1. Kinetics …or Reaction Rates

2. Change The ice melted. The Coke went flat. The nail rusted.

3. Expressing Change The ice melted. The Coke went flat. The nail rusted. mL/min g/min

4. Expressing Change Reaction Rate. Change in something divided by change in time  amount  time

5. Reaction Rate As a rxn occurs, what happens to the amount of reactant? Rate = –  reactant amt  time

6. Reaction Rate Consider: Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Rate = –  mol Mg = –  mol HCl =  t2  t  mol MgCl 2 =  mol H 2  t  t

7. Reaction Rate of a one-way reaction Will be determined by looking at the reactants only Decreases as time progresses. Why? What eventually happens to the amount of reactant?

8. Reaction Rate of an equilibrium rxn Will be determined by looking at the reactants only Decreases as time progresses. Why? What eventually happens to the rate? Does the reaction stop?

9. Reaction Rate 2NO 2 (g)  2NO(g) + O 2 (g)

10. Reaction Rate Time (s) [Concentration] (mol/L)  NO 2  NO O2O2

11. Necessities of Reaction Proper orientation 2HI  H 2 + I 2

12. Necessities of Reaction Sufficient energy--Activation energy (E a ) Rxn progress Energy EaEa 2HI H 2 + I 2 Activated Complex ΔHΔH

13. Factors that affect Reaction Rate Concentration Temperature Surface Area Presence of a catalyst Rate-determining step…or the slowest step in a rxn mechanism

14. Rate Law Relates the rate as a function of the reactant(s) Units of rate are always mol/L-s unless otherwise stated Two types: –Differential –Integrated

15. Differential Rate Law For the rxn: A  B…differential  Rate = k[A] x  k is the rate law constant…units are dependent on the order  x is the order of reactant A

16. Order may not be determined by looking at the coefficients of the reactants unless the rxn is the rate- determining step must be determined experimentally

17. Order If the rate does not change when the reactant is doubled, then the order is zero; and rate is dependent solely on the value of k  Rate = k[A] 0 or Rate = k

18. Order If the rate doubles when the reactant is doubled, then the order is first; and rate is dependent on the concentration of the reactant  Rate = k[A] 1

19. Order If the rate quadruples when the reactant is doubled, then the order is second; and rate is dependent on the square of the concentration of the reactant  Rate = k[A] 2

20. Overall Order of Reaction is the sum of the individual orders When Rate = k[A] 0, the overall order is zero When Rate = k[A] 1, the overall order is one

21. Overall Order of Reaction If a differential rate law for the following rxn:  A + B  C is Rate = k[A] 1 [B] 2 then the overall order is three –What must the units of k be in this reaction?

22. Integrated Rate Law For the rxn: 2N 2 O 5  4NO 2 + O 2 The following data were collected: [N 2 O 5 ] Time(s) 1.000 0.88200 0.78400 0.69600 0.61800

23. Integrated Rate Law If we think the reactant is zero order: Rate = k[N 2 O 5 ] 0 Integrating the rate law gives us: [N 2 O 5 ] = -kt + [N 2 O 5 ] o

24. Integrated Rate Law If we plot Time vs. [N 2 O 5 ], then we get Time (s) [N 2 O 5 ] (mol/L)

25. Integrated Rate Law Since the plot gave a curve rather than a line, the order of the N 2 O 5 cannot be zero. If we try looking at it as if it were first order, then we will need to integrate the rate law.

26. Integrated Rate Law If we think the reactant will be first order: Rate = k[N 2 O 5 ] 1 Integrating the rate law gives us: ln [N 2 O 5 ] = -kt + ln[N 2 O 5 ] o

27. Integrated Rate Law If we plot Time vs. ln[N 2 O 5 ], then we get Time (s) ln[N 2 O 5 ] (mol/L)

28. Integrated Rate Law Since the plot gave a line, the order of the N 2 O 5 is first or one.

29. Integrated Rate Law If we think the reactant will be second order: Rate = k[N 2 O 5 ] 2 Integrating the rate law gives us: [N 2 O 5 ] -1 = kt + [N 2 O 5 ] o -1

30. Integrated Rate Law If we plot Time vs. [N 2 O 5 ] -1, then we get Time (s) [N 2 O 5 ] -1 (L/ mol)

31. 1/2 Life First-order— t ½ = 0.693 k Second-order— t ½ = __1__ k[A] o Zero-order— t ½ = [A] o 2k

32. First-order Half-Life All first-order reactions have half-lives independent of the initial concentration of the reactant. All radioactive decays follow first- order kinetics.

33. First-order Half-Life Problem On November 23, 1999 I had Technetium-99 injected into my bloodstream for a bone scan. The half-life of Tc-99 is 6.0 hours. –What percentage of the original amount of Tc-99 is left in my body today?

34. Second-order Half-Life All second-order reactions have half- lives dependent on the initial concentration of the reactant. A second half-life will be longer than a first half-life because the initial concentration changes For each successive half-life, [A] o is halved; thus, for each successive half- life, the half-life is doubled

35. Second-order Half-Life Problem The decomposition of NOCl is a second- order reaction where k = 4.00 x 10 -8 s -1. –For an initial concentration of 0.50M, what is the half-life? –How much is left after 1 x 10 8 s? –What is the half-life for an initial concentration of NOCl of 0.25M?

36. Zero-order Half-Life Most often occur when a catalyst is needed for the reaction to proceed. The catalyst determines how much reactant will be used. Since the rate is constant the half-life may be determined using either the rate law constant or the rate itself.

37. Reaction Mechanism Most reactions do not occur in a single step Rather, they happen in a series of steps called elementary steps The sum of the elementary steps gives the overall reaction.

38. Reaction Mechanism Intermediates are substances that are formed in one elementary step and consumed in a subsequent elementary step. They are rarely part of the rate law. Catalysts are substances added to a step that are also produced in a subsequent step. They are rarely part of the rate law.

39. Reaction Mechanism Rate-determining step— the slowest step. If it is the first or only step, then the rate law may be written from its molecularity. –The differential rate law may be written from the reactants in the rate-determining step.

40. Reaction Mechanism Consider the following elementary steps: Step 1: OCl - + H 2 0  HOCl + OH - fast Step 2: HOCl + I -  HOI + Cl - slow Step 3: HOI + OH -  H 2 0 + OI - fast What is the overall balanced equation? What is the differential rate law? What substance(s) is an intermediate? A catalyst?

41. Reaction Mechanism What is the overall balanced equation? What is the differential rate law? What substance(s) is an intermediate? A catalyst? OCl - + I -  Cl - + OI - Rate = k[I - ] HOCl, HOI, OH - ; H 2 O

42. Reaction Mechanism A catalyst is a substance that increases the rate of reaction. –It does so by lowering the E a required for the reaction to occur.

43. A Catalyst’s Effect Rxn progress Energy EaEa Uncatalyzed Catalyzed

44. Reaction Mechanism The reaction H 2 + Cl 2  2HCl occurs in four steps. Step 1: Cl energy Cl

45. Reaction Mechanism Step 2: H Cl HHH

46. Reaction Mechanism Step 3: H Cl H

47. Reaction Mechanism Step 4: Cl

48. energy Cl HHH H H H

49. Collision Theory Molecules must move toward each other (through random motion). Molecules must hit with the proper orientation. Molecules must hit with sufficient energy. Molecules will separate after reaction occurs.

50. Collision Theory k = Ae –Ea/RT Take the ln of each side and… lnk = -Ea 1 + lnA R T y m x b

51. Collision Theory Plot 1/T vs. lnk to determine E a or A. If you have two sets of data… ln k 2 = Ea 1 1_ k 1 R T 1 T 2

52. Activation Energy for a one-way reaction

53. Activation Energy for a reversible reaction

54. Activation Energy Problem Given the following kinetics data for the reaction: NO(g) + O 3 (g)  NO 2 (g) + 0 2 (g) Determine the activation energy. T (K)k (L/mol-s) 1951.08 x 10 9 2302.95 x 10 9 2605.42 x 10 9 29812.0 x 10 9 36935.5 x 10 9

55. Reaction Mechanism A series of elementary steps must satisfy two requirements for the reaction: –The sum of the elementary steps must give the overall balanced equation for the reaction. –The mechanism must agree with the observed rate law.

56. Catalyst –provides a surface whereby the reacting molecules might position themselves more favorably for collision –lowers E a –Homogeneous (i.e. enzymes) are in the same phase as the reacting molecules –Heterogeneous (i.e. Pt or Pd pieces in the catalytic converter of a car) are in a different phase and promote adsorption

57. Enzymes are proteins in living organisms that catalyze biological reactions. –Salivary amylase –Lactase

58. Inhibitors are elements or compounds used to decrease the rate of a reaction. –Tetraethyl lead Oops! Along came catalytic converters… –Methyl t-butyl ether (MTBE) and ethanol

59. Collision Theory Concentration Temperature k is a measure of the fraction of collisions with sufficient energy to produce a reaction –k = Ae –Ea/RT (Arrhenius equation) A—frequency factor (motion/orientation) E a —activation energy (J/mol)

60. Reaction Mechanism Problem #1 The balanced equation for the reaction of nitric oxide with hydrogen is 2NO + 2H 2  2H 2 0 + N 2 The experimentally determined rate law is: rate = k[NO] 2 [H 2 ] The following mechanism has been proposed:

61. Reaction Mechanism Problem #1 (cont.) NO + H 2  N + H 2 O(slow) N + NO  N 2 O(fast) N 2 O + H 2  N 2 + H 2 O(fast) Is this mechanism consistent with the observed rate law? k1k1 k2k2 k3k3

62. Reaction Mechanism Problem #2 NO + H 2 N + H 2 O(fast, with equal rates) N + NO  N 2 O(slow) N 2 O + H 2  N 2 + H 2 O(fast) Is this mechanism consistent with the observed rate law? k1k1 k2k2 k3k3 k -1

63. Reaction Mechanism Problem #3 NO + H 2  N + H 2 O(fast) N + NO  N 2 O(fast) N 2 O + H 2  N 2 + H 2 O(slow) Is this mechanism consistent with the observed rate law? k1k1 k2k2 k3k3