Problem 51 - General Chemistry: Principles and Modern Applications

Question

The following reaction is used in some self-contained breathing devices as a source of \(\mathrm{O}_{2}(\mathrm{g})\) $$\begin{aligned} 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s}) &+3 \mathrm{O}_{2}(\mathrm{g}) \\\ K_{\mathrm{p}} &=28.5 \mathrm{at} 25^{\circ} \mathrm{C} \end{aligned}$$ Suppose that a sample of \(\mathrm{CO}_{2}(\mathrm{g})\) is added to an evacuated flask containing \(\mathrm{KO}_{2}(\mathrm{s})\) and equilibrium is established. If the equilibrium partial pressure of \(\mathrm{CO}_{2}(\mathrm{g})\) is found to be \(0.0721 \mathrm{atm},\) what are the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) and the total gas pressure?

Step-by-Step Solution

Answer

The partial pressure of \(O_2\) is \(0.10815 \, atm\) and the total pressure is \(0.18025 \, atm\).

Step 1: Understanding the stoichiometry

Looking at the balanced chemical equation, for every 2 moles of \(CO_2\) reacted, 3 moles of \(O_2\) is produced. So, the reaction ratio of \(CO_2\) to \(O_2\) is 2 to 3 or, simply, 2:3.

Step 2: Calculate the partial pressure of \(O_2\)

Since the reaction ratio of \(CO_2\) to \(O_2\) is 2:3, for every 0.0721 atm of \(CO_2\) there will be (3/2) * 0.0721 atm of \(O_2\). So solve this to find the partial pressure of \(O_2\).

Step 3: Calculate the total pressure

The total pressure will be the sum of the partial pressures of \(CO_2\) and \(O_2\), so add the values obtained in steps 1 and 2 to find this.

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In our specific exercise, we are dealing with a reaction where potassium superoxide reacts with carbon dioxide to produce potassium carbonate and oxygen. It is essential to understand the mole ratio of the reactants to the products to predict the outcomes of the reaction, which in this case is 4 moles of KO₂ reacting with 2 moles of CO₂ to produce 3 moles of O₂.

When the exercise gives us the partial pressure of CO₂ at equilibrium, using stoichiometry, we can determine the partial pressure of O₂. Since the mole ratio of CO₂ to O₂ is 2 to 3, every two parts of CO₂ correspond to three parts of O₂. This ratio allows us to perform a simple cross-multiplication to find the unknown partial pressure of O₂. Understanding stoichiometry is essential, as it is the foundation for the further calculation of partial pressures in gaseous reactions. This concept is crucial for predicting how much reactant is needed or how much product can be produced in a chemical process.

Partial Pressure

The concept of partial pressure applies to the pressure exerted by a single gas in a mixture of gases. The partial pressure is proportional to its mole fraction in the gas mixture. Dalton's law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas component.

In the exercise, understanding partial pressures assists us in calculating the equilibrium partial pressure of O₂ produced in the reaction. Given the partial pressure of CO₂ and the stoichiometry of the balanced equation, we use the mole ratio to find the partial pressure of O₂. This concept is imperative in gaseous reactions where the properties such as volume, temperature, and pressure, impact the reaction's direction and extent. To establish equilibrium partial pressures, we must account for every gas's contribution and recognize that the reaction conditions affect these pressures significantly.

Equilibrium Constant

The equilibrium constant, represented as Kp when using partial pressures, is a ratio that helps determine the extent of a chemical reaction at equilibrium at a given temperature. The value of the equilibrium constant is determined by the properties inherent to the specific reaction, including the temperature, and does not change unless these conditions are altered.

The reaction in the exercise has an established equilibrium constant, Kp, which indicates the ratio of the products' partial pressures raised to their stoichiometric coefficients to the reactants' partial pressures raised to their stoichiometric coefficients. Even though the Kp value is provided in the exercise, in this case, we are not directly using it to solve the problem. However, it's important to understand that if we had the partial pressures of all reactants and products, we could verify if the system is indeed at equilibrium. For other problems, you might have to manipulate the equilibrium constant to find unknown partial pressures and understand the dominant direction of the reaction, towards the products or reactants.