Chapter 8, Problem 8.43E

Interpretation Introduction

(a)

Interpretation:

The equilibrium expression for the given reaction is to be stated.

Concept Introduction:

The equilibrium constant of a reaction is expressed as the ratio of concentration of products and reactants each raised to the power of their stoichiometric coefficients. A general equilibrium reaction is represented as,$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The equilibrium constant for the above chemical reaction is expressed as,$K=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Where,

• $\left[\text{A}\right]$ represents the equilibrium concentration of reactant $\text{A}$.

• $\left[\text{B}\right]$ represents the equilibrium concentration of reactant $\text{B}$.

• $\left[\text{C}\right]$ represents the equilibrium concentration of product $\text{C}$.

• $\left[\text{D}\right]$ represents the equilibrium concentration of product $\text{D}$.

• $a$ represents the stoichiometric coefficient of reactant $\text{A}$.

• $b$ represents the stoichiometric coefficient of reactant $\text{B}$.

• $c$ represents the stoichiometric coefficient of product $\text{C}$.

• $d$ represents the stoichiometric coefficient of product $\text{D}$.

Answer to Problem 8.43E

The equilibrium constant for the given gaseous reaction is expressed as,$K=\frac{\left[\text{Ni}{\left({\text{NH}}_3\right)}_6{}^{2+}\right]}{\left[{\text{Ni}}^{\text{2+}}\right]{\left[{\text{NH}}_3\right]}^6}$

Explanation of Solution

The given gaseous reaction is represented as,${\text{Ni}}^{\text{2+}}+6{\text{NH}}_3\rightleftarrows \text{Ni}{\left({\text{NH}}_3\right)}_6{}^{2+}$

The concentration of the product $\text{Ni}{\left({\text{NH}}_3\right)}_6{}^{2+}$ will be in the numerator and is raised to the power $1$. The concentration of reactant ${\text{Ni}}^{\text{2+}}$ and ${\text{NH}}_3$ will be in the denominator. The concentration of reactant ${\text{Ni}}^{\text{2+}}$ is raised to the power $1$ and ${\text{NH}}_3$ is raised to the power $6$. Therefore, the equilibrium constant for the above reaction is expressed as,$K=\frac{\left[\text{Ni}{\left({\text{NH}}_3\right)}_6{}^{2+}\right]}{\left[{\text{Ni}}^{\text{2+}}\right]{\left[{\text{NH}}_3\right]}^6}$

Conclusion

The equilibrium constant for the given reaction is expressed as,$K=\frac{\left[\text{Ni}{\left({\text{NH}}_3\right)}_6{}^{2+}\right]}{\left[{\text{Ni}}^{\text{2+}}\right]{\left[{\text{NH}}_3\right]}^6}$

Interpretation Introduction

(b)

Interpretation:

The equilibrium expression for the given gaseous reaction is to be stated.

Concept Introduction:

The equilibrium constant of a reaction is expressed as the ratio of concentration of products and reactants each raised to the power of their stoichiometric coefficients. A general equilibrium reaction is represented as,$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The equilibrium constant for the above chemical reaction is expressed as,$K=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Where,

• $\left[\text{A}\right]$ represents the equilibrium concentration of reactant $\text{A}$.

• $\left[\text{B}\right]$ represents the equilibrium concentration of reactant $\text{B}$.

• $\left[\text{C}\right]$ represents the equilibrium concentration of product $\text{C}$.

• $\left[\text{D}\right]$ represents the equilibrium concentration of product $\text{D}$.

• $a$ represents the stoichiometric coefficient of reactant $\text{A}$.

• $b$ represents the stoichiometric coefficient of reactant $\text{B}$.

• $c$ represents the stoichiometric coefficient of product $\text{C}$.

• $d$ represents the stoichiometric coefficient of product $\text{D}$.

Answer to Problem 8.43E

The equilibrium constant for the given gaseous reaction is expressed as,$K=\frac{\left[{\text{Sn}}^{4+}\right]{\left[{\text{Fe}}^{2+}\right]}^2}{\left[{\text{Sn}}^{2+}\right]{\left[{\text{Fe}}^{3+}\right]}^2}$

Explanation of Solution

The given gaseous reaction is represented as,${\text{Sn}}^{2+}+2{\text{Fe}}^{3+}\rightleftarrows {\text{Sn}}^{4+}+2{\text{Fe}}^{2+}$

The concentration of the product ${\text{Sn}}^{4+}$ and ${\text{Fe}}^{2+}$ will be in the numerator. The concentration of the product ${\text{Sn}}^{4+}$ is raised to $1$ and ${\text{Fe}}^{2+}$ is raised to the power $2$. The concentration of reactant ${\text{Sn}}^{2+}$ and ${\text{Fe}}^{3+}$ will be in the denominator. The concentration of reactant ${\text{Sn}}^{2+}$ is raised to the power $1$ and ${\text{Fe}}^{3+}$ is raised to the power $2$. Therefore, the equilibrium constant for the above reaction is expressed as,$K=\frac{\left[{\text{Sn}}^{4+}\right]{\left[{\text{Fe}}^{2+}\right]}^2}{\left[{\text{Sn}}^{2+}\right]{\left[{\text{Fe}}^{3+}\right]}^2}$

Conclusion

The equilibrium constant for the given reaction is expressed as,$K=\frac{\left[{\text{Sn}}^{4+}\right]{\left[{\text{Fe}}^{2+}\right]}^2}{\left[{\text{Sn}}^{2+}\right]{\left[{\text{Fe}}^{3+}\right]}^2}$

Interpretation Introduction

(c)

Interpretation:

The equilibrium expression for the given gaseous reaction is to be stated.

Concept Introduction:

The equilibrium constant of a reaction is expressed as the ratio of concentration of products and reactants each raised to the power of their stoichiometric coefficients. A general equilibrium reaction is represented as,$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The equilibrium constant for the above chemical reaction is expressed as,$K=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Where,

• $\left[\text{A}\right]$ represents the equilibrium concentration of reactant $\text{A}$.

• $\left[\text{B}\right]$ represents the equilibrium concentration of reactant $\text{B}$.

• $\left[\text{C}\right]$ represents the equilibrium concentration of product $\text{C}$.

• $\left[\text{D}\right]$ represents the equilibrium concentration of product $\text{D}$.

• $a$ represents the stoichiometric coefficient of reactant $\text{A}$.

• $b$ represents the stoichiometric coefficient of reactant $\text{B}$.

• $c$ represents the stoichiometric coefficient of product $\text{C}$.

• $d$ represents the stoichiometric coefficient of product $\text{D}$.

Answer to Problem 8.43E

The equilibrium constant for the given gaseous reaction is expressed as,$K=\frac{\left[{\text{Cl}}_2\right]{\left[{\text{F}}^{-}\right]}^2}{{\left[{\text{Cl}}^{-}\right]}^2\left[{\text{F}}_2\right]}$

Explanation of Solution

The given gaseous reaction is represented as,${\text{F}}_2+2{\text{Cl}}^{-}\rightleftarrows 2{\text{F}}^{-}+{\text{Cl}}_2$

The concentration of the product ${\text{F}}^{-}$ and ${\text{Cl}}_2$ will be in the numerator. The concentration of the product ${\text{Cl}}_2$ is raised to $1$ and ${\text{F}}^{-}$ is raised to the power $2$. The concentration of reactant ${\text{Cl}}^{-}$ and ${\text{F}}_2$ will be in the denominator. The concentration of reactant ${\text{F}}_2$ is raised to the power $1$ and ${\text{Cl}}^{-}$ is raised to the power $2$. Therefore, the equilibrium constant for the above reaction is expressed as,$K=\frac{\left[{\text{Cl}}_2\right]{\left[{\text{F}}^{-}\right]}^2}{{\left[{\text{Cl}}^{-}\right]}^2\left[{\text{F}}_2\right]}$

Conclusion

The equilibrium constant for the given reaction is expressed as,$K=\frac{\left[{\text{Cl}}_2\right]{\left[{\text{F}}^{-}\right]}^2}{{\left[{\text{Cl}}^{-}\right]}^2\left[{\text{F}}_2\right]}$