What we learn from Chap. 15

Transcription

1 Chemical Kinetics Chapter 15 What we learn from Chap The focus of this chapter is the rates and mechanisms of chemical reactions. The applications center around pesticides, beginning with the opening chapter photo of the crop duster spraying the farm field. In the introduction, we give some statistics about pesticide use and raise the question of why the breakdown of pesticides in the environment is so slow while other chemical processes, such as combustion, are so fast. This sets the stage for kinetics. 1

2 CHAPTER OUTLINE 15.3 I. Reaction Rates A. Instantaneous Rate, Initial Rate, and Average Rate II. An Introduction to Rate Laws A. Collision Theory III. Changes in Time.The Integrated Rate Law A. Integrated First.Order Rate Law B. Half-life C. Other Rate Laws IV. Methods of Determining Rate Laws V. Looking Back at Rate Laws VI. Reaction Mechanisms A. Transition State t Theory VII. Applications of Catalysts Hydrolysis reaction of Atrazine A herbicide G<0kJ/mol, H=-35kJ/mol rapid degradation by microorganisms 1 4

3 Persistence in the Environment 15.5 Little River-Agriculture i basin --- Herbicide Pesticide Lafayette Creek-Urban basin --- Herbicide Pesticide CO(g) + NO (g) CO (g) + NO(g) E a =134 kj CO(g) + NO (g) ΔH = -6 kj CO (g) + NO(g) Kinetic vs Thermodynamic 1 6

4 Reaction Rates Depend on the concentrations of reactants Depend on temperature Affected by catalysts Instantaneous rate =dc/dt C ΔC (=C -C 1 ) C 1 average rate =C/t Δt (=t -t 1 ) t 1 t 1 8

5 Reaction Rates 15.9 Average reaction rate is the change in the concentration of a reactant or product per unit time. Initial reaction rate is the rate at the start of the reaction. Instantaneous reaction rate is the value of the rate constant at a specific c time. 1 9 Consider the reaction: N O 5 (g) 4NO (g) + O (g) 1. Rate = rate of disappearance of N O = - 5 N O 5 t Rate = rate of appearance of NO NO NO O 1 1 Rate = = = t 4 t t 3. Rate = rate of appearance of Stoichiometric Relationship 5 O = O tt NO tt 1 10

6 rate expression rate rate = C i / t aa + bb cc + dd [ C ] [ D ] [ A ] [ B ] ct dt at bt Minus sign used because concentrations of reactants decrease with time. Concentration will be expressed in molarity; time may be in seconds, minutes, years, etc Rate Law 15.1 Rate law is the relationship between the concentration and the reaction rate. Rate = k[a] n [B] m k = rate constant [A] and [B] are the molar concentrations of A and B n and m are the individual orders of the compounds Overall order = n+m 1 1

7 Sample Problem For the reaction: NO(g) + H (g) H O(l) + O (g) The rate law is Rate = k[no] [ ] [H ] ]. What is the order with respect to each reactant and what is the overall order? nd order in NO 1 st order in H 3 rd order overall 1 13 Collision Theory A reaction occurs when molecules collide with the appropriate energy and orientation. The higher the temperature the larger the rate the faster the reaction. Activation Energy (E a ) is the minimum energy to make a product. 1 14

8 Molecular oecua Collisions so s Collision o Theory

9 Collision Theory The Integrated Rate Law : Time vs. Conc. First Order Reaction A products da [ ] da [ ] da [ ] rate k[ A] : kt : k dt dt [ A] [ A] ln[ A ] [ A ] t t [ A] kt 0 0 A t -kt t 0 A ln = kt or A = A e 0 [A] t = concentration of A at time t [A] 0 = concentration of A at time t=0 k = rate constant 1 18

10 First Order Reaction Half-life of a 1 st order reaction is the time it take for a reaction to proceed to 50% completion. t 1/ = k 1 19 Half-life of a Reaction

11 Sample Problem 15.1 Cyclobutane, C 4 H 8, forms ethylene when heated. The reaction is 1 st order with k = 5.09 x 10-4 /s. How long will it take for 70% of the cyclobutane to react? If 70% reacts then 30% remains. Let [C4H 8] , [C4H 8] t 0.30 [C 4H 8] t ln kt [C4H 8] t ln s t.3710 s 1 1 Sample Problem 15. Cyclobutane, C 4 H 8, forms ethylene when heated. The reaction is 1 st order with k = 5.09 x 10-4 /s. What is the half-life? t k s s 4 1 1

12 Zero eoode Order Reaction eacto 15.3 Rate = k[a] 0 = k Does not depend on concentration of any compounds in the reaction. t = 1/ A 0 k 1 3 nd Order Reactions 15.4 Two possibilities: Rate = k[a] or Rate = k[a][b] In the case of Rate = k[a], the integrated equation is: 1 1 = kt + A A t 0 Half-life: 1 t 1/ = k A 0 1 4

13 15.4 Method of Determining Rate Law 15.5 S - +3I - - O 8 (aq) (aq) SO 4 (aq) +I 3- (aq) Experiment [S - - O 8 ] [I ] Initial Rate (M/s) x x x 10-4 Rate =k[s O 8 - ] m [I - ] n 1 5 (cont) m n rate k(0.08) 08) (0.034) 034) n ; n 1 4 m n rate k(0.08) (0.017) 4 m n rate k(008) (0.08) (0.017) 017) 1 4 ( ) m ; m m n 1 rate3.10 k(0.16) (0.017) k -4 Rate3.10 M/ s M s - - SO 8 I 0.08M0.034M -1-1 Rate =(0.081M -1 s -1 )[S O 8 - ][I - ] 1 6

14 Graphical Analysis s of Rate Equations Rate Laws

15 15.6 Reaction Mechanisms s 15.9 Reaction mechanism a series of steps by which a chemical reaction occurs. Elementary Step Each of the single steps in a chemical reaction Rate determining step the slow step in a reaction. Intermediate a compound that is formed and consumed in the course of a reaction. 1 9 Molecularity

16 Metabolism of methoxychlor NO(g)+O (g) NO (g) 15.3 k 1 NO ( g ) N O ( g ) :fast k1 NO g g NO g k ( ) +O ( ) ( ) :slow rate rate k [N O ][O ] k1[ NO] from Rxn 1, k1[ NO] k 1[ NO ] ; [ NO ] k rate kk[ NO] [O ] k '[ NO 1 k 1 ] [ O] 1 rate law (experiment) = k[no] [O ] ] 1 3

17 Ex 15.8) IBr(g) I (g)+br (g) IBr g I g Br g k ( ) 1 ( ) ( ) :slow IBr g Br g I g Br g k ( ) ( ) ( ) ( ) :fast I g I g I g k3 ( ) ( ) ( ) :fast rate rate 1 k [I Br] 1 ] 1 st order reaction 1 33 Prac NO(g)+O (g) (g) NO (g) Step1) NO(g)+O (g) NO 3 (g) fast Step) NO 3 (g)+no(g) NO (g) slow 1 34

18 Ex) I - + OCl - Cl - + OI - in basic Soln [ I ][ OCl ] experimental rate k [ OH ] mechanism k1 k OCl ( aq) H O( l) HOCl( aq) OH ( aq) :fast I aq HOCl aq OH HOI aq Cl 1 k ( ) ( ) ( ) (aq) :slow k3 OH ( aq) OH ( aq) H O( l) OI ( aq) :fast HOI 1 35 Transition State Theory Cf) Reaction coordinate 1 36

19 15.37 Transition State Theory The theory describes how bonds in reacting molecules reorganize to form bonds in products. Molecules go through a transition state and form an activated complex. Molecules must cross the energy barrier, known as the activation energy Transition State 1 38

20 Arrhenius Equation Relates the reaction rate to the activation energy, the frequency of collisions with proper orientation and the temperature. k = Ae -E a/rt k = rate constant A = frequency factor E a = activation energy 1 39 Arrhenius Equation The activation energy can be calculated if the rate of reaction at two temperatures is known. k E a 1 1 ln = k1 R T1 T k 1 and k are the rate constants at T 1 and T 1 40

21 15.41 Graphical Determination of E a 1 41 Sample Problem 15.4 The thermal decomposition of acetone was studied. It was found that k = at 650 K and k = at 800 K. What is activation energy? k E 1 1 a ln = k1 R T1 T E a ln = J/K mol Ea J/mol = 114 kj/mol 1 4

22 15.7 Catalysts Catalysts t change the mechanism of a reaction to a new reaction pathway with a lower energy of activation. Homogeneous catalyst catalyst is present in the same phase as the reacting substance. Heterogeneous catalyst catalyst is present in a different phase as the reacting substance Catalysts O( g) O( g) O ( g) Cl 3 O Cl ClO O 3 ClO O O Cl 1 44

23 Catalysts Catalysts CH( g) H( g) CH( g) Pt

24 Catalysts a s Enzyme glucose lactose galactose lactase 1 48

25 Problems ,14,4,44,5,68,9,10,