For the reaction $3A + 2B \rightleftharpoons C$,the expression for the equilibrium constant $K_c$ is:

Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$, the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.

The equilibrium constant $K_C$ of the reaction,$2 A \rightleftharpoons B + C$ is $0.5$ at $25^{\circ} C$. The reaction will proceed in the backward direction,when concentrations $[A], [B]$ and $[C]$ are,respectively:

The equilibrium constant,$K_c$ for $3 C_2H_{2(g)} \rightleftharpoons C_6H_{6(g)}$ is $4 \, L^2 \, mol^{-2}$. If the equilibrium concentration of benzene is $0.5 \, mol \, L^{-1}$,the equilibrium concentration of acetylene in $mol \, L^{-1}$ is:

When $4 \ mol$ of $A$ is mixed with $4 \ mol$ of $B$,$2 \ mol$ of $C$ is formed at equilibrium. For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant is:

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