1. Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Chapter 13 Chemical Kinetics

2. Kinetics is the study of the rates of chemical reactions. Rate is the change of concentration (c) per unit time (t): Kinetics

3. Square brackets are used to denote molar concentration. Rate is expressed either as rate of appearance of product or rate of disappearance of reactant. Rate has units of M/s or molar/s or mol/(L·s). Rate of Reaction

4. Reaction Rate

5. An average rate is a change in concentration measured over a non-zero time interval. Average rates are not very useful because they depend on the starting and ending times. Average Rates

6. The instantaneous rate of the reaction is equal to the slope of the line drawn tangent to the curve at time t. These graphs show how to determine the instantaneous rate at 10.0 s. Instantaneous Reaction Rate

7. The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry. For the reaction 2HBr (g)  H 2 (g) + Br 2 (g) two moles of HBr are consumed for every one mole of H 2 that is formed so rate of change of [HBr] is double that of [H 2 ]. Rate and Reaction Stoichiometry

8. For any reaction a A + b B  c C + d D the reaction rate is given by: Note the signs as well as the coefficients. Rate and Reaction Stoichiometry

9. For the reaction 5H 2 O 2 + 2MnO 4 - + 6H + → 2Mn 2+ + 5O 2 + 8H 2 O the experimentally determined rate of disappearance of MnO 4 - is 2.2 x 10 -3 M/s. (a) Calculate the reaction rate. (b) What is the rate of appearance of O 2 ? Example Problem

10. Experimental rate law: analysis of many experiments shows that the rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power. For a reaction aA + bB  products the rate law is the equation: rate = k[A] x [B] y Relating Rate and Concentration

11. rate = k[A] x [B] y x and y are the orders of the reaction in [A] and [B] respectively. The overall order of the reaction is x + y. x and y are usually small integers, but may be zero, negative, or fractions. k is the specific rate constant. Rate Law

12. The reaction orders are determined by noting the effect that changing the concentration of each reactant have on the rate. The rate constant, k, is evaluated once the orders in the rate law are known. Relating Rate and Concentration

13. Most often you will be given initial concentrations and rates and asked to determine the order, which is the exponent to which concentration is raised. One way to learn to predict the order is to predict the rate given the order and concentration and see the relationship. Relating Rate and Concentration

14. Dependence of Rate on Order [Concentration][Rate] First order rate law rate = k[conc] 1 123123 123123 Second order rate law rate = k[conc] 2 123123 149149 Zero order rate law rate = k[conc] 0 123123 111111

15. The initial rate method measures the time during which a known small fraction of the reactants are consumed. Experiments are performed in which initial concentrations of [A] and [B] are individually varied. The time period of measurement is small enough that the measured rate is approximately equal to the instantaneous rate. Initial Rate Method

16. Initial rates are given below for the reaction F 2 + 2ClO 2   2FClO 2 Determine the rate law and rate constant Initial Rates Trial Init. conc. [F 2 ], M Init. conc. [ClO 2 ], M Init. Rate Ms 10.100.0101.2x10 -3 20.100.0404.8x10 -3 30.200.0102.4x10 -3

17. It is helpful to express concentrations and rates on a relative scale, by dividing the entries in each column by the smallest value. Initial Rates Trial[F 2 ] Rel. conc. [ClO 2 ] Rel. conc. Initial rate Rel. rate 10.1010.01011.2x10 -3 1 20.1010.04041.8x10 -3 4 30.2020.01012.4x10 -3 2

18. In trials 1 and 2, the concentration of F 2 does not change, so the 4-fold change in [ClO 2 ] causes the 4-fold change in rate; the order of ClO 2 must be 1. In trials 1 and 3, [ClO 2 ] is the same, so the doubling of rate was caused by doubling [F 2 ]; therefore the order of F 2 is also 1. Initial Rates

19. The rate law is first order in both [F 2 ] and [ClO 2 ]. rate = k[F 2 ][ClO 2 ] Solve the equation for k, and substitute the experimental concentrations and rates. Initial Rates

20. Write the rate law for the reaction given the following data: 2NO + 2H 2  N 2 + 2H 2 O Test Your Skill Trial Init conc. [NO], M Init conc. [H 2 ], M Init. Rate, MIs 10.005700.1407.01x10 -5 20.005700.2801.40x10 -4 30.01140.1402.81x10 -4

21. A zero order rate law rate = k means that the reaction rate is independent of reactant concentration. Concentration-Time Dependence

22. A plot of concentration vs. time yields a straight line A plot of rate vs. time yields a straight line with a slope of zero Zero-Order Rate Laws

23. First order rate law Differential rate law rate = k[R] Integrated rate law Concentration-Time Dependence

24. A plot of concentration vs. time yields a curve. First order rate law: [R] = [R] o e -kt Concentration-Time Dependence

25. A plot of ln(concentration) vs. time yields a straight line. First order rate law: ln[R] = -kt + ln[R] o Concentration-Time Dependence

26. C 12 H 22 O 11 + H 2 O  C 6 H 12 O 6 + C 6 H 12 O 6 sucrose + water  glucose + fructose The reaction is 1st order, k = 6.2 x 10 -5 s -1. If [R] o = 0.40 M, what is [R] after 2 hr? First Order Rate Law

27. How long did it take for the concentration in the same experiment to drop to 0.30 M? Test Your Skill

28. Half-life, t ½, is the time required for the initial concentration to decrease by ½. Half-Life

29. Half-life, t ½, is the time required for the initial concentration to decrease by ½. For a first order rate law, the half- life is independent of the concentration. Half-Life

30. k = 6.2 x 10 -5 s -1 for the reaction C 12 H 22 O 11 + H 2 O  C 6 H 12 O 6 + C 6 H 12 O 6 Calculate the half-life. Calculating Half-Life

31. The age of objects that were once living can be found by 14 C dating, because: The concentration 14 C is a constant in the biosphere (the atmosphere and all living organisms). When an organism dies, the 14 C content decreases with first order kinetics (t ½ = 5730 years). Scientists calculate the age of an object from the concentration of 14 C in a sample. Radiocarbon Dating

32. A sample of wood has 58% of the 14 C originally present. What is the age of the wood sample? Example: 14 C Dating

33. For a second order rate law rate = k[R] 2 A plot of 1/[R] vs. t is a straight line for a system described by second order kinetics. Second Order Rate Law

34. The reaction 2NOCl  2NO + Cl 2 obeys the rate law rate = 0.020 M -1 s -1 [NOCl] 2 Calculate the concentration of NOCl after 30 minutes, when the initial concentration was 0.050 M. Example: Second Order Rate Law

35. Given the experimental data for the decomposition of 1,3-pentadiene shown below, determine the order of the reaction. Example: Order of Reaction

36. Example: Order of Reaction (cont.) Review table 13.4 on page 534.

37. Reactions proceed at faster rates at higher temperatures. Influence of Temperature on k

38. The reaction rate is proportional to the collision frequency, Z, the number of molecular collisions per second. Z depends on the temperature and the concentration of the colliding molecules. Not all molecular collisions result in the formation of products. Collision Theory

39. Activation energy (E a ): the minimum collision energy required for reaction to occur. Activated complex: the highest energy arrangement of atoms that occurs in the course of the reaction. Collision Theory

40. NO + O 3  [activated complex] *  NO 2 + O 2 The Activated Complex

41. The fraction of collisions with energy in excess of E a is given by: The collision frequency is proportional to the concentrations of colliding species. The reaction rate is proportional to the rate of collisions time the fraction of collisions with energy in excess of E a. rate = Z × f r Influence of Temperature on Kinetic Energy

42. Orientation of Reactants

43. The steric factor, p, is a number between 0 and 1 that is needed to account for factors other than energy before a reaction can occur. The reaction rate is proportional to the steric factor times the collision frequency times the fraction of collisions with energy in excess of E a : rate = p x Z x f r The Steric Factor

44. rate = p x Z x f r Combine p and Z o into a term A: Experiments show that rate = k[colliding species], so The Arrhenius Equation

45. Take the natural log of both sides of the equation: A plot of ln k vs. 1/T gives a straight line with a slope of -E a /R and an intercept of ln A. The Arrhenius Equation

46. Determine E a for the reaction, 2NO 2  2NO + O 2 given the data: Measuring Activation Energy k (M -1 ·s -1 )T (°C)In k1/T (K -1 ) 0.003500-5.82.00x10 -3 0.037550-3.301.82x10 -3 0.291500-1.2341.67x10 -3 1.666500.5071.54x10 -3 7.397002.0001.43x10 -3

47. Prepare a plot of ln k vs. 1/T Solution

48. The rate of a reaction exactly doubles, when the temperature is changed from 25.0 o C to 36.2 o C. Calculate the activation energy for this reaction. Example: Arrhenius Equation

49. A catalyst is a substance that increases the reaction rate but is not consumed in the reaction. A catalyst provides an alternate reaction path with a lower activation energy. Catalysis

50. A homogeneous catalyst is one that is present in the same phase as the reactants. Bromide ion is a homogeneous catalyst for the decomposition of hydrogen peroxide. 2H 2 O 2 (aq)  2H 2 O( l ) + O 2 (g) step 1: H 2 O 2 (aq) + 2Br - (aq) + 2H + (aq)  Br 2 (aq) + 2H 2 O( l ) step 2: H 2 O 2 (aq) + Br 2 (aq)  2Br - (aq) + 2H + (aq) + O 2 (g) Homogeneous Catalysis

51. A heterogeneous catalyst is one that is present in a different phase from the reactants. The gas phase reaction of hydrogen with many organic compounds is catalyzed by solid platinum. Heterogeneous Catalysis

52. Enzymes are large molecules (macromolecules) which catalyze specific biochemical reactions. Enzymes can increase the rates of reactions by factors as large as 10 14. Enzymes are very specific in the reactions they catalyze. Enzymes are active under mild reaction conditions. Enzyme Catalysis

53. A mechanism is a sequence of molecular-level steps that lead from reactants to products. An elementary step is an equation that describes an actual molecular level event. The concentration dependence in the rate law for an elementary step is given by the coefficients in the equation. Reaction Mechanisms

54. The molecularity of an elementary step is the number of reactant species involved in that step. Most elementary steps are either unimolecular (involving a single molecule) or bimolecular (collision of two species). Molecularity

55. The reaction 2NO + O 2  2NO 2 is believed to occur by the following sequence of elementary steps: 2NO  N 2 O 2 bimolecular reaction N 2 O 2 + O 2  2NO 2 bimolecular reaction Elementary Steps

56. The rate of an elementary step is proportional to the concentration of each reactant species raised to the power of its coefficient in the equation: step 1: 2NO  N 2 O 2 rate 1 = k 1 [NO] 2 step 2: N 2 O 2 + O 2  2NO 2 rate 2 = k 2 [N 2 O 2 ][O 2 ] Rate Laws for Elementary Reactions

57. Write the rate law for the elementary step H 2 + Cl  H 2 Cl Test Your Skill

58. The overall rate of a multistep reaction is determined by its slowest step, called the rate-limiting step. The rates of fast steps which follow the rate- limiting step have no effect on the overall rate law. The rates of fast steps that precede the rate-limiting step usually affect the concentrations of the reactant species in the rate-determining step. Rate-Limiting Steps

59. Complex Reaction Mechanisms R  P (2 steps) R  intermediatesrate 1 = k 1 [R] Intermediates  Prate 2 = k 2 [intermediates] If step 1 is slow, then it determines rate If step 2 is slow, how can we measure [intermediates]? Many fast steps prior to slow step are fast and reversible

60. Complex Reaction Mechanisms Fast reversible steps help with [intermediate] Consider the following reaction 2NO + 2H 2  N 2 + 2H 2 O rate = k[NO] 2 [H 2 ] step 1: 2NO  N 2 O 2 fast and reversible step 2: N 2 O 2 + H 2  N 2 O + H 2 O slow step 3: N 2 O + H 2  N 2 + H 2 O fast

61. Work on the board

62. Consider the two-step reaction 2NO + O 2  2NO 2 step 1: 2NO  N 2 O 2 rate 1 = k 1 [NO] 2 step 2: N 2 O 2 + O 2  2NO 2 rate 2 = k 2 [N 2 O 2 ][O 2 ] Complex Reaction Mechanisms

63. If the first step is the rate-limiting step, the rate law is: rate = k 1 [NO] 2 If the first step is rapid and the second step is the rate limiting step, the rate law is: rate = k 2 [N 2 O 2 ][O 2 ] Complex Reaction Mechanisms

64. If the first step reaches equilibrium rate 1 (forward) = rate -1 (reverse): Complex Reaction Mechanisms

65. Substituting into the expression for step 2: rate = k 2 [N 2 O 2 ][O 2 ] rate = k 2 [NO] 2 [O 2 ] Combine all the rate constants: rate = k [NO] 2 [O 2 ] The reaction is second order in NO and first order in O 2. k1k1 k -1 Complex Reaction Mechanisms

66. For the reaction 2NO 2 + O 3  N 2 O 5 + O 2 the experimentally determined rate law is rate= k[NO 2 ][O 3 ]. Identify the rate limiting step in the proposed two-step mechanism: NO 2 + O 3  NO 3 + O 2 step 1 NO 3 + NO 2  N 2 O 5 step 2 Test Your Skill

67. H 2 + I 2  2HI rate = k[H 2 ][I 2 ] For many years this reaction was believed to occur as a single bimolecular step. From more recent data, a very different mechanism is likely. I 2  2IFast, reversible I + H 2  H 2 IFast, reversible H 2 I + I  2HISlow Both mechanisms give the same rate law. The Hydrogen-Iodine Reaction

68. Most enzymes follow the Michaelis-Menten mechanism. You will see this again in biochemistry!! Enzyme Catalysis