of the line can be calculated for
the rate constant.

Integrated Rate Law
• The rate constant for the reaction
2A  B is 7.5 x 103 s1 at 110C. The
reaction is 1st order in A. How long (in
seconds) will it take for [A] to decrease
from 1.25 M to 0.71 M?

• Another form of the integrated rate law
[A]
ln
[A]
0
t kt
 
(0.71 M) 3 1
ln 7.5 10 s ( )
(1.25 M)
75 s
t
t
 
  


Your Turn!
• Consider the same first order reaction
2A  B, for which k = 7.5 x 103 s1 at
110C. With a starting concentration of
[A] = 2.25 M, what will [A] be after 2.0
minutes?

Half-Life (1st order)
• Half-life: the time
that it takes for the
reactant
concentration to
drop to half of its
original value.

Calculating First Order
Half-life
• Half-life is the time that it takes for the
reactant concentration to drop to half of
its original value.
• The expression for half-life is simplified
as
t1/ 2 
0.693
k

Half-Life
• The decomposition of ethane (C2H6) to
methyl radicals (CH3) is a first order
reaction with a rate constant of
5.36 x 104 s1 at 700 C.
C2H6  2CH3
Calculate the half-life in minutes.

Half-Life
1/2 4 1
0.693
t 1293 s
5.36 10 s
1 min
1293 s 21.5 min
60 s
 
 

 

Dependence of Reactant
Concentration on Time
Second-order reactions may be expressed in
several ways
• Example: A  product
rate = k[A]2
[A]
rate
t

 


Integrated Rate Law for
Second Order Reactions
Again, the relationships can be combined to
yield the following relationship in the form of a
straight line
   0
1 1
kt
A A
 

Integrated Rate Law for
Second Order Reactions
• For a 2nd order reaction, a plot of 1/[A]
vs time yields a straight line
• The slope = k (the rate constant)

Calculating Second-Order
Half-life
The expression for half-life is simplified as
Note: half-life for 2nd order is inversely
proportional to the initial reaction
concentration
1/2
0
1
t
[A]
k


Calculating Second Order
Half-life
I(g) + I(g)  I2(g)
The reaction is second order and has a rate
constant of 7.0 x 109 M1 s1 at 23C.
a) If the initial [I] is 0.086 M, calculate the
concentration after 2.0 min.
b) Calculate the half-life of the reaction
when the initial [I] is 0.60 M and when the
[I] is 0.42 M.

a) Use integrated rate equation for 2nd
order
   
9 1 1
11 1
12
11 1
1 1
(7.0 10 s ) 120s
A 0.086
8.4 10
1
[A] 1.2 10
8.4 10
M
M
M
M
M
 



  
 
  


b) Use the half-life formula for 2nd order
(note: half-life does not remain constant for a
2nd-order reaction!)
10
1/2 9 1 1
10
1/2 9 1 1
1
2.4 10 s
(7.0 10 s ) (0.60 )
1
3.4 10 s
(7.0 10 s ) (0.42 )
t
M M
t
M M

 

 
  

  


Zero Order
Zero Order reactions may exist but are
relatively rare
• Example: A  product
rate = k[A]0 = k
• Thus, a plot of [A] vs time yields a
straight line.

Summary of Orders

14.4 Dependence of Reaction
Rate on Temperature
• Most reactions occur faster at a higher
temperature.
• How does temperature alter rate?

Collision Theory
• Particles must collide in order to react
• The greater frequency of collisions, the
higher the reaction rate
• Only two particles may react at one time
• Many factors must be met:
– Orientation
– Energy needed to break bonds (activation
energy)

Collision Theory

Collision Theory
• Though it seems simple, not all
collisions are effective collisions
• Effective collisions: a collision that
does result in a reaction
• An activated complex (transition state)
forms in an effective collision

Activation Energy

The Arrhenius Equation
• The dependence of the rate constant of a
reaction on temperature can be expressed
Ea = activation energy
R = universal gas constant
A = frequency factor T = Kelvin temp
a/
E RT
k Ae


Arrhenius Equation
In the form of a straight line…what plot will give a straight line?
   
a 1
ln ln
E
k A
R t
 
  
 
 

Arrhenius Equation
• A plot of ln k vs 1/T will give a straight
line
• The slope of line will equal Ea/R
• The activation energy may be found by
multiplying the slope by “R”

Graphing with Arrhenius
• Rate constants for the reaction
CO(g) + NO2(g)  CO2(g) + NO(g)
Were measured at four different
temperatures. Plot the data to
determine activation energy in kJ/mol.

Graphing
318
0.332
308
0.184
298
0.101
288
0.0521
T (Kelvin)
k (M1 s1)

Graphing
Steps:
• Make a column of ln k data
• Make a column of inverse temp (1/T)
• Plot ln k vs 1/T
• Calculate the slope
• Multiply slope by R

Arrhenius Another Way!
• Another useful arrangement of the
Arrhenius equation enables calculation
of: 1) Ea (with k at two different temps)
2) the rate constant at a different
temperature (with Ea, k and temps)
1 a
2 1
2
1 1
ln
k E
T T
k R
 
  
 
 

Arrhenius Another Way!
• Use the data to calculate activation
energy of the reaction mathematically
7.0 x 101
500
6.1 x 102
450
2.9 x 103
400
k (s1)
T (Kelvin)

Arrhenius
1
2
a
2 1
a
ln
1 1
3
2.9 10
ln
2
6.1 10
8.3145J/Kmol
1 1
450 400
91 kJ/mol
k
k
E R
T T
E
 
 
  

 
 

 

 

 

  

 
 
 


14.5 Reaction Mechanisms
• Most reactions occur in a series of steps
• The balanced equation does not tell us
how the reaction occurs!
• There are often a series of steps which
add together to give the overall reaction
• The series of steps is the reaction
mechanism

Reaction Mechanisms
• Most chemical reactions occur in a
series of steps
• Energy of activation must be overcome
to form intermediates

Reaction Mechanism
Consider:
2NO(g) + O2(g)  2NO2(g)
The reaction cannot occur in a single
step.
One proposed mechanism:
Step 1: NO + NO  N2O2
Step 2: N2O2 + O2  2NO2

Reaction Mechanism
N2O2 is an intermediate in the reaction
mechanism
Intermediate: a substance that is
produced in an early step and
consumed in a later step
Elementary reaction: one that occurs in
a single collision of the reactant
molecules

Reaction Mechanism
• Molecularity: the number of reactant
molecules involved in the collision
• Unimolecular: one reactant molecule
• Bimolecular: two reactant molecules
• Termolecular: three reactant molecules
(fairly rare)

Rate Determining Step
• If the elementary reactions are known,
the order can be written from the
stoichiometric coefficients of the rate-
determining step
• Rate-determining step: the slowest
step in the mechanism

Rate-Determining Step
• Steps of a mechanism must satisfy two
requirements
– Sum of elementary steps must equal the
overall balanced equation
– The rate law must have same rate law as
determined from experimental data

• The decomposition of hydrogen
peroxide (2H2O2  2H2O + O2 )
may occur in the following two steps
Step 1: H2O2 + I H2O + IO
Step 2: H2O2 + IO H2O + O2 + I
If step 1 is the rate-determining step, then
the rate law is
rate = k1[H2O2] [I]
k1
¾ 
¾
k2
¾ 
¾

• I does not appear in the overall
balanced equation
• I serves as a catalyst in the reaction -
it is present at the start of the reaction
and is present at the end
• IO is an intermediate

Potential Energy Diagram

Reaction Mechanism
• Given overall equation:
H2(g) + I2(g)  2HI(g)
Step 1: I2 2I (fast)
Step 2: H2 + 2I 2HI (slow)
rate = k2[H2] [I]2
This rate expression does not meet the
requirement..I is an intermediate and
should not appear in the rate expression
k2
¾ 
¾
k1
k1
¾ 
¾
¬ ¾
¾

• Consider the first equilibrium step: the
forward rate is equal to the reverse rate
k1[I2] = k-1 [I]2 k1/k-1 [I2] = [I]2
If we substitute for [I]2 , the rate law becomes
rate = k[H2] [I2]
This now matches the overall balanced
equation!
(When step 2 is rate-determining this
substitution is always possible)

14.6 Catalysis
• Catalyst - a substance that increases
the rate of a chemical reaction without
being used up itself
• Provides a set of elementary steps with
more favorable kinetics than those that
exist in its absence
• Many times a catalyst lowers the
activation energy

Reaction Pathway with
Catalyst

Types of Catalysts
• Heterogeneous catalysts - reactants
and catalyst are in different phases
• Homogeneous catalysts - reactants
and catalysts are dispersed in single
phase
• Enzyme catalysts - biological catalysts

Heterogeneous Catalysts
• Most important in industrial chemistry
• Used in catalytic converters in
automobiles
– Efficient catalytic converter serves two
purposes; oxidizes CO and unburned
hydrocarbons into CO2 and H2O; converts
NO and NO2 into N2 and O2

Catalytic Converter

Homogeneous Catalysts
• Usually dispersed in liquid phase
• Acid and base catalyses are the most
important types of homogeneous
catalysis in liquid solution
• Advantages of homogeneous catalysts
– Reactions performed at room conditions
– Less expensive
– Can be designed to function selectively

Biological Catalysts
• Enzymes: large protein molecule that
contains one or more active sites where
interactions with substrates occur
• Enzymes are highly specific (lock and
key)

Reaction Pathway without and
with Enzyme-Substrate

Enzyme-Substrate Complex

Biological Molecules
(binding of glucose to
hexokinase)

Key Points
• Rate of reaction can be determined in
several ways
– Instantaneous rate
– Average rate
– Graphing using integrated rate laws
– Mechanisms

Key Points
• Write rate law expressions
• Calculate rate constant with proper units
• Distinguish orders: 1st, 2nd, 0 order
• Calculate half-life
• Collision theory and relationship to
Arrhenius equation

Key Points
• Calculate activation energy graphically
and mathematically
• Reaction mechanisms
– Elementary reactions
– Molecularity
– Rate law from slow step
– Intermediates and catalysts

Key Points
• Catalysis
– Heterogeneous
– Homogeneous
– Enzymes

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