79
13-1 CHEM 102, Spring 2012, LA TECH CTH 328 9:30-10:45 am Instructor: Dr. Upali Siriwardane e-mail: [email protected] 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

Chemistry 102(01) Spring 2012

Embed Size (px)

DESCRIPTION

Chemistry 102(01) Spring 2012. CTH 328 9:30-10:45 am Instructor : Dr. Upali Siriwardane e-mail : [email protected] 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. - PowerPoint PPT Presentation

Citation preview

Page 1: Chemistry 102(01)  Spring 2012

13-1CHEM 102, Spring 2012, LA TECH

CTH 328 9:30-10:45 am

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

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

Page 2: 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

Page 3: Chemistry 102(01)  Spring 2012

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

Page 4: Chemistry 102(01)  Spring 2012

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

Page 5: Chemistry 102(01)  Spring 2012

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

Page 6: Chemistry 102(01)  Spring 2012

13-6CHEM 102, Spring 2012, LA TECH

Reaction of cis-platin with Water

Page 7: Chemistry 102(01)  Spring 2012

13-7CHEM 102, Spring 2012, LA TECH

Disappearance of Color

Page 8: Chemistry 102(01)  Spring 2012

13-8CHEM 102, Spring 2012, LA TECH

Gas

buret

Constant temperature bath

An example reaction where gas is produced

Page 9: Chemistry 102(01)  Spring 2012

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

Page 10: Chemistry 102(01)  Spring 2012

13-10CHEM 102, Spring 2012, LA TECH

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

Page 11: Chemistry 102(01)  Spring 2012

13-11CHEM 102, Spring 2012, LA TECH

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

Page 12: Chemistry 102(01)  Spring 2012

13-12CHEM 102, Spring 2012, LA TECH

Graph

Page 13: Chemistry 102(01)  Spring 2012

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

Page 14: Chemistry 102(01)  Spring 2012

13-14CHEM 102, Spring 2012, LA TECH

Effect of Particle Size on Rate

Page 15: Chemistry 102(01)  Spring 2012

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

Page 16: Chemistry 102(01)  Spring 2012

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

Page 17: Chemistry 102(01)  Spring 2012

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

Page 18: Chemistry 102(01)  Spring 2012

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

Page 19: Chemistry 102(01)  Spring 2012

13-19CHEM 102, Spring 2012, LA TECH

Decomposition Reaction

Page 20: Chemistry 102(01)  Spring 2012

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

Page 21: Chemistry 102(01)  Spring 2012

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).

Page 22: Chemistry 102(01)  Spring 2012

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

Page 23: Chemistry 102(01)  Spring 2012

13-23CHEM 102, Spring 2012, LA TECH

Initial rate

Page 24: Chemistry 102(01)  Spring 2012

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

Page 25: Chemistry 102(01)  Spring 2012

13-25CHEM 102, Spring 2012, LA TECH

Rate data

Page 26: Chemistry 102(01)  Spring 2012

13-26CHEM 102, Spring 2012, LA TECH

Determining order

Page 27: Chemistry 102(01)  Spring 2012

13-27CHEM 102, Spring 2012, LA TECH

Determining K, Rate Constant

Page 28: Chemistry 102(01)  Spring 2012

13-28CHEM 102, Spring 2012, LA TECH

Overall order

Page 29: Chemistry 102(01)  Spring 2012

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

Page 30: Chemistry 102(01)  Spring 2012

13-30CHEM 102, Spring 2012, LA TECH

First order reactions

Page 31: Chemistry 102(01)  Spring 2012

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

Page 32: Chemistry 102(01)  Spring 2012

13-32CHEM 102, Spring 2012, LA TECH

Integrated Rate Laws

Page 33: Chemistry 102(01)  Spring 2012

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

Page 34: Chemistry 102(01)  Spring 2012

13-34CHEM 102, Spring 2012, LA TECH

Graphical Ways to get Order

Page 35: Chemistry 102(01)  Spring 2012

13-35CHEM 102, Spring 2012, LA TECH

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

Page 36: Chemistry 102(01)  Spring 2012

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]

Page 37: Chemistry 102(01)  Spring 2012

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

Page 38: Chemistry 102(01)  Spring 2012

13-38CHEM 102, Spring 2012, LA TECH

First Order ReactionsA ----> B

Page 39: Chemistry 102(01)  Spring 2012

13-39CHEM 102, Spring 2012, LA TECH

t1/2 equation 0.693 = k t1/2

0.693 t1/2 =---- k

Page 40: Chemistry 102(01)  Spring 2012

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

Page 41: Chemistry 102(01)  Spring 2012

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

Page 42: Chemistry 102(01)  Spring 2012

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

Page 43: Chemistry 102(01)  Spring 2012

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

Page 44: Chemistry 102(01)  Spring 2012

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

Page 45: Chemistry 102(01)  Spring 2012

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

Page 46: Chemistry 102(01)  Spring 2012

13-46CHEM 102, Spring 2012, LA TECH

Exothermic reaction

Endothermic reaction

Examples of reaction profiles

Page 47: Chemistry 102(01)  Spring 2012

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

Page 48: Chemistry 102(01)  Spring 2012

13-48CHEM 102, Spring 2012, LA TECH

Unimolecular Reactioncis-2-butene trans-2-butrne

Page 49: Chemistry 102(01)  Spring 2012

13-49CHEM 102, Spring 2012, LA TECH

Bimolecular Reaction

I- + CH3Br ICH3 + Br

-

Page 50: Chemistry 102(01)  Spring 2012

13-50CHEM 102, Spring 2012, LA TECH

Orientation Probability: Some Unsuccessful Collisions

I- + CH3Br ICH3 + Br

-

Page 51: Chemistry 102(01)  Spring 2012

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

Page 52: Chemistry 102(01)  Spring 2012

13-52CHEM 102, Spring 2012, LA TECH

Energy Distribution Curves:Activation Energy

Page 53: Chemistry 102(01)  Spring 2012

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

Page 54: Chemistry 102(01)  Spring 2012

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 ]

Page 55: Chemistry 102(01)  Spring 2012

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

Page 56: Chemistry 102(01)  Spring 2012

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

Page 57: Chemistry 102(01)  Spring 2012

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.

Page 58: Chemistry 102(01)  Spring 2012

13-58CHEM 102, Spring 2012, LA TECH

Effect of Concentrationon Frequency ofBimolecular Collisions

Page 59: Chemistry 102(01)  Spring 2012

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

Page 60: Chemistry 102(01)  Spring 2012

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

Page 61: Chemistry 102(01)  Spring 2012

13-61CHEM 102, Spring 2012, LA TECH

Catalysts Lowers Ea

Page 62: Chemistry 102(01)  Spring 2012

13-62CHEM 102, Spring 2012, LA TECH

Catalyzed & Uncatalyzed Reactions

Page 63: Chemistry 102(01)  Spring 2012

13-63CHEM 102, Spring 2012, LA TECH

Conversion of NO to N2 + O2

Page 64: Chemistry 102(01)  Spring 2012

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

Page 65: Chemistry 102(01)  Spring 2012

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).

Page 66: Chemistry 102(01)  Spring 2012

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

Page 67: Chemistry 102(01)  Spring 2012

13-67CHEM 102, Spring 2012, LA TECH

Relationship of Enzyme to Substrate

Page 68: Chemistry 102(01)  Spring 2012

13-68CHEM 102, Spring 2012, LA TECH

Enzyme Catalyzed Reaction

Page 69: Chemistry 102(01)  Spring 2012

13-69CHEM 102, Spring 2012, LA TECH

Maximum Velocity for an Enzyme Catalyzed Reaction

Page 70: Chemistry 102(01)  Spring 2012

13-70CHEM 102, Spring 2012, LA TECH

Enzyme Activity Destroyed by Heat

Page 71: Chemistry 102(01)  Spring 2012

13-71CHEM 102, Spring 2012, LA TECH

Reaction Mechanism

A set of elementary reactions which represent

the overall reaction

Page 72: Chemistry 102(01)  Spring 2012

13-72CHEM 102, Spring 2012, LA TECH

Mechanism Oxidation ofIodide Ion by Hydrogen Peroxide

Page 73: Chemistry 102(01)  Spring 2012

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-]

Page 74: Chemistry 102(01)  Spring 2012

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]

Page 75: Chemistry 102(01)  Spring 2012

13-75CHEM 102, Spring 2012, LA TECH

Mechanism of NO + Br2

Rate = k[NOBr2][NO]

Page 76: Chemistry 102(01)  Spring 2012

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]

Page 77: Chemistry 102(01)  Spring 2012

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]

Page 78: Chemistry 102(01)  Spring 2012

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

Page 79: Chemistry 102(01)  Spring 2012

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