Problem 25 - Chemistry: An Atoms-Focused Approach

Question

Rank the gases \(\mathrm{NO}, \mathrm{NO}_{2}, \mathrm{N}_{2} \mathrm{O}_{4},\) and \(\mathrm{N}_{2} \mathrm{O}_{5}\) in order of increasing root-mean-square speed at \(0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Question: Rank the following gases in order of increasing root-mean-square speed at 0°C: NO, NO2, N2O4, and N2O5. Answer: N2O5, N2O4, NO2, NO

Step 1: Write down given information and convert temperature to Kelvin

We are given the temperature as \(0^{\circ} \mathrm{C}\). We need to convert it to Kelvin(K) using the relation \(K=^{\circ} \mathrm{C} + 273.15\). So, \(0^{\circ} \mathrm{C} = 273.15 K\).

Step 2: Calculate the molar mass of each gas

We will find the molar mass of each gas using the molar masses of Nitrogen (N) and Oxygen (O). The molar mass of nitrogen is \(14 g/mol\), and the molar mass of oxygen is \(16 g/mol\). Molar mass of \(\mathrm{NO} = 14 + 16 = 30 g/mol\) Molar mass of \(\mathrm{NO}_{2} = 14 + 2(16) = 46 g/mol\) Molar mass of \(\mathrm{N}_{2}\mathrm{O}_{4} = 2(14) + 4(16) = 92 g/mol\) Molar mass of \(\mathrm{N}_{2}\mathrm{O}_{5} = 2(14) + 5(16) = 108 g/mol\)

Step 3: Calculate the root-mean-square speed of each gas

Using the formula \(v_{rms}=\sqrt{\frac{3RT}{M}}\), we will calculate the root-mean-square speed of each gas. Using the gas constant \(R = 8.314 \frac{J}{mol \cdot K}\). For \(\mathrm{NO}\): \(v_{rms \, NO}=\sqrt{\frac{3(8.314 J/mol.K)(273.15 K)}{0.030 kg/mol}}= \sqrt{683.72} = 26.15 m/s\) For \(\mathrm{NO}_{2}\): \(v_{rms \, NO_{2}}=\sqrt{\frac{3(8.314 J/mol.K)(273.15 K)}{0.046 kg/mol}}= \sqrt{424.40} = 20.60 m/s\) For \(\mathrm{N}_{2}\mathrm{O}_{4}\): \(v_{rms \, N_{2}O_{4}}=\sqrt{\frac{3(8.314 J/mol.K)(273.15 K)}{0.092 kg/mol}}= \sqrt{212.20} = 14.57 m/s\) For \(\mathrm{N}_{2}\mathrm{O}_{5}\): \(v_{rms \, N_{2}O_{5}}=\sqrt{\frac{3(8.314 J/mol.K)(273.15 K)}{0.108 kg/mol}}= \sqrt{184.55} = 13.59 m/s\)

Step 4: Rank the gases in order of increasing root-mean-square speed

Comparing the root-mean-square speeds of the gases, we have the ranking: \(\mathrm{N}_{2}\mathrm{O}_{5} < \mathrm{N}_{2}\mathrm{O}_{4} < \mathrm{NO}_{2} < \mathrm{NO}\) So, the gases in order of increasing root-mean-square speed are \(\mathrm{N}_{2}\mathrm{O}_{5}\), \(\mathrm{N}_{2}\mathrm{O}_{4}\), \(\mathrm{NO}_{2}\), and \(\mathrm{NO}\).

Key Concepts

Molar Mass Calculation

To calculate the molar mass of a compound, simply add up the atomic masses of all the atoms in the molecule. It's like adding apples and oranges, but for atoms!
Here's how it's done for the gases in our exercise:

• For nitrogen (N), the molar mass is 14 g/mol.
• For oxygen (O), it's 16 g/mol.

Using these, the molar masses become:

• NO: 1 nitrogen and 1 oxygen, so 14 + 16 = 30 g/mol.
• NO₂: 1 nitrogen and 2 oxygens, so 14 + 2x16 = 46 g/mol.
• N₂O₄: 2 nitrogens and 4 oxygens, so 2x14 + 4x16 = 92 g/mol.
• N₂O₅: 2 nitrogens and 5 oxygens, so 2x14 + 5x16 = 108 g/mol.

In this way, knowing the periodic table helps you quickly determine molar masses. It's essential for many calculations in chemistry!

Temperature Conversion

Temperature in scientific work is often expressed in Kelvin, especially when dealing with calculations involving gas laws.

Why use Kelvin? Because Kelvin starts at absolute zero, making it beneficial for equations, ensuring no negative temperatures that could yield puzzling results in calculations. The formula for conversion from Celsius to Kelvin is straightforward:

• \(K = °C + 273.15\)

So, for our exercise where the temperature was given as 0°C, converting it to Kelvin means adding 273.15:

• 0°C + 273.15 = 273.15K

This conversion ensures that all of our gas law calculations, like those based on the root-mean-square speed formula, use the absolute temperature scale.

Ideal Gas Law

The ideal gas law is a fan favorite in the world of chemistry equations! It combines Boyle's, Charles's, and Avogadro's laws to form a comprehensive equation: \[ PV = nRT \] Where:

• \(P\) is pressure
• \(V\) is volume
• \(n\) is the number of moles
• \(R\) is the ideal gas constant (8.314 J/mol·K)
• \(T\) is temperature in Kelvin

This formula is incredibly useful because it shows how gases behave under various pressures, temperatures, and volumes. In relation to our original exercise, while we didn't directly use the full version of the ideal gas law, it underpins the equations and theories we do use, like the expression for root-mean-square speed. Remember, in the world of gases, perfect behavior is considered "ideal," even if real-world gases sometimes beg to differ!

Kinetic Molecular Theory

This marvelous theory provides insight into the behaviors of gases, explaining how individual molecules move and interact.
The kinetic molecular theory states:

• Gases consist of very tiny particles that are in constant, random motion.
• These particles are continuously colliding with each other and the walls of their container.
• The average kinetic energy of gas particles is directly proportional to the temperature of the gas in Kelvin.
• Gas particles are assumed not to attract or repel each other, contributing to their independence.

This theory supports the calculation of root-mean-square speed, which was a central part of your original exercise. By representing an average speed of gas molecules, root-mean-square speed ties directly into how the temperature and mass impact the speed and movement of particles in a gas.
In simple terms, lighter molecules move faster than heavier ones at the same temperature, which is why NO had the highest speed in the exercise.