Both ethanol (CH3CH2OH, MM=46.07g/mol) and methanol (CH3OH, MM=32.04g/mol) have been considered as fuels for automobiles. Which is the better fuel, on a per gram basis, when burned with oxygen?
CH3CH2OH(g) DeltaHf(KJ/molrxn) = -235.10
H3OH (g) DeltaHf(KJ/molrxn) = -200.66
H2O (g) DeltaHf(KJ/molrxn) = -241.82
CO2 (g) DeltaHf(KJ/molrxn) = -393.51
To determine which fuel is better on a per gram basis when burned with oxygen, we need to compare the enthalpy changes (ΔH) of the combustion reactions for both ethanol (C2H5OH) and methanol (CH3OH).
The combustion reactions for each fuel can be written as follows:
Ethanol:
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
Methanol:
CH3OH(g) + O2(g) → CO2(g) + 2H2O(g)
Note: The coefficients in front of the compounds represent the stoichiometric coefficients required for balanced equations.
To calculate the ΔH for each reaction, we need to consider the enthalpy of formation (ΔHf) for each compound involved in the reaction. The enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
The given ΔHf values in kilojoules per mole (KJ/molrxn) are as follows:
- ΔHf for ethanol (C2H5OH): -235.10 KJ/molrxn
- ΔHf for methanol (CH3OH): -200.66 KJ/molrxn
- ΔHf for water (H2O): -241.82 KJ/molrxn
- ΔHf for carbon dioxide (CO2): -393.51 KJ/molrxn
Now, let's calculate the ΔH for each combustion reaction:
Ethanol combustion:
Calculating the ΔH for the reactants:
ΔH = (2 × ΔHf of CO2) + (3 × ΔHf of H2O) - (ΔHf of ethanol)
ΔH = (2 × -393.51 KJ/molrxn) + (3 × -241.82 KJ/molrxn) - (-235.10 KJ/molrxn)
ΔH = -787.02 KJ/molrxn + -725.46 KJ/molrxn + 235.10 KJ/molrxn
ΔH = -1276.38 KJ/molrxn
Methanol combustion:
Calculating the ΔH for the reactants:
ΔH = (ΔHf of CO2) + (2 × ΔHf of H2O) - (ΔHf of methanol)
ΔH = (-393.51 KJ/molrxn) + (2 × -241.82 KJ/molrxn) - (-200.66 KJ/molrxn)
ΔH = -393.51 KJ/molrxn + -483.64 KJ/molrxn + 200.66 KJ/molrxn
ΔH = -676.49 KJ/molrxn
Now, to compare the fuels on a per gram basis, we need to calculate the ΔH per gram of each fuel. We can do this by dividing the ΔH for each fuel by their respective molar masses.
Ethanol per gram:
ΔH per gram = ΔH per mol / Molar mass of ethanol
ΔH per gram = -1276.38 KJ/molrxn / 46.07 g/mol
ΔH per gram = -27.73 KJ/g
Methanol per gram:
ΔH per gram = ΔH per mol / Molar mass of methanol
ΔH per gram = -676.49 KJ/molrxn / 32.04 g/mol
ΔH per gram = -21.13 KJ/g
Comparing the ΔH per gram values, we can see that methanol has a lower value (more negative) than ethanol. Therefore, on a per gram basis, methanol is the better fuel when burned with oxygen.
Please note that this analysis solely considers the ΔH values of the combustion reactions and does not take into account other properties such as energy density, safety, availability, or practicality of the fuels.