MCQ
For the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$ , the partial pressure of $CO_2$ and $CO$ at equilibrium are $4$ and $8\, atm$ respectively. $K_p$ for the reaction is
- ✓$16$
- B$2$
- C$0.5$
- D$4$
atm, respectively, at equilibrium.
$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{CO}}^{2}}{\mathrm{P}_{\mathrm{CO} 2}}$
$\mathrm{K}_{\mathrm{p}}=\frac{(8.0)^{2}}{4.0}$
$\mathrm{K}_{\mathrm{p}}=16.0$
Thus, the $\mathrm{K}_{\mathrm{p}}$ of the reaction is $16.0 .$
So, the correct option is $\mathrm{A}$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2} - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - CH - CH - C{H_2} - CH - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - C{H_{3\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$