CHEMISTRY PRE-U CHEMISTRY SEM 1 CHAP 5.pdf

7/27/2019 CHEMISTRY PRE-U CHEMISTRY SEM 1 CHAP 5.pdf

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PRE-UNIVERSITY

CHEMISTRY SEMESTER 1

962 / 1CHAPTER 5 :


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CHAPTER 5 : KINETIC CHEMISTRY

5.1 Rate of reaction

5.2 Rate law, Order of reactions and rate constants

5.3 The effect of temperature on reaction kinetics

Past Year Questions Analysis

.

Topic

2007 2008 2009 2010 2011 20122013

Sem 1

2014

Sem 1

P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 AB,

CA

B,

C

5. Kinetic

Chemistry

1 5b 11b

2c

3 3 3 6 3 1 2 17


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5.1 Rate of reaction

1. Chemical kinetics is the area of chemistry concerned with thespeeds, or rates, at which a chemical reaction occurs.

Kinetics, in this case is referring to the rate of a reaction,

which is the change in the concentration of a reactant or a

product with time. Generally, rate of reaction is inverselyproportional with time, therefore, longer the time taken for a

reaction to occur, lower the rate of reaction

2. Generally in chemical reactionReactants Products

For example, when

A

B(Concentration decrease) (Concentration increase)

timerate


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b) Consider the following chemical equations :

a A (aq) + b B (aq) c C (aq) + d D (g)

Rate of reaction can also be expressed base on the stoichiometrycoefficient, where it can be written as

Rate of reaction based on A Rate of reaction based on B

td]A[d

a1Rate =

td]B[d

b1Rate =

As a conclusion,

td

]C[d

c

1Rate =

td

]D[d

d

1Rate =

td

]D[d

d

1

td

]C[d

c

1

td

]B[d

b

1

td

]A[d

a

1Rate ====


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For example, in a chemical reaction of Ostwald Process, where4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

Rate of reaction based on A Rate of reaction based on B

td]NH[d

41Rate 3=

td]O[d

51Rate 2=

Rate of reaction based on C Rate of reaction based on D

As a conclusion,

td

]NO[d

4

1Rate =

td

]OH[d

6

1Rate 2=

td

]OH[d

5

1

td

]NO[d

4

1

td

]O[d

5

1

td

]NH[d

4

1Rate 223 ====


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5.1.1 Theory of Reaction Rates

Collision Theory use to explain the effects ofconcentration and temperature on rate ofreaction.

Base on 3 main ideas

Molecules must collide to react

Molecules must collide at the minimum amount ofenergy called activation energy.


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Theory use to explain the effects of concentration and

temperature on rate of reaction.

Molecule must collide to react

In order for a reaction to occur, there must be physical interactionsthat take place, where the molecules are able to collide with each

other and form a chemical reaction. Rate of collision is directly

influenced by the following factors :

Molecule must collide to react

Concentration : As the concentration of particles increased,frequency of collision occur more rapidly. This will increase the

chances of effective collision and hence increase the rate of

reaction. From the diagram below, we can see that, as the number

of particles increase with concentration hence increase thefrequency of collision between particles. the frequency of collision

increase significantly from (a) < (b) < (c) as the number of particles

increase


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Temperature : When temperature

increase, particles absorbed theenergy supplied and stored in theform of kinetic energy. Particleswith higher kinetic energy canmove faster, and hence has ahigher frequency of collisionbetween particles. Furthermore

higher kinetic energy allows moreparticles to have energy higherthan the activation energy, henceincrease the rate of reaction.


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Molecule must collide in the right orientation inorder to form the right product.

Consider the following reaction occurs.2 AB A2 + 2 B.

For the reaction to take place, when the molecules

collide, it must collide under the correct way, inorder for that particular reaction to happen


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If the molecules collide at a wrong orientationor position is not correct, the reaction will not

occur, as illustrated in the 2 diagrams below


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Molecule must possessed certain amount ofkinetic energy called activation energy

Even if the 2 conditions above is fulfilled themolecules collide as in (a) ; it is collided in theright orientation as in (b), does it mean, thereaction will occur?

Note the diagram belowDiagram of collision with the

correct orientation. Note that,

even though it collides with theright orientation, the reaction

does not occur. This is due to

the collision is too gently, and

does not have the enoughenergy to react. So, this

minimum required for the

reaction to happened is calledactivation energy


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Activation energy is defined as ...

So, in order for a reaction to take place, there mustbe a certain amount of energy absorbed in order forthe molecules which collide at the right orientationto happen.

As discussed durin Maxwell-Boltzmann distribution

minimum amount ofenergy required to initiate a chemical reaction

graph, when temperature increase, more particleshas energy higher than activation energy, and thiswill increase the rate of reaction.

Activation is always endothermic, as heat isrequired in order for molecule to collide effectivelyand form a new compound.

The energy profile bellow shows the reaction

of both endothermic and exothermic reaction


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Endothermic process Exothermic process

Ea Ea


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Theory of transition state

Theory of transition state explained the process that takeplace during a chemical reaction.

Supposedly the reaction of BC + A C + AB is a one

step reaction where it can be described as

A + B C [A---B---C] AB + C

The height of the "barrier" in the

beginning of the graph is called the

activation energy, and theconfiguration of atoms at the

maximum in the potential energy

profile is called the transition state,

or the activated complex. In anotherwords, A--B--C formed is an activated

complex where it usually exist as an

intermediate of a chemical reaction, a

substance that appear during areaction but will not form as products


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Hydrolysis in 3o haloalkane Hydrolysis in 1o haloalkane

Equation

C(CH3)3Br + OH-

C(CH3)3OH + Br-

CH3CH2Br + OH-

CH3CH2OH +

Br-

Occur in 2 steps

Step 1 : Formation of carbocation

Occur only in 1 step

Process Step 2 : Nucleophilic attack


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Energ

y

Energy Energy / kJ

pro e

Reaction coordinate Reaction coordinate

Rate

equation

Rate = k [C(CH3)3Br] Rate = k [CH3CH2Br][OH-]


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5.2 Rate law, Order of reactions and rate constants

1. From what we had learn so far, we know that concentration isdirectly proportional to rate of reaction where

rate [concentration] ; rate = k [concentration]

The equation bolded is also known as rate equation, (also known

as rate law) a way to expresses the relationship of the rate of areaction to the rate constant and the concentrations of thereactants raised to some powers. Rate constant, k, in the otherhand, is the proportionality constant of a given chemical reaction.

2. Supposedly if a chemical reaction take placed as shown in theequation below

a A (aq) + b B (aq) c C (aq) + d D (g)

The rate equation for the reaction is expressed by

k = rate constant[A] and [B] = concentration of A and B

x and y = order of reaction withrespect to concentration of A and B

rate = k [A]x[B]y


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a) Only the concentration of reactants is taken into consideration

of rate of reaction, as the speed of the reaction largely depend

on the amount and concentration of reactants used in theprocess. Generally, greater the amount (concentration) of

reactant used, greater the rate of reaction

rate = k [A]x[B]y

b) Rate constant, k, is the proportionality constant of thereaction, in which the value remain constant under constant

temperature, regardless of the changes in concentration of

the reactants take place. However, rate constant changes

with temperature of the reaction


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c) order of reaction, x and y, is the power to which the

concentration of the reactants (in this case, [A] and [B]

respectively) is raised to in the rate equation.

i. Note that the order of reaction has no relationship with the

stoichiometry coefficient of the chemical equation. Therefore,

rate = k [A]x[B]y

.

ii. Order of reaction can only be determined from series ofexperiments carried out under different concentration.

iii. The overall order of reaction can be calculated by summing

up the order of reaction for each reactant involved. In thiscase.

Overall order = x + y


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d) Once the overall order of reaction is known, the rate constantcan be calculated accordingly by substituting the concentration

and order of reaction towards the rate of each experiment. Unit

of rate constant depend largely on the overall order of reaction.

Overall reaction

orderExample Unit of rate constant, k

= 0(mol dm-3 s-1) = k (mol dm-3)0

k = mol dm-3 s-1

First order rate = k [A](mol dm-3 s-1) = k (mol dm-3)1

k = s-1

Second order rate = k [A][B](mol dm-3 s-1) = k (mol dm-3)2

k = mol-1 dm3 s-1

Third order Rate = k [A][B]2(mol dm-3 s-1) = k (mol dm-3)3

k = mol-2 dm6 s-1

Fifth order Rate = k [A][B]2[C]2(mol dm-3 s-1) = k (mol dm-3)5

k = mol-4 dm12 s-1


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5.2.1 First order of reaction

1. A first-order reaction is a reaction whose rate depends on thereactant concentration raised to the first power. Supposedly in a

chemical reaction, where

A products

Therefore, for this reaction, rate of reaction can be expressed

or it can be written as rate = k [A]

When the rate is substituted between each other, it can be expresseddt

]A[drate =

Integration of the equation


c is a constant which can be determined when time, t = 0 (initial

time), c = ln [A]. Usually, it is indicated as ln [A]0, since it is the initial

concentration of the reaction. As a result, the equation that can be

applied to a first order reaction is :

ln [A]t = - kt + ln [A]0

]A[]A[ddtkasrearrangebecanwhich

dt]A[d]A[k ==

ckt]A[lntdk]A[

]A[d+==


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Graph in first order reaction


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3. Order of a reaction can also be determined by using half-lifemethod. Half life, t1/2 is defined as the time required for the

concentration of a reactant to decrease to half of its initial

concentration, [A]0/2. Flow below shows the decreased of

concentration of reactant for each half-life take place

]A[]A[]A[]A[ 02/1

trd302/1tnd202/1t

st1

0

-3

0.

Half life in first order of reaction

Initial concentration of a reactant is 1.00 mol dm-3

1.00 mol dm-3 0.50 mol dm-3

0.25 mol dm-3 0.125 mol dm-3

2/1tst1

2/1tnd2

2/1trd3


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Half life in first order of reaction (graph)


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c) Using the equation of the first order of reaction, we can alsocalculate the half-life of the reaction, and also rate constant, k,

if either one of these information is given. Derived from the

equation of first order of reaction :

Since after t1/2 ; concentration of the initial reactant decreasedby half. Therefore,

2

]A[]A[ 0t =

Equation of Half life in first order of reaction

substituting into the equation,

Rearranging equation

or simply ln 2 = k t1/2 or

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]A[lnkt2ln +=

2/1

20]A[

02/1

00 kt

]A[lnorkt

2

]A[ln]A[ln ==

k

693.0t 2/1 =


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5.2.2 Second order of reaction

1. A second-order reaction is a reaction whose rate depends on theconcentration of one reactant raised to the second power or on the

concentrations of two different reactants, each raised to the first power.

A + B products

Rate equation for a second order reaction can be either :rate = k [A]2 or rate = k [B]2 or rate = k [A][B]

]A[drate =

When the rate is substituted between each other, it can be expressed as


c is a constant which can be determined when time, t = 0, c = 1/[A]. Usually,

it is indicated as 1/[A]0, since it is the initial concentration of the reaction:

2

2

]A[

]A[ddtkasrearrangebecanwhich

dt

]A[d]A[k ==

ckt

]A[

1tdk

]A[

]A[d

2

+==

0t ]A[

1

kt]A[

1+=


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Graph in second order reaction


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Graph of half life in second order reaction


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The graph above clearly shows how does the half life occur for asecond order of reaction, where times taken for each stage of

half life :

For a second order of reaction : Since 1st t1/2 = 1 min

2nd t1/2 = 2 x 1st t1/2 = 2 (1 min) = 2 min

08

1

min40

4

1

min20

2

1

min10 ]A[]A[]A[]A[

3r t1/2

= 2 x 2n t1/2

= 2 (2 min) = 4 min

As for 4th t1/2 = 2 x 3rd t1/2 = 2 (4 min) = 8 min

b) Similar to first order of reaction, the equation of the half life in

second order of reaction can be derived from the second

order equation. At t1/2 ;2][][ 0AA

t =

2/1

00 ][

1

2

][

1tk

AA =

oAk

ttk

A ][

1

][

12/12/1

0

==


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1. First- and second-order reactions are the most commonreaction types. Reactions whose order is zero are rare. For a

zero-order reaction, where

A products rate equation : rate = k [A]0

rate of reaction can be expressed using the equations below,

or it can be written as rate = k]A[d

rate =

Zero order reaction

When the rate is substituted between each other, it can be

expressed as

Integration of equation

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

t

]A[dtkasrearrangebecanwhichdt

]A[dk ==

ckt]A[tdk]A[d +==

G h Z d i


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Graph Zero order reaction


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3. For a zero order of reaction, since the graph of concentrationagainst time is linear, therefore, it is impractical to determine

the half-life of zero order using practical methods. In fact, the

linear shape of the graph itself has proven that the reaction is

zero order with respect to the reactant involved.a) Though, the half-life of zero order reaction can be calculated

using equation method, where the equation is derived from the

zero order equationDuring half life, t1/2, occur ;

when [A]t is substitute in equation2

][][ 0

AA

t=

k2

]A[ttk

2

]A[]A[ 02/12/1

00 ==


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1. +

.

1.50 3 2.00 3

& 6.70 105 3 1.

) = 1

0

= ) , = /

= 6.70 105 3 1/ 1.50 3

= 4.47 105

1

) 1 , 1

= + 0

= = 4.47 10

5

3600 + 1.50= 1.28

3

= ; = 4.47 105 1.28

= 5.71 10

5

3

2 A dil t l ti f h d id b d t bl h h i It


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2. A dilute solution of hydrogen peroxide can be used to bleach hair. It

decomposes slowly in aqueous solution according to the following

equation:2 H2O2 (aq) 2 H2O (l) + O2(g)

A solution with an original concentration of 3.0 mol dm-3 was placed

in a bottle contaminated with transition metal ions, which act as

catalysts for the decomposition. The rate of decomposition wasmeasured by withdrawing 10 cm3 portions at various times and

titrating with acidified 0.10 mol dm-3 KMnO4 (aq). (5 mol of peroxide

reac w mo es o n 4. e o ow ng resu s were o a ne :

Plot a graph of volume of KMnO4 against time. Based on the graph

sketch, determine the order of reaction with respect to H2O2,

express the rate equation for the reaction and calculate the rate

constant of the reaction.

Time / min 0 5 10 15 20 25 30

V of 0.10 mol dm-3

KMnO4 / cm3

30.0 23.4 18.3 14.2 11.1 8.7 6.8


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15

20

25

30

0

5

10

0 5 10 15 20 25 30


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, 1/2 13.5 , 1/2

13.5 , , 22.

: = 22

:

1min0513.0k693.0

k;693.0

t ===.

3.


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3.

2 + 2

1/

6.00

7.00

8.00

0.00

1.00

2.00

3.00

4.00

5.00

0 50 100 150 200 250 300

a) Determine the order of reaction with respect to B.


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a) Determine the order of reaction with respect to B.

b) From the graph, determinei. the initial concentration

ii. rate constant of reaction

Second order of reaction

At t = 0 ; 1/[B] = 1.0 ; [B] = 1.0 mol dm-3

Using any two point to form tangent

c) Write the rate equation for the reaction

, . .

= 0.024 mol-1 dm3 s-1

Rate = k [B]2

4 Trichloromethane CHCl3 reacts with sodium hydroxide solution as


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4. Trichloromethane, CHCl3 reacts with sodium hydroxide solution as

represented by the following equation:

2 CHCl3 + 7 OH CO + HCOO + 6 Cl + 4 H2OThe reaction is first order with respect to each reactant.

a) Write the rate equation for the reaction above.

b) Determine the rate of production of chloride ions at 28 C when the5 1

Rate = k [CHCl3] [OH]

. .

5

53

100.1rate

100.22

1rate,so;

dt

]CHCl[d

2

1rate

=

==

5

5

100.6dt

]Cl[d

dt

]Cl[d

6

1100.1,so;

dt

]Cl[d

6

1rate

=

==

c) If the rate of reaction is r when the concentrations of both trichloromethane


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)

and sodium hydroxide are both 2.0 mol dm3, what is the rate of reaction in

terms of r when half of the hydroxide ion is reacted?

d) Sketch a graph of the rate of reaction against the concentration of

2 CHCl3 + 7 OH- products

Initial conc. 2 2

Final conc. 2 (2/7 x 1) = 12 /7 2 (2 x ) = 1

Rate, r = k [CHCl3][OH

] ; r = k [12/7 CHCl3] [1 OH

]r = 12 / 7 k [CHCl3] [OH]

So, rate is 12 / 7 r

trichloromethane if sodium hydroxide is in excess such that the hydroxide

ion concentration remains practically constant in the reaction mixture.

rate

[CHCl3]


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5.2.4 Determination of order of reaction via experiment

1. Consider the reaction between oxygen and nitrogenmonoxide, a key step in the formation of acid rain and in the

industrial production of nitric acid .

O2

(g) + 2 NO (g) 2 NO2

(g)

The rate equation, expressed in general form, is

rate = k [O2]x [NO]y

Note that the order of reaction cannot be determined directly

from the stoichiometry of the reaction. To find out the orders

of reactant with respect to each O2 and NO, we run series of

experiments, starting each one with a different set of reactant

concentrations and obtaining an initial rate in each case.


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Exp Initial concentration of reactant Initial rate(mol dm-3 s-1)O2 / mol dm

-3 NO / mol dm-3

1 1.10 x 10-2 2.50 x 10-2 2.40 x 10-3

2 2.20 x 10-2 2.50 x 10-2 4.80 x 10-3

3 1.10 x 10-2 5.00 x 10-2 9.60 x 10-3

4 3.30 x 10-2 7.50 x 10-2 x

From each experiment, the rate equations are expressed individually,where

Experiment 1 : 2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y

Experiment 2 : 4.80 x 10-3 = k (2.20 x 10-2)x (2.50 x 10-2)y

Experiment 3 : 9.60 x 10-3

= k (1.10 x 10-2

)x

(5.00 x 10-2

)y

Experiment 4 : x = k (3.30 x 10-2)x (7.50 x 10-2)y


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Comparing Experiment 2 to Experiment 1 :4.80 x 10-3 = k (2.20 x 10-2)x (2.50 x 10-2)y

2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y

2 = (2)x

Order of reaction with respect to O2 ; x = 1

Comparing Experiment 3 to Experiment 1 :

9.60 x 10-3

= k (1.10 x 10-2

)x

(5.00 x 10-2

)y

2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y

4 = (2)y

Order of reaction with respect to NO ; y = 2From the order of reaction deduced, the rate equation is

rate = k [O2][NO]2.

The overall order of reaction = 1 + 2 = 3


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Using any experiment, rate constant can be calculated. Forexample, in experiment 1

2.40 x 10-3 = k (1.10 x 10-2) (2.50 x 10-2)2

k = 349 mol-2 dm6 s-1.

Once the order of reaction and the rate constant were

determined, we can predict the rate of reaction under any

. ,

4, whererate = k [O2][NO]

2 ; rate = 349 (3.30 x 10-2) (7.50 x 10-2)2

rate = 6.48 x 10-2 mol dm-3 s-1

2. Sometimes, using a combination of graphical methods andinitial rate methods the order of reaction can be found


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initial rate methods, the order of reaction can be foundindividually. For example, in hydrolysis of bromoethane with

potassium hydroxide at different concentrations.

CH3CH2Br (l) + KOH (aq) CH3CH2OH (l) + KBr (aq)

The following results were obtained from two experiments

on such a hydrolysis. In each experiment, the overall[KOH(aq)] remained virtually constant at the value given atthe top of the column.

time /min

[CH3CH2Br] / mol dm-3 when

[KOH] = 0.10 mol dm-3[CH3CH2Br]/mol dm

-3 when

[KOH-] = 0.15 mol dm-3

0 0.0100 0.0100

40 0.0079 0.0070

80 0.0062 0.0050120 0.0049 0.0034

160 0.0038 0.0025

200 0.0030 0.0017

240 0.0024 0.0012


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0.006

0.008

0.01

0

0.002

0.004

0 50 100 150 200 250


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a) From the reaction, the rate equation can be written as

rate = k [CH3CH2Br]x [KOH]y

Therefore, to determine the order of reaction with respect to

CH3CH2Br, the half-life method is applied, since the graph of

concentration of CH3CH2Br against time are plottedaccordingly, while the order of reaction with respect to KOH

can be obtained usin initial rate methods where the initial

rate for both [CH3CH2Br] under different concentration of KOHcan be found hence calculated.

b) By the mean of half-life method, the first t1/2, second t1/2 and

third t1/2 of the concentration of CH3CH2Br under the

concentration of KOH 0.15 mol dm-3 occurred at 80 s, 160 sand 240 s. Since the 1st t1/2 = 2

nd t1/2 = 3rd t1/2 = 80 s, therefore

the order of reaction with respect to CH3CH2Br is first order of

reaction.


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d) From both methods applied, the rate equation can be written

as

rate = k [CH3

CH2

Br] [KOH]

The overall orderof reaction is 1 + 1 = 2

e) Using any experiment above, rate constant of the reaction

can e ca cu a e .

In experiment 2 : 1.00 x 10-4 = k (0.0100) (0.15)

k = 0.0667 mol-1 dm3 s-1

5.2.5 Reaction Mechanisms


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1. Chemical reactions may occur in one way reaction or a

reversible reaction. Example of a one way reaction, is the

formation of nitrogen dioxide via the reaction of nitrogenmonoxide and oxygen gas

2 NO (g) + O2 (g) 2 NO2 (g)

(symbolised one way reaction)

While the example of a reversible reaction as in production of

ammonia from nitrogen gas and hydrogen gas, which is

largely used in industrial process via Haber process.

N2 (g) + 3 H2 (g) 2 NH3 (g)

(symbolised reversible reaction)


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2. In a one way reaction, process may be taken in multiple steps,and therefore irreversible back to the reactants, as they might

involve in steps that required higher activation energies. For

example, in the reaction stated above where

2 NO (g) + O2 (g) 2 NO2 (g)

The steps (or simply mechanism) for the reaction of nitrogen

described below.

Step 1 :

Step 2 :

Overall : 2 NO + O2 2 NO2

22

slowONNO2

2

fast

222

NO2OON +


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a) In the process of the formation of nitrogen dioxide, N2O2 is formedtemporary, but it will not exist as a product in the end of reaction.

Therefore, N2O2 is also known as intermediate, a substance that

appear in the mechanism of the reaction but not in the overall

balanced equation.b) For each step of reaction, the rate equation can be described

accordingly

From the rate equation proposed for each step in the series of

mechanism proposed the order of reaction can be determinedstraight forward by the stoichiometry coefficient. From the

mechanism equation in step 1, order of reaction is second order

with respect to NO, while in step 2, order of reaction is first order

with respect to each N2O2 and O2 respectively.

Step 1 rate = k [NO]2

Step 2 rate = k [N2O2] [O2]

22slow ONNO2

2

fast

222 NO2OON +


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c) Note that in the series of step, above the arrow is written with

the word "slow" and "fast", which can be interpreted as a slow

step of reaction, and a fast step of reaction. In determining the

order with respect of each reactant involved, we can make use

the mechanism to determine the order of each reactant, sinceslow step is the rate determining step. Therefore, the rate

equation that can represent the overall equation

2 NO + O2 2 NO2 rate = k [NO]2.


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5.3 The effect of temperature on reaction kinetics


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1. Temperature often has a major effect on reaction rate. Ingeneral, increasing the temperature of a reaction increases

the average speed of particles and therefore their frequency

of collisions.

a) Arrhenius proposed that every reaction has an energy

threshold that the colliding molecules must exceed in order to

.

(EA), the energy required to activate the molecules into a statefrom which reactant bonds can change into product bonds.

b) Many of the chemical reactions near room temperature

approximately double their rates with a 10

0

C rise intemperature.


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2. The effect of temperature towards the rate of reaction can befurther explained using Maxwell-Boltzmann distribution graph.

From the graph obtained, particles at 800 K have move collision

energy compared to particles at 300 K. This is due to, as thekinetic energy increased, particles moved faster and moreparticles collides more frequently hence increased the collisionenergy. Therefore, more particles have energy greater than

activation energy, and results higher rate of reaction.


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3. Generally, the rate of reaction increased with temperature as it

affect the rate by increased the rate constant of the reaction. The

dependence of the rate constant of a reaction on temperature can

be expressed by using Arrhenius equation.

k = rate constantA = Arrhenius constant

T = tem eratureEA

Arrhenius equation shows that the rate constant is directly

proportional to A and, therefore, to the collision frequency. Inaddition, because of the minus sign associated with the exponent

EA/RT, the rate constant decreases with increasing activation

energy and increases with increasing temperature

EA = activation energyR = gas constant (8.31 J mol-1 K-1)e=

Derivation of Arrhenius Equation.


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a) Arrhenius equation expressed as natural logarithm of bothsides, where the equation can be expressed as :-

From the rearranged equation, if a graph of ln k against 1/T is

AlnT

1

R

Ekln:rearrange;

RT

EAlnkln AA +==

p o e , a nega ve gra en near ne may e o a ne ,

where the gradient, m = EA / R. Therefore, using thismethod, the activation energy of a reaction can be calculated

Example : The table below shows how the rate constant, k, varieswith the temperature for the reaction between H2 (g) and I2 (g)


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H2

(g) + I2

(g) 2 HI (g)

By plotting a suitable graph, determine the activation energy forthe reaction.

Solution : A table of 1/T and ln k is first calculated and placed

Temperature (K) 556 575 647 700 791

Rate constant,

k (mol-1 dm3 s-1)3.52 x 10-7 1.22 x 10-6 8.59 x 10-5 1.16 x 10-3 3.90 x 10-2

accordingly.

1 / T (K-1) 0.00180 0.00174 0.00155 0.00143 0.00126

ln k -14.9 -13.6 -9.36 -6.76 -3.24

1A

A

molkJ185@444,185E

31.8

E22316

2231600155.000174.0

)36.9(6.13

gradient,graphtheFrom

++=

=

=

=

0.0

0.00100 0.00110 0.00120 0.00130 0.00140 0.00150 0.00160 0.00170 0.00180 0.00190


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-8.0

-6.0

-4.0

-2.0

-16.0

-14.0

-12.0

-10.0


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b) An equation relating the rate constants k1 and k2 at temperatures

T1 and T2 can be used to calculate the activation energy or tofind the rate constant at another temperature if the activation

energy is known.

Subtracting both equation at two different temperatures above

AlnRT

EklnTAt;AlnRT

EklnTAt

2

A22

1

A11 +=+=

=

+

+=

12

A

2

1

2

A

1

A21

T1

T1

RE

kkln

AlnRTEAln

RTEklnkln

f f

;


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Example 2 : The rate constant of a first-order reaction is 3.46 x

10-2 s-1 at 298 K. What is the rate constant at 350 K if the

activation energy for the reaction is 50.2 kJ/mol?

Solution :

=12

A

2

1

T

1

T

1

R

E

k

kln

=

298

1

350

1

31.8

102.50

k

1046.3ln

3

2

2

k2 = 0.703 s-1

5.4 The role of catalysts in reactions


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1. A catalyst is a substance which alter the rate of reactionwithout changing its chemical composition. Therefore the

chemical formula will remain the same after the reaction

occur. In most of the chemical reactions, a catalyst is usually

added to speed up the reaction (increase the rate of

reaction), however, the quantity of catalyst used was only in

,

significantly increase the rate of reaction.2. A catalyst works by providing an alternative pathway for a

chemical reaction to take place, which required a lower

activation energy. However, it will not affect the enthalpy

change (H) of a chemical reaction.


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Endothermic process Exothermic processEnergy / kJ Energy / kJ

Reaction coordinate Reaction coordinate


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3. As proposed by Arrhenius' equation,

when the activation energy, EA, of a chemical reaction

decreased, the rate constant of the reaction increased,

RT

EA

eAk

=

ere ore ncrease e ra e o reac on. owever a ca a ys

do not initiate the reaction, rather it accelerate the reactionthat is already occurring

5.4.1 Autocatalysis


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1. If the product of a reaction itself acts as a catalyst for the reaction,the product is also known as autocatalyst. For example, the

reaction between manganate (VII) ions, MnO4- and ethanedioate

ions, C2O42- in the presence of sulphuric acid, H2SO4 :

2 MnO4- + 5 C2O42- + 16 H+ 2 Mn2+ + 10 CO2 + 8 H2O

Table below shows the observation from the first drop of KMnO4 is

added slowl to until the end of reaction

No of drop ofKMnO4

Observation

First two drops Purple colour of KMnO4 decolourised slowly

Following drops Purple colour of KMnO4

decolourised more rapidly.

Last few drops

before reaction

end

Purple colour of KMnO4 decoloursied less rapidly

and eventually become slower until it no longer

decolourised.


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5.8 Application of catalysis in industries

Catalysis can be categorised into 2 types, namely


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y g yp , y

heterogeneous catalyst homogeneous catalyst

5.8.1 Heterogeneous catalyst

A heterogeneous catalyst is a catalyst which has different phase withreactants. Usually it was between a solid catalyst that is used to catalyse

etween a gaseous or iqui reactants. e s a stu y speci ic

examples of heterogeneous catalyst, which are Haber Process, OstwaldProcess, Contact Process and Catalytic converter use in automobileexhaust.

5.8.1.1 Haber Process

Ammonia is an extremely valuable inorganic substance used in the fertilizeri d t th f t f l i d th li ti


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industry, the manufacture of explosives, and many other applications

The main ingredients use to synthesis ammonia are nitrogen (which can be obtainedthrough fractional distillation of liquefied air) and hydrogen (which can be obtainedeither from syn gas [C + H2O] or petroleum refining process)

In heterogeneous catalysis, the surface of the solid catalyst is usually the site of thereaction. The initial step in the Haber process involves the dissociation of N2 and H2on the metal surface.

Although the dissociated species are not truly free atoms because they are bonded to

the metal surface, they are highly reactive. The highly reactive N and H atoms combine rapidly at high temperatures to

produce NH3 molecules


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5.8.1.2 Ostwald Process

Nitric acid is one of the most important inorganic acids. It is used in the productionof fertili ers d es drugs and e plosi es The major industrial method of producing


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of fertilizers, dyes, drugs, and explosives. The major industrial method of producingnitric acid is the Ostwald process. The starting materials, ammonia and molecularoxygen, are heated in the presence of a platinum-rhodium catalyst to 8500C

Step 1

This step is the crucial step as it will determine the yield of nitric acid formed. Therest of the steps do not require catalysis and will occur at high temperature

Step 2 :The nitric oxide readily oxidizes (without catalysis) to nitrogen dioxide:

2 NO(g) + O2

(g) 2 NO2

(g)

Step 3 :When dissolved in water, NO2 forms both nitrous acid and nitric acid:

2 NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

On heating, nitrous acid (HNO2) is converted to nitric acid as follows:

3 HNO2 (aq) HNO3 (aq) + H2O (l) + 2 NO (g)

The NO generated can be recycled to produce NO2 in the second step

5.8.1.3 Contact Process

Sulphuric acid is one of the most widely use inorganic acids. Contact processis still preferable even today to synthesise high concentration of sulphuric


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is still preferable, even today, to synthesise high concentration of sulphuricacid. The following are steps in manufacturing sulphuric acid, starting fromheating sulphur with oxygen.

Step 1 : S (g) + O2 (g) SO2 (g)

After sulphur dioxide is formed and filtered, it was further oxidised to formsulphur trioxide, using vanadium (V) oxide, V2O5, as catalyst. This step iscrucial as it will influence the amount of H2SO4 formed.

Step 2

V2O5 catalyst serve as the active site and provide an alternative solution forthe formation of SO3.

Alternative Step 1 : 2 SO2 + 4V5+

+ 2 O2-

2 SO3 + 4V4+

(Oxidation ofSO2 into SO3 by V5+)

Alternative Step 2 : 4 V4+ + O2 4 V5+ + 2 O2- (Oxidation of V4+ back

into V5+ by oxygen - catalyst regenerate)

Hot sulphur trioxide passes through the heat exchanger and isdissolved in concentrated H2SO4 to form oleum


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Step 3 : H2SO4 (l) + SO3 (g) H2S2O7 (l) Oleum is reacted with water to form concentrated H2SO4Step 4 : H2S2O7 (l) + H2O (l) 2 H2SO4 (l)

The yield of sulphuric acid solution formed is around 30 - 40%. Theunreacted suphur dioxide and sulphur trioxide is then further treated

Absorption) Chamber. Through this method, the yield will bemaximised to nearly 99.8% of H2SO4

5.8.1.4 Catalytic converter

At high temperatures inside a running cars engine, nitrogen and oxygen


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gases react to form nitric oxide N2 (g) + O2 (g) 2 NO (g)

*Note that this reaction occur only when the engine is very hot. This is

due to nitrogen gas, NN has a short and strong triple bond, with a highbond energy. So, high amount of heat is required to break the bond.

Another phenomenon which caused the same reaction are when air is

surrounded by lightning.

When released into the atmosphere, NO rapidly combines with O2 to formNO2. Nitrogen dioxide and other gases emitted by an automobile, such ascarbon monoxide (CO) and various unburned hydrocarbons (CxHy), make

automobile exhaust a major source of air pollution To overcome these problem, most of cars nowadays are equipped with

catalytic converter, which contain platinum-rhodium catalyst and copper(II) oxide + chromium (III) oxide as co-catalyst


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An efficient catalytic converter serves two purposes: It oxidizes CO and

2 2 , 2 2

and O2.Oxidation : CO (g) + 1/2 O2 (g) CO2 (g)

Oxidation : CxHy + (x + y/4) O2 (g) x CO2 (g) + y/2 H2O (g)

Reduction : NOx (g) 1/2 N2 (g) + x/2 O2 (g) The suitable catalyst use is platinum / rhodium (use to oxidise CO and

CxHy) based catalyst doped with copper (II) oxide or chromium (III) oxide(use to reduce NOx). Because the catalyst serve these 3 purposes, sometimeit is also referred as three-way catalyst


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5.8.2 Homogeneous catalyst.

In homogeneous catalysis the reactants and catalyst are dispersed in a singlephase.


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p

Acid and base catalyses are the most important types of homogeneous catalysisin liquid solutions

For example, in the hydrolysis of ester

The rate equation of the reaction above is written as

However, this reaction can be catalysed by the addition of hydrogen ion (H+)from an acidic substance for example, hydrochloric acid or sulphuric acid,

where now, the rate equation can be written asrate = k [CH3COOCH2CH3] [H

+]

By this, the rate of hydrolysis of ester can be increased by the addition of

acidic substance, which may caused faster reaction, without increasing theconcentration of ester.


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Homogeneous catalysis can also take place in the gas phase. This can beexemplified by the reaction from SO2 to SO3, which is one of the majorpollutant in air. SO in the air are mainly released by the fumes of volcanic


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p2

y yactivities. However, recent papers had reported that SO2 emission could alsobe mainly contributed by the combustion of diesel oil, mining activities(sulphide based ores), and through various chemical industries processes such

as Contact Process. SO2 can be oxidised in air under the presence of nitrogen dioxide according to

the e uation below

2 SO2 (g) + 2 NO2 (g) 2 SO3 (g) + 2 NO (g) 2 NO (g) + O2 (g) 2 NO2 (g)

Overall :

SO3

produced are hygroscopic, hence they react easily with water droplet orrain to form corrosive acidic rain.

2 SO2 (g) + O2 (g) 2 SO3 (g)

5.9 Enzymes as Biological Catalysts

Chemical reactions that occur in our bodies are speeded up by enzymes, which


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act as biological cataysts.

Enzymes are the largest and most highly specialised class of proteins and areproduced by living cells from amino acids. Its molecular weight varies from

12,000 to over 1 million. Enzymes work under mild conditions and often give 100% yields and may

s eed a reaction b 106 or 1012 times.

Some enzymes require the presence of metal ions as cofactors, and these arecalled metalloenzymes. Many but not all metalloenzymes contain a transitionelement


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Type of Enzyme Metalloenzyme Function

Arginase Mn2+ Urea formation

Carboxypeptidase Zn2+ Digestion of proteins

Ferredoxin Fe2+ or Fe3+ Photosynthesis

Glutamic mutase Co Metabolism of amino acids

Nitrogenase Fe and Mo Nitrogen fixation

Tyrosinase Cu+ or Cu2+ Skin pigmentation

Compared to inorganic catalysts, enzymes are specific'in their actions.Each enzyme catalyses only one type of reaction whereas platinum


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catalyses several reactions. In the lock-and-key theory, the active site of the enzyme conforms

exactly to the substance molecule. This specificity results from the fact

that enzymes are formed from L-amino acids and therefore the activesites are asymmetrical.

Factors affecting enzyme activity: Temperature - Most enzymes have their highest activity at temperatures from

35C to 45C. Above this range, the enzymes start to denature and the reaction


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rate decreases. Above 80C, enzymes are permanently denatured. pH -The structure and geometry of an enzyme's active site changes when the pH

of the surrounding medium changes. For example, trypsin (which is active in thesmall intestine) has its maximum activity at pH 8 whereas pepsin (which is active

in the stomach) has an optimum pH of 1.5. Solvents and salt concentrations can also change the structure of a

rotein and the activit levels of enz mes.

The activity of enzymes can be inhibited by heavy metals as mercury,lead and silver. These metals are toxic because they bind irreversiblywith free sulphydryl (-SH) functional groups on enzymes, which arethen not available to bind with the necessary cofactor.

Compounds in nerve gases combine with the hydroxyl (-OH)functional group on enzymes and cause the enzymes to lose theirability to catalyse a reaction. This is why animals poisoned by nerve gasbecome paralysed.

CHEMISTRY PRE-U CHEMISTRY SEM 1 CHAP 5.pdf

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