13-1 CHEM 102, Spring 2012, LA TECH CTH 328 9:30-10:45 am Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office.

13-1CHEM 102, Spring 2012, LA TECH

CTH 328 9:30-10:45 am

Instructor: Dr. Upali Siriwardane

e-mail: upali@latech.edu

Office: CTH 311 Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;

Tu,Th,F 8:00 - 10:00 am..

Exams: 9:30-10:45 am, CTH 328.

March 21 ,  2012 (Test 1): Chapter 13

April 18 , 2012 (Test 2): Chapter 14 &15

May 14 , 2012 (Test 3): Chapter 16 &18

Optional Comprehensive Final Exam: May 17, 2012 :

Chapters 13, 14, 15, 16, 17, and 18

Chemistry 102(01) Spring 2012

13-2CHEM 102, Spring 2012, LA TECH

Chapter 13. Chemical Kinetics

13.1 Reaction Rate 13.2 Effect of Concentration on Reaction Rate 13.3 Rate Law and Order of Reaction

13.4 A Nanoscale View: Elementary Reactions

13.5 Temperature and Reaction Rate: The

Arrhenius Equation13.6 Rate Laws for Elementary Reactions13.7 Reaction Mechanisms13.8 Catalysts and Reaction Rate13.9 Enzymes: Biological Catalysts13‑10 Catalysis in Industry

13-3CHEM 102, Spring 2012, LA TECH

How do you measure rates?Rates are related to the time it required to

decay reactants or form products. The rate reaction = change in concentration of reactants/products per unit time

Average rate

rate of reaction = – D[reactant]/DtInstantaneous rate

rate of reaction = – d[reactant]/dt

13-4CHEM 102, Spring 2012, LA TECH

Rate of Appearance & Disappearance2 N2O5(g) -----> 4 NO2 (g) + O2 (g)

Disappearance is based on reactants

rate = -(D[N2O5]/ D t

Appearance is based on products

rate = D[NO2]/ D t

rate = D[O2]/ D t

Converting rates of Appearance.

rate = (D[NO2]/ D t = - 4/2 D[N2O5]/ D t

D[O2]/ D t = - 1/2 D[N2O5]/ D t

13-5CHEM 102, Spring 2012, LA TECH

Measuring Rate a A --> b B

Based on reactants

rate = -(1/a) D[A]/ D t

Based on products

rate = +(1/b) D[B]/ D t

D[A]= [A]f - [A]I Change in A

D t= tf - ti Change in t

13-6CHEM 102, Spring 2012, LA TECH

Reaction of cis-platin with Water

13-7CHEM 102, Spring 2012, LA TECH

Disappearance of Color

13-8CHEM 102, Spring 2012, LA TECH

Gas

buret

Constant temperature bath

An example reaction where gas is produced

13-9CHEM 102, Spring 2012, LA TECH

Time (s) Volume STP O2, mL

0 0

300 1.15

600 2.18

900 3.11

1200 3.95

1800 5.36

2400 6.50

3000 7.42

4200 8.75

5400 9.62

6600 10.17

7800 10.53

Here are the results for

our experiment.

Here are the results for

our experiment.

Time vs. volume of gas

13-10CHEM 102, Spring 2012, LA TECH

2 N2O5(g) -----> 4 NO2 (g) + O2 (g)

13-11CHEM 102, Spring 2012, LA TECH

Graph of 2 N2O5(g) ---> 4 NO2 (g) + O2 (g)

13-12CHEM 102, Spring 2012, LA TECH

Graph

13-13CHEM 102, Spring 2012, LA TECH

a) Temperature

b) Concentration

c) Catalysts

d) Particle size of solid reactants

Factors that affect rates of chemical reactions

13-14CHEM 102, Spring 2012, LA TECH

Effect of Particle Size on Rate

13-15CHEM 102, Spring 2012, LA TECH

Chemical Kinetics Definitions and Concepts

a) rate law

b) rate constant

c) order

d) differential rate law

c) integral rate law

13-16CHEM 102, Spring 2012, LA TECH

Every chemical reaction has a Rate Law

The rate law is an expression that relates

the rate of a chemical reaction to a constant

(rate constant-k) and concentration of

reactants raised to a power.

The power of a concentration is called the order

with respect to a particular reactant.

Rate Law

13-17CHEM 102, Spring 2012, LA TECH

Rate LawE.g. A + B -----> C

rate a [A]l[B]m

rate = k [A]l[B]m; k = rate constant

[A] = concentration of A

[B] = concentration of B

l = order with respect to A

m = order with respect to B

l & m have nothing to do with stoichiometric coefficients

13-18CHEM 102, Spring 2012, LA TECH

Rate Constant

E.g. A + B -----> C

rate a [A]l[B]m

rate = k [A]l[B]m;

k = rate constant

proportionality constant of the rate law

Larger the k faster the reaction

It is related inversely to t½

13-19CHEM 102, Spring 2012, LA TECH

Decomposition Reaction

13-20CHEM 102, Spring 2012, LA TECH

Rate Law E.g.

2 N2O5(g) -----> 4 NO2 (g) + O2 (g)

rate a [N2O5]1

rate = k [N2O5]1 ;k = rate constant

[N2O5] = concentration of N2O5

1 = order with respect to N2O5

Rate and the order are obtained by experiments

13-21CHEM 102, Spring 2012, LA TECH

Order The power of the concentrations is the order with

respect to the reactant.

E.g. A + B -----> C

If rate law: rate = k [A]1[B]2

The order of the reaction with respect to A is one (1).

The order of the reaction with respect to B is two (2).

Overall order of a chemical reaction is equal to the

sum of all orders (3).

13-22CHEM 102, Spring 2012, LA TECH

Method of initial rates

The order for each reactant is found by:

•Changing the initial concentration of that reactant.

•Holding all other initial concentrations and conditions constant.

•Measuring the initial rates of reaction

The change in rate is used to determine the order for that specific

reactant. The process is repeated for each reactant.

Finding rate laws

13-23CHEM 102, Spring 2012, LA TECH

Initial rate

13-24CHEM 102, Spring 2012, LA TECH

How do you find order? A + B -----> C

rate = k [A]l[B]m;

Hold concentration of other reactants constant

If [A] doubled, rate doubled• 1st order, [2A]1 = 2 1 x [A]1 , 2 1 = 2

b) If [A] doubled, rate quadrupled• 2nd order, [2A]2 = 2 2 x [A]2 , 2 2 = 4

c) If [A] doubled, rate increased 8 times • 3rd order, [2A]3 = 2 3 x [A]3 , 2 3 = 8

13-25CHEM 102, Spring 2012, LA TECH

Rate data

13-26CHEM 102, Spring 2012, LA TECH

Determining order

13-27CHEM 102, Spring 2012, LA TECH

Determining K, Rate Constant

13-28CHEM 102, Spring 2012, LA TECH

Overall order

13-29CHEM 102, Spring 2012, LA TECH

Units of the Rate Constant (k) 1first order: k = ─── = s-1

s L second order k = ─── mol s

L2 third order k = ─── mol2 s

13-30CHEM 102, Spring 2012, LA TECH

First order reactions

13-31CHEM 102, Spring 2012, LA TECH

Rate Law Differential Rate Law Integral Rate

rate = k [A]0 - D [A]/Dt =k ; ([A]0=1) [A]f-[A]i = -kt

rate = k [A]1 - D [A]/Dt = k [A] ln [A]o/[A]t = kt

rate = k [A]2 - D [A]/Dt = k [A]2 1/ [A]f = kt + 1/[A]i

Differential and Integral Rate Law

13-32CHEM 102, Spring 2012, LA TECH

Integrated Rate Laws

13-33CHEM 102, Spring 2012, LA TECH

Graphical methodOrder

RateLaw

Integrated Rate Law GraphX vs. time

Slope

0 rate = k [A]t = -kt + [A]0 [A]t -k

1 rate = k[A]ln[A]t = -kt + ln[A]0

ln[A]t -k

2 rate=k[A]2 = kt + k1

[A]t

1

[A]0

1

[A]t

13-34CHEM 102, Spring 2012, LA TECH

Graphical Ways to get Order

13-35CHEM 102, Spring 2012, LA TECH

First-order, Second-order,and Zeroth-order Plots

13-36CHEM 102, Spring 2012, LA TECH

Finding rate laws

0

0.05

0.1

0.15

0.2

0 2000 4000 6000 8000

-4.5

-4

-3.5

-3

-2.5

-2

-1.50 2000 4000 6000 8000

0

20

40

60

80

100

0 1000 2000 3000 4000 5000 6000 7000 8000

0 order plot

1st order plot

2nd order plot

As you can see from these

plots of the N2O5 data,

only a first order plot

results in a straight line.

As you can see from these

plots of the N2O5 data,

only a first order plot

results in a straight line.

Time (s)

Time (s)

Time (s)

[N2O

5]

1/[

N2O

5]

ln[N

2O

5]

13-37CHEM 102, Spring 2012, LA TECH

This plot of ln[cis-platin] vs.

time produces a straight line,

suggesting that the reaction

is first-order.

Comparing graphs

13-38CHEM 102, Spring 2012, LA TECH

First Order ReactionsA ----> B

13-39CHEM 102, Spring 2012, LA TECH

t1/2 equation 0.693 = k t1/2

0.693 t1/2 =---- k

13-40CHEM 102, Spring 2012, LA TECH

The half-life and the rate constant are related.

t1/2 =

Half-life can be used to calculate the first order rate constant.

For our N2O5 example, the reaction took 1900 seconds to react half way so:

k = = = 3.65 x 10-4

s-1

0.693

k

0.693

t1/2

0.693

1900 s

Half-life

13-41CHEM 102, Spring 2012, LA TECH

A Nanoscale View:Elementary Reactions

Most reactions occur through a series of simple

steps or elementary reactions.

Elementary reactions could be

unimolecular - rearrangement of a molecule

bimolecular - reaction involving the collision of

two molecules or particles

termolecular - reaction involving the collision of

three molecules or particles

13-42CHEM 102, Spring 2012, LA TECH

2NO2 (g) + F2 (g) 2NO2F (g)

If the reaction took place in a single step the rate law would be: rate = k

[NO2]2 [F2]

Observed: rate = k1 [NO2] [ F2]

If the observed rate law is not the same as if the reaction took place in a single

step that more than one step must be involved

Elementary Reactions and Mechanism

13-43CHEM 102, Spring 2012, LA TECH

Elementary ReactionsA possible reaction mechanism might be:

Step one NO2 + F2 NO2F + F (slow)

Step two NO2 + F NO2F (fast)

Overall 2NO2 + F2 2NO2F

slowest step in a multi-step mechanismthe step which determines the overall rate of the reaction

rate = k1 [NO2] [ F2]

Rate Determining Step

13-44CHEM 102, Spring 2012, LA TECH

This type of plot

shows the energy

changes during

a reaction.

This type of plot

shows the energy

changes during

a reaction.

Reaction profile of rate determining step

DH

activation

energy

Pote

nti

al

En

erg

y

Reaction coordinate

13-45CHEM 102, Spring 2012, LA TECH

What Potential Energy Curves ShowExothermic Reactions

Endothermic Reactions

Activation Energy (Ea) of reactant or the minimum

energy required to start a reaction

Effect of catalysts

Effect of temperature

13-46CHEM 102, Spring 2012, LA TECH

Exothermic reaction

Endothermic reaction

Examples of reaction profiles

13-47CHEM 102, Spring 2012, LA TECH

High activation energy (kinetic)

Low heat of reaction (thermodynamic)

Low activation energy (kinetic)

High heat of reaction (thermodynamic)

Examples of reaction profiles

13-48CHEM 102, Spring 2012, LA TECH

Unimolecular Reactioncis-2-butene trans-2-butrne

13-49CHEM 102, Spring 2012, LA TECH

Bimolecular Reaction

I- + CH3Br ICH3 + Br

-

13-50CHEM 102, Spring 2012, LA TECH

Orientation Probability: Some Unsuccessful Collisions

I- + CH3Br ICH3 + Br

-

13-51CHEM 102, Spring 2012, LA TECH

Arrhenius Equation: Dependence of Rate Constant (k) on T Rate constant (k)

k = A e-Ea/RT

A = frequency factor: A = p x z

Ea = Activation energyR = gas constantT = Kelvin temperaturep = collision factorz = Orientation factor

13-52CHEM 102, Spring 2012, LA TECH

Energy Distribution Curves:Activation Energy

13-53CHEM 102, Spring 2012, LA TECH

An alternate form of the Arrhenius equation:

k = A e-Ea/RT

ln k = + ln A

If ln k is plotted against 1/T, a straight line of slope -Ea/RT is obtained.

Activation energy - Ea

The energy that molecules must have in order to react.

( ) ( )1

T

Ea

R

-

Arrhenius Equation: ln form

13-54CHEM 102, Spring 2012, LA TECH

Calculation of Eak = A e-

Ea/RT

ln k = ln A - Ea/RT

log k = log A - Ea/ 2.303 RT

using two set of values

log k1 = log A - Ea/ 2.303 RT1

log k2 = log A - Ea/ 2.303 RT2

log k1 - log k2 = - Ea/ 2.303 RT2 + Ea/ 2.303 RT1

log k1/ k2 = Ea/ 2.303 R[ 1/T1 - 1/T2 ]

13-55CHEM 102, Spring 2012, LA TECH

Reaction rates are temperature dependent.

0

1

2

3

4

5

6

7

20 25 30 35 40 45 50

Here are rate constants

for N2O5 decomposition

at various temperatures.

T, oC k x 10

4, s

-1

20 0.235

25 0.469

30 0.933

35 1.82

40 3.62

45 6.29

k x

10

4 (

s-1

)

Temperature (o

C)

Rate vs Temperature plot

13-56CHEM 102, Spring 2012, LA TECH

y = - 1 2 3 9 2 x + 4 0 . 8 0 9

S l o p e = - 1 2 3 9 2

R = 8 . 3 5 J / m ol K

E a = 1 0 3 k J / m ol

- 2

- 1

0

1

2

3

0 . 0 0 3 1 0 . 0 0 3 2 0 . 0 0 3 3 0 . 0 0 3 4 0 . 0 0 3 5

ln k

T-1

Calculation of Ea from N2O5 data

13-57CHEM 102, Spring 2012, LA TECH

Collision ModelThree conditions must be met at the nano-scale

level if a reaction is to occur:

the molecules must collide;

they must be positioned so that the reacting

groups are together in a transition state between

reactants and products;

and the collision must have enough energy to

form the transition state and convert it into

products.

13-58CHEM 102, Spring 2012, LA TECH

Effect of Concentrationon Frequency ofBimolecular Collisions

13-59CHEM 102, Spring 2012, LA TECH

Transition State: Activated Complex or Reaction Intermediatesan unstable arrangement of atoms that has the

highest energy reached during the rearrangement

of the reactant atoms to give products of a reaction

13-60CHEM 102, Spring 2012, LA TECH

Catalyst

A substance which speeds up the rate of a

reaction while not being consumed

Homogeneous Catalysis - a catalyst which is in

the same phase as the reactants

Heterogeneous Catalysis- a catalyst which is

in the different phase as the reactants

catalytic converter• solid catalyst working on gaseous materials

13-61CHEM 102, Spring 2012, LA TECH

Catalysts Lowers Ea

13-62CHEM 102, Spring 2012, LA TECH

Catalyzed & Uncatalyzed Reactions

13-63CHEM 102, Spring 2012, LA TECH

Conversion of NO to N2 + O2

13-64CHEM 102, Spring 2012, LA TECH

Catalytic Converter catalyst

H2O(g) + HCs CO(g) + H2(g) (unbalanced)

catalyst

2 H2(g) + 2 NO(g) N2(g) + 2 H2O(g)

catalyst

HCs + O2(g) CO2(g) + H2O(g) (unbalanced)

catalyst

CO(g) + O2(g) CO2(g) (unbalanced)

catalyst = Pt-NiOHCs = unburned hydrocarbons

13-65CHEM 102, Spring 2012, LA TECH

Enzymes: Biological catalystsBiological catalysts

Typically are very large proteins.

Permit reactions to ‘go’ at conditions that the body

can tolerate.

Can process millions of molecules every second.

Are very specific - react with one or only a few types

of molecules (substrates).

13-66CHEM 102, Spring 2012, LA TECH

The active site

Enzymes are typically HUGE proteins, yet only a

small part is actually involved in the reaction. The active site has two

basic components.

catalytic site

binding site

Model of

trios-phosphate-isomerase

Model of

trios-phosphate-isomerase

13-67CHEM 102, Spring 2012, LA TECH

Relationship of Enzyme to Substrate

13-68CHEM 102, Spring 2012, LA TECH

Enzyme Catalyzed Reaction

13-69CHEM 102, Spring 2012, LA TECH

Maximum Velocity for an Enzyme Catalyzed Reaction

13-70CHEM 102, Spring 2012, LA TECH

Enzyme Activity Destroyed by Heat

13-71CHEM 102, Spring 2012, LA TECH

Reaction Mechanism

A set of elementary reactions which represent

the overall reaction

13-72CHEM 102, Spring 2012, LA TECH

Mechanism Oxidation ofIodide Ion by Hydrogen Peroxide

13-73CHEM 102, Spring 2012, LA TECH

Rate Law of Oxidation ofIodide Ion by Hydrogen Peroxide

Step 1.

HOOH + I- HOI + OH-

slow step - rate determining step, suggests that

the reaction is first order with regard to hydrogen

peroxide and iodide ion

rate = k[HOOH][I-]

13-74CHEM 102, Spring 2012, LA TECH

Mechanisms with a Fast Initial Step

2 NO(g) + Br2(g) 2NOBr(g)

rateexperimental = k[NO]2[Br2]

13-75CHEM 102, Spring 2012, LA TECH

Mechanism of NO + Br2

Rate = k[NOBr2][NO]

13-76CHEM 102, Spring 2012, LA TECH

Rate Constants for NO + Br2

Step +1(forward), rate constant k1

Step -1(backward), rate constant k-1

Step 2, rate constant k2

rateStep+1 = rateStep-1 + rateStep2

k1[NO][Br2] = k-1[NOBr2] - k2[NOBr2]

13-77CHEM 102, Spring 2012, LA TECH

Relationships of Rate Constants

k1[NO][Br2] ~ k-1[NOBr2]

thus

[NOBr2] = (k1/k-1)[NO][Br2]

substituting into

rate = k2[NOBr2][NO]

rate = k2((k1/k-1)[NO][Br2])[NO]

rate = (k2k1/k-1)[NO]2[Br2]

13-78CHEM 102, Spring 2012, LA TECH

Chain Mechanismschain initiating step • - the step of a mechanism which • starts the chain chain

propagating step(s) • the step or steps which keeps the chain going

chain terminating step(s) • the step or steps which break the chain

13-79CHEM 102, Spring 2012, LA TECH

Chain Mechanismscombustion of gasoline in an internal

combustion

engine

chain initiating step - additives which generate

free radicals, particles with unpaired electrons

chain propagating step(s) - steps which generate

new free radicals

chain terminating step(s)

- steps which do not generate new free radicals