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Chemical Kinetics Chapter 15

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					                                     1
Chemical Kinetics
        Chapter 15




  An automotive catalytic muffler.
                                              2
        Chemical Kinetics
                  Chapter 15
• We can use thermodynamics to tell if a
  reaction is product or reactant favored.
• But this gives us no info on HOW FAST
  reaction goes from reactants to products.
• KINETICS — the study of REACTION
 RATES and their relation to the way the
 reaction proceeds, i.e., its MECHANISM.
                                             3

   Thermodynamics or
        Kinetics
• Reactions can be one of the
  following:
1. Not thermodynamically favored
   (reactant favored)
2. Thermodynamically favored (product
   favored), but not kinetically favored
   (slow)
3. Thermodynamically favored (product
   favored) and kinetically favored (fast)
                              4

Thermodynamics or
     Kinetics
        • Not
          thermodynamically
          favored (reactant
          favored)
        • Sand (SiO2) will
          not decompose
          into Si and O2
                                5

Thermodynamics or
     Kinetics
        • Thermodynamically
          favored (product
          favored), but not
          kinetically favored
          (slow)
        • Diamonds will turn into
          graphite, but the
          reaction occurs very
          slowly
                                6


Thermodynamics or
     Kinetics
        • Thermodynamic
          ally favored
          (product
          favored) and
          kinetically
          favored (fast)
        • Burning of paper
          in air will turn to
          ash very quickly
                                                7

       Reaction Mechanisms
The sequence of events at the molecular
 level that control the speed and
 outcome of a reaction.
Br from biomass burning destroys
 stratospheric ozone. (See R.J. Cicerone,
 Science, volume 263, page 1243, 1994.)
Step 1:   Br + O3 ---> BrO + O2
Step 2:   Cl + O3 ---> ClO + O2
Step 3:   BrO + ClO + light ---> Br + Cl + O2
NET:      2 O3 ---> 3 O2
                                   8

       Reaction Rates
           Section 15.1
• Reaction rate = change in
  concentration of a reactant or
  product with time.
• Know about initial rate,
  average rate, and
  instantaneous rate. See
  Screen 15.2.
9
                                               10

           Determining a Reaction Rate
                          Blue dye is oxidized
                            with bleach.
                          Its concentration
                            decreases with time.
                          The rate — the
                            change in dye conc
                            with time — can be
Dye Conc




                            determined from the
                            plot.


                Time
                                11

 Factors Affecting Rates
          Section 15.2
• Concentrations and physical
  state of reactants and
  products (Screens 15.3 and
  15.4)
• Temperature (Screen 15.11)
• Catalysts (Screen 15.14)
                                              12

  Factors Affecting Rates
             Section 15.2
• Concentrations            Rate with 0.3 M HCl




                            Rate with 6.0 M HCl
                                13

  Factors Affecting Rates
              Section 15.2
• Physical state of reactants
                                      14

  Factors Affecting Rates
             Section 15.2
Catalysts: catalyzed decomp of H2O2
         2 H2O2 --> 2 H2O + O2
                            15

  Factors Affecting Rates
            Section 15.2
• Temperature
                                 16


Concentrations and Rates
To postulate a reaction
 mechanism, we study
• reaction rate and
• its concentration dependence
                                             17

 Concentrations and Rates
Take reaction
 where Cl- in     Cisplatin

 cisplatin
 [Pt(NH3)2Cl3]
 is replaced
 by H2O


     Rate of change of conc of Pt compd
        Am't of cisplatin reacting (mol/L)
      =
                 elapsed time (t)
                                                      18
  Cisplatin




                  Concentrations and Rates
              Rate of change of conc of Pt compd
                 Am't of cisplatin reacting (mol/L)
               =
                          elapsed time (t)

Rate of reaction is proportional to [Pt(NH3)2Cl2]
We express this as a RATE LAW
Rate of reaction = k [Pt(NH3)2Cl2]
where k = rate constant
k is independent of conc. but increases with T
                                        19
  Cisplatin

              Concentrations, Rates,
              and Rate Laws
In general, for
 a A + b B ---> x X with a catalyst C
Rate = k [A]m[B]n[C]p
The exponents m, n, and p
• are the reaction order
• can be 0, 1, 2 or fractions
• must be determined by experiment!!!
  Cisplatin                                          20
                  Interpreting Rate Laws
Rate = k [A]m[B]n[C]p
• If m = 1, rxn. is 1st order in A

              Rate = k [A]1
  If [A] doubles, then rate goes up by factor of ?
• If m = 2, rxn. is 2nd order in A.

              Rate = k [A]2
  Doubling [A] increases rate by ?
• If m = 0, rxn. is zero order.
              Rate = k [A]0
  If [A] doubles, rate ?
Cisplatin                                               21
                 Deriving Rate Laws
Derive rate law and k for
            CH3CHO(g) ---> CH4(g) + CO(g)
from experimental data for rate of disappearance
  of CH3CHO
Expt.               [CH3CHO]      Disappear of CH3CHO
                    (mol/L)            (mol/L•sec)
1                   0.10               0.020
2                   0.20               0.081
3                   0.30               0.182
4                   0.40               0.318
Cisplatin                                         22
              Deriving Rate Laws

Rate of rxn = k [CH3CHO]2
Here the rate goes up by ______ when initial
 conc. doubles. Therefore, we say this reaction
 is _________________ order.
Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = k (0.30 mol/L)2
            k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of
 [CH3CHO] at same T.
 Cisplatin
               Concentration/Time           23


                   Relations

Need to know what conc. of reactant is as
 function of time. Consider FIRST ORDER
 REACTIONS
For 1st order reactions, the rate law is
           - (D [A] / D time) = k [A]
 Cisplatin
               Concentration/Time                24


                   Relations

Integrating - (D [A] / D time) = k [A], we get
 Cisplatin
                        Concentration/Time               25


                            Relations

Integrating - (D [A] / D time) = k [A], we get

                            [A] = - k t
                         ln
                            [A]o
                 natural
                 logarithm            [A] at time = 0

             [A] / [A]0 =fraction remaining after time
             t has elapsed.
 Cisplatin
                        Concentration/Time               26


                            Relations

Integrating - (D [A] / D time) = k [A], we get

                            [A] = - k t
                         ln
                            [A]o
                 natural
                 logarithm            [A] at time = 0

             [A] / [A]0 =fraction remaining after time
             t has elapsed.
       Called the integrated first-order rate law.
                                      27
       Concentration/Time Relations
Sucrose               Sucrose
 decomposes to
 simpler sugars
Rate of
 disappearance of
 sucrose
 = k [sucrose]
k = 0.21 hr-1
Initial [sucrose] =
  0.010 M
How long to drop
 90% (to 0.0010 M)?
                                                                     28
     Concentration/Time Relationships
     Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
     initial [sucrose] = 0.010 M, how long to drop 90% or to
     0.0010 M?

Use the first order integrated rate law
                                                                     29
     Concentration/Time Relationships
     Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
     initial [sucrose] = 0.010 M, how long to drop 90% or to
     0.0010 M?

Use the first order integrated rate law
                                                                      30
      Concentration/Time Relationships
      Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
      initial [sucrose] = 0.010 M, how long to drop 90% or to
      0.0010 M?

Use the first order integrated rate law




ln (0.100) = - 2.3 = - (0.21 hr-1) • time
                                                                      31
      Concentration/Time Relationships
      Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
      initial [sucrose] = 0.010 M, how long to drop 90% or to
      0.0010 M?

Use the first order integrated rate law




ln (0.100) = - 2.3 = - (0.21 hr-1) • time
time = 11 hours
                                                    32
    Using the Integrated Rate Law
The integrated rate law suggests a way to tell if
  a reaction is first order based on experiment.
2 N2O5(g) ---> 4 NO2(g) + O2(g)
  Rate = k [N2O5]

Time (min)      [N2O5]0 (M)      ln [N2O5]0
  0             1.00                   0
  1.0           0.705                  -0.35
  2.0           0.497                  -0.70
  5.0           0.173                  -1.75
                                                  33
    Using the Integrated Rate Law
2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5]




Data of conc. vs.
time plot do not fit
straight line.
                                                   34
    Using the Integrated Rate Law
2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5]




Data of conc. vs.          Plot of ln [N2O5] vs.
time plot do not fit       time is a straight
straight line.             line!
                                                        35
    Using the Integrated Rate Law
                      Plot of ln [N2O5] vs. time
                      is a straight line!
                      Eqn. for straight line:
                            y = ax + b
                      ln [N 2O5] = - kt + ln [N 2O5]o

                      conc at   rate const   conc at
                      time t    = slope      time = 0

All 1st order reactions have straight line plot
  for ln [A] vs. time.
(2nd order gives straight line for plot of 1/[A]
  vs. time)
                                     36

     Graphical Methods for
   Determining Reaction Order
       and Rate Constant
• Zero-order


 [ A]t  [ A]o  kt
  y  b mx
• Straight line plot of [A]t vs. t
• Slope is –k (mol/L*s)
                                       37

     Graphical Methods for
   Determining Reaction Order
       and Rate Constant
• First-order


ln[ A]t  ln[ A]o  kt
y        b mx
• Straight line plot of ln[A]t vs. t
• Slope is – k (s-1)
                                      38

    Graphical Methods for
  Determining Reaction Order
      and Rate Constant
• Second-order

   1     1
            kt
 [ A] [ A]o
     t
 y  b mx
• Straight line plot of 1/[A]t vs t
• Slope is k (L/mol*s)
                               39


       Half-Life
Section 15.4 and Screen 15.8

HALF-LIFE is the time it
 takes for 1/2 a sample
 is disappear.
For 1st order reactions,
 the concept of HALF-
 LIFE is especially
 useful.
                          40



    Half-Life

• Reaction is 1st order
  decomposition of
  H2O2.
                        41



    Half-Life

• Reaction after 654
  min, 1 half-life.
• 1/2 of the reactant
  remains.
                        42



    Half-Life

• Reaction after
  1306 min, or 2
  half-lives.
• 1/4 of the reactant
  remains.
                        43



    Half-Life

• Reaction after 3
  half-lives, or 1962
  min.
• 1/8 of the reactant
  remains.
                        44



     Half-Life

• Reaction after 4
  half-lives, or 2616
  min.
• 1/16 of the
  reactant remains.
                                                   45
                  Half-Life
           Section 15.4 and Screen 15.8

Sugar is fermented in a 1st order process (using
 an enzyme as a catalyst).
           sugar + enzyme --> products
Rate of disappear of sugar = k[sugar]
           k = 3.3 x 10-4 sec-1
What is the half-life of this reaction?
                                                              46
                      Half-Life
             Section 15.4 and Screen 15.8

Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-
life of this reaction?
Solution
[A] / [A]0 = 1/2 when t = t1/2
Therefore, ln (1/2) = - k • t1/2
             - 0.693 = - k • t1/2

             t1/2 = 0.693 / k
So, for sugar,
t1/2 = 0.693 / k = 2100 sec = 35        min
                                                      47
                   Half-Life
            Section 15.4 and Screen 15.8

Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life
is 35 min. Start with 5.00 g sugar. How much is
left after 2 hr and 20 min?
 Solution
 2 hr and 20 min = 4 half-lives
 Half-life Time Elapsed Mass Left
 1st         35 min           5.00 g
 2nd         70               2.50 g
 3rd         105              1.25 g
 4th         140              0.625 g
                                                   48
                  Half-Life
           Section 15.4 and Screen 15.8

Radioactive decay is a first order process.
      Tritium ---> electron + helium
      3H               0 e        3He
                        -1
If you have 1.50 mg of tritium, how much is left
after 49.2 years? t1/2 = 12.3 years
Solution
                      Half-Life                              49

              Section 15.4 and Screen 15.8
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
ln [A] / [A]0 = -kt
[A] = ?      [A]0 = 1.50 mg        t = 49.2 years
Need k, so we calc k from:         k = 0.693 / t1/2
Obtain k = 0.0564 y-1
Now ln [A] / [A]0 = -kt = - (0.0564 y-1) • (49.2 y)
                    = - 2.77
Take antilog: [A] / [A]0 = e-2.77 = 0.0627
0.0627 is the fraction remaining!
                      Half-Life                              50

              Section 15.4 and Screen 15.8
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
[A] / [A]0 = 0.0627
0.0627 is the fraction remaining!
Because [A]0 = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 half-lives
 1.50 mg ---> 0.750 mg after 1
       ---> 0.375 mg after 2
       ---> 0.188 mg after 3
       ---> 0.094 mg after 4
                                                 51

 Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in
  terms of 1/2-life.
238U --> 234Th + He              4.5 x 109 y
14C --> 14N + beta               5730 y
131I --> 131Xe + beta            8.05 d
Element 106 - seaborgium
  263Sg                         0.9 s

				
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