Given: 2NH3(g) + 3N20(g) → 4N2(g) + 3H20(1) ....AH1 = -1010 kJ 4NH3(g) + 302(g) → 2N2(g) + 6H2O(1) .....AH2 = -1531 kJ The enthalpy change for the reaction: N2 + 1/202 → N20(g) is: - 510.3 kJ + 81.5 kJ - 815.0 kJ O- 303.3 kJ O+ 207.0 kJ

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 113QRT
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Given:
2NH3(g) + 3N20(g) → 4N2(g) + 3H20(1) ....AH1 = -1010 kJ
4NH3(g) + 302(g) → 2N2(g) + 6H2O(1)......H2 = -1531 kJ
The enthalpy change for the reaction:
N2 + 1/202 → N20(g) is:
- 510.3 kJ
+ 81.5 kJ
-815.0 kJ
- 303.3 kJ
+ 207.0 kJ
Transcribed Image Text:Given: 2NH3(g) + 3N20(g) → 4N2(g) + 3H20(1) ....AH1 = -1010 kJ 4NH3(g) + 302(g) → 2N2(g) + 6H2O(1)......H2 = -1531 kJ The enthalpy change for the reaction: N2 + 1/202 → N20(g) is: - 510.3 kJ + 81.5 kJ -815.0 kJ - 303.3 kJ + 207.0 kJ
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