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I've recently been reading about the different amount of energy release for different types of fuels/hydrocarbons.

Anyways, I was wondering why butanol would release more energy (kj/mol) than methanol? I'm guessing the same theory applies for methane vs butane?

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  • $\begingroup$ Consider the primary alcohol serie. Methanol as the first serie member is like half-water, while higher members converge to hydrocarbons. I missed originally you ask about the molar value, thinking about the specific value MJ/kg. The reason related to this is very trivial, butanol has bigger molecules. 3 big wooden logs in fireplace give more energy than 3 small wooden logs. $\endgroup$
    – Poutnik
    Aug 10, 2020 at 6:08

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This is essentially a question concerning the trend of molecules in the same homologus series that have differing chain lengths. When we increase chain length, the number of C-C bonds increase in proportion to C-H bonds. Looking at the bond energies, the bond energy of a C-C bond is $347 kJ/mol$, while for C-H it is $414 kJ/mol$. This means to break a C-C bond compared to a C-H bond, less energy is required. Also when increasing the number of C-C bonds, while producing the same products ($CO_{2(g)} + H_2O_{(l)}$), more carbon dioxide is produced in proportion to the amount of water. The enthalpy of formation of carbon dioxide, results in it releasing more energy than that of water. Taking this into account with the lower energy requirement to break the reactants, the chain length is generally directly proportional to the energy produced. Thus, the COMPLETE combustion of Butanol would have a higher energy out put than Methanol and likewise Butane > Methane. It is important to note that, as the compounds get more energy dense, i.e release more energy upon combustion, that it becomes harder for the compound to reach complete combustion; what I mean is that more oxygen is required for the reaction to release the full energy output. If not enough oxygen is supplied to the system soot ($C_{(s)}$) & Carbon Monoxide ($CO_{(g)}$) will be produced and the energy produced will be less than the energy produced in the complete combustion. (The degree to which this works, depends on how many moles of oxygen is supplied).

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  • $\begingroup$ (+1) Your comment on incomplete combustion being generally inefficient in oxygen with longer-chain molecules is certainly correct. However, am I correct to suspect that combustion in an atmosphere of chlorine gas where the products would be HCl and various carbon chlorides/carbon-hydrogen chlorides may actually tend to be more efficient? However, obviously the latter products are extremely problematic in an earth atmosphere, but perhaps a 3rd stage rocket in space should not be using O2 based fuel mixes. $\endgroup$
    – AJKOER
    Aug 13, 2020 at 10:08

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