Question

For the reaction, ${C_{\left( s \right)}} + {O_{2\left( g \right)}} \rightleftharpoons 2C{O_{\left( g \right)}}$ the partial pressure of $C{O_2}$ and $CO$ are $2.0$ and $4.0$ atm, respectively. At equilibrium the ${K_p}$ of the reaction is:
(A) $0.5$
(B) $4.0$
(C) $32.0$
(D) $8.0$

Answer 1

The equilibrium constant from partial pressure can be represented as ${K_p}$ It can be determined from the ratio of the partial pressure of the gaseous products and gaseous reactants. Given are the partial pressures of carbon monoxide and oxygen gas which are the gaseous products and gaseous reactants.
${K_p} = \dfrac{{{{\left( {{P_{CO}}} \right)}^2}}}{{{P_{{O_2}}}}}$
${K_p}$ is equilibrium constant
${P_{CO}}$ is partial pressure of carbon monoxide
${P_{{O_2}}}$ is the partial pressure of oxygen gas.

Complete answer:
Given that carbon on treatment with carbon dioxide forms two moles of carbon monoxide. In this reaction the carbon and carbon monoxide are gaseous reactants and products.
While calculating the equilibrium constant from partial pressure only the partial pressures of gaseous reactants and gaseous products must be considered.
Given that the partial pressure of $C{O_2}$ and $CO$ are $2.0$ and $4.0$ atm.
Substitute the above values in the formula.
${K_p} = \dfrac{{{{\left( 4 \right)}^2}}}{2}$
By simplifying the above equation, the value of the equilibrium constant will be $8$ .
Thus, for the given reaction at equilibrium the value of equilibrium constant from partial pressure ${K_p}$ is $8$
As it is the ratio of the partial pressure the units must be cancelled. Thus, it is a dimensionless quantity.
Option D is the correct one.

Note:
While calculating the equilibrium constant from partial pressure, only the partial pressure of gaseous reactants and gaseous products is considered. Thus, in the above chemical reaction only carbon dioxide and carbon monoxide are considered. As carbon is a solid reactant.