Chapter 15, Problem 1E

Based on these descriptions, write a balanced equation and the corresponding $K_z$ expression for each reversible reaction.

a. Carbonyl fluoride. ${\text{COF}}_{\text{2}}\left(\text{g}\right)$ , decomposes Write gaseous carbon dioxide and gaseous carbon tertrafluoride.

b. Copper metal displaces sliver(l) eon from aqueous solution, producing slyer metal and an aqueous solution of copper(ll) ion.

c. Peroxodisulfate ion, ${\text{S}}_2{O}_8^{2-}$ , oxidizes won(ll) ion to iron(lll) eon r aqueous solution and is itself reduced to sulfate ion

Interpretation Introduction

(a)

Interpretation:

The balanced chemical equation and the K_c expression for the following reversible reaction should be determined:

Carbonyl fluoride, COF₂(g), decomposes into gaseous carbon dioxide and carbon tetrafluoride.

Concept introduction:

The balanced reaction is that chemical reaction in which the number of atoms of each element in the product side and on reactant side are equal.

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

$\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

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

Square brackets represent the concentration.

Answer to Problem 1E

The balanced chemical equation is:

$2CO{F}_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+C{F}_4\left(g\right)$

The expression for the equilibrium constant is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[CO}}_{\text{2}}{\text{][CF}}_4\text{]}}{{{\text{[COF}}_2\text{]}}^{\text{2}}}$

Explanation of Solution

The reversible reaction for the decomposition of carbonyl fluoride, COF₂(g), into gaseous carbon dioxide, CO₂(g) and carbon tetrafluoride, CF₄(g) is:

$CO{F}_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+C{F}_4\left(g\right)$

This reaction is not balanced as:

The number of atoms of C on reactant side is 1 and on product side is 2, the number of atoms of F on reactant side is 2 and on product side is 4 and the number of atoms of O on reactant side is 1 and on product side is 2.

In order to balance the reaction, the coefficient 2 is put before COF₂ on the reactant side.

Thus, the balanced chemical equation is:

$2CO{F}_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+C{F}_4\left(g\right)$

In the balanced reversible equation, all the species that is reactant and product are in gaseous phase so, the expression for the equilibrium constant is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[CO}}_{\text{2}}{\text{][CF}}_4\text{]}}{{{\text{[COF}}_2\text{]}}^{\text{2}}}$

Interpretation Introduction

(b)

Interpretation:

The balanced chemical equation and the K_c expression for the following reversible reaction should be determined:

Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion.

Concept introduction:

The balanced reaction is that chemical reaction in which the number of atoms of each element in the product side and on reactant side are equal.

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

$\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

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

Square brackets represent the concentration.

Answer to Problem 1E

The balanced chemical equation is:

$Cu\left(s\right)+2A{g}^{+}(aq)\rightleftharpoons 2Ag\left(s\right)+C{u}^{2+}\left(aq\right)$

The expression for the equilibrium constant is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[Cu}}^{2+}\text{]}}{{{\text{[Ag}}^{+}\text{]}}^{\text{2}}}$

Explanation of Solution

The reversible reaction for the displacement of silver(I) ion, Ag⁺(aq), in aqueous solution by copper metal, Cu(s) to produce silver metal, Ag (s) and an aqueous solution of copper (II) ion, Cu²+ is:

$Cu\left(s\right)+A{g}^{+}(aq)\rightleftharpoons Ag\left(s\right)+C{u}^{2+}\left(aq\right)$

This reaction is not balanced as:

The charges on both the sides of the reaction is not balanced so in order to balance the reaction, the coefficient 2 is put before Ag⁺, the reaction becomes:

$Cu\left(s\right)+2A{g}^{+}(aq)\rightleftharpoons Ag\left(s\right)+C{u}^{2+}\left(aq\right)$

In order to balance the Ag atoms, the coefficient 2 is put before Ag on the product side.

Thus, the balanced chemical equation is:

$Cu\left(s\right)+2A{g}^{+}(aq)\rightleftharpoons 2Ag\left(s\right)+C{u}^{2+}\left(aq\right)$

In the balanced reaction, the concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[Cu}}^{2+}\text{]}}{{{\text{[Ag}}^{+}\text{]}}^{\text{2}}}$

Interpretation Introduction

(c)

Interpretation:

The balanced chemical equation and the K_c expression for the following reversible reaction should be determined:

Peroxodisulfate ion, S₂O₈^2-, oxidizes iron (II) to iron (III) ion in aqueous solution and is itself reduced to sulfate ion.

Concept introduction:

The balanced reaction is that chemical reaction in which the number of atoms of each element in the product side and on reactant side are equal.

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

$\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

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

Square brackets represent the concentration.

Answer to Problem 1E

The balanced chemical equation is:

$S_2{O}_8{}^{2-}\left(aq\right)+2F{e}^{2+}(aq)\rightleftharpoons 2F{e}^{3+}(aq)+2S{O}_4{}^{2-}\left(aq\right)$

The expression for the equilibrium constant is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[}F{e}^{3+}\text{]}}^2{\text{[}S{O}_4{}^{2-}\text{]}}^2}{{\text{[}F{e}^{2+}\text{]}}^{\text{2}}[S_2{O}_8{}^{2-}]}$

Explanation of Solution

The oxidation of iron (II) ion, Fe²+ to iron (III) ion, Fe³+ by peroxodisulfate ion, S₂O₈²- and gets reduced itself to sulfate ion, SO₄²- in aqueous solution is:

$S_2{O}_8{}^{2-}\left(aq\right)+F{e}^{2+}(aq)\rightleftharpoons F{e}^{3+}(aq)+S{O}_4{}^{2-}\left(aq\right)$

This reaction is not balanced as:

The number of atoms of S on reactant side is 2 and on product side is 1 and the number of atoms of O on reactant side is 8 and on product side is 4.

In order to balance the charge and atoms in the reaction, the coefficient 2 is put before Fe²+ on the reactant side and coefficient 2 is put before the SO₄²- and Fe³+ on the product side.

Thus, the balanced chemical equation is:

$S_2{O}_8{}^{2-}\left(aq\right)+2F{e}^{2+}(aq)\rightleftharpoons 2F{e}^{3+}(aq)+2S{O}_4{}^{2-}\left(aq\right)$

In the balanced reversible equation, all the species that is reactant and product are in gaseous phase so, the expression for the equilibrium constant is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[}F{e}^{3+}\text{]}}^2{\text{[}S{O}_4{}^{2-}\text{]}}^2}{{\text{[}F{e}^{2+}\text{]}}^{\text{2}}[S_2{O}_8{}^{2-}]}$