Chemical Kinetics

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Chemical Kinetics

• The area of chemistry that concerns reaction
  rates.

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Reaction Rate

• Change in concentration (conc) of a reactant or
  product per unit time.

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Lets look at 2 NO2?2NO O2
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• Calculate the average rate at which the
  concentration of NO2 changes over the first 50
  seconds of the reaction from the data given in
  the previous chart.
• Change in concentration / Change in time
• (.0079-.0100)/(50-0)
• -4.2x10-5 mole/Ls

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• Notice that the rate is constant but decreases
  over time. The slope of the line tangent to any
  point on the curve gives you the instantaneous
  rate of reaction (the value of the rate at a
  particular time)
• We can also define rates in terms of products.
  Stoichiometry must be considered.
• Notice that the NO2 curve is the same as the NO
  curve just flipped upside down, but the O2 curve
  is half of that. This is because O2 is produced
  at half of the rate of the NO. You can see this
  is the balanced equation

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Rate Laws

• Rate kNO2n
• k rate constant
• n rate order

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Two important points about this equation

• The concentration of the products do not appear
  in the rate law because the reaction rate is
  being studied under conditions where the reverse
  reaction does not contribute to the overall rate
  (this would make things much more difficult)
• The value of the exponent n cannot be determined
  from a balanced chemical equation it is
  determined from experiment.

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Types of Rate Laws

• Differential Rate Law expresses how rate
  depends on concentration. This is what we
  usually refer to as the rate law.
• Integrated Rate Law expresses how
  concentration depends on time.
• If we know one of these we automatically know the
  other. Each integrated rate law has a
  differential rate law that goes with it and vice
  versa

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Determining Rate Laws

• The first step is to determine the form of the
  rate law (especially its order).
• Must be determined from experimental data.
• For this reaction 2 N2O5 (aq)
  4NO2 (aq) O2(g)The reverse reaction wont play
  a role

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N2O5 (mol/L) Time
(s) 1.00 0 0.88 200 0.78 400 0.69
600 0.61 800 0.54 1000 0.48 1200 0.43
1400 0.38 1600 0.34 1800 0.30 2000

• Now graph the data

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• To find rate we have to find the slope at two
  points
• We will use the tangent method.

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At .90 M the rate is (.98 - .76) 0.22 -
5.5x 10 -4 (0-400) -400
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At .40 M the rate is (.52 - .31) 0.22 -
2.7 x 10 -4 (1000-1800) -800
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• You could then take the rate of reactions that
  you just found along with the concentrations of
  the reactants and calculate the rate law.
• This is the same thing we would do if we had
  experimental data as well.

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Determining the Form of the Rate Law

• Method of Initial Rates
• Initial Rate the instantaneous rate just
  after the reaction begins.
• The initial rate is determined in several
  experiments using different initial
  concentrations.

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An example

• For the reaction BrO3- 5 Br- 6H
  ? 3Br2 3 H2O
• The general form of the Rate Law is Rate
  kBrO3-nBr-mHp
• We use experimental data to determine the values
  of n,m,and p

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Initial concentrations (M)
Rate (M/s)
BrO3-
Br-
H
1 0.10 0.10 0.10 8.0 x 10-4 2
0.20 0.10 0.10 1.6 x 10-3 3
0.20 0.20 0.10 3.2 x 10-3 4
0.10 0.10 0.20 3.2 x 10-3

• Now we have to see how the rate changes with
  concentration

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Determining the Rate Law

• To determine the value for one variable we must
  hold the other values of initial concentration
  constant.
• For n we must hold the initial values of Br- and
  H constant. Then we can compare those rates

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Determining the Rate Law
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Determining Rate Laws
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Determining Rate Laws
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• Therefore the rate law is
• Rate kBrO3-1Br-1H2
• You can plug in any of the experimental data sets
  to determine the value for k
• K8

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Overall Reaction Order

• Sum of the order of each component in the rate
  law.
• rate kH2SeO3H2I?3
• The overall reaction order is 1 2 3 6.

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First-Order Rate Law
For aA ? Products in a 1st-order reaction,

• Integrated first-order rate law is
• lnA ?kt lnAo

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Integrated Rate Law

• The integrated rate law expresses the
  concentration of the reactant as a function of
  time.

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First Order Integrated Rate Law

• The equation shows how the concentration of A
  depends on time if the initial concentration of A
  and the rate constant k are known, the
  concentration of A at any time can be calculated
• The equation can also be looked at as ymx b,
  where ylnA, xt, m-k and blnA0. If the
  reaction is first order, plotting the natural
  logarithm of concentration versus the time will
  always give a straight line.
• The slope will be equal to -k
• The integrated rate law for a first order raction
  may also be expressed in terms of the ratio of
  A to A0.

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Practice Half-life Problem

• A first order reaction is 35 complete at the end
  of 55 minutes.
• What is the value of k?

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Solution

• t 55 min
• Ao/A .65 remember you must use the percent
  of what is left, not of what has already reacted
• k 7.8 x 10 -3 min-1

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Half-Life of a First-Order Reaction

• Half-life is the time required for a reactant to
  reach half of its original concentration
• For a first-order reaction, the half-life does
  not depend on concentration.

• t1/2 half-life of the reaction
• k rate constant

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Second-Order Rate Law

• For aA ? products in a second-order reaction,
• Integrated rate law is

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Second Order Rate Law

• A plot of 1/A versus t will produce a straight
  line with a slope that is equal to k
• The equation can be used to calculate A at any
  time if you have k and Ao.

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Half-Life of a Second-Order Reaction

• t1/2 half-life of the reaction
• k rate constant
• Ao initial concentration of A
• The half-life is dependent upon the initial
  concentration.

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Zero Order Rate Law

• A -kt Ao
• The half-life is
• A plot of A versus t will give you a straight
  line with a slope of -k

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Practice Problem

• Consider the reaction aA ? Products. A0 5.0
  M and k 1.0 x 10-2. Calculate A after 30.0
  seconds have passed, assuming the reaction is
• a) zero order.
• b) first order.
• c) second order.

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Answer

• A. 4.7M
• B. 3.7M
• C. 2.0M
• Note how the change in concentration over time
  increases with order.

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Summary of the Kinetics for Reactions of the Type
aA ? Products that are Zero, First, and Second
Order in A
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A Summary

• 1. Simplification Conditions are set such that
  only forward reaction is important.
• 2. Two types
• differential rate law
• integrated rate law
• 3. Which type? Depends on the type of data
  collected - differential and integrated forms can
  be interconverted.

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A Summary (continued)

• 4.Most common method of initial rates.
• 5. Concentration v. time used to determine
  integrated rate law, often graphically.
• 6. For several reactants choose conditions
  under which only one reactant varies
  significantly (pseudo first-order conditions).

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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.

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Reaction Mechanism (continued)

• The reaction
• has many steps in the reaction mechanism.

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Often Used Terms

• Intermediate formed in one step and used up in
  a subsequent step and so is never seen as a
  product.
• Molecularity the number of species that must
  collide to produce the reaction indicated by that
  step.
• Elementary Step A reaction whose rate law can
  be written from its molecularity.
• uni, bi and termolecular

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Rate-Determining Step

• In a multistep reaction, it is the slowest step.
  It therefore determines the rate of reaction.
• The overall reaction rate can be no faster than
  the slowest step

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• When given the elementary steps or elementary
  reactions you can write the rate law form the
  molecularity
• 2 rules must be followed
• The sum of the elementary steps must give the
  overall balanced equation for the reaction
• The mechanism must agree with the experimentally
  determined rate law

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Example

• For the elementary reaction
• NO2 C O ? NO C O2
• The following reaction steps occur
• NO2 NO2 ? NO3 NO slow step
• NO3 C O ?NO2 C O2 _fast
• The rate for this reaction would be
• kNO2NO2
• or kNO22

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• Notice that this solution fits both of our
  criteria
• Also note that this does not mean that this is
  the correct mechanism, it simple means that it is
  an acceptable mechanism
• To decide on the most probable mechanism the
  chemist doing the study would have to perform
  additional experiments

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Arrhenius Equation

• Collisions must have enough energy to produce the
  reaction (must equal or exceed the activation
  energy).
• Orientation of reactants must allow formation of
  new bonds.

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Collision Model

• Key Idea Molecules must collide to react.
• However, only a small fraction of collisions
  produces a reaction. Why?
• Arrhenius An activation energy must be overcome.

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Arrhenius Equation (continued)

• k rate constant (L/mol s)
• A frequency factor
• Ea activation energy
• T temperature in Kelvin
• R gas constant (8.3145 J/Kmol)

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Arrhenius Equation

• Plotting ln(k) vs. 1/T will give you a straight
  line
• The slope of this line is equal to Ea/R
• This equation allows you to calculate activation
  energy from a plot of ln(k) vs. 1/T

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Given the following data determine the activation
energy for the reaction 2N2O5?4NO2 O2

• K (1/s) T(C)
• 2.0 x10-5 20
• 7.3 x 10-5 30
• 2.7 x 10-4 40
• 9.1 x 10-4 50
• 2.9 x 10-3 60

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Figure 12.14 Plot of ln(k) versus 1/T for the
reaction 2N2O5(g) 4NO2(g) O2(g). The value
of the activation energy for this reaction can be
obtained from the slope of the line, which equals
-Ea/R.
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Solution

• Ea-R(slope)
• -(8.3145J/Kmol)(-1.2 x 104)
• 1.0 x 105 J/mol

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Activation Energy

• When given two values of k at different
  temperatures you may use the Arrhenius equation
  derived below

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Example

• At 550 C the rate constant for a reaction is 1.1
  L/mol s and at 625 C the rate constant is 6.4
  L/mol s. Using these values calculate the
  activation energy for this reaction.

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Answer
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Catalysis

• Catalyst A substance that speeds up a reaction
  without being consumed
• Enzyme A large molecule (usually a protein)
  that catalyzes biological reactions.
• Homogeneous catalyst Present in the same phase
  as the reacting molecules.
• Heterogeneous catalyst Present in a different
  phase than the reacting molecules.

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Heterogeneous Catalysis
Steps

• 1. Adsorption and activation of the reactants.
• 2. Migration of the adsorbed reactants on the
  surface.
• 3. Reaction of the adsorbed substances.
• 4. Escape, or desorption, of the products.