The reaction quotient $Q$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is given by $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. The reaction will proceed from right to left when:

If the equilibrium constant for $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is $K,$ the equilibrium constant for $\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{(g)}$ will be

Consider the following gaseous equilibria with equilibrium constants $K_{1}$ and $K_{2}$ respectively:
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$
The equilibrium constants are related as:

If the equilibrium constant for $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $K$,then the equilibrium constant for $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ will be :

For the reaction ${N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}}$,the reaction quotient is given by ${Q = [NH_3]^2 / ([N_2][H_2]^3)}$. The reaction will proceed from left to right when ........

In the reaction,$A + B \rightleftharpoons 2C$,at equilibrium,the concentration of $A$ and $B$ is $0.20 \ mol \ L^{-1}$ each and that of $C$ was found to be $0.60 \ mol \ L^{-1}$. The equilibrium constant of the reaction is