Consider the following reaction in a sealed vessel at equilibrium with concentrations of N_2=3.0 × 10^-3 M, O_2=4.2 × 10^-3 M and NO =2.8 × 10^-3 M.…

Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\displaystyle \mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}$ and $\displaystyle \mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}$. $\displaystyle 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ If $\displaystyle 0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ of $\displaystyle \mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation $\displaystyle (\alpha)$ of $\displaystyle \mathrm{NO}_{(\mathrm{g})}$ at equilibrium?