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

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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

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Slide 3 of 61 Contents 14-8Theoretical Models for Chemical Kinetics 14-9The Effect of Temperature on Reaction Rates 14-10Reaction Mechanisms 14-11Catalysis

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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

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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

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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

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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 + ….

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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

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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

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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

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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

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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

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Slide 13 of 61 Worked Examples Follow:

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Slide 14 of 61

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Slide 15 of 61 CRS Questions Follow:

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

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

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DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.

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