Kinetics Part V: Reaction Mechanisms

1. Kinetics Part V: Reaction Mechanisms Jespersen Chap. 14 Sec 7 Dr. C. Yau Spring 2013

2. Mechanisms Mechanism refers to the series of individual steps that add up to the overall observed reaction. It tells us on the molecular level what is happening, such as what collides with what, and which the slowest step is. Elementary reaction refers to each of the individual steps in the mechanism. The sum of the elementary reactions must give the overall reaction.

3. Reaction Mechanisms tell what happens on the molecular level, and in what order tell us which steps in a reaction are fast and slow The rate determining step is the slowest step of the reaction that accounts for most of the reaction time.

4. Example of a reaction mechanism For the reaction of NO with H2 2NO + 2H2  N2 + 2H2O The proposed mechanism is... 1. 2NO N2O2 (fast) 2. N2O2 + H2 N2O + H2O (slow) 3. N2O + H2 N2 + H2O (fast) These are the elementary rxns. Verify they add to give the overall equation. Which is the rate determining step (rds)?

5. Elementary Steps: Molecularity vs. Rate Law Molecularity refers to how many species are involved in the step. It is reflected in the overall order of reaction in the rate law. Unlike the overall rate law, rate law of the elementary rxn uses the coefficients as the order of reaction.

6. Rate Laws And Mechanisms The majority of the reaction time is taken by the rate determining step. Substances that react in this step have the greatest effect on the reaction rate. The observed rate law usually matches the rate law based on the rate determining step where the order of each reactant is its stoichiometric coefficient. Note: Observed rate law is NOT based on the stoichiometric coefficients of the overall rxn. Note: "Order" does not refer to which comes first, but to the molecularity: unimolecular? bimolecular? termolecular?

7. IntermediatesversusCatalysts Because catalystsinteract with the reactant, they will appear in the mechanism Catalysts are consumed in an early step and are regenerated in a subsequent step Intermediatesare temporary products. Intermediates are formed in an early step and consumed in a later step Catalysts are there at beginning & at the end of the rxn. Intermediates are not there at the beginning, nor at the end of the rxn. Neither will show up in the overall eqn.

8. Intermediates in a mechanism 2NO N2O2 (fast) N2O2 + H2 N2O + H2O (slow) N2O + H2 N2 + H2O (fast) Are there any catalysts or intermediates in this mechanism? Catalysts are ADDED, and they reappear at the end. Intermediates are not reactants. They appear somewhere in the elementary steps, but are consumed before the end. (No catalyst; N2O2 and N2O are intermediates.)

9. Example 1: 2H2O2 2H2O + O2Observed rate law: Rate = k[H2O2][I-] The reaction mechanism that has been proposed for the decomposition of H2O2 is H2O2 + I- H2O + IO- (slow) H2O2 + IO- H2O + O2 + I- (fast) The 1st step is slowest and therefore the rds. The rate law of the rds is Rate = k[H2O2][I-] This matches the observed rate law. Do the elementary steps shown add up to the overall reaction? The rds is said to be "bimolecular." Why is I- not in the chemical equation?

10. Example 2 The reaction: A + 3 B → D + F was studied and the following mechanism was finally determined A + B → D (fast) B + B → E (slow) E → F (very fast) What is the expected rate law of the overall reaction? rate=k[B]2Check to see that elementary steps add up to give the overall reaction.

11. Example 3 For the reaction 2NO2Cl (g) 2 NO2 (g) + Cl2 (g) the elementary steps have been proposed as… (1) NO2Cl (g) NO2 (g) + Cl (g) (2) NO2Cl (g) + Cl (g) NO2 (g) + Cl2 (g) Step 1 is "unimolecular." Step 2 is "bimolecular." Do they add up to give the overall reaction? Write the rate law for each of the elementary steps. Step 1 must be rds. Expt Rate Law: Rate = k[NO2Cl]

12. The slowest step is the rate determining step (rds). The rds limits the rate of the overall reaction so its rate law represents the rate law for the overall reaction. Example 4: NO2 (g) + CO (g) NO (g) + CO2 (g) The experimental rate law is Rate = k[NO2]2 From this we know immediately the reaction shown cannot be elementary. Can you see why not?

13. Example 4 (cont'd.)NO2 (g) + CO (g) NO (g) + CO2 (g) Proposed 2-step mechanism: (1) NO2 + NO2 NO3 + NO (2) NO3 + CO NO2 + CO2 Write the rate laws for these 2 elementary steps. Given: experimental rate law: Rate = k[NO2]2 Which is the rate determining step? How do you tell?

14. Example 52NO + 2H2 N2 + 2H2O Expt rate law: Rate = k[NO]2[H2] which suggests 2NO + H2 N2O + H2O and then N2O + H2 N2 + H2O The 1st step proposed is termolecular, which is highly unlikely. Why? It would require 3 molecules to all collide at the same time.

15. Example 5: 2NO + 2H2 N2 + 2H2O cont'd Instead: 2NO N2O2 fast N2O2 + H2 N2O + H2O slow N2O + H2 N2 + H2O fast If we look only at the rds (Step 2) Rate = k[N2O2][H2] but N2O2 is not in the overall reaction, and not in the observed rate law. This is because N2O2 is an intermediate. When this happens, we have to find an expression to replace [N2O2] in the rate law.

16. 2NO N2O2 fast N2O2 + H2 N2O + H2O slow N2O + H2 N2 + H2O fast For the equilibrium, forward reaction has… Rate = kf [NO]2 and the reverse reaction has… Rate = kr [N2O2] At equilibrium, the two rates are equal, so kf [NO]2 =kr [N2O2] which rearranges to [N2O2] =? Rate law for Step 2 is Rate = k[N2O2][H2] Substitute [N2O2] from the equilibrium step and we have a rate law that matches the observed.

17. 2NO + 2H2 N2 + 2H2O Substitute [N2O2] = kf/kr [NO]2 into Rate = k [N2O2] [H2] to give Rate = k kf/kr [NO]2[H2] Since k, kf, kr are all constants, we can combine them into a new constant, k' to give Rate = k'[NO]2[H2] and this matches the observed Rate = k[NO]2[H2] Do Practice Exercises 29, 30, 31 p.680

18. Potential Energy Diagram One-step mechanism: Potential E 1) Where is Ea? 2) Where is ΔH? 3) Is this exothermic or endothermic? 4) Where is the transition state? 5) What is the activated complex? Reaction Coordinate

19. Potential Energy Diagram Two-step mechanism: 8) Is this exo- or endothermic? 1) How do you know it has 2-steps? 2) What is at the "valley"? 3) Which step is rds? 4) Where is the ΔH of the rds? 5) How many transition states are there? 6) Which is the rds of the reverse reaction? 7) Where is the Ea of the 2nd step? Potential E Reaction Coordinate

20. Potential Energy Diagram Two-step mechanism: Potential E 1) In the reverse rxn, which is the rate determining step? 2) Where is the ΔH of the reverse rxn? 3) Is the reverse rxn exo- or endothermic? Reaction Coordinate

21. Potential Energy Diagram What can you say about this mechanism? Potential E How many steps are in the mechanism? Is it exothermic or endothermic? How many transition states are there? What is at each "valley"? Which step is rds? Reaction Coordinate

22. PE Diagram ofReversible vs. Irreversible Rxns Diagram A Diagram B Which PE diagram more closely corresponds to a reversible reaction? Why?

23. Potential Energy Diagram What would the PE graph look like if a catalyst is added? Graph for uncatalyzed rxn Graph for catalyzed rxn Potential E Note that the Ea is lowered. How does that affect the Kinetic E Diagram? Reaction Coordinate

24. Kinetic Energy Diagram: Effect of Catalyst Be able to explain why lowering the Ea in the KE diagram would affect the rate of reaction.

25. Applications of Catalytic Rxns • Oil industries: Catalysts help "crack" larger hydrocarbons into smaller pieces and reform them to more useful molecules. • Catalytic converter in your car: helps remove CO (poisonous), unburned volatile hydrocarbons and nitrogen oxides (contribute to acid rain) • Enzymes are biological catalysts: Lactase catalyzes the breakdown of lactose (Those who are lactose-intolerant lack lactase.)