Question

Using enthalpies of formation and Hess's Law, calculate the heat of combustion for each of the following : coal (c) ethanol (C2H5OH) 1-Butanol ( C4H9OH) octane (C8H18)

Answer 1

Coal

Coal + O2 --------> CO2 + H2O

Enthalpy of combustion of Coal ( delta H) = - 393.5 - 285.9 - 0 = - 679.4 kJ/mol

Ethanol

C2H5OH (l) + 3O2 (g) -----> 2CO2 (g) + 3H2O (l)

Total change in enthalpy = 2 * ( - 393.5) + 3 * ( - 285.9) = - 1644.7 kJ

Enthalpy for combustion = Total enthalpy change + Enthalpy of formation of ethanol

Enthalpy for combustion = - 1644.7 + 276.51 = - 1368 kJ/mol

Butanol

Entahlpy of combustion of butanol = ( 1g / 74 g/mol ) * 36 kJ = 2,666.67 kJ/mol

Octane

C8H18(l) +(25/2) O2(g) -----> 8CO2(g) + 9H2O (l)

delta Hf (octane) = -249.95 kJ/mol
delta Hf (CO2(g)) = -393.51 kJ/mole
delta Hf (H2O(l)) = -285.83 kJ/mole
Enthalpy for combustion of octane = 8 * ( - 393.51 ) + 9 * ( -285.83) - ( - 249.95)

where, enthalpy of formation of octane = - 249.95 kJ

Enthalpy for combustion of octane = - 5470.6 kJ/mol