chapter outline - western oregon universitypostonp/ch223/pdf/ch13-s15.pdf · 2019-04-08 ·...

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© 2014 W. W. Norton Co., Inc.

• 13.1 Cars, Trucks, and Air Quality

• 13.2 Reaction Rates

• 13.3 Effect of Concentration on Reaction Rates

• 13.4 Reaction Rates, Temperature, and the

Arrhenius Equation

• 13.5 Reaction Mechanisms

• 13.6 Catalysts

Chapter Outline


Cars, Trucks, and Air Quality

• Photochemical Smog: Mixture of gases formed when sunlight interacts with compounds produced in internal combustion engines

• Depends on chemical kinetics

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Important Reactions in Smog

1. N2(g) + O2(g) → 2 NO(g) ΔH= 180.6 kJ

2. 2 NO(g) + O2(g) → 2 NO2(g) ΔH= -114.2 kJ

3. NO2(g) NO(g) + O(g)

4. O2(g) + O(g) → O3(g)

5. O(g) + H2O(g) → 2 OH(g)

sunlight⎯⎯⎯→

Note: Products of some reactions are reactants in

other reactions.


Variations of Smog Components

Photodecomposition

of NO2 leads to high

levels of O3 in the

afternoon.

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Why Study Kinetics?

• We can control the reaction conditions to

obtain product as quickly and economically

as possible

• To understand reaction mechanisms – a

study of kinetics sheds light on how a

reaction proceeds


Factors that control reaction kinetics

• Chemical makeup of reactants and products

• Concentration of reactants

• Temperature

• Catalysts

• Surface area

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Arrhenius Equation


• 13.6 Catalysts

Chapter Outline


Rates

If driving a car, rate = change in position

change in time= slope=

∆dist∆t =

d2−d1t2−t1

Time, hours →

Dis

tan

ce

, m

iles → (t2, d2)

(t1, d1)

rise = 𝑑2 − 𝑑1

run = 𝑡2 − 𝑡1

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Reaction Rates

rate = change in conc.change in time

= slope = ∆[A]∆t =

[A]2−[A]1t2−t1

Time (s, min, hr) →

Con

ce

ntr

atio

n, M

→

(t2, [A]2)

(t1, [A]1)

rise =

[𝐴]2−[𝐴]1

run = 𝑡2 − 𝑡1


Reaction Rates:

rate of loss of reactant = negative slope

rate = change in conc.change in time

= slope = ∆[A]∆t =

[A]2−[A]1t2−t1

Time (s, min, hr) →

Con

ce

ntr

atio

n, M

→

(t2, [A]2)

(t1, [A]1)

rise =

[𝐴]2−[𝐴]1

run = 𝑡2 − 𝑡1

Here the slope is negative, so the

rate is multiplied by -1 to make

the rate positive (convention)

Rate = -∆[A]∆t

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For the reaction A → B

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8 9 10

Co

nc

en

tra

tio

n, M

Time, min

[A]t

[B]t

Note how the reaction

rate decreases over time

Here the slope is

negative


Reaction Rates and

Stoichiometry

2A B

As an example, two moles of A disappear for each

mole of B that is formed, so the rate of disappearance

of A is twice the rate of appearance of B.

Mathematically: −∆[𝐴]

∆𝑡= 2

∆[𝐵]

∆𝑡or

−1

2

∆[𝐴]

∆𝑡=

∆[𝐵]

∆𝑡

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Reaction Rates and

Stoichiometry (cont’d)

2A 3BA trickier example: two moles of A

disappear and three moles of B are

formed. First divide both sides by 3 --

Mathematically: −∆[A]∆t

= 23

∆[B]∆t

or

23

A B So now the rate of disappearance of A is

2/3 times the rate of appearance of B.

−12

∆[A]∆t

= 13

∆[B]∆t


aA + bB cC + dD

In general -

Reactants have

the minus signproducts

= 1

𝑐

∆[𝐶]

∆𝑡=

1

𝑑

∆[𝐷]

∆𝑡= -

1

𝑏

∆[𝐵]

∆𝑡-1

𝑎

∆[𝐴]

∆𝑡

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Actual Reaction Rates

N2(g) + O2(g) → 2 NO(g)Smog

reaction #1


Average Rates

N2(g) + O2(g) → 2 NO(g)

Average Rate: Change in concentration of

reactant or product over a specific time interval

average rate =∆[NO]

∆t

= ([NO]f − [NO]i)

(tf − ti)

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Practice: Reaction Rates

• Calculate the average reaction rate for the

formation of NO between 5 s and 10 s, and

then between 25 s and 30 s, using the data

below: N2(g) + O2(g) → 2 NO(g)

(a)

(b)

Δ NOAverage Rate =

Δt

(a)

(b)

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Instantaneous Reaction Rates


Instantaneous Reaction Rates

• Instantaneous Rate:

• Reaction rate at a particular instant

• Determined graphically as tangential slope

of concentration vs. time plot

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• 13.1 Cars, trucks, and Air Quality




Arrhenius Equation


• 13.6 Catalysts

Chapter Outline


Effect of Concentration on

Reaction Rates

A typical plot of reactant concentration versus time. The

rate depends on the number of molecular collisions

which depends on the concentration of reactants.

a) Initial rate (will be used in the “Method of Initial

Rates”)

b) midpoint

c) end

The rate approaches zero

as the number of reactant

molecules decreases =

concentration effect

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Reaction Order and Rate Constants

We observe in chemical reactions that rates increase as

the reactant concentrations increase. The rate is found to

be proportional to the reactant concentrations raised to

some experimentally determined power -

General Rate Law: Rate = k [A]m [B]n

o k = rate constant

o m and n are the reaction orders with respect to reactants A and B

o (m + n ) = overall order


Overall Order

The sum of the exponents for each reactant in the Rate Law.

rate = k[A] 1st order in A, 1st order overall

rate = k[A][B] 1st order in A and B, 2nd order overall

rate = k[A]2[B] 2nd order in A, 1st order in B, 3rd order overall

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The Method of Initial Rates

An experimental procedure for determining the

overall reaction order by systematically varying

the concentrations of each reactant

Rate = k [A]m [B]n

Exp. # [A], M [B], M Rate,

Ms-1

1. 0.10 0.10 1 x 10-5

2. 0.10 0.20 2 x 10-5

3. 0.20 0.20 8 x 10-5

Example experimental setup:


2 NO(g) + O2(g) → 2 NO2(g)

rate = k [NO]m [O2]n

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Systematically varying the concentrations

of each reactant, e.g., the exponent n -

Hold the concentration of one reactant while doubling,

tripling, etc the other concentration of the other reactant:

n: [NO] is held constant. [O2] doubles and the

rate doubles. Therefore n = 1



Systematically varying the concentrations

of each reactant, e.g., the exponent m -


m: [O2] is held constant. [NO] doubles and

the rate quadruples. Therefore m = 2

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2 NO(g) + O2(g) → 2 NO2(g)

Solve for the rate constant k by using the data from any of

the experiments, e.g. experiment #1 -

rate = k [NO]2 [O2]

k =rate

[NO]2[O2]

k =1 x 10−6 M s−1

0.010 𝑀 2 [0.010 M]

k = 1.0 M-2 s-1


Experimentally the rate = k [F2][ClO2]

Characteristics of 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, e.g.

1

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

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Equations for determining the exponents:

e.g. rate = k [A]m [B]n

Experiment [A], M [B], M rate

1. [A]1 [B]1 rate1

2. [A]1 [B]2 rate2

rate1rate2

=[B]1[B]2

n

Want to solve for n, so take the log of both sides first -

lograte1rate2

= log[B]1[B]2

n

rate1rate2

=[B]1[B]2

n

and since

log Xn = n log X

lograte1rate2

= n log[B]1[B]2

n=

lograte1rate2

log[𝐵]1[𝐵]2

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Sample Exercise 13.3 - Deriving a Rate

Law From Initial Reaction Rate Data

N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]

n

n

m



n

n =

lograte1rate2

log[O2]1[O2]2

=log

707500

log0.0200.010

=0.15

0.30= 1/2

Calculating n, the exponent on O2:

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Calculating m, the exponent on N2:

m =

lograte2rate3

log[N2]2[N2]3

=log

500125

log0.0400.010

=0.60

0.60= 1


n


rate = k [N2][O2]1/2

Solve for the rate constant k by using the data

from any of the experiments, e.g. experiment #1 -

k =707 M s−1

0.040 [0.020]1/2k = 1.25 x 105 M-1/2 s-1

k =rate

[N2][O2]1/2


n

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o Integrated Rate Laws: First-Order Reactions

o Reaction half-lives

o Integrated Rate Laws: Second-Order Reactions


Arrhenius Equation


• 13.6 Catalysts

Chapter Outline


Relation Between Reactant Concentration

and Time: Integrated Rate Laws

I. First Order Kinetics: rate = -∆[A]t

∆t= k1[A]

II. Second Order Kinetics: rate = -∆[A]t

∆t= k2[A]2

0.000

0.020

0.040

0.060

0.080

0.100

0.120

0 10 20 30 40 50

[A] t→

time →

2nd order

1st order

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[A]t = concentration of A at some time = t

[A]o = concentration of A at time t = 0 (initial concentration)

k1 = first order rate constant

(units = s-1, min-1, h-1, etc)

A (+ other reactants) → products

First Order Integrated Rate Equation

rate = -∆[A]t

∆t= k1[A] the solution to this equation is:


Some mathematics: ln[A]

[A]o= −k1t

exp ln[A]t[A]o

= exp −k1t

[A]t[A]o

= exp −k1t

[A]t =

Ao exp −k1t

Solving for [A]t Solving for time, t

ln[A]

[A]o= −k1t

t = −1k1

ln[A]

[A]o

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1st Order Kinetics Example – HW # 69

The following reaction is known to be 1st order with

a rate constant k = 6.32 x 10-4 s-1. How much N2O5

remains in solution after 1 hr if [N2O5]o = 0.50 M?

2 N2O5 → 4 NO2 + O2

[N2O5]t = [N2O5]o exp −k1t

[N2O5]t = (0.50 M) exp[-(6.32 x 10-4 s-1)(3600 s)]

[N2O5]t = (0.50 M)(0.103)

[A]t = 0.052 M


Linearizing the 1st Order Equation

Linearization of an equation is a common method

used in science where useful information can be

obtained from the slope and intercept (e.g. measuring

the rate constant).

ln[A]t[A]o

= −k1t And since lnab

= ln (a) − ln (b)

ln [A]t − ln [A]o = −k1t

ln [A]t = −k1t + ln [A]o

y = m x + b

m = -𝑘1

b = ln [A]o

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Example Linearization

(spreadsheet on the webpage)

0

0.002

0.004

0.006

0.008

0.01

0.012

0 20 40 60 80 100 120 140 160 180 200

[A]t

→

time, s

Example Data

-12

-10

-8

-6

-4

-2

0

0 20 40 60 80 100 120 140 160 180 200

[A]t

→

time, s

1st Order Linearization

time, s 1st order data

0 0.01

10 0.007408182

20 0.005488116

30 0.004065697

40 0.003011942

50 0.002231302

60 0.001652989

70 0.001224564

80 0.00090718

90 0.000672055

100 0.000497871

110 0.000368832

120 0.000273237

130 0.000202419

140 0.000149956

150 0.00011109

160 8.22975E-05

170 6.09675E-05

180 4.51658E-05

190 3.34597E-05

200 2.47875E-05

ln([

A] t

[A] t


Integrated Rate Law: 1st-Order

3 2 O + OhO ⎯⎯→

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y = m x + b

Plot data after

taking the log

Fit to a

straight line

m = -6.93 x 10-3 s-1

m = -k1 = -6.93 x 10-3 s-1

k1 = 6.93 x 10-3 s-1


• Half-Life (t1/2):

• The time in the course of a chemical reaction

during which the concentration of a reactant

decreases by half

• From integrated rate law, when [A]t = ½ [A]o:

• Rearrange: 1/2

0.693t =

k

0

Aln = - 0.693 = -

At kt

Half-Life: 1st Order Reactions

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Half-Life: 1st-Order Reactions

t1/2 (1) = t1/2 (2) = t1/2 (3)


Sample Exercise 13.5: Calculating the Half-Life

of a First-Order Reaction.

The rate constant for the decomposition of N2O5 at a particular

temperature is 7.8 x 10-3 s-1. What is the half-life of N2O5 at that

temperature?

t1/2 =0.693

k

t1/2 =0.693

7.8 x 10−3 s−1

t1/2 = 89 s

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Some example half lives for First Order

processes -

Process Half life, t1/2

Rate constant,

k (s-1)

Radioactive decay of 238U 4.51 X 109 yr 4.87 X 10-18

Radioactive decay of 14C 5.73 X 103 yr 1.21 X 10-4

C12H22O11(aq) + H2O →

C6H12O6(aq) + C6H12O6(aq)

(glucose) (fructose)

8.4 h 2.3 X 10-6

HC2H3O2 + H2O → H3O+ +

C2H3O2-

8.9 X 10-7 s 7.8 X 105


Second Order Kinetics


[A]t = concentration of A at some time = t

[A]o = concentration of A at time t = 0 (initial concentration)

k2 = second order rate constant (units = M-1 s-1, M-1 min-1, etc)

rate = -∆[A]t∆t

= k2[A]2 the solution to this equation is:

1[A]t

= k2t +1

[A]o

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Solving for [A]t Solving for time, t

Some mathematics:1

[A]t= k2t +

1[A]o

k2t =[A]o− [A]t[A]t[A]o

t =1

k2

[A]o− [A]t[At][A]o

k2t =1

[A]t-

1[A]o

1[A]t

= k2t +1

[A]o

1

[A]t=

[A]ok2t + 1

[A]o

[A]t=[A]o

[A]ok2t+1


Linearizing the 2nd Order Equation

y = m x + b

m = k2

b = 1

[A]o

1[A]t

= k2t +1

[A]o

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ln [NO2] plot: curved

1/[NO2] plot: straight line

slope = k2; intercept = 1/[NO2]o

How to graphically distinguish between 1st and 2nd order kinetics


Sample Exercise 13.6: Distinguishing

between First and Second order reactions

Chlorine monoxide accumulates in the stratosphere above

Antarctica each winter and plays a key role in the formation of

the ozone hole above the South pole each spring. Eventually

ClO decomposes according to the reaction:

2 ClO(g) → Cl2(g) + O2(g)

The kinetics of this reaction were studied in a laboratory

experiment at 298 K, and the data are shown in Table

13.10(b). Determine the order of the reaction, the rate law,

and the value of the rate constant k.

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2 ClO(g) → Cl2(g) + O2(g)


• because the plot of 1/[ClO]

is linear, the reaction is 2nd

order

• So the rate law = k2[ClO]2

• from the straight-line fit:

m = 7.24 x 106

y = 7.24E+06x + 6.63E+07

Results:

m = k2 = 7.24 x 106 M-1s-1

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Half Life for Second Order Reactions

When t = t½ ,

then [A]t1/2 = [A]o/2

1[A]t

= k2t +1

[A]o

k2t =1

[A]t-

1[A]o

k2t1/2 =1

[A]o/2-

1[A]o

k2t1/2 =2

[A]o-

1[A]o

k2t1/2 =1

[A]o

t1/2 =1

k2[A]o


Sample Exercise 13.7: Calculating the

half-life of Second order reactions

Calculate the half-life of the second-order decomposition of

NO2(g) if the rate constant is 0.544 M-1 s-1 at a particular

temperature; the initial concentration of NO2 = 0.0100 M.

t1/2 =1

k2[A]o=

1

(0.544 𝑀−1 𝑠−1)(1.00 x 10−2 𝑀)

t1/2 = 184 s

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Zero Order Reactions

rate = -∆[A]t

∆t

In zero-order reactions, the

concentration of the

reactant has no effect on

the rate. One important

application is measuring

enzyme kinetics.


= k0[A]0 = k0

And the integrated rate law is -

[A] t

y = m x + b

m = -ko

[A]t = −kot + [A]o


Half Life for Zero Order Reactions

When t = t½ ,

then [A]t1/2 = [A]o/2[A]t = −kot + [A]o

[A]o2

= −kot½ + [A]o

[A]o2

− [A]o = −kot½

−[A]o

2= −kot½

[A]o2ko

= t½

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Glucose bound to hexokinase

http://www.rcsb.org/pdb/explore/jmol.do?structureId=1BDG&bionumber=1

http://www.rcsb.org/pdb/explore/jmol.do?structureId=1BDG&bionumber=1

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Rate of product formation vs. substrate

concentration in an enzyme-catalyzed reaction.


Summary of the Different Rate Laws and

How to Experimentally Distinguish Them

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• 13.4 Reaction Rates, Temperature, and the Arrhenius

Equation


• 13.6 Catalysts

Chapter Outline

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Factors Affecting Rate

• Temperature:

o Increased temperature increases kinetic energy of molecules, molecular collisions

• Proper Molecular Orientation

• Activation Energy (Ea):

oMinimum energy of molecular collisions required to break bonds in reactants, leading to formation of products

• These factors are incorporated into the Arrhenius equation


Effect of Temperature

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Molecular Orientation

O3 + NO → O2 + NO2


Reaction Energy Profile

High-energy transition state (“activated complex”)

Ea =

Activation

energy

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The Transition State or “Activated Complex”

H2(g) + I2(g) → 2 HI(g) H = -13 kJ

H2 + I2

2 HI

I I

H H

Transition state or

activated complex


Transition state - bonds between reactants are

breaking at the same time that bonds are

forming between the products

H - H

I - I→ →

H H

I I

H H

I I

Bonds breaking

Bonds forming

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Arrhenius Equation

• k = rate constant

• A is a “collisional frequency factor” , includes the probability of colliding with the correct orientation

• Ea = activation energy

• R = Ideal Gas Law constant = 8.314 J/mol K

• T = temperature (in kelvin)

All of these controlling factors are combined

together into the Arrhenius Equation…..

k = Ae−Ea/RT


Ea and A can be measured graphically using

linearization again….

k = Ae-Ea/RT

ln k = ln (A▪e-Ea/RT) and since ln (a▪b) = ln a + ln b

ln k = ln A + ln (e-Ea/RT)

y = m x + b

ln k = −EaR

1T

+ ln A m = -Ea/R

b = ln A

ln k = ln A + - Ea/RT

Arrhenius Equation

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Graphical Determination of Ea

= +

- 1ln lnaEk A

R T

m = -Ea/R

b = ln A

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y = -1580x + 27.99

m = −EaR

Ea = −(8.314 J/mol K)(−1580 K) = 13,140 J = 13.1 kJ

Ea = −Rm

b = ln A A = eb

A = eb = e27.99 = 1.45 x 1012 s-1


Measuring Ea using data from two

different temperatures -

= +

1

1

- 1ln lnaEk A

R T

= +

2

2

- 1ln lnaEk A

R T

Another way to measure k if only two data points are available

ln k1 − ln k2=−Ea

R

1

T1+ ln A −

−Ea

R

1

T2+ ln A

lnk1k2

=−Ea

R

1

T1−

1

T2ln

k1k2

=Ea

R

1

T2−

1

T1

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Back to Exercise

13.9 where the first

two experiments

have been chosen.

lnk1k2

=Ea

R

1

T2−

1

T1






Arrhenius Equation


• 13.6 Catalysts

Chapter Outline

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Method of Initial

Rates: collect kinetic

data in the lab for

each reactant

General Rate Law:

analyze the kinetic data in

order to determine the

exponents on the

reactants

Propose a reasonable

reaction mechanism

that is consistent with

the rate law


Reaction Mechanisms

• Reaction Mechanism: A set of steps that describe

how a reaction occurs at the molecular level; must be

consistent with the rate law for the reaction

• Elementary Step: Molecular-level view of a single

process taking place in a chemical reaction

• Intermediate: Species produced in one step of a

reaction and consumed in a subsequent step

• Molecularity: The number of ions, atoms, or

molecules involved in an elementary step in a reaction

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Most reactions involve a

sequence of steps

Overall Br2(g) + 2 NO(g) → 2 BrNO(g)

Reaction“intermediate”

Step 1 Br2(g) + NO(g) → Br2NO(g)

Step 2 Br2NO(g) + NO(g) → 2 BrNO(g)


Each step in the mechanism is called an “elementary

step” and has it’s own Ea and rate constant, k.

Br2(g) + NO(g)Br2NO(g)

[+ NO(g)]

2 BrNO(g)

Ea1

Ea2

Reaction coordinate →

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Molecularity of Elementary Steps

1 molecule = unimolecular

2 molecules = bimolecular

3 molecules = termolecular

2. The exponents in the rate expressions for

elementary steps are the same as the

stoichiometric coefficients in each

elementary step. The reason why can be

explained by returning to Sec. 13-3…..

1. termolecular (and higher) reactions are

improbable because it's difficult to have 3 or more

molecules colliding simultaneously with the

correct energy and orientation.

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Why the rate is proportional to the product of

concentrations (Sec.13-3 revisited)

NO(g) + O3(g) → NO2(g) + O2(g) rate = k[NO][O3]

1 x 1 = 1 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 3 x 3 = 9


What if the rate law contains exponents?

2 NO(g) + O2(g) → 2 NO2(g) rate = k[NO]2[O2]

When holding O2 constant, doubling NO

should increase the rate by 22 = 4X;

tripling should increase by 32 = 9X, etc.

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equivalent

equivalent

Doubling Concentrations

Tripling Concentrationsequivalent equivalent

equivalent equivalent

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equivalent

equivalent


Reaction mechanisms have a RATE DETERMINING

STEP - the slowest step in the mechanism.

2 NH3 + OCl- → N2H4 + Cl- + H2O

Step 1 NH3 + OCl- → NH2Cl + OH- fast

Step 3 N2H5+ + OH- → N2H4 + H2O fast

Overall

Reaction2 NH3 + OCl- → N2H4 + Cl- + H2O

Step 2 NH2Cl + NH3 → N2H5+ + Cl- slow

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The overall rate expression must be

consistent with the rate law for the slow step

so the rate law = k[NH2Cl][NH3]

2 NH3 + OCl- → N2H4 + Cl- + H2O

Step 1 NH3 + OCl- → NH2Cl + OH- fast

Step 3 N2H5+ + OH- → N2H4 + H2O fast

Overall

Reaction2 NH3 + OCl- → N2H4 + Cl- + H2O

Step 2 NH2Cl + NH3 → N2H5+ + Cl- slow


• The sum of the elementary steps must give the overall

balanced equation for the reaction.

• The rate-determining step is the slowest step in the

mechanism

• The rate law for the reaction is given by the slow step,

unless…

• There is a fast equilibrium step prior to the slow step, and so

a calculation has to be made starting with ratef = rater

(coming up)

• the detection of an intermediate supports a given

mechanism

Writing plausible reaction mechanisms:

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Example Reaction Mechanisms

2 NO2 (g) → 2 NO(g) + O2(g)

Proposed mechanism:

Step 1: 2 NO2(g) → NO(g) + NO3(g)

Step 2: NO3(g) → NO(g) + O2(g)

rate = k[NO2]2

Since the observed rate is k[NO2]2, the first

step must be the RDS.

Known from exp.

2 NO2 (g) → 2 NO(g) + O2(g)

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Rate Determining Step for Mechanisms

With an Equilibrium Elementary Step

2 NO(g) + O2(g) → 2NO2(g) rate = k[NO]2[O2]

Proposed mechanism:

Step 1: NO + O2 NO3 rate1 = k1[NO][O2] fast & rev.

Step 2: NO3 + NO → 2NO2 rate2 = k2[NO3][NO]

2 NO(g) + O2(g) → 2NO2(g)

Is this mechanism consistent

with the observed rate law?


Step 1: NO + O2 NO3

2 NO(g) + O2(g) → 2 NO2(g)

at equil. ratef = rater

ratef = kf [NO][O2]

kf

rater = kr [NO3]

kr

so kf [NO][O2] = kr [NO3]

by rearrangement: [NO3] =kr

[NO][O2]kf

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[NO3] =kr

[NO][O2]kf

= k’[NO]2[O2]

Substituting into rate2 = k2[NO3][NO]

= k2kr

[NO][O2]kf [NO]

• Which is consistent with

the observed rate law

• mechanism verified


2 NO(g) + O2(g) → 2 NO2(g)

Step 1: NO + O2 NO3 Step 2: NO3 + NO → 2NO2

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HW# 107

At a given temperature, the rate law of the reaction between

NO and Cl2 is proportional to the product of concentrations of

the two gases: [NO][Cl2]. The following two-step mechanism

has been proposed for the reaction:

Step 1: NO(g) + Cl2(g) → NOCl2(g)

Step 2: NOCl2(g) + NO(g) → 2 NOCl(g)

Overall: 2 NO(g) + Cl2(g) → NOCl2(g)

Which step must be the RDS if this mechanism is correct?


HW# 108

Mechanism of Ozone Destruction:

Ozone decomposes thermally to oxygen in the following

mechanism:

O3(g) → O(g) + O2(g)

O(g) + O3(g) → 2 O2(g)

2 O3(g) → 3 O2(g)

The reaction is second order in ozone. What properties of the

two elementary steps (specifically, relative rate and

reversibility) are consistent with this mechanism?

Step 1:

Step 2:

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2 NO2(g) → N2O4(g)

N2O4(g) → N2O3(g) + O(g)

N2O3(g) + O(g) → N2O2(g) + O2(g)

N2O2(g) → 2 NO(g)

(a)slow

fast

fast

fast

HW# 110

The rate laws for the thermal and photochemical decomposition

of NO are different. Which of the following mechanisms are

possible for the thermal decomposition of NO2, and which are

possible for the photochemical decomposition of NO2? For

thermal decomposition, Rate = kT[NO2]2, and for photochemical

decomposition, Rate = kP[NO2].


2 NO2(g) → NO(g) + NO3(g)

NO3(g) → NO(g) + O2(g)

slow

fast

(b)

HW# 110, cont’d

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HW# 110, cont’d

NO2(g) → N(g) + O2(g)

N(g) + NO2(g) → N2O2(g)

N2O2(g) → 2 NO(g)

slow

fast

slow

(c)




• 13.3 Effect of Concentration on Reaction

Rates


Arrhenius Equation


• 13.6 Catalysts

Chapter Outline

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Catalysts and the Ozone Layer


Ea for Step 2 = 17.7 kJ/mol (slow step)

Natural Photodecomposition

of Ozone:

Mechanism:

⎯⎯⎯→sunlight

3 2O ( ) O ( ) + O( )g g gStep 1:

⎯⎯→3 2O ( ) + O( ) 2O ( ) g g gStep 2:

2 O3(g) → 3 O2(g)Overall:

slow

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CFC Emissions and Ozone

→3 2ClO( ) + O ( ) Cl( ) + O ( )g g g g

Ea for Cl-catalyzed reaction = 2.2 kJ/mol

Chlorofluorocarbons (CFC) e.g. CCl2F2, CCl3F, CClF3

Stable! Migrate into the stratosphere

Net: 2 O3(g) → 3 O2(g)

⎯⎯→hν

3 2CCl F( ) CCl F( ) + Cl( )g g g

→3 2Cl( ) + O ( ) ClO( ) + O ( )g g g g


Energy Profiles for O3 Decomposition

⎯⎯→ν

3 22 O 3 Oh

Cl atom:

• Not consumed

during the reaction

• Homogeneous

catalyst

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Types of Catalysts

Homogeneous

Catalyst:• One in the same

phase as the

reactants

Heterogeneous

Catalyst:• One in a different

phase from the

reactants

Platinum-rhodium gauze

catalyst used in the

production of HNO3


Automobile Catalytic Converter

Catalyst = Pt-NiO, finely divided powder. Higher

surface area provides more sites for adsorption and

reaction

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Catalytic Converters


The heterogeneous catalyst provides a surface where

the reactants can be ADSORBED on an ACTIVE SITE -

3 H2(g) + N2(g) →

2 NH3(g)

Fe

Industrial production

of ammonia (Haber

Process)-

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chapter outline - western oregon universitypostonp/ch223/pdf/ch13-s15.pdf · 2019-04-08 ·...

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