Problem 64 - Chemistry and Chemical Reactivity

Question

The gas \(\mathrm{B}_{2} \mathrm{H}_{6}\) burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{B}_{2} \mathrm{O}_{3}\) $$ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Three gases are involved in this reaction. Place them in order of increasing molecular speed. (Assume all are at the same temperature.) (b) A \(3.26-\) I. flask contains \(B_{2} H_{6}\) at a pressure of \(256 \mathrm{mm}\) Hg and a temperature of \(25^{\circ} \mathrm{C}\). Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{B}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathrm{O}_{2} ?\)

Step-by-Step Solution

Answer

The order is \( \mathrm{O}_2 < \mathrm{B}_2 \mathrm{H}_6 < \mathrm{H}_2 \mathrm{O} \) for speed; \( \mathrm{O}_2 \) partial pressure is 772.54 mm Hg.

Step 1: Understanding Molecular Speed

To determine the order of increasing molecular speed among the gases \( \mathrm{B}_{2} \mathrm{H}_{6} \), \( \mathrm{O}_{2} \), and \( \mathrm{H}_{2} \mathrm{O} \), we use the fact that molecular speed is inversely proportional to the square root of the molar mass (Graham's law). Therefore, gases with lower molar masses will have higher molecular speeds at the same temperature.

Step 2: Calculate Molar Masses

Calculate the molar masses:- \( \mathrm{B}_{2} \mathrm{H}_{6} \) has a molar mass of approximately \( 2 \times 10.81 \text{ (for B)} + 6 \times 1.01 \text{ (for H)} = 27.66 \text{ g/mol} \).- \( \mathrm{O}_{2} \) has a molar mass of \( 2 \times 16.00 = 32.00 \text{ g/mol} \).- \( \mathrm{H}_{2} \mathrm{O} \) has a molar mass of \( 2 \times 1.01 + 16.00 = 18.02 \text{ g/mol} \).

Step 3: Arrange Gases by Increasing Speed

From Step 2, we determine the order of increasing molecular speeds:1. \( \mathrm{O}_{2} \) (Molar Mass = 32.00 g/mol)2. \( \mathrm{B}_{2} \mathrm{H}_{6} \) (Molar Mass = 27.66 g/mol)3. \( \mathrm{H}_{2} \mathrm{O} \) (Molar Mass = 18.02 g/mol)Thus, the order of increasing molecular speed is \( \mathrm{O}_{2} \), \( \mathrm{B}_{2} \mathrm{H}_{6} \), \( \mathrm{H}_{2} \mathrm{O} \).

Step 4: Initial Conditions and Volume

Given the volume of the flask (\( V = 3.26 \text{ L} \)), the pressure of \( \mathrm{B}_{2} \mathrm{H}_{6} \) is \( 256 \text{ mm Hg} \), and the temperature is \( 25^{\circ} \mathrm{C} = 298 \text{ K} \).

Step 5: Apply Ideal Gas Law for \( \mathrm{B}_{2} \mathrm{H}_{6} \)

Using the ideal gas law, \( PV = nRT \), calculate the moles of \( \mathrm{B}_{2} \mathrm{H}_{6} \):\[ n = \frac{PV}{RT} = \frac{256 \text{ mm Hg} \times 3.26 \text{ L}}{62.36 \text{ L mm Hg/mol K} \times 298 \text{ K}} = 0.0452 \text{ mol} \]

Step 6: Determine Stoichiometric Moles of \( \mathrm{O}_{2} \)

According to the reaction, 1 mole of \( \mathrm{B}_{2} \mathrm{H}_{6} \) reacts with 3 moles of \( \mathrm{O}_{2} \). Therefore, \( 0.0452 \text{ mol of } \mathrm{B}_{2} \mathrm{H}_{6} \) will require:\[ 3 \times 0.0452 = 0.1356 \text{ mol of } \mathrm{O}_{2} \]

Step 7: Calculate Partial Pressure of \( \mathrm{O}_{2} \)

Using the ideal gas law, calculate the partial pressure of \( \mathrm{O}_{2} \):\[ P = \frac{nRT}{V} = \frac{0.1356 \text{ mol} \times 62.36 \text{ L mm Hg/mol K} \times 298 \text{ K}}{3.26 \text{ L}} = 772.54 \text{ mm Hg} \]

Step 8: Conclusion: Partial Pressure of \( \mathrm{O}_{2} \)

The partial pressure of \( \mathrm{O}_{2} \) when it is in the correct stoichiometric ratio with \( \mathrm{B}_{2} \mathrm{H}_{6} \) is approximately 772.54 mm Hg.

Key Concepts

Molecular Speed

When we talk about molecular speed, we're essentially discussing how fast the molecules of a gas move. This speed is affected by the molar mass of the molecules. The relationship is simple: the lighter the molecule, the faster it moves at a given temperature.

According to Graham's law, the average speed of gas molecules is inversely proportional to the square root of its molar mass. This means that if we compare gases like \(\mathrm{B}_2 \mathrm{H}_6\), \(\mathrm{O}_2\), and \(\mathrm{H}_2\mathrm{O}\), we can predict which one moves faster by looking at their molar masses.

• For \(\mathrm{B}_2 \mathrm{H}_6\), the molar mass is approximately 27.66 g/mol.
• For \(\mathrm{O}_2\), it is 32.00 g/mol.
• For \(\mathrm{H}_2\mathrm{O}\), it is 18.02 g/mol.

Thus, \(\mathrm{H}_2\mathrm{O}\) is the fastest, followed by \(\mathrm{B}_2 \mathrm{H}_6\), and \(\mathrm{O}_2\) is the slowest among them.

Ideal Gas Law

The ideal gas law is a fundamental principle that helps us understand how gases behave under various conditions. It's expressed as \( PV = nRT \), where

• \(P\) is the pressure of the gas,
• \(V\) is the volume of the container holding the gas,
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

In our given problem, we used this equation to calculate the number of moles of \(\mathrm{B}_2 \mathrm{H}_6\) in the flask. By plugging in the known values:

\[ n = \frac{256\ \mathrm{mm\ Hg} \times 3.26\ \mathrm{L}}{62.36\ \mathrm{L\ mm\ Hg/mol\ K} \times 298\ \mathrm{K}} \approx 0.0452\ \mathrm{mol} \]

This calculation shows us how many moles of \(\mathrm{B}_2 \mathrm{H}_6\) are present under the given conditions of pressure and temperature. The ideal gas law is essential for figuring out the behavior of gases and is a powerful tool in chemistry.

Stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the relationships between different substances in a chemical reaction.

In this exercise, stoichiometry helps determine how much \(\mathrm{O}_2\) is needed for the reaction with \(\mathrm{B}_2 \mathrm{H}_6\). From the balanced chemical equation:

• 1 mole of \(\mathrm{B}_2 \mathrm{H}_6\) reacts with 3 moles of \(\mathrm{O}_2\).

Given that we've calculated there are 0.0452 moles of \(\mathrm{B}_2 \mathrm{H}_6\), the stoichiometric calculation for \(\mathrm{O}_2\) would be:

\[ 3 \times 0.0452 = 0.1356 \ \mathrm{mol\ \text{of}\ } \mathrm{O}_2 \]

This tells us exactly how many moles of \(\mathrm{O}_2\) are necessary to react completely with the given amount of \(\mathrm{B}_2 \mathrm{H}_6\). Stoichiometry helps ensure that we have the correct proportions of reactants to achieve complete reaction, which is crucial in laboratory settings and industrial applications.