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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 \(\mathrm{C}_{2} \mathrm{Cl}_{2} \mathrm{F}_{4}\)

Short Answer

Expert verified
(a) F₂, (b) N₂, (c) C₂H₄, (d) CFCl₃.

Step by step solution

01

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.
02

Determine Molar Mass of COâ‚‚ and Fâ‚‚

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.
03

Determine Molar Mass of Oâ‚‚ and Nâ‚‚

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.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

You have a gas, one of the three known phosphorusfluorine compounds \(\left(\mathrm{PF}_{3}, \mathrm{PF}_{5}, \text { and } \mathrm{P}_{2} \mathrm{F}_{4}\right) .\) To find out which, you have decided to measure its molar mass. (a) First, you determine that the density of the gas is \(5.60 \mathrm{g} / \mathrm{L}\) at a pressure of \(0.971 \mathrm{atm}\) and a temperature of \(18.2^{\circ} \mathrm{C} .\) Calculate the molar mass and identify the compound. (b) To check the results from part (a), you decide to measure the molar mass based on the relative rates of effusion of the unknown gas and \(\mathrm{CO}_{2} .\) You find that \(\mathrm{CO}_{2}\) effuses at a rate of \(0.050 \mathrm{mol} / \mathrm{min}\) whereas the unknown phosphorus fluoride effuses at a rate of \(0.028 \mathrm{mol} / \mathrm{min.}\) Calculate the molar mass of the unknown gas based on these results.

You have two gas-filled balloons, one containing He and the other containing \(\mathrm{H}_{2} .\) The \(\mathrm{H}_{2}\) balloon is twice the size of the He balloon. The pressure of gas in the \(\mathrm{H}_{2}\) balloon is \(1 \mathrm{atm},\) and that in the He balloon is 2 atm. The \(H_{2}\) balloon is outside in the snow \(\left(-5^{\circ} \mathrm{C}\right)\) and the He balloon is inside a warm building \(\left(23^{\circ} \mathrm{C}\right)\) (a) Which balloon contains the greater number of molecules? (b) Which balloon contains the greater mass of gas?

In the text, it is stated that the pressure of 4.00 mol of \(\mathrm{Cl}_{2}\) in a \(4.00-\mathrm{L}\) tank at \(100.0^{\circ} \mathrm{C}\) should be 26.0 atm if calculated using the van der Waals equation. Verify this result, and compare it with the pressure predicted by the ideal gas law.

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of \(\mathrm{O}_{2}\) (L) is required for complete reaction with \(5.2 \mathrm{L}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\) What volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor ( \(\mathrm{L}\) ) is produced? Assume all gases are measured at the same temperature and pressure.

The sodium azide required for automobile air bags is made by the reaction of sodium metal with dinitrogen monoxide in liquid ammonia: $$\begin{aligned} 3 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+4 \mathrm{Na}(\mathrm{s})+\mathrm{NH}_{3}(\ell) & \rightarrow \\ & \mathrm{NaN}_{3}(\mathrm{s})+3 \mathrm{NaOH}(\mathrm{s})+2 \mathrm{N}_{2}(\mathrm{g}) \end{aligned}$$ (a) You have \(65.0 \mathrm{g}\) of sodium, a \(35.0-\mathrm{L}\). flask containing \(\mathrm{N}_{2} \mathrm{O}\) gas with a pressure of 2.12 atm at \(23^{\circ} \mathrm{C}\) and excess ammonia. What is the theoretical yield (in grams) of NaNg? (b) Draw a Lewis structure for the azide ion. Include all possible resonance structures. Which resonance structure is most likely? (c) What is the shape of the azide ion?

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