The resulting reaction is: 2CuO(s) + C(s) → 2Cu(s) + CO2(g). CuO(s) → Cu(s) + 02(g) AG° = 128 kJ/mol C(s) + O2(g) –→ CO2(g) AG° = -394 kJ/mol > 5. Calculate AG° for the overall reaction. (A) -138 kJ/mol (B) -266 kJ/mol (C) 522 kJ/mol (D) 650 kJ/mol

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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Questions 5-8 refer to the following information:
The removal of copper from metallic ores has been a challenge since
ancient times. Strong heating of copper(II) carbonate results in the
formation of copper(II) oxide.
Although the decomposition of copper(II) oxide into copper metal
and oxygen gas is not thermodynamically favorable under standard
conditions, it may be made favorable by adding carbon to the
mixture.
The resulting reaction is: 2CuO(s) + C(s) → 2Cu(s) + CO2(g).
CuO(s) → Cu(s) + 02(g)
AG° = 128 kJ/mol
>
C(s) + O2(g) → CO2(g)
AG° = -394 kJ/mol
5. Calculate AG° for the overall reaction.
(A) -138 kJ/mol
(B) -266 kJ/mol
(C) 522 kJ/mol
(D) 650 kJ/mol
Transcribed Image Text:Questions 5-8 refer to the following information: The removal of copper from metallic ores has been a challenge since ancient times. Strong heating of copper(II) carbonate results in the formation of copper(II) oxide. Although the decomposition of copper(II) oxide into copper metal and oxygen gas is not thermodynamically favorable under standard conditions, it may be made favorable by adding carbon to the mixture. The resulting reaction is: 2CuO(s) + C(s) → 2Cu(s) + CO2(g). CuO(s) → Cu(s) + 02(g) AG° = 128 kJ/mol > C(s) + O2(g) → CO2(g) AG° = -394 kJ/mol 5. Calculate AG° for the overall reaction. (A) -138 kJ/mol (B) -266 kJ/mol (C) 522 kJ/mol (D) 650 kJ/mol
Expert Solution
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The question is based on the concept of Hess law.

it states that net energy change for the reaction remains same ,  whether reaction takes place in single step or multiple Steps.

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