Chapter 12 Chemical Kinetics Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.1 Reaction Rates Average Kinetic Energy Section 12.1 Reaction Rates Section 12.1 Reaction Rates Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved 12 Section 12.1 Reaction Rates Reaction Rate Change in concentration of a reactant or product per unit time. concentration of A at time t2 concentration of A at time t1 Rate = t2 t1 A = t [A] means concentration of A in mol/L; A is the reactant or product being considered. Copyright © Cengage Learning. All rights reserved 13 Section 12.1 Reaction Rates Section 12.1 Reaction Rates The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved 15 Section 12.1 Reaction Rates The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved 16 Section 12.1 Reaction Rates Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point. Copyright © Cengage Learning. All rights reserved 17 Section 12.1 Reaction Rates Section 12.2 Rate Laws: An Introduction Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n: k = rate constant n = order of the reactant Copyright © Cengage Learning. All rights reserved 19 Section 12.2 Rate Laws: An Introduction Rate Law Rate = k[NO2]n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. Copyright © Cengage Learning. All rights reserved 20 Section 12.2 Rate Laws: An Introduction Rate Law Rate = k[NO2]n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. Copyright © Cengage Learning. All rights reserved 21 Section 12.2 Rate Laws: An Introduction Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved 22 Section 12.2 Rate Laws: An Introduction Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient. Copyright © Cengage Learning. All rights reserved 23 Section 12.2 Rate Laws: An Introduction Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Copyright © Cengage Learning. All rights reserved 24 Section 12.3 Determining the Form of the Rate Law Determine experimentally the power to which each reactant concentration must be raised in the rate law. Copyright © Cengage Learning. All rights reserved 25 Section 12.3 Determining the Form of the Rate Law Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. Copyright © Cengage Learning. All rights reserved 26 Section 12.3 Determining the Form of the Rate Law Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Copyright © Cengage Learning. All rights reserved 27 Section 12.3 Determining the Form of the Rate Law CONCEPT CHECK! How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? Copyright © Cengage Learning. All rights reserved 28 Section 12.3 Determining the Form of the Rate Law CONCEPT CHECK! How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism). Copyright © Cengage Learning. All rights reserved 29 Section 12.3 Determining the Form of the Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.2 Rate Laws: An Introduction Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved 52 Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law First-Order Rate = k[A] Integrated: ln[A] = –kt + ln[A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved 54 Section 12.4 The Integrated Rate Law Plot of ln[N2O5] vs Time Copyright © Cengage Learning. All rights reserved 55 Section 12.4 The Integrated Rate Law A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? Include units ln(0.65) = –k(55) + ln(1) k = 7.8 × 10–3 min–1 Copyright © Cengage Learning. All rights reserved 56 Section 12.4 The Integrated Rate Law A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? Include units ln[A]t - ln[A]o = –kt ln(0.65) - ln(1) = –k(55) Copyright © Cengage Learning. All rights reserved [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A 57 Section 12.4 The Integrated Rate Law A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? Include units k = 7.8 × 10–3 min–1 Copyright © Cengage Learning. All rights reserved 58 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) First order Copyright © Cengage Learning. All rights reserved 59 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) First order 3.7 M ln[A] = –(1.0×10–2)(30.0) + ln(5.0) Copyright © Cengage Learning. All rights reserved 60 Section 12.4 The Integrated Rate Law First-Order Time required for a reactant to reach half its original concentration Half–Life: t1 2 0.693 = k k = rate constant Half–life does not depend on the concentration of reactants. Copyright © Cengage Learning. All rights reserved 61 Section 12.4 The Integrated Rate Law First-Order Time required for a reactant to reach half its original concentration Half–Life: 0.693 t1 = 2 k k = rate constant Half–life does not depend on the concentration of reactants. Copyright © Cengage Learning. All rights reserved 62 Section 12.4 The Integrated Rate Law Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.4 The Integrated Rate Law Second-Order Rate = k[A]2 Integrated: 1 1 = kt + A A 0 [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved 72 Section 12.4 The Integrated Rate Law Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) Second order Copyright © Cengage Learning. All rights reserved 74 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) Second order 2.0 M (1 / [A]t) - (1 / 5.0) = (1.0×10–2)(30.0) Copyright © Cengage Learning. All rights reserved 75 Section 12.4 The Integrated Rate Law Second-Order Half–Life: t1 2 1 = k A 0 k = rate constant [A]o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Copyright © Cengage Learning. All rights reserved 76 Section 12.4 The Integrated Rate Law EXERCISE! For a second order reaction aA Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. a) Write the rate law for this reaction. Copyright © Cengage Learning. All rights reserved 77 Section 12.4 The Integrated Rate Law EXERCISE! For a second order reaction aA Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. a) Write the rate law for this reaction. 1 = kt + A Copyright © Cengage Learning. All rights reserved 1 A 0 78 Section 12.4 The Integrated Rate Law EXERCISE! For a second order reaction aA Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. If k = 8.0 x 10-3 M–1min–1 Calculate [A] at t = 525 minutes. Copyright © Cengage Learning. All rights reserved 79 Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law EXERCISE! For a second order reaction aA Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. a) Write the rate law for this reaction. rate = k[A]2 b) Calculate k. k = 8.0 × 10-3 M–1min–1 c) Calculate [A] at t = 525 minutes. [A] = 0.23 M Copyright © Cengage Learning. All rights reserved 81 Section 12.4 The Integrated Rate Law Zero-Order Rate = k[A]0 = k Integrated: [A] = –kt + [A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved 82 Section 12.4 The Integrated Rate Law Plot of [A] vs Time Copyright © Cengage Learning. All rights reserved 83 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) Zero order Copyright © Cengage Learning. All rights reserved 84 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Include units a) Second order 4.7 M [A] = –kt + [A]o [A] = –(1.0×10–2)(30.0) + 5.0 Copyright © Cengage Learning. All rights reserved 85 Section 12.4 The Integrated Rate Law Zero-Order Half–Life: t1 = 2 A 0 2k k = rate constant [A]o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Copyright © Cengage Learning. All rights reserved 86 Section 12.4 The Integrated Rate Law CONCEPT CHECK! How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? Copyright © Cengage Learning. All rights reserved 87 Section 12.4 The Integrated Rate Law Rate Laws To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 88 Section 12.4 The Integrated Rate Law Summary of the Rate Laws Copyright © Cengage Learning. All rights reserved 89 Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.4 The Integrated Rate Law Section 12.5 Reaction Mechanisms Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. Copyright © Cengage Learning. All rights reserved 93 Section 12.5 Reaction Mechanisms A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) Copyright © Cengage Learning. All rights reserved 94 Section 12.5 Reaction Mechanisms NO2(g) + CO(g) NO(g) + CO2(g) NO2(g) + NO2(g) NO3(g) + NO(g) + NO3(g) + CO(g) NO2(g) + CO2(g) ® NO3(g) + NO(g) 2(g) ® NO3(g) + NO(g) g) Section 12.5 Reaction Mechanisms Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Very rare. Copyright © Cengage Learning. All rights reserved 96 Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms Rate-Determining Step A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Copyright © Cengage Learning. All rights reserved 104 Section 12.5 Reaction Mechanisms Reaction Mechanism Requirements The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. Copyright © Cengage Learning. All rights reserved 105 Section 12.5 Reaction Mechanisms Decomposition of N2O5 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 106 Section 12.5 Reaction Mechanisms Decomposition of N2O5 2N2O5(g) 4NO2(g) + O2(g) Step 1: 2(N2O5 NO2 + NO3 ) Step 2: NO2 + NO3 → NO + O2 + NO2 Step 3: NO3 + NO → 2NO2 Copyright © Cengage Learning. All rights reserved (fast) (slow) (fast) 107 Section 12.5 Reaction Mechanisms CONCEPT CHECK! The reaction A + 2B C has the following proposed mechanism: A+B D (fast equilibrium) D+BC (slow) Write the rate law for this mechanism. rate = k[A][B]2 Copyright © Cengage Learning. All rights reserved 108 Section 12.5 Reaction Mechanisms Section 12.5 Reaction Mechanisms 2014 AP 12.5 Exam Section Reaction Mechanisms Section 12.4 The Integrated Rate Law 2014 AP 12.5 Exam Section Reaction Mechanisms Section 12.6 A Model for Chemical Kinetics Collision Model Molecules must collide to react. Main Factors: Activation energy, Ea Temperature Molecular orientations Copyright © Cengage Learning. All rights reserved 114 Section 12.6 A Model for Chemical Kinetics Activation Energy, Ea Energy that must be overcome to produce a chemical reaction. Copyright © Cengage Learning. All rights reserved 115 Section 12.6 A Model for Chemical Kinetics Transition States and Activation Energy To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 116 Section 12.6 A Model for Chemical Kinetics Change in Potential Energy Copyright © Cengage Learning. All rights reserved 117 Section 12.6 A Model for Chemical Kinetics For Reactants to Form Products Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. Copyright © Cengage Learning. All rights reserved 118 Section 12.6 A Model for Chemical Kinetics The Gas Phase Reaction of NO and Cl2 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 119 Section 12.1 Reaction Rates Section 12.1 Reaction Rates Section 12.6 A Model for Chemical Kinetics Arrhenius Equation k = Ae A Ea R T Ea / RT = = = = Copyright © Cengage Learning. All rights reserved frequency factor activation energy gas constant (8.3145 J/K·mol) temperature (in K) 122 Section 12.6 A Model for Chemical Kinetics Linear Form of Arrhenius Equation Ea 1 ln(k ) = + ln A R T Copyright © Cengage Learning. All rights reserved 123 Section 12.6 A Model for Chemical Kinetics Linear Form of Arrhenius Equation Copyright © Cengage Learning. All rights reserved 124 Section 12.6 A Model for Chemical Kinetics EXERCISE! Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? Ea = 53 kJ Copyright © Cengage Learning. All rights reserved 125 Section 12.7 Catalysis Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Copyright © Cengage Learning. All rights reserved 126 Section 12.7 Catalysis Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction Copyright © Cengage Learning. All rights reserved 127 Section 12.7 Catalysis Effect of a Catalyst on the Number of Reaction-Producing Collisions Copyright © Cengage Learning. All rights reserved 128 Section 12.7 Catalysis Heterogeneous Catalyst Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Copyright © Cengage Learning. All rights reserved 129 Section 12.7 Catalysis Heterogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 130 Section 12.7 Catalysis Heterogeneous Catalyst 1. 2. 3. 4. Adsorption and activation of the reactants. Migration of the adsorbed reactants on the surface. Reaction of the adsorbed substances. Escape, or desorption, of the products. Copyright © Cengage Learning. All rights reserved 131 Section 12.7 Catalysis Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Copyright © Cengage Learning. All rights reserved 132 Section 12.7 Catalysis Homogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 133 Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Entropy Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7 Catalysis Section 12.7Rate ? 1.0x10-3 Catalysis 2 mole 5 mole RateA = k [A] k = RateB [B] Rate = k[A] Rate = k[B] Section 12.7Rate ? 1.0x10-3 Catalysis 2 mole 5 mole Section 12.7 Catalysis Section 12.7 Catalysis

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