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					    Chemical Kinetics

 The area of chemistry concerned
with the speeds, or rates, at which a
      chemical reaction occurs.


                                        1
             Reaction Rate
The reaction rate is the change in the
 concentration of a reactant or a product
 with time, (M/s or M . s-1), where M is
 molarity and s represents seconds.

Another way to represent rate is mol . L-1 s-1



                                             2
 Factors that Influence Reaction
              Rate
Under a given set of conditions, each reaction
  has its own characteristic rate, which is
  ultimately determined by the chemical nature
  of the reactants. (You will remember this
  from Chem I - potassium and water have a
  different rate of reaction than iron and
  oxygen.)
For a given reaction (using the same reactants),
  we can control four factors that affect its
  rate: the concentration of reactants, their
  physical state, the temperature at which the
  reaction occurs, and the use of a catalyst.
                                                   3
 The Collision Theory of Chemical
              Kinetics
The kinetic molecular theory of gases postulates that gas
  molecules frequently collide with one another.
  Therefore, it seems logical to assume – and it is
  generally true - that chemical reactions occur as a
  result of collisions between reacting molecules.
In terms of the collision theory of chemical kinetics,
  then, we expect the rate of a reaction to be directly
  proportional to the frequency of the collisions (number
  of molecular collisions per second).


                    Number of collisions
     Rate =
                                 s                     4
The Collision Theory of Chemical
             Kinetics
 This simple relationship explains the
 dependence of reaction rate on
 concentration.
 •Increasing the concentration
 increases the likelihood that molecules
 will collide.

              Number of collisions
  Rate =
                       s                   5
             Concentration
 Since molecules must collide in order to
   react, the more frequently they collide, the
   more often a reaction occurs. Thus,
   reaction rate is proportional to the
   concentration of reactant
                       Means “proportional to”


Rate    collision frequency          concentration
 Therefore, if we increase the
 concentration… we increase the collision
                 frequency, which…
                                    increases
                                    the rate
                                                  6
            Physical State
Molecules must mix in order to collide. When
 reactants are in the same phase, as in aqueous
 solution, occasional stirring keeps them in
 contact. When they are in different phases,
 more vigorous mixing is needed. The more
 finely divided a solid or liquid reactant, the
 greater the surface are per unit volume, the
 more contact it makes with the other
 reactant, and the faster the reaction.


                                              7
              Temperature
Molecules must collide in order to react.
 Since the speed of a molecule depends
 on its temperature, more collisions will
 occur if the temperature is increased.


    Speed of a molecule   Number of collisions




                                                 8
             Temperature
Molecules must also collide with enough energy
 to react. Increasing the temperature
 increases the kinetic energy of the molecules,
 which in turn increases the energy of the
 collisions.

Therefore, at a higher temperature, more
 collisions occur with enough energy to react.
 Thus, raising the temperature increases the
 reaction rate by increasing the number and
 especially the energy of the collisions.
                                                 9
           Temperature
Two familiar kitchen
 appliances employ this
 effect: a
 refrigerator slows
 down chemical
 processes that spoil
 food, whereas an oven
 speeds up other
 chemical processes to
 cook it.

                          10
     Expressing Reaction Rate
Before we can deal quantitatively with the effects of
  concentration and temperature on reaction rate, we
  must be able to express the rate mathematically. A
  rate is a change in some variable per unit of time.

For example, the rate of motion of a car is the change
  of position of the car divided by time. A car that
  travels 57 miles in 60. minutes is traveling at…

     57 miles/60. minutes = .95 miles/min

In the case of chemical reactions, the
positions of the substances do not change
over time, but their concentrations do.
                                                         11
We know that any reaction can be
 represented by the general equation

         reactants  products
This equation tells us that during the
 course of a reaction, reactants are
 consumed while products are formed.
 As a result, we can follow the progress
 of a reaction by monitoring either the
 decrease in concentration of the
 reactants or the increase in
 concentration of the products.
                                           12
    The following figure shows the progress
     of a simple reaction in which A
     molecules are converted to B
     molecules:
                     A  B

                                       B




A

                                              13
The decrease in number of A molecules and the
 increase in the number of B molecules with
 time are shown below.




                                                14
In general, it is more convenient to express the
  reaction rate in terms of the change in
  concentration with time. Thus, for the
  reaction A  B we can express the rate as:

                          D[A]     [A]final – [A]initial
               Rate = -
                          Dt
Because the concentration of A decreases
during the time interval, D[A] is a negative
quantity. The rate of a reaction is a positive
quantity, so a minus sign is needed in the rate
expression to make the rate positive.
                                                           15
                        D[B]
       or     Rate =
                         Dt


The rate of product formation does not require
a minus sign because D[B] ([B]final – [B]initial) is a
positive quantity (the concentration increases
with time).



                                                         16
             D[A]                 D[B]
  Rate = -          or   Rate =
             Dt                    Dt


Where D[A] and D[B] are the changes in
 concentration (molarity) over a time
 period Dt.

These rates are average rates because
 they are averaged over a certain time
 period (Dt).
                                         17
Reaction Rates and Stoichiometry
We have seen that for stoichiometrically simple
 reactions of the type A  B, the rate can
 either be expressed in terms of the decrease
 in reactant concentration with time, -D[A]/Dt,
 or the increase in product concentration with
 time, D[B]/Dt.

For more complex reactions, we must be careful
  in writing the rate expressions.
                                 Two moles of A
Consider the reaction
                                 disappear for
                    2A  B
                                 each mole of B
                                 that forms 18
Another way to think of this is to say
 that the rate of disappearance of A is
 twice as fast as the rate of appearance
 of B. We write the rate as either
            1 D[A]                 D[B]
 Rate = -            or   Rate =
            2 Dt                   Dt

   For the reaction
                     2A  B
                                           19
In general, for the reaction

               aA + bB  cC + dD

The rate is given by

           1 D[A]         1 D[B]       1 D[C]       1 D[D]
Rate = -            = -            =            =
           a Dt           b Dt         c Dt         d Dt




                                                             20
Write the expression for the following
 reactions in terms of the disappearance
 of the reactants and the appearance of
 the products:
            3O2(g)  2O3(g)

           1 D[O2]         1 D[O3]
Rate = -             =
           3 Dt            2 Dt


                                       21
If O2 is disappearing at .25 M/s, what is
  the rate of formation of O3?


        1 D[O2]             1 D[O3]
    -              =
        3 Dt                2 Dt

        1 -.25M             1   x
    -              =
        3    s              2   s

            .50Ms = 3(x)s       x = .17M
                                            22
By definition, we know that to determine
 the rate of a reaction we have to
 monitor the concentration of the
 reactant (or product) as a function of
 time.
• For reactions in solution, the concentration of a
  species can often be measured by spectroscopic
  means.
• If ions are involved, the change in concentration can
  also be detected by an electrical conductance
  measurement.
• Reactions involving gases are most conveniently
  followed by pressure measurements.
                                                          23
    Reaction of Molecular Bromine
           and Formic Acid

In aqueous solutions, molecular bromine reacts
  with formic acid (HCOOH) as follows:
Br2(aq) + HCOOH(aq)  2Br-(aq) + 2H+(aq) + CO2(g)
Reddish-    colorless     colorless   colorless   colorless
brown

 The rate of Br2 disappearance can be determined by
 monitoring the color over time.
 As the reaction proceeds, the color of the solution …
      goes from brown to colorless
                                                          24
 Reaction of Molecular Bromine
        and Formic Acid




 Br2(aq) + HCOOH(aq)    2Br-(aq) + 2H+(aq) + CO2(g)


As the reaction proceeded, the
concentration of Br2 steadily decreased
and the color of the solution faded.
                                                      25
Measuring the change (decrease) in bromine
 concentration at some initial time ([Br2]0) and
 then at some other time, ([Br2]t) allows us to
 determine the average rate of the reaction
 during that interval:

                  D[Br2]
Average rate =
                     Dt
                       [Br2]t – [Br2]0
Average rate =
                          tfinal – tinitial
                                               26
Use the data in the following table to
 calculate the average rate over the
 first 50 second time interval.

  Time          [Br2]            Rate
   (s)           (M)            (M/s)
   0.0         0.0120         4.20 x 10-5
  50.0         0.0101         3.52 x 10-5
  100.0       0.00846         2.96 x 10-5
  150.0        0.00710        2.49 x 10-5
  200.0       0.00596         2.09 x 10-5
  250.0       0.00500         1.75 x 10-5
  300.0       0.00420         1.48 x 10-5
  350.0       0.00353         1.23 x 10-5
  400.0       0.00296         1.04 x 10-5
                                            27
                  (0.0101 – 0.0120)M
Average rate =                         = 3.80 x 10-5 M/s
                     (50.0 – 0.0)s


          Time           [Br2]            Rate
           (s)            (M)            (M/s)
           0.0          0.0120         4.20 x 10-5
          50.0          0.0101         3.52 x 10-5
          100.0         0.00846        2.96 x 10-5
          150.0         0.00710        2.49 x 10-5
          200.0         0.00596        2.09 x 10-5
          250.0         0.00500        1.75 x 10-5
          300.0         0.00420        1.48 x 10-5
          350.0         0.00353        1.23 x 10-5
          400.0         0.00296        1.04 x 10-5
                                                     28
Now use the data in the same table to
 calculate the average rate over the
 first 100 second time interval.

   Time        [Br2]            Rate
    (s)         (M)            (M/s)
   0.0         0.0120        4.20 x 10-5
   50.0        0.0101        3.52 x 10-5
  100.0       0.00846        2.96 x 10-5
  150.0       0.00710        2.49 x 10-5
  200.0       0.00596        2.09 x 10-5
  250.0       0.00500        1.75 x 10-5
  300.0       0.00420        1.48 x 10-5
  350.0       0.00353        1.23 x 10-5
  400.0       0.00296        1.04 x 10-5
                                           29
                  (0.00846 – 0.0120)M
Average rate =                       = 3.54 x 10-5 M/s
                      (100.0 – 0.0)s


          Time           [Br2]           Rate
           (s)            (M)           (M/s)
           0.0          0.0120        4.20 x 10-5
          50.0          0.0101        3.52 x 10-5
          100.0         0.00846       2.96 x 10-5
          150.0         0.00710       2.49 x 10-5
          200.0         0.00596       2.09 x 10-5
          250.0         0.00500       1.75 x 10-5
          300.0         0.00420       1.48 x 10-5
          350.0         0.00353       1.23 x 10-5
          400.0         0.00296       1.04 x 10-5
                                                    30
These calculations demonstrate that the
 average rate of the reaction depends on
 the time interval we choose.
By calculating the average reaction rate
 over shorter and shorter intervals, we
 can obtain the rate for a specific
 instant in time, which gives us the
 instantaneous rate of the reaction at
 that time.


                                       31
The figure below shows the plot of [Br2] versus time, based on
the data table given previously. Graphically, the instantaneous
rate at 100 seconds after the start of the reaction is the slope
of the line tangent to the curve at that instant.

Unless otherwise stated, we will refer to the instantaneous rate
                     as simply “the rate”.




The instantaneous rate at any other time can be determined in a
similar manner.
                                                                   32
                       k, the Rate Constant
At a specific
temperature, a rate
constant (k) is a        rate         [Br2]
constant of
                           Means “proportional to”
proportionality
between the reaction
rate and the              rate = k[Br2]
concentrations of
reactants.
 k is specific for a given reaction at a
 given temperature; it does not change as
 the reaction proceeds.
                                                     33
Rearrange the equation
               rate = k[Br2]
To solve for k
                        Since reaction
         Rate           rate has the units
    k=                  M/s, and [Br2] is in
         [Br2]          M, the unit of k
                        for this first
                        order reaction is
                        1/s or s-1.
                                           34
Calculate the rate constant for the following reaction
Br2(aq) + HCOOH(aq)  2Br-(aq) + 2H+(aq) + CO2(g)
        Time     [Br2]       Rate       k = rate/[Br2]
         (s)      (M)       (M/s)            (s-1)
         0.0    0.0120    4.20 x 10-5
        50.0    0.0101    3.52 x 10-5
        100.0   0.00846   2.96 x 10-5
        150.0   0.00710   2.49 x 10-5
        200.0   0.00596   2.09 x 10-5
        250.0   0.00500   1.75 x 10-5
        300.0   0.00420   1.48 x 10-5
        350.0   0.00353   1.23 x 10-5
        400.0   0.00296   1.04 x 10-5



                                                         35
Because k is a constant (for this reaction at this specific
temperature), it doesn’t matter which row we consider, so
let’s consider the data at time 0.0 seconds…
   Time     [Br2]       Rate                 k = rate/[Br2]
    (s)      (M)       (M/s)                      (s-1)
    0.0    0.0120    4.20 x 10-5              3.50 x 10-3
   50.0    0.0101    3.52 x 10-5
   100.0   0.00846   2.96 x 10-5
   150.0   0.00710   2.49 x 10-5
  200.0    0.00596   2.09 x 10-5
  250.0    0.00500   1.75 x 10-5
  300.0    0.00420   1.48 x 10-5
  350.0    0.00353   1.23 x 10-5
  400.0    0.00296   1.04 x 10-5

                               4.20 x 10-5
    rate = k[Br2]                              = 3.50 x 10-3
                       k=
                                   0.0120                      36
To prove that k is a constant, calculate k at time 200.0
seconds
        Time     [Br2]       Rate       k = rate/[Br2]
         (s)      (M)       (M/s)            (s-1)
         0.0    0.0120    4.20 x 10-5
        50.0    0.0101    3.52 x 10-5     3.50 x 10-3
        100.0   0.00846   2.96 x 10-5
        150.0   0.00710   2.49 x 10-5
        200.0   0.00596   2.09 x 10-5
        250.0   0.00500   1.75 x 10-5     3.51 x 10-3
        300.0   0.00420   1.48 x 10-5
        350.0   0.00353   1.23 x 10-5
        400.0   0.00296   1.04 x 10-5

  The slight variations in the values of k are due to
  experimental deviations in rate measurements.
                                                           37
Filling in the rest of the table…


    Time     [Br2]       Rate       k = rate/[Br2]
     (s)      (M)       (M/s)            (s-1)
     0.0    0.0120    4.20 x 10-5     3.50 x 10-3
    50.0    0.0101    3.52 x 10-5     3.49 x 10-3
    100.0   0.00846   2.96 x 10-5     3.50 x 10-3
    150.0   0.00710   2.49 x 10-5     3.51 x 10-3
    200.0   0.00596   2.09 x 10-5     3.51 x 10-3
    250.0   0.00500   1.75 x 10-5     3.50 x 10-3
    300.0   0.00420   1.48 x 10-5     3.52 x 10-3
    350.0   0.00353   1.23 x 10-5     3.48 x 10-3
    400.0   0.00296   1.04 x 10-5     3.51 x 10-3



                                                     38
Units of the Rate Constant k for
Several Overall Reaction Orders

 Overall    Units of k (when t is seconds)
 Reaction
  Order
    0          mol/L . s (or mol L-1 s-1)
    1                 1/s (or s-1)
    2          L/mol . s (or L mol-1 s-1)
    3         L2/mol2 . s (or L2 mol-2 s-1)

                                              39
It is important to understand that k is NOT
  affected by the concentration of Br2.

• The rate is greater at a higher
  concentration and smaller at a lower
  concentration of Br2, but the ratio of
  rate/[Br2] remains the same provided the
  temperature doesn’t change.


                                             40
                   The Rate Law
The rate law expresses the relationship of the rate of a reaction to
  the rate constant (k) and the concentrations of the reactants
  raised to a power. For the general reaction
                       aA + bB  cC + dD

The rate law takes the form

                          Rate = k[A]x[B]y
Where x and y are numbers that must be determined
  experimentally.

Note – in general, x and y are NOT equal to the stoichiometric
  coefficients a and b from the overall balanced chemical
  equation. When we know the values of x, y and k, we can use the
  rate equation shown above to calculate the rate of the reaction,
  given the concentrations of A and B.

                                                                  41
                 Rate = k[A]x[B]y

The reaction orders define how the rate is
 affected by the concentration of each
 reactant.

This reaction is xth order in A, yth order in B.




                                                   42
The following rate law was determined for
 the formation of nitrogen (VI) oxide and
 molecular oxygen from nitrogen (IV)
 oxide and ozone
           Rate = k[NO2][O3]
How would the rate of this reaction be
 affected if the concentration of NO2
 increased from 1.0 M to 2.0 M?



                                         43
                 Rate = k[NO2][O3]

This reaction is first order with respect to
 both NO2 and O3. This means that doubling
 the concentration of either reactant would
 double the rate of the reaction.
                     (2)1 = 2
                                        How many times
     How many times                     greater the rate of the
     greater the                        reaction will be
     concentration is   What the order is for
                        that reactant



                                                              44
The reaction between nitrogen monoxide
 and molecular oxygen is described by a
 different rate law.

           Rate = k [NO]2[O2]

How would the rate of this reaction be
 affected if the concentration of NO
 increased from 1.0 M to 2.0 M?



                                          45
              Rate = k[NO]2[O2]
This reaction is second order with
 respect to NO and first order with
 respect to O2.
Doubling the concentration NO would
 increase the rate by a factor of four
               (2)2 = 4 How many times
                                      greater the rate of the
    How many times                    reaction will be
    greater the
    concentration is   What the order is for
                       that reactant
                                                            46
            Rate = k [NO]2[O2]

How would the rate of this reaction be
 affected if the concentration of NO
 increased from 1.0 M to 5.0 M?
In this case, when the concentration of
NO is multiplied by 5, the rate increases
by a factor of 25!
                (5)2 = 25
                                            47
The exponents x and y specify the relationships
 between the concentrations of reactants A
 and B and the reaction rate. Added together,
 they give us the overall reaction order,
 defined as the sum of the powers to which all
 reactant concentrations appearing in the rate
 law are raised. For the equation

                 Rate = k[A]x[B]y

The overall reaction order is x + y.

                                              48
For the following reaction

          (CH3)3CBr(l) + H2O(l)  (CH3)3COH(l) + HBr(aq)

The rate law has been found to be

                             rate = k[(CH3)3CBr]

This reaction is first order in 2-bromo-2-methylpropane.

Note that the concentration of H2O does not even appear in the rate
  law. Thus, the reaction is zero order with respect to H2O. This
  means that the rate does not depend on the concentration of H2O;
  we could also write the rate law for this reaction as

                      rate = k[(CH3)3CBr][H2O]0

                                                   1st order overall (1+0=1)
What is the overall order of this reaction?

                                                                          49
Reaction orders are usually positive
 integers or zero, but they can also be
 fractional or negative. In the reaction
  CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
A fractional order appears in the rate law:
           rate = k[CHCl3][Cl2]1/2
This order means that the reaction depends on
 the square root of the Cl2 concentration. If
 the initial Cl2 concentration is increased by a
 factor of 4, for example, the rate increases by
 V4 (= 2), therefore the rate would double.
                                              50
A negative exponent means that the reaction rate
  decreases when the concentration of that
  component increases. Negative orders are often
  seen for reactions whose rate laws include
  products. For example, in the atmospheric
  reaction
                2O3(g)       3O2(g)

The rate law has been shown to be

       Rate = k[O3]2[O2]-1 ; or            [O3]2
                                  rate = k [O ]
                                             2

   If the [O2] doubles, the reaction proceeds
   half as fast.                                   51
To see how to determine the rate law of a
 reaction, let us consider the reaction
 between fluorine and chlorine dioxide:

     F2(g) + 2ClO2(g)  2FClO2(g)




                                        52
One way to study the effect of reactant
concentration on reaction rate is to determine
how the initial rate depends on the starting
concentrations. It is preferable to measure
the initial rates because as the reaction
proceeds, the concentrations of the reactants
decrease and it may become difficult to
measure the changes accurately. Also, as the
reaction continues, the product concentrations
increase,
            products  reactants
so the reverse reaction becomes increasingly
likely. Both of these complications are virtually
absent during the earliest stages of the
reaction.
                                               53
The following table shows three rate
 measurements for the formation of
 FClO2.

     [F2]0   [ClO2]0     Initial Rate
      (M)      (M)          (M/s)
     0.10     0.010       1.2 x 10-3

     0.10    0.040        4.8 x 10-3

     0.20     0.010       2.4 x 10-3


                                        54
Looking at entries 1 and 3, we see that as we
double [F2]0 while holding [ClO2]0 constant,
the reaction rate doubles. Thus the rate is
directly proportional to [F2], and the
reaction is first order with respect to F2.

    [F2]0     [ClO2]0       Initial Rate
     (M)        (M)            (M/s)
    0.10       0.010         1.2 x 10-3

    0.10      0.040          4.8 x 10-3

    0.20      0.010          2.4 x 10-3
                                                55
Similarly, the data in entries 1 and 2 show that as
we quadruple [ClO2] while holding [F2] constant,
the rate increases by four times, so the rate is
also directly proportional to [ClO2], making the
reaction 1st order with respect to [ClO2]

      [F2]       [ClO2]        Initial Rate
      (M)         (M)             (M/s)
      0.10       0.010          1.2 x 10-3

      0.10       0.040          4.8 x 10-3

      0.20       0.010          2.4 x 10-3
                                                 56
We can summarize our observations by
 writing the rate law as

             Rate = k[F2][ClO2]

Because both [F2] and [ClO2] are raised to
 the first power, the reaction is second
 order overall.



                                             57
What is the rate constant for this
 reaction at this temperature?




                                     58
From the reactant concentrations and the
 initial rate, we can calculate the rate
 constant. Using the data from the
 previous table we can write
             Rate = k[F2][ClO2]

           rate
   k=
        [F2][ClO2]

        1.2 x 10-3 M/s
                           = 1.2 L/mol.s
        (0.10M)(0.010M)
                                           59
The reaction of nitric oxide with hydrogen at
 1280 oC is
     2NO(g) + 2H2(g)  N2(g) + 2H2O(g)

From the following data that was collected experimentally
at this temperature, determine the rate law and calculate
the rate constant.
   Experiment     [NO]          [H2]       Initial Rate (M/s)

       1        5.0 x 10-3    2.0 x 10-3       1.3 x 10-5


       2        10.0 x 10-3   2.0 x 10-3       5.0 x 10-5


       3        10.0 x 10-3   4.0 x 10-3       10.0 x 10-5


                                                                60
We assume that the rate law will take the form
                         Rate = k[NO]x[H2]y
Experiments 1 and 2 show that when we double the concentration
of NO at constant concentration of H2, the rate quadruples. Thus
the reaction is second order in NO. Experiments 2 and 3 indicate
that doubling [H2] at constant [NO] doubles the rate; the reaction
is first order in H2. The rate law is therefore given by
                          Rate = k[NO]2[H2]
Which shows that it is a (1 + 2) or third-order reaction overall.

    Experiment       [NO]          [H2]        Initial Rate (M/s)

         1         5.0 x 10-3   2.0 x 10-3         1.3 x 10-5

         2        10.0 x 10-3   2.0 x 10-3         5.0 x 10-5

         3        10.0 x 10-3   4.0 x 10-3         10.0 x 10-5


                                                                     61
The rate constant can be calculated using the data from any one of
the experiments. Since


                     rate
        k=                                    5.0 x 10-5 M/s
                 [NO]2[H2]       k=
                                      (10.0 x 10-3 M)2(2.0 x 10 -3M)

                      k = 2.5 x 102/ M2 . s

    Experiment       [NO]         [H2]        Initial Rate (M/s)

         1         5.0 x 10-3   2.0 x 10-3        1.3 x 10-5

        2         10.0 x 10-3   2.0 x 10-3        5.0 x 10-5

        3         10.0 x 10-3   4.0 x 10-3        10.0 x 10-5


                                                                   62
The following points summarize our
  discussion of the rate law:
• Rate laws are ALWAYS determined
  experimentally.
• Reaction order should be defined in
  terms of reactant (not product)
  concentrations.
• The order of a reactant is not related
  to the stoichiometric coefficient of the
  reactant in the overall balanced
  equation.
                                         63
 Relationship Between Reactant
    Concentration and Time
Rate law expressions enable us to
 calculate the rate of a reaction from
 the rate constant and reactant
 concentrations.
The rate laws can also be used to
 determine the concentrations of
 reactants any time during the course of
 a reaction.

                                       64
            First-order reactions
A first order reaction is a reaction whose rate depends
  on the reactant concentration raised to the first
  power.
                      rate = k[A]
In a first-order reaction of the type
                      A  product
Rate can be expressed as

                                D[A]
                        rate = - Dt

                                                      65
Through the methods of calculus, this expression
 is integrated over time to obtain the
 integrated rate law for a first-order reaction:

        [A]0
   ln          = kt
        [A]t
               [A]0
Since    ln           = ln[A]0 – ln[A]t
               [A]t

   We can write the integrated rate
   law for a first order reaction as

         ln[A]0 – ln[A]t = kt
                                              66
Cyclobutane decomposes at 1000 oC to two moles of ethylene
   (C2H4) with a very high rate constant, 87 s-1 .
                                               -1

If the initial concentration of cyclobutane is 2.00 M, what is
   the concentration after 0.010 s?
 First order reaction, therefore kt = ln[C4H8]0 – ln[C4H8]t

 Rearranging the equation so that       ln [C4H8]t = ln[C4H8]0 - kt
 ln [C4H8]t can be solved for

Substituting the data
             ln [C4H8]t = ln(2.00 mol/L) - (87 s-1)(0.010 s)
            ln [C4H8]t = 0.69 – 0.87 = -0.18
 Taking the antilog (inverse log) of both sides
          [C4H8]t = 0.010 s = e-0.18 = 0.84 mol/L C4H8
                                                               67
What percent of the cyclobutane has
 decomposed in this time?

   2.00 M - .84 M = 1.16 M

   1.16 M/2.00 M x 100 = 58% has
   decomposed (42 % remains)




                                      68
          Reaction half-life (t1/2)

The t1/2 of a reaction is the time required
 to reach half the initial reactant
 concentration.
  For a first order reaction, the formula
  for determining t1/2 is         Note – t1/2 of a
                                   first order
       t1/2 =   ln 2   = 0.693     reaction is
                 k          k      constant, it is
                                   independent of
                                   reactant
                                   concentration! 69
  A plot of [N2O5] vs time for three half-lives.


         0.060

         0.050
[N2O5]




         0.040

         0.030

         0.020

         0.010

         0.000
                 0   24                48   72
                          Time (min)
                                                   70
Determining t1/2 for a first-order
 reaction
 Cyclopropane is the smallest cyclic
 hydrocarbon. Because its 60o bond angles
 allow only poor orbital overlap, its bonds
 are weak. As a result, it is thermally
 unstable and rearranges to propene at 100
 oC via the following reaction:



           CH2
                     CH3 – CH = CH2
     H2C      CH2
                                              71
The rate constant is 9.2 s-1. How long does
 it take for the initial concentration of
 cyclopropane to decrease by one-half?
                                  1st
  The label tells you this is a _____ order reaction.



   t1/2 =   ln 2   = 0.693 = 0.693 = 0.075 s
             k          k    9.2 s-1

  It takes 0.075 s for half the
  cyclopropane to form propene.
                                                        72
       Second-order reactions
For a general second-order rate equation, the
  expression including time can become quite
  complex, so let’s consider only the simplest case,
  one in which the rate law contains only one
  reactant
                   2A  product

  The integrated rate law for a second-order
  reaction involving one reactant:
                1 - 1
                         = kt
               [A]t [A]0
                                                 73
At 25 oC, hydrogen iodide breaks down
 very slowly to hydrogen and iodine
 according to the following:

  rate = k[HI]2   and k = 2.4 x 10-21 L/mol . s


 If 0.0100 mol HI(g) is placed in a 1.0 L
  container, how long will it take for the
  concentration of HI to reach 0.00900 mol/L?



                                                  74
rate = k[HI]2    and k = 2.4 x 10-21 L/mol . s

         1 - 1
                  = kt
        [A]t [A]0


       1   -    1
                           = (2.4 x 10-21 )(t)
   0.00900   0.0100

    111 – 100 = (2.4 x 10-21)(t)

     11/ 2.4 x 10-21 = t = 4.6 x 1021 seconds
                                                 75
In contrast to the half-life of a first-
  order reaction, the half-life of a
  second-order reaction DOES depend on
  reactant concentration:

       t1/2 =     1
                k[A]0



                                       76
In the simple decomposition reaction

          AB(g)  A(g) + B(g)

    rate = k[AB]2 and k = 0.20 L/mol . s

How long will it take for [AB] to reach half
 of its initial concentration of 1.50 M?


                                           77
t1/2 =     1
         k[A]0

t1/2 =        1
         (0.20)(1.50)

 t1/2 = 3.3 seconds




                        78
Determining the Reaction Order
 from the Integrated Rate Law
Suppose you don’t know the rate law for a
 reaction and don’t have the initial rate
 data needed to determine the reaction
 orders. Another method for finding
 reaction orders is a graphical technique
 that uses concentration-time data
 directly.


                                        79
To find the reaction order from the
  concentration-time data, some trial-
  and-error graphical plotting is required:
• If you obtain a straight line when you
  plot [reactant] vs. time, the reaction is
  zero order with respect to that
  reactant.




                                          80
Integrated rate laws
and reaction order.
  A plot of ln[A] vs t
  gives a straight line
  for a reaction that is
  first order in A

   A plot of 1/[A] vs t
   gives a straight
   line for a reaction
   that is second
   order in A              81
    Graphical determination of the reaction
    order for the decomposition of N2O5.
                          Time   [N2O5]   ln[N2O5] 1/[N2O5]

                          0      0.0165   -4.104   60.6
                          10     0.0124   -4.390   80.6
                          20     0.0093   -4.68    1.1 x 102
                          30     0.0071   -4.95    1.4 x 102
                          40     0.0053   -5.24    1.9 x 102
                          50     0.0039   -5.55    2.6 x 102

A plot of 1/[N2O5] vs t   60     0.0029   -5.84    3.4 x 102
is curved, indicating
that the reaction IS
NOT second order in
N2O5.                                                          82
     Graphical determination of the reaction
     order for the decomposition of N2O5.
                                  Time   [N2O5]   ln[N2O5] 1/[N2O5]


                                  0      0.0165   -4.104   60.6
                                  10     0.0124   -4.390   80.6
                                  20     0.0093   -4.68    1.1 x 102
                                  30     0.0071   -4.95    1.4 x 102
                                  40     0.0053   -5.24    1.9 x 102
                                  50     0.0039   -5.55    2.6 x 102

A plot of ln[N2O5] vs t gives     60     0.0029   -5.84    3.4 x 102
a straight line, indicating the
reaction is first order in
N2O5
                                                                       83
             Activation Energy
According to Collision Theory of Chemical Kinetics,
 molecules must collide to react. However, not all
 collisions lead to reactions. Energetically
 speaking, there is some minimum collision energy
 below which no reaction occurs. Any molecule in
 motion possesses kinetic energy; the faster it
 moves, the greater its kinetic energy. When
 molecules collide, part of their kinetic energy is
 converted to vibrational energy. If the initial
 kinetic energies are large enough, the colliding
 molecules will vibrate so strongly that some of
 the chemical bonds will break. This bond
 fracture is the first step toward product
 formation. If the initial kinetic energies are too
 small, the molecules will merely bounce off each
 other intact, and no change results from the
 collision.
                                                  84
We postulate that in order to react, the
 colliding molecules must have a total
 kinetic energy equal to or greater than
 the activation energy, (Ea), which is
 defined as the minimum amount of
 energy required to initiate a chemical
 reaction.



                                           85
Molecules must also be oriented in a
 favorable position – one that allows the
 bonds to break and atoms to rearrange.




                                            86
             NO(g) + NO3(g)  2NO2(g)



The picture to the right
 shows a few of the
 possible collision
 orientations for this
 simple gaseous reaction:



Of the five collisions shown, only one has an orientation in which the N
of NO collides with an O of NO3. Actually, the probability factor (p)
for this reaction is 0.006; only 6 collisions in every thousand have an
orientation that leads to a reaction.
                                                                           87
Collisions between individual atoms have p values
  near 1: almost no matter how they hit, they
  react. In such cases, the rate constant
  depends only on the frequency and energy of
  the collisions.
At the other extreme are biochemical
  reactions, in which the reactants are often
  two small molecules that can react only when
  they collide with a specific tiny region of a
  giant molecule-a protein or nucleic acid. The
  orientation factor for such reactions is often
  less than 10-6: fewer than one in a million
  sufficiently energetic collisions leads to
  product formation.
                                               88
   When reactants collide at the proper angle with energy
    equal to the activation energy (Ea), they undergo an
    extremely brief interval of bond disruption and bond
    formation called a transition state.


                                      During this transition
                                      state, the reactants form
                                      a short-lived complex that
                                      is neither reactant nor
                                      product, but has partial
                                      bonding characteristics of
                                      both. This transitional
                                      structure is called an
                                      activated complex.
Endothermic/Exothermic (choose one)
                                                             89
                           An activated complex is a
                           highly unstable species with a
                           high potential energy. (It
                           was energized by the particle
                           collision.) Once formed, it
                           will break up almost
                           immediately.
                           The activated complex exists
                           along the reaction pathway at
                           the point where the energy is
Activation energy is the
energy required to         greatest – at the peak
achieve the transition     indicated by the activation
state and form the         energy.
activated complex
                                                      90
We can think of activation energy, (Ea), as a barrier that
prevents less energetic molecules from reacting…




  …because only the molecules who have enough kinetic
  energy to exceed the activation energy can take part in
  the reaction.
                                                         91
 Maxwell Boltzmann Diagram
Because the number of reactant molecules in
an ordinary reaction is very large, the speeds,
and hence also the kinetic energies of the
molecules, vary greatly.




                                                  92
Normally, only a small fraction of the colliding
  molecules -- the fastest-moving ones – have enough
  kinetic energy to exceed the activation energy.
At higher temperature, more molecules can surpass
  the activation energy, therefore, the rate of
  product formation is greater at the higher
  temperature.

                                 As temperature
                                 increases, the
                                 probability of finding
                                 molecules at higher
                                 temperature increases.




                                                          93
With very few exceptions, reaction rates
 increase with increasing temperature.

As a general rule of thumb, you can expect a
 10 oC increase in temperature to result in
 a doubling of the reaction rate.



                                           94
            Arrhenius Equation
The dependence of the rate constant of a reaction on
  temperature can be expressed by the following equation,
  known as the Arrhenius equation:

                     k = Ae-Ea/RT
Where Ea is the activation energy (in J/mol), R the gas
constant (8.314 J/K . mol), T the absolute temperature, and
e the base of the natural logarithm scale. The quantity A
represents the collision frequency and is called the
frequency factor.

                                                        95
             Frequency Factor (A)

The frequency factor is the product of the
 collision frequency (Z) and an orientation
 factor (p) which is specific for each reaction.
 The factor p is related to the structural
 complexity of the colliding particles. You can
 think of it as the ratio of effectively oriented
 collisions to all possible collisions.
In the activation energy problems we are solving,
  the actual value of A need not be known
  because A can be treated as a constant for a
  given reacting system over a fairly wide
  temperature range.
                                                96
               k = Ae-Ea/RT
• As the activation energy increases, k
  decreases,
  and as k decreases, rate decreases

 •As the temperature increases, k
  increases
• And as k increases, rate
  increases
                                          97
You can derive the following equation
from the Arrhenius equation:

              -Ea   1
     ln k =             + ln A
               R    T




                                        98
               -Ea    1
      ln k =              + ln A
                R     T

      Y=         m
                 mx       +   b


Thus, a plot of ln k versus 1/T gives a
straight line whose
 slope
•slope (m) is equal to –Ea/R and whose
•intercept (b) with the y-axis is ln A.
                                          99
Given k and temperature, you can use your
  graphing calculator to determine the
  activation energy of a reaction.



See the following example…

                                       100
The rate constants for the decomposition
of acetaldehyde

     CH3CHO(g)  CH4(g) + CO(g)

were measured at five different
temperatures. The data are shown in the
following table.



                                       101
Plot ln k versus 1/T to determine the activation
energy (-slope) in kJ/mol for the previous reaction.

                 T (K)        k
                 700        0.011
                 730        0.035
                 760        0.105
                 790        0.343
                  810       0.789

 To determine the activation energy, we
 need to graph 1/T on the x axis and ln k
 on the y-axis.                                 102
    1/T            ln k          Graph these
 1.43 x 10-3      -4.51          values using your
                                 graphing
 1.37 x 10-3      -3.35
                                 calculator.
 1.32 x 10-3     -2.254
 1.27 x 10-3     -1.070
 1.23 x 10-3      0.237

Slope = -Ea/R
Ea = -Slope . R
       Ea = -(-2.16 x 104 K)(8.314 J/K . Mol)
               Ea = 1.80 x 105 J/mol
      Ea = 1.80 x 102 kJ/mol
                                                     103
The slope of the line can be
determined using linear
regression.




                               104
An equation relating the rate constants k1
 and k2 at temperature T1 and T2 can be
 used to calculate the activation energy
 or to find the rate constant at another
 temperature if the activation energy is
 known.

               k2   Ea 1 1
            ln    =
               k1   R T2 T1

                                        105
The rate constant of a first-order reaction is 3.46 x
10-2 s-1 at 298 K. What is the rate constant at 350 K
if the activation energy for the reaction is 50.2
kJ/mol?
    k1 = 3.46 x 10-2s-1          k2 =
  T1 = 298                     T2 = 350
               k2     Ea 1     1
            ln k =    R T
                1
                           2   T1
            k2          50,200 J/mol      1     1
   ln
       3.46 x 10-2= 8.314 J/K . mol      350 298
         k2
 ln
    3.46 x 10-2 = 3.01 Taking the antilog of both sides
         k2
                = e 3.01 = 20.3     k2 = 0.702 s-1
    3.46 x 10-2
                                                        106
      Reaction Mechanisms
As we mentioned earlier, an overall balanced
 chemical equation does not tell us much about
 how a reaction actually takes place. In many
 cases, it merely represents the sum of
 several elementary steps, or elementary
 reactions, that represent the progress of the
 overall reaction at the molecular level. The
 term for the sequence of elementary steps
 that leads to product formation is reaction
 mechanism.
The reaction mechanism is comparable to the
 route of travel followed during a trip; the
 overall chemical equation specifies only the
 origin and the destination.
                                            107
As an example of a reaction mechanism,
 let us consider the reaction between
 nitrogen monoxide and oxygen:
      2NO(g) + O2(g)  2NO2(g)

 We know that the products are not
 formed directly from the collision
 of two NO molecules with an O2
 molecule because N2O2 is detected
 during the course of the reaction.
                                      108
  Let us assume that the reaction actually takes
    place via two elementary steps as follows:
  Elementary Step 1       NO(g) + NO(g)  N2O2(g)
  Elementary Step 2       N2O2(g) + O2(g)  2NO2(g)
  Overall reaction     2NO + N2O2 + O2  N2O2 + 2NO2
Each of the elementary steps listed above is called a bimolecular
reaction because each step involves two reactant molecules. A step
that just involves one reactant molecule is a unimolecular reaction.
Very few termolecular reactions, reactions that involve the
participation of three reactant molecules in one elementary step,
are known, because the simultaneous encounter of three molecules
is a far less likely event than a bimolecular collision. (There are no
known examples of reactions involving the simultaneous encounter of
four molecules.)
                                                                109
Elementary Step 1     NO(g) + NO(g)  N2O2(g)
Elementary Step 2    N2O2(g) + O2(g)  2NO2(g)
Overall reaction    2NO + N2O2 + O2  N2O2 + 2NO2

 Species such as N2O2 are called intermediates because
 they appear in the mechanism of the reaction (that is,
 in the elementary steps) but not in the overall balanced
 equation. Keep in mind that an intermediate is always
 formed in an early elementary step and consumed in a
 later elementary step. Note – an intermediate ≠
 activated complex!


                                                     110
You can propose a mechanism for a reaction if
  you consider…
• The elementary steps in a multistep reaction
  mechanism must always add to give the
  balanced chemical equation of the overall
  process. (Any intermediates that are formed
  in earlier steps must be consumed in later
  steps.)
• Unimolecular and bimolecular reactions are
  more common than termolecular reactions.
• The rate of the overall reaction is limited by
  the rate of the slowest elementary step, (For
  that reason, the slowest elementary step is
  typically called the rate-determining step.
                                               111
Gaseous nitrogen monoxide reacts with fluorine gas to
  produce nitrogen hypofluorite (NOF). The
  intermediate product NOF2(g) has been isolated as an
  intermediate in this reaction. Propose a two step
  mechanism consistent with this intermediate product.


   Step 1: NO(g) + F2(g) NOF2(g)
   Step 2: NOF2(g) + NO(g)  2NOF(g)

  •The elementary steps add to give the overall
  balanced chemical equation
  •The first and the second step are bimolecular

                                                    112
Even when a proposed mechanism is
 consistent with the rate law, later
 experimentation may show it to be
 incorrect or only one of several
 alternatives.




                                       113
Knowing the elementary steps of a reaction enables us to
  propose a rate law.
Suppose we have the following elementary step:
                      A  products
Because there is only one reactant molecule present, this is
  a/n ___molecular reaction. It follows that the larger
       uni
  the number of A molecules present, the faster the rate
  of product formation.

Thus the rate of a unimolecular reaction is directly
  proportional to the concentration of A, or is first order
  in A:
                        Rate = k[A]
                                                         114
For a bimolecular elementary reaction
 involving A and B molecules
           A + B  product
 the rate of product formation depends
 on how frequently A and B collide, which
 in turn depends on the concentration of
 A and B. Thus we can express the rate
 as
             Rate = k[A][B]

                                        115
Similarly, for a bimolecular elementary
  reaction of the type

          A + A  products
                 or
            2A  products

the rate becomes
              Rate = k[A]2
                                          116
  Rate Laws for Elementary Steps
Elementary Step   Molecularity   Rate Law
A  product       unimolecular rate = k[A]
2A  product      bimolecular    rate = k[A]2
A + B  product   bimolecular    rate = k[A][B]
2A + B  product termolecular    rate = k[A]2[B]




                                                117
Remember, when we study a reaction that
 has more than one elementary step, the
 rate law for the overall process is given
 by the rate-determining step, which is
 the slowest step in the sequence of
 steps leading to product formation.



                                         118
Example: The gas-phase decomposition of
   dinitrogen monoxide (N2O) is believed to
   occur via two elementary steps:
Step 1: N2O N2 + O
Step 2: N2O + O  N2 + O2

Experimentally the rate law is found to be
                 Rate = k[N2O]

(a) Write the equation for the overall reaction.
(b) Identify the intermediates.
(c) What can you say about the relative rates
    of steps 1 and 2?
                                              119
(a) Adding the equations for steps 1 and 2 gives
    the overall reaction
              2N2O  2N2 + O2

(b) Since the O atom is produced in the first
    elementary step and it does not appear in the
    overall balanced equation, it is an
    intermediate.

(c) Step 1 must be the rate-determining (slower)
    step because that step is also first order
    with respect to N2O. (The rate law for the
    rate-determining step should match the rate
    law that is determined experimentally.)

                                              120
Example 2: Hydrogen Peroxide Decomposition


Does the decomposition of hydrogen
 peroxide occur in a single step?

The overall reaction is
                 I-
     2H2O2(aq)  2H2O(l) + O2(g)

By experiment, the rate law is found to be
           Rate = k[H2O2][I-]
                                             121
From this alone you can see that H2O2
 decomposition does not occur in a single
 elementary step corresponding to the
 overall balanced equation. If it did, the
 rate law would be
             Rate = [H2O2]2
 or in other words, the reaction would be
 second order in H2O2. Remember, the
 experimentally determined rate law for
 this reaction was shown to be
            Rate = k[H2O2][I-]
                                        122
                    Catalysis
                       I-
         2H2O2(aq)  2H2O(l) + O2(g)

                 Rate = k[H2O2][I-]
For the decomposition of hydrogen peroxide we see
  that the reaction rate depends on the concentration
  of iodide ions even though I- does not appear in the
  overall equation. I- is a catalyst for this reaction, a
  substance that increases the rate of a chemical
  reaction without itself being consumed.



                                                        123
A catalyst exists before the reaction
 occurs and can be recovered and reused
 after the reaction is complete. This is
 the opposite of intermediates, which
 are produced in one step of a mechanism
 and consumed in another.



                                      124
                             In many cases, a catalyst
                             increases the rate by
                             providing a set of elementary
                             steps with more favorable
                             kinetics than those that exist
                             in its absence.




In other words, a catalyst
typically lowers the
activation energy for the
reaction by forming an
activated complex that has
less potential energy.
                                                        125
Forward reaction



                                                     C+D


   C+D
                                  A+B


                  A+B
                               Reverse reaction

 Because the activation energy for the reverse
 reaction is also lowered, a catalyst enhances the
 rates of the forward and reverse reaction
 equally.                                              126
There are three general types of
 catalysis: heterogeneous catalysis,
 homogeneous catalysis, and enzyme
 catalysis.




                                       127
In heterogeneous catalysis the reactants
  and the catalyst are in different
  phases. Usually the catalyst is a solid
  and the reactants are either gases or
  liquids. Heterogeneous catalysis is by
  far the most important type of catalysis
  in industrial chemistry, especially in the
  synthesis of many key chemicals.



                                          128
Ammonia is an extremely valuable inorganic substance
  used in the fertilizer industry and many other
  applications.

N2(g) + 3H2(g)  2NH3(g)      DH = -92.6 kJ

This reaction is extremely slow at room temperature,
  and although raising the temperature accelerates the
  above reaction, it also promotes the decomposition of
  NH3 molecules into N2 and H2, thus lowering the yield
  of NH3.

In 1905, after testing literally hundreds of compounds
  at various temperatures and pressures, Fritz Haber
  discovered that iron plus a few percent of oxides of
  potassium and aluminum catalyze the reaction. This
  procedure is known as the Haber process.

                                                       129
              Haber Process
First the H2 and the
N2 molecules bind to
the surface of the
catalyst. This
interaction weakens
the covalent bonds
within the molecules
and eventually causes
the molecules to
dissociate. The highly
reactive H and N
atoms combine to
form NH3 molecules,
which then leave the
surface.                      130
Nitric acid is one of the most important
  inorganic acids. It is used in the production
  of fertilizers, dyes, drugs, and in many other
  products. The major industrial method of
  producing nitric acid is the Ostwald process.
  The starting materials, ammonia and molecular
  oxygen, are heated in the presence of a
  platinum-rhodium catalyst to about 800 oC.




                                              131
In homogeneous catalysis, the reactants
  are dispersed in a single phase, usually
  liquid. Acid and base catalyses are the
  most important types of homogeneous
  catalysis in liquid solution.
Homogeneous catalysis can also take place
  in the gas phase. A well-known example
  of catalyzed gas-phase reactions is the
  lead chamber process, which for many
  years was the primary method of
  manufacturing sulfuric acid.
                                        132
Homogeneous catalysis has several advantages
 over heterogeneous catalysis. For one thing,
 the reactions can often be carried out under
 atmospheric conditions, thus reducing
 production costs and minimizing the
 decomposition of products at high
 temperatures. In addition, homogeneous
 catalysts can be designed to function
 selectively for a particular type of reaction,
 and homogeneous catalysts cost less than the
 precious metals (for example, platinum and
 gold) used in heterogeneous catalysis.

                                              133
Of all the intricate processes that have evolved
  in living systems, none is more striking or
  more essential than enzyme catalysis.
Enzymes are biological catalysts. Enzymes can
  increase the rate of a biochemical reaction by
  a factor ranging from 106 to 1012 times!
An enzyme acts only on certain molecules, called
  substrates while leaving the rest of the
  system unaffected. It has been estimated
  that an average living cell may contain some
  3000 different enzymes, each of them
  catalyzing a specific reaction in which a
  substrate is converted into the appropriate
  products.
                                              134
An enzyme is typically a large protein molecule
 that contains one or more active sites where
 interactions with substrates takes place.
 These sites are structurally compatible with
 specific substrate molecules, in much the
 same way as a key fits a particular lock. In
 fact, the notion of a rigid enzyme structure
 that binds only to molecules whose shape
 exactly matches that of the active site was
 the basis of an early theory of enzyme
 catalysis, the so-called lock-and-key theory.


                                                  135
This theory accounts for the specificity of
enzymes, but it contradicts research evidence
that a single enzyme binds to substrates of
different sizes and shapes.




Chemists now know that an enzyme molecule (or at least its
active site) has a fair amount of structural flexibility and
can modify its shape to accommodate more than one type
of substrate.
                                                         136
The
End
      137

				
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posted:7/27/2012
language:English
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