/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A sealed glass bulb contains a m... [FREE SOLUTION] | 91Ó°ÊÓ

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A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. When the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\), what happens to these properties of the gases: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density? Assume that volume remains constant. (Hint: \(\mathrm{NO}_{2}\) is a brown gas; \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.)

Short Answer

Expert verified
With the increase in temperature, the color of the gas mixture will become browner, the pressure and the degree of dissociation will increase, while the average molar mass and the density will decrease.

Step by step solution

01

Determine the change in color

The change in color depends on the degree of dissociation. As the temperature increases, \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates into \(\mathrm{NO}_{2}\). Since \(\mathrm{NO}_{2}\) is a brown gas, the color of the gas mixture will become browner.
02

Determine the change in pressure

The pressure of a gas is directly proportional to the temperature, as per the ideal gas law. If the volume remains constant, and the moles of the gas does not change, then an increase in temperature will result in an increase in pressure.
03

Determine the change in average molar mass

The average molar mass depends upon the relative amount of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in the mixture, which changes as the \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates into \(\mathrm{NO}_{2}\). Since the molar mass of \(\mathrm{NO}_{2}\) is lower than that of \(\mathrm{N}_{2} \mathrm{O}_{4}\), the average molar mass of the gas mixture will decrease.
04

Determine the change in the degree of dissociation

The degree of dissociation increases with increase in temperature. Since the temperature increases, more of the \(\mathrm{N}_{2} \mathrm{O}_{4}\) will dissociate into \(\mathrm{NO}_{2}\). Therefore, the degree of dissociation will increase.
05

Determine the change in density

The density of a gas at constant pressure is inversely proportional to the temperature. If the pressure changes (which it does in this case), then the change in temperature and the change in molar mass both impact the density. Since both the temperature and the molar mass are increasing, the density will decrease.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental principle used to describe the behavior of gases. It states that the pressure \(P\), volume \(V\), and temperature \(T\) of a gas are related by the equation: \[ PV = nRT \] where \(n\) is the number of moles of the gas and \(R\) is the universal gas constant. When dealing with gas mixtures, this law helps us understand how changes in temperature or volume affect the gas pressure.
In our context of the sealed bulb containing \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\), when the bulb is heated from 20°C to 40°C, we focus on the pressure change. By the ideal gas law, with constant volume, increasing temperature leads directly to increased pressure. This happens because the gas molecules move faster and collide more often with the walls of the container, increasing the pressure.

By understanding the ideal gas law, we can also infer other properties of the gas, such as its density and how it changes with temperature and pressure variations. - Increase in temperature results in increased pressure if the volume is constant.- This relationship explains many of the behaviors observed in gases under heating.
Degree of Dissociation
The degree of dissociation refers to the fraction of a compound that undergoes dissociation into simpler parts. In our context, \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates into \(\mathrm{NO}_{2}\) at higher temperatures. When temperature rises, more \(\mathrm{N}_{2} \mathrm{O}_{4}\) molecules break down to form \(\mathrm{NO}_{2}\), increasing the degree of dissociation. This results in increased pressure as the number of gas molecules rises.

This process changes the properties of the gas mixture significantly:
  • The color of the gas mixture becomes darker because \(\mathrm{NO}_{2}\) is brown, whereas \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.
  • The average molar mass of the mixture decreases as more lighter \(\mathrm{NO}_{2}\) molecules are present compared to the heavier \(\mathrm{N}_{2} \mathrm{O}_{4}\).
Understanding the degree of dissociation is essential when predicting how a chemical system will change in response to environmental conditions.
Temperature Effects on Gases
The effect of temperature on gases is profound and widespread. For gaseous mixtures, such as the one with \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\), temperature changes can lead to significant shifts in physical and chemical properties.

When temperature increases:
  • The kinetic energy of the gas molecules increases, leading to more frequent and forceful collisions.
  • The dissociation rate of compounds like \(\mathrm{N}_{2} \mathrm{O}_{4}\) tends to increase, rising more \(\mathrm{NO}_{2}\) concentration.
  • Pressure increases due to increased molecular activity within the constant volume.
  • The mixture's average molar mass drops due to the higher proportion of \(\mathrm{NO}_{2}\), a lighter molecule than \(\mathrm{N}_{2} \mathrm{O}_{4}\).
Overall, recognizing these changes helps predict how a gas mixture will behave when subjected to heating or cooling, aiding in controlling processes involving gases.

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Most popular questions from this chapter

Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). What is the equilibrium constant for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) $$ at the same temperature?

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

Outline the steps for calculating the concentrations of reacting species in an equilibrium reaction.

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

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