Question

Calculate the partial pressure of carbon monoxide from the following:
$CaCO_{3(s)}\overset{\Delta}{\rightarrow}CaO_{(s)}+CO_{2(g)}\uparrow$ ${{K}_{p}}=8\times {{10}^{-2}}$
$CO_{2(s)}+C_{(s)}\rightarrow 2CO_{(s)}$ ${{K}_{p}}=2$
 (A) 0.2
(B) 0.4
(C) 1.6
(D) 4

Answer 1

Hint: When a chemical reaction reaches equilibrium, the equilibrium constant gives information about the interaction between the reactants and products. ${{K}_{p}}$ is the equilibrium constant produced by partial pressures in a reaction equation. It is used to describe the interaction between reactant and product pressures. Temperature and the stoichiometry of the change both affect ${{K}_{p}}$ values. ${{K}_{p}}$ is measured in ${{(atm)}^{\Delta n}}$ units.

Complete step by step solution:
The first equation given is
$CaCO_{3(s)}\overset{\Delta}{\rightarrow}CaO_{(s)}+CO_{2(g)}\uparrow$
As in this reaction, only $C{{O}_{2}}$is a gas. So, the ${{K}_{p}}$given in the question is the partial pressure of$C{{O}_{2}}$. ${{P}_{C{{O}_{2}}}}=8\times {{10}^{-2}}$ …(1)

Now, the second reaction given is:
$CO_{2(s)}+C_{(s)}\rightarrow 2CO_{(s)}$

The equilibrium constant is calculated as ${{K}_{p}}=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$
As ${{K}_{p}}$for this reaction given in the question is 2. So, $2=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$ … (2)

From equations (1) and (2),
$2=\frac{{{({{P}_{CO}})}^{2}}}{8\times {{10}^{-2}}}$
${{({{P}_{CO}})}^{2}}=2\times 8\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=16\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=0.16$
${{P}_{CO}}=0.4$
Hence, the partial pressure of carbon monoxide ($CO$) is 0.4.
Correct Option: (B) 0.4.

Note: Partial pressure is the pressure that one gas in a gas mixture will exert if it occupies the same volume on its own. In a mixture, every gas exerts a particular pressure. The thermodynamic activity of a gas is determined by its partial pressure. The overall pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases in the mixture, according to Dalton's equation of partial pressure. For an ideal gas mixture, Dalton's law is completely true.