chemical equilibrium for iit jee

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1. Irreversible and Reversible Reactions
Those reactions which proceeds in forward direction and reaches almost to completion are called
irreversible reactions. For example
2
3(s) (s) 2(g)MnO
2KClO 2KCl 3O
 
(Thermal decomposition)
(aq) (aq) (aq) 2NaOH HCl NaCl H O  
(Strong acid–strong base neutralisation reaction)
3 2 3 22AgNO BaCl 2AgCl Ba(NO )    (Precipitation reaction)
Whereas, those reactions which proceed in forward and backward directions both and never
reaches completion are called reversible reactions. These reactions can be initiated in any direction.
For example,
5(g) 3(g) 2(g)PCl PCl Cl (Thermal dissociation)


3 3 2
strong baseWeak acid
CH COOH NaOH CH COONa H O (Neutralisation reactions) 
2(g) 2(g) 3(g)N 3H 2NH  (synthesis reaction)
But when Fe(s) is heated and water vapour is passed over it in open vessel, it is converted to
Fe3O4(s) along with the evolution of hydrogen gas.
(s) 2 (g) 3 4(s) 2(g)3Fe 4H O 3Fe O 4H  
and when Fe3O4 is reduced with hydrogen gas, it gives Fe(s) and H2O
3 4(s) 2(g) (s) 2 (g)Fe O 4H 3Fe 4H O  
But, if the reaction is carried out in a closed container, this reaction becomes reversible.
(s) 2 (g) 3 4(s) 2(g)3Fe 4H O Fe O 4H 
2. State of Equilibrium
It has generally been observed that many changes (physical and chemical) do not proceed to
completion when they are carried out in a closed container. Consider for example vapourisation of
water,
Water Vapour
At any temperature, vapourisation of water takes place, initially the concentration of water is much
greater than the concentration of vapour, but with the progress of time, concentration of vapour
increases whereas that of water remains constant and after a certain interval of time, there is no
change in concentration of vapour, this state is known as state of physical equilibrium.
In a similar way, this has also been found for chemical reactions, for example, when PCl5(g) is
heated in a closed container, its dissociation starts with the formation of PCl3(g) and Cl2(g).
Initially, only PCl5(g) was taken, but with the progress of reaction, PCl3(g) and Cl2(g) are formed
due to dissociation of PCl5(g). After a certain interval of time, the concentration of PCl5(g),
PCl3(g) and Cl2(g) each becomes constant. It does not mean that at this point of time, dissociation
of PCl5(g) and its formation from PCl3(g) and Cl2(g) has been stopped. Actually the rate of
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dissociation of PCl5 and the rate of formation of PCl5(g) becomes equal. This state is called the
state of chemical equilibrium. So, the state of chemical equilibrium is dynamic.
5(g) 3(g) 2(g)PCl PCl Cl+
This can be shown graphically.
Equilibrium
Reactant
Time
Concentration
State of equilibrium
Time
Rateofreaction
Backward
reaction
Forward
reaction
So, ―state of chemical equilibrium can be defined as the state when the rate of forward reaction
becomes equal to rate of reverse reaction and the concentration of all the species becomes
constant‖.
3. Law of mass action
Guldberg and Waage in 1807 gave this law and according to this law, ―At constant temperature, the
rate at which a substance reacts is directly proportional to its active mass and the rate at which a
chemical reaction proceeds is directly proportional to the product of active masses of the reacting
species‖.
The term active mass of a reacting species is the effective concentration or its activity (a) which is
related to the molar concentration (C) as
a = f C
where f = activity coefficient. f  1 but f increases with dilution and as V   i.e. C  0, f 1
i.e., a  C. Thus at very low concentration the active mass is essentially the same as the molar
concentration. It is generally expressed by enclosing the formula of the reacting species in a square
bracket.
To illustrate the law of mass action, consider the following general reaction at constant
temperature,
(g) (g) (g) (g)aA bB cC dD 
Applying law of mass action,
Rate of forward reaction,
Rf  [A]a [B]b
or Rf = kf[A]a[B]b …(i)
Where kf = rate constant of forward reaction
Similarly,
Rate of reverse reaction
Rr  [C]c [D]d
or Rr = kr[C]c [D]d …(ii)
where kr = rate constant of reverse reaction.
At equilibrium,
Rate of forward reaction = Rate of reverse reaction

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i.e., Rf = Rr
So, from equations (i) and (ii), we get
kf[A]a [B]b = kr[C]c[D]d
or,
c d
f
a b
r
k [C] [D]
k [A] [B]

or
c d
f
c a b
r
k [C] [D]
K
k [A] [B]
 
Where, Kc is the equilibrium constant in terms of molar concentration.
Equilibrium Constant (Kp
) in terms of Partial Pressures: Consider the same general reaction
taking place at constant temperature,
aA(g)
+ bB(g)
 cC(g)
+ dD(g)
From law of mass action,
c d
c a b
[C] [D]
K
[A] [B]

…(1)
From ideal gas equation
PV = nRT
or,
n
P RT
V

At constant temperature,
n
P
V

Thus, we can say in a mixture of gases, Partial pressure of any component (say A)
PA
 [A]
Similarly, PB
 [B]
PC
 [C]
PD
 [D]
So, equation (1) can be rewritten as
c d
C D
p a b
A B
P P
K
P P



…(2)
4. Relationship Between Kp and Kc
From the above, for the same general reaction at constant temperature.
c d
c a b
[C] [D]
K
[A] [B]

…(1)
and,
c d
C D
p a b
A B
P P
K
P P



…(2)
From the ideal gas equation,
PV = nRT
n
P RT
V


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So, PB = [B]RT ; PD = [D]RT
Similarly, A CP [A]RT; P [C]RT 
Substituting the values of PA, PB, PC and PD in equation (2), we get
c c d d
p a a b b
[C] (RT) [D] (RT)
K
[A] (RT) [B] (RT)



or
c d (c d)
p a b (a b)
[C] [D] (RT)
K
[A] [B] (RT)


 
or
n
p cK K (RT)
 
Where,   n = (c + d) – (a + b)
i.e., n = sum of no. of moles of gaseous products – sum of no. of moles of gaseous reactants.
If  n  0, Kp  Kc
If  n = 0, Kp = Kc
and if  n  0, Kp  Kc
Illustration 1: Calculate the Kc
and Kp
for the following reactions and also deduce the relationship between Kc
and Kp
(i) 2SO2(g)
+ O2(g)
 2SO3(g)
(ii)
2(g) 2(g) 3(g)
1 3
N + H NH
2 2
Solution:
(i) 2SO2(g)
+ O2(g)
 2SO3(g)
2
3
c 2
2 2
[SO ]
K
[SO ] [O ]

…(i)
3
2 2
2
SO
p 2
SO O
P
K
P P


…(ii)
We know that
n
p cK K (RT)

n = 2 – (2+1) = – 1 
c
p
K
K
RT

(ii)
2(g) 2(g) 3(g)
1 3
N H NH
2 2
 
3
c 1/2 3/2
2 2
[NH ]
K
[N ] [H ]

…(i)
3
2 2
NH
p 1/ 2 3/ 2
N H
P
K
P P


…(ii)
Now,
n
p cK K (RT)

3 1
n 1 1
2 2
 
      
 
c
p
K
K
RT


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Illustration 2: At 400K for the gaseous reaction
2A + 3B  3C + 4D
the value of Kp is 0.05. Calculate the value of Kc (R = 0.082 dm3 atm K–1 mol–1)
Solution:
From the given data
For the reaction 2A + 3B  3C + 4D
n = (3 + 4) – (2 +3) = 2
Since R is given in dm3 atm. K–1 mol–1, we shall take P = 1 atm.
Substituting these values in the equation
n
p c
CRT
K K , we get
P

 
  
 
2
c
1 0.082 400
0.05 K
1
  
  
 
c
0.05
K
0.082 0.082 400 400

  
Kc = 4.648 × 10–5
Illustration 3: Establish a relationship between Kc and Kp for the following reactions.
(a) N2(g) + O2(g)
 2NO(g) (b)
4 2 3(s) 3(g) 2(g) 2 (g)(NH ) CO 2NH + CO + H O
Solution:
(a) ng = 0  Kp = Kc
(b) ng = 4   Kp = Kc (RT)4
5. Application of law of mass action
1. Synthesis of Hydrogen Iodide:
Suppose ‗a‘ moles of H2 and ‗b‘ moles of I2 are heated at 444°C in a closed container of volume
‗V‘ litre and at equilibrium, 2x moles of HI are formed.
H2(g) + I2(g)
 2HI(g)
Initial concentration (mol L–1)
a
V
b
V 0
Equilibrium concentration(mol L–1)
a x
V
 b x
V
 2x
V
2
c
2 2
[HI]
K
[H ][I ]

…(i)
Substituting the equilibrium concentrations of H2, I2 and HI in equation (i), we get
2
2
c
2x
4xV
K
a x b x (a x)(b x)
V V
 
 
  
     
  
   …(ii)
2 2
2
HI
p
H I
P
K
P P


…(iii)

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Suppose the total pressure of the equilibrium mixture at 444°C is P, then
PHI = mole fraction of HI × Total pressure
PHI =
2x
P
a b


Similarly,
2H
(a x)
P P
(a b)

 
 and
2I
b x
P P
(a b)

 

Substituting these values in equation (iii) we get
2
2
p
2x
P
a b
K
a x b x
P P
a b a b
 
 
 
    
     
    

2
p
4x
K
(a x)(b x)

 
…(iv)
So, one can see from equations (iii) and (iv), that
Kp = Kc
This is so, because n= 0 for the synthesis of HI from H2 and I2.
2. Thermal Dissociation of Phosphorus Pentachloride
PCl5(g) dissociates thermally according to the reaction,
PCl5(g)
 PCl3(g) + Cl2(g)
Let us consider that 1 mole of PCl5 has been taken in a container of volume V litre and at
equilibrium x moles of PCl5(g) dissociates. Thus
PCl5(g) PCl3(g) + Cl2(g)
Initial concentration (mol L–1) 1 0 0
Equilibrium concentration(mol L–1)
1 x
V
 x
V
x
V
According to law of mass action, at constant temperature,
Kc =
3 2
5
[PCl ][Cl ]
[PCl ] …(1)
and Kp =
3 2
5
PCl Cl
PCl
P P
P

…(2)
Substituing the values of equilibrium concentration, in equation (1), we have
Kc =
x x
V V
1 x
V


or
2
c
x
K
V(1 x)


Now, total number of moles at equilibrium = 1 – x + x + x = 1 + x
Mole fraction of PCl3 = Mole fraction of Cl2 =
x
1 x
and mole fraction of PCl5 =
1 x
1 x



Ph.: 0291-3291970.
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Suppose total pressure at equilibrium is P, then we have from equation (2),
2
p
x x
P
1 x 1 xK
1 x
P
1 x

 

 or
2
p 2
x P
K
(1 x )


Similarly, we can apply law of mass action on any reaction at equilibrium.
Illustration 4: Derive an expression for Kc and Kp for the reaction
N2(g) + 3H2(g)
 2NH3(g)
Assuming that in a container of volume V, initially 1 mole of N2 and 3 moles of H2 were taken and
at equilibrium 2x moles of NH3 is formed.
Solution:
N2 + 3H2 2NH3
t = 0 1 3 –
t = teq 1 – x 3 – 3x 2x = 4 – 2x
 Kc =
2
3
(2 )
(1 ) (3 3 )
x
x x  × V2 =
4
22
)x1(27
Vx

Kp =
2
3
(2 )
(1 ) (3 3 )
x
x x  ×
2
2
(4 2 )x
P

=
24
22
P)x1(27
)x2(x16


6. Equilibrium Constant for Heterogeneous Equilibria
The equilibrium which involves reactants and products in different physical states. The law of mass
action can also be applied on heterogeneous equilibria as it was applied for homogeneous equilbria
(involving reactants and products in same physical states).
(i) Thermal Dissociation of Solid Ammonium Chloride :
The thermal dissociation of NH4Cl(s) takes place in a closed container according to the equation:
NH4Cl(s)
 NH3(g) + HCl(g)
Let us consider 1 mole of NH4Cl(s) is kept in a closed container of volume ‗V‘ litre at temperature
TK and if x mole of NH4Cl dissociates at equilibrium, then
NH4Cl(s)
 NH3(g) + HCl(g)
Initial moles 1 0 0
Moles at equilibrium 1 – x x x
3
c
4
[NH ][HCl]
K
[NH Cl]

As NH4Cl is a pure solid, so there is no appreciable change in its concentration. Thus,
Kc = [NH3] [HCl]
2
c 2
x x x
K
V V V
  
and 3p NH HClK P P 

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(ii) Thermal Dissociation of Ag2CO3:
Ag2CO3(s) dissociates thermally according to the equation.
Ag2CO3(s)
 Ag2O(s) + CO2(g)
Applying law of mass action, at constant temperature, we get,
2 2
c
2 3
[Ag O][CO ]
K
[Ag CO ]

Now, let us consider that 1 mol of Ag2CO3(s) is heated in a closed container of volume V and x
mol of Ag2CO3(s) dissociates at equilibrium, then
Ag2CO3(s)
 Ag2O(s) + CO2(g)
Initial moles 1 0 0
Equilibrium moles 1 – x x x
Now, as Ag2CO3 and Ag2O are solids, so their concentration can be assumed to be constants thus
c 2
x
K [CO ]
V
 
2P COK P
Illustration 5: Calculate the partial pressure of HCl gas above solid a sample of NH4Cl(s) as a result of
its decomposition according to the reaction:
NH4Cl(s)  NH3(g) + HCl(g) Kp = 1.04 × 10–16
Solution:
NH4Cl(s)
 NH3(g) + HCl(g) Kp = 1.04 ×10–16
Kp = 3NHP × PHCl  P2 = Kp     P = pK
 P = 16
1.04 10
 = 1.02 × 10–8 atm.
7. The Le-Chatelier’s Principle
This principle, which is based on the fundamentals of a stable equilibrium, states that
“When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium
shifts in that direction in which the effect of the stress is reduced”.
Confused with ―stress‖. Well by stress here what we mean is any change of reaction conditions e.g.
in temperature, pressure, concentration etc.
This statement will be explained by the following example.
Let us consider the reaction: 2NH3 (g)
endo
exo

N2 (g) + 3H2 (g)
Let the moles of N2, H2 and NH3 at equilibrium be a, b and c moles respectively. Since the
reaction is at equilibrium,
Where,
 
 
2 2
3
3
N H
2
NH
P P
P

= Kp =
  
 
2 2
3
3
N T H T
2
NH T
X .P X .P
X .P
X terms denote respective mole fractions and PT is the total pressure of the system.

Ph.: 0291-3291970.
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 
3
T T
p2
T
a b
P P
a b c a b c
K
c
P
a b c
   
    
       
 
 
  
Here,
a
a b c  = mole fraction of N2
b
a b c  = mole fraction of H2
c
a b c  = mole fraction NH3

 
 
23
T
2 2
Pab
c a b c

 
= Kp
Since PT =
 a b c RT
V
 
( assuming all gases to be ideal)

23
2
ab RT
Vc
 
 
  = KP …(1)
Now, let us examine the effect of change in certain parameters such as number of moles, pressure,
temperature etc.
If we increase a or b, the left hand side expression becomes QP ( as it is disturbed from
equilibrium) and we can see that QP > KP
The reaction therefore moves backward to make QP = KP.
If we increase c, QP < KP and the reaction has to move forward to revert
back to equilibrium.
If we increase the volume of the container (which amounts to decreasing the pressure),
QP < KP and the reaction moves forward to attain equilibrium.
If we increase the pressure of the reaction, then equilibrium shifts towards backward direction since
in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is
reduced in backward direction.
If temperature is increased, the equilibrium will shift in forward direction since the forward
reaction is endothermic and temperature is reduced in this direction.
However from the expression if we increase the temperature of the reaction, the left hand side
increases (QP) and therefore does it mean that the reaction goes backward (since QP > KP)?. Does
this also mean that if the number of moles of reactant and product gases are equal, no change in
the reaction is observed on the changing temperature (as T would not exist on the left hand side)?.
The answer to these questions is No. This is because KP also changes with temperature. Therefore,
we need to know the effect of temperature on both QP and KP to decide the course of the reaction.
Effect of Addition of Inert Gases to A Reaction At Equilibrium
1. Addition at constant pressure
Let us take a general reaction
aA + bB cC + dD

Ph.: 0291-3291970.
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We know,
c d
C D
T T
p a b
A B
T T
n n
P P
n n
K
n n
P P
n n
   
    
   
   
   
    
   
   
 
 
Where,
nC nD, nA, nB denotes the no. of moles of respective components and PT is the total pressure and
n = total no. of moles of reactants and products.
Now, rearranging ,
n
c d
c D T
P a b
A B
n n P
K
nn n

 
   
   
Where n = (c + d) – (a + b)
Now, n can be = 0,  0 or  0
Let us take each case separately.
(a) n = 0 : No effect (b) n = ‗+ve‘ :
Addition of inert gas increases the n i.e.
TP
n
 
 
 
  is decreased and so is
n
TP
n

 
 
 
  . So products
have to increase and reactants have to decrease to maintain constancy of Kp.
So the equilibrium moves forward.
c) n = ‗–ve‘ :
In this case
TP
n
 
 
 
  decreases but
n
TP
n

 
 
 
  increases. So products have to decrease and reactants
have to increase to maintain constancy of Kp. So the equilibrium moves backward.
2. Addition at Constant Volume :
Since at constant volume, the pressure increases with addition of inert gas and at the same time n
also increases, they almost counter balance each other. So
n
TP
n

 
 
 
  can be safely approximated as
constant. Thus addition of inert gas has no effect at constant volume.
Dependence of Kp or Kc on Temperature
Now we will derive the dependence of KP on temperature.
Starting with Arrhenius equation of rate constant
af
E /RT
f fk A e

 … (i)
Where, kf = rate constant for forward reaction, Af = Arrhenius constant of forward reaction,
faE
= Energy of activation of forward reaction
ar
E /RT
r rk A e

 …(ii)
Dividing (i) by (ii) we get,
 a ar f
E E
f f RT
r r
k A
e
k A


We know that (equilibrium constant )

Ph.: 0291-3291970.
11
 K =
 a ar f
E E
f f RT
r r
k A
e
k A


At temperature T1
 a ar f
1
1
E E
RTf
T
r
A
K e
A


…(iii)
At temperature T2
 a ar f
2
2
E E
RTf
T
r
A
K e
A


… (iv)
Dividing (iv) by (iii) we get

a ar f
2 2 1
1
E E 1 1
T R T T
T
K
e
K
   
    
  

log
2 r f
1
T a a
T 2 1
K E E 1 1
K 2.303 R T T
  
  
 
The enthalpy of a reaction is defined in terms of activation energies as f ra aE E
= H
 
2
1
T
T 2 1
K H 1 1
log
K 2.303 R T T
 
  
 
  log
2
1
T
T 1 2
K ΔH 1 1
= -
K 2.303 R T T
 
 
  …(v)
For an exothermic reaction, H would be negative. If we increase the temperature of the system
(T2>T1), the right hand side of the equation (V) becomes negative.
  2 1T TK < K
, that is, the equilibrium constant at the higher temperature would be less than that at
the lower temperature.
Now let us analyse our question. Will the reaction go forward or backward?
Before answering this, we must first encounter another problem. If temperature is increased, the
new KP would either increase or decrease or may remain same. Let us assume it increases.
Now, QP can also increase, decrease or remain unchanged. If KP increases and QP decreases, then,
T T2 2
P PQ K
therefore the reaction moves forward. If KP increase and QP remains same, then
 T T T2 1 2
P P PQ Q K
. Again, the reaction moves forward. What, if KP increase and QP also increases?
Will 22 TT PP KQ 
or T T2 2
P PQ K
or T T2 2
P PQ K
? This can be answered by simply looking at the
dependence of QP and KP on temperature. You can see from the equation that KP depends on
temperature exponentially. While Q‘s dependence on T would be either to the power g,l,t……..
Therefore the variation in KP due to T would be more than in QP due to T.
  KP would still be greater than QP and the reaction moves forward again.
Therefore, to see the temperature effect, we need to look at KP only. If it increases, the reaction
moves forward, if it decreases, reaction moves backward and if it remains fixed, then no change at
all.

Ph.: 0291-3291970.
12
Illustration 6:
Under what conditions will the following reactions go in the forward direction ?
1. N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.
2. N2(g) + O2(g) 2NO(g) - 43.2 k cal.
3. C(s) + H2O(g) CO2(g) + H2(g) + X k cal.
4. N2O4(g) 2NO2(g) - 14 k cal.
Solution:
1. Low T, High P, excess of N2 and H2.
2. High T, any P, excess of N2 and O2
3. Low T, Low P, excess of C and H2O
4. High T, Low P, excess of N2O4.
Illustration 7:
Under what conditions will the following reactions go in the forward direction?
1. 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.
2. 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal.
3. PCl5(g) PCl3(g) + Cl2(g)- X k cal.
Solution :
(i) Low T, high P, excess of O2 and SO2
(ii) Low T, high P, excess of NO and O2
(ii) High T, low P, excess of PCl5
8. Le-Chatelier’s Principle and Physical Equilibria
Le Chatelier‘s principle, as already stated, is applicable to all types of equilibria involving not only
chemical but physical changes as well. A few examples of its application to physical equilbria are
discussed below.
1. Vapour pressure of a liquid: Consider the equilibrium
Liquid  Vapour
It is well known that the change of a liquid into its vapour is accompanied by absorption of heat
whereas the conversion of vapour into liquid state is accompanied by evolution of heat. According
to Le Chatelier‘s principle, therefore, addition of heat to such a system will shift the equilibrium
towards the right. On raising the temperature of the system, liquid will evaporate. This will raise
the vapour pressure of the system. Thus, the vapour pressure of a liquid increases with rise in
temperature.
2. Effect of pressure on the boiling point of a liquid: The conversion of liquid into
vapour, as represented by the above equilibrium, is accompanied by increase of pressure (vapour
pressure). Therefore, if pressure on the system is increased, some of the vapours will change into
liquid so as to lower the pressure. Thus, the application of pressure on the system tends to condense
the vapour into liquid state at a given temperature. In order to counteract it, a higher temperature is
needed. This explains the rise of boiling point of a liquid on the application of pressure.
3. Effect of pressure on the freezing point of a liquid (or melting point of a
solid): At the melting point, solid and liquid are in equilibrium:
Solid
 Liquid

Ph.: 0291-3291970.
13
Now, when a solid melts, there is usually a change, either increase or decrease, of volume. For
example, when ice melts, there is decrease in volume, or at constant volume, there is decrease in
pressure. Thus, increase of pressure on ice  water system at a constant temperature will cause
the equilibrium to shift towards the right, i.e., it will cause the ice to melt. Hence, in order to retain
ice in equilibrium with water at the higher pressure it will be necessary to lower the temperature.
Thus, the application of pressure will lower the melting point of ice.
When sulphur melts, there is increase in volume or at constant volume, there is increase in
pressure. From similar considerations, it follows that if the pressure on the system, sulphur (solid)
 sulphur (liquid) is increased, the melting point is raised.
4. Effect of temperature on solubility: In most cases, when a solute passes into solution,
heat is absorbed, i.e., cooling results. Therefore according to Le Chatelier‘s principle, when heat is
applied to a saturated solution in contact with solute, the change will take place in that direction
which absorbed heat (i.e., which tends to produce cooling). Therefore, some more of the solute will
dissolve. In other words, the solubility of the substance increases with rise in temperature.
Dissociation of a few salts (e.g., calcium salts of organic acids) is accompanied by evolution of
heat. In such cases, evidently, the solubility decreases with rise in temperature.
9. Free Energy and Chemical Equilibrium
The Gibbs free energy function is a true measure of chemical affinity under conditions of constant
temperature and pressure. The free energy change in a chemical reaction can be
defined as
G = G(products) – G(reactants)
When G = 0, there is no net work obtainable. The system is in a state of equilibrium. When G is
positive, net work must be put into the system to effect the reaction, otherwise it cannot take place.
When G is negative, the reaction can proceed spontaneously with accomplishment of the net
work. The larger the amount of this work that can be accomplished, the farther away is the reaction
from equilibrium. For this reason –G has often been called the driving force of the reaction. From
the statement of the equilibrium law, it is evident that the driving force depends on the
concentration of the reactants and products. It also depends upon the temperature and pressure
which determine the molar free energies of the reactants and products.
The reaction conducted at constant temperature (i.e., in a thermostat)
–G = – H + TS
The driving force is made up of two parts, –H term and TS term. The –H term is the heat of
reaction at constant pressure and TS is heat involved when the process is carried out reversibly.
The difference is the amount of heat of reaction which can be converted into net work (–G), i.e.,
total heat minus unavailable heat.
If the reaction is carried out at constant volume, the decrease in Helmholtz function –G =
–E + TS would be the proper measure of affinity of the reactant or the driving force of the
reaction.
Now we can see why Berthollet and Thompson were wrong in assuming that driving force of the
reaction was the heat of reaction. They neglected the TS term. The reasons for the apparent
validity of their principle was that for many reactions, H term far outweighs the TS term. This is
especially true at low temperature, since at higher temperature, TS term increases.
The fact that driving force for a reaction is large (G is large negative quantity) does not mean that
the reaction will necessarily occur under any given conditions.
For example, the reaction

Ph.: 0291-3291970.
14
2 2 2
1
H O H O; G 228.6kJ
2
    
does not occur at the laboratory temperature. The reaction mixture may be kept for years without
any detectable formation of water. Here H factor favours, but S factor disfavours the reaction.
Similarly, the reaction
2Mg + O2  2MgO; G = – 570.6kJ
is not favoured. However, the thermite reaction
2Al +
3
2 O2  Al2O3
with large value of – G proceeds favourably.
Standard Free Energy and Equilibrium Constant: The change in free energy for a
reaction taking place between gaseous reactants and products represented by the general equation.
aA bB cC dD 
According to Van‘t Hoff reaction isotherm
c d
0 C D
a b
A B
p p
G G RTln
p p

   

= G0 + RTlnQp
the condition for a system to be at equilibrium is that
G = 0
Thus at equilibrium
c d
0 0 0C D
pa b
A B
p p
0 G RTln G RTln K
p p
 
      
 
Whence G0 = – RTlnK0
p
Hence
0
0
p
G
ln K
RT


Note: 1. In the reaction, where all gaseous reactants and products; K represents Kp
2. In the reaction, where all solution reactants and products; K represents Kc
3. a mixture of solution and gaseous reactants; Kx represents the thermodynamic equilibrium
constant and we do not make the distinction between Kp and Kc.
we may conclude that for standard reactions, i.e., at 1 M or 1 atm.
When G0 = –ve or K  1: forward reaction is feasible
G0 = +ve or K  1: reverse reaction is feasible
G0 = 0 or K = 1: reaction is at equilibrium (very rare)
Illustration 8: Kc for the reaction N2O4
 2NO2 in chloroform at 291 K is 1.14. Calculate the free
energy change of the reaction when the concentration of the two gases are 0.5 mol dm–3 each at
the same temperature. (R = 0.082 lit atm K–1 mol–1.)
Solution:
From the given data
T = 291 K; R = 0.082 lit atm K–1 mol–1
Kc = 1.14; 2 2 4NO N OC C
= 0.5 mol dm–3
The reaction quotient Qc for the reaction 2 4 2N O 2NO ,

Ph.: 0291-3291970.
15
2
2
c
2 4
[NO ] 0.5 0.5
Q 0.5
[N O ] 0.5

  
Since Qp = Qc(RT)n and n = 2 – 1 = 1 in this case
Qc = 0.5 (0.082 × 291) = 11.93
Kp = Kc(RT)n = 1.14 (0.082 × 291) = 27.1
Substituting these values in the equation
 G = G0 + RTlnQp = – RT ln Kp + RT ln Qp, we get
= – 2.303 RT (log Kp – log Qp)
G = – (0.082 × 291 × 2.303) [log 27.2 – log 11.93]
= – 54.95 (1.4346 – 1.0766) = – 19.67 lit atm
Illustration 9: Calculate the pressure of CO2 gas at 700K in the heterogeneous equilibrium reaction
CaCO3(s)
 CaO(s) + CO2(g) if G0 for this reaction is 130.2 kJ mol–1.
Solution:
Here Kp = 2COP
Also,
0
pG RTlnK  

0 3 1
p 1 1
G 130.2 10 Jmol
ln K
RT (8.314JK mol )(700K)

 
 
 
 2
10
CO pp K 1.94 10 atm
  
Illustration 10:
For the equilibrium NiO(s) + CO(g)
 Ni(s) + CO2(g), G0 (J mol–1)
= – 20,700 – 11.97 T. Calculate the temperature at which the product gases at equilibrium at 1
atm will contain 400 ppm (parts per million) of carbon monoxide.
Solution: For the given reaction 2p CO COK p / p
Since 2CO COp p , hence
p 6
CO
1 1
K 2,500
p 400 10
  

0
pG RTlnK  

0
p
G 20,700 11.97T
ln K
RT RT
 
 
The equation when solved for T using R = 8.314 K–1 mol–1, gives T = 399K.
Illustration 11: Calculate the equilibrium constant of a reaction at 300 K if G0 at this temperature for
the reaction is 29.29 kJ mol–1.
Solution
(ii)  G0 = –2.303 RT log Kp
 29.29 ×103 = –2.303 × 8.31 × 300 log Kp
 Kp = 7.91 × 10–6

Ph.: 0291-3291970.
16
Illustration 12: For the formation of ammonia the equilibrium constant data at 673K and 773K
respectively are 1.64 × 10–4 and 1.44 × 10–5 respectively. Calculate heat of the reaction. Given R
= 8.314 JK–1 mol–1.
Solution: Substituting the values in the equation
2
1
p 2 1
p 2 1
K T TH
ln , we get
K R T T
 
  
 
5
4
1.44 10 H 773 673
2.303log
8.314 773 6731.64 10


   
    
H(100)
2.303log(0.0878)
8.314 773 673


 
100H = 2.303 (–1.0565) × 673 × 773 × 8.314
whence,
673 773 2.303 1.0565 8.314
H
100
    
 
= – 105216J = – 105.216kJ
Illustration 13: 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?
Solution: N2(g) + 3H2(g) 2NH3(g)
Kp = 1.6 × 10–4 at T1 = 673 K
Kp = ? at T2 = 773 K
log








 4
p
106.1
K
= 2303.2
1014.25 3


log [673 – 773]
Kp = 1.43 × 10–5
Relation between Vapour Density and Degree of Dissociation
In the following reversible chemical equation
A  yB
Initial mol 1 0
At equilibrium (1 –x) yx x = degree of dissociation
Number of moles of A and B at equilibrium = 1 – x + yx = 1 + x(y –1)
If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is
= [1 + x(y – 1)]V
Molar density before dissociation,
molecular weight m
D
volume V
 
Molar density after dissociation
m
d
[1 x(y 1)]V

 
 
D
[1 x(y 1)]1
d
  
 
D d
x
d(y 1)




Ph.: 0291-3291970.
17
Note: y is the number of moles of products from one mole of reactant.
D
d is also called Van‘t Hoff
factor.
Thus for the equilibria
I: 5(g) 3(g) 2(g)PCl PCl Cl , y = 2
II: 2 4(g) 2(g)N O 2NO , y = 2
III:
2 2 4
1
2NO N O , y =
2

 
D d
x
d


2(d D)
x
d


Also D × 2 = Molecular weight (theoretical value)
d × 2 = Molecular weight (abnormal value) of the mixture
Illustration 14: Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 40 for the
equilibrium
N2O4
 2NO2
Calculate (i) abnormal molecular weight
ii) degree of dissociation
iii) percentage of NO2 in the mixture
Solution: (i) N2O4  2NO2
Observed value of vapour density (d) = 40
Hence, abnormal molecular weight = 40 × 2 = 80
(ii) D × 2 = theoretical molecular weight = 2

92
D 46
2
 

D d 46 40
x 0.15
d 40
 
  
(iii) N2O4  2NO2
Initial mol 1 0
At equilibrium (1 – x) 2x
0.85 0.30
Total moles at equilibrium = (1 + x) = 1 + 0.15 = 1.15
 Percentage of NO2 =
2x 0.30
100 100
1 x 1.15
  
 = 26.08%
Illustration 15: N2O4(g)
 2NO2(g). In this reaction, NO2 is 20% of the total volume at
equilibrium. Calculate
(i) Vapour density (ii) abnormal molecular weight
(iii) percentage dissociation of N2O4
Solution : N2O4
 2NO2

Ph.: 0291-3291970.
18
t = 0 1 –
t = teq 1–x 2x = 1 + x
2
NO
 =
2
1
x
x
= 0.2  x =
1
9
(i)
1
9
=
46 d
d

 10d = 46 × 9
(ii) M =
92
1 x
 d = 41.4
(iii)  =
1
9
× 100 = 11.1%
Illustration 16:The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C.
Calculate the number of moles of NO2 in 100 gm mixture.
Solution:
N2O4 2NO2  =
D d
d

t = 0 1.087 0  =
46 38.3
38.3

= .2 (20%)
t = teq 1.087 (1 –) 1.087 × 2
 2NOn = 1.087 × 0.4 = 0.435
Illustration 17: The degree of dissociation at a certain or given temperature of PCl5 at 2 atm is found to
be 0.4. At what pressure, the degree of dissociation of PCl5 will be 0.6 at the same temperature?
Also calculate the equilibrium constant for the reverse reaction.
Solution:
The dissociation of PCl5 takes place according to the equation,
PCl5  PCl3(g) + Cl2(g)
Let 1 mole of PCl5 is taken in a closed container. Then
PCl5(g)  PCl3(g) + Cl2(g)
Mole(s) before dissociation 1 0 0
Mole(s) at equilibrium 1 –   
As degree of dissociation() is given at 2 atm
1 –  = 1.0 – 0.4 = 0.6
 Total moles at equilibrium = 1 –  +  +  = 1 – 0.4 + 0.4 + 0.4 = 1.4
Now,
3 2
5
PCl Cl
p
PCl
P P
K
P


=
0.4 0.4
2 2
1.4 1.4
0.6
2
1.4
  

p
0.4 0.4 4 1.4 0.32
K
0.6 2 1.4 1.4 0.84
  
 
  
Now, as temperature remains the same, Kp will also remain the same. So for
 = 0.6
PCl5(g)
 PCl3(g) + Cl2(g)
Mole before dissociation 1 0 0
Moles at equilibrium 1 – 0.6 0.6 0.6

Ph.: 0291-3291970.
19
Again,
2
p
0.6 0.6
P
1.6 1.6K
0.4
P
1.6
 



p
0.6 0.6 1.6
K P
0.4 1.6 1.6
 

 
32 9
84 16

P    P =
32 4 128
9 21 189


 = 0.677 atm
Illustration 18: (i) Calculate the percentage dissociation of H2S(g), if 0.1 mole of H2S is kept in 0.4 litre
vessel at 1000K. For the reaction, 2H2S(g)
 2H2(g) + S2(g), the value of Kc = 1.6 × 10–6
(ii) A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be
the degree of dissociation if hydrogen is added in the proportion of 1 mole for every mole of HI
originally present, the temperature and volume of the system being kept constant?
Solution:
(i) 2H2S(g)  2H2 + S2(g) Kc = 1.6 × 10–6
t = 0 0.1 – –
t = teq 0.1 – x x x/2
value of Kc suggest that (0.1 – x) =~
0.1

3
2
1
0.42(0.1)
x
 = 1.6 × 10–6  x = 2.34 ×10–3
  =
3
2.34 10
0.1


× 100 = 2.34%
(ii) 2HI H2 + I2
t = 0 1 – –
t = teq 1–  /2 /2 since  = 0.22
 Kc =
2
2
(0.11)
(0.88)
=
1
64
now 2HI H2 + I2
t = 0 1 1 0
t = teq 1–  /2 /2
1
64
= 2
(1 / 2).( / 2)
(1 )
 



=
2
2
2
4(1 )
 



 4 + 42 – 8 = 128 + 642
 602 + 136 – 4 = 0
  = 0.29

Ph.: 0291-3291970.
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Problem Set – 1
Q.1 For the given gaseous phase reaction, K1, K2, K3 are equilibrium constants respectively.
Correlate them.
2NO + O2 2NO2 4NO + 2Cl2 4NOCl
NO2 + ½ Cl2 NOCl + ½ O2
Q.2 In an equilibrium A + B C + D, the initial concentration of A was three times to that of B.
At equilibrium, the concentration of C was found to be equal to equilibrium concentration of B.
Calculate KC.
Q.3 For an equilibrium reaction A + 2B 2C + D; A and B are mixed in a reaction vessel at
300 K. The initial concentration of B was 1.5 times the initial concentration of A. After the
equilibrium, the equilibrium concentrations of A and D are same. Calculate KC.
Q.4 KC for N2O4(g) 2NO2(g) at 298°C is 5.7 ×10–9. Which species has higher
concentration at equilibrium ?
Q.5 From the given data of equilibrium constants of the following reactions :
CoO(s) + H2(g) Co(s) + H2O(g); K = 670
CoO(s) + CO(g) Co(s) + CO2(g); K = 490
Calculate KC for the reaction: CO2(g) + H2(g) CO(g) + H2O(g)
Q.6 The compound AB dissociates at 300°C of AB(g) A(g) + B(g); KC of reaction is 5 × 10–
2 and equilibrium concentration of A is 0.03 mol litre–1
(a) What was initial concentration of AB ?
(b) Calculate partial pressure of each gas and hence calculate Kp.
(c) Verify the value of Kp derived as above for its agreement with Kp = KC(RT)n
Q.7 At 25°C for the reaction A2(g) 2A(g), KC is 0.165. If one mole of A2 is placed in 0.1
litre flask at 25°C, calculate how many moles of A are formed at equilibrium ?
Q.8 A mixture of SO2 and O2 under atmospheric pressure in the ratio of 2 : 1 is passed over a
catalyst at 1170°C. After equilibrium has reached, the gas coming out has been found to contain
87% SO3 by volume. Calculate Kp for SO2 + ½ O2 SO3.
Q.9 KC for PCl5 PCl3 + Cl2 is 13.7 mol litre–1 at 546 K. Calculate the pressure developed
in 10 litre box in equilibrium if reactant concentration is 1.0 mole.
Q.10 n mole of PCl3 and n mole Cl2 are allowed to react at constant temperature T under a total
pressure P, as PCl3(g) + Cl2(g) PCl5(g)
If Y mole of PCl5 are formed at equilibrium, find Kp.
Q.11 n mole each of H2O, H2 and O2 are mixed at a suitable high temperature to attain the
equilibrium 2H2O 2H2 + O2. If y mole of H2O are dissociated and the total pressure
maintained is P, calculate the Kp.
Q.12 How many mole of PCl5 must be added to one litre vessel at 250°C in order to obtain a
concentration of 0.1 mol litre–1 of Cl2 at equilibrium. Kp for PCl5 PCl3 + Cl2 is 1.78 atm?
Q.13 The equilibrium constant of the reaction A2(g) + B2(g) 2AB(g)
at 100°C is 50. If a one litre flask containing one mole of A2 is connected to a two litre flask
containing two mole of B2, how many moles of AB will be formed at 373 K ?
Q.14 The equilibrium constant for the reaction.
CH3COOH(l)+ C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Ph.: 0291-3291970.
21
is 4. What will be the composition of the equilibrium mixture when one mole of acetic acid is
taken along with 4 moles of ethyl alcohol ?
Q.15 At 450°C, the equilibrium constant, Kp for the reaction, N2(g) + 3H2(g) 2NH3(g)
was found to be 1.6 × 10–5 at a pressure of 200 atm. If N2 and H2 are taken in 1 : 3 ratio what is
% of NH3 formed at this temperature ?
Problem Set – 2
Q.1 When limestone is heated, quick lime is formed according to the equation,
CaCO3(s) CaO(s) + CO2(g)
The experiment was carried out in the temperature range 800 – 900°C. Equilibrium constant Kp
follows the relation,
log10Kp = 7.282 – 8500/T
where T is temperature in Kelvin. At what temperature the decomposition will give CO2(g) at 1
atm?
Q.2 Consider the equilibrium :
LiCl.3NH3(s) LiCl.NH3(s) + 2NH3(g)
with Kp = 9 atm2 at 40°C. A 5 litre flask contains 0.1 mol of LiCl.NH3. How many moles of
NH3 should be added to the flask at this temperature to drive the backward reaction practically
to completion?
Q.3 A sample of CaCO3(s) is introduced into a sealed container of volume 0.654 litre and heated to
1000 K until equilibrium is reached. The equilibrium constant for the reaction.
CaCO3(s) CaO(s) + CO2(g) is 3.9 × 10–2 atm at this temperature. Calculate the mass
of CaO present at equilibrium.
Q.4 The decomposition of ammonium carbamate at 30°C is represented as
NH2COONH4(s) 2NH3(g) + CO2(g)
The equilibrium constant Kp is 2.9 × 10–5 atm3. What is the total pressure of gases in
equilibrium with NH2COONH4(s) at 300°C.
Q.5 Solid ammonium carbamate dissociates to give ammonia and carbodioxide as follows :
NH2COONH4(s) 2NH3(g) + CO2(g)
At equilibrium, ammonia is added such that partial pressure of NH3, now equals to the original
total pressure. Calculate the ratio of total pressure now to the original total pressure.
Q.6 Show that for the reaction AB(g) A(g) + B(g), the total pressure at which AB is
50% dissociated is numerically equal to three times of Kp.
Q.7 At 55°C and one atmosphere, N2O4 was found to decompose 50.3%. At what pressure and
same temperature the equilibrium, N2O4 2NO2
will give the ratio of N2O4 : NO2 at 8 : 1 ?
Q.8 Pure PCl5 is introduced into an evacuated chamber and comes to equilibrium at 250°C and
2 atmosphere. The equilibrium mixture contains 40.7% Cl2 by volume.
(a) What are the partial pressure of each constituent at equilibrium ?
(b) What are Kp and KC ?
(c) If the gas mixture is expanded to 0.200 atm at 250°C, calculate
(i) the % of PCl5 dissociated at this equilibrium.
(ii) The partial pressure of each at equilibrium

Ph.: 0291-3291970.
22
Q.9 5 mole of PCl5 and 4 mole of neon are introduced in a vessel of 110 litre and allowed to attain
equilibrium at 250°C. At equilibrium, the total pressure of reaction mixture was 4.678 atm.
Calculate degree of dissociation of PCl5 and equilibrium constant for the reaction.
Q.10 5g of PCl5 were completely vaporised at 250°C in a vessel of 1.9 litre capacity. The mixture at
equilibrium exerted a pressure of one atmosphere. Calculate , KC and Kp for the reaction PCl5
PCl3 + Cl2.
Q.11 1.0 mole of nitrogen and 3.0 mole of PCl5 are placed in 100 litre vessel heated to 227°C. The
equilibrium pressure is 2.05 atm. Assuming ideal behaviour calculate the degree of dissociation
for PCl5 and Kp for the reaction : PCl5(g) PCl3(g) + Cl2(g)
Q.12 The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 26.7°C. Calculate the
number of moles of NO2 in 100 grams of the mixture.
Q.13 For the reaction,
NH3(g) N2(g) + H2(g)
show that degree of dissociation of NH3 is given as
=
1/2
3 3
1
4 p
p
K

 
 
  
, where ‗p‘ is equilibrium pressure. If Kp of the above reaction is 78.1 atm
at 400°C, calculate KC.
Q.14 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 80 g sample of N2O4 confined to a 7
litre vessel.
(iii) What volume of above mixture will diffuse if 20 mL pure O2 diffuses in 10 minutes at same
temperature and pressure ?
Q.15 The value of KC for 2HF(g) H2(g) + F2(g) is 1.0 ×10–13 at a particular temperature. At
a certain time, the concentration of HF, H2 and F2 were found to be 0.5, 1 ×10–3 and 4 ×10–3
mol litre–1 respectively. Predict whether the reaction is in equilibrium ? If not, what is the
direction of the reaction to attain equilibrium?
Problem Set – 3
1. If for A + 2B 3C, KC = 10 and a 1 litre container contains 5 moles of C,
10 moles of B and 1.25 moles of A, then predict the direction in which the reaction
is moving.
2. A + B C + D, KC = 20. A 10 litre container contains 5 moles each of A and
B and 20 moles each of C and D.
(i) Predict direction of reaction
(ii) Calculate concentration of A, B, C, D at equilibrium.
3. A 2 litre container contains 1 moles of PCl5 2 moles of PCl3 and 2 moles of
Cl2. If KC for PCl5 PCl3 + Cl2 is 2 then, calculate concentration of PCl5,
PCl3 and Cl2 at equilibrium.

Ph.: 0291-3291970.
23
4. For the reaction , A + B 3C at 25ºC, a 3 litre vessel contains 1,2,4 mole of A, B
and C respectively . Predict the direction of reaction if :
(a) KC for the reaction is 10. (b) KC for the reaction is 15.
(c) KC for the reaction is 10.66 .
5. If Gº for the reaction given below is 1.7 kJ; the equilibrium constant of the
reaction, 2HI (g) H2(g) + I2(g) at 25 ºC is :
(A) 24.0 (B) 3.9 (C) 2.0 (D) 0.5
6. A large positive value of Gº corresponds to which of these?
(A) small positive K (B) small negative K
(C) large positive (D) large negative K
7. The equilibrium constant for a reaction is 1 × 1020 at 300 K. The standard free
energy change for this reaction is :
(A) – 115 kJ (B) +115 kJ (C) +166 kJ (D) –166 kJ
8. At equilibrium, If Kp = 1, then:
(A) Gº = 0 (B) Gº > 1 (C) Gº < 1 (D) None
9. Gº for the reaction , X + Y Z is –4.606 kcal. The equilibrium constant
for the reaction at 227 ºC is :
(A) 100 (B) 10 (C) 2 (D) 0.01
10. For the given equilibrium A + B C + D. The reaction is at equilibrium
with 0.2 M A and 0.6M B, 3M C and 4M D. Then calculate the value of Gº at
27ºC (in kcal).
11. Gº for () N2 + (3/2) H2 NH3, is –16.5 kJ mol–1. Find out Kp for the
reaction. Also report Kp and Gº for N2 + 3H2 2NH3, at 25ºC.
12. For the given equilibrium , N2O4 2NO2 total pressure at equilibrium is 5
atm and the mole ratio of N2O4 and NO2 is 3 : 2. Find the value of Gº at 227ºC
(in cal)
13. The value of KC for 2HF(g) H2(g) + F2(g) is 1.0 ×10–13 at a particular temperature. At
a certain time, the concentration of HF, H2 and F2 were found to be 0.5, 1 ×10–3 and 4 ×10–3
mol litre–1 respectively. Predict whether the reaction is in equilibrium ? If not, what is the
direction of the reaction to attain equilibrium ?
14. The equilibrium constant, Kp for the reaction
N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10–4 atm–2 at 400°C.
what will be the equilibrium constant at 500°C if heat of reaction in this range is -25.14 Kcal.
15. For a reaction X Y, heat of reaction is +83.68 kJ, energy of reactant X is 167.36 kJ and
energy of activation is 209.20 kJ. Calculate (i) threshold energy (ii) energy of product Y and (iii)
energy of activation for the reverse reaction (Y X).
Problem Set – 4
Q.1 The initial concentrations or pressure of reactants and products are given for each of the
following systems. Calculate the reaction quotient and determine the directions in which each
system will shift to reach equilibrium.
(a)2NH3(g) N2(g) + 3H2(g) K = 17
[NH3] = 0.20 M ; [N2] = 1.00 M ; [H2] = 1.00 M
(b) 2NH3(g) N2(g) + 3H2(g) Kp = 6.8 × 104 atm2

Ph.: 0291-3291970.
24
Initial pressure; NH3 = 3.0 atm; N2 = 2.0 atm ; H2 = 1.0 atm
(c) 2SO3(g) 2SO2(g) + O2(g) K = 0.230 atm
[SO3] = 0.00 M ; ]SO2] = 1.00 M; [O2] = 1.00 M
(d) 2SO3(g) 2SO2(g) + O2(g) Kp = 16.5 atm
Initial pressure : SO3= 1.0 atm ; SO2 = 1.0 atm; O2= 1.0 atm
(e)2NO(g) + Cl2(g) 2NOCl (g) K = 4.6 × 104
[NO] = 1.00 M ; [Cl2] = 1.00 M ; [NOCl] = 0M
(f) N2(g) + O2(g) 2NO(g) Kp=0.050
Initial pressure ; NO = 10.0 atm N2 = O2 = 5 atm
Q.2 Among the solubility rules is the statement that all chlorides are soluble except Hg2Cl2,, AgCl,
PbCl2 and CuCl.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation.
AgCl (s) Ag+ (aq) + CI– (aq)
Is K greater than 1, less than 1, or about equal to 1? Explain your answer.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation
Pb2+ (aq) + 2CI– (aq) PbCl2 (s)
Is K greater than 1, less than 1, or about equal to I? Explain your answer.
Q.3 Among the solubility rules is the statement that carbonates, phosphates, borates, arsenates, and
arsenites, except those of the ammonium ion and the alkali metals are insoluble.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation
CaCO3 (s) Ca2+ (aq) + CO3
2– (aq)
(b) Write the expression for the equilibrium constant for the reaction represented by the
equation. 3Ba2+ (aq) + 2PO4
3– (aq) Ba3(PO4)2 (s)
Q.4 Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is
manufactured by the catalytic conversion of acetylene to benzene.
3C2H2 C6H6
Would this reaction be most useful commercially if K were about 0.01, about 1, or about 10?
Explain your answer.
Q.5 Show the complete chemical equation and the net ionic equation for the reaction represented by
the equation
KI (aq) + I2 (aq) KI3 (aq)
give the same expression for the reaction quotient. KI3 is composed of the ions K+ and I3
–.
Q.6 Which of the following reactions goes almost all the way to completion, and which proceeds
hardly at all?
(a) N2(g) + O2(g) 2NO(g) Kc = 2.7 × 10–18
(b) 2NO(g) + O2(g) 2NO2(g) ; Kc = 6.0 × 1013
Q.7 For which of the following reactions will the equilibrium mixture contain an appreciable
concentration of both reactants and products?
(A) Cl2(g) 2Cl(g); Kc = 6.4 × 10–39
(B) Cl2(g) + 2NO(g) 2NOCl(g); Kc = 3.7 × 108
(C) Cl2(g) + 2NO2(g) 2NO2Cl(g); Kc = 1.8

Ph.: 0291-3291970.
25
Q.8 The value of Kc for the reaction 3O2(g) 2O3 (g) is 1.7 × 10–5 6 at 25ºC. Do you expect
pure air at 25ºC to contain much O3(ozone) when O2 and O3 are in equilibrium ? If the
equilibrium concentration of O2 in air at 25ºC is 8 × 10–3 M, what is the equilibrium
concentration of O3?
Q.9 At 1400 K, Kc = 2.5 × 10–3 for the reaction CH4(g) + 2H2S CS2(g) + 4H2(g) . A 10.0 L
reaction vessel at 1400 K contains 2.0 mol of CH4, 3.0 mol of CS2, 3.0 mol of H2 an 4.0 mol of
H2S. Is the reaction mixture at equilibrium? If not, in which direction does the reaction proceed
to reach equilibrium?
Q.10 The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give
water gas, a mixture of carbon monoxide and hydrogen.
H2O(g) + CH4(g) CO(g) + 3H2(g) Kc = 4.7 at 1400 K
A mixture of reactants and product at 1400 K contains 0.035 M H2O, 0.050 M CH4, 0.15 M
CO, and 0.20 M H2. In which direction does the reaction proceed to reach equilibrium?
Q.11 An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At
this temperature, Kc for the reaction N2(g) + 3H2(g) 2NH3(g) is 0.29. What is the
concentration of NH3.
Q.12 The air pollutant NO is produced in automobile engines from the high temperature reaction
N2(g) + O2(g) 2NO(g); Kc = 1.7 × 10–3 at 2300 K. If the initial concentrations of N2 and
O2 at 2300K are both 1.40 M, what are the concentration of NO, N2, and O2 when the reaction
mixture reaches equilibrium?
Q.13 At a certain temperature, the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constan
Kc = 5.8 × 10–2. Calculate the equilibrium concentrations of PCl5, PCl3 and Cl2 if only PCl5 is
present initially, at a concentration of 0.160 M.
Q.14 At 700 K Kp = 0.140 for the reaction ClF3(g) ClF(g) + F2(g). Calculate the equilibrium
partial pressure of ClF3, ClF, and F2 if only ClF3 is present initially, at a partial pressure of 1.47
atm.
Q.15 The degree of dissociation of N2O4 into NO2 at 1.5 atmosphere and 40ºC is 0.25. Calculate its
Kp at 40ºC. Also report degree of dissociation at 10 atmospheric pressure at same temperature.
Q.16 At 46ºC, Kp for the reaction N2O4(g) 2NO2(g) is 0.667 atm. Compute the percent
dissociation of N2O4 at 46ºC at a total pressure of 380 Torr.
Q.17 When 36.8 g N2O4(g) is introduced into a 1.0-litre flask at 27ºC. The following equilibrium
reaction occurs:
N2O4 (g) 2NO2(g) ; Kp = 0.1642 atm.
(a) Calculate Kc of the equilibrium reaction.
(b) What are the number of moles of N2O4 and NO2 at equilibrium?
(c) What is the total gas pressure in the flask at equilibrium?
(d) What is the percent dissociation of N2O4?
Q.18 At some temperature and under a pressure of 4 atm, PCIs is 10% dissociated. Calculate the
pressure at which PCl5 will be 20% dissociated, temperature remaining same.
Q.19 In a mixture of N2 and H2 in the ratio of 1 : 3 at 64 atmospheric pressure and 300°C, the
percentage of ammonia under equilibrium is 33.33 by volume. Calculate the equilibrium
constant of the reaction using the equation.
N2 (g) + 3H(g) 2NH3 (g).

Ph.: 0291-3291970.
26
Q.20 The system N2O 4 2 NO2 maintained in a closed vessel at 60ºC & a pressure of 5 atm has
an average (i.e. observed) molecular weight of 69, calculate Kp. At what pressure at the same
temperature would the observed molecular weight be (230/3)?
Q.21 The vapour density of N204 at a certain temperature is 30. Calculate the percentage dissociation
of N2O4 at this temperature. N2O4(g) 2NO2(g).
Q.22 In the esterfication C2H5OH (l) + CH3COOH (l) CH3COOC2H5 (l) + H2O (l) an
equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with
mole fraction = 0.333. Calculate the equilibrium constant.
Q.23 Solid Ammonium carbonate dissociates as: NH2COONH4(s) 2NH3(g) + CO2(g). In a
closed vessel solid ammonium carbonate is in equilibrium with its dissociation product. At
equilibrium, ammonia is added such that the partial pressure of NH3 at new equilibrium now
equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that
of original total. pressure.
Q.24 A sample of CaCO3(s) is introduced into a sealed container of volume 0.821 litre & heated to
1000K until equilibrium is reached. The equilibrium constant for the reaction
CaCO3(s) CaO(s) + CO2(g) is 4 × 10–2 atm at this temperature.
Calculate the mass of CaO present at equilibrium.
Q.25 Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of CaCl2, the
amount of the water taken up is governed by KP = 6.4 × 1085 for the following reaction at room
temperature, CaCl2(s) + 6H2O(g) CaCl2 .6H2O(s). What is the equilibrium vapour
pressure of water in a closed vessel that contains CaCl2(s)?
Q.26 20.0 grams of CaCO3(s) were placed in a closed vessel, heated & maintained at 727°C under
equilibrium CaCO3(s) CaO(s) + CO(g) and it is found that 75 % of CaCO3 was
decomposed. What is the value of Kp ? The volume of the container was 15 litres.
Q.27 Suggest four ways in which the concentration of hydrazine. N2H2. could be increased in an
equilibrium described by the equation
N2(g) + 2H2 (g) N2H4 (g) H = 95 kJ
Q.28 How will an increase in temperature affect each of the following equilibria? An increase in
pressure?
(a) 2NH3(g) N2(g) + 3H2(g) H = 92 kJ
(b)N2(g) + O2(g) 2NO(g) H = 181 kJ
(c)2O3(g) 3O2(g)   H = –285 kJ
(d) CaO(s) + CO2(g) CaCO3(s) H = –176 kJ
Q.29 (a) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas
and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst.
Write the expression for the equilibrium constant for the reversible reaction.
2H2(g) + CO(g) CH3OH(g) H = –90.2 kJ
(b) Assume that equilibrium has been established and predict how the concentration of H2, CO
and CH3OH will differ at a new equilibrium if (1) more H2 is added. (2) CO is removed (3)
CH3OH is added. (4) the pressure on the system is increases. (5) the temperature of the system
is increased. (6) more catalyst is added.

Ph.: 0291-3291970.
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Q.30 (a) Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction
of steam with red-hot coke, essentially pure carbon. Write the expression for the equilibrium
constant for the reversible react on.
C(s) + H2O(g) CO(g) + H2(g) H = 131.30 kJ
(b) Assume that equilibrium has been established and predict how the concentration of each
reactant and product will differ at a new equilibrium if (1) more C is added. (2) H2O is
removed. (3) CO is added. (4) the pressure on the system is increased. (5) the temperature of the
system is increased.
Q.31 Ammonia is a weak base that reacts with water according to the equation
NH3(aq) + H2O (l) NH4
+ + OH–(aq)
Will any of the following increase the percent of ammonia that is converted to the ammonium
ion in water? (a) addition of NaOH. (b) Addition of HCl. (c) Addition of NH4Cl.
Q.32 Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of
Na+, Cl–, Ag+ and NO3
–, in contact with solid AgCl.
Na+(aq) + Cl–(aq) + Ag+ (aq) + NO3
–(aq) AgCl(s) + Na+ (aq) + NO3
– (aq) H = –65.9 kJ
Q.33 Addition solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and
sulfate ion in equilibrium with solid silver sulfate. Which of the following will occur? (a) The
Ag+ and SO4
2– concentration will not change. (b) the added silver sulfate will dissolve. (c)
Additional silver sulfate will form and precipitate from solution as Ag+ ions and SO4
2– ions
combine. (d) The Ag+ ion concentration will increase and the SO4
2– ion concentration will
decreases.
Q.34 Consider a general, single-step reaction of the type A + B C. Show that the equilibrium
constant is equal to the ratio of the rate constant for the forward and reverse reaction Kc = kf/kr.
Q.35 Which of the following relative values of kf and kr results in an equilibrium mixture that
contains large amounts of reactants and small amounts of product?
(a) kf > kr (b) kf = kr (c) kf < kr
Q.36 Consider the gas-phase hydration of hexafluoroacetone, (CF3)2CO:
(CF3)2CO(g) + H2O (g) f
r
k
k
 (CF3)2C(OH)2(g)
At 76ºC, the forward and reverse rate constants are kf = 0.13 M–1s–1 and kr = 6.02 × 10–4s–1.
What is the value of the equilibrium constant Kc?
Q.37 Consider the reaction of chlorometahane with OH– in aqueous solution
CH3Cl(aq) + OH–(aq) f
r
k
k
 CH3OH (aq) + Cl–(aq)
At 25ºC, the rate constant for the forward reaction is 6 × 10–6 M–1s–1, and the equilibrium
constant Kc is 1 × 10+16. Calculate the rate constant for the reverse reaction at 25ºC.
Q.38 The progress of the reaction
A nB with time, is presented in figure. Determine
(i) the value of n.
(ii) the equilibrium constant k.
(iii) the initial rate of conversion of A. 1 3 5 7
0.1
0.3
0.5
Time/Hour

Ph.: 0291-3291970.
28
Q.39 Listed in the table are forward and reverse rate constants for the reaction 2NO(g) N2(g) +
O2(g)
Temperature (K) kf(M–1s–1) kr(M–1s–1)
1400 0.29 1.1 × 10–6
1500 1.3 1.4 × 10–5
Is the reaction endothermic or exothermic? Explain in terms of kinetics.
Q.40 Forward and reverse rate constant for the reaction CO2(g) + N2(g) CO(g) + N2O(g)
exhibit the following temperature dependence.
Temperature (K) kf(M–1s–1) kr(M–1s–1)
1200 9.1 ×10–11 1.5 × 105
1500 2.7 × 10–9 2.6 × 105
Is the reaction endothermic or exothermic? Explain in terms of kinetics.
Q.41 The equilibrium constant Kp for the reaction PCl5(g) PCl3(g) + Cl2(g) is 3.81 × 102 at
600 K and 2.69 × 103 at 700K. Calculate rH.
Q.42 As shown in figure a catalyst lowers the activation energy for the forward and reverse reactions
by the same amount, Ea.
(a) Apply the Arrhenius equation, K = aE /RT
Ae

to the forward and reverse reactions, and
show that a catalyst increases the rates of both reaction s by the same factor.
E (forward)
without catalyst
a
E (reverse)
without catalyst
a
E (reverse)
with catalyst
aE (forward)
with catalyst
a
Potentialenergy
Reaction progress
(b)Use the relation between the equilibrium constant and the forward and reverse rate constants,
Kc = kf/kr, to show that a catalyst does not affect the value of the equilibrium. constant.
Q.43 Variation of equilibrium constant ‗K‘ with temperature ‗T‘ is given by equation
log K = log A –
Hº
2.303RT

A graph between log K and 1/T was a straight line with –ve slope of 0.5 and intercept 10.
Calculate
(a)Hº.
(b)Pre exponential factor
(c)Equilibrium constant at 298 K assuming Hº to be independent of temperature.
(d)Equilibrium constant at 798 K assuming Hº to be independent of temperature.
Q.44 Rate of disappearance of the reactant A at two different temperature is given by A B
d[A]
dt

= (2 × 10–2S–1) [A] – 4 × 10–3S–1[B] ; 300 K
d[A]
dt

= ( 4 × 10–2S–1) [A] –16 × 10–4 [B] ; 400 K
Calculate heat of reaction in the given temperature range. When equilibrium is set up.

Ph.: 0291-3291970.
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Q.45 The KP for reaction A + B C + D is 1.34 at 60ºC and 6.64 at 100ºC. Determine the free
energy change of this reaction at each temperature and Hº for the reaction over this range of
temperature?
Q.46 If Kc = 7.5 × 10–9 at 1000 K for the reaction N2(g) + O2(g) 2NO(g), what is Kc at 1000
K for the reaction 2NO(g) N2(g) + O2(g)?
Q.47 An equilibrium mixture of PCl5, PCl3 and Cl2 at a certain temperature contains 8.3 × 10–3 M
PCl5, 1.5 × 10–2 M PCl3, and 3.2 × 10–2 M Cl2. Calculate the equilibrium constant Kc for the
reaction PCl5 (g) PCl3(g) + Cl2(g).
Q.48 A sample of HI2(9.30 × 10–3 mol) was placed in an empty 2.00 L container at 1000K. After
equilibrium was reached, the concentration of I2 was 6.29 × 10–4 M. Calculate the value of Kc
at 1000K for the reaction H2(g) + I2 2HI(g).
Q.49 The vapour pressure of water at 25ºC is 0.0313 atm. Calculate the values of Kp and Kc at 25ºC
for the equilibrium H2O(l) H2O(g).
Q.50 For each of the following equilibria, write the equilibrium constant expression for Kc. Where
appropriate, also write the equilibrium constant expression for Kp.
(a) Fe2O3(s) + 3CO(g) 2Fe(l) +3CO2(g) (b) 4Fe(s) + 3O2(g) 2Fe2O3(s)
(c) BaSO4(s) BaO(s) + SO3(g) (d) BaSO4(s) Ba2+ (aq) + SO4
2– (aq)
Q.51 When 0.5 mol of N2O4 is placed in a 4.00 l reaction vessel and heated at 400 K, 79.3% of the
N2O4 decomposes to NO2.
Calculate Kc and Kp at 400 K for the reaction N2O4(g) 2NO2(g)
Q.52 What concentration of NH3 is in equilibrium with 1.0 × 10–3 M N2 and 2.0 × 10–3 MH2 at 700
K? At this temperature Kc = 0.291 for the reaction N2(g) + 3H2(g) 2NH3(g).
Q.53 At 100 K, the value of Kc for the reaction C(s) + H2O(g) CO(g) + H2 (g) is 3.0 × 102.
Calculate the equilibrium concentration of H2O, CO2, and H2 in the reaction mixture obtained
by heating 6.0 mol of steam and an excess of solid carbon in a 5.0 L container. What is the
molar composition of the equilibrium mixture?
Q.54 When 1.0 mol of PCl5 is introduced into a 5.0 L container at 500 K, 78.5% of the PCl5
dissociates to give an equilibrium mixture of PCl5, PCl3, and Cl2.
PCl5 (g) PCl3(g) + Cl2(g)
(a) Calculate the values of Kc and Kp.
(b) If the initial concentration s in a particular mixture of reactants and products are [PCl5] = 0.5
M. What are the concentrations when the mixture reaches equilibrium?
Q.55 The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to
propene is 1.0 × 105at 500 K.
CH2
H C2 CH2 CH3–CH=CH2 Kc = 1.0 × 105
cyclopropane Propene
(a) What is the value of Kp at 500 K?
(b) What is the equilibrium partial pressure of cyclopropane at 500 K when the partial pressure
of propene is 5.0 atm?

Ph.: 0291-3291970.
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(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or
by decreasing the volume of the container? Explain
(d) Which has the larger rate constant, the forward reaction or the reverse reaction?
(e) Why is cyclopropane so reactive?
Q.56 -D-Glucose undergoes mutarotation to -D-Glucose in aqueous solution. If at 298 K there is
60% conversion. Calculate Gº of the reaction.
-D-Glucose -D-Glucose
Q.57 For the reaction at 298 K
A(g) + B(g) C(g) + D(g)
Hº = –29.8 kcal; Sº = – 0.1 kcal / K
Calculate Gº and K.
Q.58 The equilibrium constant of the reaction 2C3H6(g) C2H4(g) + C4H8(g) is found to fit the
expression ln K = –1.04 –
1088K
T
Calculate the standard reaction enthalpy and entropy at 400 K.
Problem Set – 5
Q.1 For an equilibrium reaction involving gases, the forward reaction is Ist order while
the reverse reaction is IInd order. The units of Kp for the forward equilibrium is:
(A) atm (B) atm2 (C) atm–1 (D) atm–2
Q.2 A chemical reaction A B is said to be at equilibrium when :
(A) Complete conversion of A to B has taken place
(B) Conversion of A to B is only 50% complete
(C) Only 10% conversion of A to B has taken place
(D) The rate of transformation of A to B is just equal to the rate of transformation
of B to A in the system.
Q.3 In the chemical reaction , N2 + 3 H2 2 NH3 at equilibrium :
(A) equal volumes of N2 & H2 are reacting
(B)equal masses of N2 & H2 are reacting
(C) the reaction has stopped
(D) the amount of ammonia formed is equal to amount of NH3 decomposed into N2 & H2
Q.4 For N2 + 3 H2 2 NH3 ; H = – ve , then :
(A) Kp = Kc (B) Kp = KcRT (C) Kp = Kc(RT)–2 (D) Kp = Kc(RT)–1
Q.5 For the reaction , H2 (g) + I2 (g) 2 HI (g) equilibrium constant , Kp changes with :
(A) Total pressure (B) Catalyst
(C) Amount of H2 and I2 present (D) Temperature
Q.6 For the reaction H2 + I2 2HI :
(A) Kc = 2Kp (B) Kc > Kp (C) Kc = Kp (D) Kc < Kp
Q.7 Which oxide of nitrogen is the most stable:
(A) 2NO2(g) N2(g) + 2O2(g); K = 6.7 × 1016 mol litre–1
(B) 2NO(g) N2(g) + O2(g); K = 2.2 × 1030 mol litre–1

Ph.: 0291-3291970.
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(C) 2N2O5(g) 2N2(g) + 5O2(g); K = 1.2 × 1034 mol–5 litre–5
(D) 2N2O(g) 2N2(g) + O2(g); K = 3.5 × 1033 mol litre–1
Q.8 The equilibrium constant for equilibria,
SO2(g) + 1
2
O2(g) SO3(g) and
2SO3(g) 2SO2(g) + O2(g) are K1 and K2 respectively. Then
(A) K2 = K1 (B) K2 = K1
2 (C) K2 = 1/K1 (D) K2 = 1/K1
2
Q.9 A higher value for equilibrium constant, K shows that :
(A) The reaction has gone to near completion towards right.
(B) The reaction has not yet started
(C) The reaction has gone to near completion towards left
(D) None of these
Q.10 If K1 and K2 are equilibrium constants for reactions (I) and (II) respectively for,
N2 + O2 2NO
1
2
N2 + 1
2
O2 NO
Then:
(A) K2 = K1 (B) K2 = 1K (C) K1 = 2K2 (D) K1 = (1/2)K2
Q.11 A reversible chemical reaction having two reactants is in equilibrium. If the concentrations of
the reactants are doubled then the equilibrium constant will:
(A) Also be doubled (B) Be halved
(C) Become one fourth (D) Remain the same
Q.12 For which reaction does the equilibrium constant depend on the units of concentration :
(A) NO(g) 1
2 N2(g) + 1
2 O2(g)
(B) Zn(s) + Cu2+ (aq.) Cu(s) + Zn2+ (aq.)
(C) C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)
(D) COCl2(g) CO(g) + Cl2(g)
Q.13 Hydrogen and oxygen were heated together in a closed vessel. The equilibrium constant is
found to decrease after 200ºC. Which is responsible for this:
(A) Backward reaction predominates (B) Forward reaction predominates
(C) Both forward and backward reaction have same rate
(D) It is a property of the system, hence no reason for lower value.
Q.14 If different quantities of ethanol and acetic acid are used in the following reversible reaction,
CH3COOH + C2H5OH CH3COOC2H5 + H2O
the equilibrium constant will have values which will be:
(A) same in all cases (B) Different in all cases
(C) Higher in cases where higher concentration of ethanol is used
(D) Higher in cases where concentration of acetic acid is used.
Q.15 Which reaction has n = 2:
(A) CaCO3(s) CaO(s) + CO2(g)
(B) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
(C) NH4Cl(g) NH3(g) + HCl (g)
(D) CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(v)

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Q.16 When a system is at equilibrium:
(A) The mass of the product is equal to the mass of the reactant
(B) The ratio of the product of the mass of the products and that of the reactants is constant
(C) The ratio of the velocity of the forward reaction and the backward reaction is one
(D) Number of moles of the product and that of the reactant are the same
Q.17 For the reaction,
CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O() which one is correct representation:
(A) Kp = 2
2
( )H Op (B) Kc = [H2O]2 (C) Kp = Kc(RT)2 (D) All
Q.18 For a hypothetical equilibrium:
4A + 5B 4X + 6Y ; The equilibrium constant Kc has the unit:
(A) mol2 litre–1 (B) litre mol–1 (C) litre2 mol–2 (D) mol litre–1
Q.19 For the gaseous phase reaction,
2NO N2 + O2, Hº = + 43.5 kcal mol–1, Which statement is correct for,
N2(g) + O2(g) 2NO(g);
(A) K is independent of temperature (B) K increases as temperature decreases
(C) K decreases as temperature decreases (D) K varies with addition of NO
Q.20 In which of the following cases, does the reaction go farthest to completion:
(A) K = 103 (B) K = 10–2 (C) K = 10 (D) K = 1
Q.21 For the homogenous gaseous phase reaction,
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) the equilibrium constant, Kc, has the units of :
(A) Unit of concentration10 (B) Unit of concentration1
(C) Unit of concentration–1 (D) It is dimensionless
Q.22 If K1 and K2 are the respective equilibrium constants for the two reactions,
XeF6(g) + H2O(g) XeOF4(g) + 2HF(g)
XeO4(g) + XeF6(g) XeOF4(g) + XeO3F2(g)
The equilibrium constant for the reaction,
XeO4(g) + 2HF(g) XeO3F2(g) + H2O(g) is:
(A) K1K2 (B) K1/K2
2 (C) K2/K1 (D) K1/K2
Q.23 If 1.0 mole of I2 is introduced into 1.0 litre flask at 1000 K, at equilibrium (Kc = 10–6), which
one is correct:
(A) [I2(g)] > [I–(g)] (B) [I2(g)] < [I–(g)] (C) [I2(g)] = [I–(g)] (D) [I2(g)] = [I–(g)]
Q.24 The active mass of 7.0 g of nitrogen in a 2.0 L container would be -
(A) 0.25 (B) 0.125 (C) 0.5 (D) 14.0
Q.25 At 2800 K, a 1.0 mole sample of CO2 in a one litre container is 50% decomposed to carbon
monoxide and oxygen at equilibrium
2CO2(g) 2CO(g) + O2(g)
The value of Kc for the reaction is -
(A) 0.25 (B) 0.50 (C) 50 (C) 25

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Problem Set – 6
Q.1 If x is the degree of dissociation of PCl5
PCl5(g) PCl3(g) + Cl2 (g)
at a given temperature and if 2 moles of PCl5 are taken in a vessel, then at equilibrium the total
number of moles of various species would be -
(A) 4 (B) 2 + x (C) 2 (1 – x) (D) 2(1 + x)
Q.2 AB dissociates as, 2AB (g) 2A(g) + B2(g)
when the initial pressure of AB is 500 mm, the pressure becomes 625 mm when the equilibrium
is attained. Calculate Kp for the reaction assuming volume remains constant -
(A) 500 (B) 125 (C) 750 (D) 375
Q.3 Starting with pure COCl2, if the initial concentration of COCl2 is 0.03 M and the equilibrium
concentration of Cl2 is 0.02 M, the equilibrium constant, KC for the reaction -
COCl2(g) CO(g) + Cl2(g) would be
(A) 4 × 10–4 (B) 4 × 10–3 (C) 4.0 (D) 4 × 10–2
Q.4 For the reaction, N2 + 3H2 2NH3
Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium 50% of each had
reacted. If the equilibrium pressure is P, the partial pressure of hydrogen at equilibrium would
be -
(A) P/2 (B) P/3 (C) P/4 (D) P/6
Q.5 For a hypothetical reaction, A(g) + B(g) X(g) + Y(g)
occurring in a single step, the specific rate constant are 2.0 × 10–2 and 5.0 × 103 respectively
for the forward and the backward reactions. The equilibrium constant is -
(A) 4.0 ×10–4 (B) 2.5 × 10–6 (C) 2.5 × 105 (D) 4.0 × 10–6
Q.6 For the equilibrium
NH4Cl(s) NH3(g) + HCl (g)
Kp = 4.9 × 10–9 atm2 at a certain temperature. The vapour pressure of NH4Cl(s) at this
temperature would be -
(A) 4.9 × 10–9 atm (B) 7 × 10–5 atm (C) 2.45 × 10–9 atm (D) 1.4 × 10–4 atm
Q.7 The equilibrium constant for the reaction,
H2 + I2 2HI, is 49 at a certain temperature. The equilibrium constant for the reaction
HI
1
2
H2 +
1
2
I2 will be -
(A) 49 (B) 7 (C) 1/7 (D) 1/49
Q.8 For an equilibrium, N2O4(g) 2NO2 (g)
the total pressure at equilibrium is P and degree of dissociation of N2O4 is x. Which one of the
following is the partial pressure of NO2 ?
(A)
2
(1 )
x
P
x
(B) 2 × P (C)
2
(1 )
P
x
(D)
2
3
× P
Q.9 For a gaseous reaction, pA + qB qC + pD
Which of the following relationship is true ?
(A) KP = KC (B) KP = KC (RT)p+ q
(C) KP = KC (RT)p–q (D) KP = KC = (RT)1/p+q
Q.10 The units of equilibrium constant for the reaction,
N2 + 3H2 2NH3 + heat, are
(A) mol–2 L2 (B) mol L–1 (C) mol2L–2 (D) L mol–1

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Q.11 One mole of SO3 was placed in a litre vessel at a certain temperature. When equilibrium was
established in the reaction
2SO3(g) 2SO2 (g) + O2(g)
The vessel was found to contain 0.4 moles of SO3. The value of equilibrium constant is
(A) 0.13 (B) 0.36 (C) 0.68 (D) 0.45
Q.12 When 1 mole of N2 and 1 mole of H2 is enclosed in 5 L vessel and the reaction is allowed to
attain equilibrium, it is found that at equilibrium there is x mole of H2. The number of moles of
NH3 formed would be
(A)
2
3
x
(B)
2(1 x)
3

(C)
2(1 )
3
x
(D)
(1 )
2
x
Q.13 A one litre vessel initially contains 2.0, 0.5 and 0.0 moles of N2, H2 and NH3 respectively. The
system after attaining equilibrium has 0.2 mole of NH3. The number of moles of N2 in the
vessel at equilibrium is
(A) 1.9 (B) 0.4 (C) 0.2 (D) 1.8
Q.14 At certain temperature, 50% of HI is dissociation into H2 and I2, the equilibrium constant of the
reaction is
(A) 1.0 (B) 3.0 (C) 0.5 (D) 0.25
Q.15 1.1 mole of A is mixed with 1.2 mole of B and the mixture is kept in a one-litre flask till the
equilibrium A + 2B 2C + D is reached. At equilibrium 0.1 mole of D is formed. The
equilibrium constant of the above reaction is
(A) 0.002 (B) 0.004 (C) 0.001 (D) 0.003
Q.16 Favourable conditions for manufacture of ammonia by the reaction
N2 + 3H2 2NH3; H = – 21.9 kcal are :
(A) Low temperature, low pressure and catalyst
(B) Low temperature, high pressure and catalyst
(C) High temperature, low pressure and catalyst
(D) High temperature, high pressure and catalyst
Q.17 For the reaction , 2X(g) + Y(g) 2Z(g) ; H = – 80 kcal the highest yield of
Z at equilibrium occurs at :
(A) 1000 atm and 500ºC (B) 500 atm and 500ºC
(C) 1000 atm and 100ºC (D) 500 atm and 100ºC
Q.18 The Haber‘s process for the manufacture of ammonia is usually carried out at about
500ºC. If a temperature of about 250ºC was used instead of 500ºC:
(A) No ammonia would be formed at all
(B) The percentage of ammonia in the equilibrium mixture would be too low
(C) A catalyst would be of no use at all at this temperature.
(D) The rate of formation of ammonia would be too slow.
Q.19 At equilibrium, If Kp = 1, then:
(A) Gº = 0 (B) Gº > 1 (C) Gº < 1 (D) None
Q.20 Does Le Chatelier‘s principle predict a change of equilibrium concentration for the
following reaction if the gas mixture is compressed :
N2O4 (g) 2 NO2 (g)
(A) Yes , backward reaction is favoured (B) Yes , forward reaction is favoured
(C) No change (D) No information
Q.21 Oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if:
(A) Temperature is increased and pressure is kept constant
(B) Temperature is reduced and pressure is increased
(C) Both temperature and pressure are increased
(D) Both temperature and pressure are reduced

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Q.22 For the equilibrium , 2NO2(g) N2O4(g) + 14.6 kcal
An increase of temperature will:
(A) Favour the formation of N2O4 (B) Favour the decomposition of N2O4
(C) Not affect the equilibrium (D) Stop the reaction
Q.23 Which equilibrium in gaseous phase would be unaffected by an increase in pressure:
(A) N2O4 2NO2 (B) N2 + O2 2NO
(C) N2 + 3H2 2NH3 (D) CO + ½O2 CO2
Q.24 For which reaction high pressure and high temperature is helpful in obtaining a high equilibrium
yield:
(A) 2NF3(g) N2(g) + 3F2(g) – 54.40 kcal
(B) N2(g) + 3H2(g) 2NH3(g) + 22.08 kcal
(C) Cl2(g) + 2O2(g) 2ClO2(g)– 49.40 kcal
(D) 2Cl2O7(g) 2Cl2(g) + 7O2(g) + 126.8 kcal
Q.25 In the gaseous equilibrium , A + 2B C + heat, the forward reaction is favoured by:
(A) Low temperature (B) Low pressure
(C) High pressure and low temperature (D) High pressure and high temperature
Problem Set – 7
Q.1 An increase in the temperature of an equilibrium system:
(A) Favours the exothermic reaction
(B) Favours the endothermic reaction
(C) Favours both the exothermic and endothermic reactions
(D) Favours neither the exothermic nor endothermic reactions.
Q.2 The degree of dissociation of PCl5 () obeying the equilibrium,
PCl5 PCl3 + Cl2, is approximately related to the pressure at equilibrium by:
(A)  P (B) 
1
P
(C)  2
1
p
(D)  4
1
p
Q.3 A reaction in which an increase in pressure will increase the yield of products:
(A) H2(g) + I2(g) 2HI(g)
(B) H2O(g) + CO(g) CO2(g) + H2(g)
(C) H2O(g) + C(s) CO(g) + H2(g)
(D) CO(g) + 3H2(g) CH4(g) + H2O(g)
Q.4 In the thermal decomposition of potassium chlorate in a closed container given as,
2KClO3(g)  2KCl (s) + 3O2(g), law of mass action:
(A) Cannot be applied
(B) Can be applied at hig temperature
(C) can be applied at low temperature
(D) Can be applied at high temperature and pressure.
Q.5 What is the direction of a reversible reaction when one of the products of the reaction is
removed:
(A) Forward direction (B) Backward direction
(C) The reaction stops (D) All the above
Q.6 The endothermic reaction (M + N P) is allowed to attain an equilibrium at 25ºC.
Formation of P can be increased by:
(A) Raising temperature (B) Lowering temperature

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(C) Keeping temperature constant (D) Decreasing the concentration of M and N
Q.7 Consider the reaction, PCl5(g) PCl3(g) + Cl2(g) in a closed container at equilibrium. At
a fixed temperature addition of more PCl5 at equilibrium will show.......in equilibrium
concentration of Cl2(g):
(A) Decrease (B) Increase (C) No change (D) None
Q.8 Vapour density of PCl5 is 104.16 but when heated to 230ºC 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%
Q.9 When pressure is applied to the equilibrium system
Ice Water
which of the following phenomenon will happen ?
(A) more ice will be formed (B) water will evaporate
(C) more water will be formed (D) equilibrium will not be disturbed
Q.10 Which of the following will not change the concentration of ammonia at the equilibrium ?
(A) increase of pressure (B) increase of volume
(C) addition of catalyst (D) decrease of temperature
Q.11 Pure ammonia is placed in a vessel at a temperature when its dissociation is appreciable. At
equilibrium
(A)  does not change with pressure
(B) concentration of ammonia does not change with pressure
(C) concentration of hydrogen is less than that of nitrogen
(D) Kp does not change significantly with pressure
Q.12 Which of the following is not favourable for the formation of SO3 ?
2SO2 (g) + O2(g) 2SO3 (g) ; H = –188 kJ
(A) high pressure (B) high temperature (C) decreasing [SO3] (D) increasing [SO2]
Q.13 For which of the following gaseous equilibrium at constant temperature, doubling the volume
would cause a shift of equilibrium to the right ?
(A) 2CO + O2 2CO2 (B) N2 + 3H2 2NH3
(C) N2 + O2 2NO (D) PCl5 PCl3 + Cl2
Q.14 The yield of the product in the reaction , A2 (g) + 2B(g) C(g) + Q kJ
would be higher at
(A) high temperature and high pressure (B) high temperature and low pressure
(C) low temperature and high pressure (D) low temperature and low pressure
Q.15 For a reaction, the value of K increases with increase in temperature. The H for the reaction
would be
(A) +ve (B) – ve (C) zero (D) cannot be predicted
Q.16 For the reaction, PCl5(g) PCl3(g) Cl2(g) at equilibrium the introduction of inert gas at
constant pressure will shift the equilibrium
(A) in forward direction (B) in backward direction
(C) neither forward nor backward direction (D) cannot be predicted
Q.17 A and B are gaseous substance which react reversibly to given two gaseous substances C and D,
accompanied by the liberation of heat. When the reaction reaches equilibrium, it is observed that
Kp = Kc. The equilibrium cannot be disturbed by
(A) adding A (B) adding D
(C) raising the temperature (D) increasing the pressure
Q.18 In a vessel containing SO3, SO2 and O2 at equilibrium some helium gas is introduced so that
the total pressure increases while temperature and volume remains constant. According to Le-
chatelier‘s principle, the dissociation of SO3

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(A) increases (B) decreases
(C) remain unaltered (D) change unpredictably
Q.19 For the reaction, PCl5(g) PCl3(g) + Cl(g), the forward reaction at constant temperature is
favaoured by
(A) introducing an inert gas at constant volume(B) introducing Cl2 gas at constant volume
(C) introducing inert gas at constant pressure(D) decreasing the volume of the container
Q.20 The reaction, SO2 + Cl2 SO2Cl2 is exothermic and reversible. A mixture of SO2(g),
Cl2(g) and SO2Cl2(l) is at equilibrium in a closed container. Now a certain quantity of extra SO2
is introduced into the container, the volume remaining the same. Which of the following are true ?
(A) the pressure inside the container will not change
(B) the temperature will not change
(C) the temperature will increase
(D) the temperature will decrease
Q.21 For the reaction, CaCO3(s) CaO(s) + CO2(g) the amount of CaO after attainment of
equilibrium can be increased by –
(A) adding some CaCO3(s) (B) removing some CaCO3(s)
(B) adding some lime water (D) decreasing the temperature
Q.22 To the system A(s) + 2X(g) + heat 2B(s) + 3Y(g), at equilibrium more X(g) is added with
temperature & volume constant. If the new steady pressure of X(g) becomes 2.828 times of the
previous steady pressure, the number of moles of Y(g) are increased by a factor of –
(A) 31/3 (B) 3 (C) 2 (D) 2
Q.23 If a chemical equilibrium Reactants
F.R.
B.R. Products is shifted to the right i.e. forming the
products both by an increasing of pressure & rise in temperature then :
(A) Forward reaction is exothermic with n < 0
(B) Forward reaction is endothermic with n > 0
(C) Backward reaction is exothermic with n < 0
(D) Backward reaction is exothermic with n > 0
Q.24 The equilibrium concentrations of (A) and (B) at a certain temperature for the reaction
2A(g) + B(g) 2C(g) + D(s)
are 15 M and 30 M respectively. When the volume was made four fold and the reaction was
allowed to reach equilibrium the concentration of B was found to be 8 M.
(i) The concentration of C at first equilibrium is -
(A) 2.89 (B) 11.55 (C) 5.77 (D) 23.1
(ii) What will be the concentration of C at Second equilibrium.
(A) 1.89 M (B) 1.50 M (C) 2.89 M (D) 2.50 M
(iii) Concentration of A at second equilibrium will be -
(A) 2.75 M (B) 3.75 M (C) 4.75 M (D) 15 M
Q.25 The equilibrium concentration of the reactants and products for the given equilibrium in a two
litre container are shown below :
PCl3(g) + Cl2(g) PCl5(g)
2M 1M 4M
(i) If 2 mole of Cl2 are added in the container, then concentration PCl3 , Cl2 & PCl5 at
newnequilibrium will be
(A) 2, 1, 4 (B) 1.5, 1.5, 4.5 (C) 2, 1.5, 4.5 (D) 2, 2, 5
(ii) If the equilibrium mixture reported initially is transferred into 4 litre vessel, the what
woulnbe the new concentration of PCl3, Cl2 & PCl5 at equilibrium
(A) 1.225, 0.725, 1.775 (B) 1.5, 0.5, 2
(C) 1.775, 0.725, 1.225 (D) none of these

Ph.: 0291-3291970.
38
Answer sheet of problem sets
Problem Set – 1
1. 22
3
1
K
K =
K
2. 0.2 3. KC = 4.0 4. N2O4 5. 1.37
6. (a) 0.048 mol lit–1 (b) PAB = 0.8467 atm, PA= PB = 1.4112 atm (c) KP = 2.35 atm
7. 0.1244 mole 8. 48.22 atm–1/2 9. 8.93 atm 10.
2
(2n - y)y
(n - y) P
11.
2
2
P(n + y / 2)(n + y)
(3n + y / 2)(n - y)
12. 0.341 mole 13. 1.868
14. CH3COOH = 0.07 mol, C2H5OH = 3.07 mol, CH3COOC2H5 = 0.93 mole, H2O = 0.93 mole
15. 17.64
Problem Set – 2
1. T = 894.26ºC 2. 0.7837 3. 0.0174 gm
4. 0.582 atm 5. 31/27 7. 97.55 atm
8. (a) PCl2 = PPCl3 = 0.814, PPCl5 = 0.372 atm (b) 1.78 atm, 0.04 mol lit–1
(c) (i) 94.8% (ii) 0.0053, 0.097 atm 9. 0.6, 1.75 atm
10. 84.6%, 5.87× 10–2 mol lit–1, 2.53 atm 11. x = 0.333, Kp = 0.204 atm
12. 0.43 mole NO2 14. (i) 64% (ii) 21.42 % (iii) V = 12.998 ml
15. Backward reaction
Problem Set – 3
1. Forward 2. Forward 3. 0.5 , 1 , 1
4. Backward , Forward , Equilibrium 5. D 6. A
7. A 8. A 9. A 10. 2.7636 kCal
11. Kp = 779.41 atm–1 G = –32.998 kJ/mole, Kp = 6.07 × 105 atm–2
12. 285.23
13. (a) [A = 0.52 M, B = C = 0.48 M] (b) A = 0.07 M, B = C = 0.18 M
14. Kp = 1.429 × 10–5 atm–2 15. (i) 376.56 kJ (ii) 251.04 kJ (iii) 125.52 kJ
Problem Set – 4
1. (a) 25, shifts left, (b) 0.22, shifts right (c)  , shifts left
(d) 1 shifts right (e) 0, shift right (f) 4, shift left
2. (a) K = [Ag+] [Cl–] is less than 1. AgCl is insoluble thus the concentration of ions are much
less than 1M
(b) K = 1/[Pb2+][Cl–]2 is greater than one because PbCl2 is insoluble and formation of the
solid will reduce the concentration of ions to a low level.

Ph.: 0291-3291970.
39
4. K about 10 6. (a) incomplete (b) almost complete
7. c 8. ~ 9 × 10–32 mol/L
9. The reaction is not an equilibrium because Qc > Kc. The reaction will proceed from right to left
to reach equilibrium
11. 5.9 × 10–3 M 12. [NO] = 0.056 M, [N2] = [O2] = 13.7 M
13. [PCl3] = [Cl2] = 0.71 M, [PCl5] = 0.089
14. PClF =
2FP = 0.389 atm,
3ClFP = 1.08 atm 15. [KP = 0.4, a 0.1]
16. 50%
17. (a)6.667 × 10–3 mol L–1; (b) n(N2O4) = 0.374 mol; n (NO2) = 0.052 mol;
(c) 10.49 atm (d) 6.44%
18. 0.97 atm 19. KP = 1.3 × 10–3 atm–2 20. KP = 2.5 atm, P = 15 atm
21. 53.33% 22. K = 4 23. 31/27
24. 22.4 mg 25.
2H OP = 5 × 10–15 atm 26. 0.821 atm
27. add N2 add H2, increase the pressure, heat the reaction
28. (a) shift right, shift left, (b) shift right, no effect, (c) shift left, shift left,
(d) shift left, shift right
29. (a) K = [CH3OH]/[H2]2[CO],
(b) 1. [H2] increase, [CO] decrease, [CH3OH] increase;
2. [H2] increase, [CO] decrease, [CH3OH] decrease;
3. [H2] increase, [CO] increase, [CH3OH] increase;
4. [H2] increase, [CO] increase, [CH3OH] increase;
5. [H2] increase, [CO] increase, [CH3OH] decrease;
6. no change]
30. (a) K = [CO[H2]/[H2O];
(b) in each of the following cases the mass of carbon will change, but its concentration
(activity) will not change
1. [H2O] no change, [CO] increase,
2. [H2] increase, [CO] decrease, [CH3OH] decrease
3. [H2] increase, [CO] increase [CH3OH] increase
4. [H2O] increase, [CO] increase [H2] increase
5. [H2O] decrease, [CO] increase [H2] increase
31. b
32. Add NaCl or some other salt that produces Cl– in the solution. Cool the solution
33. a 34. kf [A][B] = kr[C] ;
f
r
k
k
=
[ ]
[ ][ ]
C
A B
= kc 36. 216
38. (i) 2, (ii) 1.2 mol/L; (iii) 0.1 moles/hr
39. kr increase more than kf , this means that Ea (reverse) is greater than Ea(forward). The
reaction is exothermic when Ea ( reverse) > Ea (forward)
43. (a) +9.574 J/mol (b) A = 1010 (c) 9.96 × 109(d) 9.98 × 109
44. +14.06 kJ 45. –810 J/mol; – 5872 J/mol and 41.3 kJ/mol
46. 1.3 × 108 47. 0.058 48. 29.0

Ph.: 0291-3291970.
40
49. KP = 0.0313 atm, KC = 1.28 × 10–3
50. (a) KC =
3
2
3
[ ]
[ ]
CO
CO
, KP = 2
3
3
( )
( )
CO
CO
P
P
; (b) KC = 3
2
1
[ ]O
, KP =
2
3
1
( )OP
,
(c) KC = [SO3], KP = 3SOP , KC = [Ba2+][SO4
2–]
51. KC = 1.51 k, KP = 49.6 52. 1.5 × 10–6 M
53. [CO] = [H2] = 0.18 M; [H2O] = 1.02 M
54. (a) KC = 0.573 and KP = 23.5 ;
(b) to the right, [PCl5] = 0.365 M; [PCl3] = 0.285M, ; [Cl2] = 0.735M
56. –1.005 kJ/mol 57. Gº = 0 ; K = 1
58. Hº = 9.07 kJ/mol; Sº = – 8.92 J/mol–1K–1
Problem Set – 5
1. A 2. D 3. D 4. C 5. D 6. C 7. A 8. D
9. A 10. B 11. D 12. D 13. A 14. A 15. D 16. C
17. D 18. D 19. B 20. A 21. B 22. C 23. A 24. B
25. A
Problem Set – 6
1. D 2. B 3. D 4. A 5. D 6. D 7. C 8. A
9. A 10. A 11. C 12. C 13. A 14. D 15. B 16. B
17. C 18. D 19. A 20. A 21. B 22. B 23. B 24. C
25. C
Problem Set – 7
1. B 2. B 3. D 4. D 5. A 6. A 7. B 8. B
9. C 10. B 11. D 12. B 13. D 14. C 15. A 16. A
17. D 18. C 19. C 20. D 21. B 22. C 23. A
24. (i) B (ii) A (iii) A 25. (i) B (ii) A

Ph.: 0291-3291970.
41
Exercises
Exercise – 1: Single Choice Questions
Q.1 A chemical reaction, A B, is said to be in equilibrium when :
(A) rate of forward reaction is equal to rate of backward reaction
(B) conversion of A to B is only 50% complete
(C) complete conversion of A to B has taken place
(D) only 25% conversion of A to B taken place
Q.2 The equilibrium concentration of x, y and yx2 are 4, 2 and 2 respectively for the equilibrium 2x +
y yx2. The value of equilibrium constant, KC is
(A) 0.625 (B) 6.25 (C) 0.0625 (D) 62.5
Q.3 4 mole of A are mixed with 4 mole of B when 2 mole of C are formed at equilibrium, according
to the reaction, A + B C + D the equilibrium constant is :
(A) (B) 2 (C) 1 (D) 4
Q.4 HI was heated in a sealed tube at 400°C till the equilibrium was reached. HI was found to be 22%
decomposed. The equilibrium constant for dissociation is :
(A) 1.99 (B) 0.0199 (C) 0.0796 (D) 0.282
Q.5 In which of the following, the reaction proceeds towards completion ?
(A) K = 1 (B) K = 10–2 (C) K = 10 (D) K = 103
Q.6 For the following reaction at 250°C, PCl3(g) + Cl2(g) PCl5(g) the value of KC is 26 then
the value of Kp at same temperature will be
(A) 0.57 (B) 0.61 (C) 0.83 (D) 0.91
Q.7 For a reversible reaction, the rate constants for the forward and backward reactions are 2.38 ×10–
4 and 8.15 × 10–5 respectively. The equilibrium constant for the reaction is –
(A) 0.342 (B) 2.92 (B) 0.292 (D) 3.42
Q.9 If different quantities of ethanol and acetic acid was used in the following reversible reaction,
CH3COOH() + C2H5OH() CH3COOC2H5() + H2O()
the equilibrium constant will have values which will be ?
(A) different in all cases (B) same in all cases
(C) higher in cases where higher concentration of ethanol is used
(D) higher in case where higher concentration of acetic acid is used
Q.10 Which one of the following oxides is most stable? The equilibrium constants are given at the
same temperature:
(A) 2N2O5(g) 2N2(g) + 5O2(g) ; K = 1.2 × 1034
(B) 2N2O(g) 2N2(g) + O2(g) ; K = 3.5 × 1035
(C) 2NO(g) N2(g) + O2(g) ; K = 2.2 × 1030
(D) 2NO2(g) N2(g) + 2O2(g) ; K = 6.71 × 1016
Q.11 The yield of product in the reaction, A2(g) + 2B(g) C(g) + Q kJ
would be higher at :

Ph.: 0291-3291970.
42
(A) low temperature and high pressure (B) high temperature and high pressure
(B) low temperature and low pressure (D) high temperature and low pressure
Q.12 Manufacture of ammonia from the elements is represented by
N2(g) + 3H2(g) 2NH3(g) + 22.4 kcal the maximum yield of ammonia will be obtained
when the process is made to take place –
(A) at low pressure and high temperature (B) at low pressure and low temperature
(C) at high pressure and high temperature (D) at high pressure and low temperature
Q.13 In the reaction, 2SO2(g) + O2(g) 2SO3(g) + X cal, most favourable conditions of
temperature and pressure for greater yield of SO3 are :
(A) low temperature and low pressure (B) high temperature and low pressure
(C) high temperature and high pressure (D) low temperature and high pressure
Q.14 For the reaction : 2A(g) + B(g) 3C(g) + D(g)
two mole each of A and B were taken into a flask. The following must always be true when the
system attained equilibrium :
(A) [A] = [B] (B) [A] < [B] (C) [B] = [C] (D) [A] > [B]
Q.15 In a vessel containing SO2, SO3 and O2 at equilibrium, some helium gas is introduced so that
total pressure increases while temperature and volume remain the same. According to the Le
Chatelier‘s principle, the dissociation of SO3 :
(A) increases (B) decreases (C)remains unaltered (D) change unpredictably
Q.16 The equilibrium SO2Cl2(g) SO2(g) + Cl2(g) is attained at 25°C in a closed container and
an inert gas, helium, is introduced. Which of the following statements is correct ?
(A) concentrations of SO2Cl2, SO2 and Cl2 do not change
(B) more Cl2 is formed
(C) concentration of SO2 is reduced
(D) more SO2Cl2 is formed
Q.17 For the reactions,
A B, KC = 1, B C, KC = 3; C D, KC = 5
KC for the reaction A D is :
(A) 15 (B) 5 (C) 3 (D) 1
Q.18 The equilibrium constant for the reaction, N2(g) + O2(g) 2NO(g) is 4.0 × 10–4 at 2000
K. In the presence of a catalyst the equilibrium is attained ten times faster. Therefore, the
equilibrium constant in presence of the catalyst at 2000 K is –
(A) 4 × 10–4 (B) 40 × 10–4
(D) 4 × 10–2 (D) difficult to compute without more data
Q.19 The equilibrium constant for a reaction A + B C + D is 1 × 10–2 at 298 K and is 2 at 273
K. The chemical process resulting in the formation of C and D is :
(A) exothermic (B) endothermic
(C) unpredictable (D) there is no relationship between H and K
Q.20 In a flask colourless N2O4 is in equilibrium with brown coloured NO2. At equilibrium, when the
flask is heated to 100°C the brown colour deepens and on cooling, the brown colour became less
coloured. The change in enthalpy H for the system is:

Ph.: 0291-3291970.
43
(A) negative (B) positive (C) zero (D) not defined
Q.21 If 340 g of a mixture of N2 and H2 in the correct ratio gave a 20% yield of NH3, the mass
produced would be:
(A) 16 g (B) 17 g (C) 20 g (D) 68 g
Q.22 In a chemical reaction Kp is greater than Kc when
(A) the number of molecules entering into the chemical reaction is more than the number of
molecules produced.
(B) the number of molecules entering into the chemical reaction is the same as the number of
molecules produced.
(C) the number of molecules entering into the chemical reaction is less than the number of
molecules produced.
(D) the total number of moles of reactants is less than the number of moles of products.
Q.23 Which of the following statements is not correct about the equilibrium constant ?
(A) Its value does not depend upon the initial conc. of the reactants
(B) Its value does not depend upon the initial conc. of the products
(C) Its value does not depend upon nature of the reactants.
(D) Its value does not depend upon presence of catalyst.
Q.24 The equilibrium constant for the reaction, N2(g) + O2(g) 2NO(g) is 4 10–4
at 2000 K. In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the
equilibrium constant, in presence of the catalyst, at 2000 K is
(A) 40 x 10–4 (B) 4 x 10–4
(C) 4 x 10–3 (D) difficult to compute without more data.
Q.25 2XY
dissociates as 2(g) (g) (g)XY XY Y . When the initial pressure of is 600 mm of Hg,
the total pressure developed is 800 mm of Hg. for the reaction is
(A) 200 (B) 50 (C) 100 (D) 150
Q.26 The reactions 5(g) 3(g) 2(g)PCl PCl Cl and 2(g) (g) 2(g)COCl CO Cl
are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of
(g)CO
are later introduced into the vessel. After some time, the new equilibrium concentration of
(A) 5PCl
will remain unchanged (B) 2Cl
will be greater
(C) 5PCl
will become less (D) 5PCl
will become greater.
Q.27 For the reaction : (g) 2(g) 2(g)2HI H I ; the degree of dissociation () of HI(g) is related to
equilibrium constant by the expression
(A)
p1 2 K
2

(B)
p1 2K
2

(C)
p
p
2K
1 2K
(D)
p
p
2 K
1 2 K
.
Q.28 For which of the following reactions, the degree of dissociation cannot be calculated from the
vapour density data
I. (g) 2(g) 2(g)2HI H I II. 3(g) 2(g) 2(g)2NH N 3H
III. (g) 2(g) 2(g)2NO N O IV. 5(g) 3(g) 2(g)PCl PCl Cl
(A)I and III (B) II and IV (C) I and II (D) III and IV.

Ph.: 0291-3291970.
44
Exercise – 2: Subjective Type Questions
Q.1 For the reaction A + B 3C at 25ºC, a, 3 litre vessel contains 1,2,4 mole of A, B and C
respectively. Predict the direction of reaction if:
(a) KC for the reaction is 10.
(b) KC for the reaction is 15.
(c) KC for the reaction is 10.66.
Q.2 The equilibrium constant KC for A(g) B(g) is 1.1. Which gas has a molar concentration
greater than 1?
Q.3 For a gaseous phase reaction, A + 2B AB2, KC = 0.3475 litre2 mol–2 at 200ºC. When 2
mole of B are mixed with one mole of A, what total pressure is required to convert 60% of A in
AB2?
Q.4 For a gaseous phase reaction, 2HI H2 + I2, at equilibrium 7.8 g, 203.2 g and 1638.4 g of
H2, I2 and HI respectively were found in 5 litre vessel. Calculate KC. If all the reactants and
products are transferred to a 2 litre vessel., what will be the amount of reactants and product at
equilibrium?
Q.5 One mole of H2, two mole of I2 and three mole of HI are injected in one litre flask. What will be
the concentration of H2, I2 and HI at equilibrium at 500ºC? KC for reaction:
H2 + I2 2HI is 45.9.
Q.6 The degree of dissociation of HI at a particular temperature is 0.8. Calculate the volume of 2M
Na2S2O3 solution required to neutralise the iodine present in a equilibrium mixture of a reaction
when 2 mole each of H2 and I2 are heated in a closed vessel of 2 litre capacity.
Q.7 A mixture of one mole of CO2 and one mole of H2 attains equilibrium at a temperature of 250ºC
and a total pressure of 0.1 atm for the change CO2(g) + H2(g) CO(g) + H2O(g). Calculate
KP if the analysis of final reaction mixture shows 0.16 volume per cent of CO.
Q.8 At a certain temperature, KC for
SO2(g) + NO2(g) SO3(g) + NO(g)
is 16. 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 ?
Q.9 At 627ºC and one atmosphere pressure SO3 is partially dissociated into SO2 and O2 by SO3(g)
SO2(g) +
1
2
O2(g). The density of the equilibrium mixture is 0.925 g/litre. What is the
degree of dissociation?
Q.10 The equilibrium mixture for 2SO2(g) + O2(g) 2SO3(g)
present in 1 litre vessel at 600ºC contains 0.50, 0.12 and 5.0 mole of SO2, O2 and SO3
respectively.
(a) Calculate KC for the given change at 600ºC
(b)Also calculate KP
(c) How many mole of O2 must be forced into the equilibrium vessel at 600ºC in order to increase
the concentration of SO3 to 5.2 mole?

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