Chapter 5, Problem 5G.8E

(a)

Interpretation Introduction

Interpretation:

The given reaction has to be balanced using the smallest whole number coefficient, and the expression for $\text{K}$ has to be written.

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{HCl}{}_{\left(\text{g}\right)}$

Concept Introduction:

The equilibrium constant, $\text{K}$, is characteristic of the composition of reaction mixture at equilibrium at a given temperature irrespective of the initial composition of reaction mixture.

For the reaction,

  $\text{a}{\text{A}}_{\left(\text{g}\right)}\text{+}\text{b}{\text{B}}_{\left(\text{g}\right)}\rightleftharpoons \text{c}{\text{C}}_{\left(\text{g}\right)}\text{+}\text{d}{\text{D}}_{\left(\text{g}\right)}$

The equilibrium constant can be written as

  $\text{K}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}$

Explanation of Solution

The given reaction is

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{HCl}{}_{\left(\text{g}\right)}$

In the reaction, chlorine and hydrogen are unbalanced.

Chlorine can be balanced as

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}\text{2}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}\text{HCl}{}_{\left(\text{g}\right)}$

The reaction is balanced completely.  The complete balanced reaction is given as

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}\text{2}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}\text{HCl}{}_{\left(\text{g}\right)}$

The equilibrium constant of the reaction can be written by taking the partial pressures of products on numerator, raised to a power of its stoichiometric coefficient and partial pressures of reactants on denominator raised to a power of its stoichiometric coefficient.

The equilibrium constant expression is written as

  $\text{K}\text{=}\frac{\left({\text{P}}_{{\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right){\left({\text{P}}_{\text{HCl}}\right)}^{\text{2}}}{\left({\text{P}}_{{\text{CH}}_{\text{4}}}\right){\left({\text{P}}_{{\text{Cl}}_{\text{2}}}\right)}^{\text{2}}}$

(b)

Interpretation Introduction

Interpretation:

The given reaction has to be balanced using the smallest whole number coefficient, and the expression for $\text{K}$ has to be written.

  ${\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons {\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

Concept Introduction:

Refer to sub-part (a).

Explanation of Solution

The given reaction is

  ${\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons {\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

Nitrogen, hydrogen, chlorine and fluorine are unbalanced in the reaction.

Nitrogen can be balanced as

  $\text{2}{\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons {\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

Chlorine can be balanced as

  $\text{2}{\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}\text{2}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons {\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

Fluorine is balances as

  $\text{2}{\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}\text{2}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons \text{6}{\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

Now, the hydrogen’s in reaction are also balanced and the balanced reaction is given as

  $\text{2}{\text{NH}}_{\text{3}\left(\text{g}\right)}\text{+}\text{2}{\text{ClF}}_{\text{3}\left(\text{g}\right)}\rightleftharpoons \text{6}{\text{HF}}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Cl}}_{\text{2}\left(\text{g}\right)}$

The equilibrium constant expression is written as

  $\text{K}\text{=}\frac{{\left({\text{P}}_{\text{HF}}\right)}^{\text{6}}\left({\text{P}}_{{\text{N}}_{\text{2}}}\right)\left({\text{P}}_{{\text{Cl}}_{\text{2}}}\right)}{{\left({\text{P}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}{\left({\text{P}}_{{\text{ClF}}_{\text{3}}}\right)}^{\text{2}}}$

(c)

Interpretation Introduction

Interpretation:

The given reaction has to be balanced using the smallest whole number coefficient, and the expression for $\text{K}$ has to be written.

  ${\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{5}\left(\text{g}\right)}$

Concept Introduction:

Refer to sub-part (a).

Explanation of Solution

The given reaction is

  ${\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{5}\left(\text{g}\right)}$

Oxygen is unbalanced in the reaction.

Oxygen can be balanced as

  ${\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}\text{5}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}\left(\text{g}\right)}$

Now, nitrogen has to be balanced.  The balanced reaction is

  $\text{2}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}\text{5}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}\left(\text{g}\right)}$

The equilibrium constant expression is written as

  $\text{K}\text{=}\frac{{\left({\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}}\right)}^{\text{2}}}{{\left({\text{P}}_{{\text{N}}_{\text{2}}}\right)}^{\text{2}}{\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}^{\text{5}}}$