问题:
下面均是乙烯与氧气反应的热化学方程式(25℃,101kPa)
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(l)△H=-1411kJ/mol
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(g)△H=-1329kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(l)△H=-873kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(g)△H=-763kJ/mol
由此判断,乙烯的燃烧热是( )
A.1411kJ/mol
B.1329kJ/mol
C.873kJ/mol
D.763kJ/mol
下面均是乙烯与氧气反应的热化学方程式(25℃,101kPa)
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(l)△H=-1411kJ/mol
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(g)△H=-1329kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(l)△H=-873kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(g)△H=-763kJ/mol
由此判断,乙烯的燃烧热是( )
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(l)△H=-1411kJ/mol
C 2H 4(g)+3O 2(g)=2CO 2(g)+2H 2O(g)△H=-1329kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(l)△H=-873kJ/mol
C 2H 4(g)+2O 2(g)=2CO(g)+2H 2O(g)△H=-763kJ/mol
由此判断,乙烯的燃烧热是( )
A.1411kJ/mol | B.1329kJ/mol | C.873kJ/mol | D.763kJ/mol |
参考答案: