1. INTRODUCTION
It is an experimental fact that most of the process including chemical reactions, when carried out
in a closed vessel, do not go to completion. They proceed to some extent leaving considerable
amounts of reactants & products. When such stage is reached in a reaction, it is said that the
reaction has attained the state of equilibrium. Equilibrium represents the state of a process in
which the properties like temperature, pressure, concentration etc of the system do not show any
change with passage of time. In all processes which attain equilibrium, two opposing processes
are involved. Equilibrium is attained when the rates of the two opposing processes become equal.
If the opposing processes involve only physical changes, the equilibrium is called Physical
Equilibrium. If the opposing processes are chemical reactions, the equilibrium is called Chemical
Equilibrium.
Some common physical equilibria are:
Solid Liquid
Liquid Gas
Gas Solution
Generally, a chemical equilibrium is represented as
aA + bB xX + yY
Where A, B are reactants and X, Y are products.
Note:
The double arrow between the left hand part and right hand part shows that changes is
taking place in both the directions.
On the basis of extent of reaction, before equilibrium is attained chemical reactions may be
classified into three categories.
(a) Those reactions which proceed to almost completion.
(b) Those reactions which proceed to almost only upto little extent.
(c) Those reactions which proceed to such an extent, that the concentrations of reactants and
products at equilibrium are comparable.
EQUILIBRIUM IN PHYSICAL PROCESS
The different types of physical Equilibrium are briefly described below
(a) Solid – liquid Equilibrium
The equilibrium that exist between ice and water is an example of solid – liquid equilibrium. In a
close system, at 0 C ice and water attain equilibrium. At that point rate of melting of ice is equal
to rate of freezing of water. The equilibrium is represented as
   
2 2
s
H O H O
(b) Liquid-Gas Equilibrium
Evaporation of water in a closed vessel is an example of liquid – gas equilibrium. Where rate of
evaporation is equal to rate of condensation. The equilibrium is represented as
   
2 2 g
H O H O
(c) Solid – solution equilibrium
If you add more and more salt in water taken in a container of a glass and stirred with a glass rod,
after dissolving of some amount. You will find out no further salt is going to the solution and it
settles down at the bottom. The solution is now said to be saturated and in a state of equilibrium.

2. At this stage, many molecule of salt from the undissolved salt go into the solution (dissolution)
and same amount of dissolved salt are deposited back (Precipitation).
Thus, at equilibrium rate of dissolution is equal to rate of precipitation.
 
Salt(Solid) Salt in solution
(d) Gas –Solution equilibrium
Dissolution of a gas in a liquid under pressure in a closed vessel established a gas – liquid
equilibrium. The best example of this type of equilibrium is cold drink bottles. The equilibrium that
exists with in the bottle is
   
2 2
g
CO CO in solution
Equilibrium in Chemical Process (Reversible and irreversible reactions)
A reaction in which not only the reactants react to form the products under certain conditions but
also the products react to form reactants under the same conditions is called a reversible
reaction.
Examples are
(i) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
(ii) CaCO3(s) CaO(s) + CO2(g)
(iii) N2(g) + 3H2(g) 2NH3(g)
If a reaction cannot take place in the reverse direction, i.e. the products formed do not react to
give back the reactants under the same condition it is called an irreversible reaction.
Examples are:
(i) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(g)
(ii) 2M(g) + O2(g)  2MgO(s)
Note:
If any of the product will be removed from the system, reversible reaction will become
irreversible one.
CONCEPT OF CHEMICAL EQUILIBRIUM
Let we have a general reversible reaction,
A + B C + D
at time t = 0 a0 b0 0 0
as time will pass, A and B will be converted to C and D. As soon as C and D will be formed,
reaction will start in back direction also. A time will come when rate of decomposition of A and B
will be equal to the rate of formation of A & B, i.e. the rate of forward reaction will be equal to the
rate of reverse reaction. At this stage the reaction is said to be in a state of Chemical Equilibrium.
If we see variation of reactants and products with time on graph we will be having following graph.
When the equilibrium is reacted, the concentration of
reactant as well as product remains constant. At this stage
reaction seems to be stopped but in actual the reaction is
going in forward as well as reverse direction with same
rate and hence this equilibrium is called as dynamic
equilibrium.
Time
Rate
Forward Reaction
Reverse Reaction
Equilibrium point

3. Characteristics of Chemical Equilibrium
 The equilibrium is dynamic i.e. the reaction continues in both forward and reverse
directions.
 The rate of forward reaction equals to the rate of reverse reaction.
 The observable properties of the system such as pressure, concentration, density
remains invariant with time.
 The chemical equilibrium can be approached from either side.
 A catalyst can hasten the approach of equilibrium but does not alter the state of
equilibrium.
Types of Equilibria
There are mainly two types of equilibria:
(a) Homogeneous: Equilibrium is said to be homogeneous if reactants and products are in
same phase.
H2(g) + I2(g) 2HI(g)
N2(g) + 3H2(g) 2NH3(g)
N2O4(g) 2NO2(g)
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
(b) Heterogeneous: Equilibrium is said to be heterogeneous if reactants and products are in
different phases
CaCO3(s) CaO(s) + CO2(g)
NH4HS(s) NH3(g) + H2S(g)
NH2CO2NH4(s) 2NH3(g) + CO2(g)
Note:
To write the equilibrium constant expression, the concentration of pure liquid and pure
solid assumed to be unity, as the concentration of such substances remain constant, i.e.,
concentration = mole/litre  density.
LAW OF MASS ACTION
Guldberg and Waage established a relationship between rate of chemical reaction and the
concentration of the reactants or, with their partial pressure in the form of law of mass action.
According to this law, “The rate at which a substance reacts is directly proportional to its active
mass and rate of a chemical reaction is directly proportional to product of active masses of
reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in
the balanced chemical equation”.
aA bB cC dD
  
rate of reaction  [A]a.[B]b
rate of reaction = K[A]a[B]b
where K is rate constant or velocity constant of the reaction at that temperature.
Unit of rate constant (K)  
1 n 1
moles/lit time
 
 (where n is order of reaction.)
Note:
For unit concentration of reactants rate of the reaction is equal to rate constant or specific
reaction rate.
At equilibrium the rate of both forward and backward reactions become equal and after
achieving equilibria, the concentration of reactants and products remains constant as
shown in figure.

4. Time
Concentration
Mole/litre
Equilibrium
Reactant
Product
Time
Rate
Equilibrium
Forward Rate
Backward Rate
Note:
Active mass is the molar concentration of the reacting substances actually participating in
the reaction.
Hence
Active mass =
number of moles
volume in litres
Active mass of solid is taken as unity.
Illustration 1. Calculate the partial pressure of each component in the following equilibria
N2(g) + 3H2(g) 2NH3(g)
Solution: N2(g) + 3H2(g) 2NH3(g)
at t = 0 a b 0
at equilibrium a – x b – 3x 2x
ntotal = a – x + b – 3x + 2x = (a + b – 2x)
Partial pressure:
2
2
N
N T T
T
n (a x)
P P P
n (a b 2x)

   
 
2
2
H
H T T
T
n (b 3x)
P P P
n (a b 2x)

   
 
3
3
NH
NH T T
T
n 2x
P P P
n (a b 2x)
   
 
Exercise 1.
Calculate the equilibrium pressure of each reactant of given reaction with the help of
following data.
A + B C + D
Initial pressure of A and B are 6 atm and 5 atm. The equilibrium pressure of C and D
are 2 atm in each case.
Law of Chemical Equilibrium
According to this law, the ratio of product of concentration of products to the product of
concentration of reactants, with each concentration term is raised to the power by its coefficient in
overall balanced chemical equation, is a constant quantity at a given temperature and it is called
equilibrium constant.
Derivation of law of chemical equilibrium
Let us consider for the following equilibrium
aA + bB cC + dD

5. then, from Law of mass action
rate of forward reaction r1  [A]a [B]b
or r1 = Ka [A]a [B]b and rate of reverse reaction r2  [C]c [D]d
r2 = K2 [C]c [D]d
at equilibrium, r1 = r2
 K1 [A]a[B]b = K2 [C]c[D]d

c d
1
a b
2
K [C] [D]
K [A] [B]

Kc = K1/K2, an equilibrium constant in terms of active masses of reacting species.
For the reaction
SO2Cl2 SO2 + Cl2
at t = 0 a 0 0
at equilibrium a  x x x
equilibrium conc.
a x
V
 x
V
x
V
(V is volume of container)
So, 2 2 2 2
a x x
[SO Cl ] ;[SO ] [Cl ]
V V

  
So,
2
c
x x
x
V V
K
a x (a x)V
V
   

   
   
 
 
 
 
 

2
c
x
K
(a x) V

 
CHARACTERISTICS OF EQUILIBRIUM CONSTANT (Kc)
(i) Kc for a particular reaction at given temperature has a constant value.
(ii) Value of Kc always depends on nature of reactants and the temperature, but independent of
presence of catalyst or, of inert material.
(iii) Its value is always independent of the initial concentration of reactants as well as the
products.
(iv) The value of Kc indicates the proportion of products/product formed at equilibrium. Large Kc
value means large proportions of product.
(v) When the reaction is reversed, equilibrium constant for reverse reaction will also be inversed.
c
c
1
K
K
 
Let us have
A + B C + D
c
[C][D]
K
[A][B]

By reversing the reaction,
C + D A + B
c
c
[A][B] 1
K
[C][D] K
  
(vi) If the coefficients of reactants of products are halved or, doubled then accordingly, value of
Kc will change.
A + B C + D
c
[C][D]
K
[A][B]

for 2A + 2B 2C + 2D

6. 2
2 2
2
c c
2 2
[C] [D] [C][D]
K K
[A][B]
[A] [B]
 
   
 
 
(vii) When a number of equilibrium reactions are added, the equilibrium constant, for overall
reaction is the product of equilibrium constants of respective reactions.
N2(g) + O2(g) 2NO;
2
1
2 2
[NO]
K
[N ][O ]

N2O(g) N2(g) + 1/2O2(g);
1/ 2
2 2
2
2
[N ][O ]
K
[N O]

N2O(g) + 1/2O2(g) 2NO(g);
2
3 1/ 2
2 2
[NO]
K
[N O][O ]

2
3 2
1/ 2 1/ 2
2 2 2 2 2
K [N O]
[NO]
K [N O][O ] [N ][O ]
 
2
1
2 2
[NO]
K
[N ][O ]
 
i.e. 3
1
2
K
K
K

Equilibrium constant in terms of partial pressures
Let us consider a general reaction
aA + bB cC + dD
PA PB PC PD
From law of mass action
       
a b a b
1 a B 1 1 A B
r P P r K P P
  
       
c d c d
2 C D 2 2 C D
r P P r K P P
  
at equilibrium
r1 = r2 or K1 (PA)a (PB)b = K2(PC)c(PD)d
   
   
a d
C D
1
a b
2 A B
P P
K
K P P
 
   
   
c d
C D
1
p a b
2 A B
P P
K
K
K P P
  
RELATION BETWEEN KP AND KC
For a general equation,
aA + bB cC + dD
Where, a , b, c and d are coefficients of the reacting substance
 
   
d
c
C D
p a b
A B
(P ) P
K
P P

From gas equation
PV = nRT 
n
P RT
V

So, PA = [A]RT’ PB = [B]RT
PC = [C]RT; PD = [D]RT
Hence,
   
   
c d
p a b
[C]RT [D]RT
K
[A]RT [B]RT



or,   
c d
(c d) (a b)
p a b
[C] [D]
K RT
[A] [B]
  
 

7. Where ng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.
Hence, g
n
p c
K K (RT)

 
When n = 0; Kp = Kc
n  0; Kp  Kc
n  0; Kp  Kc
Note:
Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its
relation with Kp and Kc.
Illustration 2. At 27C Kp value for reaction      
3 2
CaCO s CaO s +CO g is 0.1 atm,
calculate its Kc value.
Solution: KP = Kc(RT)n
n = 1
Kc = P
K 0.1
RT 0.82 300


= 410-3
Illustration 3. Solid NH4I dissociates according to the reaction at 400 K
NH4I(s) NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI
dissociates in H2 and I2 as 2HI H2 + I2. If partial pressure of H2 at this
temp is 1 atm in the container when both the equilibrium exist simultaneously,
calculate Kp value of second equilibrium (for the dissociation of HI).
Solution: NH4I(s) NH3 (g) + HI (g) 1
P
K 16

2HI H2 + I2
P - x
x
2
x
2 2
P
K ?

Here
x
2
= 1 atm
 x = 2 atm
P(P -2) = 16
 P = 5.1
Illustration 4. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is
converted into CO on addition of graphite. What is the value of Kp if the total
pressure at equilibrium is 0.8 atm?
Solution: CO2(g) + C(s)  2CO(g)
0.5 2n
or, 0.5-n 0.5+n = 0.8
n = 0.3
Kp =
 
 
2
2 0.3
1.8 atm
0.5 0.3



Illustration 5. Given that the equilibrium constant for the reaction, H2(g) + I2(g) 2HI(g) is 50 at
700K. Calculate the equilibrium constant for the reaction.
HI(g) 2(g) 2(g)
1 1
H I
2 2

Solution: We have H2(g) + I2(g) 2HI(g)

8. Kc =
2
2 2
[HI]
50
[H ][I ]

The new equilibria is
HI(g) 2(g) 2(g)
1 1
H I
2 2

1/ 2 1/ 2
2 2
c
c
[H ] [I ] 1
K
[HI] K
  
1 2
0.141
10
50
  
Illustration 6. One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is
converted into NH3 by the following reaction
N2(g) + 3H2(g) 2NH3(g). Calculate Kc and Kp of the reaction.
(Temperature = 227°C and R = 0.08231).
Solution: We have N2(g) + 3H2(g) 2NH3(g)
Percentage N2 reacted, 25%
 x = 0.25
Now, (a – x) = 1 – 0.25 = 0.75
b – 3x = 3 – 0.75 = 2.25
(a x) 0.75
0.1875
V 4

  
b 3x 2.25
0.5625
V 4

 
 
 
 
2x 0.50
0.125
V 4
 
 
 
 

2 2
3
c 3 3
2 2
[NH ] (0.125)
K
[N ][H ] (0.1875)(0.5625)
 
= 1.48  10–5 L2 mol–2
Now, Kp = n 5 2
c
K (RT) 1.48 10 [0.0821 (227 273)]
  
     
 
5
2
1.48 10
0.082 (500)



= 8.78  10–9
Exercise 2.
The equilibrium constant for the reaction
PCl5(g) PCl3(g) + Cl2(g) is 2.4  10–3
. Determine the equilibrium constant for the
reverse reaction at same temperature.
Exercise 3.
At 21.5°C and a total pressure of 0.0787 atm, N2O4 is 48.3% dissociated into NO2.
Calculate Kp and Kc for the reaction.
N2O4(g) 2NO2(g)

9. Exercise 4.
(i) Write the expression for Kp and Kc for the following reactions.
N2 + 3H2 2NH3
(ii) Find out the heterogeneous equilibrium from the following reaction
(A) CaCO3 (s) CaO(s) + CO2 (g)
(B) H2(g) + I2(g) 2HI (g)
Exercise 5.
What is the value of equilibrium constant for the following reaction
If A D
A B K1 = 2
B C K2 = 5
C D K3 = 3
Exercise 6.
In an equilibrium A B C D
  ; A and B are mixed in a vessel at temperature T.
The initial concentration of A was twice the initial concentration of B. After the
equilibrium has established, concentration of C was thrice the equilibrium
concentration of B. Calculate Kc.
SIGNIFICANCE OF THE MAGNITUDE OF EQUILIBRIUM CONSTANT
(i) A very large value of KC or KP signifies that the forward reaction goes to completion or
very nearly so.
(ii) A very small value of KC or KP signifies that the forward reaction does not occur to any
significant extent.
(iii) A reaction is most likely to reach a state of equilibrium in which both reactants and
products are present if the numerical value of Kc or KP is neither very large nor very
small.
Units of Equilibrium Constant
As we have learned that numerical value of equilibrium constant is a function of stoichiometric
coefficient used for any balanced chemical equation.
But, what will happen in the size of units of equilibrium constant when the stoichiometric
coefficient of the reaction changes? (If sum of the exponent for product is not equal to reactant
side sum)
So for the sake of simplicity in the units of Kp for Kc the relative molarity or pressure of reactants
and products are used with respect to standard condition. (For solution standard state 1 mole/litre
and for gas standard pressure = 1 atm)
Now the resulting equilibrium constant becomes unitless by using relative molarity and pressure.
Illustration 7. The value of Kp for the reaction 2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) is
0.035 atm at 400oC, when the partial pressures are expressed in atmosphere.
Calculate Kc for the reaction,
2
1
O (g) +2HCl(g)
2
Cl2(g) + H2O(g)
Solution: KP = KC (RT)n
n = moles of product  moles of reactants = 5  4 = 1

10. R = 0.082 L atm/mol K,
T = 400 + 273 = 673 K
 0.035 = KC (0.082  673)
KC = 6.342  10-4 mol l-1
 '
C
K for the reverse reaction would be
C
1
.
K
 '
C
K = 4
1
6.342 10

= 1576.8 (mol l-1)-1
When a reaction is multiplied by any number n (integer or a fraction) the '
C
K or
'
P
K becomes (KC)n or (KP)n of the original reaction.
 KC for
1
2
O2(g) + 2HCl(g) Cl2(g) + H2O(g)
is 1576.8  39.7 (mol.l-1
)-½
Illustration 8. Kp for the equilibrium, FeO(s) + CO(g) Fe(s) + CO2(g) at 1000oC is 0.4. If
CO(g) at a pressure of 1 atm and excess FeO(s) are placed in a container at
1000oC, what are the pressures of CO(g) and CO2(g) when equilibrium is
attained?
Solution: Assuming ideal gas behaviour, partial pressures are proportional to the no. of
moles present. Since moles of CO2 formed equals moles of CO consumed, the
drop in partial pressure of CO will equal the partial pressure of CO2 produced. Let
the partial pressure of CO2 at equilibrium be ‘x’ atm. Then, partial pressure of CO
will be (1 - x) atm.
Since 2
CO
p
CO
p x
K 0.4
p 1 x
  

 x = 0.286
Hence PCO = 1-x
= 0.714 atm.
Illustration 9. At 800 K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 5
mole of SO3 at equilibrium. Kc for the equilibrium 2SO2 + O2 2SO3 is 833 lit/
mole. If the volume of the container is 1 litre, calculate how much O2 is to be
added at this equilibrium in order to get 5.2 moles of SO3 at the same
temperature.
Solution: Suppose x mole of O2 is added by which equilibrium shifts to right hand side and
y mole of O2 changes to SO3.
The new equilibrium concentration may be
2 SO2 + O2 2SO3
Moles at I equilibrium 0.5 0.12 5
II equilibrium (0.5-2y) (0.12+x-y) (5+2y)
5 + 2y = 5.2
 y = 0.1
Kc =
 
   
 
   
2 2
3
2 2
2 2
SO 5 2y
0.5 2y 0.12 x y
SO O


  

11. = 833
Substituting y = 0.1 (V = 1 lit)
We get x = 0.34 mole
Exercise 7.
Ammonium hydrogen sulphide dissociates as follows
NH4HS(s) NH3(g) + H2S(g)
If solid NH4HS is placed in an evacuated flask at certain temperature it will dissociate
until the total pressure is 600 torr.
(a) Calculate the value of equilibrium constant for the dissociation reaction
(b) Additional NH3 is introduced into the equilibrium mixture without changing the
temperature until partial pressure of NH3 is 750 torr, what is the partial pressure of
H2S under these conditions? What is the total pressure in the flask?
Exercise 8.
At 700 K hydrogen and bromine react to form hydrogen bromide. The value of
equilibrium constant for this reaction is 5  108
. Calculate the amount of H2, Br2 and
HBr at equilibrium if a mixture of 0.6 mole of H2 and 0.2 mole of Br2 is heated to 700 K.
THE REACTION QUOTIENT ‘Q’
Consider the equilibrium
PCl5 (g) PCl3(g) + Cl2 (g)
At equilibrium
   
 
2 3
5
Cl PCl
PCl
= KC. When the reaction is not at equilibrium this ratio is called ‘QC’
i.e., QC is the general term used for the above given ratio at any instant of time. And at
equilibrium QC becomes KC.
Similarly, 2 3
5
Cl PCl
PCl
P P
P
is called QP and at equilibrium it becomes KP.
 If the reaction is at equilibrium, Q = Kc
 A net reaction proceeds from left to right (forward direction) if Q  KC.
 A net reaction proceeds from right to left (the reverse direction) if Q Kc
Illustration 10. For the reaction,
A(g) + B(g) 2C(g) at 25°C, in a 2 litre vessel contains 1, 2, 3 moles of
respectively. Predict the direction of the reaction if
(a) Kc for the reaction is 3
(b) Kc for the reaction is 6
(c) Kc for the reaction is 4.5
Solution: A(g) + B(g) 2C(g)
Reaction quotient Q =
2
2
3
[C] 9
2
4.5
1 2
[A][B] 2
2 2
 
 
 
  

(a) c
Q K ,
  therefore backward reaction will be followed
(b) c
Q K


12. The forward reaction is followed
(c) Q = Kc
 The reaction is at equilibrium
Illustration 11. For the reaction : A(aq)+ B(aq) C(aq) +D(aq) , the net rate of consumption of B at
25C and at any time 't' is as given below
 
d B
-
dt
= {410-4[A] [B] – 1.3310-5 [C] [D]} mol L-1 min-1
Predict whether the reaction will be spontaneous in the direction as written in
reaction mixture in which each A, B, C and D is having a concentration of
1 mol L-1?
Solution: K = 1
2
K
K
4
5
4 10
30
1.33 10



 

Q =
  
  
C D
A B
1 1
1 1



= 1 < K
Since Q < K, so the above reaction is spontaneous in the forward direction.
Exercise 9.
The equilibrium constant is 0.403 at 1000K for the process
FeO(s) + CO(g) Fe(s)+ CO2)g)
A steam of pure CO is passed over powdered FeO at 1000K until equilibrium is
reached. What is the mole fraction of CO in the gas stream leaving the reaction
zone?
LE CHATELIER’S PRINCIPLE
“When an equilibrium is subjected to either a change in concentration, temperature or, in external
pressure, the equilibrium will shift in that direction where the effects caused by these changes are
nullified”.
This can be understood by the following example. Overall we can also predict the direction of
equilibrium by keeping in mind following theoretical assumption.
PCl5 PCl3 + Cl2
Let us assume that we have this reaction at equilibrium and the moles of Cl2, PCl3 and PCl5 at
equilibrium are a, b and c respectively, and the total pressure be PT.

T T
P
T
a b
P P
a b c a b c
K
c
P
a b c
   
 
   
   
   

 

 
 
 
= T
abP
c(a b c)
 
Since PT =
(a b c)RT
V
 
 KP =
abRT
cV

13. Now if d moles of PCl3 is added to the system, the value of Q would be,
a(b d)RT
cV

We can see that this is more than KP. So the system would move reverse to attain equilibrium.
(i) If we increase the volume of the system, the Q becomes
abRT
cV'
where V > V.
 Q becomes less, and the system would move forward to attain equilibrium.
(ii) If we add a noble gas at constant pressure, it amounts to increasing the volume of the system
and therefore the reaction moves forward.
(iii) If we add the noble gas at constant volume, the expression of Q remains as Q =
abRT
cV
and
the system continues to be in equilibrium.
 nothing happens.
(iv) Therefore for using Le-Chatlier’s principle, convert the expression of KP and KC into basic
terms and then see the effect of various changes.
Effect of Concentration
Let us have a general reaction,
aA + bB cC + dD
at a given temperature, the equilibrium constant,
c d
c a b
[C] [D]
K
[A] [B]

again if , ,  and  are the number of mole of A, B, C and D are at equilibrium
then,
c d
c a b
[ ] [ ]
K
[ ] [ ]
 

 
If any of product will be added, to keep the Kc constant, concentration of reactants will increase
i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant
side will be done, change in product’s concentration will take place to minimise the effect.
Temperature Effect
The effect of change in temperature on an equilibrium cannot be immediately seen because on
changing temperature the equilibrium constant itself changes. So first we must find out as to how
the equilibrium constant changes with temperature.
For the forward reaction, according to the Arrhenius equation,
af
E
RT
f f
k A e


And for the reverse reaction,
ar
E
RT
r r
k A e



ar af
(E E )
f f RT
r r
k A
K e
k A

 

 
ar af
1
1
E E
RT
f
T
r
A
K e
A


 
ar af
2
2
E E
RT
f
T
r
A
K e
A


It can be seen that
H ve
  
f b
H E E
  
E
Forward
E
Backward
Reaction Coordinate
Energy
P
R
Exothermic

14.  
ar af
2 2 1
1
E E 1 1
T R T T
T
K
e
K
  

 
 
 
 or
 
2
1
T ar af
T 2 1
K E E 1 1
ln
K R T T
  
 
 
 
For any reaction, H = Eaf  Ear
2
1
T
T
K
ln
K
=
H
R

2 1
1 1
T T
 

 
 
or 2
1
T
T
K
ln
K
=
H
R

1 2
1 1
T T
 

 
 
From the equation,
o
2
1 1 2
K H 1 1
log
K 2.303R T T
 

 
 
 
it is clear that
H ve
  
f b
H E E
  
E
Forward
E
Backward
Reaction Coordinate
Energy
P
R
Endothermic
R = Reactant, P = Product
(a) If Ho is +ve (endothermic), an increase in temperature (T2 > T1) will make K2 > K1, i.e., the
reaction goes more towards the forward direction and vice-versa.
(b) If Ho is -ve (exothermic), an increase in temperature (T2 > T1), will make K2 < K1 i.e., the
reaction goes in the reverse direction.
(i) Increase in temperature will shift the reaction towards left in case of exothermic reactions
and right in endothermic reactions.
(ii) Increase of pressure (decrease in volume) will shift the reaction to the side having fewer
moles of the gas; while decreases of pressure (increase in volume) will shift the reaction
to the side having more moles of the gas.
(iii) If no gases are involved in the reaction higher pressure favours the reaction to shift
towards higher density solid or liquid.
Illustration 12. Under what conditions will the following reactions go in the forward direction?
(i) N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.
(ii) 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.
(iii) N2(g) + O2(g) 2NO(g) - 43.2 k cal.
(iv) 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal.
(v) C(s) + H2O(g) CO2(g) + H2(g) + X k cal.
(vi) PCl5(g) PCl3(g) + Cl2(g)- X k cal.
(vii) N2O4(g) 2NO2(g) - 14 k cal.
Solution: (i) Low T, High P, excess of N2 and H2.
(ii) Low T, High P, excess of SO2 and O2.
(iii) High T, any P, excess of N2 and O2
(iv)Low T, High P, excess of NO and O2
(v) Low T, Low P, excess of C and H2O
(vi) High T, Low P, excess of PCl5
(vii) High T, Low P, excess of N2O4.
Illustration 13. The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g)
is 1.6  10-4 atm-2 at 400oC. What will be the equilibrium constant at 500oC
if heat of the reaction in this temperature range is 25.14 k cal?
Solution: Equilibrium constants at different temperature and heat of the reaction are
related by the equation, 2
1
P
P
K
ln
K
=
o
H
R

1 2
1 1
T T
 

 
 
2.303 log 2
1
P
P
K
K
=
o
H
R
 2 1
1 2
T T
T T
 

 
 

15. 2
p
log K =
25140 773 673
2.303 2 773 673
 
 
 
 
 
+ log 1.64  10-4
log 2
P
K 4.835
 
 2
5 2
P
K 1.462 10 atm
 
 
Effect of change of Pressure
The effect of change of pressure on chemical equilibrium can be done by the formula.
( n)
p p
K Q P 
 
Case A: When n = 0
Kp = Qp
And equilibria is independent of pressure.
Case B: When n = – ve
p
n
p p n
Q
K Q P
P


  
Here with increase in external pressure, will shift the equilibrium towards forward
direction to maintain Kp. Similarly, decrease in pressure will shift the equilibrium
in the reverse direction.
Case C: When n = +ve
n
p p
K Q P
  and, effect will be opposite to that of Case ‘B’
Exercise 10.
N2 + 3H2 2NH3;  H = negative
What are the conditions of temperature and pressure favorable for this reaction?
EFFECT OF CATALYST ON EQUILIBRIUM
Since the catalyst is associated with forward as well as reverse direction reaction. So, at
equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst
effect will be same on both forward as well as reverse. Hence catalyst never effect the point of
equilibrium but it reduces the time to attain the equilibrium.
Effect on equilibrium due to the addition of inert gases
Following are the cases, where this effect in different manner.
Case A: When n = 0 and total volume change at equilibrium remain constant.
So, addition of inert gases will increase the number of moles in the mixture
but partial pressure of each component remains constant. Hence equilibrium
remains unaffected.
Case B: When n = +ve and VT  0
The total volume will increase with addition of inert gases. This will shift the
equilibrium in forward direction.
Case C: When n = –ve and VT  0
In this case, the addition of inert gases will increase the total volume and the reaction will shift in
reverse direction.
Dependence of KP or Kc on Temperature
With the increase of temperature, equilibrium favours forward reaction in case of an endothermic
(H > 0) reaction while it favours backward reaction in the case of an exothermic (H < 0)
reaction.

16. Illustration 14. For the equilibrium
NH4I(g) NH3(g) + HI(g) H ve
  
What will be the effect on the equilibrium constant on increasing the
temperature.
Solution: Since the forward reaction is endothermic, so increasing the temperature, the
forward reaction is favoured. Thereby the equilibrium constant will increase.
Illustration 15. Kp for the reaction 2BaO2(s) 2BaO(s) + O2(g) is 1.6  10–4 atm, at 400°C.
Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas
produced at 500°C temperature, if it is carried in 2 litre reaction vessel?
Solution: We know that
2
1
p
p 1 2
K H 1 1
log
K 2.303R T T
 

 
 
 
2
p
4 3
K 25.14 773 673
log
773 673
1.6 10 2.330 2 10
 
 
 
  

    
2
5
p
K 1.46 10 atm

  
2 2
5
p o
K p 1.46 10 atm

   
Since, PV = nRT
5
1.46 10 2 n 0.0821 773

    
5
1.46 10 2
n
0.0821 773

 
 

= 4.60  107
Exercise 11.
Equilibrium constant Kp for the reaction
3
2
H2(g) +
1
2
N2(g) NH3(g) are 0.0266 and
0.0129 at 310°C and 400°C respectively. Calculate the heat of formation of
gaseous ammonia.
Standard Free Energy Change of a Reaction and its Equilibrium Constant
Let G0 be the difference in free energy of the reaction when all the reactants and products are in
the standard state and Kc or, Kp be the thermodynamic equilibrium constant of the reaction. Both
related to each other at temperature T by the following relation.
G0 = – 2.303 RT logKc and G0 = – 2.303 RT log Kp (in case of ideal gases)
Now, we know that thermodynamically,
G0 = H0 – TS0
here H0 is standard enthalpy of reaction, and S0 is standard entropy
change
(i) When G0 = 0, then, Kc = 1
(ii) When, G0  0, i.e. +ve, then Kc  1, in this case reverse reaction is feasible showing
thereby a less concentration of products at equilibrium rate.
(iii) When G0  0, i.e. ve, then, Kc  1; In this case, forward reaction is feasible showing
thereby a large concentrations of product at equilibrium state.
Illustration 16. A system is in equilibrium as
PCl5 + Heat PCl3 + Cl2
Why does the temperature of the system decreases, when PCl3 are being
removed from the equilibrium mixture at constant volume?

17. Solution: When PCl3 are being removed from the system, the reaction moved to the right.
This consumed heat and therefore, temperature is decreased.
Illustration 17. NO and Br2 at initial partial pressures of 98.4 and 41.3 torr, respectively, were
allowed to react at 300K. At equilibrium the total pressure was 110.5 torr.
Calculate the value of the equilibrium constant and the standard free energy
change at 300K for the reaction 2NO(g) + Br2(g) 2NOBr(g).
Solution: 2NO(g) + Br2(g) 2NOBr(g)
Initial pressure 98.4 41.3 0
At eqlbm. 98.4-x 41.3 -
x
2
x
The total pressure at equilibrium is 110.5 torr.
 98.4-x + 41.3-
x
2
+ x = 110.5
x = 58.4 torr
Now, 1 atm = 760.4 torr,
 x = 7.68  10-2 atm.
pNOBr = 7.68  10-2 atm ; pNO = 98.4-x = 40 torr = 5.26  10-2 atm.
pBr2
= 41.3 -
x
2
= 12.1 torr = 1.59  10-2 atm.
 
   
2
2
NOBr
p 2
NO Br
p
K
p p
 
 

   
   
=
 
   
2 2
2 2 2
7.68 10
134
5.26 10 1.59 10

 


 
atm-1
Go = -2.303 RT log K = -2.303 (1.99) (300) (log 134)
= 2.92 k cal = 12.2 kJ.
[If R is used as 1.99 cal/mol K, then Go will be in cal. If R is used as 8.314 J/mol
K, then Go will be in Joules. But KP must always be in (atm)n
.]
Exercise 12.
What is the value of R if value of GO
is in?
(i) Joules (ii) Calories
Exercise 13.
N2O4 is 25% dissociated at 27C and 1 atm pressure.
Calculate (i) Kp and (ii) the percentage dissociation at 0.1 atm and 27C.
Dependence of Degree of Dissociation on Density Measurements
The degree of dissociation of a substance is defined as the fraction of its molecules dissociating
at a given time. The following is the method of calculating the degree of dissociation of a gas
using vapour densities. This method is valid only for reactions whose KP exist, i.e., reactions
having at least one gas and having no solution.
Since PV = nRT
PV =
w
RT
M
M =
wRT RT
PV P



18.  VD =
RT
2P

Since P =
nRT
V
 VD =
RT V
V
2nRT 2n
 
 
For a reaction at equilibrium V is a constant and  is a constant.  vapour Density
1
n


Total moles at equilibrium vapour density initial D
Initial total moles vapour density at equilibrium d
  =
M
m
( molecular weight = 2  V.D)
Here M = molecular weight initial
m = molecular weight at equilibrium
Let us take a reaction
PCl5 PCl3 + Cl2
Initial moles C 0 0
At eqb. C(1) C C

C(1 ) D D
; 1
C d d
 
    =
M
m
Knowing D and d,  can be calculated and so for M and m.
Illustration 18. When PCl5 is heated it dissociates into PCl3 and Cl2. The density of the gas
mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of
dissociation at 200oC and 250oC.
Solution: PCl5(g) PCl3(g) + Cl2(g)
We are given the vapour densities at equilibrium at 200oC and 250oC.
The initial vapour density will be the same at both the temperatures as it would
be 5
PCl
M
.
2
 Initial vapour density =
 
31 5 35.5
104.25
2
 

Vapour density at equilibrium at 200oC = 70.2

Total moles at equilibrium Vapour density initial
1
Total moles initial Vapour density at equilibrium
   
=
104.25
1.485
70.2

  = 0.485
At 250oC, 1 +  =
104.25
1.8
57.9

  = 0.8
Illustration 19. The degree of dissociation of PCl5 is 60%, then find out the observed molar mass
of the mixture.
Solution: PCl5(g) PCl3(g) + Cl2(g)
Initially moles 1 0 0
At eqm. 1 –   

19. Where  = degree of dissociation = 0.6
Total moles at equilibrium = 1 +  = observed mole
observed moles molecular weight theoretical
Theoretical moles molecular weight observed

1 206.5
1 molecular weight observed
 
 
206.5 206.5
molecular weight observed
1 1.6
  
 
= 129.06
Exercise 14.
Vapour density of N2O4 which dissociates according to the equilibrium
N2O4(g) 2NO2(g) is 25.67 at 100°C and a pressure of 1 atm. Calculate the degree of
dissociation and Kp for the reaction.
Exercise 15.
How much PCl5 must be added to a one litre vessel at 250o
C in order to obtain a
concentration of 0.1 mole of Cl2? Kc for 5 3 2
PCl →PCl + Cl is 0.0414 mol/litre.

20. ANSWERS TO EXERCISES
Exercise 1:
A 
 4 atm
B 
 3 atm
Exercise 2:
2
4.16 10

Exercise 3:
p
K 0.0956 atm

3
c
K 3.96 10 mole/lit

 
Exercise 4: (i) Kc =
2
3
3
2 2
[NH ]
[N ] [H ]
Kp = 3
2 2
2
NH
3
N H
P
P P
 
 
   
   
(ii) (A) is a heterogeneous equilibrium.
Exercise 5:
K = 30
Exercise 6:
1.8
Exercise 7:
(a) 0.155 atm2
(b) 119.32 torr, 869.32 torr
Exercise 8:
0.4, 0, 0.4
Exercise 9:
0.713
Exercise 10:
Low temperature, high pressure and high concentration of N2 and H2.
Exercise 11.
12140 cals

Exercise 12:
(i) 8.314 J mole-1K-1
(ii) 2 cal mole-1K-1
Exercise 13:
(i) 0.266 atm
(ii) 93.27%
Exercise 14:
p
0.792
K 6.73 atm

Exercise 15: 0.3415 mole

21. MISCELLANEOUS EXERCISES
Exercise 1: For an equilibrium, the rate constant for the forward and the backward reaction
are 2.38  104 and 8.15  105 respectively. Calculate the equilibrium constant
for the reaction.
Exercise 2: In an experiment starting with 1 mole C2H5OH, 1 mole CH3COOH and 1 mole of
water, the equilibrium mixture on analysis shows that 54.3% of the acid is
esterified. Calculated Kc.
Exercise 3: From the values of the equilibrium constants, indicate in which case, does the
reaction go farthest to completion: K1 = 1010, K2 = 1010, K3 = 105.
Exercise 4: The value of the equilibrium constant is less than zero. What does it indicate?
Exercise 5: Would you expect equilibrium constant for the reaction    
2
I g 2I g to
increase or decrease as temperature increases? Assign reason.
Exercise 6: Acetic acid is highly soluble in water but still a weak electrolyte. Why?
Exercise 7: Which of the following reactions involve homogenous and which involve
heterogenous equilibria?
(a)      
2 2 2
2N O g 2N g +O g
(b)      
3 2 2
2NH g N g + 3H g
(c)          
3 2 2
2
2Cu NO s 2CuO s + 4NO g +O g
(d)        
3 2 5 2 3 2 5
CH COOC H + H O CH COOH +C H OH
(e)        
3+ -
3
Fe aq + 3OH aq Fe OH s
Exercise 8: For the reaction,        
4 2 2 2
CH g + 2H S g CS g + 4H g at 1173 K, the
magnitude of Kc is 3.6. For each of the following composition, decide whether the
reaction mixture is at equilibrium. If not, decide to which direction the reaction
should go?
(a)        
4 2 2 2
CH =1.07M, H S =1.20M, CS = 0.90M, H =1.78M
(b)        
4 2 2 2
CH =1.45M, H S =1.29M, CS =1.25M, H =1.75M
Exercise 9: Kp for the reaction      
2 2
H g +I g 2HI g at 460C is 49. If the partial
pressure of H2 and I2 is 0.5 atm respectively, determine the partial pressure of
each gas in the equilibrium mixture.
Exercise 10: 0.16 g N2H4 (hydrazine) are dissolved in water and the total volume is made upto
500 mL. Calculate the percentage of N2H4 that has reacted with water in solution.
The Kb for N2H4 is 4.0  106 M.

22. ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1: 2.92
Exercise 2: Kc = 4.012
Exercise 3: High value of equilibrium constant (K) means that the molar concentration of the
products is quite large. In other words, the reaction has proceeded to a large
extent in the forward direction before attaining the equilibrium. Thus for K2 = 1010,
the forward reaction has proceeded maximum before attaining the equilibrium.
Exercise 4: This indicates that the forward reaction has proceeded only to a small extent
before the equilibrium is attained.
Exercise 5: In the forward reaction, energy is needed to bring about the dissociation of I2(g)
molecules. This means that the forward reaction is endothermic in nature. The
increase in temperature will favour the forward reaction. Therefore, the
equilibrium constant will increase with rise in temperature.
Exercise 6: The strength of the electrolyte does not depend upon its amount present in
solution but on its ionization in solution. Since acetic acid is a weak acid (organic
acid), it is ionized only to small extent. Therefore, it is a weak electrolyte.
Exercise 7: Homogenous equilibria : (a), (b) (d)
Heterogenous equilibria : (c) and (e)
Exercise 8: (a) Backward direction
(b) Backward direction
Exercise 9: Partial pressure of H2 (g) = 0.112 atm
Partial pressure of I2(g) = 0.112 atm
Partial pressure of HI (g) = 0.776 atm
Exercise 10: 2.0%

23. SOLVED PROBLEMS
Subjective:
Board Type Questions
Prob 1. At 444°C, 15g mole of hydrogen are mixed with 5.2g mole of I2 vapour. When
equilibrium was established, 10g mole of HI was formed. Calculate the equilibrium
constant for the reaction.
Sol. 2(g) 2(g)
H I 2HI

a b 0
a – x b – x 2x
So,
2
2
c
2 2
2x
[HI] V
K
(a x) (b x)
[H ][I ]
V V
 
 
 
 
 

=
2 2
4x 4 5
(a x)(b x) 10 0.2


  
= 50
Prob 2. The degree of dissociation is 0.4 at 400K and 1 atm for the gaseous reaction
5 3 2
PCl PCl Cl

Assuming ideal behaviour of all the gases, calculate the density of equilibrium
mixture at 400K and 1 atm pressure.
(Atomic mass of P = 31 and Cl = 35.5)
Sol. 5 3 2
PCl PCl Cl

at t = 0 1 0 0
at eqm. 1 – 0.4 0.4 0.4
Total no. of moles after dissociation = 0.6 + 0.4 + 0.4
Now,
nRT 1.4 0.082 400
V 45.92litre
P 1
 
  
So, density =
Molecular weight 208.5
4.54g/litre
Volume 45.92
 
Prob 3. When 3.06g of solid NH4HS is introduced into a 2 litre evacuated flask at 27°C, 30%
of the solid decomposed into gaseous ammonia and hydrogen sulphide.
(i) Calculate Kc for reaction at 27°C.
(ii) What would happen to the equilibrium when more solid NH4HS is introduced into
the flask?
Sol. Moles of NH4HS introduced into flask =
3.06
0.06
51

4 (s) 3(g) 2 (g)
NH HS NH H S

0.06 (1 – x) 0.06x 0.06x
as x = 30%, so, x = 0.3
So, 0.06  0.7 0.060.3 0 0.6  0.3
So, c 3 2
K [NH ][H S]

2
0.018 0.018 3.24 10
   
Addition of NH4HS(s) does not change position of equilibrium.

24. Prob 4. Determine Kc for the reaction ½N2(g) + ½O2(g) + ½Br2(g) NOBr(g) from the
following information (at 298oK)
Kc = 2.4  1030 for 2NO(g) N2(g) + O2(g) ;
'''
C
K = 1.4 for NO(g) + ½Br2(g) NOBr(g)
Sol. N2(g) + O2(g) 2NO (g)
Kc
/ = 30
c
1 1
K 2.4 10


(i) ½N2(g) + ½O2(g ) NO(g)
Kc
// = / 30 15
c
1
K 10 0.6455 10
2.4
 
   
(ii) NO(g) + ½Br2(g) NOBr(g) '''
C
K = 1.4.
(i) + (ii) gives the net reaction:
½N2(g) + ½O2(g) + ½Br2(g) NOBr(g)
K =
 
   
 
   
 
     
1 1 1 1 1 1
2 2 2 2 2 2
2 2 2 2 2 2
NO NOBr NOBr
N O NO Br N O Br
 
= ''
C
K  '''
C
K = 0.6455  10-15  1.4 = 9.037  10-16
Prob 5. The heat of reaction at constant volume for an endothermic reaction in equilibrium is
1200 cal more than at constant pressure at 300K. Calculate the ratio of equilibrium
constant Kp and Kc.
Sol. Given that E H 1200 cal
   
Also we have H E nRT
    
 nRT 1200
  
or  
1 1
1200
n 2 R 2calK mol
2 300
 

    

Now,  
n
p c
K K RT


 
2
p 3
c
K
0.0821 300 1.648 10
K
 
   
IIT Level Question
Prob 6. Anhydrous CaCl2 is often used as dessicant. In presence of excess of CaCl2, the
amount of water taken up is governed by Kp = 1.28  1085 for the following reaction at
room temperature, CaCl2(s) + 6H2O(g) CaCl2.6H2O(s)
What is the equilibrium pressure of water in a closed vessel that contains CaCl2(s)?
Sol. 2(s) 2 (g) 2 2 (s)
CaCl 6H O CaCl .6H O

Here,
 
2
85
p 6
H O
1
K 1.28 10
p
  
 
2
6
85
H O
1
p 10
1.28

 
2
1
6
85
H O
1
p 10
1.28

 
  
 
 
2
15
H O
p 6.54 10 atm

 

25. Prob 7. NO2(g) NO(g) + 1/2O2(g)
Consider the above equilibrium where one mole of NO2 is placed in a vessel and
allowed to come to equilibrium at a total pressure of 1 atm. Experimental analysis
shows that
Temperature 700K 800K
2
NO NO
p /p 0.872 2.50
(a) Calculate Kp at 700K and 800K
(b) Calculate G0
Sol. (a) NO2(g) NO(g) +
1
2
O2(g)
at t = 0 1 mole 0 0
t eq. 1 – 2x 2x x
partial T
1 2x
P
1 x

 
 

 
T
2x
P
1 x
 
 

 
T
x
P
1 x)
 
 

 
at equilibrium pressure.
Given that
2
NO
NO
p
0.872
p
 at 700K
T
T
2x
P
1 x
0.872
1 2x
P
1 x
 
 

 
 

 
 

 
2x
0.872
1 2x
 

 x = 0.2329
 2
O T
x
p p 0.1889
1 x
 

 
1 2
2
x / 2
NO
p O
NO
p
K p
p
  
= 1/ 2
0.872 (0.1889)
 = 0.372 atm1/2
Similarly 2
p
K can be calculated
(b) 0
p
G RTlogK
  
2.303 8.314 700log0.379
   
= 5.65kJ/mol
Prob 8. At 25°C and 1 atm, N2O4 dissociates by the reaction
N2O4(g) 2NO2(g)
If it is 35% dissociated at given condition, find
(i) The percent dissociation at same temperature if total pressure is 0.2 atm.
(ii) The percent dissociation (at same temperature) in a 80g of N2O4 confined to a 7
litre vessel.
(iii) What volume of above mixture will diffuse if 20 ml of pure O2 diffuses in 10
minutes at same temperature and pressure?
Sol. N2O4(g) 2NO2(g)
t equilibrium 1 –  2
Partial pressure at equilibrium T T
1 2
P P
1 1
  
   
   
   
   

26. 2
2 4
2 2
NO
p T
2
N O
p 4
K p
p 1

  
 
=
2
2
4 (0.35)
1
1 (0.35)



= 0.56 atm
(i)
2
p 2
4
K P
1

 
 
2
2
4
0.56 0.2
1

 
 
Solving,  = 0.64
% dissociation = 64%
(ii) Since
n
p c
K K (RT)

Here n = 1
p
c
K 0.56
K
RT 0.0821 298
  

= 0.0228 mole/litre
2
c
4x
K
(1 )V


 
where x = no. of moles of N2O4(g) = 80/92
Neglecting  with respect to 1
i.e. 1 –   1
2
c
4x
K
V

 
0.0228 =
2
80
4
92
7
 
0.2142
  
% dissociation 21.42%
 
(iii) th
mix
M
1
M
  
mix
92
1 0.2142
M
 
mix
92
M 75.76
1.2142
  
Let V(ml) volume of mixture diffused in Now, from Graham’s law of diffusion.
2
2
O mix
mix O
r M
r M

20/10 75.76
V /10 32
  V = 12.998 ml
Prob 9. Solid NH4I on rapid heating in a closed vessel at 357°C develops a constant
pressure of 275 mm of Hg owing to the partial decomposition of NH4I into NH3 and HI
but the pressure gradually increases further (when excess solid residually remains in
the vessel) owing to the dissociation of HI. Calculate the final pressure developed
under equilibrium.
NH4I(s) NH3(g) + HI(g)
2HI(g) H2(g) + I2(g) Kc = 0.015 at 357°C
Sol. P =
275
760
= 0.3618 atm

27. 3
NH HI
0.3618
p p
2
  = 0.181 atm
Kp = (0.181)2 = 0.033 atm2
NH4I(s) NH3(g) + HI(g)
(0.181 + p) (0.181 + p – p) Kp = 3.3  10–2 atm2
2HI(g) H2(g) + I2(g)
(0.181 + p – p)
p
2
 p
2

Kp = 0.015
0.033 = (0.018 – p) (0.181 + p – p)
0.015 =
2
2
p
4(0.181 p p)

 
 p = 0.04 atm
p = 0.026 atm
 total pressure = 2(0.181 + p) = 0.414 = 314.64 mm Hg
Prob 10. For the gaseous reaction: C2H2 + D2O C2D2 + H2O, H is 530 cal. At 25C,
KP = 0.82. Calculate how much C2D2 will be formed if 1 mole of C2H2 and 2 moles of
D2O are put together at a total pressure of 1 atm at 100C?
Sol. Firstly calculate KP at 100C by using equation
log
 
 
o
o
P 100 C 2 1
P 1 2
25 C
K T T
H
K 2.303R T T
 


  
 
On solving we get, (KP)100C = 0.98
C2H2 + D2O C2D2 + H2O
Initial mole 1 2 0 0
At eq. mole 1– 2–  
 0.98 =
  
2
1 2

   
0.022 + 2.94 – 1.96 = 0
On solving we get,  = 0.75
Prob 11. An equilibrium mixture at 300K contains N2O4 and NO2 at 0.28 and 1.1 atm
respectively. If the volume of container is doubled, calculate the new equilibrium
pressure of two gases.
Sol. 2 4(g) 2(g)
N O 2NO
0.28 1.1
2
p
(1.1)
K 4.32atm
0.28
 
When volume is doubled pressure will reduced to half
2 4(g) 2(g)
N O 2NO
0.28
p
2
 

 
 
1.1
2p
2
 

 
 
2
p
1.1
2p
2
K 4.32
0.28
p
2
 

 
 
  
 

 
 
p 0.045
 
2 4
p(N O ) 0.14 0.045 0.095
   
2
p(NO ) 0.55 0.045 0.64atm
  
Prob 12. At 540K, 0.10 mole of PCl5 are heated in a 8 litre flask. The pressure of the
equilibrium mixture is found to be 1 atm. Calculate the Kp for the reaction.

28. Sol. 5 3 2
PCl PCl Cl

0.1 –   
Total no. of moles = 0.1 0.1
        
Now,
nRT (0.1 ) 0.082 540
P 1
V 8
   
  
(0.1 ) 0.180
   
0.08
 
So,
 
 
3 2
5
PCl Cl
p
PCl
p p
K
p


 
 
2
p p
p
0.1 0.1
0.1 (0.1 )(0.1 )
p
0.1
 
  
   
   
  
 
     
 
0.0064 1
1.78
0.18 0.02

 

atm
Prob 13. In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC, the
percentage of ammonia by volume under the equilibrium is 17.8. Calculate the
equilibrium constant (KP) of the mixture, for the reaction
N2(g) + 3H2(g) 2NH3(g)
Sol. Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as
KP will not depend on the exact no. of moles of N2 and H2. One can even start with x
and 3x).
Alternatively
N2(g) + 3H2(g) 2NH3
Initial 1 3 0
At eqb. 1x 33x 2x
Since % by volume of a NH3 gas is same as % by mole,

2x
4 2x

= 0.178

4 0.178
x 0.302
(2 2 0.178)

 
 
 Mole fraction of H2 at equilibrium =
3 3x
0.6165
4 2x



Mole fraction of N2 at equilibrium = 1  0.6165  0.178
= 0.2055
 KP =
 
   
3
2 2
2
NH T
3
N T H T
X P
X P X P

 
 
2
3
0.178 30
(0.2055 30) (0.6165 30)


 
= 7.31  10-4
atm–2
Prob 14. The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 g
dm-3. Calculate the equilibrium constant of the reaction, N2O4(g) 2NO2(g).
Sol. Let us assume that we start with C moles of N2O4(g) initially.
N2O4(g) 2NO2(g)
Initial C 0

29. At equilibrium C(1) 2C
Where  is the degree of dissociation of N2O4(g)
Since =
Total moles at equilibrium Vapour density initial
Total moles initially Vapour density at equilibrium

Initial vapour density =
92
46
2

C(1 ) 46
C d
 

Since vapour density and actual density are related by the equation,
V.D. =
RT
2P

=
1.84 0.082 348
2
 
= 26.25
 1 +  =
46
1.752
26.25

  = 0.752
 Kp =
2 2
T
T
2 C 2 0.75
P 1
C(1 ) 1.752
C(1 ) 0.248
P 1
C(1 ) 1.752
 
 
 
 
   
 
   

 
 
 
= 5.2 atm
Prob 15. At temperature T, a compound AB2(g) dissociates according to the reaction
2AB2(g) 2AB(g) + B2(g) with a degree of dissociation x, which is small compared
with unity. Deduce the expression for x in terms of the equilibrium constant, Kp and
the total pressure P.
Sol. Let the initial pressure of AB2(g) be P1
2AB2(g) 2AB(g) + B2(g)
P1(1x) P1x 1
P x
2
The total pressure is = P1 1
x
1 P ( x
2
 
 
 
 
<< 1)  P1 = P
So, KP =
 
 
2 x 3
2
2
Px P Px
2
P(1 x)



P
3
2K
x
P
 

30. Objective:
Prob 1. Reaction between iron and steam is reversible if it is carried out
2 3 4 2
3Fe 4H O Fe O 4H
 
 
(A) at constant T (B) at constant P
(C) in an open vessel (D) in a closed vessel
Sol. In open vessel H2 gas will escape.
 (D)
Prob 2. For the reaction
3(g) 2(g) 5(g)
PCl Cl PCl

The value of Kc at 250°C is 26. The value of Kp at this temperature will be
(A) 0.61 (B) 0.57
(C) 0.83 (D) 0.46
Sol. n 1
p c
K K (RT) 26 (0.082 523) 0.61
 
    
 (A)
Prob 3. In a reversible reaction, two substances are in equilibrium. If the concentration of
each one is doubled, the equilibrium constant will be
(A) Reduced to half, its original value
(B) becomes (original)/4
(C) doubled
(D) constant
Sol. Kcor Kp do not depend on concentration, but only on temperature.
 (D)
Prob 4. The equilibrium constant for the reaction,
4
2(g) 2(g)
N O 2NO is 4 10 at2000K

 
In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the
equilibrium constant, in presence of the catalyst at 2000K is
(A) 4
40 10
 (B) 4
4 10

(C) 5
4 10
 (D) difficulty to compute
Sol. Equilibrium is constant at constant temperature for a reaction
 (B)
Prob 5. 64g of HI are present in a 2 litre vessel. The active mass of HI is:
(A) 0.5 (B) 0.25
(C) 1 (D) none
Sol. Molecular mass of HI = 128
HI] =
64 1
[HI] 0.25
128 2
  
 (B)
Prob 6. For which of the following Kp may be equal to 0.5 atm
(A) 2HI H2 + I2 (B) PCl5 PCl3 + Cl2
(C) N2 + 3H2 2NH3 (D) 2NO2 N2O4

31. Sol. For Kp = 0.5 atm
n = 1 (since the unit is atm)
and PCl5 PCl3 + Cl2
n = 1
 (B)
Prob 7. The vapour density of undecomposed N2O4 is 46. When heated, vapour density
decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the
final temperature is
(A) 80 (B) 60
(C) 40 (D) 70
Sol. N2O4 2NO2
1 0 at initial
1 –  2 at equilibrium
 innitial final
final initial
V.D. n
V.D. n

46 1
25.4 1
 

1.8 = 1 +    = 0.8 or 80%
 (A)
Prob 8. If pressure is applied to the following equilibrium, liquid vapours the boiling
point of liquid
(A) will increase (B) will decrease
(C) may increase or decrease (D) will not change
Sol. Boiling point of a liquid is the temperature at which vapour pressure became equal to
atm pressure. If the pressure is applied to the above equilibrium the reaction will go
to the backward direction, i.e. vapour pressure decrease hence the boiling point
increase.
 (A)
Prob 9. For the reaction
A(g) + B(g) 3C(g) at 250°C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C
respectively. If KC for the reaction is 10, the reaction will proceed in
(A) Forward direction (B) Backward direction
(C) In either direction (D) In equilibrium
Sol.
3 3
3
[C] 4 3 3
Q
[A][B] 3 1 2
 
 
 
= 10.66
 [C] =
4
3
[A] =
1
3
 [B] =
2
3
 
 
 
 KC = 10, and Q  KC
 reaction will proceed in backward direction
 (B)
Prob 10. If CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l) Kp = 1.086  10–4 atm2 at 25°C. The
efflorescent nature of CuSO4.5H2O can be noticed when vapour pressure of H2O in
atmosphere is
(A)  7.29 mm (B)  7.92 mm
(C)  7.92 mm (D) None

32. Sol. An efflorescent salt is one that loses H2O to atmosphere.
For the reaction
CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l)
Kp = ( 2
H O
p )2 = 1.086  10–4
2
H O
p = 1.042  10–2 atm = 7.92 mm
 If 2
H O
p at 25°C  7.92 mm only then, reaction will proceed in forward direction.
 (B)
Prob 11. The enthalpies of two reaction are H1 and H2 (both positive) with
H2  H1. If the temperature of reacting system is increased from T1 to T2, predict
which of the following alternatives is correct?
(A)
 
1 2
1 2
K K
=
K K
(B)
 
1 2
1 2
K K
>
K K
(C)
 
1 2
1 2
K K
<
K K
(D) None
Sol. As the temperature of reacting system is increased the equilibrium constant of
reaction is also increased for endothermic reactions so for two reactions on
increasing the temperature by equal amounts
1 2
1 2
K K
K K
 

Hence, (C) is correct.
Prob 12. At a certain temperature 2 moles of carbonmonoxide and 3 moles of chlorine were
allowed to reach equilibrium according to the reaction CO + Cl2 COCl2 in a 5 lit
vessel. At equilibrium if one mole of CO is present then equilibrium constant for the
reaction is:
(A) 2 (B) 2.5
(C) 3.0 (D) 4
Sol. CO + Cl2 COCl2
At t = 0 2 3 0
At equilibrium (2–1) (3–1) 1
Concentrations (V = 5 lit)
1
5
2
5
1
5
Kc = 2
2
[COCl ]
[CO][Cl ]
=
1 5
1 2
5 5

=
25
10
or Kc = 2.5
Hence, (B) is correct.
Prob 13. The reaction: 3O2 2O3, H = + 69,000 calories is favoured by:
(A) high temperature and low pressure (B) high temperature and high pressure
(C) low temperature and high pressure (D) low temperature and low pressure
Sol. According to Le Chatalier principle formation of ozone is favoured by high
temperature (endothermic reaction) and high pressure.
Hence, (B) is correct.

33. Prob 14. The equilibrium constant at 323°C is 1000. What would be its value in the presence
of a catalyst in the forward reaction?
A + B C + D + 38 Kcal
(A) 1000 × concentration of catalyst (B) 1000
(C)
1000
concentration of catalyst
(D) impossible to tell
Sol. The equilibrium constant varies only with temperature. At constant temperature it will
not vary.
Hence, (B) is correct.
Prob 15. Which of the following reactions represent a heterogenous equilibrium?
(A)      
2 2
H g I g 2HI g

(B)      
3 2
CaCO s CaO s CO g

(C)    
2 4 2
N O g 2NO g
(D)    
3 2
2O g 3O g
Sol. In heterogenous equilibrium, physical state of all the reactants and products are not
same.
Hence, (B) is correct.
True/False
Prob 16. For a reversible system at constant temperature the value of Kc increase if the
concentration are changed at equilibrium.
Sol. False
Prob 17. The apparent molecular weight if PCl5 on dissociation shows lower value.
Sol. True
Prob 18. The value of Kc for the reaction 2 2 3
2SO O 2SO
 will increase if the volume of
reaction vessel is increased.
Sol. False
Fill in the Blanks
Prob 19. Free energy change at equilibrium is …………..
Sol. Zero
Prob 20. The equilibrium constant is ………. of velocity constant of forward and backward
reactions.
Sol. Ratio

34. ASSIGNMENT PROBLEMS
Subjective:
Level - O
1. What happens to rate of forward and rate of backward reaction when equilibrium is attained?
2. What happens to an equilibrium in a reversible reaction, if a catalyst is added to it?
3. The equilibrium constant of a reaction at 27°C and 127°C are 1.6  10–3 and 7.6  10–2
respectively. Is the reaction exothermic or, endothermic?
4. Ice melts slowly at higher altitudes. Explain why?
5. Predict the effect of compression on the following equilibrium reaction.
       
4 2
g g g 2 g
CH H O CO 3H
 
6. List any four important characteristics of a chemical equilibrium
7. The equilibrium constant for the reaction
N2 + 3H2 2NH3 is K. What will be the equilibrium constant for the reaction?
2 3
1 3
N NH ?
2 2

8. What are reversible and irreversible reactions? Explain with the help of some examples?
9. Why gas fizzes out when soda water bottle is opened?
10. What will happen when in the following equilibrium:
H2 + I2 2HI + Q
Temperature and pressure be increased?
11. State and explain LeChatelier’s principle. What effect will be there by addition of inert gas in
an equilibrium mixture?
12. For the reaction
CaCO3(s) CaO(s) + CO2(g), Kp = 1.16 atm at 800°C. If 20g of CaCO3 were kept in a 10
litre container and heated upto 800°C; what percentage by weight of CaCO3 would remain
unreacted at equilibrium?
13. For the reaction 2
2(g) 2(g) 3(g) c p
N 3H 2NH , show that K K (RT) .
 
Do not use the formula n
p c
K K (RT)
 . Start finding Kc and then final Kp.
14. A cylinder fitted with an air tight piston contains a small amount of a liquid at a fixed
temperature. The piston is moved out so that the volume increases.
(a) What will be the effect on the change of vapour pressure initially?
(b) How will the rates of evaporation and condensation change initially?
(c) What will happen when equilibrium is restored finally and what will be the final vapour
pressure?
15. What happens to the equilibrium?
5 3 2
PCl (g) PCl (g) Cl (g),
 if nitrogen gas is added to it (i) at constant volume (ii) at
constant pressure? Give reasons.

35. Level - I
1. An air sample containing 21:79 of O2 and N2 (molar ratio) is heated to 2400°C. If the mole
percent of NO at equilibrium is 1.8%. Calculate the Kp of the reaction
N2 + O2 2NO
2. For the reaction F2 2F. Calculate the degree of dissociation of fluorine at 4 atm and
1000K, when Kp = 1.4  10–2 atm.
3. The equilibrium constant of ester formation of propionic acid with ethyl alcohol is 7.36 at
50°C. Calculate the weight of ethyl propionate in gram existing in an equilibrium mixture when
0.5 mole of propionic acid is heated with 0.5 mole of ethyl alcohol at 50°C.
4. 0.1 mole of ethanol and 0.1 mole of butanoic acid are allowed to react. At equilibrium, the
mixture is titrated with 0.85 M NaOH solution and the titre value was 100 ml. Assuming that
no ester was hydrolysed by the base, calculate K for the reaction.
C2H 5OH + C3H7COOH C3H7COOC2H5 + H2O
5. To 500 ml of 0.150 M AgNO3 solution we add 500 ml of 1.09 M Fe2+ solution and the reaction
is allowed to reach equilibrium at 25C.
Ag+ (aq) + Fe2+ (aq) Fe3+ (aq) + Ag(s)
For 25 ml. of the solution, 30 ml. of 0.0832 M KMnO4 were required for oxidation under acidic
conditions. Calculate KC for the reaction.
6. When one mole of benzoic acid (C6H5COOH) and three moles of ethanol (C2H5OH) are
mixed and kept at 200C until equilibrium is reached, it is found that 87% of the acid is
consumed by the reaction.
C6H5COOH(l) + C2H5OH (l) C6H5COOC2H5(l) + H2O(l)
Find out the percentage of the acid consumed when one mole of the benzoic acid is mixed
with four moles of ethanol and treated in the same way.
7. N2O4 is 25% dissociated at 37C and one atmospheric pressure. Calculate (i) KP and
(ii) percent dissociation at 0.1 atmospheric pressure and 37C.
8. Kp for the reaction PCl5 PCl3 + Cl2 at 250C is 0.82. Calculate the degree of dissociation
at given temperature under a total pressure of 5 atm. What will be the degree of dissociation
if the equilibrium pressure is 10 atm at same temperature?
9. (a) Calculate the equilibrium constant for the process
N2O4 2NO2(g)
At a temperature 350 K, given that K = 0.14 at 298 K and H = 57.2 kJ
(b) The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is
1.6  10-4 atm at 400oC. What will be the equilibrium constant at 500oC if heat of the
reaction in this temperature range is 25.14 k cal?
10. Ammonium carbamate dissociates as    
2 4 3 2
NH COONH s 2NH g CO
 . In a closed
vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial
pressure of NH3 now equals to original total pressure, calculate the ratio of total pressure now
to the original pressure.

36. Level - II
1. KC for PCl5(g) PCl3(g) + Cl2(g) is 0.04 at 250°C. How many mole of PCl5 must be added to
a 3 litre flask to obtain a Cl2 concentration of 0.15 M?
2. KC for N2O4(g) 2NO2(g) is 0.00466 at 298 K. If a one litre container initially contained 0.8
mole of N2O4, what would be the concentrations of N2O4 and NO2 at equilibrium? Also
calculate equilibrium concentration of N2O4 and NO2 if the volume is halved at the same
temperature.
3. A mixture of SO2, SO3 and O2 gas is maintained in a 10 litre flask at a temperature at which
KC = 100 for the reaction
2SO2 + O2 2SO3
At equilibrium:
(a) If the number of moles of SO2 and SO3 in the flask are equal, how much O2 is present?
(b) If the number of moles of SO3 in the flask is twice the number of moles of SO2. How
much O2 is present?
4. At 800K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 5 mole of SO3 at
equilibrium. KC for the equilibrium 2SO2 + O2 2SO3 is 833 lit/mol. If the volume of the
container is 1 litre. Calculate how much O2 is to be added at this equilibrium in order to get
5.2 mole of SO3 at the same temperature.
5. At temperature T, a compound AB2(g) dissociates according to the reaction,
2AB2(g) 2AB(g) + B2(g) with degree of dissociation, , which is small compared to
unity. Deduce the expression for  in terms of the equilibrium constant KP and the total
pressure P.
6. Show that KP for the reaction
2H2S(g) 2H2 (g) + S2 (g) is given by the expression, KP =
  
3
2
P
2 1

   
Where  is the degree of dissociation and P is the total equilibrium pressure. Calculate KC of
the reaction if  at 298 K and 1 atm pressure is 0.055.
7. When PCl5 is heated it dissociates into PCl3 and Cl2. The vapour density of the gas mixture at
200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC
and 250oC.
8. When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27C, 30% of the
solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate KC and KP for
the reaction at 27C. (ii) What would happen to the equilibrium when more solid NH4HS is
introduced into the flask?
9. When 1-pentyne (A) is treated with 4N alcoholic KOH at 175C, it is converted slowly into an
equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2-pentadiene
(C). The equilibrium was maintained at 175C. Calculate G0 for the following equilibria.
B A G1
0 =? B C G2
0 = ?
From the calculated value of G1
0 and G2
0 indicate the order of stability of A, B and C. Write
a reaction mechanism showing all intermediates leading to A, B and C.
10. The density of an equilibrium mixture of N2O4 and NO2 at 1atm. and 348 K is 1.84 g dm-3.
Calculate the equilibrium constant of the reaction, N2O4(g) 2NO2(g).

37. Objective:
Level - I
1. In which of the following cases, does the reaction go furthest to completion?
(A) 3
K 10
 (B) 2
K 10

(C) 5
K 10
 (D) K = 1
2. The vapour density of undecomposed N2O4 is 46. When heated, the vapour density
decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final
temperature is
(A) 87% (B) 60%
(C) 40% (D) 70%
3. In a reaction
2 4(g) 2(g) 2 6(g)
C H H C H

the equilibrium constant can be expressed in units of
(A) 1 1
L mol
 
(B) 1
Lmol
(C) 2 2
mol litre (D) 1
mole litre
4. In a chemical reaction
A B C D
 
The concentration A, B, C and D are 0.5, 0.8, 0.4 and 1.0 at equilibrium respectively.
The equilibrium constant is:
(A) 0.1 (B) 1.0
(C) 10 (D) 
5. One mole of 3
SO was placed in a litre reaction vessel at a certain temperature. The following
equilibrium was established. 3 2 2
2SO 2SO O
 . At equilibrium, 0.6 moles of SO2 were
formed. The equilibrium constant of the reaction will be
(A) 0.36 (B) 0.45
(C) 0.54 (D) 0.675
6. HI was heated in a sealed tube at 440°C till the equilibrium was reached. HI was found to be
22% decomposed. The equilibrium constant for dissociation is
(A) 0.282 (B) 0.0769
(C) 0.0155 (D) 0.0199
7. For the reaction:      
2
C s CO g 2CO g
 , the partial pressure of CO2 and CO are 4
and 8 atm respectively. Hence Kp for the reaction is
(A) 16 (B) 2
(C) 0.5 (D) 4
8. When KOH is dissolved in water, heat is evolved. If the temperature is raised, the solubility of
KOH
(A) Increases (B) Decreases
(C) Remains the same (D) Cannot be predicted
9. A 1 litre vessel contains 2 moles each of gases A, B, C and D at equilibrium. If 1 mole each of
A and B are removed, Kc for A + B C + D; will be
(A) 4 (B) 1
(C) 1/4 (D) 2

38. 10. Consider the water gas equilibrium reaction
C(s) + H2O(g) CO(g) + H2(g)
Which of the following statement is true at equilibrium?
(A) If the amount of C(s) is increased, less water would be formed
(B) If the amount of C(s) is increased, more CO and H2 would be formed
(C) If the pressure on the system is increased by halving the volume, more water would be
formed.
(D) If the pressure on the system is increased by halving the volume, more CO and H2 would
be formed.
11. A reaction takes place in two steps with equilibrium constants 10–2 for slow step and 102 for
fast step. The equilibrium constant of the overall reaction will be
(A) 104 (B) 10—4
(C) 1 (D) 10–2
12. For the reaction PCl3(g) + Cl2(g) PCl5(g), the value of KC at 250°C is 26 litre/mole. The
value of Kp at this temperature will be
(A) 0.61 atm–1 (B) 0.57 atm–1
(C) 0.85 atm–1 (D) 0.46 atm–1
13. According to Le Chatlier’s principle adding heat to a solid and liquid in equilibrium with
endothermic nature will cause the
(A) Amount of solid to decrease (B) Amount of liquid to decrease
(C) Temperature to rise (D) Temperature to fall
14. An aqueous solution of hydrogen sulphide shows the equilibrium
H2S H+ + HS–
If dilute hydrochloric acid is added to an aqueous solution of H2S, without any change in
temperature, then
(A) The equilibrium constant will change
(B) The concentration of HS– will increase
(C) The concentration of undissociated hydrogen sulphide will decrease
(D) The concentration of HS– will decrease
15. What would happen to a reversible reaction at equilibrium when temperature is raised, given
that its H is positive
(A) More of the products are formed (B) Less of the products are formed
(C) More of the reactants are formed (D) It remains in equilibrium
16. Applying the law of mass action to the dissociation of hydrogen Iodide
2HI(g) H2 + I2
We get the following expression, 2
(a x)(b x)
K
4x
 

Where a is original concentration of H2, b is original concentration of I2, x is the number of
moles of H2 and I2 reacted with each other. If the pressure is increased in such a reaction
then
(A) K = 2
(a x)(b x)
4x
 
(B) K  2
(a x)(b x)
4x
 
(C) K  2
(a x)(b x)
4x
 
(D) none of these
17. One mole of ethanol is treated with one mole of ethanoic acid at 25C.
One–fourth of the acid changes into ester at equilibrium. The equilibrium constant for the
reaction will be
(A) 1/9 (B) 4/9
(C) 9 (D) 9/4

39. 18. The equilibrium, PCl5(g) PCl3(g) + Cl2(g) is attained at 25C in a closed container and
an inert gas He is introduced. Which of the following statements are correct?
(A) concentration of PCl5, PCl3 and Cl2 are changed
(B) more Cl2 is formed
(C) concentration of PCl3 is reduced
(D) Nothing happens
19. In which of the following equilibrium, the value of KP is less than KC?
(A) N2O4 2NO2 (B) N2 + O2 2NO
(C) N2 + 3H2 2NH3 (D) 2SO2 + O2 2SO3
20. In the reaction A2(g) + 4B2 (g) 2AB4(g) , H > 0. The decomposition of AB4 (g) will be
favoured at
(A) low temperature and high pressure
(B) high temperature and low pressure
(C) low temperature and low pressure
(D) high temperature and high pressure

40. Level - II
1. The equilibrium constant for a reaction is 10+22 at 300K, the standard free energy change for
this reaction is
(A) – 115 kJ (B) +115kJ
(C) 166kJ (D) – 166 kJ
2. The normal vapour density of PCl5 is 104.25. At 250°C, its vapour density is 57.9. The degree
of dissociation of PCl5 at this temperature will be
(A) 40% (B) 60%
(C) 80% (D) 90%
3. For the reactions,
c
A B;K 2
 , c
B C;K 4
 , c
C D;K 4

Kc for the reaction A D is
(A) 2 + 4 + 6 (B)
2 4
6

(C)
4 6
2

(D) 2 4 4
 
4. The equilibrium constant for the reaction. The reaction is 2
Br 2Br
at 500K and 700K
are 10–10 and 10–5 respectively. Hence the reaction is
(A) Endothermic (B) Exothermic
(C) Fast (D) Slow
5. If pressure is applied to the liquid vapour equilibria
liquid vapour
the boiling point of the liquid
(A) will decrease (B) will increase
(C) many increase or decrease (D) will not change
6. p c
K /K of the reaction
(g) 2(g) 2(g)
1
CO O CO
2

(A) RT (B)
1
RT
(C) RT (D) 1
7. For a system (g) (g) (g)
A B C

eqm. conc. 0.06 0.12 0.216
The value of Kc is
(A) 250 (B) 416
(C) 30 (D) 125
8. On applying pressure to the equilibrium, Ice water
Which phenomenon will happen
(A) More ice will be formed (B) More water will be formed
(C) Equilibrium will not be disturbed (D) Water will evaporate
9. Densities of diamond and graphite are 3.5 and 2.3 grams respectively. Increase of pressure
on the equilibrium Cdiamond Cgraphite
(A) Favours backward reaction (B) Favours forward reaction
(C) Have no effect (D) Increase the reaction rate

41. 10. The decomposition of N2O4 into NO2 is carried out at 280 K in chloroform. When equilibrium
has been established, 0.2 moles of N2O4 and 2  103
moles of NO2 are present in 2 litre of
the solution. The equilibrium constant for the reaction,    
2 4 2
N O g 2NO g , is
(A) 1  102
(B) 2  103
(C) 1  105
(D) 2  105
11. One mole of N2O4(g) at 300K is kept in a closed container under one atm. It is heated to 600K
when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is
(A) 1.2 atm (B) 2.4 atm
(C) 2.0 atm (D) 1.0 atm
12. At 30°C, Kp for the dissociation reaction
SO2Cl2(g) SO2(g) + Cl2(g) is 2.9  10–2 atm. If the total pressure is 1 atm, the degree of
dissociation of SO2Cl2 is
(A) 87% (B) 13%
(C) 17% (D) 29%
13. A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted
into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8
atm is
(A) 1.8 atm (B) 3 atm
(C) 0.3 atm (D) 0.18 atm
14. Vapour density of PCl5 is 104.16 but when heated at 230°C its vapour density is reduced to
62. The degree of dissociation of PCl5 at this temperature will be
(A) 6.8% (B) 68%
(C) 46% (D) 64%
15. For the reaction SO2(g) +
1
2
O2(g) SO3(g) Kp = 1.7  1012 at 20°C and 1 atm pressure.
Calculate Kc.
(A) 1.7  1012 (B) 0.7  1012
(C) 8.40  1012 (D) 1.2  1012
16. For a gaseous equilibrium
2A(g) 2B(g) + C(g) , Kp has a value 1.8 at 700K. What is the value of Kc for the
equilibrium 2B(g) + C(g) 2A at that temperature
(A)  0.031 (B)  32
(C)  44.4 (D)  1.3  10–3
17. For the reaction N2O4(g) 2NO2(g) the reaction connecting the degree of dissociation
() of N2O4(g) with its equilibrium constant KP is
(A)  = P
P
K /P
4 K /P

(B)  = P
P
K
4 K

(C)  =
1/ 2
P
P
K /P
4 K /P
 
 

 
(D)  =
1/ 2
P
P
K
4 K
 
 

 
18. In a closed container at 1 atm pressure 2 moles of SO2(g) and 1 mole of O2(g) were allowed
to react to form SO3(g) under the influence of a catalyst. Reaction
2SO2(g) + O2(g) 2SO3(g) occurred. At equilibrium it was found that 50% of SO2(g) was
converted to SO3(g). The partial pressure of O2 (g) at equilibrium will be
(A) 0.66 atm (B) 0.493 atm
(C) 0.33 atm (D) 0.20 atm

42. 19. KP for a reaction at 25C is 10 atm. The activation energy for forward and reverse reactions
are 12 and 20 kJ / mol respectively. The KC for the reaction at 40C will be
(A) 4.33 10–1 M (B ) 3.33 10–2 M
(C) 3.33 10–1 M (D) 4.33 10–2 M
20. The energy profile for the reaction:
A B C

is shown as:
The equilibrium constant for the said equilibrium
(A) increases with the increase in temperature
(B) decrease with the increase in temperature
(C) does not change with the change in temperature
(C) is equal to the rate constant of the forward reaction
Reaction path
Energy
A+B
C

43. ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level - O
1. The rate of forward reaction becomes equal to the rate of backward reaction at equilibrium.
2. When a catalyst is added, the state of equilibrium is not disturbed but equilibrium is attained
quickly.
3. As the equilibrium constant has increased with increase in temperature, the reaction is
endothermic in the forward direction.
4. (s)
Ice Water


The melting point of ice is favoured at high pressure because there is decrease in volume in
the forward reaction. Since at high altitudes, atmospheric pressure is low and therefore, ice
melts slowly.
5. The increase of pressure i.e. compression will favour the formation of reactants.
7.
2
3
3
2 2
[NH ]
K (for 1st rection)
[H ] [N ]

2
3
3
2 2
[NH ]
K (for 1st rection)
[H ] [N ]

1/ 2
2
3 3
3/ 2 1/ 2 3
2 2 2 2
[NH ] [NH ]
and K
[H ] [N ] [H ] [N ]
 
 
    
 
 
1/ 2
K K
 
12. 35%
Level – I
1. 5.131  10–4 2. 2.96%
3. 37.25 gms 4. 0.03114
5. 3.142 6. 67.57%
7. (i) 0.267 atm, (ii) 40% 8. 37.53% and 27.52%
9. (a) 4.32 (b) 1.462  10–4 atm 10. 31:27
Level – II
1. 2.1375 moles 2. (i) [N2O4] = 0.77M; [NO2] = 0.06 M
(ii) [N2O4]=1.557 M; [NO2] = 0.086 M
3. (a) 0.1 (b) 0.4
4. 0.34 5.
1/ 3
P
K
2P
 
   
 
6. 2.334  10–6 7. 0.48 and 0.79
8. Kp = 0.049 atm2 and Kc = 8.1  105
, No effect
9. 1 2
G 16.178kJ; G 12.282kJ
    , B  C  A 10. 5.14 atm

44. Objective:
Level - I
1. C 2. A 3. B
4. B 5. D 6. C
7. A 8. B 9. B
10. C 11. C 12. A
13. A 14. D 15. A
16. A 17. A 18. D
19. C & D 20. D
Level - II
1. A 2. C 3. D
4. A 5. B 6. B
7. C 8. B 9. A
10. C 11. B 12. C
13. A 14. B 15. C
16. B 17. C 18. D
19. C 20. A