For the reaction $2{\mathrm{C}}_{\left(\mathrm{s}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{CO}}_{\left(\mathrm{g}\right)},$the partial pressure of ${\mathrm{O}}_2$ and CO are 2.0 and 4.0 atm respectively at equilibrium. The ${\mathrm{K}}_{\mathrm{p}}$ for the reaction is

The equilibrium of a reaction is

$2{\mathrm{C}}_{\left(\mathrm{s}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{CO}}_{\left(\mathrm{g}\right)}$

The equilibrium constant expression for ${\mathrm{K}}_{\mathrm{p}}$ is

${\mathrm{K}}_{\mathrm{p}}=\frac{\mathrm{Partial} \mathrm{pressures} \mathrm{of} \mathrm{the} \mathrm{products}}{\mathrm{Partial} \mathrm{pressures} \mathrm{of} \mathrm{the} \mathrm{products}}$

${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\mathrm{co}}^2}{{\mathrm{P}}_{{\mathrm{O}}_2}}$

$=\frac{{\left(4.0\right)}^2}{2.0}=8.0$

Note: Pressure is not effect on solids. Hence, we consider that partial pressure for solids as unity. Therefore, the correct answer is option “ B”.