Chapter 5, Problem 5G.7E

(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.

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}\text{O}{}_{\left(\text{g}\right)}$

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

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}\text{O}{}_{\left(\text{g}\right)}$

In the reaction, oxygen and hydrogen are unbalanced.

Hydrogen can be balanced as

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}{}_{\left(\text{g}\right)}$

Oxygen can be balanced as

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}\text{2}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}{}_{\left(\text{g}\right)}$

The complete balanced reaction is given as

  ${\text{CH}}_{\text{4}\left(\text{g}\right)}\text{+}\text{2}{\text{O}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{CO}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O}{}_{\left(\text{g}\right)}$

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

  $\text{K}\text{=}\frac{\left({\text{P}}_{{\text{CO}}_{\text{2}}}\right){\left({\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\right)}^{\text{2}}}{\left({\text{P}}_{{\text{CH}}_{\text{4}}}\right){\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}^{\text{2}}}$

(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.

  ${\text{I}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{IF}}_{\text{5}\left(\text{g}\right)}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction is

  ${\text{I}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{IF}}_{\text{5}\left(\text{g}\right)}$

Iodine and fluorine are unbalanced in the reaction.

Iodine can be balanced as

  ${\text{I}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{IF}}_{\text{5}\left(\text{g}\right)}$

Fluorine can be balanced as

  ${\text{I}}_{\text{2}\left(\text{g}\right)}\text{+}\text{5}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{IF}}_{\text{5}\left(\text{g}\right)}$

Now, the complete reaction is balanced and the balanced reaction is given as

  ${\text{I}}_{\text{2}\left(\text{g}\right)}\text{+}\text{5}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{IF}}_{\text{5}\left(\text{g}\right)}$

The equilibrium constant expression is written as

  $\text{K}\text{=}\frac{{\left({\text{P}}_{{\text{IF}}_{\text{5}}}\right)}^{\text{2}}}{\left({\text{P}}_{{\text{I}}_{\text{2}}}\right){\left({\text{P}}_{{\text{F}}_{\text{2}}}\right)}^{\text{5}}}$

(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.

  ${\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{FNO}}_{\text{2}\left(\text{g}\right)}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction is

  ${\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{FNO}}_{\text{2}\left(\text{g}\right)}$

Fluorine is unbalanced in the reaction.

Fluorine can be balanced as

  ${\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{FNO}}_{\text{2}\left(\text{g}\right)}$

Now, nitrogen and oxygen has to be balanced.

  $\text{2}{\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{F}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons \text{2}{\text{FNO}}_{\text{2}\left(\text{g}\right)}$

The equilibrium constant expression is written as

  $\text{K}\text{=}\frac{{\left({\text{P}}_{{\text{FNO}}_{\text{2}}}\right)}^{\text{2}}}{{\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)}^{\text{2}}\left({\text{P}}_{{\text{F}}_{\text{2}}}\right)}$