Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 26

The element gallium (Ga) freezes at \(29.8^{\circ} \mathrm{C},\) and its molar enthalpy of fusion is \(\Delta H_{\text {fus }}=5.59 \mathrm{~kJ} / \mathrm{mol}\). (a) When molten gallium solidifies to \(\mathrm{Ga}(s)\) at its normal melting point, is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when \(60.0 \mathrm{~g}\) of \(\mathrm{Ga}(l)\) solidifies at \(29.8^{\circ} \mathrm{C}\)

Short Answer

Expert verified
(a) During the solidification process, the entropy (ΔS) decreases, making ΔS negative. (b) When 60.0 g of Ga solidifies at 29.8°C, the value of ΔS is approximately -15.67 J/K.

Step by step solution

01

Determine the sign of ΔS during the solidification process

In a phase transition where a substance changes from a liquid to a solid state (i.e., it solidifies), the substance becomes more organized and its molecules move in a more restricted way. Therefore, the entropy, which represents the degree of disorder in a system, decreases. Thus, since the entropy decreases during the solidification process, ΔS must be negative.
02

Calculate the amount of moles of Ga

To calculate the value of ΔS, we first need to find the number of moles for the given mass of Ga. We use the molar mass of Ga, which is approximately 69.72 g/mol. Moles of Ga = \( \frac{Mass}{Molar~mass} = \frac{60.0~g}{69.72~g/mol} \) Moles of Ga ≈ 0.860 mol
03

Apply Gibbs Free Energy equation to find ΔS

The Gibbs Free Energy equation is: ΔG = ΔH - TΔS Since the phase transition takes place at the melting point, the Gibbs Free Energy, ΔG, is equal to zero. Therefore, we can rewrite the equation as: 0 = ΔH - TΔS We know ΔH (the molar enthalpy of fusion) = 5.59 kJ/mol and the melting temperature (T) is 29.8°C, which needs to be converted to Kelvin (K): T(K) = T(°C) + 273.15 = 29.8 + 273.15 = 302.95 K Now, we can solve the equation for ΔS: ΔS = \( \frac{ΔH}{T} \)
04

Calculate ΔS_value for given mass of Ga

Now that we have all the information needed, we can calculate ΔS for the given amount of gallium: ΔS_value = (0.860 mol) * \( \frac{5.59~kJ/mol}{302.95~K} \) ΔS_value = (0.860 mol) * \( \frac{5500~J/mol}{302.95~K} \) (convert kJ to J by multiplying by 1000) ΔS_value ≈ -15.67 J/K The value of ΔS when 60.0 g of Ga solidifies at 29.8°C is approximately -15.67 J/K.

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

As shown here, one type of computer keyboard cleaner contains liquefied 1,1 -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{~F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1 -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1 -difluoroethane as the system, do you expect \(q_{\mathrm{sys}}\) for the process to be positive or negative? Explain. (c) Predict whether \(\Delta S\) is positive or negative for this process. (d) Given your answers to (a), (b), and (c), do you think the operation of this product depends more on heat flow or more on entropy change?

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K}\) ? (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

The following processes were all discussed in Chapter 18 , "Chemistry of the Environment." Estimate whether the entropy of the system increases or decreases during each process: (a) photodissociation of \(\mathrm{O}_{2}(g),(\mathbf{b})\) formation of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by reverse osmosis.

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K\). (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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