For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185\,^{\circ}C$. At $185\,^{\circ}C$,the value of $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is

Consider the following reactions in which all the reactants and products are present in gaseous state:
$2xy \rightleftharpoons x_2 + y_2$ $K_1 = 2.5 \times 10^5$
$xy + \frac{1}{2}z_2 \rightleftharpoons xyz$ $K_2 = 5 \times 10^{-3}$
The value of $K_3$ for the equilibrium $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$ is:

The equilibrium constant $K_{c}$ at $298 \ K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$. Starting with an equimolar solution with concentrations of $A$,$B$,$C$ and $D$ all equal to $1 \ M$,the equilibrium concentration of $D$ is $....... \times 10^{-2} \ M$. (Nearest integer)

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

If $K_1$ and $K_2$ are respective equilibrium constants for the two reactions:
$XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)}$
$XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)}$
The equilibrium constant for the reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$ will be:

The value of $K_{c}$ for the reaction $2 A \rightleftharpoons B + C$ is $2 \times 10^{-3}$. At a given time,the composition of the reaction mixture is $[ A ] = [ B ] = [ C ] = 3 \times 10^{-4} \ M$. In which direction will the reaction proceed?