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Byju's Answer
Standard XII
Chemistry
Introduction to Le Chatelier's Principle
Density of eq...
Question
Density of equilibrium mixture of
N
2
O
4
and
N
O
2
at
1
a
t
m
and
384
K
is
1.84
g
d
m
−
3
. Calculate the equilibrium constant of the reaction,
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
.
Open in App
Solution
We know that,
P
M
m
i
x
=
d
R
T
1
×
M
m
i
x
=
1.84
×
0.0821
×
384
M
=
29
×
2
Vapour density
(
d
)
at equilibrium
=
29
Initial vapour density
=
M
/
2
=
92
/
2
=
46
x
=
D
−
d
(
n
−
1
)
d
=
46
−
29
29
=
0.586
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
t
=
0
1
0
t
e
q
.
1
−
x
2
x
Total moles
=
1
+
x
p
N
2
O
4
=
(
1
−
x
)
(
1
+
x
)
×
P
;
p
N
O
2
=
2
x
(
1
+
x
)
×
P
K
p
=
4
x
2
P
1
−
x
2
=
4
×
(
0.586
)
2
×
1
1
−
(
0.586
)
2
=
2.09
a
t
m
Suggest Corrections
0
Similar questions
Q.
Gaseous
N
2
O
4
dissociates into gaseous
N
O
2
according to the reaction
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
. At
300
K
and
1
a
t
m
pressure, the degree of dissociation of
N
2
O
4
is
0.2
. If one mole of
N
2
O
4
gas is contained in a vessel, then the density of the equilibrium mixture is:
Q.
The equilibrium constant
K
p
for the reaction
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
is
4.5
. What would be the average molar mass (in g/mol) of an equilibrium mixture of
N
2
O
4
and
N
O
2
formed by the dissociation of pure
N
2
O
4
at a total pressure of
2
atm?
Q.
N
2
O
4
(
g
)
→
2
N
O
2
(
g
)
at a constant temperature, vapour density of
N
2
O
4
is found to be 30. Calculate % dissociation of
N
2
O
4
.
Q.
The decomposition of
N
2
O
4
to
N
O
2
is carried at
280
K
in chloroform. When equilirbrium has been established,
0.2
mole of
N
2
O
4
and
2
×
10
−
3
mole of
N
O
2
are present in
2
litre solution. The equilibrium constant for the reaction,
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
is:
Q.
For the reaction in equilibrium,
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
The concentrations of
N
2
O
4
and
N
O
2
at equilibrium are
4.8
×
10
−
2
and
1.2
×
10
−
2
m
o
l
L
−
1
respectively. The value of
K
c
for this reaction is:
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