利用盖斯定律解答下列各小题
(1)已知:TiO
2(s)+2Cl
2(g)═TiCl
4(l)+O
2(g)△H=+140kJ?mol
-12C(s)+O
2(g)═2CO(g)△H=-221kJ?mol
-1写出TiO
2和焦炭、氯气反应生成TiCl
4和CO气体的热化学方程式:
2C(s)+TiO2(s)+2Cl2(g)═TiCl4(l)+2CO(g)△H=-81kJ?mol-1
2C(s)+TiO2(s)+2Cl2(g)═TiCl4(l)+2CO(g)△H=-81kJ?mol-1
.
(2)25℃、101kPa下:①2Na(s)+
O
2(g)═Na
2O(s)△H
1=-414kJ?mol
-1②2Na(s)+O
2(g)═Na
2O
2(s)△H
2=-511kJ?mol
-1写出该条件下由Na
2O
2和Na生成Na
2O的热化学方程式:
2Na(s)+Na2O2(s)=2Na2O(s)△H1=-317kJ?mol-1
2Na(s)+Na2O2(s)=2Na2O(s)△H1=-317kJ?mol-1
(3)已知:C(s,石墨)+O
2(g)═CO
2(g)△H
1=-393.5kJ?mol
-1;
2H
2(g)+O
2(g)═2H
2O(l)△H
2=-571.6kJ?mol
-1;
2C
2H
2(g)+5O
2(g)═4CO
2(g)+2H
2O(l)△H
2=-2599kJ?mol
-1;
写出由C(s,石墨)和H
2(g)生成1mol C
2H
2(g)的热化学方程式
2C(s,石墨)+H2(g)=C2H2(g),△H1=226.7kJ?mol-1
2C(s,石墨)+H2(g)=C2H2(g),△H1=226.7kJ?mol-1
.