Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{~K}: \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{Cl}_{2}\), HBr. (b) Calculate and compare the \(\mathrm{ms}\) speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\).

Short Answer

Expert verified
The order of the given gases in terms of increasing average molecular speed at 300 K is CO < H₂S < Cl₂ < HBr < SF₆. The root-mean-square speeds for CO and Cl₂ at 300 K are 514.46 m/s and 324.46 m/s, respectively.

Step by step solution

01

Calculate the molar mass of each gas

In order to calculate the molecular speeds, we first need to determine the molar mass of each gas. Listed below are the atomic masses: - C: 12.01 g/mol - O: 16.00 g/mol - S: 32.07 g/mol - F: 19.00 g/mol - H: 1.01 g/mol - Br: 79.90 g/mol - Cl: 35.45 g/mol Using the atomic masses to compute the molar masses: - CO: 12.01 + 16.00 = 28.01 g/mol - SF₆: (1 × 32.07) + (6 × 19.00) = 32.07 + 114.00 = 146.07 g/mol - H₂S: (2 × 1.01) + 32.07 = 34.09 g/mol - Cl₂: 2 × 35.45 = 70.90 g/mol - HBr: 1.01 + 79.90 = 80.91 g/mol Molar masses in kg/mol: - CO: 0.02801 kg/mol - SF₆: 0.14607 kg/mol - H₂S: 0.03409 kg/mol - Cl₂: 0.07090 kg/mol - HBr: 0.08091 kg/mol
02

Calculate root-mean-square speed for CO and Cl₂ at 300 K

Using the formula for rms speed, we can calculate the speeds for CO and Cl₂: For CO: \(v_\text{rms} = \sqrt{\dfrac{3 \times 8.314 \times 300}{0.02801}} = \sqrt{\dfrac{7479.54}{0.02801}} = 514.46 \mathrm{~m/s}\) For Cl₂: \(v_\text{rms} = \sqrt{\dfrac{3 \times 8.314 \times 300}{0.07090}} = \sqrt{\dfrac{7479.54}{0.07090}} = 324.46 \mathrm{~m/s}\)
03

Place the gases in order of increasing average molecular speeds

Since root-mean-square speed is inversely proportional to the square root of molar mass, we can order the gases according to their molar masses: Lower molar mass → higher molecular speed Order of gases, based on molar mass: CO (0.02801) < H₂S (0.03409) < Cl₂ (0.07090) < HBr (0.08091) < SF₆ (0.14607) Hence, the order of gases in terms of increasing average molecular speed at 300 K is: CO < H₂S < Cl₂ < HBr < SF₆ In conclusion, the order of the given gases in terms of increasing average molecular speed at 300 K is CO < H₂S < Cl₂ < HBr < SF₆, and the root-mean-square speeds for CO and Cl₂ at 300 K are 514.46 m/s and 324.46 m/s, respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Root-Mean-Square Speed
The root-mean-square (rms) speed is a calculated value that represents the average velocity of the molecules in a gas. It's derived from the kinetic molecular theory, which explains the behavior of gas particles. According to this theory, gas particles are in constant, random motion and the rms speed provides a way to measure their motion with a single value.

To calculate the rms speed, the kinetic energy associated with the temperature of the gas is factored in, along with the molar mass of the gas. The equation used is:
\[v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\]
where:\begin{itemize}\item \( v_{\text{rms}} \) is the root-mean-square speed,\item \( R \) is the universal gas constant (8.314 J/mol⋅K),\item \( T \) is the temperature in Kelvin, and\item \( M \) is the molar mass of the gas in kilograms per mole (kg/mol).\end{itemize}
The rms speed is crucial because it allows us to compare the kinetic energies of different gases at the same temperature, assuming ideal gas behavior. When solving problems involving rms speed, it's important to use the correct mass units and to input the absolute temperature in Kelvin to ensure accurate calculations.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is a critical value in chemical calculations as it allows us to convert between the mass of a substance and the number of moles, which represent a fixed number of particles—the Avogadro number (\(6.022 \times 10^{23}\)).

For gases, molar mass influences properties such as density and diffusion rate. Molar mass can be calculated by summing the atomic masses of the elements that comprise the compound. For molecules, each atom contributes to the total molar mass proportionally to the number of atoms of that element present in the molecule. For instance, the molar mass of CO is calculated by adding the mass of one atom of carbon (\(12.01\) g/mol) to the mass of one atom of oxygen (\(16.00\) g/mol), resulting in \(28.01\) g/mol.

In the exercise we are discussing, calculating molar mass is the first step because the rms speed of gases is inversely proportional to the square root of their molar mass, making this value essential for comparisons.
Gas Laws
Gas laws are mathematical relationships between the volume, temperature, pressure, and amount of a gas. These laws arise from the kinetic molecular theory and help predict how gases will respond to changes in those conditions.

The most well-known gas laws include:
  • Boyle’s Law, which states that the pressure of a gas is inversely proportional to its volume at constant temperature,
  • Charles’s Law, which indicates that the volume of a gas is directly proportional to its absolute temperature at constant pressure,
  • Avogadro’s Law, asserting that the volume of a gas is directly proportional to the number of moles of gas at constant temperature and pressure,
  • The Ideal Gas Law, which combines the previous laws into a single equation: \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Understanding these laws is essential when working with gases, since they allow us to predict and calculate properties of a gas sample. For the calculation of molecular speed, we assume ideal gas behavior, which is a simplification that gases obey these laws under many conditions.
Kinetic Molecular Theory
The kinetic molecular theory (KMT) is a model that explains the behavior of gases at the molecular level. It's based on a few key postulates:
  • Gas is composed of a large number of small particles that are far apart relative to their size.
  • These particles are in constant, random motion and collide with each other and the walls of their container.
  • Collisions between gas particles and between particles and the container walls are elastic, meaning there is no overall loss of kinetic energy.
  • The average kinetic energy of gas particles is directly proportional to the temperature of the gas in Kelvin.

The theory provides the foundation for understanding the macroscopic properties of gases, like pressure and temperature, in terms of the microscopic behavior of the molecules. KMT helps justify the use of rms speed as a measure of molecular motion, as it relates the energy of gas particles to the temperature and allows us to compute their average speed within the gas. This theoretical framework is essential when studying the states of matter, explaining phase changes, and even calculating the rms speed in the exercise we examined.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the combustion reaction hetween \(25.0 \mathrm{mI}\). of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(12.5 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

In the United States, barometric pressures are generally reported in inches of mercury (in. \(\mathrm{Hg}\) ). On a beautiful summer day in Chicago the barometric pressure is \(30.45\) in. Hg. (a) Convert this pressure to torr. (b) A meteorologist explains the nice weather by referring to a "high-pressure area." In light of your answer to part (a), explain why this term makes sense.

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6 \mathrm{~L}\) of \(\mathrm{H}_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, will be produced?

A gas forms when elemental sulfur is heated carefully with AgF. The initial product boils at \(15^{\circ} \mathrm{C}\). Experiments on several samples yielded a gas density of \(0.803 \pm 0.010 \mathrm{~g} / \mathrm{L}\) for the gas at \(150 \mathrm{~mm}\) pressure and \(32{ }^{\circ} \mathrm{C}\). When the gas reacts with water, all the fluorine is converted to aqueous HF. Other products are elemental sulfur, \(S_{8}\), and other sulfur-containing compounds. A 480 -mL sample of the dry gas at \(126 \mathrm{~mm}\) pressure and \(28^{\circ} \mathrm{C}\), when reacted with \(80 \mathrm{~mL}\) of water, yielded a \(0.081 \mathrm{M}\) solution of HF. The initial gaseous product undergoes a transformation over a period of time to a second compound with the same empirical and molecular formula, which boils at \(-10^{\circ} \mathrm{C}\). (a) Determine the empirical and molecular formulas of the first compound formed. (b) Draw at least two reasonable Lewis structures that represent the initial compound and the one into which it is transformed over time. (c) Describe the likely geometries of these compounds, and estimate the single bond distances, given that the \(\mathrm{S}-\mathrm{S}\) bond distance in \(\mathrm{S}_{8}\) is \(2.04 \mathrm{~A}\) and the \(\mathrm{F}-\mathrm{F}\) distance in \(\mathrm{F}_{2}\) is \(1.43 \mathrm{~A}\).

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0\) L of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free