1. Chemistry 232 Chemical Kinetics

2. Chemical kinetics - speed or rate at which a reaction occurs How are rates of reactions affected by Reactant concentration? Temperature? Reactant states? Catalysts?

3. The Instantaneous Reaction Rate Consider the following reaction A + B  C Define the instantaneous rate of consumption of reactant A, A

4. Reaction Rates and Reaction Stoichiometry Look at the reaction O 3 (g) + NO(g)  NO 2 (g) + O 2 (g)

5. Another Example 2 NOCl (g)  2 NO + 1 Cl 2 (g) WHY? 2 moles of NOCl disappear for every 1 mole Cl 2 formed.

6. The General Case a A + b B  c C + d D rate = -1 d[A] = -1 d[B] = +1 d[C] = +1 d[D] a dt b dt c dt d dt Why do we define our rate in this way?  removes ambiguity in the measurement of reaction rates  Obtain a single rate for the entire equation, not just for the change in a single reactant or product.

7. Alternative Definition of the Rate Rate of conversion related to the advancement of the reaction, . V = solution volume

8. An Example Examine the following reaction 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) N2O5N2O5 NO 2 O2O2 Initial n ø 00 Change -2  +4  ++ Final n ø - 2  44

9. The N 2 O 5 Decomposition Note – constant volume system

10. The Rate Law Relates rate of the reaction to the reactant concentrations and rate constant For a general reaction a A + b B + c C  d D + e E

11. Rate Laws (Cont’d) The only way that we can determine the superscripts (x, y, and z) for a non- elementary chemical reaction is by experimentation. Use the isolation method (see first year textbooks).

12. For a general reaction  x + y + z = reaction order e.g. X = 1; Y = 1; Z = 0 2nd order reaction (x + y + z = 2) X = 0; Y = 0; Z = 1 (1st order reaction) X = 2; Y = 0; Z = 0 (2nd order)

13. Rate Laws for Multistep Processes Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism The slowest elementary step  the rate determining step (rds).

14. An Example Reaction Mechanism O atoms are intermediates in the above reaction sequence. Intermediates – generally small, indeterminate concentrations.

15. Integrated Rate Laws The rate law gives us information about how the concentration of the reactant varies with time How much reactant remains after specified period of time? Use the integrated rate laws.

16. First Order Reaction A  product Rate = v = - d[A]/d  t = k[A] How does the concentration of the reactant depend on time? k has units of s -1

17. The Half-life of a First Order Reaction For a first order reaction, the half-life t 1/2 is calculated as follows.

18. Radioactive Decay Radioactive Samples decay according to first order kinetics. This is the half-life of samples containing e.g. 14 C, 239 Pu, 99 Tc. Example

19. Second Order Reaction A  productsv = k[A] 2 A + B  products v = k[A][B] Case 1 is 2 nd order in A Case 2 is 1 st order in A and B and 2 nd order overall

20. The Dependence of Concentration on Time For a second order process where v = k[A] 2

21. Half-life for This Second Order Reaction. [A] at t = t½ = ½ [A] 0

22. Other Second Order Reactions Examine the Case 2 from above A + B  productsv = k[A][B]

23. A Pseudo-first Order Reaction Example hydration of methyl iodide CH 3 I(aq) + H 2 O(l)  CH 3 OH(aq) + H + (aq) + I - (aq) Rate = k [CH 3 I] [H 2 O] Since we carry out the reaction in aqueous solution [H 2 O] >>>> [CH 3 I]  [H 2 O] doesn’t change by a lot

24. Pseudo-first Order (cont’d) Since the concentration of H 2 O is essentially constant v = k [CH 3 I] [constant] = k`  [CH 3 I] where k` = k [H 2 O] The reaction is pseudo first order since it appears to be first order, but it is actually a second order process.

25. Sequential First Order Reactions Suppose we have two first order reactions occurring in sequence.

26. What is Our Intermediate? B is an intermediate in the above reaction sequence. Clearly B is formed in the first elementary step of the reaction mechanism and consumed in the final step.

27. The Concentration Dependencies of the Species The amounts of the reactants are related to the reaction rates as follows.

28. Sequential Reactions (3) For a set of initial conditions [A] o 0, and [B] o = [C] o =0 mol/L.

29. Sequential Reactions (4) The concentration of the intermediate can be written

30. Sequential Reactions (5)

31. The Rate Determining Step of the Reaction What happens if one of the steps in the reaction is much slower than the other reaction step? Note – assuming k a <<< k b

32. The Rate Determining Step of the Reaction (2) If k b <<< k a

33. Temperature Dependence of Reaction Rates Reaction rates generally increase with increasing temperature. Arrhenius Equation A = pre-exponential factor E a = the activation energy

34. Reactions Approaching Equilibrium Examine the concentration profiles for the following simple process. A ⇌ B

35. Approaching Equilibrium Calculate the amounts of A and B at equilibrium.

36. The Equilibrium Condition At equilibrium, v A,eq = v B,eq.

37. Elementary steps and the Molecularity Kinetics of the elementary step depends only on the number of reactant molecules in that step! Molecularity  the number of reactant molecules that participate in elementary steps

38. The Kinetics of Elementary Steps For the elementary step unimolecular step For elementary steps involving more than one reactant a bimolecular step

39. The Kinetics of Elementary Steps For the elementary step unimolecular step For elementary steps involving more than one reactant a bimolecular step

40. For the step a termolecular (three molecule) step. Termolecular (and higher) steps are not that common in reaction mechanisms.

41. The Steady-State Approximation Examine the following simple reaction mechanism Rate of product formation, v p, is proportional to the concentration of an intermediate.

42. Applying the Steady State Approximation (SSA) Look for the intermediate in the mechanism. Step 1 – B is produced. Reverse of Step 1 – B is consumed. Step 2 – B is consumed.

43. The SSA (Cont’d) The SSA applied to the intermediate B.

44. SSA – The Final Step Substitute the expression for the concentration of B into the rate law v p.

45. Competing Reactions Imagine a reaction with a competing side reaction.

46. The Reaction “Yield” We can calculate the amount of material produced from the competing reactions. k J = the rate constant for the reaction J.

47. Activated Complex Theory Consider the following bimolecular reaction Presume that the reaction proceeds through the transition state?

48. The Activated Complex The species temporarily formed by the reactant molecules – the activated complex. A small fraction of molecules usually have the required kinetic energy to get to the transition state The concentration of the activated complex is extremely small.

49. Transition State Theory (TST) TST pictures the bi-molecular reaction proceeding through the activated complex in a rapid-pre-equilibrium.

50. TST (II) From the thermodynamic equilibrium constant for the formation of AB ↕

51. TST (IV) Assume that we can obtain the Gibbs energy of activation from K ↕. The Eyring Equation!

52. TST (V) Hence, substituting for the Gibbs energy of activation in terms of the enthalpy and entropy of activation.

53. TST (VI) Expressing the enthalpy of activation in terms of the activation energy.

54. TST (VII) The final result for a bimolecular gas phase reaction.

55. TST (VIII) The final result for a unimolecular gas phase reaction (or any solution reaction).