Chapter 14, Problem 14.42QP

(a)

Interpretation Introduction

Interpretation: The equilibrium constant $K_{\text{c}}$ for the given reaction at a given temperature is $5\times {10}^{12}$. 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 $\left(K_{{\text{c}}_2}\right)$ for the given reaction at same temperature.

Answer to Problem 14.42QP

Solution

The equilibrium constant $\left(K_{{\text{c}}_2}\right)$ for the given reaction at same temperature is $\underset{\_}{2.23\times 10^6}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{NO}}_2\left(g\right)\\ \text{NO}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{NO}}_2\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $5\times {10}^{12}$.

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,

$K_{{\text{c}}_1}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^2}{{\left[\text{NO}\left(g\right)\right]}^2{\left[{\text{O}}_2\left(g\right)\right]}^1}$ (1)

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

$K_{{\text{c}}_2}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^1}{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}$ (2)

Where,

• $\left[\text{NO}\left(g\right)\right]$ is the molar concentration of $\text{NO}\left(g\right)$.
• $\left[{\text{NO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{NO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Divide equation (1) with (2).

$\begin{array}{c}\frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^2}{{\left[\text{NO}\left(g\right)\right]}^2{\left[{\text{O}}_2\left(g\right)\right]}^1}}{\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^1}{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^2}{{\left[\text{NO}\left(g\right)\right]}^2{\left[{\text{O}}_2\left(g\right)\right]}^1}\times \frac{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{NO}}_2\left(g\right)\right]}^1}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^{2-1}}{{\left[\text{NO}\left(g\right)\right]}^{2-1}{\left[{\text{O}}_2\left(g\right)\right]}^{1-\frac{1}{2}}}\\ \frac{K_{{\text{c}}_1}}{K_{{\text{c}}_2}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^1}{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}\end{array}$ (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).

$\begin{array}{c}{K}_{{\text{c}}_2}=\sqrt{5\times {10}^{12}}\\ =2.23\times {10}^6\end{array}$

Thus, the equilibrium constant $\left(K_{{\text{c}}_2}\right)$ for the given reaction at same temperature is $\underset{\_}{2.23\times 10^6}$.

(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 14.42QP

Solution

The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{2.0\times 10^{-13}}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{NO}}_2\left(g\right)\\ \text{NO}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{NO}}_2\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $5\times {10}^{12}$.

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,

$K_{{\text{c}}_{2\text{'}}}=\frac{{\left[\text{NO}\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}{{\left[{\text{NO}}_3\left(g\right)\right]}^2}$ (5)

Where,

• $\left[\text{NO}\left(g\right)\right]$ is the molar concentration of $\text{NO}\left(g\right)$.
• $\left[{\text{NO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{NO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Multiply equation (1) with (5).

$\begin{array}{c}{K}_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^2}{{\left[\text{NO}\left(g\right)\right]}^2{\left[{\text{O}}_2\left(g\right)\right]}^1}\times \frac{{\left[\text{NO}\left(g\right)\right]}^2\left[{\text{O}}_2\left(g\right)\right]}{{\left[{\text{NO}}_3\left(g\right)\right]}^2}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}}}=1\\ K_{{\text{c}}_{2\text{'}}}=\frac{1}{K_{{\text{c}}_1}}\end{array}$ (6)

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

$\begin{array}{l}{K}_{{\text{c}}_{2\text{'}}}=\frac{1}{5\times {10}^{12}}\\ K_{{\text{c}}_{2\text{'}}}=2.0\times {10}^{-13}\end{array}$

Thus, the equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ for the given reaction at same temperature is $\underset{\_}{2.0\times 10^{-13}}$.

(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 14.42QP

Solution

The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ for the given reaction at same temperature is $\underset{\_}{4.4\times 10^{-7}}$.

Explanation of Solution

Explanation

The given reactions are,

$\begin{array}{c}2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{NO}}_2\left(g\right)\\ \text{NO}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{NO}}_2\left(g\right)\end{array}$

The equilibrium constant $K_{{\text{c}}_1}$ for the first above reaction and at constant temperature is $5\times {10}^{12}$.

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,

$K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{NO}}_2\left(g\right)\right]}^1}$ (7)

Where,

• $\left[\text{NO}\left(g\right)\right]$ is the molar concentration of $\text{NO}\left(g\right)$.
• $\left[{\text{NO}}_2\left(g\right)\right]$ is the molar concentration of ${\text{NO}}_2\left(g\right)$.
• $\left[{\text{O}}_2\left(g\right)\right]$ is the molar concentration of ${\text{O}}_2\left(g\right)$.

Multiply equation (1) with (7).

$\begin{array}{c}{K}_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^2}{{\left[\text{NO}\left(g\right)\right]}^2{\left[{\text{O}}_2\left(g\right)\right]}^1}\times \frac{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}{{\left[{\text{NO}}_2\left(g\right)\right]}^1}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^{2-1}}{{\left[\text{NO}\left(g\right)\right]}^{2-1}{\left[{\text{O}}_2\left(g\right)\right]}^{1-\frac{1}{2}}}\\ K_{{\text{c}}_1}\times K_{{\text{c}}_{2\text{'}\text{'}}}=\frac{{\left[{\text{NO}}_2\left(g\right)\right]}^1}{{\left[\text{NO}\left(g\right)\right]}^1{\left[{\text{O}}_2\left(g\right)\right]}^{\frac{1}{2}}}\end{array}$ (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).

$\begin{array}{c}{K}_{{\text{c}}_{2\text{'}\text{'}}}=\sqrt{\frac{1}{5\times {10}^{12}}}\\ =\sqrt{0.2\times {10}^{-12}}\\ =4.4\times {10}^{-7}\end{array}$

Thus, the equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ for the given reaction at same temperature is $\underset{\_}{4.4\times 10^{-7}}$.

Conclusion

1. a. The equilibrium constant $\left(K_{{\text{c}}_2}\right)$ for the given reaction at same temperature is $\underset{\_}{2.23\times 10^6}$.
2. b. The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}}}\right)$ of the given reaction at same temperature is $\underset{\_}{2.0\times 10^{-13}}$.
3. c. The equilibrium constant $\left(K_{{\text{c}}_{2\text{'}\text{'}}}\right)$ for the given reaction at same temperature is $\underset{\_}{4.4\times 10^{-7}}$.