Chapter 6, Problem 29E

Interpretation Introduction

Interpretation:Moles of ${\text{O}}_2$ in indicated reaction is to be determined.

  ${\text{2H}}_2\text{O}\left(g\right)\rightleftharpoons 2{\text{H}}_{\text{2}}\left(g\right)+{\text{O}}_2\left(g\right)$

Concept introduction: The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$

Where,

• $V$ represents volume.
• $M$ represents molarity.
• $n$ represents number of moles.

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.