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Byju's Answer
Standard XII
Chemistry
Derivation of Kp and Kc
A mixture of ...
Question
A mixture of
3
moles of
S
O
2
,
4
moles of
N
O
2
and
4
moles of
N
O
is placed in a
2.0
L
vessel.
S
O
2
(
g
)
+
N
O
2
(
g
)
⇌
S
O
3
(
g
)
+
N
O
(
g
)
.
At equilibrium, the vessel is found to contain
1
mole of
S
O
2
. Calculate the value of
K
c
.
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Solution
Reaction:
S
O
2
(
g
)
+
N
O
2
(
g
)
⇌
S
O
3
(
g
)
+
N
O
(
g
)
Initial Moles
3
4
0
4
At equilibrium
3-2=1
4-2=2
2
4+2=6
So,
K
c
=
[
S
O
3
]
[
N
O
]
[
S
O
2
]
[
N
O
2
]
=
2
×
6
1
×
2
=
6
The correct answer is 6
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Similar questions
Q.
An equilibrium mixture
S
O
2
(
g
)
+
N
O
2
(
g
)
⇌
S
O
3
(
g
)
+
N
O
(
g
)
was found to contain
0.6
mole of
S
O
2
,
0.2
mole of
N
O
2
,
0.8
mole of
S
O
2
and
0.3
mole of
N
O
in a liter. vessel. How many moles of
N
O
2
per it. will have to be added to the vessel in order to increase the
N
O
concentration to
0.5
mole/liter.
Q.
The equilibrium mixture,
S
O
3
(
g
)
+
N
O
(
g
)
⇌
S
O
2
(
g
)
+
N
O
2
(
g
)
contains 0.6 mol of
S
O
3
, 0.4 mol of No, 0.8 mol of
S
O
2
and 0.1 mol of
N
O
2
per litre. If 1 mol of NO is introduced into the reaction vessel, then the number of moles of
S
O
3
,
N
O
,
S
O
2
a
n
d
N
O
2
of each gas in the new
equilibrium in moles per litre will be:
Q.
5 moles of
S
O
2
and 5 moles of
O
2
react in a closed vessel. At equilibrium 60% of the
S
O
2
is consumed. The total number of gaseous moles (
S
O
2
,
O
2
and
S
O
3
) in the vessel is:
Q.
5
moles of
S
O
2
and
5
moles of
O
2
are allowed to react to form
S
O
3
in a closed vessel. At the equilibrium,
60
% of
S
O
2
is used up. The total number of moles in the vessel at equilibrium is:
Q.
The equilibrium mixture for formation of sulphur trioxide from sulphur dioxide present in 1 litre vessel at
600
∘
C
contains 0.5, 0.12 and 5 mole of
S
O
2
,
O
2
and
S
O
3
.
A) Calculate
K
c
at
600
∘
C
B) Calculate
K
p
C) How many moles of
O
2
must be forced into reaction vessel at
600
∘
C
in order increase the concentration of
S
O
3
to 5.2 moles?
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