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Unformatted text preview: Chemical Kinetics
Chemical
Rates, Mechanisms, and
Rates,
Catalysis
Catalysis Kinetics
Chemistry is most useful when it allows us
Chemistry
to control things.
to
Thus we are interested in 2 factors...
HOW FAST.
(I feel a need for speed).
HOW FAR.
(Really, where is the equilibrium?)
Chemical Kinetics 2 Kinetics We approach kinetics in 2 major branches; 1) The macroscopic scale where we
1)
observe behavior.
observe 2) The microscopic scale where we
2)
attempt to explain behavior.
attempt Chemical Kinetics 3 Some Experiments to Think About
Lycopodium powder, in bulk vs powdered
NH3 + HCl
H2 + energy source
2 H2 + O2 2 H2O
A Simple Clock
2 N2O5 4 NO2 + O2
What would plots of concentration vs time look like
What
for these examples?
for
Chemical Kinetics 4 The rate of a chemical reaction
is affected by:
is
a.
b.
c.
d. the concentration of the reactants.
the temperature of the reaction.
the presence of a catalyst.
all of the above. Decomposition of
Dinitrogen Pentoxide Chemical Kinetics 6 The Plot for the Forward Reaction Chemical Kinetics 7 The Plot for the Reverse
The
Reaction
Reaction Chemical Kinetics 8 Consider These…
How do we normally describe the motion of a
How
vehicle?
vehicle?
How could you describe the factor that limits
How
how fast your bath tub fills?
These are common examples of the concept of
rate
rate
The reaction rate would be
The
the change in …
change Chemical Kinetics 9 Expressing Rates x
c1x Chemical Kinetics x
c2x +2x
c3+2x 10
10 N2 + 3 H2 2 NH3
The OVERALL rate is expressed as
The
x
 /∆ tmol/l.s where x = ∆ [a species]
mol/l.s
What does that mean in terms of:
What
N2,
H2,
and NH3?
and
Chemical Kinetics 11
11 Example CK.1
Give the relative rates for
Give
disappearance of reactants and
formation of products for the
production of hexamethyleneproduction
tetramine by the following reaction:
6CH2O(aq) + 4NH3(aq) (CH2)6N4(aq) + 6H2O
(aq) Chemical Kinetics 12
12 N2 + 3 H2 2 NH3
If the rate of formation of NH3 is 0.024 M/s, what
are the rates of disappearance of H2 & N2?
are Chemical Kinetics 13
13 4 NH3 + 3 O2 2 N2 + 6 H2O
NH
If the rate of formation of N2 is
0.34 M/s, what are the rates of
disappearance of NH3 & O2
disappearance
and the rate of appearance of
H2O?
Chemical Kinetics 14
14 As a chemical reaction
proceeds, the rate of the
reaction tends to:
reaction
a.
b.
c.
d. increase.
decrease.
remain constant.
oscillate. How Do You Determine a Rate?
1. Remember that the
Remember
data that you will use
to determine a rate
will look something
like one of these
curves.
curves.
2. Further, you will only
Further,
use the data from the
_______ region of
the plot.
the
3. What possibilities are
What
there to get some
sort of x/∆ t values?
sChemical Kinetics
ort 16
16 Average Rates
A+BC+D Focus on the kinetic region
Take slope of line from
Take
start to beginning of
equilibrium
equilibrium
Or 2 points in between
Note slope gets smaller,
Note
and approaches value at
one point in time.
one
Chemical Kinetics 17
17 Instantaneous Rates
Keep taking shorter
Keep
time intervals,
time
Eventually you
Eventually
reach the tangent.
reach
This is an
This
INSTANTANEOUS
rate.
rate.
Example 1
Chemical Kinetics Example 2 18
18 Initial Rate
Note the
Note
instantaneous
rate as close to
time 0 as
possible.
possible.
This is the INITIAL
This
rate, and is the
most commonly
used value.
used
Chemical Kinetics 19
19 Reconsider Concentration
• The clock reaction shows plainly that
The
concentration changes rates.
concentration
• The difficult question is always – HOW?
• Let us reason some together... What kind of
mathematical relationship
gives rise to a curve?
Chemical Kinetics 20
20 Let’s Get Mathematical
In simplest terms
Rate α[A] or Rate α [B] …
Rate
[A]
But, with the curve to consider, we include
Rate α[A]m , Rate α [B]n …
Rate α
And, of course, we can do the same for each
And,
species that alters the rate.
species
Combining, we get Rate α…
α
Combining,
Chemical Kinetics 21
21 Proportionality?
How does a mathematician make proportionality
How
easier to deal with?
easier Add a proportionality constant! Our equation becomes… Rate = k[A]m[B]n… This is known as the RATE LAW or the RATE
EQUATION
Chemical Kinetics 22
22 Rate Law (Rate Equation)
An expression of the concentration
An
dependence of the reaction rate.
dependence
For simple reactions it is the product of a
For simple
temperature dependent rate constant, k,
rate
and the reactant concentrations each
raised to an exponent – the reactant order.
order
Reactant orders MUST be determined
Reactant
experimentally.
experimentally.
Chemical Kinetics 23
23 Reactant Orders
The reactant orders are often, but NOT
The
ALWAYS, equal to the coefficients in the
balanced equation.
balanced H + I 2 HI
Rate = k[H2][I2]
2
2 2 HI H 2 + I 2
HI 2 N2O5 4 NO2 + O2 Rate = k[N2O5] Chemical Kinetics Rate = k[HI]2 24
24 Reactant Orders Essentially, data are gathered to set up 2
equations for each reactant. (i.e., at least
3 experiments) Then we use the rate law equation at each
condition. Chemical Kinetics 25
25 In Abstract
Rate1 = k[A]1m and Rate2 = k[A]2m Since k is the same in each Rate1/[A]1m = Rate2 / [A]2m = k Rate1/ Rate2 = [A]1m / [A]2m Or Rate1/ Rate2 = ([A]1 / [A]2)m
Chemical Kinetics 26
26 Rate Constant
The rate constant, k, is found experimentally.
The The rate constant is affected by temperature. The rate constant is usually different for the
forward and reverse reactions. Chemical Kinetics 27
27 Finding Orders
Here’’s a set from an other text book
s Chemical Kinetics 28
28 A+BC+D
Rate = k[A][B]
Rate
The overall order of this reaction
The
is:
is:
a.
b.
c.
d. first.
second.
third.
fourth. If tripling the concentration of
reactant A multiplies the initial
rate of the reaction by nine, the
reaction is _______ order in A.
reaction
a.
b.
c.
d.
d. zeroth
first
second
second
third Finding the Rate Constant
This is an easy task now we have
This
the orders.
the Pick any set of conditions in the
data table and solve the RATE
LAW for k. k = rate
/([A]m[B]n[C]p.
(msd)
Chemical Kinetics 31
31 A Simple Case First
The rate of decomposition of 0.10M N2O5 at
298K is 0.22 M/min. What is the first
order rate constant at 298 K?
order
2 N2O5 4 NO2 + O2
1
(of )(∆ [N2O5]/∆ t)
k[N2O5]
Rate = (rate /2 decomposition) = (first order)
k= 1
(1 )(∆ [N x / ∆
(0.22/2M/min2O51]/2)t) x 1/[N2OM
/0.10 5]
Chemical Kinetics More = 1.1 min1
32
32 Continuing…
What is the rate of formation of NO2
at 298 K when [N2O5] = 0.042 M?
at
2 N2O5 4 NO2 + O2
Rate = 1.1kmin1 x [N2O5]M
0.042 = 0.046 M/min ∆ [NO2]/∆ t = 4 x 0.046 M/min =
n
reaction rate
Chemical Kinetics 0.18 M/min
33
33 Concentration vs Time
Borrowing from calculus, we can do some
Borrowing
nifty things.
nifty Consider a 1st order reaction, for example. Rate = ∆ [R]/∆ t. (the expression) But rate = k[R]. Hence ∆ [R]t/∆ t = k[R]0. Gather terms to get ∆ [R]t/[R]0 = k∆ t
Chemical Kinetics 34
34 This is a USEFUL EQUATION
Note that ln([R]t/[R]o) = ln[R]t – ln[R]o The integrated equation is for a straight line.
Specifically: ln[R]t = ln[R]0  kt A plot of ln[R] vs time gives a line with a
slope = k and proves the reaction 1st order. A useful graphical method thus derived.
Chemical Kinetics 35
35 Other orders
Without any math… Zeroorder reactions > [R]o – [R]t = kt 2nd order > (1/[R]t) – (1/[R]o) = kt We’ll only use these occasionally, but again
they are straight line plots which give order
and k
Chemical Kinetics 36
36 Back to 1 Order
st Let’’s look at the integrated rate law at a
s
special time.
special Every reaction must reach a point where
the [R]t is exactly half the original ([R]0). If this becomes a new “original [R]0”, a new
half way point will be reached eventually.
Every one of these half way points takes
the same time. Chemical Kinetics 37
37 (Continuing)
This time is called the halflife for the
This
reaction, represented as t½
reaction, In a first order reaction this happens when
([R]t/[R]o) = 1/2 So ln(1/2) = kt½ = ln1 – ln2 Thus ln2 = kt½ = 0.693 Timerelated problems can then be dealt
with in numbers of halflives.
Chemical Kinetics 38
38 The time needed for half of the
reactant to be consumed is
called the __________ of the
reaction.
reaction.
a.
b.
c.
d. midpoint
equivalence point
halfrate
halflife The halflife of a firstorder
reaction is equal to _________,
reaction
where k is the rate constant.
where
a. 0.693/k
0.693/
b. 0.693k
0.693
c. k/2
d. 2k Why Do Reactions Happen?
Let’s review a couple of things briefly. 1. bonds are dynamic things. 2. molecules are always in motion. Chemical Kinetics 41
41 Why Do Reactions Happen? (2)
Simple Collision Collision
Effectiveness Bimolecular
Collision & Ea
Introduction to
Mechanism Chemical Kinetics 42
42 Conditions for Reaction?
Collisions MUST occur.
1. Collisions MUST have sufficient energy. 2. Collisions MUST be in correct orientation.
3. Collisions won’t necessarily succeed. Chemical Kinetics 43
43 A Microscopic View of
Microscopic
Reactions
Reactions
Activation energy barriers Nature of the reactants
Concentrations
Concentrations
Molecular orientations
Temperature and the Arrhenius equation Effect of catalysts
Catalysts in industry and the environment Chemical Kinetics 44
44 Let’s Just Play for a Minute ORA Chemical Kinetics 45
45 Factors Influencing Collisions
We’ve seen concentration affects rate… But why? What effect would temperature have? Why? Remember the Boltzmann distribution? Let’s link 2 plots!
Chemical Kinetics 46
46 Energy Relationships Chemical Kinetics 47
47 The minimum energy a collision
between molecules must have
to produce the products is called
the __________ energy.
the
a.
b.
c.
d. initiation
internal
external
activation Arrhenius
We met Arrhenius in 105 in connection
We
with acids and bases.
with But he also got into the act in kinetics. His most interesting addition was an
equation which summarizes the factors
we’ve been talking about. It is k = AeEa/RT.
Chemical Kinetics 49
49 k = Ae Ea/RT A iis called a preexponential factor and is
s
made up of 2 parts;
made
One part to account for the number of
collisions
The second to account for the fraction of
collisions which are effective The exponential term eEa/RT is always less
than one and is thought of as the fraction
with sufficient energy to react.
Chemical Kinetics 50
50 Utility of Arrhenius’ Equation
Let’s remove the exponential term. Take the natural log of the equation. Yields ln k = ln A – Ea/RT. Find k at several temperatures; A plot of ln k vs.1/T should be a straight line
with intercept ln A and slope related to Ea.
Chemical Kinetics 51
51 Reaction Mechanisms
Introduction
Introduction
Molecularity of elementary steps
Molecularity
Rate equations for elementary steps
Molecularity and reaction order
Mechanism and rate equations Chemical Kinetics 52
52 Reaction Mechanism
How does the reaction actually occur
How
on the atomic level?
on
usually a series of steps called
usually
"elementary reactions“
"elementary
Stoichiometry of these elementary
Stoichiometry
reactions IS related to the powers in the
IS
rate law.
rate
Chemical Kinetics 53
53 Let’s Just Play for a Minute ORA Chemical Kinetics 54
54 Elementary Reaction
Bimolecular Collision of 2 atoms or molecules Termolecular Collision of 3 atoms or molecules Unimolecular Reaction of 1 atom or molecule Reaction mechanism:
Sequence of Elementary Reaction Steps
Chemical Kinetics
Each step has its own rate 55
55 Rate and Mechanism
one approach:
assume one elementary reaction much
assume
slower than the rest
slower
this is the "bottleneck" that determines the
this
overall rate
overall
Called Rate Limiting Reaction
Or
Rate Determining Step
Chemical Kinetics 56
56 Simplest Case
2 NO2 + F2 2 NO2F
NO2 + F2 = NO2F + F
(slow)
NO2 + F = NO2F
(fast)
Since first step is slow:
(1/1) ∆ [F2]/∆ T = k[NO2][F2]
(1/1)
Chemical Kinetics 57
57 iif any other step is rate
f
determining,
determining,
more complicated! Chemical Kinetics 58
58 e.g., H2(g) + Br2(g) = 2HBr(g)
the rate law is very complex
The elementary reactions are
Br2 = 2 Br
fast
Br + H2 = HBr + H slow H + Br2 = HBr + Br fast H + HBr = H2 + Br
HBr fast Chemical Kinetics 59
59 2NO(g) +2H2(g) = N2(g) + 2H2O(g)
Sequence of Elementary Reaction Steps
NO + NO = N2O2
N 2O 2 + H 2 = N 2O + H 2O
N2O + H2 = N2 + H2O
each has its own equilibrium
each has forward rate constant, eg k1
each has reverse rate constant, eg k1
Chemical Kinetics 60
60 Return to Catalysts
Now we know a lot about reactions, let’’s
s
reconsider catalysts.
reconsider How does it work? provides a new "pathway" or
mechanism for the reaction. with lower activation energy. Chemical Kinetics 61
61 Catalysis Movie Chemical Kinetics 62
62 Catalyzed Pathway I Chemical Kinetics 63
63 Catalyzed Pathway II Chemical Kinetics 64
64 Catalyzed Pathway III End of Show Chemical Kinetics 65
65 A catalyst increases the rate of a
reaction by:
reaction
a. causing the molecules to move more
causing
rapidly.
rapidly.
b. causing the molecules to collide more
causing
frequently.
frequently.
c. lowering the energy of activation.
d. all of the above. Biological Catalysts
Biological Chemical Kinetics 67 Model Enzyme
Model Chemical Kinetics 68 Chemical Kinetics 69
69 THE STEPS of IRON
TRANSPORTATION
TRANSPORTATION
the first step that should be noted is that the amino acids in
the
transferrin are structured already ready to bind with iron
transferrin
once iron and transferrin join they flow through the blood stream
once
in search of a transferrin receptor (TFR) on the surface of a cell
in
transferrin and the transferrin receptor bind and create a pocket
transferrin
that holds the iron
that
this is then drawn into the cell
the cell acidifies the inside of the pocket and transferrin lets the
the
iron go
iron
the receptor and the already used transferrin are let back out of the
the
cell
cell
with the blood stream's pH the TFR releases the transferrin and
with
they start all over
they
Chemical Kinetics 70 Transferrin and Receptor
Transferrin Chemical Kinetics 71 An Indicator of Sensitivity
An Chemical Kinetics 72 Sample Rate Determination
Tangent 2 = 0.0006 0.132659
0.108703
0.084747
0.060792
0.036836 [ N 2O 5] / M Decomposition of N2O5 1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
600 560 520 480 440 400 360 320 280 240 200 160 0.3
0.086
0 0.00296 time (s) 0.00296
0.799098 0
360
500 Tangent 1 = 120 0.8
0.442
0 0.68061
0.56211
0.44362
0.32513
0.20664 80 0
120
270 2N2O5 4NO2 + O2 40 Concentration
1.000
0.993208
0.762
1
0.580
0.442
0.336
0.256
0.195
0.148
0.113
0.086
0.066
0.065
0.050
0.038
0.029
0.022
0.017
0.013
0.010
0.007
0.004
0.003
0.002
0.002
0.001
0.001 0 Time
0
40
80
120
160
200
240
280
320
360
400
440
480
520
560
600
640
680
720
800
840
880
920
960
1000 0.0006
0.300351 Chemical Kinetics Exit 73
73 Instantaneous Rate
X Chemical Kinetics 74
74 Instantaneous Rates
X Chemical Kinetics 75
75 Instantaneous Rates
X Chemical Kinetics 76
76 More Examples
X Chemical Kinetics 77
77 More Rates
X Chemical Kinetics 78
78 More Rates
X Chemical Kinetics 79
79 Halflives Chemical Kinetics 80
80 HalfLives (continued)
At time t Multiplier OR [N2O5]t [N2O5]0/1 X (1/2 ) [N2O5]t [N2O5]0 /2
/2 X (1/21) [N2O5]t [N2O5]0 /4
/4 X (1/22) [N2O5]t [N2O5]0 /8
/8 X (1/23) Amt. Left =
Chemical Kinetics 0 Init. Amt. X (1/2)0
X (1/2)1 Calc. In
X (0.5)0
X (0.5)1 X (0.5)
X (1/2)2
X (0.5)3
X (1/2)3
X (0.5)#half lives
2 81
81 Example
Iodine131 is a radionuclide that decays in
Iodine131
a firstorder process with a half life of 8.0
days.
It is a βemitter, which makes it safe for
emitter,
medical use to develop images of the
thyroid.
thyroid.
We can use our little equation to check the
We
effects of time on doses.
effects
Chemical Kinetics 82
82 Example a
We inject a patient with 1.0000 mg of I131. How much will be left after 32 days?
32 days = ? Half lives?
= 4 half lives
half
Amount left = 1.0000 mg x (0.5)4
= 0.0625 mg
0.0625 Chemical Kinetics 83
83 Example b
We use a simple inhouse generator to make 25.0
We
mg of I131 to meet expected diagnostic needs.
How much would be left after 40. Days?
How
40 days = ___ half lives?
=5
Amount left = 25.0 mg x (0.5)5
= 25.0 mg x 0.0313 = 0.78 mg
25.0 Chemical Kinetics 84
84 Example c
If we find a sample of 23.5 mg of I131, how
If
much would we expect to have left after 5.0
days?
days?
5.0 days = (5.0/8.0) half lives
5.0
= 0.625 half lives
0.625
Amount left = 23.5 mg x (0.5)0.625
= 23.5 mg x 0.648
23.5
= 15.2 mg
15.2 Chemical Kinetics 85
85 Ea calculations
T(K)
338 1.50 x 103 318 Chemical Kinetics 4.87 x 103 328
Data for the temperature
Data
dependence of the reaction
N2O5(g) 2NO2(g) + ½O2(g).
(g)
(g) ½O k(s1) 4.98 x 104 308 1.35 x 104 298 3.46 x 105 273 7.87 x 107 86
86 One Important Catalysis Chemical Kinetics 87
87 Arrhenius Equation Activity Chemical Kinetics 88
88 Surface Reactions Chemical Kinetics 89
89 Equilibrium Defined Chemical Kinetics 90
90 Half Life First Order Process Chemical Kinetics 91
91 Second Order Chemical Kinetics 92
92 Reaction Rates Chemical Kinetics 93
93 Transition State Theory Chemical Kinetics 94
94 Reaction Rates Chemical Kinetics 95
95 ...
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This document was uploaded on 11/04/2011 for the course PWS 150 at BYU.
 Fall '11
 HOPKINS

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