Chapter 17.2, Problem 5RQ

(a)

Interpretation Introduction

Interpretation: Equilibrium expression for given reactions must be written.

Concept introduction: When the rate at which forward and backward reaction occur become equal such that concentration of reactants and products remains to be constant, that state is described as equilibrium. It is indicated by double sided arrow $\rightleftharpoons$ . It can be physical or chemical in nature.

For an equilibrium reaction:

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

The expression for equilibrium constant is:

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

Where,

  $\mathrm{K}$ is equilibrium constant.

  $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are reactant concentrations for A and B.

  $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are product concentrations for C and D.

a, b, c and d are coefficients of reactants.

Answer to Problem 5RQ

  $\mathrm{K}=\frac{1}{\left[{\text{NH}}_3\right]\left[\text{HCl}\right]}$

Explanation of Solution

Concentration of Pure solids and liquids are never incorporated in the expression for equilibrium astheir concentration remains to be unchanged through the reaction.

For reaction: ${\text{NH}}_3\left(\mathrm{g}\right)+\text{HCl}\left(\mathrm{g}\right)\rightleftharpoons {\text{NH}}_{\text{4}}\text{Cl}\left(\mathrm{s}\right)$

  $\mathrm{K}=\frac{1}{\left[{\text{NH}}_3\right]\left[\text{HCl}\right]}$

(b)

Interpretation Introduction

Interpretation: Equilibrium expression for given reactions must be written.

Concept introduction: When the rate at which forward and backward reaction occur become equal such that concentration of reactants and products remains to be constant, that state is described as equilibrium. It is indicated by double sided arrow $\rightleftharpoons$ . It can be physical or chemical in nature.

For an equilibrium reaction:

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

The expression for equilibrium constant is:

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

Where,

  $\mathrm{K}$ is equilibrium constant.

  $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are reactant concentrations for A and B.

  $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are product concentrations for C and D.

a, b, c and d are coefficients of reactants.

Answer to Problem 5RQ

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

Explanation of Solution

Concentration of Pure solids and liquids are never incorporated in the expression for equilibrium as their concentration remains to be unchanged through the reaction.

For reaction: ${\text{2H}}_2\left(\mathrm{g}\right)+{\text{O}}_2\left(\mathrm{g}\right)\rightleftharpoons {\text{2H}}_2\text{O}\left(\mathrm{g}\right)$

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

(c)

Interpretation Introduction

Interpretation: Equilibrium expression for given reactions must be written.

Concept introduction: When the rate at which forward and backward reaction occur become equal such that concentration of reactants and products remains to be constant, that state is described as equilibrium. It is indicated by double sided arrow $\rightleftharpoons$ . It can be physical or chemical in nature.

For an equilibrium reaction:

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

The expression for equilibrium constant is:

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

Where,

  $\mathrm{K}$ is equilibrium constant.

  $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are reactant concentrations for A and B.

  $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are product concentrations for C and D.

a, b, c and d are coefficients of reactants.

Answer to Problem 5RQ

  $\mathrm{K}=\frac{1}{{\left[{\text{H}}_2\right]}^2\left[{\text{O}}_2\right]}$

Explanation of Solution

Concentration of Pure solids and liquids are never incorporated in the expression for equilibrium astheir concentration remains to be unchanged through the reaction.

For reaction: ${\text{2H}}_2\left(\mathrm{g}\right)+{\text{O}}_2\left(\mathrm{g}\right)\rightleftharpoons {\text{2H}}_2\text{O}\left(\mathrm{l}\right)$

  $\mathrm{K}=\frac{1}{{\left[{\text{H}}_2\right]}^2\left[{\text{O}}_2\right]}$