- ✓$\frac{{P{x^3}}}{2}$
- B$\frac{{P{x^2}}}{3}$
- C$\frac{{P{x^3}}}{3}$
- D$\frac{{P{x^2}}}{2}$
$\therefore {{\text{K}}_{\text{p}}} = \frac{{{{\text{x}}^2} \cdot {\text{x}}}}{{2{{(1 - {\text{x}})}^2}}} \times {\left[ {\frac{{\text{P}}}{{1 + \frac{{\text{x}}}{2}}}} \right]^1}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$(I)$ $\begin{array}{*{20}{c}}
{C{H_2} - C{H_2} - C{H_2} - C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - CH - C{H_2}OH} \\
{|\,} \\
{N{O_2}}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(IV)$ $CH_3-CH_2-CH_2-CH_2OH$
Select answer from codes given below :
[Atomic number of $\mathrm{Gd}=64$ ]