1. Kinetics Notes part 3 Reaction Mechanisms

2. REACTION MECHANISMS Chemical reactions involve a sequence of individual bond-making and bond-breaking steps that occurs during the conversion of reactants to products. – This series of steps is called the reaction mechanism. Chemical reactions are more complicated than it appears from the balanced equation. – The balanced equation tells us the reactants, the products, and the stoichiometry, but it doesn’t give us information about the reaction mechanism.

3. Consider the reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The rate law is Rate=k[NO 2 ] 2 The mechanism is thought to involve the steps: step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g)

4. step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) In this mechanism, NO 3 (g) is an intermediate – A species that is neither a reactant or product – It is produced and then consumed during the reaction sequence Each of these two reactions in the mechanism is called an elementary step, – A reaction whose rate law can be written from its molecularity

5. Molecularity is defined as the number of species that must collide to produce the reaction indicated by that step. – unimolecular: involves one reactant molecule – bimolecular: involves a collision between two reactant molecules – termolecular: simultaneous collision between three reactant molecules [very rare!]*

6. Remember the rate law for an elementary step can be written from its molecularity – A unimolecular step will be first order Rate = k[A] – A bimolecular step will be second order Rate=k[A] 2 or Rate=k[A][B] – Etc…

8. Better definition of reaction mechanism… A series of elementary steps that must – Add to give the overall balanced equation for the reaction – Must agree with the experimentally determined rate law

9. Let’s look at our example… step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g) +NO 3 (g) + CO(g)  NO 3 (g) + NO(g) + NO 2 (g) + CO 2 (g) Overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) so the first requirement is met (the elementary steps in the mechanism must add to give the overall balanced equation…and they do!)

10. To see whether our mechanism meets the second requirement, we must see if it agrees with the experimentally determined rate law To do so, we must determine the rate- determining step – Each individual step in the mechanism for a reaction can be fast or slow, but the overall reaction rate is dependant on the slowest step – Known as the rate determining step

11. A reaction can be no faster than its slowest, or rate determining step. – An analogy is the pouring of water rapidly into a container through a funnel. The water collects in the container at a rate that is essentially determined by the size of the funnel opening and not by the rate of pouring.

12. In our reaction between nitrogen dioxide and carbon monoxide, what is the rate determining step? – Let’s assume that it is the first step – step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) SLOW step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) FAST – Since this is an elementary step, we can write the rate law from the molecularity – This elementary step is a bimolecular step so Rate=k[NO 2 ][NO 2 ] or Rate=k[NO 2 ] 2 which agrees with the experimentally determined rate law given earlier, so this is a possible mechanism for this reaction.

13. Mechanisms with an initial slow step As in the previous example, mechanisms with an initial slow step depend on the molecularity of that first slow step, so it is pretty easy to determine the rate law from the mechanism.

14. Mechanisms with an initial fast step It is less straightforward to derive the rate law for a mechanism in which an intermediate is a reactant in the rate determining step. For example, lets consider the reaction between nitric oxide (NO) and bromine (Br 2 ) 2NO(g) + Br 2 (g)  2NOBr(g) The experimentally determined rate law is Rate=k[NO] 2 [Br 2 ]

15. One possible mechanism for this reaction would be one termolecular step where 2 molecules of NO collide with a molecule of Br 2. This is not likely because termolecular processes are very rare. Let’s look at another possible mechanism

16. This is a 2-step mechanism that does not involve a termolecular step Step 1: NO(g) + Br 2 (g)  NOBr 2 (g) FAST Step 2: NOBr 2 (g) + NO(g)  2NOBr(g) SLOW Because step 2 is the slow step, it is the rate determining step. If we write the rate law from the molecularity of the rate determining step, we get Rate=k[NOBr 2 ][NO]

17. What is wrong with this rate law? Rate=k[NOBr 2 ][NO] It contains an intermediate! Rate laws generally shouldn’t contain intermediates: they should be expressed in terms of reactants. It is easy to fix this- all we have to do is replace the intermediate with the reactants that created it.

18. Look at the proposed mechanism again: Step 1: NO(g) + Br 2 (g)  NOBr 2 (g) FAST Step 2: NOBr 2 (g) + NO(g)  2NOBr(g) SLOW Just replace the intermediate with the reactants that produced it: Rate=k[NOBr 2 ][NO] becomes Rate=k[NO][Br 2 ][NO] or Rate=k[NO] 2 [Br 2 ] which agrees with the experimentally determined rate law for this reaction

19. Energy profiles (aka reaction progress curves) and elementary steps The energy profiles we have looked at so far have just been for a single, elementary step. If a reaction occurs in a single elementary step, there will be only one “activation energy hump” on the graph.

20. For reactions that have multistep mechanisms, a reaction progress curve shows an activation- energy peak for each elementary reaction. Valleys indicate the formation of intermediates. The peaks correspond to the energies of the activated complexes. Each valley corresponds to the energy of an intermediate. How many elementary reactions are part of this reaction? How many intermediates are formed? How would a catalyst affect the energy of the reactants and the energy of the products of this reaction?

21. Exercise Nitrogen oxide is reduced by hydrogen to give water and nitrogen, 2 H 2(g) + 2 NO (g)  N 2(g) + 2 H 2 O (g) and one possible mechanism to account for this reaction is 2 NO (g)  N 2 O 2(g) N 2 O 2(g) + H 2(g)  N 2 O (g) + H 2 O (g) N 2 O (g) + H 2(g)  N 2(g) + H 2 O (g) What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net reaction.

22. REACTION MECHANISMS AND RATE EXPRESSIONS determined by experiment the rate of the overall reaction is limited by, and is exactly equal to, the combined rates of all elementary steps up to and including the slowest step in the mechanism the slowest step is the rate determining step reaction intermediate--produced in one step but consumed in another. catalyst--goes in, comes out unharmed and DOES NOT show up in the final rxn.

23. Exercise The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) The experimentally determined rate law is Rate = k [NO 2 ][F 2 ] A suggested mechanism for the reaction is NO 2 + F 2  NO 2 F + F Slow F + NO 2  NO 2 F Fast Is this an acceptable mechanism? That is, does it satisfy the two requirements? Justify.