Chapter 14 Lecture- Chemical Kinetics

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Chapter 14 Lecture on chemical kinetics

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  • Chapter 14 Lecture- Chemical Kinetics

    1. 1. <ul><li>Chapter 14- Chemical Kinetics (Reaction Rates) </li></ul><ul><li>Sections 14.1 - 14.2- Reaction Rates </li></ul><ul><ul><ul><li>1. Read pgs 575-582 </li></ul></ul></ul><ul><ul><ul><li>Pg 617 #1, 2, 11, 13, 18, 19 </li></ul></ul></ul>
    2. 2. Kinetics <ul><li>Studies the rate at which a chemical process occurs. </li></ul><ul><li>Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). </li></ul>
    3. 3. Factors That Affect Reaction Rates <ul><li>The nature of the reactants will determine many aspects of the reaction rate and humans have no control over this. </li></ul><ul><ul><li>But there are many factors that can be manipulated in order to speed up or slow down chemical reactions. </li></ul></ul>
    4. 4. Factors That Affect Reaction Rates <ul><li>Physical State of the Reactants </li></ul><ul><ul><li>In order to react, molecules must come in contact with each other. </li></ul></ul><ul><ul><li>The more homogeneous the mixture of reactants, the faster the molecules can react. </li></ul></ul>
    5. 5. Factors That Affect Reaction Rates <ul><li>Increase Concentration of Reactants </li></ul><ul><ul><li>As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. </li></ul></ul><ul><ul><ul><li>Reducing volume or increasing pressure on a gas will cause a concentration increase. </li></ul></ul></ul><ul><ul><ul><li>Increasing the surface area of a solid increases concentration of available particles. </li></ul></ul></ul>
    6. 6. Factors That Affect Reaction Rates <ul><li>Increase Temperature </li></ul><ul><ul><li>At higher temperatures, reactant molecules have more kinetic energy, move faster, and </li></ul></ul><ul><ul><li>collide more often AND with greater energy . </li></ul></ul>
    7. 7. Factors That Affect Reaction Rates <ul><li>Add a Catalyst </li></ul><ul><ul><li>Catalysts speed up reactions by changing the mechanism of the reaction. </li></ul></ul><ul><ul><li>Catalysts are not consumed during the course of the reaction. </li></ul></ul>
    8. 9. <ul><li>Increasing the partial pressures of the reactive components of a gaseous mixture increases the rate of reaction. </li></ul>
    9. 10. Reaction Rates <ul><li>Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. </li></ul>
    10. 11. Reaction Rates <ul><li>In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. </li></ul>C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq )
    11. 12. Reaction Rates <ul><li>The average rate of the reaction over each interval is the change in concentration divided by the change in time: </li></ul>C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq ) Average rate =  [C 4 H 9 Cl]  t
    12. 13. Reaction Rates <ul><li>Note that the average rate decreases as the reaction proceeds. </li></ul><ul><li>This is because as the reaction goes forward, there are fewer collisions between reactant molecules. </li></ul>C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq )
    13. 14. Reaction Rates <ul><li>A plot of concentration vs. time for this reaction yields a curve like this. </li></ul><ul><li>The slope of a line tangent to the curve at any point is the instantaneous rate at that time. </li></ul>C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq )
    14. 15. Reaction Rates <ul><li>All reactions slow down over time. </li></ul><ul><li>Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning. </li></ul>C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq )
    15. 17. <ul><li>The size of the triangle is mostly a matter of convenience; different size triangles give the same ratio (slope). </li></ul>
    16. 18. Reaction Rates and Stoichiometry <ul><li>In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. </li></ul><ul><li>Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. </li></ul>C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t
    17. 19. Reaction Rates and Stoichiometry <ul><li>What if the ratio is not 1:1? </li></ul>2 HI ( g )  H 2 ( g ) + I 2 ( g ) <ul><li>Therefore , </li></ul>Rate = − 1 2  [HI]  t =  [I 2 ]  t
    18. 20. Reaction Rates and Stoichiometry <ul><li>To generalize, then, for the reaction </li></ul>a A + b B c C + d D Rate = − 1 a  [A]  t = − 1 b  [B]  t = 1 c  [C]  t 1 d  [D]  t =
    19. 21. For the reaction: A  2 B , which of the following is a correct expression for the reaction rate? t   Rate [A] = t   Rate 2 [A] = 2. t   Rate 2 [B] = 3. 4. t   Rate 2 [B] = 1.
    20. 22. Correct Answer: In general for aA + bB  cC + dD t   Rate [A] = t   Rate 2 [A] = 2. t   Rate 2 [B] = 3. 4. t   Rate 2 [B] = 1. =  =  = =
    21. 23. For the reaction: 2 A  3 B the rate of appearance of B,  [B]/  t , is 0.30 M /s. What is the value of the rate of disappearance of A, –  [A]/  t ? <ul><li>0.05 M /s </li></ul><ul><li>0.10 M /s </li></ul><ul><li>0.20 M /s </li></ul><ul><li>0.40 M /s </li></ul><ul><li>0.60 M /s </li></ul>
    22. 24. t t     Rate 3 [B] 2 [A] = = Correct Answer: t t     Rate 3 [B] 2 [A] = = /s 0.20 3 /s) 2(0.30 [A] M M t = = =   Rate <ul><li>0.05 M /s </li></ul><ul><li>0.10 M /s </li></ul><ul><li>0.20 M /s </li></ul><ul><li>0.40 M /s </li></ul><ul><li>0.60 M /s </li></ul>
    23. 25. <ul><li>Chapter 14- Chemical Kinetics (Reaction Rates) </li></ul><ul><li>Section 14.3- Concentration & Rate </li></ul><ul><li> (rate laws & initial rates data ) </li></ul><ul><ul><ul><li>2. Read pgs 582-587 </li></ul></ul></ul><ul><ul><ul><li>#3, 21, 23, 27, 29, 30a&b, 31, 32a </li></ul></ul></ul>
    24. 26. Concentration and Rate <ul><li>One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. </li></ul>
    25. 27. Concentration and Rate <ul><li>Comparing Experiments 1 and 2, when [NH 4 + ] doubles, the initial rate doubles. </li></ul>NH 4 + ( aq ) + NO 2 − ( aq ) N 2 ( g ) + 2 H 2 O ( l )
    26. 28. Concentration and Rate <ul><li>Likewise, comparing Experiments 5 and 6, when [NO 2 − ] doubles, the initial rate doubles. </li></ul>NH 4 + ( aq ) + NO 2 − ( aq ) N 2 ( g ) + 2 H 2 O ( l )
    27. 29. Concentration and Rate <ul><li>This means </li></ul><ul><li>Rate  [NH 4 + ] </li></ul><ul><li>Rate  [NO 2 − ] </li></ul><ul><li>Rate  [NH + ] [NO 2 − ] </li></ul><ul><li>or </li></ul><ul><li>Rate = k [NH 4 + ] [NO 2 − ] </li></ul><ul><li>This equation is called the rate law , and k is the rate constant . </li></ul>
    28. 30. For the reaction: A + 2 B  B 2 A Doubling [A] doubles the rate of the reaction. However, doubling [B] increases the rate 4  . What is the correct rate law expression? [A][B] Rate k = 3. 2. 4. 1.
    29. 31. For the reaction: A + 2 B  B 2 A Doubling [A] doubles the rate of the reaction. However, doubling [B] increases the rate 4  . What is the correct rate law expression? [A][B] Rate k = 3. 2. 4. 1. The rate order on A is one, because a doubling of [A] leads to a doubling of reaction rate. The rate order on B is two, because doubling [B] leads to a 4  = (2) 2 increase in reaction rate. Rate = k [A][B] 2
    30. 32. What is the rate law for the reaction? <ul><li>Rate = k [A][B] 2 </li></ul><ul><li>Rate = k [A][B] </li></ul><ul><li>Rate = k [A] 2 [B] </li></ul><ul><li>Rate = k [A] </li></ul><ul><li>Rate = k [B] </li></ul>A + 2 B products 2.0 x 10 – 4 0.20 0.20 1.0 x 10 – 4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate M /s [B] [A]
    31. 33. What is the rate law for the reaction? <ul><li>Rate = k [A][B] 2 </li></ul><ul><li>Rate = k [A][B] </li></ul><ul><li>Rate = k [A] 2 [B] </li></ul><ul><li>Rate = k [A] </li></ul><ul><li>Rate = k [B] </li></ul>A + 2 B products 2.0 x 10 – 4 0.20 0.20 1.0 x 10 – 4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate M /s [B] [A]
    32. 34. What is the rate law for the reaction? <ul><li>Rate = k [A][B] 2 </li></ul><ul><li>Rate = k [A][B] </li></ul><ul><li>Rate = k [A] 2 [B] </li></ul><ul><li>Rate = k [A] </li></ul><ul><li>Rate = k [B] </li></ul>2 A + B products 8.0 x 10 -4 0.20 0.20 4.0 x 10 -4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate M /s [B] [A]
    33. 35. What is the rate law for the reaction? <ul><li>Rate = k [A][B] 2 </li></ul><ul><li>Rate = k [A][B] </li></ul><ul><li>Rate = k [A] 2 [B] </li></ul><ul><li>Rate = k [A] </li></ul><ul><li>Rate = k [B] </li></ul>2 A + B products 8.0 x 10 -4 0.20 0.20 4.0 x 10 -4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate M /s [B] [A]
    34. 36. Rate Laws Rate = k [NH 4 + ] [NO 2 − ] <ul><li>A rate law shows the relationship between the reaction rate and the concentrations of reactants. </li></ul><ul><li>exponents tell the order of the reaction with respect to each reactant. </li></ul><ul><li>This reaction is </li></ul><ul><li>First-order in [NH 4 + ] </li></ul><ul><li>First-order in [NO 2 − ] </li></ul>
    35. 37. <ul><li>The overall reaction order is found by adding the exponents on the reactants in the rate law. </li></ul><ul><li>This reaction is second-order overall. </li></ul><ul><ul><ul><ul><ul><li>Rate = k[A] 2 [B] </li></ul></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>2 nd order in [A] </li></ul></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>1st order in [B] </li></ul></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>3 rd order overall </li></ul></ul></ul></ul></ul>Rate Laws Rate = k [NH 4 + ] [NO 2 − ]
    36. 39. <ul><li>a.The rate law for any chemical reaction is the equation that relates concentrations of reactants to the rate of reaction. </li></ul>
    37. 40. <ul><li>a.The rate law for any chemical reaction is the equation that relates concentrations of reactants to the rate of reaction. </li></ul><ul><li>b. k in any rate law is the rate constant. </li></ul>
    38. 42. <ul><li>a. 2 nd order in NO, 1 st order in H 2 , 3 rd order overall. </li></ul><ul><li> </li></ul>
    39. 43. <ul><li>a. 2 nd order in NO, 1 st order in H 2 , 3 rd order overall. </li></ul><ul><li>b. NO </li></ul><ul><li> </li></ul>
    40. 44. What are the units of the rate constant for a reaction with a rate law of: <ul><li>M s </li></ul><ul><li>M 2 s </li></ul><ul><li>M -1 s -1 </li></ul><ul><li>M -1 s -2 </li></ul><ul><li>M -2 s -1 </li></ul>Rate = k [A][B]
    41. 45. What are the units of the rate constant for a reaction with a rate law of: <ul><li>M s </li></ul><ul><li>M 2 s </li></ul><ul><li>M -1 s -1 </li></ul><ul><li>M -1 s -2 </li></ul><ul><li>M -2 s -1 </li></ul>Rate = k [A][B]
    42. 46. <ul><li>Chapter 14- Chemical Kinetics (Reaction Rates) </li></ul><ul><li>Section 14.4- The Change of Concentration with Time (Integrated Rate Laws) </li></ul><ul><ul><ul><li>Read pgs 587-593 </li></ul></ul></ul><ul><ul><ul><li>#4, 5, 33, 34, 35, 37, 41, 43, 44 </li></ul></ul></ul>
    43. 47. Answers:   (a) rate = k [NO] 2 [H 2 ]; (b) k =1.2 M –2 s –1 ; (c) rate = 4.5  10 –4 M /s (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H 2 ] = 0.150 M . BELLWORK The following data were measured for the reaction of nitric oxide with hydrogen:
    44. 48. Rate laws (a.k.a. Differential rate laws ) are used to calculate reaction rates when given reactant concentrations. Integrated rate laws are used to calculate the amount of reactants left after a certain amount of time has passed.
    45. 49. Integrated Rate Laws <ul><li>Using calculus to integrate the rate law for a first-order process gives us </li></ul>Where [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t , during the course of the reaction. ln [A] t [A] 0 = − k t
    46. 50. Integrated Rate Laws <ul><li>Manipulating this equation produces… </li></ul>ln [A] t − ln [A] 0 = − k t ln [A] t = − k t + ln [A] 0 … which is in the form y = mx + b ln [A] t [A] 0 = − k t
    47. 51. First-Order Processes <ul><li>Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be - k . </li></ul>ln [A] t = - k t + ln [A] 0
    48. 52. First-Order Processes <ul><li>Consider the process in which methyl isonitrile is converted to acetonitrile. </li></ul>CH 3 NC CH 3 CN
    49. 53. First-Order Processes <ul><li>This data was collected for this reaction at 198.9 ° C. </li></ul>CH 3 NC CH 3 CN
    50. 54. First-Order Processes <ul><li>When ln P is plotted as a function of time, a straight line results. </li></ul><ul><li>Therefore, </li></ul><ul><ul><li>The process is first-order. </li></ul></ul><ul><ul><li>k is the negative slope: 5.1  10 -5 s − 1 . </li></ul></ul>
    51. 55. Second-Order Processes <ul><li>Similarly, integrating the rate law for a process that is second-order in reactant A, we get </li></ul>also in the form y = mx + b 1 [A] t = − kt + 1 [A] 0
    52. 56. Second-Order Processes <ul><li>So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k . </li></ul>1 [A] t = − kt + 1 [A] 0
    53. 57. Second-Order Processes The decomposition of NO 2 at 300 ° C is described by the equation and yields data comparable to this: NO 2 ( g ) NO ( g ) + 1/2 O 2 ( g ) Time ( s ) [NO 2 ], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380
    54. 58. Second-Order Processes <ul><li>Graphing ln [NO 2 ] vs . t yields: </li></ul><ul><li>The plot is not a straight line, so the process is not first-order in [A]. </li></ul>Time ( s ) [NO 2 ], M ln [NO 2 ] 0.0 0.01000 − 4.610 50.0 0.00787 − 4.845 100.0 0.00649 − 5.038 200.0 0.00481 − 5.337 300.0 0.00380 − 5.573
    55. 59. Second-Order Processes <ul><li>Graphing ln 1/[NO 2 ] vs. t , however, gives this plot. </li></ul><ul><li>Because this is a straight line, the process is second-order in [A]. </li></ul>Time ( s ) [NO 2 ], M 1/[NO 2 ] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263
    56. 60. The plot given below indicates a component whose reaction order is __. <ul><li>One </li></ul><ul><li>Two </li></ul><ul><li>Zero </li></ul><ul><li>None of the above </li></ul>
    57. 61. The plot given below indicates a component whose reaction order is __. <ul><li>One </li></ul><ul><li>Two </li></ul><ul><li>Zero </li></ul><ul><li>None of the above </li></ul>Second-order reactions will exhibit a linear plot of 1/[reactant] vs. time.
    58. 62. a) b)
    59. 63. <ul><li>a.The initial pressure of CH 3 NC </li></ul>a) b)
    60. 64. <ul><li>a.The initial pressure of CH 3 NC </li></ul><ul><li>b. The natural log of the initial pressure of CH 3 NC </li></ul>a) b)
    61. 65. Half-Life <ul><li>Half-life is defined as the time required for one-half of a reactant to react. </li></ul><ul><li>Because [A] at t 1/2 is one-half of the original [A], </li></ul><ul><li>[A] t = 0.5 [A] 0 . </li></ul>
    62. 66. Half-Life <ul><li>For a first-order process, this becomes </li></ul>ln 0.5 = − kt 1/2 − 0.693 = − kt 1/2 NOTE: For a first-order process, the half-life does not depend on [A] 0 . 0.5 [A] 0 [A] 0 ln = − kt 1/2 = t 1/2 0.693 k
    63. 67. Half-Life <ul><li>For a second-order process, </li></ul>1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 = = t 1/2 1 k [A] 0
    64. 68. For a first order reaction, k = 0.0693 s  1 and the initial concentration of the reactant is 0.10 M . Calculate the half-life of the reactant. <ul><li>0.1 s </li></ul><ul><li>1.0 s </li></ul><ul><li>10. s </li></ul><ul><li>14.4 s </li></ul><ul><li>144 s </li></ul>
    65. 69. For a first order reaction, k = 0.0693 s  1 and the initial concentration of the reactant is 0.10 M . Calculate the half-life of the reactant. <ul><li>0.1 s </li></ul><ul><li>1.0 s </li></ul><ul><li>10. s </li></ul><ul><li>14.4 s </li></ul><ul><li>144 s </li></ul>Because the reaction is first order, half-life depends only on the rate constant according to the equation: t 1/2 = 0.693/ k = 0.0693/0.0693 s  1 t 1/2 = 10. s
    66. 70. = t 1/2 1 k [A] 0
    67. 71. <ul><li>The half-life increases as the reaction proceeds. </li></ul>= t 1/2 1 k [A] 0
    68. 72. If t 1/2 = 500 s for a first-order reaction, how long will it take for the concentration of a reactant to decrease to 1/8 of its original value? <ul><li>2 x 500 = 1000 s </li></ul><ul><li>3 x 500 = 1500 s </li></ul><ul><li>4 x 500 = 2000 s </li></ul><ul><li>5 x 500 = 2500 s </li></ul><ul><li>8 x 500 = 4000 s </li></ul>
    69. 73. If t 1/2 = 500 s for a first-order reaction, how long will it take for the concentration of a reactant to decrease to 1/8 of its original value? <ul><li>2 x 500 = 1000 s </li></ul><ul><li>3 x 500 = 1500 s </li></ul><ul><li>4 x 500 = 2000 s </li></ul><ul><li>5 x 500 = 2500 s </li></ul><ul><li>8 x 500 = 4000 s </li></ul>
    70. 74. SUMMARY Order of reaction Zero First Second Rate law Rate = k Rate = k[A] Rate = k[A] 2 Integrated rate law [A] t = [A] 0 - kt ln[A] t - ln[A] 0 = -kt 1/[A] t - 1/[A] 0 = kt Units of k M/s 1/s 1/(M •s) Linear plot [A] vs. t ln[A] vs. t 1/[A] vs. t slope -k -k k Half life t 1/2 = 1/2[A] 0 k t 1/2 = 0.693/k t 1/2 = 1/(k[A] 0 )
    71. 75. <ul><li>Chapter 14- Chemical Kinetics (Reaction Rates) </li></ul><ul><li>Section 14.5- Temperature & Rate (What affects k ?) </li></ul><ul><li>Section 14.6- Reaction Mechanisms </li></ul><ul><li>Section 14.7- Catalysts </li></ul><ul><ul><ul><li>4. Read pgs 593-610 </li></ul></ul></ul><ul><ul><ul><li>#7, 8, 45, 49, 51, 53, 57, 59, 61, 65, 67, 69, 71, 79 </li></ul></ul></ul>
    72. 76. Bellwork a)What is the order of the reaction? b)What is the rate law? c) What effect will doubling the initial concentration of A have on the reaction rate?
    73. 77. Temperature and Rate <ul><li>Generally, as temperature increases, so does the reaction rate. </li></ul><ul><li>This is because k is temperature dependent. </li></ul>
    74. 78. The Collision Model <ul><li>In a chemical reaction, bonds are broken and new bonds are formed. </li></ul><ul><li>Molecules can only react if they collide with each other. </li></ul>
    75. 79. The Collision Model <ul><li>Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. </li></ul>
    76. 81. <ul><li>Molecules must collide in order to react. </li></ul>
    77. 82. Activation Energy <ul><li>In other words, there is a minimum amount of energy required for reaction: the activation energy , E a . </li></ul><ul><li>Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. </li></ul>
    78. 83. Reaction Coordinate Diagrams <ul><li>It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. </li></ul>
    79. 84. Reaction Coordinate Diagrams <ul><li>It shows the energy of the reactants and products (and, therefore,  E ). </li></ul><ul><li>The high point on the diagram is the transition state . </li></ul><ul><li>The species present at the transition state is called the activated complex . </li></ul><ul><li>The energy gap between the reactants and the activated complex is the activation energy barrier. </li></ul>
    80. 85. Reaction Coordinate Diagrams <ul><li>Comparing the activation energy ( E a ) of the forward and reverse reactions, E a is always higher for the endothermic process. </li></ul>
    81. 86. Maxwell – Boltzmann Distributions <ul><li>Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. </li></ul><ul><li>At any temperature there is a wide distribution of kinetic energies. </li></ul>
    82. 87. Maxwell – Boltzmann Distributions <ul><li>As the temperature increases, the curve flattens and broadens. </li></ul><ul><li>Thus at higher temperatures, a larger population of molecules has higher energy. </li></ul>
    83. 88. Maxwell – Boltzmann Distributions <ul><li>If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. </li></ul><ul><li>As a result, the reaction rate increases. </li></ul>
    84. 89. Maxwell – Boltzmann Distributions <ul><li>This fraction of molecules can be found through the expression </li></ul><ul><li>where R is the gas constant and T is the Kelvin temperature. </li></ul>f = e − E a / RT
    85. 91. <ul><li>Molecules must also collide both with the proper orientation and with sufficient energy to react. </li></ul>
    86. 92. Arrhenius Equation <ul><li>Svante Arrhenius developed a mathematical relationship between k and E a : </li></ul><ul><li>k = A e − E a / RT </li></ul><ul><li>where A is the frequency factor , a number that represents the likelihood that collisions would occur with the proper orientation for reaction. </li></ul>
    87. 93. Arrhenius Equation <ul><li>Taking the natural logarithm of both sides, the equation becomes </li></ul><ul><li>ln k = - E a ( ) + ln A </li></ul>y = mx + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. 1/ T . 1 RT
    88. 94. The plot below of ln k vs. 1/ T is useful because it allows determination of: <ul><li>Reaction order </li></ul><ul><li>Activation Energy </li></ul><ul><li>Activation temperature </li></ul>
    89. 95. The plot below of ln k vs. 1/ T is useful because it allows determination of: <ul><li>Reaction order </li></ul><ul><li>Activation Energy </li></ul><ul><li>Activation temperature </li></ul>The slope of the line is equal to  R / E a , where E a is the activation energy.
    90. 96. Reaction Mechanisms <ul><li>The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism . </li></ul>
    91. 97. Reaction Mechanisms <ul><li>Reactions may occur all at once or through several discrete steps. </li></ul><ul><li>Each of these processes is known as an elementary reaction or elementary process . </li></ul>
    92. 98. Reaction Mechanisms <ul><li>The molecularity of a process tells how many molecules are involved in the process. </li></ul><ul><li>For an elementary reaction, the rate law ( for that step in the mechanism only ) can be written from the equation of the elementary reaction. </li></ul>
    93. 99. <ul><li>bimolecular </li></ul><ul><li>dimolecular </li></ul><ul><li>termolecular </li></ul><ul><li>unimolecular </li></ul>
    94. 100. <ul><li>bimolecular </li></ul><ul><li>dimolecular </li></ul><ul><li>termolecular </li></ul><ul><li>unimolecular </li></ul>
    95. 101. The following mechanism has been proposed for: 2 NO + Cl 2  2 NOCl NO + Cl 2  NOCl 2 NOCl 2 + NO  2 NOCl Identify the reaction intermediate. <ul><li>NO </li></ul><ul><li>Cl 2 </li></ul><ul><li>NOCl 2 </li></ul><ul><li>NOCl </li></ul>
    96. 102. Correct Answer: NOCl 2 is not observed in the overall chemical reaction; therefore it is a short-lived reaction intermediate. <ul><li>NO </li></ul><ul><li>Cl 2 </li></ul><ul><li>NOCl 2 </li></ul><ul><li>NOCl </li></ul>
    97. 104. Multistep Mechanisms <ul><li>In a multistep process, one of the steps will be slower than all others. </li></ul><ul><li>The overall reaction cannot occur faster than this slowest, rate-determining step . </li></ul>
    98. 105. Slow Initial Step <ul><li>The rate law for this reaction is found experimentally to be </li></ul><ul><li>Rate = k [NO 2 ] 2 </li></ul><ul><li>CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. </li></ul><ul><li>This suggests the reaction occurs in two steps. </li></ul>NO 2 ( g ) + CO ( g )  NO ( g ) + CO 2 ( g )
    99. 106. Slow Initial Step <ul><li>A proposed mechanism for this reaction is </li></ul><ul><li>Step 1: NO 2 + NO 2  NO 3 + NO (slow) </li></ul><ul><li>Step 2: NO 3 + CO  NO 2 + CO 2 (fast) </li></ul><ul><li>The NO 3 intermediate is consumed in the second step. </li></ul><ul><li>As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. </li></ul>
    100. 107. Fast Initial Step <ul><li>The rate law for this reaction is found to be </li></ul><ul><li>Rate = k [NO] 2 [Br 2 ] </li></ul><ul><li>Because termolecular processes are rare, this rate law suggests a two-step mechanism. </li></ul>2 NO ( g ) + Br 2 ( g )  2 NOBr ( g )
    101. 108. Fast Initial Step <ul><li>A proposed mechanism is </li></ul>Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)
    102. 109. <ul><li>The rate of the overall reaction depends upon the rate of the slow step. </li></ul><ul><li>The rate law for that step would be </li></ul><ul><li>Rate = k [NOBr 2 ] [NO] </li></ul><ul><li>But how can we find [NOBr 2 ]? </li></ul>Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)
    103. 110. <ul><li>The rate of NOBr 2 formation is determined by [NO] and [Br 2 ] </li></ul><ul><li>Rate = k [NOBr 2 ] [NO] </li></ul><ul><li>Rate = k [NO] [Br 2 ] [NO] </li></ul><ul><li>Rate = k [NO] 2 [Br 2 ] </li></ul>Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1: NO + Br 2 NOBr 2 (fast)
    104. 111. <ul><li>The rate law is determined by the all reactants involved before the slowest step. </li></ul><ul><li>Rate laws must be determined experimentally because you can’t tell what the mechanism is by looking at the equation. </li></ul>
    105. 112. For 2 NO + Cl 2  2 NOCl, the following mechanism has been proposed. <ul><li>First </li></ul><ul><li>Second </li></ul>NO + Cl 2  NOCl 2 NOCl 2 + NO  2 NOCl The rate law is Rate = k [NO] 2 [Cl 2 ], which implies which step is the rate-limiting step?
    106. 113. Correct Answer: If the first step were rate limiting, the observed rate law would be: Rate = k [NO][Cl 2 ] The square dependence on [NO] indicates the second step must be slow relative to the first step. <ul><li>First </li></ul><ul><li>Second </li></ul>
    107. 114. <ul><li>All reactions are not elementary. </li></ul><ul><li>Some information must be known about the rate constant to determine the rate law. </li></ul><ul><li>Concentrations of reactant must be known to determine the rate law. </li></ul><ul><li>The rate law depends not on the overall reaction, but on the slowest step in the mechanism. </li></ul>
    108. 115. <ul><li>All reactions are not elementary. </li></ul><ul><li>Some information must be known about the rate constant to determine the rate law. </li></ul><ul><li>Concentrations of reactant must be known to determine the rate law. </li></ul><ul><li>The rate law depends not on the overall reaction, but on the slowest step in the mechanism. </li></ul>
    109. 116. Catalysts <ul><li>Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. </li></ul><ul><li>Catalysts change the mechanism by which the process occurs. </li></ul>
    110. 117. Catalysts <ul><li>One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. </li></ul>
    111. 118. Homogenous catalysts are in the same phase as the reactants. Heterogeneous catalysts are in a different phase than the reactants. Pt, Pd, & Rh metals are used to catalyze gas reactions in a catalytic converter.
    112. 119. <ul><li>by lowering the  H for the reaction </li></ul><ul><li>by lowering the activation energy by providing a lower energy pathway (different mechanism) for the reaction </li></ul><ul><li>by lowering the product energy </li></ul><ul><li>by lowering the reactant energy </li></ul>
    113. 120. <ul><li>by lowering the  H for the reaction </li></ul><ul><li>by lowering the activation energy by providing a lower energy pathway (different mechanism) for the reaction </li></ul><ul><li>by lowering the product energy </li></ul><ul><li>by lowering the reactant energy </li></ul>
    114. 121. <ul><li>A proposed mechanism for the decomposition of N 2 O is given below: </li></ul><ul><li>NO + N 2 O  N 2 + NO 2 </li></ul><ul><ul><li>2 NO 2  2 NO + O 2 </li></ul></ul><ul><ul><li>Which species is the catalyst? </li></ul></ul><ul><li>NO </li></ul><ul><li>NO 2 </li></ul><ul><li>N 2 O </li></ul><ul><li>N 2 </li></ul><ul><li>O 2 </li></ul>
    115. 122. Correct Answer: <ul><li>Note in the mechanism below that NO is regenerated. By definition a catalyst does not undergo permanent chemical change. </li></ul><ul><ul><li>NO + N 2 O  N 2 + NO 2 </li></ul></ul><ul><ul><li>2 NO 2  2 NO + O 2 </li></ul></ul><ul><li>NO </li></ul><ul><li>NO 2 </li></ul><ul><li>N 2 O </li></ul><ul><li>N 2 </li></ul><ul><li>O 2 </li></ul>
    116. 123. Enzymes <ul><li>Enzymes are catalysts in biological systems. </li></ul><ul><li>The substrate fits into the active site of the enzyme much like a key fits into a lock. </li></ul>
    117. 124. <ul><li>a. </li></ul><ul><li>sweet spot </li></ul><ul><li>hot spot </li></ul><ul><li>active site </li></ul><ul><li>binding site </li></ul><ul><li>b. </li></ul><ul><li>receptor </li></ul><ul><li>substrate </li></ul><ul><li>antigen </li></ul><ul><li>catalyst </li></ul>
    118. 125. <ul><li>a. </li></ul><ul><li>sweet spot </li></ul><ul><li>hot spot </li></ul><ul><li>active site </li></ul><ul><li>binding site </li></ul><ul><li>b. </li></ul><ul><li>receptor </li></ul><ul><li>substrate </li></ul><ul><li>antigen </li></ul><ul><li>catalyst </li></ul>

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