题目内容

【题目】已知下列数据: Fe(s)+ O2(g)═FeO(s)△H=﹣272kJmol﹣1
2Al(s)+ O2(g)═Al2O3(s)△H=﹣1675kJmol﹣1
则2Al(s)+3FeO(s)═Al2O3(s)+3Fe(s)的△H是(  )
A.+859 kJmol﹣1
B.﹣859 kJmol﹣1
C.﹣1403 kJmol﹣1 
D.﹣2491 kJmol﹣1

【答案】B
【解析】解:①Fe(s)+ O2(g)=FeO(s)△H=﹣272.0kJmol﹣1

②2Al(s)+ O2(g)=Al2O3(s)△H=﹣1675kJmol﹣1

将方程式②﹣①×3得2Al(s)+3FeO(s)═Al2O3(s)+3Fe(s)△H=[﹣1675kJmol﹣1﹣(﹣272.0kJmol﹣1)×3]=﹣859kJmol﹣1

故选B.

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