Chapter 17.1, Problem 1PP

(a)

Interpretation Introduction

Interpretation: The equilibrium constant of a chemical reaction quotient at chemical equilibrium needs to be determined.

Concept Introduction: The equilibrium constant expresses the relationship between product and reactants of a reaction at equilibrium.

The equilibrium constant is denoted by ${\text{'K}}_{\text{C}}\text{'}$.

Answer to Problem 1PP

${\text{K}}_{\text{C}}=\frac{{[\text{N}\text{O}(\text{g})]}^2}{[{\text{N}}_2{\text{O}}_4(\text{g})]}$

Explanation of Solution

The equilibrium constant expression is the ratio of the concentration of the products over the reactants.

The equilibrium constant is donated by ${\text{K}}_{\text{C}}$.

For example, consider a general reversible reaction.

$\text{a})\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$ .

  $\text{Reactants}\rightleftharpoons \text{Products}$

  ${\text{K}}_{\text{C}}$ for this reaction

  ${\text{K}}_{\text{C}}=\frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

$\text{b})\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$ .

  ${\text{K}}_{\text{C}}=\frac{{[\text{C}]}^{\text{c}}{[\text{D}]}^{\text{d}}}{{[\text{A}]}^{\text{a}}{[\text{B}]}^{\text{b}}}$

In this reaction, $[\text{A}][\text{B}][\text{C}][\text{D}]$ are the equilibrium concentration of the reactant and product. Therefore, ${\text{K}}_{\text{C}}=\frac{[\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}]}{[\text{R}\text{e}\text{a}\text{c}\text{t}\text{a}\text{n}\text{t}]}$

Equilibrium constant expression for the following reaction is:

${\text{N}}_2{\text{O}}_2(\text{g})\rightleftharpoons 2\text{N}{\text{O}}_2(\text{g})$

  ${\text{K}}_{\text{C}}=\frac{{[\text{N}\text{O}(\text{g})]}^2}{[{\text{N}}_2{\text{O}}_4(\text{g})]}$

(b)

Interpretation Introduction

Interpretation: The equilibrium constant of a chemical reaction quotient at chemical equilibrium needs to be determined.

Concept Introduction: The equilibrium constant expresses the relationship between product and reactants of a reaction at equilibrium.

The equilibrium constant is denoted by ${\text{'K}}_{\text{C}}\text{'}$.

Answer to Problem 1PP

  ${\text{K}}_{\text{C}}=\frac{{[{\text{H}}_2(\text{g})]}^2[{\text{S}}_2(\text{g})]}{{[{\text{H}}_2\text{S}(\text{g})]}^2}$

Explanation of Solution

The equilibrium constant expression is the ratio of the concentration of the products over the reactants.

The equilibrium constant is donated by ${\text{K}}_{\text{C}}$.

For example, consider a general reversible reaction.

$\text{a})\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$ .

  $\text{Reactants}\rightleftharpoons \text{Products}$

  ${\text{K}}_{\text{C}}$ for this reaction

  ${\text{K}}_{\text{C}}=\frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

$\text{b})\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$ .

  ${\text{K}}_{\text{C}}=\frac{{[\text{C}]}^{\text{c}}{[\text{D}]}^{\text{d}}}{{[\text{A}]}^{\text{a}}{[\text{B}]}^{\text{b}}}$

In this reaction, $[\text{A}][\text{B}][\text{C}][\text{D}]$ are the equilibrium concentration of the reactant and product. Therefore, ${\text{K}}_{\text{C}}=\frac{[\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}]}{[\text{R}\text{e}\text{a}\text{c}\text{t}\text{a}\text{n}\text{t}]}$

Equilibrium constant expression for the following reaction is:

${\text{H}}_2\text{S}(\text{g})\rightleftharpoons 2{\text{H}}_2(\text{g})+{\text{S}}_2(\text{g})$

  ${\text{K}}_{\text{C}}=\frac{{[{\text{H}}_2(\text{g})]}^2[{\text{S}}_2(\text{g})]}{{[{\text{H}}_2\text{S}(\text{g})]}^2}$

(c)

Interpretation Introduction

Interpretation: - The equilibrium constant of a chemical reaction quotient at chemical equilibrium needs to be determined.

Concept Introduction: - The equilibrium constant expresses the relationship between product and reactants of a reaction at equilibrium.

The equilibrium constant is denoted by ${\text{'K}}_{\text{C}}\text{'}$.

Answer to Problem 1PP

  ${\text{K}}_{\text{C}}=\frac{[\text{C}{\text{H}}_2(\text{g})][{\text{H}}_2\text{O}(\text{g})]}{[\text{C}\text{O}(\text{g})]{[{\text{H}}_2(\text{g})]}^3}$

Explanation of Solution

The equilibrium constant expression is the ratio of the concentration of the products over the reactants.

The equilibrium constant is donated by ${\text{K}}_{\text{C}}$.

For example, consider a general reversible reaction.

$\text{a})\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$ .

  $\text{Reactants}\rightleftharpoons \text{Products}$

  ${\text{K}}_{\text{C}}$ for this reaction

  ${\text{K}}_{\text{C}}=\frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

$\text{b})\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$ .

  ${\text{K}}_{\text{C}}=\frac{{[\text{C}]}^{\text{c}}{[\text{D}]}^{\text{d}}}{{[\text{A}]}^{\text{a}}{[\text{B}]}^{\text{b}}}$

In this reaction, $[\text{A}][\text{B}][\text{C}][\text{D}]$ are the equilibrium concentration of the reactant and product. Therefore, ${\text{K}}_{\text{C}}=\frac{[\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}]}{[\text{R}\text{e}\text{a}\text{c}\text{t}\text{a}\text{n}\text{t}]}$

Equilibrium constant expression for the following reaction is:

$\text{C}\text{O}(\text{g})+3{\text{H}}_2(\text{g})\rightleftharpoons \text{C}{\text{H}}_4(\text{g})+{\text{H}}_2\text{O}(\text{g})$

  ${\text{K}}_{\text{C}}=\frac{[\text{C}{\text{H}}_2(\text{g})][{\text{H}}_2\text{O}(\text{g})]}{[\text{C}\text{O}(\text{g})]{[{\text{H}}_2(\text{g})]}^3}$

(d)

Interpretation Introduction

Interpretation: The equilibrium constant of a chemical reaction quotient at chemical equilibrium needs to be determined.

Concept Introduction: The equilibrium constant expresses the relationship between product and reactants of a reaction at equilibrium.

The equilibrium constant is denoted by ${\text{'K}}_{\text{C}}\text{'}$.

Answer to Problem 1PP

  ${\text{K}}_{\text{C}}=\frac{{[\text{N}\text{O}(\text{g})]}^4{[{\text{H}}_2\text{O}(\text{g})]}^6}{{[\text{N}{\text{H}}_3(\text{g})]}^4{[{\text{O}}_2(\text{g})]}^5}$

Explanation of Solution

The equilibrium constant expression is the ratio of the concentration of the products over the reactants.

The equilibrium constant is donated by ${\text{K}}_{\text{C}}$.

For example, consider a general reversible reaction.

$\text{a})\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$ .

  $\text{Reactants}\rightleftharpoons \text{Products}$

  ${\text{K}}_{\text{C}}$ for this reaction

  ${\text{K}}_{\text{C}}=\frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

$\text{b})\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$ .

  ${\text{K}}_{\text{C}}=\frac{{[\text{C}]}^{\text{c}}{[\text{D}]}^{\text{d}}}{{[\text{A}]}^{\text{a}}{[\text{B}]}^{\text{b}}}$

In this reaction, $[\text{A}][\text{B}][\text{C}][\text{D}]$ are the equilibrium concentration of the reactant and product. Therefore, ${\text{K}}_{\text{C}}=\frac{[\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}]}{[\text{R}\text{e}\text{a}\text{c}\text{t}\text{a}\text{n}\text{t}]}$

Equilibrium constant expression for the following reaction is:

$4\text{N}{\text{H}}_3(\text{g})+5{\text{O}}_2(\text{g})\rightleftharpoons 4\text{N}\text{O}(\text{g})+6{\text{H}}_2\text{O}(\text{g})$

  ${\text{K}}_{\text{C}}=\frac{{[\text{N}\text{O}(\text{g})]}^4{[{\text{H}}_2\text{O}(\text{g})]}^6}{{[\text{N}{\text{H}}_3(\text{g})]}^4{[{\text{O}}_2(\text{g})]}^5}$

(e)

Interpretation Introduction

Interpretation: The equilibrium constant of a chemical reaction quotient at chemical equilibrium needs to be determined.

Concept Introduction: The equilibrium constant expresses the relationship between product and reactants of a reaction at equilibrium.

The equilibrium constant is denoted by ${\text{'K}}_{\text{C}}\text{'}$.

Answer to Problem 1PP

  ${\text{K}}_{\text{C}}=\frac{[\text{C}{\text{S}}_2(\text{g})]{[{\text{H}}_2(\text{g})]}^4}{[\text{C}{\text{H}}_4(\text{g})]{[{\text{H}}_2\text{S}(\text{g})]}^2}$

Explanation of Solution

The equilibrium constant expression is the ratio of the concentration of the products over the reactants.

The equilibrium constant is donated by ${\text{K}}_{\text{C}}$.

For example, consider a general reversible reaction.

$\text{a})\text{A}+\text{B}\rightleftharpoons \text{C}+\text{D}$ .

  $\text{Reactants}\rightleftharpoons \text{Products}$

  ${\text{K}}_{\text{C}}$ for this reaction

  ${\text{K}}_{\text{C}}=\frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

$\text{b})\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons \text{c}\text{C}+\text{d}\text{D}$ .

  ${\text{K}}_{\text{C}}=\frac{{[\text{C}]}^{\text{c}}{[\text{D}]}^{\text{d}}}{{[\text{A}]}^{\text{a}}{[\text{B}]}^{\text{b}}}$

In this reaction, $[\text{A}][\text{B}][\text{C}][\text{D}]$ are the equilibrium concentration of the reactant and product. Therefore, ${\text{K}}_{\text{C}}=\frac{[\text{P}\text{r}\text{o}\text{d}\text{u}\text{c}\text{t}]}{[\text{R}\text{e}\text{a}\text{c}\text{t}\text{a}\text{n}\text{t}]}$

Equilibrium constant expression for the following reaction is:

$\text{C}{\text{H}}_4(\text{g})+2{\text{H}}_2\text{S}(\text{g})\rightleftharpoons \text{C}{\text{S}}_2(\text{g})+4{\text{H}}_2(\text{g})$

  ${\text{K}}_{\text{C}}=\frac{[\text{C}{\text{S}}_2(\text{g})]{[{\text{H}}_2(\text{g})]}^4}{[\text{C}{\text{H}}_4(\text{g})]{[{\text{H}}_2\text{S}(\text{g})]}^2}$