1. Slide 1 of 61 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 14: Chemical Kinetics

2. Slide 2 of 61 Contents 14-1The Rate of a Chemical Reaction 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Slide 13 of 61  Worked Examples Follow:

14. Slide 14 of 61

15. Slide 15 of 61  CRS Questions Follow:

16. Slide 16 of 61 Consider the following reaction, whose rate can be expressed as Equivalent expressions are…

17. Slide 17 of 61 Consider the following reaction, whose rate can be expressed as Equivalent expressions are…