For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium $(P)$ by:

The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$. The mole fraction of $N_2O_4$ in the mixture is

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:

In a $100 \ L$ vessel,$3 \ moles$ of nitrogen and $3 \ moles$ of $PCl_5$ are taken and heated to $500 \ K$. The equilibrium pressure is $3.28 \ atm$. The percentage degree of dissociation of $PCl_5$ is : (Assume ideal behaviour for all gases).

For a reaction at equilibrium $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$,the relation between dissociation constant $(K)$,degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by?