One mole of $PCl_5$ is heated in a closed container of $1 \ L$ capacity. At equilibrium,$20\%$ of $PCl_5$ is not dissociated. What should be the value of $K_C$?

For the reversible reaction in equilibrium:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.

$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

For the reaction $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$,the value of $K_p$ is:

The equilibrium concentrations of $N_2, H_2$ and $NH_3$ in the formation of $NH_3$ at $500 \ K$ are $1.25 \times 10^{-2} \ M, 4.0 \times 10^{-2} \ M$ and $1.6 \times 10^{-2} \ M$ respectively. The equilibrium constant $K_{p}$ at the same temperature is

One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is: