$2$ moles of $PCl_5$ were heated in a closed vessel of $2 \ L$ capacity. At equilibrium,$40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is

If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.

The suitable expression for the equilibrium constant of the reaction $2NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2NOCl_{(g)}$ is

The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

The equilibrium constant $K_{c}$ at $298 \ K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$. Starting with an equimolar solution with concentrations of $A$,$B$,$C$ and $D$ all equal to $1 \ M$,the equilibrium concentration of $D$ is $....... \times 10^{-2} \ M$. (Nearest integer)