Chapter 9, Problem 9.72EP

Write equilibrium constant expressions for the following generalized reactions.

1. a. $3\text{X}(g)+2\text{Y}(g)\underset{}{\rightleftharpoons }\text{Z}(g)$
2. b. $3\text{X}(g)+2\text{Y}(s)\underset{}{\rightleftharpoons }\text{Z}(g)$
3. c. $3\text{X}(s)+2\text{Y}(s)\underset{}{\rightleftharpoons }\text{Z}(g)$
4. d. $3\text{X}(g)+2\text{Y}(g)\underset{}{\rightleftharpoons }\text{Z}(s)$

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the given reaction has to be written.

    $\text{3X(g) + 2Y(g) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    ${\text{K}}_{\begin{array}{l}\text{eq}\\ \end{array}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

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

        ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

Explanation of Solution

The given reaction is:

    $\text{3X(g) + 2Y(g) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

The equilibrium constant, ${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Z]}}{{\text{[X]}}^3{\text{[Y]}}^2}$.

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the given reaction has to be written.

    $\text{3X(g) + 2Y(s) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    ${\text{K}}_{\begin{array}{l}\text{eq}\\ \end{array}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

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

        ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

Explanation of Solution

The given reaction is:

    $\text{3X(g) + 2Y(s) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

The equilibrium constant, ${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Z]}}{{\text{[X]}}^3}$.

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the given reaction has to be written.

    $\text{3X(s) + 2Y(s) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    ${\text{K}}_{\begin{array}{l}\text{eq}\\ \end{array}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

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

        ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

Explanation of Solution

The given reaction is:

    $\text{3X(s) + 2Y(s) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(g)}$

The equilibrium constant, ${\text{K}}_{\text{eq}}\text{= [Z]}$.

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the given reaction has to be written.

    $\text{3X(g) + 2Y(g) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(s)}$

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    ${\text{K}}_{\begin{array}{l}\text{eq}\\ \end{array}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

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

        ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

Explanation of Solution

The given reaction is:

    $\text{3X(g) + 2Y(g) }\underset{}{\overset{}{\rightleftharpoons }}\text{ Z(s)}$

The equilibrium constant, ${\text{K}}_{\text{eq}}\text{=}\frac{1}{{\text{[X]}}^3{\text{[Y]}}^2}$.