CHAPTER 2: CHEMICAL KINETICS - nerdpotato · •Chemical kinetics - speed or rate at which a reaction occurs Chemical Kinetics Chemical kinetics: the study of reaction rate, a quantity

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CHAPTER 2:

CHEMICAL KINETICS

LESSON OUTCOME

• State, write and explain the term rate of reaction,

rate constant,order of reaction

• Apply the theories/concepts in chemical kinetics

• Solves the problem based on the concepts

• Determine rate of reaction

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• Chemical kinetics - speed or rate at which a

reaction occurs

Chemical Kinetics

Chemical kinetics: the study of reaction rate, a

quantity conditions affecting it,the molecular events

during a chemical reaction (mechanism), and

presence of other components (catalysis).

Factors affecting reaction rate:

Concentrations of reactants

Catalyst

Temperature

Surface area of solid reactants or catalyst

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Chemical Kinetics

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a

reactant or a product with time (M/s).

A B

rate = -D[A]

Dt

rate = D[B]

Dt

D[A] = change in concentration of A over

time period Dt

D[B] = change in concentration of B over

time period Dt

Because [A] decreases with time, D[A] is negative.

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A B

rate = -D[A]

Dt

rate = D[B]

Dt

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Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nm

light

Detector

D[Br2] a D Absorption

red-brown

t1< t2 < t3

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Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -D[Br2]

Dt= -

[Br2]final – [Br2]initial

tfinal - tinitial

slope of

tangentslope of

tangentslope of

tangent

instantaneous rate = rate for specific instance in time

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rate a [Br2]

rate = k [Br2]

k = rate

[Br2]= rate constant

= 3.50 x 10-3 s-1

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2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT

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rate = D[O2]

Dt RT

1 DP

Dt=

measure DP over time

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Reaction Rates and Stoichiometry

2A B

Two moles of A disappear for each mole of B that is formed.

rate = D[B]

Dtrate = -

D[A]

Dt

1

2

aA + bB cC + dD

rate = -D[A]

Dt

1

a= -

D[B]

Dt

1

b=

D[C]

Dt

1

c=

D[D]

Dt

1

d

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Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -D[CH4]

Dt= -

D[O2]

Dt

1

2=

D[H2O]

Dt

1

2=

D[CO2]

Dt

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Expressing reaction rates

For a chemical reaction, there are many ways to express

the reaction rate. The relationships among expressions

depend on the equation.

Note the expression and reasons for their relations for the

reaction

2 NO + O2 (g) = 2 NO2 (g)

D[O2] 1 D[NO] 1 D[NO2]

Reaction rate = – ——— = – — ———— = — ———

D t 2 D t 2 D t

Make sure you can write expressions for any reaction and

figure out the relationships. For example, give the

reaction rate expressions for

2 N2O5 = 4 NO2 + O2

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Calculating reaction rate

The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600

and 1200 s after the reactants are mixed at the appropriate

temperature. Evaluate the reaction rates for

2 N2O5 = 4 NO2 + O2

Solution:

(0.93 – 1.24)e-2 – 0.31e - 2M

Decomposition rate of N2O5 = – ————— = – ——————

1200 – 600 600 s

= 5.2e-6 M s-1.

Note however,

rate of formation of NO2 = 1.02e-5 M s-1.

rate of formation of O2 = 2.6e-6 M s-1.

The reaction rates

are expressed in 3

forms

Reaction Rates

Rates of reactions can be determined by

monitoring the change in concentration of either

reactants or products as a function of time. D[A]

vs Dt

Rxn Movie

In this reaction, the

concentration of butyl

chloride, C4H9Cl, was

measured at various

times, t.

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

[C4H9Cl] M

The average rate of the

reaction over each

interval is the change in

concentration divided

by the change in time:

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

Average Rate, M/s

• 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.

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

• A plot of concentration 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.

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

• The reaction slows down

with time because the

concentration of the

reactants decreases.

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

• In this reaction, the ratio

of C4H9Cl to C4H9OH is

1:1.

• Thus, the rate of

disappearance of C4H9Cl

is the same as the rate of

appearance of C4H9OH.

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

Rate =-D[C4H9Cl]

Dt=

D[C4H9OH]

Dt

Concentration and Rate

Each reaction has its own equation that gives

its rate as a function of reactant

concentrations.

this is called its Rate Law

To determine the rate law we measure the rate at

different starting concentrations.

Compare Experiments 1 and 2:

when [NH4+] doubles, the initial rate doubles.

Likewise, compare Experiments 5 and 6:

when [NO2-] doubles, the initial rate doubles.

This equation is called the rate law,

and k is the rate constant.

LESSON OUTCOME

• Write the rate law, rate equation & half life

for zero, first & second order of reaction

• Solves the problem involving rate equation

• Explain & describe factors that affect the

rate of reaction

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The Rate Law

The rate law expresses the relationship of the rate of a reaction

to the rate constant and the concentrations of the reactants

raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

Reaction is xth order in A

Reaction is yth order in B

Reaction is (x +y)th order overall

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F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant

(not product) concentrations.

• The order of a reactant is not related to the

stoichiometric coefficient of the reactant in the balanced

chemical equation.

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F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]x[ClO2]

y

Double [F2] with [ClO2] constant

Rate doubles

x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples

y = 1

rate = k [F2][ClO2]

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Determine the rate law and calculate the rate constant for the

following reaction from the following data:

S2O82- (aq) + 3I- (aq) 2SO4

2- (aq) + I3- (aq)

Experiment [S2O82-] [I-]

Initial Rate

(M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

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First-Order Reactions

A product rate = -D[A]

Dtrate = k [A]

k = rate

[A]= 1/s or s-1M/s

M=

D[A]

Dt= k [A]-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

[A] = [A]0e−kt ln[A] = ln[A]0 - kt

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

Graphical Determination of k

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The reaction 2A B is first order in A with a rate constant

of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease

from 0.88 M to 0.14 M ?

ln[A] = ln[A]0 - kt

kt = ln[A]0 – ln[A]

t =ln[A]0 – ln[A]

k= 66 s

[A]0 = 0.88 M

[A] = 0.14 M

ln[A]0

[A]

k=

ln0.88 M

0.14 M

2.8 x 10-2 s-1=

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First-Order Reactions

The half-life, t½, is the time required for the concentration of a

reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln 2

k=

0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant

of 5.7 x 10-4 s-1?

t½ln 2

k=

0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units of k (s-1)

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A product

First-order reaction

# of

half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

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Second-Order Reactions

A product rate = -D[A]

Dtrate = k [A]2

k = rate

[A]2= 1/M•s

M/sM2=

D[A]

Dt= k [A]2-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

1

[A]=

1

[A]0+ kt

t½ = t when [A] = [A]0/2

t½ =1

k[A]0

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Zero-Order Reactions

A product rate = -D[A]

Dtrate = k [A]0 = k

k = rate

[A]0= M/s

D[A]

Dt= k-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t = 0

t½ = t when [A] = [A]0/2

t½ =[A]02k

[A] = [A]0 - kt

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Summary of the Kinetics of Zero-Order, First-Order

and Second-Order Reactions

Order Rate Law

Concentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]0 - kt

1

[A]=

1

[A]0+ kt

[A] = [A]0 - kt

t½ln 2

k=

t½ =[A]02k

t½ =1

k[A]0

Factors That Affect Reaction Rates• Concentration of Reactants

– As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

• Temperature

– At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

• Catalysts

– Speed rxn by changing mechanism.

• Concentrations

Factors Affecting Rates

Rate with 0.3 M HCl

Rate with 6.0 M HCl

• Physical state of reactants

Factors Affecting Rates

Catalysts: catalyzed decomp of H2O2

2 H2O2 --> 2 H2O + O2

Factors Affecting Rates

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A catalyst is a substance that increases the rate of a

chemical reaction without itself being consumed.

Ea k

ratecatalyzed > rateuncatalyzed

Ea < Ea′

Uncatalyzed Catalyzed

)/( RTEaeAk

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In heterogeneous catalysis, the reactants and the catalysts

are in different phases.

In homogeneous catalysis, the reactants and the catalysts

are dispersed in a single phase, usually liquid.

• Haber synthesis of ammonia

• Ostwald process for the production of nitric acid

• Catalytic converters

• Acid catalysis

• Base catalysis

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Catalysis

A catalyst is a substance that changes

the rate of a reaction by lowing the

activation energy, Ea. It participates a

reaction in forming an intermediate, but

is regenerated.

Enzymes are marvelously selective

catalysts.

A catalyzed reaction,

NO (catalyst)

2 SO2 (g) + O2 — 2 SO3 (g)

via the mechanism

i 2 NO + O2 2 NO2

(3rdorder)

ii NO2 + SO2 SO3 + NO

Uncatalyzed

rxn

Catalyzed

rxn

rxn

Energy

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Catalyzed decomposition of ozone

The CFC decomposes in the atmosphere:

CFCl3 CFCl2 + Cl

CF2Cl3 CF2Cl + Cl.

The Cl catalyzes the reaction via the mechanism:

i O3 + h v O + O2,

ii ClO + O Cl + O2

iii O + O3 O2 + O2.

The net result or reaction is

2 O3 3 O2

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Homogenous vs. heterogeneous catalysts

A catalyst in the same phase (gases and solutions) as the

reactants is a homogeneous catalyst. It effective, but

recovery is difficult.

When the catalyst is in a different phase than reactants (and

products), the process involve heterogeneous catalysis.

Chemisorption, absorption, and adsorption cause reactions

to take place via different pathways.

Platinum is often used to catalyze hydrogenation

Catalytic converters reduce CO and NO emission.

• Temperature

Factors Affecting Rates

LESSON OUTCOME

• Explain the term of activation energy

collision theory, Arrhenius equation, reaction

mechanism, elementary step

• Calculation involving Arrhenius equation

• Write the rate law from reaction mechanism

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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.

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Exothermic Reaction Endothermic Reaction

The activation energy (Ea ) is the minimum amount of

energy required to initiate a chemical reaction.

A + B AB C + D++

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Temperature Dependence of the Rate Constant

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor

ln k = -Ea

R

1

T+ lnA

(Arrhenius equation)

)/( RTEaeAk

Alternate format:

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Alternate Form of the Arrhenius Equation

At two temperatures, T1 and T2

or

15 Chemical Kinetics 56

From k = A e – Ea / R T, calculate A, Ea, k at a specific

temperature and T.

The reaction:

2 NO2(g) -----> 2NO(g) + O2(g)

The rate constant k = 1.0e-10 s-1 at 300 K and the

activation energy

Ea = 111 kJ mol-1. What are A, k at 273 K and T when k =

1e-11?

Method: derive various versions of the same formula

k = A e – Ea / R T

A = k e Ea / R T

A / k = e Ea / R T

ln (A / k) = Ea / R T

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The reaction:

2 NO2(g) -----> 2NO(g) + O2(g)

The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy

Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?

Use the formula derived earlier:

A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1K –1*300 K))

= 2.13e9 s-1

k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K)

= 1.23e-12 s-1

T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1

= 285 K

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Importance of Molecular Orientation

effective collision

ineffective collision

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Reaction Mechanisms

The overall progress of a chemical reaction can be represented

at the molecular level by a series of simple elementary steps

or elementary reactions.

The sequence of elementary steps that leads to product

formation is the reaction mechanism.

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

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2NO (g) + O2 (g) 2NO2 (g)

Mechanism:

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Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction

mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step

and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules

reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

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Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall

balanced equation for the reaction.

• The rate-determining step should predict the same rate

law that is determined experimentally.

The rate-determining step is the slowest step in the

sequence of steps leading to product formation.

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Sequence of Steps in Studying a Reaction Mechanism

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The experimental rate law for the reaction between NO2 and CO

to produce NO and CO2 is rate = k[NO2]2. The reaction is

believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so

step 1 must be slower than step 2

Reaction Mechanisms

The sequence of events that describes the

actual process by which reactants become

products is called the reaction mechanism.

Reactions may occur all at once or through

several discrete steps.

Each of these processes is known as an

elementary reaction or elementary process.

• The molecularity of a process tells how many molecules are involved in the process.

• The rate law for an elementary step is written directly from that step.

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.

Slow Initial Step

• The rate law for this reaction is found experimentally to

be

Rate = k [NO2]2

• 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.

NO2 (g) + CO (g) NO (g) + CO2 (g)

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.

• Rate = k [NO2]2