Chemical

Equilibrium

1.

 Chemical Equilibrium

Introduction
Chemical reaction is a process in which
one or more reactants react to produce one or
more products.
Concept Ladder
Type of Chemical reaction
(a) Irreversible Reactions Physical changes can
(b) Reversible reactions also be categorised into
Reversible and Irreversible
(a) Irreversible Reaction changes.
Unidirectional reactions are known as
irreversible reactions. i.e. reactants convert
to produce products and where products
cannot convert back to the reactants.
Ex.
(1) BaCl2(aq) + H2SO4(aq)  → BaSO4(ppt)
+ 2HCl(aq)
(2) 2Mg(s) + O2(g)  → 2MgO(s)
Rack your Brain
(b) Reversible Reaction
Reversible reactions occur in both forward
Does a physical change
and backward directions and therefore
accompain a chemical change?
never go on to completion.
Ex.
(1) N2 (g) + O2 (g) 

  2 NO (g)
(2) 3 Fe (s) + 4 H2O (g) 

 Fe3O4 (s) + 4 H2 (g)

Note :
In case of Reversible Reactions, container
should be closed.
Equilibrium
It is defined as the state when measurable
properties such as position speed, pressure,
temperature and concentration do not change Concept Ladder
with time.
In mechanical equilibrium,
Equilibirum can be of a number of types.
the net forces on an object
For example: Mechanical equilibrium, thermal
is zero.
Chemical Equilibrium

equilibrium, Physical equilibrium, Chemical

equilibrium etc.

2.
Thermal Equilibrium
The type of equilibrium where the thermal
energy between two or more substances are
equal in nature.

The above figure shows two objects A and B

which are far from each other and when the come
in contact then their thermal energies become
equal. So, they are in thermal equilibrium with
each other.
Concept Ladder
Physical Equilibrium
The type of equilibrium which develops Solid-Liquid-Gaseous
between different phases and there is no change equilibrium can be seen in
in chemical composition. In physical equilibrium, it’s three states. The point
there is the existence of same substance in at which these three exist
different physical states. for H2O is known as triple
(1) Solid-liquid equilibrium point of water. Triple point
H2O (solid)  
 H2O (liquid) of water is 273.15K
Rate of transfer of ice = Rate of freezing of
water
(2) Liquid-vapour equilibrium
Chemical Equilibrium

Rate of vaporization = Rate of condensation,

H2O (liquid) 

 H2O (vapour)

3.
 Conditions necessary for a Liquid-Vapour
Equilibrium
(i) The system must be a closed system. Concept Ladder
(ii) The system must be at a constant
temperature. For any pure liquid at 1 atm
(iii) The visible properties of the system should pressure, the temperature
not change with time. at which the liquid and
(3) Solid-vapour equilibrium vapor are at equilibrium is
Certain solid substances on heating get called normal boiling point
converted directly into vapour without of liquid.
passing through the liquid phase. This
process is called sublimation. The vapour
when cooled, gives back the solid, it is called
deposition or desublimation or sublimation.
Heat
NH4Cl (solid)   NH4Cl (vapour)



For example, Ammonium chloride when.
heated sublimes.
(4) Equilibrium between a solid and its solution
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When a saturated solution is in contact
with the solid solute, there exits a dynamic
equilibrium between the solid and the When a solution becomes
solution phase. supersaturated will you be able
Solid substance  
  Solution of the to observe equilibrium?
substance
Example : Sugar and sugar solution. In a
saturated solution, a dynamic is equilibrium
is established between dissolved sugar and
solid sugar.
Sugar (solid)  
 Sugar (aqueous)
Rate of dissolution on of solid sugar =Rate
of precipitation of sugar from the solution. Concept Ladder
At the equilibrium state, the number of
sugar molecules going into the solution The number of precipitating
from the solid sugar is equal to the number molecules at equilibrium
of molecules precipitating out from the state is equal to the number
solution, i.e., at equilibrium. of molecules dissolving in
Chemical Equilibrium

(5) Equilibrium between a gas and its solution the solution.

in a liquid.
Gases dissolve in liquids. The solubility of a
gas in any liquid depends upon the,

4.
 (a) Nature of the gas and liquid.
(b) Temperature of the liquid.
(c) Pressure of the gas over the surface of Concept Ladder
the solution.
This equilibrium is governed by Henry’s When the condition is
law, which states that the mass of a gas changed at equilibrium
dissolved in a given mass of a solvent at any state, the equilibrium
temperature is proportional to the pressure changes.
of the gas above the solvent. It decreases When temperature of a
with increases of temperature. solution of gas is increased,
the rate of dissolution of
Chemical Equilibrium gas decreases.
In chemical equilibrium, concentration of
the reactants and products do not change with
time and the system does not show any further
change in properties.
In case of chemical equilibrium, rate of forward
reaction becomes equal to the rate of backward
reaction. i.e. Rate (forward reaction) = Rate
(backward reaction) Rack your Brain
Forward reaction (In an open glass container)
3 Fe (s)  4 H O (g) 

 3 Fe (s)  4 H O (g) Equilibrium state is dynamic in
2 2

If the container tube is closed, both the reactions nature. Why it is not static?
take place simultaneously and hence reaction
become reversible.
3Fe (s) + 4H2O (g) 
 Fe3O4 (s) + 4H2 (g)

Other example may include
CaCO3 (s)  CaO (s) + CO2 (g)

Characteristics of Equilibrium State
y A reaction (or a process) is said to be in
equilibrium when the rate of forward
reaction (process) becomes equal to the
rate of backward reaction (process).
y An equilibrium is dynamic in nature and not
static i.e., even after equilibrium is attained,
Chemical Equilibrium

the forward as well as the backward

reaction take place but at equal speeds.
y A chemical equilibrium can be established
only if none of the products is allowed to
escape out.

5.
 y At equilibrium, the concentration of each
of the reactants and products become
constant.

y When reaction attains equilibrium at certain

temperature and pressure, DG = 0
y Chemical equilibrium can be attained from
either direction.
Rack your Brain
y
Law of Mass Action
The rate of a reaction is proportional Does spontaneity play any role in
to the product of the active masses of the equilibrium being attained?
reactants, each raised to the power equal to its
stoichiometric coefficient as represented by the
balanced chemical equation.
aA + bB + cC + ......
Rate of reaction ∝ [A]a [B]b [C]c
Law of Chemical Equilibrium is a result obtained
by applying the Law of Mass Action to a reversible
reaction in equilibrium. Concept Ladder
A+B  
 C+D
Rate of the forward reaction ∝ [A] [B] In chemical kinetics for a
= kf [A] [B] reaction of the type
Rate of the backward reaction ∝ [C] [D] aA + bB   cC + dD
Chemical Equilibrium

= kb [C] [D] Rate of reaction = K[A]a[B]b

At equilibrium, where K depends upon
Rate of forward reaction = Rate of backward temperature.
reaction

6.

k f  A B  kb CD or
CD  k f K
A B kb
At constant temperature, kf and kb are constant,
kf
therefore, = K is also constant at constant
kb
temperature and is called ‘Equilibrium constant’.
The product of the molar concentrations
of the products, each raised to the power equal
to its stoichiometric coefficient divided by the
product of the molar concentrations of the
reactants, each raised to the power equal to its
stoichiometric coefficient is constant at constant
temperature and is called Equilibrium constant.

Q.1 The rate constant for a forward reaction in a reversible reaction (K

eq
= 108) is
10 . Calculate the rate constant for the backward reaction.
4

A.1 Keq =
K f
Kb
Where; Kf = Rate constant for forward reaction.
Kb = Rate constant for backward reaction.
Keq = Equilibrium constant.
Chemical Equilibrium

Now. Keq = 108; Kf = 104.

Therefore, from equation (1)
K 104
Kb  f  8  104
Keq 10

7.
 Importants Point about Equilibrium Constant
y If the reaction is reversed, the value of

equilibrium constant is inversed e.g.,
CH3COOH (aq) + C2 H5OH (aq) 
 CH3COOC2H5 (aq) + H2 O (l ) Kc = K,

then for
CH3COOC2H5 (aq) + H2 O (l) 
 CH3COOH (aq) + C2 H5OH (aq) , Kc = 1/K


y If the reaction is divided by 2, equilibrium

constant is the square root of the original Previous Year’s Question
e.g.,
For N2 (g) + 3 H2 (g) 
 2NH3 (g) , Kc = K

If the equilibrium constant for
then for; N2  g   O2  g  
 2 NO  g  is

K, the equilibrium constant for
1 3
N2 (g) + H2 (g) 
 NH3 (g) , Kc =
 K 1 1
2 2 N2  g   O2  g  
 NO  g 

2 2
If then reaction is multiplied by 2, will be  [NEET]
equilibrium constant is the square of the 1
Chemical Equilibrium

original. Equilibrium constatns of the step (1)  K (2) K

2
reactions e.g., if for
(3) K2 (4) K X2
N2 (g ) + 2 O2 (g) 
 2NO2 (g) ,

equilibrium constant = K

8.
 (1) NO2 (g) + O2 (g) 
 2NO2 (g) , Kc = K1

NO2 (g) + O2 (g) 
(2)  2NO2 (g) , Kc = K2
 Previous Year’s Question

then, K  K 1  K2
The equilibrium constants of the
y Effect of temperature: following are
According to Van’t Hoff equation N2 + 3 H2 
 2 NH3 ;
 K1
K2 H  T2  T1  N2 + O2 
 2 NO;
 K2
log   
K 1 2.303R  T1T2  1
N2 + O2 
 H2 O;; K3
2
Where K1 and K2 are the equilibrium
constants at temperature T1 and T2 The equilibrium constant (K) of
respectively and DH is the molar enthalpy the reaction will be:
5
change in the temperature range T1 to T2. 2 NH3 + O2  K
  2 NO + 3 H2O 
Note: 2
y For exothermic reaction, as temperature
[NEET]
increases K decreases.
y For endothermic reaction, as temperature (1)  K2K33 / K 1 (2) K2K3 / K 1
(3) K2K3 / K 1 (4) K 1K33 / K2
3
increases K increases.

Q.2 For the reactions,

A B
 KC = 1
B
C
 Kc = 3
C
D
 Kc = 5
Find KC for the reaction A 
 D is

(1) 15 (2) 5 (3) 3 (4) 1

A.2 For A
D

B  K  1;
A  1
C   K  3;
B 2
D  K 5
Chemical Equilibrium

C  3
K  K 1  K2  K3
D  1  3  5  15 [Multiplying all the three]
A 

9.
 Q.3 The equilibrium constants of the reactions
1
SO2 (g) + O2 (g)    SO3 (g)

2
and 2 SO 2 (g) + O2 (g ) 

 2SO3 (g)
are K1 and K2, respectively. The relationship between K1 and K2 is
(1) K1 = K2 (2) K22 = K1
(3) K12 = K2 (3) K2 = K 1

1 SO3 
A.3 SO2  g   O2  g  
 SO3  g  ;
  K1 
SO2 O2 
1/ 2
2

SO3 
2

2SO2  g   O2  g  
 2SO3  g  :  K2 

SO2  O2 
2

 K2  K21

Q.4 The rate at which a substance reacts, depends on its

(1) Active mass (2) Molecular mass
(3) Equivalent mass (4) Total volume

A.4 Active mass

Q.5 For the reaction:

2A(g) + B(g ) 

 3C(g) + D(g)
Two moles each of A and B were taken into a flask. The following must always
be true when the system attained equilibrium
(1) [A] = [B] (2) [A] < [B]
(3) [B] = [C] (4) [A] > [B]

A.5 For the reaction 2A(g) + B(g ) 

 3C(g) + D(g)
 2 mol of A reacts with 1 mol
of B to give products. Hence, the concentration of A will be lesser than that
Chemical Equilibrium

of B at equilibrium.

10.
Q.6 The equilibrium constant for the reaction
N2 + 2O2 

 2NO2
At a particular temperature is 100. Determine the values of equilibrium
constants for the following reactions.
2NO2 

 N2 + 2O2 ……..(1)
NO2 
 1 / 2 N2 + O2
 ……..(2)

A.6 The equilibrium constant for the reaction

N2 + 2O2 
  2 NO2


NO2 
2
Kc   100 ……..(a)
N2 O2 
2

The equilibrium constant equation for reaction (1)

N2 O2 
2

K1  ……..(b)
NO2 
2

This equation is reciprocal of equation (a)

1 1
So K 1    1  102
Kc 100
The equilibrium constant equation for reaction (2)

N2  O2 
1/ 2
K2  ……..(c)
NO2 
Comparing equation (a) and (c),

K2  K 1  102  101  0.1

Q.7 Given:
N2 (g)  3H2  g  
 2NH3 (g);K 1

N2 (g)  O2 (g ) 
 2NO  g  ; K2

1
H2 (g)  O2 (g ) 
 H2O  g  ; K3

2

The equilibrium constant for 2NH3 (g)  O2 (g) 
 2NO  g   3H3O(g )

Chemical Equilibrium

2
will be
KK K 1K23 K2K33
(1) K 1K2K3 (2) 1 2 (3) (4)
K3 K2 K1

11.
 NH3 
2
NO ; K  H2O
2

A.7 K1  ; K 
N2 H2 
3 2
N2 O2  3 H2 O2 1/2
The equilibrium constant for
5
2 NH3 (g ) + O2 (g) 
 2NO(g) + 3 H2 O(g)

2
Will be

NO H2O
2 3
K2  K33
K 
NH3  O2  /
2 5 2
K1

Q.8 
For a system, A + 2B 
 C, the equilibrium concentration are [A] = 0.06,
[B] = 0.12, and [C] = 0.216. The Kc for the reaction is
(1) 120 (2) 400
(3) 4 × 10–3 (4) 250

A.8 
A + 2B 
C

K
C  
(0.216)
 250
A B
2
(0.06)(0.012)2

Homogeneous Equilibrium
In a homogeneous system, all the
reactants and products are in the same phase.
For example, in the gaseous reaction,
CaCO3  solid 

 CaO  solid  CO2  gas  , reactants and

products are in the homogeneous phase.

Similarly, for the reactions,
CH COOC H  aq  H O l   CH COOH  aq  C H OH  aq
3 2 5 2  3 2 5

3
Fe  aq  SCBN  aq 
 Fe SCN

2
  aq Concept Ladder
Equilibrium constant in gaseous reaction
Is case of Homogeneous
For gaseous reactions, equilibrium
equilibrium of gases
constant is expressed in terms of partial pressure.
Chemical Equilibrium

both the pressure and

For example the decomposition of N2O4 into
concentration of gases can
NO2 in close container is gaseous reaction at
be used to represent liquid
equilibrium.
and equilibrium states.

12.
Ideal gas equation; PV = nRT
P = n RT ……(1)
v
Where, P = Pressure of gas in “Bar”
n = number of moles of gas in mol
V = volume of gas in liter
Concept Ladder
T = absolute temperature in Kelvin “K”
R = 0.0831 bar L mol- K-1 R can have different units
Now form equation (1) also.
P = C.R.T. or P = [gas] RT; where C = n R = 0.0821 atm. L mol-1K-1
v
8.314 J
At constant temperature, pressure of gas is =
mol K
proportional to its concentration i.e.,
P ∝ [gas] 2 cal
=
mol K
for general reaction in equilibrium
aA (g) + bB (g) 
 cC (g) + dD (g)

Chemical Equilibrium

PC  PD 
c d
or K 
p a b
PA  PB 

13.
 Equilibrium constant in terms of mole fraction
Consider a general reaction in equilibrium

aA + bB  
 cC + dD Concept Ladder
Amount of A = XA (mole fraction of A)
B = XB (mole fraction of B) Active mass of a solid
Amount of
substance is taken as unity
Amount of C = XC (mole fraction of C) during calculations.
Amount of D = XD (mole fraction of D)
Now equilibrium constant in term of mole fraction

X  X 
c d

KX  C a D b
 XA   XB 
Relationship between Kc, Kp and KX
(a) Relation between Kp and KX
Consider a general reaction in equilibrium
aA + bB 
  cC + dD

according to Raoult’s Law
pA = XA .P Previous Year’s Question
Where, pA = partial pressure of gas A
Equilibrium constant KP for the
XA = mole fraction of gas A
following reaction
P = Total pressure.
MgCO3  s  
 MgO  s   CO2  g 

similarity; pB = XB P
pC = XC P  [NEET]
pD = XD P
(1)  KP = PCO2
p  p 
c d

Now; Kp  C a D b  PCO2 PMgO 

pA  pB  (2) KP  PCO2  
 PMgCO3 
 XC   XD  P(c d)(a b)  PCO PMgO 
Kp 
 XA   XB  (3) KP   2 
 PMgCO
 3 
Kp  K X .P
n

PMgCO3
where, ng   c  d   a  b  (4)  KP 
PCO2  PMgO
Chemical Equilibrium

(b) Relation between Kp and Kc

Consider a general reaction in equilibrium
aA + bB 
 cC + dD ……… (1)
n
= p = RT CRT where C = n
v v

14.
 for an ideal ‘A’, pA = CA.RT ……… (2)
where; pA = Partial pressure of gas ‘A’
CA = concentration of gas ‘A’
Concept Ladder
R = gas constant
T = absolute temperature ng
Similarly for gas ‘B’ KP  KC RT 
pB = CB .RT .......(3) where Dng = no. of moles
pC = CC .RT .......(4) of gaseous product no. of
moles of gaseous reactant.
pD = CD .RT .......(5)
If Dng > 0, then KP > KC
Now Kp for the reaction (1) is If Dng < 0, then KP < KC
(p )c (p )d If Dng = 0, then KP = KC
Kp = C a D b .....(6)
(pA ) (pB )
Now put the values of (2), (3), (4) and (5) in
the given equation (6), will get
C D RT (c d)
c d

Kp  Previous Year’s Question

A  B RT (a b)
a b

Here R and T are constants In which of the following

equilibrium KC and KP are not
C D RT cda b
c d

Now, Kp  …….(7) equal? [NEET]

A  B
a b
(1) 2 NO  g   N2  g   O2g


C D
c d
(2) SO  g   NO2  g  
 SO3  g   NO  g 

KC 
where; (3) H2  g   I2  g    2H g 

A  B
a b

(4) 2 C  s   O2  g  
 2 CO2  g 

Now above equation (7) becomes

Chemical Equilibrium

15.
 Units of equilibrium constant
n
Unit of Kp   atm g Rack your Brain
Unit of Kc = (mol L–1)Dn
A reaction which is unimolecular
Unit of K XC = unit less
w.r.t to forward reaction and
Heterogeneous Equilibrium bimolecular w.r.t backward
In a heterogeneous system, all the reaction has 2 and 4 moles
reactants and products are in different respectively what would be the
phases. For example, in the given reaction, relationship between KP & KC?
CaCO3  solid   CaO  solid  CO2  gas  ,



reactants and products are in the heterogeneous
phase.
Chemical Equilibrium

Similarly, for the reactions,

NH4HS  s  

 NH3  g   H2S  g 
NH2COONH4  s  
 2 NH3  g   CO3  g 


16.
Factors influencing equilibrium
(a) Concentration

Note:
Increasing concentration of one side favours the other side.
(b) Temperature

Chemical Equilibrium

Note :
y If the forward reaction exothermic then backward reaction will be endothermic and
vice-versa.

17.
 y or exothermic reaction as temperature
F
increases K decreases; for endothermic Concept Ladder
reaction as temperature increases K
increases. When the value of ethalpy
According to van’t Hoff equation change for the reaction
K2 H  T2  T1  would be zero than the
log   
K 1 2.303R  T1T2  value of K is independent
of temperature.
Where K1 and K2 are the equilibrium
constants at temperature T1 and T2
respectively and DH is the molar enthalpy
change in the temperature T1 to T2.

(c) Pressure

Note:
If the number of gas molecules is the same
on either side, then changing of pressure
will have no effect.
Chemical Equilibrium

18.
Q.9 The following concentrations were obtained for the formation of NH3 from N2
and H2 at equilibrium at 500K. [N2] = 1.5 × 10–2M, [H2] = 3.0 × 10–2M, and [NH3] =
1.2 × 10–2M. calculate the equilibrium constant.

A.9 The equilibrium constant for the reaction, N2(g) + 3H2(g)  
 2NH3(g) can
be written as
 
2

NH3 (g) 1.2  102
2

Kc    3.55  102
  
3 2 3
N2 (g) H2  g  1.5  10 2
3.0  10

Q.10 For the equilibrium,

2NOCl(g)  
 2NO(g) + Cl2(g)Kc is 3.75 × 10–6 at 1069K. Calculate
the Kp for this reaction at this temperature.

A.10 Kp  Kc RT 
n

n  (2  1)  2  1
Kp  3.75  106 (0.0831  1069)1 = 0.033

Q.11 The equation constants of the dissociation of various oxides of an elements A

are given at constant temperature:
(1) 2A 2O(g) 
 2A 2 (g)  O2 (g);KC  4.0  10

30

(2) 2AO(g)   A 2 (g)  O2 (g);KC  2.0  10

27


(3) 2AO2 (g)   A 2 (g)  2 O2 (g);KC  7.0  10
13


(4) 2A 2O5 (g) 
 2A 2 (g)  5 O2 (g);KC  1.0  10

31

Write the stability of these oxides in increasing order.

A.11 A2O5 < A2O < AO < AO2

At higher rate constant, the reaction will have more tendency to go on
completion.
Chemical Equilibrium

In this case, metal oxide dissociates in forward direction. Therefore, the metal
oxide with lesser value of equilibrium constant will be stable.

19.
 Q.12 At a certain temperature, the equilibrium constant (Kc) is 16 for the reaction:
SO2 (g) + NO2 (g) 
 SO3 (g) + NO(g)

If we take one mole of each of the four gases in one litre container, what
would be the equilibrium concentration of NO and NO2?

A.12 SO2 (g) + NO2 (g) 

  SO3 (g) + NO(g)


Initial 1 1 1 1
conc.
Equilibrium 1-x 1-x 1+x 1+x
conc.
Applying the law of mass action,
SO3  NO  (1  x)(1  x)  16
KC 
SO2 NO2  (1  x)(1  x)
1x
4 or 1  x  4  4x
1x
Concentration of NO2 at equilibrium = (1 – 0.6) = 0.4 mol.
Concentration of NO at equilibrium = (1 + 0.6) = 1.6 mol.

Q.13 For an ideal gas reaction

2A + B 

C +D
the value of Kp will be
nCnD V nCnD V
(1) Kp = . (2) Kp = .
n2AnB RT2 n2AnB RT
nCnD RT nCnD V
(3) Kp = . (4) Kp = .
n2AnB V 4n2AnB RT

A.13 2A + B 
C +D

nCRT nDRT
pC .pD . nn V
Kp  2  V V  C2 D .
2
pA .pB  nART   nBRT  nAnB RT
 V   V 
   
Chemical Equilibrium

20.
Q.14 Prove that the pressure necessary to obtain 50% dissociation of PCl5 at 250°C
is numerically three times of Kp.

A.14 PCl 5 

  PCl 3
  Cl 2

(Inital mole) 1 0 0
(Mole at equilibrium) (11  0.5) 0.5 0.5
Total mole of equilibrium = (1-0.5) + 0.5 + 0.5 = 1.5
 0.5   0.5 
 P  P
pPCl3 .pCl2  1.5   1.5 
Kp   (where; P  inital pressure)
pPCl5  0.5 
 P
 1.5 
1
or Kp = 3 P

or P = 3 Kp

Q.15 At 700K, the equilibrium constant Kp for the reaction

2SO3 (g) 

 2SO2 (g) + O2 (g)
is 1.80 × 10–3 kPa. What is the numerical value of Kc in moles per litre for this
reaction at the same temperature?

A.15 Given
2SO3 (g) 
 2SO2 (g) + O2 (g)

1 atm = 105 Pa; Dn = 1
Kp  1.8  103 KPa  1.8 Pa  1.8  105 atm
T = 700 K
R = 0.082 lit. atm
ng
Kp  Kc RT 
Kp
Kc =
RT
1.8  105
Chemical Equilibrium

Kc   0.031  105  3.1  107 mol L1

700  0.0821

21.
 Q.16 Write the expressions for equilibrium constant for the following reactions. If
the concentrations are expressed in mol L–1, give the units in each case.

(1) N2O4 (g) 

 2NO2 (g)

(2) 4NH3 (g) + 5 O2 (g ) 
 4NO(g) + 6 H2 O(g)

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


(4) 2HI(g) 
 H2 (g ) + I2 (g)

(5) 2N2O5 (g) 
 4NO2 (g) + O2 (g)


NO2 
2

A.16 (1) Kc 
N2O4 
, units = mol L

NO H2O
4 6

(2) Kc  , units = mol L–1

NH3  O3 
4

NO H2O
4 6

(3) Kc  , units = mol-2L2

NH3  O3 
4

(4) Kc 
H2 I2  , no units
HI
2

NO2  O2 
4

(5) Kc  ; units = mol3L–3

N2O4 
2
Previous Year’s Question

Applications of Equilibrium constant

If the value of equilibrium
(1) Predicting the extent of a reaction on the
constant for a particular reaction
basis of its magnitude.
is 1.6 × 1012, then at equilibrium
(2) Predicting the extent of a reaction
the system will contain [NEET]
High value of equilibrium constant indicates
(1) Mostly products
that products concentration is high and its
(2) Similar amount of reactants
low value indicates that concentration of
and products
the products in equilibrium mixture is low
(3) All reactants
H (g) + Br (g) 

 2HBr(g)
2 2 (4) Mostly reactants.
Chemical Equilibrium

(pHBr )2
KP   5.4  1018
(pH2 )(pBr2 )
The large volume of equilibrium constant
indicates that concentration of the product,

22.
 HBr is very high and reaction goes nearly to
completion. Rack your Brain
Similarly, equilibrium constant for [H2(g)
An equilibrium reaction has the
+ Cl4(g) 
 2 HCI (g)] is very high and
value of rate constant as 3.8
reaction goes virtually to completion.
× 1021. What will happen to the

HCl   4.0  1031
2
reaction.
KC 
H2 Cl 2 

y he reaction of H2 with O2 at 500 K has a

T
very large equilibrium constant, Previous Year’s Question
Kc = 2.4 × 1047 h
y The decomposition of H2O into H2 and O2 at
In Haber process, 30 litres of
500K has a very small equilibrium constant,
dihydrogen and 30 litres of
Kc = 4.1 × 10–48
dinitrogen were taken for reaction
which yielded only 50% of the
Note:
expected product. What will
large value of KP, or KC (larger than about
be the composition of gaseous
103), favour the products strongly. For
mixture under the aforesaid
intermediated values of K (approximately in
condition in the end?  [AIIMS]
the range of 10-3 to 103), the concentrations
(1) 20 litres ammonia, 20 litres
of reactants and products are comparable.
nitrogen, 20 litres hydrogen
Small values of equilibrium constant
(2) 10 litres ammonia, 25 litres
(smaller than 10-3)., favour the reactants
nitrogen, 15 litres hydrogen
strongly.
(3) 20 litres ammonia, 10 litres
Chemical Equilibrium

At 298 K for reaction,

 nitrogen, 30 litres hydrogen
N2,(g) + O2(g)  
 2NO(g)
(4) 20 litres ammonia, 25 litres

NO  4.8  1031
2
nitrogen, 15 litres hydrogen
KC 
N2 O2 

23.
 The very small value of. implies that reactants
N2, and 02, will be the predominant species
in the reaction mixture at equilibrium
(ii) Predicting the direction of the reaction.
The equilibrium constant is also used
to find in which direction the reaction
will proceed fora given concentration of
reactants and products. For this purpose,
we calculate the Reaction Quotient (Q).
The reaction quotient is defined in the
same way as the equilibrium constant (with
Previous Year’s Question
molar concentrations to give Qc, or with h
partial pressure to give QP) at any stage of
reaction. Fora general reaction: The reaction quotient (Q) for the
n1A + n2B  reaction

 m1C + m2D

C D
m1 m2
N2g   3 H2g  
 2 NH3 g 

QC 
A  B
1n n
2

NH3  .
2

is given by Q  The
Note: N2 H2 
3

If QC > KC the reaction will proceed in the

reaction will proceed from right
backward direction
to left if.  [AIPMT]
If Qc < Kc, the reaction will move in the
(1) Q = KC (2) Q < KC
forward direction
(3) Q > KC (4) Q = 0
If Qc = KC then reaction will be in equilibrium.
Chemical Equilibrium

24.
For example,
In the reaction H2 (g) + I2 (g) 
 2 HI (g) , if
 Rack your Brain
the molar concentration of H2, I2 and HI are The value of QC for a reaction is
0.1 mol L-1 respectively at 783K, then reaction determined on the basis of active
quotient at this stage of the reaction is mass. Why not some other term

HI  (0.4)2  8 is considered to calculate it?
2

QC 
H2 I2  (0.1)(0.2)
KC for this reaction at 783 K at 46 and we
find that QC < KC. The reaction, therefore,
will move to right i.e. more H2(g) and I2(g)
will react to form more HI(g) and their
concentration will decrease till QC = Kc.
Degree of Dissociation
Degree of dissociation is the fraction of a
mole of the reactant that undergoes dissociation. Previous Year’s Question
It is represented by ‘a’
no. of moles of reactant dissociated If a is dissociation constant, then

no. of moles of reacttant present initially the total number of moles for the
reaction, 2 HI   H2  I2 will be
For example,
 [AIPMT]
For reversible reaction
(1) 1 (2) 1-a
1 3
NH3 (g) 
 N2 (g) + H2 (g)
 (3) 2 (4) 2-a
2 2
Moles initially a 0 0
aa 3aa
At equilibrium; a(1-a)
2 2
Here, a represented the degree of dissociation.
The process of formation of ammonia takes place
by combination of nitrogen and hydrogen gases in
the presence of iron as catalyst and molybdenum
Definitions
as promoter.
This process is known as Haber’s process. The Degree of dissociation is
reactants and products are in equilibrium phase the fraction of molecules
dissociated at any given time.
Chemical Equilibrium

with each other.

This process takes place at a temperature range
400 - 450 °C and pressure range of 150 - 200
atm.

25.
Chemical Equilibrium

26.
Q.17 2.56 gm of sulphur S8(s) is taken which is in equilibrium with its vapour
according to reaction, If vapour occupies 960 mL at 1 atm and 273 K then the
degree of dissociation of S8(s) will be
(1) 0.5 (2) 0.55
(3) 0.4 (4) 0.44

2.56
nS8   0.01
A.17 8  32
S8 (s)  8S(g)
0.01(1  ) 8  0.1  
960
1  (0.01  8  )  0.08  273
1000
  0.55

Q.18 For the reaction 2 HI  g  

 H2  g   I2  g 

The degree of dissociation (a) of HI(g) is related to equilibrium constant Kp by
the expression.
1 + 2 Kp 1 + 2 Kp 2 Kp 2 Kp
(1) (2) (3) (4)
2 2 1 + 2 Kp 1 + 2 Kp

A.18
2 HI  g  

 H2  g   I2  g 
Initial 1 0 0
At equilibrium 1– a a/2 a/2
2
   PI  PH 
 PT    KP  2 2 2 
 2  
Kp  2  PHI  
 1    PT2
 2 KP
 2 KP  
1 1  2 KP
Chemical Equilibrium

27.
 Q.19 For the dissociation reaction N2O 

 2 NO2  g  , the equilibrium constant
KP is 0.120 atm at 298K and total pressure of system is 2 atm. Calculate the
degree of dissociation of N2O4.

A.19 For the reaction N2O 


  2 NO2  g 

Initial 1 0
At equilibrium (1-a) 2a
Let a be the degree of dissociation and P is the total pressure, then
Total number of moles  2   1    1  
 PN2O  moles of N2O  Ptotal

1
 P
1
 2 
and PNO2   P
1
2
 2  
P   1   P 
2
  
 
NO2
 KP 
PN2O  1  
 P
1 
4 2P2 1 4 2P ……. (i)
  
1    1    P 1  2
Given atm; KP = 0.120 atm; P = 2 atm

Substituting all the values in equation (i), we get

4 2  2  8 2
0.120  
1    2
 1  2 
 0.120  1     8
2 2
1/ 2
 0.120 
Degree of dissociation,      0.121
 8.12 
Chemical Equilibrium

28.
Calculation of KP and KC
(a) Homogeneous equilibrium in gaseous phase
Concept Ladder
(b) Homogeneous equilibrium in solution phase
(c) Equilibrium constant for various
The numerical value of
heterogeneous equilibrium
equilibrium constant KC is
(a) Homogeneous equilibrium in gaseous phase
not influenced by catalyst.
Formation of Nitric Oxide : (Dn = 0)
Presence of a catalyst
1. Calculation of KC :–
simply helps in attaining the
Suppose the initial concentration of N2 and
equilibrium earlier
O2 is a and b respectively. x is the degree of
dissociation.

N2 + O2 

 2NO
Initial moles a b 0
Moles at equilibrium (a – x) (b – x) 2x

(a − x) (b − x) 2x
Active mass (mol l–1) V V
V
Here, V is the volume of container in litre.
According to the law of mass action
Previous Year’s Question

NO
2

KC  The reaction,
N2 O2  2A(g) + B(g)  
  3C(g) + D(g)
is begun with the concentrations
Substitution the values in the above of A and B both at an initial value
equation of 1.00 M. When equilibrium is
 2x 
2 reached, the concentration of D is
  measured and found to be 0.25M.
 v 
KC  The value for the equilibrium
a  x b  x 
   constant for this reaction is given
 v  v 
by the expression [AIPMT]
3 2
(1)  0.75  0.25    1.00   1.00  
3 2
x2 (2)  0.75  0.25  0.50  0.75
KC 
(a  x)(b x) (3)  0.75  0.25  0.50  0.25 
3 2
Chemical Equilibrium

(4)  0.75  0.25   0.75 0.25 

3 2
KC for this reaction is independent of V of
the reaction container.

29.
 2. Calculation of KP
All the thing being same as above, except
pressure. Let P atmosphere is the pressure
at equilibrium
N2 + O2 
 2NO
Initial moles a b 0
At equilibrium (a – x) (b–x) 2x
Total no. of moles = (a – x) + (b – x) + 2x = (a + b)

a  x)P
PN2 
(a  b) Rack your Brain

The value of KP depends upon

b  x)P
PO2  partial pressure of gases at
(a  b)
equilibrium. Would it depend
upon degree of dissociation?
2x)P
PNO 
(a  b)

According to the law of mass action

PNO 
2

KP 
PN2  PO2 

Substituting the value of PNO , PN2 , PO2 in the

above equation of KP. Concept Ladder

KP and KC are equilibrium

2
 (2 x)P  constant of ideal gas
 (a  b)  mixture considered under
KP   
 (a  x)P   (b  x)P  reversible reaction KP is an
 (a  b)   (a  b)  equilibrium constant wrt. to
  
atmospheric pressure and
the KC is the equilibrium
4x2 constant wrt. concentration
KP 
Chemical Equilibrium

(a  x)(b x) expressed in molarity.

Thermal Dissociation of Phosphorus

pentachloride (Dn > 0)

30.
Dissociation of PCl5
1. Calculation of KC : Suppose one mole
of PCl5 is take in a closed container of V
litre. Further at equilibrium x mol of PCl5
dissociated

PCl 5 
 PCl 3 + Cl 2
Initial moles 1 0 0
Moles at equilibrium (1 – x) x x
1−x x x
Concentration (mol l–1)
v v v
According to law of mass action Previous Year’s Question

KC 
PCl 3  Cl 2  The dissociation equilibrium of a
PCl5  gas AB2 can be represented as:

2AB2(g)    2AB(g) + B2(g)
Substituting the values in the above the degree of dissociation is x
equation. and is small compared to 1. the
expression relating the degree of
 x  x 
   dissociation (x) with equilibrium
v v
KC      constant KP and total pressure P
1x
  is  [AIPMT]
 v 
(1) (2KP/P)1/2
x2 (2) (KP/P)
KC  (3) (2KP/P)
(1  x) v
(4) (2KP/P)1/3

The formula of KC has V in the denominator,

hence the equilibrium will be affected by
V of the reaction container for the given
reaction.
If x << 1, then, 1 – x ≈ 1
Concept Ladder
x2
So, KC =
v
x 2
= K C .v
The numerical value of KC
is independent of initial
Chemical Equilibrium

x ∝ v 2

concentration of reactants
x∝ v and product.

If we increase the volume, the dissociation
x is also increased.

31.
 2. Calculation of Kp
PCl 5  
  PCl 3 + Cl 2
Initial moles 1 0
0
Moles at equilibrium (1 – x) x
x Rack your Brain
total no. of moles at equilibrium,
(1  x)  x  x  (1  x)moles Why the concentration of pure
solid and liquid should be ignored
According to law of mass action
while writing the equilibrium

PPCl3  PCl2 constant expression?
KP 
PPCl5
x P
At equilibrium PPCl3 
(1  x)
x P
PCl2 
(1  x)
(1  x)P
PPCl5 
(1  x)
Substituting the values in the above
equation of KP –

 x P  x P 
  
1  x  1  x 
KP  
(1  x)  P
(1  x) Previous Year’s Question
The equation of KP is not independent of
pressure. In which of the following
Suppose, x<<1, then, 1 – x2 ≈ 1 equilibrium KC and KP are not
K = x2P equal? [AIPMT]
P
(1) 2 NOg   N2  g   O2 (g)

KP
x2 = (2) SO2 g   NO2 g  
 2 HIg 

P
1 (3) H2 g   I2 g  
x∝  2 HIg 

P
Chemical Equilibrium

(4) 2 Cs  O2 g  
 2 CO2 g 

The degree of dissociation of PCl5 is
inversely proportional to the square root of
pressure so, decrease of pressure increases
dissociation of PCl5.

32.
Formation of Ammonia – (Dn < 0)
1. Calculation of KC :

N2 + 3 H2  2NH3
Initial moles 1 3 0
moles at equilibrium (1 – x) (3 – 3x) 2x

1x  3  3x   2x 
Active mass (mol l–1)      
 v   v   v 
According to law of mass action

NH3 
2

KC 
N2 H2 
3

Substituting the values in the above

equation
Previous Year’s Question
2
 2x 
 
KC   v  Equilibrium constant KP for
3
 1  x   3  3x  following reaction
  
 v  v  MgCO3 s 
  MgOs  CO2 g  

4x2 v 2
KC  [AIPMT]
(1  x)(3  3 x)3 (1) KP = PCO
2

4x2 v 2
KC 
27(1  x)4 PCO2  PMgO
(2) KP  KCO 
The formula of KC has V in the numerator;
2
PMgCO3
hence the equilibrium will be affected by V
of the reaction container. PCO2  PMgO
(3) KP 
Dependence If, x<<1 then, (1  x)  1
4
PMgCO3

x 2 v 2
KC  PMgCO3
27 (4) KP 
PCO2  PMGo

27KC
x2 =
4V 2
1 1
x2 ∝ x∝
Chemical Equilibrium

v2 v
If we increase the volume of container the
degree of dissociation x is decreased.

33.
 2. Calculation of KP:
N2 + 3 H2    2NH3
Initial concentration 1 3 0
moles at equilibrium (1-x) (3-3x) 2x
Total number of moles at equilibrium  ( 1  x)  (3  3x)  2 x  (4  2 x)
According to the law of mass action

 
2
PNH3
KP 
   
3
PN2  PH2
(2 x).P
At equilibrium; pNH3 
(4  2 x)
(1  x).P
pN2 
(4  2 x)
(3  3 x).P
pH2 
(4  2 x)
Substituting the values in the above
equation of KP.
Previous Year’s Question
2
 2x 
 .P 
KP   4  2x  A 20 litre container at 400K
 1  x   3  3x  contains CO2(g) at pressure
 .P   .P 
 4  2x   4  2x  0.4 atm and an excees of SrO
(neglect the volume of solid SrO).
4x2 (4  2 x)2
KP  the volume of the container is
(1  x)(3  3 x)3 P2
now decreased by moving the
movable piston fitted in the
16x2 (2  x)2
KP  container. The maximum volume
27(1  x)4 P2
of the container, when pressure
The equation of KP is not independent of of CO2 attains its maximum value,
pressure will be
Suppose, x << 1 then, (Given by: SrCO3s 
 SrOs  CO2 g 

(1  x)4  1 KP = 1.6 atm) [NEET]
(1) 10 litre (2) 4 litre
And (2  x)2  4
(3) 2 litre (4) 5 litre
64x2
KP =
Chemical Equilibrium

27P2
x2 ∝ P2 x ∝ P
If we increase the pressure the degree of
dissociation x is also increased.

34.
(b) Homogeneous equilibrium is solution phase
Formation of ethyl acetate
Equilibrium is represented as

C2H5  OH     CH3COOH    
 CH3COOC2H5     H2O   

Initial moles 1 1 0 0
Moles at equilibrium 1-x 1-x x x

KC 
CH3COOC2H5  H2O
C2H5OHCH3COOH
x x

V V x2
KC  
1  x  1  x 1  x 1  x
v v Previous Year’s Question
(c) Equilibrium constant for various
heterogeneous equilibrium If the concentration of HO- ions
Heterogenous equilibrium results from in the reaction
a reversible reaction involving reactants Fe(OH)3(s) 

 Fe (aq) + 3OH (aq)
3+ -

and product that are in different phases. is decreased by 1/4 times, then
The law of mass action is applicable to equilibrium concentration of Fe3+
a homogeneous equilibrium and is also will increase by  [AIPMT]
applicable to a heterogeneous system. (1) 64 times (2) 4 times

1. Decomposition of solid CaCO3 into solid (3) 8 times (4) 16 times

CaO and gaseous CO2
Let ‘a’ moles of CaCO3 are taken in a vessel
of volume ‘V’ litre at temperature ‘T’ K.
CaCO3 (s) 
 CaO(s) + CO2 (g)


Moles initially a 0 0
Moles at equilibrium a – x x x

Keq 
CaOCO2 
CaCO3 
As CaCO3 and CaO(s) are pure solids, so
Chemical Equilibrium

their concentration is unity

PCO2
 KC 
RT
K C (RT) = PCO2

35.
 Since KC R and T are constant, their product
will also be a constant referred as KP.

xRT
 KP  PCO2 
V
Simultaneous Equilibrium
In simultaneous equilibrium more than one
equilibrium are established in a vessel at the Concept Ladder
same time and any one of the reactant or product
is common in more than one equilibrium, then Equilibrium concentration
the equilibrium concentration of the common of common species at
species in all the equilibrium would be same. equilibrium are always same
For example, if we take CaCO3(s) and C(s) in case of simultaneous
together in a vessel of capacity ‘V’ litre and heat it equilibrium.
at temperature ‘T’ K, then CaCO3 decomposes to
CaO(s) and CO2(g). further, evolved CO2 combines
with the C(s) to give carbon monoxide. Let the
moles of CaCO3 and carbon taken initially be ‘a’
and ‘b’ respectively.
CaCO3 (s) 
 CaO(s) + CO2 (g)

Moles at equilibrium a – x x (x –y)
CO2 (s) + C(s)  2 CO(g)


Moles at equilibrium (x – y) (b - y) 2y
Thus, as CO2 is common in both the equilibrium so
its concentration is same in both the equilibrium
constant expression.
Equiliibrium constant for first equilibrium, Rack your Brain
xy
KC1  CO2  
V s
Equilibrium constant for second equilibrium,
.
 CO
2 2
 2y  V 4y 2
K C2   
CO2  V2 (x y) V(x y)
Calculation of Degree of dissociation by vapour
Chemical Equilibrium

density measurement
Reactions in which there is a change in the
number of moles after dissociation, the extent of
dissociation can be determined by vapour density

36.
measurement.
PCl 5    PCl 3 + Cl 2


Initially 1 0 0
Moles at equilibrium (1 – a) a a (‘a’ is the degree of dissociation)
Total number of moles at equilibrium;  1          1   

V is the volume occupied by 1 mol of PCl5(s) which

have vapour density is ‘D’ before dissociation
and after dissociation is ‘d’. Under the same
conditions, the volume occupied by (1 + a) moles
at equilibrium would be (1 + a)V litre.
1
Density ∝
Volume
Concept Ladder
1 1
D∝ d
V 1   V For polymerization reaction
1 nA  g  
 An  g  ,

D V
or   1   where n  2
d 1 Dd Mt  M0
= 
1   V 1  1
d   1  M0   1 

n  n 

D Dd
or   1
d d
Molecular mass = 2 × Vapour density
Mt  M0
So  
M0
Where, Mt = calculated molecular mass
M0 = observed molecular mass Rack your Brain

In which case the degree

of dissociation constant be
determined by measuring vapour
density?
Chemical Equilibrium

37.
 Q.20 At 627°C and 1 atm SO 3
is partially dissociated into SO2 and O2 by the reaction
SO3  g  
 SO2  g   1 / 2 O2  g 

The density of the equilibrium mixture is 0.925g L–1. What is the degree of
dissociation.

A.20 Let the molecular mass of the mixture at equilibrium be

Applying the reaction
.

dRT 0.925  0.0821  900

Mmix    63.348
P 1
Molecular mass of SO3 = 80

80
Vapour density of SO3=
,D = 40
2
68.348
Vapour density of mixture,
= d = 34.174
2
Let the degree of dissociation be x.
Dd 40  34.174 5.826  2
x    0.34
n  1 d  3  1   34.174 34.174
 
2 
or x = 34% dissociated
i.e. SO3 is 34% dissociated.

Q.21 The vapour density of N2O4 at a certain temperature is 30. Calculate the
percentage dissociation of N2O4 at this temperature.

A.21 N2O4  g  
 2 NO2  g 

Molecular mass of N2O4   28  64   92

Vapour density, = 92
D = 46
2
Let the degree of dissociation be x.
Given, d = 30
Applying the relationship,
Chemical Equilibrium

D  d  46  30  16
x    0.533
d 30 30
Degree of dissociation = 53.3%

38.
LE-CHAPTELIER’S PRINCIPLE
How an equilibrium state relieves the external
stress?
Concept Ladder
How a state in equilibrium (a stable state) adjusts
to the external stress (change in Temperature,
According to Le-Chatelier,
pressure or concentration of reactants/products) to get better yield of NH3-
is generalised in Le-Chatelier’s principle. (a) High pressure
(b) Low Pressure
Le-Chatelier’s Principal states that : (c) High concentration of
If a stress is applied to a system in reactants
equilibrium, the equilibrium condition is upset; A (d) Low concentration of
net reaction occurs in that direction which tends product
to relieve the external stress and finally a new
equilibrium is attained.
To understand its application to a system, let us
consider following example :
N2(g) + 3H2(g)  
 2NH3(g)
DH = -92 kJ for Forward Reaction
DH = +92 kJ for Backward Reaction
Note :
DH : Enthalpy Change is a measure of heat
evolved or heat absorbed in a chemical
Previous Year’s Question
reaction. It is negative when heat is evolved
and positive when heat is absorbed during a
For the reversible reaction,
chemical change. You will learn details of it
N2g   3 H2g  
 2 NH3 g   heat

later in the Chapter on Chemical Energetics.
Note that in the above reaction : The equilibrium shifts in forward
(a) Forward reaction is exothermic (favours direction [NEET]
formation of NH3) and backward reaction (1) by increasing the
is endothermic (favours decomposition of concentration of NH3(g)
NH3) (2) by decreasing the pressure
(b) Formation of NH3 results in decrease in (3) by decreasing the
number of moles (from 4 total moles of N2 concentrations of N2(g) and
and H2 to 2 total moles of NH3) is a decrease H2(g)
in volume to right (in forward reaction) (4) by increasing pressure and
Chemical Equilibrium

(c) Both reactants and products are gases and decreasing temperature.
they will be influenced by changes in P, T
and changing concentrations.

39.
 Effect of Temperature
Temperature can be increased by adding heat
and can be decreased by taking out heat from Previous Year’s Question
the system.
y Increase the temperature by supplying
Which of the following information
heat: According to Le-Chatelier’s principle can be obtained on the basis of
the disturbed equilibrium state will move Le-Chatelier’s principle. [AIIMS]
in that direction where heat is being (1) entropy change in a reaction
absorbed (where stress is relieved) i.e., (2) Dissociation constant of a
in endothermic direction. In case of given weak acid
situation, reverse direction will be favoured (3) Equilibrium constant of a
(that being endothermic) till whole of extra chemical reaction
heat in consumed. so NH3 will decompose (4) Shift in equilibrium position
on increasing temperature. on changing value of a
y Decrease the temperature by extracting
constant
heat: According to Le Chatelier’s principle,
the system will go in the direction where
heat is evolved i.e. in exothermic direction. Rack your Brain
In given situation, forward reaction will be
favoured (i.e. formation of NH3) till the new
Why formation of NO requires
equilibrium is again established.
high temperature whereas
Note:
formation of NH3 is preferred at
The new equilibrium state has a new value
low temperature?
of equilibrium constant K on changing the
temperature.

Effect of Pressure
y Increase in pressure would result in

decrease in volume thereby increasing Previous Year’s Question
the concentration (mol/L). The system will
shift in a direction where number of moles According to Le-Chatelier’s
decreases (decreasing concentration). In principle, adding heat to a solid
given situation, there is decrease in number and liquid in equilibrium will
of moles in forward direction, so increasing cause the  [NEET]
pressure favours forward reaction (i.e, (1) temperature to increase
formation of NH3) (2) temperature to decrease
Chemical Equilibrium

y Decreasing the pressure would mean lower (3) amount of liquid to decrease
number of moles/L. The system will shift (4) amount of solid to decrease
in a direction which will produce more
moles. In given situation, there is increase

40.
 is number of moles in reverse direction,
so decreasing pressure favours backward
Rack your Brain
reaction (i.e., decomposition of NH3)
Effect of Concentration:
What is the effect of a catalyst in
The Concentration can be changed in two ways :
a system at equilibrim?
(a) By removing some of a component or
(b) By adding some more of a component.

According to Le-Chatelier’s principle :

y The addition of any component to a side
(reactants and products) of a reaction
in equilibrium shifts the equilibrium in Concept Ladder
the direction away from that side or one
can say that equilibrium shifts in that
Solubility of NaOH increases
direction which consumes the increased
with temperature although
concentration.
dissolution is exothermic
y In given situation ; increasing amount of

(an exception).
pure N2 and H2 would favour formation of
NH3 . increasing amount of NH3 would favour
decomposition of NH3
y The removal of any component from a side
(reactants and products) of a reaction in
equilibrium shifts the equilibrium in the
direction towards that side, or one can say
the equilibrium shifts in that direction which
produces the decreased concentration.
y In given situation : Decreasing the amount Previous Year’s Question
of NH3 from right side drives the equilibrium
to forward direction i.e. favours formation Which one of the following
of NH3 . Decreasing amount of N2 or H2 inforation can be obtained on the
from left drives the equilibrium to reverse basis of Le-Chatelier principle?
direction i.e. favours decomposition of NH3  [AIPMT]
(1) dissociation constant of a
Effect of Catalyst: weak acid
y Catalyst increases the rate of both forward (2) Entropy change in a reaction
and backward reactions simultaneously and (3) Equilibrium constant of a
Chemical Equilibrium

to the same extent in a reversible reaction. chemical reaction

By increasing both rates, catalyst reduces (4) Shift in equilibrium position
the time to reach equilibrium state. on changing value of a
constraint

41.
 y
Catalyst does not change the relative
amounts of either reactants or products, Rack your Brain
hence it has no effect on equilibrium
constant
Water boil at high temperature in
pressure cooker why?
Evaluation of Keq at different temperatures:
If K1 be the equilibrium constant at T1 (in
Kelvin) and K2 be the equilibrium constant at T2
(in Kelvin) (T2 > T1), the two constants are related
by Van’t Hoff equation as follows:
K2 H  T2  T1  Concept Ladder
log 10   
K 1 2.303R  T1T2 
Where r is gas constant and DH° is the standard Liquid phase of H2O, ie.,
heat of reaction. water cannot exist above
647.15K and 218 atm. these
Le-Chatelier’s principle and Physical equilibrium: values are called critical
Consider the physical equilibrium of change of temperature and critical
state: pressure for water.
solid  
  liquid
Effect of pressure on melting:
When a solid melt, there is a decrease
in volume for some solids (ice, diamond,
carborundum, magnesium nitride, quartz etc.)
and there is an increase in volume for some Previous Year’s Question
solids (sulphur, iron, copper, silver, gold etc.).
When ice melts, there is a decrease in For a given exothermic reaction,
volume.On this kind of system where there is a KP and K’P are the equilibrium
decrease in volume due to melting, increasing constants at temperature T1 and
pressure will reduce the melting point of the T2, respectively. Assuming that
system. Thus, increasing the pressure will favor heat of reaction is constant in
the melting of ice. temperature range between T1
Similarly, when sulphur melts, there is an and T2 it is readily observed that
increase in volume. On this kind of system where  [NEET]
there is an increase in volume due to melting, (1) KP > K’P
decreasing pressure will reduce the melting point (2) KP < K’P
Chemical Equilibrium

of the system. Thus, decreasing the pressure will (3) KP = K’P

favor the melting of sulphur.
1
Vapour pressure of liquids: (4) KP =
K 'P

42.
Consider liquid vapour equilibrium:
This is an endothermic reaction in forward
direction. Hence rise in temperature will favour
evaporation. This means, increase in temperature
Rack your Brain
results in increase in vapour pressure of the
system.
Why sealed soda water bottle on
Effect of pressure on boiling point: opening shows the evolution of
As we increase pressure on the system : gas with effervescences?
Liquid vapour ; vapours condense lowering the
vapour pressure of system. This means boiling
point rises on increasing pressure as to reach the
pressure required for the liquid to start boiling,
needs to be increased.

Effect of temperature on solubility:

In most cases, formation of solution (solute
in solvent) is an endothermic process. In such Concept Ladder
cases, increasing temperature, increases the
solubility of solutes. In cases, where dissolution The solubility of NH3 in water
of solute is followed by evolution of heat; also involves H-bonding
increasing temperature lowers the solubility of and thus, solubiliy of NH3 in
solutes. water depends on pressure
as well as tendency to show
Solubility of gases in liquid : H-bonding.
As the temperature increases, the solubility
of a gas decreases. This means that more gas is
present in a solution with a lower temperature as
compared to a solution with higher temperature.
Effect of pressure on solubility :
When a gas dissolves in liquid, there is a decrease
in volume of the gas (X(g)  
 X(aq.)). Thus,
increase of pressure will favour the dissolution of
gas in liquid.
Rack your Brain
Chemical Equilibrium

Ice melts slowly at higher

altitude, Why?

43.
Chemical Equilibrium

44.
Q.22 What should be the respective active masses of methyl alcohol and carbon
tetrachloride, if their densities are 0.5 and 1.2 g/mL?
(1) 15.62 and 7.79 (2) 16.65 and 7.40
(3) 15.46 and 7.80 (4) 15.40 and 6.50

A.22 (1)

Density × 1000
Active mass = (when density in g/mL)
Molecular weight
0.5 × 1000
(a) [CH3OH] = = 15.62 g mol L–1
32
[Molecular weight of CH3OH = 12 + 3 + 16 + 1 = 32]
1.2 × 1000
(b) [CCl4] = = 7.79 g mol L–1
154
[Molecular weight of CCl4 = 12 + 35.5 × 4 = 12 + 142 = 154]

Q.23 Write down the equilibrium constant for the following reactions:

(a) N2 + 3H2  
 2NH3 
(b) PCl5  PCl3 + Cl2

(c) 3A + 2B  C + 4D (d) CaCO3(s) 
 
 CaO(s) + CO2(g)

(e) 2KClO3(s)  2KCl(s) + 3O2(g)

(f) CH3COOH(l) + C2H5OH(l)  
 CH¬3COOC2H5(l) + H2O(l)
 
(g) NH3(aq) + H2O  NH4 (aq) + OH–(aq)
+


(h) H2O(l)  
 H2O(g)

[NH3 ]2 [PCl 3 ][Cl 2 ]

A.23 (a) K =
[N2 ][H2 ]3
(b) K =
[PCl5 ]

[C][D]4
(c) K = (d) K = [CO2] (Active mass of solid is 1)
[A]3 [B]2
(e) K = [O2]3

[CH3COOC2H5 ][H2O]
(f) K = (here H2O is not in excess)
[CH3COOH][C2H5OH]

[NH4 ][OH ]
Chemical Equilibrium

(g) K  (here H2O is in excess (solvent) so its concentration doesn’t change.)

[NH3 ]
(h) K = [H2O](g)

45.
 Q.24 If 0.5 mol of H 2
is reacted with 0.5 mol of I2 in a 10 L container at 444°C and
at same temperature value of equilibrium constant KC is 49, the ratio of [HI]
and [I2] will be
1 1
(1) 7 (2) (3) (4) 49
7 7

A.24 (1)

H2 + I2 
 2HI
[HI]
KC = if [H2] = [I2]
[H2 ][I2 ]
[HI]2
KC =
[I2 ]2
[HI]
or = KC = 49 = 7
[I2 ]

Q.25 One mole of ammonium carbamate dissociate as shown below at 500 K.


NH2COONH4(s) 
 2NH3(g) + CO2(g)

If the pressure exerted by the released gases is 3.0 atm, the value of KP is

(1) 7 atm (2) 3 atm (3) 4 atm (4) 8 atm

A.25 (3)
Applying the law of chemical equilibrium, we get
KP = (PNH3 )2 (PCO2 )
Since total pressure is 3 atm, the partial pressure of NH3(g) and CO2(g) are

2
PNH3  = 3 × = 2 atm
3

PCO2  = 3 × 1
Chemical Equilibrium

= 1 atm
3
KP = [2.0]2 [1.0] = 4.0 atm

46.
Q.26 K for the reaction A(g) + 2B(g) 
P
 3C(g) + D(g); is 0.05 atm. What will be

its KC at 1000 K in terms of R:

5  105 R
(1) (2) (3) 5 × 10–5R (4) None of these
R 5  105

A.26 (1)
We know that

KP = KC (RT)∆n
KP
or, KC =
(RT)Dn
Here, ∆n = 4 – 3 = 1

T = 1000 K, KP = 0.05
0.05 5  105
KC  1

(R  1000) R

Q.27 For N O  NO + NO2, if total pressure is P atm and amount of dissociation
2 3 
is 50%, the value of KP will be:

P P
(1) 3 P (2) 2 P (3) (4)
3 2

A.27 (3)

N2O3 

 NO + NO2

Initial mol 1 0 0

at equilibrium 1 – 0.5 0.5 0.5

2 2
xP xP
KP  
( 1  x) ( 1  x) 1  x2
Chemical Equilibrium

0.5  0.5  P P
KP  
1.5  0.5 3

47.
 Q.28 The equilibrium constant for the reaction H (g) + S(s)
2

 H2S(g); is 18.5
at 925 K and 9.25 at 1000 K respectively. The enthalpy of the reaction will be:
(1) –68000.05 J mol–1 (2) –71080.57 J mol–1
(3) –80071.75 J mol–1 (4) 57080.75 J mol–1

A.28 (2)
Using the relation,

K2 H  T2  T1 
log   
K1 2.303 R  T1T2 

9.25 H 75
log  
18.5 2.303  8.314 925  1000

H
–0.301 =  ∆H = –71080.57 J mol–1
2.303  8.314  925  1000

Q.29 In reaction,
CO(g) + 2H2(g)  
 CH3OH(g) ∆H° = –92 kJ/mol–1
concentrations of hydrogen, carbon monoxide and methanol become constant
at equilibrium. What will happen if:
(1) Volume of the reaction vessel in which reactants and products are
contained is suddenly reduced to half?
(2) Partial pressure of hydrogen is suddenly doubled?
(3) An inert gas is added to the system at constant pressure?
(4) The temperature is increased?

A.29 For the equilibrium


CO(g) + 2H2(g)   CH3OH(g)
PCH3OH
[CH3OH]
KC  2
 KP 
[CO][H2 ] PCO  PH22
(1) When the volume of the vessel is suddenly reduced to half, the partial
pressure of various species gets doubled. Therefore,
2PCH3OH 1
Chemical Equilibrium

QP  2
 KP
2PCO  (2PH2 ) 4
Since QP is less than KP, the equilibrium shift in the forward direction
producing more CH3COH.

48.
 (2) When partial pressure of hydrogen is suddenly, QP changes and is no
longer equal to KP.
2PCH3OH 1
QP  2
 KP
PCO  (2PH2 ) 4
Equilibrium will shift from left to right.
(3) When an inert gas is added to the system at constant pressure,
equilibrium shifts from lower number of moles to higher number of
moles (in backward direction).
(4) By increasing the temperature, KP will decrease the equilibrium will
shift from right to left.

Q.30 A mixture of 4.2 moles of N , 2.0 moles of H

2 2
and 10.0 moles of NH3 is introduced
into a 10.0 L reaction vessel at 500 K. At this temperature, equilibrium constant
KC is
1.7 × 102, for the reaction N2(g) + 3H2(g) 
 2NH3(g)
(i) is the reaction mixture at equilibrium?
(ii) if not, what is the direction of the reaction?

A.30 [N ] = 4.2
= 0.42 M
2
10
2.0
[H2] = = 0.2 M
10
10
[NH3] = = 0.1 M
10
For the concentration, reaction quotient (Q) for the reaction


N2(g) + 3H2(g) 
 2NH3(g) is
[NH 3 ]2 (0.1)2
Q   2.976
[N2 ][H2 ]3 (0.42)  (0.2)3
But KC = 1.7 × 102

(i) Since Q ≠ KC, hence reaction is not at equilibrium.

Chemical Equilibrium

(ii) Also Q < KC, the reaction will proceed from left to right.

49.
 Chapter Summary

1. Equilibrium is defined as the point at which the rate of forward reaction is equal to
the rate of backward reaction.

2. Chemical equilibrium is dynamic in nature and equilibrium state can be approached

from both sides.

3. Active mass is molar concentration of the substance. Active mass of solid and pure
liquid is taken as unity.

4. Equilibrium constant has definite value for every chemical reaction at a given
temperature. It is independent of concentration and catalyst.

5. If a reaction can be expressed as the sum of two or more reactions then overall KC
will be equal to the product of the equilibrium constant of individual reaction.

6. Non-affecting factors of the equilibrium constant : The value of equilibrium constant

is independent of the following factors :
(a) Concentration of reactants and products. (b) Pressure
(c) Volume (d) The presence of a catalyst.
(e) Presence of inert materials

7. Degree of Dissociation
x x
  %    100
a a
Where a = Degree of dissociation in percentage
x = Number of dissociates moles
a = Initial no. of moles

8. It is not necessary that all the types of equilibrium constants are defined for every
reaction for eg. for a reaction involving only solutions Kp is not defined.

9. If inert gas mixed at constant temperature and constant volume in an equilibrium

Chemical Equilibrium

chemical reaction then total number of moles of gases are present in a container,
increases i.e. total pressure of gases increases but concentration in terms of moles/
litre and partial pressure of reacting substances are unchanged so dissociation (x)
unchanged.

50.
 1 
10. SO2 (g)  O2 (g)  SO3 (g)  K 1
2
 1
NO2 (g)   NO(g)  O2 (g)  K2
2

then SO2 (g)  NO2 (g) 
 SO3 (g)  NO(g)  K
So K  K 1  K2
11. Change in temperature, pressure or concentration favours one of the reactions
and thus shift the equilibrium point in one direction.

12. A catalyst allows the system to reach a state of equilibrium more quickly.

13. Pressure and volume has no effect on the reaction in which there is no change in
the number of moles.

14. If the concentration of reactants is increased and product is removed, the
reaction will take place in forward direction.

15. Free energy change
DG = DG° + 2.303 RT log Q At equilibrium DG = 0, (T is in Kelvin), Q = K so
DG° = – 2.303 RT log K, where K is equilibrium constant.

Chemical Equilibrium

51.