14 Chemical Kinetics Chemistry 140 Fall 2002

14 Chemical Kinetics Chemistry 140 Fall 2002
Rocket fuel is designed to give a rapid release of gaseous products and energy to provide a rocket maximum thrust. Milk is stored in a refrigerator to slow down the chemical reactions that cause it to spoil. Current strategies to reduce the rate of deterioration of the ozone layer try to deprive the ozone-consuming reaction cycle of key intermediates that come from chlorofluorocarbons (CFCs). Catalysts are used to reduce the harmful emissions from internal combustion engines that contribute to smog. These examples illustrate the importance of the rates of chemical reactions. Moreover, how fast a reaction occurs depends on the reaction mechanism—the step-by-step molecular pathway leading from reactants to products. Thus, chemical kinetics concerns how rates of chemical reactions are measured, how they can be predicted, and how reaction rate data are used to deduce probable reaction mechanisms.

General Chemistry: Chapter 14
Chemistry 140 Fall 2002 Chemical Kinetics 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 Although stable at room temperature, ammonium dichormate decomposes very rapidly once ignited: The rates of chemical reactions and the effect of temperature on those rates are among several key concepts explored in this chapter. (NH4)2Cr2O7  N2(g) + 4H2O(g) + Cr2O3(s) General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Chemistry 140 Fall 2002 Chemical Kinetics Contents 14-8 Theoretical Models for Chemical Kinetics 14-9 The Effect of Temperature on Reaction Rates 14-10 Reaction Mechanisms 14-11 Catalysis The dissolving of a cube of sugar (sucrose) is seen here as swirls of a higher-density sucrose solution falling through the lower-density water. In this chapter we will explore several solution properties whose values depend on solution concentration. Our emphasis will be on describing solution phenomena and their applications and explaining these phenomena at the molecular level. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

How fast a reaction occurs depends on the reactions mechanism- the step by step molecular pathway leading from reactants to products. Chemical kinetics concerns how rate of chemical reactions are measured, how they can be predicted and how reaction rate data are used to deduce probable reaction mechanism. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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. General Chemistry: Chapter 12 Copyright © 2011 Pearson Canada Inc.

(NH4)2Cr2O7  N2(g) + 4H2O(g) + Cr2O3(s)
Although stable at room temperature, ammonium dichormate decomposes very rapidly once ignited: The rates of chemical reactions and the effect of temperature on those rates are among several key concepts explored in this chapter. (NH4)2Cr2O7  N2(g) + 4H2O(g) + Cr2O3(s) General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Rate, or speed, refers to something that happens in a unit of time
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.

The Rate of a Chemical Reaction
Chemistry 140 Fall 2002 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+] = M Δt = 38.5 s Δ[Fe2+] = ( – 0) M 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. Rate of formation of Fe2+= = = 2.610-5 M s-1 Δ[Fe2+] Δt M 38.5 s

2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
Δt Δ[Fe2+] Δt = 1 2 Δ[Fe3+] Δt = - 1 2

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

Chemistry 140 Fall 2002 Measuring Reaction Rates by measuring the changes in concentration over time H2O2(aq) → H2O(l) + ½ O2(g) Measure the rate by monitoring volume of O2 formed, Or, from time to time by chemical analysis of samples of the reaction mixture for their H2O2 content 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ → 2 Mn H2O(l) + 5 O2(g) Experimental setup for determining the rate of decomposition of H2O2

2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ → 2 Mn2+ + 8 H2O(l) + 5 O2(g)
or by chemical analysis of aliquots 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ → 2 Mn H2O(l) + 5 O2(g)

Initial Rate of Reaction
Chemistry 140 Fall 2002 Initial Rate of Reaction Rate = -Δ[H2O2] Δt -(-2.32 M / 1360 s) = 1.71  10-3 M s-1 Rate of Reaction at time t -(-1.7 M / 2800 s) = 6.1  10-4 M s-1 Graphical representation of kinetic data for the reaction H2O2(aq) → H2O(l) + ½ O2(g);The reaction rate is not constant; the lower the remaining concentration of H2O2, the more slowly the reaction proceeds. 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 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 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. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Chemistry 140 Fall 2002 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. 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: A) overall second order B) overall first order C) overall third order D) zero order in A E) second order in B General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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 to increase the most? A) doubling [A] B) lowering temperature C) tripling [B] D) quadrupling [A] E) doubling [B] General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

If increasing the concentration of A in a chemical reaction causes no increase in the rate of the reaction, then we may say: A) A is a catalyst B) the reaction rate is zero order in A C) the reaction rate is zero order in [A] D) the reaction rate is first order in [A] E) A is not involved in the reaction General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Establishing the Order of a reaction Method of Initial Rates
Chemistry 140 Fall 2002 Establishing the Order of a reaction Method of Initial Rates Use the data provided establish the order of the reaction with respect to HgCl2 and C2O42- and also the overall order of the reaction. 2 HgCl2(aq) + C2O42-(aq) Cl-(aq) + 2 CO2(g) + Hg2Cl2(s) rate of reaction = k[HgCl2]m[C2O42-]n As its name implies, this method requires us to work with initial rates of reaction. As an example, let’s look at a specific reaction: that between mercury(II) chloride and oxalate ion. We can follow the reaction by measuring the quantity of HgCl2 formed as a function of time. Some representative data are given in Table 14.3, which we can assume are based on either the rate of formation of Hg2Cl2 or the rate of disappearance of C2O42-. Example 4.3 uses these data to illustrate the 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 HgCl2 and C2O2 2- and also the overall order of the reaction.

Chemistry 140 Fall 2002 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. We made an important observation in Example 14-3: If a reaction is first order in one of the reactants, doubling the initial concentration of that reactant causes the initial rate of reaction to double. Following is the general effect of doubling the initial concentration of a particular reactant (with other reactant concentrations held constant). General Chemistry: Chapter 14

Chemistry 140 Fall 2002 For 2NO + O2 → 2NO2, initial rate data are: [NO] M [O2] M rate mM/sec The rate law is Rate = k[NO]x[O2]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 B General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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
dt = k -d[A] Move to the infinitesimal = k Δt And integrate from 0 to time t - dt = k d[A]  [A]0 [A]t t -[A]t + [A]0 = kt [A]t = - kt + [A]0 Copyright © 2011 Pearson Canada Inc.

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 FIGURE 14-3 A zero-order reaction: A products

14-5 First-Order Reactions
Chemistry 140 Fall 2002 14-5 First-Order Reactions H2O2(aq) → H2O(l) + ½ O2(g) d[H2O2 ] = -k[H2O2] [k] = s-1 dt An Integrated rate Law for a First-Order Reaction = - k dt [H2O2] d[H2O2 ]  [A]0 [A]t t = -kt ln [A]t [A]0 ln[A]t = -kt + ln[A]0 General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Chemistry 140 Fall 2002 Test for a first-order reaction: Decomposition of H2O2(aq) This figure is used in Example 14-5 An easy test for a first-order reaction is to plot the natural logarithm of a reactant concentration versus time and see if the graph is linear. The data from Table 14.1 are plotted in Figure 14-4, and the rate constant k is derived from the slope of the line. An alternative, nongraphical approach, illustrated in Practice Example 14-5B, is to substitute data points into equation (14.13) and solve for k. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

14-6 Second-Order Reactions
Chemistry 140 Fall 2002 14-6 Second-Order Reactions Rate law where sum of exponents m + n +… = 2 A → products dt = -k[A]2 d[A] [k] = M-1 s-1 = L mol-1 s-1 dt = - k d[A] [A]2  [A]0 [A]t t = kt + 1 [A]0 [A]t General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

A straight-line plot for the second order reaction A products
Chemistry 140 Fall 2002 The reciprocal of the concentration, 1/[A], is plotted against time. As the reaction proceeds, [A] decreases and 1/[A] increases in a linear fashion. The slope of the line is the rate constant k. FIGURE 14-6 A straight-line plot for the second order reaction A products General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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 1/[A] vs t.

2 N2O5(g) → 2 N2O4(g) + O2(g) Time (sec) [N2O5] (M)

This reaction has been shown to be First Order in N2O5; meaning the Rate Law can be written as:
Rate = k [N2O5] or in Integral form as: ln [N2O5] = ln [N2O5]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 Quench the reaction, measure concentrations For gas phase, measure pressure vs. time 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 … General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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

the time taken for one-half of a reactant to be consumed.
Chemistry 140 Fall 2002 Half-Life, t½ the time taken for one-half of a reactant to be consumed. = -kt ln [A]t [A]0 = -kt½ ln ½[A]0 [A]0 - ln 2 = -kt½ The decomposition reaction is described through equation (14.16). In this graph of the partial pressure of DTBP as a function of time, three successive half-life periods of 80 min each are indicated. This constancy of the half-life is proof that the reaction is first order. t½ = ln 2 k 0.693 =

14-9 Effect of Temperature on Reaction Rates
Chemistry 140 Fall 2002 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 ln k = lnA R -Ea T 1 on this slide

Temperature dependence of the rate constant k for a reaction
N2O5(CCl4) → N2O4(CCl4) + ½ O2(g) = -1.2104 K R -Ea -Ea = 1.0102 kJ mol-1 Temperature dependence of the rate constant k for a reaction

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: 2CH3CH2OH → CH3COOH + C2H4 + H2 Ea= KJ/mol, k = ◦ C Fermentation can only produce wine with a 3M of ethanol. The bottle was submerged in 4 ◦ C water for centuries. How many years was the ship submerged? General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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 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).

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. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

An analogy for a reaction profile and activation energy
Chemistry 140 Fall 2002 FIGURE 14-11 An analogy for a reaction profile and activation energy

A) low, low, molecules are so far apart
Chemistry 140 Fall 2002 According to the collision theory in gaseous molecules, collision frequency is ________ and reaction rate is ________ because ________. A) low, low, molecules are so far apart B) high, high, each collision results in a reaction C) low, 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 Answer: D

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. As temperature increases, reaction rate increases. Orientation of molecules may be important.

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

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

In the Arrhenius equation, ln k = -Ea/RT + ln A, the symbol A denotes:
Chemistry 140 Fall 2002 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 Answer: D

Transition State Theory
Chemistry 140 Fall 2002 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 activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state. A reaction profile for the reaction N2O(g) + NO(g) N2(g) + NO2(g)

Step-by-step description of a reaction.
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: Stoichiometry for the overall reaction. The experimentally determined rate law.

A mechanism with a Slow Step Followed by a Fast Step
d[P] H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g) = k[H2][ICl] dt Postulate a mechanism: dt = k[H2][ICl] d[HI] slow H2(g) + ICl(g) HI(g) + HCl(g) dt = k[HI][ICl] d[I2] fast HI(g) + ICl(g) I2(g) + HCl(g) dt = k[H2][ICl] d[P] H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

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

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; 1st order rate = k[A] Unimolecular 2A → products; 2nd order rate = k[A]2 Bimolecular A + B → prod.; 2nd order rate = k[A][B] Bimolecular A + B + C → prod. 3rd order rate = k[A][B][C] Termolecular

Chemistry 140 Fall 2002 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. 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. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

What is the rate law for the following mechanism?
Chemistry 140 Fall 2002 What is the rate law for the following mechanism? N2O + NO → N2ONO (Slow) N2ONO → N2 + NO2 (Fast) A) Rate = k[N2O] B) Rate = k[NO] C) Rate = k[N2O][NO] D) Rate = k[N2][NO2] E) Rate = k[N2ONO] Answer: C

What is the rate law for the following mechanism?
Chemistry 140 Fall 2002 What is the rate law for the following mechanism? CH3COOC2H5 + H2O → CH3COOC2H6+ + OH- (Slow) CH3COOC2H6+ → CH3COOH + C2H5+ (Fast) C2H5+ + OH- → C2H5OH (Fast) A) Rate = k[CH3COOC2H5][H2O]2 B) Rate = k[C2H5OH] C) Rate = k[CH3COOH] D) Rate = k[CH3COOC2H5] E) Rate = k[CH3COOC2H5][H2O] Answer: E

What is the rate law for the following reaction and its mechanism?
Chemistry 140 Fall 2002 What is the rate law for the following reaction and its mechanism? 2O3 → 3O (overall reaction) O3 → O2 + O (Slow) O∙ + O3 → 2O2 (Fast) A) Rate = k[O3] B) Rate = k[O3]2 C) Rate = k[O3]2/[O2] D) Rate = k[O3]/[O2] E) Rate = k[O3][O2] Answer: A

Substitution: One functional group replaces another
Alkyl halides react with a nucleophile to give a substituted product CH3–Cl + NaOH → CH3–OH + NaCl Nucleophile: A species with an unshared electron pair; a reagent that seeks a positive charge. Examples: HO-, RO-, CN-, CH3S-, :NH3 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-, H2O

Nucleophilic Substitution Reactions
Chemistry 140 Fall 2002 Nucleophilic Substitution Reactions Nucleophiles, often denoted by the abbreviation Nu, can be negatively charged or neutral, but every nucleophile contains at least one pair of unshared electrons. Nucleophiles are, in fact, Lewis bases. The nucleophilic substitution of a haloalkane is described by either of two general equations: 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. The electron pair from the bond that is broken ends up as a lone pair on the leaving group. In light of the discussion of acid–base chemistry in Section 27A on equilibrium favors forming the substitution product if the leaving group is a weaker base than the nucleophile. 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 1890s, 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! General Chemistry: Chapter 27

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.

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) General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Kinetic and Thermodynamic Control
At low temperature, the reaction is under kinetic control (rate, irreversible conditions) and the major product is that from fastest reaction. At high temperature, the reaction is under thermodynamic control (equilibrium, reversible conditions) and the major product is the more stable system Energy Reaction Coordinate 8/8/2018

1. At low temperature, the reaction preferentially proceeds along the green path to P1 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 P1 and P2, k1: k2. 2. At some slightly higher temperature, reaction 1 will become reversible while reaction 2 remains irreversible. So although P1 may form initially, over time it will revert to SM and react to give the more stable P2. 3. At high temperature, both reaction 1 and 2 are reversible and the product ratio of the reaction is dictated by the equilibrium constants for P1 and P2, K1 : K2. 8/8/2018

Reaction profile for an SN2 reaction

Consider the nucleophilic substitution of Br- by Cl- in the series of compounds to the right.
If the reaction proceeds via an SN2 mechanism, predict the order of increasing rate of reaction. A) B) C) D) A < B < C < D D < C < B < A B < C < D < A A < D < C < B Slide 71 of 33

Consider the nucleophilic substitution of Br- by Cl- in the series of compounds to the right.
If the reaction proceeds via an SN2 mechanism, predict the order of increasing rate of reaction. A) B) C) D) A < B < C < D D < C < B < A B < C < D < A A < D < C < B Slide 72 of 33

The SN1 Mechanism

Consider the nucleophilic substitution of Br- by OH- occurring in an aqueous solution for the series of compounds to the right. If the reaction proceeds via an SN1 mechanism, predict the order of increasing rate of reaction. A) B) C) D) A < B < C < D D < C < B < A B < C < D < A A < D < C < B Slide 77 of 33

A) B) C) A < B < C < D D < C < B < A D)
Consider the nucleophilic substitution of Br- by OH- occurring in an aqueous solution for the series of compounds to the right. If the reaction proceeds via an SN1 mechanism, predict the order of increasing rate of reaction. A) B) C) D) A < B < C < D D < C < B < A B < C < D < A A < D < C < B Slide 78 of 33

Relative rates of reaction
SN1 reaction

Alternative reaction pathway of lower energy. Homogeneous catalysis.
Chemistry 140 Fall 2002 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

Catalysis on a Surface Figure 14-19 Reaction profile for a surface-catalyzed reaction Heterogeneous catalysis in the reaction 2 CO + 2 NO CO2 + N2

Chemistry 140 Fall 2002 The decomposition of hydrogen peroxide, H2O2 , to H2O and O2 is a highly exothermic reaction that is catalyzed by platinum metal. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

I) lowers activation energy II) provides an alternate reaction pathway
Chemistry 140 Fall 2002 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, II, IV, and V Answer: E

If a catalyst is added to a reaction: I) the value of k is increased
Chemistry 140 Fall 2002 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 D

General Chemisry: Chapter 14
Chemistry 140 Fall 2002 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 Answer: E General Chemisry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Enzymes as Catalysts t Chemistry 140 Fall 2002
A computer graphics representation of the enzyme phosphoglycerate kinase (carbon backbone shown as blue ribbon). A molecule of ATP, the substrate, is shown in green.

Effect of substrate concentration on the rate of an enzyme reaction.
Chemistry 140 Fall 2002 E + S ES k1 k-1 → E + P k2 dt = k2[ES] d[P] dt = k1[E][S] – k-1[ES] – k2[ES]= 0 d[P] k1[E][S] = (k-1+ k2 )[ES] [E] = [E]0 – [ES] k1[S]([E]0 –[ES]) = (k-1+ k2 )[ES] (k-1+ k2 ) + k1[S] k1[E]0 [S] [ES] = FIGURE 14-21 Effect of substrate concentration on the rate of an enzyme reaction. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Effect of substrate concentration on the rate of an enzyme reaction.
Chemistry 140 Fall 2002 E + S ES k1 k-1 → E + P k2 dt = d[P] (k-1+ k2 ) + k1[S] k1k2[E]0 [S] dt = d[P] k2[E]0 dt = d[P] (k-1+ k2 ) + [S] k2[E]0 [S] k1 dt = d[P] KM k2 [E]0 [S] Two kinetic assumptions can be made. KM>>[S] at low substrate concentration and [S] >> KM at high substrate concentration, this gives rise to two limiting kinetic expressions characteristic of saturation kinetics. dt = d[P] KM + [S] k2[E]0 [S] FIGURE 14-21 Effect of substrate concentration on the rate of an enzyme reaction. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

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. General Chemistry: Chapter 14 Copyright © 2011 Pearson Canada Inc.

Kinetic equations [A]t = - kt + [A]0 = kt + 1 [A]0 [A]t k = Ae-Ea/RT
Chemistry 140 Fall 2002 Kinetic equations [A]t = - kt + [A]0 R = L atm mol-1 K-1 ln[A]t = -kt + ln[A]0 R = J mol-1 K-1 R = m3 Pa mol-1 K-1 = kt + 1 [A]0 [A]t k = Ae-Ea/RT ln = R -Ea T1 1 k2 k1 T2

14 Chemical Kinetics Chemistry 140 Fall 2002

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14 Chemical Kinetics Chemistry 140 Fall 2002

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Presentation on theme: "14 Chemical Kinetics Chemistry 140 Fall 2002"— Presentation transcript:

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About project

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