题目内容
【题目】肼(N2H4)是火箭发动机的一种燃料,反应时N2O4为氧化剂,生成N2和H2O(g),已知: N2(g)+2O2(g)═N2O4(g),△H=+8.7kJ/mol;
N2H4(g)+O2(g)═N2(g)+2H2O(g),△H=﹣534.0kJ/mol;
下列表示肼跟N2O4反应的热化学反应方程式,正确的是( )
A.2N2H4(g)+N2O4(g)═3N2(g)+4H2O(g);△H=﹣542.7 kJ/mol
B.2N2H4(g)+N2O4(g)═3N2(g)+4H2O(g);△H=﹣1059.3 kJ/mol
C.N2H4(g)+ N2O4(g)═ N2(g)+2H2O(g);△H=﹣1076.7 kJ/mol
D.2N2H4(g)+N2O4(g)═3N2(g)+4H2O(g);△H=﹣1076.7 kJ/mol
【答案】D
【解析】解:已知①N2(g)+2O2(g)═N2O4(g),△H=+8.7kJ/mol,②N2H4(g)+O2(g)═N2(g)+2H2O(g),△H=﹣534.0kJ/mol,
利用盖斯定律将②×2﹣①可得2N2H4(g)+N2O4(g)═3N2(g)+4H2O(g),△H=(﹣534.0kJ/mol)×2﹣(+8.7kJ/mol)=﹣1076.7kJ/mol,或N2H4(g)+ N2O4(g)═ N2(g)+2H2O(g)△H=﹣1076.7 kJ/mol,
故选D.
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