Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 53

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\) a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process $\mathrm{S}_{\text { rhombic}}(s)\longrightarrow \mathrm{S}_{\text { monoclinic }}(s)$ b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Short Answer

Expert verified
a. For the phase change from rhombic to monoclinic crystal structures, ∆H is positive (endothermic process) and ∆S is positive (increased disorder). b. The rhombic crystal structure has a more ordered crystalline structure (smaller positional probability) than the monoclinic crystal structure.

Step by step solution

01

Understanding the relationship between enthalpy and entropy

Enthalpy (H) represents the heat content of a system, while entropy (S) represents the disorder or randomness of a system. The change in enthalpy (∆H) and change in entropy (∆S) depict the changes in heat content and disorder during a process or reaction. Positive ∆H indicates an endothermic process (absorption of heat), while negative ∆H indicates an exothermic process (release of heat). Positive ∆S indicates increased randomness or disorder, while negative ∆S indicates increased order.
02

Analyzing the phase change

During the phase transition, sulfur transforms from the rhombic crystal structure to the monoclinic crystal structure at temperatures above 95°C. As the temperature increases, this phase change occurs to minimize the overall energy of the system, implying that heat is absorbed in this process.
03

Predicting the signs of ∆H and ∆S

Since heat is absorbed during the phase change, ∆H will be positive, indicating an endothermic process. The crystal structure transformation suggests a change in the positional probability, or order, of the system. As the temperature increases, the increased kinetic energy of the atoms usually results in a more disordered system. Therefore, we can predict that ∆S is positive, indicating increased disorder. #b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?#
04

Comparing the entropy values

A more ordered crystalline structure will have lower entropy or smaller positional probability. Since we predicted that ∆S is positive for the phase transition, this implies that the initial rhombic crystal structure has a lower entropy value compared to the final monoclinic crystal structure.
05

Identifying the more ordered crystalline structure of sulfur

Based on the provided information, the rhombic crystal structure has a lower entropy value or smaller positional probability, and thus, it possesses the more ordered crystalline structure.

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

Consider the system $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\) a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol},\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C} .\) c. Show by calculations that \(\Delta G=0\) at equilibrium.

True or false: High temperatures are favorable to a reaction both kinetically and thermodynamically. Explain.

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?

Consider the reactions $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ $$\mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text { reaction } 2}>K_{\text { reaction }}\) . Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

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