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AIEEE 2008 Solved Paper

Section: Chemistry

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Question : 7 of 104

Marks: +1, -0

The equilibrium constants

K_{P 1}

K_{P 2}

for the reactions

X \leftrightharpoons 2 Y

Z \rightleftharpoons P+Q

, respectively are in the ratio of

1: 9

. If the degree of dissociation of

X

Z

be equal then the ratio of total pressure at these equilibria is

$1: 36$
$1: 1$
$1: 3$
$1: 9$

Solution:

Let the initial moles of

X

be

{}^\prime a^\prime

and that of

Z

be

{}^\prime b^\prime

the for the given reactions,

we have

X \rightleftharpoons 2 Y

Initial	a moles	0
At equi. (moles)	$\; a(1-\alpha)$	$2 a \alpha$

Total no. of moles

=a(1-\alpha)+2 a \alpha

=a-a\alpha+2 a\alpha=a(1+\alpha)

Now,

K_{p_1}= \frac{(n_y)^2}{n_x} \times \left( \frac{P_{T_1}}{\sum n} \right)^{\Delta n}

or,

K_{p_1}= \frac{(2 a \alpha)^2 \cdot P_{T_1}}{[a(1-\alpha)][a(1+\alpha)]}

	$Z\;\;\;\;\rightleftharpoons$	$P\;\;+\;\;Q$
Initial	b moles	$0\;\;\;\;\;\;\;\;\;0$
Ar equi.	$b(1-\alpha)$	$b\alpha\;\;\;\;b\alpha$

Total no. of moles

=b(1-\alpha)+b\alpha+b\alpha

=b-b\alpha+b\alpha+b\alpha

=b(1+\alpha)

Now

K_{P_2}= \frac{n_Q \times n_P}{n_z} \times \left[ \frac{P_{T_2}}{\sum_{n}} \right]^{\Delta n}

or

K_{P_2}= \frac{(b\alpha)(b\alpha) \cdot P_{T_2}}{[b(1-\alpha)][b(1+\alpha)]}

Or

\;\; \frac{K_{P_1}}{K_{P 2}} = \frac{4 \alpha^2 \cdot P_{T_1}}{(1-\alpha^2)} \times \frac{(1-\alpha)^2}{P_{T_2} \cdot \alpha^2}

= \frac{4 P_{T_1}}{P_{T_2}}

or

\frac{P_{T_1}}{P_{T 2}} = \frac{1}{9} \;\;

[ as

\frac{K_{P_1}}{K_{P_1}} = \frac{1}{9}

given ]

or

\frac{P_{T_1}}{P_{T 2}} = \frac{1}{36}

or

1: 36

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