Answer to Question #114281 in General Chemistry for Cesca

Question #114281
Calculate the enthalpy of formation of ethanol (C2H5OH) given the following enthalpies of combustion.
ΔcH C(s) = -393, H2(g) = -286, C2H5OH(l) = -1371 kJ/mol
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Expert's answer
2020-05-07T14:16:02-0400

4C+6H2+O22C2H5OH4C+6H_2+O_2\to 2C_2H_5OH ΔfH=?\Delta fH=?

C+O2CO2C+O_2\to CO_2 ΔcH(C)=393kJ/mol\Delta cH(C)=-393kJ/mol

2H2+O22H2O2H_2+O_2\to2H_2O ΔcH(H2)=286kJ/mol\Delta cH (H_2)=-286 kJ/mol

C2H5OH+3O22CO2+3H2OC_2H_5OH+3O_2\to 2CO_2+3H_2O ΔcH(C2H5OH)=1371kJ/mol\Delta cH(C_2H_5OH)=-1371kJ/mol

ΔfH=4ΔcH(C)+6ΔcH(H2)2ΔcH(C2H5OH)\Delta fH=4\Delta cH(C)+6\Delta cH(H_2)-2\Delta cH(C_2H_5OH)

ΔfH=\Delta fH= 4*(-393kJ/mol)+6*(-286kJ/mol)-2(-1371kJ/mol)=-546kJ/mol

Answer:-546kJ/mol


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