In a reaction $A + B \rightleftharpoons C + D$,the concentrations of $A$,$B$,$C$ and $D$ (in $mol/L$) are $0.5$,$0.8$,$0.4$ and $1.0$ respectively. The equilibrium constant is

At $T \ (K)$,$K_{p}$ value for the reaction,$2 \ AO_{2(g)} + O_{2(g)} \rightleftharpoons 2 \ AO_{3(g)}$ is $4 \times 10^{10}$. What is the $K_{p}^{\prime}$ value for $3 \ AO_{2(g)} + \frac{3}{2} \ O_{2(g)} \rightleftharpoons 3 \ AO_{3(g)}$ at $T \ (K)$?

The equilibrium constant for the given reaction $H_2 + I_2 \rightleftharpoons 2HI$ is correctly given by the expression:

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. 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 $M$)

State the law of chemical equilibrium for an equilibrium process.

For the equilibrium $N_2 + 3H_2 \rightleftharpoons 2NH_3$,$K_c$ at $1000 \ K$ is $2.37 \times 10^{-3}$. If at equilibrium $[N_2] = 2 \ M$ and $[H_2] = 3 \ M$,the concentration of $NH_3$ is: (in $M$)