Chapter 8 Chemical Equilibrium by TEACHING CARE Online coaching and tuition classes

 

Whenever we hear the word Equilibrium immediately a picture arises in our mind an object under the influence of two opposing forces. For chemical reactions also this is true. A reaction also can exist in a state of equilibrium balancing forward and backward reactions.

• Definition : “Equilibrium is the state at which the concentration of reactants and products do not change with i.e. concentrations of reactants and products become constant.”
• Characteristics : Following are the important characteristics of equilibrium state,
  • Equilibrium state can be recognised by the constancy of certain measurable properties such as pressure, density, colour, concentration

by changing these conditions of the system, we can control the extent to which a reaction proceeds.

• Equilibrium state can only be achieved in close vessel, but if the process is carried out in an open vessel equilibrium state cannot be attained because in an open vessel, the reverse process will not take
• Equilibrium state is reversible in
• Equilibrium state is also dynamic in nature. Dynamic means moving and at a microscopic level, the system is in motion. The dynamic state of equilibrium can be compared to water tank having an inlet and outlet. Water in tank can remain at the same level if the rate of flow of water from inlet (compared to rate of forward reaction) is made equal to the rate of flow of water from outlet (compared to rate of backward reaction). Thus, the water level in the tank remains constant, though both the inlet and outlet of water are working all the
• At equilibrium state,

Rate of forward reaction = Rate of backward reaction

• At equilibrium state, DG = 0, so that DH = TDS.

• Types : Equilibrium in a system implies the existence of the following types of equilibria simultaneously,
  • Thermal equilibrium : There is no flow of heat from one part to another e. T = constant.
  • Mechanical equilibrium : There is no flow of matter from one part to another e. P = constant.
  • Physical equilibrium : There is the substance exist in three states: solid, liquid and
  • Chemical equilibrium : There is no change in composition of any part of the system with

 

The physical equilibrium is a state of equilibrium between the same chemical species in different phases (solid, liquid and gaseous). The various equilibria which can exist in any physical system are,

 

 

 

Solid	⇌	Liquid
Liquid	⇌	Vapour
Solid	⇌	Gas(vapour)
Solid	⇌	Saturated solution of solid in a liquid
Gas(vapour	⇌	Saturated solution of gas in a liquid
)

(1)  Solid-liquid equilibrium

Rate of transfer of molecules from ice to water       =       Rate of transfer of molecules from water to ice Rate of melting of ice                                         =       Rate of freezing of water

Free energy change and solid-liquid equilibrium in water : For ice-water system free energy change

 

(DG), at 273 K, and one atmosphere pressure is zero i.e., DG = 0; Ice ⇌ Water;

H ₂O(s) ⇌

H 2 O(l)

 

• At temperature higher than 273 K, and 1 atm pressure, DG < Thus, the process in the forward direction would become favourable and ice will melt to give more water.
• At temperature less than 273 K, and 1 atm pressure, DG > 0. Thus, the reverse reaction will become favourable, and more ice will be formed from liquid
  • Liquid-vapour equilibrium : A liquid placed in an open container disappears completely. After some time vapours of the liquid held in an open container can escape out to the atmosphere. Thus, when vapour of liquid exists in equilibrium with the liquid, then

Rate of vaporisation = Rate of condensation,

 

H ₂O(l) ⇌

H 2 O(v)

 

Conditions necessary for a liquid-vapour equilibrium

• The system must be a closed system e., the amount of matter in the system must remain constant.
• The system must be at a constant
• The visible properties of the system should not change with
  • Solid-vapour equilibrium : Certain solid substances on heating get

 

 

 

 

 

converted directly into vapour

 

without passing through the liquid phase. This process is called sublimation. The vapour when cooled, gives back the solid, it is called disposition.

Solid ⇌ Vapour

The substances which undergo sublimation are camphor, iodine, ammonium chloride etc.

cool

For exampie, Ammonium chloride whenhehaet ated sublimes.

NH ₄ Cl (s)                 NH ₄ Cl (v)

• Equilibrium between a solid and its solution : When a saturated solution is in contact with the solid solute, there exists a dynamic equilibrium between the solid and the solution

 

 

 

Solid substance ⇌ Solution of the substance

Example : Sugar and sugar solution. In a saturated solution, a dynamic equilibrium is established between dissolved sugar and solid sugar.

Sugar (s) ⇌ Sugar (aq)

At the equilibrium state, the number of sugar molecules going into the solution from the solid sugar is equal to the number of molecules precipitating out from the solution, i.e., at equilibrium,

Rate of dissolution of solid sugar = Rate of precipitation of sugar from the solution.

• Equilibrium between a gas and its solution in a liquid : Gases dissolve in The solubility of a gas in any liquid depends upon the,

• Nature of the gas and
• Temperature of the
• Pressure of the gas over the surface of the

Henry’s law : “At a certain temperature, the mass of a gas which dissolves in a definite volume of a liquid is proportional to the pressure of the gas in equilibrium with the solution.”

m µ P or m = KP ; (where K is the proportionality constant)

Thus, at a constant temperature, the ratio of the molar concentration of the gas in the solution and into the atmosphere is constant.

Limitations of Henry’s law

• Henry’s law is applicable to ideal gases Henry’s law should be applied only at low pressures because real gases behave like ideal gases at low pressures.
• Henry’s law is not applicable to gases which react chemically with the

 

• Henry’s law will not apply to the solution of gases like ammonia water because these gases react chemically with

(NH 3 )

and hydrogen chloride (HCl) in

 

Note : ®             A chilled soda water bottle fizzes out when opened because, soda water is a solution of carbon

 

dioxide gas,

CO2 (g)

in water at high pressure. As soon as the bottle is opened under normal

 

atmospheric conditions, the dissolved gas escapes out to reach a new equilibrium state, so that the pressure of the gas inside the bottle becomes equal to the atmospheric pressure. At low pressure, the solubility of the gas in water decreases.

The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium.

• Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium

 

Example :     (i)

C2 H5 OH (l) + CH3 COOH (l) ⇌

CH3 COOC2 H5 (l) +  H 2 O(l)

liquid phase

 

(ii)

 

(iii)

N ₂ (g) + 3H ₂ (g) ⇌

2SO₂ (g) + O₂ (g) ⇌

2NH 3 (g)

gas phase

2SO3 (g)

gas phase

 

• Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are present in different phases are called heterogeneous equilibrium

 

 

 

 

Example :  (i)

CaCO₃ (s) ⇌

CaO(s) + CO₂ (g)

 

(ii)

2NaHCO₃ (s) ⇌

Na2 CO3 (s) + CO2 (g) + H 2 O(g)

 

(iii)

H 2 O(l) ⇌

H 2 O(g)

 

Ca(OH)₂ (s) + H ₂O(l) ⇌

Ca 2+ (aq) + 2OH – (aq)

 

Note : ® The equilibrium expression for heterogeneous reactions does not include the concentrations of pure solids because their concentrations remain constant.

A chemical reaction is said to have taken place when the concentration of reactants decreases, and the concentration of the products increases with time. The chemical reactions are classified on the basis of the extent to which they proceed, into the following two classes;

• Reversible reactions : Reactions in which only a part of the total amount of reactants is converted into products are termed as reversible reactions.
• Characteristics of reversible reactions
• These reactions can be started from either side,
• These reactions are never complete,
• These reactions have a tendency to attain a state of equilibrium,
• This sign (⇌) represents the reversibility of the reaction,
• Free energy change in a reversible reaction is zero (DG = 0),
• Examples of reversible reactions
• Neutralisation between an acid and a base either of which or both are weak g.,

CH 3 COOH + Na OH ⇌ CH3 COONa + H 2 O

 

• Salt hydrolysis, e.g.,

Fe Cl₃ + 3H ₂O

⇌ Fe (OH)3  + 6HCl

 

• Thermal decomposition, g.,

PCl5 (g)

⇌ PCl 3 (g) + Cl2(g) – Q

 

CaCO3(s)

⇌ CaO(s) + CO2(g)

; 2HI (g) ⇌ H 2(g) + I 2 (g)

 

• Esterification, g., CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

 

• Evaporation of water in a closed vessel, g., H 2 O(l) ⇌

H 2 O(g) – Q

 

• Other reactions, g.,

N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)

• Q ;

2SO2(g)  + O2(g)  ⇌ 2SO3 (g)   + Q

 

• Irreversible reactions : Reactions in which the entire amounts of the reactants are converted into products are termed as irreversible reactions.
• Characteristics of irreversible reactions
• These reactions proceed only in one direction (forward direction),
• These reactions can proceed to completion,
• The arrow (®) is placed between reactants and products,
• In an irreversible reaction, DG < 0,
• Examples of irreversible reactions

 

 

 

 

• Neutralisation between strong acid and strong base g.

NaOH + HCl ® NaCl + H ₂O + 13.7 kcal

 

 

• Double decomposition reactions or precipitation reactions g.

BaCl 2 (aq) + H 2 SO4 (aq) ® BaSO4 (s) ¯ +2HCl(aq) ; AgNO3(aq) + NaCl(aq) ® AgCl(g) ¯ + NaNO3(aq)

 

(s)

• Thermal decomposition, e.g. 2KClO3 ¾¾^MnO¾2¾^,D ® 2KCl(s)  + 3O2  ­

 

2Pb(NO3 )2  ¾¾^he¾^at ® 2PbO + 4 NO2  + O2  ­ ;   NH 4 NO2  ¾¾^he¾^at ® N 2  ­ +2H 2 O ­

• Redox reactions, g., SnCl2 (aq) + 2FeCl 3 (aq) ® SnCl4 (aq) + 2FeCl 2 (aq)

 

 

• Other reactions, g.,

NaCl + H 2 SO4  ® NaHSO4  + HCl ­

 

 

On the basis of observations of many equilibrium reactions, two Norwegian chemists Guldberg and Waage suggested (1864) a quantitative relationship between the rates of reactions and the concentration of the reacting substances. This relationship is known as law of mass action. It states that

“The rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants at a constant temperature at any given time.”

The molar concentration i.e. number of moles per litre is also called active mass. It is expressed by enclosing the symbols of formulae of the substance in square brackets. For example, molar concentration of A is expressed as [A].

Let us consider a simple reaction between the species A and B : A + B ® Products According to law of mass action, rate of reaction, r µ [A] [B] = k [A] [B]

Where [A] and [B] are the molar concentrations of the reactants A and B respectively, k is a constant of proportionality for the forward reaction and is known as rate constant. The rate constant is also called velocity constant. Now, if the concentration of each of the reactants involved in the reaction is unity, i.e., [A] = [B] = 1, then, rate of reaction, r = k ´ 1 ´ 1 or r = k

Thus, the rate constant of a reaction at a given temperature may be defined as “the rate of the reaction when the concentration of each of the reactants is unity.”

For a general reaction, aA + bB + cC ® Products

The law of mass action may be written as : Rate of reaction, r = k[A]^a [B]^b [C]^c

Thus, the law of mass action may be restated as, “The rate of a chemical reaction at any particular temperature is proportional to the product of the molar concentrations of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactants taking part in the reaction.”

The number of molecules of a reactant taking part in a reaction is also called its stoichiometric coefficient. For example, a, b and c…. in the above equation are called stoichiometric coefficients of A, B and C…. respectively.

 

 

 

• Equilibrium constant in terms of law of mass action : The law of mass action may be applied to a reversible reaction to derive a mathematical expression for equilibrium constant known as law of chemical equilibrium.

Let us consider a simple reversible reaction, A + B ⇌ X + Y in which an equilibrium exists between the reactants (A and B) and the products (X and Y). The forward reaction is,

A + B ® X + Y

According to law of mass action,

Rate of forward reaction µ [A][B] = k f [A][B]

 

Where k f

is the rate constant for the forward reaction and [A] and [B] are molar concentrations of reactants A

 

and B respectively.

Similarly, the backward reaction is ; X + Y ® A + B

Rate of backward reaction µ [X][Y] = kb [X][Y]

 

Where kb

is the rate constant for the backward reaction and [X] and [Y] are molar concentrations of products

 

X and Y respectively.

At equilibrium, the rates of two opposing reactions become equal. Therefore, at equilibrium,

Rate of forward reaction = Rate of backward reaction

k f [A][B] = kb [X][Y]

 

k f	= [X][Y]	or	K	=	[X][Y]
kb	[A][B]	[A][B]

 

The combined constant K, which is equal to k f / kb

is called equilibrium constant and has a constant value

 

for a reaction at a given temperature. The above equation is known as law of chemical equilibrium.

For a general reaction of the type : aA + bB ⇌ cC + dD

 

The equilibrium constant may be represented as :

K = [C]^c [D]^d

[A]^a [B]^b

 

where the exponents a, b, c and d have the same values as those in the balanced chemical equation. Thus,

the equilibrium constant may be defined as,

“The ratio between the products of molar concentrations of the products to that of the molar concentrations of the reactants with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation at a constant temperature.”

(2)  Characteristics of equilibrium constant

• The value of equilibrium constant is independent of the original concentration of reactants.

For example, the equilibrium constant for the reaction,

 

 

Fe ³⁺ (aq) + SCN ^– (aq) = FeSCN ²⁺ (aq) ;

K = [FeSCN 2+ ] [Fe 3+ ][SCN – ]

= 138.0 Lmol -1 (at 298 K)

 

Whatever may be the initial concentrations of the reactants, to be 138.0 L mol^–1 at 298 K.

Fe 3+

and

SCN ^–

ions, the value of K comes out

 

 

 

• The equilibrium constant has a definite value for every reaction at a particular However, it varies with change in temperature.

For example, the equilibrium constant for the reaction between hydrogen and iodine to form hydrogen iodide is 48 at 717 K.

H (g) + I (g) = 2HI(g);     K =   [HI]2     = 48

 

2

2             2                                         [H

][I 2 ]

 

For this reaction, the value of K is fixed as long as the temperature remains constant.

• For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward

For example, if equilibrium constant, K, for the reaction of combination between hydrogen and iodine at 717 K is 48

 

 

H (g) + I

• ⇌ 2HI (g);

K ¢ =

[HI]²   = 48

 

2

2             2                                            [H

][I 2 ]

 

Then, the equilibrium constant for the decomposition of hydrogen iodide is the inverse of the above equilibrium constant.

 

2HI(g) ⇌ H

(g) + I

(g);

K = [H 2 ][I 2 ] = 1 = 1 = 0.02

 

 

 

In general,

^{2        2}              [HI]2

K¢    48

 

 

• The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction. If value of K is large, the reaction proceeds to a greater extent in the forward direction and if it is small, the reverse reaction proceeds to a large extent and the progress in the forward direction is
• The equilibrium constant is independent of the presence of catalyst. This is so because the catalyst affects the rates of forward and backward reactions
• The value of equilibrium constant changes with the change of temperature. Thermodynamically, it can be

 

shown that if

K1 and

K ₂ be the equilibrium constants of a reaction at absolute temperatures T₁

and

T₂ . If DH is

 

the heat of reaction at constant volume, then

 

(Van’t Hoff equation)

 

The effect of temperature can be studied in the following three cases

• When DH = 0 e., neither heat is evolved nor absorbed

 

log K₂ – log K₁ = 0

or log K₂ = log K₁ or

K2 = K1

 

Thus, equilibrium constant remains the same at all temperatures.

• When DH = +ve e., heat is absorbed, the reaction is endothermic. The temperature T₂ is higher than T₁ .

 

log K2 – log K1 = +ve or

log K2  > log K1 or

K2 > K1

 

The value of equilibrium constant is higher at higher temperature in case of endothermic reactions.

• When DH = –ve, e., heat is evolved, the reaction is exothermic. The temperature T₂ is higher than T₁ .

 

log K2 – log K1 = –ve

or log K1 > log K2 or

K1  > K2

 

The value of equilibrium constant is lower at higher temperature in the case of exothermic reactions.

• The value of the equilibrium constant depends upon the stoichiometry of the chemical equation.

 

 

 

Examples :

• If the equation (having equilibrium constant K) is divided by 2, then the equilibrium constant for the new

 

equation is the square root of K i.e.

ways as follows,

. For example, the thermal dissociation of

SO₃ can be represented in two

 

2SO3 (g) ⇌

2SO2 (g) + O2 (g) and

SO3 (g) ⇌

SO₂ (g) + 1 / 2O₂ (g)

 

K = [SO2 ]²[O2 ]

[SO3 ]²

and

K¢ = [SO2 ][O2 ]^{1 / 2} ;

[SO3 ]

 

K¢ = K 2

• Similarly, if a particular equation is multiplied by 2, the equilibrium constant for the new reaction (K¢) will be the square of the equilibrium constant (K) for the original reaction e.,

 

 

and

• If the chemical equation for a particular reaction is written in two steps having equilibrium constants K₁

K = K1 ´ K2

K₂ , then the equilibrium constants are related as

 

For example, the reaction N ₂ (g) + 2O₂ (g) ⇌ 2NO₂ (g) with equilibrium constant (K) can be written in two steps :

 

N 2 (g) + O2 (g) ⇌ 2NO (g) ; (Equilibrium constant =

K1 )

 

2NO(g) + O₂ (g) ⇌

2NO2 (g) ; (Equilibrium constant = K2 )

 

 

Now, K   =

[NO]²

and K   =

[NO2 ]²

 

 

1

 

 

2

Therefore,

[N 2 ][O2 ]

2

 

[NO]²

[NO]²[O2 ]

[NO2 ]²

 

[NO2 ]²

 

K1 ´ K2 = [N

´

][O2 ]

[NO]²

[O2 ]

=                   = K

[N 2 ][O2 ]²

 

• Types of equilibrium constant : Generally two types of equilibrium constants are used,
  • K_c ® It is used when the various species are generally expressed in terms of moles/litre or in terms of molar
  • K_p ® It is used when in gaseous reactions, the concentration of gases expressed in terms of their partial

Kp  = Kc (RT)Dn

K_p is not always equal to K_c. K_p and K_c are related by the following expression,

where, R = Gas constant = 0.0831 bar dm³ mol ^-1 k ^-1 ; T = Temperature in Kelvin

Dn = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation

• Unit of equilibrium constant : Equilibrium constant K has no units e., dimensionless if the total number of moles of the products is exactly equal to the total number of moles of reactants. On the other hand if the number of mioles of products and reactants are not equal, K has specific units.

Units of K_p and K_c and the value of Dn

 

Value of	Dn	Relation between Kp and Kc	Units of Kp	Units of Kc
0	Kp = Kc		No unit	No unit
>0	Kp > Kc		(atm)Dn	(mole l–1)Dn
<0	Kp < Kc		(atm)Dn	(mole l–1)Dn

 

 

 

• Applications of equilibrium constant : Knowing the value of the equilibrium constant for a chemical reaction is important in many ways. For example, it judge the extent of the reaction and predict the direction of the
• Judging the extent of reaction

We can make the following generalisations concerning the composition of equilibrium mixture.

 

• If

K  > 103 , products predominate over reactants. If K

is very large, the reaction proceeds almost all

 

c

c

the way to completion.

 

• If

K < 10 -3 , reactants predominate over products. If K

is very small, the reaction proceeds hardly at all.

 

c

c

• If Kc

is in the range 10 -3

to 103 , appreaciable concentration of both reactants and products are present.

 

This is illustrated as follows,

• Reaction quotient and predicting the direction of reaction : The concentration ratio, e., ratio of the product of concentrations of products to that of reactants is also known as concentration quotient and is denoted by Q.

Concentration quotient, Q = [X][Y] .

[A][B]

It may be noted that Q becomes equal to equilibrium constant (K) when the reaction is at the equilibrium

 

state. At equilibrium,

Q = K = Kc

= Kp . Thus,

 

• If Q > K, the reaction will proceed in the direction of reactants (reverse reaction).
• If Q < K, the reaction will proceed in the direction of the products (forward reaction).
• If Q = K, the reaction mixture is already at equilibrium.

Thus, a reaction has a tendency to form products if Q < K and to form reactants if Q > K. This has also been shown in figure,

 

 

• Calculation of equilibrium constant : We have studied in the characteristics of the equilibrium constant that its value does not depend upon the original concentrations of the reactants and products involved in the However, its value depends upon their concentrations at the equilibrium point. Thus, if the equilibrium concentrations of the species taking part in the reaction be known, then the value of the equilibrium constant and vice versa can be calculated.

Homogeneous equilibria and equations for equilibrium constant (Equilibrium pressure is P atm in a V L flask)

Heterogeneous equilibria and equation for equilibrium constant (Equilibrium pressure is P atm)

 

NH4 HS(s) ⇌ NH3(g) + H2S(g)        C(s) + CO2(g) ⇌ 2CO(g)    NH2CO2 NH4 (s) ⇌ 2NH3(g) + CO2(g)

Initial mole

0   x

1

0   x 2x    x   = 1 2x     2        P   2        P 2 4

1              1   (1–x) 2x (1+x)   æ 1 – x ö ç 1 + x ÷ è          ø æ   2x   ö ç 1 + x ÷ è          ø Pæ 1 – x ö ç 1 + x ÷ è          ø Pæ  2x   ö ç 1 + x ÷ è          ø 4 P 2x2 (1 – x2)

0

1

0   2x   3x   2 3       2P   3

0             1 3

Mole at equilibrium

(1–x)

(1–x)

(1–x)

x

Total moles  at  equilibrium

(solid not included)

Mole fraction

1

2

Partial pressure

P

2

P

3

Kp

4P 3

27

 

 

 

• Equilibrium constant and standard free energy change : Standard free energy change of a reaction and its equilibrium constant are related to each other at temperature T by the following relation,

 

when, when,

DG^o = –ve , the value of equilibrium constant will be large positive quantity and

DG^o = +ve , the value of equilibrium constant is less than 1 i.e., low concentration of products at

 

equilibrium state.

 

 

Example : 1        In the reversible reaction A + B ⇌ C + D, the concentration of each C and D at equilibrium was 0.8

 

mole/litre, then the equilibrium constant Kc

will be                                                                [MP PET 1986]

 

(a) 6.4                          (b) 0.64                        (c) 1.6                          (d) 16.0

Solution: (d)     Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each formed remaining concentration of A and B will be (1 – 0.8) = 0.2 mole/litre each.

K = [C][D] = 0.8 ´ 0.8 = 16.0

 

^c       [A][B]     0.2 ´ 0.2

Example : 2       For the system A(g) + 2B(g) ⇌ C(g), the equilibrium concentrations are (A) 0.06 mole/litre (B) 0.12 mole/litre

• 216 mole/litre. The Keq for the reaction is                                                                              [CPMT 1983]

 

(a) 250                         (b) 416                         (c)

4 ´ 10 ^–3

(d) 125

 

Solution: (a)      For reaction A + 2B ⇌ C;

Keq   =     [C]    =           0.216         = 250

 

[A][B]²

0.06 ´ 0.12 ´ 0.12

 

Example : 3       Molar concentration of O₂ is 96 gm, it contained in 2 litre vessel, active mass will be

• 16 mole/litre (b) 5 mole/litre                   (c) 4 mole/litre      (d) 24 mole/litre

weight

 

 

Solution: (b)     Active mass = M.wt. =

Volume

weight        =

M.wt. ´ Volume

96

 

32 ´ 2

= 3 = 1.5 mol / litre

2

 

Example : 4 2 moles of PCl₅ were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl₅ is dissociated into PCl₃ and Cl₂. The value of equilibrium constant is                                        [MP PMT 1989] (a) 0.266            (b) 0.53                              (c) 2.66                        (d) 5.3

 

Solution: (a)     At start,

PCl5  ⇌ PCl3 + Cl2

 

 

At equilibrium.

2

2 ´ 60

 

100

0

2 ´ 40

 

100

0

2 ´ 40

 

100

 

Volume of cantainer = 2 litre

2 ´ 40 ´ 2 ´ 40

 

 

\ Kc

= 100 ´ 2    100 ´ 2

2 ´ 60

 

100 ´ 2

= 0.266

 

Example : 5      A mixture of 0.3 mole of

H ₂ and 0.3 mole of

I 2 is allowed to react in a 10 litre evacuated flask at 500^o C .

 

The reaction is H₂ + I₂ ⇌ 2HI the Kc

is found to be 64. The amount of unreacted I ₂ at equilibrium is

[KCET 1990]

 

• 15 mole (b) 0.06 mole                      (c) 0.03 mole                     (d) 0.2 mole

 

 

 

 

 

Solution: (b)

K   =   [HI]²

; 64 =          x 2

 

2

^c      [H ][I ]

2

0.03 ´ 0.03

 

x ² = 64 ´ 9 ´ 10 ^–4 ; x = 8 ´ 3 ´ 10 ^–2 = 0.24

 

x is the amount of HI at equilibrium. Amount of I ₂ at equilibrium will be

0.30 – 0.24 = 0.06 mole

Example : 6       The rate constant for forward and backward reactions of hydrolysis of ester are 1.1 ´ 10 ^–2

minute respectively. Equilibrium constant for the reaction,

 

 

and 1.5 ´ 10 ^–3 per

 

CH 3 COOC2 H 5  + H 2 O ⇌ CH 3 COOH + C2 H 5 OH

is                                                             [AIIMS   1999]

 

(a) 4.33                        (b) 5.33                        (c) 6.33                        (d) 7.33

 

 

Solution: (d)

K f  = 1.1 ´ 10 ^–2 ,

K f

K

=

Kb = 1.5 ´ 10 ^–3 ; Kc

b

= 1.1 ´ 10 -2  = 7.33

1.5 ´ 10 ^–3

 

Example : 7        For the reaction

PCl₃ (g) + Cl₂ (g) ⇌ PCl ₅

at 250^o C , the value of Kc

is 26, then the value of K p

on the

 

same temperature will be                                                                                         [MNR  1990; MP PET 2001]

(a) 0.61                        (b) 0.57                        (c) 0.83                        (d) 0.46

Solution: (a)      Dn_g = 1 – 2 = –1

\ Kp  = Kc (RT)^Dn ;  Kp  = Kc (RT)^–1

Since R = 0.0821 litre atm k^–1 mol^–1, T = 250^oC = 250 + 273 = 523 Kp = 26(0.0821 ´ 523)^–1 = 0.61

 

Example : 8       If K_p for reaction

A(g) + 2B(g) ⇌

3C(g) + D(g) is 0.05 atm at 1000K its K_c in term of R will be       [CBSE PMT 1989]

 

 

(a)

5 ´ 10^–4

R

5                           (c)

R

5 ´ 10^–5

R

(d) None of these

 

 

Solution: (c)

K    = K (RT)^Dn Þ 5 ´ 10 ^–2 = K (R ´ 1000)¹ Þ K

= 5 ´ 10 ^–5

 

p              c                                                                  c                                                 c                   R

Example : 9       If the equilibrium constant of the reaction 2HI ⇌ H₂ + I₂ is 0.25, then the equilibrium constant of the reaction

H₂ + I₂ ⇌ 2HI would be                                                                                                 [MP PMT 1989, 95]

(a) 1.0                          (b) 2.0                          (c) 3.0                          (d) 4.0

 

Solution: (d)

Kc      for the IInd reaction is reverse of Ist for reaction

2HI ⇌

H 2 + I 2

is  0.25Kc¢

for reaction,

H 2 + I 2   ⇌

 

 

2HI

will be

Kc¢

=   1 =    1   = 4 .

 

Kc            0.25

Example : 10 If equilibrium constant for reaction 2AB ⇌ A₂ + B₂ , is 49, then the equilibrium constant for reaction

 

AB ⇌

1 A   + 1 B

, will be                                                                            [MP PMT 2003; EAMCET 1998]

 

2   2      2 2

(a) 7                            (b) 20                           (c)  49                           (d) 21

Solution: (a)      2AB ⇌ A2 + B2

K   = [A2 ][B2 ] = 49

^c         [AB]²

 

 

 

 

For reaction AB ⇌

1 A   + 1 B

 

 

Kc¢

2   2      2   2

= [A2 ]1 / 2 [B2 ]1 / 2

[AB]

 

Kc¢  =

=           = 7

 

Example : 11  For the reaction

2NO2(g)   ⇌

2NO(g)   + O2(g)

Kc   = 1.8 ´ 10 ^–6

at 185^o C . At 185^o C , the value of

Kc¢   for the

 

reaction NO      + 1 O

⇌ NO         is                                                                                    [UPSEAT  2000]

 

 

(a)

(g)

 

0.9 ´ 10⁶

2    2(g)

(b)

2(g)

 

7.5 ´ 10 ²

(c)

1.95 ´ 10 ^–3

(d)

1.95 ´ 10³

 

Solution: (b)     Reaction is reversed and halved.

 

\  Kc¢

=    1   ;

Kc¢  =          1       

= 7.5 ´ 10²

 

 

 

Example : 12   In an equilibrium reaction for which DG⁰ = 0 the equilibrium constant Kp

should be                    [BHU 1987]

 

(a) 0                            (b) 1                            (c) 2                            (d) 10

 

Solution: (b)     If DG⁰ = 0

and DG⁰ = -2.303RT log Kp

 

log Kp = 0 , Kp  = 1

 

Example : 13

DG ⁰

for HI(g) @ +1.7kJ / mole

what is the equilibrium constant at 25 ^o C for 2HI (g)

⇌ H 2(g) + I 2(g)

[KCET 1992]

 

 

Solution: (d)

(a) 24.0                        (b) 3.9                          (c) 2.0                          (d) 0.5

DG⁰ = -2.303RT log Kp = -2.303 ´ 8.314 ´ 10^–3 ´ 298 log K p

1.7 = -2.303 ´ 8.314 ´ 10-3 ´ 298 log Kp

K p = 0.5

 

Example : 14 It is found that the equilibrium constant increases by a factor of four when the temperature is increased from

25 ^o C to 40 ^o C . The value of DH^o  is

 

(a)

25.46 kJ mol ^–1

(b)

171.67 kJ mol ^–1

(c)

89.43 kJ mol ^–1

(d)

71.67 kJ mol ^–1

 

Solution: (d)     Using the equation,

 

(K_p )40o C

DH      æ 1      1 ö

 

log (K

)25^o C

=                 ç

2.303R è T1

–       ÷ ,

T2 ø

 

p

we get log 4 =              DH           æ       1      –        1      ö

ø

2.303 ´ 8.314 ç 273 + 25         273 + 40 ÷

è

\ DH = 71.67 kJ mol ^–1

 

Example : 15

Kc for the reaction

A (g) + B (g) ⇌ 2C(g) is 3.0 at 400K. In an experiment “a” mol of A is mixed with 3 mol

 

of B in a 1-L vessel. At equilibrium 3 mol of C is formed. The value of ‘a’ will be,

(a) 4.5 mol                          (b) 9.5 mol                          (c) 2.5 mol                          (d) 3.5 mol

 

 

 

 

Solution: (d)

A(g) + B(g) ⇌

2C(g)

 

a a– x

3                0

3- x                     2x

 

From the equation, 2x = 3 Þ x = 1.5

 

 

Kc  =

4 x ²

 

(a – x)(b – x)

; 3 =

4 ´ (1.5)²

 

(a – 1.5)(3 – 1.5)

; 3 =  4 ´ 2.25

(a – 1.5)(1.5)

Þ a = 3.5

 

Example : 16   For the reaction

AB(g) ⇌

A(g) + B(g), AB is 33% dissociated at a total pressure of P. Then

 

(a)

P = K p

(b)

P = 4 K p

(c)

P = 3K p

(d)

p = 8K p

 

 

 

Solution: (d)

AB(g) ⇌ A(g) + B(g)

 

1

-1 / 3

2 / 3

 

(å n)eq / m   =

0

+1 / 3

1 / 3

 

2 +

0

+1 / 3

1 / 3

 

1 + 1 = 4

 

3    3    3     3

1 / 3 P 1 / 3 P

K     = PA PB   = 4 / 3     4 / 3    = 1 P

 

^p         PAB

2 / 3 P                8

4 / 3

 

P = 8Kp

Example : 17 The total pressure observed at equilibrium in the dissociation of solid ammonium carbamate at a certain temperature is 2.0 atm. The equilibrium constant K p is

(a) 4.185                      (b) 1.185                      (c) 2.276                      (d) 1.072

 

Solution: (b)

NH ₄ COONH ₂ (s) ⇌

2NH ₃ (g) + CO₂ (g)

 

 

 

₂               æ 2

 

2 P                        1 P

3                  3

 

ö ² æ 1    ö       4     ₃      4         ₃

 

Kp  = PNH3 PCO2

= ç 3 P ÷ ç 3 P ÷ = 27 P

= 27 ´ (2)

= 1.185

 

è        ø è        ø

 

Example : 18   At the equilibrium of the reaction

N 2 O4 (g) ⇌ 2NO2 (g) , the observed molar mass of

N 2 O4

is 77.70 g. The

 

percentage dissociation of

N 2 O4  is

 

(a) 28.4                        (b) 46.7                        (c) 22.4                        (d) 18.4

 

 

Solution: (d)

a = MTh – M obs ; Molar mass of

M obs (n – 1)

N 2 O4

= 92 g mol ^–1

 

 

Here, n = 2;

a = 92.00 – 77.70 = 0.184 = 18.4%

77.70(2 – 1)

 

 

 

 

Example : 19  In the equilibrium

H ₂O(g) ⇌

H 2 (g) + 1 O2 (g) , the extent of dissociation of water when p = 1 atm and

2

 

K = 2.08 ´ 10 ^–3

is approximately

 

(a) 2%                         (b) 0.2%                       (c) 20%                        (d) 1%

 

 

 

Solution: (a)     For the equilibrium H 2 O (g) ® H 2 (g) + 1 O2 (g)

2

p

K     = a 3 / 2 P 1 / 2 P = 1 atm

 

 

a = (

2.Kp )^{2 / 3} = 0.0205 » 2%

 

 

Le-Chatelier and Braun (1884), French chemists, made certain generalizations to explain the effect of changes in concentration, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system readjusts itself until it returns to equilibrium. The generalization is known as Le-Chatelier’s principle. It may stated as :

“Change in any of the factors that determine the equilibrium conditions of a system will shift the equilibrium in such a manner to reduce or to counteract the effect of the change.”

The principle is very helpful in predicting qualitatively the effect of change in concentration, pressure or temperature on a system in equilibrium. This is applicable to all physical and chemical equilibria.

• Effect of change of concentration : According to Le-Chatelier’s principle, “If concentration of one or all the reactant species is increased, the equilibrium shifts in the forward direction and more of the products are Alternatively, if the concentration of one or all the product species is increased, the equilibrium shifts in the backward direction forming more reactants.”

Thus,

• Effect of change of temperature : According to Le-Chatelier’s principle, “If the temperature of the system at equilibrium is increased (heat is supplied), the equilibrium will shift in the direction in which the added heat is absorbed. In other words, the equilibrium will shift in the direction of endothermic reaction with increase in temperature. Alternatively, the decrease in temperature will shift the equilibrium towards the direction in which heat is produced and, therefore, will favour exothermic ”

Thus,

 

 

 

 

 

• Effect of change of pressure : Pressure has hardly effect on the reactions carried in solids and liquids. However, it does influence the equilibrium state of the reactions that are carried in the gases. The effect of pressure depends upon the number of moles of the reactants and products involved in a particular reaction. According to Le- Chatelier’s principle, “Increase in pressure shifts the equilibrium in the direction of decreasing gaseous

 

 

 

Alternatively, decrease in pressure shifts the equilibrium in the direction of increasing gaseous moles and pressure has no effect if the gaseous reactants and products have equal moles.”

Thus,

 

• Effect of volume change : We know that increase in pressure means decrease in volume, so the effect of change of volume will be exactly reverse to that of Thus, “decreasing the volume of a mixture of gases at equilibrium shifts the equilibrium in the direction of decreasing gaseous moles while increasing the volume shifts the equilibrium in the direction of increasing gaseous moles.”

Thus,

 

 

 

 

 

• Effect of catalyst : Catalyst has no effect on This is because, catalyst favours the rate of forward and backward reactions equally. Therefore, the ratio of the forward to reverse rates remains same and no net change occurs in the relative amount of reactants and products present at equilibrium. Thus, a catalyst does not affect the position of the equilibrium. It simply helps to achieve the equilibrium quickly. It may also be noted that a catalyst has no effect on the equilibrium composition of a reaction mixture.

Thus,

• Effect of addition of inert gas : The addition of an inert gas (like helium, neon, etc.) has the following effects on the equilibrium depending upon the conditions :
  • Addition of an inert gas at constant volume : When an inert gas is added to the equilibrium system at constant volume, then the total pressure will increase. But the concentrations of the reactants and products (ratio of their moles to the volume of the container) will not change. Hence, under these conditions, there will be no effect on the
  • Addition of an inert gas at constant pressure : When an inert gas is added to the equilibrium system at constant pressure, then the volume will As a result, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift in a direction in which there is increase in number of moles of gases.

Thus,

 

 

 

The Le-Chateliers principle has a great significance for the chemical, physical systems and in every day life in a state of equilibrium. Let us discuss in brief a few applications.

• Applications to the chemical equilibrium : With the help of this principle, most favourable conditions for a particular reaction can be

 

• Synthesis of ammonia (Haber’s process):

N 2 + 3H 2

⇌ 2NH₃ + 23kcal

(exothermic)

 

1 vol        3 vol

2 vol

 

• High pressure forward

(Dn < 0)

• Low temperature (c) Excess of

N 2 and

H 2  (d) Removal of

NH 3

favours

 

• Formation of sulphur trioxide :

2SO2 + O2

⇌ 2SO₃ + 45 kcal

(exothermic)

 

2 vol

1 vol

2 vol

 

• High pressure

(Dn < 0)

• Low temperature (c) Excess of

SO2

and

O2 , favours the reaction in forward

 

• Synthesis of nitric oxide :

N 2 + O2

⇌ 2NO

• 2 kcal

(endothermic )

 

1 vol        1 vol

2 vol

 

• High temperature (b) Excess of

N 2 and O2

• Since reaction takes place without change in volume e.,

 

Dn = 0 , pressure has no effect on equilibrium.

• Formation of nitrogen dioxide :

2NO + O2

⇌ 2NO2 + 27.8 Kcal

 

2 vol

1 vol

2 vol

 

• High pressure (b) Low temperature (c) Excess of NO and O2

favours the reaction in forward direction.

 

• Dissociation of phosphours pentachloride :

PCl5   ⇌

PCl 3 + Cl2 – 15 kcal

 

1 vol

1 vol

1 vol

 

• Low pressure or high volume of the container, Dn > 0

• High temperature (c) Excess of

PCl5 .

 

• Applications to the physical equilibrium : Le-Chatelier’s principle is applicable to the physical equilibrium in the following manner;

 

• Melting of ice (Ice – water system) :

Ice

(Greater Volume)

Water

⇌

(Lesser Volume)

• x kcal

 

(In this reaction volume is decreased from 1.09 c.c. to 1.01 c.c. per gm.)

(a) At high temperature more water is formed as it absorbs heat.(b) At high pressure more water is formed as it is accompanied by decrease in volume.(c) At higher pressure, melting point of ice is lowered, while boiling point of water is increased.

 

• Melting of sulphur : S(s) ⇌

S(l ) – x kcal

 

(This reaction accompanies with increase in volume.)

(a) At high temperature, more liquid sulphur is formed. (b) At higher pressure, less sulphur will melt as melting increases volume.(c) At higher pressure, melting point of sulphur is increased.

 

• Boiling of water (water- water vapour system) :

Water

(Low volume)

Water Vapours- x kcal

⇌

(Higher volume)

 

(It is accompanied by absorption of heat and increase in volume.)

(a) At high temperature more vapours are formed.(b) At higher pressure, vapours will be converted to liquid as it decreases volume.(c) At higher pressure, boiling point of water is increased (principle of pressure cooker).

 

 

 

• Solubility of salts : If solubility of a salt is accompanied by absorption of heat, its solubility increases

 

with rise in temperature; e.g.,

NH 4 Cl, K2 SO4 , KNO3  etc.

KNO3(s) + (aq) ¾¾® KNO3(aq) – x kcal

 

On the other hand if it is accompanied by evolution of heat, solubility decreases with increase in temperature;

e.g., CaCl2 , Ca(OH)2 , NaOH, KOH etc. Ca(OH)2( s) + (aq) ¾¾® Ca(OH)2 (aq) + x kcal

• Application in every day life : We have studied the application of the Le-Chatelier’s principle to some equilibria involved in the physical and chemical In addition to these, the principle is also useful to explain certain observations which we come across in every day life. A few out of them are discussed below,
  • Clothes dry quicker in a windy day : When wet clothes are spread on a stand, the water evaporates and the surrounding air tends to get saturated thus hampering the process of drying. On a windy day when breeze blows, the nearby wet air is replaced by dry air which helps the process of evaporation Thus, clothes dry quicker when there is a breeze.
  • We sweat more on a humid day : The explanation is the same as given In a humid day, the air is already saturated with water vapours. This means that the water that comes out of the pores of the body as sweat does not vaporise. This will result in greater sweating in a humid day.
  • Carriage of oxygen by haemoglobin in blood : The haemoglobin (Hb) in red corpuscles of our blood

 

carries oxygen to the tissues. This involves the equilibrium,

Hb(s) + O2 (g) ⇌

HbO2 (s)

 

The blood that is in equilibrium with the oxygen of the air in the lungs finds a situation in the tissues where the partial pressure of oxygen is low. According to Le-Chatelier’s principle, the equilibrium shifts towards the left so that some of the oxyhaemoglobin changes to haemoglobin giving up the oxygen. When the blood returns to the lungs, the partial pressure of the oxygen is higher and the equilibrium favours the formation of more oxyhaemoglobin.

 

• Removal of carbon dioxide from the tissues by blood : Blood removes

CO2

from the tissues. The

 

equilibrium is,

CO₂(g) + H₂O(l) ⇌

H₂CO₃ (aq) ⇌

H ⁺ (aq) + HCO^– (aq)

 

3

Carbon dioxide dissolves in the blood in the tissues since the partial pressure of lungs, where the partial pressure of CO₂ is low, it is released from the blood.

CO2

is high. However in the

 

• Sweet substances cause tooth decay : Tooth enamel consists of an insoluble substance called

 

hydroxyapatite,

Ca5 (PO4 )3 OH.

The dissolution of this substance from the teeth is called demineralization and its

 

formation is called remineralization. Even with the healthy teeth, there is an equilibrium in the mouth as

 

Ca5

(PO4 )3

OH(s)

exothermic

 

endothermic

5Ca²⁺ (aq) + 3PO^3– (aq) + OH ^– (aq)

 

4

When sugar substances are taken, sugar is absorbed on teeth and gets fermented to give H⁺ ions. The H⁺ ions

produced distrub the equilibrium by combining with OH^– to form water and with PO^3– to form HPO^2– . Removal of

4                               4

products cause the equilibrium to shift towards right and therefore, Ca5 (PO4 )3 OH dissolves causing tooth decay.

In the following reversible chemical equation.

A     ⇌    yB

Initial mol         1            0

At equilibrium    (1–x)       yx                     x = degree of dissociation

Number of moles of A and B at equilibrium = 1 – x + yx = 1 + x(y – 1)

D = molecular weight = volume m V

If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is, = [1 + x(y – 1)]V

Molar density before dissociation,

 

 

 

 

Molar density after dissociation, d =                   m            ;

[1 + x(y – 1)]V

D = [1 + x(y – 1)] ;

d

x =  D – d  

d(y – 1)

 

y is the number of moles of products from one mole of reactant. In terms of molecular mass,

D is also called Van’t Hoff factor.

d

 

x = M – m

(y – 1) m

; Where

M = Initial molecular mass, m = molecular mass at equilibrium

 

Thus for the equilibria

 

(I)

PCl5(g )   ⇌

PCl

 

3(g )

• Cl

 

2(g )

, y = 2 (II)

N 2 O

4(g ) ⇌

2NO

 

2(g )

, y = 2 (III) 2NO2 ⇌ N

2 O4

, y = 1

2

 

\ x = D – d

d

(for I and II) and

x = 2(d – D)

d

(for III)

 

Also D ´ 2 = Molecular weight (theoretical value)

d ´ 2 = Molecular weight (abnormal value) of the mixture

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