题目内容
已知热化学方程式:
4Al(s)+3O2(g)═2Al2O3(s)△H1=-3288.6kJ?mol-1,
4Fe(s)+3O2(g)═2Fe2O3(s)△H2=-1631.8kJ?mol-1,
则铝粉与氧化铁发生铝热反应的热化学方程式为______.
4Al(s)+3O2(g)═2Al2O3(s)△H1=-3288.6kJ?mol-1,
4Fe(s)+3O2(g)═2Fe2O3(s)△H2=-1631.8kJ?mol-1,
则铝粉与氧化铁发生铝热反应的热化学方程式为______.
①4Al(s)+3O2(g)═2Al2O3 (s)△H1=-3288.6kJ?mol-1;
②4Fe(s)+3O2(g)═2Fe2O3 (s)△H2=-1631.8kJ?mol-1;
由盖斯定律计算①-②得到:4Al(s)+2Fe2O3 (s)═2Al2O3 (s)+4Fe(s)△H=-1656.8KJ/mol
热化学方程式为:2Al(s)+Fe2O3 (s)═Al2O3 (s)+2Fe(s)△H=-828.4KJ/mol
故答案为:2Al(s)+Fe2O3 (s)═Al2O3 (s)+2Fe(s)△H=-828.4KJ/mol.
②4Fe(s)+3O2(g)═2Fe2O3 (s)△H2=-1631.8kJ?mol-1;
由盖斯定律计算①-②得到:4Al(s)+2Fe2O3 (s)═2Al2O3 (s)+4Fe(s)△H=-1656.8KJ/mol
热化学方程式为:2Al(s)+Fe2O3 (s)═Al2O3 (s)+2Fe(s)△H=-828.4KJ/mol
故答案为:2Al(s)+Fe2O3 (s)═Al2O3 (s)+2Fe(s)△H=-828.4KJ/mol.
练习册系列答案
相关题目