Question

You have an equimolar mixture of the gases \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2},\) along with some \(\mathrm{He}\), in a container fitted with a piston. The density of this mixture at STP is 1.924 \(\mathrm{g} / \mathrm{L}\) . Assume ideal behavior and constant temperature and pressure. a. What is the mole fraction of He in the original mixture? b. The \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react to completion to form \(\mathrm{SO}_{3} .\) What is the density of the gas mixture after the reaction is complete?

Short answer

a. The mole fraction of He in the original mixture is approximately 0.0924. b. The density of the gas mixture after the reaction between SO₂ and O₂ is complete and SO₃ is formed is 3.14 g/L.

Step-by-step solution

Use the density formula

Recall that density is mass per unit volume. For an ideal gas mixture, we can use the formula: \[Density = \frac{M * P}{R * T}\] Where \(Density\) is given by the problem (1.924 g/L), \(M\) is the molar mass of the gas mixture, \(P\) is the pressure (STP) of 1 atm, \(R\) is the ideal gas constant (0.0821 L * atm/(mol * K)), and \(T\) is the temperature (STP) of 273 K. Plug in the relevant values: \[1.924 \frac{g}{L} = \frac{M * 1 \ atm}{0.0821 \frac{L \cdot atm}{mol \cdot K} * 273 \ K}\] Now, solve for the molar mass of the original gas mixture. #Step 2: Write the equimolar reaction between SO₂ and O₂#

Write the balanced chemical equation

The reaction between SO₂ and O₂ is given by the balanced chemical equation: \[SO_{2} + \frac{1}{2} O_{2} \rightarrow SO_{3}\] Because the mixture is equimolar, for every mole of SO₂, there is half a mole of O₂. #Step 3: Calculate the mole fraction of He#

Calculate the molar masses and mole fractions

To find the mole fraction of He, we first need the molar masses of all three gases in the mixture: - SO₂: 32 (S) + 2 * 16 (O) = 64 g/mol - O₂: 2 * 16 (O) = 32 g/mol - He: 4 g/mol Define the mole fractions of SO₂, O₂, and He using 1 mol in total: \[x_{SO_{2}} = \frac{64}{M}\] \[x_{O_{2}} = \frac{32}{M}\] \[x_{He} = \frac{4}{M}\] We are looking for the molar mass of the original gas mixture. From step 1, we can calculate \(M\): \[M = \frac{1.924 \times (0.0821) * 273}{1} \approx 43.3 \ g/mol\] Now calculate the mole fraction of He: \[x_{He} = \frac{4}{43.3} \approx 0.0924\] This means there are approximately 0.0924 moles of He in every mole of the mixture. #Step 4: Find the density of the gas mixture after the reaction is complete#

Calculate the molar mass of the new mixture

After the reaction between SO₂ and O₂ is complete, the remaining mixture in the container is made of SO₃ and He. Calculate the molar mass of SO₃: - SO₃: 32 (S) + 3 * 16 (O) = 80 g/mol Now, calculate the molar mass of the new mixture considering the mole fraction of He (0.0924): \[Mnew = (1 - 0.0924) * 80 + 0.0924 * 4 = 75.04 \ g/mol\]

Calculate the density of the new gas mixture

Now we can use this molar mass to find the density of the new mix containing only He and SO₃ at STP, using the density formula: \[Density = \frac{Mnew * P}{R * T} = \frac{75.04 \times 1}{0.0821 * 273} = 3.14 \frac{g}{L}\] The density of the gas mixture after the reaction between SO₂ and O₂ is complete and SO₃ is formed is 3.14 g/L.

Key concepts

Mole Fraction

When discussing the composition of a gas mixture, the term "mole fraction" is commonly used. The mole fraction indicates the proportion of a component in the total mixture. It is a way to express concentrations and is useful in calculations involving gases. To calculate the mole fraction, you divide the number of moles of the substance of interest by the total number of moles in the mixture.

For example, in the problem given, we need to calculate the mole fraction of helium in a gas mixture consisting of \ \(\mathrm{SO}_{2}\), \ \(\mathrm{O}_{2}\), and helium (He). By knowing the molar masses of these gases, we use the molar mass of the entire gas mixture to find the fraction for helium:

• \(M_{SO_{2}} = 64 \, \text{g/mol} \)
• \(M_{O_{2}} = 32 \, \text{g/mol} \)
• \(M_{He} = 4 \, \text{g/mol} \)

Using the calculated total molar mass of the mixture \(M = 43.3 \, \text{g/mol}\), we found that the mole fraction of helium \(x_{He}\) is given by \(x_{He} = \frac{4}{43.3} \approx 0.0924\). This tells us that 9.24% of the gas mixture is helium.

Chemical Reaction

A crucial part of understanding this exercise is recognizing the chemical reactions that occur between the components of the gas mixture. The problem statement involves the reaction between sulfur dioxide (\(\mathrm{SO}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) to form sulfur trioxide (\(\mathrm{SO}_{3}\)):

• The balanced chemical equation is \(\mathrm{SO}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3}\).
• This reaction is exothermic, meaning it releases energy.
• Equimolar conditions imply that every mole of \(\mathrm{SO}_{2}\) reacts with half a mole of \(\mathrm{O}_{2}\) to produce one mole of \(\mathrm{SO}_{3}\).

The importance of this reaction being complete means that no \(\mathrm{SO}_{2}\) or \(\mathrm{O}_{2}\) remains post-reaction. This reaction changes the gas mixture composition, requiring a new density calculation based on the presence of \(\mathrm{SO}_{3}\) and the non-reactive \(\mathrm{He}\).

Density Calculation

Density calculations are essential when dealing with gases, especially under the assumptions of ideal behavior. Density (\(\rho\)) is defined as mass divided by volume. For gases, it can be expressed using the relation:

• \(\rho = \frac{M \cdot P}{R \cdot T}\)

where:

• \(M\) is the molar mass of the gas,
• \(P\) is the pressure,
• \(R\) is the ideal gas constant,
• \(T\) is the temperature.

For our gas mixture after the chemical reaction, we recalculated the molar mass \(M_{\text{new}}\) considering the produced \(\mathrm{SO}_{3}\) (80 g/mol) and helium's mole fraction (0.0924). Using these values, the density of the new gas mixture was determined:

• \(M_{\text{new}} = (1 - 0.0924) \times 80 + 0.0924 \times 4 = 75.04 \, \text{g/mol}\)

Applying the density formula:

• \(\rho = \frac{75.04 \times 1}{0.0821 \times 273} = 3.14 \, \text{g/L}\)

This calculation reveals the increased density of the new gas mixture compared to the initial one.