10. CHALLENGE. Calculate AHrxn. N2H4(1) + O2(g) → N2(g) + 2 H₂O(1)given:2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(D)N2O(g) + 3 H2(g) → N2H4(D) + H₂O(1)->>2 NH3(g) + 12 O2(g) → N2H4(1) + H₂O(1)H₂(g) + O2(g) → H₂O(n)AH = -1013 kJΔΗ = -317 kJAH = -142.9 kJΔΗ = -285.8 kJ

Answered: Image /qna-images/answer/ec60e1a9-a202-4bfd-8f54-eb8c114a1c16.jpg

10. CHALLENGE. Calculate AHrxn. N2H4(1) + O2(g) → N2(g) + 2 H₂O(1) given: 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H₂0(D) N2O(g) + 3 H2(g) → N2H4(D) + H₂0 (1) 2 NH3(g) + 1/2O2(g) → N2H4(D) + H₂O(1) H2(g) + 2 O2(g) → H₂0(D) AH = -1013 kJ ΔΗ = -317 kJ AH = -142.9 kJ ΔΗ = -285.8 kJ

10. CHALLENGE. Calculate AHrxn. N2H4(1) + O2(g) → N2(g) + 2 H₂O(1) given: 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H₂0(D) N2O(g) + 3 H2(g) → N2H4(D) + H₂0 (1) 2 NH3(g) + 1/2O2(g) → N2H4(D) + H₂O(1) H2(g) + 2 O2(g) → H₂0(D) AH = -1013 kJ ΔΗ = -317 kJ AH = -142.9 kJ ΔΗ = -285.8 kJ

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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10. CHALLENGE. Calculate AHrxn. N2H4(1) + O2(g) → N2(g) + 2 H₂O(1)
given:
2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(D)
N2O(g) + 3 H2(g) → N2H4(D) + H₂O(1)
->>
2 NH3(g) + 12 O2(g) → N2H4(1) + H₂O(1)
H₂(g) + O2(g) → H₂O(n)
AH = -1013 kJ
ΔΗ = -317 kJ
AH = -142.9 kJ
ΔΗ = -285.8 kJ

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