Chapter 14, Problem 14.41QP

(a)

Interpretation Introduction

Interpretation: The equilibrium constant $K_{\text{c}}$ for the given reaction at a given temperature is $2.4\times {10}^{-3}$. On the basis of this information given questions are to be answered.

Concept introduction: The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

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

To determine: The equilibrium constant $K_{\text{c}}$ for the given reaction at same temperature.

Answer to Problem 14.41QP

Solution

The equilibrium constant of the given reaction at same temperature is $\underset{\_}{4.8\times 10^{-2}}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{SO}}_{\text{3}}\left(g\right)\\ {\text{SO}}_2\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{SO}}_{\text{3}}\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $2.4\times {10}^{-3}$.

For a simple reaction,

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

The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

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

Where,

• $\left[\text{C}\right]$ is the molar concentration of $\text{C}$.
• $\left[\text{D}\right]$ is the molar concentration of $\text{D}$.
• $\left[\text{B}\right]$ is the molar concentration of $\text{B}$.
• $\left[\text{A}\right]$ is the molar concentration of $\text{A}$.
• $c$ is the molar coefficient of $\text{C}$.
• $d$ is the molar coefficient of $\text{D}$.
• $b$ is the molar coefficient of $\text{B}$.
• $a$ is the molar coefficient of $\text{A}$.

The expression for the equilibrium constant of first $\left(K_{{\text{c}}_1}\right)$ given reaction is,

$K_{{\text{c}}_1}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^2}{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}$ (1)

The expression for the equilibrium constant of second $\left(K_{{\text{c}}_2}\right)$ given reaction is,

$K_{{\text{c}}_2}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^1}{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}$ (2)

Where,

• $\left[{\text{SO}}_3\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_3\left(g\right)$.
• $\left[{\text{SO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Divide equation (1) with (2).

$\begin{array}{c}\frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^2}{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}}{\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^1}{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^2}{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}\times \frac{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{SO}}_3\left(g\right)\right]}^1}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^{2-1}}{{\left[{\text{SO}}_2\left(g\right)\right]}^{2-1}{\left[{\text{O}}_2\left(g\right)\right]}^{1-\frac{1}{2}}}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^1}{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}\end{array}$ (3)

Compare equation (3) with (2).

$\begin{array}{c}\frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=K_{{\text{c}}_2}\\ {\left(K_{{\text{c}}_2}\right)}^2=K_{{\text{c}}_1}\\ K_{{\text{c}}_2}=\sqrt{K_{{\text{c}}_1}}\end{array}$ (4)

Substitute the value of $K_{{\text{c}}_1}$ in equation (4).

$\begin{array}{c}{K}_{{\text{c}}_2}=\sqrt{2.4\times {10}^{-3}}\\ =4.8\times {10}^{-2}\end{array}$

Thus, the equilibrium constant of the given reaction at same temperature is $\underset{\_}{4.8\times 10^{-2}}$.

(b)

Interpretation Introduction

To determine: The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ for the given reaction at same temperature.

Answer to Problem 14.41QP

Solution

The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{4.1\times 10^2}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{SO}}_{\text{3}}\left(g\right)\\ {\text{2SO}}_3\left(g\right)\rightleftharpoons 2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $2.4\times {10}^{-3}$.

For a simple reaction,

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

The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

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

Where,

• $\left[\text{C}\right]$ is the molar concentration of $\text{C}$.
• $\left[\text{D}\right]$ is the molar concentration of $\text{D}$.
• $\left[\text{B}\right]$ is the molar concentration of $\text{B}$.
• $\left[\text{A}\right]$ is the molar concentration of $\text{A}$.
• $c$ is the molar coefficient of $\text{C}$.
• $d$ is the molar coefficient of $\text{D}$.
• $b$ is the molar coefficient of $\text{B}$.
• $a$ is the molar coefficient of $\text{A}$.

The expression for the equilibrium constant of second $\left(K_{{\text{c}}_{2\text{'}}}\right)$ given reaction is,

$K_{{\text{c}}_{2\text{'}}}=\frac{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}{{\left[{\text{SO}}_3\left(g\right)\right]}^2}$ (5)

Where,

• $\left[{\text{SO}}_3\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_3\left(g\right)$.
• $\left[{\text{SO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Multiply equation (1) with (5).

$\begin{array}{c}{K}_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^2}{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}\times \frac{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}{{\left[{\text{SO}}_3\left(g\right)\right]}^2}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_2\text{'}}=1\\ K_{{\text{c}}_{2\text{'}}}=\frac{1}{K_{{\text{c}}_1}}\end{array}$ (6)

Substitute the value of $K_{{\text{c}}_1}$ in equation (6).

$\begin{array}{l}{K}_{{\text{c}}_{2\text{'}}}=\frac{1}{2.4\times {10}^{-3}}\\ K_{{\text{c}}_2\text{'}}=4.1\times {10}^2\end{array}$

Thus, the equilibrium constant $\left(K_{{\text{c}}_2\text{'}}\right)$ of the given reaction at same temperature is $\underset{\_}{4.1\times 10^2}$.

(c)

Interpretation Introduction

To determine: The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ for the given reaction at same temperature.

Answer to Problem 14.41QP

Solution

The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{2.02}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{SO}}_{\text{3}}\left(g\right)\\ {\text{SO}}_3\left(g\right)\rightleftharpoons {\text{SO}}_2\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $2.4\times {10}^{-3}$.

For a simple reaction,

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

The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

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

Where,

• $\left[\text{C}\right]$ is the molar concentration of $\text{C}$.
• $\left[\text{D}\right]$ is the molar concentration of $\text{D}$.
• $\left[\text{B}\right]$ is the molar concentration of $\text{B}$.
• $\left[\text{A}\right]$ is the molar concentration of $\text{A}$.
• $c$ is the molar coefficient of $\text{C}$.
• $d$ is the molar coefficient of $\text{D}$.
• $b$ is the molar coefficient of $\text{B}$.
• $a$ is the molar coefficient of $\text{A}$.

The expression for the equilibrium constant of second $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ given reaction is,

$K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{SO}}_3\left(g\right)\right]}^1}$ (7)

Where,

• $\left[{\text{SO}}_3\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_3\left(g\right)$.
• $\left[{\text{SO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{SO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Multiply equation (1) with (7).

$\begin{array}{c}{K}_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^2}{{\left[{\text{SO}}_2\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}\times \frac{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{SO}}_3\left(g\right)\right]}^1}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^{2-1}}{{\left[{\text{SO}}_2\left(g\right)\right]}^{2-1}{\left[{\text{O}}_2\left(g\right)\right]}^{1-\frac{1}{2}}}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{SO}}_3\left(g\right)\right]}^1}{{\left[{\text{SO}}_2\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}\end{array}$ (8)

Compare equation (8) with (7).

$\begin{array}{c}{K}_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{1}{K_{{\text{c}}_2}}\\ {\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)}^2=\frac{1}{K_{{\text{c}}_1}}\\ K_{{\text{c}}_{2\text{'}\text{'}}}=\sqrt{\frac{1}{K_{{\text{c}}_1}}}\end{array}$ (9)

Substitute the value of $K_{{\text{c}}_1}$ in equation (9).

$\begin{array}{c}{K}_{{\text{c}}_{2\text{'}\text{'}}}=\sqrt{\frac{1}{2.4\times {10}^{-3}}}\\ =\sqrt{4.1\times {10}^2}\\ =2.02\end{array}$

Thus, the equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{2.02}$.

Conclusion

1. a. The equilibrium constant $\left(K_{{\text{c}}_2}\right)$ of the given reaction at same temperature is $\underset{\_}{4.8\times 10^{-2}}$.
2. b. The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{4.1\times 10^2}$.
3. c. The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{2.02}$.