Chapter 5.10, Problem 5.37P

(a)

Interpretation Introduction

Interpretation:

The forward and reverse reaction in the given equilibrium reaction has to be calculated.

  ${\text{2SO}}_{\text{2}}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2SO}}_{\text{3}}\left(\text{g}\right)$

Concept introduction:

A reversible reaction can occur in either direction from reactants to products or form products to reactants.  The forward reaction proceeds from left to right as drawn.  The reverse reaction proceeds from right to left as drawn.

Example:

  ${\text{CO}}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\rightleftarrows {\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}\left(\text{g}\right)}$

The forward reaction is,

  ${\text{CO}}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}\left(\text{g}\right)}$

The reverse reaction is,

  ${\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }{\text{CO}}_{\left(\text{g}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(b)

Interpretation Introduction

Interpretation:

The forward and reverse reaction in the given equilibrium reaction has to be calculated.

  ${\text{N}}_{\text{2}}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows \text{2NO}\left(\text{g}\right)$

Concept introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The forward and reverse reaction in the given equilibrium reaction has to be calculated.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{\text{+CH}}_{\text{4}}\text{O}\rightleftarrows {\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

Concept introduction:

Refer to part (a).