题目内容
由金红石(TiO2)制取单质Ti,涉及到的步骤为:TiO2→TiCl4
Ti已知:
①
②
③
则(1)TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g)的△H=______.
(2)碳在氧气中不完全燃烧生成CO的热化学方程式为______.
| ||
| 800℃.Al |
①
|
②
|
③
|
则(1)TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g)的△H=______.
(2)碳在氧气中不完全燃烧生成CO的热化学方程式为______.
(1)已知:①
②
③
利用盖斯定律将③+2×①-②可得:
TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g);△H=(+141kJ/mol)+2×(-393.5kJ/mol)-(-566kJ/mol)=-80kJ/mol,故答案为:-80KJ/mol;
(2)利用盖斯定律将①×2-②可得:
2C(s)+O2(g)═2CO(g);△H=2×(-393.5kJ/mol)-(-566kJ/mol)=-221kJ/mol,
或C(s)+
O2(g)═CO(g);△H=-110.5kJ/mol
故答案为:2C(s)+O2(g)═2CO(g);△H=-221kJ/mol或C(s)+
O2(g)═CO(g);△H=-110.5kJ/mol.
|
②
|
③
|
利用盖斯定律将③+2×①-②可得:
TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g);△H=(+141kJ/mol)+2×(-393.5kJ/mol)-(-566kJ/mol)=-80kJ/mol,故答案为:-80KJ/mol;
(2)利用盖斯定律将①×2-②可得:
2C(s)+O2(g)═2CO(g);△H=2×(-393.5kJ/mol)-(-566kJ/mol)=-221kJ/mol,
或C(s)+
| 1 |
| 2 |
故答案为:2C(s)+O2(g)═2CO(g);△H=-221kJ/mol或C(s)+
| 1 |
| 2 |
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