AIEEE 2008 Solved Paper

Let the initial moles of X be ′a′
and that of Z be ‌′b′ the for the given reactions,
we have X⇌2Y

Initial	a moles	0
At equi. (moles)	‌‌a(1−α)	2aα

Total no. of moles =a(1−α)+2aα
=a−aα+2aα=a(1+α)
Now, Kp1=‌

(ny)2

nx

×(‌

PT1

∑n

)∆n
or, Kp1=‌

(2aα)2⋅PT1

[a(1−α)][a(1+α)]

	Z‌‌‌‌⇌	P‌‌+‌‌Q
Initial	b moles	0‌‌‌‌‌‌‌‌‌0
Ar equi.	b(1−α)	bα‌‌‌‌bα

Total no. of moles
=b(1−α)+bα+bα
=b−bα+bα+bα
=b(1+α)
Now KP2=‌

nQ×nP

nz

×[‌

PT2

∑n

]∆n
or KP2=‌

(bα)(bα)⋅PT2

[b(1−α)][b(1+α)]

Or ‌‌‌

KP1

KP2

=‌

4α2⋅PT1

(1−α2)

×‌

(1−α)2

PT2⋅α2

=‌

4PT1

PT2

or ‌

PT1

PT2

=‌

1

9

‌‌ [ as ‌

KP1

KP1

=‌

1

9

given ]
or ‌

PT1

PT2

=‌

1

36

or 1:36