/*! 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 90 Some water is placed in a coffee... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 90

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 \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surt}},\) and \(\Delta S_{\mathrm{univ}} ?\)

Short Answer

Expert verified
In the dissolving process, the signs for the entropy changes are as follows: \( \Delta S_{sys} \) (system) is positive since entropy increases with increasing disorder. \( \Delta S_{surr} \) (surroundings) is positive since heat is released to the surroundings during the exothermic process, and temperature is always positive. Finally, \( \Delta S_{univ} \) (universe) is positive because it is the sum of the positive entropy changes for the system and surroundings.

Step by step solution

01

Determine the sign of 鈭哠_sys

To determine the sign of 鈭哠_sys, we must analyze the molecular states involved in the dissolving process. The solid ionic compound is in a highly ordered state. When it dissolves in the water, the ions disperse, and the overall system becomes less ordered. The change in entropy of the system (鈭哠_sys) will be positive, since entropy increases with increasing disorder.
02

Determine the sign of 鈭哠_surr

Since the dissolving process is exothermic, it releases heat to the surroundings. The sign of 鈭哠_surr depends on the sign of the heat (q) released and the temperature (T) at which the process occurs. Here, the heat is released to the surroundings, making q a negative value. The temperature is always positive. Therefore, 鈭哠_surr = -q/T, and the negative sign of q changes to a positive sign when multiplied by the constant negative sign. So, the change in entropy of the surroundings (鈭哠_surr) is positive.
03

Determine the sign of 鈭哠_univ

To find the sign of 鈭哠_univ, we sum the change in entropy for the system and the surroundings. Since both 鈭哠_sys and 鈭哠_surr are positive, their sum (鈭哠_univ) will also be positive. In conclusion, the signs for the entropy changes are as follows: 鈭哠_sys: Positive 鈭哠_surr: Positive 鈭哠_univ: 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!

Key Concepts

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

Entropy
Entropy is a measure of disorder or randomness in a system. When we talk about a system becoming more disordered, we mean that the particles, like atoms or molecules, are moving and spread out in a less organized way.
During the dissolving process of an ionic solid in water, the molecular structure transforms. The ionic solid, initially in an ordered state as a lattice, shifts to a disordered state when its ions are dispersed in the solution.
This increase in disorder leads to a positive \( \Delta S_{\text{sys}} \) because entropy increases.
  • A solid ionic compound is well-ordered and structured.
  • When dissolved, it disperses into separate ions.
  • This dispersion signifies a higher state of randomness.
This understanding of entropy helps explain phenomena where systems evolve towards greater disorder, following the second law of thermodynamics.
Exothermic Process
An exothermic process is one where energy is released in the form of heat during a reaction. In our context of calorimetry, when an ionic solid dissolves in water and the temperature increases, the process is exothermic.
This means heat is transferred from the system (the dissolving ionic solid) to the surroundings, like the water in the calorimeter.
  • The system releases energy by decreasing its enthalpy (energy content).
  • This released energy heats up the surroundings.
  • The temperature rise confirms the exothermic nature of the process.
Consequently, the entropy change of the surroundings, \( \Delta S_{\text{surr}} \), is positive due to the heat flow into the environment, demonstrating how heat distributions contribute to entropy.
Thermal Equilibrium
Thermal equilibrium occurs when two objects interacting in a system reach the same temperature and no further heat transfer occurs. In calorimetry, reaching thermal equilibrium is key for accurate measurements.
Once the ionic solid has entirely dissolved in water, thermal equilibrium is reached when the solution stabilizes at the increased temperature.
  • Heat flows naturally from hotter to cooler regions.
  • Equilibrium is achieved when temperatures balance in the system.
  • This state allows us to conclude the heat exchange process.
Understanding thermal equilibrium is essential in calorimetry experiments since it ensures accurate assessments of energy changes and supports the analysis of entropic changes within a system.

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

For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for \(\mathrm{CS}_{2}(g) .\)

Consider the following energy levels, each capable of holding two particles: $$\begin{array}{l}{E=2 \mathrm{kJ}} \\ {E=1 \mathrm{kJ}} \\ {E=0 \quad \underline{X X}}\end{array}$$ Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K}\) , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ},\) for a reaction at \(25^{\circ} \mathrm{C}\) . How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C} ?\) What assumptions are made?

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical 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.