Chapter 14 Chemical Kinetics

Transcription

1 7/10/003 Chapter 14 Chemical Kinetics 14-1 Rates of Chemical Reactions 14- Reaction Rates and Concentrations 14-3 The Dependence of Concentrations on Time 14-4 Reaction Mechanisms 14-5 Reaction Mechanism and Rate Laws 14-6 Effect of Temperature on Reaction Rates 14-7 Kinetics of Catalysis OFB Chapter 14 1

2 Chapter 14 Chemical Kinetics Thus far in the course, we have been using and taling about chemical reactions Reactants, products, and how much is involved Chemical Kinetics deals with how fast a reaction proceeds Kinetics Rates of Chemical Reactions And how to deduce reaction mechanisms from observed rates of reactions Activation Energy Catalysts 7/10/003 OFB Chapter 14

3 average reaction rate units are mol/l s average rate [X] t mol mol L s [X] f [X] i t t f i L [X] t 1 1 s To measure rates we could monitor the disappearance of reactants or appearance of products e.g., NO + CO NO + CO 7/10/003 OFB Chapter 14 3

4 To measure rates we could monitor the disappearance of reactants or appearance of products e.g., NO + F NO F 7/10/003 OFB Chapter 14 4

5 OFB Chapter /10/003 Generalized Reaction aa + bb cc + dd t D d t C c t B b t A a rate ] [ 1 ] [ 1 ] [ 1 ] [ 1 +

6 Order of a Reaction N O + decomposition 5 1 NO O e.g., rate aa [A] n products 7/10/003 OFB Chapter 14 6

7 e.g., aa rate [ A] products n n does not have to be an integer n 3/ three halves order reaction Note that n does not equal the coefficient of the reactant. It is related to the reaction mechanism and determined experimentally. 7/10/003 OFB Chapter 14 7

8 aa + bb rate [A] m products [B] n 7/10/003 OFB Chapter 14 8

9 Chapter 14 Chemical Kinetics Example 14- At elevated temperatures, HI reacts according to the chemical equation HI H + I at 443 C, the rate of reaction increases with concentration of HI, as shown in this table. Data Point 1 3 [HI] (mol L -1 ) Rate (mol L -1 s -1 ) 7.5E E-03 1.E-0 a) Determine the order of the reaction with respect to HI and write the rate expression b) Calculate the rate constant and give its units c) Calculate the instantaneous rate of reaction for a [HI] 0.000M 7/10/003 OFB Chapter 14 9

10 HI H + I Data Point 1 3 [HI] (mol L -1 ) Rate (mol L -1 s -1 ) 7.5E E-03 1.E-0 tae the ratio of any two data points 3.0x10 7.5x n 7/10/003 rate second OFB Chapter [HI] order in HI and answer to part A

11 Example 14- b) Calculate the rate constant and give its units rate [X] answer to part A Example 14- c) Calculate the instantaneous rate of reaction for a [HI] 0.000M 7/10/003 OFB Chapter 14 11

12 Similarly for two or more concentrations rate [ A] m [ B] n order is p m + n units of are mol ( p 1) ( L p 1) s 1 Example A + B C initial Rate [A] [B] mol L -1 s E E10-4.8E E E E10-6.0E E E10-5 7/10/003 OFB Chapter 14 1

13 For Example A + B C initial Rate [A] [B] mol L -1 s E E10-4.8E E E E10-6.0E E E st When A is constant (1.0E -4), B increases X3 and rate increases X3 7/10/003 OFB Chapter 14 13

14 For Example A + B C initial Rate [A] [B] mol L -1 s E E10-4.8E E E E10-6.0E E E10-5 nd When B is constant (3.0E -4), A increases X and rate increases X4 7/10/003 OFB Chapter 14 14

15 Can now solve for rate [A] [B] 1 [A] rate [B] 1 [1x10.8x10 4 ] 6 [1x10 4 ] 1.8x10 6 L mol - s -1 7/10/003 OFB Chapter 14 15

16 A 14-3 The Dependence of Concentrations on Time First Order Reactions in general, for first order products [ A] rate t [ A] For now, just accept this important formula 7/10/003 OFB Chapter 14 16

17 If we tae the ln of both sides Recall y mx + b or y b + mx A plot of ln [A] vs t will be a straight line with the Intercept ln[a] 0 Slope - 7/10/003 OFB Chapter 14 17

18 ln[ A] ln[ A] 0 t First Order Reaction ln [A} (in mol / L 0.0E+00.0E E E Slope -t Time (in seconds) 7/10/003 OFB Chapter 14 18

19 First Order Reactions A useful concept for 1 st order reactions (only) is the halflife i.e., the time for ½ [A] 0 (see boo for the derivation) Applies only to First order reactions TRY Examples 14-5 and Exercise 7/10/003 OFB Chapter 14 19

20 Second Order Reactions In General, for second order A From calculus products 1 A [ ] rate - t [ A] This is also the equation for a straight line Y mx +b 7/10/003 OFB Chapter 14 0

21 7/10/003 OFB Chapter 14 1

22 1 [A] t + [ 1 A] o Second Order Reaction 1/[A] (L /mol) Slope Time (in seconds) Second Order Reactions Half-life is NOT A useful concept for nd order reactions but, 7/10/003 t 1/ 1/[A] 0 OFB Chapter 14

23 Summary Integrated Rate Laws First Order Reactions ln[ A] ln[ A] 0 t Second Order Reactions 1 1 t + [A] [ A] o 7/10/003 OFB Chapter 14 3

24 In the real world, if we don t now the order of the reaction [A] n plot both ln [A} (in mol / L First Order Reaction 0.0E+00.0E E E Time (in seconds) 1/[A] (L /mol) Second Order Reaction Time (in seconds) 7/10/003 OFB Chapter 14 4

25 Chapter 14-4 Reaction Mechanisms Most reactions proceed not thru a single step but through a series of steps Each Step is called an elementary reaction Types of elementary Reactions 1. Unimolecular (a single reactant) E.g., A B + C (a decomposition). Bimolecular (most common type) E.g., A + B products 3. Termolecular (less liely event) E.g., A + B + C products 7/10/003 OFB Chapter 14 5

26 Chapter 14 Chemical Kinetics Reaction Mechanism is a detailed series of elementary steps and rates which are combined to yield the overall reaction One goal of chemical inetics is to use the observed rate to chose between several possible reaction mechanisms. 7/10/003 OFB Chapter 14 6

27 1.).) NO NO NO CO NO NO NO (slow) CO (fast) Notice that NO 3 is formed and consumed. This is called a reactive intermediate Notice also that Step 1 is bimolecular Step is bimolecular Reactive intermediate is NO 3 7/10/003 OFB Chapter 14 7

28 7/10/003 Chemical Equilibrium A direct connection exists between the equilibrium constant of a reaction that taes place in a sequence of steps and the rate constants in each step. a.) at equil: forward rate reverse rate b.) K eq f / r aa 1.) f.) f & r + bb cc + dd Forward reaction rate Reverse reaction rate f r [A] a [B] [C] [A] c a b [D] [B] d b r [C] K c eq r [D] d f [A] [C] OFB Chapter 14 8 c a [B] [D] d b

29 A direct connection exists between the equilibrium constant of a reaction that taes place in a sequence of steps and the rate constants in each step. a.) at equil: forward rate reverse rate b.) K eq f / r (sometimes n / -n ) The boo uses a general reaction to illustrate this principle. A (g) + B (g) C (g) + D (g) Suppose the reaction proceeds by the following two step mechanism K K 7/10/ ) A (g) + A (g) A (g) Rate 1 1 [A] -1 Rate -1-1 [A ].) A (g) + B (g) C (g) + D (g) Rate [A ][B] - Rate - - [C][D] 1 K1K 1 [ A ][ C][ D] [ C][ D] [ A] [ A ][ B] [ A] [ B] 1 OFB Chapter

30 14-5 Reaction Mechanism & Rate Laws Typically with a reaction one of several elementary step reaction is the slowest step. This is called the Rate Determining Step (RDS) Case #1: When the RDS occurs first, the first step is slow and determines the rate of the overall reaction. 1.).) NO NO + F + F NO Step 1 is the RDS rate 1 NO 1 F [NO F + F ][F ] (slow) (fast) 7/10/003 OFB Chapter 14 30

31 1.).) NO NO + F + F rate 1 NO NO F Step 1 is the RDS 1[NO F + F ][F ] (slow) (fast) E n e r g y F + NO NO + F NO F Reaction Progress 7/10/003 OFB Chapter 14 31

32 Case #: When the RDS occurs after one or more Fast steps, mechanisms are often signaled by a reaction order greater than two. The slow step determines the overall rate of the reaction. 1.).) NO N Overall NO O + NO N + O reaction + O NO 3 molecule reaction. Is it A Termolecular or Bimolecular reactions? Three way collisions are rare. Try a two step mechanism. 1-1 O NO (fast) (slow) rate [N O ][O But N O is a reactive intermediate ] 7/10/003 OFB Chapter 14 3

33 1.) NO +.) N O NO N + O 1-1 O NO (fast) (slow) rate [N O ][O ] E n e r g y N O + O NO + O NO Reaction Progress 7/10/003 OFB Chapter 14 33

34 OFB Chapter /10/003 ] ][O O [N rate (slow) NO O O N.) (fast) O N NO NO 1.) [NO] ] O [N K ] O [N [NO] at equilibrium

35 1.) NO +.) N O NO N + O 1-1 O NO (fast) (slow) rate [N O ][O ] O ] K [NO] 1 [N Substituting for [N O ] in the rate expression above 7/10/003 OFB Chapter 14 35

36 1.) NO +.) N O NO N + O 1-1 O NO (fast) (slow) rate [N O ][O ] rate K [NO] [O ] 1 E slow n e fast r g y N O + O NO + O NO Reaction Progress 7/10/003 OFB Chapter 14 36

37 14-6 The Effect of Temperature on Reaction Rates where E a is the Activation energy units are energy per mole and A is the "pre - exponential factor" is a constant and has the units of 7/10/003 OFB Chapter 14 37

38 Arrenhius Equation A Ea RT y mx + b 18.5 An Arrenhius Plot Slope -Ea/R 18 ln /T (K -1 ) 7/10/003 OFB Chapter 14 38

39 Transition State 7/10/003 OFB Chapter 14 39

40 Transition State E n e r g y E a f E a r Reaction Progress 7/10/003 OFB Chapter 14 40

41 7/10/003 OFB Chapter 14 41

42 E n e r g y slow fast fast Reaction Progress 7/10/003 OFB Chapter 14 4

43 slow E n e r g y fast fast Reaction Progress 7/10/003 OFB Chapter 14 43

44 Catalyst Chapter 14 Chemical Kinetics provides a lower energy path, but it does not alter the energy of the starting material and product rather it changes the energy of the transition (s), in the reaction A catalyst has no effect on the thermodynamics of the overall reaction Inhibitor is a negative catalyst. It slows the rate of a reaction frequently by barring access to path of low Ea and thereby forcing the reaction to process by a path of higher Ea. 7/10/003 OFB Chapter 14 44

45 7/10/003 OFB Chapter 14 45

46 Kinetics of Catalysis A catalyst has no effect on the thermodynamics of the overall reaction It only provides a lower energy path Examples Pt and Pd are typical catalysts for hydrogenation reactions (e.g., ethylene to ethane conversion) Enzymes act as catalysts Phases Homogenous catalysis the reactants and catalyst are in the same catalyst (gas or liquid phase) Heterogeneous catalysis reaction occurs at the boundary of two different phases (a gas or liquid at the surface of a solid) 7/10/003 OFB Chapter 14 46

47 Chapter 14 Chemical Kinetics Example / exercise 14-1, 14-, 14-3, 14-4, 14-5, 14-6, 14-7, 14-8, 14-9 Problems 7, 9, 11, 15, 19, 1, 3, 5, 37, 41, 43, 45, 51, 53 7/10/003 OFB Chapter 14 47