chemistry 102(60) summer 2001 n instructor: dr. upali siriwardane n e-mail:upali@chem.latech n...

Chemistry 102(60) Summer 2001 Instructor: Dr. Upali Siriwardane e-mail:upali@chem.latech Office: CTH 311 Phone 257-4941 Office Hours: 8:30-10:30 a.m., M, W, Tu,Th, F

(Test 1): Chapter 12(Test 2): Chapter 13. (Test 3): Chapter 14 (Test 4): Chapter 15. (Test 5): Chapter 17.

Chapter 13. Rates of Reactions Chemical Kinetics

the branch of chemistry dealing with the

rates of chemical reactions.

Using chemical kinetics we can find

time to complete a reaction, the effect of

temperature on the rate

Effect of other substances (catalysts or

inhibitors) on the reactions

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

An example reaction

Gasburet

Constant temperature bath

An example reaction 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.

Rate a A --> b B based on reactants rate = -(1/a) [A]/ t Based on products rate = +(1/b) [B]/ t [A]= [A]f - [A]I Change in A

t= tf - ti Change in t

Rate of Reaction

2 N2O5(g) -----> 4 NO2 (g) + O2 (g) based on reactants

rate = -(1/2) [N2O5]/ t Based on products rate = +(1/4) [NO2]/ t

rate = +(1/1) [O2]/ t

Rate of Appearance & disappearance

2 N2O5(g) -----> 4 NO2 (g) + O2 (g) Disappearance is based on reactants

rate = -([N2O5]/ t Appearance is based on products rate = [NO2]/ t

rate = [O2]/ t converting rates of Appearance.. rate = ([NO2]/ t = - 4/2 [N2O5]/ t

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

Chemical Kinetics Definitions and Concepts

a) rate law b) rate constant c) order d) differential rate law c) integral rate law

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 the particular reactant.

Rate Law

Rate Law E.g. A + B -----> C rate [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

Rate Constant

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

rate [A]l[B]m rate = k [A]l[B]m; k = rate constant proportionality constant of the rate law

Rate Law E.g. 2 N2O5(g) -----> 4 NO2 (g) + O2 (g)

rate [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

Order The power of the concentrations is the

order with respect to the reactant. E.g. A + B -----> C 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).

Finding rate laws Method of initial rates.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.

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

Method of Initial Rates A + B ----> C The rate law : rate = k [A]x [B]y

[A],mol/L [B],mol/L rate, mol/LS 4.6 x 10-4 3.1 x 10-5 2.08 x 10-3 4.6 x 10 -4 6.2 x 10 -5 4.16 x 10-3 9.2 x 10 -4 6.2 x 10 -5 1.664 x 10-2

Order wrt A

1.664 x 10-2 k (9.2 x 10-4)x (6.2 x 10-5)y --------------- = -------------------------------------- 4.16 x 10-3 k (4.6 x 10-4)x (6.2 x 10-5)y 1.664 x 10-2 (9.2 x 10-4)x --------------- = -------------- 4.16 x 10-3 (4.6 x 10-4)x

4 = (9.2 x 10-4)/(4.6 x 10-4)x = 2x 4 = 2x

x = 2

Order wrt B 4.16 x 10-3 k (4.6 x 10-4)x (6.2 x 10-5)y

----------- = ------------------------------ 2.08 x 10-3 k (4.6 x 10-4)x (3.1 x 10-5)y 4.16 x 10 -3 (6.2 x 10-5) y ----------- = ------------- 2.08 x 10 -3 (3.1 x 10-5) y

2 = (6.2 x 10-5/3.1 x 10-5)y = 2y 2 = 2y

y = 1 The rate law : rate = k [A]2 [B]1

Units of the Rate Constant

1 first order: k = ─── = s-1

s L second order k = ─── mol s

Rate Law Differential Rate Law Integral Rate

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

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

rate k [A]2 [A]/t = k [A]2 1/ [A]f = kt + 1/[A]i

Differential Rate Law

Normal form Differential form

zero order rate = k [A]0 rate = - [A]/ t = k ( [A]0=1) first order rate law rate = k [A]1 rate = - [A]/ t = k [A]1

Integral Rate Law

Differential form Integral form

zero order rate = - [A]/ t = [A]f-[A]i = -kt = k ( [A]0=1) first order rate law rate = - [A]/ t ln [A]o/[A]t = kt = k [A]1

Finding rate lawsGraphical method.Graphical method.

Rate integrated Graph Slope

Order law rate law vs. time

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]0

1[A]t

1[A]t

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 2000 4000 6000 8000

0 order plot

1st order plot

2nd order plot

As you can see from theseplots of the N2O5 data, only a first order plotresults in a straight line.

As you can see from theseplots of the N2O5 data, only a first order plotresults in a straight line.

Time (s)

Time (s)

Time (s)

[N2O

5]

1/[

N2O

5]

ln[N

2O

5]

First order reactions Reactions that are first order with respect

to a reactant are of great importance. Describe how many drugs pass into the

blood stream or used by the body. Often useful in geochemistry Radioactive decay Half-life (Half-life (tt1/21/2)) The time required for one-half of the

quantity of reactant originally present to react.

First Order Reactions A ----> B Differential rate law [A] - ───── = k [A] t [A]t [A]0

ln ─── = - k t or ln ─── = k t [A]0 [A]t

Half-life form of 1st order

t2 is defined as time for [A]0, the initial concentration to decay half the original value

ie 1/2 x [A]0 = [A]t.

t2 equation

0.693 = k t2

0.693 t2 = ---- k

Half-life

The half-life and the rate constant are related. tt1/21/2 = = 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

0.6930.693kk

0.693

t1/2

0.6931900 s

Half-life

0

0.05

0.1

0.15

0.2

0 2000 4000 6000 8000

From our N2O5 data, we cansee that it takes about 1900seconds for the concentrationto be reduced in half.

It takes another 1900 secondsto reduce the concentration inhalf again.

From our N2O5 data, we cansee that it takes about 1900seconds for the concentrationto be reduced in half.

It takes another 1900 secondsto reduce the concentration inhalf again.

Time (s)

[N2O

5]

Theories of reaction rates Collision theoryCollision theory Based on kinetic-molecular theory. It assumes that reactants must collide for

a reaction to occur. They must hit with sufficient energy and

with the proper orientation so as to break the original bonds and form new ones.

As temperature is increased, the average kinetic energy increases - so will the rate.

As concentration increases, the number of collisions will also increase, also increasing the rate.

Effective collision

Reactants must have sufficient energy and the proper orientation for a collision to result in a reaction.

Transition state theory As reactants collide, they initially form an activated

complex. The activated complex is in the transition state. It lasts for approximately 10-100 fs. It can then form products or reactants. Once products are formed, it is much harder to return

to the transition state, for exothermic reactions.

Reaction profiles can be used to show this process.

What are the factors that affect rates of chemical reactions?

a) Temperature b) Concentration c) Catalysts d) Particle size of solid reactants

This type of plotshows the energychanges during

a reaction.

This type of plotshows the energychanges during

a reaction.

Reaction profile

Hactivation

energyPote

nti

al

En

erg

y

Reaction coordinate

Potential Energy Curves

Exothermic Reactions Endothermic Reactions Effect of catalysts Effect of temperature

Examples of reaction profiles

Exothermic reaction

Endothermic reaction

Examples of reaction profiles

High activation energyLow heat of reaction

Low activation energyHigh heat of reaction

Arrhenius Equation

Rate constant (k) k = A e-Ea/RT

A = frequency factor Ea = Activation energy R = gas constant T = Kelvintemperature

Rate and temperature

Reaction rates are temperature dependent.

0

1

2

3

4

5

6

7

20 25 30 35 40 45 50

0

1

2

3

4

5

6

7

20 25 30 35 40 45 50

Here are rate constantsfor N2O5 decompositionat various temperatures.

T, oC k x 104, 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 (oC)

Rate and temperature

The relationship between rate constant and temperature is mathematically described by the Arrhenius equationArrhenius equation.

k = A e A constant Ea activation energy T temperature, Kelvin R gas law constant

-Ea / RT

Rate and temperature

An alternate form of the Arrhenius equation is:

ln k = + ln A

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

Activation energy - Activation energy - EEaa The energy that molecules must have in

order to react.

( )( ) 1

T

Ea

R-

Calculation of Ea

k = 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 ]

Calculation of Ea from N2O5 data

y = -12392x + 40.809

Slope = -12392R = 8.35 J/mol KEa = 103 kJ / mol

-2

-1

0

1

2

3

0.0031 0.0032 0.0033 0.0034 0.0035

ln k

T-1

Reaction mechanisms

A detailed molecular-level picture of how a reaction might take place.

ON

OO

O+ O

NO

OO

ON

OO

O

activatedcomplex

= bonds in the process of breaking or being formed

Reaction mechanisms

Elementary processElementary process Each step in a mechanism. MolecularityMolecularity The number of particles that come

together to form the activated complex in an elementary process.

1 - unimolecular 2 - bimolecular 3 - termolecular

Reaction Mechanisms

Consider the following reaction. 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] However, the experimentally observed

rate law is: rate = k [NO2] [F2]

Reaction Mechanisms Since the observed rate law is not the same as

if the reaction took place in a single step, we know two things.• More than one step must be involved• The activated complex must be produced from two

species. A possible reaction mechanism might be: Step oneStep one NO2 + F2 NO2F + F

Step twoStep two NO2 + F NO2F

OverallOverall 2NO2 + F2 2NO2F

Reaction Mechanism

Elementary Reactions: NO2 + F2 --> NO2F + F (slow)

F + NO2 --> NO2F (fast) Molecularity? Of Elementary Reactions unimolecular, bimolecular,

termolecular?

Reaction Mechanisms

Rate-determining step.Rate-determining step. When a reaction occurs in a series

of steps, with one slow step, it is the slowslow step that determines the overall rate.

Step oneStep one NO2 + F2 NO2F + F Expected to be slow. It involves breaking

an F-F bond. Step twoStep two NO2 + F NO2F Expected to be fast. A fluorine atom is

very reactive.

Reaction Mechanisms Since step one is slow, we can expect this

step to determine the overall rate of the reaction.

NO2 + F2 NO2F + F This would give a rate expression of: rate = k1 [NO2] [ F2] This agrees with the experimentally

observed results.

Catalysis CatalystCatalyst

A substance that changes the rate of a reaction without being consumed in the reaction.

Provides an easier way to react.

Lower activation energy.

Still make the same products. EnzymesEnzymes are biological catalysts. InhibitorInhibitor

A substance that decreases the rate of reaction.

Catalysts Lowers ECatalysts Lowers Eaa

Catalysis

Types of catalystsTypes of catalysts

HomogeneousHomogeneous - same phase

Catalyst is uniformly distributed throughout the reaction mixture

Example - I- in peroxide.

HeterogeneousHeterogeneous - different phase

Catalyst is usually a solid and thereactants are gases or liquids

Example - Automobile catalytic converter.

Heterogeneous catalysis

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 (substratessubstrates).

Classification of enzymes Based on type of reactionBased on type of reaction

OxireductaseOxireductase catalyze a redox reaction

TransferaseTransferase transfer a functional group

HydrolaseHydrolase cause hydrolysis reactions

LyaseLyase break C-O, C-C or C-N bonds

IsomerasesIsomerases rearrange functional groups

LigaseLigase join two molecules

The active site Enzymes are typically HUGE proteins, yet only

a small part is actually involved in the reaction.

The active site has twobasic components.

catalytic sitecatalytic site

binding sitebinding site

Model oftrios-phosphate-isomerase

Model oftrios-phosphate-isomerase

Characteristics of enzyme active sites

Catalytic siteCatalytic site Where the reaction actually occurs.

Binding siteBinding site Area that holds substrate in proper place. Enzymes uses weak, non-covalent interactions to hold

the substrate in place based on R groups of amino acids.

Shape is complementary to the substrate and determines the specificity of the enzyme.

Sites are pockets or clefts on the enzyme surface.