The decomposition of crystalline N2O5 N2O5(s)?2NO2(g)+12O(g) is an example of a reaction that is thermodynamically favored even though it absorbs heat. At 25 ?C we have the following values for the standard state enthalpy and free energy changes of the reaction: Delta H = +109.6 kJ/mol Delta Standard Gibbs Free Energy = -30.5 kJ/mol C) Calculate delta U for this reaction at 25 C.
The decomposition of crystalline N2O5 N2O5(s)?2NO2(g)+12O(g) is an example of a reaction that is thermodynamically favored even though it absorbs heat. At 25 ?C we have the following values for the standard state enthalpy and free energy changes of the reaction: Delta H = +109.6 kJ/mol Delta Standard Gibbs Free Energy = -30.5 kJ/mol C) Calculate delta U for this reaction at 25 C.
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
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The decomposition of crystalline N2O5
N2O5(s)?2NO2(g)+12O(g)
is an example of a reaction that is
Delta H = +109.6 kJ/mol
Delta Standard Gibbs Free Energy = -30.5 kJ/mol
C) Calculate delta U for this reaction at 25 C.
![The decomposition of crystalline N¿O5
NeOs(S)→ 2NO2(g) + 어(미)
is an example of a reaction that is thermodynamically favored even
though it absorbs heat. At 25 °C we have the following values for the
standard state enthalpy and free energy changes of the reaction:
AH° = +109.6 kJ/mol
AG° = -30.5 kJ/mol
(a) Calculate AS° at 25 °C.
(b) Why is the entropy change so favorable for this reaction?
(c) Calculate AU° for this reaction at 25 °C.
(d) Why is AH° greater than AU?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe05ba2f0-76a3-41b9-bc6b-4d089d545847%2Ff3731633-0b31-4a78-ac41-c6a1ff16bbd9%2F7kxe6t_processed.png&w=3840&q=75)
Transcribed Image Text:The decomposition of crystalline N¿O5
NeOs(S)→ 2NO2(g) + 어(미)
is an example of a reaction that is thermodynamically favored even
though it absorbs heat. At 25 °C we have the following values for the
standard state enthalpy and free energy changes of the reaction:
AH° = +109.6 kJ/mol
AG° = -30.5 kJ/mol
(a) Calculate AS° at 25 °C.
(b) Why is the entropy change so favorable for this reaction?
(c) Calculate AU° for this reaction at 25 °C.
(d) Why is AH° greater than AU?
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