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Ch12 z5e kinetics

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Ch12 z5e kinetics

1. 1. Kinetics pp The study of reaction rates . Spontaneous reactions are reactions that will happen - but we can’t tell how fast. Diamond will spontaneously turn to graphite – eventually. Reaction mechanism - the steps by which a reaction takes place.
2. 2. 12.1 Reaction Rate Rate = Conc. of A at t 2 - Conc. of A at t 1 t 2 - t 1 Rate =   [A]  t Change in concentration per unit time For this reaction N 2 + 3H 2 2NH 3
3. 3. N 2 + 3H 2 2NH 3 As the reaction progresses the concentration H 2 goes down Concentration Time [H 2 ]
4. 4. As the reaction progresses the concentration N 2 goes down 1/3 as fast as for H 2 3 moles H 2 being consumed for every 1 mole of N 2 Concentration Time [H 2 ] [N 2 ]
5. 5. As the reaction progresses the concentration NH 3 goes up “mole proportionally” Concentration Time [H 2 ] [N 2 ] [NH 3 ]
6. 6. Calculating Rates Average rates are taken over long intervals Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time We take the derivative.
7. 7. Average slope method Concentration Time  [H 2 ]  t
8. 8. Instantaneous slope method. Concentration Time  [H 2 ]  t
9. 9. Defining Rate pp We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product . In our example N 2 + 3H 2 2NH 3 -  [ N 2 ] = - 3  [ H 2 ] = 2  [ NH 3 ]  t  t  t Since [ reactants ] always decreases with time, any rate expression includes (-) sign
10. 10. 12.2 Rate Laws: An Introduction Reactions are reversible . As products accumulate they can begin to turn back into reactants. Early on the rate will depend on only the amount of reactants present. So, we want to measure the reactants as soon as they are mixed . This is called the Initial rate method .
11. 11. Two key points: The concentration of the products do not appear in the rate law because this is an initial rate (only the [reactants]). The order must be determined experimentally . Cannot be obtained from the equation’s coefficients (remember this!). Rate Laws pp
12. 12. Rate Laws N 2 + 3H 2  2NH 3 When forward &amp; reverse reaction rates are equal then there is no change in concentration of either. The reaction is at equilibrium . Rate = k [N 2 ] m [H 2 ] n Note: “m” and “n” are not coefficients of the balanced equation, must be determined experimentally .
13. 13. You will find the rate depends only on the concentration of the reactants . Rate = k [ NO 2 ] n This is called a rate law expression . k is called the rate constant. n is the order of the reactant - usually a positive integer (1st, 2nd, etc.). It is not the coefficient of the balanced reaction (e.g., it could be “1”) We can instead define Rate in terms of production of O 2 vs. consumption of NO 2 2 NO 2 2 NO + O 2
14. 14. The rate of appearance of O 2 is . . . Rate &apos; =  [ O 2 ] = k &apos; [ NO 2 ] n  t Because there are 2 NO 2 consumed for each O 2 produced . . . Rate = k[NO 2 ] n = 2 x Rate ’ = 2 x k &apos; [NO 2 ] n So k[NO 2 ] n = 2 x k &apos; [NO 2 ] n So k = 2 x k&apos; 2 NO 2 2 NO + O 2
15. 15. Figure 12.1 p. 562 Definition of Rate. 2NO 2  2NO + O 2
16. 16. Types of Rate Laws Differential Rate law - describes how rate depends on concentration . Integrated Rate Law - Describes how concentration depends on time . For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms.
17. 17. 12.3 Determining Rate Laws The first step is to determine the form of the rate law (especially its order ). Must be determined from experimental data ( not from the coefficients!!). For this reaction: 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role. Next slide shows experimental results (like you’ll do in lab 12).
18. 18. Now graph the data [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
19. 19. To find rate we have to find the slope at two points We will use the tangent method.
20. 20. At .90 M the rate is - (.98 - .76) = -0.22 = 5.5x 10 -4 M• s -1 (0-400) -400
21. 21. At .45 M the rate is - (.52 - .31) = -0.22 = 2.7 x 10 -4 (1000-1800) -800
22. 22. Since the rate at twice the concentration is twice as fast , the rate law must be.. Rate = -  [N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ]  t We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment. “ n” is not same number as coefficient (only a coincidence). 1st order means doubling the [ ] doubles the rate (know this). Rate = k[A] 1
23. 23. A Plot of [N 2 O 5 ] as f(t) for 2N 2 O 5  4NO 2 + O 2 pp Note that the reaction rate at 0.90 M is twice that at 0.45 M So, the reaction is 1st order because doubling the [ ] doubles the rate
24. 24. The method of Initial Rates pp This method requires that a reaction be run several times (do in Lab 12). The initial concentrations of the reactants are varied . The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction and makes it easier to calculate.
25. 25. Here’s an AP-type Question pp Write general form of rate law for: BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O Answer is . . . Rate = k [BrO 3 - ] n [Br - ] m [H + ] p Determine n , m , p by comparing rates Add n , m , p values to get overall order Find value of k by plugging in values from any one of the experiment rates
26. 26. An AP Test Example pp For the reaction BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p Be able to write the above as the 1st question, given the above reaction. We use experimental data to determine the values of n,m, and p
27. 27. Now we have to see how the rate changes with concentration 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
28. 28. Find n by ratio of Rate 2/Rate 1, in which only [BrO 3 - ] changes n = 1 because (next slide) 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
29. 29. The math pp Rate 2 = 1 .6 x 10 -3 mol/Ls = k (0.20 mol/L) n (0.10 mol/l) m (0.10 mol/L) p Rate 1 8.0 x 10 -4 mol/Ls k (0.10 mol/L) n (0.10 mol/l) m (0.10 mol/L) p = 2.0 = (0.20 mol/L ÷ 0.10 mol/L) n = (2.0) n So, n = 1
30. 30. Find m by ratio of Rate 3/Rate 2, in which only [Br - ] changes m = 1 (ditto on math) 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
31. 31. Find p by ratio of Rate 4/Rate 1, in which only [H + ] changes p = 2 (ditto on math) 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
32. 32. AP-type question pp So, rate of reaction is first order in BrO 3 - and Br - and second order in H + Overall rate = 1 + 1 + 2 = 4 Rate law can be written as: Rate = k [BrO 3 - ][Br - ][H + ] 2
33. 33. AP-type question pp Rate = k [BrO 3 - ][Br - ][H + ] 2 Try calculating k from Expt. 2 BrO 3 1- Br 1- H 1+ Rate ( M /s ) 0.20 0.10 0.10 1.6 x 10 -3 The answer is . . . k = 8.0 L 3 /mol 3 •s (or 8.0 M -3 s -1 ) The complete rate law is . . . Rate = 8.0 L 3 /mol 3 •s [BrO 3 - ][Br - ][H + ] 2 Be sure to express the correct units!
34. 34. AP-type question Units are very important!!! Let’s try Text # 27, p. 568 . . . Answer is . . . 25 a. Rate = K[NOCl] 2 b. k= 6.6 x 10 -29 cm 3 /molecules•s c. k = 4.0 x 10 -8 M -1 s -1 Do # 25 at home. (Quiz soon!)
35. 35. 12.4 Integrated Rate Law Expresses the reaction concentration as a function of time (vs. rate as f(conc.)). Form of the equation depends on the order of the rate law (from differential). Changes Rate = -  [A] n = k [A] n  t We will only work with n = 0, 1, and 2 for reactions with only a single reactant (decomposition)
36. 36. First Order For the reaction 2N 2 O 5 4NO 2 + O 2 We found the (differential) Rate = k [ N 2 O 5 ] 1 I.e. , if concentration doubles rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law ln[N 2 O 5 ] = - k t + ln[N 2 O 5 ] 0 (y = mx + b) ln is the natural log [N 2 O 5 ] 0 is the initial concentration.
37. 37. General form is Rate = -  [A] /  t = k[A] ln[A] = - k t + ln[A] 0 In the form y = mx + b y = ln[A] m = - k x = t b = ln[A] 0 A graph of ln[A] vs. time is a straight line if it is first order (otherwise, a different graph is a straight line, see following slides). First Order
38. 38. By getting a straight line you prove it is first order. Conversely, if a plot of ln[A] vs . t (as opposed to plotting other values) is not a straight line, the reaction is not first order in [A]. So, you have to plot several possibilities to determine which order it is. First Order
39. 39. A Plot of [N 2 O 5 ] as f(t) for 2N 2 O 5  4NO 2 + O 2 . Decomposition reaction Note that the reaction rate at 0.90 M is twice that at 0.45 M So, the reaction is 1st order because doubling the [ ] doubles the rate This is not a plot of ln [N 2 O 5 ], which would be a straight line if it was first order . . .
40. 40. ln [A] = - k t + ln[A] 0 By getting the straight line (plot ln [A] vs. t) you can prove it is first order. Often expressed in a ratio First Order
41. 41. Figure 12.4 A Plot of In (N 2 O 5 ) Versus Time (1st order)
42. 42. Half Life The time required to reach half the original concentration. If a bacterium doubles every minute, and takes 40 minutes to fill the jar, how long to fill half the jar? 39 minutes. If the reaction is first order, then: [A] = [A] 0 /2 when t = t 1/2
43. 43. Figure 12.5: A Plot of (N 2 O 5 ) vs. Time for the Decomposition Reaction of N 2 O 5
44. 44. Half Life ln(2) = k t 1/2 Solving for t 1/2 : The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2
45. 45. Half Life t 1/2 = ln(2)/k = 0.693/ k So, the time to reach half the original concentration does not depend on the starting concentration. This is an easy way to find k if you know or can determine the t 1/2 and the reaction is 1st order.
46. 46. Second Order Rate = -  [A] /  t = k[A] 2 integrated rate law has the form: 1/[A] = k t + 1/[A] 0 , where y = mx + b, so: y= 1/[A] m = k x= t b = 1/[A] 0 Get a straight line if 1/[A] vs t is graphed (instead of ln[A] = - k t + ln[A] 0 ) Knowing k and [A] 0 you can calculate [A] at any time t by graphing y = mx + b
47. 47. Second Order Half Life [A] = [A] 0 /2 at t = t 1/2
48. 48. Z7e 545 Fig 12.6 (a) A Plot of In(C 4 H 6 ) Versus t (not straight line, so not 1st order) (b) A Plot of 1/(C 4 H 6 ) Versus t (straight line so must be 2nd order)
49. 49. Determining Rate Laws Since we get a straight line with (b) and not with (a) in previous slide: The reaction must be second order in C 4 H 6
50. 50. Zero Order Rate Law Rate = k[A] 0 = k Rate does not change with concentration Important for drug pharmacokinetics Integrated rate law: [A] = -kt + [A] 0 y = mx + b format When [A] = [A] 0 /2 t = t 1/2 t 1/2 = [A] 0 /2k for zero order.
51. 51. Z7e 546 Figure 12.7 A Plot of [A] Versus t for a Zero-Order Reaction
52. 52. Occurs most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry and other reactions involving catalysts . Zero Order Rate Law
53. 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. 54. Zero Order Graph Time Concentration
55. 55. Zero Order Graph Time Concentration  A]/  t = slope  t -k =  A] t 1/2 = [A 0 ]/2k
56. 56. More Complicated Reactions Reactions with more than one reactant . BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O For this reaction we experimentally found the (differential) rate law to be Rate = k[BrO 3 - ][Br - ][H + ] 2 To investigate this reaction rate we need to control the conditions.
57. 57. Rate = k[BrO 3 - ][Br - ][H + ] 2 We set up the experiment so two of the reactants are in large excess (Lab 12). [BrO 3 - ] 0 = 1.0 x 10 -3 M [Br - ] 0 = 1.0 M (1000 x as great) [H + ] 0 = 1.0 M (ditto) As the reaction proceeds [BrO 3 - ] changes noticeably ( limiting reactant ). [Br - ] and [H + ] do not (since in large excess)
58. 58. This rate law can be rewritten . . . Rate = k[Br - ] 0 [H + ] 0 2 [BrO 3 - ] = k ’ [BrO 3 - ] because, [Br - ] 0 and [H + ] 0 are constant (since in large excess). Solving for k ’ . . . k ’ = k[Br - ] 0 [H + ] 0 2 and substituting k ’ into the original rate law . . . Rate = k ’ [BrO 3 - ] This is called a pseudo first order rate law. k = k ’ [Br - ] 0 [H + ] 0 2 Rate = k[BrO 3 - ][Br - ][H + ] 2
59. 59. 12.5 Summary of Rate Laws Assume only the forward reaction is important so can produce rate laws that only contain reactant concentrations Differential rate law ( aka “rate law”) shows how rate depends on concentration. Integrated rate law shows how concentration depends on time . Choice of using either method depends on data being experimentally derived.
60. 60. Summary of Rate Laws Most common method to determine the differential rate law is method of initial rates (get t as close to zero as possible) We will use this method in Lab 12 If instead using the integrated rate law, measure concentration at various times. You’ll graph this in Lab 12 and plot the values to get a straight line as follows:
61. 61. Summary of Rate Laws pp Memorize the following rate laws (for AP questions) and use the t 1/2 for our lab 12. (rf. Table 12.6 p. 548) - use for online HW! Reaction order is found by whichever of the following gives straight line: Zero order when [A] vs . t yields straight line Rate = k and t 1/2 = [A] o /2 k 1st order when ln [A] vs. t yields straight line Rate = k[A] and t 1/2 = 0.693/ k 2nd order when 1/ [A] vs. t yields straight line Rate = k[A] 2 and t 1/2 = 1 /k[ A] o
62. 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. 63. 12.6 Reaction Mechanisms The series of steps that actually occur in a chemical reaction. A balanced equation does not tell us how the reactants become products. Kinetics can tell us something about the mechanism.
64. 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. 65. 2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ][F 2 ] The proposed mechanism is: NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F (fast) F is called an intermediate. It is formed then consumed in the reaction Reaction Mechanisms
66. 66. Each of the two reactions is called an elementary step . The rate for each reaction can be written from its molecularity . Molecularity is the number of pieces that must come together (collide) to produce the reaction indicated by that step. Reaction Mechanisms
67. 67. Unimolecular step involves one molecule - Rate is first order. Bimolecular step - requires two molecules - Rate is second order Termolecular step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule. With molecularity the coefficients can become the exponents.
68. 68. A products A+A products 2A products A+B products A+A+B Products 2A+B Products A+B+C Products This is also in Table 12.7, p. 550. Know. Rate = k[A] Rate= k[A] 2 Rate= k[A] 2 Rate= k[A][B] Rate= k[A] 2 [B] Rate= k[A] 2 [B] Rate= k[A][B][C]
69. 69. Finding the Reaction Mechanism pp Must satisfy two requirements : Sum of the elementary steps does give the overall balanced equation for the reaction The proposed mechanism must agree with the experimentally derived rate law.
70. 70. How to get rid of intermediates pp Use the reactions that form them. If the reactions are fast and irreversible the concentration of the intermediate is based on stoichiometry . (Use spider). If it is formed by a reversible reaction set the reversible reaction rates equal to each other.
71. 71. Formed in reversible reactions pp 2 NO + O 2 2 NO 2 Overall reaction Given : Expt-derived Rate = k [NO] 2 [O 2 ] Proposed Mechanism : 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) 2nd step is rate determining (slow), so need to get this one to look like the expt.-derived rate, k [NO] 2 [O 2 ] 1st step is reversible so set both rates in this step equal to each other: k 1 [NO] 2 = k -1 [N 2 O 2 ]
72. 72. Formed in reversible reactions pp 2 NO + O 2 2 NO 2 overall reaction Given: Expt-derived Rate = k [NO] 2 [O 2 ] Proposed Mechanism 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) 2nd step rate comes from molecularity A+B products Rate= k[A][B] Rate = k 2 [N 2 O 2 ][O 2 ]
73. 73. Formed in reversible reactions pp Given: Expt-derived Rate = k [NO] 2 [O 2 ] 2 NO + O 2 2 NO 2 overall reaction Proposed Mechanism 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) Rate = k 1 [NO] 2 = k -1 [N 2 O 2 ] (from 1st step) Rate = k 2 [N 2 O 2 ][O 2 ] (from 2nd step) Solve for [N 2 O 2 ] from (1) &amp; sub into (2) rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]= k[NO] 2 [O 2 ]
74. 74. Formed in reversible reactions pp 2 NO + O 2 2 NO 2 overall reaction Proposed Mechanism 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) Since model-derived Rate = k[NO] 2 [O 2 ] agrees with the expt.-derived rate . . . And sum of the steps = overall reaction: The 2-step proposed mechanism works!
75. 75. Formed in reversible reactions pp In other words: Since this last equation, which we derived by eliminating the intermediate is the same as the experimentally derived rate: And , the sum of the 2 steps add up to the overall balanced reaction: The proposed 2-step mechanism is correct. There are AP questions that have you do this for.
76. 76. Formed in fast, irreversible reactions pp 2 IBr I 2 + Br 2 overall reaction Experimentally-derived Rate = k [IBr] 2 Proposed Mechanism IBr I + Br (fast) IBr + Br I + Br 2 ( slow ) I + I I 2 (fast) Step 2 is slow, so it’s rate determining Slow Rate = k[ I Br][Br] but [Br]= [ I Br] So, Rate = k[ I Br][ I Br] = k[ I Br] 2
77. 77. Formed in fast, irreversible reactions pp 2 IBr I 2 + Br 2 overall reaction Proposed Mechanism IBr I + Br (fast) IBr + Br I + Br 2 (slow) I + I I 2 (fast) Sum of steps = overall balanced reaction Rate = k[IBr][IBr] = k[IBr] 2 is same as experimentally derived So, the mechanism works (But this doesn’t prove it’s the correct mechanism)
78. 78. 12.7 A Model for Chemical Kinetics Collision Theory Molecules must collide to react. Concentration affects rates because collisions are more likely with more [ ]. Must collide hard enough . Temperature and rate are related. Only a small number of collisions produce reactions ( orientation , force ).
79. 79. Terms Activation energy - the minimum energy needed to make a reaction happen. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier. Write rate laws for CH 3 NC (g)  CH 3 CN (g) O 3(g) + O (g)  2O 2(g) Try #53 p. 571 ( similar to next slide):
80. 80. Potential Energy Reaction Coordinate Reactants Products
81. 81. Potential Energy Reaction Coordinate Reactants Products Activation Energy E a
82. 82. Potential Energy Reaction Coordinate Reactants Products Activated complex
83. 83. Potential Energy Reaction Coordinate Reactants Products  E }
84. 84. Potential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State
85. 85. Arrhenius Said that reaction rate should increase with temperature . At high temperature more molecules have the energy required to get over the barrier (i.e . the activation energy). So a higher E a means slower reaction at a given temperature. The number of collisions with the necessary energy increases exponentially with higher temperature.
86. 86. Arrhenius Number of collisions with the required energy = ze -E a /RT z = total collisions e is Euler’s number (opposite of ln) E a = activation energy R = ideal gas constant = 8.314 J /K mol T is temperature in Kelvin
87. 87. Problems Observed rate is less than the number of collisions that have the minimum energy. This is due to Molecular orientation Written into equation as p (the steric factor ). k = z p e -E a /RT
88. 88. Figure 12.13 Several Possible Orientations for a Collision Between Two BrNO Molecules. (a) &amp; (b) can lead to a collision, (c) cannot.
89. 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. 90. Arrhenius Equation k = zpe -E a /RT = Ae -E a /RT A is called the frequency factor = zp ln k = -(E a /R)(1/T) + ln A Another line !!!! (y = mx + b) That is, ln k vs 1/T is a straight line Remember, T is temperature in Kelvin! Another form: ln(k 2 /k 1 ) = (E a /R)(1/T 1 - 1/T 2 ) You will do this in Lab 12
91. 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. 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. 93. Activation Energy and Rates The final saga
94. 94. Mechanisms and rates There is an activation energy for each elementary step. Activation energy determines k . k = Ae - (E a /RT) k determines rate So, slowest step (rate determining) must have the highest activation energy.
95. 95. This reaction takes place in three steps
96. 96. First step is fast Low activation energy E a
97. 97. Second step is slow High activation energy E a
98. 98. E a Third step is fast Low activation energy
99. 99. Second step is rate determining
100. 100. Intermediates are present
101. 101. Activated Complexes or Transition States
102. 102. 12.8 Catalysts Speed up a reaction without being used up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.
103. 103. How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy (so, has a faster rate). More molecules will have this activation energy. Does not change overall  E
104. 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. 105. Figure 12.16 Effect of a Catalyst on the Number of Reaction-Producing Collisions (increases even though no temperature increase).
106. 106. Hydrogen bonds to surface of metal. Break H-H bonds Heterogenous Catalysts Pt surface H H H H H H H H
107. 107. Heterogenous Catalysts Pt surface H H H H C H H C H H
108. 108. Heterogenous Catalysts The double bond breaks and bonds to the catalyst. Pt surface H H H H C H H C H H
109. 109. Heterogenous Catalysts The hydrogen atoms bond with the carbon Pt surface H H H H C H H C H H
110. 110. Heterogenous Catalysts Pt surface H C H H C H H H H H
111. 111. Heterogenous Catalysts Summary: usually in 4 steps: Adsorption &amp; activation of reactants onto the catalytic surface. Migration of the adsorbed reactants across the surface. Reaction of the adsorbed reactants. Escape (desorption) of the products .
112. 112. Homogenous Catalysts Chlorofluorocarbons catalyze the decomposition of ozone. Both are in same phase (gas) Another example: Enzymes regulating the body processes. (Protein catalysts)
113. 113. Catalysts and rate Catalysts will speed up a reaction but only to a certain point . Past a certain point adding more reactants won’t change the rate. Becomes Zero Order (saturation kinetics) See saturation kinetics with renal disease &amp; some drugs (aminoglycosides).
114. 114. Catalysts and rate. Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration Concentration of reactants Rate