Answer:
ΔH° = + 6.2 kJ
Explanation:
Fe₂O₃(s) + 3CO(g) –––––> 2 Fe(s) + 3 CO₂(g); ΔH° = –26.8 kJ ......... ( 1 )
FeO(s) + CO(g) –––––> Fe(s) + CO₂(g); ΔH° = –16.5 kJ .........( 2 )
Multiplying equation ( 2 ) by 2 and subtracting it from ( 1 )
Fe₂O₃(s) + 3CO(g) -2 FeO(s) - 2CO(g) ––> 2 Fe(s) + 3 CO₂(g) - 2Fe(s) - 2CO₂(g) ΔH° = –26.8 kJ - ( 2 x –16.5 kJ )
Fe₂O₃(s) + CO(g) -2 FeO(s)––> CO₂(g) ΔH° = –26.8 kJ + 33 kJ
Fe₂O₃(s) + CO(g) ––>2 FeO(s) +CO₂(g) ΔH° = + 6.2 kJ
Required ΔH° = + 6.2 kJ Ans .
