1. Chemical Kinetics Tro, Chemistry: A Molecular Approach

6. Tro, Chemistry: A Molecular Approach at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1

7. Tro, Chemistry: A Molecular Approach at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2

8. Tro, Chemistry: A Molecular Approach at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3

9. Hypothetical Reaction Red  Blue in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time Time (sec) Number Red Number Blue 0 100 0 5 84 16 10 71 29 15 59 41 20 50 50 25 42 58 30 35 65 35 30 70 40 25 75 45 21 79 50 18 82

10. Hypothetical Reaction Red  Blue Tro, Chemistry: A Molecular Approach

11. Hypothetical Reaction Red  Blue Tro, Chemistry: A Molecular Approach

14. Hypothetical Reaction Red  Blue Avg. Rate Avg. Rate Avg. Rate Time (sec) Number Red Number Blue (5 sec intervals) (10 sec intervals) (25 sec intervals) 0 100 0 5 84 16 3.2 10 71 29 2.6 2.9 15 59 41 2.4 20 50 50 1.8 2.1 25 42 58 1.6 2.3 30 35 65 1.4 1.5 35 30 70 1 40 25 75 1 1 45 21 79 0.8 50 18 82 0.6 0.7 1

15. H 2 I 2 HI Stoichiometry tells us that for every 1 mole/L of H 2 used, 2 moles/L of HI are made. Assuming a 1 L container, at 10 s, we used 0.181 moles of H 2 . Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles At 60 s, we used 0.699 moles of H 2 . Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles The average rate is the change in the concentration in a given time period. In the first 10 s, the  [H 2 ] is -0.181 M, so the rate is Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 10.000 0.819 20.000 0.670 30.000 0.549 40.000 0.449 50.000 0.368 60.000 0.301 70.000 0.247 80.000 0.202 90.000 0.165 100.000 0.135 Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 0.000 10.000 0.819 0.362 20.000 0.670 0.660 30.000 0.549 0.902 40.000 0.449 1.102 50.000 0.368 1.264 60.000 0.301 1.398 70.000 0.247 1.506 80.000 0.202 1.596 90.000 0.165 1.670 100.000 0.135 1.730 Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 0.000 1.000 0.000 10.000 0.819 0.362 0.0181 20.000 0.670 0.660 0.0149 30.000 0.549 0.902 0.0121 40.000 0.449 1.102 0.0100 50.000 0.368 1.264 0.0081 60.000 0.301 1.398 0.0067 70.000 0.247 1.506 0.0054 80.000 0.202 1.596 0.0045 90.000 0.165 1.670 0.0037 100.000 0.135 1.730 0.0030 Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 0.000 10.000 0.819 0.362 0.0181 0.0181 20.000 0.670 0.660 0.0149 0.0149 30.000 0.549 0.902 0.0121 0.0121 40.000 0.449 1.102 0.0100 0.0100 50.000 0.368 1.264 0.0081 0.0081 60.000 0.301 1.398 0.0067 0.0067 70.000 0.247 1.506 0.0054 0.0054 80.000 0.202 1.596 0.0045 0.0045 90.000 0.165 1.670 0.0037 0.0037 100.000 0.135 1.730 0.0030 0.0030

16. Tro, Chemistry: A Molecular Approach average rate in a given time period =  slope of the line connecting the [H 2 ] points; and ½ +slope of the line for [HI] the average rate for the first 10 s is 0.0181 M/s the average rate for the first 40 s is 0.0150 M/s the average rate for the first 80 s is 0.0108 M/s

18. H 2 ( g ) + I 2 ( g )  2 HI ( g ) Using [H 2 ], the instantaneous rate at 50 s is: Using [HI], the instantaneous rate at 50 s is:

19. Ex 13.1 - For the reaction given, the [I  ] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the  [H + ]. H 2 O 2 ( aq ) + 3 I  ( aq ) + 2 H + ( aq )  I 3  ( aq ) + 2 H 2 O ( l ) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value

29. Sample Rate Laws Tro, Chemistry: A Molecular Approach The reaction is autocatalytic, because a product affects the rate. Hg 2+ is a negative catalyst, increasing its concentration slows the reaction.

30. Reactant Concentration vs. Time A  Products Tro, Chemistry: A Molecular Approach

34. Tro, Chemistry: A Molecular Approach ln[A] 0 ln[A] time slope = − k

35. Half-Life of a First-Order Reaction Is Constant Tro, Chemistry: A Molecular Approach

36. Rate Data for C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl Tro, Chemistry: A Molecular Approach Time (sec) [C 4 H 9 Cl], M 0.0 0.1000 50.0 0.0905 100.0 0.0820 150.0 0.0741 200.0 0.0671 300.0 0.0549 400.0 0.0448 500.0 0.0368 800.0 0.0200 10000.0 0.0000

37. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach

38. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach

39. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach slope = -2.01 x 10 -3 k = 2.01 x 10 -3 s -1

41. Tro, Chemistry: A Molecular Approach l/[A] 0 1/[A] time slope = k

42. Rate Data For 2 NO 2  2 NO + O 2 Time (hrs.) Partial Pressure NO 2 , mmHg ln(P NO2 ) 1/(P NO2 ) 0 100.0 4.605 0.01000 30 62.5 4.135 0.01600 60 45.5 3.817 0.02200 90 35.7 3.576 0.02800 120 29.4 3.381 0.03400 150 25.0 3.219 0.04000 180 21.7 3.079 0.04600 210 19.2 2.957 0.05200 240 17.2 2.847 0.05800

43. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach

44. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach

45. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach

47. Tro, Chemistry: A Molecular Approach

48. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach

49. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach

50. Tro, Chemistry: A Molecular Approach

51. Tro, Chemistry: A Molecular Approach

52. Tro, Chemistry: A Molecular Approach

53. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach the reaction is second order, Rate = -  [A]  t = 0.1 [A] 2

54. Ex. 13.4 – The reaction SO 2 Cl 2( g )  SO 2( g ) + Cl 2( g ) is first order with a rate constant of 2.90 x 10 -4 s -1 at a given set of conditions. Find the [SO 2 Cl 2 ] at 865 s when [SO 2 Cl 2 ] 0 = 0.0225 M the new concentration is less than the original, as expected [SO 2 Cl 2 ] 0 = 0.0225 M, t = 865, k = 2.90 x 10 -4 s -1 [SO 2 Cl 2 ] Check: Solution: Concept Plan: Relationships: Given: Find: [SO 2 Cl 2 ] [SO 2 Cl 2 ] 0 , t, k

56. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Comparing Expt #1 and Expt #2, the [NO 2 ] changes but the [CO] does not Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

57. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Determine by what factor the concentrations and rates change in these two experiments. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

58. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Determine to what power the concentration factor must be raised to equal the rate factor. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

59. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Repeat for the other reactants Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

60. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach n = 2, m = 0 Substitute the exponents into the general rate law to get the rate law for the reaction Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

61. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033

62. Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Tro, Chemistry: A Molecular Approach Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 1 0.0200 0.200 10.8 2 0.0600 0.200 32.3 3 0.200 0.0202 10.8 4 0.200 0.0404 21.6

63. Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Rate = k [NH 4 + ] n [NO 2  ] m Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 1 0.0200 0.200 10.8 2 0.0600 0.200 32.3 3 0.200 0.0202 10.8 4 0.200 0.0404 21.6

65. Tro, Chemistry: A Molecular Approach

67. Isomerization of Methyl Isonitrile Tro, Chemistry: A Molecular Approach methyl isonitrile rearranges to acetonitrile in order for the reaction to occur, the H 3 C-N bond must break; and a new H 3 C-C bond form

68. Energy Profile for the Isomerization of Methyl Isonitrile the collision frequency is the number of molecules that approach the peak in a given period of time the activation energy is the difference in energy between the reactants and the activated complex the activated complex is a chemical species with partial bonds As the reaction begins, the C-N bond weakens enough for the C  N group to start to rotate

70. Tro, Chemistry: A Molecular Approach

72. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach Temp, K k, M -1 ∙s -1 Temp, K k, M -1 ∙s -1 600 3.37 x 10 3 1300 7.83 x 10 7 700 4.83 x 10 4 1400 1.45 x 10 8 800 3.58 x 10 5 1500 2.46 x 10 8 900 1.70 x 10 6 1600 3.93 x 10 8 1000 5.90 x 10 6 1700 5.93 x 10 8 1100 1.63 x 10 7 1800 8.55 x 10 8 1200 3.81 x 10 7 1900 1.19 x 10 9

73. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach use a spreadsheet to graph ln( k ) vs. (1/T)

74. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach E a = m∙(- R ) solve for E a A = e y -intercept solve for A

76. Ex. 13.8 – The reaction NO 2( g ) + CO ( g )  CO 2( g ) + NO ( g ) has a rate constant of 2.57 M -1 ∙ s -1 at 701 K and 567 M -1 ∙ s -1 at 895 K. Find the activation energy in kJ/mol most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable T 1 = 701 K, k 1 = 2.57 M -1 ∙s -1 , T 2 = 895 K, k 2 = 567 M -1 ∙s -1 E a , kJ/mol Check: Solution: Concept Plan: Relationships: Given: Find: E a T 1 , k 1 , T 2 , k 2

79. Effective Collisions Kinetic Energy Factor Tro, Chemistry: A Molecular Approach for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex

80. Effective Collisions Orientation Effect Tro, Chemistry: A Molecular Approach

88. Rate Laws of Elementary Steps Tro, Chemistry: A Molecular Approach

93. An Example Tro, Chemistry: A Molecular Approach 2 H 2 (g ) + 2 NO ( g )  2 H 2 O ( g ) + N 2( g ) Rate obs = k [H 2 ][NO] 2 2 NO ( g )  N 2 O 2( g ) Fast H 2 (g ) + N 2 O 2( g )  H 2 O ( g ) + N 2 O ( g ) Slow Rate = k 2 [H 2 ][N 2 O 2 ] H 2 (g ) + N 2 O ( g )  H 2 O ( g ) + N 2( g ) Fast k 1 k - 1 for Step 1 Rate forward = Rate reverse

94. Ex 13.9 Show that the proposed mechanism for the reaction 2 O 3( g )  3 O 2( g ) matches the observed rate law Rate = k [O 3 ] 2 [O 2 ] -1 Tro, Chemistry: A Molecular Approach O 3( g )  O 2( g ) + O ( g ) Fast O 3 (g ) + O ( g )  2 O 2( g ) Slow Rate = k 2 [O 3 ][O] k 1 k - 1 for Step 1 Rate forward = Rate reverse

96. Ozone Depletion over the Antarctic Tro, Chemistry: A Molecular Approach

97. Energy Profile of Catalyzed Reaction Tro, Chemistry: A Molecular Approach polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

99. Types of Catalysts Tro, Chemistry: A Molecular Approach

100. Catalytic Hydrogenation H 2 C=CH 2 + H 2 -> CH 3 CH 3 Tro, Chemistry: A Molecular Approach

102. Enzyme-Substrate Binding Lock and Key Mechanism Tro, Chemistry: A Molecular Approach

103. Enzymatic Hydrolysis of Sucrose Tro, Chemistry: A Molecular Approach

104. Chapter 13 Chemical Kinetics 2008, Prentice Hall Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA