For the following equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ in gaseous phase,$NO_2$ is $50\%$ of the total volume when equilibrium is set up. Hence,the percentage of dissociation of $N_2O_4$ is.......$\%$

For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

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:

The degree of dissociation of a weak acid is $7.2 \times 10^{-4}$. What is the value of its percent dissociation in a $0.025 \ M$ solution (in $\%$)?

$3.2$ moles of hydrogen iodide were heated in a sealed bulb at $444\,^{\circ}C$ till the equilibrium was reached. The degree of dissociation of $HI$ at this temperature was found to be $22\%$. The number of moles of hydrogen iodide present at equilibrium are

If $D_T$ and $D_0$ are the theoretical and observed vapour densities at a definite temperature and $\alpha$ is the degree of dissociation of a substance,then $\alpha$ in terms of $D_0, D_T$ and $n$ (number of moles of product formed from $1 \, \text{mole}$ of reactant) is calculated by the formula: