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Kinetics  pp ,[object Object],[object Object],[object Object],[object Object]
12.1 Reaction Rate ,[object Object],[object Object],[object Object],[object Object]
[object Object],Concentration Time [H 2 ]
[object Object],[object Object],Concentration Time [H 2 ] [N 2 ]
[object Object],Concentration Time [H 2 ] [N 2 ] [NH 3 ]
Calculating Rates ,[object Object],[object Object],[object Object]
[object Object],Concentration Time  [H 2 ]  t
[object Object],Concentration Time  [H 2 ]    t
Defining Rate  pp ,[object Object],[object Object],[object Object],[object Object]
12.2 Rate Laws:  An Introduction ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],Rate Laws  pp
Rate Laws ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],2 NO 2  2   NO  +  O 2
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],2 NO 2  2   NO  +  O 2
Figure 12.1 p. 562 Definition of Rate. 2NO 2     2NO + O 2
Types of Rate Laws ,[object Object],[object Object],[object Object],[object Object]
12.3 Determining Rate Laws ,[object Object],[object Object],[object Object],[object Object]
[object Object],[N 2 O 5 ] (mol/L)    Time (s) 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000
[object Object],[object Object]
At  .90  M   the rate is - (.98 - .76)  =   -0.22  =  5.5x 10  -4   M• s -1     (0-400)  -400
At  .45 M  the rate is - (.52 - .31)  =   -0.22  = 2.7 x 10  -4   (1000-1800)  -800
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
A Plot of [N 2 O 5 ] as f(t) for 2N 2 O 5    4NO 2  + O 2   pp ,[object Object],[object Object]
The method of Initial Rates  pp ,[object Object],[object Object],[object Object],[object Object]
Here’s an AP-type Question  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
An AP Test Example  pp ,[object Object],[object Object],[object Object],[object Object]
[object Object],Initial concentrations (M)  pp Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
[object Object],[object Object],Initial concentrations ( M )  pp Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
The math  pp ,[object Object],[object Object],[object Object]
[object Object],[object Object],Initial concentrations ( M )  pp Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
[object Object],[object Object],Initial concentrations ( M )  pp Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
AP-type question  pp ,[object Object],[object Object],[object Object],[object Object]
AP-type question  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
AP-type question ,[object Object],[object Object],[object Object],[object Object],[object Object]
12.4 Integrated Rate Law ,[object Object],[object Object],[object Object],[object Object]
First Order ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],First  Order
[object Object],[object Object],[object Object],First Order
A Plot of [N 2 O 5 ] as f(t) for 2N 2 O 5    4NO 2  + O 2 . ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],First Order
Figure 12.4 A Plot of  In (N 2 O 5 ) Versus Time (1st order)
Half Life ,[object Object],[object Object],[object Object],[object Object]
Figure 12.5:  A Plot of (N 2 O 5 ) vs. Time for the  Decomposition  Reaction of N 2 O 5
Half Life ,[object Object],[object Object],[object Object],[object Object],[object Object]
Half Life ,[object Object],[object Object],[object Object]
Second Order ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Second Order Half Life ,[object Object]
Z7e  545 Fig 12.6 ,[object Object],[object Object]
Determining Rate Laws ,[object Object],[object Object]
Zero Order Rate Law ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Z7e 546 Figure 12.7 A Plot of  [A] Versus  t   for a  Zero-Order  Reaction
[object Object],[object Object],Zero Order Rate Law
Fig. 12.8 Decomposition Reaction N 2 O    2N 2  + O 2 Even though [N 2 O] is   twice as great in (b) as in (a)   the reaction occurs on a   saturated   platinum surface, so rate is   zero order
Zero Order Graph Time Concentration
Zero Order Graph Time Concentration  A]/  t = slope  t -k =   A] t 1/2  = [A 0 ]/2k
More Complicated Reactions ,[object Object],[object Object],[object Object],[object Object]
Rate = k[BrO 3 - ][Br - ][H + ] 2 ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Rate = k[BrO 3 - ][Br - ][H + ] 2
12.5 Summary of Rate Laws ,[object Object],[object Object],[object Object],[object Object]
Summary of Rate Laws ,[object Object],[object Object],[object Object],[object Object]
Summary of Rate Laws  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Use for Online HW  pp Use for online HW Hint:  on one problem you will have to first determine the  order , then use the  half-life equation  (5 th  line) to solve for k, then use the  Integrated rate law  (2 nd  line) to solve for time (t).
12.6 Reaction Mechanisms ,[object Object],[object Object],[object Object]
Figure 12.9 A Molecular Representation of the Elementary Steps in the Reaction of NO 2  and CO   Overall:  NO 2  + CO    NO + CO 2 Step 1:  NO 2  + NO 2      NO 3  + NO  (k 1 ) Step 2:  NO 3  + CO    NO 2  + CO 2   (k 2 ) NO 3  is an  intermediate
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Reaction Mechanisms
[object Object],[object Object],[object Object],Reaction Mechanisms
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Finding the Reaction Mechanism  pp   ,[object Object],[object Object],[object Object]
How to get rid of intermediates  pp ,[object Object],[object Object],[object Object]
Formed in  reversible  reactions  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Formed in  reversible  reactions  pp ,[object Object],[object Object],[object Object],[object Object]
Formed in  reversible  reactions  pp   Given:  Expt-derived Rate =  k [NO] 2 [O 2 ] ,[object Object],[object Object],[object Object]
Formed in  reversible  reactions  pp ,[object Object],[object Object],[object Object],[object Object]
Formed in  reversible  reactions  pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Formed in  fast, irreversible  reactions  pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Formed in fast, irreversible reactions  pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
12.7 A Model for Chemical Kinetics ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Terms ,[object Object],[object Object],[object Object],[object Object]
Potential Energy Reaction Coordinate Reactants Products
Potential Energy Reaction Coordinate Reactants Products Activation Energy E a
Potential Energy Reaction Coordinate Reactants Products Activated complex
Potential Energy Reaction Coordinate Reactants Products  E }
Potential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State
Arrhenius ,[object Object],[object Object],[object Object],[object Object]
Arrhenius ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Problems ,[object Object],[object Object],[object Object],[object Object]
Figure 12.13 Several Possible Orientations for a Collision Between Two BrNO Molecules.   (a) & (b) can lead to a collision, (c) cannot.
No Reaction O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br
Arrhenius Equation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Figure 12.14  pp   Plot of In( k )  vs.  1/ T  for   2N 2 O 5( g )      g )  + O 2( g ). Can get E a  from slope = -E a /R Do in lab 12.
Figure 12.14 Plot of In( k )  vs.  1/ T  for   2N 2 O 5( g )      g )  + O 2( g ). Can get E a  from slope = -E a /R Do in lab 12.
Activation Energy and Rates The final saga
Mechanisms  and rates   ,[object Object],[object Object],[object Object],[object Object]
[object Object]
[object Object],[object Object],E a
Second step is slow High activation energy E a
E a Third step is fast Low activation energy
Second step is rate determining
Intermediates are present
Activated Complexes or Transition States
12.8 Catalysts ,[object Object],[object Object],[object Object],[object Object]
How Catalysts Work ,[object Object],[object Object],[object Object],[object Object]
Figure 12.15 Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction. ∆ E  is the same in both cases.
Figure 12.16 Effect of a Catalyst on the  Number  of Reaction-Producing  Collisions   (increases even though no temperature increase).
[object Object],[object Object],Heterogenous Catalysts Pt surface H H H H H H H H
Heterogenous Catalysts Pt surface H H H H C H H C H H
Heterogenous Catalysts ,[object Object],Pt surface H H H H C H H C H H
Heterogenous Catalysts ,[object Object],Pt surface H H H H C H H C H H
Heterogenous Catalysts Pt surface H C H H C H H H H H
Heterogenous Catalysts ,[object Object],[object Object],[object Object],[object Object],[object Object]
Homogenous Catalysts ,[object Object],[object Object],[object Object]
Catalysts and rate ,[object Object],[object Object],[object Object],[object Object]
Catalysts and rate. ,[object Object],[object Object],Concentration of reactants Rate
 

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Ch12 z5e kinetics

  • 1.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15. Figure 12.1 p. 562 Definition of Rate. 2NO 2  2NO + O 2
  • 16.
  • 17.
  • 18.
  • 19.
  • 20. At .90 M the rate is - (.98 - .76) = -0.22 = 5.5x 10 -4 M• s -1 (0-400) -400
  • 21. At .45 M the rate is - (.52 - .31) = -0.22 = 2.7 x 10 -4 (1000-1800) -800
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.
  • 27.
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  • 33.
  • 34.
  • 35.
  • 36.
  • 37.
  • 38.
  • 39.
  • 40.
  • 41. Figure 12.4 A Plot of In (N 2 O 5 ) Versus Time (1st order)
  • 42.
  • 43. Figure 12.5: A Plot of (N 2 O 5 ) vs. Time for the Decomposition Reaction of N 2 O 5
  • 44.
  • 45.
  • 46.
  • 47.
  • 48.
  • 49.
  • 50.
  • 51. Z7e 546 Figure 12.7 A Plot of [A] Versus t for a Zero-Order Reaction
  • 52.
  • 53. Fig. 12.8 Decomposition Reaction N 2 O  2N 2 + O 2 Even though [N 2 O] is twice as great in (b) as in (a) the reaction occurs on a saturated platinum surface, so rate is zero order
  • 54. Zero Order Graph Time Concentration
  • 55. Zero Order Graph Time Concentration  A]/  t = slope  t -k =  A] t 1/2 = [A 0 ]/2k
  • 56.
  • 57.
  • 58.
  • 59.
  • 60.
  • 61.
  • 62. Use for Online HW pp Use for online HW Hint: on one problem you will have to first determine the order , then use the half-life equation (5 th line) to solve for k, then use the Integrated rate law (2 nd line) to solve for time (t).
  • 63.
  • 64. Figure 12.9 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO Overall: NO 2 + CO  NO + CO 2 Step 1: NO 2 + NO 2  NO 3 + NO (k 1 ) Step 2: NO 3 + CO  NO 2 + CO 2 (k 2 ) NO 3 is an intermediate
  • 65.
  • 66.
  • 67.
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  • 71.
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  • 78.
  • 79.
  • 80. Potential Energy Reaction Coordinate Reactants Products
  • 81. Potential Energy Reaction Coordinate Reactants Products Activation Energy E a
  • 82. Potential Energy Reaction Coordinate Reactants Products Activated complex
  • 83. Potential Energy Reaction Coordinate Reactants Products  E }
  • 84. Potential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State
  • 85.
  • 86.
  • 87.
  • 88. Figure 12.13 Several Possible Orientations for a Collision Between Two BrNO Molecules. (a) & (b) can lead to a collision, (c) cannot.
  • 89. No Reaction O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br
  • 90.
  • 91. Figure 12.14 pp Plot of In( k ) vs. 1/ T for 2N 2 O 5( g )    g ) + O 2( g ). Can get E a from slope = -E a /R Do in lab 12.
  • 92. Figure 12.14 Plot of In( k ) vs. 1/ T for 2N 2 O 5( g )    g ) + O 2( g ). Can get E a from slope = -E a /R Do in lab 12.
  • 93. Activation Energy and Rates The final saga
  • 94.
  • 95.
  • 96.
  • 97. Second step is slow High activation energy E a
  • 98. E a Third step is fast Low activation energy
  • 99. Second step is rate determining
  • 101. Activated Complexes or Transition States
  • 102.
  • 103.
  • 104. Figure 12.15 Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction. ∆ E is the same in both cases.
  • 105. Figure 12.16 Effect of a Catalyst on the Number of Reaction-Producing Collisions (increases even though no temperature increase).
  • 106.
  • 107. Heterogenous Catalysts Pt surface H H H H C H H C H H
  • 108.
  • 109.
  • 110. Heterogenous Catalysts Pt surface H C H H C H H H H H
  • 111.
  • 112.
  • 113.
  • 114.
  • 115.  
  • 116.