Calculate AH for the reaction: 2NH3 (g) + O2(9) → N½H4(1) + H20(!) given the following data: 2NH3 (9) + 3N20(g) → 4N2(g) + 3H2O(1) AH = -1010. kJ %3D N20(g) + 3H29)→ N,H4(1) + H2O(1) AH = -317 kJ %3D N2H4 (1) + O2(9) N2(9) + 2H,0(1) AH = -623 kJ %3D H2 (g) + O2(g) → H20(1) AH = -286 kJ %3D AH= kJ %3D

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Chapter1: Chemical Foundations
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Calculate AH for the reaction:
2NH3 (9) +O2(9) → N2H4 (1) + H20(1)
given the following data:
2NH3(g) + 3N,0(g) → 4N,(g) + 3H,O(1) AH = - 1010. kJ
N20(g) + 3H, (g) → N,H4 () + H,O(1) AH = -317 kJ
N2 H4 (1) + O2(g) →
N2 (g) + 2H2O(1)
AH =-623 kJ
H2 (g) + 02(g)
→ H,0(1)
AH =
==286 k.J
AH =
kJ
Transcribed Image Text:Calculate AH for the reaction: 2NH3 (9) +O2(9) → N2H4 (1) + H20(1) given the following data: 2NH3(g) + 3N,0(g) → 4N,(g) + 3H,O(1) AH = - 1010. kJ N20(g) + 3H, (g) → N,H4 () + H,O(1) AH = -317 kJ N2 H4 (1) + O2(g) → N2 (g) + 2H2O(1) AH =-623 kJ H2 (g) + 02(g) → H,0(1) AH = ==286 k.J AH = kJ
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