CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
14th Edition
ISBN: 9781259137815
Author: Julia Burdge
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 23, Problem 49QP
Interpretation Introduction

Interpretation:

The minimum voltage required to produce one mole of aluminum at the temperature at which aluminum produced by the Hall process and the energy required to produce 1.00kg

of the metal is to be calculated.

Concept Introduction:

The Hall–Heroult process is considered an enormous industrial process in which the smelting of aluminum takes place.

The decomposition of chemical process takes place by the electrolytic cell that contains electrical energy.

The number of moles can be calculated as:

moles=givenmassmolarmass.

The relationship between cell voltage and free energy difference is represented as:

ΔG=nFE.

Expert Solution & Answer
Check Mark

Answer to Problem 49QP

Solution:

(a) The minimum voltage required to produce one mole of aluminum is 1.03V.

(b) The energy required to produce 1.00kg

of the metal is 3.32×104 kJ/mol.

Explanation of Solution

Given information: The overall reaction for the electrolytic production of aluminum by the Hall process is represented as:

AL2O3(s)+3C(s)2Al(l)+3CO(g).

At 1000°C, the standard free-energy change for this process is 594kJ/mol.

a) The minimum voltage required to produce 1 mol of aluminum at this temperature

The relationship between cell voltage and free energy difference is represented as:

ΔG=nFE.

Here,

ΔG=594×103j/mol,

n=6.

F=96500j/V×mol.

All the given values are to be put in the formula as:

E=ΔGnF=(594×103J/mol6×96500J/V×mol)=1.03V.

Aluminum contains two moles in the balanced equation so, the whole given equation is divided by two.

Thus, the new equation is as

12AL2O3(s)+32C(s)Al(l)+32CO(g).

For the new equation,

n=3

The Gibbs energy is calculated as

ΔG=(594×103 J/mol2)=297kJ/mol.

These values are put in the formula as

E=ΔGnF=(297×103J/mol3×96500J/V×mol)=1.03EV.

Hence, the required voltage would be the same, whether one mole or 1000moles of aluminum are produced, but the amount of current used in each case would be different.

b) The energy required to produce 1.00 kg of the metal.

Primarily, 1.00kg (1000g) of Al

change into moles takes place.

moles=givenmassmolarmass

(1.00×103g Al)×1mol Al26.98gAl=37.1mol Al.

The electrical energy can be found by using the same formula taking other voltage as

ΔG=nFE=(37.1mol Al)(3mol e1mol Al×96500C1mol e)(3.09J1C)=3.32×107J/mol=3.32×104kJ/mol.

The electrical work can be obtained by the multiplication of the voltage and the amount of charge transported through the circuit as

Joules=Volts×Coulombs

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
You are trying to decide if there is a single reagent you can add that will make the following synthesis possible without any other major side products: xi 1. ☑ 2. H₂O хе i Draw the missing reagent X you think will make this synthesis work in the drawing area below. If there is no reagent that will make your desired product in good yield or without complications, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. There is no reagent that will make this synthesis work without complications. : ☐ S ☐
Predict the major products of this organic reaction: H OH 1. LiAlH4 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. G C टे
For each reaction below, decide if the first stable organic product that forms in solution will create a new C-C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 CI MgCl ? Will the first product that forms in this reaction create a new CC bond? Yes No MgBr ? Will the first product that forms in this reaction create a new CC bond? Yes No G टे

Chapter 23 Solutions

CHEMISTRY >CUSTOM<

Ch. 23 - Prob. 11QPCh. 23 - Prob. 12QPCh. 23 - Prob. 13QPCh. 23 - Prob. 14QPCh. 23 - Prob. 15QPCh. 23 - Prob. 16QPCh. 23 - Prob. 17QPCh. 23 - Prob. 18QPCh. 23 - Prob. 19QPCh. 23 - Although iron is only about two-thirds as abundant...Ch. 23 - Prob. 21QPCh. 23 - Prob. 22QPCh. 23 - Prob. 23QPCh. 23 - Prob. 24QPCh. 23 - Prob. 25QPCh. 23 - Prob. 26QPCh. 23 - Prob. 27QPCh. 23 - Prob. 28QPCh. 23 - Prob. 29QPCh. 23 - Prob. 30QPCh. 23 - Prob. 31QPCh. 23 - Prob. 32QPCh. 23 - Prob. 33QPCh. 23 - Prob. 34QPCh. 23 - Prob. 35QPCh. 23 - Prob. 36QPCh. 23 - Prob. 37QPCh. 23 - Prob. 38QPCh. 23 - Prob. 39QPCh. 23 - Describe two ways of preparing magnesium chloride.Ch. 23 - Prob. 41QPCh. 23 - Prob. 42QPCh. 23 - Prob. 43QPCh. 23 - Prob. 44QPCh. 23 - Prob. 45QPCh. 23 - Prob. 46QPCh. 23 - Prob. 47QPCh. 23 - With the Hall process, how many hours will it take...Ch. 23 - Prob. 49QPCh. 23 - Prob. 50QPCh. 23 - Prob. 51QPCh. 23 - Prob. 52QPCh. 23 - Prob. 53QPCh. 23 - Prob. 54QPCh. 23 - Prob. 55QPCh. 23 - Prob. 56QPCh. 23 - Prob. 57QPCh. 23 - Prob. 58APCh. 23 - Prob. 59APCh. 23 - Prob. 60APCh. 23 - Prob. 61APCh. 23 - 23.62 A 0.450-g sample of steel contains manganese...Ch. 23 - Given that Δ G ( Fe 2 O 3 ) f o = − 741.0 kJ/mol...Ch. 23 - Prob. 64APCh. 23 - Prob. 65APCh. 23 - Prob. 66APCh. 23 - Prob. 67APCh. 23 - Write balanced equations for the following...Ch. 23 - Prob. 69APCh. 23 - Prob. 70APCh. 23 - Prob. 71APCh. 23 - Prob. 72APCh. 23 - Prob. 73APCh. 23 - Prob. 74APCh. 23 - Prob. 75APCh. 23 - Prob. 76APCh. 23 - Prob. 77APCh. 23 - Prob. 78APCh. 23 - Prob. 79APCh. 23 - 23.80 The electrical conductance of copper metal...Ch. 23 - Prob. 81APCh. 23 - Prob. 82APCh. 23 - Prob. 1SEPPCh. 23 - Prob. 2SEPPCh. 23 - Prob. 3SEPPCh. 23 - Prob. 4SEPP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY