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Jeffrey Mack California State University, Sacramento Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions.

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1 Jeffrey Mack California State University, Sacramento Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

2 Chemical Kinetics will now provide information about the arrow! HOW This gives us information on HOW a reaction occurs! Reactants Products Chemical Kinetics: The Rates of Chemical Reactions Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed?

3 Kinetics is the study of how fast (Rates) chemical reactions occur. Important factors that affect the rates of chemical reactions: reactant concentration or surface area in solids temperature action of catalysts Our goal: Our goal: Use kinetics to understand chemical reactions at the particle or molecular level. Chemical Kinetics

4 Reactants go away with time. Products appear with time. The rate of a reaction can be measured by either. In this example by the loss of color with time. Rate of Reactions

5 Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc. with time — can be determined from the plot. Dye Conc Time Determining a Reaction Rate A B

6 aA + bB  cC + dD In general for the reaction: reactants go away with time therefore the negative sign… Reaction Rate & Stoichiometry Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

7 Determining a Reaction Rate

8 concentrationtime The rate of appearance or disappearance is measured in units of concentration vs. time. Rate = time There are three “types” of rates 1.initial rate 2.average rate 3.instantaneous rate = M  s  1 or M  min  1 etc... Determining a Reaction Rate

9 Reactant concentration (M) The concentration of a reactant decreases with time. Time Reaction Rates

10 Reactant concentration (M) Time Initial rate Reaction Rates

11 Reactant concentration (M) Time Reaction Rates

12 Reactant concentration (M) Time Instantaneous rate (tangent line) Reaction Rates

13 Reactant concentration (M) Time Instantaneous rate (tangent line) Initial rate Reaction Rates During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction.

14 Problem: Consider the reaction: Over a period of 50.0 to s, the concentration of NO(g) drops from M to M. a) What is the average rate of disappearance of NO(g) during this time?

15 a) Problem: Consider the reaction: Over a period of 50.0 to s, the concentration of NO(g) drops from M to M. a)What is the rate of rxn? b)What is the average rate of disappearance of NO(g) during this time? = 1.50  10  4 Ms  1 (0.0100M  M) s  50.0 s RATE= -  [NO] tt 1 2 b)  [NO]/ tt = /50.0 M/s = X M/s

16 Practice example Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt

17 There are several important factors that will directly affect the rate of a reaction: Temperature The physical state of the reactants Addition of a catalyst All of the above can have a dramatic impact on the rate of a chemical process. Reaction Conditions & Rate

18 Bleach at 54 ˚CBleach at 22 ˚C Reaction Conditions & Rate: Temperature

19 Reaction Conditions & Rate: Physical State of Reactants

20 Catalyzed decomposition of H 2 O 2 2 H 2 O 2  2 H 2 O + O 2 Reaction Conditions & Rate: Catalysts

21 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq)  MgCl 2 (aq) + H 2 (g) Effect of Concentration on Reaction Rate: Concentration

22 concentration The rate of reaction must be a function of concentration: collisionsAs concentration increases, so do the number of collisions… probabilityAs the number of collisions increase, so does the probability of a reaction. rateThis in turn increases the rate of conversion of reactants to products. Rate Law ExpressionThe relationship between reaction rate and concentration is given by the reaction Rate Law Expression. The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. The Rate Law Expression

23 aA + bB  cC + dD x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms  1 ). For the general reaction: Reaction Order

24 A reaction order can be zero, or positive integer and fractional number. OrderNameRate Law 0zerothrate = k[A] 0 = k 1firstrate = k[A] 2secondrate = k[A] 2 0.5one-halfrate = k[A] 1/2 1.5three-halfrate = k[A] 3/ two-thirds rate = k[A] 2/3 Reaction Orders

25 Overall order: or seven  halves order note: when the order of a reaction is 1 (first order) no exponent is written. 1 + ½ + 2 = 3.5 =7/2 Reaction Order

26 To find the units of the rate constant, divide the rate units by the Molarity raised to the power of the overall reaction order. k = if (x+y) = 1k has units of s -1 if (x+y) = 2k has units of M -1 s -1 Reaction Constant

27 The rate constant, k, is a proportionality constant that relates rate and concentration. It is found through experiment that the rate constant is a function temperature. Rate constants must therefore be reported at the temperature with which they are measured. The rate constant also contains information about the energetics and collision efficiency of the reaction. Reaction Constant

28 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O (g) first orderthird order is experimentally found to be first order in H 2 and third order in NO a) Write the rate law.

29 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O (g) first orderthird order is experimentally found to be first order in H 2 and third order in NO a) Write the rate law. Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the overall order of the reaction?

30 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O (g) first orderthird order is experimentally found to be first order in H 2 and third order in NO a) Write the rate law. Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the overall order of the reaction? = 4Overall order =1+ 3 “4 th order” c) What are the units of the rate constant?

31 EXAMPLE: The reaction, 2 NO (g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O (g) first orderthird order is experimentally found to be first order in H 2 and third order in NO a) Write the rate law. Rate(Ms -1 ) =k[H 2 ][NO] 3 b) What is the overall order of the reaction? = 4Overall order =1+ 3 “4 th order” c) What are the units of the rate constant?

32 doublesdoubles If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? Rate Law Expression

33 doublesdoubles If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one… 1 Rate Law Expression

34 doublesdoubles If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one… 1 Rate Law Expression

35 Instantaneous rate (tangent line) Reactant concentration (M) During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction. Initial rate Determining a Rate Equation

36 The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H 2 (g)  N 2 (g) + 2H 2 O (g) From the following experimental data, determine the rate law and rate constant. Determining Reaction Order: The Method of Initial Rates

37 The reaction of nitric oxide with hydrogen at 1280 °C is as follows: Notice that in Trial 1 and 3, the initial concentration of NO is held constant while H 2 is changed. Determining Reaction Order: The Method of Initial Rates 2NO (g) + 2H 2 (g)  N 2 (g) + 2H 2 O (g)

38 Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H 2 (g)  N 2 (g) + 2H 2 O (g) This means that any changes to the rate must be due to the changes in H 2 which is related to the concentration of H 2 & its order!

39 Rate(M/min) = k [NO] x [H 2 ] y The rate law for the reaction is given by: 2NO(g) + 2H 2 (g)  N 2 (g) + 2H 2 O(g) Taking the ratio of the rates of Trials 3 and 1 one finds: Plugging in the values from the data: Rate (Trial 3) Rate (Trial 1) = Determining Reaction Order: The Method of Initial Rates

40 2.00 log Take the log of both sides of the equation: log(2.00) y = 1 Rate(M/min) = k [NO] x [H 2 ] Determining Reaction Order: The Method of Initial Rates

41 x = 3 Similarly for x: Rate(M/min) = k [NO] x [H 2 ] y k k Determining Reaction Order: The Method of Initial Rates

42 The Rate Law expression is: The order for NO is 3 The order for H 2 is 1 The over all order is =4 2NO(g) + 2H 2 (g)  N 2 (g) + 2H 2 O(g) Determining Reaction Order: The Method of Initial Rates

43 The Rate constant Rate(M/min) = k [NO] 3 [H 2 ] To find the rate constant, choose one set of data and solve: Determining Reaction Order: The Method of Initial Rates

44 It is important know how long a reaction must proceed to reach a predetermined concentration of some reactant or product. We need a mathematical equation that relates time and concentration: This equation would yield concentration of reactants or products at a given time. It will all yield the time required for a given amount of reactant to react. Concentration–Time Relationships: Integrated Rate Laws

45 zero For a zero order process where “A” goes onto products, the rate law can be written: A  products = k k has units of Ms  1 For a zero order process, the rate is the rate constant! Integrated Rate Laws

46 This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate” Zero order kinetics A  products Integrated Rate Laws = k

47 [A] t  [A] o =  k  (t  0)=  k  t and [A] = [A] at time t = t where [A] = [A] o at time t = 0 [A] t  [A] o =  kt Zero order kinetics Integrated Rate Laws

48 [A] o  [A] t = kt what’s this look like? y = mx + b [A] t =  kt + [A] o rearranging… a plot of [A] t vs t looks like… [A] t (mols/L) t (time) y-intercept  the y-intercept is [A] o Conclusion: If a plot of reactant concentration vs. time yields a straight line, then the reactant order is ZERO!  slope =  k k has units of M×(time)  1 Zero order kinetics Integrated Rate Laws

49 first order For a first order process, the rate law can be written: A  products This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate” Integrated Rate Laws

50 Taking the exponent to each side of the equation: or Conclusion: Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time. First order kinetics Integrated Rate Laws

51 Taking the natural log of both sides… ln[A] t = rearranging… = ln[A] o  kt  kt + ln[A] o First order kinetics Integrated Rate Laws

52 y = mx + bso a plot of ln[A] t vs t looks like… ln[A] t t (time) y-intercept  the y-intercept is ln[A] o Conclusion: Conclusion: If a plot of natural log of reactant concentration vs. time yields a straight line, then the reactant order is FIRST!  slope =  k ln[A] t =  kt + ln[A] o k has units of (time)  1 First order kinetics Integrated Rate Laws

53 Problem: The decomposition of N 2 O 5 (g) following 1 st order kinetics. If 2.56 mg of N 2 O 5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s  1 ?

54 Problem: The decomposition of N 2 O 5 (g) following 1 st order kinetics. If 2.56 mg of N 2 O 5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s  1 ? Begin with the integrated rate law for a 1 st order process: Wait… what is the volume of the container??? Do we need to convert to moles?

55 Problem: The decomposition of N 2 O 5 (g) following 1 st order kinetics. If 2.56 mg of N 2 O 5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s  1 ? Check it out! You don’t need the volume of the container!

56 Problem: The decomposition of N 2 O 5 (g) following 1 st order kinetics. If 2.56 mg of N 2 O 5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s  1 ? taking the natural log and substituting time in seconds: k = 9.3  10  5 s  1

57 Rate = k[A] 2 Integrating as before we find: A  Products k has units of M  1 s  1 Second order kinetics Integrated Rate Laws

58 y = mx + bso a plot of 1/[A] t vs t looks like… 1/[A] t t (time) y-intercept  the y-intercept is 1/[A] o Conclusion: Conclusion: If a plot of one over reactant concentration vs. time yields a straight line, then the reactant order is second!  slope = k Second order kinetics k has units of M  1 s  1 Integrated Rate Laws

59 Summary of Integrated Rate Laws

60 Half-life of a reaction is the time taken for the concentration of a reactant to drop to one-half of the original value. Half-life & First-Order Reactions

61 For a first order process the half life (t ½ ) is found mathematically from: Start with the integrated rate law expression for a 1 st order process Bring the concentration terms to one side. Express the concentration terms as a fraction using the rules of ln. at time = t ½ Reaction Half-Life: 1 st Order Kinetics

62 exchange [A] with [A] 0 to reverse the sign of the ln term and cancel the negative sign in front of k Substitute the value of [A] at the half-life Reaction Half-Life: 1 st Order Kinetics

63 So, knowing the rate constant for a first order process, one can find the half-life! The half-life is independent of the initial concentration! Reaction Half-Life: 1 st Order Kinetics

64 Problem: A certain reaction proceeds through first order kinetics. The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s?

65 Problem: A certain reaction proceeds through first order kinetics. The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find:

66 Problem: A certain reaction proceeds through first order kinetics. The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s -1

67 Problem: A certain reaction proceeds through first order kinetics. The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s -1 =

68 Problem: A certain reaction proceeds through first order kinetics. The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s -1 = Since the ratio of [A] t to [A] 0 represents the fraction of [A] that remains, the % is given by: 100  = 3.12%

69 Elements that decay via radioactive processes do so according to 1 st order kinetics: Element:Half-life: 238 U  234 Th +  14 C  14 N +  131 I  131 Xe +  4.5  10 9 years 5730 years 8.05 days Reaction Half-Life: 1 st Order Kinetics

70 Tritium decays to helium by beta (  ) decay: The half-life of this process is 12.3 years Starting with 1.50 mg of 3 H, what quantity remains after 49.2 years. Solution: Begin with the integrated rate law expression for 1 st order kinetics. Reaction Half-Life: 1 st Order Kinetics

71 Recall that that the rate constant for a 1 st order process is given by: [ 3 H] t = 1.50 mg  = mg Reaction Half-Life: 1 st Order Kinetics

72 Notice that 49.2 years is 4 half-lives… After 1 half life:= 0.75 mg remains After 2 half life's:= 0.38 mg remains After 3 half life's:= 0.19 mg remains After 4 half life's:= mg remains Reaction Half-Life: 1 st Order Kinetics

73 Arrhenius: Arrhenius: Molecules must posses a minimum amount of energy to react. Why? (1) In order to form products, bonds must be broken in the reactants. (2) Bond breakage requires energy. (3) Molecules moving too slowly, with too little kinetic energy, don’t react when they collide. The Activation energy, E a, is the minimum energy required to initiate a chemical reaction. E a is specific to a particular reaction. A Microscopic View of Reaction Rates

74 When one writes a reaction all that is seen are the reactants and products. This details the overall reaction stoichiometry. mechanism How a reaction proceeds is given by the reaction mechanism. A Microscopic View of Reaction Rates

75 The reaction of NO 2 and CO (to give NO and CO 2 ) has an activation energy barrier of 132 kJ/mol-rxn. The reverse reaction (NO + CO 2  NO2 + CO) requires 358 kJ/mol-rxn. The net energy change for the reaction of NO 2 and CO is  226 kJ/mol-rxn. Activation Energy

76 reactants products Reaction Coordinate The progress of a chemical reaction as the reactants transform to products can be described graphically by a Reaction Coordinate. Potential Energy Reaction Progress reactants products  H RXN E act In order for the reaction to proceed, the reactants must posses enough energy to surmount a reaction barrier. Transition State Activation Energy

77 The temperature for a system of particles is described by a distribution of energies. At higher temps, more particles have enough energy to go over the barrier. E > E a E < E a Since the probability of a molecule reacting increases, the rate increases. Activation Energy

78 Molecules need a minimum amount of energy to react. activation energy, E a. Visualized as an energy barrier - activation energy, E a. Reaction coordinate diagram Activation Energy

79 Orientation factors into the equation The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs. reactive or effective collision When the green atom collides with the green atom on the molecule, a reactive or effective collision occurs. The reaction occurs only when the orientation of the molecules is just right… Activation Energy

80 Orientation factors into the equation non-reactive or ineffective collision When the green atom collides with the red atom on the molecule, this leads to a non-reactive or ineffective collision occurs. In some cases, the reactants must have proper orientation for the collision to yield products. This reduces the number of collisions that are reactive! Activation Energy

81 Arhenius discovered that most reaction-rate data obeyed an equation based on three factors: (1) The number of collisions per unit time. (2) The fraction of collisions that occur with the correct orientation. (3) The fraction of the colliding molecules that have an energy greater than or equal to E a. Arrhenius equation From these observations Arrhenius developed the aptly named Arrhenius equation. The Arrhenius Equation

82 Both A and E a are specific to a given reaction. k is the rate constant E a is the activation energy R is the ideal-gas constant (8.314 J/K  mol) T is the temperature in K Afrequencypre–exponential factor A is known the frequency or pre–exponential factor In addition to carrying the units of the rate constant, “A” relates to the frequency of collisions and the orientation of a favorable collision probability The Arrhenius Equation

83 Temperature Dependence of the Rate Constant: Increasing the temperature of a reaction generally speeds up the process (increases the rate) because the rate constant increases according to the Arrhenius Equation. Rate (Ms -1 ) = k[A] x [B] y As T increases, the value of the exponential part of the equation becomes less negative thus increasing the value of k. The Arrhenius Equation

84 Reactions generally occur slower at lower T. Iodine clock reaction. H 2 O I H +  2 H 2 O + I 2 Room temperature In ice at 0 o C Effect of Temperature

85 Determining the Activation Energy E a may be determined experimentally. First take natural log of both sides of the Arrhenius equation: ln ln k y = mx + b The Arrhenius Equation

86 One can determine the activation energy of a reaction by measuring the rate constant at two temperatures: Writing the Arrhenius equation for each temperature: Subtracting k 1 from k 2 we find that: Determining the Activation Energy The Arrhenius Equation

87 Knowing the rate constants at two temps yields the activation energy. or Knowing the E a and the rate constant at one temp allows one to find k(T 2 ) Determining the Activation Energy The Arrhenius Equation

88 Problem: The activation energy of a first order reaction is 50.2 kJ/mol at 25 °C. At what temperature will the rate constant double? (1) (2) (3)

89 Problem: (4) (5) T 2 = 308 K A 10 °C change of temperature doubles the rate!! algebra!

90 Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. What is a catalyst? Catalysts and society Catalysts and the environment Catalysis

91 In auto exhaust systems — Pt, NiO 2 CO + O 2  2 CO 2 2 NO  N 2 + O 2 Catalysis

92 2.Polymers: H 2 C=CH 2  polyethylene 3.Acetic acid:CH 3 OH + CO  CH 3 CO 2 H 4. Enzymes — biological catalysts Catalysis

93 Catalysis and activation energy Uncatalyzed reaction Catalyzed reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2  2 H 2 O + O 2 Catalysis

94 Iodine-Catalyzed Isomerization of cis-2-Butene

95

96 The overall stoichiometry of a chemical reaction is most often the sum of several steps: (1) 2AB  A 2 B 2 (2) A 2 B 2 + C 2  A 2 B + C 2 B (3) A 2 B + C 2  A 2 + C 2 B 2AB + A 2 B 2 + A 2 B + 2C 2  A 2 B 2 + A 2 B + A 2 + 2C 2 B Net: 2AB + 2C 2  A 2 + 2C 2 B The sequence of steps (1-3) describes a possible “reaction mechanism”. Reaction Mechanisms

97 The overall stoichiometry of a chemical reaction is most often the sum of several steps: (1) 2AB  A 2 B 2 (2) A 2 B 2 + C 2  A 2 B + C 2 B (3) A 2 B + C 2  A 2 + C 2 B 2AB + A 2 B 2 + A 2 B + 2C 2  A 2 B 2 + A 2 B + A 2 + 2C 2 B Net: 2AB + 2C 2  A 2 + 2C 2 B The species that cancel out (not part of the overall reaction) are called “reaction intermediates”. Reaction Mechanisms

98 (1) 2AB  A 2 B 2 (2) A 2 B 2 + C 2  A 2 B + C 2 B (3) A 2 B + C 2  A 2 + C 2 B 2AB + 2C 2  A 2 + 2C 2 B “elementary step”. Each step in the mechanism is called an “elementary step”. molecularity The number of reactants in an elementary step is called the “molecularity”. bimolecular process In this example each step (1-3) is a bimolecular process. (2 reactants) unimolecular process A 2  2A is a unimolecular process (1 reactant) termolecular process 2A + B  is a termolecular process (3 reactants) Reaction Mechanisms

99 The rate of the overall reaction can never be faster than the “slowest step” in the mechanism. If reaction (1) is the slowest of the three steps in the mechanism… Then it is known as the “rate determining step” slow fast (1) 2AB  A 2 B 2 (2) A 2 B 2 + C 2  A 2 B + C 2 B (3) A 2 B + C 2  A 2 + C 2 B 2AB + 2C 2  A 2 + 2C 2 B Reaction Mechanisms

100 slow fast (1) 2AB  A 2 B 2 (2) A 2 B 2 + C 2  A 2 B + C 2 B (3) A 2 B + C 2  A 2 + C 2 B 2AB + 2C 2  A 2 + 2C 2 B The slowest step controls the rate of the reaction. It determines the rate law! Rate (Ms -1 ) = k[AB] 2 The rate law is not based on the overall reaction: Rate (Ms -1 ) = k[AB] 2 [C 2 ] 2 Reaction Mechanisms

101 Consider the following reaction: 2NO 2 (g) + F 2 (g)  2FNO 2 (g) If the reaction proceeded by the overall reaction, the rate law for the reaction would be 3 rd order overall. The actual rate law is found to be: Rate = k[NO 2 ][F 2 ] Indicating that the slowest step in the mechanism is a bimolecular reaction between NO 2 and F 2. Reaction Mechanisms

102 2NO 2 (g) + F 2 (g)  2FNO 2 (g) Rate = k[NO 2 ][F 2 ] To explain the observed kinetics, a possible mechanism is proposed: Reaction Mechanisms

103 2NO 2 (g) + F 2 (g)  2FNO 2 (g) Rate = k[NO 2 ][F 2 ] Since the 1 st step in the reaction is the slow step, it determines the kinetics and rate law for the reaction: Reaction Mechanisms

104 Validating a Reaction Mechanism: A mechanism is a proposal of how the reaction proceeds at the molecular level. 1.The individual elementary steps must sum to yield the overall reaction with correct stoichiometry. 2.The predicted reaction rate law must be in agreement with the experimentally determined rate law. 3.Note that there may be more than one mechanism that is in agreement with the reaction stoichiometry and kinetics. Reaction Mechanisms


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