题目内容

已知:Fe2O3(s)3C(石墨)=2Fe(s)3CO(g)ΔH=+489.0 kJ·mol1

CO(g)O2(g)=CO2(g)ΔH=-283.0 kJ·mol1

C(石墨)O2(g)=CO2(g)ΔH=-393.5 kJ·mol1

4Fe(s)3O2(g)=2Fe2O3(s)ΔH(  )

A.+1 164.1 kJ·mol1   B.-1 641.0 kJ·mol1

C.-259.7 kJ·mol1 D.-519.4 kJ·mol1

 

B

【解析】根据盖斯定律×6×2×64Fe(s)3O2(g)=2Fe2O3(s)ΔH=-393.5 kJ·mol1×6489.0 kJ·mol1×2283.0 kJ·mol1×6=-1 641.0 kJ·mol1

 

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