Petrucci Harwood Herring Madura GENERAL CHEMISTRY Ninth Edition

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

Contents 14 -1 The Rate of a Chemical Reaction 14 -2 Measuring Reaction Rates 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 2 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

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 Δt = 38. 5 s [Fe 2+] = 0. 0010 M Δ[Fe 2+] = (0. 0010 – 0) M Rate of formation of Fe 2+= 4 Δ[Fe 2+] Δt = 0. 0010 M 38. 5 s General Chemistry: Chapter 14 = 2. 6 10 -5 M s-1 Prentice-Hall © 2007

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

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

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 + …. 7 General Chemistry: Chapter 14 Prentice-Hall © 2007

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 Hg. Cl 2 and C 2 O 22 - and also the overall order of the reaction. 8 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 9 General Chemistry: Chapter 14 Prentice-Hall © 2007

EXAMPLE 14 -3 R 3 = k [Hg. Cl 2]3 m [C 2 O 42 -]3 n R 2 = k [Hg. Cl 2]2 m [C 2 O 42 -]2 n = k (2[Hg. Cl 2]3)m [C 2 O 42 -]3 n R 2 = R 3 k [Hg. Cl 2]3 m [C 2 O 42 -]3 n k 2 m [Hg. Cl 2]3 m [C 2 O 42 -]3 n R 2 2 m. R 3 = = = 2. 0 m 2 n R 3 k [Hg. Cl 2]3 [C 2 O 4 ]3 R 3 2 m = 2. 0 therefore m = 1. 0 10 General Chemistry: Chapter 14 Prentice-Hall © 2007

EXAMPLE 14 -3 R 2 = k [Hg. Cl 2]21 [C 2 O 42 -]2 n = k (0. 105) (0. 30)n R 1 = k [Hg. Cl 2]11 [C 2 O 42 -]1 n = k (0. 105) (0. 15)n R 2 = R 1 k (0. 105) (0. 30)n k (0. 105) (0. 15)n R 2 = R 1 (0. 30)n (0. 15)n = 2 n 7. 1 10 -5 = = 3. 94 -5 1. 8 10 2 n = 3. 94 therefore n = 2. 0 11 General Chemistry: Chapter 14 Prentice-Hall © 2007

EXAMPLE 14 -3 R 2 = k [Hg. Cl 2]21 [C 2 O 42 -]2 2 First order + 12 Second order General Chemistry: Chapter 14 = Third Order Prentice-Hall © 2007

14 -4 Zero-Order Reactions A → products Rrxn = k [A]0 Rrxn = k [k] = mol L-1 s-1 13 General Chemistry: Chapter 14 Prentice-Hall © 2007

Integrated Rate Law -Δ[A] Δt = k Move to the -d[A] infinitesimal dt = k And integrate from 0 to time t t [A]t - d[A] [A]0 = k dt 0 -[A]t + [A]0 = kt [A]t = [A]0 - kt 14 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

First-Order Reactions 16 General Chemistry: Chapter 14 Prentice-Hall © 2007

Half-Life ¨ t½ is the time taken for one-half of a reactant to be consumed. ln ln [A]t [A]0 = -kt ½[A]0 = -kt½ [A]0 - ln 2 = -kt½ ln 2 0. 693 t½ = = k k 17 General Chemistry: Chapter 14 Prentice-Hall © 2007

Half-Life But. OOBut(g) → 2 CH 3 CO(g) + C 2 H 4(g) 18 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

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

Second-Order Reaction 21 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

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

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. 24 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

14 -8 Theoretical Models for Chemical Kinetics Collision Theory ¨ Kinetic-Molecular theory can be used to calculate the collision frequency. · In gases 1030 collisions per second. · If each collision produced a reaction, the rate would be about 106 M s-1. · Actual rates are on the order of 104 M s-1. ◦ Still a very rapid rate. · Only a fraction of collisions yield a reaction. 26 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 27 General Chemistry: Chapter 14 Prentice-Hall © 2007

Activation Energy 28 General Chemistry: Chapter 14 Prentice-Hall © 2007

Kinetic Energy 29 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 30 General Chemistry: Chapter 14 Prentice-Hall © 2007

Collision Theory 31 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 32 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

Arrhenius Plot N 2 O 5(CCl 4) → N 2 O 4(CCl 4) + ½ O 2(g) -Ea R = -1. 2 104 K -Ea = 1. 0 102 k. J mol-1 34 General Chemistry: Chapter 14 Prentice-Hall © 2007

Arrhenius Equation k= ln R T + ln A -Ea 1 ln k 2– ln k 1 = 35 -Ea 1 ln k = Ae-Ea/RT R k 2 k 1 = 1 -E a + ln A - ln A T 2 R T 1 -Ea 1 R T 2 - 1 T 1 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 36 General Chemistry: Chapter 14 Prentice-Hall © 2007

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. 37 General Chemistry: Chapter 14 Prentice-Hall © 2007

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

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

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

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

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. 42 General Chemistry: Chapter 14 Prentice-Hall © 2007

14 -5 Catalysis 43 General Chemistry: Chapter 14 Prentice-Hall © 2007

Catalysis on a Surface 44 General Chemistry: Chapter 14 Prentice-Hall © 2007

Enzyme Catalysis E+S 45 k 1 k-1 ES k 2 ES → E + P General Chemistry: Chapter 14 Prentice-Hall © 2007