Chapter 9, Problem 9.78EP

Use the given K_eq value and the terminology in Table 9-2 to describe the relative amounts of reactants and products present in each of the following equilibrium situations.

1. a. $2\text{NO}(g)\rightleftharpoons {\text{N}}_2(g)+{\text{O}}_2(g)K_{\text{eq}}(25°\text{C})=1\times {10}^{30}$
2. b. ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons {\text{2NH}}_3(g)K_{\text{eq}}(25°\text{C})=1\times {10}^9$
3. c. ${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)K_{\text{eq}}(127°\text{C})=1\times {10}^{-2}$
4. d. ${\text{2Na}}_2\text{O}(s)\rightleftharpoons 4\text{Na}(l)+{\text{O}}_2(g)K_{\text{eq}}(427°\text{C})=1\times {10}^{-25}$

Interpretation Introduction

Interpretation:

For the reaction $\text{2NO(g) }\underset{}{\rightleftharpoons }{\text{ N}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}$, the relative amount of reactants and products present has to be described using the given ${\text{K}}_{\text{eq}}$ value.

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:

    $\text{2NO(g) }\underset{}{\rightleftharpoons }{\text{ N}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}$

Given,

  ${\text{K}}_{\text{eq}}{\text{(25}}^{\text{o}}{\text{C) = 1×10}}^{30}$

The value of ${\text{K}}_{\text{eq}}$ is very large, so the products formed will be more than the reactants.  Equilibrium will be far to the right.

Hence the product formed will be more.

Interpretation Introduction

Interpretation:

For the reaction ${\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ 2NH}}_{\text{3}}\text{(g)}$, the relative amount of reactants and products present has to be described using the given ${\text{K}}_{\text{eq}}$ value.

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:

    ${\text{ N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ 2NH}}_{\text{3}}\text{(g)}$

Given the,

  ${\text{K}}_{\text{eq}}{\text{(25}}^{\text{o}}{\text{C) = 1×10}}^{\text{9}}$

The value of ${\text{K}}_{\text{eq}}$ is very large, so the products formed will be more than the reactants.  Equilibrium will shift to the right.

Hence the product formed will be more.

Interpretation Introduction

Interpretation:

For the reaction ${\text{PCl}}_{\text{5}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ PCl}}_{\text{3}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$, the relative amount of reactants and products present has to be described using the given ${\text{K}}_{\text{eq}}$ value.

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:

    ${\text{PCl}}_{\text{5}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ PCl}}_{\text{3}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$

Given,

  ${\text{K}}_{\text{eq}}{\text{(127}}^{\text{o}}{\text{C) = 1×10}}^{\text{-2}}$

The value of ${\text{K}}_{\text{eq}}$ is near to unity, so there are significant amounts of reactants and products in the reaction mixture.

Interpretation Introduction

Interpretation:

For the reaction ${\text{2Na}}_{\text{2}}\text{O(s) }\underset{}{\rightleftharpoons }{\text{ 4Na(l) + O}}_{\text{2}}\text{(g)}$, the relative amount of reactants and products present has to be described using the given ${\text{K}}_{\text{eq}}$ value.

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:

    ${\text{2Na}}_{\text{2}}\text{O(s) }\underset{}{\rightleftharpoons }{\text{ 4Na(l) + O}}_{\text{2}}\text{(g)}$

Given,

  ${\text{K}}_{\text{eq}}{\text{(427}}^{\text{o}}{\text{C) = 1×10}}^{\text{-25}}$

The value of ${\text{K}}_{\text{eq}}$ is very small, essentially the reaction mixture will contain more reactants.  The equilibrium is far to the left.