Download Chemical Kinetics

Chemical Kinetics
Rates at which chemical
reactions occur.
Main goal is to understand the steps by
which a reaction takes place:
Reaction Mechanism
1
As we will see, there are several factors which
affect the rate of a chemical reaction
including:
1. the nature and concentrations of the
reactants.
2. the temperature of the reaction system.
3. the presence of a catalyst and
4. the surface area of the reactants or catalyst.
2
Reaction Rate
Change in concentration of a reactant or product per
change in time.
∆[ ] = [final] – [initial]
Rate = ∆[reactants]
∆t
Rate = ∆[products]
∆t
concentration
concentration
( )Sign is used if the [ ] is decreasing
Units usually molar/second
3
2NO2(g)  2NO(g) + O2(g)
in a flask at 300C, measure change in
concentration as it decomposes.
Calculate the average rate it changes in the first 50
sec.
.0079mol / L  .0100mol / L
50. s  0s
  4.2 x10  5 mol / L  s
(-)means decreasing
4
Instantaneous rates
Value at a particular time: slope of the line tangent
to the curve at that point.
 y  [ NO2 ]

x
t
 [ NO2 ]
Rate  
t
Rate   ( slope of tan gent )
  0.0026mol / L 




110s
 2.4 x10  5 mol / L s
5
Rates and Stoichiometry
Relative rates
(need to look at the coefficients)
2NO2(g)  2NO(g) + O2(g)
Must include reference:
 disappearance of reactant
 appearance of product
6
2NO2(g)  2NO(g) + O2(g)
Rate of consumption = Rate of production =2(rate of production
of NO2
of NO
of O2)
 

 NO 2
t

=
∆[NO]
∆t
2
 
 O2
t
Rate of O2 production is half that of NO
4PH3  P4 + 6H2
7
Example:
2A + 3B C + 4D
What is the rate of B, C and D in reference to one mole of A?
- Δ[A]
Δt
3 -Δ[B] = 1 Δ[C]
= /
/
2 Δt
2 Δt
= 2
Δ[D]
Δt
8
Rate Laws
2NO2(g)   2NO(g) + O2(g)
Reactions are reversible
Choose conditions where the reverse has
negligible contributions.
Study at a point soon after they are mixed
before product builds up.
Reaction rate will depend only on
concentration of the reactants.
9
Rate Laws
Two key points
1. The concentration of the products do not
appear in the rate law because this is an
initial rate.
2. The order (exponent) must be determined
experimentally, can’t be obtained from the
equation
So, what is a Rate Law?
• Algebraic expression of the relationship
between concentration and the rate of a
reaction at a particular temperature.
• Constant of proportionality in the
expression is given the symbol k and is
referred to as the specific rate constant for
the reaction
11
Rate Laws
( ignore reverse)
Rate = k[NO2]n
k = rate constant (constant of proportionality)
n = rate order (must be determined
experimentally not from a balanced equation)
can be an integer including 0.
[product] does not appear in the rate law
12
Types of Rate Laws
Differential Rate Law: expresses how rate
depends on concentration. (rate law)
Integrated Rate Law: expresses how
concentration depends on time.
13
Determining the Form of the Rate Law
*how a reaction occurs*
1. Need to determine n (order)
2N2O5(aq)  4NO2(aq) + O2(g)
N2O5 = 0.90M
= 0.45M
5.4 x 10 -4mol/L·s
2.7 x 10-4mol/L·s
½ the concentration , ½ the rate
14
Slope of the
tangent to the
curve
15
This means the rate of the reaction depends on
concentration of N2O5 to the first power.
The differential rate law is:
 [ N 2 O5 ]
Rate  
 k [ N 2 O5 ]1  k [ N 2 O5 ]
t
***first order reaction
doubling the concentration doubles the
reaction rate.
16
Reaction Rate and Concentration
Click in this box to enter notes.
Go to Slide Show View (press F5) to play the video or animation. (To exit, press Esc.)
This media requires PowerPoint® 2000 (or newer) and the Macromedia Flash Player (7 or higher).
[To delete this message, click inside the box, click the border of the box, and then press delete.]
Copyright © Houghton Mifflin Company. All rights reserved.
Method of Initial Rates
(experimentally determining the form of the rate law)
Initial Rate: the “instantaneous rate” just after
the reaction begins before the [initial] of
reactants have changed significantly. Just after
t=0
1. Run the reaction several times
2. Vary initial concentrations
3. Measure reaction rate just after reaction is
mixed
18
NH4+(aq) + NO2-(aq) →N2(g) + 2H2O(l)
Double the NO2 ; double the initial rate
Rate  k[ NH4  ]n [ NO2  ]m
19
Rate 2 = 2.7 x 10-7 = (.0100)m = (2.0)m
Rate 1 1.35 x 10-7 (.0050)m
=2.00 =
(2.0)m
In terms of
NO2-
Value of m=1
First order reaction in terms of NO2-.
20
Rate 3 = 5.4 x 10-7 = (.200)n = (2.0)n
Rate 2 2.7 x 10-7 (.100)n
=2.00 = (2.0)n
In terms of
NH4+
Value of n = 1 and the value of m=1
First order in both NH4 and NO2
Rate law is : Rate = k[NH4+][NO2-]
21
Overall Reaction Order
Sum of the order of each component in the
rate law.
rate = k[NH4+][NO2-]
The overall reaction order is 1 + 1=2.
22
Calculate k
1.35x10-7=k(0.100mol/L) (0.0050mol/L)
k=2.7 x 10-4L/mol.s
23
Sample 12.1
The reaction
between bromate
ions and bromide
ions in acidic
aqueous solution is
given by the
equation
BrO3-(aq) + 5Br-(aq) + 6H+(aq) →3Br2(l)+ 3H2O(l)
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
24
Determine the order for all three reactants, the overall reaction
Order, and the value of the rate constant.
Rate = k[BrO3-]n[Br-]m[H+]p
Determine the values of n, m, and p by comparing the
rates from the various experiments.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
25
n= 1
m= 1
p= 2
The rate is 1st order in BrO3- and br-, and 2nd order in H+.
The overall reaction order is n + m + p = 4
Rate law is written as;
Rate = k[BrO3-][Br-][H+]2
K= 8.0 L3/mol3•s
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
26
Integrated Rate Laws
So far looked at rate as a funx of the
[reactant].
Also useful to express the [reactant] as a funx
of time, give the (differential) rate law for
the reaction.
27
2N2O5(soln) → 4NO2(soln) + O2(g)
 [ N 2 O5 ]
Rate  
 k [ N 2 O5 ]
t
Since the rate depends on the [N2O5]
to the 1st power,
1st order reaction
This means if the [N2O5] doubled, the
rate would double.
Put into different form using calculus
28
First-Order Rate Law
For aA  Products in a 1st-order reaction,
   A
Rate =
 k A
t
Integrated first-order rate law is
ln[A] = kt + ln[A]o
**Concentration as a function of time**
29
Integrated first-order rate law is
ln[A] = kt + ln[A]o
1. The equation shows how [A] depends on
time.
2. Is in the form y=mx + b, where a plot of y
vs. x is a straight line with slope m and
intercept b.
y=ln[A]
x=t
m=-k
b=ln[A]0
30
For the reaction: aA→products
The reaction is first order in A if a plot of ln[A] versus t is
a straight line.
If it is not a straight line it is not 1st order.
3.
This integrated rate law for a first-order reaction also can
be expressed in terms of ratio of [A] and [A]0 as follows;
 [ A]0 
ln
  kt
 [ A] 
31
2N2O5(soln) → 4NO2(soln) + O2(g)
Using data, verify that the rate law is first order in [N2O5],
And calculate the value of the rate constant.
[N2O5] (mol/L)
.1000
0.707
0.0500
0.0250
0.0125
0.00625
Time (s)
0
50
100
200
300
400
32
Figure 12.4: A plot of ln[N2O5]
versus time.
This verifies
that the
reaction is
1st order.
33
ln[N2O5] = -kt + ln[N2O5]0
y =
mx + b
Since the reaction is 1st order, the slope of the line equals
-k, where
change y  (ln[ N 2 O5 ])
Slope 

change x
t
1
 5075
.  (2.303)  2772
3
slope 

  6.93x10 s
400s  0s
400
k   ( slope)  6.39 x10  3 s  1
34
Using the data given, calculate [N2O5] at 150s
after the start of the reaction.
[N2O5]
(mol/L)
.1000
0.707
0.0500
0.0250
0.0125
0.00625
Time (s)
ln[N2O5] = -kt + ln[N2O5]0
k =6.39x10-3s-
0
50
100
200
300
400
ln([ N 2 O5 ]) t 150   (6.93x10  3 s 1 )(150s)  ln(010
. )
  1040
.
 2.303   3.343
[ N 2 O5 ]t 150  anti log(  3.343)  0.0353mol / L
35
Half-Life of a First-Order
Reaction
Time required to reach [1/2]
t1/2
0.693

k
t1/2 = half-life of the reaction
k = rate constant
For a first-order reaction, the half-life does
not depend on concentration.
36
Figure 12.5: A plot of [N2O5] versus time for the
decomposition reaction of N2O5.
t1/2=100s
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
37
Formula for t1/2 is derived from integrated rate law:
aA→ product
If the reaction
is 1st order in [A]
 [ A]0 
ln
  kt
 [ A] 
[ A]0
[ A] 
2
By definition
when t=t1/2,
 [ A]0 
  kt1/ 2
ln
 [ A]0 / 2 
ln(2)  kt1/ 2
t1/ 2
0.693

k
38
Half-life first order reaction
A certain first-order reaction has a half-life of
20.0min.
a. Calculate the rate constant for this
reaction.
b. how much time is required for this
reaction to be 75% complete.
39
1.
0.693
k
 3.47 x10  2 min
20.0 min
2. Using the integrated rate law in the form
 [ A]0 
ln
  kt
 [ A] 
If the reaction is 75% complete, 75% of reactant has been
consumed, leaving 25% in the original form.
[ A]
x100%  25%
[ A0 ]
This means
that
[ A]0
[ A]
1
 .25 or

 4.0
[ A]0
[ A]
.25
40
 [ A] 
 347
. x10  2 
  ln 4.0  kt  
then ln
t
 min 
 [ A]0 
and
ln 4.0
t
 2  40.min
3.47 x10
min
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
41
Another way to solve..
• t1/2 is at 50% completion
• If the [1mol/L]0, after one half-life the
[0.50mol/L].
• One more half-life produces [0.25mol/L].
• 25% of reactant is left.
• Two half-lives = 2(20.0min)=40.0min
42
Second-Order Rate Law
For aA  products in a second-order reaction,
 A
Rate =
k A
t
Integrated rate law is
2
*Doubling the
concentration
quadruples the
reaction rate.
1
1
 kt +
A
Ao
43
1
1
 kt +
A
Ao
1. A plot of 1/[A] versus t will produce a
straight line w/slope = k
2. Shows how [A] depends on time and can
be used to calculate [A] at any time t,
provided k and [A]0 are known.
44
Half-Life of a Second-Order
Reaction
t1/2
1

kA
o
t1/2 = half-life of the reaction
k = rate constant
Ao = initial concentration of A
The half-life is dependent upon the initial
concentration.
45
Half life of second order reaction
At t=t1/2,
[ A]0
[ A] 
2
Then the equation becomes
1
1
 kt 1/ 2 +
 A 0
 A o
2
2
1

 kt 1/ 2


[ A]0
Ao
1
 kt 1/ 2
 A o
Solve for t1/ 2
1

k [ A]0
46
Butadiene reacts to form its dimer according to the equation:
2C4H6(g)→C8H12(g)
(two identical molecules
combine)
The following data were collected for this reaction at a given temp :
[C4H6](mol/L)
Time (+,- s)
0.01000
0.00625
0.00476
0.00370
0.00313
0.00270
0.00241
0.00208
0
1000
1800
2800
3600
4400
5200
6200
a. Is the reaction 1st or 2nd order?
b. What is the value of the rate
constant for the reaction?
c. What is the half-life for the
reaction under the conditions
of this experiment?
47
Figure 12.6: (a) A plot of
ln[C4H6] versus t. (b) A plot of
1n[C4H6] versus t.
48
a. Is it first or second order?
Write the rate law.
Rate =
2nd

  C 4 H6
t
 k C H
 
4
2
6
49
b. What is the rate constant?
2nd order: plot 1/[C4H6] vs. t and get a
straight line with slope of k
1
(
)
change y
C4 H6
Slope 

change x
t
1
(481  100) L / mol
381
slope 

 614
. x10  2 s
6200s  0s
6200
50
c. Half-life?
t1/2
1

kA
o
t1/2=1/(6.14x10-2L/mol.s)(1.000x10-2mol/L)
= 1.63x103s
51
Difference between ½-life of a 1st
order and 2nd order reaction
2nd order
•Depends on both k and
[A]0
•Each ½ life is double of
the preceding one
1st order
•Depends only on k
•Constant time for each
½-life
52
Zero order reaction
Rate =k[A]0=k(1)=k
The rate is constant (does not change with
concentration)
Integrated rate law :
[A]=-kt+[A]0
Plot of [A] versus t gives a straight line of slope -k
53
Half-life
[A]=[A]0/2 when t=t1/2 , so
[ A]0
  kt1/ 2  [ A]0
2
[ A]0
t1/ 2 
2k
54
Figure
12.7: A
plot of [A]
versus t for
a zeroorder
reaction.
Zero order
Metal surface or an enzyme is required for the
reaction to occur.
56
The decomposition reaction
2N2O(g) → 2N2(g) + O2(g)
takes place on a platinum surface.
Concentration has no effect on the rate
57
More Complicated Reactions
• BrO3- + 5 Br- + 6H+  3Br2 + 3 H2O
• For this reaction we found the rate law to be
• Rate = k[BrO3-][Br-][H+]2
• To investigate this reaction rate we need to
control the conditions
Rate = k[BrO3-][Br-][H+]2
• We set up the experiment so that two of the
reactants are in large excess.
• [BrO3-]0= 1.0 x 10-3 M
• [Br-]0 = 1.0 M
• [H+]0 = 1.0 M
• As the reaction proceeds [BrO3-] changes
noticeably
• [Br-] and [H+] don’t
Rate = k[BrO3-][Br-][H+]2
• Assume that through the reaction:
[Br-] = [Br-]0
[H+] = [H+]0
Therefore the rate law can be written:
60
Rate = k[BrO3-][Br-][H+]2
• Rate = k[BrO3-][Br-]0[H+]02
• Rate = k[BrO3-][Br-]0[H+]02
k’ = k[Br-]0[H+]02
Rate = k’[BrO3-]
This is called a pseudo first order rate law.
k’
k=
[Br-] [H+] 2
A Summary
1.
Simplification: Conditions are set such
that only forward reaction is important.
2.
Two types:
3.
Which type? Depends on the type of
data collected - differential and
integrated forms can be interconverted.
differential rate law
integrated rate law
62
A Summary
(continued)
4.
Most common: method of initial rates.
5.
Concentration v. time: used to
determine integrated rate law, often
graphically.
6.
For several reactants: choose
conditions under which only one
reactant varies significantly (pseudo
first-order conditions).
63
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
64
Reaction Mechanism
 The
series of steps by which a chemical
reaction occurs. (bond making and bond
breaking)
 Must
be determined by experiment!
 Must
agree with overall stoichiometry
AND the experimentally determined rate
law.
A chemical equation does not tell us how reactants become
products - it is a summary of the overall process.
65
Elementary steps: the steps in the reaction mechanism that
add up to the balance equation
Molecularity: how the molecules collide
The rate for a reaction can be written from its molecularity .
66
Reaction Mechanisms
2NO2 + F2  2NO2F
Rate = k[NO2][F2]
The proposed mechanism is:
Elementary steps
NO2 + F2  NO2F + F (slow)
F + NO2  NO2F
(fast)
2NO2 + F2  2NO2F
F is called an intermediate it is formed then consumed in the reaction
Formed in reversible reactions
2 NO + O2  2 NO2
Mechanism
2 NO  N2O2 (fast and reversible)
N2O2 + O2  2 NO2
(slow) Determining!
rate = k[N2O2][O2] No intermediates
Substitute [NO]2 for [N2O2]
What is the rate determining step in this video clip
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of
collisions produces a reaction. Why?
Arrhenius: An activation energy must be
overcome.
Mechanism Sim
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
69
Factors that Affect Reaction Rate
1. Concentrations of reactants
• More reactants mean more collisions if
enough energy is present
2. Temperature
• Collision Theory: When two chemicals
react, their molecules have to collide
with each other with sufficient energy
for the reaction to take place.
• Kinetic Theory: Increasing temperature
means the molecules move faster.
Collision Theory
•
•
•
•
•
•
Particles have to collide to react.
Have to hit hard enough
Things that increase this increase rate
High temp – faster reaction
High concentration – faster reaction
Small particles = greater surface area means
faster reaction
Rates and collisions
However…
• Has been found that the rate of a reaction is
much smaller than the calculated collision
frequency in a collection of gas particles.
• Therefore only a small fraction of the
collisions produces a reaction. WHY?
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
72
Arrhenius Equation
 Collisions
must have enough energy to
produce the reaction (must equal or exceed
the activation energy (threshold energy)).
 Orientation
of reactants must allow
formation of new bonds.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
73
Figure 12.11: (a) The change in potential energy as a function of reaction
progress for the reaction 2BrNO  2NO + Br2. The activation energy Ea
represents the energy needed to disrupt the BrNO molecules so that they
can form products. The quantity ∆E represents the net change in energy in
going from reactant to products. (b) A molecular representation of the
reaction.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
74
Figure 12.12: Plot showing the
number of collisions with a
particular energy at T1 and T2,
where T2 > T1.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
75
Temperature Dependence of the Rate Constant
**10 increase will double the rate**
k = A • exp( -Ea/RT )
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
-Ea 1
Ln k = + lnA
R T
13.4
ln(k) = -Ea/RT + ln(A)
Catalysts
• Speed up a reaction without being used up
in the reaction.
• Enzymes are biological catalysts.
• Homogenous Catalysts are in the same
phase as the reactants.
• Heterogeneous Catalysts are in a different
phase as the reactants.
Ways to speed up a reaction:
1. Catalysts
Speed up reactions by lowering activation energy
2. Surface area of a solid reactant
Bread and Butter theory: more area for reactants to be in contact
3. Pressure of gaseous reactants or
products
Increased number of collisions
79
Catalysts allow
reactions to proceed
by a different
mechanism - a new
pathway with a
lower activation
energy.
More molecules will
have this activation
energy.
Does not affect ∆E
Show up as a
reactant in one step
and a product in a
later step
80
Figure 12.16: Effect of a catalyst on
the number of reaction-producing
collisions.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
81
Heterogeneous Catalysis
(gas on a solid surface)
Steps:
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.
82
Heterogenous Catalysts
H H
• Hydrogen bonds to surface
of metal.
• Break H-H bonds
H
H H
H H
H
Pt surface
Hydrogenation of unsaturated fats
Heterogenous Catalysts
H
H
H
C
C
H
H H
H
H
Pt surface
Heterogenous Catalysts
• The double bond breaks and bonds to the
catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H
Heterogenous Catalysts
• The hydrogen atoms bond with the carbon
H
H
H
C
H
C
H
H
Pt surface
H H
Heterogenous Catalysts
H
H
H
H
C
C
H
H
H
Pt surface
catalyst sim
H
Homogenous Catalysts
(same phase)
• Chlorofluorocarbons (CFCs) catalyze the
decomposition of ozone.
• Enzymes regulating the body processes.
(Protein catalysts)
Ostwald Process
Pt catalyst
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
2NO (g) + O2 (g)  2NO2 (g)
2NO2 (g) + H2O (l)  HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Hot Pt wire
over NH3 solution
13.6
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
13.6
Enzyme Catalysis
13.6
Catalysts and rate
• Catalysts will speed up a reaction but only
to a certain point.
• Past a certain point adding more reactants
won’t change the rate.
• Zero Order
Catalysts and rate.
R
a
t
e
• Rate increases until the active sites of
catalyst are filled.
• Then rate is independent of
concentration
Concentration of reactants
•
This reaction takes place in three steps

Ea
First step is fast
Low activation energy

Ea
Second step is slow
High activation energy

Ea
Third step is fast
Low activation energy
Second step is rate determining
Intermediates are present
Activated Complexes or
Transition States
Catalysts
• Speed up a reaction without being used up
in the reaction.
• Enzymes are biological catalysts.
• Homogenous Catalysts are in the same
phase as the reactants.
• Heterogeneous Catalysts are in a different
phase as the reactants.
How Catalysts Work
• Catalysts allow reactions to proceed by a
different mechanism - a new pathway.
• New pathway has a lower activation energy.
• More molecules will have this activation
energy.
• Does not change E
• Show up as a reactant in one step and a
product in a later step
Figure 12.15: Energy plots for a
catalyzed and an uncatalyzed
pathway for a given reaction.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
103
Chapter 12(b)
Chemical Kinetics
(cont’d)
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
104
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
105
Figure 12.9: A molecular
representation of the elementary
steps in the reaction of NO2 and
CO.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
106
12.10:
A plot
showing the
exponential
dependence
of the rate
constant on
absolute
temperature.
orientations for a collision
between two BrNO molecules.
Orientations (a) and (b) can lead
to a reaction, but orientation (c)
cannot.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
108
for the
reaction
2N2O5(g)
4NO2(g) +
O2(g). The
value of the
activation
energy for
this reaction
can be
obtained
12.17:
Heterogeneo
us catalysis
of the
hydrogenatio
n
of ethylene.
Figure 12.18: The exhaust gases
from an automobile engine are passed
through a catalytic converter to
minimize environmental damage.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
111
removal of
the end
amino acid
from a
protein by
reaction with
a molecule
of water.
The products
are an amino
acid and a
enzyme carboxypeptidase-A, which
contains 307 amino acids. The zinc ion is
shown above as a black sphere in the
center. (b) Carboxy-peptidase-A with a
substrate (pink) in place.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
113
interaction. The substrate is shown in
black and red, with the red representing
the terminal amino acid. Blue indicates
side chains from the enzyme that help
bind the substrate.