10.Hafta Chemical Kinetics

8/3/2019 10.Hafta Chemical Kinetics

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TOPIC 10-CHEMICAL KINETICS

CONTENTS

The rate of a Chemical Reaction

Measuring Reaction Rates

The Effect of Concentrations on

Rates of reaction:The Rate LawZero-Order Reactions

First-order Reactions

Second-Order Reactions

Reaction Kinetics:Summary



The Rate of a Chemical Reaction



Measuring Reaction RatesIn the decomposition of H2O2, O2(g) escapes from the

reaction mixture and the reaction goes to completion

2 H2O2(aq) 2 H2O + O2(g)

We can follow the progress of reaction by focusing either on

the formation of O2(g) or the disappearance of H2O2(aq).

That is, we can

• Measure the volumes of O2(g) and relate these volumes

decrease in concentration of H2O2

• Remove small samples of the reaction mixture from time

to time and analyze these samples of H2O2 content. This

can be done, for example, by titration of KMnO4 in acidisolution. The net ionic equation for the oxidation-

reduction reaction is

• 2 MnO4- + 5 H2O2+ 6 H+ 2 Mn2+ + 8 H2O + 5 O2(g




Table 1. Decomposition of H 2O2 – Measurement of the rate of reaction

I II

Δt(s)

400

400

400

400

400

III IV

Δ[H2O2], M

V

REACTION RATE= -

Δ[H2O2]/ Δt, Ms-1 TIME(s) [H2O2], M

0 2,32

400 1,72

800 1,30

1200 0,98

1600 0,73

2000 0,54

2400 0,39

2800 0,28

-0,42

- 0,32

- 0,25

- 0,60

-0,19

-0,15

-0,11

15,0X 10-4

10,5X10-4

8,0X10-4

6,3X10-4

4,8X10-4

3,8X10-4

2,8X10-4




The rate of this reaction is not

constant. In order to calculate

the instantaneous rate ofreaction a tangent line must be

drawn. Reaction rate is

determined from the slopes of

the tangent lines. The slope of

the first tangent line is given

below:

s M v

s

M v

/ 107,5

2850

)63,1(

4

Graphical Representation of kinetic data for the

reaction 2 H2O2(aq) 2 H2O + O2(g)




=

11

11

Example

From the data in the previous Table and Figure for the decomposition

of H2O2 a) determine the initial rate of decomposition of H2O2 and b)[H2O2]t at t=100 s.

initial rate 1,74x10-3Ms-1




An alternate method is to use data from the table given : [H2O2]=2,32

M at t=0 and [H2O2]=2,01 M at t=200 s

The agreement between two methods is fairly good, although it

would probably better if the time interval were less than 200 s. Of

the two results, that based on the tangent line is presumably more

reliable because it is expressed with more significant figures. On the

other hand, the reliability of the tangent line depends on howcarefully the line is constructed.



The Effect of Concentrations on Rates of

Reactions: The Rate Law

One of the goals in chemical kinetics study is to derive an equation that can be

used to predict how a rate of reaction depends on the concentrations of

reactants. Such an experimentally determine equation is called a rate law or

rate equation.

Consider the hypothetical reaction

a A + b B + …… g G + h H + ……

Where a,b…. Stand for coefficients in the balanced equation. We can express

the rate of reaction as

rate of reaction = k [A]m [B]n

The terms [A] and [B] represent molarities. The exponents,m,n are generally

small whole numbers, although in some cases they may be zero, fractional

and/or negative.





Establishing the Order of a Reaction by the Method of Initial Rates

INITIAL CONCENTRATIONS, M INITIAL RATE OF REACTION,Ms-1, R

EXPT [S2O8-2] [I-] R

1 0,038 0,060 R1= 1,4 x 10-5

2 0,076 0,060 R2= 2,8 x 10-5

3 0,060 0,120 R3= 4,4 x 10-5

Use data from the table above to establish the order of reaction

S2O8

2-

+ 3 I-

2 SO4

2-

+ I3

-

with respect to S2O82- and I- and also the overall order of the reaction




Reactions: The Rate LawÖrnek





This is not a problem because we now know the value m=1 and we can

substitute actual concentrations for [S2O82-].

R3 = k x (0,060 M)1 x (2 x [I-]2)n

R2 = k x (0,076 M)1 x [I-]2n

R3 / R2 = 4,4 x 10-5 / 2,8 x 10-5 = 0,060 x 2n / 0,076

2n = 2 n=1 .

The reaction is first order in S2O82- (m=1), first order in I- (n=1) andsecond order overall ( m + n= 2)





In general, the effect of the initial rate caused by doubling of the

concentration of a reactant is related to the reaction order in the following

way:

For a zero-order reaction the initial rate is unaffected(20=1);

For a first order reaction the initial rate doubles(21=2)

With a second order reaction the initial rate increases fourfold(22=4)

With a third order reaction,eightfold(23=8)

h ff f C i f





Practice: Use the results in the previous example and data from the table testablish the value of k in the rate equation

113101,6 s M k



Zero-Order ReactionsA zero-order reaction has a rate equation in which the

sum of the exponents(m+n+….) is equal to 0. In a

reaction in which a single reactant A decomposes to

products

A Products, if the reaction is zero-order, the rate

equation is

Rate of reaction= k [A]0=k=constant

Other features of the zero-order reactions are

• The concentration-time graph(plotted above) is a

straight line with negative slope

• The rate of reaction, which remains constant

throughout the reaction, is the negative of the slope of this line

• The units of k are the same as as the units of the rate

of reaction: molL-1s-1 or Ms-1



First-Order Reactions

A first order reaction has a rate

equation in which the sum of theexponents (m+n+…) is equal to 1. A

particularly common type of first order

reaction and the only type we will

consider, is that in which a single

reactant decomposes into products.



First Order Reactions

A A t k

t 0

lnln

Equation of straight line y= mx + b



First-Order ReactionsExample:

H2O2(aq), initially at a concentration of 2,32 M, is allowed to decomposWhat will [H2O2] be at t=1200 s? Use k= 7,30x10-4 s-1 for the first order

decomposition.



First-Order Reactions



Reactions Involving Gases

For gaseous reactions, rates are often measured in terms of gas

pressures. For the hypothetical reaction, A(g) products, the

initial partial pressure (PA)0 , and the partial pressure at time t, (PA)t,

are related to the expression:

To see how this equation is derived, start with the ideal gas equation

written for reactant A: PAV= nART. Note that the ratio nA /V is the same

as [A]. So, [A]0 = (PA)0 /RT and [A]t= (PA)t /RT. Substitute these terms in

the equation of the first order rate of reaction ,then you will get the

equation above.



Reactions Involving Gases

From the figure we see that t1/2= 8,0 x 101 min. For a first order

reaction : 13min107,8

min80

693,0693,0

21

t

k

a.

b.

mmHgP

mmHgP

s DTBP

DTBP

50][

800][ 0

[

16

1][ DTBs DTBP

PP

min320min80442

1 t t

The reaction must go through four half-lives

C8H12O2(g) 2 C3H6O(g) + C2H6(g)

DTBP Aceton Ethane

Reaction is started with pure DTBP at 147˚C and 800,0 mm Hg pressure in

a flask of constant volume.a) What is the value of the rate constant k? b ) A

what time will the partial pressure of DTBP be 50,0 mm Hg?



First Order Reactions

SOME TYPICAL FIRST-ORDER PROCESSES

PROCESS HALF-LIFE,t1/2 RATE CONSTANT k,s-1

Radioactive decay of uranium isotope(MA=238) 4,51 x 109 years 4,87 x 10-18

Radioactive decay of carbon isotope(MA=14) 5,73 x 103 years 3,83 x 10-12

Radioactive decay of phosphorus(MA=32) 14,3 days 5,61 x 10-7

C12H22O11(aq) + H2O C6H12O6(aq) + C6H12O6(aq)

sucrose fructose glucose

8,4 h 2,3 x 10-5

(CH2)2O (g) CH4(g) +CO(g)

Ethylene oxide

56,3 min 2,05 x 10-4

2 N2O5 2 N2O4 + O2(g) 18,6 min 6,21 x 10-4

HC2H3O2(aq) H+(aq) + C2H3O2(aq) 8,9 x 10-7 s 7,8 x 105

15C

415˚C

in

CCl4

45˚C




][

][ 0

2

0][

][

A

A

t

dt k A Ad

A second order reaction has a rate equation in which the sum of

exponents(m+n+….) is equal to 2. The reaction of peroxidisulfate-iodide is

second order overall:

rate of reaction= k [S2O82-] [I-]

Some reactions involving a single reactant, that is of the type A products, ar

also second order and their rate equations are

rate of reaction= -(rate of change of [A] = k [A]2

To convert the rate law into an integrated equation, a

calculus derivation is required:




The straight line plot is obtained for1/[A] vs. t,(graph on the right hand) ,so the reaction is second ord

Testing for the order of a reaction



Second- Order Reactions

Graphing Data to determine the Order of a Reaction-The data

listed in the table below refers to the decomposition of Aproducts. a) Establish the order of the reaction b) What is the rate

constant k? c) What is the half-life, t1/2 , if [A]0=1,00 M.

KINETIC DATA

TIME,min [A], M ln [A] 1/[A]

0 1,00 0,00 1,00

5 0,63 -0,46 1,6

10 0,46 -0,78 2,2

15 0,36 -1,02 2,8

25 0,25 -1,39 4,0



Establishing the Order of a Reaction

•



Reaction Kinetics (Summary)1. To calculate a rate of reaction when the rate law is known, use the

expression: rate of reaction= k [A]m [B]n

2. To determine a rate of reaction when the rate law is not given, use

• The slope of an appropriate tangent line to the graph of [A] vs. t.

• The expression –Δ[A]/ Δt, with a short time interval Δt

3. To determine the reaction order, depending on the data given,

• Use the method of initial rates;

• Find the graph of rate data that yields a straight line;

• Test for the constancy of the half-life(good only for first-order)

• Substitute data into integrated rate equations to find the equation that givesa constant k,

4. To relate reactant concentrations and times, use the appropriated integrated

rate equation after first determining k



Theoretical Models For Chemical KineticsCOLLISION THEORY: With the kinetic molecular theory it is possible to calculatethe number of molecular collisions per unit time ( νc= COLLISION FREQUENCY). Ina typical reaction involving gases ν

c

= 1030 collisions per second. If each collisionyielded product molecules, the rate of reaction would be about υrxn ≈106 M/s ,anextremely rapid rate. Gas phase reactions generally proceed at much slower rate, υrxn

≈10-4 M/s . This means that only a fraction of the collisions among the gaseousmolecules lead to a chemical reaction. The minimum total kinetic energy that moleculesmust bring to their collisions for a chemical reaction to occur is called ActivationEnergy=Ea. The rate of a reaction depends on the product of the collision frequency

and the fraction of these activated molecules. Moreover the higher the Ea of a reactionthe smaller the fraction of energetic collisions and the slower the reaction.



Theoretical Models for Chemical Kinetics



Theoretical Models for Chemical Kinetics• Another factor that affects the rate of reaction is the orientation of

the molecules at time of the collision. A particular orientation of the

molecules may be required if a collision is to be effective inproducing a chemical reaction. The number of unfavourable collisions

often exceeds the number of favorable ones. (The collision of two

slow moving automobiles bumper to bumper may produce no damage

at all, whereas if the bumper of one car strikes the fender of the other,

the damage may be considerable.• TRANSITION STATE THEORY: A hypothetical species with

properties intermediate to those of the reactants and products, called

transition state or activated complex.

• The enthalpy change of a reaction is equal to the difference in

activation energies of the forward and reverse reactions ΔΗrxn=Ea(forward)- Ea(reverse)

• For an endothermic reaction the activation energy must be equal to

or greater than the enthalpy of reaction



The Effect of Temperature on Reaction Rates

To speed up the biochemical reactions involved in cooking we raise the

temperature. To slow down other reaction we lower the temperature, asin refrigerating milk to prevent it from souring.

In 1889 Svante Arrhenius demonstrated the rate constants of manychemical reactions vary with temperature in accordance with theexpression:



The Effect of Temperature on Reaction Rates

Data are plotted as follows for the

representative point in black

To evaluate Ea,

Slope of line =



The Effect of Temperature on Reaction RatesExercise: Determine the temperature (T2) at which the decompositon

constant of N2O5 k 2 was found 9,63x10-5 considering the first condition

T1 = 298 K and k 1 = 3,46x10-5 s-1 ; (Ea = 106 kJ/mol)?



CATALYSIS

A reaction can generally be made to go faster by raising the temperature.

Another way to speed up a reaction is to use a catalyst. A catalyst provides

an alternate reaction pathway of lower activation energy. It participates in a

chemical reaction without undergoing permanent change. As a result the

formula of a catalyst does not appear in the net chemical equation(Its

formula is placed over the reaction arrow.)

Some examples for the catalysts can be given as follows:

1-) Pt-Rh catalyst speed up the oxidation of NH3 (g)

2-) Iodide ion (I-) is a good catalyst for the decomposition of H2O2(aq)

There are also some important biological enzymes such as lactase whichhelps to convert milk sugar( lactose) into glucose and galactose

10.Hafta Chemical Kinetics

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