题目内容
18.发射卫星用N2H4气体为燃料,NO2气体为氧化剂,两者反应生成N2和水蒸气,已知:N2 (g)+2O2 (g)═2NO2(g)△H1=+67.7kJ•mol-1;
N2H4 (g)+O2 (g)═N2(g)+2H2O(g)△H3=-534kJ•mol-1;
H2 (g)+F2 (g)═2HF(g)△H4=-538kJ•mol-1;
2H2 (g)+O2 (g)═2H2O(g)△H2=-484kJ•mol-1;
H2O(1)═H2O(g)△H5=+44.0kJ•mol-1.
(1)写出N2H4气体与NO2气体反应生成氮气和液态水的热化学方程式:2N2H4(g)+2NO2(g)=3N2(g)+4H2O(l)△H=-1311.7kJ/mol.
(2)写出N2H4气体与F2气体反应的热化学方程式:N2H4(g)+2F2(g)=N2(g)+4HF(g)△H=-1126KJ/mol.
分析 (1)已知:①N2(g)+2O2(g)=2NO2(g)△H1=+67.7kJ•mol-1;
②N2H4(g)+O2(g)=N2(g)+2H2O(g)△H2=-534kJ•mol-1;
⑤H2O(l)=H2O(g)△H3=+44kJ•mol-1.
2N2H4(g)+2NO2(g)=3N2(g)+4H2O(l)可以是反应②×2-①-⑤×4得到,由此分析解答;
(2)①N2H4(g)+O2(g)=N2(g)+2H2O(g)△H2=-534kJ/mol;
③H2(g)+F2(g)=2HF(g)△H4=-538kJ/mol
④2H2(g)+O2(g)=2H2O (g)△H2=-484kJ/mol;
依据盖斯定律③×2-④+①得到,由此分析解答.
解答 解:(1)已知:①N2(g)+2O2(g)=2NO2(g)△H1=+67.7kJ•mol-1;
②N2H4(g)+O2(g)=N2(g)+2H2O(g)△H2=-534kJ•mol-1;
⑤H2O(l)=H2O(g)△H3=+44kJ•mol-1.
2N2H4(g)+2NO2(g)=3N2(g)+4H2O(l)可以是反应②×2-①-⑤×4得到,
所以该反应的焓变=(-534kJ•mol-1)×2-67.7kJ•mol-1-(+44kJ•mol-1)×4=-1311.7kJ/mol,
故答案为:2N2H4(g)+2NO2(g)=3N2(g)+4H2O(l)△H=-1311.7kJ/mol;
(2)①N2H4(g)+O2(g)=N2(g)+2H2O(g)△H2=-534kJ/mol;
③H2(g)+F2(g)=2HF(g)△H4=-538kJ/mol
④2H2(g)+O2(g)=2H2O (g)△H2=-484kJ/mol;
依据盖斯定律③×2-④+①得到热化学方程式为:N2H4(g)+2F2(g)=N2(g)+4HF(g)△H=-1126KJ/mol;
故答案为:N2H4(g)+2F2(g)=N2(g)+4HF(g)△H=-1126KJ/mol.
点评 本题考查了热化学方程式的书写方法和盖斯定律的计算应用,题目较简单.
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