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JEE Advanced Model Paper 7 with solutions for online practice
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© examsnet.com
Question : 66
Total: 66
The equilibrium pressure of
N
H
4
C
N
(
s
)
⇌
N
H
3
3
(
g
)
+
H
C
N
(
g
)
is 0.298 atm. If
N
H
4
C
N
is allowedto decompose in presence of
N
H
3
at 0.25 atm, calculate the partial pressure of HCN atequilibrium. It partial pressure of HCN at equilibrium is y ×
10
−
2
atm , then ehat is y
Your Answer:
Validate
Solution:
P
N
H
4
=
P
H
C
N
= 0.149 atm ⇒
K
p
= 0.149 × 0.149 = 0.022 atm
In second case :
N
H
4
C
N
→
N
H
3
+ HCN
Initially : (0.25) , 0
At equation : (0.25 + x) , x
K
p
= (0.25 + x) x = 0.022
or
x
2
+ 0.25x - 0.022 = 0
∴ x =
−
0.25
+
√
(
0.25
)
2
−
4
(
1
)
(
0.022
)
2
(
1
)
=
−
0.25
+
√
0.0625
−
0.088
2
=
−
0.25
+
√
0.1505
2
=
−
0.25
+
0.388
2
= 0.07
∴ [HCN] = 7 ×
10
−
2
Hence y = 7
© examsnet.com
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