Chemical Kinetics http://www.chem1.com/acad/webtext/dynamics/dynamics-3.html http://ibchem.com/IB/ibnotes/brief/kin-sl.htm

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Chemical Kinetics http//www.chem1.com/acad/we
btext/dynamics/dynamics-3.html
http//ibchem.com/IB/ibnotes/brief/kin-sl.htm
http//www.docbrown.info/page03/3_31rates.htmThee
ffectofaCatalysthttp//www.practicalchemistry.or
g/experiments/the-rate-of-reaction-of-magnesium-wi
th-hydrochloric-acid,100,EX.html
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What do we mean by kinetics?

• Kinetics is the study of
• the rate at which chemical reactions occur.
• The reaction mechanism or pathway through which a
  reaction proceeds.

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How an Airbag Works http//www.youtube.com/wa
tch?vdZfLOnXoVOQ

• When sodium azide, NaN3, is ignited by a spark,
  it releases nitrogen gas which can instantly
  inflate an airbag.(one 25th of a second) 
• SiO2
• 10NaN3 (s) 2KNO3 (s)gt K2O (s) 5 Na2 O(s)
  16N2 (g)

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Reaction Rate

• The change in the concentration of a reactant or
  a product with time (M/s).
• Reactant ? Products
• A ? B
• Since reactants go away with time
• Rate - ?A/?t ?B/?t mol
  dm-3 s-1
• Simulation
• http//www.chm.davidson.edu/vce/kinetics/ReactionR
  ates.html

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A gt B Rate - ?A/?t
?B/?t until completion
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http//www.avogadro.co.uk/kinetics/rate_equatio
n.htm

• When the rate does not depend on the
  concentration

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Reaction Mechanism

• Most chemical reactions consist of a sequence of
  two or more simpler reactions.
• For example, recent evidence indicates that the
  reaction
•  2O3 ---gt 3O2  
• occurs in three steps after intense ultraviolet
  radiation from the sun liberates chlorine atoms
  from certain compounds in Earths stratosphere.

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For example

• 1. Chlorine atoms decompose ozone according to
  the equation
• Cl O3 --- gt  O2 ClO
• 2. Ultraviolet radiation causes the decomposition
  reaction
• O3 --- gt O2 O
• 3. ClO produced in the reaction in step 1 reacts
  with O produced in step 2 according to the
  equation
• ClO O --- gt Cl O2

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• A complex reaction is one that consists of two or
  more elementary steps. The complete sequence of
  elementary steps that make up a complex reaction
  is called a reaction mechanism.

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Rate Determining Step

• Chemical reactions, too, have a weakest link in
  that a complex reaction can proceed no faster
  than the slowest of its elementary steps. In
  other words, the slowest elementary step in a
  reaction mechanism limits the instantaneous rate
  of the overall reaction.

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Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

• The reaction slows down with time because the
  concentration of the reactant decreases.

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Collision Theory

• http//www.saskschools.ca/curr_content/chem30_05/2
  _kinetics/kinetics2_1.htm
• For a reaction to occur Particles must collide
  with each other and these collisions must be
  effective
• Sufficient energy(Activation energy) and proper
  orientation.

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Factors That Affect Reaction Rateshttp//www.doc
brown.info/page03/3_31rates.htmhttp//www.blackgo
ld.ab.ca/ict/Divison3/reactionrates/reactionrates.
htm

• The Nature of the Reactants
• Chemical compounds vary considerably in their
  chemical reactivities.
• Concentration of Reactants
• As the concentration of reactants increases, so
  does the likelihood that reactant molecules will
  collide.
• Temperature
• At higher temperatures, reactant molecules have
  more kinetic energy, move faster, and collide
  more often and with greater energy.
• Catalysts
• Change the rate of a reaction by changing the
  mechanism.
• Surface Area
• http//www.physchem.co.za/OB12-che/rates.htm

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Maxwell Boltzman

• As you increase the temperature the rate of
  reaction increases. As a rough approximation, for
  many reactions happening at around room
  temperature, the rate of reaction doubles for
  every 10C rise in temperature.

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Catalyst and Ea

• A catalyst provides an alternative route for the
  reaction. That alternative route has a lower
  activation energy.

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http//www.gcsescience.com/rc4-sodium-thiosulfa
te-hydrochloric.htm

• investigate the effect that the temperature of
  water has on the reaction rate of antacid
  tablets.
• investigate the change in reaction rate when
  varying sizes of zinc particles are reacted with
  sulfuric acid.
• explore the relationship between concentration
  and reaction rate.
• observing how quickly reactants are used up or
  how quickly products are forming. This can be
  calculated in three ways of precipitation - When
  the products of the reaction is a precipitate,
  which will cloud the solution. Observing a marker
  through the solution and timing how long it takes
  for this marker to disappear will determine the
  rate of reaction.
• Hydrochloric Acid and Sodium Thiosulphate
• HCl(aq)       Na2S2O3(aq)             
  NaCl(aq)            SO2(g)              S(s) 
    H2O(l)

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The graph lines W, X, original, Y and Z on the
left diagram are typical of when a gaseous
product is being collected

• X ,a catalyst was added, forming the same amount
  of product, but faster.
• Z could represent taking half the amount of
  reactants or half a concentration. The reaction
  is slower and only half as much gas is formed
• W might represent taking double the quantity of
  reactants, forming twice as much gas e.g. same
  volume of reactant solution but doubling the
  concentration, so producing twice as much gas,
  initially at double the speed (gradient twice as
  steep).
• W gt X gt original gt Y gt Z
• See SG graphs

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Measuring the rate of a reaction

• The change in concentration can be measured by
  using any property that changes during conversion
  of reactants to products
• CaCO3(s) HCl(aq) gt CaCl2(aq) H2O(l)
  CO2(g)
• 1. mass and volume changes for gaseous reactions
• 2. change in pH for acids and bases
• 3. change in conductivity for metals
• 4. colorimetry for color changes

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Measuring the volume of gas produced with time
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Mass Loss Method
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Graphs

• The graph below shows the volume of carbon
  dioxide gas produced against time when excess
  calcium carbonate is added to x cm3 of 2.0 mol
  dm-3 hydrochloric acid.
• (i)      Write a balanced equation for the
  reaction.
• (ii)      State and explain the change in the
  rate of reaction with time.  Outline how you
  would determine the rate of the reaction at a
  particular time.
• (iii)     Sketch the above graph on an answer
  sheet. On the same graph, draw the curves you
  would expect if
• I.       the same volume (x cm3) of 1.0 mol dm-3 
  HCl is used.
• II.      double the volume (2x cm3) of 1.0 mol
  dm-3  HCl is used.
•           Label the curves and explain your
  answer in each case.

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Graph mass x time
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SG page 36
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Example 1 Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
C4H9Cl M

• In this reaction, the concentration of butyl
  chloride, C4H9Cl, was measured at various times,
  t.

?C4H9Cl ?t
Rate
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Reaction Rates Calculation
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl (aq)
Average Rate, M/s

• The average rate of the reaction over each
  interval is the change in concentration divided
  by the change in time

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Reaction Rate Determination
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

• Note that the average rate decreases as the
  reaction proceeds.
• This is because as the reaction goes forward,
  there are fewer collisions between the reacting
  molecules.

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Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

• A plot of concentration vs. time for this
  reaction yields a curve like this.
• The slope of a line tangent to the curve at any
  point is the instantaneous rate at that time.

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Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

• The reaction slows down with time because the
  concentration of the reactants decreases.

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Reaction Rates and Stoichiometry
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

• In this reaction, the ratio of C4H9Cl to C4H9OH
  is 11.
• Thus, the rate of disappearance of C4H9Cl is the
  same as the rate of appearance of C4H9OH.

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Reaction Rates Stoichiometry

• Suppose that the mole ratio is not 11?

Example H2(g) I2(g) ??? 2 HI(g)
2 moles of HI are produced for each mole of H2
used.
The rate at which H2 disappears is only half of
the rate at which HI is generated
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Concentration and Rate

• Each reaction has its own equation that gives its
  rate as a function of reactant concentrations.
• This is called its Rate Law
• The general form of the rate law is
• Rate kAxBy
• Where k is the rate constant, A and B are the
  concentrations of the reactants. X and y are
  exponents known as rate orders that must be
  determined experimentally
• To determine the rate law we measure the rate at
  different starting concentrations.

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Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)

• Compare Experiments 1 and 2when NH4 doubles,
  the initial rate doubles.

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Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)

• Likewise, compare Experiments 5 and 6 when
  NO2- doubles, the initial rate doubles.

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Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
This equation is called the rate law, and k is
the rate constant.
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The Rate Law

• A rate law shows the relationship between the
  reaction rate and the concentrations of
  reactants.
• For gas-phase reactants use PA instead of A.
• k is a constant that has a specific value for
  each reaction.
• The value of k is determined experimentally.
• Rate K NH4 NO2-
• Constant is relative here-
• the rate constant k is unique for each reaction
• and the value of k changes with temperature

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The Rate Law

• Exponents tell the order of the reaction with
  respect to each reactant.
• This reaction is
• First-order in NH4
• First-order in NO2-
• The overall reaction order can be found by adding
  the exponents on the reactants in the rate law.
• This reaction is second-order overall.
• Rate K NH4 1NO2- 1

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Determining the Rate constant and Order

• The following data was collected for the
  reaction of substances A and B to produce
  products C and D.
• Deduce the order of this reaction with
  respect to A and to B. Write an expression for
  the rate law in this reaction and calculate the
  value of the rate constant.

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Graphical Representation

• http//www.chem.purdue.edu/gchelp/howtosolveit/Kin
  etics/IntegratedRateLaws.html
• A ? B

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Zero Order

• For a zero order reaction
• A versus t (linear for a zero order
  reaction)
• rate k (k - slope of
  line)

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First Order

• For a 1st order reaction
• rate kA (k - slope of line)

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First-Order Processes

• Consider the process in which methyl isonitrile
  is converted to acetonitrile.

How do we know this is a first order reaction?
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First-Order Processes

• This data was collected for this reaction at
  198.9C.

Does ratekCH3NC for all time intervals?
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First-Order Processes

• When Ln P is plotted as a function of time, a
  straight line results.
• The process is first-order.
• k is the negative slope 5.1 ? 10-5 s-1.

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Half-Life of a Reaction

• Half-life is defined as the time required for
  one-half of a reactant to react.
• Because A at t1/2 is one-half of the original
  A,
• At 0.5 A0.

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Half-Life of a First Order Reaction

• For a first-order process, set At0.5 A0 in
  integrated rate equation

NOTE For a first-order process, the half-life
does not depend on the initial concentration,
A0.
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First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?

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First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?
LnAt LnAo -kt
LnAt Ln0.80 - (0.010 s-1 )(90 s) LnAt
- (0.010 s-1 )(90 s) Ln0.80 LnAt
-0.90 - 0.2231 LnAt -1.1231 At
0.325 mol dm-3
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First Order Rate Calculations

Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.



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First Order Rate Calculations

Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.


Solution k 0.693/t1/2 0.693/120s 0.005775
s-1 LnA Ln(2.00) -0.005775 s-1 (80 s)
-0.462 Ln A - 0.462 0.693 0.231 A
1.26 mol dm-3
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First Order Rate Calculations

Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?



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First Order Rate Calculations

Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?


Solution k 0.693/t1/2 0.693/28.8 yr 0.02406
yr-1 Ln1 Ln(100) - (0.02406 yr-1)t -
4.065 t - 4.062 . - 0.02406
yr-1 t 168.8 years
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• A versus t (linear for a zero order
  reaction)
• ln A versus t (linear for a 1st order
  reaction)
• 1 / A versus t (linear for a 2nd order
  reaction)

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Second Order

• For a 2nd order reaction
• rate kA2
  (k slope of line)

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Determining Reaction OrderDistinguishing Between
1st and 2nd Order
The decomposition of NO2 at 300C is described by
the equation
A experiment with this reaction yields this data
Time (s) NO2, M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
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Determining Reaction OrderDistinguishing Between
1st and 2nd Order
Graphing ln NO2 vs. t yields

• The graph is not a straight line, so this process
  cannot be first-order in A.

Time (s) NO2, M ln NO2
0.0 0.01000 -4.610
50.0 0.00787 -4.845
100.0 0.00649 -5.038
200.0 0.00481 -5.337
300.0 0.00380 -5.573
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Second-Order Reaction Kinetics
A graph of 1/NO2 vs. t gives this plot.
Time (s) NO2, M 1/NO2
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263

• This is a straight line. Therefore, the process
  is second-order in NO2.
• The slope of the line is the rate constant, k.

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Half-Life for 2nd Order Reactions

• For a second-order process, set
• At0.5 A0 in 2nd order equation.

In this case the half-life depends on the initial
concentration of the reactant A.
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Sample Problem 1 Second Order

• Acetaldehyde, CH3CHO, decomposes by second-order
  kinetics with a rate constant of 0.334
  mol-1dm3s-1 at 500oC. Calculate the amount of
  time it would take for 80 of the acetaldehyde
  to decompose in a sample that has an initial
  concentration of 0.00750 M.

The final concentration will be 20 of the
original 0.00750 M or 0.00150
1 . .00150
1 . .00750
0.334 mol-1dm3s-1 t
666.7 0.334 t 133.33 0.334 t 533.4
t 1600 seconds
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Sample Problem 2 Second Order

• Acetaldehyde, CH3CHO, decomposes by second-order
  kinetics with a rate constant of 0.334
  mol-1dm3s-1 at 500oC. If the initial
  concentration of acetaldehyde is 0.00200 M. Find
  the concentration after 20 minutes (1200 seconds)

Solution
1 . At
1 . 0.00200 mol dm-3
0.334 mol-1dm3s-1 (1200s)
1 . At
0.334 mol-1dm3 s-1 (1200s) 500 mol-1dm3
900.8 mol-1dm3
1 _____. 900.8 mol-1dm3
At
0.00111 mol dm-3
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Summary of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
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Temperature and Rate

• Generally speaking, the reaction rate increases
  as the temperature increases.
• This is because k is temperature dependent.
• As a rule of thumb a reaction rate increases
  about 10 fold for each 10oC rise in temperature

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The Collision Model

• In a chemical reaction, bonds are broken and new
  bonds are formed.
• Molecules can only react if they collide with
  each other.
• These collisions must occur with sufficient
  energy and at the appropriate orientation.

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The Collision Model

• Furthermore, molecules must collide with the
  correct orientation and with enough energy to
  cause bonds to break and new bonds to form

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Activation Energy

• In other words, there is a minimum amount of
  energy required for reaction the activation
  energy, Ea.
• Just as a ball cannot get over a hill if it does
  not roll up the hill with enough energy, a
  reaction cannot occur unless the molecules
  possess sufficient energy to get over the
  activation energy barrier.

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Reaction Coordinate Diagrams

• It is helpful to visualize energy changes
  throughout a process on a reaction coordinate
  diagram like this one for the rearrangement of
  methyl isonitrile.

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Reaction Coordinate Diagrams

• It shows the energy of the reactants and products
  (and, therefore, ?E).
• The high point on the diagram is the transition
  state.

• The species present at the transition state is
  called the activated complex.
• The energy gap between the reactants and the
  activated complex is the activation energy
  barrier.

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MaxwellBoltzmann Distributions

• Temperature is defined as a measure of the
  average kinetic energy of the molecules in a
  sample.

• At any temperature there is a wide distribution
  of kinetic energies.

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MaxwellBoltzmann Distributions

• As the temperature increases, the curve flattens
  and broadens.
• Thus at higher temperatures, a larger population
  of molecules has higher energy.

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MaxwellBoltzmann Distributions

• If the dotted line represents the activation
  energy, as the temperature increases, so does the
  fraction of molecules that can overcome the
  activation energy barrier.

• As a result, the reaction rate increases.

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MaxwellBoltzmann Distributions

• This fraction of molecules can be found through
  the expression
• where R is the gas constant and T is the
  temperature in Kelvin .

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Arrhenius Equation

• Svante Arrhenius developed a mathematical
  relationship between k and Ea
• where A is the frequency factor, a number that
  represents the likelihood that collisions would
  occur with the proper orientation for reaction.
  Ea is the activation energy. T is the Kelvin
  temperature and R is the universal thermodynamics
  (gas) constant.
• R 8.314 J mol-1 K-1 or 8.314 x 10-3 J mol-1
  K-1

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Arrhenius Equation

• Taking the natural logarithm of both sides, the
  equation becomes

y mx b
When k is determined experimentally at several
temperatures, Ea can be calculated from the slope
of a plot of ln k vs. 1/T.
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Arrhenius Equation for 2 Temperatures
When measurements are taken for two different
temperatures the Arrhenius equation can be
symplified as follows

Write the above equation twice, once for each of
the two Temperatures and then subtract the lower
temperature conditions from the higher
temperature. The equation then becomes
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Arrhenius Equation Sample Problem 1
The rate constant for the decomposition of
hydrogen iodide was determined at two different
temperatures 2HI ? H2 I2. At 650 K, k1
2.15 x 10-8 dm3 mol-1s-1 At 700 K, k2
2.39 x 10-7 dm3 mol-1s-1 Find the activation
energy for this reaction.

2.39 x 10-7 Ea Ln ---------------- -
------------------------ x 2.15 x 10-8
(8.314 J mol-1 K-1) Ea 180,000
J mol-1 180 kJ mol-1
1 1 ------ -- ------ 700K 650K
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Overview of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
Rate and Temp (T)
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Reaction Mechanisms

• The sequence of events that describes the actual
  process by which reactants become products is
  called the reaction mechanism.

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Reaction Mechanisms

• Reactions may occur all at once or through
  several discrete steps.
• Each of these processes is known as an elementary
  reaction or elementary process.

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Reaction Mechanisms

• The molecularity of a process tells how many
  molecules are involved in the process.
• The rate law for an elementary step is written
  directly from that step.

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Multistep Mechanisms

• In a multistep process, one of the steps will be
  slower than all others.
• The overall reaction cannot occur faster than
  this slowest, rate-determining step.

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Slow Initial Step
NO2 (g) CO (g) ??? NO (g) CO2 (g)

• The rate law for this reaction is found
  experimentally to be
• Rate k NO22
• CO is necessary for this reaction to occur, but
  the rate of the reaction does not depend on its
  concentration.
• This suggests the reaction occurs in two steps.

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Slow Initial Step

• A proposed mechanism for this reaction is
• Step 1 NO2 NO2 ??? NO3 NO (slow)
• Step 2 NO3 CO ??? NO2 CO2 (fast)
• The NO3 intermediate is consumed in the second
  step.
• As CO is not involved in the slow,
  rate-determining step, it does not appear in the
  rate law.

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Fast Initial Step

• The rate law for this reaction is found
  (experimentally) to be
• Because termolecular ( trimolecular) processes
  are rare, this rate law suggests a two-step
  mechanism.

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Fast Initial Step

• A proposed mechanism is

Step 1 is an equilibrium- it includes the
forward and reverse reactions.
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Fast Initial Step

• The rate of the overall reaction depends upon the
  rate of the slow step.
• The rate law for that step would be
• But how can we find NOBr2?

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Fast Initial Step

• NOBr2 can react two ways
• With NO to form NOBr
• By decomposition to reform NO and Br2
• The reactants and products of the first step are
  in equilibrium with each other.
• Therefore,
• Ratef Rater

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Fast Initial Step

• Because Ratef Rater ,
• k1 NO Br2 k-1 NOBr2
• Solving for NOBr2 gives us

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Fast Initial Step

• Substituting this expression for NOBr2 in the
  rate law for the rate-determining step gives

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Catalysts

• Catalysts increase the rate of a reaction by
  decreasing the activation energy of the reaction.
• Catalysts change the mechanism by which the
  process occurs.
• Some catalysts also make atoms line up in the
  correct orientation so as to enhance the reaction
  rate

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Catalysts

• Catalysts may be either homogeneous or
  heterogeneous
• A homogeneous catalyst is in the same phase as
  the substances reacting.
• A heterogeneous catalyst is in a different phase

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Catalysts

• One way a catalyst can speed up a reaction is by
  holding the reactants together and helping bonds
  to break.
• Heterogeneous catalysts often act in this way

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Catalysts

• Some catalysts help to lower the energy for
  formation for the activated complex or provide a
  new activated complex with a lower activation
  energy

AlCl3 Cl2 ? Cl AlCl4- Cl C6H6 ? C6H5Cl
H H AlCl4- ? HCl AlCl3 Overall
reaction C6H6 Cl2 ? C6H5Cl HCl
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Catalysts Stratospheric Ozone
In the stratosphere, oxygen molecules absorb
ultraviolet light and break into individual
oxygen atoms known as free radicals The
oxygen radicals can then combine with ordinary
oxygen molecules to make ozone. Ozone can also
be split up again into ordinary oxygen and an
oxygen radical by absorbing ultraviolet light.

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Catalysts Stratospheric Ozone
The presence of chlorofluorcarbons in the
stratosphere can catalyze the destruction of
ozone. UV light causes a Chlorine free radical
to be released

The chlorine free radical attacks ozone and
converts it Back to oxygen. It is then
regenerated to repeat the Process. The result is
that each chlorine free radical can Repeat this
process many many times. The result is
that Ozone is destroyed faster than it is formed,
causing its level to drop
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Enzymes

• Enzymes are catalysts in biological systems.
• The substrate fits into the active site of the
  enzyme much like a key fits into a lock.

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96.