Chemistry Form 6 Sem 1 06

8/13/2019 Chemistry Form 6 Sem 1 06

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CHEMISTRY LOWER 6SEMESTER 1

HAPTER 6CHEMICAL EQUILIBRIA


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6.0 Introduction

Among the reactionsdiscussed so far, most of

them are irreversible

reactions. For example, the

dissociation of hydrogenperoxide, H2O2 is an

irreversible process.

Equation :

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

So, if the graph of

concentration of hydrogen

peroxide and hydrogen gas

against time is plotted


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6.1 Reversible reaction and dynamic equilibria

Actually, the example cited above, is one of a few chemical

reactions proceed in only one direction, Most of the chemicalreaction occur around us are usually reversible.

At the start of a reversible process, the reaction proceeds toward

the formation of products. As soon as some product molecules are

formed, the reverse process begins to take place and reactant

molecules are formed from product molecules.

Chemical equilibrium is achieved when the rates of the

forward and reverse reactions are equal and the concentrations ofthe reactants and products remain constant.

An example of a reversible reaction, is the dissociation of

dinitrogen tetraoxide, N2O4, into nitrogen dioxide, NO2, where

the chemical equation can be written asN2O4 (g) 2 NO2 (g)

Diagram below shows 3 different graphs of changes in the

concentration of N2O

4and NO

2. In each case, equilibrium is

established to the right of the vertical line


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a) Initially only NO2present b) Initially only N2O4presentc) Initially a mixture of NO2and N2O4 is present.

Observation : Observation : Observation :Since there are only NO2, reaction

will move to backward to form

some N2O4. at the same time

concentration of NO2 decrrease

Since there are only N2O4,

reaction will move to forward to

dissociate some N2O4 and form

NO2. At the same time

concentration of N2O4 decrrease

Depending on the situation,

concentration of N2O4 and NO2may varies. From the graph, the

concentration of NO2 decrease as

it form more N2O4.


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6.2 Homogeneous Equilibria and Equilibrium

Constant

The term homogeneous equilibrium applies to reactions in

which all reacting species are in the same phase. Example of

the dissociation of N2O4 above is one of homogeneous gas-

phase equilibrium since both reactant and product are gases.

N2O4 (g) 2 NO2 (g)

As mentioned earlier, a reversible reaction is to say achieved a

chemical equilibrium, when the rate of forward reaction is the

same as rate of backward reaction. Consider the following equilibrium reaction.

aA + bB cC + dD

Since a reversible reaction only take one single steps, hence

the stoichiometry coefficient shall be the order of reaction in

which it can be expressed as the following

rate of forward reaction ;

rate of backward reaction ;


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When reaction achieved equilibrium

rate of forward reaction = rate of backward reaction

ba

dc

BA

DC

k

k

][][

][][

1 =

So, k [A]a[B]b = k1 [C]c[D]d

tconsk

Since tan1

=

1

=kkKcOr


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6.2.1 Equilibrium constant of concentration, Kc

According to equilibrium law (mass law action) states that,

at a given temperature, when the amount of

concentration of reactants in a chemical reaction, is

divided by the amount of concentration of products is a

constant where the concentration of each substance israised to the power equal to coefficient in the balanced

reversible chemical equation

,

pA (aq) + qB (aq) rC (aq) + sD (aq)

whereA and B are reactants ; C and D are products ;p , q ,

r , s are stoichiometry coefficient of reaction

Kc of the reaction above can be written as


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Note that the [X] is used to express concentration for both

products and reactants. Hence, in terms of equilibrium

constant, KC, is known as the equilibrium constant ofconcentration. The c stands for concentration. Since the

unit of concentration for products / reactants is mol dm-3.

So, the unit of KC

depend on the total concentration of

products over the reactants.

Kc can also be used to express substances in gaseous

hase


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Chemical equation Equilibrium constant, K C

a) CH3COOH (aq) + C3H7OH (aq) CH3COOC3H7 (aq) + H2O

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

]OHHC][COOHCH[]OH][HCOOCCH[K

733

2733c=

2232

22

c]I[]Fe[]I[]Fe[K

+

+

=

4

c) 2 N2O5 (g) 4 NO2 (g) + O2 (g)

d) AgNO3 (aq) Ag+

(aq) + NO3-

(aq)

e) H2SO4 (aq) + 8 HI (aq)

4 I2 (g) + H2S (g) + 4 H2O (l)

2

52

22c]ON[K =

]AgNO[

]NO][Ag[

K3

3

c

+

=

8

42

4

2

4

22c

]HI][SOH[

]OH[]I][SH[K =


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Example 1 : Consider the following equation

3 A + 2 B 2 C + D

Given the concentration of A, B, C and D are

1.20 mol dm-3 ; 0.920 mol dm-3 ; 1.10 mol dm-3 ; 1.30 mol dm-3 respectively. Calculate the

KC for the reaction above

Example 2 : Given the KC for following

equation 2 E + F G + 2 H

is 4.30. Given the concentration of E, F and G

are 0.35 mol dm-3 ; 0.46 mol dm-3 ; 0.30 moldm-3 respectively. Calculate the concentration

of H in the reaction above.

Example 3 :When 60.0 g of sulphur dioxide,

SO react with 46.0 of ox en to form 58.7

Example 4 : An equilibrium mixture contains

1.25 mol of h dro en, 2.00 mol of bromine

KC = [C]2[D] / [A]3[B]2

Kc = (1.10)

2

(1.30) / (1.20)

3

(0.920)

2

KC = 1.08 mol-2 dm6

KC = [H]2[G] / [F] [E]2

4.30 = (H)2(0.30) / (0.46)(0.35)2[H] = 0.90 mol dm-3

g of sulphur trioxide in 1 dm3

vessel. Calculatethe KC of system.[RAM S = 32 , O = 16]

and 0.50 hydrogen bromide in a 4.0 dm3

ofvessel. Calculate the KC of the system

H2 + Br2 2 HBr2 SO2 + O2 2 SO3

mol = 60.0 / 64 46.0 / 32 58.7 / 80

= 0.9375 = 1.438 = 0.7338

KC = [SO3]2 / [SO 2]

2[O2]

Kc = (0.7338/1)2 / (0.9375/1)2(1.438/1)

KC = 0.426 mol-1 dm3

KC = [HBr]2 / [H2] [Br2]

Kc = (0.50/4.0)2 / (1.25/4.0) (2.00 /4.0)

KC = 0.100


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6.2.2 Equilibrium constant of partial pressure, Kp

Different from Kc, which can be use to express the

equilibrium constant of both aqueous and gaseous phase,

Kp (equilibrium constant of partial pressure) is specifically

an equilibrium constant used to expressed for gaseous

substance. For example, in the reaction of production of ammonia,

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

Equilibrium constant of concentration,

Kc

Equilibrium constant of partial pressure,

Kp

We can expressed equilibrium constant as both Kc and Kpas followIn terms of KC, it is written as

3

22

2

3c

]H][N[

]NH[K =

3

HN

2

NH

P)P)(P(

)P(K

22

3=


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a) H2 (g) + I2 (g) 2 HI (g)

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

c) PCl5 (g) PCl3 (g) + Cl2 (g)

)P)(P(

)P(K

22 IH

2

HIP =

)P()P(

)P(K

22

3

O

2

SO

2

SO

P =

)P)(P(23 ClPCl

d) 2 N2O5 (g) 4 NO2 (g) + O2 (g)

e) N2O4 (g) 2 NO2 (g)

)P( 5PClP

2ON

O

4

NO

P )P(

)P()P(K

52

22=

)P(

)P(K

42

2

ON

2

NO

P=


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Example 5 : The conversion of oxygen to

ozone at high altitude can be represented

by the following equation.3 O2 (g) 2 O3 (g).

In an equilibrium mixture, the pressure of

oxygen gas and ozone are 0.120 atm and

0.200 atm respectively. Calculate the KPof reaction.

Example 6 : A mixture of hydrogen and

nitrogen gas is made up in the ratio of 3 :

1 by volume, and is left to attainequilibrium at 1.01 kPa. The temperature

is kept at 650 K. At constant temperature

and pressure, the equilibrium mixture

contains 20% of ammonia. Calculate the

KP at equilibrium.

3

2

O

PP

)P(K 3= N2 + 3 H2 2 NH3

KP = (0.200)2/ (0.120)3

KP = 23.1 atm-1

2

At eq : 1 : 3 = 80% 20%20% 60%

Since Pa = xa . Ptot(0.20)(1.01) (0.60)(1.01) (0.20)(1.01)

Pgas : 0.202 kPa 0.606 kPa 0.202 kPa

Kp = (PNH3)2 / (PN2) (PH2)

3

KP = (0.202)2/ (0.202) (0.606)3

KP = 0.908 kPa-2


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Example 7 : Chlorine and carbon monoxide react to form phosgene,

COCl2, according to the equation.

Cl2 (g) + CO (g) COCl2 (g).Chlorine gas was placed in a reaction vessel connected to a manometer and

its pressure was found to be 0.69 atm. Carbon monoxide, with an initial

pressure of 0.48 atm was then allowed to react with Cl2 in the vessel. At

equilibrium, the total pressure was found to be 0.82 atm. Calculate KP.

CO (g) + Cl2 (g) COCl2 (g)

Initial : 0.48 atm 0.69 atm 0 atm

At equilibrium : (0.48 x) (0.69 x) xPtotal = 0.82atm

So, (0.48 x) + (0.69 x) + x = 0.82atm

x = 0.35 atm

At equilibrium : 0.13 atm 0.34 atm 0.35 atmKP = (PCOCl2) / ( PCl2) (PCO)

KP = (0.35) / (0.13)(0.34)

KP = 7.9 atm-1


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6.2.3 Relationship between KC and KP.

In general,KC is not equal toKP, because the partial

pressures of reactants and products are not equal to their

concentrations expressed in moles per liter. A simple

relationship betweenKP andKc can be derived as follows.

Let us consider the following equilibrium in the gas phasewA (g) + xB (g) y C (g) + z D (g)

Equilibrium constant of partial Equilibrium constant of

Expressing

Equilibrium

constant

pressure, Kp concentration, Kc

Kp


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Assuming ideal gas behaviour, where PV = nRT, where V is

the volume of container. Rearrange the equation

@ P = [conc]RT

For each partial pressure of the gas

Substituting the partial pressure of each gas into the

equilibrium constant of partial pressure

PA = [A] RT PB = [B] RT PC = [C] RT PD = [D] RT

Factorised RT from equation will make the equation

become :


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The general equation relating KC and KP can be expressed

KP = KC (RT)n

where n = (total mol of gaseous products) - (total mol of

gaseous reactants)

From the equation, there may be a possibility where KP =

KC is when the total amount of mol of product is equal to

the total amount of mol of reactant in a balanced chemicalequation.


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6.3 Heterogeneous equilibria

Heterogeneous equilibrium results from a reversible

reaction involving reactants and products that are indifferent phases

Consider a heterogeneous equilibrium system of calcium

carbonate, when it is heated in a vesselCaCO3 (s) CaO (s) + CO2 (g)

In the heterogeneous system, the equilibirum constant can

be expressed as

*KC' is used in this case to distinguish it from the final form

of equilibrium constant derived shortly


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The concentration of a solid, like its density, is an intensive

property and does not depend on how much of the substance is

present. For example, the molar concentration of calciumcarbonate (density: 2.83 g/cm3) at 20C is the same, whether we

have 1 gram or 1 ton of the metal

[CaCO3] =

For this reason, the terms [CaCO3] and [CaO] are themselves

constants and can be combined with the equilibrium constant.

We can rearran e and sim lif e uation above b writin

where KC, the new equilibrium constant, is conveniently

expressed in terms of a single concentration, that of CO2. Notethat the value of KC does not depend on how much CaCO3 and

CaO are present, as long as some of each is present at

equilibrium


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Another example of heterogeneous equilibria can be

explained using hydrolysis of ester.

CH3COOCH3 (aq) + H2O (l) CH3COOH (aq) + CH3OH (aq)

In the heterogeneous system, the equilibirum constant can

be expressed as

Concentration of a liquid, similar to solid, is another

intensive property and does not depend too, on ow much thesubstance presence. In 1 liter (1 dm3) of pure water, the

molarity can be calculated accordingly

[H2O] =

*Similar to the example above, Kc does not depend on how

much the concentration of water used during hydrolysis as

long as water is present in equilibrium


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From both of example above, in expressing equilibrium

constant of a heterogeneous equilibria, pure solid and pure

liquid can be ignored from the expression of the equilibriumconstant


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Reversible reaction K C KP

1. CO2 (g) + C (s) 2 CO (g) ]CO[

]CO[K

2

2

c=

)P(

)P(K

2CO

2

COP =

2. 3 Fe (s) + 4 H2O (g)

Fe3O4 (s) + 4 H2 (g)

3. Ca(OH)2 (s) + 2 NH4Cl (aq)

CaCl2 (aq) + 2 NH3 (g) + H2O (l)

4

2

42

c]OH[

]H[K =

2

NHP )P(K 3=

2

4

2

32c

]ClNH[]NH][CaCl[K =

4

OH

4

HP

)P()P(K

2

2=


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9.3.1 Dissociation reactions

Dissociation reaction ~ ..

.

For example,

N2O4 (g)

2 N2O5 (g)2 HI (g)

PCl5 (g)

Reaction where large molecule decomposed

to form smaller molecules

2 NO2 (g)

4 NO2 (g) + O2 (g)H2 (g) + I2 (g)

PCl3 (g) + Cl2 (g)

Most dissociation reactions are reversible reaction, meaningthat the dissociation process proceed until a certain extent

before reaching equilibrium.

The extent of dissociation of the compound is measured by

If = 1 it indicates molecules are ... dissociated. If

< 1, molecules are dissociated.

Example, if = 0.20, % of molecules are dissociated.

degree of dissociationcompletely

partially

20


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The following example shows the method how is usually

Kc is calculated when a system with single reactant is

dissociated into its component until it achievedequilibrium.

PCl5 (g) PCl3 (g) + Cl2 (g)

Initial mol 1 mol 0 0

degree of dissociation, - + +

At equilibrium 1 -


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Example 8 : At a temperature of 398oC, hydrogen iodide (HI)has a

degree of dissociation of 25.0%

a) Calculate the KC of the above reaction at 398o

C.a) 2 HI (g) H2 (g) + I2 (g)

Initial : 1.00 0.00 0.00

When 25% dissociate : 0.25 + 0.25 / 2 + 0.25 / 2

At equilibrium 0.75 0.125 0.125

2

22C

]HI[

]I][H[K = 028.0

)(

))((275.0

125.0125.0

==V

VVCK


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Example 9 : At 300 K and 1.0 atm, dinitrogen tetraoxide, N2O4, is 20%

dissociated.

a) Calculate the equilibrium constant of pressure for the reactionb) Calculate the degree of dissociation at 300 K and the total pressure of

0.20 atm.

c) Calculate the pressure that must be applied so that the degree of

dissociation is lowered to 15%a) N2O4 (g) 2 NO2 (g)

Initial 1.00 mol 0 mol

en ssoc a e . + .

At equilibrium 0.80 mol 0.40 mol Total mol = 1.2 mol

atm3

2atm00.1

20.1

80.0P

42ON == atm

3

1atm00.1

20.1

40.0P

2NO ==

)P(

)P(K

42

2

ON

2

NO

P= atmKP 17.0)(

)(

32

2

31

==


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Calculate the degree of dissociation at 300 K and the total pressure of

0.20 atm.

b) N2O4 (g) 2 NO2 (g)Initial 1.00 mol 0 mol

When dissociated + 2

At equilibrium 1.00 + 2 Total mol = 1.00 +

atm200.01

1P

42ON

+

= atm20.0

1

2P

2NO +

=

= 0.416

atm167.0)P(

)P(K

42

2

ON

2NO

P ==

)atm200.01

1(

)atm20.01(K

2

P

+

+

=


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Calculate the pressure that must be applied so that the degree of

dissociation is lowered to 15%

c) N2O4 (g) 2 NO2 (g)Initial 1.00 mol 0 mol

When dissociated 0.15 + 2 (0.15)

At equilibrium 0.85 mol 0.300 mol Total mol = 1.15

atmP15.1

85.0P

42ON = atmP

15.1

30.0P

2NO =

P = 1.8 atm

atmPPK

ON

NO

P 17.0)()(

42

2

2 == atmatmP

atmPKP 17.0

)15.1

85.0(

)15.1

.(

2

=

=


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Example 10 : At 500oC, the KP for the reaction

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

is 1.5 x 10-5 atm-2. Calculate the pressure that has to be applied to thesystem containing a mixture of N2 and H2 in the molar ratio 1 : 3 so

that half of the reactants are converted to ammonia at equilibrium

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

Initial 1.00 mol 3.00 mol 0 mol

When dissociated 0.50 3 (0.50) + 2 (0.50)

At equilibrium 0.50 mol 1.50 mol 1.00 mol

Total mol = 0.50 mol + 1.50 mol + 1.00 mol = 3.00 mol

P = 5.96 x 10-3 atm

atmP00.3

50.0P

2N = atmP

00.3

50.1P

2H = atmP

00.3

00.1P

3NH =

25

3

HN

2

NH

P atm10x5.1)P)(P(

)P(

K22

3

== 25

3

21

61

2

3

1

P atm10x5.1)P)(P(

)P(K

==


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Example 11 : Consider the heterogeneous equilibrium

MO (s) + H2 (g) M (g) + H2O (g)

At ToC and pressure of 120 kPa, 1.5 mol of metal oxide, MO and 1.0mol of H2 reacted to produce 0.012 mol of M and 0.012 mol H2O at

equilibrium. Calculate a value for the equilibrium constant, KP at ToC.

MO (s) + H2 (g) M (g) + H2O (g)

Initial : 1.5 mol 1.0 mol 0 mol 0 mol

Mol dissociate : - 0.012 0.012 + 0.012 mol + 0.012 mol

At equilibrium : 1.488 mol 0.988 mol 0.012 mol 0.012 mol

total mol = 0.988 mol + 0.012 mol + 0.012 mol= 1.012 mol

kPa117kPa120012.1

988.0P

2H == kPa42.1kPa120

012.1

012.0PM ==

kPa42.1kPa120012.1012.0P OH2 ==

kPa117

kPa42.1xkPa42.1

)P(

)P)(P(K

2

2

H

MOH

P == KP

= 0.017 kPa


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9.4 Position of Equilibrium

In general, the equilibrium constant helps us to predict the direction

in which a reaction mixture will proceed to achieve equilibrium and tocalculate the concentrations of reactants and products once

equilibrium has been reached. Since

For example, The equilibrium constantKc for the formation of

hydrogen iodide from molecular hydrogen and molecular iodine in the

as hase

]reactatns[

]products[KC

H2 (g) + I2 (g) 2HI(g) KC = 54.3 at 430C. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146

mole of I2, and 1.98 moles of HI all in a 1.00 L container at 430C.

Inserting the starting concentrations in the equilibrium constant

expression, we write

111)146.0)(243.0(

)98.1(

][][

][ 2

0202

20 === cCC QQIH

HIQ

Reaction quotient (Qc) is a value obtained by


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Reaction quotient (Qc) is a value obtained by

substituting the initial concentrations into the equilibrium

constant expression in order to determine position ofequilibrium compare to equilibrium constant (KC) .

When value of Qc is large, we shall have more products

over reactants while, when value of Qc is small, we shall

have more reactants over products.

To determine the direction in which the net reaction will

roceed to achieve e uilibrium we com are the values of

QC and KC.

QC < KC

The ratio of initial concentrations of products to

reactants is too small. To reach equilibrium,

reactants must be converted to products. Thesystem proceeds from left to right (consuming

reactants, forming products) to reach equilibrium.

At this moment, we described it as equilibrium

shift to right


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QC =

KC

The initial concentrations are equilibrium

concentrations. The system is at equilibrium.

At this moment, we described it as

equilibrium remain unchanged

QC >

KC

The ratio of initial concentrations of products

to reactants is too large. To reach equilibrium,

products must be converted to reactants. The

system proceeds from right to left (consumingproducts, forming reactants) to reach

equilibrium. At this moment, we described it as

equilibrium shift to left


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6.6 Factors that Affect the Position of Equilibrium

Once a system has achieved equilibrium, it is possible to change

the chemical composition of equilibrium mixture by changing the

condition of the reaction. In other words, position of equilibrium

can be altered.

2 important factors which must be considered in chemicalequilibria are

the position of equilibrium

e ra e a w c equ r um s ac eve .

The factors which affect the position of equilibrium can be

deduced using Le Chatelier's principle.

Le Chateliers Principle stated that if an external stress is

applied to a system at equilibrium, the system adjusts in such away that the stress is partially offset as the system reaches a new

equilibrium position


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The factors that affected the position of equilibrium

are

the concentration of reactants and products

the temperature of the experiment

t e pressure n react ons nvo v ng gases

9 4 1 The Effect of Concentration on Equilibrium


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9.4.1 The Effect of Concentration on Equilibrium.

If the concentration of the reactants are increased in a reaction at

equilibrium, the equilibrium will .., formingmore By this way, some of the reactants will .....,

and vice versa. The equilibrium constant will remain ..

Changing the concentration of one or more of the reactants or products

of a reaction at equilibrium disturbed the original equilibrium, but the

value of .. remain the same and the system

changes until equilibrium is re-established.

Example, in the reaction of the formation of nitrogen dioxide

shift its position to rightproducts be decreased

the same

equilibrium constant

2 NO (g) + O2 (g) 2 NO2 (g) Equilibrium is established at one point.


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[NO]

[O2]

[NO2]

Then, the equilibrium is disturbed by adding nitrogen monoxide gas

into the gas vessel. When this happened, concentration of NO . According to Le Chateliers Principle, equilibrium will shift to

to decrease the concentration of

At the same time, concentration of NO2 : .

concentration of O2 :

increaserightNO

increasedecrease


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Equation Substance alteredEquilibrium

positionConcentration of other substances

1. H2 + I2 2 HI [H2] increased [I2] [HI]

2. 2 Fe3+ + 2 I- 2 Fe2+ + I2 [Fe2+] decreased [Fe3+] [I2]

Shift to

rightdecrease increase

Shift to


3. N2 + 3 H2 2 NH3 [H2] decrease [N2] [NH3]

4. 2 SO2 + O2 2 SO3 [SO3] increased [SO2] [O2]

5. N2O4 2 NO2 [NO2] decreased [N2O4]

Shift toleft

increase decrease

Shift to

leftincrease increase

Shift to

rightdecrease

9 4 2 The effect of Pressure on a system in equilibrium


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9.4.2 The effect of Pressure on a system in equilibrium

Change in pressure of system will only affect the position

of equilibrium of reaction involving ..

Result in a change in the of both forward and

backward reactions.

Result in a change in the of equilibrium mixture Will not change . or . at constant

temperature.

gas

pressure

positionrate, k Equilibrium, K C

Le Chateliers principle stated that, When the pressure of a gaseous system increased, equilibrium will

shift to the position with .. total amount of gaseous mole

When the pressure of a gaseous system decreased, equilibrium will

shift to the position with .... total amount of gaseous mole Equilibrium constant, KC and KP will remain

less

moreconstant

Consider the reaction ; N O (g) 2 NO (g)


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Consider the reaction ; N2O4 (g) 2 NO2 (g).

The pressure in the vessel increased

According to Le Chateliers principle, the system will react to(i) decrease the pressure of the system

(ii) decrease the concentration of all species involved.

The results are :

Position will shift to the position with the least total number of

moles. From the equation above, equilibrium shift to

Concentration of N O : where are NO :

left

increase decrease

KC and KP remain constant

F t E ilib i Ch f t ti f


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EquationFactor

altered

Equilibriu

m position

Changes of concentration of

substances

2 SO2 (g) + O2 (g)

2 SO3 (g)P

[SO2] [SO3]

4 NH3 (g) + 5 O2 (g)

4 NO (g) + 6 H2O (g)P

[NH3] [NO]

Shift to

right decrease increase

Shift to


CO (g) + SO3 (g) CO2 (g) + SO2 (g)

P [CO] [SO3]

CaCO3 (s) CaO (s) + CO2(g) P

[CO2]

2 N2O5 (g) 4 NO2 (g) + O2(g) V [N2O5] [NO2]

Nochange No change No change

Shift to

left decrease

Shift to


Adding a noble gas (inert gas) on an equilibrium.


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Adding a noble gas (inert gas) on an equilibrium.

(i) Adding an inert gas into the system at a constant

pressure and the volume of vessel remain.. but .. the partial pressure

of the gases involved in the system.

(ii) This will cause the equilibrium to shift to the directionwith the .. total number of moles.

Factor Equilibriu

Changes of concentration of

unchanged lowered

more

altered m position substances

PCl5 (g) PCl3 (g) + Cl2 (g) + Ne[PCl5] [PCl3]

2 H2 (g) + O2 (g) 2 H2O (g) + Ar[H2] [H2O]

2 HI (g) H2

(g) + I2

(g) + Kr[I2] [HI]

Shift to


Shift toleft increase decrease

No

change remain remain

9.4.3 Effect of Temperature on an Equilibrium


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p q

In a reversible reaction, the changes in temperature will affect

. . . .

In kinetic chemistry earlier, we used . equation to

explain the effect on temperature toward the rate of reaction

Rate of forward reactionRate of backward reaction

k, rate constantK, Equilibrium constant

Arrhenius

RT

EA

eAk

= n equ r um, t e e ect o temperature towar an equ r um system

can be explained by using equation

At different temperature, Vant Hoff equation can be derived to become

=

211

2 11lnTTR

H

K

K

cRT

HKln +

=

Vant Hoff

K = equilibrium constant, KCH = enthalpy change

R = gas constant, 8.31 J mol-1 K-1

T = temperature in Kelvin, K

Exothermic reaction Endothermic reaction


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KC

T

KC

T

From the graph, the equilibrium

constant ...... withtemperature.

From the graph, the equilibrium

constant ......withtemperature.

decrease increase

According to Le Chateliers principle, the changes of temperature will affect

the position of equilibrium by

o When temperature increased, equilibrium will shift to the position wherethe system is countered by .. the temperature, which is toward

the position of . process

o When temperature decreased, equilibrium will shift to the position where

the system is countered by .. the temperature, which is toward

the position of . Process

decreaseendothermic

increase

exothermic

Factor Equilibriu Rate Equilibriu


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Equation ProcessFactor

altered

Equilibriu

m position

Rate

constant

Equilibriu

m constant

2 HI (g) H2 (g) + I2 (g) Endothermic T

2 H2O2 (l) 2 H2O (l) + O2 (g) Exothermic T

Shift to

right

increase increase

Shift to

leftincrease decrease

CH4(g) + 2O2(g)

CO2 (g) + 2H2OExothermic T

2 NO2 (g) N2O4 (g) Endothermic T

4 NO2 (g) + O2 (g)

2 N2O5 (g) Exothermic T

Shift to


Shift toleft decrease decrease

Shift to

left

increase decrease

Example 12 At a specific temperature T, a 1.00 dm3 flask was found to

t i ilib i i t f 0 500 l f l h di id 0 100 l f


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contain an equilibrium mixture of 0.500 mol of sulphur dioxide, 0.100 mol of

oxygen and 4.60 mol of sulphur trioxide.

a) Calculate the equilibrium constant, KC of the reaction.

b) Calculate the number of moles of oxygen that must be forced into the mixture

to increase the number of moles of sulphur trioxide at equilibrium to 4.70 mol.

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


]O[]SO[

]SO[K

2

2

2

2

3c =

31

C100.02500.0

2

00.160.4

c dmmol846K][][

][K ==

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


Amount of mol insert 0.100 mol 0.0500 + x + 0.100

At new equilibrium 0.400 mol 0.0500 + x 4.70 mol

..

]O[]SO[

]SO[K

2

2

2

2

3c=

31

00.1

X0500.02

00.1400.0

2

00.170.4

c dmmol846][][

][K

+ ==

X = 0.113 mol

Example 13 Consider the equilibrium N2O4 (g) 2 NO2 (g).

A 30 C d 1 0 30% f N O i di i d


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At 30oC and 1.0 atm., 30% of N2O4 is dissociated

a) calculate the equilibrium constant, KP, for the reaction

b) calculate the pressure needed to increase the dissociation to 40% at

fixed temperature.

a) N2O4 (g) 2 NO2 (g)

Initial 1.00 mol 0 molWhen dissociated 0.30 + 2 (0.30)


atm538.0atm00.130.1

70.0P 42ON == atm462.0atm00.130.1

60.0P 2NO ==

)P(

)P(K

42

2

ON

2

NO

P = atmKP 40.0

)538.0(

)462.0( 2==

b) calculate the pressure needed to increase the dissociation to 40% at

fi d t t


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fixed temperature

N2O

4(g) 2 NO

2(g)

Initial 1.00 mol 0 mol

When dissociated 0.40 + 2 (0.40)


atmP4286.0atmP40.160.0P

42ON == atmP5714.0atmP

40.180.0P

2NO ==

P 2 2

)P(K

42

2

ON

P = atm396.0)P4286.0(

.KP == P = 0.53 atm

Example 14 Consider the reaction N2 (g) + 3 H2 (g) 2 NH3 (g)


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The KP values at different temperatures are given below.

Plot a graph of ln KP against 1/T. From your graph, determinethe value of H of the reaction

T (oC) 25 127 227 327 427

Kp (atm-2) 6.76 x 105 4.07 x 101 3.55 x 10-2 1.66 x 10-3 7.76 x 10-5

1/T K-1 3.36 x 10-3 2.50 x 10-3 2.00 x 10-3 1.67 x 10-3 1.43 x 10-3

ln Kp (atm-2) 13.4 3.71 3.34 6.40 9.46

Gradient = 3.71 (6.40) / 2.50 x 10-3 1.67 x 10-3

Gradient = H / R = 12180

H = 101 kJ / mol

6.5 The Industrial Application of Chemical Equilibria

i id d b Ch i l I d i i d i


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3 main aspects considered by Chemical Industries in producing

chemical substance

i) ii) iii)

In this Chapter, there are 2 industrial processes which are discussed.

1. Haber process process of synthesizing . in industry.

N2 (g) + 3 H2 (g) 2 NH3 (g) H = 92 kJ / mol

Consider the Haber reaction using Le Chateliers Principle

High yield Short time Low cost

ammonia

In order to produce large amount (high yield) equilibrium must shift

to .

Pressure : to make equilibrium shift to right, the pressure must be

as right side of the equation has total amount of moles

Temperature : to make equilibrium shift to right, the temperature must

be as the forward reaction is an . reaction.

Adding catalyst (Iron) will not change the . but it will

only increase the

right

high lesser

low exothermicyield of product

rate of reaction

2. Contact Process process of synthesizing . in industriessulphuric acid


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2 SO2 (g) + O2 (g) 2 SO3 (g) H = -90 kJ/mol

Consider the Contact process reaction using Le Chateliers Principle

In order to produce large amount (high yield) equilibrium must shift

to

Pressure : to make equilibrium shift to right, the pressure must be

right

. as r g t s e o t e equat on as tota amount o mo es

Temperature : to make equilibrium shift to right, the temperature

must be as the forward reaction is an reaction.

Adding catalyst will not change the . but it will onlyincrease the

low exothermic

yield of productrate of reaction

Chemistry Form 6 Sem 1 06

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