Chapter Thirteen

Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Chemical Kinetics: A Preview

•

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).

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 2

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.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 3

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 )

General

rate of reaction = rate of disappearance of reactant or of formation of product stoichiometric coefficient of that reactant or product in the balanced equation Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 4

5

If the rate of consumption of H 2 O 2 is 4.6 M/h, then … … the rate of formation of H 2 O must also be 4.6 M/h, and … … the rate of formation of O 2 is 2.3 M/h

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.1

Consider the hypothetical reaction A + 2 B  3 C + 2 D Suppose that at one point in the reaction, [A] = 0.4658 M and 125 s later [A] = 0.4282 M. During this time period, what is the average

(a)

rate of reaction expressed in M s –1 and

(b)

rate of formation of C, expressed in M min –1 .

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

2 H

2

O

2



2 H

2

O + O

2

1 L

2.960 g O 2 (0.09250 mole) produced in 60 s means …

Prentice Hall © 2005 Rate = 0.1850 mol H 2 O 2 /L 60 s

… 0.1850 mol H 2 O 2 reacted in 60 s.

=

0.00131 M H 2 O 2 s –1

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 7

Average vs. Instantaneous Rate

8

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.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.2

Use data from Table 13.1 and/or Figure 13.5 to (a) determine the initial rate of reaction and (b) calculate [H 2 O 2 ] at

t

= 30 s. 9 Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

The Rate Law of a Chemical Reaction

• The

rate law

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

aA + bB + cC …



products rate = k[A]

n

[B]

m

[C]

p

…

• 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

.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 10

The Rate Law

Rate =

k

[A]

1

=

k

[A] Rate =

k

[A]

2

Rate =

k

[A]

3 Prentice Hall © 2005

Reaction is

first

order in A Reaction is

second

order in A Reaction is

third

order in A

If we triple the concentration of A in a second-order reaction, the rate increases by a factor of ________.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 11

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.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 12

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.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 13

14

The concentration of NO was held the same in Experiments 1 and 2 … … while the concentration of Cl 2 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 Cl 2 .

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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 Cl 2 ?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.3

For the reaction 2 NO(g) + Cl 2 (g)  2 NOCl(g) described in the text and in Table 13.2,

(a)

what is the initial rate for a hypothetical Experiment 4, which has [NO] = 0.0500 M and [Cl 2 ] = 0.0255 M?

(b)

What is the value of

k

for the reaction?

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

First-Order Reactions

• In a

first-order reaction

, the exponent in the rate law is 1.

• Rate =

k

[A] 1 =

k

[A] • The

integrated rate law

describes the concentration of a reactant as a function of time. For a first-order process: [A]

t

ln = –

kt

ln [A]

t

ln [A]

t

[A] 0 – ln [A] 0 = –

kt

= –

kt

+ ln [A] 0

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

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

proportional

to concentration.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 16

Decomposition of H

2

O

2

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.4

For the first-order decomposition of H 2 O 2 (aq), given

k

3.66 x 10 –3 s –1 and [H 2 O 2 ] 0 = 0.882 M, determine

(a)

= the time at which [H 2 O 2 ] = 0.600 M and

(b)

[H 2 O 2 ] after 225 s. 18 Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Half-life of a Reaction

• The

half-life (t

½

)

of a reaction is the time required for one half of the reactant originally present to be consumed.

• At

t

½ , [A]

t

= ½[A] 0 , and for a first order reaction: ln ½[A] 0 [A] 0 = –

kt

½ ln (½) = –

kt

½ –0.693 = –

kt

½

t

½ = 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.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Initial amount

Decomposition of N

2

O

5

at 67 °C

20

After one half-life, half the N 2 O 5 has reacted.

After two half-lives, half of the remaining N 2 O 5 has reacted—three-fourths has been consumed.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.5 A Conceptual Example

Use data from Figure 13.7 to evaluate the

(a)

half-life and

(b)

rate constant for the first-order decomposition of N 2 O 5 at 67 °C.

Example 13.6 An Estimation Example

Which seems like a probable approximate time for 90% of a sample of N 2 O 5 to undergo decomposition at 67 °C: (a) 200 s, (b) 300 s, (c) 400 s, or (d) 500 s?

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

A Zero-Order Reaction

rate = k[A] 0 = k Rate is independent of initial concentration

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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: 1 –––– =

kt

[A]

t

1 + –––– [A] 0

What do we plot vs. time to get a straight line?

• The

half-life

of a second-order reaction depends on the initial concentration as well as on the rate constant

k

:

t

½ 1 = –––––

k

[A] 0 Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 23

24

Example 13.7

The second-order decomposition of HI(g) at 700 K is represented in Figure 13.9.

HI(g)  ½ H 2 (g) + ½ I 2 (g) Rate =

k

[HI] 2 What are the:

(a)

rate constant and

(b)

half-life of the decomposition of 1.00 M HI(g) at 700 K?

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.8 A Conceptual Example

Shown here are graphs of [A] versus time for two different experiments dealing with the reaction A  products. What is the order of this reaction?

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Summary of Kinetic Data

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 (E

a

)

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.

• The

spatial orientations

of the colliding species may also determine whether a collision is effective.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 27

Distribution of Kinetic Energies

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry

At higher temperature (red), more molecules have the necessary activation energy.

Chapter Thirteen

Importance of Orientation

One hydrogen atom can approach another from any direction … Effective collision; the I atom can bond to the C atom to form CH 3 I

29

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

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Ineffective collision; orientation is important in this reaction.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 30

A Reaction Profile

CO (g) + NO 2 (g)  CO 2 (g) + NO (g) 31 Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

An Analogy for Reaction Profiles and Activation Energy

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Effect of Temperature on the Rates of Reactions

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

k

:

k

=

A

e –

E

a /

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 –

E

a /

RT

represents the fraction of molecular collisions

sufficiently energetic

to produce a reaction.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 33

Example 13.9

Estimate a value of

k

at 375 K for the decomposition of dinitrogen pentoxide illustrated in Figure 13.15, given that

k

= 2.5 x 10 –3 s –1 at 332 K.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 35

Molecularity

The

molecularity

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

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4 th edition, Hill, Petrucci, McCreary, Perry

Termolecular processes are unusual, for the same reason that three basketballs shot at the same time are unlikely to collide at the same instant …

Chapter Thirteen

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.

• 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.

Fast Mechanism for 2 NO + O 2  2 NO 2 Slow Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 37

Example 13.10

For the reaction H 2 (g) + I 2 (g)  mechanism is: 2 HI(g), a proposed

Fast step:

I 2

k

1

k

–1 2 I

k

2

Slow step:

2 I + H 2 2 HI What is the net equation for the overall reaction, and what is the order of the reaction according to this mechanism?

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Catalysis

• A

catalyst

increases the reaction rate without itself being used up in a chemical reaction.

• In general, a catalyst works by changing the mechanism of a chemical reaction.

• Often the catalyst is consumed in one step of the mechanism, but is regenerated in another step.

• The pathway of a catalyzed reaction has a lower activation energy than that of an uncatalyzed reaction, so more molecules at a fixed temperature have the necessary activation energy.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 39

Effect of Catalyst on Reaction Profile and Activation Energy

A catalyst lowers the activation energy, making it easier for the reactants to “climb the energy hill” and form the products.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Homogeneous Catalysis

Ozone decomposition catalyzed by chlorine atoms has a much lower activation energy and proceeds much more rapidly than the uncatalyzed reaction

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Heterogeneous Catalysis

• Many reactions are catalyzed by the surfaces of appropriate solids.

• A good catalyst provides a higher frequency of effective collisions.

• Four steps in heterogeneous catalysis: – Reactant molecules are

adsorbed

.

– Reactant molecules diffuse along the surface.

– Reactant molecules react to form product molecules.

– Product molecules are

desorbed

(released from the surface).

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 42

Heterogeneous Catalysis

Hydrogen is adsorbed onto the surface of a nickel catalyst. A C=C approaches …

43

… and is adsorbed.

Hydrogen atoms attach to the carbon atoms, and the molecule is desorbed.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

A Surface-Catalyzed Reaction

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Enzyme Catalysis

•

Enzymes

are high-molecular-mass proteins that usually catalyze one specific reaction, or a set of similar reactions.

• The reactant substance, called the

substrate

(S), attaches itself to an area on the enzyme (E) called the

active site

, to form an enzyme-substrate complex (ES).

• The enzyme–substrate complex decomposes to form products (P), and the enzyme is regenerated.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Factors Influencing Enzyme Activity

The rates of enzyme-catalyzed reactions are influenced by factors such as concentration of the substrate, concentration of the enzyme, acidity of the medium, and temperature.

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4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Enzyme Activity as a Function of Temperature

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Mercury Poisoning: An Example of Enzyme Inhibition

When Hg reacts with an enzyme …

48

… the Hg binds to sulfur atoms … … changing the shape of the active site, so that it no longer “fits” the substrate.

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Action of Cholinesterase and Its Inhibition

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General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Cumulative Example

In acidic aqueous solutions, nitroacetic acid decomposes to nitromethane and carbon dioxide gas: CH 2 (NO 2 )COOH  CH 3 NO 2 + CO 2 (g) Nitroacetic acid Nitromethane In one experiment, 0.1051 g nitroacetic acid was allowed to decompose, and the evolved CO 2 (g) was collected at 25.0 °C over CaCl 2 (aq) saturated with CO 2 (g) and having a water vapor pressure of 9.7 Torr. The barometric pressure was 758.2 Torr. Use the following data on collected gas volume as a function of time to determine the half-life of this reaction at 25.0 °C.

50

Time, min Volume, mL

Prentice Hall © 2005 0 0 1.64 3.96 3.64

7.93 6.14

9.64 15.19

11.92 15.93 19.93

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen