Estimate the enthalpy change for the combustion of one mole of ethanol (C2H5OH)

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C2H5OH + 3(O2) ==> 2(CO2) + 3(H2O) enthalpy of combustion = enthalpy of above reaction = enthalpy of formation of product - enthalpy of formation of reactant enthalpy of product = 2*(122+122) + 3*(463+463) = 3226 enthalpy of reactant = (5*413 + 1*154 + 1*358 + 1*463) + 3*(498) = 4534 => enthalpy of combustion = 3226 - 4534 = -1308 KJ/mol formation reation.. 2(CO2) + 3(H2O) ==> C2H5OH + 3(O2) heat of formation of C2H5OH = 2*(-394) + 3*(-242) = -1514 KJ/mol

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