MCQ
$HI$ was heated in a closed tube at ${440\,^o}C$ till equilibrium is obtained. At this temperature $22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be
- A$0.282$
- B$0.0796$
- ✓$0.0199$
- D$1.99$
Initial conc. $2$ moles $0\,\,\,\, 0$
at equilibrium $\frac{{22}}{{100}} \times 2$ $0.22\,\,\,\,\, 0.22$
$ = 2 - 0.44 = 1.56$
$K = \frac{{[{H_2}]\,\,[{I_2}]}}{{{{[HI]}^2}}} = \frac{{0.22 \times 0.22}}{{{{[1.56]}^2}}} = 0.0199$.
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${C_6}{H_5} - H\mathop {\xrightarrow{{[X]}}}\limits_{AlC{l_3}} $ $\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
{{C}_{6}}{{H}_{5}}-C-C{{H}_{2}}-{{C}_{6}}{{H}_{5}} \\
\end{matrix}$
$[X]$ will be :
$(I) HCHO$,
$(II)$ $C{H_3}CHO$,
$(III)$ $C{H_3}COC{H_3}$ is
${H_3}C - C \equiv CH\,\xrightarrow[{HgS{O_4}}]{{{H_2}O,\,{H_2}S{O_4}}}\,\mathop {Intermediate}\limits_{(A)} \,$$ \to \,\mathop {\Pr oduct}\limits_{(B)} $