At $444\,^oC$,the equilibrium constant $K$ for the reaction $2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $\frac{1}{64}$. The degree of dissociation of $AB$ will be .....$\%$

At temperature,$T$,a compound $AB_{2(g)}$ dissociates according to the reaction; $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is small compared with unity. The expression for $K_p$,in terms of $x$ and the total pressure,$P$ is

For the reaction $2A \rightleftharpoons B + 2C$,the degree of dissociation $(x)$ in terms of vapour density is:

For the reaction $A \rightleftharpoons nB$,if $a$ moles of $A$ are taken initially and $x$ moles of $A$ dissociate at equilibrium,find the value of the degree of dissociation.

For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_p$. The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as:

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be: