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Rates of Reaction
       Reaction Rates

 Chemical reactions require varying lengths
  of time for completion.
   – This reaction rate depends on the
     characteristics of the reactants and
     products and the conditions under which the
     reaction is run. (see Figure 14.1)
   – By understanding how the rate of a reaction is
     affected by changing conditions, one can learn
     the details of what is happening at the
     molecular level.

                                                      2
       Reaction Rates

 The questions posed in this chapter will be:

  – How is the rate of a reaction measured?
  – What conditions will affect the rate of a
    reaction?
  – How do you express the relationship of rate to
    the variables affecting the rate?
  – What happens on a molecular level during a
    chemical reaction?


                                                     3
          Reaction Rates

 Chemical kinetics is the study of reaction rates,
  how reaction rates change under varying
  conditions, and what molecular events occur
  during the overall reaction.
   – What variables affect reaction rate?
      •   Concentration of reactants.
      •   Concentration of a catalyst
      •   Temperature at which the reaction occurs.
      •   Surface area of a solid reactant or catalyst.



                                                          4
        Reaction Rates

 Chemical kinetics is the study of reaction rates,
  how reaction rates change under varying
  conditions, and what molecular events occur
  during the overall reaction.

   – What variables affect reaction
     rate?
         Let’s look at each in more detail.


                                                      5
       Factors Affecting Reaction
       Rates
 Concentration of reactants.
 – More often than not, the rate of a reaction increases
   when the concentration of a reactant is increased.
 – Increasing the population of reactants increases the
   likelihood of a successful collision.
 – In some reactions, however, the rate is unaffected by
   the concentration of a particular reactant, as long as
   it is present at some concentration.




                                                       6
       Factors Affecting Reaction
       Rates
 Concentration of a catalyst.
  – A catalyst is a substance that increases
    the rate of a reaction without being
    consumed in the overall reaction.
  – The catalyst generally does not appear in
    the overall balanced chemical equation
    (although its presence may be indicated by
    writing its formula over the arrow).

                      
       2 H 2O2( aq )  2 H 2O(l )  O2( g )
                  HBr ( aq )


                                                7
       Factors Affecting Reaction
       Rates
 Concentration of a catalyst.
  – Figure 14.2 shows the HBr catalyzed
    decomposition of H2O2 to H2O and O2.

                      )
       2H 2O 2( aq ) HBr 2H 2O( l )  O 2( g )
                        ( aq


  – A catalyst speeds up reactions by reducing the
    “activation energy” needed for successful reaction.
  – A catalyst may also provide an alternative
    mechanism, or pathway, that results in a faster
    rate.

                                                      8
      Factors Affecting Reaction
      Rates
 Temperature at which a reaction occurs.

  – Usually reactions speed up when the
    temperature increases.
  – A good “rule of thumb” is that reactions
    approximately double in rate with a 10 oC
    rise in temperature.



                                                9
       Factors Affecting Reaction
       Rates
 Surface area of a solid reactant or
  catalyst.
     – Because the reaction occurs at the
       surface of the solid, the rate
       increases with increasing surface
       area.
     – Figure 14.3 shows the effect of
       surface area on reaction rate.


                                            10
       Definition of Reaction Rate

 The reaction rate is the increase in molar
  concentration of a product of a reaction
  per unit time.
  – Always a positive value.
  – It can also be expressed as the decrease in
    molar concentration of a reactant per unit
    time.



                                               11
       Definition of Reaction Rates

 Consider the gas-phase decomposition of
 dintrogen pentoxide.
      2N 2O5 (g )  4NO2 (g )  O 2 (g )
  – If we denote molar concentrations using brackets,
    then the change in the molarity of O2 would be
    represented as

                     D[O 2 ]           where the
     symbol, D (capital Greek delta), means
  “the change in.”
                                                    12
        Definition of Reaction Rates

 Then, in a given time interval, Dt , the molar
  concentration of O2 would increase by
  D[O2].
   – The rate of the reaction is given by:
                                D[O 2 ]
  Rate of formation of oxygen 
                                 Dt
 – This equation gives the average rate over the time
   interval, Dt.
 – If Dt is short, you obtain an instantaneous rate, that
   is, the rate at a particular instant. (Figure 14.4)
                                                        13
Figure 14.4 The instantaneous rate of reaction
                                                 In the reaction

                                                 2N2O5 (g)  4NO2 (g)  O2 (g)

                                                 The concentration of
                                                 O2 increases over time.
                                                 You obtain the
                                                 instantaneous rate
                                                 from the slope of the
                                                 tangent at the point of
                                                 the curve
                                                 corresponding to that
                                                 time.
       Definition of Reaction Rates

 Figure 14.5 shows the increase in
  concentration of O2 during the
  decomposition of N2O5.
• Note that the rate decreases as the
  reaction proceeds.




                                        15
Figure 14.5 Calculation of the average rate.

                                               When the time
                                               changes from 600 s
                                               to 1200 s, the
                                               average rate is
                                               2.5 x 10-6 mol/(L.s).
                                               Later when the time
                                               changes from 4200 s
                                               to 4800 s, the
                                               average rate has
                                               slowed to
                                               5 x 10-7 mol/(L.s).
                                               Thus, the rate of a
                                               reaction decreases
                                               as the reaction
                                               proceeds.
         Definition of Reaction Rates

   Because the amounts of products and
    reactants are related by stoichiometry, any
    substance in the reaction can be used to
    express the rate.
                                   D[N 2O5 ]
Rate of decomposition of N 2O5  -
                                     Dt
 • Note the negative sign. This results in a
   positive rate as reactant concentrations
   decrease.
                                                  17
          Definition of Reaction Rates
 The rate of decomposition of N2O5 and the formation of O2
  are easily related.
   2N 2O5 (g )  4NO2 (g )  O 2 (g )
           D[O 2 ] 1         D[N 2O 5 ]
            Dt
                  
                     2
                         ( -
                               Dt
                                        )
– Since two moles of N2O5 decompose for each
  mole of O2 formed, the rate of the
  decomposition of N2O5 is twice the rate of the
  formation of O2.

                                                         18
       Experimental Determination
       of Reaction Rates
 To obtain the rate of a reaction you must
  determine the concentration of a reactant or
  product during the course of the reaction.

  – One method for slow reactions is to withdraw
    samples from the reaction vessel at various times
    and analyze them.
  – More convenient are techniques that continuously
    monitor the progress of a reaction based on some
    physical property of the system.

                                                    19
       Experimental Determination
       of Reaction Rates
 Gas-phase partial pressures.
  – When dinitrogen pentoxide crystals are sealed in a
    vessel equipped with a manometer (see Figure 14.6)
    and heated to 45oC, the crystals vaporize and the
    N2O5(g) decomposes.

        2N 2O5 (g )  4NO2 (g )  O 2 (g )

  – Manometer readings provide the concentration of
    N2O5 during the course of the reaction based on
    partial pressures.

                                                      20
This device indicates the difference between two pressures
(differential pressure), or between a single pressure and
atmosphere (gage pressure), when one side is open to atmosphere.
If a U-tube is filled to the half way point with water and air pressure
is exerted on one of the columns, the fluid will be displaced. Thus
one leg of water column will rise and the other falls. The difference
in height "h" which is the sum of the readings above and below the
half way point, indicates the pressure .




                                                                21
         Experimental Determination
         of Reaction Rates
 Colorimetry
  – Consider the reaction of the hypochlorite ion with
    iodide.
                                               
ClO (aq)  I (aq)  IO (aq)  Cl (aq)
  – The hypoiodate ion, IO-, absorbs near 400 nm. The
    intensity of the absorbtion is proportional to [IO-], and
    you can use the absorbtion rate to determine the
    reaction rate.


                                                         22
       Dependence of Rate on
       Concentration
 Experimentally, it has been found that the
  rate of a reaction depends on the
  concentration of certain reactants as well as
  catalysts.
  – Let’s look at the reaction of nitrogen dioxide with
    fluorine to give nitryl fluoride.
         2NO2 (g )  F2 (g )  2NO2F(g )
  – The rate of this reaction has been observed to be
    proportional to the concentration of nitrogen dioxide.

                                                          23
     Dependence of Rate on
     Concentration
• When the concentration of nitrogen
  dioxide is doubled, the reaction rate
  doubles.
   – The rate is also proportional to the
     concentration of fluorine; doubling the
     concentration of fluorine also doubles the
     rate.
   – We need a mathematical expression to
     relate the rate of the reaction to the
     concentrations of the reactants.

                                                  24
       Dependence of Rate on
       Concentration
 A rate law is an equation that relates the
  rate of a reaction to the concentration of
  reactants (and catalyst) raised to various
  powers.
          Rate  k[NO2 ][F2 ]
  – The rate constant, k, is a proportionality
    constant in the relationship between rate
    and concentrations.

                                                 25
       Dependence of Rate on
       Concentration
 As a more general example, consider the
  reaction of substances A and B to give D
  and E. C
        
aA  bB  dD  eE                       C  catalyst
  – You could write the rate law in the form

          Rate  k[A] [B] [C]m      n      p

  – The exponents m, n, and p are frequently, but not
    always, integers. They must be determined
    experimentally and cannot be obtained by simply
    looking at the balanced equation.
                                                        26
           Find the rate law expression and evaluate k for
           2H2 + 2NO  2H2O + N2 @800K; all gases



    H2       NO     Rate               Rate = k [A]x[B]y
                    (atm/min
                                          [A]x [B]y
1   .001     .006   .025
                               Trial 2 = .002 x x .006 y= .050
2   .002     .006   .050       Trial 1 .001 .006 .025
3   .003     .006   .075                    2x x 1y = 2
4   .009     .001   .0063                           2x = 2
5   .009     .002   .025                             x = 1
6   .009     .003   .056

                                                         27
           Find the rate law expression and evaluate k for
           2H2 + 2NO  2H2O + N2 @800K; all gases


    H2       NO      Rate                   [A]x [B]y
                     (atm/min   Trial 6 = .009 1 x .003 y= .056
1   .001     .006    .025       Trial 4 .009       .001 .0063
2   .002     .006    .050                            3y = 8.89
                                                 ylog3 =log 8.89
3   .003     .006    .075
                                                     y= 2
4   .009     .001    .0063      Rxn is first order for x
5   .009     .002    .025       second order for y , and
6   .009     .003    .056       (2 + 1 =) third order overall.


                                                                 28
             Rate = k[H2][NO]2
From Trial 1: .025atm = k(.001M)(.006M)2
                 min
              .025atm = k(.001M)(.006M)2
                 min
          6.94 x 105 atm = k
                 min M3




                                           29
    Find the rate law expression and evaluate
    k for
    H2 + 2NO  2H2O + N2O
                               @1100K; all gases

       PNO [atm]   PH2 [atm]        Rate
                                    [atm/min]

1      .150        .400             .020

2      .075        .400             .005

3      .150        .200             .101


                                                   30
      Find the rate law expression and
      evaluate k for
      H2 + 2NO  2H2O + N2O
                            @1100K; all gases
 ANSWERS:
x = 2
y = 1
 k = 2.22(min atm)-1




                                            31
       Dependence of Rate on
       Concentration
 Reaction Order
  – The reaction order with respect to a given
    reactant species equals the exponent of the
    concentration of that species in the rate law, as
    determined experimentally.

  – The overall order of the reaction equals the sum
    of the orders of the reacting species in the rate
    law.



                                                    32
        Dependence of Rate on
        Concentration
 Reaction Order
   – Consider the reaction of nitric oxide with
     hydrogen according to the following equation.
   2NO(g )  2H 2 (g )  N 2 (g )  2H 2O(g )
   – The experimentally determined rate law is
            Rate  k[NO] [H 2 ]
                             2

   – Thus, the reaction is second order in NO, first
     order in H2, and third order overall.
                                                 33
       Dependence of Rate on
       Concentration
 Reaction Order
   – Although reaction orders frequently have
     whole number values (particularly 1 and 2),
     they can be fractional.
   – Zero and negative orders are also possible.
   – The concentration of a reactant with a zero-
     order dependence has no effect on the rate
     of the reaction.


                                              34
       Dependence of Rate on
       Concentration
 Determining the Rate Law.
  – One method for determining the order of a
    reaction with respect to each reactant is the
    method of initial rates.
  – It involves running the experiment multiple
    times, each time varying the concentration of
    only one reactant and measuring its initial
    rate.
  – The resulting change in rate indicates the
    order with respect to that reactant.
                                              35
       Dependence of Rate on
       Concentration
 Determining the Rate Law.
  – If doubling the concentration of a reactant has a
    doubling effect on the rate, then one would deduce
    it was a first-order dependence. 2:2 correspondence
  – If doubling the concentration had a quadrupling
    effect on the rate, one would deduce it was a
    second-order dependence. 2:4 correspondence
  – A doubling of concentration that results in an
    eight-fold increase in the rate would be a third-
    order dependence. 2:8 correspondence

                                                    36
           A Problem to Consider

   Iodide ion is oxidized in acidic solution to
    triiodide ion, I3- , by hydrogen peroxide.
                                         
H 2O2 (aq)  3I (aq)  2 H (aq)  I 3 (aq)  2 H 2O(l )
    – A series of four experiments was run at different
      concentrations, and the initial rates of I3- formation
      were determined.
    – From the following data, obtain the reaction orders
      with respect to H2O2, I-, and H+.
    – Calculate the numerical value of the rate constant.

                                                           37
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6
  – Comparing Experiment 1 and Experiment 2, you see
    that when the H2O2 concentration doubles (with other
    concentrations constant), the rate doubles.
  – This implies a first-order dependence with respect
    to H2O2.

                                                            38
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6

  – Comparing Experiment 1 and Experiment 3, you see
    that when the I- concentration doubles (with other
    concentrations constant), the rate doubles.
  – This implies a first-order dependence with respect
    to I-.
                                                            39
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6
   – Comparing Experiment 1 and Experiment 4, you see
     that when the H+ concentration doubles (with other
     concentrations constant), the rate is unchanged.
   – This implies a zero-order dependence with
     respect to H+.

                                                            40
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6
   – Because [H+]0 = 1, the rate law is:
                                         
               Rate  k[H 2O 2 ][I ]
   – The reaction orders with respect to H2O2, I-, and H+, are
     1, 1, and 0, respectively.
                                                            41
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6
   – You can now calculate the rate constant by substituting
     values from any of the experiments. Using Experiment 1
     you obtain:

                 6mol             mol         mol
         1.15  10      k  0.010      0.010
                   Ls              L           L
                                                            42
 A Problem to Consider
         Initial Concentrations (mol/L)
          H2O2          I-          H+   Initial Rate [mol/(L.s)]
Exp. 1    0.010       0.010      0.00050        1.15 x 10-6
Exp. 2    0.020       0.010      0.00050        2.30 x 10-6
Exp. 3    0.010       0.020      0.00050        2.30 x 10-6
Exp. 4    0.010       0.010      0.00100        1.15 x 10-6
   – You can now calculate the rate constant by substituting
     values from any of the experiments. Using Experiment 1
     you obtain:
               1.15 106 s 1
         k                       1.2 102 L /( mol  s )
            0.010  0.010mol / L
                                                              43
       Change of Concentration with
       Time
 A rate law simply tells you how the rate of
  reaction changes as reactant concentrations
  change.
  – A more useful mathematical relationship
    would show how a reactant concentration
    changes over a period of time.




                                                44
       Change of Concentration with
       Time
 A rate law simply tells you how the rate of
  reaction changes as reactant concentrations
  change.
  – Using calculus we can transform a rate law
    into a mathematical relationship between
    concentration and time.
  – This provides a graphical method for
    determining rate laws.


                                                45
        Concentration-Time
        Equations
 First-Order Integrated Rate Law
   – You could write the rate law in the form

               D[ A]
      Rate          k[ A ]
                Dt




                                                46
        Concentration-Time
        Equations
 First-Order Integrated Rate Law
  – Using calculus, you get the following equation.

                     [ A]t
                  ln        - kt
                     [ A]o
  – Here [A]t is the concentration of reactant A at time t,
    and [A]o is the initial concentration.
  – The ratio [A]t/[A]o is the fraction of A remaining at
    time t.

                                                        47
         A Problem to Consider
 The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.8 x 10-4 s-1. If the
  initial concentration of N2O5 is 1.65 x 10-2 mol/L,
  what is the concentration of N2O5 after 825
  seconds?
   – The first-order time-concentration equation for this
     reaction would be:

                [N 2O 5 ]t
             ln             - kt
                [N 2O 5 ]o
                                                         48
         A Problem to Consider
 The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.8 x 10-4 s-1. If the
  initial concentration of N2O5 is 1.65 x 10-2 mol/L,
  what is the concentration of N2O5 after 825
  seconds?
   – Substituting the given information we obtain:

        [N 2O 5 ]t
ln                      - (4.80  10-4 s -1 )  (825 s)
   1.65  102 mol / L

                                                         49
         A Problem to Consider
 The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.8 x 10-4 s-1. If the
  initial concentration of N2O5 is 1.65 x 10-2 mol/L,
  what is the concentration of N2O5 after 825
  seconds?
   – Substituting the given information we obtain:

                 [N 2O 5 ]t
         ln          2
                               - 0.396
            1.65  10 mol / L

                                                         50
         A Problem to Consider
 The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.8 x 10-4 s-1. If the
  initial concentration of N2O5 is 1.65 x 10-2 mol/L,
  what is the concentration of N2O5 after 825
  seconds?
   – Taking the inverse natural log of both sides we
     obtain:
            [N 2O 5 ]t
                2
                         e -0.396
                                    0.673
       1.65  10 mol / L

                                                         51
          A Problem to Consider
 The decomposition of N2O5 to NO2 and O2 is first
   order with a rate constant of 4.8 x 10-4 s-1. If the
   initial concentration of N2O5 is 1.65 x 10-2 mol/L,
   what is the concentration of N2O5 after 825
   seconds?
    – Solving for [N2O5] at 825 s we obtain:


[N 2O5 ]  (1.65  10-2 mol / L)  (0.673)  0.0111 mol / L


                                                          52
        Concentration-Time
        Equations
 Second-Order Integrated Rate Law
   – You could write the rate law in the form

               D[ A]
      Rate          k[ A ]2
                Dt




                                                53
        Concentration-Time
        Equations
 Second-Order Integrated Rate Law
   – Using calculus, you get the following equation.
             1     1
                      kt
           [ A]t [A]o
   – Here [A]t is the concentration of reactant A
     at time t, and [A]o is the initial concentration.

                                                   54
       Concentration-Time
       Equations
 Zero-Order Integrated Rate Law
   – The Zero-Order Integrated Rate Law equation
     is:.

          [ A]o  [ A]  kt


                                            55
          Concentration-Time
          Equations
 First order integrated rate law


                      - kt
               [ A]t
            ln
               [ A]o
 Second order integrated rate law
           1      1
                      kt
         [ A]t   [A]o
 Zero order integrated rate law

         [ A]o  [ A]  kt
                                     56
        Concentration-Time
        Equations
 k values
General k formula:
           Units k = (L/mol)order-1
                      unit of time
 First order:      s-1
 Second order: L/mol•sec, or L•mol-1•sec-1
 Zero order:       L2/mol2•sec, or L2•mol-2•sec-1


                                                     57
          Half-life
 The half-life of a reaction is the time required for the
  reactant concentration to decrease to one-half of its initial
  value.

      – For a first-order reaction, the half-life is
        independent of the initial concentration of reactant.

      – In one half-life the amount of reactant decreases
                              ln(   1
                                    2   )     kt   1
                                                     2




        by one-half. Substituting into the first-order
        concentration-time equation, we get:
                        ln( 1 / 2)  .693  kt1
                                                         2
                                                                  58
         Half-life

 The half-life of a reaction is the time required for
  the reactant concentration to decrease to one-half
  of its initial value.
     – Solving for t1/2 we obtain:


                        0.693
                    t 
                      1
                      2
                          k
     – Figure 14.8 illustrates the half-life of a first-order
       reaction.
                                                            59
        Half-life

 Sulfuryl chloride, SO2Cl2, decomposes in a
  first-order reaction to SO2 and Cl2.
      SO 2Cl 2 (g )  SO 2 (g )  Cl 2 (g )
    – At 320 oC, the rate constant is 2.2 x 10-5 s-1.
      What is the half-life of SO2Cl2 vapor
      at this temperature?
    – Substitute the value of k into the relationship
      between k and t1/2.
                             0.693
                         t 
                          1
                          2
                               k                        60
       Half-life

 Sulfuryl chloride, SO2Cl2, decomposes in a
  first-order reaction to SO2 and Cl2.
      SO 2Cl 2 (g )  SO 2 (g )  Cl 2 (g )
    – At 320 oC, the rate constant is 2.20 x 10-5 s-1.
    What is the half-life of SO2Cl2 vapor at
    this temperature?
    – Substitute the value of k into the relationship
    between k and t1/2.
                          0.693
                  t           5 -1
                      2.20  10 s
                    1
                    2
                                                         61
       Half-life

 Sulfuryl chloride, SO2Cl2, decomposes in a
  first-order reaction to SO2 and Cl2.
      SO 2Cl 2 (g )  SO 2 (g )  Cl 2 (g )
      – At 320 oC, the rate constant is 2.20 x 10-5 s-1.
      What is the half-life of SO2Cl2 vapor at this
      temperature?
      – Substitute the value of k into the relationship
      between k and t1/2.
                  t  3.15  10 s
                    1
                    2
                                       4

                                                       62
        Half-life

 For a second-order reaction, half-life
  depends on the initial concentration and
  becomes larger as time goes on.
    – Again, assuming that [A]t = ½[A]o after one half-life, it
      can be shown that:
                           1
                    t 
                      1
                      2
                        k[ A]o
    – Each succeeding half-life is twice the length of its
      predecessor.
                                                         63
       Half-life

 For Zero-Order reactions, the half-lite is
  dependent upon the initial concentration of
  the reactant and becomes shorter as the
  reaction proceeds.
                   [ A ]o
              t1 
               2
                     2k

                                                64
        Half-life

 First order reactions
       ln( 1 / 2)  .693  kt1
                                              2
 Second order reactions
              1
       t 
        1
        2
           k[ A]o
 Zero order reactions
             [ A]o
        t1               ln( 1 )  kt
                              2           1
                                          2




         2
               2k                                 65
        Graphing Kinetic Data

 In addition to the method of initial rates,
  rate laws can be deduced by graphical
  methods.
    – If we rewrite the first-order concentration-
      time equation in a slightly different form, it
      can be identified as the equation of a straight
      line.
           ln[ A]t  kt  ln[ A]o
              y    = mx + b

                                                66
        Graphing Kinetic Data

 In addition to the method of initial rates,
  rate laws can be deduced by graphical
  methods.
     – If we rewrite the first-order concentration-time
       equation in a slightly different form, it can be
       identified as the equation of a straight line.
     – This means if you plot ln[A] versus time, you will
       get a straight line for a first-order reaction. (see
       Figure 14.9)


                                                          67
        Graphing Kinetic Data

 In addition to the method of initial rates,
  rate laws can be deduced by graphical
  methods.
     – If we rewrite the second-order concentration-time
       equation in a slightly different form, it can be identified
       as the equation of a straight line.
                  1            1
                       kt 
                [ A]t        [ A]o
                   y = mx + b
                                                          68
        Graphing Kinetic Data

 In addition to the method of initial rates,
  rate laws can be deduced by graphical
  methods.
     – If we rewrite the second-order concentration-time
       equation in a slightly different form, it can be identified
       as the equation of a straight line.
     – This means if you plot 1/[A] versus time, you will
       get a straight line for a second-order reaction.

     – Figure 14.10 illustrates the graphical method of
       deducing the order of a reaction.
                                                          69
70
71
       Collision Theory

 Rate constants vary with temperature.
  Consequently, the actual rate of a reaction is
  very temperature dependent.
     • Why the rate depends on temperature
       can by explained by collision theory.




                                               72
       Collision Theory

 Collision theory assumes that for a reaction
  to occur, reactant molecules must collide
  with sufficient energy and the proper
  orientation.
      • The minimum energy of collision
        required for two molecules to react is
        called the activation energy, Ea.


                                             73
       Transition-State Theory

 Transition-state theory explains the
  reaction resulting from the collision of two
  molecules in terms of an activated
  complex.
  – An activated complex (transition state) is
    an unstable grouping of atoms that can
    break up to form products.
  – A simple analogy would be the collision of
    three billiard balls on a billiard table.
                                                 74
       Transition-State Theory

 Transition-state theory explains the
  reaction resulting from the collision of two
  molecules in terms of an activated
  complex.
   – Suppose two balls are coated with a
     slightly stick adhesive.
   – We’ll take a third ball covered with an
     extremely sticky adhesive and collide it
     with our joined pair.
                                                 75
       Transition-State Theory

 Transition-state theory explains the
  reaction resulting from the collision of two
  molecules in terms of an activated
  complex.
    – At the instant of impact, when all three spheres
      are joined, we have an unstable transition-state
      complex.
    – The “incoming” billiard ball would likely stick to
      one of the joined spheres and provide sufficient
      energy to dislodge the other, resulting in a new
      “pairing.”
                                                     76
       Transition-State Theory

 Transition-state theory explains the
  reaction resulting from the collision of two
  molecules in terms of an activated
  complex.
    – If we repeated this scenario several times, some
      collisions would be successful and others
      (because of either insufficient energy or improper
      orientation) would not be successful.
    – We could compare the energy we provided to the
      billiard balls to the activation energy, Ea.
                                                    77
        Potential-Energy Diagrams for
        Reactions
 To illustrate graphically the formation of a
  transition state, we can plot the potential
  energy of a reaction versus time.
    – Figure 14.13 illustrates the endothermic reaction of
      nitric oxide and chlorine gas.
    – Note that the forward activation energy is the
      energy necessary to form the activated complex.
    – The DH of the reaction is the net change in energy
      between reactants and products.

                                                      78
Figure 14.13 Potential-energy curve for the endothermic reaction of nitric
oxide and chlorine.
       Potential-Energy Diagrams for
       Reactions
 The potential-energy diagram for an
  exothermic reaction shows that the products
  are more stable than the reactants.
     – Figure 14.14 illustrates the potential-energy
       diagram for an exothermic reaction.
     – We see again that the forward activation energy is
       required to form the transition-state complex.
     – In both of these graphs, the reverse reaction must
       still supply enough activation energy to form the
       activated complex.
                                                   80
Figure 14.14 Potential-energy curve for an exothermic reaction.
       Collision Theory and the
       Arrhenius Equation
 Collision theory maintains that the rate
  constant for a reaction is the product of
  three factors.
    1. Z, the collision frequency
    2. f, the fraction of collisions with sufficient energy to
       react
    3. p, the fraction of collisions with the proper
       orientation to react

                   k  Zpf
                                                         82
       Collision Theory and the
       Arrhenius Equation
 Z is only slightly temperature dependent.

    – This is illustrated using the kinetic theory
      of gases, which shows the relationship
      between the velocity of gas molecules and
      their absolute temperature.
           3RTabs
velocity                or    velocity  Tabs
            Mm
                                               83
       Collision Theory and the
       Arrhenius Equation
 Z is only slightly temperature dependent.

    – This alone does not account for the
      observed increases in rates with only
      small increases in temperature.
    – From kinetic theory, it can be shown
      that a 10 oC rise in temperature will
      produce only a 2% rise in collision
      frequency.

                                              84
       Collision Theory and the
       Arrhenius Equation
 On the other hand, f, the fraction of
  molecules with sufficient activation energy,
  turns out to be very temperature dependent.
       – It can be shown that f is related to Ea by the
         following expression.


                       f e
                                  - Ea
                                  RT

        – Here e = 2.718… , and R is the ideal gas
          constant, 8.31 J/(mol.K).
                                                     85
       Collision Theory and the
       Arrhenius Equation
 On the other hand, f, the fraction of
  molecules with sufficient activation energy
  turns out to be very temperature dependent.
      – From this relationship, as temperature
        increases, f increases.

                 f e
                           - Ea
                           RT

         Also, a decrease in the activation energy, Ea,
         increases the value of f.
                                                   86
       Collision Theory and the
       Arrhenius Equation
 On the other hand, f, the fraction of
  molecules with sufficient activation energy
  turns out to be very temperature dependent.

    – This is the primary factor relating
      temperature increases to observed rate
      increases.

                  f e
                           - Ea
                           RT


                                                87
       Collision Theory and the
       Arrhenius Equation
 The reaction rate also depends on p, the
  fraction of collisions with the proper
  orientation.
    – This factor is independent of temperature
      changes.
    – So, with changes in temperature, Z and p
      remain fairly constant.
    – We can use that fact to derive a mathematical
      relationship between the rate constant, k, and
      the absolute temperature.

                                                       88
       The Arrhenius Equation

 If we were to combine the relatively
  constant terms, Z and p, into one constant,
  let’s call it A. We obtain the Arrhenius
  equation:

               k  Ae
                                - Ea
                                RT


   – The Arrhenius equation expresses the
     dependence of the rate constant on absolute
     temperature and activation energy.

                                                   89
       The Arrhenius Equation

 It is useful to recast the Arrhenius equation
  in logarithmic form.
   – Taking the natural logarithm of both sides
     of the equation, we get:


           ln k  ln A -     Ea
                             RT



                                                  90
       The Arrhenius Equation

 It is useful to recast the Arrhenius equation
  in logarithmic form.
       We can relate this equation to the (somewhat
       rearranged) general formula for a straight line.
             ln k  ln           Ea 1
                             A - R (T)
                y =        b      +mx
       A plot of ln k versus (1/T) should yield a straight line
       with a slope of (-Ea/R) and an intercept of ln A.
       (see Figure 14.15)
                                                          91
        The Arrhenius Equation

 A more useful form of the equation emerges if we
  look at two points on the line this equation
  describes that is, (k1, (1/T1)) and (k2, (1/T2)).
   – The two equations describing the relationship at
     each coordinate would be

         ln k 1  ln          Ea 1
                          A - R ( T1 )
                      and
         ln k 2  ln           Ea 1
                           A - R ( T2 )
                                                        92
        The Arrhenius Equation

 A more useful form of the equation emerges if we
  look at two points on the line this equation
  describes that is, (k1, (1/T1)) and (k2, (1/T2)).
   – We can eliminate ln A by subtracting the two
     equations to obtain




            ln   k2
                 k1      Ea 1
                          R ( T1       1)
                                        T2
                                                      93
        The Arrhenius Equation

 A more useful form of the equation emerges if we
  look at two points on the line this equation
  describes that is, (k1, (1/T1)) and (k2, (1/T2)).
    – With this form of the equation, given the
      activation energy and the rate constant k1 at a
      given temperature T1, we can find the rate
      constant k2 at any other temperature, T2.


            ln   k2
                 k1      Ea 1
                          R ( T1       1)
                                        T2
                                                        94
        A Problem to Consider

 The rate constant for the formation of
  hydrogen iodide from its elements
           H 2 (g )  I 2 (g )  2HI(g )
       is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3
       L/(mol.s) at 650 K. Find the activation energy, Ea.
    – Substitute the given data into the Arrhenius
      equation.

       3.5  10-3
    ln
       2.7  10 4
                   
                            Ea
                                     (  1
                                          
                                             1
                     8.31 J/(mol  K) 600K 650K
                                                       )
                                                       95
        A Problem to Consider

 The rate constant for the formation of
  hydrogen iodide from its elements
           H 2 (g )  I 2 (g )  2HI(g )
        is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3
        L/(mol.s) at 650 K. Find the activation energy, Ea.

     – Simplifying, we get:
                              Ea                  4
ln (1.30  10 )  1.11 
            1
                                       (1.28  10 )
                         8.31 J/(mol)

                                                        96
        A Problem to Consider

 The rate constant for the formation of
  hydrogen iodide from its elements
          H 2 (g )  I 2 (g )  2HI(g )
       is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3
       L/(mol.s) at 650 K. Find the activation energy, Ea.

   – Solving for Ea:

           1.11  8.31J / mol
      Ea                      1.66  105 J
              1.28  104
                                                       97
      Reaction Mechanisms

 Even though a balanced chemical equation may
  give the ultimate result of a reaction, what actually
  happens in the reaction may take place in several
  steps.
  – This “pathway” the reaction takes is referred to
    as the reaction mechanism.
  – The individual steps in the larger overall reaction
    are referred to as elementary reactions. (See
    animation: Decomposition of N2O5 Step 1)


                                                     98
           Elementary Reactions

    Consider the reaction of nitrogen dioxide
     with carbon monoxide.
      NO2 (g )  CO(g )  NO(g )  CO2 (g )
      – This reaction is believed to take place in
        two steps.
NO2 (g )  NO2 (g )  NO3 (g )  NO(g )   (elementary reaction)

NO3 (g )  CO(g )  NO2 (g )  CO2 (g )   (elementary reaction)

                                                         99
       Elementary Reactions

 Each step is a singular molecular event
  resulting in the formation of products.
   – Note that NO3 does not appear in the
     overall equation, but is formed as a
     temporary reaction intermediate.




                                            100
                Elementary Reactions

      Each step is a singular molecular event
        resulting in the formation of products.
           – The overall chemical equation is obtained by
             adding the two steps together and canceling any
             species common to both sides.
                NO2 (g )  NO2 (g )  NO3 (g )  NO(g )

                NO3 (g )  CO(g )  NO2 (g )  CO2 (g )

NO2 (g )  NO2 (g )  NO3 (g )  CO(g )  NO3 (g )  NO(g )  NO2 (g )  CO2 (g )


                                                                        101
       Molecularity

 We can classify reactions according to their
  molecularity, that is, the number of
  molecules that must collide for the
  elementary reaction to occur.
  – A unimolecular reaction involves only one
    reactant molecule.
  – A bimolecular reaction involves the collision of
    two reactant molecules.
  – A termolecular reaction requires the collision of
    three reactant molecules.
                                                    102
       Molecularity

 We can classify reactions according to their
  molecularity, that is, the number of
  molecules that must collide for the
  elementary reaction to occur.
  – Higher molecularities are rare because of
    the small statistical probability that four or
    more molecules would all collide at the
    same instant.


                                                 103
       Rate Equations for
       Elementary Reactions
 Since a chemical reaction may occur in
  several steps, there is no easily stated
  relationship between its overall reaction and
  its rate law.
    • For elementary reactions, the rate is
       proportional to the concentrations of all
       reactant molecules involved.


                                            104
       Rate Equations for
       Elementary Reactions
 For example, consider the generic equation
  below.
                A  products
      The rate is dependent only on the
      concentration of A; that is,

                  Rate  k[A]


                                           105
       Rate Equations for
       Elementary Reactions
 However, for the reaction
             A  B  products
        the rate is dependent on the
        concentrations of both A and B.


             Rate  k[A][B]


                                          106
       Rate Equations for
       Elementary Reactions
 For a termolecular reaction
          A  B  C  products
     the rate is dependent on the populations of all
     three participants.

              Rate  k[A][B][C]



                                                       107
       Rate Equations for
       Elementary Reactions
 Note that if two molecules of a given
  reactant are required, it appears twice in the
  rate law. For example, the reaction
            2A  B  products
        would have the rate law:

Rate  k[A][A][B] or Rate  k[A] [B]     2



                                               108
        Rate Equations for
        Elementary Reactions
 So, in essence, for an elementary reaction, the
  coefficient of each reactant becomes the power
  to which it is raised in the rate law for that
  reaction.
     – Note that many chemical reactions occur in
       multiple steps and it is, therefore,
       impossible to predict the rate law based
       solely on the overall reaction.



                                                    109
       Rate Laws and Mechanisms

 Consider the reaction below.
     2 NO2 (g)  F2 (g)  2 NO2F(g)
  – Experiments performed with this reaction
    show that the rate law is
          Rate  k[NO2 ][F2 ]
  – The reaction is first order with respect to
    each reactant, even though the coefficient
    for NO2 in the overall reaction is 2.
                                             110
       Rate Laws and Mechanisms

 Consider the reaction below.
     2 NO2 (g)  F2 (g)  2 NO2F(g)
  – Experiments performed with this reaction show that
    the rate law is

           Rate  k[NO2 ][F2 ]
  – This implies that the reaction above is not an
    elementary reaction but rather the result of
    multiple steps.

                                                     111
       Rate-Determining Step

 In multiple-step reactions, one of the
  elementary reactions in the sequence is
  often slower than the rest.
   – The overall reaction cannot proceed any
     faster than this slowest rate-determining
     step.




                                             112
        Rate-Determining Step

 In multiple-step reactions, one of the
  elementary reactions in the sequence is
  often slower than the rest.
   – Our previous example occurs in two elementary
     steps where the first step is much slower.
                          k
                          
        NO 2 (g)  F2 (g) 1 NO 2F(g)  F(g) (slow)
                       k2
         NO 2 (g)  F(g)  NO 2F(g)           (fast)

       2 NO2 (g)  F2 (g)  2 NO2F(g)

                                                   113
        Rate-Determining Step

 In multiple-step reactions, one of the
  elementary reactions in the sequence is
  often slower than the rest.
   – Since the overall rate of this reaction is determined
     by the slow step, it seems logical that the observed
     rate law is Rate = k1[NO2][F2].
                          k1
                          
        NO 2 (g)  F2 (g)  NO 2F(g)  F(g) (slow)



                                                      114
       Rate-Determining Step

 In a mechanism where the first elementary
  step is the rate-determining step, the overall
  rate law is simply expressed as the
  elementary rate law for that slow step.

   – A more complicated scenario occurs
     when the rate-determining step contains
     a reaction intermediate, as you’ll see in
     the next section.
                                              115
       Rate-Determining Step

 Mechanisms with an Initial Fast Step
   – There are cases where the rate-
     determining step of a mechanism contains
     a reaction intermediate that does not
     appear in the overall reaction.
   – The experimental rate law, however, can
     be expressed only in terms of
     substances that appear in the overall
     reaction.

                                          116
Rate-Determining Step
  Consider the reduction of nitric oxide with H2.
      2NO(g )  2H 2 (g )  N 2 (g )  2H 2O(g )
   – A proposed mechanism is:
                     k1
            2NO            N 2O 2          (fast, equilibrium)
                     k-1

                     N 2O  H 2O
      N 2O 2  H 2 k  2                   (slow)

                  
       N 2O  H 2  N 2  H 2O
                     k3                    (fast)

   – It has been experimentally determined that the rate
     law is Rate = k [NO]2[H2]
                                                      117
        Rate-Determining Step

 The rate-determining step (step 2 in this
  case) generally outlines the rate law for the
  overall reaction.
            Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)
   – As mentioned earlier, the overall rate law can be
     expressed only in terms of substances
     represented in the overall reaction and cannot
     contain reaction intermediates.

                                                    118
       Rate-Determining Step

 The rate-determining step (step 2 in this
  case) generally outlines the rate law for the
  overall reaction.
            Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)

 – It is necessary to reexpress this proposed rate law
   after eliminating [N2O2].



                                                    119
       Rate-Determining Step

 The rate-determining step (step 2 in this
  case) generally outlines the rate law for the
  overall reaction.
            Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)

 – We can do this by looking at the first step,
   which is fast and establishes equilibrium.


                                                  120
       Rate-Determining Step

 The rate-determining step (step 2 in this
  case) generally outlines the rate law for the
  overall reaction.
            Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)

 – At equilibrium, the forward rate and the
   reverse rate are equal.
            k 1[NO]  k 1[N 2O 2 ]
                     2


                                                 121
       Rate-Determining Step

 The rate-determining step (step 2 in this
  case) generally outlines the rate law for the
  overall reaction.
            Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)
   – Therefore,
         [N 2O 2 ]  (k 1 / k 1 )[NO]  2

 – If we substitute this into our proposed rate
   law we obtain:
                                                 122
        Rate-Determining Step

 The rate-determining step (step 2 in this case)
  generally outlines the rate law for the overall
  reaction.
             Rate  k 2 [N 2O 2 ][H 2 ]
      (Rate law for the rate-determining step)
                   k 2k 1
            Rate             2
                          [NO] [H 2 ]
                    k 1
       – If we replace the constants (k2k1/k-1)
         with k, we obtain the observed rate law:
         Rate = k[NO]2[H2].
                                                    123
       Rate-Determining Step

 Determine the overall reaction and the rate
  expression that corresponds to the following
  reaction mechanism.
                 2A + B ↔ D
               D+B→E+F
                    F→G
                                     Whitten 6th Ed. Ch 16 # 66a




                                                           124
       Rate-Determining Step

 Determine the overall reaction and the rate
  expression that corresponds to the following
  reaction mechanism.
                 2A + B ↔ D
               D+B→E+F
                    F→G
 Answer:
  2A + 2B → E + G
  Rate = k[A]2[B]2
                                                125
       Catalysts

 A catalyst is a substance that provides a
  good “environment” for a reaction to occur,
  thereby increasing the reaction rate without
  being consumed by the reaction.

  – To avoid being consumed, the catalyst must
    participate in at least one step of the
    reaction and then be regenerated in a later
    step.
                                              126
       Catalysts

 A catalyst is a substance that provides a
  good “environment” for a reaction to occur,
  thereby increasing the reaction rate without
  being consumed by the reaction.
   – Its presence increases the rate of
     reaction by either increasing the
     frequency factor, A (from the Arrhenius
     equation) or lowering the activation
     energy, Ea.
                                               127
       Catalysts

 Homogeneous catalysis is the use of a
 catalyst in the same phase as the reacting
 species.
   – The oxidation of sulfur dioxide using
     nitric oxide as a catalyst is an example
     where all species are in the gas phase.

                             
      2SO 2 (g )  O 2 (g )   2SO 3 (g )
                          NO ( g )



                                                128
       Catalysts

 Heterogeneous catalysis is the use of a
  catalyst that exists in a different phase from
  the reacting species, usually a solid catalyst
  in contact with a liquid or gaseous solution
  of reactants.
   – Such surface catalysis is thought to occur by
     chemical adsorbtion of the reactants onto the
     surface of the catalyst.
   – Adsorbtion is the attraction of molecules to a
     surface.
                                                      129
      Enzyme Catalysis

 Enzymes have enormous catalytic
 activity.
   – The substance whose reaction the
     enzyme catalyzes is called the
     substrate. (see Figure 14.20)
   – Figure 14.21 illustrates the reduction in
     acivation energy resulting from the
     formation of an enzyme-substrate
     complex.

                                                 130
        Operational Skills
 Relating the different ways of expressing reaction
  rates
 Calculating the average reaction rate
 Determining the order of reaction from the rate
  law
 Determining the rate law from initial rates
 Using the concentration-time equation for first-
  order reactions
 Relating the half-life of a reaction to the rate
  constant
                                                     131
        Operational Skills

 Using the Arrhenius equation
• Writing the overall chemical equation from a
  mechanism
• Determining the molecularity of an elementary
  reaction
• Writing the rate equation for an elementary
  reaction
• Determining the rate law from a mechanism


                                             132
Figure 14.1:
Combining
product of
formaldehyde
with hydrogen
sulfite ion.
Photo courtesy of
James Scherer.




Return to Slide 2
                    133
Figure 14.9: A plot of ln [N2O5]
versus time.




       Return to Slide 58
                                   134
Figure
14.15: Plot
of ln k
versus 1/T.




Return to Slide 80
                     135
  Animation: Decomposition of
  N2O5



(Click here to open QuickTime animation)




           Return to Slide 87
                                           136
Figure 14.20: Enzyme action
(lock-and-key model).




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                               137
Figure 14.21 Potential-energy curves
for the reaction of substrate, S, to
products, P.




      Return to Slide 117
                               138

								
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