Atkins & de Paula: Elements of Physical Chemistry: 5e

1. Atkins & de Paula: Elements of Physical Chemistry: 5e Chapter 11: Chemical Kinetics: Accounting for the Rate Laws

2. End of chapter 11 assignments Discussion questions: • 2, 5 Exercises: • 5, 13, 14, 18, 25 Use Excel if data needs to be graphed

3. Time for me to “nag” you! • Did you: • Read the chapter? • Work through the example problems? • Connect to the publisher’s website & access the “Living Graphs”? • Examine the “Checklist of Key Ideas”? • Work assigned end-of-chapter exercises? • Review terms & concepts that you should remember from previous chapters or courses

4. Reaction schemes • All chemical rxns proceed toward a state of equilibrium • The reverse rxn becomes increasingly important as the rxn proceeds • Many rxns proceed through a series of intermediates • These intermediates may be important

5. Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. 14.2

6. F2(g) + 2ClO2(g) 2FClO2(g) aA + bB cC + dD [C]c[D]d [A]a[B]b Kc = Rate Laws • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. • The stoichiometric coefficient of the reactant in the balanced chemical equation is the exponent in the K expression. 1 rate = k [F2][ClO2] 14.2

7. Digression: Rate law and Keq • Keq is concerned with the initial and final conditions of the rxn (overall) • Rate law, order, “kinetics” are concerned with everything in between • Let me show you what IUPAC says about the rate law and Keq

8. Rate law and Keq The equilibrium law or the equilibrium constant are in the first instance derived from the thermodynamics of the reaction, the value of the equilibrium constant Kc depending entirely upon the characteristics of the reactants and these of the reaction products. --continued

9. Rate law and Keq Rate of reaction and reaction mechanism are kinetic aspects of the reaction. They have nothing to do with the expression for Kc or with the value of Kc. It would be simplistic and imprudent to assume all sorts of unrealistic reaction equations to obtain an expression for Kc.

10. Rate law and Keq • Quoted from IUPAC’s web site • http://www.iupac.org/ • Specifically: http://www.iupac.org/didac/Didac%20Eng/Didac02/Content/E06%20-%20E07%20-%20E08.htm

11. Order & molecularity The overall reaction order is the sum of the powers of the concentrations of all of the substances appearing in the experimental rate law for the reaction; hence, it is the sum of the individual orders (exponents) associated with a given reactant (or product). Reaction order is an experimentally determined, not theoretical, quantity, although theory may attempt to predict it. Quoted from another book by Atkins…

12. Order & molecularity Molecularity is the number of reactant molecules partici-pating in an elementary reaction. This concept has meaning only for an elementary reaction, but reaction order applies to any reaction. In general, reaction order bears no necessary relation to the stoichiometry of the reaction, with the exception of elementary reactions, where the order of the reaction corresponds to the number of molecules participating in the reaction; that is, to its molecularity. Thus for an elementary reaction, overall order and molecularity are the same and are determined by the stoichiometry. Quoted from another book by Atkins…

13. Profile of an exothermic rxn • Fig 11.1 (258) • Ea is greater for the reverse rxn than for the forward rxn • So the Rf increases less sharply (than the Rr) as T increases • As the T is raised, the products are favored

14. The approach to equilibrium • Derivation 11.1 (259) • You should work through this derivation

15. The approach to equilibrium • Fig 11.2 (259) • The approach to equilibrium of a rxn that is 1st order in both directions • In this example, k = 2k’, so at equilibrium the ratio of b:a is 2:1 AB

16. Relaxation methods • Some externally applied influence shifts the rxn equilibrium abruptly • Temperature jump • Pressure jump • The adjustment back to equilibrium is called “relaxation”

17. Relaxation methods Fig 11.3 (259) Relaxation to a new equilibrium following a sud-den change of temperature from T1 to T2

18. 1  Relaxation methods • When a sudden temperature jump shifts the equilibrium of a 1st order rxn: • x = x0e–t/ and = ka + kb • x is the shift in equilibrium at the new t • x0 is the shift in equilibrium immediately after the temperature change •  is the relaxation time • t is the time (s)

19. Consecutive reactions Common examples are radioactive decay chains and biochemical rxn sequences (e.g., gylcoysis, Krebs cycle, etc)

20. The Uranium Decay Series

21. Glycolysis

22. Consecutive reactions • Fig 11.4 (263) • Concentrations in consecutive rxns, A  I  P • A = reactant • I = intermediate • P = product • At any time the sum [A] + [I] + [P] is a constant

23. Elementary reactions • Many reactions occur as a series of steps called “elementary reactions” • Molecularity is the number of molecules coming together to react • Unimolecular • Bimolecular • What’s next?

24. Elementary reactions Fig 11.5 (263) A unimolecular elementary rxn

25. Elementary reactions Fig 11.6 (264) A bimolecular elementary rxn

26. The formulation of rate laws • Assuming that a chemical rxn is a series of elementary rxns… • One of these elementary rxns is the rate determining step • An acceptable rate law for an overall rxn is expressed only in terms of the concentrations of the species in the overall rxn, not in terms of an intermediate (264)

27. The formulation of rate laws • Consider this rxn: • 2 NO(g) + O2(g) d 2 NO2(g) • Experimentally: rate = k[NO]2[O2] (which is 3rd order overall) • How can we explain this? Is it a termolecular rxn (2 NOs and 1 O2)? • Let’s see… (264)

28. Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2 Step 1. NO + NO dN2O2 Rate of formation of N2O2 = ka[NO]2 Step 2. N2O2 dNO + NO Rate of decomposition of N2O2 = k’a[N2O2] Step 3. N2O2 + O2dNO2 + NO2 Rate of consumption of N2O2 = 2kb[N2O2][O2] Thus the net rate of formation of N2O2 is = ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2] Does text have an error in this last eqn?

29. Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2 Net rate of formation of N2O2 = ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2] • Solving this eqn involves solving a very difficult differential eqn and gives a complex expression • This drives us to consider…“The Steady-State Approximation” (265)

30. The steady-state approximation • Our assumption: the concentrations of all intermediates remain constant and small throughout the rxn (except at the very beginning and at the very end) • So, the net rate of formation of N2O2 = 0 • So, ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2] = 0 • Solving for [N2O2] • Solving for [NO2] gives 11.10, which accounts for overall 3rd order rxn (265)

31. The rate-determining step • If step 3 is much faster than steps 1 and 2, (this is the case if the [O2] is high) then k’a can be ignored • Now the rate law simplifies to 11.12 • [NO2] = = 2ka[NO]2 • Now the rxn is 2nd order and [O2] does not appear in the rate law 2kakb[NO]2[O2] kb [NO]2[O2]

32. The rate-determining step The rate determining step (RDS) must have two characteristics: • It must be the slowest step, and • It must be a crucial part of the path from reactants to products

33. The rate-determining step • Fig 11.7 (266) • Diagrammatic illustration of the RDS • Heavy lines = fast steps • Thin lines = slow steps

34. The rate-determining step • Fig 11.7 (266) • Heavy lines = fast steps • Thin lines = slow steps • In (b), the thin line is not a bottleneck because there is an alternate route

35. [P2] k2 = k1 [P1] Kinetic control Consider these competing rxns: • A + B d P1 Rate of formation of P1 = k1[A][B] • A + B d P2 Rate of formation of P2 = k2[A][B]  Represents kinetic control

36. Kinetic control If a rxn is allowed to reach equilibrium, then thermodynamics (rather than kinetic considerations) determines the proportions of products, and the ratio of concentrations is controlled by standard Gibbs energies of reactants and products (267)

37. Reaction profile Fig 11.7 (267) Multistep mechanism in which the first step is the RDS (highest Ea)

38. Unimolecular reactions • Many gas-phase rxns proceed by two gas molecules colliding but are 1st order • These collisions are bimolecular events and the rxn should be 2nd order • Explain! • The Lindemann mechanism posits that the RDS is unimolecular (267f)

39. Activation control & diffusion control Now we consider reactions in solution

40. Activation control & diffusion control • kd is the rate constant (of diffusion) • Diffusion-controlled limit—the rate of the rxn is controlled by the rate at which reactants diffuse (kd) • Activation-controlled limit—the rate of the rxn depends on the rate at which energy accumulates in the “encounter pair” (ka)

41. Activation control & diffusion control • Now read next-to-last paragraph, p.269 • “A lesson to learn from this analysis is the concept of the rate-determining step is rather subtle. Thus, in the…”

42. 8RT 3 Activation control & diffusion control • Vis-a-vis the rate of diffusion in a liquid, the rate constant (kd) is related to the coefficient of viscosity,  • How are kd and  related? kd = •  is the Greek lowercase letter eta

43. Diffusion This section relates directly to the lab experiment you’ve worked on this semester (270)

44. Number of particles passing through a window Area of window  Time interval Diffusion • Terms: diffusion, random walk, flux (J) • Rate of diffusion  Concentration gradient • J = • J = – D  Concentration gradient (11.18b) • D is the diffusion coefficient(area/time) • This is Fick’s first law of diffusion

45. Diffusion • Fig 11.9 (271) • The flux of solute particles is proportional to the concen-tration gradient • Gradient is negative; flux is positive

46. Diffusion • Fig 11.18 (281) Further Information

47. Diffusion • Diffusion is often aided by convection • The diffusion eqn below lets us predict the rate at which the concentration of a solute changes in a “non-uniform” solution • Fick’s second law: Rate of change of concentration in a region = D  curvature of the concentration in the region (11.19) • So, what’s a “wrinkle”? 271

48. Diffusion in a non-uniform solution • Fig 11.10 (272) • Nature abhors a wrinkle • The phrase “non-uniform solution” is an oxymoron

49. Table 11.1 Diffusion coefficients at 25°C, D/(10-9 m2 s-1) (270)

50. Diffusion in water • Fig 11.11 (273) • Temperature dependence of the viscosity of water • Water becomes more fluid as temperature increases • How does this affect diffusion of solute particles?