Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 14: Chemical Kinetics

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 2 of 55 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

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 3 of 55 Contents 14-8Theoretical Models for Chemical Kinetics 14-9The Effect of Temperature on Reaction Rates 14-10Reaction Mechanisms 14-11Catalysis Focus On Combustion and Explosions

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 4 of 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+ ] = M Δt = 38.5 sΔ[Fe 2+ ] = ( – 0) M Rate of formation of Fe 2+ = = = 2.6x10 -5 M s -1 Δ[Fe 2+ ] ΔtΔt M 38.5 s

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 5 of 55 Rates of Chemical Reaction Δ[Sn 4+ ] ΔtΔt 2 Fe 3+ (aq) + Sn 2+ (aq) → 2 Fe 2+ (aq) + Sn 4+ (aq) Δ[Fe 2+ ] ΔtΔt = 1 2 Δ[Fe 3+ ] ΔtΔt = - 1 2

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 6 of 55 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

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 7 of Measuring Reaction Rates H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) 2 MnO 4 - (aq) + 5 H 2 O 2 (aq) + 6 H + → 2 Mn H 2 O(l) + 5 O 2 (g)

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 8 of 55 H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) Example (-1.7 M / 2800 s) = 6 x M s -1 -(-2.32 M / 1360 s) = 1.7 x M s -1 Determining and Using an Initial Rate of Reaction. Rate = -Δ[H 2 O 2 ] ΔtΔt

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 9 of 55 Example Δ[H 2 O 2 ] = -([H 2 O 2 ] f - [H 2 O 2 ] i )= 1.7 x10 -3 M s -1 x Δt Rate = 1.7x M s -1 ΔtΔt = - Δ[H 2 O 2 ] [H 2 O 2 ] 100 s – 2.32 M =-1.7 x M s -1 x 100 s = 2.15 M = 2.32 M M [H 2 O 2 ] 100 s What is the concentration at 100s? [H 2 O 2 ] i = 2.32 M

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 10 of 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 + ….

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 11 of 55 Example 14-3 Method of Initial Rates 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 4 2- and also the overall order of the reaction.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 12 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 13 of 55 Example 14-3 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

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 14 of 55 Example 14-3 R 2 = k [HgCl 2 ] 2 1 x[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 = 7.1x x10 -5 = 3.94 R2R2 R1R1 (0.30) n (0.15) n = = 2 n = 2 n = 3.98 therefore n = 2.0

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 15 of 55 + = Third Order R 2 = k [HgCl 2 ] 2 [C 2 O 4 2- ] 2 First order Example Second order 2

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 16 of Zero-Order Reactions A → products R rxn = k [A] 0 R rxn = k [k] = mol L -1 s -1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 17 of 55 Integrated Rate Law -[A] t + [A] 0 = kt [A] t = [A] 0 - kt ΔtΔt -Δ[A] dt = k -d[A] Move to the infinitesimal = k And integrate from 0 to time t

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 18 of 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 ] dt = -kt ln [A] t [A] 0 ln[A] t = -kt + ln[A] 0 k = [s -1 ]

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 19 of 55 First-Order Reactions

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 20 of 55 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 k =

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 21 of 55 Half-Life Bu t OOBu t (g) → 2 CH 3 CO(g) + C 2 H 4 (g)

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 22 of 55 Some Typical First-Order Processes

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 23 of Second-Order Reactions Rate law where sum of exponents m + n + … = 2. A → products = kt + 1 [A] 0 [A] t 1 dt = -k[A] 2 d[A] [k] = M -1 s -1 = L mol -1 s -1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 24 of 55 Second-Order Reaction

Summary of the orders Zero-orderFirst-orderSecond-order General Chemistry: Chapter 15Slide 25 of 55

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 26 of 55 Pseudo First-Order Reactions Simplify the kinetics of complex reactions Rate laws become easier to work with. CH 3 CO 2 C 2 H 5 + H 2 O → CH 3 CO 2 H + C 2 H 5 OH If the concentration of water does not change appreciably during the reaction. –Rate law appears to be first order. Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 27 of 55 Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 28 of 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 29 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 30 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 31 of Theoretical Models for Chemical Kinetics Kinetic-Molecular theory can be used to calculate the collision frequency. –In gases 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

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 32 of 55 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 is: –The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 33 of 55 Activation Energy

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 34 of 55 Kinetic Energy

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 35 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 36 of 55 Collision Theory

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 37 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 38 of 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 = + ln A R -Ea T 1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 39 of 55 Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2 x 10 4 K R -Ea-Ea -E a = 1.0 x10 2 kJ mol -1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 40 of 55 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 k2k2 k1k1 T1T1 1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 41 of 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 42 of 55 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 43 of 55 A Rate Determining Step

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 44 of 55 Slow Step Followed by a Fast Step H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) dt = 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) dt = k[H 2 ][ICl] d[HI] dt = k[HI][ICl] d[I 2 ] dt = k[H 2 ][ICl] d[P]

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 45 of 55 Slow Step Followed by a Fast Step

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 46 of 55 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: dt = 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 dt = k 2 [NO] 2 [O 2 ] d[NO 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 ]

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 47 of 55 The Steady State Approximation dt = k 1 [NO] 2 – k 2 [N 2 O 2 ] – k 3 [N 2 O 2 ][O 2 ] = 0 d[N 2 O 2 ] N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 2NO(g) N 2 O 2 (g) k -1 k1k1 2NO(g) N 2 O 2 (g) N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 N 2 O 2 (g) 2NO(g) k2k2 k1k1 2NO(g) N 2 O 2 (g) dt = k 3 [N 2 O 2 ][O 2 ] d[NO 2 ]

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 48 of 55 The Steady State Approximation dt = k 1 [NO] 2 – k 2 [N 2 O 2 ] – k 3 [N 2 O 2 ][O 2 ] = 0 d[N 2 O 2 ] k 1 [NO] 2 = [N 2 O 2 ](k 2 + k 3 [O 2 ]) k 1 [NO] 2 [N 2 O 2 ] = (k 2 + k 3 [O 2 ]) dt = k 3 [N 2 O 2 ][O 2 ] d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = (k 2 + k 3 [O 2 ])

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 49 of 55 Kinetic Consequences of Assumptions dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = (k 2 + k 3 [O 2 ]) N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 N 2 O 2 (g) 2NO(g) k2k2 k1k1 2NO(g) N 2 O 2 (g) dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = ( k 3 [O 2 ]) k 1 [NO] 2 = dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = ( k 2 ) [NO] 2 [O 2 ] = k1k3k1k3 k2k2 Let k 2 << k 3 Let k 2 >> k 3 Or

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 50 of 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.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 51 of Catalysis

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 52 of 55 Catalysis on a Surface

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 53 of 55 Enzyme Catalysis E + S = ES k1k1 k -1 ES → E + P k2k2

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 54 of 55 Saturation Kinetics E + S = ES k1k1 k -1 → E + P k2k2 dt = k 1 [E][S] – k -1 [ES] – k 2 [ES]= 0 d[P] dt = k 2 [ES] d[P] k 1 [E][S] = (k -1 +k 2 )[ES] [E] = [E] 0 – [ES] k 1 [S]([E] 0 –[ES]) = (k -1 +k 2 )[ES] (k -1 +k 2 ) + k 1 [S] k 1 [E] 0 [S] [ES] =

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 55 of 55 Michaelis-Menten dt = d[P] (k -1 +k 2 ) + k 1 [S] k 1 k 2 [E] 0 [S] dt = d[P] (k -1 +k 2 ) + [S] k 2 [E] 0 [S] k1k1 dt = d[P] K M + [S] k 2 [E] 0 [S] dt = d[P] k 2 [E] 0 dt = d[P] KMKM k2k2 [E] 0 [S]

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 56 of 55 Chapter 15 Questions Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.