Determining the Components of the Rate Equation aa + bb yy + zz

Transcription

1 Determining the Components of the Rate Equation aa + bb yy + zz Rate k[a] [B] The coefficients and components of the rate equation Must be found by experiment Cannot be deduced from stoichiometry Do not necessarily tell us the physical mechanism Also, the reaction orders can be fractional, negative, zero, or even not-definable

2 Rate Laws We have seen how to obtain the differential form of rate laws based upon experimental observation. As they involve derivatives, we must integrate the rate equations to obtain the time dependence of concentrations. We will do this for a few cases, all involving these empirical rate laws. Here the rate law need not bear any relationship to the stoichiometry of the reaction. Our next step will be to understand the origin of the empirical laws. This leads to the concept of elementary reactions where the reaction is direct and occurs (or nor) in a single encounter. We will see how the more complex rate laws arise from multiple elementary processes. Understanding the nature of the elementary reactions and the role of elementary reactions in complex processes will occupy the remainder of the semester.

3 First Order Reactions, n= Differential form: Integrated form: A X d[a] dt d [A] dt k[a] k[a] or = k[a] d[a] [A] kdt Integrating, [ A] t d [A] [A] kdt or ln [A] [A] 0 which says that A A e [ A] kt kt

4 [A] ConcepTest What is X the X order X X X of X this X reaction? X X X X X X A. XZero order X X X X X X B. XFirst order C. Second order X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X

5 Different Types of Composite Reactions Simultaneous: A Y and A Z Consecutive: A X Y Opposing: A B Z Consecutive with opposition: A X Y In order for the system to be at complete equilibrium, the time derivative of all species concentrations must be zero!

6 Kinetics for a Simple Consecutive Reaction k 2 A X and X Z k with A A e kt 0 d X k X k A (2) 2 dt becomes d X kt k X k A e 2 dt 0 d X kt k X k A e a "standard form" o.d.e. with solution 2 dt 0 k 2 kt k2t X e e A and k k 0 2 Z A A X A ke k 2 kt ke k kt

7 Quasi Steady State Approximation Consider the differential rate equations for [A], [X] and [Z]. d A k A dt d X k X k A 0 2 dt k k X A A e k k 0 or ( d Z dt 2 A X and X Z 2 k X 2 k 2 kt () (2) (3) 2.5) k Let [A] t=0 = [A] 0 and [X] t=0 = [Z] t=0 = 0 For (2) and (2.5) to be compatible, we must have (k /k 2 ) << A A e kt 0 Make QSS approx Z k A e dt t 0 0 kt kt e A 0

8 Compare Exact and Steady State Solutions for k 2 = 20 k k t

9 Sample Problem Given reaction with stoichiometry: A + B Y + Z The reaction follows the mechanism: k A X k 2 k X + B Y + Z Obtain an expression for the rate using the steady state treatment a. What is the rate determining step if the second reaction is slow relative to the initial equilibrium? b. What is the rate determining step if the second reaction is rapid relative to the initial equilibrium?

10 ConcepTest Reaction: 2 O 3 3 O 2 k Mechanism: O 3 O + O 2 k k2 (fast) O + O 3 2 O 2 (slow) What is the rate equation for O 3 loss? A. Rate = k [O 3 ] 2 B. Rate = k [O 3 ] [O 2 ] C. Rate = k [O 3 ] 2 [O 2 ] B. Rate = k [O 3 ] [O 2 ]

11 Solution Reaction: 2 O 3 3 O 2 k Mechanism: O 3 O + O 2 k k2 (fast) O + O 3 2 O 2 (slow) What is the rate equation for O 3 loss? Rxn equilibrium : k O 3 k O O 2 k O 3 O k O 2 with a slow second step, the equilibrium is maintained d O Rate dt 2 3 k O O 3 3 k O O k O k k O O 2 2

12 Lindemann Hinshelwood Mechanism A + B AB* Complex with excess energy Given enough time, it will fall apart AB* A + B Unless a collision removes the excess energy AB*+ M AB + M*

13 Lindemann Hinshelwood Mechanism Consider only one species A. This is easy to generalize. d[a*] 2 A + A A* + A ka[a] dt d[a*] dt d[a*] A* P kb[a*] dt ' A* + A A + A ka[a*][a] The rate law for formation of A* is (with s.s. approximation) d[a*] dt 2 ' a a b k A k A A* k A* 0

14 Lindemann Hinshelwood Mechanism With the result Given A* k b ka A k ' a 2 A 2 d[p] kk[ ] A* a b A kb dt k k A d [P] kk[ A] a b k A where k eff eff ' dt k k A b a ' ka k k k eff a b k A a This is not a first order reaction. However, if the rate for stabilization of A* is much greater than the unimolecular decay, A*A, k a [A*][A] >> k b [A*] or just k a [A] >> k b, then Such a reaction goes from 2 nd order to st order as the pressure increases. This is a very common gas phase mechanism, as collisions supply (take away) energy. b ' a

15 Lindemann Hinshelwood Mechanism O + NO NO 2 * NO 2 * + M NO 2 + M* Normally, bath gas dominates

16 Example 3. The effective rate constant for a gaseous reaction which has a Lindemann Hinshelwood mechanism is s at.30 kpa and s at 2 Pa. Calculate the rate constant for the activation step in the mechanism. First, we check to see if the reaction is in the Lindemann mechanism high pressure limit. In that limit, the rate constant would be simply k B and would be pressure independent. Is it?it is not, so we use the full Lindemann expression: Now

17 Exam 3 Subject matter: Kinetics Rates, rate laws, mechanisms, method of initial rates Arrhenius rate law physical interpretation Reaction paths, potential energy surfaces Molecular basis of reactions Elementary reactions, composite reactions Fast and slow rate approximations; steady state rate determining steps; catalyzed reactions

18 Exam 3 ~ 5 problems (weights given) budget your time Closed book Don t memorize formulas/constants You will be given things you need Specifically, you will be given the integrated rate laws. However, you may well be asked to conjure a mechanism that is with a given set of observations. You might well be asked to use the steady state approximation, and to explain why/when it is useful. This exam will have some numeric problems, similar to the homework in complexity. If a problem seems lengthy, do another problem & come back later Understanding homework will be quite useful EXAM held in JILA Auditorium, Dec, 0 AM