Chapter 5.6, Problem 83P

To determine

The equilibrium conditions.

Explanation of Solution

Given:

The chemical reactions are

  $2\text{H}\leftrightarrow {\text{H}}_{\text{2}}$ ……. (1)

  $2\text{CO}+{\text{O}}_{\text{2}}\leftrightarrow 2{\text{CO}}_{\text{2}}$ …… (2)

  ${\text{CH}}_{\text{4}}+2{\text{O}}_{\text{2}}\leftrightarrow 2{\text{H}}_{\text{2}}\text{O}\text{+}{\text{CO}}_2$ ……. (3)

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\leftrightarrow 2{\text{H}}^{\text{+}}+{\text{SO}}_{\text{4}}^{\text{2-}}$ …… (4)

  $2\text{p}\text{+}\text{2n}\leftrightarrow 4\text{He}$ …… (5)

Introduction:

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction. In other words, there is no net change in concentrations of reactants and products. This kind of equilibrium is also called dynamic equilibrium.

In other word, chemical equilibrium for a chemical equation can be stated when the sum of chemical potentials of the reactants is equal to the sum of the chemical potential of the products.

The equilibrium condition for reaction (1) is:

  $2{\mu }_{\text{H}}={\mu }_{{\text{H}}_{\text{2}}}$

Here, ${\mu }_{\text{H}}$ is the chemical potential of $\text{H}$ and ${\mu }_{{\text{H}}_{\text{2}}}$ is the chemical potential of ${\text{H}}_{\text{2}}$ .

The equilibrium condition for reaction (2) is:

  $2{\mu }_{\text{CO}}+{\mu }_{{\text{O}}_{\text{2}}}=2{\mu }_{{\text{CO}}_{\text{2}}}$

Here, ${\mu }_{\text{CO}}$ is the chemical potential of $\text{CO}$ , ${\mu }_{{\text{O}}_{\text{2}}}$ is the chemical potential of ${\text{O}}_{\text{2}}$ and ${\mu }_{{\text{CO}}_{\text{2}}}$ is the chemical potential of ${\text{CO}}_{\text{2}}$ .

The equilibrium condition for reaction (3) is:

  ${\mu }_{{\text{CH}}_4}+2{\mu }_{{\text{O}}_{\text{2}}}=2{\mu }_{{\text{H}}_{\text{2}}\text{O}}+{\mu }_{{\text{CO}}_{\text{2}}}$

Here, ${\mu }_{{\text{CH}}_4}$ is the chemical potential of ${\text{CH}}_4$ , ${\mu }_{{\text{O}}_{\text{2}}}$ is the chemical potential of ${\text{O}}_{\text{2}}$ , ${\mu }_{{\text{CO}}_{\text{2}}}$ is the chemical potential of ${\text{CO}}_{\text{2}}$ and ${\mu }_{{\text{H}}_{\text{2}}\text{O}}$ is the chemical potential of ${\text{H}}_{\text{2}}\text{O}$ .

The equilibrium condition for reaction (4) is:

  $2{\mu }_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}=2{\mu }_{{\text{H}}^{\text{+}}}+{\mu }_{{\text{SO}}_{\text{4}}^{\text{2-}}}$

Here, ${\mu }_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ is the chemical potential of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ , ${\mu }_{{\text{H}}^{\text{+}}}$ is the chemical potential of ${\text{H}}^{\text{+}}$ and ${\mu }_{{\text{SO}}_{\text{4}}^{\text{2-}}}$ is the chemical potential of ${\text{SO}}_{\text{4}}^{\text{2-}}$ .

The equilibrium condition for reaction (5) is:

  $2{\mu }_{\text{p}}+2{\mu }_{\text{n}}=4{\mu }_{\text{He}}$

Here, ${\mu }_{\text{p}}$ is the chemical potential of $\text{p}$ , ${\mu }_{\text{n}}$ is the chemical potential of $\text{n}$ and ${\mu }_{\text{He}}$ is the chemical potential of $\text{He}$ .

Conclusion:

Thus, the conditions for the chemical equilibrium of the reactions are stated.