Problem 44 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

In a closed container maintained at 1 atm pressure and \(25^{\circ} \mathrm{C}, 2\) moles of \(\mathrm{SO}_{2}(\mathrm{~g})\) and 1 mole of \(\mathrm{O}_{2}(\mathrm{~g})\) were allowed to react to form \(\mathrm{SO}_{3}(\mathrm{~g})\) under the influence of a catalyst. \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) At equilibrium, it was found that \(50 \%\) of \(\mathrm{SO}_{2}(\mathrm{~g})\) was converted to \(\mathrm{SO}_{3}(\mathrm{~g})\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) at equilibrium will be (a) \(0.17\) atm (b) \(0.5 \mathrm{~atm}\) (c) \(0.33\) atm (d) \(0.20 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The partial pressure of O2 at equilibrium is 0.2 atm.

Step 1: Understand the Chemical Equation

The chemical equation for the reaction is given as: \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\). We are told that at equilibrium, 50% of \(\text{SO}_2\) is converted into \(\text{SO}_3\). This means we started with 2 moles of \(\text{SO}_2\) and ended up with 1 mole of \(\text{SO}_2\) at equilibrium because half of it reacted.

Step 2: Calculate Moles of SO2 and SO3 at Equilibrium

At the beginning there were 2 moles of \(\text{SO}_2\). Since 50% has reacted, 1 mole of \(\text{SO}_2\) is left. This corresponds to the creation of 1 mole of \(\text{SO}_3\), as the stoichiometry is 1:1.

Step 3: Determine Moles of O2 at Equilibrium

Half a mole of \(\text{O}_2\) would have been required to react with 1 mole of \(\text{SO}_2\) according to the stoichiometry of the reaction. This means at equilibrium, there would be \(1 - 0.5 = 0.5\) moles of \(\text{O}_2\) left.

Step 4: Calculate the Partial Pressure of O2

If the total pressure in the container is 1 atm, and there are a total of 2.5 moles of gas present at equilibrium (1 mole of \(\text{SO}_2\), 1 mole of \(\text{SO}_3\), and 0.5 moles of \(\text{O}_2\)), the partial pressure of \(\text{O}_2\) can be calculated using Dalton's Law of Partial Pressures (\(P_{\text{O}_2} = P_{total} \times (\frac{n_{\text{O}_2}}{n_{total}})\)).

Step 5: Perform the Calculation

Using Dalton's Law, the partial pressure of \(\text{O}_2\) is calculated as follows:\[P_{\text{O}_2} = 1 \,\text{atm} \times (\frac{0.5 \,\text{moles}}{2.5 \,\text{moles}}) = 0.2 \,\text{atm}\].

Key Concepts

Partial Pressure Calculations

In the context of chemistry and gas laws, partial pressure calculations are essential for understanding the behavior of gases in mixtures. This concept plays a critical role when dealing with chemical reactions involving gases, as seen in the provided exercise. Partial pressure refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume of the mixture alone at the same temperature.

To calculate the partial pressure of a specific gas, we use the formula:\begin{align*}P_{i} = P_{total} \times \left(\frac{n_{i}}{n_{total}}\right),\end{align*}where \(P_{i}\) represents the partial pressure of gas \(i\), \(P_{total}\) is the total pressure of the gas mixture, \(n_{i}\) is the number of moles of gas \(i\), and \(n_{total}\) is the total number of moles of all gases present. Understanding this calculation is pivotal for predicting the outcome of reactions and determining gas concentrations under different conditions.

Reaction Stoichiometry

Understanding reaction stoichiometry is a fundamental aspect of chemistry that enables us to calculate the proportions of reactants and products in chemical reactions. In simple terms, stoichiometry is the 'recipe' for a chemical reaction, telling you how much of each substance is needed to react completely.

The exercise demonstrates reaction stoichiometry through the conversion of \(\mathrm{SO}_{2}(\mathrm{~g})\) to \(\mathrm{SO}_{3}(\mathrm{~g})\) with the presence of \(\mathrm{O}_{2}(\mathrm{~g})\). Students are often challenged by stoichiometry because it requires careful balance and proportion calculations. For every two moles of \(\mathrm{SO}_{2}\), one mole of \(\mathrm{O}_{2}\) is needed to produce two moles of \(\mathrm{SO}_{3}\) according to the balanced chemical equation. When the problem states that 50% of \(\mathrm{SO}_{2}\) is converted, it directly impacts the amount of \(\mathrm{O}_{2}\) used. Recognizing and applying these stoichiometric relationships is vital for solving chemical equilibrium problems.

Dalton's Law of Partial Pressures

When dealing with gas mixtures, Dalton's Law of Partial Pressures is a useful principle. It states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Dalton's Law is symbolically represented as:\begin{align*}P_{total} = P_{1} + P_{2} + P_{3} + ... + P_{n},\end{align*}where \(P_{total}\) is the total pressure, and \(P_{1}, P_{2}, P_{3}, ..., P_{n}\) are the partial pressures of the gases.

In the exercise, we apply this law to find the partial pressure of oxygen at equilibrium. It shows how the total pressure in the container and the mole fraction of oxygen contribute to its partial pressure. Understanding how Dalton's Law is invoked in these situations allows students to draw connections between theoretical gas laws and practical chemical problems. It's crucial to understand that this law is applicable only when the gases involved do not react with each other, although it can be used in the context of chemical reactions at equilibrium to find the partial pressures of reactants and products if they are gaseous.