Problem 47 - Chemistry and Chemical Reactivity

Question

In each pair of gases below, tell which will effuse faster: (a) \(\mathrm{CO}_{2}\) Or \(\mathrm{F}_{2}\) (b) \(\mathrm{O}_{2}\) or \(\mathrm{N}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}\) or \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) two chlorofluorocarbons: CFCl \(_{3}\) or \(\mathrm{C}_{2} \mathrm{Cl}_{2} \mathrm{F}_{4}\)

Step-by-Step Solution

Answer

(a) F₂, (b) N₂, (c) C₂H₄, (d) CFCl₃.

Step 1: Understanding the Concept of Effusion

Effusion is the process by which gas molecules escape through a tiny hole into a vacuum. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that a gas with a lower molar mass will effuse faster.

Step 1: Determine Molar Mass of CO₂ and F₂

Calculate the molar mass of \(\text{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}\)and for \(\text{F}_2 = 2 \times 19.00 = 38.00 \, \text{g/mol}\). Since the molar mass of \(\text{F}_2\)is less than that of \(\text{CO}_2\), \(\text{F}_2\)will effuse faster.

Step 2: Determine Molar Mass of O₂ and N₂

Calculate the molar mass of \(\text{O}_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}\)and for \(\text{N}_2 = 2 \times 14.01 = 28.02 \, \text{g/mol}\). Since the molar mass of \(\text{N}_2\)is less than that of \(\text{O}_2\), \(\text{N}_2\)will effuse faster.

Step 3: Determine Molar Mass of C₂H₄ and C₂H₆

Calculate the molar mass of \(\text{C}_2\text{H}_4 = 2 \times 12.01 + 4 \times 1.01 = 28.06 \, \text{g/mol}\)and for \(\text{C}_2\text{H}_6 = 2 \times 12.01 + 6 \times 1.01 = 30.08 \, \text{g/mol}\). Since the molar mass of \(\text{C}_2\text{H}_4\)is less than that of \(\text{C}_2\text{H}_6\), \(\text{C}_2\text{H}_4\)will effuse faster.

Step 4: Determine Molar Mass of CFCl₃ and C₂Cl₂F₄

Calculate the molar mass of \(\text{CFCl}_3 = 12.01 + 19.00 + 3 \times 35.45 = 137.36 \, \text{g/mol}\)and for \(\text{C}_2\text{Cl}_2\text{F}_4 = 2 \times 12.01 + 2 \times 35.45 + 4 \times 19.00 = 170.92 \, \text{g/mol}\). Since the molar mass of \(\text{CFCl}_3\)is less than that of \(\text{C}_2\text{Cl}_2\text{F}_4\), \(\text{CFCl}_3\)will effuse faster.

Key Concepts

Graham's Law

Graham's Law is a key principle in understanding gas effusion. It explains how gases behave when they pass through a small opening into a vacuum. According to this law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. In simpler terms, this means that lighter gases effuse faster than heavier gases. This relationship can be mathematically represented as: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]Here, \(r_1\) and \(r_2\) represent the effusion rates of gas 1 and gas 2, while \(M_1\) and \(M_2\) are their respective molar masses. Graham's law is particularly useful for comparing the effusion rates of different gases, which can help in identifying the lighter gas that will effuse faster.

Molar Mass

Molar mass is a fundamental concept in chemistry that is essential for calculating the effusion rate of gases. The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all atoms in a molecule. * For example, the molar mass of carbon dioxide (\(\text{CO}_2\)) is calculated as: \[ 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \] * Similarly, for fluorine (\(\text{F}_2\)), the molar mass is: \[ 2 \times 19.00 = 38.00 \, \text{g/mol} \] Understanding molar mass allows us to determine which gas will effuse faster when comparing two gases. The one with the lower molar mass will effuse at a higher rate, as dictated by Graham's law.

Gas Molecules

Gas molecules are the tiny particles that make up gases. They are always in motion, and their movement is a key factor in processes like effusion. Characteristics of gas molecules include:

• High kinetic energy, which causes them to move rapidly and randomly.
• Negligible interactions with each other due to the significant space between them.

When gas molecules effuse, their behaviors follow certain predictable patterns due to physical laws like Graham's Law. This behavior makes gases excellent subjects for studying molecular movement and kinetic theory.

Effusion Rate

The effusion rate of a gas is a measure of how quickly gas molecules escape through a small opening into a vacuum. This rate is influenced by several factors, such as:

• The molar mass of the gas, where lighter gases effuse quicker than heavier ones.
• Temperature, since higher temperatures increase the speed of gas molecules, potentially increasing the effusion rate.
• The size of the opening, with larger openings allowing more molecules to pass through simultaneously.

Effusion rate is crucial in applications such as separating isotopes for nuclear power and understanding natural processes like the earth's atmosphere leaks gases into space. Graham's Law provides a quantitative way to compare effusion rates and predict the behavior of different gases under similar conditions.