Chapter 13, Problem 38E

Write expressions for K_p for the following reactions.

a. $2\text{Fe}\left(S\right)+\frac{3}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{Fe}}_2{\text{O}}_3\left(S\right)$

b. ${\text{CO}}_2\left(g\right)+\text{M}gO\left(s\right)\rightleftharpoons {\text{MgCO}}_3\left(s\right)$

c. $\text{C}\left(s\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$

d. $4{\text{KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 4\text{KOH}\left(s\right)+3{\mathrm{O}}_2\left(g\right)$

Interpretation Introduction

Interpretation: The expression for $K$ and $K_{\text{p}}$ for the given reactions is to be determined.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$. When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$.

To determine: The expression for $K$ and $K_{\text{p}}$ for the given reaction.

Answer to Problem 38E

The expression for $K$ and $K_{\text{p}}$ is as follows.

Explanation of Solution

Given

The stated reaction is,

$\text{2Fe}\left(s\right)+\frac{3}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{Fe}}_2{\text{O}}_3\left(s\right)$

At equilibrium, the equilibrium constant expression $\left(K\right)$ is stated as,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The solid components do not affect the value of the equilibrium constant.

For the given reaction, the equilibrium constant expression $\left(K\right)$ is,

$K=\frac{1}{{\left[{\text{O}}_2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

The expression for $K_{\text{p}}$ the given reaction.

The expression for $K_{\text{p}}$ is as follows.

Given

The stated reaction is,

$\text{2Fe}\left(s\right)+\frac{3}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{Fe}}_2{\text{O}}_3\left(s\right)$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The solid components do not affect the value of the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{\text{1}}{{\left({\text{P}}_{{\text{O}}_2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Conclusion

When the equilibrium constant is expressed in terms of concentration, it is represented $K$ When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$. The solid components do not affect the value of the equilibrium constant.

Interpretation Introduction

Interpretation: The expression for $K$ and $K_{\text{p}}$ for the given reactions is to be determined.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$. When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$.

To determine: The expression for $K$ and $K_{\text{p}}$ for the given reaction.

Answer to Problem 38E

The expression for $K$ and $K_{\text{p}}$ is as follows.

Explanation of Solution

Given

The stated reaction is,

${\text{CO}}_2\left(g\right)+\text{MgO}\left(s\right)\rightleftharpoons {\text{MgCO}}_3\left(s\right)$

At equilibrium, the equilibrium constant expression $\left(K\right)$ is stated as,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The solid components do not affect the value of the equilibrium constant.

For the given reaction, the equilibrium constant expression $\left(K\right)$ is,

$K=\frac{1}{\left[{\text{CO}}_2\right]}$

The expression for $K_{\text{p}}$ the given reaction.

The expression for $K_{\text{p}}$ is as follows.

Given

The stated reaction is,

${\text{CO}}_2\left(g\right)+\text{MgO}\left(s\right)\rightleftharpoons {\text{MgCO}}_3\left(s\right)$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The solid components do not affect the value of the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{1}{\left({\text{P}}_{{\text{CO}}_2}\right)}$

Conclusion

When the equilibrium constant is expressed in terms of concentration, it is represented $K$. When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$. The solid components do not affect the value of the equilibrium constant.

Interpretation Introduction

Interpretation: The expression for $K$ and $K_{\text{p}}$ for the given reactions is to be determined.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$. When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$.

To determine: The expression for $K$ and $K_{\text{p}}$ for the given reaction.

Answer to Problem 38E

The expression for $K$ and $K_{\text{p}}$ is as follows.

Explanation of Solution

Given

The stated reaction is,

$\text{C}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$

At equilibrium, the equilibrium constant expression $\left(K\right)$ is stated as,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The solid components do not affect the value of the equilibrium constant.

For the given reaction, the equilibrium constant expression $\left(K\right)$ is,

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

The expression for $K_{\text{p}}$ the given reaction.

The expression for $K_{\text{p}}$ is as follows.

Given

The stated reaction is,

$\text{C}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The solid components do not affect the value of the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{\left({\text{P}}_{\text{CO}}\right)\left({\text{P}}_{{\text{H}}_2}\right)}{\left({\text{P}}_{{\text{H}}_2\text{O}}\right)}$

Conclusion

When the equilibrium constant is expressed in terms of concentration, it is represented as $K$. When the equilibrium constant is expressed in terms of pressure, it is represented as $K_{\text{p}}$. The solid components do not affect the value of the equilibrium constant.

Interpretation Introduction

Interpretation: The expression for $K$ and $K_{\text{p}}$ for the given reactions is to be determined.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$. When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$.

To determine: The expression for $K$ and $K_{\text{p}}$ for the given reaction.

The expression for $K$ the given reaction.

Answer to Problem 38E

The expression for $K$ and $K_{\text{p}}$ is as follows.

Explanation of Solution

Given

The stated reaction is,

${\text{4KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons \text{4KOH}\left(s\right)+{\text{O}}_2\left(g\right)$

At equilibrium, the equilibrium constant expression $\left(K\right)$ is stated as,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The solid and liquid components do not affect the value of the equilibrium constant.

For the given reaction, the equilibrium constant expression $\left(K\right)$ is,

$K=\frac{\left[{\text{O}}_2\right]}{{\left[{\text{H}}_2\text{O}\right]}^2}$

The expression for $K_{\text{p}}$ the given reaction.

The expression for $K_{\text{p}}$ is as follows.

Given

The stated reaction is,

${\text{4KO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons \text{4KOH}\left(s\right)+{\text{O}}_2\left(g\right)$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The solid components do not affect the value of the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{\left({\text{P}}_{{\text{O}}_2}\right)}{{\left({\text{P}}_{{\text{H}}_2\text{O}}\right)}^2}$

Conclusion

When the equilibrium constant is expressed in terms of concentration, it is represented $K$ When the equilibrium constant is expressed in terms of pressure, it is represented $K_{\text{p}}$. The solid components do not affect the value of the equilibrium constant