a
\({N_2}(g) + {O_2}(g) \leftrightarrow 2NO(g)\)

At equilibrium, we have  \([{N_2}] = 0.79\,(1 - \alpha );\)

\([{O_2}] = 0.21\,(1 - \alpha );\,[NO] = 2\alpha \)

Total number of moles

\( = 0.79(1 - \alpha ) + 0.21(1 - \alpha )9 + 2\alpha  = 1 + \alpha \)

\({P_{{N_2}}} = \frac{{0.79(1 - \alpha )}}{{1 + \alpha }} \times 1;\)

\({P_{{O_2}}} = \frac{{0.21(1 - \alpha )}}{{1 + \alpha }} \times 1;\,{P_{NO}} = \frac{{2\alpha }}{{1 + \alpha }} \times 1\)

\({K_P} = \frac{{P_{NO}^2}}{{{P_{{N_2}}}.{P_{{O_2}}}}}\)

\(1.1 \times {10^{ - 3}} = \frac{{4{\alpha ^2}}}{{0.79 \times 0.21{{(1 - \alpha )}^2}}}\)

or \(\alpha  = 0.0067\)

\( \Rightarrow \) vol \(\%\) of \(NO = 2\alpha  \times 100\)

                        \( = 2 \times 0.0067 \times 100 = 1.33\,\% \)