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
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

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(0.622) The equilibrium constant $K_{c}$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is given by the expression:
$K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(2.8 \times 10^{-3})^{2}}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_{c} = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}}$
$K_{c} = 0.622$

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Consider the following gaseous equilibrium reactions $(I)$,$(II)$ and $(III)$ with equilibrium constants $K_1$,$K_2$ and $K_3$ respectively:
$I$) $\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$
$II$) $2 NO \rightleftharpoons N_2 + O_2$
$III$) $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O$
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For the reactions:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is:

Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:

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