己知298K时下列热化学方程式: 1 2NH 3 (g) → N 2 (g) + 3H 2 (g) Δ r H m Θ =92.2 kJ·mol -1 2 H 2 (g) + 1/2O 2 (g) → H 2 O(g) Δ r H m Θ =-241.8 kJ·mol -1 3 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) Δ r H m Θ =-905.6 kJ·mol -1 计算: Δ f H m Θ (NH 3 ,g,298K)= kJ·mol -1 ; Δ f H m Θ (H 2 O,g,298K)= kJ·mol -1 ; Δ f H m Θ (NO,g,298K)= kJ·mol -1 ; 由NH 3 (g)生产1.00 mol NO(g),放出热量为 kJ。