Problem 50 - Chemistry The Central Science In Si Units

Question

For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\) at \(700 \mathrm{~K}\). In a 2.00-L vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

Step-by-Step Solution

Answer

Approximately 0.1634 grams of \( \mathrm{SO}_2 \) are present in the vessel.

Step 1: Determine Moles of Given Substances

First, calculate the moles of each given substance using their molecular weights. \( \text{Molar mass of } \mathrm{SO}_3 = 80.07 \text{ g/mol} \), thus moles of \( \mathrm{SO}_3 = \frac{1.17 \text{ g}}{80.07 \text{ g/mol}} = 0.0146 \text{ mol} \). For \( \mathrm{O}_2\), with molar mass \( 32.00 \text{ g/mol} \), moles of \( \mathrm{O}_2 = \frac{0.105 \text{ g}}{32.00 \text{ g/mol}} = 0.00328 \text{ mol} \).

Step 2: Use Ideal Gas Law for Moles of SO3 and O2

Calculate the partial pressures using the ideal gas law: \( PV = nRT \), rearranged to \( P = \frac{nRT}{V} \). Using \( R = 0.0821 \text{ L atm K}^{-1}\text{mol}^{-1} \) and \( T = 700 \text{ K} \).. The volumes cancel since we know \( V = 2.00 \text{ L} \) for each, leading to \( P_{\mathrm{SO}_3} = \frac{0.0146 \times 0.0821 \times 700}{2.00} \) and \( P_{\mathrm{O}_2} = \frac{0.00328 \times 0.0821 \times 700}{2.00} \).

Step 3: Calculate Partial Pressures

Find the partial pressures: \( P_{\mathrm{SO}_3} = 0.420 \text{ atm} \) and \( P_{\mathrm{O}_2} = 0.0938 \text{ atm} \).

Step 4: Apply Equilibrium Expression Formula

Use the equilibrium expression for the reaction: \( K_p = \frac{(P_{\mathrm{SO}_3})^2}{(P_{\mathrm{SO}_2})^2 (P_{\mathrm{O}_2})} = 3.0 \times 10^2 \).

Step 5: Solve for SO2 Pressure

Rearrange to solve for \( P_{\mathrm{SO}_2} \): \( P_{\mathrm{SO}_2} = \sqrt{\frac{(P_{\mathrm{SO}_3})^2}{K_p \times P_{\mathrm{O}_2}}} \). Substitute in known values to get \( P_{\mathrm{SO}_2} = \sqrt{\frac{(0.420)^2}{3.0 \times 10^2 \times 0.0938}} = 0.073 \text{ atm} \).

Step 6: Calculate Moles of SO2

Back-calculate the moles of \( \mathrm{SO}_2 \) using \( P_{\mathrm{SO}_2} = \frac{nRT}{V} \), thus: \( n_{\mathrm{SO}_2} = \frac{P_{\mathrm{SO}_2} \times V}{R \times T} = \frac{0.073 \times 2.00}{0.0821 \times 700} = 0.00255 \text{ mol} \).

Step 7: Convert Moles of SO2 to Grams

Finally, convert moles of \( \mathrm{SO}_2 \) to grams using its molar mass \( 64.07 \text{ g/mol} \): \( \text{grams of } \mathrm{SO}_2 = 0.00255 \times 64.07 = 0.1634 \text{ g} \).

Key Concepts

Equilibrium Constant (Kp)

In chemical equilibrium, the equilibrium constant, denoted as \( K_p \) for gases, is a crucial concept that indicates the ratio of the product's partial pressures to the reactant's partial pressures at equilibrium. This ratio is specific to a given temperature. In our reaction scenario, the equilibrium equation is represented as \( K_p = \frac{(P_{\mathrm{SO}_3})^2}{(P_{\mathrm{SO}_2})^2 (P_{\mathrm{O}_2})} \). By understanding \( K_p \), we can determine whether the reaction mixture is at equilibrium, or whether the forward or reverse reaction is favored. If \( K_p \) is much greater than 1, as in our example, it suggests that the products are favored, meaning more \( \mathrm{SO}_3 \) is present at equilibrium compared to \( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \). This information is pivotal when predicting the outcome and direction of gas-phase reactions.

Molar Mass Calculations

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. It is a critical conversion factor in chemistry, especially when dealing with gases at equilibrium. To start the calculations in our problem, we first needed the molar masses of the compounds involved: \( \mathrm{SO}_3 \) had a molar mass of 80.07 g/mol, \( \mathrm{O}_2 \) was 32.00 g/mol, and \( \mathrm{SO}_2 \) came in at 64.07 g/mol. Thus, determining the moles of each substance using the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \) is essential to use stoichiometry and the ideal gas law correctly. It also helps convert mass to moles and vice versa, which is vital in chemical reactions, particularly in achieving and analyzing equilibrium.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas in terms of pressure, volume, temperature, and moles. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature. This equation is critical for calculating the unknown variable when knowing the other three. In our solution, after determining the moles of \( \mathrm{SO}_3 \) and \( \mathrm{O}_2 \), we used the Ideal Gas Law to convert these moles into partial pressures. By rearranging to \( P = \frac{nRT}{V} \), this allowed us to determine the pressure exerted by each gas in the vessel, using the known values of \( R = 0.0821 \text{ L atm K}^{-1}\text{mol}^{-1} \) and temperature \( T = 700 \text{ K} \). This understanding is essential for utilizing chemical equilibrium data in gas reactions.

Partial Pressure

Partial pressure is the pressure that one component of a mixture of gases would exert if it were alone in the container. Understanding partial pressures is vital for interpreting and predicting the behavior of gas mixtures, particularly at equilibrium. In our reaction involving \( \mathrm{SO}_3 \), \( \mathrm{SO}_2 \), and \( \mathrm{O}_2 \), the partial pressures are required to apply the equilibrium law \( K_p \). By using the rearranged Ideal Gas Law formula \( P = \frac{nRT}{V} \), we calculate the partial pressure of \( \mathrm{SO}_3 \) and \( \mathrm{O}_2 \). From these values, the remaining pressure attribute for \( \mathrm{SO}_2 \) can be deduced using \( K_p \) and the equilibrium expression. This analysis shows the role of partial pressures in determining the extent of the reaction at a given state.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on balanced chemical equations and allows the quantification of substances involved. In our given reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), stoichiometry helps to maintain the proportion of moles that react and form products. After computing the equilibrium pressures and applying the equilibrium constant \( K_p \), stoichiometry aids in determining how much \( \mathrm{SO}_2 \) is expected to react to reach equilibrium. It is calculated using the relation of moles within the balanced equation as well as using the determined partial pressures, allowing us to find the specific mass or moles of a participating substance derived from the starting or equilibrium quantities. This ensures the chemical equation follows the law of conservation of mass.