• 14-1 rates of chemical • 14-2 reaction rates and...

OFB Chapter 14 13/27/2003

Chapter 14Chemical Kinetics

• 14-1 Rates of Chemical Reactions

• 14-2 Reaction Rates and Concentrations

• 14-3 The Dependence of Concentrations on Time

• 14-4 Reaction Mechanisms• 14-5 Reaction Mechanism and

Rate Laws• 14-6 Effect of Temperature on

Reaction Rates• 14-7 Kinetics of Catalysis



• Thus far in the course, we have been using and talking about chemical reactions– Reactants, products, and how much is

involved

• Chemical Kinetics deals with how fast a reaction proceeds– Kinetics = Rates of Chemical Reactions– And how to deduce reaction mechanisms

from observed rates of reactions– Activation Energy– Catalysts


∆t∆[X]

tt[X][X] rate average

sLmolsL

mols

mol/L are units

∆t∆[X] rate reaction average

if

if

11

=−−

=

••=•

=

=

−−

To measure rates we could monitor the disappearance of reactants or appearance of products

e.g., NO2 + CO → NO + CO2


To measure rates we could monitor the disappearance of reactants or appearance of products

e.g., 2NO2 + F2 → 2NO2F


Generalized Reaction

aA + bB → cC + dD

tD

dtC

c

tB

btA

arate

∆∆

∆∆

∆∆

∆∆

][1][1

][1][1

=

+=

−=

−=


Order of a Reaction

→ 221

252 2NO OkON + decomposition

→

k[A]rate

k aA n

productse.g.,

=


→

][

k aA productse.g.,

nAkrate =

n does not have to be an integer

n = 3/2 “three halves order” reaction

Note that n does not equal the coefficient of the reactant. It is related to the reaction mechanism and determined experimentally.


→nm

k

[B]k[A] rate

products bB aA

=

+



• Example 14-2At elevated temperatures, HI reacts according to the chemical equation

2HI → H2 + I2at 443°C, the rate of reaction increases with concentration of HI, as shown in this table.

Data Point 1 2 3[HI] (mol L-1) 0.005 0.01 0.02Rate (mol L-1 s-1) 7.5E-04 3.0E-03 1.2E-02

a) Determine the order of the reaction with respect to HI and write the rate expressionb) Calculate the rate constant and give its unitsc) Calculate the instantaneous rate of reaction for a [HI] = 0.0020M

OFB Chapter 14 103/27/2003

part A toanswer 2

n

4

3

points datatwoanyofratio the take

k[HI] rate

and HI inorder second

0.00500.010

7.5x103.0x10

=

∴

=−

−

2HI → H2 + I2

Data Point 1 2 3[HI] (mol L-1) 0.005 0.01 0.02Rate (mol L-1 s-1) 7.5E-04 3.0E-03 1.2E-02

OFB Chapter 14 113/27/2003

Example 14-2b) Calculate the rate constant and give its units

Example 14-2c) Calculate the instantaneous rate of

reaction for a [HI] = 0.0020M

part A toanswer 2 k[X] rate =

OFB Chapter 14 123/27/2003

Similarly for two or more concentrations

nmp isorder ][][ +== nm BAkrate1)1()1(

arek of units −−−− sLmol pp

C B AExample

→+initial Rate

[A] [B] mol L-1 s-1

1.0E10-4 1.0E10 -4 2.8E10 -6

1.0E10-4 3.0E10 -4 8.4E10 -6

2.0E10-4 3.0E10 -4 3.4E10 -5

OFB Chapter 14 133/27/2003

C B AExampleFor →+ initial Rate

[A] [B] mol L-1 s-1

1.0E10-4 1.0E10 -4 2.8E10 -6

1.0E10-4 3.0E10 -4 8.4E10 -6

2.0E10-4 3.0E10 -4 3.4E10 -5

1st When A is constant (1.0E -4),

B increases X3 and rate increases X3

OFB Chapter 14 143/27/2003

C B AExampleFor →+ initial Rate

[A] [B] mol L-1 s-1

1.0E10-4 1.0E10 -4 2.8E10 -6

1.0E10-4 3.0E10 -4 8.4E10 -6

2.0E10-4 3.0E10 -4 3.4E10 -5

2nd When B is constant (3.0E -4),

A increases X2 and rate increases X4

OFB Chapter 14 153/27/2003

Can now solve for k

1-2- 26

1424

6

12

12

s molL2.8x10

][1x10][1x102.8x10

[B][A]ratek

[B]k[A]rate

=

=

=

=

−−

−

OFB Chapter 14 163/27/2003

14-3 The Dependence of Concentrations on Time

→

][][ k A products

orderfirst for general, in

AktArate =−=∆∆

First Order Reactions

For now, just accept this

important formula

OFB Chapter 14 173/27/2003

If we take the ln of both sides

Recall y = mx + b or

y = b + mxA plot of ln [A] vs t will be a straight line with the

Intercept = ln[A]0

Slope = -k

OFB Chapter 14 183/27/2003

ktAA −= 0]ln[]ln[First Order Reaction

-6

-5

-4-3

-2

-1

00.0E+00 2.0E+04 4.0E+04 6.0E+04

Time (in seconds)

ln [A

} (in

mol

/ L

Slope = -kt

OFB Chapter 14 193/27/2003


A useful concept for 1st order reactions (only) is the half=life

i.e., the time for ½ [A]0

(see book for the derivation)

Applies only to First order reactions

TRY

Examples 14-5 and Exercise

OFB Chapter 14 203/27/2003

Second Order Reactions

This is also the equation for a straight line

Y = mx +b

From calculus

→

2

k

ordersecondfor General, In

][][21- rate

products A2

AktA

==∆∆

OFB Chapter 14 213/27/2003

OFB Chapter 14 223/27/2003

Second Order Reaction

0

50

100

150

200

250

0 500 1000

Time (in seconds)

1/[A

] (L

/mol

)oA

kt][

12[A]1

+=

Slope = 2k


Half-life is NOT A useful concept for 2nd order reactions but,

t1/2 = 1/2k[A]0

OFB Chapter 14 233/27/2003

ktAA −= 0]ln[]ln[



oAkt

][12

[A]1

+=

OFB Chapter 14 243/27/2003

In the real world, if we don’t know the order of the reaction

[A]n plot both

First Order Reaction

-6-5-4-3-2-10

0.0E+00 2.0E+04 4.0E+04 6.0E+04

Time (in seconds)

ln [A

} (in

mol

/ L

Second Order Reaction

050

100150200250

0 500 1000

Time (in seconds)

1/[A

] (L

/mol

)

OFB Chapter 14 253/27/2003

Chapter 14-4Reaction Mechanisms

• Most reactions proceed not thru a single step but through a series of steps

• Each Step is called an elementary reaction

• Types of elementary Reactions1. Unimolecular (a single reactant)

E.g., A → B + C (a decomposition)

2. Bimolecular (most common type)E.g., A + B → products

3. Termolecular (less likely event)E.g., A + B + C → products

OFB Chapter 14 263/27/2003


• Reaction Mechanism– is a detailed series of elementary

steps and rates which are combined to yield the overall reaction

– One goal of chemical kinetics is to use the observed rate to chose between several possible reaction mechanisms.

OFB Chapter 14 273/27/2003

(fast) CO NO CO NO 2.)(slow) NONONO NO 1.)

223

322

+→+

+→+

Notice that NO3 is formed and consumed. This is called a reactive intermediate

Notice also that

Step 1 is bimolecular

Step 2 is bimolecular

Reactive intermediate is NO3

OFB Chapter 14 283/27/2003

A direct connection exists between the equilibrium constant of a reaction that takes place in a sequence of steps and the rate constants in each step.

Suppose the reaction proceeds by the following mechanism

][][]][[

]][[][]][][[

22

22

2

2

1

1

2

2

1

121

BADC

BAADCAK

kk

kk

kk

kkKKK

==

=

==

−−−−

The book uses a general reaction to illustrate this principle.

2 A (g) + B (g) ↔ C (g) + D (g)

1.) A (g) + A (g) ↔ A2 (g) rate = k1[A]2

2.) A2 (g) + B (g) ↔ C (g) + D (g) rate = k2[A2][B]

k1

k2

k-1

k-2

rate = k-1[A2]

rate = k-2[C][D]

a.) forward rate = reverse rateb.) Keq = kf / kr

OFB Chapter 14 293/27/2003

14-4 Kinetics and Chemical Equilibrium

E.g., a 3 step mechanism (not in book) similar to the previous general reaction

→← 22

1-k

1 k

ON NO NO 1.) +

→←

OH ON H ON 2.) 222-k

2 k

222 ++

→←

OH N H ON 3.) 223-k

3 k

22 ++

→←

O2H N 2H 2NO 222overall ++

OFB Chapter 14 303/27/2003

→←

O2H N 2H 2NO 222overall ++

From earlier this semester,

at equilibrium

Notice that N2O and N2O2 are reactive intermediates and not in the Equilibrium Constant (K)

Now, let’s derive the same Keq from its elementary steps.

OFB Chapter 14 313/27/2003

A.) at equilibrium

Forward rate = reverse rate

(called “The Principle of Detailed Balance”)

B.) Keq = kf / kr

→← 22

1-k

1 k

ON NO NO 1.) +

]O[Nk[NO]k 2212

1 −=

222

1

1eq1 ][

][KNO

ONkk

==−

OFB Chapter 14 323/27/2003

→←

OH ON H ON 2.) 222-k

2 k

222 ++

O]O][H[Nk]][HO[Nk

222

2222

−=

]][HO[NO]O][H[N

kkK

222

22

2

22 ==

−

eq

A.) at equilibrium


B.) Keq = kf / kr

OFB Chapter 14 333/27/2003

→←

OH N H ON 3.) 223-k

3 k

22 ++

O]][H[Nk]O][H[Nk

223

223

−=

]O][H[NO]][H[N

kkK

22

22

3

3eq3 ==

−

A.) at equilibrium


B.) Keq = kf / kr

OFB Chapter 14 343/27/2003

We added the elementary steps to get the Overall Reaction

O2H N 2H 2NO 222overall +←→

+

=

]O][HNO]][H[N

]][HO[NO]O][H[N

[NO]]O[N

22

22

222

22222

22

22

22

eq ][H[NO]][NO][HK =

OFB Chapter 14 353/27/2003

22

22

22

eq ][H[NO]][NO][HK =

Proven. This is the same K as that written directly from the overall reaction

Notice that the two reactive intermediates N2O2 and N2O cancelled out

OFB Chapter 14 363/27/2003

14-5 Reaction Mechanism & Rate Laws

Typically with a reaction one of several elementary step reaction is the slowest step. This is called the Rate Determining Step (RDS)

Case #1: When the RDS occurs first, the first step is slow and determines the rate of the overall reaction.

→

→

]][F[NOk rateRDS the is 1 Step

(fast) FNO F NO 2.)

(slow) FFNO F NO 1.)

221

2k2

2

2k1

22

=

+

++

OFB Chapter 14 373/27/2003

Reaction Progress

Energy

→

→

]][Fk1[NO rateRDS the is 1 Step

(fast) FNO F NO 2.)

(slow) FFNO F NO 1.)

22

2k2

2

2k1

22

=

+

++

NO2F

F + NO2

NO2+ F2

OFB Chapter 14 383/27/2003

Case #2: When the RDS occurs after one or more Fast steps, mechanisms are often signaled by a reaction order greater than two. The slow step determines the overall rate of the reaction.

But N2O2 is a reactive intermediate

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

22 2NO O 2NOreactionOverall→+

3 molecule reaction. Is it A Termolecular or Bimolecular reactions? Three way collisions are rare. Try a two step mechanism.

OFB Chapter 14 393/27/2003

Reaction Progress

Energy

2NO + O2

2NO2

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

N2O2 + O2

OFB Chapter 14 403/27/2003

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

222

1-

11

221-2

1

[NO]]O[N

kk K

]O[Nk [NO]k

mequilibriuat

==

=

OFB Chapter 14 413/27/2003

2122 [NO]K ]O[N =∴

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

Substituting for [N2O2] in the rate expression above

OFB Chapter 14 423/27/2003

Reaction Progress

Energy N2O2 + O2

slow

fast

2NO + O2

2NO2

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

][O[NO]Kk rate 22

12=

OFB Chapter 14 433/27/2003

→←

→←

→ (slow) OI OH HOI OH 3.)

(fast) Cl HOI HOCl I 2.)

(fast)OH HOCl OH OCl 1.)

1-2

k31-

1-

2-k

k21-

1-

1-k

k1

21-

++

++

++

Exercise 14-8 Suppose the mechanism for the reaction given in the Example 14-8 were

What rate law is now predicted?

OFB Chapter 14 443/27/2003

→←

→←




1-2

k31-

1-

2-k

k21-

1-

1-k

k1

21-

++

++

++

Notice that [OH-1] and [HOI] are both reactive intermediates. Need to express them in other species

1-1-1-1 OIClOCl I

solutionaqinreaction Overall

+→+−

OFB Chapter 14 453/27/2003

O]][H[OCl] [HOCl][OH K

21-

1-

1 = [H2O]=1

][HOCl][I][HOI][Cl K 1-

1-

2 =

→←

→←




1-2

k31-

1-

2-k

k21-

1-

1-k

k1

21-

++

++

++

][HOI][OHkrate 13

−=

OFB Chapter 14 463/27/2003

[HOCl]] [OClK ][OH

1-11- =

] [Cl][HOCl] [IK [HOI] 1-

1-2=

][HOI][OHkrate 13

−=Can now substitute [OH-1] and [HOI] in the original rate expression.

=

] [Cl][HOCl] [IK

[HOCl]] [OClKkrate 1-

1-

2

1-

13

OFB Chapter 14 473/27/2003

Reaction Progress

Energy

slowfast

fast

intermediates

→←

→←




1-2

k31-

1-

2-k

k21-

1-

1-k

k1

21-

++

++

++

1-1-1-1 OI Cl OCl Isolution aqin reaction Overall

+→+−

] [Cl] ][I [OClKKkrate 1-

-1-1

213=

[HOI]

[OH-1]

-11 OCl I +−

-1-1 OI Cl +

OFB Chapter 14 483/27/2003

Chain ReactionsA reaction that proceeds through a series of

elementary steps, in polymerization reactions these can be repeated several or many times.

1. Initiation, generation of a reactive intermediate.

• Free Radical

2. Propagation

• A reactive intermediate reacts with starting material or others to generate a new reactive intermediate

• Chain growth in polymers

3. Termination

• Reactive intermediate is “quenched in some manner to give a stable product.

OFB Chapter 14 493/27/2003

→

→ •+•

•

M M R

radical) free ( R Initiator

Initiation

ki

kd

•→+•

•→+•

•→+•

+1)(X kp

x

3 kp

2

2 kp

1

M M M

M M M

M M M

nPropagatio

Coupling M M MnTerminatio

y)(xyx +→•+•

OFB Chapter 14 503/27/2003

• Fiber Optics– UV curable polymers

• Fiber coating• Coloring• Ribbon materials

• Many Polymers– Free-radical polymerization

OFB Chapter 14 513/27/2003

14-6 The Effect of Temperature on Reaction Rates

k of unitsthehasandconstant a isfactor" lexponentia-pre" the is A

andmoleper energy are unitsenergy n Activatiothe is E

where

a

OFB Chapter 14 523/27/2003

Ak RTEa

EquationArrenhius

−=

An Arrenhius Plot

16

16.5

17

17.5

18

18.5

0.0025 0.0027 0.0029 0.0031 0.0033 0.0035

1/T (K-1)

ln k

y = mx + b

Slope = -Ea/R

OFB Chapter 14 533/27/2003

Transition State

OFB Chapter 14 543/27/2003

Reaction Progress

Energy

Ear

Eaf

Transition State

OFB Chapter 14 553/27/2003

OFB Chapter 14 563/27/2003

Reaction Progress

Energy

slowfast

fast

OFB Chapter 14 573/27/2003

Reaction Progress

Energy

slow

fast fast

OFB Chapter 14 583/27/2003


• Catalyst– provides a lower energy path, but it

does not alter the energy of the starting material and product

– rather it changes the energy of the transition (s), in the reaction

– A catalyst has no effect on the thermodynamics of the overall reaction

• Inhibitor– is a negative catalyst. It slows the rate

of a reaction frequently by barring access to path of low Ea and thereby forcing the reaction to process by a path of higher Ea.

OFB Chapter 14 593/27/2003

OFB Chapter 14 613/27/2003

OFB Chapter 14 623/27/2003

Kinetics of Catalysis• A catalyst has no effect on the

thermodynamics of the overall reaction

• It only provides a lower energy path• Examples

– Pt and Pd are typical catalysts for hydrogenation reactions (e.g., ethylene to ethane conversion)

– Enzymes act as catalysts• Phases

– Homogenous catalysis – the reactants and catalyst are in the same catalyst (gas or liquid phase)

– Heterogeneous catalysis – reaction occurs at the boundary of two different phases (a gas or liquid at the surface of a solid)

OFB Chapter 14 633/27/2003


• Example / exercise14-1, 14-2, 14-3, 14-4, 14-5, 14-6, 14-7, 14-8, 14-9

• Problems7, 9, 11, 15, 19, 21, 23, 25, 37, 41, 43, 45, 51, 53

• 14-1 rates of chemical • 14-2 reaction rates and...

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