Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 52

In each of the following processes, is any work done when the reaction is carried out at constant pressure in a vessel open to the atmosphere? If so, is work done by the reacting system or on it? (a) Reaction of nitrogen monoxide and oxygen gases to form gaseous nitrogen dioxide; (b) precipitation of magnesium hydroxide by the reaction of aqueous solutions of \(\mathrm{NaOH}\) and \(\mathrm{MgCl}_{2} ;\) (c) reaction of copper(II) sulfate and water vapor to form copper(II) sulfate pentahydrate.

Short Answer

Expert verified
In (a) and (b) no work is done as there is no change in volume. In (c), work is done on the system due to contraction of gas.

Step by step solution

01

Evaluate work done in gas-gas reaction

In (a), Nitrogen monoxide and oxygen gases react to form gaseous nitrogen dioxide. The volumes of reacting and produced gases are the same, so there is no change in volume. Hence, in this scenario, no work is done by or on the system.
02

Determine work done in precipitation reaction

In (b), a precipitation reaction occurs where magnesium hydroxide is formed by the reaction of sodium hydroxide (\(NaOH\)) and magnesium chloride (\(MgCl_2\)). As this reaction happens in solution, no gas is formed, and hence there is no volume change for work to be done. Therefore, no work is done by or on the system in this case.
03

Assess work for hydrate formation

In (c), Copper(II) sulfate reacts with water vapor to form Copper(II) sulfate pentahydrate. Here, water vapor is a gaseous reactant, and the product is a solid hydrate. Hence, contraction of gas takes place. According to the thermodynamic definition, for a system, work done is positive if it is done on the system and negative if it is done by the system. Here, the system contracts so work is done on the system and it is positive.

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

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

\(\mathrm{CCl}_{4},\) an important commercial solvent, is prepared by the reaction of \(\mathrm{Cl}_{2}(\mathrm{g})\) with a carbon compound. Determine \(\Delta H^{\circ}\) for the reaction $$ \mathrm{CS}_{2}(1)+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1)+\mathrm{S}_{2} \mathrm{Cl}_{2}(1) $$ Use appropriate data from the following listing. $$\begin{aligned} \mathrm{CS}_{2}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{SO}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-1077 \mathrm{kJ} \end{aligned}$$ $$2 \mathrm{S}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=-58.2 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1) \quad \Delta H^{\circ}=-135.4 \mathrm{kJ}$$ $$\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-296.8 \mathrm{kJ}$$ $$\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=+97.3 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-393.5 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{CCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g})+\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) & \\ \Delta H^{\circ}=&-5.2 \mathrm{kJ} \end{aligned}$$

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear 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