Problem 83 - Chemistry: The Central Science

Question

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\). (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C}\). (c) Calculate the most probable speed of an ozone molecule in the stratosphere, where the temperature is \(270 \mathrm{~K}\).

Step-by-Step Solution

Answer

The order of increasing average molecular speed of the given gases is Ne < CO < \(\mathrm{SO_2}\) < \(\mathrm{NF_3}\) < HBr. The rms speed of \(\mathrm{NF}_3\) at \(25^{\circ} \mathrm{C}\) is approximately \(670.6 \mathrm{~m/s}\), and the most probable speed of an ozone molecule in the stratosphere at \(270 \mathrm{~K}\) is approximately \(402.5 \mathrm{~m/s}\).

Step 1: Find the formula for molecular speed

There are a few ways to think about molecular speed, but the most useful for this problem is the root mean square speed, defined as: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] where \(v_{rms}\) is the root-mean-square speed, \(R\) is the ideal gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kg/mol. As the temperature is constant, we can sort the molecules by their inverse molar masses, since the larger the molar mass, the slower the rms speed.

Step 2: Find the molar masses of the gases

To order the gases, we need to find their molar masses, which can be done by adding up the atomic masses of the elements in each compound: 1. Ne: 20.18 g/mol (1 helium atom) 2. HBr: 1.01 g/mol (H) + 79.9 g/mol (Br) = 80.91 g/mol 3. \(\mathrm{SO_2}\): 32.06 g/mol (S) + 2 * 16.00 g/mol (O) = 64.06 g/mol 4. \(\mathrm{NF_3}\): 14.01 g/mol (N) + 3 * 19.00 g/mol (F) = 71.01 g/mol 5. CO: 12.01 g/mol (C) + 16.00 g/mol (O) = 28.01 g/mol

Step 3: Arrange the gases in order of increasing average molecular speed

Using the inverse relationship between molar mass and rms speed at constant temperature, we can now arrange the molecules in the order of increasing average molecular speed: Ne < CO < \(\mathrm{SO_2}\) < \(\mathrm{NF_3}\) < HBr

Step 4: Calculate the rms speed of \(\mathrm{NF}_3\) at 25°C

Now, we will use the rms speed formula to calculate the rms speed of \(\mathrm{NF}_3\) at 25°C (298.15 K): \[v_{rms} = \sqrt{\frac{3RT}{M}}\] For \(\mathrm{NF}_3\), \(M = 71.01 \times 10^{-3}\) kg/mol (converted from g/mol to kg/mol) and \(T = 298.15\) K. Using these values and the ideal gas constant, \(R = 8.314 \mathrm{~J/mol\cdot K}\), we get: \[v_{rms} = \sqrt{\frac{3 \times 8.314 \times 298.15}{71.01 \times 10^{-3}}}\] \[v_{rms} \approx 670.6 \mathrm{~m/s}\]

Step 5: Calculate the most probable speed of an ozone molecule in the stratosphere

The most probable speed of a molecule is given by the formula: \[v_p = \sqrt{\frac{2RT}{M}}\] For an ozone molecule, O\(_3\), the molar mass is \(3 \times 16.00 = 48.00\) g/mol, which we will convert to kg/mol: \(M = 48.00 \times 10^{-3}\) kg/mol. The temperature in the stratosphere is given as \(270\) K. Now we can use these values to compute the most probable speed of an ozone molecule in the stratosphere: \[v_p = \sqrt{\frac{2 \times 8.314 \times 270}{48.00 \times 10^{-3}}}\] \[v_p \approx 402.5 \mathrm{~m/s}\] To summarize, the order of increasing average molecular speed for the gases is Ne < CO < \(\mathrm{SO_2}\) < \(\mathrm{NF_3}\) < HBr, the rms speed of \(\mathrm{NF}_3\) at 25°C is approximately 670.6 m/s, and the most probable speed of an ozone molecule in the stratosphere is approximately 402.5 m/s.

Key Concepts

Root Mean Square Speed

When we discuss the motion of molecules in a gas, one key concept is the root mean square (rms) speed, which represents the average speed of these molecules. It is derived from the kinetic theory of gases which provides insights into the microscopic behavior of gas particles.

The formula for root mean square speed is given by:
\[v_{rms} = \sqrt{\frac{3RT}{M}}\]
where:

• \(v_{rms}\) is the root mean square speed of the molecules in m/s,
• \(R\) is the ideal gas constant which has a value of 8.314 J/mol·K,
• \(T\) is the absolute temperature of the gas in Kelvin, and
• \(M\) is the molar mass of the gas in kg/mol.

Given the same temperature and pressure conditions, lighter molecules will move faster than heavier ones, thus having a higher rms speed. This is why when solving problems involving molecular speed, understanding molar mass is crucial to determine how fast molecules are moving relative to one another. Translating this concept into practical use, as in the textbook exercise, can essentially help students comprehend why certain gases diffuse faster in a mixture.

Molar Mass

To truly understand molecular speeds, we must first grasp the concept of molar mass. Molar mass is fundamental to many areas of chemistry and is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

Calculating the molar mass of a chemical compound involves summing up the atomic masses of its constituent elements in their respective ratios. For example:

• \(\mathrm{H_2O}\) has a molar mass of approximately 18.02 g/mol, derived from twice the atomic mass of hydrogen plus the atomic mass of oxygen.

The molar mass is crucial when calculating the rms speed, as gases with larger molar masses move slower at a given temperature. In classroom settings, emphasizing the method of calculating molar mass helps students understand how to predict the behavior of gases and use it in equations involving the ideal gas law and molecular speeds.

Ideal Gas Law

A cornerstone of gas behavior is captured in the ideal gas law, which provides a clear relationship between the volume, pressure, temperature, and number of moles of a gas. The law is elegantly expressed by the equation:
\[PV = nRT\]
where:

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

The ideal gas law assumes that gases are composed of particles that exert no forces on each other and occupy no volume of their own. This is a simplification, but it works well under many conditions and is particularly useful for predicting and calculating the properties of gases in reactions and containers. When gases deviate from ideal behavior, real gas equations are used, but for most high school and initial university-level courses, the ideal gas law provides adequate descriptions and is a building block for understanding gas behavior, including concepts like molecular speed.