Chapter 12, Problem 12QAP

Write a chemical equation for an equilibrium system that would lead to the following expressions $\left(\text{a}-\text{d}\right)$ for K.

(a) $K=\frac{{\left(P_{{\text{H}}_2\text{S}}\right)}^2{\left(P_{{\text{O}}_2}\right)}^3}{{\left(P_{{\text{SO}}_2}\right)}^2{\left(P_{{\text{H}}_2\text{O}}\right)}^2}$(b) $K=\frac{{\left(P_{{\text{F}}_2}\right)}^{1/2}{\left(P_{{\text{I}}_2}\right)}^{1/2}}{P_{\text{IF}}}$(c) $K=\frac{{\left[{\text{Cl}}^{-}\right]}^2}{\left(P_{{\text{cl}}_2}\right){\left[{\text{Br}}^{-}\right]}^2}$(d) $K=\frac{{\left(P_{\text{NO}}\right)}^2{\left(P_{{\text{H}}_2\text{O}}\right)}^4{\left[{\text{Cu}}^{2+}\right]}^3}{{\left[{\text{NO}}_3{}^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^8}$

Interpretation Introduction

(a)

Interpretation:

For the given expression of equilibrium constant, the equation for the equilibrium reaction needs to be determined.

$K=\frac{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2{\left(P_{{\text{O}}_{\text{2}}}\right)}^3}{{\left(P_{{\text{SO}}_{\text{2}}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^2}$

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 12QAP

${\text{2SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_{\text{2}}^{}\left(g\right)$

Explanation of Solution

The given expression for the equilibrium constant, K is as follows:

$K=\frac{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2{\left(P_{{\text{O}}_{\text{2}}}\right)}^3}{{\left(P_{{\text{SO}}_{\text{2}}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^2}$

Here, ${\text{H}}_{\text{2}}\text{S}\left(g\right)$ and ${\text{O}}_{\text{2}}^{}$ (g) will be the products and ${\text{SO}}_2$ (g) and ${\text{H}}_{\text{2}}\text{O}$ (g) will be the reactants.

The equation for the equilibrium system will be:

${\text{2SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_{\text{2}}^{}\left(g\right)$

Interpretation Introduction

(b)

Interpretation:

For the given expression of equilibrium constant, the equation for the equilibrium reaction needs to be determined.

$K=\frac{{\left(P_{{\text{F}}_2}\right)}^{1/2}{\left(P_{{\text{I}}_{\text{2}}}\right)}^{1/2}}{\left(P_{\text{IF}}\right)}$

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 12QAP

$\text{IF}\left(g\right)\rightleftharpoons \frac{1}{2}{\text{F}}_2\left(g\right)+\frac{1}{2}{\text{I}}_2\left(g\right)$

Explanation of Solution

The given expression for the equilibrium constant, K is as follows:

$K=\frac{{\left(P_{{\text{F}}_2}\right)}^{1/2}{\left(P_{{\text{I}}_{\text{2}}}\right)}^{1/2}}{\left(P_{\text{IF}}\right)}$

Here, ${\text{F}}_2$ (g) and ${\text{I}}_2$ (g) will be the products and IF (g) will be the reactant.

The equation for the equilibrium system will be:

$\text{IF}\left(g\right)\rightleftharpoons \frac{1}{2}{\text{F}}_2\left(g\right)+\frac{1}{2}{\text{I}}_2\left(g\right)$

Or,

$\text{2IF}\left(g\right)\rightleftharpoons {\text{F}}_2\left(g\right)+{\text{I}}_2\left(g\right)$

Interpretation Introduction

(c)

Interpretation:

For the given expression of equilibrium constant, the equation for the equilibrium reaction needs to be determined.

$K=\frac{{\left[{\text{Cl}}_{}^{-}\right]}^2}{\left(P_{{\text{Cl}}_{\text{2}}}\right){\left[{\text{Br}}_{}^{-}\right]}^2}$

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 12QAP

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{Br}}_{}^{-}\left(aq\right)\rightleftharpoons 2{\text{Cl}}_{}^{-}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)$

Explanation of Solution

The given expression for the equilibrium constant, K is as follows:

$K=\frac{{\left[{\text{Cl}}_{}^{-}\right]}^2}{\left(P_{{\text{Cl}}_{\text{2}}}\right){\left[{\text{Br}}_{}^{-}\right]}^2}$

Here, ${\text{Cl}}_{}^{-}\left(aq\right)$ will be the product and ${\text{Cl}}_{\text{2}}$ (g) and ${\text{Br}}_{}^{-}\left(aq\right)$ will be the reactants.

The equation for the equilibrium system will be:

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{Br}}_{}^{-}\left(aq\right)\rightleftharpoons 2{\text{Cl}}_{}^{-}\left(aq\right)$

To balance the above reaction:

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{Br}}_{}^{-}\left(aq\right)\rightleftharpoons 2{\text{Cl}}_{}^{-}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)$

Interpretation Introduction

(d)

Interpretation:

For the given expression of equilibrium constant, the equation for the equilibrium reaction needs to be determined.

$K=\frac{{\left(P_{\text{NO}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^4{\left[{\text{Cu}}_{}^{2+}\right]}^3}{{\left({\text{NO}}_3^{-}\right)}^2{\left[{\text{H}}_{}^{+}\right]}^8}$

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 12QAP

$2{\text{NO}}_3^{-}\left(aq\right)+8{\text{H}}_{}^{+}\left(aq\right)+\text{Cu}\left(s\right)\rightleftharpoons \text{2NO}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)+3{\text{Cu}}^{2+}\left(aq\right)$

Explanation of Solution

The given expression for the equilibrium constant, K is as follows:

$K=\frac{{\left(P_{\text{NO}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^4{\left[{\text{Cu}}_{}^{2+}\right]}^3}{{\left({\text{NO}}_3^{-}\right)}^2{\left[{\text{H}}_{}^{+}\right]}^8}$

Here, $\text{NO}\left(g\right)$, ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ and ${\text{Cu}}^{2+}\left(aq\right)$ will be the products and ${\text{NO}}_3^{-}\left(aq\right)$ (g) and ${\text{H}}_{}^{+}\left(aq\right)$ will be the reactants.

The equation for the equilibrium system will be:

$2{\text{NO}}_3^{-}\left(aq\right)+8{\text{H}}_{}^{+}\left(aq\right)\rightleftharpoons \text{2NO}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)+3{\text{Cu}}^{2+}\left(aq\right)$

To balance the above reaction:

$2{\text{NO}}_3^{-}\left(aq\right)+8{\text{H}}_{}^{+}\left(aq\right)+\text{Cu}\left(s\right)\rightleftharpoons \text{2NO}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)+3{\text{Cu}}^{2+}\left(aq\right)$