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

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg},\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

Short Answer

Expert verified
(a) NO_2, O_2, NO; (b) 75 mm Hg; (c) 150 mm Hg.

Step by step solution

01

Understand Root Mean Square (RMS) Speed

The RMS speed of a gas molecule at a given temperature can be found using the formula: \( \text{RMS speed} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. This implies that RMS speed is inversely proportional to the square root of the molar mass.
02

Determine Molar Mass of Each Gas

Calculate the molar masses: NO has a molar mass of \( 30 \ g/mol \), \( O_2 \) has a molar mass of \( 32 \ g/mol \), and \( NO_2 \) has a molar mass of \( 46 \ g/mol \).
03

Order Gases by Increasing RMS Speed

Using Step 1, the gas with the highest molar mass has the lowest RMS speed. Therefore, the order of increasing RMS speed is \( NO_2 < O_2 < NO \).
04

Determine Oxygen Partial Pressure

From the stoichiometry of the balanced equation, 2 moles of NO react with 1 mole of \( O_2 \). Thus, the partial pressure of \( O_2 \) must be half that of \( NO \), which is \( 75 \ mm Hg \).
05

Calculate Final Pressure of \( NO_2 \)

Assuming the reaction goes to completion, all the \( NO \) and \( O_2 \) convert to \( NO_2 \). Since 2 moles of \( NO \) produce 2 moles of \( NO_2 \), the initial partial pressures of NO will result entirely in \( NO_2 \) pressure. Thus, \( NO_2 \) will have a pressure of \( 150 \ mm Hg + 0 \ mm Hg \) (as the oxygen was added just enough according to stoichiometry). Therefore, \( P_{NO_2} = 150 \ mm Hg \).

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.

Gas Laws
Gas laws are essential in understanding how gases behave under various conditions. Three key gas laws are Boyle's Law, Charles's Law, and Avogadro's Law, which describe relationships between volume, pressure, and temperature of gases.
Boyle's Law explains that the pressure of a gas is inversely proportional to its volume, provided the temperature and number of moles remain constant.
  • This means as the volume increases, the pressure decreases and vice versa.
  • The equation used is: \(P_1V_1 = P_2V_2\).
Charles's Law states that the volume of a gas is directly proportional to its Kelvin temperature when the pressure is constant.
  • This implies as the temperature increases, so does the volume.
  • The formula is: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
Finally, Avogadro's Law shows that the volume of a gas is directly proportional to the number of moles, given constant pressure and temperature.
  • This can be represented as \(V_1/n_1 = V_2/n_2\).
Together, these laws can be combined in the ideal gas law: \(PV = nRT\), where \(R\) is the gas constant. Understanding these relationships helps in predicting the behavior of gases during chemical reactions under different conditions.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict how much of a substance is needed or produced in a reaction. One fundamental aspect of stoichiometry is understanding and using the balanced chemical equation. This indicates the mole ratio of reactants and products.
In the given reaction, \(2 \ ext{NO}(g) + \text{O}_2(g) \rightarrow 2 \ ext{NO}_2(g)\), stoichiometry tells us:
  • 2 moles of \(NO\) react with 1 mole of \(O_2\).
  • If 1 mole of \(O_2\) is used, it will completely react with 2 moles of \(NO\) to produce 2 moles of \(NO_2\).
This process can also be used to calculate the necessary amounts of reactants needed for a reaction to proceed. For example, if we start with a certain pressure of \(NO\), we can use stoichiometry to find the required pressure of \(O_2\). This ensures the reactants are in the correct proportions, leading to a complete reaction and accurate product amounts. Understanding stoichiometry is crucial for efficiency and cost-effectiveness in chemical reactions.
RMS Speed
Root Mean Square (RMS) speed is the measure of the speed of particles in a gas. It is related to the average kinetic energy of gas particles. RMS speed can be calculated using the formula: \[ \text{RMS speed} = \sqrt{\frac{3RT}{M}} \]where:
  • \(R\) is the universal gas constant, \(8.314 \ J/(mol\cdot K)\).
  • \(T\) is the absolute temperature in Kelvin.
  • \(M\) is the molar mass of the gas in kilograms per mole.
This equation shows that the speed of gas molecules increases with temperature and decreases with increased molar mass.
For instance, in the given exercise:
  • \(NO\) has the lowest molar mass (30 g/mol) and hence the highest RMS speed.
  • \(O_2\) follows with a slighter greater molar mass.
  • \(NO_2\) has the highest molar mass (46 g/mol) and thus the lowest RMS speed.
Understanding RMS speed is crucial in calculating and predicting the behavior of gases in reactions by analyzing their molecular speeds at different temperatures.
Partial Pressure
Partial pressure is a concept that helps us understand the contribution of each gas in a mixture to the total pressure. According to Dalton's Law, in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.
  • This can be mathematically represented as \(P_{total} = P_1 + P_2 + \ldots + P_n\).
To find the partial pressure of a gas, you need its mole fraction and the total pressure. The formula is: \[ P_i = \chi_i \times P_{total} \]where \(P_i\) is the partial pressure, \(\chi_i\) is the mole fraction, and \(P_{total}\) is the total pressure.
In the exercise, stoichiometry helps determine that since 2 moles of \(NO\) react with 1 mole of \(O_2\), if \(NO\) has a partial pressure of 150 mm Hg, \(O_2\)'s partial pressure must be half, 75 mm Hg. Understanding partial pressures is significant in predicting how gases will behave when mixed, ensuring proper conditions for reactions.

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 sample of nitrogen gas has a pressure of \(67.5 \mathrm{mm} \mathrm{Hg}\) in a \(500 .\) -mL flask. What is the pressure of this gas sample when it is transferred to a \(125-\mathrm{mL}\) flask at the same temperature?

Argon gas is 10 times denser than helium gas at the same temperature and pressure. Which gas is predicted to effuse faster? How much faster?

You have a gas, one of the three known phosphorus-fluorine compounds \(\left(\mathrm{PF}_{3}, \mathrm{PF}_{5},\right.\) and \(\mathrm{P}_{2} \mathrm{F}_{4}\) ). 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 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 mol/ min. Calculate the molar mass of the unknown gas based on these results.

You have a sample of \(\mathrm{CO}_{2}\) in flask \(\mathrm{A}\) with a volume of \(25.0 \mathrm{mL}\). At \(20.5^{\circ} \mathrm{C},\) the pressure of the gas is \(436.5 \mathrm{mm}\) Hg. To find the volume of another flask, B, you move the \(\mathrm{CO}_{2}\) to that flask and find that its pressure is now \(94.3 \mathrm{mm}\) Hg at 24.5 "C. What is the volume of flask B?

Under which set of conditions will \(\mathrm{CO}_{2}\) deviate most from ideal gas behavior? (a) \(1 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (b) 0.1 atm, \(100^{\circ} \mathrm{C}\) (c) \(10 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (d) \(1 \mathrm{atm}, 100^{\circ} \mathrm{C}\)

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