lu 4 slides reaction kinetics

8/2/2019 LU 4 Slides Reaction Kinetics

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4.1 rate of reaction4.2 collision theory of chemical reaction

LU4 Reaction kinetics

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. rans n s a e e ry4.4 rate law

4.5 Factors affecting rate & rate constant

4.6 Homogeneous & heterogeneous catalyst

PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



-Involve chemical reactant(s)

-

A chemical reaction


- any relation on their rates?


FOR SALE



2N2O5 (g)→ 4NO2(g) + O2(g)

A chemical reaction


getting less over time

Products are getting more andmore over time


FOR SALE



Rates of Chemical ReactionsConcentration ofreactant

Concentration ofproduct


Time

As the reactionproceeds, the

[reactant] decreases.

Time

As the reactionproceeds, the

[product] increases.PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Reaction Rates & Stoichiometry

• For the general reaction

where a, b, c & d are the stoichiometric coefficients

a A + b B c C + d D


the rate expression is given by

t d t ct bt a ∆

∆⋅=

∆

∆⋅=

∆

∆⋅−=

∆

∆⋅−=

D][1C][1B][1A][1rate


FOR SALE



2N2O5 (g)→ 4NO2(g) + O2(g)

A chemical reaction

= −


∆t ∆t

∆[O2] = −½ ∆[N2O5]∆t ∆t


FOR SALE



2N2

O5

(g)→ 4NO2

(g) + O2

(g)

A chemical reaction

=− −1 −1


∆t

∆[NO2] = −2(−2.2 mol L−1s−1)∆t =4.4 mol L−1s−1


FOR SALE



2N2

O5

(g)→ 4NO2

(g) + O2

(g)

A chemical reaction

=− −1 −1


∆t

∆[O2] = −½(−2.2 mol L−1s−1)∆t =1.1 mol L−1s−1


FOR SALE



2H2

O2

(l) → O2

(g) + 2H2

O(l)

Consider the reaction

if ∆ O = 0.21 mol s-1


∆t

determine ∆[H2O2]∆t


FOR SALE



Rate of a reaction

Time involved

what is the difference between


Average rateinstantaneous rate

initial rate


FOR SALE



the reactionBr2(aq)+ HCOOH(aq) →2Br−(aq) +2H+(aq) + CO2(g)

Time / s [Br2]/ mol L−1

0.0 0.0120

50.0 0.0101


100.0 0.00846150.0 0.00710

200.0 0.00596

300.0 0.00420400.0 0.00296

What do you notice [Br2] ?PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE




Average rate = ∆[Br2]

∆t


. .

Average rate = [Br2]final − [Br2]initial

= 0.0101 − 0.0120 mol L−1

50.0 − 0.0 s

= −3.80x10−5 mol L−1 s−1

tfnal − tinitial

PRK 1026 CHEMISTRI II UNIMAS (C) NOTFOR SALE




InstantaneousRateThe rate of


a react onat a particular time





When t=0,a special instant

Such rate known


as initial rateof the reaction




If involve gas, measure gasvolume against time

How to determine rate experimentally?





Examples of gaseous reactions

Zn(s) +H2SO4(aq) →H2(g) + ZnSO4(aq)

CaCO3(s) + 2HCl(aq) →CO2(g) + CaCl2(aq) + H2O(l)


2H2O2(aq) →O2(g) + H2O(l)

As gas has pressure, can also measurepressure changes over time




If involve colour changes, usecolourimetric method

How to determine rate experimentally?





To result in chemical productsreactant molecules must collide

A chemical reaction


( The Collision Theory )

Minimum energy needed to

bring about product formation

( Activation energy )PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Any chemical reaction


Collision is a necessity




A reaction needs minimum energy


A collision may have no reaction




Overcome activation energy


Collisions > Ea have reactions




The Boltzman distribution of

Molecules at any T

r o f

l e s


Molecular energy

N u m b

m o l e c




Molecules with energy ≥ Eaonly can react

r o f

e s


Molecular energy

N u m b

m o l e c

Ea




Kinetic energy of molecules

T2

T1

T2 > T1


More molecules > Ea at higher T




Right orientation of collision


Effective collisions only can result

in productsPRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Right orientation of collision

CH3 CHCl3→

CH4 +

CCl3

CH H-C Cl → CH + CClCl


Cl

CH3

Cl-C Cl → no product

Cl

H

orientation

Wrong

orientation




Frequency of effective collisions

Effect of increasing Temperature

Reactant molecules moving faster


increased

Rate increased




When molecules collide

effectively,

Transition state before product


A B A B A B

Effective collision activated complex product

in transition state




Transition state before product

Transition state is

an activated


maximum energyto rearrange itself

to the product




Reaction kinetics

•How fast a reaction can take place?

•Find out experimentally

•What relation between rate of


reaction and concentration of reactants?




Reaction kinetics

A + B →→→→ C + D

How fast C is produced?





Reaction kinetics

A + B → C + D

Rate of C roduced de ends on A & B


Rate = k[A]x[B]y

Order wrt A

Order wrt B

Rate constant




Order of a reaction

A + B → C + D

Rate = k A x B y

Order wrt A

Order wrt B


If x + y = 0 0th order

If x + y = 1 1st

orderIf x + y = 2 2nd order




Rate constant, k

A constant at a fixed TemperatureIf T , then k will


Once order of reaction known, kcan be calculated.




The hydrolysis of CH3Br by OH- isgiven by the rate law

Rate = k[CH3Br][OH-]


r er o reac on =1st order wrt CH3Br

1st order wrt OH-




The reaction A + B + C → D + E

has a rate law as below:-

Rate = k[A][B]2[C]0


a) What is the order wrt A?

b) What is the order of thereaction?




Consider another reactionCH3COOCH3 + OH−→CH3COO− + CH3OH

Write the rate law for the reaction.


Answer:

rate of reaction =k [CH3COOCH3]x [OH−]y




Order of a reaction

A + B → C + D

Order must find from ex eriment onl


Using different concentrations of A & B

Get different initial rate obtained




Expt [X] /

mol L-1

[Y] /

mol L-1

Initial rate /

Mol L-1s-1

1 2.20 1.50 0.00700

2 2.20 3.00 0.0140

3 6.60 1.50 0.0210


Rate = k[X]x[Y]y

Step 1


Step 2



From expt 1 & 3

Initial rate 3 =

Initial rate 1

Step 2


From expt 1 & 30.0210 = k(6.60)x(1.50)y

0.00700 k(2.20)x(1.50)y

X=1




From expt 1 & 2Initial rate 2

Initial rate 1


From expt 1 & 2

0.0140 =k(2.20)x(3.00)y

0.00700 k(2.20)x(1.50)y

Y=1PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Step 3: Conclude the rate law

Rate= k[X]1[Y]1





Rate = k[X]1[Y]1

Order of reaction = 2nd

Step 4: Conclude order of chemical reaction


, u e exOr any other expt

k=2.12x10-3 mol-1Ls-1




More example on Order of Reaction

• Consider the reaction

F2 (g) + 2ClO2 (g) 2FClO2 (g)


• The rate law is expressed as

• To determine the values of x and y , we have to

look at the experimental data for this reaction.

y xk ]ClO[]F[rate 22=




Order of Reaction

• Experimental data

Exp

#[F2] (M) [ClO2] (M) Initial Rate (M/s)

1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

-3


. . .

Doubling [F2] while holding [ClO2] constant

will double the rate of reaction.

∴The rate is directly proportional to [F 2 ] The reaction is first order in F 2

x = 1PRK 1026 CHEMISTRI II UNIMAS (C) NOTFOR SALE



Order of Reaction

• Experimental data

Exp


1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

3 0.20 0.010 2.4 x 10-3


Quadrupling [ClO2] while holding [F2]constant

will quadruple the rate of reaction.

∴The rate is directly proportional to [ClO 2 ]

The reaction is first order in ClO 2

y = 1 PRK 1026 CHEMISTRI II UNIMAS (C) NOTFOR SALE



Order of Reaction

Exp


1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

3 0.20 0.010 2.4 x 10-3


• Since the reaction is first order for both [F2]and [ClO2], the rate law is written as

• The reaction is (1+1) or second order overall.

]ClO][F[rate 22k =


QUESTION



Expt [A] /

mol L− 1

[B] /

mol L− 1

Initial rate /

mol L− 1

s− 1

1 0.100 0.100 5.50x10− 6

− 5

Consider the reaction at 25 ° C

A + 2B → C


. . .

3 0.400 0.100 8.80x10− 5

4 0.100 0.300 1.65x10− 5

a) Determine the rate law for the reaction.

b) What is the rate constant for the reaction?




Try 2

In the reaction A → B , the initial rate is 1.7x10− 4 mol

L− 1

s− 1

when the concentration of A is 0.25 mol L− 1

a) What will be the initial rate of reaction if [A] is 0.75


mol L− with zero order of reaction?

b) What will be the initial rate of reaction if [A] is 0.75

mol L− 1

with 1st

order of reaction?


FOR SALE



Rate constant

•A constant at a particular temperature

•Always bigger at a higher temperature

•K and Ea related as in Arrhenius equation


k=AeWhere A is the frequency factor

Ea

is the activation energy

R is the universal gas constant

T is the reaction temperature.

−


FOR SALE



Rate constant

•If many k s are done at different temperature

•Can determine Ea of a reaction


k=Ae

−

•Plot graph of ln k against 1/T and Ea can be

determined from gradient of the linear graph


FOR SALE



Arrhenius Equation

ln A

A

T R

E k a ln

1ln +

−= RT

E a

Aek −

=



FOR SALE

l n

k

1 /T

R

E a−=slope



1st order reaction

Ad ][

A→ productRate of reaction = k[A]

Rate of product = −


t

= k[A]

So, k[A] = -dt

Ad ][

A

Ad ][−


FOR SALE



1st order reaction

= k[A]A

Ad ][By integration of −

Obtained


ln( ) = kt or][

][ 0

A

A

ln [A]0 − ln[A] = kt


FOR SALE



Gradient of the linear

graph equals −k ln[A]

t


[A]

t


FOR SALE



The reaction 2A→ B is a first order reaction

with a rate constant of 2.8x10−2 s−1 at 80 °C.

How long (in seconds) will it take to decrease from 0.88

mol L−1

to 0.14 mol L−1 ?


answerln [A]

0 −ln[A] = kt

t = ( ln [A]0

- ln[A] ) /k

= ln0.88 − ln0.14

2.8x10−2 s−1

= 66 sPRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Try 1

The reaction 2N 2O5 → 4NO2 + O2 is a first order reaction with a rate constant of


. x − s− a .

If the initial concentration of N 2O5 is 0.25 mol L− 1 ,

what is the concentration after 3.2 minutes?


FOR SALE



Half-life of a reaction

Time required for the initialamount of reactant to decrease


to half its initial amount

10 g t 5 g8 g t 4 g


FOR SALE



Half-life of a reaction

For a 1st order reaction


kt1/2 = ln2


FOR SALE



examplethe decomposition of cyclopropane to propene

is a first order reaction with a rate constant

of 6.7x10−

4 s−

1 at 500 °C.

Calculate the half-life of the reaction.


ln2 = kt1/2hence, t

1/2= ln2/k

= ln2/6.7x10−4 s−1

= 1.0x103 s


FOR SALE



Try

The half-life of first order reactionis 84.1 minute.


Calculate the rate constant of the reaction.


FOR SALE



Half-life of 1st

order reaction isindependent of amount of active substance



FOR SALE



what fraction of active materialis left after 4 half-lives?

Half-life


½ ¼ 1/8

answer

1/16


FOR SALE



a) Most reaction goes to productthrough a number of steps

Reaction mechanism


b)always has a slowest step

c)rate determining step


FOR SALE



Reaction mechanism

A reacts with B to form products X and Y


Step 2 M + A → Y fast

Step 1 + 2A + B → X + Y

Step 2

So, rate = k[A][B]PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE



Reaction mechanism

NO reacts with O 2 in 2 elementary steps

Step 1 NO + O 2 → NO 3


Step 2 NO 3 + NO → 2NO 2

Given that, rate = k[NO] 2 [O 2 ] Which step is rate determining? Why?


FOR SALE



Helps to increase rate of a

A catalyst


,

How a catalyst works?


FOR SALE



How catalyst increases rate


Ea reduced in presence of a catalyst∆H not changed


FOR SALE

Effect of Catalyst on E U t l d



Effect of Catalyst on E a

e n e r g y

E a uncatalysed

Uncatalysed

reactionCatalysedreaction

Normalreaction

mechanism


P o t e n t

i a l

Progress of reaction

A + B

C + D

E a catalysed


FOR SALE

Alternate

reactionmechanism



Catalyst reduces Ea



FOR SALE



How catalyst increases rate

•Catalyst provides alternative routewith a lower Ea


•ata yst ta es part n react on

•Catalyst not used up

Note: catalyst does not changethe products nor its quantity


FOR SALE

How a catalyst works



How a catalyst works

• Keep in mind that:

– The normal uncatalysed reaction mechanism still

exists with E a = E a uncatalysed .


– Catalyst provides an alternative reaction withE a = E a catalysed .

– Because E a catalysed

< E a uncatalysed

, most reactants

would follow the alternative reaction mechanism

as it is easier.


FOR SALE



Types of catalysts

Homogeneous catalyst


Heterogeneous catalyst


FOR SALE




2- -

Catalyst and reactants of same phases


2 8

I2 (aq) and SO42-(aq) using

Fe3+(aq) as catalyst


FOR SALE




2- - 3+


2 8

I2 (aq) +SO42-(aq)

Reaction takes place in 2 stepsPRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE




2Fe3+(aq) + 2I-(aq)→2Fe2+(aq) + I2 (aq)


More reasonable for reaction tooccur with opposite charges

S2O82-(aq) +2Fe2+(aq)→2Fe3+(aq)+ 2SO42-(aq)


FOR SALE




In the reaction , Fe3+(aq) is a catalyst,


an e + aq s an nterme ate


FOR SALE

Intermediates



X and Y reacts to produce Z in the presence

of a catalyst, C.

The reaction steps are as:


.........

Y + XC → XYC ........(2) XYC → CZ .........(3)

CZ →

C + Z .........(4)

State the intermediates.PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE




The Haber Process of makin NH

Catalyst and reactants of different phases


from H2(g) and N2(g) using Fe ascatalyst


FOR SALE




3H2(g) +N2(g) Fe as catalyst 2NH3(g)


Use surface adsorption theory to explain therole of Fe in increasing the rate of reaction


FOR SALE

Surface adsorption theory




H2 and N2 molecules adsorbed on surfaceDissociation happens to give H and N atoms

H and N atoms form bonds to become NH3


FOR SALE

E.g.: Hydrogenation of C H



2 4• The hydrogenation of ethene to ethane iscatalysed by nickel:

)(CHCH)(H)(CHCH 33222 ggg −→+=



FOR SALE

C CH H

H H

Ethene H2

Nickel(catalyst)

Adsorption



C C

H

H

H

H

H H


• Both ethene & hydrogen bonds to nickel.


FOR SALE

HH

Formation of •CH -CH radical



2 3

H H

•CH2-CH3 radical

H H


• The radical uses the unpaired electron to

bond to the surface.PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE

C CH H

HC CH H

H

Reaction with another H atom



H H


• The radical reacts with another H atom.


FOR SALE

HC CH H

H

Desorption



H HEthane


• Desorption of ethane from the surface of

nickel occurs.PRK 1026 CHEMISTRI II UNIMAS (C) NOT

FOR SALE

HC CH H

H

Summary on reaction kinetics



•Rate of reaction can change

•Collision theory to explain reaction•A reaction needs activation ener


•Reaction has a mechanism•Catalyst changes route of reaction

but not ∆H nor quantity of products•Catalyst is specific in action


FOR SALE

lu 4 slides reaction kinetics

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