Question

The minerals hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) and magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)\) exist in equilibrium with atmospheric oxygen: \(4 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad K_{\mathrm{p}}=2.5 \times 10^{87}\) at \(298 \mathrm{~K}\) (a) Determine \(P_{\mathrm{O}_{2}}\) at equilibrium. (b) Given that \(P_{\mathrm{O}_{2}}\) in air is 0.21 atm, in which direction will the reaction proceed to reach equilibrium? (c) Calculate \(K_{c}\) at \(298 \mathrm{~K}\).

Short answer

(a) 4 \(\times 10^{-88}\) atm \(P_{\mathrm{O}_2}\); (b) Reaction proceeds left; (c) \(K_c = 6.1165 \times 10^{88}\)

Step-by-step solution

- Understand the given equation and constants

The equilibrium reaction is given as: \[4 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) + \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\] Here, \(K_p\) at 298 K is \(2.5 \times 10^{87}\). We need to find the equilibrium partial pressure \(P_{\mathrm{O}_2}\).

- Identify the expression for the equilibrium constant \(K_p\)

The equilibrium constant for the reaction concerning \(P_{\mathrm{O}_2}\) is: \[K_p = \frac{[\mathrm{Fe}_{2} \mathrm{O}_{3}]^6}{[\mathrm{Fe}_{3} \mathrm{O}_{4}]^4 \cdot [\mathrm{O}_{2}]} = 2.5 \times 10^{87}\]Since \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) are solids, their concentrations are constant and can be omitted from the expression.

- Simplify the expression for \(K_p\)

The equilibrium expression then simplifies to: \[K_p = \frac{1}{P_{\mathrm{O}_2}} = 2.5 \times 10^{87}\]Therefore, solving for \(P_{\mathrm{O}_2}\), we get: \[P_{\mathrm{O}_2} = \frac{1}{2.5 \times 10^{87}}\]

- Calculate \(P_{\mathrm{O}_2}\) at equilibrium

Now calculating: \[P_{\mathrm{O}_2} = 4 \times 10^{-88} \text{ atm}\]

- Determine the direction of reaction with given air's \(P_{\mathrm{O}_2}\)

Given \(P_{\mathrm{O}_2}\) in air is 0.21 atm, the reaction will shift to the left (towards forming \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and \(\mathrm{O}_{2}\)) to reach equilibrium because 0.21 atm is much greater than 4 \(\times 10^{-88}\) atm.

- Relationship between \(K_p\) and \(K_c\)

The relationship between \(K_p\) and \(K_c\) is given by the equation: \[K_p = K_c (RT)^{\Delta n}\]where \(R = 0.0821 \text{ L·atm·K}^{-1} \text{·mol}^{-1}\) and \(\Delta n\) is the change in moles of gas.

- Calculate \(\Delta n\)

\[\Delta n = \text{(moles of gas products) - (moles of gas reactants)} = 0 - 1 = -1\]

- Solve for \(K_c\)

Using the relationship: \[K_c = K_p \cdot (RT)^{-\Delta n} = 2.5 \times 10^{87} \left(0.0821 \times 298\right)^{-(-1)}\]Calculating further: \[K_c = 2.5 \times 10^{87} \times 24.466 = 6.1165 \times 10^{88}\]

Key concepts

Partial Pressure

Partial pressure refers to the individual pressure exerted by a particular gas in a mixture of gases. In this exercise, partial pressure plays a critical role in determining the concentration of oxygen at equilibrium. The partial pressure of a gas can be calculated using the ideal gas law and is an essential component when dealing with gases in equilibrium.

Reaction Direction

Determining the direction of a reaction is crucial when understanding how the system will move towards equilibrium. In this exercise, the reaction shifts based on the comparison between the given partial pressure of oxygen in air and the calculated partial pressure at equilibrium. If the partial pressure of oxygen in air is greater than the equilibrium partial pressure, the reaction will shift towards the formation of the reactants to decrease the concentration of oxygen.

Kp and Kc Relationship

Understanding the relationship between the equilibrium constants Kp and Kc is vital for converting between different forms of the constant based on the nature of the reaction.

The relationship between Kp and Kc is given by the equation:

Kp = Kc (RT)^{Delta n}

Here, 'R' refers to the gas constant (0.0821 L·atm·K^{-1}·mol^{-1}), 'T' is the temperature in Kelvin, and 'Delta n' is the change in the number of moles of gas from reactants to products.

In the exercise provided, determining 'Delta n' helped us convert Kp into Kc.

Solid and Gas Phases

In chemical equilibrium, the states of the reactants and products are important. Solids and gases behave differently in equilibrium conditions. In the given exercise, Fe_2O_3 and Fe_3O_4 are solids, while O_2 is a gas. This distinction allows us to omit solids from the equilibrium constant expression, simplifying the calculations.

Understanding which phases are involved helps to formulate a correct expression for the equilibrium constant and makes the problem-solving process more manageable.