The equilibrium constant K_c for the reaction ${\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_3\left(\mathrm{g}\right)+\mathrm{NO}\left(\mathrm{g}\right)\text{ is }16$. If 1 mole of each of all the four gases is taken in 1dm³ vessel, the equilibrium concentration of NO would be :

$$\begin{gathered} \mathrm{Q}=1<{\mathrm{k}}_{\text{eq }} \\ \mathrm{reaction} \mathrm{moves} \mathrm{forward} \\ \mathrm{i}.\mathrm{e}., {\mathrm{SO}}_2 + {\mathrm{NO}}_2 \rightleftharpoons {\mathrm{SO}}_3 + \mathrm{NO} \\ \mathrm{t}=0 ; 1 1 1 1 \\ {\mathrm{t}}_{\mathrm{eq}.}; (1-\mathrm{x}) (1-\mathrm{x}) (1+\mathrm{x}) (1+\mathrm{x}) \\ {\mathrm{k}}_{\mathrm{c}}=\frac{\left[{\mathrm{SO}}_3\right]\left[\mathrm{NO}\right]}{\left[{\mathrm{SO}}_2\right]\left[{\mathrm{NO}}_2\right]}=16 \\ 4=\frac{1+\mathrm{x}}{1-\mathrm{x}}\Rightarrow \mathrm{x}=0.6 \\ \left[\mathrm{NO}\right]=1+\mathrm{x}=1.6\mathrm{M} \end{gathered}$$