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Chapter 14: Chemical Kinetics

Petrucci • Harwood • Herring • Madura. GENERAL. Ninth Edition. CHEMISTRY. Principles and Modern Applications. Chapter 14: Chemical Kinetics. Contents. 14-1 The Rate of a Chemical Reaction 14-3 Effect of Concentration on Reaction Rates: The Rate Law 14-4 Zero-Order Reactions

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Chapter 14: Chemical Kinetics

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

  2. Contents 14-1 The Rate of a Chemical Reaction 14-3 Effect of Concentration on Reaction Rates: The Rate Law 14-4 Zero-Order Reactions 14-5 First-Order Reactions 14-6 Second-Order Reactions 14-7 Reaction Kinetics: A Summary

  3. Contents 14-8 Theoretical Models for Chemical Kinetics 14-9 The Effect of Temperature on Reaction Rates 14-10 Reaction Mechanisms 14-11 Catalysis

  4. Δ[Fe2+] 0.0010 M Rate of formation of Fe2+= = = 2.610-5 M s-1 Δt 38.5 s 14-1 The Rate of a Chemical Reaction • Rate of change of concentration with time. 2 Fe3+(aq) + Sn2+→ 2 Fe2+(aq) + Sn4+(aq) t = 38.5 s [Fe2+] = 0.0010 M Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0) M

  5. 1 Δ[Fe3+] Δ[Sn4+] 1 Δ[Fe2+] = - = Δt Δt Δt 2 2 Rates of Chemical Reaction 2 Fe3+(aq) + Sn2+→ 2 Fe2+(aq) + Sn4+(aq)

  6. Δ[B] Δ[A] 1 1 = - = - b a Δt Δt Δ[D] Δ[C] 1 1 = = d c Δt Δt General Rate of Reaction a A + b B → c C + d D Rate of reaction = rate of disappearance of reactants = rate of appearance of products

  7. 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. EXAMPLE 14-3 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 HgCl2 and C2O22- and also the overall order of the reaction.

  9. EXAMPLE 14-3 Notice that concentration changes between reactions are by a factor of 2. Write and take ratios of rate laws taking this into account.

  10. EXAMPLE 14-3 k(2[HgCl2]3)m[C2O42-]3n R2 = R3 k[HgCl2]3m[C2O42-]3n k2m[HgCl2]3m[C2O42-]3n R2 2mR3 = = = 2.0 R3 k[HgCl2]3m[C2O42-]3n R3 R3 = k[HgCl2]3m[C2O42-]3n = k(2[HgCl2]3)m[C2O42-]3n R2 = k[HgCl2]2m[C2O42-]2n 2m = 2.0therefore m = 1.0

  11. EXAMPLE 14-3 k(0.105)(0.30)n R2 = R1 k(0.105)(0.15)n (0.30)n R2 7.110-5 = 2n = = = 3.94 R1 (0.15)n 1.810-5 R2 = k[HgCl2]21[C2O42-]2n = k(0.105)(0.30)n R1 = k[HgCl2]11[C2O42-]1n = k(0.105)(0.15)n 2n= 3.94 therefore n = 2.0

  12. EXAMPLE 14-3 2 1 R2 = k[HgCl2]2[C2O42-]2 First order + = Third Order Second order

  13. Worked Examples Follow:

  14. CRS Questions Follow:

  15. Consider the following reaction, whose rate can be expressed as Equivalent expressions are…

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

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