4NH3+502 → 4NO + 6H20 The standard enthalpies of formation for NH3, NO and H2O are 294.1 kj/mol, 90.250 kj/mol and -241.82 kj/mol, respectively. Calculate the enthalpy of reaction at 389 K using the specific heat capacity equations of the components depending on temperature. j j Cp(NH3 mol. R) = 33,573 - 0, 012581T + 0,0000889067 Cp(0, - 29,526 0,00889 2' mol. K K =
4NH3+502 → 4NO + 6H20 The standard enthalpies of formation for NH3, NO and H2O are 294.1 kj/mol, 90.250 kj/mol and -241.82 kj/mol, respectively. Calculate the enthalpy of reaction at 389 K using the specific heat capacity equations of the components depending on temperature. j j Cp(NH3 mol. R) = 33,573 - 0, 012581T + 0,0000889067 Cp(0, - 29,526 0,00889 2' mol. K K =
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter4: Energy And Chemical Reactions
Section: Chapter Questions
Problem 99QRT
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![4NH3+502 → 4NO + 6H20 The standard enthalpies of formation for NH3, NO and H2O
are 294.1 kj/mol, 90.250 kj/mol and -241.82 kj/mol, respectively. Calculate the enthalpy of
reaction at 389 K using the specific heat capacity equations of the components depending
on temperature.
j
j
Cp(NH3 mol. R) = 33,573 - 0, 012581T + 0,0000889067 Cp(0,
-
29,526 0,00889
2' mol. K
K
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fb6b972-a7a3-4a7f-b019-87bbdbd309ae%2Fc2736470-795a-4937-9f8f-454652fc61a9%2Fatu90ee_processed.png&w=3840&q=75)
Transcribed Image Text:4NH3+502 → 4NO + 6H20 The standard enthalpies of formation for NH3, NO and H2O
are 294.1 kj/mol, 90.250 kj/mol and -241.82 kj/mol, respectively. Calculate the enthalpy of
reaction at 389 K using the specific heat capacity equations of the components depending
on temperature.
j
j
Cp(NH3 mol. R) = 33,573 - 0, 012581T + 0,0000889067 Cp(0,
-
29,526 0,00889
2' mol. K
K
=
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