4C+6H2+O2→2C2H5OH ΔfH=?
C+O2→CO2 ΔcH(C)=−393kJ/mol
2H2+O2→2H2O ΔcH(H2)=−286kJ/mol
C2H5OH+3O2→2CO2+3H2O ΔcH(C2H5OH)=−1371kJ/mol
ΔfH=4ΔcH(C)+6ΔcH(H2)−2ΔcH(C2H5OH)
ΔfH= 4*(-393kJ/mol)+6*(-286kJ/mol)-2(-1371kJ/mol)=-546kJ/mol
Answer:-546kJ/mol
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