$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$
જો પ્રક્રિયાનું $K_p$ $1.1\times10^{-3}$ છે, તોકદના ટકાની દ્રષ્ટિએ ઉત્પન્ન થયેલ નાઇટ્રિક ઓકસાઈડની માત્રાની ગણતરી કરો.
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\,\% $
$\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} \text {. }$
(આપેલ : R = $\left.8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$