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# Ch36.ppt

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1. 1. Chapter 36 Elementary Chemical Kinetics Engel & Reid
2. 2. Figure 36.1 Figure 36.1 Concentration as a function of time for the conversion of reactant A into product B. The concentration of A at time 0 is [A]0, and the concentration of B is zero. As the reaction proceeds, the loss of A results in the production of B.
3. 3. 36.2 reaction rate 0 A B C+ D The number of moles os a species during any point of the reaction :advancement of the reaction 1 Reaction rate Reaction rate in intensive expression R= i i i i i i i a b c d n n v dn d v dt dt dn d Rate dt v dt             rate 1 1 1 [ ] V i i i dn d i V v dt v dt         2 5 2 2 2 (g) 2 (g) 2 5 (g) N O NO O 4NO O 2N O 1 1 4 2 dn dn dn Rate dt dt dt        Example
4. 4. 36.3 rate laws Rate law : R = k [A]a [B]b ... Rate Constant Reaction order • The rate of reaction is often found to be proportional to the molar concentrations of the reactants raised to a simple power. • It cannot be overemphasized that reaction orders have no relation to stoichiometric coefficients, and they are determined by experiment.
5. 5. 36.3 Rate laws
6. 6. 36.3 Rate laws 36.3.1 Measuring Reaction Rates      -1 1 A B first order reaction ; 40 A 40 A k k s R k s            1 0 1 30 A A 40Ms A 12Ms t t ms d R dt d R initial rate dt d R dt             
7. 7. 36.3.2 Determining Reaction Orders A + B C [ ] [ ] k R k A B a b    Strategy 1. Isolation method The reaction is performed with all species but one in excess. Under these conditions, only the concentration of one species will vary to a significant extent during the reaction. R = k’[B]b
8. 8. 36.3.2 Determining Reaction Orders Strategy 2. Method of initial rates The concentration of a single reactant is changed while holding all other concentrations constant, and the initial rate of the reaction is determined. A + B C [ ] [ ] k R k A B a b        1 0 1 1 2 2 0 2 1 1 2 2 A [A] [B] [A] [B] [A] A ln ln [A] k R R k R R a a b a b a                         
9. 9. Example Problem 36.2 Using the following data for the reaction illustrated in equation 36.13. Determine the order of the reaction with respect to A and B, and the rate constant for the reaction [A] (M) [B] (M) Initial Rate (Ms-1) 2.3010-4 3.1010-5 5.2510-4 4.6010-4 6.2010-5 4.2010-3 9.2010-4 6.2010-5 1.7010-2
10. 10.   3 4 2 4 2 1 1 1 2 2 2 2 2 4 4 5 3 4 5 2 2 4 -1 4 4.20 10 4.60 10 ln ln 2 1.70 10 9.20 10 [A] [B] [A] [B] 5.25 10 2.30 10 3.10 10 1 4.20 10 4.60 10 6.20 10 [A] [B] 5.2 10 Ms 2.3 10 M 3.1 R k R k R k k b b b a a b                                                                   5 8 -2 1 8 -2 1 2 10 M 3.17 10 M 3.17 10 M [A] [B] k s R s        Solution Example 36.2
11. 11. 36.3 Rate law Determining the rate of a chemical reaction, experimentally •Chemical methods •Physical methods •Stopped-flow techniques •Flash photolysis techniques •Perturbation-relaxation methods
12. 12. Figure 36.3 Figure 36.3 Schematic of a stopped-flow experiment. Two reactants are rapidly introduced into the mixing chamber by syringes. After mixing chamber, the reaction kinetics are monitored by observing the change in sample concentration versus time, in this example by measuring the absorption of light as a function of time after mixing.
13. 13. 36.5 Integrated Rate Law Expressions                   kt e e kt kdt d k dt d dt d R k R P kt kt t k                               0 0 0 0 0 0 A A A ln A ln 1 ] A [ A] [ ] A [ ] P [ A ] A [ ; A A ln A ] A [ ]; A [ ] A [ ] A [ ]; A [ A reaction Order - First 0
14. 14. 36.5 Integrated Rate Law Expressions   kt   0 A ln ] A ln[   kt e  0 A ] A [
15. 15. 36.5 Integrated Rate Law Expressions     k k kt 2 ln 2 ln A 2 / A ln Reaction Order - First and life - Half 2 / 1 0 0 2 / 1             
16. 16. Example 36.3 Example Problem 36.3 The decomposition of N2O5 is an important process in tropospheric chemistry. The half-life for the first order decomposition of this compound is 2.05×104 s. How long will it take for a sample of N2O5 to decay to 60% of its initial value? Solution               s 10 51 . 1 s 10 38 . 3 6 . 0 ln 6 . 0 O N O N 6 . 0 O N O N O N s 10 38 . 3 s 10 05 . 2 2 ln 2 ln 4 1 - 5 s 10 38 . 3 s 10 38 . 3 0 5 2 0 5 2 5 2 0 5 2 5 2 1 - 5 4 2 / 1 1 - 5 1 - 5                        t e e e t k t t kt
17. 17. Example 36.3 Example Problem 36.4 Catbon-14 is a radioactive nucleus with half-life of 5760 years. Living matter exchange carbon with its surroundings (for example, through CO2) so that s constant level of C14 is maintained, corresponding to 15.3 decay events per minute. Once living matter has died, carbon contained in the matter is not exchanged with the surroundings, and the amount of C14 that remains in the dead material decreases with time due to radioactive decay. Consider a piece of fossilized wood that demonstrates 2.4 C decay events per minute. How old is the wood.
18. 18. Example 36.4               s 10 86 . 4 157 . 0 ln s 10 81 . 3 1 C C ln 1 C C 157 . 0 min 3 . 15 min 4 . 2 C C s 10 81 . 3 s 10 82 . 1 2 ln years 5760 2 ln 2 ln 11 1 - 12 0 14 14 0 14 14 1 - 1 - 0 14 14 1 - 12 11 2 / 1                             k t e t k kt Solution
19. 19. 36.5 Integrated Rate Law Expressions           t k dt k d k dt d dt d R k R P eff t eff k               0 0 A A 2 2 2 A 1 A 1 A ] A [ ; ] A [ 2 ] A [ ] A [ 2 1 ; ] A [ A 2 I) (Type reaction Order - Second 0
20. 20. Figure 36.5a     t keff   0 A 1 A 1
21. 21. 36.5 Integrated Rate Law Expressions        0 2 / 1 2 / 1 0 2 / 1 0 A 1 A 1 ; A 1 A 2 I) (Type Reaction Order - Second and life - Half eff eff eff k k t k t k    
22. 22. 36.5 Integrated Rate Law Expressions                           A B A B let ; A A B B B B A A ] B [ ] A [ ; B ] A [ B A II) (Type reaction Order - Second 0 0 0 0 0 0                     dt d dt d R k R P k
23. 23.                 kt kt kt kt kdt d k k dt d t                                                                                           0 0 0 0 0 0 0 0 A A 0 A A ] A /[ ] A [ ] B /[ ] B [ ln A B 1 ] B [ ] B [ ln ] A [ ] B [ ln 1 ] A [ ] A [ ln ] A [ ] A [ ln 1 ] A [ ] A [ ln 1 ] A [ ] A [ ] A [ ] A [ ] A [ ] B ][ A [ ] A [ 0 0 36.5 Integrated Rate Law Expressions
24. 24. Figure 36.6                      t t t t t k t k k t k t k dt d A A A A A A A ; A A A A P A                     Figure 36.6 Schematic representation of the numerical evaluation of a rate law.
25. 25. Figure 36.7 Figure 36.7 Comparison of the numerical approximation method to the integrated rate law expression for a first-order reaction. The rate constant for the reaction is 0.1 m s-1. The time evolution in reactant concentration determined by the integrated rate law expression
26. 26. 36.7 Sequential First-Order Reaction               I P I A I A A P I A I I A A I A k dt d k k dt d k dt d k k           At t = 0                                       0 A I A 0 0 0 A I A 0 A A 0 A 1 P I A A P P I A A A I I A I A I A A I A A A                                 k k e k e k e e k k k k e k k k dt d e t k I t k t k t k t t k t t k A I Sequential First-Order Reaction
27. 27. Figure 36.8a Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (a) kA=2kf =0.1 s-1.
28. 28. Figure 36.8b Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (b) kA=8kf =0.4 s-1. Notice that both the maximal amount of I in a ddition to the time for the maximum is changed relative to the first channel.
29. 29. Figure 36.8c Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (c) kA=0.025kf=0.0125 s-1. In this case, very little intermediate is formed, and the maximum in [I] is delayed relative to the first two examples.
30. 30. 36.7 Sequential First-order Reaction Maximum Intermediate Concentration                     I A I A t t k k k k t dt d ln 1 0 I max max
31. 31. Example problem 36.5 Example Problem 36.5 Determine the time at which [I] is at a maximum for kA = 2kI = 0.1 s-1. Solution s 9 . 13 s 05 . 0 s 1 . 0 ln s 05 . 0 s 1 . 0 1 ln 1 1 - -1 1 - 1 - max                      I A I A k k k k t
32. 32. Rate-Determining Steps              0 0 0 0 A 1 A 1 lim P lim A 1 A 1 lim P lim t k A I t k I t k A k k t k t k I A t k A I t k I t k A k k t k t k I A IA A I A A I A I A I A A I A e k k e k e k e e k k e k k e k e k e e k k                                                                     36.7 Sequential First-order Reaction
33. 33. Figure 36.9a Figure 36.9 Rate-limiting step behavior in sequential reactions. (a) kA=20kf =1 s-1 such that the rate-limiting step is the decay of intermediate I. In this case, the reduction in [I] is reflected by the appearances of [P]. The time evolution of [P] predicted by the sequential mechanism is given by the yellow line, and the corresponding evolution assuming rate-limiting step behavior, [P]rl, is given by the red curve.
34. 34. Figure 36.9b Figure 36.9 Rate-limiting step behavior in sequential reactions. (b) The opposite case from part (a) kA=0.04kf = 0.02 s-1 such that the rate-limiting step is the decay of reactant A.
35. 35. 36.7 Sequential First-order Reaction                     2 2 2 2 1 1 2 1 1 1 2 1 P A A A P A 2 1 A I k dt d I k I k dt I d I k k dt I d k dt d I I A A k k k              The steady-State Approximation This approximation is particularly god when the decay rate of the intermediate is greater than the rate of production so that the intermediates are present at very small concentrations during the reaction. In Steady-State approximation, the time derivative of intermediate concentrations is se to zero.                                     t k ss t k A ss t k A ss ss ss ss ss t k A A ss ss A ss A A A A e e k I k dt d e k k k k I I k I k dt I d e k k k k I I k k dt I d                      1 A P A P A I 0 A A A 0 0 0 2 2 0 2 1 2 1 2 2 2 1 1 2 0 1 1 1 1 1 1
36. 36. Figure 36.10 Figure 36.10 Concentration determined by numerical evolution of the sequential reaction scheme presented in Equation (36.44) where kA= 0.02 s-1 and kf = k2 = 0.2 s-1.
37. 37. Figure 36.11 Figure 36.11 Comparison of the numerical and steady-state concentration profiles for the sequential reaction scheme presented in Equation (36.44) where kA= 0.02 s-1 and kf = k2 = 0.2 s-1. Curves corresponding to the steady- state approximation are indicated by the subscript ss.
38. 38. 36.8 Parallel Reactions Parallel Reactions                                                  A B t k k C B C t k k C B B t k k C B C B C B k k e k k k e k k k e condition initial k dt d k dt d k k k k dt d C B C B C B                           C B 1 A C 1 A B A A 0 C B , 0 A C C ; A A A A A A 0 0 0 0
39. 39. Figure 36.12 Figure 36.12 Concentration for a parallel reaction where kB = 2kC = 0.1 s-1.
40. 40. 36.8 Parallel Reactions Yield, F, is defined as the probability that given product will be formed by decay of the reactant.   F n n i k k
41. 41. Example Problem 36.7 In acidic conditions, benzyl peniciline (BP) undergoes the following paralll reaction: In the molecular structure, R1 and R2 indicate alkyl substitutions. In a solution where pH=3, the rat constants for the processes at 22 ℃are k1=7.0×10-4 s-1, k2=4.1×10-3 s-1, and k2=5.7×10-3 s-1. what is the yield for P1 formation? Example Problem 36.7
42. 42. Example Problem 36.7 067 . 0 s 10 7 . 5 s 10 1 . 4 s 10 0 . 7 s 10 0 . 7 1 3 1 3 1 4 1 4 3 2 1 1 P1            F         k k k k Solution
43. 43. 36.9 Temperature Dependence of Rate Constants RT E A k Ae k rrehenius a RT Ea     ln ln expression A / A frequency factor or Arrhenius pre-exponential factor Ea activation energy
44. 44. Example Problem 36.8 Example Problem 36.8 The temperature dependence of the acid-catalyzed hydrolysis of penicillin (illustrated in Example problem 36.7) is investigated, and the dependece of k1 on temperature is iven in the following table. What is the activation eergy and Arrhenius preexponential factor for this branch of hydrolysis reaction? Temperature ℃ k1 （s-1） 22.2 7.0×10-4 27.2 9.8×10-4 33.7 1.6×10-3 38.0 2.0×10-3
45. 45. Example Problem 36.8 Draw a plot of ln (k1) versus 1/T     1 1 - 1 6 1 . 14 mole K J 314 . 8 K 3 . 6306 10 33 . 1 1 . 14 ln 1 . 14 T 1 K 3 . 6306 ln                a a E R E slope s e A A k Solution
46. 46. 36.9 Temperature Dependence of Rate Constant Figure 36.13 A schematic drawing of the energy profile for a chemical reaction. Reactants must acquire sufficient energy to overcome the activation energy, Ea, for the reaction. The reaction coordinate represents the binding and geometry changes hat occur in the transformation of reactants into products.
47. 47. 36.10 Reversible Reactions and Equilibrium                                                                  t k k dt k k k d k k k k -k k -k dt d k - k dt d k -k dt d 0 A A 0 B B A 0 B B A 0 B A B A 0 B A B A - A - A A A A - A - A A B A A B A A Since B A B B A A B A 0 A B                                                               B A B 0 eq B A B 0 eq B A A B 0 B A A B 0 1 A B lim B A A lim A 1 A B A A B A B A k k k k k k k k e k k k k e k k t t t k k t k k Reversible Reactions
48. 48. 36.10 Reversible Reactions and Equilibrium             C eq eq eq eq eq eq K k k k k dt d dt d        A B B A 0 B A B A B A At equilibrium
49. 49. 36.10 Reversible Reactions and Equilibrium Figure 36.14 Reaction coordinate demonstrating the activation energy for reactants to form products, Ea, and the back reaction in which products form reactants, E’a
50. 50. 36.10 Reversible Reactions and Equilibrium Figure 36.15 Time-dependent concentrations in which both forward and back reactions exist between reactant A and product B. In this example, kA=2kB=0.06 s-1. Note that the concentration reach a constant value at longer times (t > teq) at which pint the reaction reaches equilibrium.
51. 51. 36.10 Reversible Reactions and Equilibrium Figure 36.16 Methodology for determining fprward and back rate constants. The apparent rate constant for reactant dacay is equal to the sum of forward, kA, and back, kB, rate constnats. The equilibrium constant is equal to kA / kB. These two measurements provide a system of two equations and two unknowns that can be readily evaluated to produce kA, and kB.
52. 52. Example Problem 36.9 Example Problem 36.9 Consider the interconversion of the boat and chair conformation of cyclohexane: The reaction is first order in each direction, with a equilibrium constant of 104. The activation enegy for the conversion of the chair conformer to the boat conformer is 42 kJ/mol. Assuming an Arrhenius preexponential factor of 1012 s-1, what is the expected observed reaction rate constant at 298 K if one were to initiate this reaction starting with only the boat conformer?
53. 53. Example Problem 36.9    1 - 8 1 - 8 4 4 1 - 4 1 - 1 - 1 - 1 - 12 / B s 10 34 . 4 s 10 34 . 4 10 10 s 10 34 . 4 K 298 K mol J 8.314 mol J 000 , 42 exp s 10                       B A app B A B A C RT E k k k k k k k K Ae k a Solution
54. 54. Figure 36.17 Figure 36.17 Example of a temperature-jump experiment for a reaction in which the forward and back rate pocesses are first order. The yellow and blue portions of the graph indicate times before and after the temperature jump, respectively. After the temperature jump, [A] decrease with a time constant related to the sum of he forward and back rate constants. The change between the pre-jump and post-jump equilibrium concentrations is given by 0
55. 55. 36.13 Potential Energy Surface Figure 36.18 Definition of geometric coordinates for the AB + CA+BC reaction.
56. 56. 36.13 Potential Energy Surface Figure 36.19 Illustration of a potential surface for the AB+C reaction at a colinear geometry (=180° in Figure 36.18). (a,b) Three dimensional views of the surface.
57. 57. 36.13 Potential Energy Surface Figure 36.19 Illustration of a potential surface for the AB+C reaction at a colinear geometry (=180° in Figure 36.18). (c) Counter plot of the surface with contours of equipotential energy. The curved dashes line represents the path of a reactive event, corresponding to the reaction coordinate. The transition state for this coordinate is indicated by the symbol ‡.
58. 58. 36.13 Potential Energy Surface Figure 36.19 Illustration of a potential surface for the AB+C reaction at a colinear geometry (=180° in Figure 36.18). (d,e) Cross sections of the potential energy surface along the lines a-a and b-b, respectively. These two graphs corresponds to the potential for two-body interactions of B with C, and A with B.
59. 59. Figure 36.20 Figure 36.20 Reaction coordinates involving an activated complex and a reactive intermediate. The graph corresponds to the reaction coordinate derived from the dashed line between points c and d on the contour plot of Figure 36.19c. The maximum in energy along this coordinate corresponds to the transition state, and the species at this maximum is referred to as an activated complex.
60. 60. Figure 36.21 Figure 36.21 Illustration of transition state theory. Similar to reaction coordinates depicted previously, the reactants (A and B) and product (P) are separated by an energy barrier. The transition state is an activated reactant complex envisioned to exist at the free-energy maximum along the reaction coordinate.