1. Unit 6: Kinetics IB Topics 6 & 16 Part 3: Reaction Mechanisms & Activation Energy

2. Reaction Mechanisms Reaction Mechanisms: most reactions that occur at a measurable rate occur as a series of simple steps, each involving a small number of particles. This sequence of steps is known as the reaction mechanism.

3. Reaction Mechanisms Individual steps are called elementary steps  usually cannot be observed directly  mechanism is in effect a theory about the sequence of events as the overall reaction proceeds from reactants to products.

4. Catalyst speeds up a reaction without being used up (should not appear in overall reaction because it was present initially and then reforms as a product somewhere in the process. Example: which one is the catalyst? step 1: H 2 O 2 + I -  H 2 O + IO - step 2: H 2 O 2 + IO-  H 2 O + O 2 + I -

5. Intermediate / Activated complex product of one elementary step that is used up in a subsequent step. Example: which one is the intermediate? step 1: H 2 O 2 + I -  H 2 O + IO - step 2: H 2 O 2 + IO-  H 2 O + O 2 + I -

6. Chemical reactions take place as a result of collisions between molecules. Each type of collision in a given reaction is called a single event or an elementary step.

7. Molecularity: # of species participating as reactants in a given step of a reaction. Unimolecular: A single reactant particle is involved in the elementary step Bimolecular: Two particles are involved as reactants Termolecular: Three particles are involved.  Termolecular steps are far less probable than unimolecular or bimolecular processes and are rarely encountered. The chances that four or more molecules will collide simultaneously with any regularity is extremely remote, consequently, such collisions are never proposed as a part of a reaction mechanism.

8. Rate Laws of Elementary Steps A  ProductsRate = k[A] A+A  ProductsRate = k[A] 2 A+B  ProductsRate = k[A][B] A+A+A  ProductsRate = k[A] 3 A+A+B  ProductsRate = k[A] 2 [B] A+B+C  ProductsRate = k[A][B][C] unimolecular bimolecular termolecular

9. The reaction between NO 2 and CO has the overall reaction: NO 2 + CO  NO + CO 2 A study of the kinetics of this reaction revealed the rate law for the reaction is. Rate = k[NO 2 ] 2 This Rate Law requires that the slow step of the reaction involves a collision between two NO 2 molecules. How can this be a step in the seemingly simple reaction above? Example:

10. Example: NO 2 + CO  NO + CO 2 NO 2 + NO 2  NO 3 + NO NO 3 + CO  NO 2 + CO 2 2NO 2 + NO 3 + CO  NO 3 + NO + NO 2 + CO 2 NO 2 + CO  NO + CO 2 Further study of this reaction established that two NO 2 molecules can react as follows: NO 3 is a highly reactive material which is capable of transferring an oxygen atom.

11. NO 2 + NO 2  NO 3 + NO NO 3 + CO  NO 2 + CO 2 NO 2 + CO  NO + CO 2 slow fast The rate law only provides information about the slowest reaction in the mechanism. The rate law is based off of the first elementary step, so it must be the slow/rate-determining one. Thus, the reaction mechanism can be expressed as follows: Which molecule is an intermediate in the reaction mechanism above? Example: NO 2 + CO  NO + CO 2 NO 3 Rate = k[NO 2 ] 2 Step 1: Step 2: Overall: So if you were given this mechanism, you should be able to produce the rate law ( rate = k[NO 2 ] 2 )

12. Determining the rate expression for an overall reaction from the reaction mechanism The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step, which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law & order as is determined experimentally. For elementary steps (since they are true single- step reactions), the exponents in the rate laws are the same as the coefficients of the reactants in the chemical equation.

13. Determining the rate expression for an overall reaction from the reaction mechanism Equation for rate- determining step MolecularityRate law A → productsunimolecular rate=k[A] 2A → productsbimolecular rate=k[A] 2 A + B→ productsbimolecular rate=k[A][B] The rate law for the rate-determining step, predictable from its equation, leads to the rate expression for the overall reaction. If the rate-determining step is the first step (or only step) in the mechanism, then its rate law is the rate expression for the overall reaction.

14. Example: The overall reaction 2NO 2 Cl(g)  2NO 2 (g) + Cl 2 (g) is believed to have the following mechanism NO 2 Cl(g)  NO 2 (g) + Cl(g) NO 2 Cl(g) + Cl(g)  NO 2 (g) + Cl 2 (g) 2NO 2 Cl(g)  2NO 2 (g) + Cl 2 (g) slow: rate-determining stepStep 1: Step 2: Overall: fast The rate expression for the overall reaction is that of the rate-determining step: rate = k[NO 2 Cl] This is a first-order reaction.

15. What if the rate- determining step is not the first step in the mechanism?

16. It’s a little more complicated because the reactant concentrations depend on an earlier step, so this must be taken into account.

17. Example: For the reaction NO(g) + O 2 (g)  2NO 2 (g), the following reaction mechanism has been proposed NO(g) + NO(g)  N 2 O 2 (g) N 2 O 2 (g) + O 2 (g)  2NO 2 (g) 2NO(g) + O 2 (g)  2NO 2 (g) slow: rate-determining step Step 1: Step 2: Overall: fast So the rate depends on step 2, for which… rate = k[N 2 O 2 ][O 2 ] This is a third-order reaction. But N 2 O 2 is a product of step 1, so the concentration of this intermediate depends on [NO] 2. Therefore, we substitute this into the equation above and the rate expression for the overall reaction is: rate = k[NO] 2 [O 2 ]

18. Activation Energy The rate constant, k, is temperature dependent Rxn rate increases as temperature increases  faster particles = increased collision rate

19. Activation Energy Rule of thumb: a 10  C increase in temperature leads to a doubling of the rate.  Note: this is only a rule of thumb, not a law of nature. Thus, it is not true for all reactions.

21. Activation Energy When activation energy is large, temperature has a greater effect on the number of particles with sufficient energy to react (and thus rate of reaction). This can be shown by studying Maxwell-Boltzman distributions.

22. Arrhenius Equation: mathematical relationship between temperature, the rate constant and the activation energy. 1859-1927 Svante Arrhenius The fraction of molecules with energy greater than activation energy, E a, at temperature T is proportional to the expression e -Ea/RT

23. Arrhenius Equation: mathematical relationship between temperature, the rate constant and the activation energy. 1859-1927 Svante Arrhenius Thus, here is my famous equation… k  e -Ea/RT or k = Ae -Ea/RT In the IB data booklet

24. k = Ae -Ea/RT  k = rate constant  A = Arrhenius constant (a.k.a. frequency factor, pre-exponential factor, and steric factor) Takes into account the frequency with shich successful collisions occur based on collision geometry and energy requirements Constant for a rxn & has units the same as k (and so varies with the order of the reaction)  E a = activation energy (in Joules mol -1 )  R = 8.31 J mol -1 K -1 (ideal gas constant)  T = absolute temperature (in Kelvin)

25. k = Ae -Ea/RT Taking the natural logarithm (ln base e) of both sides of the equation above: or Notice that this is a form of y = mx + b (straight line) In the IB data booklet

26. Arrhenius Plot graph of ln k v. 1/T ln k 1/T (K -1 ) slope = y-intercept =

27. The next few slides contain additional examples to help explain these concepts further… include only if time allows Examples

28. Example: What happens to the rate of a reaction if the temperature increases from 20 o C to 30 o C? The frequency factor, a, is approximately constant for such a small temperature change. We need to look at how e -(Ea/RT) changes --- the fraction of molecules with energies equal to or in excess of the activation energy. Assume an activation energy of 50 kJ/mol. In the equation, we have to write that as 50,000 J/mol. The value of the gas constant R is 8.31 J/K·mol Now raise the temperature just a little bit to 30 o C (303k)

29. At 20 o CAt 30 o C The fraction of the molecules able to react has almost doubled by increasing the temperature by 10 o C. That causes the rate of the reaction to almost double. This is the basis for the old rule of thumb that a reaction rate doubles for every 10 o C rise in temperature.

30. The effect of a catalyst A catalyst provides an alternate reaction mechanism, or route for the reaction. This alternate route necessarily has a lower activation energy. The overall effect of a catalyst is to lower the activation energy. Continuing our example of a reaction with an activation energy of 50,000 J/mol… What is the effect on the rate of lowering the activation to 25,000 J/mol? With Catalyst Without Catalyst No wonder catalysts speed up reactions!

31. H-C-C-O-C-H + H 2 O  H-C-C-OH + H-C-OH H H H H H H H H O Methyl Acetate + water  Acetic Acid + Methanol Reaction mechanism example: Experimental evidence for reaction mechanisms --- Consider the reaction of methyl acetate with water…. The methyl acetate molecule can break in one of two places to complete the reaction…

32. H-C-C-O-C-H + H 2 O  H-C-C-OH + H-C-OH H H H H H H H H O Methyl Acetate + water  Acetic Acid + Methanol H-C-C- O-C-H O-H H H H O H H -OR-

33. H-C-C-O-C-H + H 2 O  H-C-C-OH + H-C-OH H H H H H H H H O Methyl Acetate + water  Acetic Acid + Methanol H-C-C- O-C-H O-H H H H O H H H-C-C-O- -C-H H H H H O O-H H -OR- How can we determine experimentally which is correct?

34. H-C-C-O 18 -H H-O 18 H H H Use a radioactive isotope of oxygen, O 18 To make up the water O

35. H-C-C-O-C-H + H 2 O  H-C-C-OH + H-C-OH H H H H H H H H O Methyl Acetate + water  Acetic Acid + Methanol H-C-C- O-C-H O-H H H H O H H H-C-C-O- -C-H H H H H O O-H H -OR-

36. Measuring the rate of a reaction. Formulating the Rate Law Postulating a reasonable reaction mechanism Sequence of steps in the study of a reaction mechanism

37. Decomposition of Hydrogen Peroxide The decomposition of Hydrogen Peroxide is facilitated by iodide ions. The overall reaction is… 2H 2 O 2  2H 2 O + O 2 By experiment, the rate law is found to be… rate = k[H 2 O 2 ][I - ] Thus the reaction is first order with respect to both H 2 O 2 and I -. Example:

38. rate = k[H 2 O 2 ][I - ] You can see that H 2 O 2 decomposition does not occur in a single elementary step corresponding to the overall balanced equation. If it did, the reaction would be second order in H 2 O 2 (as a result of the collision of two H 2 O 2 molecules). What’s more, the I - ion, which is not even part of the overall equation, appears in the rate law expression. We can account for the observed rate law by assuming that the reaction takes place in two separate elementary steps, each of which is bimolecular: step 1: H 2 O 2 + I -  H 2 O + IO - step 2: H 2 O 2 + IO-  H 2 O + O 2 + I - 2H 2 O 2  2H 2 O + O 2

39. step 1: H 2 O 2 + I -  H 2 O + IO - step 2: H 2 O 2 + IO -  H 2 O + O 2 + I - The first step must be the rate determining step, because it matches the rate law. Thus, the rate of the reaction can be determined from the first step alone: Note that the IO - ion is an intermediate because it does not appear in the overall balanced equation. Although the I - ion also does not appear in the overall equation, I - differs from IO - in that the former is present at the start of the reaction and at its completion. The function of I - is to speed up the reaction --- that is, it is a catalyst. rate = k[H 2 O 2 ][I - ] Slow

40. Another Example : Consider the Reaction… 2NO 2 Cl  2NO 2 + Cl 2 This is found to be a first order reaction, it’s experimentally determined rate law is… rate = k[NO 2 Cl] Could the overall reaction occur in a single step by the collision of two NO 2 Cl molecules? This would lead to… rate = k[NO 2 Cl] 2 Chemists believe the actual mechanism is… NO 2 Cl  NO 2 + Cl NO 2 Cl + Cl  NO 2 + Cl 2 Slow Fast