$AB_{3(g)}$ dissociates as:
$AB_{3(g)} \rightleftharpoons AB_{2(g)} + \frac{1}{2} B_{2(g)}$. When the initial pressure of $AB_3$ is $800 \ torr$ and the total pressure developed at equilibrium is $900 \ torr$,what percentage of $AB_{3(g)}$ is dissociated?

Before equilibrium is set up for the chemical reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the vapour density $d$ of the gaseous mixture was measured. If $D$ is the theoretical value of vapour density,the variation of $\alpha$ with $D/d$ is given by the graph. What is the value of $D/d$ at point $A$?

At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is

At $T$ $K$,consider the following gaseous reaction,which is in equilibrium: $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$. What is the fraction of $N_2O_5$ decomposed at constant volume and temperature,if the initial pressure is $300 \ mm \ Hg$ and pressure at equilibrium is $480 \ mm \ Hg$? (Assume all gases as ideal)

For the reaction $A_{3(g)} \rightleftharpoons 3A_{(g)}$,the initial moles of $A_3$ is $a$. If $\alpha$ is the degree of dissociation of $A_3$,then the total number of moles at equilibrium will be:

The vapour density of $N_2O_4$ in the equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ is $40$. The degree of dissociation is: