Chapter 12. Chemical Kinetics
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1 Chapter 12 Chemical Kinetics
2 Section 12.1 Reaction Rates Reaction Rate Change in concentration of a reactant or product per unit time. Rate = concentration of A at time t t concentration of A at time t t = A t [A] means concentration of A in mol/l; A is the reactant or product being considered. Copyright Cengage Learning. All rights reserved 2
3 Section 12.1 Reaction Rates The Decomposition of Nitrogen Dioxide Copyright Cengage Learning. All rights reserved 3
4 Section 12.1 Reaction Rates The Decomposition of Nitrogen Dioxide Copyright Cengage Learning. All rights reserved 4
5 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 5
6 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: 2NO 2 (g) 2NO(g) + O 2 (g) Rate = k[no 2 ] n : k = rate constant n = order of the reactant Copyright Cengage Learning. All rights reserved 6
7 Section 12.2 Rate Laws: An Introduction Rate Law Rate = k[no 2 ] 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 7
8 Section 12.2 Rate Laws: An Introduction Rate Law Rate = k[no 2 ] 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 8
9 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 9
10 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 10
11 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 11
12 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 12
13 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 13
14 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 14
15 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 15
16 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 16
17 Section 12.4 The Integrated Rate Law Plot of ln[n 2 O 5 ] vs Time Copyright Cengage Learning. All rights reserved 17
18 Section 12.4 The Integrated Rate Law First-Order Time required for a reactant to reach half its original concentration Half Life: t 1 = 2 k k = rate constant Half life does not depend on the concentration of reactants. Copyright Cengage Learning. All rights reserved 18
19 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? k = min 1 Copyright Cengage Learning. All rights reserved 19
20 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 20
21 Section 12.4 The Integrated Rate Law Plot of ln[c 4 H 6 ] vs Time and Plot of 1/[C 4 H 6 ] vs Time
22 Section 12.4 The Integrated Rate Law Second-Order Half Life: t = k 1 A 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 22
23 Section 12.4 The Integrated Rate Law EXERCISE! For a 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 = M 1 min 1 c) Calculate [A] at t = 525 minutes. [A] = 0.23 M Copyright Cengage Learning. All rights reserved 23
24 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 24
25 Section 12.4 The Integrated Rate Law Plot of [A] vs Time Copyright Cengage Learning. All rights reserved 25
26 Section 12.4 The Integrated Rate Law Zero-Order Half Life: t = 2 k 1 2 A 0 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 26
27 Section 12.4 The Integrated Rate Law CONCEPT CHECK! How can you tell the difference among 0 th, 1 st, and 2 nd order rate laws from their graphs? Copyright Cengage Learning. All rights reserved 27
28 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 28
29 Section 12.4 The Integrated Rate Law Summary of the Rate Laws Copyright Cengage Learning. All rights reserved 29
30 Section 12.4 The Integrated Rate Law EXERCISE! Consider the reaction aa Products. [A] 0 = 5.0 M and k = (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: a) Zero order b) First order c) Second order 4.7 M 3.7 M 2.0 M Copyright Cengage Learning. All rights reserved 30
31 Section 12.5 Reaction Mechanisms Reaction Mechanism 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 31
32 Section 12.5 Reaction Mechanisms A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 (g) + CO(g) NO(g) + CO 2 (g) Copyright Cengage Learning. All rights reserved 32
33 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 33
34 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 34
35 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 35
36 Section 12.5 Reaction Mechanisms Decomposition of N 2 O 5 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 36
37 Section 12.5 Reaction Mechanisms Decomposition of N 2 O 5 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) 2( ) Step 1: N 2 O 5 NO 2 + NO 3 (fast) Step 2: NO 2 + NO 3 NO + O 2 + NO 2 Step 3: NO 3 + NO 2NO 2 (slow) (fast) Copyright Cengage Learning. All rights reserved 37
38 Section 12.5 Reaction Mechanisms CONCEPT CHECK! The reaction A + 2B C has the following proposed mechanism: A + B D (fast equilibrium) D + B C (slow) Write the rate law for this mechanism. rate = k[a][b] 2 Copyright Cengage Learning. All rights reserved 38
39 Section 12.6 A Model for Chemical Kinetics Collision Model Molecules must collide to react. Main Factors: Activation energy, E a Temperature Molecular orientations Copyright Cengage Learning. All rights reserved 39
40 Section 12.6 A Model for Chemical Kinetics Activation Energy, E a Energy that must be overcome to produce a chemical reaction. Copyright Cengage Learning. All rights reserved 40
41 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 41
42 Section 12.6 A Model for Chemical Kinetics Change in Potential Energy Copyright Cengage Learning. All rights reserved 42
43 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 43
44 Section 12.6 A Model for Chemical Kinetics The Gas Phase Reaction of NO and Cl 2 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 44
45 Section 12.6 A Model for Chemical Kinetics Arrhenius Equation k E a = / RT Ae A = frequency factor E a = activation energy R = gas constant ( J/K mol) T = temperature (in K) Copyright Cengage Learning. All rights reserved 45
46 Section 12.6 A Model for Chemical Kinetics Linear Form of Arrhenius Equation ln( ) = E 1 R T k a + ln A Copyright Cengage Learning. All rights reserved 46
47 Section 12.6 A Model for Chemical Kinetics Linear Form of Arrhenius Equation Copyright Cengage Learning. All rights reserved 47
48 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? E a = 53 kj Copyright Cengage Learning. All rights reserved 48
49 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 49
50 Section 12.7 Catalysis Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction Copyright Cengage Learning. All rights reserved 50
51 Section 12.7 Catalysis Effect of a Catalyst on the Number of Reaction-Producing Collisions Copyright Cengage Learning. All rights reserved 51
52 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 52
53 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 53
54 Section 12.7 Catalysis Heterogeneous Catalyst 1. Adsorption and activation of the reactants. 2. Migration of the adsorbed reactants on the surface. 3. Reaction of the adsorbed substances. 4. Escape, or desorption, of the products. Copyright Cengage Learning. All rights reserved 54
55 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 55
56 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 56
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