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

Ammonia is heated in a closed vessel in the presence of a catalyst at $15 \ atm$ pressure and a temperature range of $27 \ ^\circ C$ to $347 \ ^\circ C$. Under these conditions,ammonia undergoes partial dissociation according to the reaction $2NH_3 \rightleftharpoons N_2 + 3H_2$. When the pressure is increased to $50 \ atm$ while keeping the volume of the vessel constant,calculate the percentage of $NH_3$ dissociated.

For the reaction $P \rightleftharpoons Q + R$,initially $2 \, \text{mol}$ of $P$ is taken. At equilibrium,$0.5 \, \text{mol}$ of $P$ dissociates. The degree of dissociation will be ...... .

The dissociation equilibrium of a gas $AB_{2}$ can be represented as:
$2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$
The degree of dissociation is $x$ and is small compared to $1$. The expression relating the degree of dissociation $(x)$ with equilibrium constant $K_P$ and total pressure $P$ is:

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 $A$?

Total number of moles for the reaction $2HI \rightleftharpoons H_2 + I_2$ at equilibrium,if $\alpha$ is the degree of dissociation,is: