Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 14

Consider these changes. (a) \(\operatorname{Hg}(l) \longrightarrow \operatorname{Hg}(g)\) (b) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Short Answer

Expert verified
In reactions (a) and (c), work is done by the system on the surroundings due to an increase in volume. In reaction (b), work is done by the surroundings on the system due to a decrease in volume. And in reaction (d), no work is done since there is no change in volume.

Step by step solution

01

Reaction (a)

Reaction (a) is \(\operatorname{Hg}(l) \longrightarrow \operatorname{Hg}(g)\). It involves the conversion of liquid mercury to its gaseous form. Hence, there is an increase in volume. Since there's an expansion of gases, work is done by the system on the surroundings.
02

Reaction (b)

Reaction (b) is \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g)\). Here, three moles of O2 gas convert into two moles of O3 gas. This means a decrease in volume. Hence, work is done by the surroundings on the system.
03

Reaction (c)

Reaction (c) is $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)$. Here, the hydrated salt loses its water molecules to form gas, which leads to an increase in volume. Hence, in this case, work is done by the system on the surroundings.
04

Reaction (d)

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) describes the reaction. Here, two moles of H2 and F2 gas react to produce two moles of HF gas. The total number of moles of reactants is equal to the total number of moles of products. This implies no change in volume, hence no work is done in this reaction.

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

From the following heats of combustion, $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-726.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: \(\mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\)

Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? (The specific heats of water vapor and air are approximately \(1.9 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively.

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Define these terms: thermochemistry, exothermic process, endothermic process.

The average temperature in deserts is high during the day but quite cool at night, whereas that in regions along the coastline is more moderate. Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

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