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已知298K时下列热化学方程式:①2NH₃(g)→N₂(g)+3H₂(g)rHmΘ=92.2kJ·mol—1②H₂(g)+ 1/2O₂(g)→H₂O(g)rHmΘ=-241.8 kJ·mol-1③4NH₃(g)+5O₂(g)→4NO(g)+6H₂O(g)rHmΘ=-905。6 kJ·mol-1试确定fHmΘ(NH₃,g,298K)=()kJ·mol-1;fHmΘ(H₂O,g,298K)=()kJ·mol—1;fHmΘ(NO,g,298K)=()kJ·mol—1。由NH₃(g)生产1.00kgNO(g)则放出热量为()kJ。
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