VITEEE 2010 Paper

For the given reaction,
2A(g)+B2(g)⇌2AB2(g)
The equilibrium constant,
Kp=

P2AB2

P2A.PB2

=16...(i)
For the other given reaction,
AB2(g)⇌A(g)+

1

2

B2(g)
The equilibrium constant,
K′p=

PA.PB21∕2

PAB2

...(ii)
On squaring Eq.(ii), we obtain,
(K′p)2=

P2A.PB2

P2AB2

...(iii)
Now, from Eq.(i) and (ii), we obtain,
Kp.(K′p)2=1
⇒16.(K′p)2=1‌‌‌‌‌‌‌‌‌(∵Kp=16.0)
⇒(K′p)=[

1

16

]1∕2
⇒K′p=

1

4

=0.25