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

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Presentation on theme: "Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

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

2 1. Distinguish between average and instantaneous rates of chemical reactions. 2. Define rate constant. 3. Use the concepts of stoichiometry to write reaction rate expressions in terms of the disappearance of reactants and the appearance of products. 4. Sketch the rate of reaction versus concentration of reactant for zero and first order reactions. 5. Use rate data to detemine rate laws and rate constants. 6. Show mathematically that λn[A] = -k t + λn[A] 0 for first order reactions. 7. Determine the time required for the concentration of a reactant to change a desired amount given the initial concentration and the rate constant for a first order reaction. 8. Show that half-life is independent of initial concentration of the reactant in a first order reaction. 9. Use the concept of half-life to determine concentration of reactants over time. 10. Show mathematically that t ½ = 1/ k[A] 0 for second order reactions.

3 11.Describe the Collision Theory of Chemical Kinetics using the terms activation energy, activated complex (transition state), potential energy profiles, endothermic and exothermic reactions. 12. Use Arrhenius equation to determine the activation energy of a reaction. 13.Show that the sum of elementary steps is the overall reaction for a reaction mechanism and that intermediates appear in the reaction mechanism but not in the overall reaction. 14. Define molecularity of unimolecular, bimolecular, and termolecular reactions. 15. Relate the importance of the rate-determining step in determination of reaction mechanisms. 16. Suggest how the use of isotopes can be used to experimentally determine reaction mechanisms. 17. Describe what a catalysis does, how it effects activation energy and the difference between homogeneous and heterogeneous catalysis. 18. Relate the importance of enzymes as biological catalysts.

4 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. 13.1

5 A B 13.1 rate = -  [A] tt rate = [B][B] tt time

6 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption 393 nm Br 2 (aq) 13.1

7 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 13.1

8 rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] 13.1 = rate constant = 3.50 x 10 -3 s -1

9 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 13.1

10 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) 13.1

11 Reaction Rates and Stoichiometry 13.1 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

12 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 13.1

13 The Rate Law 13.2 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

14 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 ] 13.2

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 13.2

16 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I-][I-] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.081/M s 13.2 rate = k [S 2 O 8 2- ][I - ]

17 First-Order Reactions 13.3 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 exp(-kt) ln[A] = ln[A] 0 - kt

18 The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 = 13.3

19 First-Order Reactions 13.3 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 ln2 k = 0.693 k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln2 k = 0.693 5.7 x 10 -4 s -1 = = 1216 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

20 A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16 13.3

21 Second-Order Reactions 13.3 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

22 Zero-Order Reactions 13.3 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

23 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 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 ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 13.3

24 A + B C + D Exothermic Reaction Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. 13.4

25 Temperature Dependence of the Rate Constant k = A exp( -E a /RT ) 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 lnk = - EaEa R 1 T + lnA (Arrhenius equation) 13.4

26 lnk = - EaEa R 1 T + lnA

27 13.5 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 +

28 13.5 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

29 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps 13.5 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.

30 The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 What is the intermediate? NO 3 What can you say about the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2 13.5

31 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a /RT )EaEa k uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed E a < E a ‘ 13.6

32 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 catalyses Base catalyses 13.6

33 N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process 13.6

34 Ostwald Process Hot Pt wire over NH 3 solution 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) 13.6

35 Catalytic Converters 13.6 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter NO + NO 2 N 2 + O 2 catalytic converter

36 Enzyme Catalysis 13.6

37 uncatalyzed enzyme catalyzed 13.6


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