题目内容
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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＝＋8.7 kJ·mol－1 N2H4(g)＋O2(g)＝N2(g)＋2H2O(g)；△H＝－534.0 kJ·mol－1 下列表示肼跟N2O4反应的热化学反应方程式，正确的是 | |
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A． |
2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)；△H＝－542.7 kJ·mol－1 |
B． |
2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)；△H＝－1059.3 kJ·mol－1 |
C． |
N2H4(g)＋ |
D． |
2N2H4(g)＋N2O4(g)＝3N2(g)＋4H2O(g)；△H＝－1076.7 kJ.mol－1 |
N2O4(g)＝
N2(g)＋2H2O(g)；△H＝－1076.7 kJ·mol－1
答案：D
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相关题目
N2O4(g) ΔH=+8.7 kJ·mol-1
N2(g)+2H2O(g) ΔH=-534.0 kJ·mol-1
3N2(g)+4H2O(g) ΔH=-542.7 kJ·mol-1
3N2(g)+4H2O(g) ΔH=-1 059.3 kJ·mol-1
N2O4(g) 
N2(g)+2H2O(g) ΔH=-1 076.7 kJ·mol-1
3N2(g)+4H2O(g) ΔH=-1 076.7 kJ·mol-1
N2O4(g) ══
N2(g)+2H2O(g);ΔH=-1076.7 kJ·mol -1??
N2O4(g) ====
N2(g)+2H2O(g)ΔH="-1076.7" kJ·mol-1