/*! 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 87 For an explosion in an open vess... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 87

For an explosion in an open vessel, one would expect a. \(\Delta \mathrm{H}\) to be negative and \(\Delta \mathrm{E}\) to be greater than \(\Delta \mathrm{H}\) b. \(\Delta \mathrm{H}\) to be negative and \(\Delta \mathrm{E}\) to be less than \(\Delta \mathrm{H}\) c. \(\Delta \mathrm{H}\) to be positive and \(\Delta \mathrm{E}\) to be less than \(\Delta \mathrm{H}\) d. \(\Delta \mathrm{H}\) to be positive and \(\Delta \mathrm{E}\) to be greater than \(\Delta \mathrm{H}\)

Short Answer

Expert verified
b. 9 H is negative and 9 E is less than 9 H

Step by step solution

01

Understanding Enthalpy Change

In an explosion, a reaction releases energy to the surroundings. Enthalpy change (9 H) measures the heat exchange at constant pressure. For an explosion in an open vessel, which involves releasing heat, 9 H should be negative, indicating an exothermic reaction.
02

Analyzing Internal Energy Change

The change in internal energy (9 E) is affected by both heat transfer and work done (particularly expansion work). For reactions involving gas expansion, such as explosions in open vessels, 9 E is typically less than 9 H because some energy is used to do work on the surroundings (e.g., pushing air).
03

Application to Answer Choices

Given the prior analysis, correct statements should reflect 9 H as negative and 9 E less than 9 H. This makes choice b correct: 9 H is negative and 9 E is less than 9 H.

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 91影视!

Key Concepts

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

Internal Energy Change
In thermodynamics, the internal energy of a system is the total energy contained within it. This includes both kinetic and potential energies of particles. When we talk about the change in internal energy, denoted as \(\Delta E\), we're referring to how this total energy shifts during a process.
Different factors can influence \(\Delta E\), notably heat exchange with the environment and work done by or on the system. The formula to calculate internal energy change is:
  • \(\Delta E = q + w\)
Here, \(q\) represents the heat exchange, and \(w\) stands for work done. In the context of an explosion in an open vessel, where gases expand, work is typically done on the surroundings. As a result, not all energy content remains within the system, leading to a smaller value of \(\Delta E\) compared to the enthalpy change \(\Delta H\).
Exothermic Reactions
Exothermic reactions are characterized by the release of heat. In such processes, the stored energy of reactants is greater than that of the products. This excess energy gets released to the environment, hence we observe a heat outflow.
The enthalpy change, \(\Delta H\), is a key indicator of whether a reaction is exothermic. For exothermic reactions, \(\Delta H\) is negative. This negative sign signifies that energy is being released, rather than absorbed.
Explosions, as seen in the exercise, are classic examples of exothermic reactions. They generate heat and release it fast, reflecting in the negative \(\Delta H\) value. Understanding this helps explain why choice \(b\) from the exercise is accurate.
Gas Expansion Work
Gas expansion work occurs when a system expands against an external pressure. In thermodynamics, this kind of work is crucial, as it helps determine the energy changes in reactions involving gases. It forms part of the total work \(w\) done by or on the system.
The amount of gas expansion work can be calculated using the formula:
  • \(w = -P_{ext} \times \Delta V\)
Here, \(P_{ext}\) is the external pressure, and \(\Delta V\) is the change in volume of the gas. When the volume increases, work is performed on the surroundings, which often happens during chemical reactions like explosions.
This work transfers some of the internal energy to the environment, resulting in \(\Delta E\) being lower than \(\Delta H\) in cases involving significant gas expansion. Recognizing this dynamic is key to solving the given exercise correctly.

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

The correct statement/s among the following is/are a. mass plus energy of the universe remains always constant while entropy of the universe remains increasing continuously b. an exothermic reaction with \(\Delta \mathrm{S}\) being positive, will be spontaneous only at high temperature c. in a reversible process, the system always in equilibrium with surroundings d. in any cyclic process \(\Delta \mathrm{X}=0\) where \(\mathrm{X}\) is a state fuction.

The lattice energy of \(\mathrm{NaCl}\) is \(780 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The enthalpies of hydration of \(\mathrm{Na}^{+}(\mathrm{g})\) and \(\mathrm{Cl}^{-}(\mathrm{g})\) ions are \(-406\) \(\mathrm{kJ} \mathrm{mol}^{-1}\) and \(-364 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy of solution of \(\mathrm{NaCl}(\mathrm{s})\) is a. \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\) b. \(-10 \mathrm{~kJ} \mathrm{~mol}^{-1}\) c. \(736 \mathrm{~kJ} \mathrm{~mol}^{-1}\) d. \(100 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Which of the following is/are true? a. A spontaneous process is one that can proceed on its own. b. The reverse of a non-spontaneous process is always spontaneous. c. A spontaneous process need not occur immediately. d. A spontaneous process must be exothermic and must have an increase in entropy.

The heat evolved during the combustion of \(46 \mathrm{gm}\) of ethanol in a bomb calorimeter wad determined to be \(670.5 \mathrm{kcal} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). The value of \(\Delta \mathrm{U}\) of the reaction at the same temperature is a. \(-335.25 \mathrm{Kcal}\) b. \(-660.3 \mathrm{Kcal}\) c. \(-670.5 \mathrm{Kcal}\) d. \(-2802.6 \mathrm{~kJ}\)

(A): In the case of an ideal gas, the changes in Gibbs and Helmholtz free energies are equal to each other \((\Delta \mathrm{G}=\Delta \mathrm{A})\) for isothermal reversible processes. (R): There is no change in internal energies and enthalpies for ideal gases at constant temperature.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Organic 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.