Chapter 5, Problem 5.31E

(a)

Interpretation Introduction

Interpretation:

The proper expression for the equilibrium constant of the reaction is to be stated.

Concept introduction:

Equilibrium constant for a chemical reaction defines the extent of reaction. It is the ratio of concentration of products formed to the concentration of reactants with time which is expressed in terms of activity and molarity. The equilibrium constant is constant at a specified temperature and pure solids and pure liquids are not taken in the expression for equilibrium constant. In order to get a unitless quantity, the whole product is divided by the standard molal concentration.

Answer to Problem 5.31E

The expression for equilibrium constant for the reaction ${\text{PbCl}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{2+}}\left(\text{aq}\right)+{\text{2Cl}}^{\text{-}}\left(\text{aq}\right)$ is given below.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{Pb}}^{\text{2+}}}{\text{m}}_{{\text{Pb}}^{\text{2+}}}}{\text{m°}}\right]{\left[\frac{{\text{γ}}_{{\text{Cl}}^{-}}{\text{m}}_{{\text{Cl}}^{-}}}{\text{m°}}\right]}^{\text{2}}}{\frac{{\text{γ}}_{{\text{PbCl}}_{\text{2}}}{\text{m}}_{{\text{PbCl}}_{\text{2}}}}{{\text{m}}^{\text{ο}}}}$

Explanation of Solution

The equilibrium constant for the given reaction can be expressed as the product of the concentration of products as given below.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{Pb}}^{\text{2+}}}{\text{m}}_{{\text{Pb}}^{\text{2+}}}}{\text{m°}}\right]{\left[\frac{{\text{γ}}_{{\text{Cl}}^{-}}{\text{m}}_{{\text{Cl}}^{-}}}{\text{m°}}\right]}^{\text{2}}}{\frac{{\text{γ}}_{{\text{PbCl}}_{\text{2}}}{\text{m}}_{{\text{PbCl}}_{\text{2}}}}{{\text{m}}^{\text{ο}}}}$

In the above expression, ${\text{K}}_{\text{eq}}$ is the equilibrium constant and the square brackets represent the concentrations of products (${\text{Pb}}^{\text{2+}}$ and ${\text{Cl}}^{\text{-}}$). The power of $2$ at ${\left[{\text{Cl}}^{-}\right]}^{\text{2}}$ is due to its stoichiometric coefficient in the reaction. The term ${\text{m}}^{\text{ο}}$ represents the standard molal concentration. Since ${\text{PbCl}}_{\text{2}}$ is in the solid form, so it will be taken as $1$ in the equilibrium constant for the reaction.

Conclusion

The equilibrium constant for the given reaction is ${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{Pb}}^{\text{2+}}}{\text{m}}_{{\text{Pb}}^{\text{2+}}}}{\text{m°}}\right]{\left[\frac{{\text{γ}}_{{\text{Cl}}^{\text{-}}}{\text{m}}_{{\text{Cl}}^{\text{-}}}}{\text{m°}}\right]}^{\text{2}}}{\frac{{\text{γ}}_{{\text{PbCl}}_{\text{2}}}{\text{m}}_{{}_{{\text{PbCl}}_{\text{2}}}}}{{\text{m}}^{\text{ο}}}}$.

(b)

Interpretation Introduction

Interpretation:

The proper expression for the equilibrium constant of the reaction is to be stated.

Concept introduction:

Equilibrium constant for a chemical reaction defines the extent of reaction. It is the ratio of concentration of products formed to the concentration of reactants with time which is expressed in terms of activity and molarity. The equilibrium constant is constant at a specified temperature and pure solids and pure liquids are not taken in the expression for equilibrium constant.

Answer to Problem 5.31E

The expression for equilibrium constant for the reaction ${\text{HNO}}_{\text{2}}\left(\text{aq}\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)$ is given below.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{m}}_{{\text{H}}^{\text{+}}}}{\text{m°}}\right]\left[\frac{{\text{γ}}_{{\text{NO}}_{\text{2}}^{\text{-}}}{\text{m}}_{{\text{NO}}_{\text{2}}^{\text{-}}}}{\text{m°}}\right]}{\left[\frac{{\text{γ}}_{{\text{HNO}}_{\text{2}}}{\text{m}}_{{\text{HNO}}_{\text{2}}}}{\text{m°}}\right]}$.

Explanation of Solution

The equilibrium constant for the given reaction can be expressed as the ratio of the product of the concentration of products to the product of the concentration of reactants that is:

${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{m}}_{{\text{H}}^{\text{+}}}}{\text{m°}}\right]\left[\frac{{\text{γ}}_{{\text{NO}}_{\text{2}}^{-}}{\text{m}}_{{\text{NO}}_{\text{2}}^{-}}}{\text{m°}}\right]}{\left[\frac{{\text{γ}}_{{\text{HNO}}_{\text{2}}}{\text{m}}_{{\text{HNO}}_{\text{2}}}}{\text{m°}}\right]}$

In the above expression, ${\text{K}}_{\text{eq}}$ is the equilibrium constant and the square brackets represent the concentrations of products and reactants, that is ${\text{H}}^{\text{+}}$, ${\text{NO}}_{\text{2}}^{-}$ and ${\text{HNO}}_{\text{2}}$. The term ${\text{m}}^{\text{ο}}$ represents the standard molal concentration. Since the stoichiometric coefficient of the reactants and the product is $1$ therefore, it is taken as the direct product.

Conclusion

The equilibrium constant for the given reaction is ${\text{K}}_{\text{eq}}\text{=}\frac{\left[\frac{{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{m}}_{{\text{H}}^{\text{+}}}}{\text{m°}}\right]\left[\frac{{\text{γ}}_{{\text{NO}}_{\text{2}}^{\text{-}}}{\text{m}}_{{\text{NO}}_{\text{2}}^{\text{-}}}}{\text{m°}}\right]}{\left[\frac{{\text{γ}}_{{\text{HNO}}_{\text{2}}}{\text{m}}_{{\text{HNO}}_{\text{2}}}}{\text{m°}}\right]}$.

(c)

Interpretation Introduction

Interpretation:

The proper expression for the equilibrium constant of the reaction is to bestated.

Concept introduction:

Equilibrium constant for a chemical reaction defines the extent of reaction. It is the ratio of concentration of products formed to the concentration of reactants with time which is expressed in terms of activity and molarity. The equilibrium constant is constant at a specified temperature and pure solids and pure liquids are not taken in the expression for equilibrium constant.

Answer to Problem 5.31E

The expression for equilibrium constant for the given reaction ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{aq}\right)\rightleftharpoons {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$ is given below.

${\text{K}}_{\text{eq}}\text{=}\frac{\frac{{\text{pCO}}_{\text{2}}}{\text{p°}}}{\left[\frac{{\text{γ}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{m}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{m°}}\right]}$

Explanation of Solution

The equilibrium constant for the given reaction can be expressed as the ratio of the product of the concentration of products to the product of the concentration of reactants that is,

${\text{K}}_{\text{eq}}\text{=}\frac{\frac{{\text{pCO}}_{\text{2}}}{\text{p°}}}{\left[\frac{{\text{γ}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{m}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{m°}}\right]}$

In the above expression, ${\text{K}}_{\text{eq}}$ is the equilibrium constant, ${\text{pCO}}_{\text{2}}$ is the partial pressure of carbon dioxide gas evolved as a product in the reaction and $\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right]$ is the concentration of reactant. The term ${\text{m}}^{\text{ο}}$ represents the standard molal concentration. In case of gaseous reactants or products, their partial pressure is taken instead of molar concentrations.

Conclusion

The equilibrium constant for the given reaction is ${\text{K}}_{\text{eq}}\text{=}\frac{\frac{{\text{pCO}}_{\text{2}}}{\text{p°}}}{\left[\frac{{\text{γ}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{m}}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{m°}}\right]}$.