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2N203(g) → 2N2(g) + 302(g) from the equations: AH = 39.7 kJ a. N203(g) – NO(g) + NO2(g) b. ½N2(g) + ½O2{g) → NO(g) c. ½N2(g) + O2{g) – NO2(g) %3D AH = 90.4 kJ %3D AH 33.8 kJ %3!

2N203(g) → 2N2(g) + 302(g) from the equations: AH = 39.7 kJ a. N203(g) – NO(g) + NO2(g) b. ½N2(g) + ½O2{g) → NO(g) c. ½N2(g) + O2{g) – NO2(g) %3D AH = 90.4 kJ %3D AH 33.8 kJ %3!

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Calculate value of delta H. Show your complete solution and write readable. Thank you.
2N203(g) - 2N2(g) + 302(g) from the equations: - NO(g) + NO2(g) AH = 39.7 kJ a. N2O3(g) b. ½N2(g) + ½O2(g) → NO(g) c. ½N2(g) + O2(g) AH = 90.4 kJ NO2(g) AH = 33.8 kJ
Transcribed Image Text:2N203(g) - 2N2(g) + 302(g) from the equations: - NO(g) + NO2(g) AH = 39.7 kJ a. N2O3(g) b. ½N2(g) + ½O2(g) → NO(g) c. ½N2(g) + O2(g) AH = 90.4 kJ NO2(g) AH = 33.8 kJ
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