由氢气和氧气反应生成1 mol水蒸气放热241.8 kJ,写出该反应的热化学

由氢气和氧气反应生成1 mol水蒸气放热241.8 kJ,写出该反应的热化学方程式:                                                   。若1 g水蒸气转化成液态水放热2.444 kJ,则反应2H2(g)+O2(g)2H2O(l);ΔH =             kJ·mol—1

 

答案

H2(g)+1/2O2(g)H2O(g);ΔH = -241.8kJ·mol1,-571.6 kJ·mol—1

解析:

 

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