The equilibrium constants Kp1 and Kp2 for the reactions, X⇌2Y and Z⇌P+Q respectively, are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal than the ratio of total pressures at these equilibrium is

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The equilibrium constants $K_{p 1}$ and $K_{p 2}$ for the reactions, $X ⇌ 2 Y$ and $Z ⇌ P + Q$ respectively, are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal than the ratio of total pressures at these equilibrium is

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$1 : 9$

$1 : 36$

1 : 3

1 : 1

answer is B.

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Detailed Solution

We have

$X ⇌ 2 Y p_{X} (1 - α) p_{X} (2 α) p_{eql} = p_{X} (1 - α) + p_{X} (2 α) = p_{X} (1 + α)$

$K_{p 1} = (\frac{p_{Y}^{2}}{p_{X}}) = \frac{{\{p_{X} (2 α)\}}^{2}}{p_{X} (1 - α)} = \frac{4 p_{X} α^{2}}{1 - α}$

$Z ⇌ P + Q$

$p_{Z} (1 - α) p_{Z} α p_{Z} α p_{eq 2} = p_{Z} (1 - α) + p_{Z} α + p_{Z} α = p_{Z} (1 + α)$

$K_{p 2} = (\frac{p_{P} p_{Q}}{p_{Z}}) = \frac{(p_{Z} α) (p_{Z} α)}{p_{Z} (1 - α)} = \frac{p_{Z} α^{2}}{1 - α}$ . Now $\frac{K_{p 1}}{K_{p 2}} = \frac{4 p_{X} α^{2} / (1 - α)}{p_{Z} α^{2} / (1 - α)} = \frac{4 p_{X}}{p_{Z}}$

It is given $K_{p 1} / K_{p 2}$ that= 1/9. Hence $\frac{p_{X}}{p_{Z}} = \frac{1}{36}$ Also $\frac{p_{eq 1}}{p_{eq 2}} = \frac{p_{X}}{p_{Z}} = \frac{1}{36}$