Logout Open In App
Open In App
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

Chapter 13: Problem 66

What will happen to the number of moles of \(\mathrm{SO}_{3}\) in equilibrium with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ in each of the following cases? a. Oxygen gas is added. b. The pressure is increased by decreasing the volume of the reaction container. c. In a rigid reaction container, the pressure is increased by adding argon gas. d. The temperature is decreased (the reaction is endothermic). e. Gaseous sulfur dioxide is removed.

Short Answer

Expert verified
a. The number of moles of \(\mathrm{SO}_{3}\) will increase. b. The number of moles of \(\mathrm{SO}_{3}\) will increase. c. There will be no change in the number of moles of \(\mathrm{SO}_{3}\). d. The number of moles of \(\mathrm{SO}_{3}\) will decrease. e. The number of moles of \(\mathrm{SO}_{3}\) will decrease.

Step by step solution

01

a. Oxygen gas is added.

Adding more oxygen gas to the system will increase the concentration of \(\mathrm{O}_{2}\). To counteract this, the reaction will shift to the left to consume the excess \(\mathrm{O}_{2}\). This results in an increase in the concentration of \(\mathrm{SO}_{3}\).
02

b. The pressure is increased by decreasing the volume.

When the volume of the container is decreased, the pressure inside increases. According to Le Chatelier's Principle, the system will respond by shifting towards the side with fewer moles of gas. In this case, the left side of the reaction (2 moles of \(\mathrm{SO}_{3}\)) has fewer moles than the right side (3 moles - 2 moles of \(\mathrm{SO}_{2}\) and 1 mole of \(\mathrm{O}_{2}\)). Therefore, the reaction will shift to the left, and the concentration of \(\mathrm{SO}_{3}\) will increase.
03

c. In a rigid container, the pressure is increased by adding argon gas.

Since argon is an inert gas and does not participate in the reaction, adding it does not affect the equilibrium directly. However, when the pressure is increased due to the addition of argon, the partial pressures of the components will not change. Therefore, there will be no change in the number of moles of \(\mathrm{SO}_{3}\).
04

d. The temperature is decreased (the reaction is endothermic).

In an endothermic reaction, heat can be considered a reactant. Decreasing the temperature is equivalent to removing heat from the system. According to Le Chatelier's Principle, the reaction will shift towards the side that absorbs the heat, in this case, the right side. The shift to the right will result in a decrease in the concentration of \(\mathrm{SO}_{3}\).
05

e. Gaseous sulfur dioxide is removed.

When \(\mathrm{SO}_{2}\) is removed, its concentration decreases. To counteract the change, the reaction will shift towards the right side to produce more \(\mathrm{SO}_{2}\). This means that more \(\mathrm{SO}_{3}\) will be consumed, and its concentration will decrease.

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.

Start your free trial

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

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

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium: $$ \underset{\text { Purple }}{\mathrm{CoCl}_{2}(s)}+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \underset{\text { Pink }}{\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)} $$ What color will such an indicator be if rain is imminent?

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is \(1.5\) atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\). a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\). b. If the volume of the cylinder is increased until the total pressure is \(1.0 \mathrm{~atm}\) (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\). c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{~atm}\) )?

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ At equilibrium the total pressure and the density of the gaseous mixture were \(1.80 \mathrm{~atm}\) and \(1.60 \mathrm{~g} / \mathrm{L}\), respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2\) \(\begin{array}{ll}\text { Acetic acid } & \text { Ethanol } & \text { Ethyl acetate }\end{array}\) For the following mixtures \((\mathrm{a}-\mathrm{d})\), will the concentration of \(\mathrm{H}_{2} \mathrm{O}\) increase, decrease, or remain the same as equilibrium is established? a. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 \mathrm{M}\) b. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 M\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 \mathrm{M}\) c. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 \mathrm{M}\) d. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 \mathrm{M}\) e. What must the concentration of water be for a mixture with \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]\) \(=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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