Chapter 12, Problem 10QAP

Interpretation Introduction

(a)

Interpretation:

A balanced chemical equation for the given description and the equilibrium constant for the expression should be determined.

Concept introduction:

For a general equilibrium reaction as follows:

$aA+bB\rightleftharpoons cC+dD$

The expression for equilibrium constant of a reaction can be calculated as follows:

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

Here, A and B are reactants with stoichiometric coefficient a and b respectively. Similarly, C and D are products with stoichiometric coefficient c and d respectively

In the expression for equilibrium constant, the species in pure solid and liquid state are not written as equilibrium constant does not depend on them.

Answer to Problem 10QAP

$K=P_{{\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}\left(g\right)}$

Explanation of Solution

According to the given description, the liquid acetone, ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}\left(l\right)$ is in equilibrium with its vapor state.

The reaction can be written as follows:

${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}\left(g\right)$

The above reaction is already balanced.

The expression for equilibrium constant will be as follows:

$K=P_{{\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}\left(g\right)}$

Interpretation Introduction

(b)

Interpretation:

A balanced chemical equation for the given description and the equilibrium constant for the expression should be determined.

Concept introduction:

For a general equilibrium reaction as follows:

$aA+bB\rightleftharpoons cC+dD$

The expression for equilibrium constant of a reaction can be calculated as follows:

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

Here, A and B are reactants with stoichiometric coefficient a and b respectively. Similarly, C and D are products with stoichiometric coefficient c and d respectively

In the expression for equilibrium constant, the species in pure solid and liquid state are not written as equilibrium constant does not depend on them.

Answer to Problem 10QAP

$K=\frac{{\left(P_{{\text{NH}}_{\text{3}}\left(g\right)}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}\right)}^4}{{\left(P_{{\text{H}}_{\text{2}}\left(g\right)}\right)}^7{\left(P_{{\text{NO}}_{\text{2}}\left(g\right)}\right)}^2}$

Explanation of Solution

According to the given description, the reaction of hydrogen gas ${\text{H}}_{\text{2}}\left(g\right)$ with nitrogen dioxide gas ${\text{NO}}_{\text{2}}\left(g\right)$ results in the production of ammonia ${\text{NH}}_{\text{3}}\left(g\right)$ and steam ${\text{H}}_{\text{2}}\text{O}\left(g\right)$.

The reaction can be written as follows:

${\text{H}}_{\text{2}}\left(g\right)+{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{NH}}_{\text{3}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

To balance the number of oxygen atoms, give coefficient 2 to ${\text{H}}_{\text{2}}\text{O}\left(g\right)$.

${\text{H}}_{\text{2}}\left(g\right)+{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{NH}}_{\text{3}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Give coefficient 7/2 to ${\text{H}}_{\text{2}}\left(g\right)$ to balance the number of hydrogen atoms.

${\text{7/2H}}_{\text{2}}\left(g\right)+{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{NH}}_{\text{3}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Or,

${\text{7H}}_{\text{2}}\left(g\right)+2{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons 2{\text{NH}}_{\text{3}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The expression for equilibrium constant will be as follows:

$K=\frac{{\left(P_{{\text{NH}}_{\text{3}}\left(g\right)}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}\right)}^4}{{\left(P_{{\text{H}}_{\text{2}}\left(g\right)}\right)}^7{\left(P_{{\text{NO}}_{\text{2}}\left(g\right)}\right)}^2}$

Interpretation Introduction

(c)

Interpretation:

A balanced chemical equation for the given description and the equilibrium constant for the expression should be determined.

Concept introduction:

For a general equilibrium reaction as follows:

$aA+bB\rightleftharpoons cC+dD$

The expression for equilibrium constant of a reaction can be calculated as follows:

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

Here, A and B are reactants with stoichiometric coefficient a and b respectively. Similarly, C and D are products with stoichiometric coefficient c and d respectively

In the expression for equilibrium constant, the species in pure solid and liquid state are not written as equilibrium constant does not depend on them.

Answer to Problem 10QAP

$K=\frac{{\left[{\text{H}}^{+}\left(aq\right)\right]}^2}{P_{{\text{H}}_{\text{2}}\text{S}\left(g\right)}\left[{\text{Pb}}^{2+}\left(aq\right)\right]}$

Explanation of Solution

According to the given description, the reaction of hydrogen sulfide gas ${\text{H}}_{\text{2}}\text{S}\left(g\right)$ and aqueous solution of lead (II) ion ${\text{Pb}}^{2+}\left(aq\right)$ results in the formation of lead sulfide precipitate $\text{PbS}\left(s\right)$ and hydrogen ions ${\text{H}}^{+}\left(aq\right)$.

Thus, the reaction is as follows:

${\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{Pb}}^{2+}\left(aq\right)\rightleftharpoons \text{PbS}\left(s\right)+{\text{H}}^{+}\left(aq\right)$

Give coefficient 2 to ${\text{H}}^{+}\left(aq\right)$ to balance the number of hydrogen atoms.

${\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{Pb}}^{2+}\left(aq\right)\rightleftharpoons \text{PbS}\left(s\right)+2{\text{H}}^{+}\left(aq\right)$

The expression for the equilibrium constant will be:

$K=\frac{{\left[{\text{H}}^{+}\left(aq\right)\right]}^2}{P_{{\text{H}}_{\text{2}}\text{S}\left(g\right)}\left[{\text{Pb}}^{2+}\left(aq\right)\right]}$