The values of $K_{P_1}$ and $K_{P_2}$ for the reactions
$X \rightleftharpoons Y + Z$ --- $(1)$
$A \rightleftharpoons 2B$ --- $(2)$
are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio:

The equilibrium constants for the following reactions are $K_1$ and $K_2$,respectively.
$2 P_{(g)} + 3 Cl_{2(g)} \rightleftharpoons 2 PCl_{3(g)}$
$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$
Then,the equilibrium constant for the reaction,$2 P_{(g)} + 5 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)}$ is

$\Delta _f G^o$ at $500 \, K$ for substance '$S$' in liquid state and gaseous state are $+100.7 \, kcal \, mol^{-1}$ and $+103 \, kcal \, mol^{-1}$,respectively. The vapour pressure of liquid '$S$' at $500 \, K$ is approximately equal to $(R = 2 \, cal \, K^{-1} \, mol^{-1}) \dots \dots \text{atm}$.

$15 \ mol$ of $H_2$ and $5.2 \ mol$ of $I_2$ are mixed and allowed to attain equilibrium at $773 \ K$. At equilibrium,the number of moles of $HI$ is found to be $10$. The equilibrium constant for the dissociation of $HI$ is

The amount of $PCl_5$ (in moles) that needs to be added to a $1\,L$ vessel at $250\,^oC$ in order to obtain $0.1\,mol$ of $Cl_2$ for the given reaction is:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

$A$ $20 \, L$ container at $400 \, K$ contains $CO_{2(g)}$ at a pressure of $0.4 \, atm$ and an excess of $SrO$ (neglect the volume of solid $SrO$). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container,when the pressure of $CO_2$ attains its maximum value,will be........$L$
(Given: $SrCO_{3(s)} \rightleftharpoons SrO_{(s)} + CO_{2(g)}, K_p = 1.6 \, atm$)