(1)2CH 3OH(1)+3O 2(g)===2CO 2(g)+4H 2O(l) △H=-1452.8 kJ/mol
(2)-93(3)△H=+226.7 kJ/mol
(1)甲醇燃烧化学方程式为:2CH 3OH+3O 2(g)
2CO 2(g)+4H 2O,故2mol甲醇燃烧放出的热量为
1452.8kJ,故热化学方程式为:2CH 3OH(1)+3O 2(g)===2CO 2(g)+4H 2O(l) △H=-1452.8 kJ/mol
(2)△H=反应物的键能之和—生成物的键能之和=945+3×436—2×3×391=-93kJ/mol
(3)依据盖斯定律可知,2×①+②/2 —③/2:
2C(s,石墨)+H 2(g)=C 2H 2(g)△H=+226.7 kJ/mol