Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of formation of the reactants and products. C2H5OH (l) + 3 O2 (g) = 2CO2 (g) + 3 H2O

For the reaction

C2H5OH (l) + 3 O2 (g) = 2CO2 (g) + 3 H2O

We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O (l) - (ΔHºf C2H5OH (l) - 3ΔHºfO2 (g))

(we were not give the water state but we know we are at standard conditions so it is in its liquid state)

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O (l) - (ΔHºf C2H5OH (l) - + 3ΔHºfO2 (g))

= [ 2 (-393.52) + 3 (-285.83) ] - [ (-276.2 + 0) ] kJ

ΔHºc = - 1368.33 kJ