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Chemical Kinetics 14 Chemical Kinetics 14

Chemical Kinetics CONTENTS 14 -1 14 -2 Measuring Reaction Rates 14 -3 Effect of Chemical Kinetics CONTENTS 14 -1 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 Slide 2 of 56 The Rate of a Chemical Reaction Kinetics: A Summary General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Chemical Kinetics CONTENTS 14 -8 14 -9 The Effect of Temperature on Reaction Rates Chemical Kinetics CONTENTS 14 -8 14 -9 The Effect of Temperature on Reaction Rates 14 -10 Reaction Mechanisms 14 -11 Slide 3 of 56 Theoretical Models for Chemical Kinetics Catalysis General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

How fast a reaction occurs depends on the reactions mechanism- the step by step How fast a reaction occurs depends on the reactions mechanism- the step by step molecular events leading from reactants to products. . Slide 4 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Elementary Reactions A mechanism is a description of the actual molecular events that occur Elementary Reactions A mechanism is a description of the actual molecular events that occur during a chemical reaction. Each such event is an elementary reaction. Elementary reactions involve one, two or occasionally three (rare) reactants molecules or atoms.

The nucleophile approaches the carbon from the side directly opposite the leaving group : The nucleophile approaches the carbon from the side directly opposite the leaving group : a backside attack. As the reaction progresses the bonding orbital between the carbon and leaving group weakens. As the leaving group is pushed away, the carbon atom has its configuration turned inside out, becoming inverted. This mechanisms only involves one step, proceeding through a high energy, short lasting transition state.

Chemical kinetics concerns how rate of chemical reactions are measured. Some of the most Chemical kinetics concerns how rate of chemical reactions are measured. Some of the most important information about a mechanism comes from experiments that determine how fast a chemical reaction occurs under various conditions. Kinetic studies are essential to explorations of reaction mechanisms. Both Concentration and Temperature affect the rate of a chemical reaction. 7

 • Reaction rates and reaction order are determined experimentally. • Reaction rates and • Reaction rates and reaction order are determined experimentally. • Reaction rates and reaction order are critical to understanding how the reaction progresses and can be controlled.

Rate, or speed, refers to something that happens in a unit of time. A Rate, or speed, refers to something that happens in a unit of time. A car traveling at 60 mph, for example, covers a distance of 60 miles in one hour. For chemical reactions, the rate of reaction describes how fast the concentration of a reactant or product changes with time. Slide 9 of 56

The Rate of a Chemical Reaction Rate of change of concentration with time. 2 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 s Δ[Fe 2+] = (0. 0010 – 0) M Δ[Fe 2+] 0. 0010 M Rate of formation of Fe 2+= = 2. 6 10 -5 M s-1 Δt 38. 5 s

2 Fe 3+(aq) + Sn 2+ → 2 Fe 2+(aq) + Sn 4+(aq) 1 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

General Rate of Reaction a A + b B → g G + h General Rate of Reaction a A + b B → g G + h H Rate of reaction = negative of rate of disappearance of reactants 1 Δ[B] 1 Δ[A] = = b Δt a Δt = rate of appearance of products 1 Δ[H] 1 Δ[G] = = h Δt g Δt

Measuring Reaction Rates by measuring the changes in concentration over time H 2 O Measuring Reaction Rates by measuring the changes in concentration over time H 2 O 2(aq) → H 2 O(l) + ½ O 2(g) Measure the rate by monitoring volume of O 2 formed, Or, from time to time by chemical analysis of samples of the reaction mixture for their H 2 O 2 content 2 Mn. O 4 -(aq) + 5 H 2 O 2(aq) + 6 H+ → 2 Mn 2+ + 8 H 2 O(l) + 5 O 2(g) EXPERIMENTAL SETUP FOR DETERMINING THE RATE OF DECOMPOSITION OF H 2 O 2

or by chemical analysis of aliquots 2 Mn. O 4 -(aq) + 5 H or by chemical analysis of aliquots 2 Mn. O 4 -(aq) + 5 H 2 O 2(aq) + 6 H+ → 2 Mn 2+ + 8 H 2 O(l) + 5 O 2(g)

Initial Rate of Reaction -(-2. 32 M / 1360 s) = 1. 71 10 Initial Rate of Reaction -(-2. 32 M / 1360 s) = 1. 71 10 -3 M s-1 -(-1. 7 M / 2800 s) = Rate = -Δ[H 2 O 2] Δt of Reaction at time t 6. 1 10 -4 M s-1 THE RATE OF THE REACTION IS DETERMINED FROM THE SLOPE OF A TANGENT LINE TO A CONCENTRATION-TIME CURVE, THIS IS INSTANTANEOUS RATE OF REACTION.

The Rate at which a chemical reaction proceeds is typically influenced by the amount The Rate at which a chemical reaction proceeds is typically influenced by the amount of each reactant present and the temperature of the reaction vessel. And, typically, this relationship between the Reaction Rate and Reagent Concentration is known as the Rate Law.

Effect of Concentration on Reaction Rates: The Rate Law or Rate Equation a A Effect of Concentration on Reaction Rates: The Rate Law or Rate Equation 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 + …. The terms [A] and [B] represent reactant molarities. The required exponents, m, n, …are generally small, positive whole numbers, although in some cases they may be zero, fractional, or negative. They must be determined by experiment and are generally not related to stoichiometric coefficients a, b, …. That is, often m ≠ a, n ≠ b, and so on.

If a reaction has a rate equation of rate = k[A][B][C] then it is: If a reaction has a rate equation of rate = k[A][B][C] then it is: A) overall second order B) overall first order C) overall third order D) zero order in A E) second order in B Slide 18 of 56

For a reaction Rate = k[A][B]2, what factor will keep k unchanged? A) raising For a reaction Rate = k[A][B]2, what factor will keep k unchanged? A) raising temperature B) adding inhibitor C) increasing [A] D) adding catalyst

The reaction has the rate law Rate = k[A][B]2. Which will cause the rate The reaction has the rate law Rate = k[A][B]2. Which will cause the rate to increase the most? A) doubling [A] B) lowering temperature C) tripling [B] D) quadrupling [A] E) doubling [B]

Establishing the Order of a reaction Method of Initial Rates Use the data provided Establishing the Order of a reaction Method of Initial Rates Use the data provided establish the order of the reaction with respect to Hg. Cl 2 and C 2 O 42 - and also the overall order of the reaction. 2 Hg. Cl 2(aq) + C 2 O 42 -(aq) 2 Cl-(aq) + 2 CO 2(g) + Hg 2 Cl 2(s) rate of reaction = k[Hg. Cl 2]m[C 2 O 42 -]n

General effect of doubling the initial concentration of a particular reactant (with other reactant General effect of doubling the initial concentration of a particular reactant (with other reactant concentrations held constant). • Zero order in the reactant—there is no effect on the initial rate of reaction. • First order in the reactant—the initial rate of reaction doubles. • Second order in the reactant—the initial rate of reaction quadruples. • Third order in the reactant—the initial rate of reaction increases eightfold. Slide 22 of 56 General Chemistry: Chapter 14

For 2 NO + O 2 → 2 NO 2, initial rate data are: For 2 NO + O 2 → 2 NO 2, initial rate data are: [NO] 0. 010 0. 030 M [O 2] 0. 010 0. 020 M rate 2. 5 5. 0 45. 0 m. M/sec The rate law is Rate = k[NO]x[O 2]y: A) x = 1, y = 2 B) x = 2, y = 1 C) x = 1, y = 1 D) x = 2, y = 2 E) x = 0, y = 2

The rate of a specific chemical reaction is independent of the concentrations of the The rate of a specific chemical reaction is independent of the concentrations of the reactants. Thus the reaction is: A) first order in A B) second order C) first order in the product D) catalyzed E) overall zero order

Another useful equation is: Integrated Rate Law -Δ[A] Δt Move to the infinitesimal = Another useful equation is: Integrated Rate Law -Δ[A] Δt Move to the infinitesimal = k -d[A] 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 = - kt + [A]0

14 -4 Zero-Order Reactions A → products Rrxn = k [A]0 (0 - [A]0) 14 -4 Zero-Order Reactions A → products Rrxn = k [A]0 (0 - [A]0) [A]0 Rrxn = k = = (tf – 0) tf [k] = mol L-1 s-1 A ZERO-ORDER REACTION: A PRODUCTS

2 N 2 O(g) → 2 N 2(g) + O 2(g) The platinum surface 2 N 2 O(g) → 2 N 2(g) + O 2(g) The platinum surface is completely covered with N 2 O molecules. • If we increase the concentration, it has no effect on the rate. • Since only those N 2 O molecules on the surface can react. • An Example of a zero-order reaction. Slide 27 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

14 -5 First-Order Reactions H 2 O 2(aq) → H 2 O(l) + ½ 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 [k] = s-1 An Integrated rate Law for a First-Order Reaction [A]t [A]0 ln Slide 28 of 56 [A]t [A]0 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 Copyright © 2011 Pearson Canada Inc.

Test for a first-order reaction: Decomposition of H 2 O 2(aq) Slide 29 of Test for a first-order reaction: Decomposition of H 2 O 2(aq) Slide 29 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

14 -6 Second-Order Reactions Rate law where sum of exponents m + n +… 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 Slide 30 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

FIGURE 14 -6 A straight-line plot for the second order reaction A Slide 31 FIGURE 14 -6 A straight-line plot for the second order reaction A Slide 31 of 56 General Chemistry: Chapter 14 products Copyright © 2011 Pearson Canada Inc.

Integrating our other simple Rate Laws into their “Linear” form provides us with: If Integrating our other simple Rate Laws into their “Linear” form provides us with: If we do not know the order of a given reaction, we can simply plot the data in all three “Linear” forms and see which results in a straight-line.

Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t.

2 N 2 O 5(g) → 2 N 2 O 4(g) + O 2(g) 2 N 2 O 5(g) → 2 N 2 O 4(g) + O 2(g) Time (sec) [N 2 O 5] (M) 0 1. 40 200 1. 24 400 1. 10 600 0. 98 800 0. 87 1000 0. 77 1200 0. 68 1400 0. 60 1600 0. 53 1800 0. 47

This reaction has been shown to be First Order in N 2 O 5; This reaction has been shown to be First Order in N 2 O 5; meaning the Rate Law can be written as: Rate = k [N 2 O 5] or in Integral form as: ln [N 2 O 5] = ln [N 2 O 5]o - k t Thus, a plot of the Natural Log of the above Concentration data vs. Time should give us a straight line with a slope = k. This is in fact the case.

Determining Orders of Reactions I) Getting the data a)Quench the reaction, measure concentrations b)For Determining Orders of Reactions I) Getting the data a)Quench the reaction, measure concentrations b)For gas phase, measure pressure vs. time c)Spectroscopically follow reactants/products Etc… II) Analyzing the data A) Reactions with one reactant: A → products a) Plot or analyze [A] vs. t ln[A] vs. t 1/[A] vs. t … Slide 36 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Zero-Order Reactions [A]t = - kt + [A]0 [k] = mol L-1 s-1 First Zero-Order Reactions [A]t = - kt + [A]0 [k] = mol L-1 s-1 First -Order Reactions ln[A]t = -kt + ln[A]0 [k] = s-1 Second -Order Reactions 1 [A]t = kt + 1 [A]0 [k] = M-1 s-1 = L mol-1 s-1

Half-Life, t½ the time taken for one-half of a reactant to be consumed. ln Half-Life, t½ 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

Plutonium 239 is a radioactive isotope used as fuel in nuclear reactors. For the Plutonium 239 is a radioactive isotope used as fuel in nuclear reactors. For the decay below, the rate constant K=5. 5× 10 -11 239 Pu → 235 U + 4 He a. Calculate the half life for 239 Pu. b. How long would it take for 60% of a 1. 35 g sample to decay? Slide 40 of 56 General Chemistry: Chapter 14

14 -9 Effect of Temperature on Reaction Rates Svante Arrhenius demonstrated that many rate 14 -9 Effect of Temperature on Reaction Rates Svante Arrhenius demonstrated that many rate constants vary with temperature according to this equation: k = Ae-Ea/RT -Ea 1 ln k = + ln. A R T

N 2 O 5(CCl 4) → N 2 O 4(CCl 4) + ½ O 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 Temperature dependence of the rate constant k for a reaction

k = Ae-Ea/RT -Ea 1 ln k = + ln A R T -Ea k = Ae-Ea/RT -Ea 1 ln k = + ln A R T -Ea 1 ln k 2– ln k 1 = + ln A - - ln A R T 2 R T 1 k 2 1 -Ea 1 ln = - k 1 T 1 R T 2

An ancient shipwreck has been discovered off Greenland. An unopened 1 liter bottle of An ancient shipwreck has been discovered off Greenland. An unopened 1 liter bottle of wine brought to the surface is opened and found to smell strongly like vinegar. The contents are analyzed later and also found to contain 1. 2 × 10 -4 moles of ethanol. It is known that ethanol decomposes to acetic acid as below: 2 CH 3 CH 2 OH → CH 3 COOH + C 2 H 4 + H 2 Ea= 121. 3 KJ/mol, k = 40. 9 M-1 month-1 @ 25 ◦ C Fermentation can only produce wine with a 3 M of ethanol. The bottle was submerged in 4 ◦ C water for centuries. How many years was the ship submerged? Slide 44 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

The active ingredient in Clearasil BP is benzoyl peroxide, (C 6 H 5 COO)2, The active ingredient in Clearasil BP is benzoyl peroxide, (C 6 H 5 COO)2, in fighting bacteria, a first order reaction occurs in which benzoyl peroxide decomposes into two high energy radicals. The t 1/2 of benzoyl peroxide at 37◦C has been shown to be about 898 min-1. a murder victim is found at 11: 30 PM on a warm summer’s night. The medical examiner isolates 1. 63× 10 -1 grams of benzoyl peroxide from the victim’s face. It is know that the victim usually put on Clearasil BP just before leaving the house. It is further determined that an average application of Clearasil BP contains about 2. 13× 10 -1 grams of benzoyl peroxide. The detectives have asked you to estimate when the murder victim left home.

14 -8 Theoretical Models for Chemical Kinetics Collision Theory In gases 1030 collisions per 14 -8 Theoretical Models for Chemical Kinetics Collision Theory In gases 1030 collisions per liter per second. If each collision produced a reaction, the rate would be about 106 M s-1, extremely rapid! Actual rates are on the order of 10 -4 M s-1. So, Only a fraction of collisions yield a reaction.

Activation Energy: The minimum energy that molecules must bring to their collisions for a Activation Energy: The minimum energy that molecules must bring to their collisions for a chemical reaction to occur. For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).

FIGURE 14 -11 An analogy for a reaction profile and activation energy FIGURE 14 -11 An analogy for a reaction profile and activation energy

For a chemical reaction to proceed at a reasonable rate, there should exist an For a chemical reaction to proceed at a reasonable rate, there should exist an appreciable number of molecules with energy equal to or greater than the activation energy. Slide 49 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Collision Theory The rate of the reaction depends on how often the molecules with Collision Theory The rate of the reaction depends on how often the molecules with sufficient kinetic energy are likely to collide with each other. If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower. Orientation of molecules may be important. As temperature increases, reaction rate increases.

Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the molecules whose molecular collisions lead to chemical reaction

Orientation of molecules may be important Molecular collisions and chemical reactions Orientation of molecules may be important Molecular collisions and chemical reactions

Transition State Theory The activated complex is a hypothetical species lying between reactants and 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. The formation of the activated Complex is a reversible process. A REACTION PROFILE FOR THE REACTION N 2 O(G) + NO(G) NO 2(G) N 2(G) +

The potential outcome of a reaction is usually influenced by two factors: the relative The potential outcome of a reaction is usually influenced by two factors: the relative stability of the products (i. e. thermodynamic factors) the rate of product formation (i. e. kinetic factors) Slide 54 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Kinetic and Thermodynamic Control • At low temperature, the reaction is under kinetic control Kinetic and Thermodynamic Control • At low temperature, the reaction is under kinetic control (rate, irreversible conditions) and the major product is that from Energy fastest reaction. • At high temperature, the reaction is under thermodynamic control (equilibrium, reversible conditions) and the major product is the more stable one. Reaction Coordinate P 1 is Favored under Low Temp. and short time P 2 is Favored under High Temp. and long time 3/19/2018

Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the molecules whose molecular collisions lead to chemical reaction

According to the collision theory in gaseous molecules, collision frequency is ____ and reaction According to the collision theory in gaseous molecules, collision frequency is ____ and reaction rate is ____ because ____. A) low, molecules are so far apart B) high, each collision results in a reaction C) low, molecules must collide before they can react D) high, relatively low, only a fraction of the collisions lead to a reaction E) low, high, molecules are moving so fast that each reaction causes many others

In the Arrhenius equation, ln k = -Ea/RT + ln A, the symbol A In the Arrhenius equation, ln k = -Ea/RT + ln A, the symbol A denotes: A) the initial concentration of A B) the activation energy C) the rate constant D) a constant that represents the frequency of collisions with the proper orientation and other steric conditions favorable for a reaction E) the absolute temperature

14 -10 Reaction Mechanisms Step-by-step description of a reaction. Each step is called an 14 -10 Reaction Mechanisms Step-by-step description of a reaction. Each step is called an elementary process. Any molecular event that significantly alters a molecule’s energy or geometry or produces a new molecule. Reaction mechanism must be consistent with: 1. Stoichiometry for the overall reaction. 2. The experimentally determined rate law.

A mechanism with a Slow Step Followed by a Fast Step H 2(g) + A mechanism with a 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: slow H 2(g) + ICl(g) HI(g) + HCl(g) fast HI(g) + ICl(g) I 2(g) + HCl(g) H 2(g) + 2 ICl(g) → I 2(g) + 2 HCl(g) d[HI] dt d[I 2] dt d[P] dt = k[H 2][ICl] = k[HI][ICl] = k[H 2][ICl]

FIGURE 14 -14 A reaction profile for a two-step mechanism Slide 61 of 56 FIGURE 14 -14 A reaction profile for a two-step mechanism Slide 61 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Molecularity: Molecularity is the number of molecules that need to collide, and in one Molecularity: Molecularity is the number of molecules that need to collide, and in one step form the products. For single step elementary reactions, Molecularity = Order A → products; 1 st order rate = k[A] Unimolecular 2 A → products; 2 nd order rate = k[A]2 Bimolecular A + B → prod. ; 2 nd order rate = k[A][B] Bimolecular A + B + C → prod. 3 rd order rate = k[A][B][C] Termolecular

Elementary Processes Unimolecular or bimolecular. Exponents for concentration terms are the same as the Elementary Processes Unimolecular or bimolecular. Exponents for concentration terms are the same as the stoichiometric factors for the elementary process. Intermediates are produced in one elementary process and consumed in another. Intermediates must not appear in the overall equation or the overall rate law. One elementary step is usually slower than all the others and is known as the rate determining step.

Choose the INCORRECT statement. A) The rate-determining step is always the first step. B) Choose the INCORRECT statement. A) The rate-determining step is always the first step. B) A unimolecular process is one in which a single molecule dissociates. C) A bimolecular process is one involving a collision of two molecules. D) A reaction mechanism is a step-by-step detailed description of a chemical reaction. E) An elementary process is a step in the mechanism. Slide 64 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

What is the rate law for the following mechanism? N 2 O + NO What is the rate law for the following mechanism? N 2 O + NO → N 2 ONO (Slow) N 2 ONO → N 2 + NO 2 (Fast) A) Rate = k[N 2 O] B) Rate = k[NO] C) Rate = k[N 2 O][NO] D) Rate = k[N 2][NO 2] E) Rate = k[N 2 ONO]

What is the rate law for the following mechanism? CH 3 COOC 2 H What is the rate law for the following mechanism? CH 3 COOC 2 H 5 + H 2 O → CH 3 COOC 2 H 6+ + OH- (Slow) CH 3 COOC 2 H 6+ → CH 3 COOH + C 2 H 5+ + OH- → C 2 H 5 OH (Fast) A) Rate = k[CH 3 COOC 2 H 5][H 2 O]2 B) Rate = k[C 2 H 5 OH] C) Rate = k[CH 3 COOH] D) Rate = k[CH 3 COOC 2 H 5] E) Rate = k[CH 3 COOC 2 H 5][H 2 O] (Fast)

What is the rate law for the following reaction and its mechanism? 2 O What is the rate law for the following reaction and its mechanism? 2 O 3 → 3 O 2 (overall reaction) O 3 → O 2 + O. (Slow) O∙ + O 3 → 2 O 2 (Fast) A) Rate = k[O 3] B) Rate = k[O 3]2 C) Rate = k[O 3]2/[O 2] D) Rate = k[O 3]/[O 2] E) Rate = k[O 3][O 2]

 What is the rate law for the following reaction and its mechanism? 2 What is the rate law for the following reaction and its mechanism? 2 Hg. Cl 2 + C 2 O 42 - → 2 Cl- + 2 CO 2 + Hg 2 Cl 2 (overall reaction) Hg. Cl 2 + C 2 O 42 - ⇌ Hg. Cl 2 C 2 O 42(Fast) Hg. Cl 2 C 2 O 42 - + C 2 O 42 - → Hg + 2 C 2 O 4 Cl 2 - (Slow) Hg + Hg. Cl 2 → Hg 2 Cl 2 (Fast) 2 C 2 O 4 Cl 2 - → C 2 O 42 - + 2 Cl- + 2 CO 2 (Fast) A) Rate = k[Hg. Cl 2][C 2 O 42 -] B) Rate = k[Hg. Cl 2]2[C 2 O 42 -] C) Rate = k[Hg 2 Cl 2] D) Rate = k[Hg. Cl 2][C 2 O 42 -]2 E) Rate = k[Hg. Cl 2]2[C 2 O 42 -]2 Slide 68 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Substitution: One functional group replaces another Alkyl halides react with a nucleophile to give Substitution: One functional group replaces another Alkyl halides react with a nucleophile to give a substituted product CH 3–Cl + Na. OH → CH 3–OH + Na. Cl • Nucleophile: A species with an unshared electron pair; a reagent that seeks a positive charge. Examples: HO-, RO-, CN-, CH 3 S-, : NH 3

 • Leaving Group: For a molecule to be reactive to substitution the leaving • Leaving Group: For a molecule to be reactive to substitution the leaving group must be a good one. • A good leaving group must be able to leave as a relatively stable molecule, or ion. Examples: Cl-, Br-, I-, H 2 O Slide 70 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Nucleophilic Substitution Reactions In a nucleophilic substitution reaction, the bond (in blue) between carbon Nucleophilic Substitution Reactions In a nucleophilic substitution reaction, the bond (in blue) between carbon and the leaving group is broken and a new bond (in red) is formed by using a lone pair from the nucleophile.

Chemical research, starting in the 1890 s, has shown that nucleophilic substitution reactions can Chemical research, starting in the 1890 s, has shown that nucleophilic substitution reactions can involve two types of mechanisms. 1. Carbon–Halogen bond breaks at the same time the carbon–nucleophile bond is formed. Everything happens in one step! Slide 72 of 69 General Chemistry: Chapter 27

The nucleophile approaches the carbon from the side directly opposite the leaving group : The nucleophile approaches the carbon from the side directly opposite the leaving group : a backside attack. As the reaction progresses the bonding orbital between the carbon and leaving group weakens. As the leaving group is pushed away, the carbon atom has its configuration turned inside out, becoming inverted. This mechanisms only involves one step, proceeding through a high energy, short lasting transition state. Slide 73 of 69

Reaction profile for an SN 2 reaction Reaction profile for an SN 2 reaction

Consider the nucleophilic substitution of Br- by Cl- in the series of compounds to Consider the nucleophilic substitution of Br- by Cl- in the series of compounds to the right. If the reaction proceeds via an SN 2 mechanism, predict the order of increasing rate of reaction. A) B) 1. A < B < C < D 2. D < C < B < A C) 3. B < C < D < A 4. A < D < C < B Copyright © 2011 Pearson Canada Inc. D) Slide 76 of 33

Consider the nucleophilic substitution of Brby Cl- in the series of compounds to the Consider the nucleophilic substitution of Brby Cl- in the series of compounds to the right. If the reaction proceeds via an SN 2 mechanism, predict the order of increasing rate of reaction. A) B) 1. A < B < C < D 2. D < C < B < A C) 3. B < C < D < A 4. A < D < C < B D) Copyright © 2011 Pearson Canada Inc. Slide 77 of 33

Carbon–Halogen bond breaks first, then the nucleophile attacks. Happens in 2 steps! Slide 79 Carbon–Halogen bond breaks first, then the nucleophile attacks. Happens in 2 steps! Slide 79 of 69 General Chemistry: Chapter 27 3/19/2018 Copyright © 2011 Pearson Canada Inc.

The SN 1 Mechanism The SN 1 Mechanism

Reaction profile for the reaction between t-butyl bromide and water Slide 81 of 69 Reaction profile for the reaction between t-butyl bromide and water Slide 81 of 69 General Chemistry: Chapter 27 Copyright © 2011 Pearson Canada Inc.

Consider the nucleophilic substitution of Br- by OH- occurring in an aqueous solution for Consider the nucleophilic substitution of Br- by OH- occurring in an aqueous solution for the series of compounds to the right. A) If the reaction proceeds via an SN 1 mechanism, predict the order of increasing rate of reaction. B) 1. A < B < C < D C) 2. D < C < B < A 3. B < C < D < A 4. A < D < C < B Copyright © 2011 Pearson Canada Inc. D) Slide 82 of 33

Consider the nucleophilic substitution of Br- by OHoccurring in an aqueous solution for the Consider the nucleophilic substitution of Br- by OHoccurring in an aqueous solution for the series of compounds to the right. If the reaction proceeds via an SN 1 mechanism, predict the order of increasing rate of reaction. A) B) 1. A < B < C < D 2. D < C < B < A C) 3. B < C < D < A 4. A < D < C < B D) Copyright © 2011 Pearson Canada Inc. Slide 83 of 33

Relative rates of reaction SN 1 reaction Slide 85 of 69 Relative rates of reaction SN 1 reaction Slide 85 of 69

SN 2 reaction Slide 86 of 69 General Chemistry: Chapter 27 Copyright © 2011 SN 2 reaction Slide 86 of 69 General Chemistry: Chapter 27 Copyright © 2011 Pearson Canada Inc.

Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the Distribution of molecular kinetic energy Distribution of molecular kinetic energies These are “activated”; the molecules whose molecular collisions lead to chemical reaction

14 -5 Catalysis • Alternative reaction pathway of lower energy. • Homogeneous catalysis. All 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.

14 -5 Catalysis AN EXAMPLE OF HOMOGENEOUS CATALYSIS 14 -5 Catalysis AN EXAMPLE OF HOMOGENEOUS CATALYSIS

Catalysis on a Surface Heterogeneous catalysis in the reaction 2 CO + 2 NO Catalysis on a Surface Heterogeneous catalysis in the reaction 2 CO + 2 NO 2 CO 2 + N 2 Figure 14 -19 Reaction profile for a surface-catalyzed reaction

Slide 91 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc. Slide 91 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

k 1 E + S ES k-1 k 2 ES → E + P k 1 E + S ES k-1 k 2 ES → E + P Lock-and-key model of enzyme action Slide 92 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

A catalyst: I) lowers activation energy II) provides an alternate reaction pathway III) is A catalyst: I) lowers activation energy II) provides an alternate reaction pathway III) is consumed in the reaction and therefore does not appear in the chemical equation of each mechanism IV) speeds a reaction V)is heterogeneous if it is in a different phase than the reactants A) I, III, and IV B) I, IV, and V C) II, III, and IV D) II and IV E) I, IV, and V

If a catalyst is added to a reaction: I) the value of k is If a catalyst is added to a reaction: I) the value of k is increased II) the value of k is decreased III) the rate is increased IV) the rate is decreased V)neither rate nor the constant are changed, only the order A) I and IV B) II and IV C) II and III D) I and III E) V only

A factor that decreases the activation energy for a reaction: I) decreases the rate A factor that decreases the activation energy for a reaction: I) decreases the rate constant II) increases the rate constant III) has no effect on the rate constant IV) makes the product yield increase V) might be a catalyst A) I and IV B) II and IV C) I, IV, and V D) IV and III E) II and V Slide 95 of 56 General Chemisry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Enzymes as Catalysts t Enzymes as Catalysts t

k 1 k 2 E + S ES → E + P d[P] k-1 k 1 k 2 E + S ES → E + P d[P] k-1 dt d[P] dt = k 2[ES] = k 1[E][S] – k-1[ES] – k 2[ES]= 0 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[E]0 [S] (k-1+ k 2 ) + k 1[S] FIGURE 14 -21 Effect of substrate concentration on the rate of an enzyme reaction. Slide 97 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

k 1 k 2 E + S ES → E + P k-1 d[P] k 1 k 2 E + S ES → E + P k-1 d[P] dt d[P] = k 2[E]0 dt d[P] dt = = k 1 k 2[E]0 [S] (k-1+ k 2 ) + k 1[S] k 2[E]0 [S] (k-1+ k 2 ) + [S] k 1 d[P] dt = k 2 [E]0 [S] KM d[P] dt = k 2[E]0 [S] KM + [S] FIGURE 14 -21 Effect of substrate concentration on the rate of an enzyme reaction. Slide 98 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

End of Chapter Questions Dimensional Analysis is your friend. Never leave units off of End of Chapter Questions Dimensional Analysis is your friend. Never leave units off of a number. You are better off leaving off the numerical part of the number and working ONLY with the units. The units must correctly cancel out. The units left after that process must be the correct units for your answer. Only then should you calculate. Slide 99 of 56 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Kinetic equations [A]t = - kt + [A]0 ln[A]t = -kt + ln[A]0 1 Kinetic equations [A]t = - kt + [A]0 ln[A]t = -kt + ln[A]0 1 1 = kt + [A]t [A]0 R = 0. 082057 L atm mol-1 K-1 R = 8. 3145 m 3 Pa mol-1 K-1 R = 8. 3145 J mol-1 K-1 k = Ae-Ea/RT k 1 1 -Ea 1 ln = - k 2 T 2 R T 1

1. At low temperature, the reaction preferentially proceeds along the green path to P 1. At low temperature, the reaction preferentially proceeds along the green path to P 1 and stops since they lack sufficient energy to reverse to SM, i. e. it is irreversible, so the product ratio of the reaction is dictated by the rates of formation of P 1 and P 2, k 1: k 2. At some slightly higher temperature, reaction 1 will become reversible while reaction 2 remains irreversible. So although P 1 may form initially, over time it will revert to SM and react to give the more stable P 2. 3. At high temperature, both reaction 1 and 2 are reversible and the product ratio of the reaction is 3/19/2018