1. Chapter 14 Chemical Kinetics Chemical Kinetics

2. Kinetics The study of reaction rates – the speed at which reactants are converted to products. The study of reaction rates – the speed at which reactants are converted to products. Explosions happen in seconds- very fast. Explosions happen in seconds- very fast. Coal (carbon) will turn into diamond  eventually. Coal (carbon) will turn into diamond  eventually. Spontaneous reactions are reactions that will happen  but we can’t tell how fast. Spontaneous reactions are reactions that will happen  but we can’t tell how fast. Kinetics is very important because it allows us to manipulate and control reactions. Kinetics is very important because it allows us to manipulate and control reactions.

3. Reactions occur when... Molecules find each other Molecules find each other Effective collisions happen between molecules Effective collisions happen between molecules Frequency of collisions is high Frequency of collisions is high

4. Factors that affect Reaction Rate: 1. Temperature Hot = fast Cold = slowHot = fast Cold = slow 2. Concentration of Reactants more concentrated = fastermore concentrated = faster 3. Physical conditions of reactants. more surface area = fastermore surface area = faster 4. Presence of a Catalyst – increases rate without being used up increases rate without being used up

5. Reaction Rate The change in concentration of a reactant or a product per unit of time. Rates decrease with time Typically rates are expressed as positive values Instantaneous rate can be determined by finding the slop of a line tangent to a point representing a particular time

6. Reaction Rate - “speed of the reaction” Rate = Conc. of A at t 2  Conc. of A at t 1 t 2  t 1 Rate = Conc. of A at t 2  Conc. of A at t 1 t 2  t 1 Rate = [A] D t Rate = [A] D t Units: Change in concentration per time Units: Change in concentration per time For this reaction For this reaction N 2 + 3H 2 2NH 3 N 2 + 3H 2 2NH 3

7. As the reaction progresses the concentration of H 2 goes down N 2 + 3H 2 2NH 3 As the reaction progresses the concentration of H 2 goes down N 2 + 3H 2 2NH 3 ConcentrationConcentration Time [H 2 ]

8. As the reaction progresses the concentration N 2 goes down 1/3 as fast. Why? N 2 + 3H 2 2NH 3 As the reaction progresses the concentration N 2 goes down 1/3 as fast. Why? N 2 + 3H 2 2NH 3 ConcentrationConcentration Time [H 2 ] [N 2 ]

9. As the reaction progresses the concentration NH 3 goes up. N 2 + 3H 2 2NH 3 As the reaction progresses the concentration NH 3 goes up. N 2 + 3H 2 2NH 3 ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ]

10. Calculating Rates 1. Average rates are taken over long intervals 2. 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 (Finding the Derivative.)

11. Average slope method Average slope method ConcentrationConcentration Time D[H 2 ] DtDtDtDt

12. Instantaneous slope method. Instantaneous slope method. ConcentrationConcentration Time  [H 2 ] D t D t

13. Find the Instantaneous rate of disappearance: In order to find the instantaneous rate of disappearance at time 0, a tangent line was drawn and the following information was obtained: the concentration of the unknown changed from 0.100 M to 0.060 M from 0 seconds to 200 seconds. What is the instantaneous rate of disappearance at time 0? 2.0 x 10 -4 M/s

14. Reaction Rates and Stoichiometry Remember, the rate of disappearance is a  slope and the rate of appearance is a + slope Remember, the rate of disappearance is a  slope and the rate of appearance is a + slope For general equation: For general equation: aA + bB cC + dD aA + bB cC + dD General rate equation: General rate equation: Rate = -1[A] = -1[B] = 1[C] = 1∆[D] a t b t c t d∆t

15. Defining Rate We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. 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 In our example N 2 + 3H 2 2NH 3 -[N 2 ] = -[H 2 ] = [NH 3 ] t 3 t 2 t -[N 2 ] = -[H 2 ] = [NH 3 ] t 3 t 2 t

16. Practice How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O 3  3O 2 ? If the rate of appearance of O 2, is 6.0x10 -5 M/s at a particular instant, what is the value of the rate of disappearance of O 3 at this same time? 4.0x10 -5 M/s

17. The decomposition of N 2 O 5 proceeds according to the following equation: 2N 2 O 5  4NO 2 + O 2 If the rate of decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2x10 -7 M/s, what is the rate of appearance of NO 2 ? O 2 ? NO 2 = 8.4x10 -7 M/s O 2 = 2.1 x 10 -7 M/s

18. Rate Laws To study the rate of a reaction and determine the rate law, a series of experiments must be done varying the initial concentrations of reactants. This is called the Initial rate method. The products do not appear in the rate law because we are only concerned with the very start of the experiment where conc. of reactants are carefully chosen.

19. The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD aA + bB cC + dD Rate = k [A] x [B] y } The rate law expression Rate = k [A] x [B] y } The rate law expression k is the rate constant (temperature dependent) k is the rate constant (temperature dependent) Units of k vary Units of k vary x and y are integers, usually 0, 1 or 2 x and y are integers, usually 0, 1 or 2 reaction is x th order in A, reaction is y th order in B reaction is x th order in A, reaction is y th order in B reaction is (x +y) th order overall reaction is (x +y) th order overall

20. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Initial [ ] Double [F 2 ] with [ClO 2 ] constant: rate doubles Quadruple [ClO 2 ] with [F 2 ] constant: rate quadruples Therefore, rate law is: rate = k [F 2 ] 1 [ClO 2 ] 1 We say, “first order with respect to F 2, first order with respect to ClO 2, and 2nd order overall” (All readings recorded at same temp.)

21. 1. Rate laws are always determined experimentally. 2. Reaction order is always defined in terms of reactant (not product) concentrations. 3. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. Rate Laws - Key Points F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] 1 [ClO 2 ] 1 4. Once you obtain the Rate Law and the value of k, you can calculate the rate of reaction for any set of concentrations.

22. Try this. A + B ---> C + D Run # Initial [A] Initial [B] Initial Rate (v 0 ) ([A] 0 ) ([B] 0 ) 1 1.00 M 1.00 M 1.25 x 10 -2 M/s 1 1.00 M 1.00 M 1.25 x 10 -2 M/s 2 1.00 M 2.00 M 2.5 x 10 -2 M/s 2 1.00 M 2.00 M 2.5 x 10 -2 M/s 3 2.00 M 2.00 M 2.5 x 10 -2 M/s 3 2.00 M 2.00 M 2.5 x 10 -2 M/s What is the order with respect to A? What is the order with respect to B? What is the overall order of the reaction? Write the rate law expression 0 1 1 Rate = k [A] 0 [B] 1

23. Try this. 2NO (g) + Cl 2 (g) ---> 2NOCl [NO (g) ] (M) [Cl 2(g) ] (M) Initial Rate(Ms -1 ) [NO (g) ] (M) [Cl 2(g) ] (M) Initial Rate(Ms -1 ) 0.250 0.250 1.43 x 10 -6 0.250 0.500 2.86 x 10 -6 0.500 0.500 1.14 x 10 -5 What is the order with respect to Cl 2 ? What is the order with respect to NO? What is the overall order of the reaction? Write the rate law expression Calculate the value of k 1 2 3 Rate = k [NO] 2 [Cl 2 ] 1 9.15 x 10 -5 1/M 2 s

24. Units of k 1st Order overall units: 1/s 2nd Order overall units: 1/M s 3rd Order overall units: 1/ M 2 s What the Order means: Zero order means doubling or tripling the [reactant] o has NO effect on the rate. First order means doubling the [reactant] o doubles the rate, tripling the [reactant] o triples the rate, etc… Second order means doubling the [reactant] o quadruples the rate (raises to a power of 2), etc...

25. Initial concentrations (M) Rate (M/s) [BrO 3 - [BrO 3 - ] [Br - [Br - ] [H + [H + ] 0.100.100.10 8.0 x 10 -4 0.100.100.10 8.0 x 10 -4 0.200.100.10 1.6 x 10 -3 0.200.100.10 1.6 x 10 -3 0.200.200.10 3.2 x 10 -3 0.200.200.10 3.2 x 10 -3 0.100.100.20 3.2 x 10 -3 0.100.100.20 3.2 x 10 -3 Determine the rate law expression Calculate the value of k Try this: BrO 3 - + 5 Br - + 6H + ----> 3Br 2 + 3 H 2 O Rate = k [BrO 3 ] 1 [Br - ] 1 [H + ] 2 fourth order overall 8.0 1/M 3 s

26. Determine the Rate Law Expression Determine the value for k [NH 4 + ] [NO 2 - ] Initial Rate (M/s) 0.100 M 0.0050 M 1.35 x 10 -7 0.100 M 0.010 M 2.70 x 10 -7 0.200 M 0.010 M 5.40 x 10 -7 NH 4 + + NO 2 -  N 2 + 2H 2 O

27. Types of Rate Laws Differential Rate Law – describes how rate depends on concentration Rate = k [A] x [B] y Integrated Rate Law – describes how concentration depends on time For each type of differential rate law, there is an integrated rate law and vise versa. We can use these laws to help us better understand reaction mechanisms

28. Integrated Rate Law Equations allow us to: 1. Determine the concentration of reactant remaining at any time after the reaction has started. 2. Determine the time required for a fraction of a sample to react. 3. Determine the time for a reactant concentration to fall to a certain level. 4. Verify whether a reaction is first order and determine the rate constant.

29. Integrated Rate Law Expresses the reaction concentration as a function of time. Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Form of the equation depends on the order of the rate law (differential). A Changing Rate = [A] n t A Changing Rate = [A] n t We will only work with n = 0, 1, and 2 We will only work with n = 0, 1, and 2

30. First Order For the reaction 2N 2 O 5 4NO 2 + O 2 For the reaction 2N 2 O 5 4NO 2 + O 2 Rate = k[ N 2 O 5 ] 1 Rate = k[ N 2 O 5 ] 1 If concentration doubles the rate doubles. If concentration doubles the rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law If we integrate this equation with respect to time we get the Integrated Rate Law ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln is the natural log ln is the natural log [N 2 O 5 ] 0 is the initial concentration. [N 2 O 5 ] 0 is the initial concentration.

31. General form-- Rate = [A] / t = k[A] General form-- Rate = [A] / t = k[A] ln[A] = - kt + ln[A] 0 ln[A] = - kt + ln[A] 0 In the form y = mx + b In the form y = mx + b y = ln[A] ; m = -k ; x = t ;b = ln[A] 0 y = ln[A] ; m = -k ; x = t ;b = ln[A] 0 A graph of ln[A] vs time is a straight line. A graph of ln[A] vs time is a straight line. First Order

32. A graph of ln[A] vs time is a straight line. A graph of ln[A] vs time is a straight line. By getting the straight line you can prove it is first order By getting the straight line you can prove it is first order First Order

33. For the reaction 2N 2 O 5 4NO 2 + O 2 For the reaction 2N 2 O 5 4NO 2 + O 2 Rate = k[N 2 O 5 ] 1 Rate = k[N 2 O 5 ] 1 ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 First Order

34. Try this The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = - kt + ln[A] 0 ln[A] = - kt + ln[A] 0 ln[A] - ln[A] 0 = - kt ln[A] - ln[A] 0 = - kt ln[0.14 M] - ln[0.88M] 0 = - (2.8 x 10 -2 s -1 ) t - 1.97  - 0.128 = - (2.8 x 10 -2 s -1 ) t - 1.84 = - (2.8 x 10 -2 s -1 ) t 66 s = t

35. Second Order (Overall) Rate = -[A] / t = k[A] 2 Rate = -[A] / t = k[A] 2 integrated rate law: 1/[A] = kt + 1/[A] 0 integrated rate law: 1/[A] = kt + 1/[A] 0 y= 1/[A] m = k y= 1/[A] m = k x= tb = 1/[A] 0 x= tb = 1/[A] 0 A straight line if 1/[A] vs time is graphed A straight line if 1/[A] vs time is graphed Knowing k and [A] 0 you can calculate [A] at any time t Knowing k and [A] 0 you can calculate [A] at any time t

36. Second Order Looking at the data alone, you cannot tell if 1 st or 2 nd order. To tell the difference you must graph the data, or look at the units of k.

37. Zero Order Rate Law Rate = k[A] 0 = k Rate = k[A] 0 = k Rate does not change with concentration. Rate does not change with concentration. Integrated [A] = -kt + [A] 0 Integrated [A] = -kt + [A] 0 In the form y = mx + b In the form y = mx + b Most often when reaction happens on a surface because the surface area stays constant. Most often when reaction happens on a surface because the surface area stays constant.

38. Time ConcentrationConcentration Zero Order

39. Time ConcentrationConcentration  A]/  t tt k =  A] Zero Order

41. Half Life The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 t ½ = t when [A] = [A] 0 /2

42. Half Life The time required to reach half the original concentration. The time required to reach half the original concentration. If the reaction is first order If the reaction is first order t ½ = t when [A] = [A] 0 /2 t ½ = t when [A] = [A] 0 /2 If the reaction is 2nd order: If the reaction is 2nd order: t ½ = t when [A] = [A] 0 /2 t ½ = t when [A] = [A] 0 /2 If the reaction is zero order: If the reaction is zero order: t ½ = t when [A] = [A] 0 /2 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 ln [A] 0 [A] 0 /2 k = t½t½ ln 2 k =.693 k =

43. Try This What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? Which order is it? How do you know? Which order is it? How do you know? t½t½ Ln 2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes

44. Summary of Rate Laws Order Rate Law Integrated Rate Law Equation Half-Life

45. Reaction Rates So, WHY do some reactions go fast and others are slow? Reaction Mechanisms Collision Theory and Activation Energy

46. Reaction Mechanisms The series of steps that actually occur in a chemical reaction. The series of steps that actually occur in a chemical reaction. Each step is called an elementary step. Each step is called an elementary step. Kinetics can tell us something about the mechanism Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products. A balanced equation does not tell us how the reactants become products.

47. 2NO 2 + F 2 2NO 2 F 2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ] [F 2 ] (determined experimentally) Rate = k[NO 2 ] [F 2 ] (determined experimentally) The proposed mechanism is The proposed mechanism is NO 2 + F 2 NO 2 F + F (slow) NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F(fast) F + NO 2 NO 2 F(fast) F is called an intermediate It is formed then consumed in the reaction F is called an intermediate It is formed then consumed in the reaction The slow step is the rate determining step! The slow step is the rate determining step! Reaction Mechanisms -- Example

48. Each of the two reactions is called an elementary step. Each of the two reactions is called an elementary step. The rate for a reaction can be written from its molecularity The rate for a reaction can be written from its molecularity Molecularity is the number of pieces that must collide to produce the reaction indicated by the step. Molecularity is the number of pieces that must collide to produce the reaction indicated by the step. Molecularity is the number of molecules which must collide effectively for reaction to occur. Molecularity is the number of molecules which must collide effectively for reaction to occur. Reaction Mechanisms

49. Unimolecular step involves one molecule - Rate is first order. Unimolecular step involves one molecule - Rate is first order. Bimolecular step - requires two molecules - Rate is second order Bimolecular step - requires two molecules - Rate is second order Termolecular step- requires three molecules - Rate is third order Termolecular step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules colliding effectively at the same time are miniscule. Termolecular steps are almost never heard of because the chances of three molecules colliding effectively at the same time are miniscule.

50. A products Rate = k[A] A products Rate = k[A] A+A productsRate= k[A] 2 A+A productsRate= k[A] 2 2A productsRate= k[A] 2 2A productsRate= k[A] 2 A+B productsRate= k[A][B] A+B productsRate= k[A][B] A+A+B Products Rate= k[A] 2 [B] A+A+B Products Rate= k[A] 2 [B] 2A+B Products Rate= k[A] 2 [B] 2A+B Products Rate= k[A] 2 [B] A+B+C Products; Rate= k[A][B][C] A+B+C Products; Rate= k[A][B][C] Rate Laws for Elementary Steps (these ARE based on the stoichiometry)

51. Going back to our example: 2NO 2 + F 2 2NO 2 F (overall reaction) 2NO 2 + F 2 2NO 2 F (overall reaction) Rate = k[NO 2 ] [F 2 ] (determined experimentally) Rate = k[NO 2 ] [F 2 ] (determined experimentally) The proposed mechanism consists of The proposed mechanism consists of 2 elementary steps: 2 elementary steps: NO 2 + F 2 NO 2 F + F (slow) NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F(fast) F + NO 2 NO 2 F(fast) F is called an intermediate - It is formed then consumed in the reaction. F is called an intermediate - It is formed then consumed in the reaction. The slow bimolecular elementary step determines the rate of the reaction and sets the rate law as: The slow bimolecular elementary step determines the rate of the reaction and sets the rate law as: 2nd order overall.A + B P rate = k[A] [B]

52. Multistep Example It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps: –O 3  O 2 + O –O 3 + O  2O 2 Describe the Molecularity of each step Write the equation for the overall reaction. Identify the intermediate(s)

53. For the Equation: Mo(CO) 6 + P(CH 3 ) 3  Mo(CO) 5 P(CH 3 ) 3 + CO The proposed mechanism is: –Mo(CO) 6  Mo(CO) 5 + CO –Mo(CO) 5 + P(CH 3 ) 3  Mo(CO) 5 P(CH 3 ) 3 Is the proposed mechanism consistent with the overall reaction? Identify the intermediate(s)

54. Consider the following reaction: 2NO+ Br 2  2NOBr Write the rate law for the reaction assuming it involves a single step. Is a single-step mechanism likely for this reaction? Explain.

55. In summary Reaction mechanisms describe what really happens during a reaction: how the molecules collide and how many are involved. Reaction mechanisms describe what really happens during a reaction: how the molecules collide and how many are involved. There is a bit of guessing as to what the mechanism might be. There is a bit of guessing as to what the mechanism might be. Chemists try to match a proposed mechanism to an actual experimentally determined rate law. Chemists try to match a proposed mechanism to an actual experimentally determined rate law. The rate of the reaction is the slowest elementary step of the mechanism. The overall rate order is set by this step. The rate of the reaction is the slowest elementary step of the mechanism. The overall rate order is set by this step.

56. What about the role of temperature? Temperature, Rate, and Temperature, Rate, and Effective Collisions Effective Collisions

57. Collision Theory Molecules must collide to react. Molecules must collide to react. Concentration affects rates because collisions are more likely. Concentration affects rates because collisions are more likely. Must collide hard enough. Must collide hard enough. Temperature and rate are related. Temperature and rate are related. Only a small number of collisions produce reactions. Only a small number of collisions produce reactions. In order to react, colliding molecules must have a total KE equal or greater than some minimum value. In order to react, colliding molecules must have a total KE equal or greater than some minimum value.

58. O N Br O N O N O N O N O N O N O N O N O N No Reaction 2 NOBr 2 NO + Br 2 No reaction an effective collision

59. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Energy Diagram

60. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activation Energy E a Energy Diagram

61. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex Energy Diagram

62. Potential EnergyPotential Energy Reaction Coordinate Reactants Products EE } Energy Diagram

63. Potential EnergyPotential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO 2 Transition State

64. Terms Activation energy - the minimum kinetic energy needed to make a reaction happen. Activation energy - the minimum kinetic energy needed to make a reaction happen. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

65. A + B C + D ExothermicEndothermic The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction.

66. Exothermic and the Reverse Endothermic (a)Activation energy (Ea) for the forward reaction (a)Activation energy (Ea) for the reverse reaction (c) ∆H 50 kJ/mol 300 kJ/mol 150 kJ/mol 100 kJ/mol -100 kJ/mol +200 kJ/mol Exothermic Endothermic

67. Arrhenius Said that molecules must possess a certain minimum energy in order to react. Said that molecules must possess a certain minimum energy in order to react. Said the reaction rate should increase with temperature. Said the reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier. At high temperature more molecules have the energy required to get over the barrier. The number of collisions with the necessary energy increases exponentially. The number of collisions with the necessary energy increases exponentially.

68. Arrhenius Equation Number of collisions with the required energy, Number of collisions with the required energy, worked into an equation: worked into an equation: k = Ae -E a /RT k = Ae -E a /RT k = rate constant k = rate constant A = frequency factor (total collisions) A = frequency factor (total collisions) e is Euler’s number (opposite of ln) e is Euler’s number (opposite of ln) E a = activation energy E a = activation energy R = ideal gas law constant R = ideal gas law constant T is temperature in Kelvin T is temperature in Kelvin

69. To find the activation energy: Rearrange Arrhenius Equation by taking the natural log of each side Rearrange Arrhenius Equation by taking the natural log of each side ln k = Ea + ln A ln k = Ea + ln A RT RT

70. Activation Energy and Rates The final saga

71. Mechanisms and rates There is an activation energy (hill to get over) for each elementary step. There is an activation energy (hill to get over) for each elementary step. Activation energy determines k. Activation energy determines k. k = Ae - (E a /RT) k = Ae - (E a /RT) k determines rate k determines rate Slowest step (rate determining) must have the highest activation energy. Slowest step (rate determining) must have the highest activation energy.

72. + This reaction takes place in three steps E Reaction Progress

73. 1 EaEa First step is fast Low activation energy

74. Second step is slow High activation energy 2 EaEa

75. 3 EaEa Third step is fast Low activation energy

76. Second step is rate determining

77. Intermediates are present

78. Activated Complexes or Transition States

79. Catalysts - How do they work? Speed up a reaction without being used up in the reaction. Speed up a reaction without being used up in the reaction. Enzymes are biological catalysts. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.

80. How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy. New pathway has a lower activation energy. More molecules will have this activation energy. More molecules will have this activation energy. Do not change  E (∆H) of the reaction. Do not change  E (∆H) of the reaction.

81. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. No catalyst With a catalyst No catalyst With a catalyst uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed E a, k

82. Pt surface HHHH HHHH Hydrogen bonds to surface of metal. Break H-H bonds Heterogeneous Catalysts H 2 + C 2 H 4 -----> C 2 H 6

83. Pt surface HHHH Heterogeneous Catalysts C HH C HH H 2 + C 2 H 4 -----> C 2 H 6

84. Pt surface HHHH Heterogeneous Catalysts C HH C HH The double bond breaks and bonds to the catalyst. H 2 + C 2 H 4 -----> C 2 H 6

85. Pt surface HHHH Heterogeneous Catalysts C HH C HH The hydrogen atoms bond with the carbon H 2 + C 2 H 4 -----> C 2 H 6

86. Pt surface H Heterogeneous Catalysts C HH C HH HHH H 2 + C 2 H 4 -----> C 2 H 6

87. Catalysts and rate Catalysts will speed up a reaction but only to a certain point. Catalysts will speed up a reaction but only to a certain point. Past a certain point adding more reactants won’t change the rate. Past a certain point adding more reactants won’t change the rate. Zero Order Zero Order

88. Catalysts and Rate Concentration of reactants RateRate Rate increases until the active sites of catalyst are filled. Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration Then rate is independent of concentration