Chapter 13. Kinetics: Mechanisms and Rates of Reactions 13.1 What is a Reaction Mechanism? 13.2...

Chapter 13. Kinetics: Mechanisms and Rates of Reactions

13.1 What is a Reaction Mechanism?13.2 Rates of Chemical Reactions13.3 Concentration and Reaction Rates13.4 Experimental Kinetics13.5 Linking Mechanisms and Rate Laws13.6 Reaction Rates and Temperature13.7 Catalysis

Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.

13.1 What is a Reaction Mechanism?

Learning objective:

Explain the concepts of a mechanism and a rate-determining step in a chemical reaction


13.1 What is a Reaction Mechanism?

Reaction mechanisms – the exact molecular pathway that starting materials follow on their way to becoming products.

Each step in a multi-step reaction is called an elementary step.


Formation of N2O4


Elementary Reactions

Elementary reactions are classified by their molecularity – the number of molecules (or atoms) on the reactant side of the chemical equation for the elementary reaction.

A → Products UnimolecularA + B → Products, 2A → Products BimolecularA + B + C → Products, 2A + B → Products Termolecular


Step 1: NO2 (g) + NO2 (g) → NO (g) + NO3 (g) Elementary Reaction

Step 2: NO3 (g) + CO (g) → NO2 (g) + CO2 (g) Elementary Reaction

NO2 (g) + CO (g) → NO (g) + CO2 (g) Overall Reaction

For Example…

Elementary reaction – describes the behavior of the individual molecules

Overall reaction – describes the reaction stoichiometry

Reaction intermediate: species only present in the elementary reactions, formed in one step and consumed in another (NO3 in this case).


The Rate – Determining Step

Rate – the speed of a reaction. Every elementary reaction has a characteristic rate.

Rate-determining step (RDS) – the slowest elementary step in a mechanism.

The RDS governs the rate of the overall chemical reaction because no net chemical reaction can go faster than its slowest step.


13.2 Rates of Chemical Reactions

Learning objective:

Determine the rate of a reaction based on the rate of change of concentration of a reactant or product.


13.2 Rates of Chemical Reactions

Kinetics: the study of the rates of chemical reactions

Molecular View → deals with the actual collisions theory, what pathways are taken by the atoms and molecules

Macroscopic view → deals with rates of the reactions, how to determine the rate and how factors affect it


A Molecular View


Collisions must give the cis-2-butene enough energy to break the C-C -bond

A Macroscopic View

Rate: dependent on how the concentration of a reactant or product changes over time.

Units of rate: mol L-1 s-1 (if concentration is expressed as molarity)

Rate is dependent on three things: concentration of reactants, temperature, and catalysts.

ConcentrationRate =

Time


2 NO2 → 2 NO + O2


Reaction Rate

aA + bB → dD + eE

The relationship among reaction rates and stoichiometric coefficients can be applied to any reaction.

Δt

EΔ

e

1

Δt

DΔ

d

1

Δt

BΔ

b

1

Δt

AΔ

a

1RateReaction


Example 13–1 Relative Rates of Reaction

Acrylonitrile is produced from propene, ammonia, and oxygen by the following balanced equation

2 C3H6 + 2 NH3 + 3 O2 → 2 CH2CHCN + 6 H2O

Relate the rates of reaction of starting materials and products


13.3 Concentration and Reaction Rates


Learning objective:

Determine the rate law, given the mechanism and knowledge of the relative rates of steps of a reaction

13.3 Concentration and Reaction Rates


Example 13–2 Rates and Number of Molecules

The container on the left contains O3 and NO. Compared with the container on the right (which has the same volume), how fast will the reaction proceed?


Rate Laws

Rate Law: the effect of concentration on the rate of a particular chemical reaction.

Rate laws for elementary reactions can be written directly from the stoichiometry of the reactants. (But only for elementary reaction steps)

The experimentally observed rate law for an overall reaction depends on the reaction mechanism.


The rate–law for the rate determining step is the rate– law for the overall reaction.


Rate Laws

Rate laws have a general form

Rate Constant: k (for a given temperature)y and z: exponents which relate rate to the change in

concentrations of the reactants (not related to the stoichiometric coefficients).

Reaction Order = y+z

y zRate = k[A] [B]

y and z must be determined experimentallyy and z must be determined experimentally


Rate Laws

Stoichiometry: 2 NO2 → 2 NO + O2

Experimentally, rate = k[NO2]2

Possible Mechanism I:NO2 → NO + O (slow, i.e. rate determining)

O + NO2 → O2 + NO (fast)

Possible Mechanism II:2 NO2 → NO3 + NO (slow, i.e. rate determining)

NO3 → NO + O2 (fast)

Which mechanism gives the observed rate law?Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.

2 NO2 → 2 NO + O2

Observed rate law: rate = k[NO2]2

Possible Mechanism I:NO2 → NO + O (slow, i.e. rate determining)

O + NO2 → O2 + NO (fast)

Doubling [NO2] will double the rate, hence the predicted rate law for this mechanism is rate = k[NO2]. This cannot be the correct mechanism.


2 NO2 → 2 NO + O2

Observed rate law: rate = k[NO2]2

Possible Mechanism II:2 NO2 → NO3 + NO (slow, i.e. rate determining)

NO3 → NO + O2 (fast)

Doubling [NO2] will quadruple the rate, hence the predicted rate law for this mechanism is rate = k[NO2]2. This may be the correct mechanism.

(Would have to verify the presence of NO3 to be sure!)Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.

Example 13–3 Units of a Rate Constant

Reactions in aqueous solution can have complicated kinetics. An example is the reaction between arsenic acid and iodide ions:

H3AsO4 + 3 I- + 2 H3O+ → H3AsO3 + I3- + 3 H2O

The rate law for this reaction has been found experimentally to be as follows:

Rate = k[H3AsO4][I-][H3O+]What are the units of the rate constant when the time is

expressed in minutes?


Reaction Order for A→ products

Zero-order 0ARate = - = k[A] = k

t

ARate = - = k[A]

t

2ARate = - = k[A]

t

First – order

Second – order


13.4 Experimental Kinetics

Learning objective:

Determine rate laws from concentration versus time data


13.4 Experimental Kinetics

We have learned how to determine the rate of a reaction.

Often it is important to know how long a reaction must proceed in order to obtain a certain concentration of some reagent.


Zero-order Integrated Rate Law

• [A]0 and [A]t are the concentrations of the reactant at t = 0 and a later time, t

• Notice that the integrated equation is a straight line.

t 0[A] = -kt + [A]

(y = mx + b)


First – order Integrated Rate Law

First-order integrated rate law for the reaction aA → products

[A]0 and [A]t are the concentrations of the reactant at t=0 and a later time, t

oA

ln = aktA


First Order Plot for C5H11Br → C5H10 + HBr


First Order Plot for C5H11Br → C5H10 + HBr


Linearity impliesfirst order!

Example 13–4 First Order Kinetic Analysis

According to proposed Mechanism I, the decomposition of NO2 should follow first order kinetics. Do the experimental data of Figure 13–6 support this mechanism?


First-Order Rate Laws and Half-Life

Half-life – the time required for the reactant concentration to drop to one-half its original value, t1/2

The half-life is constant because it depends only on the rate constant and not the reactant concentrations.

1/2

0.693t =

ak


Example 13-5 Half Lives

Carbon-14 (14C) is a radioactive isotope with a half life of 5.73 x 103 years. The fractional amount of 14C present in an object can be used to determine its age. Calculate the rate constant for decay of 14C and determine how long it takes for 90% of the 14C in a sample to decompose.


Second – Order Reactions

2nd Order Integrated Rate Law for the reaction 2 A → products Notice it has a positive slope equal to the rate constant.

Half-life – notice it is dependent on the initial concentration

t 0

1 1 - = 2kt

[A] [A]

1/20

1t =

2k[A]


Example 13-6 Analysis of Rate Data

The Diels–Alder reaction, in which two alkenes combine to give a new product, is one of the most frequently used reactions for the synthesis of organic compounds. Thousands of examples are found in the chemical literature. The reaction of butadiene is a simple example.

Use the data in the table to determine the rate law and the rate constant for the Diels–Alder reaction of butadiene:Time (min) 0 4.0 8.0 12.0 16.0 20.0 30.0[C4H6] (M) 0.130 0.0872 0.0650 0.0535 0.0453 0.0370 0.0281


Isolation Method

A method for determining the order of a reaction by adjusting the conditions.

The initial concentration of one starting material is much smaller that the initial concentrations of the others.

Let’s examine the reaction of ozone with a terpene, isoprene which is produced by the plants and trees in the forest.

O3 + isoprene → products Rate=k[O3]y[isoprene]z


O3 + Isopene → Products


O3 + Isopene → Products


Linearity implies first order in [isoprene]

Example 13–7 Rate Law from Isolation Experiments

The reaction of hydrogen and bromine produces hydrogen bromide, a highly corrosive gas

H2 + Br2 → 2 HBrTo determine the rate law for this reaction, a chemist performed

two isolation experiments using different initial concentrations. Both experiments gave linear graphs of ln([H2]o/[H2]) vs. t, but with different slopes. Here are the details:[Br2]o [H2]o Slope of graph

3.50 x 10-5M 2.50 x 10-7M 8.87 x 10-4 s-1

2.00 x 10-5M 2.50 x 10-7M 6.71 x 10-4 s-1

Determine the rate law and the rate constant for the reaction.


Initial Rates

This method measures the rate at the very beginning of the reaction for different concentrations.

Consider the following data and determine the rate2 H2 + 2 NO → N2 + 2 H2O

[H2] [NO] Initial Rate

0.0010 M 0.0020 M 1.2 x 10-4 M/s0.0010 M 0.0030 M 1.8 x 10-4 M/s0.0020 M 0.0020 M 4.8 x 10-4 M/s


Initial Rates 2 H2 + 2 NO → N2 + 2 H2O

Initial rate = k[H2]xinitial[NO]y

initial

Initial rate1 = 1.2 x 10-4 M/s = k(0.0010 M)x(0.0020)y


Thus, y = 1


4 y1

4 y2

Initial rate 1.8 10 M / s (0.0030M)Initial rate 1.2 10 M / s (0.0020M)

Initial Rates 2 H2 + 2 NO → N2 + 2 H2O

Initial rate = k[H2]xinitial[NO]y

initial



Thus, x = 2Thus, rate = k[H2]2[NO]


x41

4 x2

Initial rate (0.0020M)4.8 10 M / sInitial rate 1.2 10 M / s (0.0010M)

Example 13–8 Rate Law from Initial Rates

The reaction of nitrogen dioxide with fluorine generates nitryl fluoride

2 NO2 + 2 F2 → 2 NO2FTo determine the rate law for this reaction, a chemist performed

several initial rate experiments using different initial concentrations. Determine the rate law and the rate constant for the reaction.

[NO2] [F2] [NO2F] Initial Rate

1.0 mM 5.0 mM 0.10 mM 2.0 x 10-4 M/s2.0 mM 5.0 mM 0.10 mM 4.0 x 10-4 M/s2.0 mM 5.0 mM 1.0 mM 4.0 x 10-4 M/s2.0 mM 7.5 mM 0.10 mM 6.0 x 10-4 M/s


Rate Parameters for Simple Reaction Orders


13.5 Linking Mechanisms and Rate Laws

Learning objective:

Show that the mechanism and rate law are closely related


13.5 Linking Mechanisms and Rate Laws

1. The mechanism is one or more elementary reactions describing how the chemical reaction occurs.

2. The sum of the individual steps in the mechanism must give the overall balanced chemical equation.

3. The reaction mechanism must be consistent with the experimental rate law.


Rate Laws and Reaction Mechanisms

Bimolecular Elementary ReactionA + B → Products Elementary Rate = k[A][B]

Unimolecular Elementary ReactionC → Products Elementary Rate = k[C]

When the first step of a mechanism is rate-determining, the predicted rate law for the overall reaction is the rate law for that first step.


Some Mechanisms and Rate Laws


Example 13–9 Predicted Rate Laws

At elevated temperature, NO2 reacts with CO to produce CO2 and NO. Below are 2 possible mechanisms. What is the rate law for each mechanism?

A.

B.


Rate-Determining Later Step

H2 + Br2 → 2 HBr Rate = k[H2][Br2]1/2


Example 13–10 Reaction Between NO and O3

Nitrogen oxide converts ozone into molecular oxygen, as follows:O3 + NO → O2 + NO2

The experimental rate law is rate = k[O3][NO]. Which of the following mechanisms are consistent with the experimental rate law?

Mechanism I Mechanism II

Mechanism III


1

2

2

k3 3

k3 2

k3 2

O NO O NO (slow)

O O 2O (fast)

NO NO 2NO (fast)

1

1

2

k

3 2k

k2

O O O (fast,reversible)

NO O NO (slow)

3k3 2 2O NO O NO (slow)

13.6 Reaction Rates and Temperature

In order for reactions to occur, molecules need a certain amount of energy to overcome an energy barrier - activation energy, Ea


Activation Energy

The reaction of O3 and NO to produce O2 and NO2

Is the forward reaction exo- or endothermic?

What about the reverse?


Molecular Orientation and Reaction Rate

Though there must be sufficient collision energy in order for a reaction to occur, there must also be the correct orientation.

The “steric-factor” is important in determining how fast a reaction is and it affects the value of the rate constant k.

The lower the probability of alignment, the lower the value of k and the slower the reaction.


The Arrhenius Equation

All the requirements: energy, orientation and collision frequency are summarized in the Arrhenius equation

1. Can be used to calculate the value of Ea

2. Can be used to calculate the rate constant at a given temperature if all else is known.

a-ERTk = Ae


Example 13–11 Graphing to Determine Ea

At high temperature cyclopropane isomerizes to propene.

When this reaction is studied at different temperatures, the following rate constants are obtained:

T(°C) 477 523 577 623K (s-1) 1.8 x 10-4 2.0 x 10-4 2.7 x 10-4 2.3 x 10-4

What is the activation energy for the isomerization of cyclopropane?


The Arrhenius Equation

The ratio of rates at two different temperatures is

a2

1 1 2

Ek 1 1ln

k R T T


Example 13–12 Calculating an Activation Energy

The reactions of NO2 have been studied as a function of temperature. For the following decomposition reaction, the rate constant is 2.7 x 10-2 M-1 s-1 at 227 °C and 2.4 x 10-1 M-1 s-1 at 277 °C:

2 NO2 → 2 NO + O2

Studies of the conversion of NO2 to N2O4 give k = 5.2 x 108 M-1 s-1 at both 25 °C and 87 °C:

2 NO2 → N2O4

Calculate the activation energies of these two reactions.


Example 13–13 Calculating k from Ea

When two NO2 molecules collide and react, they can form a bond or exchange an oxygen atom. For the oxygen atom exchange, Ea = 1.0 x 102 kJ/mol, and the rate constant at 250°C is 8.6 x 10-2 M-1 s-1. Estimate the rate constant for oxygen exchange at room temperature (25°C).


Values of Activation Energy

Most reactions between stable molecules have activation energies of 100 kJ/mol or greater, even when the overall reaction is exothermic.

For example, a violent explosion occurs when a spark is applied to a mixture of H2 and O2. The spark gives some of the molecules enough energy to overcome the activation barrier.


Kinetics of SN1 and SN2 Reactions

SN1 Mechanism:

Step 1: R-X R+ + X- (slow)Step 2: R+ + Nu Product (faster)

Rate = k[R–X]

SN2 Mechanism:

R–X + Nu R–Nu + LGRate = k[R–X][Nu] ()


13.7 Catalysis

Learning objective:

Explain the mechanisms by which catalysts function.


13.7 Catalysis

Catalysts – substances that speed up the rate of a chemical reaction

Catalysts are not consumed in a chemical reactionCatalyst Function: to provide a different pathway with a

lower activation energy.


Catalysis and the Ozone Problem


Types of Catalysts

Homogeneous catalysts – when the catalyst is in the same phase as the reacting substance

Heterogeneous catalysts – when the catalyst is in a different phase than the reacting substance.


Biocatalysis: Enzymes

Enzyme (E) – specialized proteins that catalyze specific biochemical reactions.

Substrate (S) – a reactant moleculeFirst the substrate binds to the enzyme

E + S Ý [ES]Then the binding distorts the structure of the substrate

[ES] Ý E---SFinally the distortion allows the susbtrate to react with

another reactant (R) and for the desired product (P).E---S + R → E --- P → E + P



Chapter 13 Visual Summary













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Chapter 13. Kinetics: Mechanisms and Rates of Reactions 13.1 What is a Reaction Mechanism? 13.2...

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