$X \rightleftharpoons Y + Z - - - - - - \left( 1 \right)$
$A \rightleftharpoons 2B - - - - - - \left( 2 \right)$
are in ratio of $9 :1$. If degree of dissociation of $X$ and $A$ be equal, then total pressure at equilibrium $(1)$ and $(2)$ are in the ratio
$P_{x}=\left(\frac{1-\alpha}{1+\alpha}\right) P_{1} \& P_{y}=\left(\frac{\alpha}{1-\alpha}\right) P_{1} \& P_{z}=\left(\frac{\alpha}{1+\alpha}\right) P_{1}$
$\mathrm{K}_{\mathrm{R}}=\frac{\mathrm{P}_{\mathrm{y}} \times \mathrm{P}_{\mathrm{z}}}{\mathrm{P}_{\mathrm{x}}}$
$\mathop {\mathop A\limits_1 }\limits_{1 - \alpha } \rightleftharpoons \mathop {\mathop {2B}\limits_0 }\limits_{2\alpha } $
$P_{A}=\left(\frac{1-\alpha}{1+\alpha}\right) P_{2} \& P_{B}=\left(\frac{2 \alpha}{1+\alpha}\right) P_{2}$
$\mathrm{K}_{\mathrm{P}_{2}}=\frac{\mathrm{P}_{\mathrm{B}}^{2}}{\mathrm{P}_{\mathrm{A}}}$
$\frac{\mathrm{K}_{\mathrm{R}}}{\mathrm{K}_{\mathrm{p}_{2}}}=\frac{9}{1}$
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$[A]$ With increase in temperature, the value of $K$ for exothermic reaction decreases because entropy change of the system is positive
$[B]$ With increase in temperature, the value of $K$ for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases
$[C]$ With increase in temperature, the value of $K$ for endothermic reaction increases because the entropy change of the system is negative
$[D]$ With increase in temperature, the value of $K$ for exothermic reaction decreases because favourable change in entropy of the surrounding decreases
if one mole of each of $X$ and $Y$ with $0.05 mol$ of $Z$ gives compound $XYZ _{3}$. (Given : Atomic masses of $X , Y$ and $Z$ are 10,20 and $30 amu$, respectively). The yield of $XYZ _{3}$ is $.........g$.(Nearest integer)