Question

Hydrogen and methanol have both been proposed as alternatives to hydrocarbon fuels.

1. Use standard enthalpies of formation to calculate the amount of heat released per kilogram of the fuel

2. How does the energy of these fuels compare to that of octane (C8H18)?

Answer 1

I am assuming that they are producing H2O(l) @ dHf = -285.8 kJ/mol, rather than H2O(g) @ dHf = - –241.8

H2 (g) & 1/2O2 (g) --> H2O(l)
dH = -285.8 kJ/mol of H2O produced, & per mole of H2 burned
so, using molar mass:
(-285.8 kJ/mol) / (2.016 gram H2 / mol) = - 142.9 kJ / gram H2 times 1000 grams per kilogram =
dH = - 142,900 kJ / kg H2 = 142MJ/kg

methanol:
CH3OH(l) & 2 O2 --> CO2 & 2H2O(l)
dH rxn = dHf products - dHf reactants

dH rxn = (dHf CO2 & 2 dHf H2O) - (dHf CH3OH & 2 dHf O2)

dH rxn = (-393.5 & 2 X -285.8) - (–238.7 & zero)
dH rxn = (-393.5 & - 571.6) - (–238.7 )
dH rxn = (-965.1) - (–238.7 )
dH rxn = (-965.1 + 238.7 )
dH rxn = -726.4 kJ / mole methanol burned
using molar mass:
(-726.4 kJ / mole) / (32.042 g methanol / mole) = -22.67 kJ / gram methanol times 1000 grams per kilogram= dH = -22,670 kJ/kg methanol burned = 22.67MJ/kg

Using the above method, heat of combustion of octane comes out to be 5460 KJ/mol. Converting it into kg,

= (5460/114)*1000 = 47894 KJ/kg = 47.894 MJ/kg

Hydrogen has the highest heat of combustion and is thus used a rocket fuel.