Problem 97 - Chemistry and Chemical Reactivity

Question

A \(1.0-\) L flask contains \(10.0 \mathrm{g}\) each of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) (a) Which gas has the greater partial pressure, \(\mathbf{O}_{2}\) or \(\mathbf{C O}_{2}\) or are they the same? (b) Which molecules have the greater average speed, or are they the same? (c) Which molecules have the greater average kinetic energy, or are they the same?

Step-by-Step Solution

Answer

(a) \\(O_2\\) has greater partial pressure. (b) \\(O_2\\) molecules have greater average speed. (c) Both gases have the same average kinetic energy.

Step 1: Calculate Moles of Each Gas

First, determine the number of moles for each gas using their molecular weights. The molecular weight of \( ext{O}_2\) is \( ext{32.00 g/mol}\) and for \( ext{CO}_2\) it is \( ext{44.01 g/mol}\). Compute the moles: \[ n_{O_2} = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.3125 \text{ mol} \]\[ n_{CO_2} = \frac{10.0 \text{ g}}{44.01 \text{ g/mol}} = 0.2273 \text{ mol} \]

Step 2: Calculate Partial Pressure of Each Gas Using the Ideal Gas Law

Use the ideal gas law \(PV = nRT\) to calculate partial pressures. Assume \(R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}\) and \(T = 298\, \text{K (since } 25^\circ \text{C} = 298 \text{ K)}\):\[ P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{0.3125 \cdot 0.0821 \cdot 298}{1.0} = 7.64 \text{ atm} \] \[ P_{CO_2} = \frac{n_{CO_2}RT}{V} = \frac{0.2273 \cdot 0.0821 \cdot 298}{1.0} = 5.55 \text{ atm} \]

Step 3: Compare Partial Pressures

From Step 2, the partial pressure of \( ext{O}_2\) is greater than that of \( ext{CO}_2\). Therefore, \( ext{O}_2\) has a greater partial pressure than \( ext{CO}_2\).

Step 4: Calculate Root Mean Square Speed for Each Gas

For average speed, use the root mean square speed formula: \(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(M\) is the molar mass in kg/mol. Convert molar mass from g/mol to kg/mol:\[ M_{O_2} = 32.00 \times 10^{-3} \text{ kg/mol} \]\[ M_{CO_2} = 44.01 \times 10^{-3} \text{ kg/mol} \]Calculate the speeds:\[ v_{rms,O_2} = \sqrt{\frac{3 \times 8.314 \times 298}{32.00 \times 10^{-3}}} = 483.78 \text{ m/s} \]\[ v_{rms,CO_2} = \sqrt{\frac{3 \times 8.314 \times 298}{44.01 \times 10^{-3}}} = 394.44 \text{ m/s} \]\( ext{O}_2\) molecules have a greater average speed.

Step 5: Compare Average Kinetic Energy

The average kinetic energy of a gas molecule depends only on the temperature and is given by: \( \frac{3}{2}kT\), where \(k\) is the Boltzmann constant. As both gases are at the same temperature, their average kinetic energies are the same.

Key Concepts

Partial Pressure

In the context of gases, partial pressure refers to the pressure exerted by a specific gas in a mixture of gases. Each gas within a mixture behaves as if it is alone, contributing its own pressure independently of the others. The total pressure in a container is the sum of the partial pressures of all gases present. This concept forms a crucial component of Dalton's Law of Partial Pressures.

To calculate the partial pressure of a single gas within the mixture, we use the Ideal Gas Law in the format:

• \( PV = nRT \)

where:

• \( P \) = pressure in atmospheres
• \( V \) = volume in liters
• \( n \) = number of moles
• \( R \) = ideal gas constant (0.0821 L atm K^-1 mol^-1)
• \( T \) = temperature in Kelvin

For each gas in the provided example, after determining the number of moles, the partial pressures were obtained by applying the Ideal Gas Law. Oxygen \(O_2\) had more moles (0.3125 mol) than carbon dioxide \(CO_2\) (0.2273 mol) in the same container, resulting in a higher partial pressure of 7.64 atm compared to 5.55 atm for \(CO_2\). This illustrates that the higher the moles of a gas in a fixed volume at a constant temperature, the greater its contribution to the total pressure.

Root Mean Square Speed

When considering the speed of gas molecules, the root mean square speed (\(v_{rms}\)) is highly informative. It provides an average velocity value that depicts the movement of particles in a gaseous state. Calculating this speed involves the formula:

• \[v_{rms} = \sqrt{\frac{3RT}{M}}\]

where:

• \(R\) is the gas constant, 8.314 J mol^-1 K^-1
• \(T\) is temperature in Kelvin
• \(M\) is the molar mass of the gas in kg mol^-1

Importantly, the root mean square speed depends inversely on the molar mass of the gas—meaning lighter gases move faster. For the gases in the exercise, \(O_2\) (molar mass = 32.00 g/mol or 0.032 kg/mol) has a higher speed of 483.78 m/s compared to \(CO_2\) (molar mass = 44.01 g/mol or 0.044 kg/mol) which moves at 394.44 m/s. This is due to the fact that lighter \(O_2\) molecules can travel faster than relatively heavier \(CO_2\) molecules, even at the same temperature.

Average Kinetic Energy

Average kinetic energy in gases quantifies the energy distributed among gas molecules due to their motion. Remarkably, in an ideal gas, this quantity does not depend on the type of gas but solely on its temperature. The expression for average kinetic energy \(E_k\) is given by:

• \(E_k = \frac{3}{2}kT\)

where:

• \(k\) is the Boltzmann constant, 1.38 x 10^-23 J/K
• \(T\) is the absolute temperature in Kelvin

For both \(O_2\) and \(CO_2\), since they are at the same temperature (298 K), the average kinetic energy of their molecules is identical. This stems from the principle that at a given temperature, all ideal gas particles irrespective of type have the same kinetic energy distribution. Therefore, the energy due to movement, characterized by kinetic energy, remains constant for both gases as it's entirely dependent on temperature, not molecular identity or mass. This emphasizes the idea that temperature is a measure of the average kinetic energy in gas samples.