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

The reaction of \(\mathrm{SO}_{2}\) with \(\mathrm{Cl}_{2}\) gives dichlorine oxide, which is used to bleach wood pulp and to treat wastewater: $$\mathrm{SO}_{2}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{OSCl}_{2}(\mathrm{g})+\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})$$ All of the compounds involved in the reaction are gases. List them in order of increasing rms speed.

Short Answer

Expert verified
\(\mathrm{OSCl_{2}}\), \(\mathrm{Cl_{2}O}\), \(\mathrm{Cl_{2}}\), \(\mathrm{SO_{2}}\)

Step by step solution

01

Understand RMS Speed

The root-mean-square (rms) speed of a gas is given by the formula \(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kilograms per mole. Since \(R\) and \(T\) are constant for all gases at a given temperature, the rms speed is inversely proportional to the square root of the molar mass (\(M\)).
02

Identify Gases and Their Molar Masses

The gases involved in the reaction are \(\mathrm{SO_{2}}\), \(\mathrm{Cl_{2}}\), \(\mathrm{OSCl_{2}}\), and \(\mathrm{Cl_{2}O}\). Calculate their molar masses: \(\mathrm{SO_{2}}\) = 32 + 2(16) = 64 g/mol, \(\mathrm{Cl_{2}}\) = 2(35.5) = 71 g/mol, \(\mathrm{OSCl_{2}}\) = 16 + 32 + 2(35.5) = 102 g/mol, \(\mathrm{Cl_{2}O}\) = 2(35.5) + 16 = 86 g/mol.
03

Rank Gases by Molar Mass

The order of molar masses from highest to lowest is: \(\mathrm{OSCl_{2}\) (102 g/mol), \(\mathrm{Cl_{2}O}\) (86 g/mol), \(\mathrm{Cl_{2}}\) (71 g/mol), \(\mathrm{SO_{2}}\) (64 g/mol). Since rms speed is inversely proportional to the square root of molar mass, the order from lowest to highest speed is the reverse of molar mass.
04

Order in Increasing RMS Speed

Given that the rms speed is inversely proportional to molar mass, the gases are ordered by increasing rms speed as follows: \(\mathrm{OSCl_{2}}\), \(\mathrm{Cl_{2}O}\), \(\mathrm{Cl_{2}}\), \(\mathrm{SO_{2}}\).

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.

Molar Mass
Molar mass is an important factor when discussing the behavior of gases. It is the mass of one mole of a substance, usually expressed in grams per mole. For an ideal gas, the molar mass directly affects the root-mean-square (rms) speed.

In a chemical equation, each element's atomic mass contributes to the total molar mass of a compound. For example, sulfur dioxide (\(\mathrm{SO}_2\)) has a molar mass calculated by adding the atomic mass of sulfur (32 g/mol) and twice the atomic mass of oxygen (16 g/mol per oxygen), which results in 64 g/mol.
Similarly, dichlorine oxide (\(\mathrm{OSCl}_2\)) combines the masses of one oxygen (16 g/mol), one sulfur (32 g/mol), and two chlorine atoms (each 35.5 g/mol), totaling 102 g/mol.

The concept of molar mass becomes particularly significant when comparing the speeds of different gases. Since the rms speed depends largely on molar mass, knowing this weight allows for deductions about how atoms and molecules move in gaseous states.
Understanding these weight differences helps students predict behaviors in reactions involving gases, like their speeds and how quickly they might react.
Gas Constant
The gas constant (\(R\)) is a universal constant that plays a critical role in various equations, such as the Ideal Gas Law. In the context of root-mean-square speed, it helps define the energy and pressure relationships in a gas. The gas constant is generally given as 8.314 J/(mol·K).

It acts as the bridge allowing conversion between the energy scale and the molar scale. When plugged into the rms speed equation \(v_{rms} = \sqrt{\frac{3RT}{M}}\), \(R\) ensures that quantities like temperature (which is in Kelvin) and molar mass (in kg/mol) are properly accounted for, providing accurate speed values.

This constant is fixed and universal for all ideal gas calculations. Therefore, at a given temperature, \(R\) remains unchanged, allowing the rms speed to vary only with changes in molar mass. Understanding the role of the gas constant is crucial when working out problems involving gas behavior and reactions.
Inversely Proportional Relationship
An inversely proportional relationship occurs when two variables behave in such a way that as one increases, the other decreases. This is seen in many facets of chemistry, especially in gas behaviors.

For the rms speed of a gas, the formula \(v_{rms} = \sqrt{\frac{3RT}{M}}\) shows that speed is inversely proportional to the square root of molar mass \(M\).
This means that gases with larger molar masses will have slower rms speeds when compared to lighter gases, as long as the temperature and gas constant remain constant.

This concept is vital in predicting how different gases will move under identical conditions. It helps explain why gases such as \(\mathrm{SO}_2\) with a smaller molar mass will travel faster than gases like \(\mathrm{OSCl}_2\). The inversely proportional relationship provides a clue into kinetic behaviors and is a fundamental concept in thermodynamics and kinetic theory.

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

A You have \(1.249 \mathrm{g}\) of a mixture of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) You find that \(12.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{HCl}\) is required to convert the sample completely to \(\mathrm{NaCl}\), \(\mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{CO}_{2}\) \(\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ What volume of \(\mathrm{CO}_{2}\) is evolved at \(745 \mathrm{mm}\) Hg and \(25^{\circ} \mathrm{C} ?\)

On combustion, \(1.0 \mathrm{L}\) of a gaseous compound of hydrogen, carbon, and nitrogen gives \(2.0 \mathrm{L}\) of \(\mathrm{CO}_{2}\) \(3.5 \mathrm{L}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor, and \(0.50 \mathrm{L}\) of \(\mathrm{N}_{2}\) at STP. What is the empirical formula of the compound?

One of the cylinders of an automobile engine has a volume of \(400 . \mathrm{cm}^{3} .\) The engine takes in air at a pressure of 1.00 atm and a temperature of \(15^{\circ} \mathrm{C}\) and compresses the air to a volume of \(50.0 \mathrm{cm}^{3}\) at \(77^{\circ} \mathrm{C}\). What is the final pressure of the gas in the cylinder? (The ratio of before and after volumes-in this case, 400: 50 or \(8: 1-\) is called the compresion ratio.)

Chlorine trifluoride, \(\mathrm{CIF}_{3}\), is a valuable reagent because it can be used to convert metal oxides to metal fluorides: \(6 \mathrm{NiO}(\mathrm{s})+4 \mathrm{ClF}_{3}(\mathrm{g}) \rightarrow 6 \mathrm{NiF}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})\) (a) What mass of NiO will react with CIF \(_{3}\) gas if the gas has a pressure of \(250 \mathrm{mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) in a \(2.5-\mathrm{L}\) flask? (b) If the CIF a described in part (a) is completely consumed, what are the partial pressures of \(\mathrm{Cl}_{2}\) and of \(\mathrm{O}_{2}\) in the 2.5 -I. flask at \(20^{\circ} \mathrm{C}\) (in \(\mathrm{mm}\) Hg)? What is the total pressure in the flask?

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.

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