For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression $K_c$ is:

At equilibrium,the concentrations of $N_{2} = 3.0 \times 10^{-3} \, M$,$O_{2} = 4.2 \times 10^{-3} \, M$ and $NO = 2.8 \times 10^{-3} \, M$ in a sealed vessel at $800 \, K$. What will be $K_{c}$ for the reaction
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

The dissociation of $SO_{3(g)}$ into $SO_{2(g)}$ and $O_{2(g)}$ is carried out in a closed container at a constant temperature $T$. The equilibrium constant for the reaction is $K_P = x \ atm$. The partial pressure variation for the three gases is as shown in the graph. The value of $x$ is

Given the reaction between $2$ gases represented by $A_2$ and $B_2$ to give the compound $AB_{(g)}$:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be:

At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)