Using the equations; N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol...

Question

Question

Using the equations;

N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

Determine the enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).

answer in kJ/mol

science chemistry

Answer 1

Answer #1

297 +0249) -→2009) AM = 1 80.6 kJ/mol 0 N269+34, 9) -> 2N4 (9) AN = -91.8 kJ/mol - © 24₂ (g) + O2 lg) 240 (g) .An= -489, 7 kJ

Answer 2

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