/*! 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 84 Some water is placed in a coffee... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 84

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{~g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\text {sys }}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) ?

Short Answer

Expert verified
In conclusion, for the dissolving process, the signs for the entropy changes are: 螖S_sys is positive because the ionic solid dissolves into its respective ions, increasing randomness in the system, 螖S_surr is positive because the exothermic process releases heat into the surroundings, causing an increase in randomness, and 螖S_univ is positive because it is the sum of the positive 螖S_sys and 螖S_surr values.

Step by step solution

01

1. Determine the sign for 螖S_sys

To determine the sign for 螖S_sys, consider the dissolving process of the solid. Since it dissolves into its respective ions in the solution, the overall number of particles increases, leading to an increase in the randomness of the system. Therefore, the entropy of the system (螖S_sys) is positive.
02

2. Determine the sign for 螖S_surr

Now we need to determine the sign of 螖S_surr. We know that the dissolving process is exothermic, which means it releases heat into the surroundings. The released heat causes the temperature of the surroundings to rise and increase the randomness of the surroundings. Therefore, the entropy of the surroundings (螖S_surr) is positive.
03

3. Determine the sign for 螖S_univ

Since both 螖S_sys and 螖S_surr are positive, the change in entropy of the universe (螖S_univ) will be the sum of the two: \(螖S_{univ} = 螖S_{sys} + 螖S_{surr}\) As both terms are positive, the overall change in entropy of the universe (螖S_univ) will also be positive. In conclusion, the signs for the entropy changes are: 螖S_sys is positive, 螖S_surr is positive, and 螖S_univ 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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 71 .

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

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.