Reaction Kinetics. An Introduction

Transcription

1 Reaction Kinetics An Introduction

2 A condition of equilibrium is reached in a system when opposing changes occur simultaneously at the same rate. The rate of a chemical reaction may be defined as the # of mols of a substance which disappear or are formed by the reaction per unit volume in a unit of time.

3 Example I C Rate forward I t I t Δ I Δ t 3

4 The previous rate is for the DISAPPEARANCE of I, therefore: rate Δ I Δ t where the negative sign means disappearing or " loss of " 4

5 Bacwards Example Rate bward where the C t positive C t sign Δ C Δ t means when both reaction rates study I Δ C + Δ t for min g 5

6 More Complex Reactions X 3 Z Rate Δ Δ X t Δ Δ Z t this is due to Z appearing 3X as fast as X is disappearing Rate Δ Δ X t 3 Δ Δ Z t 6

7 In General For the Reaction: pp + qq rr + ss Rate p Δ Δ P t q Δ Q Δ t r Δ Δ R t s Δ Δ S t 7

8 Reaction Order 8

9 In General For the Reaction: pp + qq rr + ss Rate p Δ Δ P t q Δ Q Δ t r Δ Δ R t s Δ Δ S t and is proportional to P n Q m or Rate P n Q m where proportionality cons tan t or rate cons tan t 9

10 A reaction with an incredibly large rate constant is faster than a reaction with an incredibly small rate constant. 0

11 For the Reaction: pp + qq rr + ss The reaction is n order in P and m order in Q OR Is OVERALL (n + m) order

12 KEY!!!!!! n and m DO NOT necessarily equal p, q, r or s

13 3 Example 4 ) ( ) ( 4 ) ( O N or t O t NO t O N Rate g O g NO g O N Δ Δ Δ Δ Δ Δ +

14 This reaction is FIRST order in N O 5, NOT SECOND order as one might intuit from the stoichiometry. 4

15 The Order of The Reaction the specification of the empirical (experimentallydetermined) dependence of the rate of the reaction on CONCENTRATIONS The order may 0, a whole number or a non-whole number, e.g., 0 ½ We ll focus on whole numbers and 0 (zero) for reactions of the type: ff + gg Products AND! The order of the reaction is defined in terms of REACTANTS not the products, therefore, products do not need to be specified 5

16 6 Zero-Order Reactions form b mx y line Straight t at F t time at F Chem P here not Integrate t F arrange t F and f t F coefficient a only is f G F t F f Rate o t o F F + Δ Δ Δ Δ Δ Δ Δ Δ : 0 ; : ) ( : Re ' ' 0 0

17 For Zero-Order Reactions These reactions are relatively rare Occur on metal surfaces Reaction rate is INDEPENDENT of concentration of reactants 7

18 First Order Reactions Assume reaction is st order in F and zero order in G: Rate Δ f Δt Δ F f ' F Δt Δ F Re arrange : F F ln F F o F 0 Δt OR F o ' F e and int egrate : t t G 8

19 Many radioactive decays fit st order reactions: 6 Ra 88 Rn He 38 U 9 34 Th He The rate is proportional to F 9

20 The Rate is Proportional to F Δ Δt F F Double F F F F 3 Rate Change X X 4 X 8 0

21 Second Order Reactions: Type t F F egrate and arrange G to respect with order Zero F to respect with order Second F t F and f G F t F f Rate o Δ Δ Δ Δ : int Re ' ' 0

22 Rate Second Order Reactions: Type Re arrange G Re action f o is f Δ F Δt ' Δ F Δt and F FIRST and, F G int egrate ( by G F ln F G order SECOND order o g ΔG Δt hence, in o o F t overall ' F G parts) AND in G, Second order reactions are the most common reactions The rate is proportional to G, G o, F, F o

23 Pseudo-First Order Reactions A special ind of d order reaction: Example M acetyl chloride (AcCl) reacted with 56 M water (XSSV amount) to for HOAC and HCl AcCl + H O HOAc + HCl 3

24 4 Order First PSEUDO Hence order first be to APPEARS action AcCl Rate O H so ect to small too is O H in change The O H AcCl d t AcCl d Rate :, " " Re ' ',, det

25 5 Empirical Method in Determining Reaction Orders studied being are A in ONLY differing reactions different Where A A t t X Step Second A reaction for order unnown the X A A t t rate rate Step First X log log : " " : Δ Δ Δ Δ

26 6 Example A in order ond is reaction M M A A t t X t M A and B A action t M A and B A action sec log.734 log log 5' 5' log log log 5' ; 0.73 : Re 5' ; 0.3 : Re Δ Δ Δ Δ

27 Practical Application Reaction rates are proportional to some power of reactant Determined by using initial reaction rate method 7

28 Compare Reaction with Reaction 8

29 Compare Reaction 3 with Reaction 9

30 Fuse both effects and the rate equation for this reaction Rate Zn HCl REMEMBER: The exponent in a rate equation generally does NOT match the chemical equation coefficients. The exponent MUST be determined experimentally. 30

31 Another way to do this 3

32 Compare Reaction with Reaction 3

33 Compare Reaction 3 with Reaction 33

34 Reaction Order Half-Lives 34

35 35 : ln : : s M F t Order Second CONCENTRATION of INDEPENDENT is Order t First s t Order First M s F t Order Zero o o

36 Example For the reaction: C D + E, Half of the C is used up in 60 seconds. Calculate the fraction of C used up after 0 minutes reaction is first order in C. 36

37 37 Solution Up C Used or C USED UP Hence LEFT C C C C C C Key t C C onds and t o o o 99.9% : ln (0.055)(600) ln ln : ln sec ln

38 Example 4 C is present at about.*0-3 mol% naturally in living matter. A bone dug up showed 9*0-5 mol% 4 C. The half life of 4 C is 570 years. How old is the bone? 38

39 39 Solution old years t t C C yrs t o *0.503 ) (.5*0.*0 9*0 ln ln.* ln

40 Example A decomposition reaction occurs in a fixed-volume container at 460 C. Its rate constant is 4.5*0-3 seconds -. At t 0, P 0.75 atm. What is the pressure (P) after 8 minutes? 40

41 Solution ln ln P P P ln (4.5* P ln 0.75 o t 3 (4.5*0 ln P P atm )(480) 3 )(480) 4

42 Q 0 Effect 4

43 Example If the rate of a chemical reaction doubles for every 0 C rise in temperature, how much faster would the reaction proceed at 55 C than at 5 C? 43

44 Q 0 Effect Q 0 Effect on Reaction Rate 70, 8 Relative Reaction Rate , 64 50, 3 40, 6 30, 8 0, 0, 0, Temperature (Celsius) 44

45 Solution Temperature increased 30 C, therefore, reaction rate increases 8-fold Example: What if the temperature was increased to 05 C from 5 C? Temperature increased 80 C, therefore reaction rate increases 56-fold 45

46 Example How much faster would a reaction go at 00 C than at 5 C? For every 0 C increase in temperature, the reaction rate doubles. The change in temperature is 75 C. This is C increases. Hence times faster 46

47 Example In an experiment, a sample of NaOCl was 85% decomposed in 64 minutes. How long would it have taen if the temperature was 50 C higher? For every 0 C increase, the reaction rate doubles. 50 C increase is 5 0 C increases. Hence: 5 3 times faster So: (64 minutes)/(3 times faster) minutes 47

48 There are 4 types of temperature dependence for reaction rates -- Rate increases with increasing temperature NORMAL 48

49 There are 4 types of temperature dependence for reaction rates -- Rate increases to a point, then reduces with increasing temperature E.g., enzymes being denatured 49

50 There are 4 types of temperature dependence for reaction rates -- 3 Rate decreases with increasing temperature VERY RARE Known only for a few reactions that are multi-step reactions: A B Fast step B C Rate limiting step 50

51 There are 4 types of temperature dependence for reaction rates -- 4 Rate increases with increasing temperature Odd behavior Explosive reaction when temperature shoots up Gradual rise in temperature due to chain reactions 5

52 The Q 0 effect is not entirely perfect There is another way to study temperature dependence: Mathematically with Energy of Activation 5

53 E Ae a / RT rate cons tan t A frequency factor ( total of collisions between reac tan E a Energy of Activation R Gas cons tan t T absolute temperature frequency t molecules) 53

54 Tae Natural Log (ln) of Above Equation ln ln A E a RT 54

55 Previous Equation Can Be Manipulated If you now the rate constants for reactions at different temperatures you can calculate the E a : ln E a R T T 55

56 Example At T of 308 K, 0.36 s - ; at T of 38 K,.5 s -. R 8.34 J/mol-K. Determine E a Ea ln R T T.5 Ea ln Ea.6 (.0* (.6)(8.34) J Ea 0, *0 mol ) 0.6 J mol 56

57 E a -- Transition Top of Energy curve ( hump ) transition state The smaller the E a, the easier it is for the reaction to go 57

58 E a thermic E a is related to ΔG 58

59 Reaction Mechanisms 59

60 Unimolecular Steps A single molecule or atom breas down or rearranges itself into another species. 60

61 Bimolecular Steps Collision of two molecules or atoms; relatively common 6

62 Termolecular Steps 3 molecules or atoms collide simultaneously; relatively rare and SLOW 6

63 Elementary Steps each step in a mechanism; a specific occurrence in the reaction sequence The rate of elementary steps are proportional to the number of collisions per step The number of collisions per step are proportional to every reactant concentration Hence the rate of elementary steps are proportional to the concentration of every reactant 63

64 The exponents of an elementary step rate equation are equal to the coefficients in the step s equation 64

65 65 E.g., Bi Elementary Steps Order st SO Step for equation Rate Order d O SO Step for equation Rate Step O SO SO Step SO O SO Rxn Overall O SO O SO : :

66 66 E.g., Termolecular Elementary Steps Order st Cl H Step for equation Rate Order d H Cl Step for equation Rate Order st Cl Step for equation Rate Step HCl Cl H Step Cl H H Cl Step Cl Cl Rxn Overall HCl H Cl 3 : : :

67 Reaction Mechanisms TWO requirements of postulated mechanisms: ) MUST account for the postulated mechanism ) MUST explain the experimental rate equation ) INHERENT in this is the assumption that ) step is incredibly SLOWER than the others, hence, it determines the speed of the reaction. 3) This step is called the rate limiting step. 4) Therefore, the overall rate equation is determined by rate equations for the rate limiting step. 5) ALSO: when a catalyst is present, there MUST BE at least steps in the mechanism! 67

68 Example The empirically derived rate equation for: NO (g) + Cl (g) NO Cl is NO Cl Write a two-step mechanism with a slow first step 68

69 69 slow step the is Step Hence Equation Rate Empirical Step for Equation Rate BUT Cl NO Cl Equation Rate Step NO Cl Cl NO Cl Cl NO Equation Rate Step Cl NO Cl Cl NO Rxn Overall NO Cl Cl NO,

70 Comments Any steps after Step will NOT effect the rate because step is the rate limiting step Add up the steps (just lie in thermodynamics) and you get the overall reaction 70

71 7 Illustrative Example ), Re ( purposes illustrative for was this STEP is ality In Limiting Rate is Step Equation Step Equals Equation Rate Empirical that Note HI ICl Equation Rate Step HCl I HI ICl H ICl Equation Rate Step HCl HI H ICl Equation Rate Derived Empirically HI ICl Rxn Overall I HCl H ICl

72 Example The empirical rate equation for ClO + F FClO Is: ClO F Write the reaction mechanism (-step) consistent with the rate equation. 7

73 73 Limiting Rate is Step Hence Equation Rate d Dtn Empirically Matches Equation Rate Step F ClO Equation Rate Step FClO F ClO F ClO Equation Rate Step F FClO F ClO Rxn Overall FClO F ClO, '

74 Steady State Approximation The majority of mechanisms are NOT as easy as previous examples. Let s return to our Bimolecular Reaction Step Example: 74

75 Focus SO 5 This is an intermediate a transition An intermediate (or transition) is a compound that is neither reactant nor product, rather in between the two. Intermediates are very difficult to isolate in many cases. They are highly reactive and, more often than not, are used up as quicly as they are formed. It is because of this characteristic that we have the Steady State Approximation 75

76 Steady State Approximation The concentration of intermediate is a constant I.e., the rate of formation of the intermediate is equal to the rate of disappearance of the intermediate: d intermediate d t d intermediate d t 76

77 Example O 3 3O Empirically derived rate equation O 3 /O I prefer O O 3 O + + catalyst O + O + catalyst rate for bacwards reaction ( some sources call it as it re min ds O 3 3 rate for forward reaction O MECHANISM : + O + catalyst Step O 3 + catalyst Step FAST FAST you that it' s for the bacwards reaction clearly Step 3 SLOW ) 77

78 The constant of a reverse reaction is indicated by (-), e.g., - The intermediate in this mechanism is the oxygen ATOM, hence: d O d O d t d t 78

79 Knowing this, we need to show that the empirical rate equation and mechanism are consistent with each other. 79

80 80 ' ' O O O O for Solve O O O catalyst out Cancel catalyst O O catalyst O Two The Equate Now catalyst O O t d O d catalyst O t d O d Rxn Bcw d of Rate Rxn d For of Rate

81 Stop temporarily REMEMBER: the rate limiting step is #3; the rate equation for this step 3 O O 3 So, substitute the solution for O from the first for d and bw d reactions in to the rate equation for Step 3: 8

82 8 equation rate derived empirically to identical Is O O equation substitute and Let O O O : * *

83 This series of inetic equations fulfills the requirements: ) Accounts for the products, and, ) Explains the observed rate law 83

84 Second Example For NO + O NO, the empirical rate equation is: NO O The postulated mechanism follows: NO + O NO 3 Step FAST NO 3 + NO NO Step SLOW 84

85 Show that the empirical rate equation and mechanism are consistent with each other. Use the methodology we used in the previous example. 85

86 NO + NO NO 3 3 O + NO NO NO + O 3 NO and d NO 3 d NO 3 d t d t 86

87 87 temporarily here Stop NO O NO NO for Solve NO O NO two the Equate NO t d NO d O NO t d NO d Equation Rate Bw d Equation Rate d For : : ' '

88 The rate limiting step is step 3. The rate equation for step 3 is: NO 3 NO Substitute the solution for NO 3 into the rate equation for step 3. 88

89 NO O NO NO O 89

90 Apply This to Enzymes Enzymes are, with a couple of exceptions, proteins Enzymes are biological catalysts Enzymes speed up biological reactions incredibly For this discussion: E enzyme, S substrate and P product 90

91 E + S ES FAST ES E + S FAST ES E + P SLOW AND d ES d ES d t d t 9

92 Short Method E S ES Solve for ES : E S ES Stop temporarily 9

93 Rate limiting step is step 3 Rate equation is: ES Substitute as before: E S E S Write the overall reaction : E + S ES E + P This is a Uni Uni Rxn 93

94 Uni-Uni Reaction Cleland Plot 94

95 Bi-molecular Reactions An enzyme catalyzed reaction may utilize substrates. This reaction is always SEQUENTIAL, however, May be ORDERED or RANDOM 95

96 E.g., Ordered Sequential Reaction E + X + Y E + R + S E is still enzyme X and Y are substrates R and S are products 96

97 Putative Mechanism E + X EX EX + Y EXY EXY ERS SLOW STEP ERS ER + S ER E + S And: d ERS d EXY d t d t 97

98 98 temporarily here Stop EXY EX Y X E EXY for Solve EXY Y EX X E Equate EXY t d EXY d Y EX X E d t ERS d 3 3 3

99 Rate Limiting Step is: 3 EXY Substitute: 3 3 E X Y EX E X Y EX E X Y EX 99

100 Kinetic Data Tells us: The following sequence MUST be taing place: E + X + Y EX (FIRST!) EXY (SECOND!) ERS ES + R E + S And is an Ordered Bi Bi Reaction 00

101 Ordered Bi Bi Reaction Cleland Plot 0

102 In the case where separate experiments about the same system give different rate equations, e.g., E X Y EX And E X Y EY 0

103 Mechanism Random Sequential Cleland Plot 03

104 What, though, if an enzyme catalyzed a reaction that bound one substrate, released its product, then binds a SECOND substrate and releases ITS product? Overall Reaction is: E + X + Y E + R + S 04

105 Mechanism E + X EX EX ER st rate limiting step ER E + R E + Y EY EY ES d rate limiting step ES E + S 05

106 Note: while written unidirectionally, in many cases the reactions are reversible With rate limiting steps, this reaction and its inetics get ugly fast. This sort of reaction between substrates and the enzyme act lie a ping pong game. 06

107 Ping Pong Mechanism Cleland Plot 07

108 Function of Kinetics To Determine Reaction Mechanisms 08