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 \(\mathrm{K}\) .

Short answer

The gases in order of increasing average molecular speed at 25°C are HBr, NF3, SO2, CO, and Ne. The rms speed of NF3 molecules at 25°C is 485.92 m/s, and the most probable speed of an ozone molecule in the stratosphere at 270 K is 461.30 m/s.

Step-by-step solution

Part (a): Arrange the gases in order of increasing average molecular speed at 25°C

First, we need to understand that the average molecular speed is inversely proportional to the square root of the molar mass of the gas. We can write it as: \(v_{avg} \propto \frac{1}{\sqrt{M}}\) where v_avg is the average molecular speed and M is the molar mass of the gas. Now, we'll find the molar mass (in grams/mole) for each gas: Ne - 20.18 g/mol HBr - 80.91 g/mol (1 * H + 79.9 * Br) SO2 - 64.07 g/mol (32.07 * S + 2 * 16 * O) NF3 - 71.00 g/mol (1* N + 3 * 19 * F) CO - 28.01 g/mol (12.01 * C + 16 * O) Now, we can arrange these gases in order of increasing average molecular speed as the lighter gases move faster: 1. HBr 2. NF3 3. SO2 4. CO 5. Ne

Part (b): Calculate the root-mean-square (rms) speed of NF3 molecules at 25°C

Now we will calculate the rms speed of NF3 using the following formula: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) where v_rms is the rms speed, R is the universal gas constant (8.314 J/mol·K), T is the temperature (in kelvin) and M is the molar mass (in kg/mol). First, convert 25°C to Kelvin: T = 25 + 273.15 = 298.15 K Second, convert the molar mass of NF3 to kg/mol: M = 71.00 g/mol * (1 kg/1000 g) = 0.071 kg/mol Now, plug in the values: \(v_{rms} = \sqrt{\frac{3(8.314)(298.15)}{0.071}} = 485.92 \, m/s\) So the rms speed of NF3 molecules at 25°C is 485.92 m/s.

Part (c): Calculate the most probable speed of an ozone molecule at 270 K

Next, we will calculate the most probable speed of an ozone (O3) molecule at 270 K using the following formula: \(v_{mp} = \sqrt{\frac{2RT}{M}}\) where v_mp is the most probable speed, R is the universal gas constant (8.314 J/mol·K), T is the temperature (in kelvin) and M is the molar mass (in kg/mol). First, find the molar mass of O3: O3 - 48 g/mol (3 * 16 * O) Now, convert the molar mass to kg/mol: M = 48 g/mol * (1 kg/1000 g) = 0.048 kg/mol Plug in the values into the formula: \(v_{mp} = \sqrt{\frac{2(8.314)(270)}{0.048}} = 461.30 \, m/s\) So the most probable speed of ozone molecules in the stratosphere at 270 K is 461.30 m/s.

Key concepts

Root-Mean-Square Speed

Understanding the root-mean-square (rms) speed of gas molecules is crucial for students studying thermodynamics and kinetic theory of gases. It represents a type of average speed of the particles in a gas, more precisely a measure of the speed of particles that takes into account both their mass and their energy.

When we look at the movement of particles in a gas, they move at a variety of speeds, but the rms speed gives us a single value representing the speed of a 'typical' particle. The formula to calculate the rms speed is given by: \[v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\]where \(v_{\text{rms}}\) is the root-mean-square speed, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kilograms per mole.

Importantly, the rms speed is sensitive to temperature; as the temperature increases, the average kinetic energy of the gas molecules increases, leading to a higher rms speed. This is why in our exercise, the rms speed of NF3 molecules increases when converted from the temperature in Celsius to Kelvin.

Most Probable Speed

The most probable speed (\(v_{\text{mp}}\)) is distinct from the average or rms speed of molecules in a gas. It refers to the speed at which the greatest number of particles in a gas sample are moving. Hence, it's the speed most likely to be possessed by any one molecule at a given temperature.

To determine the most probable speed of gas molecules, we can use the formula: \[v_{\text{mp}} = \sqrt{\frac{2RT}{M}}\].

The most probable speed is always lower than the rms speed because it doesn't take into account the square of the velocities. Also, like rms speed, it is temperature dependent and will increase as the gas gets hotter. The concept of most probable speed has pivotal implications in fields such as chemical kinetics and is foundational for understanding the distribution of molecular speeds in a gas, known as the Maxwell-Boltzmann distribution. When calculating the most probable speed for ozone molecules in the stratosphere as per the given exercise, students can observe how the unique molar mass of the gas influences its most probable speed at a constant temperature.

Molar Mass

Molar mass is a property of a substance that links the mass of a sample to the number of particles contained in that sample. It can be defined as the mass of one mole of a pure substance. When dealing with gases, the molar mass becomes a critical factor in understanding their kinetic properties, such as diffusion, effusion, and their speeds.

The molar mass is usually expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of all the atoms present in a molecule. For instance, in our exercise, we calculated the molar mass of HBr (hydrogen bromide) by adding the atomic mass of hydrogen (1 g/mol) to that of bromine (79.9 g/mol), yielding 80.91 g/mol.

Understanding molar mass is not only fundamental in stoichiometric calculations but also in predictive applications; it allows us to predict how a gas will behave under certain conditions by using the molar mass in conjunction with the ideal gas law. Additionally, considering the relationship between molecular speed and molar mass, students can determine the order of speeds for different gases at a given temperature, as demonstrated in the exercise provided.