Problem 16 - General Chemistry: Principles and Modern Applications

Question

Rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}),\) is caused by the oxidation of iron by oxygen. Write the equilibrium constant expression first in terms of activities, and then in terms of concentration and pressure.

Step-by-Step Solution

Answer

The equilibrium constant expression in terms of activities is \(K_{A} = a(\mathrm{O_{2}})^{3}\) and in terms of concentration and pressure is \(K_{P} = P(\mathrm{O_{2}})^{3}\).

Step 1: Formulate the chemical equation

The reaction represented can be written as: \(4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \leftrightharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\).

Step 2: Write the equilibrium constant expression in terms of activities

The equilibrium constant expression (K) in terms of activities is given by the product of the activities (more precisely, the effective concentrations) of the products divided by the product of the activities of the reactants, each raised to the power of its stoichiometric coefficient. Pure solids and liquids do not appear in the expression. Thus, \(K_{A} = a(\mathrm{O_{2}})^{3}\) since the activity of pure solids is one, and the other reactant, iron, is a pure solid.

Step 3: Write the equilibrium constant expression in terms of concentration and pressure

For a gas, the activity is equal to its partial pressure divided by a standard pressure (usually 1 bar or 1 atm). Thus, in this case, the equilibrium constant expression in terms of concentration and pressure, \(K_{P}\), will also be equal to the partial pressure of oxygen cubed. As such, \(K_{P} = P(\mathrm{O_{2}})^{3}\), which is the same as \(K_{A}\) since solids and pure liquids do not appear in the expression.

Key Concepts

Activity

When discussing equilibrium reactions, the term "activity" can often be encountered. It refers to the "effective concentration" of a substance in a mixture. In simpler terms, activity measures how potential a component is in influencing the progression of a reaction. This is particularly useful in determining equilibrium constant expressions in chemistry.

Unlike concentrations, which may change due to external conditions like pressure and temperature, activity adjusts for these variables. For pure solids and pure liquids, their activities are always considered to be 1. This means they don’t affect the equilibrium constant as their effective concentration remains constant regardless of the amount present.

• For gases, activity is often related to partial pressures.
• For dissolved substances, activity can often be represented with molarity and activity coefficients.

Essentially, the concept of activity allows chemists to accurately predict reaction behavior in real-world conditions that aren't ideal. It provides a corrected value that makes it immensely easier to apply equilibrium constants to non-ideal systems.

Partial Pressure

Partial pressure plays a significant role in gases involved in chemical reactions. It describes the pressure that a single component of a mixture of gases would exert if it occupied the entire volume by itself. In the context of equilibrium problems, the partial pressure is pivotal as it is directly utilized when expressing the equilibrium constant in terms of pressure, known as \( K_P \).

For example, in the rusting process, oxygen is the gas interacting with solid iron. Its activity or contribution to the reaction can be expressed using its partial pressure. The partial pressure of a gas in a mixture is denoted \( P(X) \), where \( X \) represents the gas in question. The formula to calculate it is:

• \( P(X) = \frac{n(X)}{n_{ ext{total}}} \times P_{ ext{total}} \)

The ability to express equilibrium constants using partial pressures is especially beneficial because it takes into account how gases behave under different conditions of temperature and pressure. Calculating \( K_P \) makes use of these principles for predicting how equilibrium positions will shift given a change in external conditions.

Chemical Equation

The chemical equation represents the fundamental starting point for understanding reactions. It shows the reactants and products involved in a chemical process and their stoichiometric relationships. In the context of equilibrium, the chemical equation provides the roadmap for constructing equilibrium constant expressions.

In the rusting of iron, the equation is:
\[ 4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \leftrightharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \]
This balanced equation reveals important information like the stoichiometric coefficients, which are crucial when formulating the equilibrium constant. These coefficients tell you the ratio of substances reacting and forming products and are used as exponents in the equilibrium constant expressions.

• Simplifies the interpretation of which substances are in excess or limiting.
• Assists in determining catalyst approach or yield percentage in real-world reactions.

Understanding this balanced form lets students calculate the progression and extent of reactions, as well as appreciate the chemical behavior under a variety of conditions.