Chapter 12 Chemical Kinetics

Chapter 12 Table of Contents Copyright © Cengage Learning. All rights reserved Reaction Rates 12.2 Rate Laws: An Introduction 12.3 Determining the Form of the Rate Law 12.4 The Integrated Rate Law 12.5 Reaction Mechanisms 12.6A Model for Chemical Kinetics 12.7Catalysis

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 3 Objectives To define Chemical Kinetics To define Reaction Rate and illustrate how it is determined To illustrate that the Instantaneous Rate is calculated by determining the slope of the tangent line at a given point on a curve.

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 4 Chemical Kinetics Chemical Kinetics – the area of chemistry that studies the rate at which a chemical reaction takes place. The following reaction is considered spontaneous (favorable) and will release energy: N 2 (g) + 3H 2 (g)  2NH 3 (g) + energy However, the two gas reactants may coexist indefinitely under normal conditions. Not so good if you are in the fertilizer business!

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 5 Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered.

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 6 The Decomposition of Nitrogen Dioxide 2NO 2(g)  2NO (g) + O 2(g) 300 O C)

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 7 The Decomposition of Nitrogen Dioxide 2NO 2  2 NO + O 2 Rate =  [NO 2 ]/  t Slope! Average if you use the data points or instantaneous if you use the tangent line at a given point for a given time.

Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 8 Objectives Review To define Chemical Kinetics To define Reaction Rate and illustrate how it is determined using a slope calculation To illustrate that the Instantaneous Rate is calculated by determining the slope of the tangent line at a given point on a curve. Work Session: page 566 #7, 19, 21

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 9 Objectives To define the Rate Law Equation and its variables To determine the form (order) of the rate law experimentally

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 10 Rate Law The Rate Law shows how the rate depends on the [concentrations] of reactants. [Reactant] units of M 2NO 2 (g) → 2NO(g) + O 2 (g) Rate = k[NO 2 ] n  k = rate constant (coefficient of parabola or m if linear)  [NO 2 ] = Concentration of reactant (M = mol/L)  n = order of the reactant (form) This form of the rate law allows you to model the continuous curve regardless of the order (exponent).

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 11 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. 2NO 2 (g) 2NO(g) + O 2 (g) The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 12 Rate Law Rate = k[NO 2 ] n Rate =  [NO 2 ]/  t

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 13 Rate Law Rate = k[NO 2 ] n (How to Calculate n experimentally) ExperimentInitial [NO 2 ] Initial Rate(mol/L*s) M1.25 X M2.50 X Rate = k[NO 2 ] n write the expression for both exp’ts the bigger value in the numerator…. Rate 2 = k[0.9] n = 2.50 X Rate1 = k[0.45] n = 1.25 X Simplify and solve for n [2] n = 2

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 14 Rate Law Rate 2 = k[0.9] n = 2.50 X Rate1 = k[0.45] n = 1.25 X Simplify and solve for n [2] n = 2 n = 1 or in other words, The rate law for this reaction is first order with respect to the reactant NO 2. (Important to define) 2NO 2 (g)  2NO(g) + O 2 (g)

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 15 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. We are usually looking for the slow step in a process.

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 16 Objectives Review To define the Rate Law Equation and its variables To determine the form (order) of the rate law experimentally

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 17 Objectives To determine the overall reaction order of the rate law experimentally using data from several reactants

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 18 Overall Reaction Order The sum of the exponents in the reaction rate equation. aA + bB  cC + …? Rate = k[A] n [B] m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 19 Let’s use the following EXPERIMENTAL data to determine the order (form) for each reactant. NH NO 2 -  N 2 + 2H 2 O First, write the general rate law……Rate = ?? Exp’t Initial Initial Initial Rate [NH 4 + ] [NO 2 - ] (mol/L*s) M M 1.35 X M M 2.70 X M M 5.40 X Remember the trick to find rate order = bigger numerator For order w/respect to [NH 4 + ], [other] must stay constant

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 20 Overall Reaction Order Remember bigger numerator! NH NO 2 -  N 2 + 2H 2 O Exp’t Initial Initial Initial Rate [NH 4 + ] [NO 2 - ] (mol/L*s) M M 1.35 X M M 2.70 X M M 5.40 X Rate 2 = k[0.100] n [0.010] m = 2.70 X Rate1 = k[0.100] n [0.0050] m = 1.35 X Simplify and solve for m

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 21 NH NO 2 -  N 2 + 2H 2 O Exp’t Initial Initial Initial Rate [NH 4 + ] [NO 2 - ] (mol/L*s) M M 1.35 X M M 2.70 X M M 5.40 X Rate 2 = k[0.100] n [0.010] m = 2.70 X Rate1 = k[0.100] n [0.0050] m = 1.35 X Simplify and solve for m [2] m = 2 The rate law for this reaction is first order in the reactant NO 2 -. What about the other reactant?

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 22 With respect to NH 4 + (keep other [conc] same) NH NO 2 -  N 2 + 2H 2 O Exp’t Initial Initial Initial Rate [NH 4 + ] [NO 2 - ] (mol/L*s) M M 1.35 X M M 2.70 X M M 5.40 X Rate 3 = k[0.200] n [0.010] m = 5.40 X Rate2 = k[0.100] n [0.010] m = 2.70 X Simplify and solve for n Also first order with respect to NH 4 +

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 23 Overall Reaction Order The sum of the exponents in the reaction rate equation. aA + bB  cC + …? Rate = k[A] n [B] m Overall reaction order = n + m m = 1, n = 1 Overall Reaction order = 2

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 24 Turn to page 567 and work on #25.

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 25 Concept Check How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why?

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 26 Objectives Review To determine the overall reaction order of the rate law experimentally using data from several reactants Work Session: # 26, 27, 28 (just find the order) To Slide 43 to skip Integrated Rate LawSlide 43

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 27 Half-Life of Reactions

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 28 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 First-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 29 Plot of ln[N 2 O 5 ] vs Time

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 30 Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. First-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 31 Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? k = 7.8 x 10 –3 min –1

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 32 Rate = k[A] 2 Integrated: [A] = concentration of A at time t k = rate constant t = time [A] o = initial concentration of A Second-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 33 Plot of ln[C 4 H 6 ] vs Time and Plot of 1/[C 4 H 6 ] vs Time

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 34 Half–Life: 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. Second-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 35 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 = 8.0 x M –1 min –1 c)Calculate [A] at t = 525 minutes. [A] = 0.23 M

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 36 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 Zero-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 37 Plot of [A] vs Time

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 38 Half–Life: k = rate constant [A] o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Zero-Order

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 39 Concept Check How can you tell the difference among 0 th, 1 st, and 2 nd order rate laws from their graphs?

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 40 Rate Laws

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 41 Summary of the Rate Laws

Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 42 Exercise Consider the reaction aA  Products. [A] 0 = 5.0 M and k = 1.0 x 10 –2 (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

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 43 To define reaction mechanism To determine if a mechanism is a possible explanation of observed behavior Objectives

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 44 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 reactant or product in the overall balanced reaction. Reaction Mechanism

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 45 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) NO 2 (g) + NO 2 (g) → NO(g) + NO 3 (g). NO 3 (g) + CO(g) → NO 2 (g) + CO 2 (g) intermediate?

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 46 Molecularity – the number of species that must collide to produce the reaction in that step. Unimolecular – reaction involving one molecule; first order. A  products Rate = k[A] Bimolecular – reaction involving the collision of two species; second order. A + A  products Rate = k[A] 2 A + B  products Rate=k[A][B] Elementary Steps (Molecularity) Page 550

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 47 Termolecular – reaction involving the collision of three species; third order. A + A + B  products Rate = [ ][ ][ ] = [ ] [ ] A + B + C  products Rate = Elementary Steps (Molecularity) Page 550

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 48 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. Focus on the slow step….. Rate-Determining Step

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 49 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. If these two conditions are met, the mechanism is possibly correct as a mechanism can never be proved absolutely correct. Reaction Mechanism Requirements

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 50 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) NO 2 (g) + NO 2 (g) → NO(g) + NO 3 (g) (slow) NO 3 (g) + CO(g) → NO 2 (g) + CO 2 (g) (fast) Does the sum of the elementary steps give the overall balanced equation for the reaction? Does the mechanism agree with the experimentally determined rate law given as Rate = k[NO 2 ] 2 ? Is this a possible mechanism for this reaction? The overall rate can not be faster than the slow step.

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 51 Decomposition of N 2 O 5

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 52 Decomposition of N 2 O 5 Is this a possible Mechanism? 1?2? 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Step 1: N 2 O 5 NO 2 + NO 3 (fast) Step 2: NO 2 + NO 3 → NO + O 2 + NO 2 (slow) Step 3: NO 3 + NO → 2NO 2 (fast) Does the sum of the elementary steps give the overall balanced equation for the reaction? Does the mechanism agree with the experimentally determined rate law?? Is this a possible mechanism for this reaction? 2( )

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 53 Write the balanced reaction for the synthesis of gases nitrogen dioxide and fluorine that forms 1 product. ____NO 2 + ____F 2  ____ NO 2 F The experimentally determined rate law is: Rate = k[NO 2 ][F 2 ] A suggested mechanism is: NO 2 + F 2  NO 2 F + F(slow) F + NO 2  NO 2 F(fast) Is this an acceptable mechanism?

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 54 The experimentally determined rate law is: Rate = k[NO 2 ][F 2 ] A suggested mechanism is: NO 2 + F 2  NO 2 F + F(slow) F + NO 2  NO 2 F(fast) Is this an acceptable mechanism? Does the sum of the elementary steps give the overall balanced equation for the reaction? Does the mechanism agree with the experimentally determined rate law??

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 55 Concept Check (sometimes the Rate Law Looks Tricky.. The reaction A + 2B  C has the following proposed mechanism: A + B D(fast equilibrium) D + B  C(slow) Does this rate law make sense? rate = k[A][B] 2 A has to collide with B, then with another B Why wouldn’t rate = [B] work? Slow step.

Section 12.6 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 56 To define reaction mechanism To determine if a mechanism is a possible explanation of observed behavior Work Session: 49, 51 Objectives Review

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 57 To define activation energy E A To understand E A on a Potential Energy Graph Objectives

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 58 A=frequency factor (Molecular Orientations E a =activation energy R=gas constant ( J/K·mol) T =temperature (in K) Arrhenius Equation Molecules must collide to react. Main Factors:

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 59 Energy that must be overcome to produce a chemical reaction. Activation Energy, E a

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 60 Transition States and Activation Energy

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 61 Change in Potential Energy

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 62 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. For Reactants to Form Products

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 63 The Gas Phase Reaction of NO and Cl 2

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 64 To define activation energy E A To understand E A on a Potential Energy Graph Work Session: Page 571 #53 Objectives Review

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 65 To define a catalyst To identify catalysis on a Potential Energy Graph Objectives

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 66 A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Catalyst

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 67 Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 68 Effect of a Catalyst on the Number of Reaction-Producing Collisions

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 69 Heterogeneous Catalysis

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 70 Homogeneous Catalysis

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 71 To define a catalyst To identify catalysis on a Potential Energy Graph Work Session Page 572 # 65(a, b) #68a Catalyst Video 1:40Catalyst Video End of Chapter 12! Objectives Review

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 72 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. Heterogeneous Catalyst

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 73 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. Heterogeneous Catalyst

Section 12.8 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 74 Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Homogeneous Catalyst

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 75 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.

Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 76 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.

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 77 Determine experimentally the power to which each reactant concentration must be raised in the rate law.

Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 78 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.

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 79 Linear Form of Arrhenius Equation

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 80 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

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 81 Linear Form of Arrhenius Equation

Section 12.7 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 82 Molecules must collide to react. Main Factors:  Activation energy, E a  Temperature  Molecular orientations Collision Model