Equimolar mixture of two gases $X_{2}$ and $Y_{2}$ is taken in a rigid vessel at temperature 300K. The gases reacts according to given equations

If the initial pressure in the container was 2 atm and final pressure developed at equilibrium is $2.75 a t m$ in which equilibrium partial pressure of gas XY was $0.5 a t m,$ calculate the ratio of $\frac{K_{P_{2}}}{K_{P_{1}}}$ ?

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$P e q = P_{A_{2}} + P_{B_{2}} + P_{B} + P_{A} + P_{A B}$
$y + 2 x + 2 y + 2$

$= 2 + x + y = 2.75 þ x + y = 0.75$
and $2 z = 0.5 þ z = 0.25$
$K_{P_{3}} = \frac{{(P_{A B})}^{2}}{{P_{A_{2}}}^{'} P_{B_{2}}} = \frac{{(0.5)}^{2}}{(0.75 - x) (0.75 - y)}$
$K_{P_{1}} = \frac{{(P_{A})}^{2}}{P_{A_{2}}} = \frac{{(2 x)}^{2}}{(0.75 - x)}$

$K_{P_{2}} = \frac{{(P_{B})}^{2}}{P_{B_{2}}} = \frac{{(2 y)}^{2}}{(0.75 - y)}$
${K_{P_{1}}}^{'} K_{P_{2}} = \frac{{(2 x)}^{2} {(2 y)}^{2}}{(0.75 - x) (0.75 - y)}$

${K_{P_{1}}}^{'} K_{P_{2}} = \frac{16 x^{2} y^{2}}{(0.5)}' 2$
$\frac{K_{P_{1}}}{K_{P_{2}}} = \frac{(1 / 4) (1 / 2)}{1 (1 / 4)} = \frac{1}{8}$