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
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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
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