1. Unit 7 (Chp 14): Chemical Kinetics (rates) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Chemical Kinetics is the study of: 1)Rates at which reactants are consumed or products are produced in a chemical rxn. 2)Factors that affect the rate of a reaction according to Collision Theory (temperature, concentration, surface area, & catalyst). 3)Mechanisms, or sequences of steps, for how a reaction actually occurs. 4)Rate Laws (equations) used to calculate rates, rate constants, concentrations, & time. Chemical Kinetics

3. Reaction Rates Reaction rates are described by the change in concentration (  M) of reactants (consumed) or of products (produced) per change in time.

4. Reaction Rates C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) [C 4 H 9 Cl] measured at various times recall: [ ] brackets represent concentration in molarity (M)

5. Reaction Rates C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)  [C 4 H 9 Cl]  t “change in”

6. Reaction Rates fewer reactant collisions average rate slows WHY? C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

7. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. Reaction Rates C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)  [C 4 H 9 Cl]  t (rise) (run)

8. All reactions slow down over time. The initial rate of reaction is commonly chosen for analysis and comparison. Reaction Rates C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

9. rate of consumption of reactant = rate of production of product. (IF 1:1 mol ratio) –  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t ↓ reactant = ↑ product Reaction Rates and Stoichiometry C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rate :

10. Reaction Rates: 2 NO 1 O 2 2 NO 2  2 NO + O 2 –  [NO 2 ] 2  t =  [NO] 2  t = [O2]t[O2]t –  [NO 2 ]  t +  [NO]  t  [NO 2 ] tt time (t) change in conc. change in time concentration ( M )

11. Initial rate of production of H 2 is 0.050 M∙s –1. What is the rate of consumption of HI? aA + bBcC + dD Rate Ratios Rate 1a1a  [A]  t = 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t = 2 HI (g)  H 2 (g) + I 2 (g) − −= 0.050 M∙s –1 H 2 x mol H 2 mol HI 1 = 2 0.10 M∙s –1 HI –0.10 M∙s –1 HI mol L∙s mol ratio =?=? HW p. 619 #20

12. Rate equations (or rate laws) have the form: Rate Laws rate = k[A] x [B] y [C] z rate constant order with respect to reactants A, B, & C …or… number of each particle involved in collision that affects the rate overall order of reaction = x + y +… rate = k[BrO 3 – ][Br – ][H + ] 2 Example: overall order = ___ order 4 th (4 particles in collision)

13. reactant bonds break, then product bonds form. Reaction rates depend on collisions between reactant particles with: 1) greater frequency 2) enough energy 3) proper orientation The Collision Model reactantsproductsCollision activation energy: minimum E required to start reaction (E act ) successful unsuccessful

14. Reaction Coordinate Diagram transition state ∆E∆E E act …aka… Energy Profile Potential Energy  Reaction progress  demo

15. 4 Factors that Affect Reaction Rates: 1) Concentration 2) Temperature 3) (exposed) Surface Area (particle size) 4) Catalyst Factors That Affect Reaction Rates

16. 1)Concentration of Reactants  ↑ concentration, ↑ collision frequency  increase pressure of gases Factors That Affect Reaction Rates Fe (s) + O 2 (g)  Fe 2 O 3 (s) 20% of air is O 2 (g) 100% O 2 (g)

17. E act at higher Temp more particles over E act collisions of…  greater frequency  greater energy unsuccessful collisions (bounce off) Factors That Affect Reaction Rates 2) Temperature successful collisions (react) ↑ Temp, ↑ rate

18. k is temp. dependent (k changes with temp) Temperature and k (rate constant) rate = k [A] x

19. 3)Surface Area (particle size)  smaller pieces, more exposed surface area for collision. Factors That Affect Reaction Rates

20.  consumed, then produced (not used up) 4) Catalyst Factors That Affect Reaction Rates Uncatalyzed …lowering the E act. reaction progress potential energy 2 H 2 O 2 2 H 2 O + O 2 2 H 2 O 2 2 H 2 O + O 2 + 2 Br – + 2 H + Br 2 (intermediate) Catalyzed ↑↑ rate by changing the reaction mechanism by… Br –, H + demo

21. Surface Catalysts Catalysts can orient reactants to help bonds break and form. H 2 + H 2 C=CH 2  H 3 C–CH 3 H 2 + H 2 C=CH 2 CH 3

22. biological catalysts in living systems. A substrate fits into the active site of the enzyme much like a key fits into a lock. (IMAFs work here) Enzymes HW p. 621 #50,51,64

23. Rate equations (or rate laws) have the form: Rate Laws rate = k[A] x [B] y [C] z rate constant order with respect to reactants A, B, & C …or… number of each particle involved in collision that affects the rate overall order of reaction = x + y +… rate = k[BrO 3 – ][Br – ][H + ] 2 Example: overall order = ___ order 4 th (4 particles in collision) recall…

24. Reaction Mechanisms The sequence of molecular collisions and changes by which reactants become products is called the reaction mechanism. Rxns may occur in separate elementary steps. The overall reaction occurs only as fast as the slowest, rate-determining step. (RDS)

25. Slow First Step A proposed mechanism for this reaction is: Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) NO 3 intermediate is produced then consumed. NO 2 (g) + CO (g)  NO (g) + CO 2 (g) CO is not involved in the slow RDS, so it does not appear in the rate law. (OR…the order w.r.t. CO is ___) Rate law depends on the slow 1 st step: rate = k [NO 2 ] 2 0 th

26. Slow Second Step A proposed mechanism is: Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast) 2 NO (g) + Br 2 (g)  2 NOBr (g) NOBr 2 intermediate is produced then consumed. Rate law depends on the slow 2 nd step: rate = k 2 [NOBr 2 ] [NO] But cannot include intermediate [NOBr 2 ] (b/c it’s difficult to control conc.’s of intermediates).

27. Slow Second Step b/c step 1 is in equilibrium Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast) From Step 1: rate forward = rate reverse k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ] From RDS Step 2: rate = k 2 [NOBr 2 ] [NO] solve for [NOBr 2 ] substitute for [NOBr 2 ]

28. Slow Second Step Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast) k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ] rate = k 2 [NOBr 2 ] [NO] substitute for [NOBr 2 ] k2k1k−1k2k1k−1 rate =[NO] [Br 2 ] [NO] rate = k [NO] 2 [Br 2 ] HW p. 625 #92, 94 WS Reaction Mechanisms

29. Rate equations (or rate laws) have the form: Rate Laws rate = k[A] x [B] y [C] z rate constant order with respect to reactants A, B, & C …or… number of each particle involved in collision that affects the rate overall order of reaction = x + y +… rate = k[BrO 3 – ][Br – ][H + ] 2 Example: overall order = ___ order 4 th (4 particles in collision) recall…

30. Orders from Experimental Data: Rate varies with Concentration Initial Concentrations Rate in M per unit time Experiment[BrO 3 – ], M[Br – ], M[H + ], M 10.00500.250.3010 20.0100.250.3020 30.0100.500.3040 40.0100.500.60160

31. Initial Concentrations Rate in M per unit time Experiment[BrO 3 – ], M[Br – ], M[H + ], M 10.00500.250.3010 20.0100.250.3020 30.0100.500.3040 40.0100.500.60160 In experiments A & B: doubling [BrO 3 – ] doubles the rate. (other conc.’s kept constant) rate is 1 st order w.r.t. [BrO 3 – ] Orders from Experimental Data: Rate varies with Concentration rate = k [BrO 3 – ] x 2 = [2] x “w.r.t.” (with respect to) 1 = x

32. Initial Concentrations Rate in M per unit time Experiment[BrO 3 – ], M[Br – ], M[H + ], M 10.00500.250.3010 20.0100.250.3020 30.0100.500.3040 40.0100.500.60160 In experiments B & C: doubling [Br – ] doubles the rate (other conc.’s kept constant) rate is 1 st order w.r.t. [Br – ] Orders from Experimental Data: Rate varies with Concentration rate = k [Br – ] y 2 = [2] y 1 = y

33. Initial Concentrations Rate in M per unit time Experiment[BrO 3 – ], M[Br – ], M[H + ], M 10.00500.250.3010 20.0100.250.3020 30.0100.500.3040 40.0100.500.60160 In experiments C & D: doubling [H + ] quadruples the rate (other conc.’s kept constant) rate is 2 nd order w.r.t. [H + ] Orders from Experimental Data: Rate varies with Concentration rate = k [H + ] z 4 = [2] z 2 = z

34. Initial Concentrations Rate in M per unit time Experiment[BrO 3 – ], M[Br – ], M[H + ], M 10.00500.250.3010 20.0100.250.3020 30.0100.500.3040 40.0100.500.60160 Rate law is: rate = k [BrO 3 – ] [Br – ] [H + ] 2 Orders from Experimental Data: Rate varies with Concentration Orders found by experiment are: x = 1, y = 1, z = 2

35. 1)…only found experimentally (from data). 2)…do NOT come from the coefficients of reactants of an overall reaction. 3)…represent the number of reactant particles (coefficients) in the RDS of the mechanism. 4)…zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS). 5)…typically 0, 1, 2, but can be any # or fraction Orders in Rate Laws

36. Order Rate Law k Units M = ?∙M 3 s M = ?∙M 2 s M = ?∙M s M = ? s Units of k (rate constant) Units of k give info about order. Rate is usually (M∙s –1 ), or (M∙min –1 ), etc. M∙s –1 s –1 M –1∙ s –1 M –2∙ s –1 1 M 2 ∙s 1 M∙s 1 s MsMs 0 th 1 st 2 nd 3 rd rate = k[A] 0 rate = k[A] rate = k[A] 2 rate = k[A] 2 [B] HW p. 618 #22,24,28

37. Determine the rate law for the reaction (from experimental data). HW p. 619 #28 rate = k [ClO 2 ] x [OH – ] y rate 1 rate 2 = (0.0248) = (0.00276) = k (0.060) x (0.030) y = k (0.020) x (0.030) y 0.060 x 0.020 0.030 y 0.030 (3) x (1) y 9 = (3) x (0.0248) = (0.00276) = 9 = x = 2 (a)

38. Determine the rate law for the reaction (from experimental data). HW p. 619 #28 rate = k [ClO 2 ] 2 [OH – ] y rate 3 rate 2 = (0.00828) = (0.00276) = k (0.020) 2 (0.090) y = k (0.020) 2 (0.030) y 3 = (3) y y = 1 rate = k [ClO 2 ] 2 [OH – ] 1 OR rate = k [ClO 2 ] 2 [OH – ] (a)

39. Calculate the rate constant (with units). HW p. 619 #28 rate = k [ClO 2 ] 2 [OH – ] Exp 1:(0.0248)= k (0.060) 2 (0.030) (b) (0.0248) (0.060) 2 (0.030) k = 230 k = M –2 ∙s –1 M = ?∙M 2 ∙M s

40. Calculate the rate when [ClO 2 ] = 0.010 M and [OH – ] = 0.025 M. HW p. 619 #28 rate = k [ClO 2 ] 2 [OH – ] rate= (230)(0.010) 2 (0.025) (c) rate = 0.00058 M∙s –1

41. Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Integrated rate laws express (reveal) the relationship between concentration of reactants and time. The differential rate law is usually just called “the rate law.” Rate Laws

42. Integrated Rate Laws ln [A] t [A] 0 = −kt ln [A] t – ln [A] 0 = –kt ln [A] t = –kt + ln [A] 0 y = mx + b given on exam Using calculus to integrate a first-order rate law gives us: [A] 0 = initial conc. of A at t = 0. [A] t = conc. of A at any time, t.

43. If a reaction is first-order, a plot of ln [A] vs. t is a straight line, and the slope of the line will be –k. First-Order Processes ln [A] t = –kt + ln [A] 0 y = mx + b ln [A] t m = –k first-order

44. First-Order Processes CH 3 NCCH 3 CN at 198.9 o C

45. First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore,  The process is first-order.  k is the negative slope: 5.1  10 –5 s −1.

46. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get… 1 [A] t = kt + 1 [A] 0 y = mx + b 1 [A] t – 1 [A] 0 = kt given on exam

47. Second-Order Processes 1 [A] t = kt + 1 [A] 0 y = mx + b If a reaction is second-order, a plot of 1/[A] vs. t is a straight line, and the slope of the line is k. 1 [A] t m = k second-order

48. Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields the following data: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380

49. Second-Order Processes Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000−4.610 50.00.00787−4.845 100.00.00649−5.038 200.00.00481−5.337 300.00.00380−5.573 The plot is NOT a straight line, so the process is NOT first-order in [A].

50. Second-Order Processes Graphing 1/[NO 2 ] vs. t, however, gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 Because this IS a straight line, the process is second-order in [A].

51. If a reaction is zero-order, a plot of [A] vs. t is a straight line, with a slope = rate. Zero-Order Processes [A] t zero-order

52. [A] t ln [A] t = –kt + ln [A] 0 ln [A] t m = –k first-order 1 [A] t m = k second-order Summary of Integrated Rate Laws and Straight-Line Graphs 1 [A] t = kt + 1 [A] 0 HW p. 620 #33, 38, 43

53. Half-life (t 1/2 ): time at which half of initial amount remains. Half-life (t 1/2 ) is constant for 1 st order only. [A] t = 0.5 [A] 0 concentration at time, t initial concentration at time, t = 0

54. For a 1 st order process: 0.5 [A] 0 [A] 0 ln = −kt 1/2 ln 0.5 = − kt 1/2 −0.693 = − kt 1/2 = t 1/2 0.693 k ln [A] t – ln [A] 0 = –kt ln 0.5 [A] 0 – ln [A] 0 = –kt 1/2 [A] t 1/2 = 0.5 [A] 0 given on exam Half-life (t 1/2 ) depends on k:

55. A wooden object from an archeological site is subjected to radiometric dating by carbon-14. The activity of the sample due to 14 C is measured to be 11.6 disintegrations per second (current amount of C-14). The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14). The half-life of 14 C is 5715 yr. What is the age of the archeological sample? Half-life & Radiometric Dating

56. A wooden object from an archeological site is subjected to radiometric dating by carbon-14. The activity of the sample due to 14 C is measured to be 11.6 disintegrations per second (current amount of C-14). The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14). The half-life of 14 C is 5715 yr. What is the age of the archeological sample? t 1/2 = 5715 yr [A] 0 = 15.2 initial dis/s at time, t 0 = 0 s [A] t = 11.6 current dis/s at time, t t = ? yr

57. = 5715 yr 0.693 k = k 0.693 5715 yr First, determine the rate constant, k. k = 1.21  10 −4 yr −1 ln N t − ln N 0 = – kt ln 0.5N 0 − ln N 0 = – k(5715) 0.5 N 0 N 0 ln = −k(5715) –0.693 –5715 = k easy way = t 1/2 0.693 k Half-life & Radiometric Dating 0.5 ln = −k(5715) –0.693 = −k(5715) (given: t 1/2 = 5715 yr)

58. Now we can determine t: = −(1.21  10 −4 ) t ln(11.6) – ln(15.2) = −(1.21  10 −4 ) t –0.270 = t 2230 yr ln N t − ln N 0 = – kt Half-life & Radiometric Dating

59. Summary0 th Order1 st Order2 nd Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = –kt + [A] 0 ln[A] = –kt + ln[A] 0 Linear plot [A]ln[A] k & slope of line Slope = –k Slope = k M∙s –1 s –1 M –1∙ s –1 Half-Life depends on [A] 0 HW p. 620 #34,36,40,32 M = ?∙M 2 s M = ?∙M s M = ? s 1 [A] Units of k ttt y = mx + b