Problem: Calculate ΔH for the reaction2NH3 (g) + 1/2 O2 (g) → N2H4 (l) + H2O (l) given the following data:2NH3 (g) + 3N2O (g) → 4N2 (g) + 3H2O (l)  ΔH = -1010 kJN2O (g) + 3H2 (g) → N2H4 (l) + H2O (l)      ΔH = -317 kJN2H4(l) + O2 (g) → N2(g) + 2H2O (l)      ΔH = -623 kJH2(g) + 1/2O2 (g) → H2O (l)          ΔH = -286 kJ

FREE Expert Solution

2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (l)             ΔH = -1010 kJ

N2O (g) + 3 H(g) → N2H4 (l) + H2O (l)                      ΔH = -317 kJ

(N2H4 (l) + H2O (l) → N2O (g) + 3 H(g)) x3               ΔH = (+317 kJ)x3 

3 N2H4 (l) + 3 H2O (l) → 3 N2O (g) + 9 H(g)           ΔH = 951 kJ

N2H(l) + O2 (g) → N(g) + 2 H2O (l)                         ΔH = -623 kJ

(N(g) + 2 H2O (l) → N2H(l) + O2 (g) )x4                      ΔH = (+623 kJ)x4

4 N(g) + 8 H2O (l) → 4 N2H(l) + 4 O2 (g)                 ΔH = +2492 kJ

(H(g) + 1/2 O2 (g) → H2O (l))x9                                ΔH = (-286 kJ) x 9 = -2574 kJ

9 H(g) + 4.5 O2 (g) → 9 H2O (l)                                 ΔH = -2574 kJ   


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Problem Details

Calculate ΔH for the reaction

2NH3 (g) + 1/2 O2 (g) → N2H(l) + H2O (l) 

given the following data:

2NH3 (g) + 3N2O (g) → 4N2 (g) + 3H2O (l)  ΔH = -1010 kJ

N2O (g) + 3H(g) → N2H4 (l) + H2O (l)      ΔH = -317 kJ

N2H4(l) + O2 (g) → N2(g) + 2H2O (l)      ΔH = -623 kJ

H2(g) + 1/2O2 (g) → H2O (l)          ΔH = -286 kJ

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