7:19 AM Tue Nov 7 1 4 What is Sº for A in the reaction 3 A → 2 Bif AS°(rxn) =-132.9 J/mol K? [S° (B) = (240.0 J/mol K)] 7 ● Question 18 of 40 +/- Tap here or pull up for additional resources ● 115.7 J/mol K 2580 369 O X C 97% x 100 Submit
7:19 AM Tue Nov 7 1 4 What is Sº for A in the reaction 3 A → 2 Bif AS°(rxn) =-132.9 J/mol K? [S° (B) = (240.0 J/mol K)] 7 ● Question 18 of 40 +/- Tap here or pull up for additional resources ● 115.7 J/mol K 2580 369 O X C 97% x 100 Submit
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question:**
What is S° for A in the reaction 3 A → 2 B if ΔS°(rxn) = -132.9 J/mol·K? [S°(B) = 240.0 J/mol·K]
**Answer Explanation:**
The problem involves calculating the standard molar entropy (S°) for substance A given the reaction entropy change (ΔS°) and the standard molar entropy for substance B. The reaction can be expressed as:
3 A → 2 B
The entropy change for the reaction can be described by the equation:
ΔS°(rxn) = Σ S°(products) - Σ S°(reactants)
Substituting the given values:
-132.9 J/mol·K = [2 * 240.0 J/mol·K] - [3 * S°(A)]
Solving for S°(A):
S°(A) = [(2 * 240.0) + 132.9] / 3
S°(A) = (480.0 + 132.9) / 3
S°(A) = 612.9 / 3
S°(A) = 204.3 J/mol·K
The text field shows the entered value as "115.7 J/mol·K" and options to input using a keypad.
**Keypad Display:**
A numeric keypad is available below the text box, indicating that the user can input answers directly. The current numerical answer displayed is 115.7 J/mol·K, which seems to be incorrect based on the calculation above.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8099e47-3d0e-448d-afd1-356c22244100%2F622714a8-f618-4333-8fa6-41a2a32caf31%2Fxk9jm0q_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What is S° for A in the reaction 3 A → 2 B if ΔS°(rxn) = -132.9 J/mol·K? [S°(B) = 240.0 J/mol·K]
**Answer Explanation:**
The problem involves calculating the standard molar entropy (S°) for substance A given the reaction entropy change (ΔS°) and the standard molar entropy for substance B. The reaction can be expressed as:
3 A → 2 B
The entropy change for the reaction can be described by the equation:
ΔS°(rxn) = Σ S°(products) - Σ S°(reactants)
Substituting the given values:
-132.9 J/mol·K = [2 * 240.0 J/mol·K] - [3 * S°(A)]
Solving for S°(A):
S°(A) = [(2 * 240.0) + 132.9] / 3
S°(A) = (480.0 + 132.9) / 3
S°(A) = 612.9 / 3
S°(A) = 204.3 J/mol·K
The text field shows the entered value as "115.7 J/mol·K" and options to input using a keypad.
**Keypad Display:**
A numeric keypad is available below the text box, indicating that the user can input answers directly. The current numerical answer displayed is 115.7 J/mol·K, which seems to be incorrect based on the calculation above.
![**Question 28 of 40**
What is the maximum amount of work that is possible for an electrochemical cell where \( E = 1.10 \, \text{V} \) and \( n = 2 \)? (\( F = 96,500 \, \text{J/(V} \cdot \text{mol)} \))
**Answer:**
-212.300 J
---
**Explanation:**
This question involves calculating the maximum work done by an electrochemical cell, which can be determined using the formula:
\[ \text{Work} = -n \cdot F \cdot E \]
Where:
- \( n \) is the number of moles of electrons transferred (2 in this case),
- \( F \) is Faraday's constant (96,500 J/(V·mol)),
- \( E \) is the electromotive force (1.10 V).
**Calculation:**
\[ \text{Work} = -2 \cdot 96,500 \cdot 1.10 \]
\[ \text{Work} = -212,300 \, \text{J} \]
The result is negative, indicating the energy extracted from the system. The answer given is -212.300 J, which appears to be a typing or calculation error in the interface as the correct result should be -212,300 J. The decimal placement may need correction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8099e47-3d0e-448d-afd1-356c22244100%2F622714a8-f618-4333-8fa6-41a2a32caf31%2Frmyz7b8_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 28 of 40**
What is the maximum amount of work that is possible for an electrochemical cell where \( E = 1.10 \, \text{V} \) and \( n = 2 \)? (\( F = 96,500 \, \text{J/(V} \cdot \text{mol)} \))
**Answer:**
-212.300 J
---
**Explanation:**
This question involves calculating the maximum work done by an electrochemical cell, which can be determined using the formula:
\[ \text{Work} = -n \cdot F \cdot E \]
Where:
- \( n \) is the number of moles of electrons transferred (2 in this case),
- \( F \) is Faraday's constant (96,500 J/(V·mol)),
- \( E \) is the electromotive force (1.10 V).
**Calculation:**
\[ \text{Work} = -2 \cdot 96,500 \cdot 1.10 \]
\[ \text{Work} = -212,300 \, \text{J} \]
The result is negative, indicating the energy extracted from the system. The answer given is -212.300 J, which appears to be a typing or calculation error in the interface as the correct result should be -212,300 J. The decimal placement may need correction.
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