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Chemistry 142 Chapter 13 Chemica

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					       Chemistry 142
Chapter 13: Chemical Kinetics
            Chapter 13 Examples – Rate


                                                          at t = 0
at t = 0
                                                          [X] = 8
[A] = 8
                                                           [Y] = 8
[B] = 8
                                                           [Z] = 0
[C] = 0


at t = 16                                                 at t = 16
[A] = 4                                                     [X] = 7
[B] = 4                                                     [Y] = 7
[C] = 4                                                     [Z] = 1




                   Tro, Chemistry: A Molecular Approach               2
Concentration as a function of time
   2 N2O5 (g)  4 NO2 (g) + O2 (g)
                                Continuous Monitoring




                                       Sampling




                                                        5
Tro, Chemistry: A Molecular Approach
                         First Order Reactions
                      ln[A]0




                   ln[A]




                                 time
Tro, Chemistry: A Molecular
                                   6
Approach
                                        Rate Data for
         C4H9Cl               (aq) +   H2O (l)  C4H9OH (aq) + HCl (aq)
                               Time (sec) [C4H9Cl], M
                                        0.0       0.1000
                                          50.0        0.0905
                                         100.0        0.0820
                                         150.0        0.0741
                                         200.0        0.0671
                                         300.0        0.0549
                                         400.0        0.0448
                                         500.0        0.0368
                                         800.0        0.0200
                                       10000.0        0.0000
Tro, Chemistry: A Molecular
                                                 7
Approach
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl (aq)
                                          Concentration vs. Time for the Hydrolysis of C 4H9Cl
                               0.12


                                0.1


                               0.08
          concentration, (M)




                               0.06


                               0.04


                               0.02


                                 0
                                      0           200          400                   600   800   1000
                                                                         time, (s)
Tro, Chemistry: A Molecular
                                                                     8
Approach
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl
                                                    (aq)
                                     Rate vs. Time for Hydrolysis of C 4H9Cl

                       2.5E-04



                       2.0E-04



                       1.5E-04
         Rate, (M/s)




                       1.0E-04



                       5.0E-05



                       0.0E+00
                                 0   100    200     300         400       500   600   700   800
                                                              time, (s)
Tro, Chemistry: A Molecular
                                                          9
Approach
                     C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl (aq)
                                   LN([C 4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl
                      0

                    -0.5                                                                                       slope =
                                                                                                               -2.01 x 10-3
                     -1
                                                                                                               k=
                    -1.5                                                                                       2.01 x 10-3 s-1
LN(concentration)




                     -2

                    -2.5
                                                                                                                       0.693
                     -3
                                                                                                               t1 
                                                                             y = -2.01E-03x - 2.30E+00             2
                                                                                                                         k
                    -3.5
                                                                                                                        0.693
                                                                                                               
                     -4
                                                                                                                 2.01 10 3 s -1
                    -4.5                                                                                        345 s
                           0       100         200   300     400       500        600       700          800
                                                           time, (s)

                     Tro, Chemistry: A Molecular
                                                                       10
                     Approach
     Chapter 13 – Examples
        Reaction Order

Using the data below verify that the
                                          Time (s)   [N2O5] (mol/L)
decomposition       of       dinitrogen
                                             0           0.100
pentaoxide is first order and
                                            50          0.0707
calculate the value of the rate
                                            100         0.0500
constant, where rate = -Δ[N2O5]/Δt
                                            200         0.0250
     2 N2O5 (g)  4 NO2 (g) + O2 (g)        300         0.0125
                                            400         0.00625
                        Chapter 13 – Examples
                           Reaction Order

Time (s)   [N2O5] (mol/L)                                              Concentration [N2O5] versus
   0           0.100                                                              Time
                                                                   0.120
  50          0.0707
  100         0.0500                                               0.100




                                    Concentration [N2O5] (mol/L)
  200         0.0250
                                                                   0.080

  300         0.0125
                                                                   0.060
  400         0.00625
                                                                   0.040


                                                                   0.020


                                                                   0.000
                                                                           0   100   200              300   400   500
                                                                                           time (s)
                        Chapter 13 – Examples
                           Reaction Order

Time (s)   [N2O5] (mol/L)   ln[N2O5]                          ln[N2O5] versus Time
   0           0.100        -2.303                0.000
                                                          0   100   200              300      400       500
  50          0.0707        -2.649
                                              -1.000
  100         0.0500        -2.996
  200         0.0250        -3.689            -2.000

  300         0.0125        -4.382                                              y = -0.0069x - 2.3026




                                       ln[N2O5]
  400         0.00625       -5.075            -3.000



                                              -4.000



                                              -5.000



                                              -6.000
                                                                          time (s)
Half-Life of a First-Order Reaction Is Constant




Tro, Chemistry: A Molecular
                              14
Approach
              Chapter 13 – Examples
                     Half-life
A certain first order reaction has a half-life of
20.0 minutes.
  a. Calculate the rate constant.
  b. How much time is required for this reaction to be
      75% complete?
Second Order Reactions


1/[A]



 l/[A]0



                     time
          Tro, Chemistry: A Molecular Approach   16
                     Chapter 13 – Examples
                          Rate Laws
Butadiene reacts to form its         time (s)   [C4H6] (mol/L)
  dimer according to:                    0         0.0100
                                       1000        0.00625
      2 C4H6 (g)  C8H12 (g)           1800        0.00476
                                       2800        0.00370
a) Is this reaction first order or     3600        0.00313
                                       4400        0.00270
   second order?
                                       5200        0.00241
b) What is the value of the rate       6200        0.00208
   constant function of the
   reaction?
c) What is the half-life for the
   reaction under the conditions
   of this experiment?
                                                    Chapter 13 – Examples
                                                         Rate Laws
                                time (s)              [C4H6] (mol/L)                ln[C4H6]                  1/[C4H6] (L/mol)
                                     0                   0.0100                            -4.605                   100
                                    1000                0.00625                            -5.075                   160
                                    1800                0.00476                            -5.348                   210
                                    2800                0.00370                            -5.599                   270
                                    3600                0.00313                            -5.767                   319
                                    4400                0.00270                            -5.915                   370
                                    5200                0.00241                            -6.028                   415
                                    6200                0.00208                            -6.175                   481

                       ln[C4H6] versus time                                                                   1/[C4H6] versus time
           -4.500                                                                                600
           -4.700 0   1000   2000    3000      4000     5000   6000    7000   1/[C4H6] (L/mol)
                                                                                                 500
           -4.900
                                                                                                 400
           -5.100
                                                                                                 300
ln[C4H6]




           -5.300
           -5.500                                                                                200
           -5.700
                                                                                                                                  y = 0.0612x + 99.368
                                                                                                 100
           -5.900
                                                                                                   0
           -6.100                                                                                      0   1000   2000    3000    4000   5000   6000   7000
           -6.300
                                         time (s)                                                                           time (s)
Half-life for a second order reaction
Zero Order Reactions
 [A]0




[A]




                   time

        Tro, Chemistry: A Molecular Approach   20
           Chapter 13 – Examples
             Initial rate method
Describe the general form of the rate law for
the reaction given the results of four
experiments below:
    NH4+ (aq) + NO2- (aq)  N2 (g) + 2 H2O (l)
   Experiment   [NH4+ ]0   [NO2- ]0   Initial rate (mol/L s)
       1        0.100 M    0.0050 M        1.35 x 10-7
       2        0.100 M    0.010 M         2.70 x 10-7
       3        0.200 M    0.010 M         5.40 x 10-7
           Chapter 13 – Examples
             Initial rate method
Describe the general form of the rate law for
the reaction given the results of three
experiments below:
CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + CH3OH (aq)

   Experiment   [CH3COOH ]0   [OH- ]0   Initial rate (mol/L s)
       1         0.040 M      0.040 M         0.00022
       2         0.040 M      0.080 M         0.00045
       3         0.080 M      0.080 M         0.00090
              Chapter 13 – Examples
                Initial rate method
Describe the general form of the rate law for
the reaction given the results of three
experiments below:
 CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + CH3OH (aq)

Experiment   [CH3COOH ]0   [OH- ]0   Initial rate (mol/L s)   k (L/mol s)
    1         0.040 M      0.040 M         0.00022             0.0138
    2         0.040 M      0.080 M         0.00045             0.0141
    3         0.080 M      0.080 M         0.00090             0.0141
           Chapter 13 – Examples
            Initial Rate Method
The reaction between bromate ions and bromide ions in an
acidic solution is given by the equation:
       BrO3- (aq) + 5 Br- (aq) + 6 H+ (aq)  3 Br2 (l) + 3 H2O (l)
Describe the general form of the rate law for this reaction
given the results of four experiments below.


  Experiment   [BrO3- ]0   [Br- ]0   [H+ ]0   Initial rate (mol/L s)
      1        0.10 M      0.10 M    0.10 M        8.0 x 10-4
      2        0.20 M      0.10 M    0.10 M        1.6 x 10-3
      3        0.20 M      0.20 M    0.10 M        3.2 x 10-3
      4        0.10 M      0.10 M    0.20 M        3.2 x 10-3
Tro, Chemistry: A Molecular
                              25
Approach
Tro, Chemistry: A Molecular
                              26
Approach
Tro, Chemistry: A Molecular
                              27
Approach
     Energy Profile for the
Isomerization of Methyl Isonitrile




                28
                              Effective Collisions
                              Orientation Effect




Tro, Chemistry: A Molecular
                                       29
Approach
Arrhenius Plot
                   Chapter 13 – Examples
                       Arrhenius Plot
For the reaction: 2 N2O5 (g)  4 NO2 (g) + O2 (g)
Several values of k were obtained at different temperatures.
Calculate the activation energy for the reaction.

                     T (°C)   k (s-1)
                     20       2.0 x 10-5
                     30       7.3 x 10-5
                     40       2.7 x 10-4
                     50       9.1 x 10-4
                     60       2.9 x 10-3
                               Chapter 13 – Examples
                                  Arrhenius Plot
                                   2 N2O5 (g)  4 NO2 (g) + O2 (g)
T (°C)   T (K)   1/T (1/K)     k (s-1)               ln k
20       293     3.41 x 10-3   2.0 x 10-5            -10.82
30       303     3.30 x 10-3   7.3 x 10-5            -9.53
40       313     3.19 x 10-3   2.7 x 10-4            -8.22
50       323     3.10 x 10-3   9.1 x 10-4            -7.00
60       333     3.00 x10-3    2.9 x 10-3            -5.84
                                                                              ln k versus 1/T
                                                       0
                                                       2.90E-03   3.00E-03   3.10E-03   3.20E-03    3.30E-03   3.40E-03   3.50E-03
                                                       -2

                                                       -4

                                                                                              y = -12174x + 30.694
                                            ln (k)




                                                       -6
                                                                                                   R² = 0.9998
                                                       -8

                                                      -10

                                                      -12
                                                                                        1/T (1/K)
              Chapter 13 – Examples
               Arrhenius Equation
For the gas phase reaction:
              CH4 (g) + 2 S2 (g)  CS2 (g) + 2 H2S (g)
The rate constant at 550 °C is 1.1 L/mol s and at 625 °C the
value of k is 6.4 L/mol s. Calculate the activation energy for
the reaction.
Catalysis
                         Enzyme-Substrate Binding
                         Lock and Key Mechanism




Tro, Chemistry: A Molecular
                                    35
Approach
     Enzymatic Hydrolysis of Sucrose




Tro, Chemistry: A Molecular
                              36
Approach
                              Types of Catalysts




Tro, Chemistry: A Molecular
                                      37
Approach
                              Catalytic Hydrogenation
                              H2C=CH2 + H2 → CH3CH3




Tro, Chemistry: A Molecular
                                         38
Approach
        Rate Laws of Elementary Steps




Tro, Chemistry: A Molecular
                              41
Approach
                Chapter 13 – Examples
                Mechanism Validation
The balanced equation for the reaction of nitrogen dioxide
and fluorine gases is:
                2 NO2 (g) + F2 (g)  2 NO2F (g)
the experimentally determined rate law is
                       rate = k[NO2][F2]
A suggested mechanism is
                          k1
          NO2 (g) + F2 (g)  NO2F (g) + F (g)   slow
                         k2
           F (g) + NO2 (g)  NO2F (g)           fast
Is this an acceptable mechanism?
             Chapter 13 – Examples
             Mechanism Validation
The balanced equation for the reaction of nitrogen dioxide
and fluorine gases is:
                   2 NO2 (g) + F2 (g)  2 NO2F (g)
the experimentally determined rate law is
                           rate = k[NO2][F2]
Another suggested mechanism is
          NO2 (g) + F2 (g)  NOF2 (g) + O (g)      slow
          NO2 (g) + O (g)  NO3 (g)                fast
          NOF2 (g) + NO2 (g)  NO2F (g) + NOF (g)  fast
          NO3 (g) + NOF (g)  NO2F (g) + NO2 (g)   fast
Is this an acceptable mechanism?
                Chapter 13 – Examples
                Mechanism Validation
The balanced equation for the reaction of nitrogen monoxide and
chlorine gases is:
                   2 NO (g) + Cl2 (g)  2 NOCl (g)
the experimentally determined rate law is
                       rate = k[NO]2[Cl2]
Another suggested mechanism is
                       k1
        NO (g) + Cl2 (g)  NOCl2 (g)             fast
                       k-1
                             k2
       NOCl2 (g) + NO (g)  2 NOCl (g)           slow

Is this an acceptable mechanism?

				
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