$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$
If the $K_p$ of the reaction is $1.1\times10^{-3}$, calculate the amount of nitric oxide produced in terms of volume percent.
- A$1.33$
- B$1.12$
- C$1.02$
- ✓$1.44$
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\,\% $
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$2A{B_3}(g) \rightleftharpoons {A_2}(g) + 3{B_2}(g)$. At equilibrium, $2\, mol$ of $A_2$ are found to be present. The equilibrium constant of this reaction is
$\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - CH - C{H_3}\xrightarrow{{C{H_3}{O^\Theta }}}X} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{F\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ product
The alkene formed in major amount