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$C{H_3} - \mathop {\mathop {\mathop {\mathop C\limits^| }\limits^{Cl} - }\limits_{|\,\,\,\,} }\limits_{OH} C{H_2}CH = CHC{H_3}$
$(i)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,OC{H_3}}
\end{array}$
$(ii)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iii)\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus {H_2} - NH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv)\,C{H_3} - \mathop C\limits^ \oplus {H_2}$
$2 NO _{( g )}+ O _{2}( g ) \rightleftarrows 2 NO _{2}( g )$
The reaction occurring as above comes to equilibrium under a total pressure of 1 atom. Analysis of the system shows that $0.6 mol$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is $.........$(Nearest integer).
Given :
Molar mass $N =14\,g\,mol ^{-1} ; O =16\,g\,mol ^{-1} ; C =12\,g\,mol ^{-1} ; H =1\,g\,mol ^{-1}$;