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Gc Chemical Kinetics

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1. Study of speed with which a chemical reaction occurs and the factors affecting that speed
2. Provides information about the feasibility of a chemical reaction
3. Provides information about the time it takes for a chemical reaction to occur
4. Provides information about the series of elementary steps which lead to the formation of product

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Gc Chemical Kinetics

  1. 1. Chemical Kinetics<br />Study of speed with which a chemical reaction occurs and the factors affecting that speed<br />Provides information about the feasibility of a chemical reaction<br />Provides information about the time it takes for a chemical reaction to occur<br />Provides information about the series of elementary steps which lead to the formation of product <br />
  2. 2. Rate Data for A + B -> C<br />
  3. 3. A + B -> C<br />C<br />A<br />B<br />
  4. 4. The Rate of a Chemical Reaction<br />The speed of a reaction can be examined by the decrease in reactants or the increase in products.<br />a A + b B -> c C + d D<br />Where m and n are determined experimentally, and not necessarily<br />Equal to the stiochiometry of the reaction<br />
  5. 5. Reaction A -> 2 B<br />A<br />A<br />B<br />B<br />B<br />B<br />A<br />B<br />B<br />B<br />B<br />A<br />A<br />A<br />B<br />B<br />B<br />B<br />A<br />A<br />A<br />B<br />B<br />B<br />B<br />A<br />B<br />B<br />B<br />B<br />22<br />22<br />=<br />6.022 x 10<br /> molecules<br />B<br />6.022 x 10<br /> molecules<br />=<br />A<br />in a 1.00 L container<br />in a 1.00 Liter container<br />1 mol/L<br />2 mol/L<br />
  6. 6. Average Rate<br />Rate of A disappearing is<br />Let’s suppose that after 20 seconds ½ half of A disappears.<br />Then <br />And Rate of B appearing is <br />Then<br />
  7. 7. Average Rate Law for the General Equationa A + b B -> c C + d D<br />For Example:<br />N2O5 (g) -> 2 NO2 (g) + ½ O2 (g)<br />
  8. 8. Determination of the Rate Equation<br />Determined Experimentally<br />Can be obtained by examining the initial rate after about 1% or 2% of the limiting reagent has been consumed.<br />
  9. 9. Consider the Reaction:CH3CH2CH2CH2Cl (aq) + H2O (l) -> CH3CH2CH2CH2OH (aq) + HCl (aq)<br />
  10. 10. Average Rates, <br />
  11. 11. Average Rates, <br />
  12. 12. Average Rates, <br />
  13. 13. Average Rates, <br />
  14. 14. Instantaneous Rate or initial rate at t=0 s<br />Instantaneous Rate at t = 500 s<br />
  15. 15.
  16. 16. Order of Reaction<br />Zero order –independent of the concentration of the reactants, e.g, depends on light<br />First order - depends on a step in the mechanism that is unimolecular<br />Pseudo first order reaction – one of the reactants in the rate determining step is the solvent<br />Second order – depends on a step in the mechanism that is bimolecular<br />Rarely third order – depends on the step in the mechanism that is termolecular<br />
  17. 17. Data from the hydrolysis of n-butyl chloride<br />
  18. 18. IF Zero Order<br />
  19. 19. Therefore, the reaction is not zero order<br />
  20. 20. If Second Order<br />
  21. 21. Therefore, the reaction is not second order<br />
  22. 22. IF First Order Reaction<br />
  23. 23. First Order Plot<br />
  24. 24. First Order Plot<br />
  25. 25. Slope<br />
  26. 26. Slope<br />
  27. 27. Rate of the Reaction<br />
  28. 28. For the ReactionN2O5 (g) -> 2 NO2 (g) + ½ O2 (g)<br />The rate can be used to explain the mechanism<br />
  29. 29. (1) Slow Step<br />(2) Fast Step<br />
  30. 30. Sum of the two steps:<br />N2O5 -> 2 NO2 + ½ O2<br />or<br />2 N2O5 -> 4 NO2 + O2<br />
  31. 31. ApplicationMechanism of a Chemical Reaction<br />(a)<br />Suggest a possible mechanism for<br />NO2 (g) + CO (g) -> NO (g) + CO2 (g)<br />Given that <br />(b)<br />Suggest a possible mechanism for<br />2 NO2 (g) + F2 (g) -> 2 NO2F (g)<br />Given that <br />
  32. 32. Factors Affecting the Rate of a Chemical Reaction<br />The Physical State of Matter<br />The Concentration of the Reactants<br />Temperature<br />Catalyst<br />
  33. 33. For A Reaction to Occur<br />Molecules Must Collide<br />Molecules must have the Appropriate Orientation<br />Molecules must have sufficient energy to overcome the energy barrier to the reaction-<br />Bonds must break and bonds must form<br />
  34. 34. A Second Order Reaction<br />
  35. 35. Rate Constant “k”<br />Must be determined experimentally<br />Its value allows one to find the reaction rate for a new set of concentrations<br />
  36. 36. The following data were collected for the rate of the reaction <br />Between A and B, A + B -> C , at 25oC. Determine the rate law <br />for the reaction and calculate k.<br />
  37. 37. From Experiments 1 and 2<br />Solution A:<br />Divide equation (1) into equation (2)<br />
  38. 38. Solution B:<br />Subtract equation (2) from equation (1)<br />
  39. 39. From Experiments 4 and 5<br />Solution A:<br />Divide equation (1) into equation (2)<br />
  40. 40. Solution B:<br />Subtract equation (2) from equation (1)<br />
  41. 41. Rate Constant k<br />
  42. 42. Rate Constant kFrom Experiment 3<br />
  43. 43. Rate Constant kFrom Experiment 1<br />
  44. 44. Your Understanding of this ProcessConsider the Data for the Following Reaction:<br />Determine the Rate Law Expression and the value of k consistent<br />With these data.<br />
  45. 45. From Experiments 1 and 2<br />Solution :<br />Divide equation (1) into equation (2)<br />
  46. 46. From Experiments 2 and 3<br />Solution :<br />Divide equation (1) into equation (2)<br />
  47. 47. Rate Expression<br />
  48. 48. AssignmentDetermine the Rate Law for the following reaction from the given data:<br />2 NO (g) + O2 (g) -> 2 NO2 (g)<br />
  49. 49. Relationship Between Concentration and Time<br />First Order Reaction<br />
  50. 50. A plot of<br />Gives a Straight line<br />The Following Reaction is a First Order Reaction:<br />Plot the linear graph for concentration versus time<br />and obtain the rate constant for the reaction.<br />
  51. 51. Data for the Transformation of cylcpropane to propene<br />
  52. 52. slope = 2 x 10-5 s-1<br />
  53. 53. Data for the Transformation of cylcpropane to propene<br />
  54. 54. -slope = -2 x 10-5s-1<br />Slope = 2 x 10-5 s-1<br />
  55. 55. Relationship Between Concentration and Time<br />Second Order Reaction<br />
  56. 56. A plot of<br />Gives a Straight line<br />The Following Reaction is a Second Order Reaction:<br />2 HI (g) -> H2 (g) + I2 (g)<br />Plot the linear graph for concentration versus time<br />and obtain the rate constant for the reaction.<br />
  57. 57. Data for the Transformation of hydrogen iodide gas to hydrogen and iodine<br />
  58. 58. slope = 30. L mol-1 min-1<br />
  59. 59. Graphical Method for Determining the Order of a Reaction<br />First Order Reaction: y = ln (ao – x); x = t; slope = -k; and the intercept is lnao<br /> or<br /> y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0 <br />Second Order Reaction: y = 1/ (ao – x); x = t; slope = k; and the intercept = 1/ ao<br />Zero Order Reaction: y = x; x = t; slope = k and the intercept = 0<br />or y = ao – x ; x = t; slope = -k and the intercept = ao<br />
  60. 60. Zero Order Reaction<br />
  61. 61. Application of the Graphical Method for Determining the Order of a Reaction<br />N2O5 (g) -> 2 NO2 (g) + ½ O2 (g)<br />Tabulate the data so that each order may be tested<br />
  62. 62. Data tabulation to determine which order will give a linear graph<br />
  63. 63. Test for zero order reaction<br />Not linear; therefore, the reaction is not zero order<br />
  64. 64. Test for first order reaction<br />Linear; therefore, the reaction is first order<br />
  65. 65. Test for second order reaction<br />Non-linear; therefore, the reaction is not second order<br />
  66. 66. Half Life for a First Order Reaction<br />
  67. 67. Half Life for a Second Order Reaction<br />
  68. 68. Half Life for a Zero Order Reaction<br />
  69. 69. Application of Half Life<br />The rate constant for transforming cyclopropane into propene is 0.054 h-1<br />Calculate the half-life of cyclopropane.<br />Calculate the fraction of cyclopropane remaining after 18.0 hours.<br />Calculate the fraction of cyclopropane remaining after 51.5 hours.<br />
  70. 70. Half-Life<br />Fraction of cyclopropane<br />Remaining after 18.0 hours<br />Fraction of cyclopropane<br />Remaining after 51.5 hours<br />
  71. 71. Effect of Temperature on the Reaction Rate<br />Arrhenius Equation<br />
  72. 72. Use the Following Data to Determine the Eactfor<br />
  73. 73.
  74. 74. Effect of a Catalyst on the Rate of a Reaction<br />Lowers the energy barrier to the reaction via lowering the energy of activation<br />Homogeneous catalyst- in the same phase as the reacting molecules<br />Herterogeneous catalyst – in a different phase from the reacting molecules<br />
  75. 75. Example of a Homogeneous Catalyst<br />
  76. 76.
  77. 77. Example of Heterogeneous Catalyst<br />
  78. 78. An interesting problem:<br />The reaction between propionaldehyde and hydrocyanic acid <br />have been observed by Svirbely and Roth and reported in the <br />Journal of the American Chemical Society. Use this data to <br />ascertain the order of the reaction and the value of the rate <br />constant for this reaction.<br />
  79. 79.
  80. 80. Check to determine first order in HCN<br />
  81. 81. Check to determine first order in propionaldehyde<br />
  82. 82. So close; therefore, let’s take another approach. Let [HCN] = ao and <br />[propionaldehyde] = bo<br />Then,<br />
  83. 83.
  84. 84.
  85. 85. Let’s construct the data in a different format<br />
  86. 86. Slope = 0.678; therefore, k = 0.678 M-1min-1<br />
  87. 87. Mechanism:<br />
  88. 88. Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step<br />
  89. 89.
  90. 90.
  91. 91. Revisit the kinetics for<br />2 NO (g) + O2 (g) -> 2 NO2 (g)<br />

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