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
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

3
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.

4
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.

5
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.
• It is found that the reaction initially occurs
  very rapidly but gradually slows down.

6
An example reaction
Gas buret
Constant temperature bath
7
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 hypothetical
experiment.
8
An example reaction
Volume, mL O2
The rate of O2 production slows down with time.
Time, s
9
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.
10
Examining Rates of Reaction

• Because the graph forms a curve, 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.
• Recall that the slope of a line or the slope of a
  line tangent to a curve is often a meaningful
  value
• Steeper slope faster rate
• In higher math the slope of a line tangent to a
  curve can be calculated using the derivative
• Initial rate of reaction
• The rate of formation at time zero when the
  reactants are initially mixed.

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Instantaneous Rate
Volume, mL O2
Time, s
12
Initial Rate
Volume, mL O2
Time, s
13
Our 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
14
Our example reaction, as a graph
Volume, mL O2
N2O5
Time, s
15
General reaction rates

• For the general reaction
• a A b B . . . e E f F . .
  .
• The rate of reaction can be expressed as
• Rate

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Remember silly sign?

• aA bB ? cC dD
• In stoichiometry
• In kinetics
• Notice that there is a - in front of the two
  reactants. This is because we are describing the
  rate at which they are disappearing. The rate at
  which the reactants are disappearing is related
  in stoichiometric ratio to the rate at which the
  products are appearing.

17
Section 15.1

• Review Examples 15.1 and 15.2 (pg 606)
• Do Exercise 15.1 and 15.2 (pg 606-607)

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Conditions That Affect Rate

• Concentrations of reactants
• In general, increasing the concentrations of
  reactants results in a faster rate
• Temperature
• A rule of thumb is that rate doubles for each
  increase of 10?C (10K)
• Catalysts
• Substances that increase the rate of reaction,
  but are not themselves changed

19
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.

20
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.

21
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
• N.B. - The order is NOT the same as the
  coefficients for the balanced reaction.

22
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.

23
N2O5 example

• 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.

24
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 gives 2.01 (2.00)x
• and x 1 (first order reaction)

25
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.

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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

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A more complex example

• Order for B
• Use experiments one and three.
• 4.9 x 10-8 M/s (0.020 M)y
• 1.7 x 10-8 M/s (0.010 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

Fear not! It is rare to have fractional orders
of reaction!
3 2
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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

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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

30
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
• Notice that all of the integrated rate laws are
  in the form y mx b.

31
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
32
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.

33
Half-life
From our N2O5 data, we can see that it takes
about 1900 seconds for the concentration to be
reduced in half. It takes another 1900
seconds to reduce the concentration in half again.
N2O5
Time (s)
34
Half-life

• The half-life and the rate constant are related.
• t1/2
• Half-life can be used to calculate the first
  order rate constant.
• For our N2O5 example, the reaction took 1900
  seconds to react half way so
• k 3.65 x 10-4 s-1

0.693 k
0.693 t1/2
0.693 1900 s
35
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)
36
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
37
Rate and temperature

• An alternate form of the Arrhenius equation is
• We arrive at this equation if we take the ln of
  both sides of the Arrhenius equation as shown in
  the previous slide
• Notice that this is also in the form y mx b
• 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.

38
Calculation of Ea from N2O5 data
ln k
T-1
39
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.

40
Effective collision

• Reactants must have sufficient energy and the
  proper orientation for a collision to result in a
  reaction.

41
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
42
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.

43
Effective collision
Activated Complex
A temporary state where bonds are in the process
of breaking and forming.
44
Reaction profile
This type of plot shows the energy changes
during a reaction.
Potential Energy
activation energy
?H
Reaction coordinate
45
Examples of reaction profiles
Exothermic reaction
Endothermic reaction
46
Examples ofreaction profiles
High activation energy Low heat of reaction
Low activation energy High heat of reaction
47
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
48
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
• Even termolecular processes are very rare. WHY?
• Think KMT Collision Theory

49
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

50
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.
• Lets look at N2O5 - again!

51
Reaction mechanisms

• Earlier we found that for
• 2N2O5 2N2O4 O2
• The rate law was
• rate k N2O5
• According to the equation, it should be second
  order but the data shows it to be first order.
• The reaction must involve more than one step.

52
Reaction Mechanisms

• Consider the following reaction.
• 2 NO2 (g) F2 (g) 2 NO2F (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

53
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

54
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.

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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.

56
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.

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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.
58
Heterogeneous catalysis
59
Enzymes

• Biological catalysts
• Typically are very large proteins.
• Permit reactions to go at conditions that the
  body can tolerate.
• In some cases enzymes process millions of
  molecules every second.
• Are very specific - react with one or only a few
  types of molecules (substrates).

60
Classification of enzymes

• Based on type of reaction
• Oxireductase catalyze a redox reaction
• Transferase transfer a functional group
• Hydrolase cause hydrolysis reactions
• Lyase break C-O, C-C or C-N bonds
• Isomerases rearrange functional groups
• Ligase join two molecules

61
The active site

• Enzymes are typically HUGE proteins, yet only a
  small part is actually involved in the reaction.

The active site has two basic components. catalyt
ic site binding site
Model of trios-phosphate-isomerase
62
Characteristics of enzyme active sites

• Catalytic site
• Where the reaction actually occurs.
• Binding site
• Area that holds substrate in proper place.
• Enzymes uses weak, non-covalent interactions to
  hold the substrate in place based on R groups of
  amino acids.
• Shape is complementary to the substrate and
  determines the specificity of the enzyme.
• Sites are pockets or clefts on the enzyme surface.

63
Characteristics of enzyme active sites
Binding site
Catalytic site
Enzyme
Substrate
64
Enzyme-substrate reaction