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KEAM 2015 Physics and Chemistry Paper
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© examsnet.com
Question : 108
Total: 120
Equilibrium constants are given for the following two equilibria
(i)
A
2
(
g
)
+
B
2
(
g
)
⇌
2
A
B
(
g
)
;
K
=
2
×
10
−
4
(ii)
2
A
B
(
g
)
+
C
2
(
g
)
⇌
2
A
B
C
(
g
)
;
K
=
2
×
10
−
2
L
m
o
l
−
1
Calculate the equilibrium constant for the following equilibrium
A
B
C
(
g
)
⇌
1
2
A
2
(
g
)
+
1
2
B
2
(
g
)
+
1
2
C
2
(
g
)
500
m
o
l
1
2
L
1
2
4
×
10
−
6
m
o
l
1
2
L
1
2
500
m
o
l
−
1
2
L
1
2
200
m
o
l
1
2
L
−
1
2
500
m
o
l
1
2
L
−
1
2
Validate
Solution:
A
2
(
g
)
+
B
2
(
g
)
⇌
2
A
B
(
g
)
K
1
=
2
×
10
−
4
L
/
m
o
l
K
1
=
[
A
B
]
2
[
A
2
]
[
B
2
]
....(i)
2
A
B
(
g
)
+
C
2
(
g
)
⇌
2
A
B
C
(
g
)
K
2
=
2
×
10
−
2
L
∕
m
o
l
So,
K
2
=
[
A
B
C
]
2
[
A
B
]
2
[
C
2
]
....(ii)
A
B
C
⇌
1
2
A
2
(
g
)
+
1
2
B
2
(
g
)
+
1
2
C
2
(
g
)
So.
K
3
=
|
A
2
|
1
/
2
[
B
2
|
1
/
2
|
C
2
|
1
/
2
|
A
B
C
|
.....(iii)
Eq. (i) and Eq. (ii) are added, reversed and divided by 2 then we get Eq. (iii).
Therefore,
K
3
=
1
√
K
1
×
K
2
=
1
√
2
×
10
−
4
×
2
×
10
−
2
=
1000
2
=
500
mol
1
/
2
L
−
1
/
2
© examsnet.com
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