Chapter 6, Problem 21E

(a)

Interpretation Introduction

Interpretation:The expression for $K$ and $K_{\text{P}}$ for indicated reaction should bewritten.

  ${\text{2NH}}_3\left(g\right)+{\text{CO}}_2\left(g\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{CH}}_{\text{4}}\text{O}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept introduction: Forthe equilibrium as indicated below.

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

Law of mass action states that for such equilibrium that involves solutions and gases expression of equilibrium constant in principle is written in accordance with law of mass action as follows:

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

Where,

• $\left[\text{C}\right]$ denotes equilibrium concentration of C.
• $\left[\text{D}\right]$ denotes equilibrium concentration of D.
• $\left[\text{A}\right]$ denotes equilibrium concentration of A.
• $\left[\text{B}\right]$ denotes equilibrium concentration of B.
• $K$ denotes equilibrium constant.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

The expression for $K_{\text{p}}$ is written as follows:

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

Where,

• $\left[P_{\text{C}}\right]$ denotes partial pressure of C.
• $\left[\text{D}\right]$ denotes partial pressure of D.
• $\left[\text{A}\right]$ denotes partial pressure of A.
• $\left[\text{B}\right]$ denotes partial pressure of B.
• $K_{\text{p}}$ denotes equilibrium constant for gaseous equilibrium.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

Explanation of Solution

The indicated reaction is given as follows:

  ${\text{2NH}}_3\left(g\right)+{\text{CO}}_2\left(g\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{CH}}_{\text{4}}\text{O}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The corresponding equilibrium constant expression with solid $\left[{\text{N}}_{\text{2}}{\text{CH}}_{\text{4}}\text{O}\right]$ assumed as constant is written as follows:

  $K=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]}{{\left[{\text{NH}}_3\right]}^2\left[{\text{CO}}_2\right]}$

In terms of partial pressure corresponding $K_{\text{P}}$ expression is written as follows:

  $K_{\text{P}}=\frac{\left[{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\right]}{{\left[{\text{P}}_{{\text{NH}}_3}\right]}^2\left[{\text{P}}_{{\text{CO}}_2}\right]}$

(b)

Interpretation Introduction

Interpretation: The expression for $K$ and $K_{\text{P}}$ for indicated reaction should bewritten.

  ${\text{2NBr}}_{\text{3}}\left(s\right)\rightleftharpoons {\text{N}}_{\text{2}}\left(s\right)+3{\text{Br}}_{\text{2}}\left(g\right)$

Concept introduction: For the equilibrium as indicated below.

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

Law of mass action states that for such equilibrium that involves solutions and gases expression of equilibrium constant in principle is written in accordance with law of mass action as follows:

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

Where,

• $\left[\text{C}\right]$ denotes equilibrium concentration of C.
• $\left[\text{D}\right]$ denotes equilibrium concentration of D.
• $\left[\text{A}\right]$ denotes equilibrium concentration of A.
• $\left[\text{B}\right]$ denotes equilibrium concentration of B.
• $K$ denotes equilibrium constant.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

The expression for $K_{\text{p}}$ is written as follows:

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

Where,

• $\left[P_{\text{C}}\right]$ denotes partial pressure of C.
• $\left[\text{D}\right]$ denotes partial pressure of D.
• $\left[\text{A}\right]$ denotes partial pressure of A.
• $\left[\text{B}\right]$ denotes partial pressure of B.
• $K_{\text{p}}$ denotes equilibrium constant for gaseous equilibrium.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

Explanation of Solution

Decomposition of ${\text{NBr}}_{\text{3}}$ occurs as follows:

  ${\text{2NBr}}_{\text{3}}\left(s\right)\rightleftharpoons {\text{N}}_{\text{2}}\left(s\right)+3{\text{Br}}_{\text{2}}\left(g\right)$

Equilibrium constant expression with solid $\left[{\text{NBr}}_{\text{3}}\right]$ assumed as constant is written as follows:

  $K=\left[{\text{N}}_{\text{2}}\right]{\left[{\text{Br}}_{\text{2}}\right]}^3$

The corresponding $K_{\text{P}}$ expression with solid $\left[{\text{NBr}}_{\text{3}}\right]$ assumed as constant is written as follows:

  $K_{\text{P}}={\left[{\text{P}}_{{\text{Br}}_{\text{2}}}\right]}^3\left[{\text{P}}_{{\text{N}}_{\text{2}}}\right]$

(c)

Interpretation Introduction

Interpretation:The expression for $K$ and $K_{\text{P}}$ for indicated reaction should bewritten.

  ${\text{2KClO}}_{\text{3}}\left(s\right)\rightleftharpoons 2\text{KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

Concept introduction: For the equilibrium as indicated below.

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

Law of mass action states that for such equilibrium that involves solutions and gases expression of equilibrium constant in principle is written in accordance with law of mass action as follows:

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

Where,

• $\left[\text{C}\right]$ denotes equilibrium concentration of C.
• $\left[\text{D}\right]$ denotes equilibrium concentration of D.
• $\left[\text{A}\right]$ denotes equilibrium concentration of A.
• $\left[\text{B}\right]$ denotes equilibrium concentration of B.
• $K$ denotes equilibrium constant.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

The expression for $K_{\text{p}}$ is written as follows:

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

Where,

• $\left[P_{\text{C}}\right]$ denotes partial pressure of C.
• $\left[\text{D}\right]$ denotes partial pressure of D.
• $\left[\text{A}\right]$ denotes partial pressure of A.
• $\left[\text{B}\right]$ denotes partial pressure of B.
• $K_{\text{p}}$ denotes equilibrium constant for gaseous equilibrium.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

Explanation of Solution

Decomposition of ${\text{KClO}}_{\text{3}}$ occurs as follows:

  ${\text{2KClO}}_{\text{3}}\left(s\right)\rightleftharpoons 2\text{KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

Equilibrium constant expression with solid $\left[{\text{KClO}}_{\text{3}}\right]$ and $\left[\text{KCl}\right]$ assumed as constant is written as follows:

  $K={\left[{\text{O}}_2\right]}^3$

The corresponding $K_{\text{P}}$ expression in terms of partial pressure is written as follows:

  $K_{\text{P}}={\left[{\text{P}}_{{\text{O}}_2}\right]}^3$

(d)

Interpretation Introduction

Interpretation:The expression for $K$ and $K_{\text{P}}$ for indicated reaction should be written.

  $\text{CuO}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)\rightleftharpoons \text{Cu}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept introduction: For the equilibrium as indicated below.

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

Law of mass action states that for such equilibrium that involves solutions and gases expression of equilibrium constant in principle is written in accordance with law of mass action as follows:

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

Where,

• $\left[\text{C}\right]$ denotes equilibrium concentration of C.
• $\left[\text{D}\right]$ denotes equilibrium concentration of D.
• $\left[\text{A}\right]$ denotes equilibrium concentration of A.
• $\left[\text{B}\right]$ denotes equilibrium concentration of B.
• $K$ denotes equilibrium constant.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

The expression for $K_{\text{p}}$ is written as follows:

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

Where,

• $\left[P_{\text{C}}\right]$ denotes partial pressure of C.
• $\left[\text{D}\right]$ denotes partial pressure of D.
• $\left[\text{A}\right]$ denotes partial pressure of A.
• $\left[\text{B}\right]$ denotes partial pressure of B.
• $K_{\text{p}}$ denotes equilibrium constant for gaseous equilibrium.
• $a$ , $b$ , $c$ and $d$ are stoichiometric coefficients.

Explanation of Solution

The indicated reaction is given as follows:

  $\text{CuO}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)\rightleftharpoons \text{Cu}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The corresponding equilibrium constant expression with solid $\left[\text{CuO}\right]$ and $\left[\text{Cu}\right]$ assumed as constant is written as follows:

  $K=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]}{\left[{\text{H}}_{\text{2}}\right]}$

In terms of partial pressure corresponding $K_{\text{P}}$ expression is written as follows:

  $K_{\text{P}}=\frac{\left[{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\right]}{\left[{\text{P}}_{{\text{H}}_2}\right]}$