The equilibrium constant $\left(K_c\right)$ for the reaction ${\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_3\left(\mathrm{g}\right)+\mathrm{NO}\left(\mathrm{g}\right)$ is 16. If 1 mol of each of all the four gases is taken in $1{\mathrm{dm}}^3$ vessel, the equilibrium concentration of NO would be

$\begin{array}{l}{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_3\left(\mathrm{g}\right)+\mathrm{NO}\left(\mathrm{g}\right)\\ 1\mathrm{mol}-x1\mathrm{mol}-x1\mathrm{mol}+x1\mathrm{mol}+x\end{array}$

$K_c=\frac{(1\mathrm{mol}+x{)}^2}{(1\mathrm{mol}-x{)}^2}$ i.e. $16={\left(\frac{1\mathrm{mol}+x}{1\mathrm{mol}-x}\right)}^2$

or $(1\mathrm{mol}+x)=4(1\mathrm{mol}-x)$ or $x=0.6\mathrm{mol}$

Amount of NO= 1 mol + x = 1.6 mol and Concentration of NO = 1.6 M