The minerals hematite ${}^{\left({\mathrm{Fe}}_2{O}_3\right)}$and magnetite ${}^{\left({\mathrm{Fe}}_3{O}_4\right)}$exist in equilibrium with atmospheric oxygen:role="math" localid="1654925446427" ${}^{{\mathrm{Fe}}_3{O}_4\left(s\right)+O_2\left(g\right)\rightleftharpoons {\mathrm{Fe}}_6{O}_3\left(s\right)K_P=2.5\times {10}^{87}}$(a) Determine ${}^{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}}$at equilibrium. (b) Given that ${}^{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}}$ in air is 0.21 atm, in which direction will the reaction proceed to reach equilibrium? (c) Calculate${}^{{\mathbf{K}}_{\mathbf{C}}}$ at 298 K.

The equilibrium constant gives the relative concentration of products and reactants at equilibrium of a reversible reaction.In the case of reactions involving gases, the equilibrium constant can be determined as the ratio of partial pressure of gaseous products to that of gaseous reactants.

The equilibrium constant determines the direction favoured by a reaction at the given temperature. A higher value of the equilibrium constant implies a higher rate for forward reaction, while a lower value implies that the backward reaction is favoured.

The equilibrium constant can be expressed in terms of concentration or partial pressure. The partial pressures of solids are considered unity. Therefore, only gases are considered to determine the equilibrium constant in terms of partial pressure.

The relation between the two equilibrium constants is derived using the ideal gas equation and is given as:

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{K}}_{\mathbf{C}}\mathbf{}{\left(\mathrm{RT}\right)}^{\mathbf{∆}\mathbf{n}}$

Where ${}^{{\mathbf{K}}_{\mathbf{C}}}$is the equilibrium constant in terms of concentration, and ${}^{{\mathbf{K}}_{\mathbf{P}}}$is that in terms of partial pressures. R is the universal gas constant with a value of 0.0821 Liter atmosphere per mole per Kelvin. ${}^{\mathbf{∆}\mathbf{n}}$is the difference between the number of moles of gases in products to that in the reactant.

(a)

In the given equilibrium, hematite and magnetite are solids. Therefore, their partial pressures are 1.

The equilibrium constant in terms of partial pressures can then be defined as:

$$\begin{gathered} {\mathbf{K}}_{\mathbf{P}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{P}}_{{\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}}}{{\mathbf{P}}_{{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\times }{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{1}}{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}} \end{gathered}$$

The equilibrium constant value is ${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{87}}}$at 298 K. Therefore, the partial pressure of oxygen can be calculated as:

$$\begin{gathered} {\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{K}}_{\mathbf{P}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{87}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{88}} \end{gathered}$$

The partial pressure of oxygen is${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{88}}}$ .

(b) The partial pressure of oxygen in the air is 0.21 atm, and the partial pressure of oxygen at equilibrium is${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{88}}}$ . Therefore, the partial pressure of oxygen in the air is greater.

Oxygen is on the reactant side, and therefore, a higher concentration of reactant shifts the reaction to the product side. The equilibrium is reached when the partial pressure of oxygen in the air and that at equilibrium becomes equal.

Therefore, the reaction proceeds in the right direction (forward direction) till the partial pressure of oxygen in the air is equal to that at equilibrium.

(c) The reactant side contains 1 mole of oxygen gas, and the product side does not contain any gases.

Therefore,

$$\begin{gathered} \mathbf{\Delta n}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{n}}_{\mathbf{products}}\mathbf{}\mathbf{-}{\mathbf{n}}_{\mathbf{reactants}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{-}\mathbf{1} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Therefore, the equilibrium constant in terms of concentration at 298 K can be calculated as:

$$\begin{gathered} {\mathbf{K}}_{\mathbf{c}}\mathbf{}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{P}}}{{\mathbf{\left(}\mathbf{RT}\mathbf{\right)}}^{\mathbf{\Delta n}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{87}}}{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0821}\mathbf{\times }\mathbf{298}\mathbf{)}}^{\mathbf{-}\mathbf{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{88}} \end{gathered}$$

The equilibrium constant in terms of concentration is${}^{\mathbf{}\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{88}}}$ .