14-1 Rates of Chemical • 14-2 Reaction Rates and...

OFB Chapter 14 17/10/2003

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 14 107/10/2003

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 14 117/10/2003

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 14 127/10/2003

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 14 137/10/2003

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 14 147/10/2003

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 14 157/10/2003

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 14 167/10/2003

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 14 177/10/2003

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 14 187/10/2003

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 14 197/10/2003

First Order Reactions

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 14 207/10/2003

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 14 217/10/2003

OFB Chapter 14 227/10/2003

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 14 237/10/2003

ktAA −= 0]ln[]ln[First Order Reactions


oAkt

][12

[A]1

+=

Summary

Integrated Rate Laws

OFB Chapter 14 247/10/2003

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 14 257/10/2003

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 14 267/10/2003


• 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 14 277/10/2003

(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 14 287/10/2003

Chemical EquilibriumA 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.

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

eqba

dc

r

f

dcr

baf

dcr

baf

rf

K[B][A][D][C]

kk 2.)

[D][C]k[B][A]k 1.)

[D][C]krate reaction Reverse

[B][A]krate reaction Forward

dDcCk&k

bBaA

==

=

=

=

+ →+ ←

OFB Chapter 14 297/10/2003

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 two step 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)

2.) A2 (g) + B (g) ↔ C (g) + D (g)

k1

k2

k-1

k-2

Rate-1 = k-1[A2]

Rate-2 = k-2[C][D]

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

Rate1 = k1[A]2

Rate2 = k2[A2][B]

(sometimes kn / k-n)

OFB Chapter 14 307/10/2003

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 14 317/10/2003

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 14 327/10/2003

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 14 337/10/2003

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 14 347/10/2003

→←

→

]][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 14 357/10/2003

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 14 367/10/2003

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 14 377/10/2003

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 14 387/10/2003

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 14 397/10/2003

Transition State

OFB Chapter 14 407/10/2003

Reaction Progress

Energy

Ear

Eaf

Transition State

OFB Chapter 14 417/10/2003

OFB Chapter 14 427/10/2003

Reaction Progress

Energy

slowfast

fast

OFB Chapter 14 437/10/2003

Reaction Progress

Energy

slow

fast fast

OFB Chapter 14 447/10/2003


• 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 14 457/10/2003

OFB Chapter 14 467/10/2003

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 14 477/10/2003


• 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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