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Chemistry: A Molecular Approach | 5th Edition | ISBN: 9780134874371 | Authors: Nivaldo J. Tro ISBN: 9780134874371 2047

Solution for problem 69 Chapter 19

Chemistry: A Molecular Approach | 5th Edition

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Chemistry: A Molecular Approach | 5th Edition | ISBN: 9780134874371 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 5th Edition

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Problem 69

Consider the sublimation of iodine at \(25.0^{\circ} \mathrm{C}\) :

\(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\)

a. Find \(\Delta G_{\mathrm{rxn}}^{\circ}\) at \(25.0^{\circ} \mathrm{C}\).

b. Find \(\Delta G_{\mathrm{rxn}}^{\circ}\) at \(25.0^{\circ} \mathrm{C}\) under the following nonstandard conditions:

i. \(P_{\mathrm{l}_{2}}=1.00 \mathrm{mmHg}\)

ii. \(P_{\mathrm{I}_{2}}=0.100 \mathrm{mmHg}\)

c. Explain why iodine spontaneously sublimes in open air at \(25.0^{\circ} \mathrm{C}\).

Text Transcription:

25^circ C

I2(s) longrightarrow I2( g)

Delta G_rxn^circ

PI2 = 1.00 mmHg

PI2 = 0.100 mmHg

Step-by-Step Solution:
Step 1 of 3

Chem 222 notes Week 13 Sample problem: Calculate ΔG for the following reaction at standard conditions: 2Ag +(aq) Cu (s)> 2Ag (s)+ Cu 2+(aq) The standard reduction potential for silver is 0.799, and the standard reduction potential for copper is 0.337. First step: figure out what you have, what you need, and what relationship gets you from what you have to what you need. You have the E for your two reactions, and 0 you need ΔG. ΔG is related to E by the following equation: ΔG=-nFE. E for the complete reaction is the sum of the E of the two half-reactions. One half reaction is going forward, and the other is going backward. In the equation at the top, silver is written as a reduction; thus, its reduction potential is the same as given in the table. DO NOT multiply by the balancing coefficient. Copper is being oxidized in the above reaction, not reduced, meaning that its E will have an opposite sign of the number given in the table. Add the two half-reactions to get the E for the full reaction: Ecell +0.799 + (-0.337) = 0.462V. Plug this value into the expression for ΔG (n is 2 because you have 2 electrons being transferred in the above reaction, and F is Faraday’s constant, 96485): ΔG=-nFE = -(2)(96485)(.462)= -89000 J In an electrolytic cell, an electric current drives a nonspontaneous reaction. Electrolytic cells have a negative E, this is how you distinguish an electrolytic cell from a galvanic cell. Positive E means spontaneous, negative E means nonspontaneous. (Notice that this is the opposite of what the sign means when

Step 2 of 3

Chapter 19, Problem 69 is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 5
Author: Nivaldo J. Tro
ISBN: 9780134874371

This full solution covers the following key subjects: . This expansive textbook survival guide covers 24 chapters, and 162 solutions. Chemistry: A Molecular Approach was written by Aimee Notetaker and is associated to the ISBN: 9780134874371. Since the solution to 69 from 19 chapter was answered, more than 201 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 69 from chapter: 19 was answered by Aimee Notetaker, our top Chemistry solution expert on 06/03/22, 04:20PM. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 5. The answer to “?Consider the sublimation of iodine at \(25.0^{\circ} \mathrm{C}\) : \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\)a. Find \(\Delta G_{\mathrm{rxn}}^{\circ}\) at \(25.0^{\circ} \mathrm{C}\).b. Find \(\Delta G_{\mathrm{rxn}}^{\circ}\) at \(25.0^{\circ} \mathrm{C}\) under the following nonstandard conditions:i. \(P_{\mathrm{l}_{2}}=1.00 \mathrm{mmHg}\)ii. \(P_{\mathrm{I}_{2}}=0.100 \mathrm{mmHg}\)c. Explain why iodine spontaneously sublimes in open air at \(25.0^{\circ} \mathrm{C}\).Text Transcription:25^circ CI2(s) longrightarrow I2( g)Delta G_rxn^circPI2 = 1.00 mmHgPI2 = 0.100 mmHg” is broken down into a number of easy to follow steps, and 56 words.

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