1 Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative.

3 A B rate = -  [A] tt rate = [B][B] tt

4 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption red-brown t 1 < t 2 < t 3

5 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time

6 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time

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8 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d

Example 13.1 Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products:

Example 13.2 Consider the reaction Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of M/s. (a)At what rate is N 2 O 5 being formed? (b)At what rate is NO 2 reacting?

Review of Concepts Write a balanced chemical equation for a gas- phase reaction whose rate is given by 11

RATE LAW 12 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g)

13 rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x s -1

14 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. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall

15 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1

16 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ]

Example 13.3 The reaction of nitric oxide with hydrogen at 1280°C is From the following data collected at this temperature, determine (a) the rate law (b) the rate constant (c) the rate of the reaction when [NO] = 12.0 × 10 −3 M and [H 2 ] = 6.0 × 10 −3 M

Example 13.3

Review of Concepts The relative rates of the reaction 2A + B → products shown in the diagrams (a)-(c) are 1:2:4. The red spheres represent A molecules and the green spheres represent B molecules. Write a rate law for this reaction. 19

INTEGRATED RATE LAW 20

21 First-Order Reactions A product rate = -  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 [A] = [A] 0 e −kt ln[A] = -kt + ln[A] 0

Example 13.4 The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 × 10 −4 s −1 at 500°C. (a)If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b)How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c)How long (in minutes) will it take to convert 74 percent of the starting material?

23 2N 2 O 5 4NO 2 (g) + O 2 (g) Graphical Determination of k

Example 13.5 The rate of decomposition of azomethane (C 2 H 6 N 2 ) is studied by monitoring the partial pressure of the reactant as a function of time: The data obtained at 300°C are shown in the following table: Are these values consistent with first-order kinetics? If so, determine the rate constant.

Example 13.5

26 First-Order Reactions 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 ln [A] 0 [A] 0 /2 k = t½t½ ln 2 k = k =

27 A product First-order reaction # of half-lives [A] = [A] 0 /n

Example 13.6 The decomposition of ethane (C 2 H 6 ) to methyl radicals is a first- order reaction with a rate constant of 5.36 × 10 −4 s −1 at 700°C: Calculate the half-life of the reaction in minutes.

Review of Concepts Consider the first-order reaction A B in which A molecules (blue spheres) are converted to B molecules (orange spheres). (a) What are the half-life and rate constant for the reaction? (b) How many molecules of A and B are present at t = 20s and t = 30s? 29

30 Second-Order Reactions A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0

Example 13.7 Iodine atoms combine to form molecular iodine in the gas phase This reaction follows second-order kinetics and has the high rate constant 7.0 × 10 9 /M · s at 23°C. (a)If the initial concentration of I was M, calculate the concentration after 2.0 min. (b)Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.

32 Zero-Order Reactions A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t = 0 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt

33 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln 2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0

Rate Laws Summary 1.Assume rxn studied under conditions where only forward rxn is important 2.Two types of rate laws a.Differential rate law b.Integrated rate law 3.Use method of initial rates to obtain rate law 34

TEMPERATURE DEPENDENCE OF THE RATE CONSTANT 35

Collision Theory Parameters 1.Collision frequency 2.Energy of collisions 3.Orientation of collisions Temp velocities frequency 36

37 Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +

38 Kinetic Energy # of collisions

39 Importance of Molecular Orientation effective collision ineffective collision

40 E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor ln k = - EaEa R 1 T + lnA Arrhenius equation Alternate format:

Example 13.8 The rate constants for the decomposition of acetaldehyde were measured at five different temperatures. The data are shown in the table. Plot ln k versus 1/T, and determine the activation energy (in kJ/mol) for the reaction. Note that the reaction is “3/2” order in CH 3 CHO, so k has the units of 1/M ½ ·s.

42 Alternate Form of the Arrhenius Equation At two temperatures, T 1 and T 2 or

Example 13.9 The rate constant of a first-order reaction is 3.46 × 10 −2 s −1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

44 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. EaEa k rate catalyzed > rate uncatalyzed E a < E a ′ UncatalyzedCatalyzed

45 In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis

46 N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process

47 Ostwald Process Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq)

48 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

49 Enzyme Catalysis

50 Binding of Glucose to Hexokinase

51 rate =  [P] tt rate = k [ES] Enzyme Kinetics

Factors that affect rxn rate 1. Temperature 2. Concentration 3. Surface area / physical state 4. presence of a catalyst 52

What does a chemical equation tell you? 53 CO (g) + NO 2 (g) CO 2 (g) + NO (g)

54 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +

55 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:

56 Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

57 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation.

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59 Sequence of Steps in Studying a Reaction Mechanism

Example The gas-phase decomposition of nitrous oxide (N 2 O) is believed to occur via two elementary steps: Experimentally the rate law is found to be rate = k[N 2 O]. (a)Write the equation for the overall reaction. (b)Identify the intermediate. (c)What can you say about the relative rates of steps 1 and 2?

Example NO 3 is detected Experimental rate law: rate=k[NO 2 ] 2 Give a plausible mechanism 61 CO (g) + NO 2 (g) CO 2 (g) + NO (g)

Review of Concepts The rate law for the reaction H 2 + 2IBr I 2 + 2HBr is rate=k[H 2 ][IBr]. Given that HI is and intermediate, write a two-step mechanism for the reaction. 62