Chapter 12, Problem 11QAP

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

(a) $K=\frac{{\left(P_{\text{co}}\right)}^2{\left(P_{{\text{H}}_2}\right)}^5}{\left(P_{{\text{C}}_2{\text{H}}_6}\right){\left(P_{{\text{H}}_2\text{O}}\right)}^2}$(b) $K=\frac{{\left(P_{{\text{NH}}_3}\right)}^4{\left(P_{{\text{O}}_2}\right)}^5}{{\left(P_{\text{NO}}\right)}^4{\left(P_{{\text{H}}_2\text{O}}\right)}^6}$(c) $K=\frac{{\left[{\text{ClO}}_3{}^{-}\right]}^2{\left[{\text{Mn}}^{2+}\right]}^2}{\left(P_{{\text{cl}}_2}\right){\left[{\text{MNO}}_4{}^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^4}$; liquid water is a product

Interpretation Introduction

(a)

Interpretation:

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

$K=\frac{{\left(P_{\text{CO}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}}\right)}^5}{\left(P_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right){\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^2}$

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 11QAP

${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+5{\text{H}}_{\text{2}}^{}\left(g\right)$

Explanation of Solution

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

$K=\frac{{\left(P_{\text{CO}}\right)}^2{\left(P_{{\text{H}}_{\text{2}}}\right)}^5}{\left(P_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right){\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^2}$

Here, CO (g) and ${\text{H}}_{\text{2}}^{}$ (g) will be the products and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ (g) and ${\text{H}}_{\text{2}}\text{O}$ (g) will be the reactants.

The equation for the equilibrium system will be:

${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+5{\text{H}}_{\text{2}}^{}\left(g\right)$

Interpretation Introduction

(b)

Interpretation:

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

$K=\frac{{\left(P_{{\text{NH}}_3}\right)}^4{\left(P_{{\text{O}}_{\text{2}}}\right)}^5}{{\left(P_{\text{NO}}\right)}^4{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^6}$

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 11QAP

$\text{4NO}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 4{\text{NH}}_{\text{3}}\left(g\right)+5{\text{O}}_2\left(g\right)$

Explanation of Solution

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

$K=\frac{{\left(P_{{\text{NH}}_3}\right)}^4{\left(P_{{\text{O}}_{\text{2}}}\right)}^5}{{\left(P_{\text{NO}}\right)}^4{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)}^6}$

Here, ${\text{NH}}_{\text{3}}$ (g) and ${\text{O}}_2$ (g) will be the products and $\text{NO}$ (g) and ${\text{H}}_{\text{2}}\text{O}$ (g) will be the reactants.

The equation for the equilibrium system will be:

$\text{4NO}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 4{\text{NH}}_{\text{3}}\left(g\right)+5{\text{O}}_2\left(g\right)$

Interpretation Introduction

(c)

Interpretation:

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

$K=\frac{{\left[{\text{ClO}}_3^{-}\right]}^2{\left[{\text{Mn}}^{2+}\right]}^2}{\left(P_{{\text{Cl}}_{\text{2}}}\right){\left[{\text{MnO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^4}$

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 11QAP

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{MnO}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\rightleftharpoons 2{\text{ClO}}_3^{-}\left(aq\right)+2{\text{Mn}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

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

$K=\frac{{\left[{\text{ClO}}_3^{-}\right]}^2{\left[{\text{Mn}}^{2+}\right]}^2}{\left(P_{{\text{Cl}}_{\text{2}}}\right){\left[{\text{MnO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^4}$

It is also given that ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is a product.

Here, ${\text{ClO}}_3^{-}\left(aq\right)$ and ${\text{Mn}}^{2+}\left(aq\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ will be the products and ${\text{Cl}}_{\text{2}}$ (g), ${\text{MnO}}_4^{-}\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ will be the reactants.

The equation for the equilibrium system will be:

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{MnO}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\rightleftharpoons 2{\text{ClO}}_3^{-}\left(aq\right)+2{\text{Mn}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$