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# NCERT Solution - Equilibrium Class 11 Notes | EduRev

## Class 11 : NCERT Solution - Equilibrium Class 11 Notes | EduRev

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CHAPTER-7 NCERT SOLUTION

Question 7.1:
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume
of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the volume of the container is suddenly increased, then the vapour pressure would
decrease initially. This is because the amount of vapour remains the same, but the volume
increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the
volume of the container is increased, the density of the vapour phase decreases. As a result, the
rate of collisions of the vapour particles also decreases. Hence, the rate of condensation
decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of
condensation. In this case, only the volume changes while the temperature remains constant.
The vapour pressure depends on temperature and not on volume. Hence, the final vapour
pressure will be equal to the original vapour pressure of the system.

Question 7.2:
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is:
[SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?

The equilibrium constant (K
c
) for the give reaction is:

Hence, K
c
for the equilibrium is .

Question 7.3:
At a certain temperature and total pressure of 10
5
Pa, iodine vapour contains 40% by volume of I
atoms

Calculate K
p
for the equilibrium.

Partial pressure of I atoms,

Page 2

CHAPTER-7 NCERT SOLUTION

Question 7.1:
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume
of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the volume of the container is suddenly increased, then the vapour pressure would
decrease initially. This is because the amount of vapour remains the same, but the volume
increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the
volume of the container is increased, the density of the vapour phase decreases. As a result, the
rate of collisions of the vapour particles also decreases. Hence, the rate of condensation
decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of
condensation. In this case, only the volume changes while the temperature remains constant.
The vapour pressure depends on temperature and not on volume. Hence, the final vapour
pressure will be equal to the original vapour pressure of the system.

Question 7.2:
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is:
[SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?

The equilibrium constant (K
c
) for the give reaction is:

Hence, K
c
for the equilibrium is .

Question 7.3:
At a certain temperature and total pressure of 10
5
Pa, iodine vapour contains 40% by volume of I
atoms

Calculate K
p
for the equilibrium.

Partial pressure of I atoms,

Partial pressure of I
2
molecules,

Now, for the given reaction,

Question 7.4:
Write the expression for the equilibrium constant, K
c
for each of the following
reactions:
(i)
(ii)
(iii)
(iv)
(v)

Page 3

CHAPTER-7 NCERT SOLUTION

Question 7.1:
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume
of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the volume of the container is suddenly increased, then the vapour pressure would
decrease initially. This is because the amount of vapour remains the same, but the volume
increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the
volume of the container is increased, the density of the vapour phase decreases. As a result, the
rate of collisions of the vapour particles also decreases. Hence, the rate of condensation
decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of
condensation. In this case, only the volume changes while the temperature remains constant.
The vapour pressure depends on temperature and not on volume. Hence, the final vapour
pressure will be equal to the original vapour pressure of the system.

Question 7.2:
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is:
[SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?

The equilibrium constant (K
c
) for the give reaction is:

Hence, K
c
for the equilibrium is .

Question 7.3:
At a certain temperature and total pressure of 10
5
Pa, iodine vapour contains 40% by volume of I
atoms

Calculate K
p
for the equilibrium.

Partial pressure of I atoms,

Partial pressure of I
2
molecules,

Now, for the given reaction,

Question 7.4:
Write the expression for the equilibrium constant, K
c
for each of the following
reactions:
(i)
(ii)
(iii)
(iv)
(v)

Question 7.5:
Find out the value of K
c
for each of the following equilibria from the value of K
p
:

The relation between K
p
and K
c
is given as:
K
p
= K
c
(RT)
?n

(a) Here,
?n = 3 – 2 = 1
R = 0.0831 barLmol
–1
K
–1

T = 500 K
K
p
= 1.8 × 10
–2

Now,
K
p
= K
c
(RT)
?n

(b) Here,
?n = 2 – 1 = 1
Page 4

CHAPTER-7 NCERT SOLUTION

Question 7.1:
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume
of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the volume of the container is suddenly increased, then the vapour pressure would
decrease initially. This is because the amount of vapour remains the same, but the volume
increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the
volume of the container is increased, the density of the vapour phase decreases. As a result, the
rate of collisions of the vapour particles also decreases. Hence, the rate of condensation
decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of
condensation. In this case, only the volume changes while the temperature remains constant.
The vapour pressure depends on temperature and not on volume. Hence, the final vapour
pressure will be equal to the original vapour pressure of the system.

Question 7.2:
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is:
[SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?

The equilibrium constant (K
c
) for the give reaction is:

Hence, K
c
for the equilibrium is .

Question 7.3:
At a certain temperature and total pressure of 10
5
Pa, iodine vapour contains 40% by volume of I
atoms

Calculate K
p
for the equilibrium.

Partial pressure of I atoms,

Partial pressure of I
2
molecules,

Now, for the given reaction,

Question 7.4:
Write the expression for the equilibrium constant, K
c
for each of the following
reactions:
(i)
(ii)
(iii)
(iv)
(v)

Question 7.5:
Find out the value of K
c
for each of the following equilibria from the value of K
p
:

The relation between K
p
and K
c
is given as:
K
p
= K
c
(RT)
?n

(a) Here,
?n = 3 – 2 = 1
R = 0.0831 barLmol
–1
K
–1

T = 500 K
K
p
= 1.8 × 10
–2

Now,
K
p
= K
c
(RT)
?n

(b) Here,
?n = 2 – 1 = 1
R = 0.0831 barLmol
–1
K
–1

T = 1073 K
K
p
= 167
Now,
K
p
= K
c
(RT)
?n

Question 7.6:
For the following equilibrium,

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions.
What is K
c
, for the reverse reaction?

It is given that  for the forward reaction is
Then, for the reverse reaction will be,

Question 7.7:
Explain why pure liquids and solids can be ignored while writing the equilibrium constant
expression?

For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always
fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances
are not mentioned in the equilibrium constant expression.

Question 7.8:
Reaction between N
2
and O
2
takes place as follows:
Page 5

CHAPTER-7 NCERT SOLUTION

Question 7.1:
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume
of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the volume of the container is suddenly increased, then the vapour pressure would
decrease initially. This is because the amount of vapour remains the same, but the volume
increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the
volume of the container is increased, the density of the vapour phase decreases. As a result, the
rate of collisions of the vapour particles also decreases. Hence, the rate of condensation
decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of
condensation. In this case, only the volume changes while the temperature remains constant.
The vapour pressure depends on temperature and not on volume. Hence, the final vapour
pressure will be equal to the original vapour pressure of the system.

Question 7.2:
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is:
[SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?

The equilibrium constant (K
c
) for the give reaction is:

Hence, K
c
for the equilibrium is .

Question 7.3:
At a certain temperature and total pressure of 10
5
Pa, iodine vapour contains 40% by volume of I
atoms

Calculate K
p
for the equilibrium.

Partial pressure of I atoms,

Partial pressure of I
2
molecules,

Now, for the given reaction,

Question 7.4:
Write the expression for the equilibrium constant, K
c
for each of the following
reactions:
(i)
(ii)
(iii)
(iv)
(v)

Question 7.5:
Find out the value of K
c
for each of the following equilibria from the value of K
p
:

The relation between K
p
and K
c
is given as:
K
p
= K
c
(RT)
?n

(a) Here,
?n = 3 – 2 = 1
R = 0.0831 barLmol
–1
K
–1

T = 500 K
K
p
= 1.8 × 10
–2

Now,
K
p
= K
c
(RT)
?n

(b) Here,
?n = 2 – 1 = 1
R = 0.0831 barLmol
–1
K
–1

T = 1073 K
K
p
= 167
Now,
K
p
= K
c
(RT)
?n

Question 7.6:
For the following equilibrium,

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions.
What is K
c
, for the reverse reaction?

It is given that  for the forward reaction is
Then, for the reverse reaction will be,

Question 7.7:
Explain why pure liquids and solids can be ignored while writing the equilibrium constant
expression?

For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always
fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances
are not mentioned in the equilibrium constant expression.

Question 7.8:
Reaction between N
2
and O
2
takes place as follows:

If a mixture of 0.482 mol of N
2
and 0.933 mol of O
2
is placed in a 10 L reaction vessel and
allowed to form N
2
O at a temperature for which K
c
= 2.0 × 10
–37
, determine the composition of
equilibrium mixture.

Let the concentration of N
2
O at equilibrium be x.
The given reaction is:

Therefore, at equilibrium, in the 10 L vessel:

The value of equilibrium constant i.e., = 2.0 × 10
–37
is very small. Therefore, the amount of
N
2
and O
2
reacted is also very small. Thus, x can be neglected from the expressions of molar
concentrations of N
2
and O
2
.
Then,

Now,

Question 7.9:
Nitric oxide reacts with Br
2
and gives nitrosyl bromide as per reaction given below:

When 0.087 mol of NO and 0.0437 mol of Br
2
are mixed in a closed container at constant
temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO
and Br
2
.

The given reaction is:
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