Question

Sulfur dioxide is used to make sulfuric acid. One method of producing it is by roasting mineral sulfides, for example, $$\operatorname{FeS}_{2}(s)+\mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} \mathrm{SO}_{2}(g)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad[\text {unbalanced }]$$ A production error leads to the sulfide being placed in a \(950-\mathrm{L}\) vessel with insufficient oxygen. Initially, the partial pressure of \(\mathrm{O}_{2}\) is 0.64 atm, and the total pressure is 1.05 atm, with the balance due to \(\mathrm{N}_{2}\). The reaction is run until \(85 \%\) of the \(\mathrm{O}_{2}\) is consumed, and the vessel is then cooled to its initial temperature. What is the total pressure in the vessel and the partial pressure of each gas in it?

Short answer

Total pressure: 0.902 atm. Partial pressures: \( \mathrm{O}_{2} \) - 0.096 atm, \( \mathrm{N}_{2} \) - 0.41 atm, \( \mathrm{SO}_{2} \) - 0.396 atm.

Step-by-step solution

Write the balanced chemical equation

Starting with the unbalanced equation \ \[ \operatorname{FeS}_{2}(s)+ \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} \mathrm{SO}_{2}(g)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \] \ Balance the equation: \[ 4 \operatorname{FeS}_{2}(s) + 11 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{SO}_{2}(g) + 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \]

Calculate the initial pressures of gases

Given the partial pressure of \( \mathrm{O}_{2} \) is 0.64 atm and the total pressure is 1.05 atm. The balance includes nitrogen, so the partial pressure of \( \mathrm{N}_{2} \) is: \[ 1.05 \, \text{atm} - 0.64 \, \text{atm} = 0.41 \, \text{atm} \]

Determine the amount of \( \mathrm{O}_{2} \) consumed

85% of the \( \mathrm{O}_{2} \) is consumed: \ \[ 0.85 \times 0.64 \,\text{atm} = 0.544 \, \text{atm consumed} \] Remaining \( \mathrm{O}_{2} \) is: \ \[ 0.64 \, \text{atm} - 0.544 \, \text{atm} = 0.096 \, \text{atm remaining} \]

Determine the partial pressure of \( \mathrm{SO}_{2} \)

From the equation, 11 moles of \( \mathrm{O}_{2} \) produce 8 moles of \( \mathrm{SO}_{2} \). According to the stoichiometry: \ For every 1 mole of \( \mathrm{O}_{2} \): \[ \text{Partial pressure of} \; \mathrm{SO}_{2} = \frac{8}{11} \times 0.544 \, \text{atm} = 0.396 \, \text{atm} \]

Total pressure in the vessel

Add up the partial pressures of the remaining gases: \ Partial pressure of \( \mathrm{O}_{2} \): \ \[ 0.096 \, \text{atm} \] Partial pressure of \( \mathrm{SO}_{2} \): \ \[ 0.396 \, \text{atm} \] Partial pressure of \( \mathrm{N}_{2} \): \ \[ 0.41 \, \text{atm} \] Total pressure: \ \[ 0.096 \, \text{atm} + 0.396 \, \text{atm} + 0.41 \, \text{atm} = 0.902 \, \text{atm} \]

Key concepts

Chemical Reaction Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It is essential for understanding how much of each substance is involved in a reaction. The initial step in stoichiometry is to balance the chemical equation. By balancing, you ensure that the number of atoms for each element is the same on both sides of the equation. This is crucial for accurately calculating the amounts of reactants and products.

In the example provided, the unbalanced equation is: $$ \text{FeS}_{2}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) + \mathrm{Fe}_{2}\mathrm{O}_{3}(s) \quad [\text {unbalanced}] $$ The balanced equation is: $$ 4 \text{FeS}_{2}(s) + 11 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{SO}_{2}(g) + 2 \mathrm{Fe}_{2}\mathrm{O}_{3}(s) $$ Balancing equations allows you to see the direct relationship between reactants and products. For instance, 11 moles of \( \mathrm{O}_{2} \) react with 4 moles of \( \text{FeS}_{2} \) to produce 8 moles of \( \mathrm{SO}_{2} \) and 2 moles of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \). This stoichiometric relationship is fundamental for calculating the partial pressures of gases involved.

Partial Pressure Calculation

Partial pressure is the pressure exerted by an individual gas in a mixture. Understanding partial pressures is vital in predicting the behavior of gases during chemical reactions. For the given problem, the initial total pressure in the vessel is 1.05 atm, consisting of \( \mathrm{O}_{2} \) and \( \mathrm{N}_{2} \).

The partial pressure of \( \mathrm{O}_{2} \) is 0.64 atm, and the remainder is \( \mathrm{N}_{2} \). Using the equation: \[ \text{Partial pressure of} \; \mathrm{N}_{2} = 1.05 - 0.64 = 0.41 \; \text{atm} \] When 85% of the \( \mathrm{O}_{2} \) is consumed, you calculate the amount consumed and the remaining \( \mathrm{O}_{2} \): \[ 0.85 \times 0.64 = 0.544 \; \text{atm consumed} \] \[ 0.64 - 0.544 = 0.096 \; \text{atm remaining} \] Next, calculate the partial pressure of \( \mathrm{SO}_{2} \) formed using stoichiometry: For every 11 moles of \( \mathrm{O}_{2} \), 8 moles of \( \mathrm{SO}_{2} \) are produced. Therefore, the partial pressure of \( \mathrm{SO}_{2} \) is: \[ \text{Partial pressure of} \; \mathrm{SO}_{2} = \frac{8}{11} \times 0.544 = 0.396 \; \text{atm} \] Partial pressure calculations make use of the ideal gas law and stoichiometric relationships to determine the contributions of each gas to the total pressure.

Gas Laws

Gas laws describe the behavior of gases in various conditions. These laws are fundamental in understanding how gases interact in chemical reactions. The key laws relevant to this problem are Dalton's Law of Partial Pressures and the Ideal Gas Law.

Dalton's Law states that the total pressure exerted by a gas mixture is the sum of the partial pressures of each individual gas. This is applied in the final step to determine the total pressure in the vessel: \[ \text{Total pressure} = \text{Partial Pressure of} \; \mathrm{O}_{2} + \text{Partial Pressure of} \; \mathrm{SO}_{2} + \text{Partial Pressure of} \; \mathrm{N}_{2} \] Thus: \[ 0.096 + 0.396 + 0.41 = 0.902 \; \text{atm} \] The Ideal Gas Law, given by \( PV = nRT \), connects pressure (P), volume (V), number of moles (n), the ideal gas constant (R), and temperature (T). Although it's not directly used in this particular problem, it's crucial when dealing with gas reactions and predicting the behavior of gases under different conditions.

Understanding these gas laws helps interpret and predict gas behavior, which is essential for calculating pressures and preparing for real-world chemical reactions involving gases.