Question

Diborane, \(\mathrm{B}_{2} \mathrm{H}_{6},\) reacts with oxygen to give boric oxide and water vapor. $$ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If you \(\operatorname{mix} \mathrm{B}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio, and if the total pressure of the mixture is \(228 \mathrm{mm} \mathrm{Hg}\), what are the partial pressures of \(\mathrm{B}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) ? If the temperature and volume do not change, what is the pressure of the water vapor?

Short answer

Partial pressures: \(\mathrm{B}_{2} \mathrm{H}_{6}\) is 57 mm Hg, \(\mathrm{O}_{2}\) is 171 mm Hg. Water vapor pressure: 171 mm Hg.

Step-by-step solution

Write the Balanced Chemical Equation

First, confirm the balanced chemical reaction:\[ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g}) + 3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s}) + 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]This equation is already balanced, showing that one mole of diborane reacts with three moles of oxygen gas to form one mole of boric oxide and three moles of water vapor.

Determine the Stoichiometric Ratio

The stoichiometric ratio can be found by examining the coefficients in the balanced equation:For every 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\), 3 moles of \(\mathrm{O}_{2}\) are needed. This means the ratio of \(\mathrm{B}_{2} \mathrm{H}_{6}\) to \(\mathrm{O}_{2}\) is 1:3.

Calculate Partial Pressures from Total Pressure

Using the stoichiometric ratio and the total pressure of 228 mm Hg:- Let \(P_{\mathrm{B}_{2} \mathrm{H}_{6}}\) be the partial pressure of diborane.- Let \(P_{\mathrm{O}_{2}}\) be the partial pressure of oxygen.Since the ratio is 1:3, you can let \(P_{\mathrm{B}_{2} \mathrm{H}_{6}} = x\) and \(P_{\mathrm{O}_{2}} = 3x\). The total pressure is the sum of these pressures:\[ x + 3x = 228 \, \mathrm{mm} \, \mathrm{Hg} \]\[ 4x = 228 \, \mathrm{mm} \, \mathrm{Hg} \]

Solve for Partial Pressures

Solve the equation from the previous step to find the value of \(x\):\[ 4x = 228 \, \mathrm{mm} \, \mathrm{Hg} \]\[ x = \frac{228}{4} = 57 \, \mathrm{mm} \, \mathrm{Hg} \]Thus, the partial pressure of \(\mathrm{B}_{2} \mathrm{H}_{6}\) is 57 mm Hg and the partial pressure of \(\mathrm{O}_{2}\) is \(3 \times 57 = 171 \, \mathrm{mm} \, \mathrm{Hg}\).

Calculate Pressure of Water Vapor

From the balanced equation, 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\) produces 3 moles of \(\mathrm{H}_{2} \mathrm{O}\). Since all the reactants react completely:The pressure of water vapor \(P_{\mathrm{H}_2 \mathrm{O}}\) is equivalent to the initial partial pressure of \(\mathrm{O}_{2}\) because 3 times the amount of \(\mathrm{B}_{2} \mathrm{H}_{6}\) forms 3 times \(\mathrm{H}_{2} \mathrm{O}\):Therefore, \(P_{\mathrm{H}_2 \mathrm{O}} = 171 \, \mathrm{mm} \, \mathrm{Hg}\).

Key concepts

Partial Pressure

In a gas mixture, each component gas exerts its own pressure. This is known as its partial pressure. Even though the gases are together in a mixture, each gas behaves independently and has its own pressure contributing to the total pressure. The total pressure is simply the sum of all individual partial pressures in the mixture. This concept is crucial when dealing with reactions involving gases, like the one in the exercise provided.

When calculating partial pressures, it's important to remember that they are proportional to the amount or number of moles of each gas in the mixture. Using the stoichiometry from the balanced chemical equation, you determine how much of each substance is present. In the exercise, diborane (\(\mathrm{B}_2\mathrm{H}_6\)) and oxygen (\(\mathrm{O}_2\)) are mixed in a 1:3 ratio, which dictates their individual partial pressures. Thus, the individual partial pressures can be calculated using the formula:

• \(P_{\text{total}} = P_{\mathrm{B}_2\mathrm{H}_6} + P_{\mathrm{O}_2}\)
• Using the known conditions and stoichiometric ratio, solve for each: \(P_{\mathrm{B}_2\mathrm{H}_6} = x\) and \(P_{\mathrm{O}_2} = 3x\)

Balanced Chemical Equation

A balanced chemical equation accurately represents a chemical reaction. It shows the reactants and products in terms of moles, following the Law of Conservation of Mass. This means that for any chemical reaction, the mass of the reactants must equal the mass of the products.

This is why it's important to balance chemical equations; it ensures that the number of each type of atom is the same on both sides of the equation. In the exercise, the balanced equation presented is:\[\mathrm{B}_2\mathrm{H}_6(\mathrm{g}) + 3\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{B}_2\mathrm{O}_3(\mathrm{s}) + 3\mathrm{H}_2\mathrm{O}(\mathrm{g})\]This equation tells us that 1 mole of diborane reacts with 3 moles of oxygen gas to make 1 mole of boric oxide and 3 moles of water vapor. The coefficients of the substances in the balanced equation indicate the stoichiometric ratios used to determine how substances interact in the reaction.

• Reactants: Diborane and oxygen.
• Products: Boric oxide and water vapor.
• Ratio: 1:3 of diborane to oxygen.

Gas Laws

Gas laws are fundamental principles that describe how gases behave under different conditions of pressure, volume, and temperature. They're rooted in kinetic molecular theory, which states that gas molecules are in constant, random motion. Understanding gas laws helps us to predict how gases will respond to changes in these conditions.

One such crucial law is Dalton’s Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. When all gases in a mixture behave ideally, the pressure they exert depends solely on their temperature, volume, and number of moles. This means that each gas in a mixture behaves as if it were the only gas present in the container.

• Formula: \(P_{\text{total}} = P_{a} + P_{b} + \ldots\)
• Where \(P_{a}, P_{b}, \ldots\) represent partial pressures of individual gases.
• Helps calculate individual contributions in a gas reaction, like calculating water vapor pressure as demonstrated in the solution.

Using these principles, students can understand how to determine the pressure of each individual gas during reactions just like in the exercise where boranes, oxygen, and water vapor are accounted for in their pressure calculations.