2) Calculate AH for the reaction: 2N;H,(1) + 202(g) → 2N2(g) + 4H2O(1I) given the following data: 2NH3(g) + 3N½O(g) → 4N2(g) + 3H20(1); AH = -1010 kJ N20(g) + 3H2(8) → N2H«(I) + H20(1); AH = -317 kJ 2NH3(g) + ½Oz(g) → N2H«(l) + H2O(1); AH = -143 kJ H2(8) + ¼O2(g) → H2O(1); AH = -286 kJ

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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2) Calculate AH for the reaction: 2N2H4(1) + 202(g) → 2N2(g) + 4H2O(1) given the following data:
2NH3(g) + 3N20(g) → 4N2(g) + 3H20(1); AH = -1010 kJ
N20(g) + 3H2(g) → N2H«(1) + H2O(1); AH = -317 kJ
2NH3(g) + ½O2(g) → N2H«(I) + H2O(I); AH = -143 kJ
H2(g) + ½O2(g) → H2O(1); AH = -286 kJ
Transcribed Image Text:2) Calculate AH for the reaction: 2N2H4(1) + 202(g) → 2N2(g) + 4H2O(1) given the following data: 2NH3(g) + 3N20(g) → 4N2(g) + 3H20(1); AH = -1010 kJ N20(g) + 3H2(g) → N2H«(1) + H2O(1); AH = -317 kJ 2NH3(g) + ½O2(g) → N2H«(I) + H2O(I); AH = -143 kJ H2(g) + ½O2(g) → H2O(1); AH = -286 kJ
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