If the degree of dissociation for the reaction,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is $20\%$ at $1 \ atm$ pressure,calculate $K_c$.

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:

Consider the reaction $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$. The equation representing the correct relationship between the degree of dissociation $(x)$ of $X_2Y_{(g)}$ with its equilibrium constant $Kp$ is. . . . . . . Assume $x$ to be very very small.

In a closed vessel,the dissociation of phosphorus pentachloride occurs as: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If the total pressure at equilibrium is $P$ and the degree of dissociation of $PCl_5$ is $x$,then the partial pressure of $PCl_3$ will be:

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

At $NTP$,$5.6 \ L$ of a gas weighs $8 \ g$. The vapour density of the gas is: