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Chapter 14Chemical Kinetics

14.1 Factors that Affect Reaction Rates

Kinetics

- In kinetics we study the rate at which a chemical

process occurs. - Besides information about the speed at which

reactions occur, kinetics also sheds light on the

reaction mechanism (exactly how the reaction

occurs).

Factors That Affect Reaction Rates

- Physical State of the Reactants
- In order to react, molecules must come in contact

with each other. - The more homogeneous the mixture of reactants,

the faster the molecules can react.

Factors That Affect Reaction Rates

- Concentration of Reactants
- As the concentration of reactants increases, so

does the likelihood that reactant molecules will

collide.

Factors That Affect Reaction Rates

- Temperature
- At higher temperatures, reactant molecules have

more kinetic energy, move faster, and collide

more often and with greater energy.

Factors That Affect Reaction Rates

- Presence of a Catalyst
- Catalysts speed up reactions by changing the

mechanism of the reaction. - Catalysts are not consumed during the course of

the reaction. - P.575 GIST How does increasing the partial

pressures of the reactive components of a gaseous

mixture affect the rate at which the components

react with one another?

14.2 Reaction Rates

Reaction Rates

- Rates of reactions can be determined by

monitoring the change in concentration of either

reactants or products as a function of time.

Sample Exercise 14.1 Calculating an Average Rate

of Reaction

From the data given in the caption of Figure

14.3, calculate the average rate at which A

disappears over the time interval from 20 s to 40

s.

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- In this reaction, the concentration of butyl

chloride, C4H9Cl, was measured at various times.

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- The average rate of the reaction over each

interval is the change in concentration divided

by the change in time

p.577 GIST Why do the rates of reactions

decrease as concentrations decrease?

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- Note that the average rate decreases as the

reaction proceeds. - This is because as the reaction goes forward,

there are fewer collisions between reactant

molecules.

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- A plot of C4H9Cl vs. time for this reaction

yields a curve like this. - The slope of a line tangent to the curve at any

point is the instantaneous rate at that time.

p.578 GIST This figure shows two triangles used

to determine the slope of the curve at two

different times. How do you determine how large

to draw the triangle when determining the slope

of a curve at a particular point?

Reaction Rates

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- All reactions slow down over time.
- Therefore, the best indicator of the rate of a

reaction is the instantaneous rate near the

beginning of the reaction.

Sample Exercise 14.2 Calculating an Instantaneous

Rate of Reaction

Using Figure 14.4, calculate the instantaneous

rate of disappearance of C4H9Cl at t 0 (the

initial rate).

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Reaction Rates and Stoichiometry

C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)

- In this reaction, the ratio of C4H9Cl to C4H9OH

is 11. - Thus, the rate of disappearance of C4H9Cl is the

same as the rate of appearance of C4H9OH.

Reaction Rates and Stoichiometry

- What if the ratio is not 11?

2 HI(g) ??? H2(g) I2(g)

- In such a case,

Reaction Rates and Stoichiometry

- To generalize, then, for the reaction

Sample Exercise 14.3

- How is the rate at which ozone disappears related

to the rate at which oxygen appears in the

reaction 2O3(g) ? 3O2(g)? - If the rate at which O2 appears is 6.0 x 10-5 M/s

at a particular instant, at what rate is O3

disappearing at the same time?

Practice

- The decomposition of N2O5 proceeds according to

the following equation - 2N2O5(g) ? 4NO2(g) O2(g)
- If the rate of decomposition of N2O5 at a

particular instant in a reaction vessel is 4.2 x

10-7 M/s, what is the rate of appearance of NO2?

Of O2?

14.3 Concentration and Rate

Concentration and Rate

- One can gain information about the rate of a

reaction by seeing how the rate changes with

changes in concentration.

Concentration and Rate

- If we compare Experiments 1 and 2, we see that

when NH4 doubles, the initial rate doubles.

Concentration and Rate

- Likewise, when we compare Experiments 5 and 6,

we see that when NO2- doubles, the initial rate

doubles.

Concentration and Rate

- This means
- Rate ? NH4
- Rate ? NO2-
- Rate ? NH4 NO2-
- which, when written as an equation, becomes
- Rate k NH4 NO2-
- This equation is called the rate law, and k is

the rate constant.

Therefore,

Rate Laws

- A rate law shows the relationship between the

reaction rate and the concentrations of

reactants. - The exponents tell the order of the reaction with

respect to each reactant. - Since the rate law is
- Rate k NH4 NO2-
- the reaction is
- First-order in NH4 and
- First-order in NO2-.

Rate Laws

- Rate k NH4 NO2-
- The overall reaction order can be found by adding

the exponents on the reactants in the rate law. - This reaction is second-order overall.

14.3 GIST

- What is a rate law?
- What is the name of the quantity k in any rate

law? - The experimentally determined rate law for the

reaction 2NO 2H2 ? N2 2H2O is rate

kNO2H2. What are the reaction orders in this

rate law? Does doubling the concentration of NO

have the same effect on rate as doubling the

concentration of H2?

Sample Exercise 14.4

- Consider a reaction A B ? C for which rate

kAB2. Each of the boxes represent a reaction

mixture (A is red, B is purple). Rank these

mixtures in order of increasing rate of reaction.

Practice

- Assuming the rate kAB, rank the mixtures in

order of increasing rate.

Units of Rate Constant

- Units of rate (units of rate constant)(units of

concentration)x So, - Units of rate constant units of rate / (units

of concentration)x - Units of k MEMORIZE
- Zero order M/s
- First order s-1
- Second order M-1s-1
- Third order M-2s-1

Sample Exercise 14.5 Determining Reaction Order

and Units of Rate Constants

(a) What are the overall reaction orders for the

reactions described in Equations 14.9 and 14.10?

(b) What are the units of the rate constant for

the rate law in Equation 14.9?

14.9 2 N2O5 ? 4 NO2 O2 Rate kN2O5

14.10 CHCl3 Cl2 ? CCl4 HCl Rate

kCHCl3Cl21/2

14.11 H2 I2 ? 2HI Rate kH2I2

Method of Initial Rates

- Used to find the form of the rate law
- Write the general form of the rate law, then

determine order by doing the following. - Choose one reactant to start with
- Find two experiments where the concentration of

that reactant changes but all other reactants

stay the same - Write the rate laws for both experiments
- Divide the two rate laws
- Use log rules to solve for the order
- Follow the same technique for other reactants

Example

- Choose one reactant to start with
- Find two experiments where the concentration of

that reactant changes but all other reactants

stay the same

Example

Sample Exercise 14.6 Determining a Rate Law from

Initial Rate Data

Sample Exercise 14.6 Determining a Rate Law from

Initial Rate Data

14.4 The Change of Concentration With Time

Integrated Rate Laws

- Using calculus to integrate the rate law for a

first-order process gives us

Where

A0 is the initial concentration of A, and At

is the concentration of A at some time, t, during

the course of the reaction.

Integrated Rate Laws

- Manipulating this equation produces

ln At - ln A0 - kt

ln At - kt ln A0

which is in the form

y mx b

First Order

- lnA -kt lnA0
- can calculate the concentration at any time
- can be determined by a graph
- where
- y lnA
- x t
- m -k
- b lnA0

lnA vs. t is a straight line

First Order

- half-life
- time is takes for a reactant concentration to

reach half of its original concentration - does not depend on the concentration

First-Order Processes

ln At -kt ln A0

- Therefore, if a reaction is first-order, a plot

of ln A vs. t will yield a straight line, and

the slope of the line will be -k.

First-Order Processes

- Consider the process in which methyl isonitrile

is converted to acetonitrile.

First-Order Processes

- This data was collected for this reaction at

198.9 C.

First-Order Processes

- When ln P is plotted as a function of time, a

straight line results. - Therefore,
- The process is first-order.
- k is the negative of the slope 5.1 ? 10-5 s-1.
- P.587 GIST what do the y-intercepts in the above

graphs represent?

Sample Exercise 14.7 Using the Integrated

First-Order Rate Law

The decomposition of a certain insecticide in

water follows first-order kinetics with a rate

constant of 1.45 yr1 at 12 C. A quantity of

this insecticide is washed into a lake on June 1,

leading to a concentration of 5.0 107 g/cm3.

Assume that the average temperature of the lake

is 12 C. (a) What is the concentration of the

insecticide on June 1 of the following year? (b)

How long will it take for the concentration of

the insecticide to decrease to 3.0 107 g/cm3?

Sample Exercise 14.7 Using the Integrated

First-Order Rate Law

Second-Order Processes

- Similarly, integrating the rate law for a

process that is second-order in reactant A, we get

also in the form

y mx b

Second-Order Processes

- So if a process is second-order in A, a plot of

vs. t will yield a straight line, and the

slope of that line is k.

Second-Order Processes

The decomposition of NO2 at 300C is described by

the equation

and yields data comparable to this

Time (s) NO2, M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

Second-Order Processes

- Plotting ln NO2 vs. t yields the graph at the

right.

- The plot is not a straight line, so the process

is not first-order in A.

Time (s) NO2, M ln NO2

0.0 0.01000 -4.610

50.0 0.00787 -4.845

100.0 0.00649 -5.038

200.0 0.00481 -5.337

300.0 0.00380 -5.573

Second-Order Processes

- Graphing ln vs. t, however, gives this

plot.

- Because this is a straight line, the process is

second-order in A.

Time (s) NO2, M 1/NO2

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

Second Order

- 1/A kt 1/A0
- 1/A vs. t is a straight line with a slope of k

Sample Exercise 14.8 Determining Reaction Order

from the Integrated Rate Law

Sample Exercise 14.8 Determining Reaction Order

from the Integrated Rate Law

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Half-Life

- Half-life is defined as the time required for

one-half of a reactant to react. - Because A at t1/2 is one-half of the original

A, - At 0.5 A0.

Half-Life

- For a first-order process, this becomes

ln 0.5 -kt1/2

-0.693 -kt1/2

NOTE For a first-order process, then, the

half-life does not depend on A0.

Half-Life

- For a second-order process,

Second Order

- half-life does depend on the concentration for

second order

14.5 GIST

- If a solution containing 10.0 g of a substance

reacts by first-order kinetics, how many grams

remain after 3 half-lives? - How does the half-life of a second-order reaction

change as the reaction proceeds?

Sample Exercise 14.9 Determining the Half-Life of

a First-Order Reaction

The reaction of C4H9Cl with water is a

first-order reaction. Figure 14.4 shows how the

concentration of C4H9Cl changes with time at a

particular temperature. (a) From that graph,

estimate the half-life for this reaction. (b) Use

the half-life from (a) to calculate the rate

constant.

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Zero Order

- half-life does depend on the concentration

Example

- A certain first order reaction has t1/2 of 20.0

minutes - Find k
- How much time is required for this reaction to be

75 complete?

14.5 Temperature and Rate

Temperature and Rate

- Generally, as temperature increases, so does the

reaction rate. - This is because k is temperature dependent.

The Collision Model

- In a chemical reaction, bonds are broken and new

bonds are formed. - Molecules can only react if they collide with

each other.

The Collision Model

- Furthermore, molecules must collide with the

correct orientation and with enough energy to

cause bond breakage and formation.

Activation Energy

- In other words, there is a minimum amount of

energy required for reaction the activation

energy, Ea. - Just as a ball cannot get over a hill if it does

not roll up the hill with enough energy, a

reaction cannot occur unless the molecules

possess sufficient energy to get over the

activation energy barrier.

Reaction Coordinate Diagrams

- It is helpful to visualize energy changes

throughout a process on a reaction coordinate

diagram like this one for the rearrangement of

methyl isonitrile.

Reaction Coordinate Diagrams

- The diagram shows the energy of the reactants and

products (and, therefore, ?E). - The high point on the diagram is the transition

state.

- The species present at the transition state is

called the activated complex.

- The energy gap between the reactants and the

activated complex is the activation energy

barrier.

MaxwellBoltzmann Distributions

- Temperature is defined as a measure of the

average kinetic energy of the molecules in a

sample.

- At any temperature there is a wide distribution

of kinetic energies.

MaxwellBoltzmann Distributions

- As the temperature increases, the curve flattens

and broadens. - Thus at higher temperatures, a larger population

of molecules has higher energy.

MaxwellBoltzmann Distributions

- If the dotted line represents the activation

energy, then as the temperature increases, so

does the fraction of molecules that can overcome

the activation energy barrier.

- As a result, the reaction rate increases.

MaxwellBoltzmann Distributions

- This fraction of molecules can be found through

the expression - where R is the gas constant and T is the Kelvin

temperature.

f e

Arrhenius Equation

- Svante Arrhenius developed a mathematical

relationship between k and Ea - k A e
- where A is the frequency factor, a number that

represents the likelihood that collisions would

occur with the proper orientation for reaction.

Arrhenius Equation

- Taking the natural logarithm of both sides, the

equation becomes - ln k - ( ) ln A

y m x b

Therefore, if k is determined experimentally at

several temperatures, Ea can be calculated from

the slope of a plot of ln k vs. .

14.5 GIST

- What is the central idea of the collision model?
- Why isnt collision frequency the only factor

affecting a reaction rate?

Sample Exercise 14.10 Relating Energy Profiles to

Activation Energies and Speed of Reaction

Sample Exercise 14.11 Determining the Energy of

Activation

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Sample Exercise 14.11 Determining the Energy of

Activation

14.6 Reaction Mechanisms

Reaction Mechanisms

- The sequence of events that describes the actual

process by which reactants become products is

called the reaction mechanism.

Reaction Mechanisms

- Reactions may occur all at once or through

several discrete steps. - Each of these processes is known as an elementary

reaction or elementary process. - Rate laws can be determined directly from

elementary reactions. - The rate law for the overall reaction is

determined from the slowest elementary step in

the mechanism

Reaction Mechanisms

- Example
- Overall NO2(g) CO(g) ? NO(g) CO2(g)
- 2 Steps NO2(g) NO2(g) ? NO3(g) NO(g)
- NO3(g) CO(g) ? NO2(g) CO2(g)

Reaction Mechanisms

- intermediate
- species that is formed and consumed in the steps

so does not appear in overall reaction - appears as a product in one reaction and then

reactant in following reaction - catalyst
- species used to speed up the reaction
- is not in overall reaction
- appears as reactant and then as a product
- Intermediates and catalysts do not appear in the

rate law.

Reaction Mechanisms

- The molecularity of a process tells how many

molecules are involved in the process.

How do you know if the proposed mechanism is

correct?

- 1. The series of elementary steps must add up to

give overall balanced equation - 2. must agree with the experimentally determined

rate law

p.598 GIST

- What is the molecularity of this elementary

reaction? - NO(g) Cl2(g) ? NOCl(g) Cl(g)

Sample Exercise 14.12 Determining Molecularity

and Identifying Intermediates

Sample Exercise 14.12 Determining Molecularity

and Identifying Intermediates

Sample Exercise 14.13 Predicting Rate Law for an

Elementary Reaction

Multistep Mechanisms

- In a multistep process, one of the steps will be

slower than all others. - The overall reaction cannot occur faster than

this slowest, rate-determining step.

p.601 GIST

- Why cant the rate law for a reaction generally

be deduced from the balanced equation for the

reaction?

Slow Initial Step

NO2 (g) CO (g) ??? NO (g) CO2 (g)

- The rate law for this reaction is found

experimentally to be - Rate k NO22
- CO is necessary for this reaction to occur, but

the rate of the reaction does not depend on its

concentration. - This suggests the reaction occurs in two steps.

Slow Initial Step

- A proposed mechanism for this reaction is
- Step 1 NO2 NO2 ??? NO3 NO (slow)
- Step 2 NO3 CO ??? NO2 CO2 (fast)
- The NO3 intermediate is consumed in the second

step. - As CO is not involved in the slow,

rate-determining step, it does not appear in the

rate law.

Sample Exercise 14.14 Determining the Rate Law

for a Multistep Mechanism

The reaction is believed to occur in two

steps The experimental rate law is rate

kO3NO2. What can you say about the relative

rates of the two steps of the mechanism?

Fast Initial Step

2 NO (g) Br2 (g) ??? 2 NOBr (g)

- The rate law for this reaction is found to be
- Rate k NO2 Br2
- Because termolecular processes are rare, this

rate law suggests a two-step mechanism.

Fast Initial Step

- A proposed mechanism is

Step 2 NOBr2 NO ??? 2 NOBr (slow)

Step 1 includes the forward and reverse reactions.

Fast Initial Step

- The rate of the overall reaction depends upon the

rate of the slow step. - The rate law for that step would be
- Rate k2 NOBr2 NO
- But how can we find NOBr2?

Fast Initial Step

- NOBr2 can react two ways
- With NO to form NOBr
- By decomposition to reform NO and Br2
- The reactants and products of the first step are

in equilibrium with each other. - Therefore,
- Ratef Rater

Fast Initial Step

- Because Ratef Rater ,
- k1 NO Br2 k-1 NOBr2
- Solving for NOBr2 gives us

Fast Initial Step

- Substituting this expression for NOBr2 in the

rate law for the rate-determining step gives

k NO2 Br2

Sample Exercise 14.15 Deriving the Rate Law for a

Mechanism with a Fast Initial Step

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14.7 Catalysis

Catalysts

- Catalysts increase the rate of a reaction by

decreasing the activation energy of the reaction,

more collisions have the minimum Ea. - Catalysts change the mechanism by which the

process occurs. - is never consumed in the reaction so it is not

considered a reactant

Catalysts

- One way a catalyst can speed up a reaction is by

holding the reactants together and helping bonds

to break.

Catalysis

- heterogeneous
- in different phases
- homogeneous
- same phase

- enzyme
- catalysts in body

Enzymes

- Enzymes are catalysts in biological systems.
- The substrate fits into the active site of the

enzyme much like a key fits into a lock.

14.7 GIST

- How does a catalyst increase the rate of a

reaction? - How does a homogeneous catalyst compare with a

heterogeneous one regarding the ease of recovery

of the catalyst from the reaction mixture? - What names are given to the following aspects of

enzyme catalysis - The place on the enzyme where catalysis occurs
- The substances that undergo catalysis

Sample Integrative Exercise Putting Concepts

Together

Formic acid (HCOOH) decomposes in the gas phase

at elevated temperatures as follows HCOOH(g) ?

CO2(g) H2(g) The uncatalyzed decomposition

reaction is determined to be first order. A graph

of the partial pressure of HCOOH versus time for

decomposition at 838 K is shown as the red curve

in Figure 14.28. When a small amount of solid ZnO

is added to the reaction chamber, the partial

pressure of acid versus time varies as shown by

the blue curve in Figure 14.28. (a) Estimate the

half-life and first-order rate constant for

formic acid decomposition. (b) What can you

conclude from the effect of added ZnO on the

decomposition of formic acid? (c) The progress of

the reaction was followed by measuring the

partial pressure of formic acid vapor at selected

times. Suppose that, instead, we had plotted the

concentration of formic acid in units of mol/L.

What effect would this have had on the calculated

value of k? (d) The pressure of formic acid vapor

at the start of the reaction is 3.00 102 torr.

Assuming constant temperature and ideal-gas

behavior, what is the pressure in the system at

the end of the reaction? If the volume of the

reaction chamber is 436 cm3, how many moles of

gas occupy the reaction chamber at the end of the

reaction? (e) The standard heat of formation of

formic acid vapor is ?Hf 378.6 kJ/mol.

Calculate ?H for the overall reaction. If the

activation energy (Ea) for the reaction is 184

kJ/mol, sketch an approximate energy profile for

the reaction, and label Ea, ?H, and the

transition state.

Figure 14.28