Chapter 15, Problem 18QP

Write equilibrium constant expressions for $K_c\text{, and for }Kp$, if applicable, for the following processes:

$\begin{array}{l}(a)\\ 2C{O}_2(g)\rightleftarrows \text{ 2CO}(g)+O_2(g)\\ (b)\\ 3O_2(g)\rightleftarrows {\text{ 2O}}_3(g)\\ (C)\\ \text{CO(}g{\text{) + Cl}}_2(g)\rightleftarrows COC{l}_2(g)\\ (d)\\ H_2\text{O(}g\text{) + C(}s\text{) }\rightleftarrows \text{ CO(}g{\text{) +H}}_2(g)\\ (e)\text{ HCOOH(}aq\text{)}\rightleftarrows {\text{ H}}^{+}(aq)\text{ HCOO(}aq\text{)}\\ \text{(f)}\\ \text{2HgO(}s\text{) }\rightleftarrows \text{2Hg(}l{\text{) + O}}_2(g)\end{array}$

Interpretation Introduction

Interpretation:

Theexpressions for equilibrium constant $K_p$ and $K_c$ for the given equilibrium reactions are to be determined.

Concept introduction:

Equilibrium constants of the gas phase reaction are written in terms of partial pressure because the concentration of gases is directly proportional to the partial pressure. The equilibrium constant in terms of partial pressure is denoted by $K_P$ and the reactants present in the solid or liquid phase are not present in the $K_P$ expression.

Answer to Problem 18QP

Solution:

$K_c=\frac{{\left[\text{CO}\right]}^2\left[{\text{O}}_2\right]}{{\left[{\text{CO}}_2\right]}^2}$ and $K_p=\frac{P_{\text{CO}\left(g\right)}^2{P}_{{\text{O}}_2\left(g\right)}}{P_{{\text{CO}}_2\left(g\right)}^2}$

$K_c=\frac{{\left[{\text{O}}_3\right]}^2}{{\left[{\text{O}}_2\right]}^3}$ and $K_p=\frac{P_{{\text{O}}_3\left(g\right)}^2}{P_{{\text{O}}_2\left(g\right)}^3}$

$K_c=\frac{\left[{\text{COCl}}_2\right]}{\left[\text{CO}\right]\left[{\text{Cl}}_2\right]}$ and $K_p=\frac{P_{{\text{COCl}}_2}}{P_{\text{CO}}{P}_{{\text{Cl}}_2}}$

$K_c=\frac{\left[\text{CO}\right]\left[{\text{H}}_2\right]}{\left[{\text{H}}_2\text{O}\right]}$ and $K_p=\frac{P_{\text{CO}}{P}_{{\text{H}}_2}}{P_{{\text{H}}_2\text{O}}}$

$K_c=\frac{\left[{\text{HCOO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HCOOH}\right]}$

$K_c=\left[{\text{O}}_2\right]$ and $K_p=P_{{\text{O}}_2}$

Explanation of Solution

a) ${\text{2CO}}_{\text{2}}\left(g\right)\rightleftharpoons \text{2CO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

In the given reaction, all reactants are present in the gaseous phase. Therefore, the equilibrium constant of the reaction is written in terms of both partial pressure and concentration of gases.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constantexpression for the reaction is:

$K_c=\frac{{\left[\text{CO}\right]}^2\left[{\text{O}}_2\right]}{{\left[{\text{CO}}_2\right]}^2}$

The general formula for writing the equilibrium constant expression in terms of partial pressure is:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, W and X are reactants, Y and Z are products, and $a,b,c$ and $d$ are their respective stoichiometric coefficients.

The expression of the equilibrium constant for the given reaction is:

$K_p=\frac{P_{\text{CO}\left(g\right)}^2{P}_{{\text{O}}_2\left(g\right)}}{P_{{\text{CO}}_2\left(g\right)}^2}$

b) ${\text{3O}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{2O}}_{\text{3}}\left(g\right)$

In the given reaction, all reactants are present in the gaseous phase. Therefore, the equilibrium constant of the reaction is written in terms of both partial pressure and concentration of gases.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constantexpression for the reaction is:

$K_c=\frac{{\left[{\text{O}}_3\right]}^2}{{\left[{\text{O}}_2\right]}^3}$

The general formula for writing the equilibrium constant expression in terms of partial pressure is:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, W and X are reactants, Y and Z are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The expression of the equilibrium constant for the reaction is:

$K_p=\frac{P_{{\text{O}}_3\left(g\right)}^2}{P_{{\text{O}}_2\left(g\right)}^3}$

c) $\text{CO}\left(g\right)+{\text{Cl}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{COCl}}_{\text{2}}\left(g\right)$

In the given reaction, all reactants are present in the gaseous phase. Therefore, the equilibrium constant of the reaction is written in terms of both partial pressure and concentration of gases.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constantexpression for the reaction is:

$K_c=\frac{\left[{\text{COCl}}_2\right]}{\left[\text{CO}\right]\left[{\text{Cl}}_2\right]}$

The general formula forwriting the equilibrium constant expression in terms of partial pressure is:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, W and X are reactants, Y and Z are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The expression of the equilibrium constant for the reaction is:

$K_p=\frac{P_{{\text{COCl}}_2}}{P_{\text{CO}}{P}_{{\text{Cl}}_2}}$

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

In the given reaction, gaseous reactants are present. Therefore, the equilibrium constant of the reaction is written in terms of both partial pressure and concentration of gases.

The species thatis present in the solid or liquid phase is not considered in the equilibrium constant expression because its concentration remains nearly constant during the course of the reaction. Since, carbon is present in the solid state in this reaction, it will not be written in the equilibrium constant expression.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constant expression for the reaction is:

$K_c=\frac{\left[\text{CO}\right]\left[{\text{H}}_2\right]}{\left[{\text{H}}_2\text{O}\right]}$

The general formula for writing the equilibrium constant expression in terms of partial pressure is:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, W and X are reactants, Y and Z are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The expression of the equilibrium constant for the reaction is:

$K_p=\frac{P_{\text{CO}}{P}_{{\text{H}}_2}}{P_{{\text{H}}_2\text{O}}}$

e) $\text{HCOOH}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HCOO}}^{-}\left(aq\right)$

In the given reaction, all reactants are present in the liquid phase. Therefore, the equilibrium constant of the reaction is written only in terms of concentration of gases and not in terms of partial pressure.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constantexpression for the reaction is:

$K_c=\frac{\left[{\text{HCOO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HCOOH}\right]}$

f) $\text{2HgO}\left(s\right)\rightleftharpoons \text{2Hg}\left(l\right)+{\text{O}}_2\left(g\right)$

In the given reaction, gaseous reactants are present. Therefore, the equilibrium constant of the reaction is written in terms of both partial pressure and concentration of gases.

The species that is present in the solid or liquid phase is not considered in the equilibrium constant expression because its concentration remains nearly constant during the course of the reaction. Since, mercury is present in the liquid state and $\text{HgO}$ in solid state in this reaction, they will not be written in the equilibrium constant expression.

The general formula for writing the equilibrium constant expression for the reaction is:

$K_c=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here, A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The equilibrium constantexpression for the reaction is:

$K_c=\left[{\text{O}}_2\right]$

The general formula for writing the equilibrium constant expression in terms of partial pressure is:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, W and X are reactants, Y and Z are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

The expression of the equilibrium constant for the reaction is:

$K_p=P_{{\text{O}}_2}$