Chemical Kinetics Ch t ap 1 er

Transcription

1 Chemical Kinetics Chapter 13 1

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 rate = [B] 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 time rate = - [A] t rate = [B] t

4 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time Br 2 (aq) m 393 nm light Detector [Br 2 ] Absorption 393 n

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

6 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) slope of tangent slope of tangentt slope of tangent instantaneous rate = rate for specific instance in time

7 rate [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant =350x s

8 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

9 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F x 2 ] [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 ]

10 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. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ]

11 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 2-8 (aq) + 3I - (aq) 2SO 2-4 (aq) + I - 3 (aq) Experiment [S 2 O 8 2- ] [I - ] 2 8 Initial Rate (M/s) x x x

12 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 2-8 (aq) + 3I - (aq) 2SO 2-4 (aq) + I - 3 (aq) Experiment [S 2 O 8 2- ] [I - ] Initial Rate (M/s) x 10-4 y = 1 rate = k [S 2 O 2-8 ] x [I - ] y x = x x 10-4 Double [I - ], rate doubles (experiment 1 & 2) Double [S 2 O 2-8 ], rate doubles (experiment 2 & 3) rate = k [S 2 O 8 2- ][I - ] rate k = [S2 O 2-8 ][I - ] 2.2 x 10-4 M/s = = 0.08/M s08/m (0.08 M)(0.034 M)

13 Example # 1 [A] [B] Rate Write the rate law Calculate k 13

14 Example # 2 [A] [B] Rate Write the rate law Calculate k 14

15 You Try [A] [B] Rate Write the rate law Calculate k 15

16 A product k = rate [A] = M/s M First-Order Reactions = 1/s or s -1 rate = - [A] t - rate = k [A] [A] 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

17 Decomposition of N 2 O 5 ln[a] = ln[a] 0 - kt

18 Decomposition of N 2 O

19 The reaction 2A B is first order in A with a rate constant of x 10-2 s -1 at t80 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 = [A] ln 0 [A] k = 0.88 M ln 0.14 M [A] 0 = 0.88 M [A] = 0.14 M 2.8 x 10-2 s -1 = 66 s

20 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 ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 ln = = = k k 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? t ½ = ln2 k = = 1200 s = 20 minutes 5.7 x 10-4 s -1 How do you know decomposition is first order? units of k (s -1 )

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

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23 Second-Order Reactions A product rate = - [A] t rate = k [A] 2 k = rate [A] 2 = M/s M 2 = 1/M s - [A] t = k [A] 2 1 [A] = 1 [A] 0 + kt [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 = 1 k[a]

24 Zero-Order Reactions A product rate = - [A] t rate = k [A] 0 = k k = rate [A] = M/s [A] - 0 t = k [A] = [A] 0 - kt [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 2k

25 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order Rate Law Concentration-Time Equation Half-Life 0 rate = k [A] = [A] 0 - kt t ½ = [A] 0 2k 1 rate = k [A] ln[a] = ln[a] 0 - kt t ln2 ½ = k rate = k [A] 2 [A] = 1 + kt t [A] ½ = 0 k[a]

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

27 Temperature Dependence of the Rate Constant k = A exp( -E a /RT ) (Arrhenius equation) E a is the activation energy (J/mol) R is the gas constant (8.314 J/K mol) T is the absolute temperature A is the frequency factor lnk = - E a R E 1 1 T + lna

28 lnk = - E a R 1 T + lna

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30 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

31 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. 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 The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular l reaction elementary step with 1 molecule l Bimolecular reaction elementary step with 2 molecules Termolecular reaction elementary step with 3 molecules

32 Rate Laws and Elementary Steps Unimolecular reaction A products rate = k [A] Bimolecular reaction A + B products rate = k [A][B] Bimolecular reaction A + A products rate = k [A] 2 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

33 The experimental rate law for the reaction between NO 2 and dcot to produce NO and dco 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? What is the intermediate? NO 2 + CO NO + CO 2 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

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

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

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

37 Practice The thermal decomposition of phosphine: hi 4 PH 3(g) P 4(g) + 6 H 2(g) is first order kinetics. If the half-life is 35.0 s at 680 o C, calculate: 1. the rate constant 2. the time required for 95% to decompose