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Chemical kinetics

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Chemical kinetics The rate of a chemical reaction is dependent on: reactant concentrations state of reactants (solid, liquid, powder, etc.) temperature (e.g., eggs ... – PowerPoint PPT presentation

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Title: Chemical kinetics


1
Chemical kinetics
The rate of a chemical reaction is dependent on
  • reactant concentrations
  • state of reactants (solid, liquid, powder, etc.)
  • temperature (e.g., eggs cook faster at higher
    temperatures)
  • catalyst (e.g., catalytic converter in your car
    speeds the formation of less polluting products
    from your engine)

Kinetics predicts on how fast you reach
equilibrium... not the extent to which the
reaction proceeds!
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2
Reaction rates are determined experimentally
NOTE Rate varies with time
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3
Average Reaction Rate
4
Relative Rates
  • aA bB ? cC dD
  • Rate

5
Example
  • The reaction of hydrogen with nitrogen to produce
    ammonia is 3H2 N2 ?? 2NH3
  • If the rate of appearance of NH3 is 3.0 x 10-6
    M/s, what is the rate of disappearance of H2?

6
Write relative rate expressions for the following
reactions in terms of the disappearance of the
reactants and the appearance of the products
  • 3O2(g) ? 2O3(g)
  • 4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)

7
Types of Rate Laws
  • Differential Rate Law expresses how rate
    depends on concentration.
  • Integrated Rate Law expresses how concentration
    depends on time.

8
Differential Rate Law
  • This rate law shows how the rate of a reaction
    depends on concentrations of various species in
    the reaction.

A and B are reactants
The Rate Law is determined experimentally does
not necessarily reflect the stoichiometry of the
reaction
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9
How does value of k affect reaction rate?
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10
Rate Law Example
  • The reaction2NO(g) 2H2(g) ? N2(g)
    2H2O(g)has the rate law R k NO2 H2
  • Therefore the reaction is 2nd order in NO, 1st
    order in H2 and 3rd order overall.

11
Rate Constant Units
  • The units for the rate constant depend on the
    rate law.
  • For example, if the rate law is R k NO2
    Br2then the units are
  • R ? M sec-1
  • therefore, k ?

12
Measured Reaction Rates
  • The rate of reaction changes with time, as the
    concentrations of reactants and products change.
  • The rate constant, but not the rate, is
    independent of time.
  • The initial rate is the reaction rate measured
    before the initial concentrations have had time
    to change.

13
Determining Reaction Orders
  • The rate law for a given reaction must be
    determined experimentally. This means
    determining the order with respect to each
    species involved.
  • The two major methods are
  • Method of Initial Rates
  • Integrated Rate Law Method

14
Method of Initial Rates
15
Method of Initial Rates
  • .involves measuring the rate of reaction at very
    short times before any significant changes in
    concentration occur

16
The initial rate is determined in several
experiments using different initial
concentrations.
e.g.
BrO3- 5Br- 6H3O ? 3Br2 _ 9H2O
initial concentrations in starting solution
Expt BrO3- Br- H3O Initial rate (M/s)
1 0.10 0.10 0.10 1.2x10-3
2 0.20 0.10 0.10 2.4x10-3
3 0.10 0.30 0.10 3.5x10-3
4 0.20 0.10 0.15 5.5x10-3
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17
How to use the initial rate data
Note k does not change
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18
return to problem...
Expt BrO3- Br- H3O Initial rate (M/s)
1 0.10 0.10 0.10 1.2x10-3
2 0.20 0.10 0.10 2.4x10-3
3 0.10 0.30 0.10 3.5x10-3
4 0.20 0.10 0.15 5.5x10-3
BrO3-
Find expt pair where only ONE conc changes
Find similar pair of expts for other reactants
Br-
H3O
RatekBrO3- Br- H3O2
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19
Integrated Rate Laws
20
First-Order Rate Law
  • For aA ? products in a second-order reaction,

21
Can we predict concentration with time?
We will consider only first order reaction...
Rate kA
A?B
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22
Example problem radioactive decay (a first
order process)
14C has a half-life of 5,730 years. If the
concentration of 14C originally in an artifact
was 1.3x10-9 moles/g and the current
concentration is 0.87x10-9 moles/g,. How old is
the artifact?
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23
Recognizing First-Order
  • Plot the experimental ln A vs. time
  • If the graph is linear, the reaction is
    first-order.
  • The rate constant is k -slope

ln A - kt ln A0
y m t b
24
SO2Cl Example
  • The reaction SO2Cl2(g) ? SO2(g) Cl2(g) gives
    the following experimental data

Time (s) Pressure SO2Cl2 (atm)
0 1.000
2,500 0.947
5,000 0.895
7,500 0.848
10,000 0.803
25
SO2Cl Example
ln P vs time
slope 2.19 x 10-5 sec-1
ln(p/atm)
time/sec
26
Second-Order Rate Law
  • For aA ? products in a second-order reaction,

27
Second-Order Rate Law (Type I)
k
2A ? P
28
Recognizing Second-Order
  • Plot the experimental 1/A vs. time
  • If the graph is linear, the reaction is
    second-order.
  • The rate constant is k slope

y m t b
29
NO2 Example
  • The following data were obtained for the
    gas-phase decomposition of NO2(g) at 300C.

Time (s) NO2 ln NO2 1/NO2
0 0.01000 -4.610 100
50 0.00787 -4.845 127
100 0.00649 -5.038 154
200 0.00481 -5.337 208
300 0.00380 -5.573 263
30
First Order Plot
ln NO2
time / s
31
Second Order Plot
1/ NO2
time / s
32
Zero Order Rate Law
  • Reaction rate is does not depend on concentration
  • Rate k
  • Plot of A vs. time produces straight line

33
Zero Order Rate Law
34
Reaction Mechanisms
35
Reaction Mechanism
  • The series of steps by which a chemical reaction
    occurs.
  • A chemical equation does not tell us how
    reactants become products - it is a summary of
    the overall process.

36
Reaction Mechanism
  • The mechanism of a reaction is the sequence of
    individual collisions, known as elementary steps,
    that take the reactant molecule(s) to the product
    molecule(s).

37
Often Used Terms
  • Elementary Step A reaction whose rate law can
    be written from its chemical equation.

38
Often Used Terms
  • Intermediate formed in one step and used up in
    a subsequent step and so is never seen as a
    product.

39
Often Used Terms
  • Molecularity the number of species that must
    collide to produce the reaction indicated by that
    step.

40
Ozone Example
  • The conversion of ozone O3 to oxygen O2 has the
    overall balanced equation
  • 2O3(g) ? 3O2(g).
  • A possible mechanism for this reaction has two
    elementary steps
  • O3(g) ? O(g) O2(g)
  • O3(g) O(g) ? 2O2(g)
  • Net result 2O3(g) ? 3O2(g)
  • O(g) is an intermediate

41
Reaction Mechanisms
  • Criteria
  • Elementary steps must add to give overall
    balanced equation
  • Rate law of slow step must agree with actual rate
    law

42
Rate Determining Steps
  • Some reactions have a mechanism in which one of
    the elementary steps is much slower than the
    others.
  • Such a slow elementary step is called the rate
    determining step, since it acts as the bottleneck
    for the overall reaction.
  • The overall rate law is then simply the rate law
    for the rate determining step.

43
Schematic of rate of reaction for2NO O2 ? 2NO2
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44
When the slow step isnt the first
stepintermediates in the rate law
  • Step 1 NO NO ? N2O2
  • Step 2 N2O2 O2?2NO2 (slow)

45
Collision Model
  • Molecules must collide with sufficient energy to
    react.
  • only a small fraction of collisions produces a
    reaction.
  • Arrhenius An activation energy, Ea, must be
    overcome.

46
EFFECT OF TEMPERATURE ON REACTION
KINETICSMinimum energy needed for effective
collision, i.e., product formed
no reaction
Products
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47
Temperature dependence of the number of molecules
that have the minimum energy to react.
Activation energy
At higher temperatures, more molecules have the
minimum energy and the rate of reaction increases.
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48
Activation Energy and rate of reaction
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49
Arrhenius Equation
  • Collisions must have enough energy to produce the
    reaction (must equal or exceed the activation
    energy Ea).
  • Orientation of reactants must allow formation of
    new bonds. (frequency factor, A)

50
The frequency factor (A) can be visualized as a
geometry factor. Not all collisions (even with
enough energy) will result in a successful
reaction. What if it hits in the wrong spot?
51
Lets deal with this concept quantitativelyArrhe
nius equation
k rate constant Ea activation energy A
frequency factor R 8.314 J/mol-K
Useful form of equation for comparing rates when
temperature is changed (Ea stays the same)
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52
How much faster is a reaction if the temperature
is raised from 25 to 35oC when Ea 58.1 kJ/mol?
53
Arrhenius Equation
Therefore, ln k vs. 1/T is linear with slope Ea/R
54
CH3NC Example
  • The table below shows the rate constant for the
    CH3NC rearrangement at various temperatures.
    Find the activation energy.

Temperature / C k / s-1
189.7 2.52 x 10-5
198.9 5.25 x 10-5
230.3 6.30 x 10-4
251.2 3.16 x 10-3
55
Slope -Ea/R -19050 -Ea/8.3145 J/Kmol Ea
158391 J 158.391 kJ
56
Catalysis
  • Catalyst A substance that speeds up a reaction
    without being consumed
  • Enzyme A large molecule (usually a protein)
    that catalyzes biological reactions.

57
Catalysts increase reaction rate
  • Catalysts change the reaction mechanism to lower
    Ea
  • Catalysts are involved in reaction but are
    neither created nor consumed
  • Homogeneous catalyst Same phase as reactants
    (e.g., a soluble solution species)
  • Heterogeneous catalyst Separate phase from
    reactants(e.g., catalytic muffler on your
    automobile)

58
Catalysis
  • Homogeneous catalyst Present in the same phase
    as the reacting molecules.
  • Heterogeneous catalyst Present in a different
    phase than the reacting molecules.

59
Heterogeneous Catalysis
Steps
  • 1. Adsorption and activation of the reactants.
  • 2. Migration of the adsorbed reactants on the
    surface.

60
Heterogeneous Catalysis
Steps
  • 3. Reaction of the adsorbed substances.
  • 4. Escape, or desorption, of the products.

61
Example of Heterogeneous catalysis
C2H4 H2 ? C2H6
Catalytic converter
C2H5? (radical)
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