1. Introduction to Chemical Kinetics

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1 1. Introduction to Chemical Kinetics objectives of chemical kinetics 1) Determine empirical rate laws H 2 + I 2 2HI How does the concentration of H 2, I 2, and HI change with time? 2) Determine the mechanism of the reaction H 2 + I 2 2HI is a stoichiometric statement reactants intermediates products elementary reaction: the stoichiometric equation reflects what actually happens to the molecules mechanism: series of elementary reactions that constitute the stoichiometric equation Adv PChem 1.1

2 example 2H 2 + O 2 2H 2 O some intermediate steps: H 2 + O 2 HO 2 + H H 2 + HO 2 OH + H 2 O OH + H 2 H 2 O + H O 2 + H OH + O H 2 + O OH + H 3) Empirical study of elementary reactions important intermediates are often unstable, shortlived species: H, OH,... 4) Predict rates of reactions theoretical calculations 5) Chemical dynamics study the dynamics of collisions and reactions of individual molecules; high vacuum Adv PChem 1.2

3 reaction rates 2NO + O 2 2NO 2 time dependence: d[no], d[o 2], d[no 2 ] ambiguity: NO reacts twice as fast as O 2 chemical reaction ν J J = 0 ν J positive for products ν J negative for reactants extent of the reaction ξ dn J ν J dξ Using numbers of moles in the definition of the extent of the reaction, we ensure that it is general and that it applies even to non-constant volume reactions. dn NO = 2dξ dn O2 = dξ dn NO2 = 2dξ Adv PChem 1.3

4 reaction rate: ξ = dξ (rate of conversion) dn J = ν J dξ = dn J = ν J ξ ξ = 1 ν J dn J single rate for reaction can follow the time-dependence of any component If V = const, then the reaction rate can be expressed by the rate of concentration change v 1 V ξ for V = const v = reaction velocity dn J = ν J ξ 1 V Adv PChem 1.4

5 1 V dn J = d(n J/V ) = ν J V v J d[j] = dc J = ν J ξ = ν J v V ξ = V ν J v J ξ if V = const reaction velocity of species J v J = ν J v rate law: v = f (concentrations) If v depends on the concentration of some species that does not appear in the stoichiometric equation, then that species is called a catalyst if v as [catalyst] and an inhibitor if v as [inhibitor] If v depends on one or more products, then the reaction is autocatalytic if v as [product], and the reaction is self-inhibiting if v as [product]. Adv PChem 1.5

6 often v = k[a] x [B] y [C] z k: rate constant (depends on T ), k > 0 x, y, z,...: order of the reaction, typically but not always integers x + y + z + = overall order For elementary reactions, the x, y,... are integers. A D v = k[a] unimolecular A + B D v = k[a][b] bimolecular A + B + C D v = k[a][b][c] termolecular (rare) Reactions that behave kinetically just like a singlestep, elementary reaction are called kinetically simple: order of reactant J i = νi. Such reactions are also said to obey the Law of Mass Action action or to have Adv PChem 1.6

7 mass action kinetics. That means, the stoichiometric equation looks like an elementary reaction, but generally it is not elementary. Example: H 2 + I 2 2HI empirical rate law: v = k[h 2 ][I 2 ], i.e., the equation is kinetically simple and looks like a bimolecular elementary reaction but H 2 + Br 2 2HBr empirical rate law: v = k [H 2][Br 2 ] 3/2 [Br 2 ] + k [HBr] The reaction is first order in H 2 ; no order is definable for Br 2 or HBr, and no overall order is definable. The reaction is self-inhibiting. Adv PChem 1.7

8 Further examples of empirical rate laws: C 2 H 4 O CH 4 + CO v = k[c 2 H 4 O] CH 3 CHO CH 4 + CO v = k[ch 3 CHO] 3/2 2O 3 N 2 O 5 3O 2 v = k[o 3 ] 2/3 [N 2 O 5 ] 2/3 2SO 2 + O 2 2N 2 O Pt 2SO 3 v = k[so 2 ][SO 3 ] 1/2 Pt 741 C 2N 2 + O 2 v = k[n 2O] 1 + k [N 2 O] Experimental determination of rate laws One cannot measure v J directly. measurements: [J](t); make connection between [J](t) and [J](t) two possibilities: differentiate the data or integrate the rate law 1) Differential method A P Adv PChem 1.8

9 [A](t 2) [A](t 1 ) t 2 t 1 v A ([A]) = [A] [A] = 1 2 t { [A](t 1 ) + [A](t 2 ) = [A] t average velocity } If the rate law has a simple order, say v A = k[a] n, then ln v A = lnk + n ln[a]; so to determine n, plot ln v A versus ln[a]. Method has a major inherent difficulty: [A] t if t is small but then [A] is small also: division of two small numbers = need very high precision in the concentration measurement 2) Method of initial velocities (variant of the differential method) example: say v A = k[a] x [B] y Adv PChem 1.9

10 t = 0 : v A,0 = k[a] x 0[B] y 0 ln v A,0 = lnk + x ln[a] 0 + y ln[b] 0 keep [B] 0 fixed and vary [A] 0 : slope of ln v A,0 versus ln[a] 0 is equal to x keep [A] 0 fixed and vary [B] 0 : slope of ln v A,0 versus ln[b] 0 is equal to y problems with this method: H 2 + Br 2 2HBr v 0 = k [H 2] 0 [Br 2 ] 3/2 0 [Br 2 ] 0 + k [HBr] }{{} 0 =0 = k[h 2 ] 0 [Br 2 ] 1/2 0 method does not work for multistep mechanisms method does not work for complex reactions with initial transients and induction periods 3) Method of integrated rate laws d[j] = v J = ν J f (concentrations) Adv PChem 1.10

11 rate law = differential equation solve (integrate) the differential equation (analytically or numerically); then fit data Zero-order reaction A k P v = k = ν A v = v = k = k [A](t) [A] 0 = k t [A](t) [A] 0 = kt [A](t) = [A] 0 kt 0 For a zero-order reaction, a plot of [A](t) versus t yields a straight line. Note: Reactions cannot be of zero order for all times there are no true zero-order reactions because the reactant concentration cannot become less than zero, which would occur at t = [A] 0 /k. A zero-order is Adv PChem 1.11

12 an approximation, which is valid only as long as the reactant is in excess. First-order reaction A k P v = k[a] = ν A v = v = k[a] [A] = k [A](t) t [A] 0 [A] = ( ) [A](t) ln = kt [A] 0 0 k [A](t) = [A] 0 exp( kt) For a first-order reaction, a plot of ln([a](t)/[a] 0 ) versus t results in a straight line with slope k. Second-order reaction A k P v = k[a] 2 Adv PChem 1.12

13 = ν A v = v = k[a] 2 [A] 2 = k [A](t) [A] 0 [A] 2 = t 0 k 1 [A](t) + 1 [A] 0 = kt 1 [A] 0 [A](t) = = kt[a] + kt 0 [A] 0 For a second-order reaction, a plot of 1/[A](t) versus t results in a straight line with slope k. Note: If the reaction is A + A k P v = k[a] 2 then = ν A v = 2v = 2k[A] 2 Adv PChem 1.13

14 and k has to be replaced by 2k in the expressions above. A + B k P v = k[a][b] = ν A v = v = k[a][b] [A](t) = [A] 0 z(t), = dz dz = k([a] 0 z) ([B] 0 z) dz ([A] 0 z) ([B] 0 z) = k z(t) 0 z(t) [B](t) = [B] 0 z(t) dz ([A] 0 z) ([B] 0 z) = kt ( ) [B] 0 [A] 0 [A] 0 z 1 dz = kt [B] 0 z [ ( ) ( )] 1 [A]0 z(t) [B]0 z(t) ln + ln = kt [B] 0 [A] 0 [A] 0 [B] 0 Adv PChem 1.14

15 ( ) 1 [A]0 [B](t) ln = kt [B] 0 [A] 0 [A](t) [B] 0 If [B] 0 = [A] 0, then = k[a] 2 Reactions of other orders A k P v = k[a] x = ν A v = v = k[a] x [A] x = k [A](t) [A] 0 [A](t) [A] x 1 x 1 [A] x = t 0 [A] x = k t 0 k { [A](t) 1 x [A] 1 x } 0 = kt, x 1 { } 1 [A](t) x 1 1 [A] x 1 = kt, x 1 0 Adv PChem 1.15

16 For more complicated rate laws, use the method of isolation or method of flooding. Focus on one species; prepare all other species in large excess. Example: A + B k P v = k[a][b] 3/2 [B] 0 [A] 0 : = k[a][b] 3/2 k[b] 3/2 0 [A] = k eff [A] pseudo-first order [A] 0 [B] 0 : = k[a][b] 3/2 k[a] 0 [B] 3/2 = k eff [B] 3/2 pseudo-one-and-a-half order Adv PChem 1.16

17 Information on rate laws from half-life data half-life t 1/2 : [A](t 1/2 ) = 1 2 [A] 0 First-order reaction, A P: ( 1 2 ln [A] ) 0 = kt 1/2 [A] 0 t 1/2 = ln2 k = 0.693k 1 Second-order reaction, A P: 1 2 [A] 0 = kt 1/2 [A] 0 = 1 t 1/2 = 1 k[a] 0 [A] kt 1/2 [A] 0 Reactions of order x (x 1), A P: t 1/2 = 2 x 1 1 (x 1)k[A] x 1 0 Adv PChem 1.17

18 Temperature dependence of reaction rates k = A exp ( E a RT ) Arrhenius law E a activation energy; A pre-exponential factor In the Arrhenius law, E a and A are considered to be constants, though in applications they are often not strictly constant but vary slowly with temperature. if E a is large, then k is very sensitive to changes in temperature if E a is zero, e.g., radical recombination reactions, then the rate is largely independent of temperature Include backward reactions Reversible first-order reaction A k 1 B k 1 Reversible has a different meaning in kinetics than in thermodynamics! Adv PChem 1.18

19 All chemical reactions are reversible to some extent, but if K > 100, this fact can usually be ignored for purposes of chemical kinetics. ν f A = 1, ν b A = +1, = ν f A v f + ν b A v b v f = k 1 [A] v b = k 1 [B] = k 1 [A] + k 1 [B] d[b] = ν f B v f + ν b B v b d[b] = k 1 [A] k 1 [B] Add the two rate equations: d + d[b] ( [A] + [B] = k 1 [A] + k 1 [B] + k 1 [A] k 1 [B] ) = 0 [A](t) + [B](t) = const = [A](0) + [B](0) = C Adv PChem 1.19

20 [B](t) = C [A](t) = k 1 [A] + k 1 {C [A](t)} = k 1 C (k 1 + k 1 )[A] equilibrium: eq = 0 0 = k 1 [A] eq + k 1 [B] eq [B] eq [A] eq = k 1 k 1 = K c C [A] eq [A] eq = k 1 k 1 = K c [A] eq = k 1 k 1 + k 1 C = k 1 C (k 1 + k 1 )[A] { = (k 1 + k 1 ) [A] k } 1 C k 1 + k 1 Adv PChem 1.20

21 = (k 1 + k 1 ) { } [A] [A] eq [A](t) = (k 1 + k 1 )t [A] 0 [A] [A] eq ( ) [A](t) [A]eq ln = (k 1 + k 1 )t [A] 0 [A] eq [A](t) [A] eq [A] 0 [A] eq = exp { (k 1 + k 1 )t } [A](t) = [A] eq + ( [A] 0 [A] eq ) exp { (k1 + k 1 )t } relaxation methods: relaxation = return of a system to equilibrium Relaxation methods are used to measure rate constants. Apply a sudden perturbation to the equilibrium, e.g., a temperature jump or a pressure jump, and monitor the relaxation. Let x be the deviation from equilibrium. Then for a reversible first-order reaction A B, our results show Adv PChem 1.21

22 that x(t) = [A](t) [A] eq = x(0)exp( t/τ) with τ = 1 k 1 + k 1 i.e., after a temperature jump at t = 0, the system relaxes exponentially with a rate 1/τ = k 1 + k 1. Combine the measurement of the equilibrium constant K c = k 1 /k 1 with a measurement of the relaxation time τ to obtain the forward and backward rate constants k 1 and k 1. Parallel first-order reactions Assume the reactions are irreversible for the moment. A k 1 B A k 2 C = k 1 [A] k 2 [A] Adv PChem 1.22

23 d[b] = k 1 [A] d[c] = k 2 [A] add the three equations: d ( ) [A] + [B] + [C] = 0 [A](t) + [B](t) + [C](t) = const = [A](0) + [B](0) + [C](0) = [A] 0 if initially only the reactant A is present conversion of reactant into products: [A](t) = [A] 0 exp{ (k 1 + k 2 )t} d[b] = k 1 [A] = k 1 [A] 0 exp{ (k 1 + k 2 )t} [B](t) = k 1[A] 0 k 1 + k 2 { 1 exp[ (k1 + k 2 )t] } [C](t) = k 2[A] 0 k 1 + k 2 { 1 exp[ (k1 + k 2 )t] } Adv PChem 1.23

24 yield (t ) Y B irr = [B]( ) = k 1 [A] 0 k 1 + k 2 Y C irr = [C]( ) = k 2 [A] 0 k 1 + k 2 selectivity S irr = [B](t) [C](t) = k 1 k 2 for all times What happens if the reactions are reversible? A k 1 k 1 A k 2 k 2 B C = k 1 [A] + k 1 [B] k 2 [A] + k 2 [C] d[b] = k 1 [A] k 1 [B] Adv PChem 1.24

25 d[c] = k 2 [A] k 2 [C] [A](t) + [B](t) + [C](t) = const = [A] 0 can eliminate on variable, say [C]: [C](t) = [A] 0 [A](t) [B](t) Consider the case t. The reaction reaches equilibrium, i.e., the concentrations no longer change with time: = k 1 [A] + k 1 [B] k 2 [A] + k 2 [C] = 0 d[b] = k 1 [A] k 1 [B] = 0 d[c] = k 2 [A] k 2 [C] = 0 k 1 [A] eq = k 1 [B] eq k 2 [A] eq = k 2 [C] eq [B] eq [A] eq = k 1 k 1 = K c,1 Adv PChem 1.25

26 [C] eq [A] eq = k 2 k 2 = K c,2 [A] eq + [B] eq + [C] eq = [A] 0 [A] eq + K c,1 [A] eq + K c,2 [A] eq = [A] 0 [A] eq = [B] eq = [C] eq = [A] K c,1 + K c,2 K c,1 [A] K c,1 + K c,2 K c,2 [A] K c,1 + K c,2 yield (t ) Y B rev = [B] eq [A] 0 = Y C rev = [C] eq [A] 0 = K c,1 1 + K c,1 + K c,2 K c,2 1 + K c,1 + K c,2 Adv PChem 1.26

27 selectivity S rev = [B] eq [C] eq = K c,1 K c,2 = k 1 k 1 k 2 k 2 = k 1 k 2 k 2 k 1 S rev = S irr k 2 k 1 The selectivity S = [B] [C] = k 1 k 2 = S irr will be observed for short times, as along as the backward reactions are negligible: The reaction is under kinetic control. For long times, the selectivity approaches the equilibrium value S = [B] [C] t K c,1 K c,2 = S rev The reaction is under thermodynamic control. Adv PChem 1.27

28 Example: k 1 = 1 s 1, k 1 = 0.01 s 1, k 2 = 0.1 s 1, k 2 = s 1 Initially B is formed ten times faster than C, however K c,1 = K c,2 = 1 s 1 = s 0.1 s 1 = s and C is the more stable product. conc t Figure 1: Parallel first-order reversible reactions. [A]: solid line, [B]: dashed line, [C]: dot-dash line. Adv PChem 1.28

29 conc t Figure 2: Parallel first-order reversible reactions: Initial stage. [A]: solid line, [B]: dashed line, [C]: dot dash line. Adv PChem 1.29

30 conc t Figure 3: Parallel first-order reversible reactions: concentrations versus log t. [A]: solid line, [B]: dashed line, [C]: dot-dash line. In the above parallel-reaction mechanism C can be converted into B only via A. What happens if there is the additional reaction C k 3 k 3 B (An example are the isomerization reactions between ortho-xylene, meta-xylene, and para-xylene: Adv PChem 1.30

31 o-xylene m-xylene, m-xylene p-xylene, p-xylene o-xylene.) Let us assume that the three steps are the mechanism of the reaction, i.e., all reactions are elementary. Is it then possible to establish equilibrium by the cyclic reaction mechanism below? A B C = k 1 [A] + k 2 [C] d[b] = k 1 [A] k 3 [B] d[c] = k 3 [B] k 2 [C] [A](t) + [B](t) + [C](t) = const = [A] 0 stationary state: k 1 [A] = k 2 [C] = [C] [A] = k 1 k 2 Adv PChem 1.31

32 k 1 [A] = k 3 [B] = [B] [A] = k 1 k 3 k 3 [B] = k 2 [C], which follows from the previous two equations [B] = k 1 k 3 [A] [C] = k 1 k 2 [A] [A] + k 1 k 3 [A] + k 1 k 2 [A] = [A] 0 [A] = [A] k 1 k 3 + k 1 k 2 We have determined the stationary concentrations [A], [B], and [C] in terms of [A] 0 and the three rate constants. This stationary state is not an equilibrium state!! It violates the principle of detailed balance. The principle of microscopic reversibility: In a system at equilibrium, any molecular pro- Adv PChem 1.32

33 cess and the reverse of that process occur, on the average, at the same rate. implies the principle of detailed balance: In a system at equilibrium, each collision has its exact counterpart in the reverse direction, so that the rate of every chemical process is exactly balanced by that of the reverse process. = In a system at equilibrium, every elementary forward reaction is balanced by its own backward reaction. In other words, in a system at equilibrium each elementary reaction must be individually at equilibrium. Principle of detailed balance implies that all elementary reactions must be reversible. Application to the scheme A k 1 k 1 B Adv PChem 1.33

34 A k 2 k 2 C k 3 k 3 C B equilibrium: k 1 [A] eq = k 1 [B] eq = K c,1 = [B] eq [A] eq = k 1 k 1 k 2 [A] eq = k 2 [C] eq = K c,2 = [C] eq [A] eq = k 2 k 2 k 3 [C] eq = k 3 [B] eq = K c,3 = [B] eq [C] eq = k 3 k 3 The product of the second and third equilibrium constants is K c,2 K c,3 = [C] eq [A] eq [B] eq [C] eq = [B] eq [A] eq = K c,1 = Adv PChem 1.34

35 k 1 k 1 = k 2 k 2 k 3 k 3 or k 1 k 2 k 3 = k 1 k 2 k 3 This condition on the six rate constants has been derived by considering chemical equilibrium, but rate constants are concentration independent. Therefore, this relation is general for this mechanism; it also holds for all nonequilibrium conditions. We see that the rate constants can never be chosen independently, and the cyclic mechanism is impossible, since it requires k 1 = k 2 = k 3 = 0, which is forbidden by the relation. Consecutive first-order reactions (series reactions) A k 1 B k 2 C Assume the reactions are first order and irreversible. = k 1 [A] Adv PChem 1.35

36 d[b] = k 1 [A] k 2 [B] d[c] = k 2 [B] Assume again [A](0) = [A] 0, [B](0) = 0, [C](0) = 0, and add the three equations: [A](t) + [B](t) + [C](t) = const = [A] 0 [A](t) = [A] 0 exp( k 1 t) d[b] = k 1 [A] 0 exp( k 1 t) k 2 [B] [B](t) = k 1[A] 0 k 2 k 1 {exp( k 1 t) exp( k 2 t)} [C] = [A] 0 [A](t) [B](t) { = [A] } [k 1 exp( k 2 t) k 2 exp( k 1 t)] k 2 k 1 Consider the case: k 2 k 1. Then exp( k 2 t) exp( k 1 t) Adv PChem 1.36

37 and k 2 k 1 k 2 [C] [A] 0 {1 + 1 } ( k 2 )exp( k 1 t) k 2 = [A] 0 {1 exp( k 1 t)} The formation of C depends only on the rate constant of the first step A B. A B is the rate-determining step. Motivation for the Steady-State Approximation: After an initial induction period, the concentration and the rate of change of all reaction intermediates are negligibly small. Illustrate for the consecutive scheme: [B](t) [A](t) = k 1 [A] 0 k 2 k 1 {exp( k 1 t) exp( k 2 t)} [A] 0 exp( k 1 t) = k 1 k 2 k 1 {1 exp[ (k 2 k 1 )t]} Adv PChem 1.37

38 For k 2 k 1 : [B](t) [A](t) k 1 {1 exp[ k 2 t] } = k 2 }{{} 0 for t 1/k 2 k 1 k 2 [B](t) = k 1 k 2 [A](t) [A](t) d[b] = k 1 k 2 If we apply directly the steady-state approximation (SSA), also known as the quasisteady-state approximation (QSSA), to the intermediate B, we obtain 0 d[b] = = k 1 [A] k 2 [B] [B] = k 1 k 2 [A] d[c] = k 2 [B] = k 2 k 1 k 2 [A] = k 1 [A] Adv PChem 1.38

39 = k 1 [A] 0 exp( k 1 t) [C](t) = [A] 0 {1 exp( k 1 t)} (same as above) Application: decomposition of ozone 2O 3 3O 2 empirical rate law (M is an inert gas): v = 1 2 d[o 3 ] = 1 3 d[o 2 ] = k[o 3 ] 2 [M] k [O 2 ][M] + [O 3 ] proposed mechanism: stoichiometric number O 3 + M k 1 O 2 + O + M 2 O 2 + O + M k 1 O 3 + M 1 O + O 3 k 2 2O 2 1 Adv PChem 1.39

40 2O 3 + 2M + O 2 + O + M + O + O 3 2O 3 3O 2 2O 2 + 2O + 2M + O 3 + M + 2O 2 intermediate O: steady-state approximation d[o] = 0 Intermediates do not appear in the rate law!! d[o] [O] = d[o 3 ] = k 1 [O 3 ][M] k 1 [O 2 ][O][M] k 2 [O][O 3 ] = 0 k 1 [O 3 ][M] k 1 [O 2 ][M] + k 2 [O 3 ] = k 1 [O 3 ][M] + k 1 [O 2 ][O][M] k 2 [O][O 3 ] = k 1 [O 3 ][M] + { } k 1 [O 2 ][M] k 2 [O 3 ] [O] Adv PChem 1.40

41 d[o 3 ] = k 1 [O 3 ][M]+ { } k 1 [O 3 ][M] + k 1 [O 2 ][M] k 2 [O 3 ] k 1 [O 2 ][M] + k 2 [O 3 ] = 1 k 1 [O 2 ][M] + k 2 [O 3 ] { k 1 [O 3 ][M]k 1 [O 2 ][M] k 1 [O 3 ][M]k 2 [O 3 ]+ } + k 1 [O 2 ][M]k 1 [O 3 ][M] k 2 [O 3 ]k 1 [O 3 ][M] = 2k 1k 2 [O 3 ] 2 [M] k 1 [O 2 ][M] + k 2 [O 3 ] = 2 k 1 [O 3 ] 2 [M] k 1 k 2 [O 2 ][M] + [O 3 ] = 2v empirical rate law:k = k 1, k = k 1 /k 2 If initially we have pure ozone, then k 1 [O 2 ][M] k 2 [O 3 ] Adv PChem 1.41

42 and d[o 3 ] = 2k 1 [O 3 ][M] = the first step is rate-determining The reaction is self-inhibiting in O 2, v as [O 2 ] Another approximation method Pre-equilibrium or fast-equilibrium approximation A + B k 1 X k 2 C k 1 k 1 k 2, k 1 k 2 A, B, and X are in equilibrium: k 1 [A][B] = k 1 [X] [X] = k 1 k 1 [A][B] = K [A][B] (Note: K = K c!!) Adv PChem 1.42

43 d[c] = k 2 [X] = k 2 K [A][B] = k[a][b] k = k 2 K = k 2k 1 k 1 enzyme reactions Michaelis-Menten mechanism E + S k 1 (ES) k 2 P + E k 1 d[p] = v = k 2 [(ES)] Apply the steady-state approximation: d[(es)] = k 1 [E][S] k 1 [(ES)] k 2 [(ES)] = 0 [(ES)] = k 1[E][S] k 1 + k 2 = [E][S] K M K M = k 1 + k 2 k 1 Michaelis constant Adv PChem 1.43

44 d[p] = k 2 K M [E][S] conservation of total enzyme concentration: [E] 0 = [E] + [(ES)] = [E] + [E][S] = [E] [E] = [E] [S] K M d[p] K M ( 1 + [S] ) K M = k 2 K M [E][S] d[p] = k 2 [E] 0 [S] K M 1 + [S] K M = v = k 2[E] 0 [S] K M + [S] Michaelis-Menten rate law Adv PChem 1.44

45 At high substrate concentrations, [S] K M, d[p] k 2[E] 0 [S] [S] = k 2 [E] 0, the reaction is zero order, and the rate reaches its maximum: v max = k 2 [E] 0 [Figure: rate versus substrate concentration; Atkins 9th ed., Fig. 23.3] Adv PChem 1.45

46 The Michaelis-Menten rate law can be written in terms of v max : d[p] = v = v max 1 + K M /[S] We rearrange this expression into a form that is amenable to data analysis by linear regression: 1 v = 1 ( ) KM 1 + v max v max [S] A Lineweaver-Burk plot is a plot of the reciprocal of the velocity v versus the reciprocal of the substrate concentration [S] at fixed enzyme concentration. If the enzyme obeys Michaelis-Menten kinetics, then the plot should yield a straight line, with slope K M /v max, that intercepts the ordinate axis at 1/v max and the abscissa at 1/K M. Adv PChem 1.46

47 [Figure: Lineweaver-Burk plot; Atkins 9th ed., Fig. 23.4] At low substrate concentrations, [S] K M, d[p] k 2[E] 0 [S] K M = k [S], the reaction is first order. The turnover number or catalytic constant of an enzyme, k cat, is the number of catalytic cycles (turnovers) performed by the active site in a given Adv PChem 1.47

48 time interval divided by the duration of that interval. It has units of a first-order rate constant, and for the Michaelis-Menten mechanism is given by k 2, the rate constant for release of product from the enzymesubstrate complex. k cat = k 2 = v max [E] 0 The catalytic efficiency, η, of an enzyme is given by η = k cat K M The higher the value of η, the more efficient is the enzyme. For the Michaelis-Menten mechanism η = k 2 = k 1 +k 2 k 1 k 1 k 2 k 1 + k 2 η η max = k 1 if k 2 k 1 = rate of formation of the enzyme-substrate complex; diffusion-limited: < M 1 s 1 for enzyme-sized molecules at room temperature Adv PChem 1.48

49 catalase (decomposition of hydrogen peroxide) η = M 1 s 1 : catalytic perfection Adv PChem 1.49

Advanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS

Experimental Kinetics and Gas Phase Reactions Advanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS Professor Angelo R. Rossi http://homepages.uconn.edu/rossi Department of Chemistry, Room