CHAPTER 12: KINETICS

Dr. Aimée TomlinsonChem1212

Reaction Rates

Section 12.1

Reaction RateThe rates at which products are formed and reactants are consumed are connected

They are always represented as concentration change / time change

For the general case aA + bB cC + dD the relationships are:

where t = tf - ti, [A] is the concentration of A (moles/L) and [A] = [A]f - [A]i

we loose reactants as products are formed which is why the rates of A & B are negative

1 [ ] 1 [ ] 1 [ ] 1 [ ]A B C DRatea t b t c t d t

Example Application of Definition1. What is the rate relationship between the production of O2 and O3?

2O3(g) 3O2(g)

the rate of O2 production is 1.5 times faster than the ate of consumption of O3the rate of O3 consumption is 2/3 times the production of O2

3 3 32 2 2[ ] [ ] [ ][ ] [ ] [ ]1 1 3 2 or 2 3 2 3

O O OO O Ot t t t t t

2. The decomposition of N2O5 proceeds according to the equation: 2N2O5(g) 4NO2(g) + O2(g)

If the rate of decomposition of N2O5 at a particular instant is 4.2 x 10-7 M/s, what is the rate of production of NO2 and O2?

2 5 2 2

7 72 52

72 52

[ ] [ ] [ ]1 12 4

[ ][ ] 4 2 4.2 10 8.4 102

[ ][ ] 1 2.1 102

N O NO ORatet t tN ONO M M

s st tN OO M

st t

Example: 2 5( ) 2( ) 2( )2 4g g gN O NO O

52 52 5

622

22

0.0101 0.0120[ ]rate of decomposition of N O 1.9 10400 300

0.00049 0.0040[ ]rate of formation of O 9 10400 300

0.0197 0.0160[ ]rate of formation of NO 3.7400 300

MN O Mst s

MO Mst s

MNOt s

510 M

s

Different Types of Rates

Average RateConcentration change over a time interval (colored region in plot)

Instantaneous RateSlope of the tangent line at a given time (purple line in plot)

Rate Law & Reaction Order

Section 12.2

Rate Law

Relates the rate directly to reactant concentrations

For the General case aA + bB cC + dD, the rate law is:

Rate = k[A]m[B]n

where m and n are determined experimentallyk is called the rate constant

CAUTION!!! Rate law is NOT related to stoichiometry!2 5( ) 2( ) 2( ) 2 5

12

3( ) 2( ) 4( ) ( ) 3 2

2 4 [ ]

[ ][ ]

g g g

g g g g

N O NO O Rate k N O

CHCl Cl CCl HCl Rate k CHCl Cl

Reaction OrderThe power to which each reactant is raised

For the General case aA + bB cC + dD, with

Rate = k[A]m[B]n

“m” is the order of reactant A and “n” is the order of reactant BOverall reaction order is the sum of all or m+n in this caseName the reactant orders and overall reaction order for

It is 1st order in CHCl3 and ½ order for Cl2 with 1 ½

order overall

12

3 2[ ][ ]Rate k CHCl Cl

Experimental Determination of Rate

Law

Section 12.3

Illustrative Example

Experiment [A]0 [B]0 Initial Rate M/s

1 0.100

0.100 4.0 x 10-5

2 0.100

0.200 4.0 x 10-5

3 0.200

0.100 16.0 x 10-5

Determine the rate law, the rate constant and the reaction orders for each reactant and the overall reaction order using the data given below.

Rate constant k & Overall OrderIt is an indicator of the overall rate order of the equation

?

1

?

1?

n

M Ms

n

n n

rate k A

M Ms

Ms M s M

zeroth 0 /first 1 1/second 2 1/

order m k unitsM s

sM s

Final Thoughts

What to do when finding ‘m’ isn’t obvious

1 1 1 1ln ln4 2 4 2

1ln4 21ln2

m

m

m

First-Order Integrated Rate Law & its Half-Life

Section 12.4 & 12.5

First-Order Integrated Rate Law• Final equation is the integrated rate law• Change in reactant concentration over time

may be plotted to get rate law• We plot ln[A]t versus t:

0

0

[ ]

0

[ ]

[ ][ ] 0

0

[ ][ ]

[ ][ ]

ln[

[ ][ ]

] | |

ln[

ln[

]

] ln[ ]

ln[ ]

tA t

A o

A t t

t

A

t

Arate k A

A k tA

d A kdtA

A kdt

A

A t

t

t

k

A

A

k

0ln[ ] ln[ ]tm xy b

A k t A

Half-Life for First-Order Reaction

0

102

120

12

12

12

[ ]ln

[ ][ ]

ln[ ]1ln2

0.693

0.693

tAkt

AA

ktA

kt

kt

t k

Second-Order Reactions

Section 12.6

Second-Order Integrated Rate Law• Final equation is the integrated rate law• This time we plot 1/[A]t versus t to get

linearity

0

0

2

[ ]

2[ ]

[

2

0

][ ] 0

0

[

1 1

][ ]

[ ][ ]

1 |

][ ]

|

1 1

[

tA t

A

t

o

A t tA

t

ktA A

A k tA

d A kdtA

kdtA

ktA A

Arate k At

1 1[ ] [ ]t m x

by

k tA A

Zeroth-Order Reactions

Section 12.7

Zeroth-Order Integrated Rate Law

0

0

[ ][ ][ ] 0 0

[ ]

0

[ ]

[

[ ] [

] [ ] | | [ ]

[

]

]

]

[tA t

A t t

o

t

A tA

A k t

d A kdt A

A

kdt A A

A k

rate kt

t

kt

A

Summary of Integrated Equations

Reaction Mechanisms

Section 12.8

Definitions

Reaction Mechanisms: how electrons move during a reaction

IntermediateSpecies that is produced then consumedNever partakes or appears in the rate law

Elementary StepA step that occurs in a reactionMost reactions have multiple elementary stepsThe stoichiometry of these steps CAN be used to get the reaction orderThe sum of these steps leads to the overall reaction

MolecularityNumber of reacting particles in an elementary stepUni- (1 species), Bi- (2 species), Ter- (3 species)

Example ApplicationGiven the mechanism below state the overall reaction, rate law and molecularity of each step as well as the intermediate.

2 2 3

3 2 2

2 2

1:2 :

step NO NO NO NO elementarystep NO CO NO CO elementary

NO CO NO CO overall

Rate law for step 1: rate1 = k1[NO2]2

Rate law for step 2: rate2 = k2[NO3][CO]Both steps have 2 interacting species so bimolecularIntermediate is NO3 which is produced then consumed

Rate Laws & Reaction Mechanisms

Section 12.9

Examples of Molecularity

Molecularity Elementary step Rate LawUnimolecular A products Rate = k[A]Bimolecular A + A products Rate = k[A]2

Bimolecular A + B products Rate = [A][B]Termolecular A + A + A products Rate = k[A]3

Termolecular A + A + B products Rate = k[A]2[B]Termolecular A + B + C products Rate = k[A][B][C]

Example ProblemWhat is the molecularity & rate law for each of the following?

2 2H NO N H O

2N NO N O

2 2H O H O

2[ ][ ]rate k H NObimolecular

bimolecular

bimolecular

[ ][ ]rate k N NO

2[ ][ ]rate k H O

molecularity Rate law

Rate Laws for Overall Reactions

Section 12.10

Rate Determining StepThe slowest step in a mechanism

Just like the slowest person in a relay race – the slowest step in the mechanism will ruin the speed/time of the reaction

If the slow step is first it is very easy to get the rate law:1

2

2 3

3 2

1: 2 ( )

2 : ( )

k

k

step NO NO NO slow

step NO NO O fast

21 2the rate law would then be: [ ]rate k NO

Example for Fast Step FirstWhat is the rate law for the mechanism below?

1

2

2 3

3 2

1: ( )

2 : 2 ( )

k

k

step NO O NO fast

step NO NO NO slow

Fast Equilibrium Applied ExampleThe rate laws for the thermal and photochemical decomposition of NO2 are different. Which of the following mechanisms are possible for thermal and photochemical rates given the information below?

Thermal rate = k[NO2]2

Photochemical rate = k[NO2] 2

2 2

.)

slow

fast

a NO NO O

O NO NO O

2 2 2 4

2 4 3

3 2

.)

fast

slow

fast

b NO NO N O

N O NO NO

NO NO O

2 2 3

3 2

.)

slow

fast

c NO NO NO NO

NO NO O

Reaction Rates & Temperatures –

Arrhenius Equation

Section 12.11

Arrhenius EquationRelates rate to energy and temperature

aERTk Ae

Frequency Factor AIndicative of the number of successful collisions

Energy of Activation, Ea

Energy that must be overcome to form products

Using the Arrhenius Equation

Section 12.12

Finding the Ea through plotting

Mechanisms & Energy Profiles

14_19.jpg

2 2 3

3 2

slow

fast

NO NO NO NO

NO NO O

The slow step has the larger Ea since it takes longer to generate more energy

Transition StateDefn: state at which reactant bonds are broken and product bonds begin to form

Located at the top of the hump for each elementary step in the reaction profile

Catalysts

Section 12.13 & 12.14

CatalystA species that lowers the activation energy of a chemical reaction and does not undergo any permanent chemical change

it is not present in the overall reaction expression

it is not present in the rate law

it must be consumed and produced in the elementary steps

Two types of catalysts:Homogeneous: in the same phase as the reactantsHeterogeneous: a different phase from reactants

Example of Catalysis

14_21.jpg

Uncatalyzed mechanism - blue line in the figure

Cl Catalyzed mechanism - red line

3 2

3 2

3 2

1:

2 : 2

2 317.7 /

hv

a

step O O O

step O O O

O OE kJ mol

3 2

3 2

3 2

1:

2 : 2

2 32.2 /a

step Cl O O ClO

step ClO O Cl O

O OE kJ mol