1. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 1 of 61 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 14: Chemical Kinetics

2. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 2 of 61 Contents 14-1The Rate of a Chemical Reaction 14-2Measuring Reaction Rates 14-3Effect of Concentration on Reaction Rates: The Rate Law 14-4Zero-Order Reactions 14-5First-Order Reactions 14-6Second-Order Reactions 14-7Reaction Kinetics: A Summary

3. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 3 of 61 Contents 14-8Theoretical Models for Chemical Kinetics 14-9The Effect of Temperature on Reaction Rates 14-10Reaction Mechanisms 14-11Catalysis

4. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 4 of 61 14-1 The Rate of a Chemical Reaction  Rate of change of concentration with time. 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s [Fe 2+ ] = 0.0010 M Δt = 38.5 sΔ[Fe 2+ ] = (0.0010 – 0) M Rate of formation of Fe 2+ = = = 2.6  10 -5 M s -1 Δ[Fe 2+ ] ΔtΔt 0.0010 M 38.5 s

5. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 5 of 61 Rates of Chemical Reaction Δ[Sn 4+ ] ΔtΔt 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) Δ[Fe 2+ ] ΔtΔt = 1 2 Δ[Fe 3+ ] ΔtΔt = - 1 2

6. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 6 of 61 General Rate of Reaction a A + b B → c C + d D Rate of reaction = rate of disappearance of reactants = Δ[C] ΔtΔt 1 c = Δ[D] ΔtΔt 1 d Δ[A] ΔtΔt 1 a = - Δ[B] ΔtΔt 1 b = - = rate of appearance of products

7. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 7 of 61 14-3 Effect of Concentration on Reaction Rates: The Rate Law a A + b B …. → g G + h H …. Rate of reaction = k[A] m [B] n …. Rate constant = k Overall order of reaction = m + n + ….

8. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 8 of 61 Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respect to HgCl 2 and C 2 O 2 2- and also the overall order of the reaction. EXAMPLE 14-3

9. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 9 of 61 Notice that concentration changes between reactions are by a factor of 2. Write and take ratios of rate laws taking this into account. EXAMPLE 14-3

10. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 10 of 61 R 2 = k  [HgCl 2 ] 2 m  [C 2 O 4 2- ] 2 n R 3 = k  [HgCl 2 ] 3 m  [C 2 O 4 2- ] 3 n R2R2 R3R3 k  (2[HgCl 2 ] 3 ) m  [C 2 O 4 2- ] 3 n k  [HgCl 2 ] 3 m  [C 2 O 4 2- ] 3 n = 2 m = 2.0 therefore m = 1.0 R2R2 R3R3 k  2 m  [HgCl 2 ] 3 m  [C 2 O 4 2- ] 3 n k  [HgCl 2 ] 3 m  [C 2 O 4 2- ] 3 n == 2.0= 2mR32mR3 R3R3 = k  (2[HgCl 2 ] 3 ) m  [C 2 O 4 2- ] 3 n EXAMPLE 14-3

11. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 11 of 61 R 2 = k  [HgCl 2 ] 2 1  [C 2 O 4 2- ] 2 n = k  (0.105)  (0.30) n R 1 = k  [HgCl 2 ] 1 1  [C 2 O 4 2- ] 1 n = k  (0.105)  (0.15) n R2R2 R1R1 k  (0.105)  (0.30) n k  (0.105)  (0.15) n = 7.1  10 -5 1.8  10 -5 = 3.94 R2R2 R1R1 (0.30) n (0.15) n = = 2 n = 2 n = 3.94 therefore n = 2.0 EXAMPLE 14-3

12. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 12 of 61 + = Third Order R 2 = k  [HgCl 2 ] 2  [C 2 O 4 2- ] 2 First order 1 Second order 2 EXAMPLE 14-3

13. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 13 of 61 14-4 Zero-Order Reactions A → products R rxn = k [A] 0 R rxn = k [k] = mol L -1 s -1

14. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 14 of 61 Integrated Rate Law -- dtdt= k d[A]  [A] 0 [A] t 0 t -[A] t + [A] 0 = kt [A] t = [A] 0 - kt ΔtΔt -Δ[A] dtdt = k -d[A] Move to the infinitesimal = k And integrate from 0 to time t

15. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 15 of 61 14-5 First-Order Reactions H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) = -k[H 2 O 2 ] d[H 2 O 2 ] dtdt = - k dt [H 2 O 2 ] d[H 2 O 2 ]  [A] 0 [A] t  0 t = -kt ln [A] t [A] 0 ln[A] t = -kt + ln[A] 0 [k] = s -1

16. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 16 of 61 First-Order Reactions

17. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 17 of 61 Half-Life  t ½ is the time taken for one-half of a reactant to be consumed. = -kt ln [A] t [A] 0 = -kt ½ ln ½[A] 0 [A] 0 - ln 2 = -kt ½ t ½ = ln 2 k 0.693 k =

18. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 18 of 61 Half-Life Bu t OOBu t (g) → 2 CH 3 CO(g) + C 2 H 4 (g)

19. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 19 of 61 Some Typical First-Order Processes

20. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 20 of 61 14-6 Second-Order Reactions  Rate law where sum of exponents m + n + … = 2. A → products dtdt= - k d[A] [A] 2  [A] 0 [A] t  0 t = kt + 1 [A] 0 [A] t 1 dtdt = -k[A] 2 d[A] [k] = M -1 s -1 = L mol -1 s -1

21. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 21 of 61 Second-Order Reaction

22. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 22 of 61 Testing for a Rate Law Plot [A] vs t.Plot ln[A] vs t.Plot 1/[A] vs t.

23. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 23 of 61 14-7 Reaction Kinetics: A Summary  Calculate the rate of a reaction from a known rate law using:  Determine the instantaneous rate of the reaction by: Rate of reaction = k [A] m [B] n …. Finding the slope of the tangent line of [A] vs t or, Evaluate –Δ[A]/Δt, with a short Δt interval.

24. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 24 of 61 Summary of Kinetics  Determine the order of reaction by: Using the method of initial rates. Find the graph that yields a straight line. Test for the half-life to find first order reactions. Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.

25. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 25 of 61 Summary of Kinetics  Find the rate constant k by:  Find reactant concentrations or times for certain conditions using the integrated rate law after determining k. Determining the slope of a straight line graph. Evaluating k with the integrated rate law. Measuring the half life of first-order reactions.

26. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 26 of 61 14-8 Theoretical Models for Chemical Kinetics  Kinetic-Molecular theory can be used to calculate the collision frequency.  In gases 10 30 collisions per second.  If each collision produced a reaction, the rate would be about 10 6 M s -1.  Actual rates are on the order of 10 4 M s -1. ◦Still a very rapid rate.  Only a fraction of collisions yield a reaction. Collision Theory

27. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 27 of 61 Activation Energy  For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).  Activation Energy:  The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

28. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 28 of 61 Activation Energy

29. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 29 of 61 Kinetic Energy

30. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 30 of 61 Collision Theory  If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.  As temperature increases, reaction rate increases.  Orientation of molecules may be important.

31. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 31 of 61 Collision Theory

32. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 32 of 61 Transition State Theory  The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

33. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 33 of 61 14-9 Effect of Temperature on Reaction Rates  Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = Ae -E a /RT ln k = + lnA R -Ea T 1

34. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 34 of 61 Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2  10 4 K R -Ea-Ea -E a = 1.0  10 2 kJ mol -1

35. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 35 of 61 Arrhenius Equation k = Ae -E a /RT ln k = + ln A R -Ea-Ea T 1 ln k 2 – ln k 1 = + ln A - - ln A R -Ea-Ea T2T2 1 R -Ea-Ea T1T1 1 ln = - R -Ea-Ea T2T2 1 k1k1 k2k2 T1T1 1

36. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 36 of 61 14-10 Reaction Mechanisms  A step-by-step description of a chemical reaction.  Each step is called an elementary process.  Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule.  Reaction mechanism must be consistent with:  Stoichiometry for the overall reaction.  The experimentally determined rate law.

37. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 37 of 61 Elementary Processes  Unimolecular or bimolecular.  Exponents for concentration terms are the same as the stoichiometric factors for the elementary process.  Elementary processes are reversible.  Intermediates are produced in one elementary process and consumed in another.  One elementary step is usually slower than all the others and is known as the rate determining step.

38. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 38 of 61 Slow Step Followed by a Fast Step H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) dtdt = k[H 2 ][ICl] d[P] Postulate a mechanism: H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) slow H 2 (g) + ICl(g) HI(g) + HCl(g) fast HI(g) + ICl(g) I 2 (g) + HCl(g) dtdt = k[H 2 ][ICl] d[HI] dtdt = k[HI][ICl] d[I 2 ] dtdt = k[H 2 ][ICl] d[P]

39. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 39 of 61 Slow Step Followed by a Fast Step

40. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 40 of 61 Fast Reversible Step Followed by a Slow Step 2NO(g) + O 2 (g) → 2 NO 2 (g) dt = -k obs [NO 2 ] 2 [O 2 ] d[P] Postulate a mechanism: dtdt = k 2 [N 2 O 2 ][O 2 ] d[NO 2 ] fast 2NO(g) N 2 O 2 (g) k1k1 k -1 slow N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k2k2 dtdt = k 2 [NO] 2 [O 2 ] d[I 2 ] k -1 k1k1 2NO(g) + O 2 (g) → 2 NO 2 (g) K = k -1 k1k1 = [NO] [N 2 O 2 ] = K[NO] 2 k -1 k1k1 = [NO] 2 [N 2 O 2 ]

41. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 41 of 61 Catalytic Converters  Dual catalyst system for oxidation of CO and reduction of NO. CO 2 + N 2 CO + NO cat

42. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 42 of 61 14-5 Catalysis  Alternative reaction pathway of lower energy.  Homogeneous catalysis.  All species in the reaction are in solution.  Heterogeneous catalysis.  The catalyst is in the solid state.  Reactants from gas or solution phase are adsorbed.  Active sites on the catalytic surface are important.

43. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 43 of 61 14-5 Catalysis

44. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 44 of 61 Catalysis on a Surface

45. Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 45 of 61 Enzyme Catalysis ES → E + P k2k2 E + S ES k1k1 k -1