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Chemical Kinetics

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Title: Chemical Kinetics


1
Chemical Kinetics
  • Rates of Reactions
  • Rate and Concentration
  • Finding Rate Laws
  • First-Order Reactions
  • Rate and Temperature
  • Theory of Reaction Rates
  • Reaction Mechanisms
  • Catalysts
  • Steady-State Approximations

2
Reaction Rate The Central Focus of Chemical
Kinetics
3
Some Reactions are Fast and Others are Slow. Why?
Reactions of gases can be very fast, or very
slow, even though the gases may be able to
diffuse very fast. Why is this?
H2 (g) F2 (g) 2 HF(g)
(very fast!)
3 H2 (g) N2 (g) 2 NH3 (g)
(very slow!)
The reaction of metals with water can also occur
fast or slow, and the speed of these chemical
reactions depends upon the chemistry of the metal
and the reactivity of the water molecule. Why ?
Mg(s) 2 H2O(l) Mg2(aq) 2 OH-(aq)
H2 (g) (slow)
Ba(s) 2 H2O(l) Ba2(aq) 2 OH-(aq)
H2 (g) (fast)
4
There are Several Important Reasons that
Different Reactions, and Even the Same Chemical
Reaction May Go at Different Rates
1) Concentration molecules must collide in order
to react. Reaction rate is proportional to
the concentration of reactants. Rate k
(collision frequency) k (concentration)
2) Physical state molecules must mix in order to
collide. The physical state (solid, liquid,
gas) will affect frequency of collisions, as
well as the physical size of droplets (liquid) or
particles in the case of solids.
3) Temperature molecules must collide with
enough energy and
correct orientation to react. Raising the
temperature increases the reaction rate by
increasing the number of collisions, and
especially, the energy of the collisions.
5
The Effect of Surface Area on Reaction Rate
Hot Steel wool in O2
Hot Nail in O2
6
Collision Energy and Reaction Rate Bimolecular
7
Rates of reactions
  • Rate of a chemical reaction.
  • The change in the quantity of a reactant or
    product that takes place in a period of time.
  • rate

8
Rates of reactions
  • To study rates of reaction, you must
  • Identify the reactants and products.
  • Carry out the reaction.
  • Measure the concentrations of one of the
    reactants or products at known intervals.
  • You must have a way to measure at least one of
    the species involved.
  • Continuous monitoring methods should be used
    whenever possible.

9
An example reaction
  • Decomposition of N2O5.
  • Dinitrogen pentoxide is known to decompose
    completely by the following reaction.
  • 2N2O5 (g) ? 2N2O4 (g) O2 (g)
  • This reaction can be conducted in an inert
    solvent like carbon tetrachloride.
  • When N2O5 decomposes, N2O4 remains in solution
    and O2 escapes and can be measured.

10
An example reaction
  • We can easily measure the oxygen as dinitrogen
    pentoxide decomposes.
  • Temperature must be maintained to within 0.01
    oC.
  • The reaction flask must be shaken to keep oxygen
    from forming a supersaturated solution.

11
An example reaction
Gas buret
Constant temperature bath
12
An example reaction
  • Time (s) Volume STP O2, mL
  • 0 0
  • 300 1.15
  • 600 2.18
  • 900 3.11
  • 1200 3.95
  • 1800 5.36
  • 2400 6.50
  • 3000 7.42
  • 4200 8.75
  • 5400 9.62
  • 6600 10.17
  • 7800 10.53

Here are the results for our experiment.
13
An example reaction
Volume, mL O2
The rate of O2 production slows down with time.
Time, s
14
Average rates
  • We can calculate the average rate of oxygen
    formation during any time interval as
  • Average rate of
  • O2 formation


The rates shown here have units of mL O2 at
STP/s. Note how the rate decreases with time.
15
Instantaneous rates
  • We know that the rate of our reaction is
    constantly changing with time.
  • Instantaneous rate
  • The rate of reaction at any point in time.
  • It can be found by taking the tangent of our
    earlier plot.
  • Initial rate of reaction
  • The rate of formation at time zero when the
    reactants are initially mixed.

Volume, mL O2
Time, s
16
An example reaction, again
  • Since we know what the stoichiometry for our
    reaction is, we can calculate the concentration
    of N2O5 during our reaction.
  • 2N2O5 (g) 2N2O4 (g) O2 (g)
  • For each mole of O2 produced, two moles of N2O5
    will have decomposed.
  • The rate of reaction will be
  • rate of reaction -

1 2
17
An example reaction, again
Volume, mL O2
N2O5
Time, s
18
The Rate of Reaction
  • Consider the hypothetical reaction,
  • A(g) B(g) C(g) D(g)
  • equimolar amounts of reactants, A B, will be
    consumed and products, C D, will be formed as
    indicated in this graph

19
A(g) B(g) C(g) D(g)
  • A concentration of A in M ( mol/L)
  • Note that rxn does not go to completion

20
Generic Rate Equation
21
The Rate of Reaction
  • Rate of a simple one-step reaction is directly
    proportional to the concentration of the reacting
    substance
  • R is the rate of the reaction
  • k specific rate constant

22
The Rate of Reaction
  • For a simple rate-law expression like
  • R kA
  • doubling the initial concentration of A doubles
    the initial rate of reaction
  • halving the initial concentration of A halves the
    initial rate of reaction

23
The Rate of Reaction
  • Very often, more than one reactant molecule
    appears in the equation for a one-step reaction
  • experimentally determine that reaction rate

?
24
The Rate of Reaction
  • Rate Law Expressions must be determined
    experimentally
  • cannot be determined from balanced equations
  • most chemical reactions are not one-step
    reactions
  • Rate law expressions are also called
  • rate laws
  • rate equations
  • rate expressions

25
Rate and concentration
  • We can develop a quantitative relationship
    between instantaneous rates and concentration.
  • By drawing tangents along the curve for N2O5,
    we can measure the following rates of reaction.

26
Rate and concentration
  • The earlier data indicated that the rate is
    directly proportional to concentration.
  • rate k N2O5
  • We can verify this by calculating the value for
    k for the various rates we measured.

27
2H2O2(aq) ---gt 2H2O(l) O2(g)
  • How can we find the rate of this reaction?
  • Decrease of H2O2
  • Increase of H2O or O2
  • Evolution of a gas - moles/liter per unit time

28
Average Rate
29
Average Rate
30
Reaction rates
  • The rate of formation of a product is always a
    positive quantity
  • The rate of disappearance of a reactant is always
    a negative quantity
  • The rate of appearance and disappearance of
    reactants and products is stoichiometric

2H2O2(aq) ---gt 2H2O(l) O2(g)
31
Rate of reaction
2H2O2(aq) ---gt 2H2O(l) O2(g)
  • Rate of Reaction?
  • Positive quantity
  • remove the stoichiometry (based on per mole)

32
The Rate of Reaction
  • Order of a reaction
  • expressed in terms of either
  • each reactant
  • overall reaction
  • For example

33
The Rate of Reaction
If the rate law does not match the
stoichiometry? Not a single-step reaction!
34
The Rate of Reaction
35
The Rate of Reaction
  • Look at the following one step reaction and its
    experimentally determined rate-law expression
  • because it is a second order rate-law expression
  • doubling the A increases the rate of reaction
    by a factor of 4
  • 22 4
  • halving the A decreases the rate of reaction
    by a factor of 4
  • (½)2 ¼

36
Rate and concentration
  • For the general reaction
  • a A b B . . . e E f F .
    . .
  • the rate expression will often have the form of
  • rate k Ax By . . .
  • k rate constant
  • x, y order for A B, respectively
  • x y order for the reaction
  • Note The order is NOT the same as the
  • coefficients for the balanced reaction.

37
Reaction mechanisms
  • Elementary process
  • Each step in a mechanism.
  • Molecularity
  • The number of particles that come together to
    form the activated complex in an elementary
    process.
  • 1 - unimolecular
  • 2 - bimolecular
  • 3 - termolecular

38
Elementary Reactions and Molecularity Rate Laws
and Mechanism
39
Fig 14-2
Pg 646
A unimolecular elementary reaction may involve
bond breakage. If an N2O4 molecule possesses
enough energy, the vibration can break the N-N
bond to produce two NO2 molecules.
40
Reaction mechanisms
  • For elementary processes, the exponents for each
    species in the rate law are the same as the
    coefficients in the equation for the step.
  • For our earlier example,
  • the rate law is
  • rate k NO O3

41
Reaction mechanisms
  • In general, the rate law gives the composition of
    the activated complex.
  • The power of a species in the rate law is the
    same as the number of particles of the species in
    the activated complex.
  • If the exponents in the rate law are not the same
    as the coefficients of the equation for the
    reaction, the overall reaction must consist of
    more than one step.

42
Pg 645A
Bimolecular reaction
If this was an elementary reaction rate k
NO22
43
Pg 645B
44
Proposed Mechanism
45
Alternative Mechanism
46
rate k NO22 Slow first Step Matches
Stoichiometry
47
Fig 14-14
Pg 674
H2 Br2 ? 2 HBr
A molecular view of the accepted mechanism for
the accepted mechanism for the reaction of H2 and
Br2 to produce 2 HBr.
48
Pg 675
49
Example 14-9
Pg 676
50
Pg 675
What would the proposed rate law be?
51
Finding rate laws
  • Method of initial rates.
  • The order for each reactant is found by
  • Changing the initial concentration of that
    reactant.
  • Holding all other initial concentrations and
    conditions constant.
  • Measuring the initial rates of reaction
  • The change in rate is used to determine the order
    for that specific reactant. The process is
    repeated for each reactant.

52
N2O5 example
2N2O5 (g) ? 2N2O4 (g) O2 (g)
  • The following data was obtained for the
    decomposition of N2O5.
  • Experiment N2O5 Initial rate, M/s
  • 1 0.100 3.62 x 10-5
  • 2 0.200 7.29 x 10-5
  • We know that the rate expression is
  • rate k N2O5x
  • Our goal is to determine what x (the order) is.

53
N2O5 example
  • For exp. 2 7.29 x 10-5 M/s k (0.200 M)x
  • For exp. 1 3.62 x 10-5 M/s k (0.100 M)x
  • We can now divide the equation for experiment
    two by the one for experiment one.
  • 7.29 x 10-5 M/s k (0.200 M)x
  • 3.62 x 10-5 M/s k (0.100 M)x
  • which give 2.01 (2.00)x
  • and x 1 (first order reaction)


54
A more complex example
  • The initial rate of reaction was obtained for the
    following reaction under the conditions listed.
  • A B C . . .
  • Exp. A B C
    Initial rate, M/s
  • 1 0.030 0.010 0.050 1.7 x 10-8
  • 2 0.060 0.010 0.050 6.8 x
    10-8
  • 3 0.030 0.020 0.050 4.9 x 10-8
  • 4 0.030 0.010 0.100 1.7 x 10-8
  • With this series of experiments, the
    concentration of a single reactant is doubled in
    concentration for experiments 2-4, compared to
    experiment one.

55
A more complex example
  • While this type of problem is more time
    consuming, its not any more difficult than the
    previous example.
  • Order for A
  • Use experiments one and two.
  • 6.8 x 10-8 M/s (0.060 M)x
  • 1.7 x 10-8 M/s (0.030 M)x
  • 4.0 (2.0)x
  • By inspection, x 2


56
A more complex example
  • Order for B
  • Use experiments one and three.
  • 4.9 x 10-8 M/s (0.010 M)y
  • 1.7 x 10-8 M/s (0.020 M)y
  • 2.9 (2.0)y
  • The order is not obvious by inspection. You
    must take the logarithm of both sides and solve
    for y.
  • ln 2.9 y ln 2.0
  • y 1.54 or


3 2
57
A more complex example
  • Order for C
  • Use experiments one and four.
  • Experiment C Initial Rate
  • 1 0.050 1.7 x 10-8
  • 4 0.100 1.7 x 10-8
  • Here the rate did not change when C was
    doubled. This is an example of a zero order
    reaction.
  • z 0

58
A more complex example
  • We can now write the overall rate law.
  • rate A2 B3/2 C0
  • or since C has no effect on the rate
  • rate A2 B3/2
  • The overall order for the reaction is
  • x y z 2 3/2 0 3 1/2

59
Determining Reaction Order from the Rate Law
Problem For each of the following reactions,
determine the reaction order with respect to
each reactant and the overall order of the given
rate law a) 5 Br-(aq) BrO3-(aq) 6 H(aq)
3 Br2 (aq) 3 H2O(aq)
b) 2 NO(g)
2 H2 (g) N2 (g) 2 H2O(g)
Rate kNO2H2 PlanWe inspect the exponents
in the rate law, not the coefficients of
the balanced equation, to find the individual
orders, and then take their sum to obtain the
overall reaction order.
Rate kBrBrO3-H2
Solution a) The reaction is first order with
respect to Br- and BrO3- and second
order with respect to H. The reaction is fourth
order overall. b) The reaction is
second order in NO and first order in H2 and
third order overall.
60
Integrated rate laws
  • Integrate this eqn. over time from zero to
    infinity

61
Finding Rate Laws
  • Using graphical method
  • based on integrated rate laws

zero order for A
first order for A
second order for A
62
2H2O2(aq) ---gt 2H2O(l) O2(g)
  • How can we find the rate of this reaction?
  • Decrease of H2O2
  • Increase of H2O or O2
  • Evolution of a gas - moles/liter per unit time

63
Zero Order?
64
First Order?
65
Second Order?
66
First Order
67
Finding rate laws
  • Graphical method.
  • Using integrated rate laws, one can produce
    straight line plots. The order for a reactant if
    verified if the data fits the plot.
  • Rate integrated Graph
    Slope
  • Order law rate law
    vs. time
  • 0 rate k At -kt A0
    At -k
  • 1 rate kA lnAt -kt lnA0
    lnAt -k
  • 2 ratekA2 kt
    k

68
Finding rate laws
0 order plot
2nd order plot
N2O5
1/N2O5
Time (s)
Time (s)
Time (s)
As you can see from these plots of the N2O5 data,
only a first order plot results in a straight
line.
1st order plot
lnN2O5
69
Reaction profile
This type of plot shows the energy changes
during a reaction.
Potential Energy
activation energy
?H
Reaction coordinate
70
Effective collision
Activated Complex
A temporary state where bonds are in the process
of breaking and forming.
71
Examples of reaction profiles
Exothermic reaction
Endothermic reaction
72
Reaction mechanisms
  • A detailed molecular-level picture of how a
    reaction might take place.

activated complex
bonds in the process of breaking or
being formed
73
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75
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76
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77
First order reactions
  • Reactions that are first order with respect to a
    reactant are of great importance.
  • Describe how many drugs pass into the blood
    stream or used by the body.
  • Often useful in geochemistry
  • Radioactive decay
  • Half-life (t1/2)
  • The time required for one-half of the quantity
    of reactant originally present to react.

78
Half-Life
79
Using the Integrated rate law
  • Find value of k from the slope
  • Given the rate constant you can calculate
  • time to reach a certain concentration
  • concentration reached in a given time
  • the initial concentration, given some conc. with
    time
  • Half-Life time for reactant to decrease by
    1/2

80
Use of 1/2 Life
  • C4H8(g) --gt 2C2H4 (g) First Order Reaction
  • At 1000C k 87 s-1
  • What is the 1/2 life?
  • So 50 is gone after 8ms
  • After 24ms how much is gone?

81
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Use of 1/2 Life
of t1/2s 0 1 2 3 4
Percent Left 100 50 25 12.5 6.25
83
Theories of reaction rates
  • Collision theory
  • Based on kinetic-molecular theory.
  • It assumes that reactants must collide for a
    reaction to occur.
  • They must hit with sufficient energy and with the
    proper orientation so as to break the original
    bonds and form new ones.
  • As temperature is increased, the average kinetic
    energy increases - so will the rate.
  • As concentration increases, the number of
    collisions will also increase, also increasing
    the rate.

84
Rate and temperature
  • Reaction rates are temperature dependent.

Here are rate constants for N2O5 decomposition at
various temperatures. T, oC k x 104, s-1
20 0.235 25 0.469 30
0.933 35 1.82 40 3.62 45
6.29
k x 104 (s-1)
Temperature (oC)
85
Rate and temperature
  • The relationship between rate constant and
    temperature is mathematically described by the
    Arrhenius equation.
  • k A e
  • A constant
  • Ea activation energy
  • T temperature, Kelvin
  • R gas law constant

-Ea / RT
86
Rate and temperature
  • An alternate form of the Arrhenius equation is
  • ln k ln A
  • If ln k is plotted against 1/T, a straight line
    of slope -Ea/RT is obtained.
  • Activation energy - Ea
  • The energy that molecules must have in order to
    react.

87
Calculation of Ea from N2O5 data
ln k
T-1
88
Temperature and Ea
  • As the temperature is increased, a higher
    fraction of molecules will have a kinetic energy
    that is greater that the activation energy.

T1 lt T2 lt T3
89
Transition state theory
  • As reactants collide, they initially form an
    activated complex.
  • The activated complex is in the transition state.
  • It lasts for approximately 10-100 fs.
  • It can then form products or reactants.
  • Once products are formed, it is much harder to
    return to the transition state, for exothermic
    reactions.
  • Reaction profiles can be used to show this
    process.

90
Reaction Mechanisms
  • Consider the following reaction.
  • 2NO2 (g) F2 (g) 2NO2F (g)
  • If the reaction took place in a single step the
    rate law would be
  • rate k NO22 F2
  • However, the experimentally observed rate law is
  • rate k NO2 F2

91
Reaction Mechanisms
  • Since the observed rate law is not the same as if
    the reaction took place in a single step, we know
    two things.
  • More than one step must be involved
  • The activated complex must be produced from two
    species.
  • A possible reaction mechanism might be
  • Step one NO2 F2 NO2F F
  • Step two NO2 F NO2F
  • Overall 2NO2 F2 2NO2F

92
Reaction Mechanisms
  • Rate-determining step.
  • When a reaction occurs in a series of steps,
    with one slow step, it is the slow step that
    determines the overall rate.
  • Step one NO2 F2 NO2F F
  • Expected to be slow. It involves breaking an
    F-F bond.
  • Step two NO2 F NO2F
  • Expected to be fast. A fluorine atom is very
    reactive.

93
Reaction Mechanisms
  • Since step one is slow, we can expect this step
    to determine the overall rate of the reaction.
  • NO2 F2 NO2F F
  • This would give a rate expression of
  • rate k1 NO2 F2
  • This agrees with the experimentally observed
    results.

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Catalysis
  • Catalyst A substance that changes the rate of a
    reaction without being consumed in the
    reaction.
  • Provides an easier way to react.
  • Lower activation energy.
  • Still make the same products.
  • Enzymes are biological catalysts.
  • Inhibitor A substance that decreases the rate
    of reaction.

96
Catalysis
Types of catalysts Homogeneous - same
phase Catalyst is uniformly distributed
throughout the reaction mixture Example - I-
in peroxide. Heterogeneous - different
phase Catalyst is usually a solid and
the reactants are gases or liquids Example -
Automobile catalytic converter.
97
End-of-Chapter 14 Exercises 9, 12, 13, 15, 17,
23, 24, 25, 27, 29, 31, 35, 36, 41, 42, 45, 47,
52, 57, 61, 62, 71, 77, 78, 96  
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