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Chemical Kinetics Chapter 15

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Chapter 15 An automotive catalytic muffler. Chemical Kinetics Chapter 15 We can use thermodynamics to tell if a reaction is product or reactant favored. – PowerPoint PPT presentation

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Title: Chemical Kinetics Chapter 15


1
Chemical KineticsChapter 15
An automotive catalytic muffler.
2
Chemical KineticsChapter 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
  • Not thermodynamically favored (reactant favored)
  • Thermodynamically favored (product favored), but
    not kinetically favored (slow)
  • 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
  • Thermodynamically 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 ---gt BrO O2
  • Step 2 Cl O3 ---gt ClO O2
  • Step 3 BrO ClO light ---gt Br Cl O2
  • NET 2 O3 ---gt 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
(No Transcript)
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 determined from the plot.

Dye Conc
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 --gt 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
    Pt(NH3)2Cl3 is replaced by H2O

18
Concentrations and Rates
  • 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
Concentrations, Rates, and Rate Laws
  • In general, for
  • a A b B ---gt x X with a catalyst C
  • Rate k AmBnCp
  • The exponents m, n, and p
  • are the reaction order
  • can be 0, 1, 2 or fractions
  • must be determined by experiment!!!

20
Interpreting Rate Laws
  • Rate k AmBnCp
  • If m 1, rxn. is 1st order in A
  • Rate k A1
  • If A doubles, then rate goes up by factor of
    ?
  • If m 2, rxn. is 2nd order in A.
  • Rate k A2
  • Doubling A increases rate by ?
  • If m 0, rxn. is zero order.
  • Rate k A0
  • If A doubles, rate ?

21
Deriving Rate Laws
  • Derive rate law and k for
  • CH3CHO(g) ---gt CH4(g) CO(g)
  • from experimental data for rate of disappearance
    of CH3CHO
  • Expt. CH3CHO Disappear of CH3CHO (mol/L) (
    mol/Lsec)
  • 1 0.10 0.020
  • 2 0.20 0.081
  • 3 0.30 0.182
  • 4 0.40 0.318

22
Deriving Rate Laws
  • Rate of rxn k CH3CHO2
  • 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/Ls k (0.30 mol/L)2
  • k 2.0 (L / mols)
  • Using k you can calc. rate at other values of
    CH3CHO at same T.

23
Concentration/Time 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

24
Concentration/Time Relations
  • Integrating - (D A / D time) k A, we get

25
Concentration/Time Relations
  • Integrating - (D A / D time) k A, we get

A / A0 fraction remaining after time t has
elapsed.
26
Concentration/Time Relations
  • Integrating - (D A / D time) k A, we get

A / A0 fraction remaining after time t has
elapsed.
Called the integrated first-order rate law.
27
Concentration/Time Relations
  • 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)?

Sucrose
28
Concentration/Time RelationshipsRate 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 RelationshipsRate 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 RelationshipsRate 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 RelationshipsRate 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) ---gt 4 NO2(g) O2(g)
  • Rate k N2O5
  • Time (min) N2O50 (M) ln N2O50
  • 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) ---gt 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) ---gt 4 NO2(g) O2(g) Rate k
    N2O5

Plot of ln N2O5 vs. time is a straight line!
Data of conc. vs. time plot do not fit straight
line.
35
Using the Integrated Rate Law
Plot of ln N2O5 vs. time is a straight line!
Eqn. for straight line y ax b
  • 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
  • Straight line plot of At vs. t
  • Slope is k (mol/Ls)

37
Graphical Methods for Determining Reaction Order
and Rate Constant
  • First-order
  • Straight line plot of lnAt vs. t
  • Slope is k (s-1)

38
Graphical Methods for Determining Reaction Order
and Rate Constant
  • Second-order
  • Straight line plot of 1/At vs t
  • Slope is k (L/mols)

39
Half-LifeSection 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-LifeSection 15.4 and Screen 15.8
  • Sugar is fermented in a 1st order process (using
    an enzyme as a catalyst).
  • sugar enzyme --gt products
  • Rate of disappear of sugar ksugar
  • k 3.3 x 10-4 sec-1
  • What is the half-life of this reaction?

46
Half-LifeSection 15.4 and Screen 15.8
Rate ksugar and k 3.3 x 10-4 sec-1. What
is the half-life of this reaction?
  • Solution
  • A / A0 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-LifeSection 15.4 and Screen 15.8
Rate ksugar 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-LifeSection 15.4 and Screen 15.8
Radioactive decay is a first order process.
Tritium ---gt electron helium
3H 0-1e 3He If you have 1.50 mg of tritium,
how much is left after 49.2 years? t1/2 12.3
years
  • Solution

49
Half-LifeSection 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 / A0 -kt
  • A ? A0 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 / A0 -kt - (0.0564 y-1)
    (49.2 y)
  • - 2.77
  • Take antilog A / A0 e-2.77 0.0627
  • 0.0627 is the fraction remaining!

50
Half-LifeSection 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 / A0 0.0627
  • 0.0627 is the fraction remaining!
  • Because A0 1.50 mg, A 0.094 mg
  • But notice that 49.2 y 4.00 half-lives
  • 1.50 mg ---gt 0.750 mg after 1
  • ---gt 0.375 mg after 2
  • ---gt 0.188 mg after 3
  • ---gt 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 --gt 234Th He 4.5 x 109 y
  • 14C --gt 14N beta 5730 y
  • 131I --gt 131Xe beta 8.05 d
  • Element 106 - seaborgium263Sg 0.9 s

52
  • Darleane Hoffman of UC-Berkeley studies the
    newest elements.
  • Hoffman is Director of Seaborg Institute. Recd
    ACS Award in Nuclear Chem and the Garvan Medal.
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