Chapter 12 Chemical Kinetics. 2 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen ASSIGNMENTS

Chapter 12

Chemical Kinetics

2

General Chemistry 4th edition, Hill, Petrucci, McCreary, PerryPrentice Hall © 2005Hall © 2005 Chapter Thirteen

ASSIGNMENTS 3-7-13 AP Chem

• Stop here Thursday - 3-7-13• Work on problem sets from handout AND

the separate page...• HW: ch. 12.1 to 12.4 #1-13 due Monday.• HW: Read ch. 12 by Mon. • HW: Read lab report• TEST next WED. ch. 12• TURN in last week’s lab report.• Yesterday’s lab report due next Thursday.• Kinetics LAB tomorrow - switch rooms

3

Chapter Twelve

Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

Chapter 12Chapter 12

4

Chapter 12

Table of Contents

Copyright © Cengage Learning. All rights reserved

12.1 Reaction Rates

12.2 Rate Laws: An Introduction

12.3 Determining the Form of the Rate Law

12.4 The Integrated Rate Law

12.5 Reaction Mechanisms

12.6A Model for Chemical Kinetics

12.7Catalysis

5

Chapter 12

Table of Contents


• AP CHEM - WED - 1/29/14• CW: Notes 12.1 to 12.3• TURN in Clue Mystery Lab Report• CW/HW: Assignment - 2 handouts #1-6 problems & 1-13

(but only have to complete 1-6 for sect. 12.1 to 12.3 by Tuesday 2-4

• HW: Read over Kinetics Lab for Friday• On Friday, you will receive the ch.10-11 take home test to

complete by Monday, Feb. 10th

6


• Chemical kinetics is the study of:– the rates of chemical reactions

– factors that affect these rates

– the mechanisms by which reactions occur

• Reaction rates vary greatly – some are very fast (burning, precipitation) and some are very slow (rusting, disintegration of a plastic bottle in sunlight, production of a diamond).

Chemical Kinetics: A Preview

7


Variables in Reaction Rates

• Concentrations of reactants: Reaction rates generally increase as the concentrations of the reactants are increased.

• Temperature: Reaction rates generally increase rapidly as the temperature is increased.

• Surface area: For reactions that occur on a surface rather than in solution, the rate increases as the surface area is increased.

• Catalysts: Catalysts speed up reactions and inhibitors slow them down.

8


The Meaning of Rate• The rate of a reaction is the change in concentration

of a product per unit of time (rate of formation of product).

• Rate is also viewed as the negative of the change in concentration of a reactant per unit of time (rate of disappearance of reactant).

• The rate of reaction often has the units of moles per liter per unit time (mol L–1 s–1 or M s–1)

rate of disappearance of reactant or of formation of product

stoichiometric coefficient of that reactant or product in the balanced equation

General rate of reaction =

9

Section 12.1

Reaction Rates


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Reaction Rate

• Change in concentration of a reactant or product per unit time.

[A] means concentration of A in mol/L; A is the reactant or product being considered.

Rate

10


If the rate of consumption of H2O2 is 4.6 M/h, then …

… the rate of formation of H2O must also be 4.6 M/h, and …

… the rate of formation of O2 is 2.3 M/h

11


2 H2O2 2 H2O + O2

… 0.1850 mol H2O2 reacted in 60 s.

Rate =0.1850 mol H2O2/L

60 s

= 0.00131 M H2O2 s–1

1 L

2.960 g O2 (0.09250 mole) produced in 60 s means …

12

Section 12.1

Reaction Rates


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The Decomposition of Nitrogen Dioxide

13

Section 12.1

Reaction Rates


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The Decomposition of Nitrogen Dioxide

http://college.cengage.com/chemistry/zumdahl/chemistry/7e/assets/students/protected/fae/index.html?layer=act&src=qtiworkflowflash_12_34.xml&w=750&h=434

14

Section 12.1

Reaction Rates


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Instantaneous Rate

• Value of the rate at a particular time.• Can be obtained by computing the slope of a

line tangent to the curve at that point.

15


Average vs. Instantaneous Rate

Instantaneous rate is the slope of the tangent to the curve at a particular time.

We often are interested in the initial instantane-ous rate; for the initial concentrations of reactants and products are known at this time.

16

Section 12.3

Atomic Masses


Rate Laws: An Introduction

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Rate Law

• Shows how the rate depends on the concentrations of reactants.

• For the decomposition of nitrogen dioxide:

2NO2(g) → 2NO(g) + O2(g)

Rate = k[NO2]n:

k = rate constant n = order of the reactant

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Section 12.3

Atomic Masses



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Rate Law

Rate = k[NO2]n

• The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

18

Section 12.3

Atomic Masses



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Rate Law

Rate = k[NO2]n

• The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

19

Section 12.3

Atomic Masses



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Types of Rate Laws

• Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations.

• Integrated Rate Law – shows how the concentrations of species in the reaction depend on time.

20

Section 12.3

Atomic Masses



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Rate Laws: A Summary

• Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants.

• Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient.

21

Section 12.3

Atomic Masses



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Rate Laws: A Summary

• Experimental convenience usually dictates which type of rate law is determined experimentally.

• Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

22

Section 12.3

Atomic Masses



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Videoclip Summary Kinetic #1 - 7 minutes

• Click below to watch videoclip.• Link for Lesson 12.1-12.2 7 minute video• CosmoLearning Videoclips

http://www.cosmolearning.com/video-lectures/kinetics-i-rate-laws/

23

Section 12.3

The Mole Determining the Form of the Rate Law


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• Determine experimentally the power to which each reactant concentration must be raised in the rate law.

24

Section 12.3



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Method of Initial Rates

• The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible.

• Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run.

• The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

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Section 12.3



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Overall Reaction Order

• The sum of the exponents in the reaction rate equation.

Rate = k[A]n[B]m

Overall reaction order = n + m

k = rate constant

[A] = concentration of reactant A

[B] = concentration of reactant B

26


The Rate Law of a Chemical Reaction

• The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants.

• The exponents (m, n, p…) are determined by experiment.

• Exponents are not derived from the coefficients in the balanced chemical equation, though in some instances the exponents and the coefficients may be the same.

• The value of an exponent in a rate law is the order of the reaction with respect to the reactant in question.

• The proportionality constant, k, is the rate constant.

aA + bB + cC … products rate = k[A]n[B]m[C]p …

27


The Rate Law

Rate = k[A]1 = k[A] Reaction is first order in A

Rate = k[A]3 Reaction is third order in A

Rate = k[A]2 Reaction is second order in A

If we triple the concentration of A in a second-order reaction, the rate

increases by a factor of ________.

9

28


More About the Rate Constant k

• The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants.

• The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants.

• The rate and the rate constant have the same numerical values and units only in zero-order reactions.

• For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.

29


Method of Initial Rates

• The method of initial rates is a method of establishing the rate law for a reaction—finding the values of the exponents in the rate law, and the value of k.

• A series of experiments is performed in which the initial concentration of one reactant is varied. Concentrations of the other reactants are held constant.

• When we double the concentration of a reactant A, if:

– there is no effect on the rate, the reaction is zero-order in A.

– the rate doubles, the reaction is first-order in A.

– the rate quadruples, the reaction is second-order in A.

– the rate increases eight times, the reaction is third-order in A.

30


The concentration of NO was held the

same in Experiments 1 and

2 …

… while the concentration of

Cl2 in Experiment 2 is

twice that of Experiment 1.

The rate in Experiment 2 is twice that in

Experiment 1, so the reaction must be first

order in Cl2.

Which two experiments are used to find the order

of the reaction in NO?

How do we find the value of k after obtaining the

order of the reaction in NO and in Cl2?

31

Section 12.3

Atomic Masses



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Videoclip Finding Rate Laws from data - 3.5 minutes

• Click below to watch videoclip.• Link for Lesson 12.3 Rate Law 3.5 minutes video• CosmoLearning Videoclips

http://www.cosmolearning.com/video-lectures/kinetics-ii-determining-rate-law-of-reaction/

32

Section 12.3

Atomic Masses



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Problems

• Day 1 - Stop here and work practice problems assigned.

• 2 handouts - #1-6 practice problems (Dr. paper)• #1-13 12.1 to 12.4 assigned problems (but only have to

do section 12.1 to 12.3 for now but should read ahead for section 12.4 in textbook reading assignment due Tues Feb. 4th -

• Friday - Kinetics Lab (Read and look over pre-lab)• Turn in Clue Lab report today

33


Assignments -

• Notes 12.3 revisited & 12.4 problems -handoutsCW/HW: #7-12 Bradley worksheet Chem 201B

• CW/HW: 12.4 #7-13 worksheet chemical “Kinetics p.279”

• CW/HW: problems D & E new handout with notes

• HW: Ch. 10-11 Take Home Test• Get help on any other problems previously

assigned

34

Section 12.3

Atomic Masses



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Videoclip introduction similar to our Kinetic Lab data• http://www.cosmolearning.com/video-lectures/kinetics-iii-

integrated-rate-law-in-0th-1st-and-2nd-order/

http://www.cosmolearning.com/video-lectures/kinetics-iii-integrated-rate-law-in-0th-1st-and-2nd-order/

http://www.cosmolearning.com/video-lectures/kinetics-iii-integrated-rate-law-in-0th-1st-and-2nd-order/

35

Section 12.3

Atomic Masses



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Kinetics III Integrated Rate Law - 0,1,2 order

• Example from videoclip

• 2CO2 --> 2CO + O2

• t (time) vs concentration• 0 0.10 M• 5 0.050 M• 10 0.025 M• 15 0.0125 M

36

Section 12.4

The Integrated Rate Law


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• Rate = k[A]• Integrated:

ln[A] = –kt + ln[A]o

[A] = concentration of A at time tk = rate constant

t = time

[A]o = initial concentration of A

First-Order

37


First-Order Reactions

• In a first-order reaction, the exponent in the rate law is 1.

• Rate = k[A]1 = k[A]

Look! It’s an equation for a straight line!

ln [A]t – ln [A]0 = –kt

ln [A]t = –kt + ln [A]0

• At times, it is convenient to replace molarities in an integrated rate law by quantities that are proportional to concentration.

• The integrated rate law describes the concentration of a reactant as a function of time. For a first-order process:

ln[A]t

[A]0

= –kt

y = mx+b

38

Section 12.4



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Half-Life of Reactions

Click above when hand appearsto see visual. Be in play mode.Use my log-in as needed to access. [email protected] “Bryant123”

mailto:[email protected]

39

Section 12.4



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Plot of ln[N2O5] vs Time

40


Half-life of a Reaction• The half-life (t½) of a reaction is the time required for

one-half of the reactant originally present to be consumed.

• At t½, [A]t = ½[A]0, and for a first order reaction:

ln½[A]0

[A]0

= –kt½

ln (½) = –kt½

–0.693 = –kt½

t½ = 0.693/k

• Thus, for a first-order reaction, the half-life is a constant; it depends only on the rate constant, k, and not on the concentration of reactant.

41

Section 12.4



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• Half–Life:

k = rate constant

• Half–life does not depend on the concentration of reactants.

First-Order

42

Section 12.4



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Exercise

A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?

43

Section 12.4



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Exercise


1st order so look at ln (xo/x) = kt If 35% is completed then 65% remains - original would 100% or 1 so

ln (1/.65) = kt where t is 55 minutes then solve for k

k = 7.8 x 10–3 min–1

44


Second-Order Reactions

• A reaction that is second order in a reactant has a rate law in which the exponent for that reactant is 2.

Rate = k[A]2

• The integrated rate law has the form:

• The half-life of a second-order reaction depends on the initial concentration as well as on the rate constant k:

1 1–––– = kt + –––– [A]t [A]0

What do we plot vs. time to get a straight

line?

1t½ = ––––– k[A]0

45

Section 12.4



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• Rate = k[A]2

• Integrated:


t = time


Second-Order

46

Section 12.4



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Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time

47

Section 12.4



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• Half–Life:

k = rate constant


• Half–life gets longer as the reaction progresses and the concentration of reactants decrease.

• Each successive half–life is double the preceding one.

Second-Order

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Section 12.4



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Exercise

For a reaction aA Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively.

a) Write the rate law for this reaction.

? rate = k[A]2

b) Calculate k.

k = 8.0 x 10-3 M–1min–1

– Calculate [A] at t = 525 minutes.

[A] = 0.23 M

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Section 12.4



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• Rate = k[A]0 = k• Integrated:

[A] = –kt + [A]o


t = time


Zero-Order

50


Rate is independent of

initial concentration

A Zero-Order Reaction

rate = k[A]0

= k

51

Section 12.4



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Plot of [A] vs Time

52

Section 12.4



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• Half–Life:

k = rate constant


• Half–life gets shorter as the reaction progresses and the concentration of reactants decrease.

Zero-Order

53

Section 12.4



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Concept Check

How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs?

54

Section 12.4



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Rate LawsClick below to watch visual.

55


Summary of Kinetic Data

56


Assignments 2014 - Thurs 2-6

• CW/HW: #7-12 Bradley worksheet Chem 201B

• CW/HW: 12.4 #7-13 worksheet chemical “Kinetics p.279”

• CW/HW: problems D & E new handout with notes

• HW: Ch. 10-11 Take Home Test• Get help on any other problems previously

assigned

57


ASSIGNMENTS 3-7-13 AP Chem

• Stop here Thursday - 3-7-13• Work on problem sets from handout AND

the separate page...• HW: ch. 12.1 to 12.4 #1-13 due Monday.• HW: Read ch. 12 by Mon. • HW: Read lab report• TEST next WED. ch. 12• TURN in last week’s lab report.• Yesterday’s lab report due next Thursday.• Kinetics LAB tomorrow - switch rooms

58


ASSIGNMENTS 3-11-13 AP Chem• HW ch. 12 handout section 12.1-12.4 due -

check answers; make corrections with WORK shown to get credit. HW: ch. 12.1 to 12.4 #1-13 due today.

• Notes 12.4-12.8• TEST ch. 12 WEDNESDAY.• CW: Quiz ch. 12 (12 problems - 20 min.)• If you haven’t read ch. 12, you need to along with packet

handout.

• HW: Lab report due FRIDAY. The other lab report not

formal lab report from EOC day. (11th will do Thurs.)

• Turn in previous labs, etc. this WEEK!

59


Assignments Mon 2-10-14

• Notes - Kinetics starting with 12.6 packet with reaction mechanisms then the math/science packet.

• Turn in ch. 10-11 Test ANSWERS - We will put in the computer later.

• HW: Kinetics I tutorial packet for grade.• HW: Kinetics checklist packet with

provided answers for practice and test preparation - will not count as grade

60


Integrated Rate Law Summary

61

Section 12.4



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Exercise


62

Section 12.4



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Exercise


1st order so look at ln (xo/x) = kt If 35% is completed then 65% remains - original would 100% or 1 so

ln (1/.65) = kt where t is 55 minutes then solve for k

k = 7.8 x 10–3 min–1

63

Section 12.4



Return to TOC

Exercise

Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is:

– Zero order– First order – Second order

4.7 M3.7 M2.0 M

64

Section 12.5

Reaction Mechanisms


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• Most chemical reactions occur by a series of elementary steps.

• An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction.

Reaction Mechanism

65


Reaction Mechanisms

• Analogy: a banana split is made by steps in sequence: slice banana; three scoops ice cream; chocolate sauce; strawberries; pineapple; whipped cream; end with cherry.

• A chemical reaction occurs according to a reaction mechanism—a series of collisions or dissociations—that lead from initial reactants to the final products.

• An elementary reaction represents, at the molecular level, a single step in the progress of the overall reaction.

• A proposed mechanism must:

– account for the experimentally determined rate law.

– be consistent with the stoichiometry of the overall or net reaction.

66


Molecularity

The molecularity of an elementary reaction refers to the number of free atoms, ions, or molecules that collide or dissociate in that step.

Termolecular processes are

unusual, for the same reason that three

basketballs shot at the same time are

unlikely to collide at the same instant …

67

Section 12.5

Reaction Mechanisms


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• Unimolecular – reaction involving one molecule; first order.

• Bimolecular – reaction involving the collision of two species; second order.

• Termolecular – reaction involving the collision of three species; third order.

Elementary Steps (Molecularity)

68

Section 12.5

Reaction Mechanisms


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A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO

NO2(g) + CO(g) → NO(g) + CO2(g)

69


The Rate-Determining Step

• The rate-determining step is the crucial step in establishing the rate of the overall reaction. It is usually the slowest step. A reaction is only as fast as its slowest step.

• Some two-step mechanisms have a slow first step followed by a fast second step, while others have a fast reversible first step followed by a slow second step.

Slow

Fast

Mechanism for

2 NO + O2 2 NO2

• The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.

70

Section 12.5

Reaction Mechanisms


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• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The mechanism must agree with the experimentally determined rate law.

Reaction Mechanism Requirements

71


After two half-lives, half of the remaining N2O5 has reacted—three-

fourths has been consumed.

After one half-life, half the N2O5 has reacted.

Initial amount Decomposition of N2O5 at 67 °C

72

Section 12.5

Reaction Mechanisms


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Decomposition of N2O5Click below to watch visual.

73

Section 12.5

Reaction Mechanisms


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Decomposition of N2O5

2N2O5(g) 4NO2(g) + O2(g)

Step 1: N2O5 NO2 + NO3 (fast)

Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)

Step 3: NO3 + NO → 2NO2 (fast)

2( )

74

Section 12.5

Reaction Mechanisms


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Concept Check

The reaction A + 2B C has the following proposed mechanism:

A + B D (fast equilibrium)

D + B C (slow)

Write the rate law for this mechanism.

rate = k[A][B]2

75


Theories of Chemical Kinetics: Collision Theory

• Before atoms, molecules, or ions can react, they must first collide.

• An effective collision between two molecules puts enough energy into key bonds to break them.

• The activation energy (Ea) is the minimum energy that must be supplied by collisions for a reaction to occur.

• A certain fraction of all molecules in a sample will have the necessary activation energy to react; that fraction increases with increasing temperature. (general rule: 10 degree increase in temp. (Celsius/Kelvin) = doubles rate)

• The spatial orientations of the colliding species may also determine whether a collision is effective.

76


Importance of OrientationOne hydrogen atom

can approach another from any

direction …

… and reaction will still occur; the spherical symmetry of the atoms means that orientation does not

matter.

Effective collision; the I atom can bond

to the C atom to form CH3I

Ineffective collision; orientation is

important in this reaction.

77

Section 12.6

A Model for Chemical Kinetics


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• Molecules must collide to react.• Main Factors:

Activation energy, Ea

Temperature Molecular orientations

Collision Model

78


Distribution of Kinetic Energies

At higher temperature (red),

more molecules have the necessary

activation energy.

79

Section 12.6



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• Energy that must be overcome to produce a chemical reaction.

Activation Energy, Ea

80


Transition State Theory

• The configuration of the atoms of the colliding species at the time of the collision is called the transition state.

• The transitory species having this configuration is called the activated complex.

• A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction.

• Reactant molecules must have enough energy to surmount the energy “hill” separating products from reactants.

81

Section 12.6



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Transition States and Activation EnergyClick below to watch visual animation.

82

Section 12.6



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Change in Potential Energy

83


CO(g) + NO2(g) → CO2(g) + NO(g)

A Reaction Profile

84


An Analogy for Reaction Profiles and Activation Energy

85

Section 12.6



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• Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy).

• Relative orientation of the reactants must allow formation of any new bonds necessary to produce products.

For Reactants to Form Products

86

Section 12.6



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The Gas Phase Reaction of NO and Cl2

Click below to watch visual animation.

87


Effect of Temperature onthe Rates of Reactions

• In 1889, Svante Arrhenius proposed the following expression for the effect of temperature on the rate constant, k:

k = Ae–Ea/RT

• The constant A, called the frequency factor, is an expression of collision frequency and orientation; it represents the number of collisions per unit time that are capable of leading to reaction.

• The term e–Ea/RT represents the fraction of molecular collisions sufficiently energetic to produce a reaction.

88

Section 12.6



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A = frequency factor

Ea = activation energy

R = gas constant (8.3145 J/K·mol)

T = temperature (in K)

Arrhenius Equation

Note the division sign “/”.

89

Section 12.6



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Linear Form of Arrhenius Equation

90

Section 12.6



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Linear Form of Arrhenius Equation

91

Section 12.6



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Exercise

Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C?

Ea = 53 kJ

92

Section 12.7

Catalysis


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• A substance that speeds up a reaction without being consumed itself.

• Provides a new pathway for the reaction with a lower activation energy.

Catalyst

93

Section 12.7

Catalysis


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Catalysis Teaching Videoclip & Example

• Optional• Click below. Must be logged in cengage to watch.

• http://college.cengage.com/chemistry/discipline/thinkwell/2944.html

• Good example to show Elephant Toothpaste experiment demonstration.

94

Section 12.7

Catalysis


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Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction

95

Section 12.7

Catalysis


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Effect of a Catalyst on the Number of Reaction-Producing Collisions

96

Section 12.7

Catalysis


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• Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst.

• Adsorption – collection of one substance on the surface of another substance.

Heterogeneous Catalyst

97

Section 12.7

Catalysis


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Heterogeneous Catalysis Click picture below to watch animation.

98

Section 12.7

Catalysis


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1. Adsorption and activation of the reactants.

2. Migration of the adsorbed reactants on the surface.

3. Reaction of the adsorbed substances.

4. Escape, or desorption, of the products.

Heterogeneous Catalyst

99

Section 12.7

Catalysis


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• Exists in the same phase as the reacting molecules.

• Enzymes are nature’s catalysts.

Homogeneous Catalyst

100

Section 12.7

Catalysis


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Homogeneous Catalysis - Click below for visual. Requires log-in

Documents

Chapter 12 Chemical Kinetics. 2 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen ASSIGNMENTS