由金红石(TiO2)制取单质Ti,涉及的步骤为:已知:①C(s)+O2(g)====CO2(g) ΔH=-393.5 kJ·mol-1②2CO(g)+O2(g) ====2CO2(g) ΔH=-566 kJ·mol-1③TiO2(s)+2Cl2(g) ====TiCl4(s)+O2(g) ΔH=+141 kJ·mol-1则(1)TiO2(s)+2Cl2(g)+2C(s) ====TiCl4
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 16:20:22
![由金红石(TiO2)制取单质Ti,涉及的步骤为:已知:①C(s)+O2(g)====CO2(g) ΔH=-393.5 kJ·mol-1②2CO(g)+O2(g) ====2CO2(g) ΔH=-566 kJ·mol-1③TiO2(s)+2Cl2(g) ====TiCl4(s)+O2(g) ΔH=+141 kJ·mol-1则(1)TiO2(s)+2Cl2(g)+2C(s) ====TiCl4](/uploads/image/z/4740576-24-6.jpg?t=%E7%94%B1%E9%87%91%E7%BA%A2%E7%9F%B3%28TiO2%29%E5%88%B6%E5%8F%96%E5%8D%95%E8%B4%A8Ti%2C%E6%B6%89%E5%8F%8A%E7%9A%84%E6%AD%A5%E9%AA%A4%E4%B8%BA%EF%BC%9A%E5%B7%B2%E7%9F%A5%EF%BC%9A%E2%91%A0C%28s%29%2BO2%28g%29%3D%3D%3D%3DCO2%28g%29+%CE%94H%3D%EF%BC%8D393.5+kJ%C2%B7mol-1%E2%91%A12CO%28g%29%2BO2%28g%29+%3D%3D%3D%3D2CO2%28g%29+%CE%94H%3D%EF%BC%8D566+kJ%C2%B7mol-1%E2%91%A2TiO2%28s%29%2B2Cl2%28g%29+%3D%3D%3D%3DTiCl4%28s%29%2BO2%28g%29+%CE%94H%3D%2B141+kJ%C2%B7mol-1%E5%88%99%281%29TiO2%28s%29%2B2Cl2%28g%29%2B2C%28s%29+%3D%3D%3D%3DTiCl4)
由金红石(TiO2)制取单质Ti,涉及的步骤为:已知:①C(s)+O2(g)====CO2(g) ΔH=-393.5 kJ·mol-1②2CO(g)+O2(g) ====2CO2(g) ΔH=-566 kJ·mol-1③TiO2(s)+2Cl2(g) ====TiCl4(s)+O2(g) ΔH=+141 kJ·mol-1则(1)TiO2(s)+2Cl2(g)+2C(s) ====TiCl4
由金红石(TiO2)制取单质Ti,涉及的步骤为:已知:①C(s)+O2(g)====CO2(g) ΔH=-393.5 kJ·mol-1
②2CO(g)+O2(g) ====2CO2(g) ΔH=-566 kJ·mol-1
③TiO2(s)+2Cl2(g) ====TiCl4(s)+O2(g) ΔH=+141 kJ·mol-1
则(1)TiO2(s)+2Cl2(g)+2C(s) ====TiCl4(s)+2CO(g)的ΔH=____________________.
求详细步骤最好说明原因
由金红石(TiO2)制取单质Ti,涉及的步骤为:已知:①C(s)+O2(g)====CO2(g) ΔH=-393.5 kJ·mol-1②2CO(g)+O2(g) ====2CO2(g) ΔH=-566 kJ·mol-1③TiO2(s)+2Cl2(g) ====TiCl4(s)+O2(g) ΔH=+141 kJ·mol-1则(1)TiO2(s)+2Cl2(g)+2C(s) ====TiCl4
ΔH=-80kJ/mol
利用盖斯定律做.
将2改变为2CO2(g)=2CO(g)+O2(g) ΔH=+566 kJ·mol-1
把1乘以2得到 2C(s)+2O2(g)====2CO2(g) ΔH=-787 kJ·mol-1
与2合并得到
2C(s)+O2(g)=2CO(g) ΔH=-221 kJ·mol-1【+566+(-787)=-221】
与3合并得到
TiO2(s)+2Cl2(g)+2C(s) ====TiCl4(s)+2CO(g)的ΔH=-221+(+141)=-80kJ/mol
2x(1)-(2)+(3)=-2x393.5+566+141=120kj.mol
qwrfq