Question

【题目】肼(N2H4)是火箭发动机的燃料，它与N2O4反应时，N2O4为氧化剂，生成氮气和水蒸气。已知：①N2(g)＋2O2(g)＝N2O4(g)ΔH＝＋8.7kJ·mol－1；②N2H4(g)＋O2(g)＝N2(g)＋2H2O(g)ΔH＝－534.0kJ·mol－1；下列表示肼跟N2O4反应的热化学方程式正确的是

A.2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)ΔH=－542.7kJ·mol－1

B.2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)ΔH=－1059.3kJ·mol－1

C.N2H4(g)＋N2O4(g)＝N2(g)＋2H2O(g)ΔH=－1076.7kJ·mol－1

D.2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)ΔH=－1076.7kJ·mol－1

Answer 1

【答案】D

【解析】

肼(N2H4)是火箭发动机的燃料，它与N2O4反应生成氮气和水蒸气，已知：①N2(g)＋2O2(g)＝N2O4(g) ΔH＝＋8.7kJ·mol－1、②N2H4(g)＋O2(g)＝N2(g)＋2H2O(g) ΔH=－534.0kJ·mol－1，将方程式②×2-①得N2H4跟N2O4反应的热化学方程式，2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g) ΔH=－1076.7kJ·mol－1或N2H4(g)＋N2O4(g)＝N2(g)＋2H2O(g) ΔH=-538.35kJ·mol－1，D符合题意；

答案选D。