announcements, nov. 19 - georgia institute of...

OFB Chapter 14 111/30/2004

Announcements, Nov. 19th

• PRS Quiz results through July 9are posted on the course website.

• July 19, 21– Review Exam 3 and exam analysis– Chapter 14

• Friday July 23 PRS Course Review– Bring PRS unit and calculator– Wrap-up course

• Monday July 26 Final Exam– 2:50-5:40 (this room)– 4 Crib sheets from Exam 1, 2, 3 and

Chapter 14– Part A (1-6), B (7-9), C (10-13), D (14)

Lecture topic: Chemical Kinetics– Check against Kinetics LabCheck

against Kinetics Lab– On Final Exam, NOT 3rd Exam

WebCT set posted (Homework 11) – Ch. 14.16(b), 24(b)– Due Wed. 24th, 1700

• Monday, Nov 22– 3rd Exam , usual format; topics:

Chs. 9-13 (omit 9.5-6, 12.4-5, 13.6-7)- Practice Exam, posted

• Wednesday, Nov 24– RSVP your attendance

• This announcement to be posted

OFB Chapter 14 211/30/2004

Chapter 14: Chemical Kinetics

• 14-1 Rates of Chemical Reactions

• 14-2 Rates & 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

OFB Chapter 14 311/30/2004

Chapter 14: Chemical Kinetics

• Thus far, we have been using and talking about chemical reactions– Reactants, products, and how much is involved– Chemical equilibrium is “dynamic”

• Chemical Kinetics* concerns how fast a reaction proceeds– Kinetics = Rates of Chemical Reactions– And how to deduce reaction mechanisms from

observed rates of reactions– Activation Energy (Role of Catalysts)

How to analyze, quantify, and predict observed reaction rates?

*Kinetics, Kinematics, Cinematics, Kino

OFB Chapter 14 411/30/2004

Reaction Mechanisms

• Most reactions, as written, actually proceed through a series of steps

• Each Step is called an elementary reaction

• Classes of elementary reactions:1. Unimolecular (a single reactant)

A → B + C (a decomposition)2. Bimolecular (very common)

A + B → products3. Termolecular (less likely event)

A + B + C → products

OFB Chapter 14 511/30/2004

tX

ttXX

sLmolsL

molsLmol

tX

if

if

∆∆

∆∆

][][][ rate average

/ are units

][ rate reaction average

11

=−−

=

••=•

=

=

−−

To measure rates, monitor disappearance of reactants or appearance of products

e.g., NO2 + CO → NO + CO2

∆t]∆[CO

∆t∆[NO]

∆t∆[CO]

∆t]∆[NOrate rxn

2

2

+=+=

−=−=

OFB Chapter 14 611/30/2004

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

e.g., 2NO2 + F2 → 2NO2F

tFNO

tF

tNOrate

∆∆

∆∆

∆∆

][21

][][21

2

22

+=

−=

−=

OFB Chapter 14 711/30/2004

For a Generalized Reaction

aA + bB → cC + dD

tD

dtC

c

tB

btA

arate

∆∆

∆∆

∆∆

∆∆

][1][1

][1][1

=

+=

−=

−=

tX

ttXX

if

if

∆∆ ][][][

rate average =−−

=

NO2 + CO → NO + CO2

OFB Chapter 14 811/30/2004

Order of a Reaction

→ 221

252 2NO OkON + decomposition

][ 52ONkrate =Called a Rate Expression

Or Rate Law

Called a Rate constant f(Temperature)

→

k[A]rate

k aA n

productse.g.,

=An “nth order” reaction

OFB Chapter 14 911/30/2004

Summary

aA + bB → cC + dD

tD

dtC

c

tB

btA

arate

∆∆

∆∆

∆∆

∆∆

][1][1

][1][1

=

+=

−=

−=

nk[A]rate

products kaA

=

→ Called a Rate Expression

Or Rate Law

nth order reaction

→nm

k

[B]k[A] rateproducts bB aA

=

+

Overall order = m + n

mth order in [A] and nth order in [B]

OFB Chapter 14 1011/30/2004

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 [HI] RatePoint (mol L-1) (mol L-1 s-1)

1 0.005 7.5 x 10-4

2 0.01 3.0 x 10-3

3 0.02 1.2 x 10-2

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 1111/30/2004

( )( )n

n

HIkrate

HIkrate

22

11

][

][

=

=22

k I H 2HI +→

( )( )

( )

part A toanswer 2

n

n

4

3

n1

n2

1

2

pointsdatatwoanyofratio the take

k[HI] rate

and HI inorder second 2n24

0.00500.010

7.5x103.0x10

[HI]k[HI]k

raterate

=

∴==

=

=

−

−

Data [HI] RatePoint (mol L-1) (mol L-1 s-1)

1 0.005 7.5 x 10-4

2 0.01 3.0 x 10-3

3 0.02 1.2 x 10-2

a) Determine the order of the reaction with respect to HI and write the rate expression

OFB Chapter 14 1211/30/2004

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

part A toanswer 2 k[HI] rate =

B toanswer 1-1-

21-

1- 1-4

2

smol L 30k

)L mol (0.0050s L mol 7.5x10

[HI]ratek

=

==−

Example 14-2c) Calculate the instantaneous rate of

reaction for a [HI] = 0.0020M

( )( )C toanswer

1-1-4-

21-1-1-

2

sLmol 10x 1.2

L mol 0.0020s mol L 30

k[HI]rate

=

=

=

OFB Chapter 14 1311/30/2004

Similarly for two or more concentrations

nmp isorder ][][ +== nm BAkrate→products bB aA k+

1)1()1(arek of units −−−− sLmol pp

C B AExample

→+( )( )n1

n2

1

2

pointsdata two anyofratiothetake

[X]k[X]k

raterate

=

initial Rate

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

1.0x10-4 1.0x10 -4 2.8x10 -6

1.0x10-4 3.0x10 -4 8.4x10 -6

2.0x10-4 3.0x10 -4 3.4x10 -5

OFB Chapter 14 1411/30/2004

C B AExampleFor →+ initial Rate

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

1.0x10-4 1.0x10 -4 2.8x10 -6

1.0x10-4 3.0x10 -4 8.4x10 -6

2.0x10-4 3.0x10 -4 3.4x10 -5

nm BAkrate ][][= ( )( )n1

n2

1

2

pointsdatatwo any ofratiothetake

[X]k[X]k

raterate

=

1st When [A] is constant (1.0x 10-4),

When [B] increases times 3 and rate increases times 3

1

1

][][

][][

BAkrate

BBn

n

=

=

∴

OFB Chapter 14 1511/30/2004

C B AExampleFor →+ initial Rate

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

1.0x10-4 1.0x10 -4 2.8x10 -6

1.0E10-4 3.0x10 -4 8.4x10 -6

2.0x10-4 3.0x10 -4 3.4x10 -51][][ BAk m=rate

2nd When [B] is constant (3.0x 10-4),

[A] increases times 2 & rate increases times 4

12

2

][][

24]2[

4][

BAkrate

m

A m

=∴

==

=( )( )m1

m2

1

2

pointsdata two anyofratiothetake

[X]k[X]k

raterate

=

What is the rxn order with respect to

A? B? Overall?

OFB Chapter 14 1611/30/2004

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

=

=

=

=

−−

−

1)1()1(are k of units −−−− smolL pp

Actually Example 13-4

2 NO + O2 → 2 NO2

][][ 22 ONOkrate =

OFB Chapter 14 1711/30/2004

14-3 The Dependence of Concentrations on Time

First Order Reactions

→

k[A]∆t∆[A]1rate

aA

n

productsk

orderfirst for general, in

=−=a

kt0

A of conc initial theis0if

e[A][A]

[A] −= For now, just

accept this important formula

a term that diminishes [A]0

OFB Chapter 14 1811/30/2004

kt0e[A][A]

ReactionOrder 1st afor LawRateIntegrated

−=If we take the ln of both sides

ktAA −= 0]ln[]ln[

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 1911/30/2004

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 = -k

k = 1.72 x 10-5 s-1

OFB Chapter 14 2011/30/2004

Second Order Reactions

n

k

k[A]∆t∆[A]

21- rate

products A 2order secondfor General,In

==

→

From calculus

o[A]12kt

[A]1

ReactionOrder 2nd afor Law Rate Integrated

+=

0][1intercept

2slope][

12[A]1

A

kA

kto

=

=

+=

This is also the equation for a straight line

y = mx +b

OFB Chapter 14 2111/30/2004

Slope = 2k

Or

k = ½ slope

OFB Chapter 14 2211/30/2004

oAkt

][12

[A]1

+=

Second Order Reaction

0

50

100

150

200

250

0 500 1000

Time (in seconds)

1/[A

] (L

/mol

)

Slope = 2k

OFB Chapter 14 2311/30/2004

Summary

Integrated Rate Laws


ktAA −= 0]ln[]ln[

Second Order Reactions

oAkt

][12

[A]1

+=

OFB Chapter 14 2411/30/2004

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

[A]n plot both

If a plot of ln [A] vs t is a straight line, then the reaction is 1st order

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

)

If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order

OFB Chapter 14 2511/30/2004

In a study of the reaction of pyridine (C5H5N) with methyl iodide (CH3I) in a benzene solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.

[C5H5N] [CH3I] Rate(mol L-1) (mol L-1) (mol L-1 s-1)

1.0 x 10-4 1.0 x 10-4 7.5 x 10-7

2.0 x 10-4 2.0 x 10-4 3.0 x 10-6

2.0 x 10-4 4.0 x 10-4 6.0 x 10-6

a) Write the rate law for this reactionb) Calculate the rate constant and give its unitsc) Predict the initial reaction rate that would be seen in a solution in which [C5H5N] = 5.0 x 10-5 M and [CH3I] = 2.0 x 10-5 M

OFB Chapter 14 2611/30/2004


1.0 x 10-4 1.0 x 10-4 7.5 x 10-7

2.0 x 10-4 2.0 x 10-4 3.0 x 10-6

2.0 x 10-4 4.0 x 10-4 6.0 x 10-6

a) Write the rate law for this reaction

( )( )n

23

n33

2

3

I][CHkI][CHk

raterate

=

( )( )

122

10 x 210 x 4

10 x 310 x 6

n4-

n4-

6-

6-

==

=

n

n

nNHCICHkrate ][][ 551

3=∴

OFB Chapter 14 2711/30/2004


1.0 x 10-4 1.0 x 10-4 7.5 x 10-7

2.0 x 10-4 2.0 x 10-4 3.0 x 10-6

2.0 x 10-4 4.0 x 10-4 6.0 x 10-6


n1N]5H5[C

11I]3[CHk

n2N]5H5[C

12I]3[CHk

raterate

1

2

=

( ) ( )( ) ( )

122

)2()2(410 x 110 x 110 x 210 x 2

10 x 7.510 x 3

1

n4-14-

n4-14-

7-

6-

==

=

=

n

n

n

155

13 ][][ NHCICHkrate=∴

OFB Chapter 14 2811/30/2004

In a study of the reaction of pyridine (C5H5N) with methyl iodide (CH3I) in a benzene solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.


1.0 x 10-4 1.0 x 10-4 7.5 x 10-7

2.0 x 10-4 2.0 x 10-4 3.0 x 10-6

2.0 x 10-4 4.0 x 10-4 6.0 x 10-6


b) Calculate the rate constant and give its unitsc) Predict the initial reaction rate that would be seen in a solution in which [C5H5N] = 5.0 x 10-5 M and [CH3I] = 2.0 x 10-5 M

]][[ 553 NHCICHkrate =∴

OFB Chapter 14 2911/30/2004

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.

dDcCrkfk

bBaA +←

→+

a.) at equil: forward reaction rate = reverse reaction rate

b.) Keq = kf / kr

eqba

dc

r

f

dcr

baf

dcr

baf

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

kk 2.)

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

[D][C]kratereaction Reverse

[B][A]kratereaction Forward

==

=

=

=

OFB Chapter 14 3011/30/2004

a.) at equil: forward rate = reverse rateb.) Keq = kf / kr (sometimes kn / k-n)

A general reaction to illustrate this principle.

2 A (g) + B (g) ↔ C (g) + D (g)

Suppose the reaction proceeds by the following two step mechanism

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]

Rate1 = k1[A]2

Rate2 = k2[A2][B]

22

1

11 ][

][AA

kkK ==− ]][[

]][[

22

22 BA

DCkkK ==−

][][]][[

]][[]][[

][][

22

22

2

2

1

121

BADC

BADC

AAK

kk

kkKKK

=

=

==

−−

OFB Chapter 14 3111/30/2004

PRS Quiz, Question 1In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.

[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)

1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q1 What is the order of the reaction with respect to compound A?

1. 02. 13. 24. 3

OFB Chapter 14 3211/30/2004

OFB Chapter 14 3311/30/2004

PRS Quiz, Solution 1In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.


1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q1 What is the order of the reaction with respect to compound A?

1. 02. 13. 24. 3

( )( )

( )( )

224

1x10

2x109x10

36x10

[A]k[A]k

raterate

n4-

n4-

6-

6-

n2

n3

2

3

==

=

=

n

n

OFB Chapter 14 3411/30/2004



1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q2 What is the order of the reaction with respect to compound B?

1. 02. 13. 24. 3

OFB Chapter 14 3511/30/2004

OFB Chapter 14 3611/30/2004

( )( )

( )( )

133

1x10

3x103x109x10

[B]k[B]k

raterate

n4-

n4-

6-

6-

n1

n2

1

2

==

=

=

n

n



1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q2 What is the order of the reaction with respect to compound B?

1. 02. 13. 24. 3

OFB Chapter 14 3711/30/2004



1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q3 What is the rate expression for the above reaction?

1. Rate = k [A]2. Rate = k [A][B]3. Rate = k [A]2[B]2

4. Rate = k [A]2[B]

OFB Chapter 14 3811/30/2004

OFB Chapter 14 3911/30/2004



1.0 x 10-4 1.0 x 10-4 3.0 x 10-6

1.0 x 10-4 3.0 x 10-4 9.0 x 10-6

2.0 x 10-4 3.0 x 10-4 3.6 x 10-5

Q3 What is the rate expression for the above reaction?

1. Rate = k [A]2. Rate = k [A][B]3. Rate = k [A]2[B]2

4. Rate = k [A]2[B]

OFB Chapter 14 4011/30/2004

dDcCrkfk

bBaA +←

→+

tD

dtC

ctB

btA

arate

∆∆

=

∆∆

+=

∆∆

−=

∆∆

−=

][1][1][1][1

nmf [B][A]k rate =

Called a Rate Expression Or Rate Law

kf called rate constant

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

eqba

dc

r

f

dcr

baf

dcr

baf

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

kk 2.)

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

[D][C]kratereaction Reverse

[B][A]kratereaction Forward

==

=

=

=

OFB Chapter 14 4111/30/2004

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 theis 1 Step

(fast) FNO F NO 2.)

(slow) FFNO F NO 1.)

221

2k

2

2k

22

2

1

=

→+

+→+

F22NO 2F 22NO →+

OFB Chapter 14 4211/30/2004

]][F[NOk rateRDS theis 1 Step

(fast) FNO F NO 2.)

(slow) FFNO F NO 1.)

221

2k2

2

2k1

22

=

→+

+→+

Reaction Progress

Energy

F + NO2

NO2+ F2

slow fast

NO2F

F22NO 2F 22NO →+

OFB Chapter 14 4311/30/2004

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.

22 2NO O 2NOreactionOverall→+

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

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

→+

←→

+

But N2O2 is a reactive intermediate

OFB Chapter 14 4411/30/2004

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

→+

←→

+

Reaction Progress

Energy N2O2 + O2

slow

fast

2NO2NO2

22 2NO O 2NO →+

OFB Chapter 14 4511/30/2004

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

Need to express [intermediates] in terms of other reactants

]O[Nk rate reverse[NO]k rate forward

221-

21

==

2[NO]

]2O2[N

1-k1k

1K

]2O2[N1-k 2[NO]1k

rate reverse rate forward that recall mequilibriuat

==

=

=

2122 [NO]K ]O[N =∴

OFB Chapter 14 4611/30/2004

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

][O[NO]Kk rate 22

12=

OFB Chapter 14 4711/30/2004

→←

→

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

221-k

1k

=

+

+

Reaction Progress

Energy N2O2 + O2

slow

fast

2NO2NO2

][O[NO]Kk rate 22

12=

22 2NO O 2NO →+

1

11 k

kK−

=

OFB Chapter 14 4811/30/2004

Reaction Mechanism

• Intermediates

•Transition states

Reaction Progress

Energy

slowfast

fast

intermediates

Transition States

OFB Chapter 14 4911/30/2004

Transition States

OFB Chapter 14 5011/30/2004

14-6 The Effect of Temperature on Reaction Rates

RTE a

eA k

Equation Arrenhius−

=

k of units the has andconstant a isfactor" lexponentia-pre" the is A

andmoleper energy are units

energy n Activatiothe is aE where

OFB Chapter 14 5111/30/2004

Aek RTEa

EquationArrenhius

−=

RTElnA lnk a−=

y = mx + bAn 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

Slope = - Ea / R

OFB Chapter 14 5211/30/2004

Reaction Progress

Energy

∆E= minus

exothermic

Ear

Eaf

Transition State

∆E = Eaf - Ea

r

The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products

OFB Chapter 14 5311/30/2004

Reaction Progress

Energy

∆E= positive

endothermic

Ear

Eaf

Transition State

∆E = Eaf - Ea

r

The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products

OFB Chapter 14 5411/30/2004

Chapter 14Chemical Kinetics

• 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 Eaand thereby forcing the reaction to process by a path of higher Ea.

OFB Chapter 14 5511/30/2004

OFB Chapter 14 5611/30/2004

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 5711/30/2004

Chapter 14Chemical Kinetics

• 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

OFB Chapter 14 5811/30/2004


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)

2/1

2/1

6931.0

6931.0

tk

kt

=

=

Applies only to First order reactions

TRY

Examples 14-5 and Exercise

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