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Class 11
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Chemistry
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Equilibrium
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Homogeneous and Heterogeneous Equilibria
Homogeneous and Heterogeneous Equilibria
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Quick Summary With Stories
Classification of Chemical Equilibrium
2 mins
Important Questions
Derive the relation between
K
p
and
K
c
for a general chemical equilibrium reaction.
Medium
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>
The dissociation equilibrium of a gas
A
B
2
can be represented as:
2
A
B
2
(
g
)
⇌
2
A
B
(
g
)
+
B
2
(
g
)
The degree of dissociation is
x
and is small compared to
1
. The expression relating the degree of dissociation
x
with equilibrium constant
K
p
and total pressure
p
is:
Hard
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>
Write the expression for the equilibrium constant,
K
c
for each of the following reactions.
(i)
2
N
O
C
l
(
g
)
⇌
2
N
O
(
g
)
+
C
l
2
(
g
)
(ii)
2
C
u
(
N
O
3
)
2
(
s
)
⇌
2
C
u
O
(
s
)
+
4
N
O
2
(
g
)
+
O
2
(
g
)
(iii)
C
H
3
C
O
O
C
2
H
5
(
a
q
)
+
H
2
O
(
1
)
⇌
C
H
3
C
O
O
H
(
a
q
)
+
C
2
H
5
O
H
(
a
q
)
(iv)
F
e
3
+
(
a
q
)
+
3
O
H
−
(
a
q
)
⇌
F
e
(
O
H
)
3
(
s
)
(v)
I
2
(
s
)
+
5
F
2
⇌
2
I
F
5
Hard
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>
2
moles of
P
C
l
5
were heated in a closed vessel of 2 litre capacity. At equilibrium,
4
0
%
of
P
C
l
5
is dissociated into
P
C
l
3
and
C
l
2
. The value of equilibrium constant is-
Hard
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>
For the reaction
P
C
l
3
(
g
)
+
C
l
2
(
g
)
⇌
P
C
l
5
(
g
)
the value of
K
c
at
2
5
0
o
C
is 26. The value of
K
p
at this temperature will be :
Hard
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>
For the reaction equilibrium
2
N
O
B
r
(
g
)
⇋
2
N
O
(
g
)
+
B
r
2
(
g
)
, if
P
B
r
2
=
9
P
at equilibrium and P is the initial total pressure, then the ratio
P
K
P
is equal to :
Hard
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>
At certain temperature compound
A
B
2
(
g
)
dissociates according to the reaction:
2
A
B
2
(
g
)
⇌
2
A
B
(
g
)
+
B
2
(
g
)
With the degree of dissociation
α
, which is small compared with unity. The expression of
K
p
, in terms of
α
and initial pressure P is:
Hard
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>
lo
g
K
c
K
p
+
lo
g
R
T
=
0
is true relationship for the following reaction:
Hard
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>
In which of the following equilibrium,
K
c
=
K
p
?
Medium
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>
For the reaction
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
, the degree of dissociation at equilibrium is
0
.
2
at
1
atm pressure.
The equilibrium constant
K
p
will be:
Hard
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>